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2025-03-21T14:48:30.046651
| 2020-03-10T20:25:27 |
354646
|
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|
Stack Exchange
|
Henselization and completions of local rings & schemes
That's the second part of my coarse becoming acquainted with Henselizations of fields and local rings. (in this question we focus on local rings as it is more algebro geometric motivated). So let $(R, \mathfrak m, \kappa= R/\mathfrak m )$ be a local ring with max ideal $m$.
We can obtain two new rings $R^h$ (the Henselization) and $\widehat{R}_m$ the completion wrt $m$. Consider $R$ as a stalk of a nice enough scheme $S$ we can use these two constructions to obtain new new objects stalkwise: $S^h$ (here we have to differ between strict and "weak" Henselization) and the completion $\widehat{S}$. (recall $\widehat{S}$ is not more a scheme but a ringed space: localizations and completions not behave well to each other).
in would like to compare the main differences & (dis)advantages of completions & Henselizations from viewpoint of commutative algebra and (as well possible) geometric intuition.
The main motivation is that I often read comments like "in pracise it's often nicer to work with Henselisations than with completions" in order to study the ring $R$ itself.
Question: Could anybody point out what are the advantages making Henselizations from certain viewpoint nicer to handle with then with completions?
In many comments the hand weavy arguments apearing in this context are like $\widehat{R}_m$ is much "bigger" that $R^h$ making it not "so easy handable like $R^h$". Could anybody bring more light in this formulation? When is mean by "bigger" (the added limits ofCauchy sequences I guess) but much more intersting what makes $R^h$ more "handable"?
The only point that I found out is that $Frac(R)=K \subset K^h$ stays algebraic and in many situations even finite. Is $R \to R^h$ also a finite $R$-module. In general that's not true for completions $ R \to \widehat{R}_m$.
Is this the only point making $R^h$ more handable than $ R \to \widehat{R}_m$?
What can we say about the geometrical part? The completion $\widehat{S}$ gives in certain way "analytic structure" to an (algebraic) scheme $S$ (very hand weavy; I know). About what kind of "geometry" one can think when one consider a henselization of a scheme (as for completion: local ring wise)? Some sources refer to "etale topology". It's a starting point of a hige mashinery cumulating in stack theory.
Is there an geometric intuition how one can draw comparisons between endowings of $S$ "analytical structure" (as for completions) and with "etale topology" for $S^h$?
I know that there are a couple of questions here with similar titles (eg henselization and completion , Completion versus henselization. , comparison of completion and Henselization in class field theory ) but none of them deal with question of pure comparison of tqo constructions in the way I explained above.
|
2025-03-21T14:48:30.046932
| 2020-03-10T20:55:54 |
354649
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Alexey Milovanov",
"Aryeh Kontorovich",
"fedja",
"https://mathoverflow.net/users/1131",
"https://mathoverflow.net/users/12518",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/354649"
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|
Stack Exchange
|
A possible generalization of Solomonoff's theorem
Assume that $P$ and $Q$ are probability distribution on the binary tree,
i.e. $P$ and $Q$ are functions $\{0,1\}^{*} \to \mathbb{R}$ such that:
for every $x$: $P(x)=P(x0)+P(x1)$ and $P( \text{empty word})= 1$.
Theorem(Solomonoff)
Assume that for some $c>0$ it holds that for every $x$ $Q(x) \ge c P(x)$ .
Then:
$$ \sum_{x \in \{0,1\}^*} P(x) (P(0 | x) - Q(0 | x))^2 = O(- \log c).$$
(the same is true for $(P(1 | x) - Q(1 | x))^2$ since it is equal to
$(P(0 | x) - Q(0 | x))^2$).
Usually this theorem is formulated in terms of computable (semi-)measures.
For the reader's convenience I represent the
sketch of the proof.
Denote the restrictions of $P$ and $Q$ on $\{0,1\}^n$ as $P_n$ and $Q_n$.
The Kullback–Leibler divergence $D(P_n | Q_n)$ is defined as $\sum_{x \in \{0,1\}^n} P_n(x) \log \frac{P_n(x)}{Q_n(x)}$.
Lemma 1 $D(P_1 | Q_1) \ge (P(0) - Q(0))^2$.
For $b \in \{0,1\}, x \in \{ 0,1 \}^n$ denote by $P_{n+1}( \cdot | x)$ the
probabilty distribution on $\{0,1\}$ with probabilities $P_{n+1}(0|x)$ and $P_{n+1}(1|x)$.
Lemma 2
$$D(P_{n+1} | Q_{n+1}) =D(P_n | Q_n) + \sum_{x \in \{0,1\}^n}D(P_n(\cdot |x), Q_n(\cdot |x)).$$
Hence, $D(P_n | Q_n) = \sum_{k=1}^{n} \sum_{x \in \{0,1\}^k} D(P_k(\cdot |x), Q_k(\cdot |x))$.
By using this equality and the first lemma we obtain:
$$D(P_n | Q_n) \ge \sum_{k=1}^{n} \sum_{x \in \{0,1\}^k} P(x) (P(0|x) - Q(0|x))^2.$$
Therefore,
$$\sum_{x \in \{0,1\}^*} P(x) (P(0 | x) - Q(0 | x))^2 = \lim_{n \to \infty}D(P_n | Q_n) = \lim_{n \to \infty}P_n(x) \log \frac{P_n(x)}{Q_n(x)} \le \log c.$$
End of the proof.
Now assume that the condition $Q(x) \ge c P(x)$ only at some subset $T\subseteq \{0,1\}^*$. Assume also that we also have $P(x) \ge c Q(x)$ for every $x \in T$.
Question: Is it true that then
$$ \sum_{x \in T} P(x) (P(0 | x) - Q(0 | x))^2 = O(- \ln c) ? $$
Can you state Solomonoff's full result, including what it means to "predict well", and give a reference?
Just to make sure I understand the notation right: is $P(0|x)=P(x0)/P(x)$?
@fedja, yes. If $P(x)=0$ then $P(0 | x):=0$.
@AryehKontorovich The Solomonoff's theorem is usually formulated for computable measures and semi-measures. You can find it the book of Li and Vitanyi: https://www.springer.com/gp/book/9781489984456.
But in fact the computability is not important: if $Q(x) \ge c P(x)$ for every $x$ then the corresponding sum is less than $O(\log c)$. This result means that $P(0|x)$ is close to $Q(0|x)$ for random (according to measure P) $x$. So $Q$ can predict $P$ well.
|
2025-03-21T14:48:30.047144
| 2020-03-10T21:58:01 |
354655
|
{
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"ACL",
"Fedor Petrov",
"Iosif Pinelis",
"Maximilian Janisch",
"Shahrooz",
"Tony Huynh",
"Vladimir Dotsenko",
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|
Stack Exchange
|
How to prove $e^x\left|\int_x^{x+1}\sin(e^t) \,\mathrm d t\right|\le 1.4$?
Related question asked by me on Math SE a few days ago: How to prove $e^x\left|\int_x^{x+1}\sin(e^t) \,\mathrm d t\right|\le 1.4$?
A few days ago, somebody asked How to prove $ \mathrm{e}^x\left|\int_x^{x+1}\sin\mathrm e^t \mathrm d t\right|\leqslant 2$? on Math StackExchange.
However, this bound does not appear to be sharp so I was wondering how to find the maxima/minima of $$f(x)=e^x\int_x^{x+1}\sin(e^t) \,\mathrm d t$$
or at least how to prove $-1.4\le f(x)\le 1.4$.
Some observations, using the substitution $y=e^t$:
$$f(x)=e^x \int_{e^x}^{e^{x+1}} \frac{\sin(y)}y\,\mathrm dy=g(e^x),$$
where I have defined $$g(z)=z \int_z^{e z} \frac{\sin(y)}y\,\mathrm dy = z (\operatorname{Si}(e z)-\operatorname{Si}(z)).$$
($\operatorname{Si}$ is the Sine integral.)
So the question reduces to: What are the maxima/minima of $g(z)$ for $z\geq 0$ ?
Using the series of $\mathrm{Si}(z)$, we get
$$g(z)=\sum_{k=1}^\infty (-1)^{k-1} \frac{z^{2k}(e^{2k-1}-1)}{(2k-1)!\cdot(2k-1)}$$
and here is a plot of $g(z)$:
Also, notice that $g$ is analytic and $g'(z)=\sin (e z)-\sin (z)+\text{Si}(e z)-\text{Si}(z)$ which might help for the search of critical points (although I don't think that $g'(z)=0$ has closed form solutions).
Your initial trick does not seem to work because $\sin(e^t)$ does not have constant sign.
The proof of the upper bound on Stackexchange gives $\cos(e^x)+1-(cos(e^{x+1}+1)/e$, which is $(\cos(e^x)-e^{-1}\cos(e^x))+(1-1/e)$.
The first term is bounded above by $\sqrt{1+1/e^2}\approx 1.07$, and the second is $\approx 0.63$. This gives an upper bound of $1.70$. Not yet $1.40$.
Did you worked on the first and second derivative of $f(x)$?
It would be better to suppress the erroneous "trick" to not mislead the subsequent readers, probably ?
Thanks to @ACL and @ VladimirDotsenko for noting that the trick is erroneous (now removed)
Integrate by parts:
\begin{align}
\int_x^{x+1}\sin(e^t)dt
& =\int_x^{x+1}e^{-t}d(-\cos(e^t)) \\
& =e^{-x}\cos e^x-e^{-x-1}\cos e^{x+1}-\int_x^{x+1}e^{-t}\cos e^{t}dt\\
& =e^{-x}\cos e^x-e^{-x-1}\cos e^{x+1}-\int_x^{x+1}e^{-2t}d\sin e^{t}\\
& =e^{-x}\cos e^x-e^{-x-1}\cos e^{x+1}-e^{-2(x+1)}\sin e^{x+1}\\
& \hphantom{={}}+e^{-2x}\sin e^x+2\int_x^{x+1}e^{-2t}\sin e^tdt.\end{align}
From here we see that $e^x \int_x^{x+1}\sin(e^t)dt$ is bounded by $1+1/e+O(e^{-x})$ and $1+1/e\approx 1.368$ can not be improved, since both $\cos e^x$ and $-\cos e^{x+1}$ may be almost equal to 1:
if $e^x=2\pi n$ for large integer $n$, then $e^{x+1}=2\pi e n$, we want this to be close to $\pi+2\pi k$, i.e., we want $en$ to be close to $\frac12+k$.
This is possible since $e$ is irrational. Moreover, $e$ is so special number that you may find explicit $n$ for which $en$ is nearly half-integer: $n=m!/2$ for large even $m$ works. Indeed, $e=\sum_{i=0}^{m-1}1/i!+1/m!+o(1/m!)$ yields $em!/2=\text{integer}+1/2+\text{small}$.
I think you switched integration variables from $t$ to $x$, which is somewhat confusing.
Very nice! Integration by parts, of course, to alleviate the oscillations. Somehow, I had not noticed your answer before completing mine. By the way, we both seem to be using that $e/\pi$ is irrational -- do you know if that is so?
@IosifPinelis I think this is still not known to be true, although it follows from Schanuel's conjecture. See https://mathoverflow.net/questions/33817/work-on-independence-of-pi-and-e
@IosifPinelis hm, it looks that I need that $e$ is irrational, not $e/\pi$.
@TonyHuynh : Thank you for the information.
@FedorPetrov: Right, my bad.
The answer https://math.stackexchange.com/questions/3573887/how-to-prove-ex-left-int-xx1-sinet-mathrm-d-t-right-le-1-4 is more accurate.
Here is a method that will allow one to find the exact upper and lower bounds on $g(z)$ over $z>0$ with any degree of accuracy.
Take any real $z>0$. Since
\begin{equation*}
\frac1y=\int_0^\infty dt\,e^{-y t}
\end{equation*}
for any real $y>0$, we have
\begin{align*}
\frac{g(z)}z
&=\int_z^{e z} dy\, \frac{\sin y}y \\
&=\int_0^\infty dt\,\int_z^{e z} dy\,e^{-y t}\sin y \\
&=\int_0^\infty dt\,
\Big( \frac{e^{-t z} (\cos z+t \sin z)}{t^2+1}
-\frac{e^{-e t z} (\cos ez+t \sin ez)}{t^2+1}\Big) \\
&=I_1(z) \cos z+I_2(z)\sin z -I_1(ez) \cos ez-I_2(ez)\sin ez, \tag{1}
\end{align*}
where
\begin{align*}
I_1(z)&:=\int_0^\infty dt\,\frac{e^{-t z}}{t^2+1}, \\
I_2(z)&:=\int_0^\infty dt\,\frac{e^{-t z}t}{t^2+1}.
\end{align*}
Next, letting $c_1$ and $c_2$ denote functions with values in $(0,1)$, we have
\begin{align*}
I_1(z)&=\frac1z\,\int_0^\infty du\,\frac{e^{-u}}{1+u^2/z^2} \\
&=\frac1z\,\int_0^\infty du\,e^{-u}
-\frac1z\,\int_0^\infty du\,\frac{u^2e^{-u}}{z^2+u^2} \\
&=\frac1z-\frac{2c_1(z)}{z^3};
\end{align*}
at the last step here, we used the inequality $z^2+u^2>z^2$ for $u>0$;
similarly,
\begin{align*}
I_2(z)&=\frac1{z^2}-\frac{3c_2(z)}{z^4}.
\end{align*}
So, by (1),
\begin{equation*}
g=h+r,
\end{equation*}
where
\begin{equation*}
h(z):=\cos z-\tfrac1e\,\cos ez
\end{equation*}
and
\begin{equation*}
r(z):=-\frac{2c_1(z)}{z^2}\, \cos z-\frac{3c_2(z)}{z^3}\,\sin z
+\frac{2c_1(ez)}{e^3z^2}\, \cos ez+\frac{3c_2(2z)}{e^4z^3}\,\sin ez
\end{equation*}
is the "remainder",
so that
\begin{equation*}
|r(z)|<\frac{2.1}{z^2}+\frac{3.1}{z^3},
\end{equation*}
which can be made however small if $z$ is large enough.
On the other hand, since $e$ is irrational, we will have
\begin{equation*}
\sup_{z>0}h(z)=-\inf_{z>0}h(z)=1+1/e=1.367\dots
\end{equation*}
(which is somewhat close to your value $1.4$).
So, to compute $\sup_{z>0}g(z)$ and $\inf_{z>0}g(z)$ with any degree of accuracy, it suffices to be able to compute $\sup_{z\in(0,a]}g(z)$ and $\inf_{z\in(0,a]}g(z)$ with any degree of accuracy for any given real $a>0$, which can be done by (say) the interval arithmetic method, using the formula $g(z)=z(\text{Si}(e z)-\text{Si}(z))$ and the monotonicity of the function $\text{Si}$ on each of the intervals of the form $[k\pi,(k+1)\pi]$ for $k=0,1,\dots$.
This can be done with help of Maple in such a way. First, we find the estimated expression explicitly by
a := (exp(x)*int(sin(exp(t)), t = x .. x + 1) assuming x::real;
$ {{\rm e}^{x}} \left( -{\it Si} \left( {{\rm e}^{x}} \right) +{\it Si}
\left( {{\rm e}^{x+1}} \right) \right)
$
In fact, the integral is reduced to another integrals. Next, the asymptotics of $a$ is found. Maple is not able to find this asymptotics directly so the change $x=\log y$ should be used:
asympt(simplify(eval(a, x = log(y))), y, 2);
$-{\frac {\cos \left( y{\rm e} \right) }{{\rm e}}}+\cos \left( y
\right) +O \left( {y}^{-1} \right)
$
Now we return to $x$ by
eval(%, y = exp(x));
$-{\frac {\cos \left( {{\rm e}^{x}}{\rm e} \right) }{{\rm e}}}+\cos
\left( {{\rm e}^{x}} \right) +O \left( \left( {{\rm e}^{x}} \right)
^{-1} \right) $
The rest is as in the Fedor Petrov's answer.
|
2025-03-21T14:48:30.047622
| 2020-03-10T21:58:44 |
354656
|
{
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"authors": [
"Will Sawin",
"https://mathoverflow.net/users/18060"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/354656"
}
|
Stack Exchange
|
Galois action on $\overline{k}$-valued points extends to action on $k$-scheme $X$
Let $X$ be $k$ variety or more genrally a $k$-scheme. Denote the algebraic closure of $k$ by $\overline{k}$. it's a fact that $X(\overline{k}):=Hom(\operatorname{Spec} \ \overline{k}, X)$ as set is dense in the underlying topological space of $X$.
Obviously the Galois group $Gal(\overline{k}/k)$ acts on $X(\overline{k})$ by composition: let $\sigma \in Gal(\overline{k}/k)$ then $\alpha: X(\overline{k}) \to X(\overline{k}), \alpha \mapsto \alpha \circ \operatorname{Spec} \ \sigma$ where $\alpha \in X(\overline{k})$ and $\operatorname{Spec} \ \sigma$ is the spec morphism induced by $\alpha$.
Question: Is it true and if yes then why that the action of $\alpha$ as described above on $\overline{k}$-valued points $X(\overline{k})$ extends in appropriate way to an action on the whole scheme $k$-scheme $X$. By density $X(\overline{k})$ of it's clear that if such extension of action exists it's unique. That is the question is if such extension always or under "weak" assumptions on $X$ always exist and how does it look like. Is there a concrete description of it?
If we consider $X$ as contravariant functor $X: (Sch/k) \to (Set), Y \mapsto Hom_k(Y,X)=X(Y)$ then obviously it suffice to show that the action of $Gal(\overline{k}/k)$ on $X(\overline{k})$ extends to an action on $X(Y)$ for every $k$-scheme $Y$. That's also not obvious.
Does somebody know when under extra assumptions on $X$ such Galois action on $X(\overline{k})$ as described above extends to $X$? My hope is that the density condition could "somehow" allow a continuation (in what sense ever) on the action.
If you're interested in only a single element of the Galois group, and not the full Galois group, you can do this in characteristic $p$ for any power of Frobenius. I suspect this is the only example but wasn't able to prove it (except for $\mathbb A^1$).
No, the Galois action on $X(\overline k)$ corresponds to a trivial action on the scheme $X$. This you can already see if $X=\mathbf A^1_k=\mathop{\rm Spec}(k[T])$ is the affine line.
Then $X(\overline k)=\overline k$, with its obvious Galois action. However, the scheme $X$ has two kind of points: the generic point, and closed points corresponding to maximal ideals of $k[T]$, that is, to irreducible monic polynomials $P(T)$. The $\overline k$-roots of such a polynomial are exchanged by the Galois group, but the maximal ideal is itself invariant.
You have, however, a Galois action on $X_{\overline k}=X\otimes_k \overline k$, which is inherited by the Galois action on $\overline k$.
|
2025-03-21T14:48:30.047832
| 2020-03-11T00:05:14 |
354662
|
{
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"Andrej Bauer",
"Caleb Stanford",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/354662"
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|
Stack Exchange
|
How strong is this set theory?
In the spirit of this related question, consider a set theory with the following axioms:
Axiom of extension:
$$ \forall x \forall y (\forall z (z \in x \leftrightarrow z \in y) \rightarrow x = y) $$
Axiom schema of comprehension:
$$ \exists x \forall y (y \in x \leftrightarrow (\phi y \land C y)) $$
Axiom of construction:
$$ \forall x (\forall y (y \in x \rightarrow C y) \leftrightarrow C x) $$
where $C$ is a new symbol like $\in$ that intuitively represents a kind of "constructibility". Some properties I've deduced are
Every hereditarily finite set exists and is constructible (in particular, the empty set $\varnothing$ exists and is constructible, as are all the finite ordinals).
The quasi-universal set $U$ with $\phi y \equiv \top$ exists, is constructible, and contains itself.
The quasi-Russell set $R$ with $\phi y \equiv y \not\in y$ exists, is not constructible (since that yields the Russell contradiction), and does not contain itself.
The quasi-co-Russell set $R^\ast$ with $\phi y \equiv y \in y$ exists and is constructible iff it contains itself, which seems to be independent of the theory.
Let $y^+ \equiv y \cup \{y\}$ and let $\textsf{inductive } x \equiv \varnothing \in x \land \forall y (y \in x \rightarrow y^+ \in x)$ denote that a set is inductive. $C \varnothing$ and $C y \rightarrow C y^+$, so $\textsf{inductive } U$.
Let $\textsf{natural } x \equiv \forall y (\textsf{inductive } y \rightarrow x \in y)$ denote that a set belongs to every inductive set (i.e. is a natural number). Since $\textsf{inductive } U$, $\textsf{natural } x \rightarrow x \in U$. But $x \in U \rightarrow C x$. Thus $\textsf{natural } x \rightarrow C x$. Instantiate the axiom schema of comprehension with $\phi \equiv \textsf{natural}$. Then $\exists x \forall y (y \in x \leftrightarrow \textsf{natural } y)$. That is, the set of natural numbers $\mathbb{N}$ exists, and moreover is constructible.
My question is this: How strong is this set theory compared to ZFC or other alternative set theories?
Edit: In light of the contradiction pointed out below, one might consider restricting the formulas allowed in the axiom schema of comprehension. For example, we might consider restricting to positive formulas, as in positive set theory.
Edit 2: An interesting alternative system is presented here.
Link to related mathSE post
Hmm if you restrict to positive formulae, you cannot use implications. So do you want to do something less restrictive? Maybe positive formulae but allowing restricted quantifiers (which can capture some implications)?
@user21820 That's certainly an option, too.
Well you'd have to decide what flavour of positive formulae you like. Note that your definition of natural numbers relies on implications, and that this issue is solved in positive set theory via the closure axiom, which you don't have. I'm not sure that adding restricted quantifiers is strong enough, but at least it feels safe enough. =)
Even if $φ$ is restricted to not use $C$, the theory is inconsistent! Here is the simple 2-line proof.
Let $R$ be such that $∀x\ ( x∈R ⇔ C(x) ∧ x∉x )$ by Comprehension.
Then $C(R)$ by Construction. Thus $R∈R ⇔ R∉R$. Contradiction.
Do you mean $C(x)$ in your second line? Otherwise, I'm a bit confused as to how you used comprehension.
@user44191: Yes it was a silly typo because the original axiom had a "$y$". Thanks!
I've added a statement on the possibility of restricting comprehension (e.g. to positive formulas), in case you have any thoughts on that.
@user76284 i think what needs to be done is to remove $Cy$ form comprehension, ban the use of $C$ in $\phi$ and place the pre-condition that $\phi$ only holds of $C$ objects in comprehension.
Suppose first that $\phi$ in the comprehension schema is allowed to use the symbol $C$. Then I claim your theory is inconsistent. Indeed, suppose we had a model $M$ of this theory, and let $M'$ be the submodel consisting of the elements of $M$ that satisfy the predicate $C$ in $M$. Then every definable-in-$M'$ family of elements of $M'$ is also definable in $M$ (by a formula that may need $C$ to restrict quantifiers) and therefore, according to comprehension, given by an element $x$ of $M$. By the axiom of construction, $x$ satisfies $C$ in $M$ and is therefore in $M'$. Thus $M'$ satisfies unrestricted comprehension, which is impossible by Russell's paradox.
Now suppose the $\phi$ in comprehension is not allowed to use $C$, so it's a formula in the language of ZF. That causes a problem in the preceding paragraph, because a definable-in-$M'$ family of elements of $M'$ might not be definable in $M$ without using $C$. Specifically, if the definition in $M'$ involved a quantifier, the corresponding definition in $M$ would need to refer to $C$ in order to make the quantification range only over $M'$ and not all of $M$. Fortunately, that doesn't matter, simply because the formula involved in Russell's paradox doesn't contain any quantifiers.
So even with the weaker, $C$-less-$\phi$ version of comprehension, the theory is inconsistent.
Thanks for this insightful answer. I am wondering if this trick also works to show this theory is inconsistent? I can re-post that question on MathOverflow if preferred (I asked it over 2 years ago).
Thanks for your answer. I've added a statement on the possibility of restricting comprehension (e.g. to positive formulas), in case you have any thoughts on that.
So, in other words, it's the strongest set theory.
Regarding the latter suggestion to restrict $\phi$ to positive formulas. The resulting theory would be consistent relative to positive set theory, but it would be just a redundant re-exposition of a fragment of it. Simply take $C$ to be the predicate of being "equal to itself" i.e.:
$C(x) \leftrightarrow x=x$,
and comprehension would just be positive comprehension, and axiom of construction would be trivially true! Same would apply if for example we've restricted $\phi$ to stratified formulas, we'd only get Quine's New foundations set theory.
Actually that argument would hold for any kind of a theory that has a naive like comprehension axiom with only syntactical restrictions on the defining formula.
I think the only reasonable adjustments to salvage your approach are those you've had with your original theory, the one you raised this theory in connection with, which I think is as strong as $PA$, if you want to strengthen it, you add an axiom of infinity, or even a more risky approach is to add a universal class of all $C$ objects but with changing the bound on parameters in comprehension (the schema of construction) for all of them to be elements of that universal class. But I don't know if those are consistent, and what would be their consistency strength?
|
2025-03-21T14:48:30.048316
| 2020-03-11T03:57:45 |
354673
|
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|
Stack Exchange
|
Middle cohomology of very general hyperplane sections
Let $X$ be a smooth, projective variety over $\mathbb{C}$ of dimension $n$ satisfying the property that for every $i \ge 0$, $H^{i,i}(X,\mathbb{C}) \cap H^{2i}(X,\mathbb{Q})=\mathbb{Q}c_1(\mathcal{O}_X(1))^i$. Is it then true for a very general smooth hyperplane section $Y \subset X$, we have $H^{i,i}(Y,\mathbb{C}) \cap H^{2i}(Y,\mathbb{Q})=\mathbb{Q}c_1(\mathcal{O}_Y(1))^i$? In particular, is it true that for a fixed integer $n$ and a very general complete intersection subvariety $Y$ of $\mathbb{P}^n$, we have $H^{i,i}(Y,\mathbb{C}) \cap H^{2i}(Y,\mathbb{Q})=\mathbb{Q}c_1(\mathcal{O}_Y(1))^i$? We know that this holds true for very general hypersurfaces in $\mathbb{P}^n$.
Any hint/reference will be most welcome.
The question would make more sense with $\Bbb{Z}$ replaced by $\Bbb{Q}$; then the answer (for the last question) would be positive, with some exceptions. See Deligne's Exposé 19 in SGA 7.
@abx Thank you, I have edited the question by replacing $\mathbb{Z}$ by $\mathbb{Q}$.
Now Theorem 1.4 in Deligne's Exposé 19 gives you the best general result you can hope for.
I guess that this is not true even for hypersurfaces. Take a smooth quadric in $Y=Q \subset \mathbb{P}^{3} = X$. Then $h^{1,1}(Q) = b_{2}(Q) = 2$, so the second cohomology cannot be generated by $1$ element.
But, very general hypersurfaces in $\mathbb{P}^3$ of degree at least 4 has Picard number $1$. This is the Noether-Lefschetz theorem.
|
2025-03-21T14:48:30.048438
| 2020-03-11T04:08:40 |
354675
|
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|
Stack Exchange
|
Certain Fourier transforms involving Whittaker function and Bessel functions
I recently meet the following two weird "Fourier transform" questions.
(I), Suppose that $F$ is a $p$-adic field (the same question can be asked over any local field, including $\mathbb{R}$ and $\mathbb{C}$) and $\psi$ be a fixed nontrivial additive character of $F$. Let $W$ be a function on $GL_2(F)$ which satisfies the following 2 conditions:
(1), $W\left(\begin{pmatrix} 1& x \\ &1 \end{pmatrix}
g\right)=\psi(x)W(g)$, for all $x\in F, g\in GL_2(F)$, and
(2), there exists an open subgroup $K$ of $GL_2(F)$ such that
$W(gk)=W(g)$ for all $g\in GL_2(F), k\in K$.
One can think $W$ as a Whittaker function of a smooth representation of $GL_2(F)$. One extend the definition of $W$ to all $2\times 2$ matrices by zero extension, i.e., if $g\in \mathrm{Mat}_{2\times 2}(F)$ and $\det(g)=0$, then define $W(g)=0$.
Let $\mathcal{S}(F^2)$ be the space of Bruhat-Schwartz functions on $F^2$. Fix a nonzero Whittaker function $W$ as above. For $\phi\in \mathcal{S}(F^2)$, define
$$\widehat{\phi}(x_1,x_2)=\int_{F^2}W\left( I_2+\begin{pmatrix} x_1\\ x_2 \end{pmatrix}\begin{pmatrix} y_1 & y_2\end{pmatrix}\right)\phi(y_1,y_2)dy_1 dy_2.$$
Here $I_2$ is the $2\times 2$ identity matrix and $W$ is omitted from the notation $\widehat{\phi}$.
Question (I): Do we know that $\widehat \phi\in \mathcal{S}(F^2)$ and $\phi\mapsto \widehat\phi$ defines an isomorphism?
This looks like a Fourier transform. Also, it is easy to see that the function $x_1\mapsto \widehat{\phi}(x_1,0)$ has compact support. But I don't know how to prove the above statement in general.
(II) The second question involves a Fourier transform in a similar flavor, but not the same with the above one. Suppose that $k$ is a finite field and $\psi$ is a nontrivial additive character of $k$. Let $\mathscr{B}$ be the set of functions $B$ on $GL_2(k)$ such that
$$B\left(\begin{pmatrix}1&x\\ &1 \end{pmatrix} g \begin{pmatrix}1&y\\ &1 \end{pmatrix}\right)=\psi(x)\psi(-y)B(g), \forall x,y\in k, g\in GL_2(k).$$
One can think such a function $B$ as a Bessel function.
For $B\in \mathscr{B}$, define
\begin{align*}&\mathcal{F}_B\left(\begin{pmatrix} x_{11} & x_{12}\\ x_{21}& x_{22} \end{pmatrix}\right)\\
&\qquad :=\sum_{s_1,s_2,r_1,r_2\in k} \psi(x_{22}r_1+x_{21}r_2+s_1)B\left(\begin{pmatrix} x_{11} & x_{12}\\ x_{21}& x_{22} \end{pmatrix} \left(I_2+\begin{pmatrix} s_2\\ -s_1 \end{pmatrix}\begin{pmatrix} -r_1 & r_2\end{pmatrix} \right) \right),
\end{align*}
where if $\det(g)=0$, we view $B(g)=0.$
It is not hard to show that $\mathcal{F}_B\in \mathscr{B}$. The negative signs in the definition of $\mathcal{F}_B$ is to make sure $\mathcal{F}_B\in \mathscr{B}.$
Question (II): Is it true that $B\mapsto \mathcal{F}_B$ is a bijection?
If the answer of (II) is affirmative, I am also wondering if there is a local field analogue.
Question (III): Are there generalizations to $GL_n$?
Thanks in advance.
|
2025-03-21T14:48:30.048741
| 2020-03-11T04:23:33 |
354677
|
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|
Stack Exchange
|
Is every simplicial spectrum equivalent to an abelian group simplicial spectrum?
I wonder if every simplicial $S^1$-spectrum stable equivalent to an abelian group simplicial $S^1$-spectrum? It seems that I could use the stable Dold-Kan on the heart and use Postnikov towers?
If I understand correctly your question, the category of simplicial $S^{1}$-spectrum is the stable category with respect to the suspension functor i.e., it is the category of spectra. The category of abelian group simplicial $S^{1}$-spectrum is the stabilization category of the model category of simplicial abelian groups with respect to the "suspension" functor, this category is the category of $HZ$-spactrum i.e the category of spectrum which have the structure of $HZ$-module. The answer to your question (if my interpretation is correct) is no.
More precisely (if GSM's interpretation of your question is correct, I'm also not 100% sure what your objects are), what goes wrong with your argument sketch is that the objects in the heart are $HZ$-modules, but the way they're glued together (the k-invariants in the postnikov tower) doesn't respect that, i.e. there are non-$HZ$-linear maps between $HZ$-modules.
@Achim Krause I completely agree with you.
|
2025-03-21T14:48:30.048849
| 2020-03-11T09:07:13 |
354683
|
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|
Stack Exchange
|
Integrating the Riemann curvature tensor over a singular 2-disc
There's a classic characterization of the Riemann curvature tensor. Say, take a Riemann metric on an open subset $U$ of $\mathbb R^n$. Given a point $p \in U$ and two vectors $v,w \in T_p U$ you compute the holonomy around the boundary of the parallelogram $\{p + t_1 v + t_2 w : 0 \leq t_1 \leq a, 0 \leq t_2 \leq b\}$. The second-order Taylor expansion of this holonomy, as a function of $a$ and $b$ (centred at $a=b=0$) is given by:
$$Id_{T_p U} + ab R_p(v,w)$$
If you think carefully about this, what it tells you is that if you have any map $D^2 \to N$ where $N$ is a Riemann manifold, you can compute the holonomy around the boundary of the disc by an appropriate "integral" on the interior of the disc, of the Riemann curvature tensor. It's not an integral in the traditional sense, as you are performing a limit of a system of composites of functions. But in spirit, it is reasonably-close to an integral.
Other than being a limit of a composite of a large number of functions, a technical issue is the choice of basepoint and ensuring you are using an appropriate transport of the Riemann tensor back to one tangent space.
I imagine this observation has been written up somewhere in the literature. Who has proven this theorem, and where might it appear? Does this type of groupoid-y integration have a standard name?
I think this can be extracted from section 6.2 of "Gromov's almost flat manifold" by Buser and Karcher. They give an upper bound on holonomy in terms of norm of the curvature tensor and the area of the disk.
There is "How the curvature generates the holonomy of a connection in an arbitrary fibre bundle" by Helmut Reckziegel and Eva Wilhelmus in Results in Mathematics 46 (2006). I'm not sure that is the first place where this is proved properly, but it does do so nicely (in a more general context).
Thanks, I'll have a look when I'm back in my office and (hopefully) inside their paywall.
@RyanBudney: Googling the title of the paper gave me a pdf version.
Everything I find on Google is behind a paywall, unfortunately. Anyhow, I'll head into the office today.
|
2025-03-21T14:48:30.049033
| 2020-03-11T09:50:15 |
354686
|
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|
Stack Exchange
|
antiderivative always exists?
It is well known that many real valued real functions are not Riemann integrable on subsets of $\mathbb{R}$, but formally an antiderivative may still exist. May I see an example of a function having no antiderivative? Or does any function (without additional hypothesis) always have an antiderivative?
Edit: Does a continuous function always satisfy Choquet's characterization? If yes, take $f$ differentiable and such that $f'$ is bijective. In order to explicitly compute the antiderivative of $f'^{-1}$ I find myself needing that $f'$ is also differentiable. Could such an additional hypothesis be ruled out somehow?
(1) $f(x) = 0$ for $x \ne 0$, $f(0) = 1$ has no antiderivative (and it is Riemann-integrable, by the way). (2) Every continuous function clearly has an anti-derivative. (3) Using the Riemann–Stieltjes integral and integration by parts, we have $$\begin{aligned} \int (f')^{-1}(s) ds & = \int x df'(x) \& = xf'(x) - \int f'(x) dx = xf'(x) - f(x) + C \& = s(f')^{-1}(s)-f((f')^{-1}(s)) + C.\end{aligned}$$
Writing $df'(s)$ in the first line does not imply that $f'$ is differentiable?
No, it is perfectly sufficient to have $f'$ strictly monotone (by assumption) and continuous (by monotonicity and intermediate value property). This becomes fairly straightforward if written as the limit of Riemann–Stieltjes sums.
This is a good question, but more appropriate at MSE.
There are elementary necessary conditions on real functions to be the derivative of a real function. For instance, Darboux's theorem states that a derivative must satisfy the intermediate value theorem. The Baire category theorem implies that a derivative is continuous on a dense set. It allows to find functions with no antiderivative easily.
There is also a characterization due to Choquet :
A real function $f$ admits an antiderivative if and only if there exists an homeomorphism $\phi$ such that $f \circ \phi$ is of Baire class 1 and satisfies the intermediate value theorem.
Reference : French Wikipedia page on Gustave Choquet
Does a continuous function always satisfy Choquet's characterization? If yes, take $f$ differentiable and such that $f'$ is bijective. In order to explicitly compute the antiderivative of $f'^{-1}$ I find myself needing that $f'$ is also differentiable. Could such an additional hypothesis be ruled out somehow?
@Hair80 Assuming "Baire first category" means "Baire class 1", then yes, by definition continuous function have Baire class 0 and hence also class 1. Of course, continuous functions satisfy IVT, so we can take $\phi$ the identity.
Any distribution $T$ on the real line has an anti-derivative, i.e. there exists a distribution $S$ such that $$S'=T\tag{$\ast$}.$$ Here is a constructive proof: with a given $T$,
define the distribution $S$ by
$$
\langle S, \phi\rangle_{\mathscr D',\mathscr D }=-\langle T, \psi_\phi\rangle_{\mathscr D',\mathscr D },
\quad \text{with}\quad (\psi_\phi)(x)=\int_{-\infty}^x \phi(t) dt-\chi_0(x)\int_{\mathbb R} \phi(t) dt,
$$
where the function $\chi_0$ smooth, equal to 1 on $[\max\text{supp} \phi,+\infty)$ and equal to $0$ on $(-\infty, \min \text{supp}\phi]$.
This makes sense since the function $\psi_\phi$ is smooth (obvious) and also compactly supported: if $x\ge \max\text{supp} \phi$, then
$$
(\psi_\phi)(x)=(1-\chi_0(x))\int_{\mathbb R} \phi(t) dt=0.
$$
If $x\le \min\text{supp} \phi$, then
$
(\psi_\phi)(x)=-\chi_0(x)\int_{\mathbb R} \phi(t) dt=0.
$
We check now $S'=T$. Indeed we have
$$
\langle S', \phi\rangle_{\mathscr D',\mathscr D }=
-\langle S, \phi'\rangle_{\mathscr D',\mathscr D }
=\langle T, \psi_{\phi'}\rangle_{\mathscr D',\mathscr D }.
$$
We note that $\int_\mathbb R\phi'(t) dt =0$, so that
$
(\psi_{\phi'})(x)=\phi(x)
$
and thus, we find
$$
\langle S', \phi\rangle_{\mathscr D',\mathscr D }=\langle T, \phi\rangle_{\mathscr D',\mathscr D },\quad \text{i.e.}\quad S'=T.
$$
A simple addendum: if $S_1, S_2$ are two antiderivatives, the $S_2-S_1=\text{constant}.$
|
2025-03-21T14:48:30.049278
| 2020-03-11T09:52:55 |
354687
|
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|
Stack Exchange
|
Well-spread (weak Sidon) sets
Does anybody have access to the paper
A. Kotzig: On well spread sets of integers, 1972,
or does anybody know the proof of $\sigma^*(n)\geq 4+\binom{n-1}2$ for $n\geq7$
(as cited in Marr,Wallis: Magic Graphs, Springer 2013, Th. 2.15)
Notions: A set $A=\{a_1,\dots,a_n\}$ of integers is a well-spread (weak Sidon) set of cardinality $n$, if $a_i+a_j \neq a_k+a_\ell$ whenever these four elements of $A$ are distinct. Then $\sigma(A):= \max A - \min A +1$ is the size of $A$, and $\sigma^*(n)$ is the minimal $\sigma(A)$ taken over all well-spread sets $A$ of cardinality $n$.
Bibliographic info: A. Kotzig, On well spread sets of integers. Tech. Report CRM-161, Centre Res. Math., Université de Montréal (1972), 83 pp. – maybe it's worth contacting the U of Montreal.
For $n=3$ we have a weak Sidon set of spread 3, which looks to be less than 5.
The lower bound on $\sigma^*(n)$ holds for $n\geq7$.
Suppose that $A\subset[1,l]$ is weak Sidon; we show that $l>\binom{n-1}2$, which is very close to your estimate.
There are $n(n-1)/2$ positive differences of the form $a_1-a_2$ with $a_1,a_2\in A$. All these differences are in $[1,l-1]$, but we cannot conclude that $l-1\ge n(n-1)/2$ because some of the differences can coincide; namely, we have $a_1-a_2=a_3-a_4$ if and only if exactly one of the following holds: (i) $a_1=a_4$ and $a_2,a_1=a_4,a_3$ is a three-term arithmetic progression; (ii) $a_2=a_3$ and $a_1,a_2=a_3,a_4$ is a three-term arithmetic progression. We associate with every equality of the form $a_1-a_2=a_3-a_4$ the middle term of the corresponding progression.
The crucial observation is that for every element $a\in A$, the integer $2a$ is the middle term of at most one progression: for an equality of the form $a_1+a_2=b_1+b_2=2a$ would contradict the weak Sidon property. Therefore, there are at most $n$ equalities of the form $a_1-a_2=a_3-a_4$. Removing for every such equality the corresponding difference from the count, we get $n(n-1)/2-n\le l-1$; that is,
$$ l \ge \frac12(n^2-3n+2) = \frac12(n-1)(n-2)=\binom{n-1}2. $$
Middle terms of these 3-term progressions may take only $n-2$ possible values, not $n$. Therefore $l\geqslant (n-1)(n-2)/2+2$.
Yes, but this lower bound misses Kotzig's Claim by 2. How to get the improvement?
@BerndM: see Fedor Petrov's comment.
@Seva: I saw it, it is still 2 less than Kotzig's claim.
|
2025-03-21T14:48:30.049456
| 2020-03-11T10:20:15 |
354690
|
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|
Stack Exchange
|
Formality and symplectic forms on a smooth manifold
I saw one paper which asks this question.
"Let $(M,\omega,J)$ be a Kähler manifold. Then does $M$ admit a symplectic structure $\sigma$ of non-hard Lefschetz type?".
I was wondering whether I could ask the similar question but instead of "hard Lefschetz property" I will replace with "formality".
That is:
Let $(M,\omega, J)$ be a Kähler manifold. Is there a symplectic structure $\sigma$ on $M$ which is of non-formal type?
Note the first question was answered by Yunhyung Cho https://arxiv.org/pdf/1403.1418.pdf.
|
2025-03-21T14:48:30.049529
| 2020-03-11T10:41:07 |
354692
|
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|
Stack Exchange
|
The Yoneda pairing, hypercohomology, and cup product
Let $\mathcal{F}$ and $\mathcal{G}$ be coherent analytic sheaves on $\mathbb{P}^n$. Let $\mathcal{F}_\bullet$ be a locally free resolution of $\mathcal{F}$. In Principles of Algebraic Geometry by Griffiths and Harris, $\text{Ext}^p(\mathcal{F},\mathcal{G})$ is defined as the hypercohomology of the complex $\mathcal{Hom}(\mathcal{F}_\bullet,\mathcal{G})$, i.e., the cohomology of the complex $\bigoplus_{p=k+\ell} C^k(\mathfrak{U},\mathcal{Hom}(\mathcal{F}_\ell,\mathcal{G}))$, see pages 705 and 446. Here $C^\bullet(\mathfrak{U},\mathcal{Hom}(\mathcal{F}_\ell,\mathcal{G}))$ denotes the Čech complex with respect to some affine open cover $\mathfrak{U}$ of $\mathbb{P}^n$.
If I understand correctly, the Yoneda pairing $$\text{Ext}^p(\mathcal{F},\mathcal{G}) \times \text{Ext}^q(\mathcal{G},\mathcal{H}) \rightarrow
\text{Ext}^{p+q}(\mathcal{F},\mathcal{H})$$ should then be induced by the cup product in Čech cohomology. However, I fail to see precisely how this works out.
Edit: To clarify:
What I am primarily interested in is how the Yoneda pairing can be expressed in terms of the concrete representatives when $\text{Ext}$ is realized as hypercohomology. In [HL, Section 10.1.1], there is an explicit cup product on hypercohomology, which then is said to induce a product $Ext^i(F^\bullet,G^\bullet)\otimes Ext^j(E^\bullet,F^\bullet) \to Ext^{i+j}(E^\bullet,G^\bullet)$. It is then also stated that
"If we interpret $Ext^i(E^\bullet,F^\bullet)$ as $Hom_{\mathcal D}(E^\bullet,F^\bullet[i])$, where $\mathcal{D}$ is the derived category of quasi-coherent sheaves, then the cup product for Ext-groups is simply given by composition." To me, it is not clear neither how the cup product induces the product on $\text{Ext}$, nor why this product coincides with composition in the derived category. Any references to where this is discussed in more detail, or hints on how to prove this would be welcome.
[HL] Huybrechts, Lehn: The Geometry of Moduli Spaces of Sheaves
I think these questions become clearer if you go full derived and see $\mathrm{Ext}^p(\mathcal{F},\mathcal{G})$ as the cohomology of the internal hom in the derived category of quasi-coherent analytic sheaves.
Suppose Č(U)→Hom(Q(F),G)[p] and Č(V)→Hom(Q(G),H)[q] represent
elements in Ext^p(F,G) and Ext^q(G,H).
Here Č(U) denotes the Čech chains for an open cover U and Q(-) is the functor
that computes appropriate replacements (often cofibrant), e.g., locally free resolutions.
Choose a common refinement W of U and V and restrict the above maps
along the morphisms Č(W)→Č(U) and Č(W)→Č(V).
Tensor the two resulting maps together and restrict along the diagonal map,
obtaining a map Č(W)→Hom(Q(F),G)⊗Hom(Q(G),H)[p+q]
The remaining problem is the mismatch between G and Q(G).
The canonical map Q(G)→G goes in the wrong direction,
and so does the induced map Hom(Q(F),Q(G))→Hom(Q(F),G).
This prevents us from composing things in the most obvious way.
If one is not interested in specific models for representatives,
one can observe that Hom(Q(F),Q(G))→Hom(Q(F),G) becomes an isomorphism
in the derived category, so can be tensored with Hom(Q(G),H)
and then inverted and composed with the other map,
yielding a map Č(W)→Hom(Q(F),H)[p+q], as desired.
Otherwise, working in (say) the projective or flat model structure
one can explicitly lift the cofibration 0→Č(U)
against the acyclic fibration
Hom(Q(F),Q(G))[p]→Hom(Q(F),G)[p],
obtaining a new representative Č(U)→Hom(Q(F),Q(G))[p],
and then apply the second paragraph above to get
the desired map Č(W)→Hom(Q(F),H)[p+q].
|
2025-03-21T14:48:30.049772
| 2020-03-11T10:45:35 |
354693
|
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"Jochen Glueck",
"YCor",
"https://mathoverflow.net/users/102946",
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"sort": "votes",
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|
Stack Exchange
|
Continuity of local spectral radius
Let $H$ be a complex Hilbert space and let $T \in \mathcal{B}(H)$ be a linear, bounded operator. Given $x \in H$ we define its local spectral radius as $$r_T(x) = \limsup\limits_{n\rightarrow\infty} \|T^nx\|^{1/n}.$$
Now, let $(x_n)_n$ be a sequence converging to $x$. I've been looking for any results to assure the convergence of the local spectral radius, id est, $r_T(x_n) \rightarrow r_T(x)$. Obviously this convergence cannot happen for every operator and for every vector, but I would like to know if anyone knows any condition to assure this convergence.
Thank you very much.
I don't know if the terminology is standard but if not, I'd rather call it "pointwise spectral radius".
@YCor: Actually, the terminology is standard. There's an entire field called "local spectral theory" that deals with local - in the sense of "pointwise" - spectral properties.
|
2025-03-21T14:48:30.049867
| 2020-03-11T11:24:39 |
354694
|
{
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"authors": [
"Aleksei Kulikov",
"Lucia",
"https://mathoverflow.net/users/104330",
"https://mathoverflow.net/users/38624"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/354694"
}
|
Stack Exchange
|
Bounding a Fourier coefficient of a non-negative periodic function in terms of its $L^2$-norm
This question is motivated by the earlier MO question: Show that $(\sum_{k=1}^{n}x_{k}\cos{k})^2+(\sum_{k=1}^{n}x_{k}\sin{k})^2\le (2+\frac{n}{4})\sum_{k=1}^{n}x^2_{k}$ .
It is a cleaned up asymptotic version of that question.
Let $f$ be a non-negative function, periodic with period $1$, and square integrable on ${\Bbb R}/{\Bbb Z}$. Is it true that
$$
|{\widehat f}(1)|^2 = \Big| \int_0^1 f(x) e^{-2\pi ix} dx \Big|^2 \le \frac 14 \int_0^1 f(x)^2 dx \ \ ?
$$
Equality is attained for example when $f(x) = \max(0, \cos(2\pi x))$.
Note that $|\widehat f(1)| =|\widehat f(-1)|$ and, since $f$ is non-negative, $|\widehat f(1)| \le \widehat f(0)$. Therefore
$$
\int_0^1 f(x)^2 dx = \sum_n |\widehat f(n)|^2 \ge 3 |\widehat f(1)|^2,
$$
so that the estimate holds with $1/3$ in place of $1/4$. There is a lot of scope to improve this argument, and with a more careful application of Bessel's inequality I could get the constant $1/4+1/4\pi$. But the claimed inequality looks very clean, and I wonder if (i) it is true!, (ii) is known in some context, and (iii) (hopefully) has an elegant proof?
Your conjecture is indeed correct. The proof is based on the following simple yet powerful trick I learned many years ago: $|z| = \sup_{|v| = 1} \Re (zv)$. Therefore it is enough to only bound from above $\Re \left(\int_0^1 vf(x)e^{-2\pi ix}dx\right)$ for all $v\in \mathbb{T}$. And now it is clear by Cauchy-Schwarz that $f$ should be proportional to $\max(0, \Re(ve^{-2\pi ix}))$, and all such functions gives us $\frac{1}{4}$ ($f$ from the OP corresponds to the choice $v = 1$).
Cool! Missed that completely; nice proof!
And your proof works for the other problem too of course.
@Lucia hmm, are you sure?.. I can see it only in asymptotic sense, with $n(1/4 + \varepsilon)$
Right, I did have in mind asymptotic version ...
|
2025-03-21T14:48:30.050272
| 2020-03-11T11:55:11 |
354695
|
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"F.Abellan",
"Gerrit Begher",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/354695"
}
|
Stack Exchange
|
On equivalences of cartesian fibrations
Let $p:X^{\natural} \to S$, $q:Y^{\natural} \to S$ be two cartesian fibrations of simplicial sets (where the marking is given by cartesian edges) and assume that we are given an equivalence of cartesian fibrations $X^{\natural} \to Y^{\natural}$.
Given a marking on $S$ we define the marked simplicial set $X^{\dagger}$ (resp. $Y^{\dagger}$) by declaring and edge to be marked if and only if it is marked in $X^{\natural}$ and its image under $p$ is marked in $S$.
My question is the following: Is the induced map $X^{\dagger} \to Y^{\dagger}$ a weak equivalence of marked simplicial sets (over the point)?
What is a marking?
@GerritBegher a choice of 1-simplices that contains all degenerate edges.
Yes. Since $X^{\natural} \to S$ and $Y^{\natural} \to S$ are both cartesian fibrations they are fibrant and cofibrant objects in the cartesian model structure over $S$, which is a simplicial model structure. If two such objects are weakly equivalent then there must exist maps $f:X^{\natural} \to Y^{\natural}$ and $g: Y^{\natural} \to X^{\natural}$ over $S$ with homotopies $\eta:(\Delta^1)^{\sharp} \times X^{\natural} \to X^{\natural}$ and $\tau:(\Delta^1)^{\sharp} \times Y^{\natural} \to Y^{\natural}$ (again over $S$) from $g \circ f$ and $f \circ g$ to the respective identities. These maps and homotopies imply in particular that $X^{\natural}$ and $Y^{\natural}$ are equivalent as marked simplicial sets over the point. This remains true if we remove some of the marking, as long as we do it in a way that depends only on some marking of $S$. Indeed, as long as $f,g,\eta$ and $\tau$ remain marking-preserving they will continue to determine an equivalence.
|
2025-03-21T14:48:30.050401
| 2020-03-11T12:50:51 |
354696
|
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"Iosif Pinelis",
"T. Tharoor",
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"https://mathoverflow.net/users/36721"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/354696"
}
|
Stack Exchange
|
Convergence of estimator given by a fixed point
Let $X$ be a non-negative random variable with cdf $F$ and define
$$G(s) = E[\max(0,u(X)-sX)],$$ where $u$ is some real function.
Let $s_0$ be the unique fixed point of $G$.
Now let $X_1,\dots,X_t$ be a sequence of samples drawn independantly from $F$.
Let
$$G_t(s) = \frac{1}{t}\sum_{i=1}^t \max(0,u(X_i)-sX_i)$$
and let $s_t$ be the unique fixed point of $G_t$.
Is it true that $s_t$ converges almost surely to $s_0$, and is it possible to say something about the speed of convergence? If not, how could I construct a converging estimator of $s_0$?
$\newcommand\N{\{1,2,\dots\}}
\newcommand\NN{\{1,2,\dots,\infty\}}
\newcommand\si{\sigma}
\newcommand{\ep}{\varepsilon}$
Suppose that $E u(X)_+<\infty$, where $x_+:=\max(0,x)$. Then the convergence takes place, with the rate $O(1/\sqrt t)$ (as $t\to\infty$) if we also assume that $EX<\infty$, $E u(X)_+^a<\infty$ for some $a>2$, and
$$\si(s):=\sqrt{Var\,(u(X)-s X)_+}\ne0$$
for any real $s$.
Indeed, let
$$G_\infty(s):=G(s)=E(u(X)-sX)_+,$$
and then let
$$H_t(s):=G_t(s)-s$$
for all $t\in\NN$, so that $EH_t(s)=H_\infty(s)$. For each $t$, the function $G_t$ is nonincreasing and hence $H_t$ is strictly decreasing. Moreover, by dominated convergence, $G_\infty$ is continuous and hence $H_\infty$ is continuous. Also, $H_t$ is obviously continuous for $t\in\N$. Further, $H_t(0)\ge0>-\infty=H_t(\infty-)$. So, for each $t\in\NN$, the function $H_t$ has a unique zero in $[0,\infty)$, that is, the function $G_t$ has a unique fixed point, say $s_t$, in $[0,\infty)$.
Also, using the condition $E u(X)_+^a<\infty$ with $a>2$ and, again, the dominated convergence theorem, we see that, for each $b\in[0,a]$, $E(u(X)-sX)_+^b$ is continuous in $s$ and hence bounded in $s$ in any compact interval.
So, by the Lyapunov--Bikelis inequality (see e.g. inequality (1)),
\begin{equation}
P(Z_t(s)>z)-P(Z>z)\to0\tag{1}
\end{equation}
uniformly over all real $z$ and all $s$ in any compact interval, where $Z\sim N(0,1)$ and
$$Z_t(s):=\frac{H_t(s)-H_\infty(s)}{\si(s)/\sqrt t}.$$
So, for any real $z$ and any $t\in\N$
\begin{align*}
P(s_t>s_\infty+z/\sqrt t)&=P(H_t(s_t)<H_t(s_\infty+z/\sqrt t)) \\
&=P(H_t(s_\infty+z/\sqrt t)>0) \\
& =P\Big(Z_t(s_\infty+z/\sqrt t)>\frac{-H_\infty(s_\infty+z/\sqrt t)}{\si(s_\infty+z/\sqrt t)/\sqrt t}\Big) \\
& =P\Big(Z>\frac{-H_\infty(s_\infty+z/\sqrt t)}{\si(s_\infty+z/\sqrt t)/\sqrt t}\Big)+o(1),
\end{align*}
by (1).
Next, again by dominated convergence, the right derivative of $H_\infty$ at any real $s$ is $-\mu^+(s)-1$, where
$$\mu^+(s):=EX1_{u(X)-sX>0}.$$
So, for any real $z>0$, $H_\infty(s_\infty+z/\sqrt t)=H_\infty(s_\infty+z/\sqrt t)-H_\infty(s_\infty)\sim-(1+\mu^+(s_\infty))z/\sqrt t$.
Also, using again the mentioned continuity of $E(u(X)-sX)_+^b$ in $s$ for each $b\in[0,a]$, we see that $\si(s_\infty+z/\sqrt t)\to\si(s_\infty)$.
We conclude that for any real $z>0$
\begin{align*}
P(s_t-s_\infty>z/\sqrt t)&\to P\Big(Z>\frac{(1+\mu^+(s_\infty))z}{\si(s_\infty)}\Big).
\end{align*}
Similarly, for any real $z>0$
\begin{align*}
P(s_t-s_\infty<-z/\sqrt t)&\to P\Big(Z<-\frac{(1+\mu^-(s_\infty))z}{\si(s_\infty)}\Big),
\end{align*}
where $\mu^-(s):=EX1_{u(X)-sX\ge0}$.
Thus indeed, $s_t$ converges to $s_\infty$ at the rate $O(1/\sqrt t)$.
Let us also show that $s_t\to s_\infty$ almost surely (a.s). Take any real $\ep>0$. Since the function $H_\infty$ is strictly decreasing, we have
\begin{equation*}
H_\infty(s_\infty-\ep)>H_\infty(s_\infty)>H_\infty(s_\infty+\ep).
\end{equation*}
By the strong law of large numbers, $H_t(s_\infty\pm\ep)\to H_\infty(s_\infty\pm\ep)$ a.s. So, for some random variable $T_\ep$ such that $P(T_\ep\in\N)=1$ and all natural $t$, on the event $\{t\ge T_\ep\}$ we have
\begin{equation*}
H_t(s_\infty-\ep)>H_\infty(s_\infty)=0=H_t(s_t)=H_\infty(s_\infty)>H_t(s_\infty+\ep),
\end{equation*}
which yields
\begin{equation*}
s_\infty-\ep<s_t<s_\infty+\ep,
\end{equation*}
because the function $H_t$ is strictly decreasing. Thus indeed, $s_t\to s_\infty$ a.s.
Thank you for your answer. Correct me if I'm wrong but this does not imply almost sure convergence, right?
Moreover, as far as I understand, $(1)$ is the central limit theorem, why do you need a stronger result?
@pyth : (i) It is significantly easier to get almost sure convergence (by using the strong law of large numbers) than the rate of convergence (for which you need a central limit theorem, and then the rate can be only given for the convergence in distribution). I decided to focus on the harder part of the problem. (ii) (1) is not a simple central limit theorem: we need uniformity in $s$, which is not provided by the standard central limit theorem.
From the strong law of large numbers, for all $s$ $H_t(s)\to H_\infty(s)$ almost surely. But from there I do not see how to pursue.
I have added a proof of the almost sure convergence.
|
2025-03-21T14:48:30.050791
| 2020-03-11T13:17:02 |
354699
|
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"url": "https://mathoverflow.net/questions/354699"
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|
Stack Exchange
|
Starting point of roundtrip coloring in connected graphs
This is a subquestion for an older question about a certain kind of greedy coloring.
Let $G = (V,E)$ be a finite, connected, simple, undirected graph. By a roundtrip of $G$ we mean a map $r:\{0,\ldots,n\} \to V$ for some $n\in\mathbb{N}$ with the following properties:
$r$ is surjective,
$\{r(k), r(k+1)\} \in E$ for all $k \in \{0, \ldots, n-1\}$, and
$r(0) = r(n)$.
An easy inductive argument shows that we can select $n$ such that $n \leq 2|G|$.
Given a roundtrip $r$ and a vertex $v\in V$, we assign a roundtrip coloring $c_{r,v}:V\to\mathbb{N}$ of $G$, starting at $v$ in the following manner: Start at $v$, proceed along the roundtrip $r$ and greedy-color every uncolored vertex along the way, until every vertex has been colored. Let $\chi(r,v)$ be the number of colors used.
Question. What is an example of a connected graph $G=(V,E)$ and $v\neq v'\in V$ such that $\chi(r,v)\neq \chi(r,v')$?
My answer to your other question can be modified to give a family of graphs and a roundtrip such that the difference $\chi(r,v)-\chi(r,v')$ is linear in the number of vertices. Let $G$ be a graph with vertex set $\{v_i,v_i'\mid 1 \leq i \leq n\}$ where $n$ is an even integer and edges $v_iv_j'$ for $i \neq j$ and $v_iv_{i+1}$ and $v_i'v_{i+1}'$ for $1 \leq i < n$.
In other words, take a complete bipartite graph, delete a perfect matching, and add the edges of a spanning path in each of the bipartite parts.
Let $r$ be the roundtrip
$$
v_1,v_2,v_1',v_2',v_3,v_4,v_3',v_4', \dots, v_{n-1},v_n,v_{n-1}',v_n',v_1.
$$
Then $\chi(r,v_1) = n$ (like in my answer to your other question) whereas $\chi(r,v_1') = 4$.
|
2025-03-21T14:48:30.050922
| 2020-03-11T13:28:26 |
354700
|
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|
Stack Exchange
|
Relative degree of a prime over a number field (notation from Algebraic Number Fields from Gerald J. Janusz)
I´m working with "Algebraic Number Fields" from Gerald J. Janusz (1. edition from 1973) and I have a question about his notation.
In chapter IV proposition 4.5 he states if K is an algebraic number field and S is the set of primes of K which have relative degree one over Q then S is an infinte set.
Up to this point Janusz just defined the relative degree of a prime over R, with R as a ring of integers.
Can someone tell me, if there is a difference between the relative degree of a prime over a number field and the relative degree of a prime over a ring of integers?
Thanks in advance
Julian
No, Janusz clearly refers to the relative degree over the ring of integers of $\mathbb Q$, that is $\mathbb Z$. It is very common in algebraic number theory to speak of something relative to a number field, when in reality it means relative to its ring of integers. For instance, we commonly talk of ideals of $K$, which should only mean $(0)$ and $K$, when we actually mean ideals of $\mathcal O_K$.
|
2025-03-21T14:48:30.051025
| 2020-03-11T13:50:49 |
354702
|
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"André Henriques",
"YCor",
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"user43326"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/354702"
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|
Stack Exchange
|
Differential version of $G\mapsto H^3(G,\mathbb Z)$?
Let $\mathit{cLieGrp}^{\mathrm{inj}}$ be the category of compact connected Lie groups, and injective continuous group homomorphisms.
Is there a reasonable functor (some kind of degree $3$ differential cohomology?)
$$
F:\mathit{cLieGrp}^{\mathrm{inj}} \to \mathit{Ab}
$$
to the category of abelian groups with the following features:
(1) On the full subcategory of compact semisimple Lie groups, this functor agrees with $G\mapsto H^3(G,\mathbb Z)$ (integral 3rd cohomology of the underlying manifold of $G$).
(2) On the full subcategory of compact abelian Lie groups (tori), this functor agrees with $T\mapsto H^4(BT,\mathbb R)$.
Motivation for the question:
The functor $$G \mapsto \{ \text{full WZW models with gauge group } G\}$$ seems to behave like the functor $F$ above (except that the set of full WZW models with gauge group $G$ needs to be group-completed in order to make it into an abelian group).
Does "injective maps" mean "injective continuous group homomorphisms"? or "injective continuous maps" (or anything else)?
Well, $H^3(T,Z)=Z$ and $H^4(BT,R)=R$, this looks impossible...
By the way I meant $T=S^1$ in the above.
@YCor. Yes: I meant "injective continuous group homomorphisms".
@user43326. For $T:=S^1$, we have $H^3(T,\mathbb Z)=0$. It's just the cohomology of the underlying manifold (and as the manifold is 1-dimensional, there is no $H^3$).
OK, I thought you meant "group cohomology". But in that case it still doesn't work since $H^4(BT,R)$ is still different from $H^3(T,Z)$. Or here does $B$ mean something other than the classifying space?
@user43326. The classes of groups covered under cases (1) and (2) are disjoint. So there is no contradiction. The group $S^1$ is not semisimple.
Oh, I.see. There still is a problem. The inclusion of the maximal torus of a semi-simple Lie group is a morphism in your category, right? So such a cohomology theory can't possess Becker-Gottlieb type transfer.
@user43326 Yes, the inclusion $\iota:T \to G$ of a torus into a semi-simple Lie group is a morphism in $\mathit{cLieGrp}^{\mathrm{inj}}$. In that case, I would expect $F(\iota)$ to be $H^3(G,\mathbb Z) \to H^3(G,\mathbb R)\cong H^4(BG,\mathbb R) \to H^4(BT,\mathbb R)$.
The point is when the source is torsion-free, if there is transfer, the image should be the invariant of the target under the action of the Weil group.
|
2025-03-21T14:48:30.051206
| 2020-03-11T15:56:03 |
354714
|
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|
Stack Exchange
|
Wavelet momentum identity
I am reading an article on wavelet connection coefficients (G. Beylkin, "On the representation of operators in bases of compactly supported wavelets", 1992 (MSN)) and I came across Equation (3.31):
\begin{equation}
\sum_{l=-\infty}^\infty l^m\phi(x-l) = x^m + \sum_{l=1}^m (-1)^l
\begin{pmatrix} m\\l \end{pmatrix}
M_l^\phi x^{m-l}
\end{equation}
where $\phi(x)$ is the scaling function and
\begin{equation}
M_l^\phi = \int_{-\infty}^\infty x^l\phi(x)\,dx
\end{equation}
is the $l$-th momentum of $\phi$.
The author claims that the equation is well-known if $M_l^\phi = 0$ for $l=0,\dotsc,m$, and the general case follows from taking Fourier transforms. However, I could not find it, and trying to prove it myself is not working.
I recognize that both sides are kinds of convolutions, but when taking the Fourier transform the expressions (apparently) lead nowhere. Is there some trick I need to be aware of, or is it simply lack of practice/knowledge?
I do not know if this information should help, but the ultimate goal is to prove that
\begin{equation}
\sum_{l=-\infty}^\infty lr_l = -1
\end{equation}
where
\begin{equation}
r_l = \int_{-\infty}^\infty \phi(x-l)\phi'(x)\,dx.
\end{equation}
EDIT: Using the Poisson summation as @Nemo suggested in a comment, I was able to find that
\begin{equation}
\sum_{l=-\infty}^\infty l^m\phi(x-l) = \sum_{k=0}^m \sum_{l=-\infty}^\infty (-1)^k
\begin{pmatrix} m\\k \end{pmatrix}
e^{-ilx} i^k \frac{d^k\hat{\phi}}{d\xi^k}(-l) x^{m-k}.
\end{equation}
Now, I know that $i^k\frac{d^k\hat{\phi}}{d\xi^k}(0) = M_l^\phi$ but I'm still stuck with the terms $i^k e^{-ilx} \frac{d^k\hat{\phi}}{d\xi^k}(-l)$ for $l\ne0$. Is there any identity I am not aware of?
https://mathworld.wolfram.com/PoissonSumFormula.html
I used it by defining $f(y) = y^m\phi(x-y)$ and almost found the result. However, instead of $M_\phi^l$, I got $i^k e^{i l x} \frac{d^k\hat{\phi}}{d\xi^k} (-l)$, which (I think) I cannot turn into $M_\phi^l$...
The identity in the OP does not hold for any $m$, but only for $m< N$ where $N$ is the number of vanishing moments of the wavelet function. To complete the Poisson-summation derivation, one needs the socalled Strang-Fix condition, which says that $\frac{d^k\hat{\phi}}{d\xi^k}(-l)=0$ for integer $l$ unequal to 0 and $k< N$.
The identity says that integer shifts of the scaling function can reproduce polynomials of order $N$. For a proof, see theorems 4.26 and 4.27 in this book. (The identity is equivalent to the recursion relation 4.51.)
I couldn't have a look at it during the weekend, but this does look like what I was needing. I'll check the theorems and award the bounty later.
|
2025-03-21T14:48:30.051392
| 2020-03-11T16:19:22 |
354717
|
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"Will Sawin",
"curious math guy",
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|
Stack Exchange
|
L-function in p-adic spaces
I've been learning more about different $p$-adic geometries, namely Berkovich spaces, Huber's Adic spaces and ridgid analytic spaces. In arithmetic geometry, it is often very interesting to assoicate and study the L-function of an space. My question is if this has been studied for such $p$-adic spaces. One way one could naively hope to get such L-functions is to take the L-function associated to the cohomology of such a space and study that. However, I've been unable to find a reference for this.
The issue is that google searches lead straight to $p$-adic L-functions, which as far I can tell is not what I want.
Because $p$-adic spaces lie over $p$-adic fields, they could only ever have $L$-functions that are local at $p$ - there would be no way to define $L$-factors at other primes. One can define these $L$-factors whenever one has a finite-dimensional cohomology theory with Frobenius acting on it (which I think is fine for at least proper rigid analytic spaces), but it's not clear what they get you.
Thanks @Will Sawin. On a related note, does there exist a notion of a zeta function for such spaces?
Again, there exists a notion of local zeta factor (which would be a simple product of local $L$-factors).
|
2025-03-21T14:48:30.051508
| 2020-03-11T16:40:13 |
354718
|
{
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"Evgeny Shinder",
"Mikhail Bondarko",
"Sasha",
"THC",
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"naf"
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}
|
Stack Exchange
|
Field extensions in Grothendieck rings
Let $k$ be a field, and consider the Grothendieck ring of $k$-varieties, $K_0(V_k)$. Let $K/k$ and $K'/k$ be field extensions of $k$. We view $\mathrm{Spec}(K)$ and $\mathrm{Spec}(K')$ as $k$-schemes and consider their classes $[\mathrm{Spec}(K)]$ and $[\mathrm{Spec}(K')]$ in $K_0(V_k)$.
I have two general questions:
(A) what are the most interesting properties/criteria which lead to the equality $[\mathrm{Spec}(K)] = [\mathrm{Spec}(K')]$ ?
(B) when can we decide that $[\mathrm{Spec}(K)] \ne [\mathrm{Spec}(K')]$ ?
I understand that some of the first properties will be pretty easy or expected (but still interesting to mention), but also that the "next generation" of properties could be highly interesting.
You're using the notation $[]$ without defining it, and you're defining $K_0$ without using it.
@YCor : I have no idea what you mean ...
@THC: What do you mean by the equalities? Do you consider $Spec(K)$ and $Spec(K')$ as $k$-schemes and compare their classes in $K_0(V_k)$? This is quite unclear...
@Sasha : yes, I consider them as $k$-schemes, and compare the classes in $K_0(V_k)$. (Did I miss something ?)
I removed the tag "spectral theory" and you added it again. Have you seen what questions are asked with this tag? Have you seen what are the arxiv papers with this tag? The usage guidance of this tag is Schrodinger operators, operators on manifolds, general differential operators, numerical studies, integral operators, discrete models, resonances, non-self-adjoint operators, random operators/matrices. Putting this tag just spams people not interested with this stuff...
PS the tag [arithmetic-geometry] would perfectly fit this question in lieu of the wrong one. FYI at this moment, [arithmetic-geometry]: 224 watchers, 1426 questions; [sp.spectral-theory]: 81 watchers, 683 questions.
@YCor : I indeed understand the tag. And some of the most interesting open cases (related to my question) I know of, arise from / give rise to isospectral manifolds.
Fine. Anyway if you have in mind arithmetic manifolds [nt.number-theory] or [analytic-number-theory] will probably be more useful. If you have in mind Kähler manifolds, [complex-geometry].
I can try to explain why the equality of classes implies that the (Artin) motives of these fields over k are isomorphic. Unfortunately, it does not follow that the fields itself are isomorphic; thus I doubt that the converse implication is vald. Are you interested?
Actually, if k is of characteristic 0 then my claim is an easy application of the results of Gillet and Soule. Would you like me to turn this into an answer? I can also prove the statement if k is of characteristic $p>0$; yet then I will have to cite my own results.:)
@MikhailBondarko : Yes, very interested !
Since Shinder's answer is better, I will only say that in the case char(k)=0 one can apply Theorem 4 of Gillet H., Soulé C., Descent, motives and K-theory// J. f. die reine
und ang. Math. v. 478, 1996, 127-176, to numerical motives (that coincide with Chow ones for varieties of dimension 0).
In characteristic zero $[\mathrm{Spec}(K)] = [\mathrm{Spec}(K')]$ for finite field extensions of $k$ implies that $K$ and $K'$ are isomorphic.
Indeed, by the Larsen-Lunts theorem for smooth projective connected schemes of finite type $[X] = [Y]$ implies that $X$ and $Y$ are stably birational; this applies to $\mathrm{Spec}(K)$, $\mathrm{Spec}(K')$ as they are connected smooth and projective. Now if $\mathrm{Spec}(K)$, $\mathrm{Spec}(K')$ are stably birational, then they are isomorphic. This is because if $X$ is smooth projective and stably birational to $\mathrm{Spec}(K)$, then $K = \Gamma(X, \mathcal{O}_X)$. Same argument applies to products of fields that is to reduced zero-dimensional finite $k$-schemes: here the Larsen-Lunts theorem will match up the stable birational types of the connected components.
In characteristic $p > 0$ the result may be still true but hopeless to prove without resolution of singularities.
UPDATE: the result above about fields is given in Proposition 5 in a paper by Liu and Sebag where they study what $[X] = [Y]$ implies in general in characteristic zero, using Larsen-Lunts Theorem: https://link.springer.com/article/10.1007/s00209-009-0518-7
One question I would like to know the answer myself is: what if $\mathbf{L}^n\cdot([\mathrm{K}]-[\mathrm{K'}]) = 0$? Are $K$, $K'$ isomorphic then? Here $\mathbf{L} = [\mathbf{A}^1]$.
I would guess that this follows by looking at the etale Euler characteristic.
@ulrich: What exactly do you mean? I think all the `cohomological' data (Artin motives, Galois modules) which can be extracted from $[X]$ will not be able to distinguish fields.
OK, I suppose I was wrong. What I was suggesting (looking at the Artin motive) works if one of the fields is Galois, but not otherwise.
|
2025-03-21T14:48:30.051822
| 2020-03-11T16:46:37 |
354719
|
{
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"Gerry Myerson",
"Michael Hardy",
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"darij grinberg",
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|
Stack Exchange
|
Journal of serious opinions on weighty matters for mathematicians
Is there any scholarly journal
of opinion on professional matters,
for mathematicians, that would exclude things just as relevant to other fields as to mathematics,
that is not just $\text{“}\,$leisure reading$\text{,''}$ (i.e., it is on weighty consequential matters)?
"Professional matters" would include research, philosophy, and pedagogy, and probably other things.
I don't know any journal that is exclusively opinion, but many publications often print such opinion pieces, e.g. AMS Notices.
I posted this on m.s.e. without getting any replies: https://math.stackexchange.com/questions/3574020/journal-of-serious-opinions-on-weighty-matters-for-mathematicians
@NateEldredge : "Many" is an interesting word.
and let me guess, you want it not to degrade to Blue-Red mudslinging in a few issues? ;)
Another journal that may interest you is Journal of Humanistic Mathematics.
Are you exclusively interested in journals in English language? (A lot of non-native speakers easily write mathematical papers, but would have a hard time writing a well-written text with opinions etc.)
@YCor : Not necessarily. $\qquad$
@darijgrinberg : Anyone who disagrees with me about which versions of the Axiom of Choice should be allowed in a certain narrow area of research is a vile Nazi!! $${}$$ yeah, I guess exculding that sort of material would be nice. Or would be required.
The post to mathstack was deleted a few years ago.
This question was asked some time ago but remains on the unanswered queue. I want to echo Nate's comment regarding the Journal of Humanistic Mathematics. I've published two papers there. The first one included plenty of "serious opinions on weighty matters" such as the balance between writing research papers concisely vs welcoming new mathematicians into the field, and our collective habit of obfuscating the thought process that led to a result, leading to the impression that doing math successfully requires an inexplicable stroke of genius. We also opined on the importance of finding ways to share mental imagery with readers/listeners, and finding ways to explain math research to the general public. The second paper included opinions (and data from surveys) about many topics, like the importance of play as part of the research process, the value of expository writing, the importance of sharing personal hardships so that junior people know it's possible to bounce back from setbacks and still have a productive professional career (I have done this myself a few times), improving upon the culture of spreading knowledge via word-of-mouth (especially with an eye towards making math more inclusive, including to people who can't easily attend conferences), and more.
I'm a big fan of the Journal of Humanistic Mathematics and would love to see more people submitting articles there. They certainly seem to welcome (well thought-out, potentially research driven) opinions on matters that many mathematicians care about.
|
2025-03-21T14:48:30.052064
| 2020-03-11T17:01:15 |
354720
|
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"url": "https://mathoverflow.net/questions/354720"
}
|
Stack Exchange
|
Marcenko-Pastur and Tracy-Widom laws for sample covariance and Gram matrices when the "features" are correlated: references
Let us assume we've a rectangular data matrix $X=[x_1 \dots x_n] \in \mathbb{R}^{p \times n}$, where the $x_i \in \mathbb{R}^{p \times 1}$ are iid column vectors. I'm not assuming here that the entries of the matrix $X$ are iid, or also that $x_i$ are of the form $x_i = C^{1/2}z_i$, where $Z:=[z_1 \dots z_n]$ have iid entries (I think the work of Baik and Silverstein includes this case for ESD's). I'm really assuling that the co-ordinates ("features") of each $x_i$ are corelated.
I'm wondering (some references will be enough, but detailed answers appreciated!) if there're equivalents of Marcenko-Pastur and Tracy-Wisdom theorems in this case? More precisely:
(1) What's the limiting empirical spectral distribution of sample covariance $\frac{1}{p}XX'$ and Gram matrix $\frac{1}{n}X'X$?
(2) How are the largest eigenvalues of the above two matrices distributed?
Thank you for your answer!
If the correlations can be described by a multivariate Gaussian there exist results for the spectral density in the limit of large matrices, see for example Spectral Moments of Correlated Wishart Matrices (2005).
For the distribution of the largest eigenvalue I only know results for complex matrix elements, see Large complex correlated Wishart matrices: Fluctuations and asymptotic independence at the edges (2014).
Thanks - just accepted your answer!
|
2025-03-21T14:48:30.052197
| 2020-03-11T17:28:51 |
354723
|
{
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"Yangong Wu",
"fedja",
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"https://mathoverflow.net/users/153484"
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"url": "https://mathoverflow.net/questions/354723"
}
|
Stack Exchange
|
Optimal function existence? what is it?
It's a problem abstracted from a real engineering project.
I want to find the best curve $y=y(x)$, $x \in [0,1]$: $y$ doesn't have to be a continuous function.
The constraint is
$$
L=\int_{0}^{1} \frac{1}{y^3} dx =\text{const.},
$$ which represents the energy consumption of the system.
The objective function is
$$
J[y]=\int_{0}^{1} \left[ L \int_{0}^{z} \frac{1}{y^4} dx - \int_{0}^{z} \frac{1}{y^3} dx \int_{0}^{1} \frac{1}{y^4} dx \right] dz
$$
which instead represents the thrust force of the system.
The meaning of this question is find out what's the maximum force exerted by a specific system when its energy consumption is limited.
Questions
Does the objective function has a finite maximum?
If the objective function has a finite maximum, what's the expression of $y$?
If the objective function does not have a finite maximum, what's the expression of $y$?
A possible condition can be added $y(1)=1$.
Is $y$ positive, or not necessarily?
From the view of engineering, y must be greater than 0.
By introducing a few more variables and constraints you can also write it into a form which can partially be solved with Pontryagin's maximum principle:
\begin{align}
\max_y & \int_0^1 L_1\,x_1(t) - L_2\,x_2(t)\,dt \\
\text{s.t.}\, & \frac{d\,x_1(t)}{dt} = \frac{1}{y^4(t)},\ \frac{d\,x_2(t)}{dt} = \frac{1}{y^3(t)}, \\
& x_1(0) = 0,\ x_2(0) = 0, \\
& x_1(1) = L_2,\ x_2(1) = L_1.
\end{align}
After solving this problem one would still be free to choose $L_2$, but this would just be a scalar optimization problem, which should be easier to solve. The resulting Hamiltonian associated with the problem can be written as
$$
H(x,y,\lambda) = \frac{\lambda_1}{y^4} + \frac{\lambda_2}{y^3} + L_2\,x_2 - L_1\,x_1
$$
The dynamics of the co-states are given by
$$
\frac{d\,\lambda_1(t)}{dt} = L_1, \\ \frac{d\,\lambda_2(t)}{dt} = -L_2.
$$
The expression for $y$ can be found by maximizing the Hamiltonian. However, depending on the values of $\lambda_1$ and $\lambda_2$ this does not necessarily have to satisfy the solution $H_y = 0$, with $H_y$ the partial derivative of $H$ with respect to $y$.
Thank you, this is the right answer. I checked it with numerical methods.
|
2025-03-21T14:48:30.052361
| 2020-03-11T17:49:54 |
354726
|
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|
Stack Exchange
|
A question about homogeneous distribution
A distribution in $\mathscr{S}^{\prime}\left(\mathbb{R}^{n}\right)$ is called homogeneous of degree $\gamma \in \mathbb{C}$ if for all $\lambda>0$ and for all $\varphi \in \mathscr{S}\left(\mathbb{R}^{n}\right),$ we have
$$
\left\langle u, \delta^{\lambda} \varphi\right\rangle=\lambda^{-n-\gamma}\langle u, \varphi\rangle.
$$
where $\delta^{\lambda} \varphi(x)=\varphi(\lambda x)$. Now suppose that $u \in C^{\infty}\left(\mathbb{R}^{n} \backslash\{0\}\right)$ is homogeneous of degree $-n+i \tau, \tau \in \mathbb{R} .$ How to prove that the operator given by convolution with $u$ maps $L^{2}\left(\mathbb{R}^{n}\right)$ to $L^{2}\left(\mathbb{R}^{n}\right)$.
Let $u$ be a smooth function on $\mathbb R^n\backslash\{0\}$ homogeneous with degree $\lambda$ (on $\mathbb R^n\backslash\{0\}$). If $\lambda$ is not an integer $\le -n$, then $u$ can be uniquely extended to a tempered distribution homogeneous with degree $\lambda$. Moreover, the Fourier transform of an homogeneous distribution with degree $\lambda$ is an homogeneous distribution of degree $-\lambda-n$.
As a result, the Fourier transform of your $u$ is homogeneous with degree $n-i\tau-n=-i\tau$ when $\tau\in \mathbb R^*$, so is in one dimension a linear combination of $\xi_\pm^{-i\tau}$ which is thus bounded, proving the sought $L^2$ boundedness.
If $\tau =0$, then $u$ is homogeneous of degree $-1$ in one dimension. You need $u$ to be odd for the $L^2$ boundedness to hold. Take for instance (still in one dimension)
$u(x)=1/\vert x\vert$, obviously homogeneous with degree $-1$ and smooth on $\mathbb R^*$. The singular integral with kernel $1/\vert x-y\vert$ is not bounded on $L^2$, but the Hilbert transform with kernel $1/(x-y)$ is bounded on $L^2$ (with norm π).
Thanks a lot ! I already know how to prove it.
By the Proposition 2.4.8. of L. Grafakos(GTM249), we have $\hat{u}\in C^{\infty}\left(\mathbb{R}^{n} \backslash\{0\}\right)$. It follows from $u$ is homogeneous of degree $-n+i\tau$ that $\hat{u}$ is a homogeneous distribution of degree $-i\tau$.Then,
$$\hat{u}(\xi)=|\xi|^{-i\tau}\hat{u}(\frac{\xi}{|\xi|}),$$
which implies that $|\hat{u}(\xi)|=\sup_{|x|=1}|\hat{u}(x)|<\infty$, i.e. $\hat{u}\in L^{\infty}(\mathbb{R}^{n})$. Thus,
$$\|u*f\|_{L^{2}}=\|\hat{u}\hat{f}\|_{L^{2}}\leq \|\hat{u}\|_{L^{\infty}}\|\hat{f}\|_{L^{2}}=\|\hat{u}\|_{L^{\infty}}\|f\|_{L^{2}}$$
for all $f \in L^{2}(\mathbb{R}^{n})$.
The meaning of $\hat u$ is not clear, take for instance $u(x)=1/\vert x\vert$ and see the constraints in my answer above.
" $$ is homogeneous of degree $-n+i\tau$" indicates that $u$ is a tempered distributions, so fourier transform is meaningful.
Well, the extension of an homogenous functions on $\mathbb R^n\backslash{0}$ to an homogeneous distribution on $\mathbb R^n$ could be impossible for integer indices $\le -n$.
|
2025-03-21T14:48:30.052544
| 2020-03-11T18:34:35 |
354728
|
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|
Stack Exchange
|
Orbits of irreducible representations, the isotropy group H, the double-coset space $H\backslash G/H$
I first thought that this would be a simple question and placed it on Mathematics StackExchange
https://math.stackexchange.com/questions/3575564/orbits-of-irreducible-representations-the-isotropy-group-h-the-double-coset-sp
To my surprise, no answers have been offered; so I now put it here.
If $\,D(K)\,$ is an irreducible of $\,K\vartriangleleft G\,$, then $\,D_g(K)\equiv D(g^{-1}Kg)\,$ are irreducibles of $\,K\,$ for $\,\forall~g\in G\,$.
The elements $\,h\,$ generating representations equivalent to $\,D\,$, constitute an isotropy group:
$$
D_h(K)~\simeq~D(K)\quad\Longleftrightarrow\quad h\in H~<~G
$$
Likewise, the elements $\,h^{\,\prime}\in G\,$, which satisfy
$$
\left(D_g\right)_{h'}(K)~\simeq~D_g(K)~~,
$$
constitute the isotropy group $\,H_g\,$ of $\,D_g(K)\,$. The groups $\,H_g\,$ and $\,H\,$ are conjugate:
$$
H_g~=~g~H~g^{-1}~~.
$$
So the set of $\,D_g\,$ splits into disjoint classes (orbits) of mutually equivalent irreducibles.
Lemma 1
$\phantom{qqi}$ If $\,H\,$ is the isotropy group of $\,D(K)\,$, then, for $~\forall\,w,\,w^{\,\prime}\,\in\, G\,$,
$$
D_{w^{~\prime}}(K)~\simeq~D_w(K)\quad\iff\quad w^{~\prime}~=~w~h~~,~~h\in H~~.
$$
This can be easily proven from the definition of $\,H\,$.
Lemma 2
$\phantom{qqi}$ For $~\forall~w,~w^{~\prime} \in G~$ and $~\forall h \in H\,$,
$$
w^{~\prime}~=~w~h\quad\Longrightarrow\quad H~w^{~\prime}~H~=~H~w~H~~.
$$
$\phantom{qqi}$ In the special case of $H \lhd G $,
$$
w^{~\prime}~=~w~h\quad\iff\quad H~w^{~\prime}~H~=~H~w~H~~.
$$
The former statement is self-evident. To prove the latter one, i.e. to show that for $H\lhd G$ the implication becomes equivalence, we notice that in this situation, for $\forall h_1,~h_2\in H$ and $\forall w \in G$, a product $ h_1 w h_2 $ can be cast as
$$
h_1~ w~h_2~=~w~(w^{-1}~h_1~ w)~h_2~=~w~h~~,~~~ h\in H~~.
$$
Hence,
$$
H~w^{~\prime}~H~=~H~w~H
\quad\iff\quad
w^{~\prime}~=~h_1~w~h_2
\quad\iff\quad
w^{~\prime}~=~w~h~~.
$$
QED
Combined, the former and latter lemmas amount to the following
Theorem
$$
D_{w^{\,\prime}}(K)~\simeq~D_w(K)\quad\iff\quad w^{~\prime}~=~w~h\quad\Longrightarrow\quad H\,w^{~\prime}~H~=~H~w~H~~.
$$
$\phantom{qqi}$ In the special case of $H \lhd G $, this works in both directions:
$$
D_{w^{~\prime}}(K)~\simeq~D_w(K)\quad\iff\quad w^{~\prime}~=~w~h\quad\iff\quad H\,w^{~\prime}~H~=~H~w~H~~.\quad
$$
Thus, if for some $\,K\lhd G\,$ we start out with a representation $\,D(K)\,$ and its isotropy group $\,H\,$, and build the space $\,H\backslash G/H\,$, then distinct double cosets $~H w H~$ and $~H w^{~\prime} H~$ parameterise inequivalent representations $D_w(K)$ and $D_{w^{~\prime}}(K)$ $-$ and, thereby, different orbits.
While in general it is not prohibited for one double coset to correspond to several inequivalent irreducibles (i.e. to several orbits),
in the case of $H \lhd G $ the mapping becomes bijective, so one coset corresponds to one orbit.
QUESTION:
Can you please offer me a simple example of a situation where $w$ ranges within one coset $HwH$ and generates several inequivalent $D_w(K)~$?
I think you'll get an example by taking $H$ to be as far from normal as possible. In finite group theory, such a subgroup $H$ is called a Frobenius complement. See here for examples: https://en.wikipedia.org/wiki/Frobenius_group .
@HJRW Could you please also advise me on the best notation for the set of all distinct orbits of $,D_g(K),$? The notation $,\left{,D_g(K),\right},,~g\in G,$, looks awkward, because it fails to emphasise that we factorise. So it looks like the set of all representations, not all orbits.
A set of orbits is a quotient. If I have understood you correctly, you want to think about the orbit of $D(K)$ under conjugation by elements of $G$, and then factor out the action of $H$. So I think the following would be good notation: $H\backslash (G.D(K))$.
@HJRW A silly question, if I may. Why do you say H∖(G, D(K)) and not (G, D(K))/H ?
Because, if I understand correctly, H acts the left. The former is for a left action, the latter is for a right action. (In either case, it’s a full stop, not a comma. It can be omitted in any case.)
@HJRW What exactly do you mean by saying that H acts from the left? I introduce $,D_h(K),=,D(h^{-1},K,h),$ and define $,H,=,\left{,h;|;D_h(K),\simeq,D \right},$. Similarly, for some $,D_w(K),\not\simeq,D(K),$, I define $,H_w,=,\left{,h^{,\prime};|;(D_w)_{h^{,\prime}}(K),\simeq,D_w \right},$. In what sense does this imply that "H acts from the left"?
(I am not a mathematician by training, so each step is difficult to me.)
Sorry, there's a small mistake in your question that confused me. I think I have it right now. (It's easy to get these things muddled up!) The action where $g$ sends $D(K)$ to $D_g(K)=D(g^{-1}Kg)$ is a right action of $G$. It follows that the stabiliser (ie isotropy subgroup $H_g$) of $D_g(K)$ is $g^{-1}Hg$, not $gHg^{-1}$ as you wrote. Therefore I should have written $D(K)G/H$, not $H\backslash GD(K)$. Anyway, these questions are about elementary group theory, so I suggest you ask any more of them that you may have on math.stackexchange.com .
Thank you, @HJRW . A lot of food for thought...
|
2025-03-21T14:48:30.053004
| 2020-03-11T19:53:17 |
354730
|
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"Jochen Glueck",
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|
Stack Exchange
|
Positive solutions for semilinear parabolic equations
Let $X$ be a Banach lattice. Consider the system
$$y'(t)=Ay(t)+f(t,y(t)) \qquad \text{in } (0,T) , \qquad y(0)=y_0, \qquad (*)$$
where $T>0$, $A$ generates an analytic positive semigroup $S(t)$ on $X$ and $f$ is a locally Lipschitz function.
Question:
Let $\eta \in X$. Assuming that $y_0 \ge \eta$, what are some minimal conditions on $f$ to ensure that the solution $y$ of $(*)$ satisfies:
$$y(t)\ge \eta, \qquad \forall t\in (0,T).$$
Is there any theory or theorem that answers such question in a general framework?
Are you interested to the case where $\eta \neq 0$?
@GiorgioMetafune: I can't speak for the OP, of course; but I would also be interested in references that deal with the special case $\eta=0$.
If $\eta=0$, then $y\geq 0$ whenever $f \ge 0$, by the variation of constants formula. For more general $f(y)$, let us assume that $f(y)=o(y)$ as $y \to 0$, otherwise we incorporate the linear part in $A$. Then we modify $f$ putting $0$ for $y<0$ and obtain $g$. If we assume that $g \ge 0$, then we solve the above problem with $g$ obtaining a positive soluition which is a solution for the initial problem. So a sufficient condition would be that $f(y) \ge 0$ when $y\ge 0$.
@GiorgioMetafune I think the condition $f(y) \ge 0$ is too much. For example, for $y'=f(y)$, $y(0)\ge 0$, I found that a necessary and sufficient condition to ensure $y(t)\ge 0$ is $f(0) \ge 0$. This can be generalized to the heat equation. See this.
Proposition 2.1 and Remark 2.2.
True, I agree, but in both cases you can use subsolutions or the maximum principle. In general maybe it is difficult to find precise conditions. Probably other criteria come out when $A$ is given by a form, by the proof of Beurling-Deny conditions.
Yes, I would be interested to the form framework, since in my special case $A$ is given by a form, say in $L^2$.
Theorem 3.3 in the paper of Haraux that you linked suggests that the analogous condition on a general Banach lattice $X$ might be that the semigroup generated by $A$ be not only positive but also "positivity improving" and that $f$ be cross-positive in the sense that we have $\langle x', f(x)\rangle \ge 0$" whenever $0 \le x' \in X'$ and $0 \le x \in X$ with $\langle x', x \rangle = 0$. However, I don't know which precise technical conditions are needed to make all the details work.
Anyway, note that Theorem 3.3 alone does not give a condition purely in terms of $A$ and $f$; it also requires to already know that the given trajectory is positive for small times. In the applications to heat equations in Sections 4 and 5 Haraux checks this by a perturbation argument together with lower estimates for the heat semigroup.
Assume only that $f(0) \ge 0$ and choose $\lambda$ so that $\lambda y+f(y)$ is positive for positive $y$. Then rewriting the equation as $y'=(A-\lambda)y+(\lambda y+f(y)$ allows to apply my previous comment. Any mistake?
I find that the condition $A\eta+f(\eta) \ge 0$ is necessary and sufficient. Let us assume that $f$ is globally Lipschitz and depends only on $y$, so that there is no problem about global existence. Let us have in mind that $X$ is an $L^p$ space or a space of continuous functions.
1) Assume first that $\eta=0$. If we take $y_0=0$, then $y'(0)=f(0)$; if $f(0) <0$, then $y$ would be negative for small $t$ (pointwise in spaces of continuous functions, against a positive functional in $L^p$...).
2) If $f \ge 0$ everywhere, then $y \ge 0$ by the variation of constants formula.
3) If $f$ is positive only for positive $y$, then we modify $f$ to $g$ by setting $g(y)=f(0)$ for $y \le 0$ and $f=g$ for positive $y$. Then we solve the problem with $g$ instead of $f$ and we get a positive solution, by the previous argument. This solution is then the solution of the given problem, by uniqueness.
4) Assume now that $f(0) \ge 0$ and take $\lambda$ such that $\lambda y+f(y) \ge 0$ for $y \ge 0$. We rewrite the problem in the equivalent form
$$
y'=(A-\lambda )y+(\lambda y+f(y)).$$ Since $A-\lambda$ generates a positive semigroup, as well, we may apply point 3) and $y \ge 0$.
5) The general case reduces to $\eta=0$ writing $y=\eta+z, y_0=\eta+z_0$. The equation for $z$ becomes $z'=Az+A\eta+f(\eta+z), z_0 \ge 0$ and the condition $A\eta+f(\eta) \ge 0$.
If I understand correctly, you are implicitly assuming that $f: X \to X$ is not a general mapping, but the composition with a function from $\mathbb{R}$ to $\mathbb{R}$ that you also denote by $f$ (which is why you assumed $X$ to be a function space). However, I think the OP is actually interested in general functions $f$, as indicated by the linked paper of Haraux.
Yes true. I thought the question was on reaction diffusion problems.
@GiorgioMetafune Thank you for the answer. For the scalar valued case (in $\mathbb{R}$), no problem with the existence of $\lambda$ s.t $\lambda y + f(y) \ge 0$. I'm stuck on this condition in general case, for example for product space of $L^2$ as mentioned by Jochen.
Obviously, one should have $\eta \in D(A)$.
Yes $\eta \in D(A)$. For systems I do not know. In case of systems of ODE's $y'=f(y)$ one has simple criterion for the invariance of a set $D$ (the inner product of the outer normal and $f$ must be nonpositive on the boundary). For PDE's I do not know. I see that the question was another. Should I delete the answer or leave so that everybody can follow the discussion?
I found your answer very useful. Maybe a generalization of the result you mentioned is this.
It's a geometric condition of set invariance by flow. The problem I have now is how to link between this and the positivity assumption on $f$.
@S.Euler: For $\eta = 0$, I think one can use the paper of Pavel that you linked above to show that cross-positivity is indeed the right assumption on $f$ (as suggested in my comment under your question). If this observation is of any help for you, I can try to write down the details in an answer.
Good! Unfortunately I do not have access to the paper from home and, as you know, universities are closed in Italy. By the way, probably I found and example of a pure second order operator with non real eigenvalues; we discussed in another topic.
@S.Euler: Hmm, I've been thinking about it for a while now, and it seems that the relation of Pavel's result to cross positivity is much more subtle than I suspected at first glance. I'll have to think about it for a bit longer (and I have to admit I'm not even sure any longer whether it really works out well).
|
2025-03-21T14:48:30.053459
| 2020-03-11T20:27:56 |
354733
|
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|
Stack Exchange
|
Can Ackermann theory minus foundation minus class comprehension permit allowing every proper subclass of $V$ to be a set?
Lets add a constant symbol $V$ to the signature of the language of set theory. So working in first order logic with equality, add the following axioms about $\in $ and $V$.
Extensionality: $\forall x \forall y (\forall z (z \in x \leftrightarrow z \in y) \to x=y)$
Set construction (reflection): if $\phi$ is a formula in which all and only symbols $y,x_1,..,x_n$ occur free, and non of them occur bound, in which the symbol $V$ doesn't occur, then: $$\forall x_1 \in V,...,\forall x_n \in V \\ \forall y (\phi \to y \in V) \to \exists x \in V \forall y (y \in x \leftrightarrow \phi)$$; is an axiom.
Set-hood: $\forall x \ (x \subsetneq V \leftrightarrow x \in V)$
Now this theory is formulated in the same langauge of Ackermann's, it share with it the first two axioms and the right to left implication of the third axiom. However the left to right implication of the third axiom is in some sense daunting! The idea here is that this theory doesn't have comprehension axioms about proper classes, clearly by the third axiom all classes the second axiom constructs are sets! If we just add the class comprehension schema of Ackermann, then we immediately get a contradiction, since the Russell class would be a set. Now this theory easily prove all axioms of Zermelo set theory, and I'd think that (if consistent) it might even be equi-interpretable with the full Ackermann's set theory itself. Its also to be noted that if instead of adding the symbol $V$ as a constant, we added it as a one place predicate symbol (and so every formula $x \in V$ would be turned to $V(x)$) [as it is the case in the original formulation of Ackermann's where he actually used the symbol $\mathcal M$ for that], then it appears that Infinity would not be provable, which proves that the way how $V$ is added as a primitive does matter as regards the consistency strength of extensions of fragments of Ackermann's set theory.
Question: What's the exact consistency strength of this theory?
Is $x \subsetneq V$ necessary instead of the simpler $x \subseteq V$?
Might a parameter-free version of comprehension be sufficient, like it is for ZFC?
@user76284 if we allow $x \subseteq V$ then we'd have $V \in V$ therefore easily getting Russell's paradox by formula $x \in x_1 \land x \not \in x$ [just substitute$ x_1$ by $V$] and if you unleash parameters then you will encode $V$ in an indirect manner, so it would invoke Russell's paradox again. Yes parameter free would work but with the restriction of not containing $V$, of course. But I don't know its strength.
Doesn’t that mean your formula contains $V$ somewhere, though? Do you have an example for the indirect encoding?
Wait, why did you edit the construction schema?
@user76284, to have elements in $V$
But you can already prove $\varnothing \in V$ without that assumption.
@user76284 how?
I think like this: Consider $\phi y = \bot$. Then by the original schema $\exists x \forall y (y \in x \leftrightarrow \bot)$. That is, $\varnothing$ exists. Now it is trivially true that any element in $\varnothing$ belongs to $V$ (since there are no elements), so by the set-hood axiom $\varnothing \in V$.
@user76284 No to be an element of $V$ you need to be a proper subset of $V$, and there is no proof that the empty set is a proper subset of $V$. So you need to enforce that by modifying the set construction schema. Or simply add an axiom that $V$ is non empty.
Hmm, I see. That takes us back to the previous question of whether $x \subsetneq V$ is necessary.
@user76284, of course its necessary otherwise you'll have $V \in V$ and then you can substitute $V$ for any parameter $x_1$ and get the contradiction. As for an indirect way of coding $V$ if we unleash parameter (i.e. let the parameters range over all classes), then clearly the same would apply by substitution of parameters. The parameter free approach would work of course (provided that $V$ is shunned from being used) but its weak, I don't think it can prove pairing over all elements of $V$ unless you stipulate that all elements of $V$ are parameter free definable.
Sorry, how exactly are you getting the Russell contradiction from $V \in V$? Can you post a complete formal derivation? I think I might be misreading the axioms.
@user76284, OK, lets take $\phi$ to be the formula $ y \in x_1 \land y \not \in y$, now let $x_1=V$, since you have $V \in V$ then this would fulfill the antecedent of set construction, and the rest of the proof is easy.
@user76284 you get $R \in R \leftrightarrow R \not \in R$.
Makes sense. I'll think about the parameter-free version.
@user76284, parameter free version is OK if you don't use the constant symbol $V$, but the problem is that it is very weak, how can you prove pairing within $V$ for example?
The schema of https://math.stackexchange.com/questions/2499051/is-there-a-contradiction-hiding-in-this-alternative-set-theory-with-3-axioms has a different structure since the parameters for the antecedent and consequent are separate.
But in that case don't you need to show $\forall x_1 \forall x (x \not\in x \land x \in x_1 \rightarrow x \in V)$, even for those $x_1$ that are not $V$, before getting to the consequent?
Oh, I forgot it assumes all the parameters like $x_1$ are in $V$.
Since $x_1 \in V$ then by set-hood, this wold be proved.
Tricky. And making $V$ a predicate again would make us lose infinity.
@user76284 your structure would offer no protection, the same argument applies, actually if you take the formula $y \in x_1 \land y \not \in y$ you get the Russell set by letting $x_1=V$ in the consequent. You need to let $V$ be a predicate for that structure to work, and you lose infinity.
In that case, can we add the axiom of infinity I proposed to the predicate version you posted at https://math.stackexchange.com/questions/3251691/is-this-simple-class-theory-equi-interpretable-with-zfc? Or would it also be inconsistent?
@user76284, if you add a class of all constructible sets, then that class would act like $V$ here, so you need to amend 'set-hood' axiom as to require only proper subsets of $V$ to be constructible. And you need to change the parameter structure to the one given here (so the consequent of set construction scheme doesn't have unleashed parameters) since otherwise $V$ can seep through those unleashed parameters in the consequent. All in all you'll come back to this theory here.
The theory is inconsistent.
Let ZG(x) be the formula ∀u∈x∀v∈u(v∈x)∧∀t(x∈t→∃s∈t(∀∈s(y∉t)))∧(∀t(∃s∈x(s∉t)→∃y∈x(y∉t∧∀u∈(x-t)(u∉y))))∧∀t∈x∃s∀v(v∈s↔(v=tνv=x))∧∀u∈x∃t∈x∀s(s∈t↔(s∈u∧∃r(r∈s)))∧∃t∀s(s∈t↔(s∈x∧∃r(r∈s)))
(That is x is transitive; if x is in t, then t has an ∈-minimal element; if x is not contained in t, then there is an ∈-minimal element of x which is not in t; if t is in x then
the pair {t,x} exists; if t is in x then t-{0} exists; and x-{0} exists.)
We note some simple properties of x for which ZG(x) holds:
If x∈ then x⊆.(If x were not contained in V, then there is an ∈-minimal element m of x which is not in V. If m is not V, then by Set-hood m is in V. If m=V,then {V,x} exists
and has no ∈-minimal element.)
If x∈, then there is a W∈ such that t is in W iff t is contained in x.(By the above property x⊆, and so anything contained in x is properly contained in V since x is not in
x. Therefore by Set construction such a W must exist.) We will denote such a W by Px.
If x∈ then ZG(Px).
Suppose that ZG(x) implies x∈. By Set construction, there is a z∈ which consists of all x for which ZG(x) holds. Let y be the union of z. Then y is in and ZG(y). Therefore ZG(Py).
Then Py is contained in y. But then y∈y which is impossible. Therefore there must be an x such that ZG(x) and x∉. If x is contained in then x-{0} is in and so x is in . If x is
not contained in then there is an ∈-minimal element m of x which is not in . But then m-{0} is in and so m is in .
Forgive me if I'm missing something, but doesn't $\phi y = \bot$ prove the empty set $\varnothing$ exists, and then $\varnothing \in V$, and then $\phi y = \forall x (x \in y \leftrightarrow x = \varnothing)$ shows the unit set ${\varnothing}$ exists and is also in $V$?
@ZuhairAl-Johar I don't understand - user76284's comment seems correct to me.
@NoahSchweber, you came late. Kirmayer is responding to my old presentation of the theory in which I missed stipulating the asserted class in set construction to be a set. the user 76284 comment in that old formulation fails because the empty class need not be a proper subclass of $V$.
@ I think you are right! I think this proof can be simplified to require any object x to fulfill ZG if and only if x-{0} exists and whatever class t that x is not a subset of then x has a minimal element m that is not in t and for which m-{0} exists. This would enforce every ZG object to be a member of V. Clearly every von Neumann ordinal in V fulfills ZG, then the property "ZG(x) and x is a von Neumann ordinal" would enact Burali-Forti. Great! Thanks!
I wonder if the theory presented at the below mentioned link would be subject to the same argument of inconsistency? https://math.stackexchange.com/questions/3251691/is-this-simple-class-theory-equi-interpretable-with-zfc
|
2025-03-21T14:48:30.054136
| 2020-03-11T21:18:11 |
354735
|
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|
Stack Exchange
|
Bayesian posterior consistency when prior distribution is induced by a diffusion
Let $\Pi_{b,\sigma}$ be a prior distribution on $\{z_t\}_{t<T}\in C_0[0,T]$ induced by the following diffusion:
\begin{align}
d\tilde z_t&=b(\tilde z_t,t)dt+\sigma(\tilde z_t,t) dW_t, ~\text{ for $t<T$}.
\end{align}
Let we observe $n$ i.i.d. samples at time $T$, denoted as $X_n^T$, from the model $P(\cdot|\{z_t\}_{t<T})$. Define the posterior distribution as the following Radon-Nikodym derivative
\begin{align}
\frac{d\Pi_{b,\sigma_0}(\{z_t\}_{t<T}|X_n^T)}{d\Pi_{b,\sigma}(\{z_t\}_{t<T})}= \frac{ P(X_n^T|\{z_t\}_{t<T}) }{\int d\Pi_{b,\sigma}(\{z_t\}_{t<T}) P(X_n^T|\{z_t\}_{t<T})}.
\end{align}
Question
1. What is the notion of consistency of posterior distribution on $\{z_t\}_{t<T}$ as number of samples $n\to \infty$?
2. References to read more about such posteriors and their asymptotic properties.
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2025-03-21T14:48:30.054228
| 2020-03-11T21:19:58 |
354736
|
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|
Stack Exchange
|
Are $0, 1, 4, 7, 8$ the only dimensions in which a bivector-valued cross product exists?
It is a well-known mathematical curiosity that ordinary (vector-valued) cross products over $\mathbb{R}$ exist only in dimensions $0, 1, 3$ and $7$ (this fact is related to Hurwitz's theorem that real composition algebras only exist in dimensions $1, 2, 4, 8$). I am interested in a generalization of this result to bivector-valued cross products.
A bivector-valued cross product (see here for the terminology) is a bilinear map $\times : V^2 \to \Lambda^2 V$ with the following two properties:
The resulting bivector is orthogonal to all inputs (i.e. the left contractions $\mathbf{x} \, \lrcorner\, (\mathbf{x}\times \mathbf{y})$ and $\mathbf{y} \, \lrcorner\, (\mathbf{x}\times \mathbf{y})$ vanish). If we write the bivector as a skew-symmetric matrix $M_{\mathbf{x},\mathbf{y}}$, this condition is simply $M_{\mathbf{x},\mathbf{y}} \mathbf{x} = M_{\mathbf{x},\mathbf{y}} \mathbf{y} = 0$.
In analogy with ordinary cross products, the resulting bivector's norm (the Frobenius norm of the previous matrix) equals the area of the paralellogram formed by the inputs. That is, we have $$|\mathbf{x}\times \mathbf{y}|^2 = \det \begin{pmatrix} \mathbf{x}\cdot\mathbf{x} & \mathbf{x}\cdot\mathbf{y}\\ \mathbf{y}\cdot\mathbf{x} & \mathbf{y}\cdot\mathbf{y} \end{pmatrix}.$$
The following dimensions are known to admit a bivector-valued cross product:
In dimensions $0$ and $1$ the identically zero product trivially satisfies both conditions.
In dimension $4$, we can take the four-dimensional volume form, raise two indices and multiply by the appropriate normalization constant.
Finally, in dimensions $7$ and $8$ bivector-valued cross products can be defined in terms of a coassociative form (the Hodge dual of an associative form) and a Cayley form respectively, by raising two indices and normalizing.
On the other hand:
In dimensions $2$ and $3$ such a product is impossible: there is enough room that some parallelograms are nontrivial, but not enough room to fit in a nonzero orthogonal bivector.
In dimensions $5$ and $6$ there aren't any bivector-valued cross products either, though the proof I have is less straightforward (it involves Hodge duality and appealing to the known classifications of scalar-valued and vector-valued $k$-ary products, as I mention in the MSE question below).
My question is:
Does there exist any bivector-valued cross product in dimension $9$ or higher?
Note: this is the last remaining case of a more general question I asked in Math StackExchange. My hope is that there is already a reference somewhere dealing with this problem, or with something equivalent to it.
Update (02/01/2021):
An equivalent way to state the defining conditions is in terms of a totally antisymmetric tensor $f_{abcd}$ (in abstract index notation) of rank $4$ such that $(x \times y)_{ab} = f_{abcd} x^c y^d$. This tensor satisfies
$$f_{ab}{}^{ef}f_{cdfe}+f_{da}{}^{ef}f_{bcfe}=2g_{ac}g_{bd}-g_{ab}g_{cd}-g_{ad}g_{bc},$$
where $g_{ab}$ denotes the standard inner product, which I also implicitly used to raise and lower indices. This identity comes from polarizing the area condition. Note that the orthogonality condition follows from the full antisymmetry of $f$.
A related problem has been studied in Chapter 20 of the book Group Theory by P. Cvitanović. Here $f$ and $g$ (of the same rank and symmetry) are assumed primitive invariant tensors, meaning roughly that every tensor invariant under $G = \operatorname{Aut}(V,f,g)$ can be decomposed as a linear combination of certain contractions of $f$ and $g$, namely those that are tree-like in Penrose's graphical notation.${}^{(*)}$
This primitiveness condition turns out to imply that $f$ is (almost always) a bivector-valued cross product: adding to Equation (20.6) its $90{}^\circ$-rotated version and using antisymmetry of $f$, we obtain precisely the polarized area condition modulo relabeling and raising/lowering indices, except that there is an overall factor of $2b(b-1)$ multiplying the right hand side, where $b$ is a nonnegative constant. This factor can be absorbed by redefining $f\mapsto f/\sqrt{2b(b-1)}$ provided $b>1$.
The author classifies the possible Lie groups $G$ arising in this way (or rather their associated Lie algebras), as well as the corresponding dimension $n$ of the defining representation, by means of Diophantine conditions. He obtains the known nontrivial cross products in dimensions $4, 7$ and $8$, acted on by $SO(4), G_2$ and $\mathrm{Spin}(7)$ respectively (the case $n=10$ acted on by $SO(10)$ has $b=1$, so it does not produce a cross product). Thus we don't find any new examples of bivector valued cross products.
If the reverse implication somehow held (being a cross product implies being primitive), that would constitute a proof of the nonexistence of bivector valued cross products in dimension $9$ or higher. If I understand correctly, primitiveness can fail in two ways:
The tensor $f_{abcd}$ can fail to be primitive itself, in which case it is a linear combination of tree contractions of primitive invariant tensors of lower rank. I'm not entirely sure, but I think these should be able to be ruled out manually. For example, if $f_{abcd} = f'_{[ab} f'_{cd]}$ for some antisymmetric primitive $f'$ of rank $2$, a short graphical manipulation implies that there is no way to satisfy the area condition unless $n=-4$, which is impossible (but see the remark below). The case of rank $3$ seems more involved, and I haven't been able to rule it out yet.
There may exist more primitive invariant tensors. The group preserving them would presumably have to be a Lie subgroup of $SO(4), G_2$ or $\mathrm{Spin}(7)$, by the argument in Section 4.7 of the book. I'm not sure of the details, but in principle one only needs to consider the finitely many types of extra primitive invariants that would spoil the derivation of the Diophantine conditions on the dimension.
I don't really know how plausible this whole approach is, but a priori it looks like a viable way to attack the problem.
$(*)$ This sort of graphical calculus, known under many other names (birdtracks, string diagrams, trace diagrams, etc.) is very useful for manipulating tensor equations with many indices. Another reference where these diagrams are extensively applied is L. Cadorin's thesis, where it is used to classify, among other things, ternary vector valued cross products; in particular, Equation (27.10) is the same as the area condition up to a factor of $d-2$ multiplying the right hand side (note she uses a different graphical convention for antisymmetric tensors of even rank). As above, one can absorb this factor into $f$ provided the dimension $d$ is not $2$. What this tells us is that given a ternary vector valued cross product ($d=0, 1, 2, 4, 8$) in any dimension other than $2$, we obtain a corresponding binary bivector valued cross product. Again we obtain no new examples.
Interestingly, both Cadorin and Cvitanović obtain negative even solutions for the dimension, which may be interpreted as evidence for some sort of cross product-like structures in Grassmann coordinates. As Cvitanović points out, his negative solutions contain the third row of Freudenthal's magic square.
Related question (but not duplicate) https://mathoverflow.net/questions/314613
Fun fact: this sequence is invariant under the involution $n\mapsto 8-n$.
|
2025-03-21T14:48:30.054728
| 2020-03-11T21:57:49 |
354741
|
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|
Stack Exchange
|
Comprehension factorization for locally small categories
It is well-known that there is an orthogonal factorization system on Cat (the category of small categories and functors) where the final functors are left-orthogonal to the discrete fibrations. This factorization is called "comprehensive" in the original paper by Street and Walters.
However, in that paper the authors do not address any size issue.
Can this factorization system be extended to the category of locally small categories?
Discrete fibrations are faithful functors, so any discrete fibration to a locally small category is locally small.
In particular if you take any comprehension factorization:
$$ C \rightarrow D \rightarrow T $$
If $T$ is locally small, then $D$ is also locally small. so the answer to your question is yes.
Though you have to be aware that the kind of discrete fibrations we are talking about here can have large fibers.
|
2025-03-21T14:48:30.054933
| 2020-03-11T22:23:37 |
354742
|
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|
Stack Exchange
|
"Strengthening" the mean value theorem for the sine function
The present discussion arises from this MO question. Below, $e$ stands for Euler's number and let
$$\tau:=\arccos\left(\frac{\sin e-\sin 1}{e-1}\right)\approx 1.82\cdots.$$
An application of the Mean Value Theorem (for derivatives) to the function $f(t)=\sin t$ leads to
$$\frac{\sin(e\,t)-\sin(t)}{e\,t-t}=\cos(\xi_tt) \qquad \text{for some $1\leq\xi_t\leq e$}. \tag1$$
QUESTION. Is it true that for each $t>0$, one can always find some $\xi_t\geq\tau$ such that (1) holds? Example: $\xi_1=\tau$.
You found such $\xi_t\in[1,e]$ in (1). Now add to it $2\pi N/t$ where $N$ is large, and you obtain new $\xi_t>\tau$ satisfying (1).
Given that the answers (1 2) to your question indicate that it's not what you meant, you should probably change it so that it says what you mean.
In view of the answer by Carlo Beenakker and the comment by Alexandre Eremenko, it appears that what you actually had in mind is the following question:
By the mean value theorem, for each $t\in(0,1]$ there is some $\xi_t\in(1,e)$ such that
\begin{equation*}
r(t):=\frac{\sin et-\sin t}{(e-1)t}=\cos(\xi_t t). \tag{2}
\end{equation*}
(Since $\cos u$ is strictly decreasing in $u\in[0,e]$, the value of $\xi_t$ is unique for each $t\in(0,1]$.)
Is it true that $\xi_t\ge\tau$ for all $t\in(0,1]$?
The answer to this question is yes. Indeed, for $t\in(0,1)$ we have $\xi_t t\in(0,e)\subset[0,\pi]$ and $\tau t\in(0,\tau]\subset[0,\pi]$. Therefore, in view of (2) and because $\cos$ is strictly decreasing on $[0,\pi]$, we see that
\begin{equation*}
\xi_t>\tau\iff d(t):=\cos\tau t-r(t)>0; \tag{3}
\end{equation*}
here and in what follows, $t\in(0,1)$.
Next,
\begin{equation*}
d_1(t):=(e-1) d(t)/t^2=\sum_{j=1}^\infty(-1)^jb_j t^{2j-2}-\sum_{j=1}^\infty(-1)^ja_j t^{2j-2},
\end{equation*}
where
\begin{equation*}
a_j:=\frac{e^{2 j+1}-1}{(2 j+1)!},\quad b_j:=\frac{(e-1) \tau^{2 j}}{(2 j)!}.
\end{equation*}
It is easy to see that $0<a_j<a_{j+1}$ and $0<b_j<b_{j+1}$ for all natural $j$. So,
\begin{equation*}
d_1(t)>-b_1+b_2t^2-b_3t^4+a_1-a_2t^2>0 \quad\text{if}\quad 0<t\le4/5.
\end{equation*}
It remains to prove that
\begin{equation}
d_2(t):=(e-1)t d(t)>0\quad\text{if}\quad 4/5<t<1.
\end{equation}
Since $d_2(1)=0$, it suffices to show that
\begin{equation}
d_2'(t)=\cos t-e \cos et+(e-1) \cos \tau t-(e-1) \tau t \sin \tau t<0
\end{equation}
for $t\in(4/5,1)$.
Since $\cos t,\cos et,\cos \tau t$ are decreasing in $t\in(4/5,1)$ and $\sin \tau t$ is concave in $t\in(4/5,1)$, the desired result follows because for $t\in[t_j,t_{j+1}]$ and $j=0,\dots,n-1$
\begin{equation}
d_2'(t)\le\cos t_j-e \cos et_{j+1}+(e-1) \cos \tau t_j
-(e-1) \tau t_j \min(\sin \tau t_j,\sin \tau t_{j+1})<0,
\end{equation}
where $n:=20$ and $t_j:=4/5+j/(5n)$.
To illustrate the above proof, here is the graph $\{(t,d(t))\colon0<t<1\}$:
This is indeed what I meant, although I missed specifying the range $0<t<1$. Cool job!
this is a plot of $\frac{\sin e\,t-\sin t}{e\,t-t}-\cos\xi \,t$ as a function of $\xi$ for $t=1.1$; the curve does not cross zero in the interval $[\tau,e]$, so I would conclude that (1) does not hold.
This is true due to my mistake in not saying $0<t<1$. Thanks.
|
2025-03-21T14:48:30.055174
| 2020-03-11T22:29:17 |
354744
|
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|
Stack Exchange
|
Are the very hyperlows closed under join?
Call $X$ very hyperlow if $\mathcal{O}^X \le_T \mathcal{O}$, where $\mathcal{O}$ is your favorite $\Pi^1_1$-complete set. Note: Turing reducibility, not hyp-reducibility. Observe that this is a (Turing) degree invariant notion.
Are the very hyperlow Turing degrees closed under join?
I don't think so. There is an $\mathscr{O}$-recursive sequence $\{D_i\}_{i}$ of dense open sets so that for any real $g$ meeting every member of the sequence, $\mathscr{O}^g\leq_T g\oplus \mathscr{O}$. Now it is simple to construct two such reals $g_1$ and $g_2$ so that $g_1\oplus g_2\equiv_T \mathscr{O}$.
Nice, thank you.
|
2025-03-21T14:48:30.055255
| 2020-03-11T23:06:32 |
354745
|
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|
Stack Exchange
|
Are modular representations isomorphic if they're isomorphic after raising to the pth power?
Consider algebraic representations of a reductive group $G$ over a field in characteristic $p$. I even want to allow potentially disconnected reductive groups, i.e. $G$ could be a finite group. (However I'm also interested if the behavior in the connected case is different.)
If $V$ and $W$ are two such representations with $V^{\otimes p}$ ismomorphic to $W^{\otimes p}$, are $V$ and $W$ necessarily isomorphic?
In characteristic $\neq p$ this is obviously false because you can tensor with a character of order $p$, but my question is in characteristic $p$.
Over some rings, there are (non-finitely generated) modules $M \neq 0$ so that $M \otimes M = 0$. Can this happen in this situation?
Here the tensor products are over the field $k$, so no: for this question we would never be worried about something like that. Also, when I said "algebraic representation" I meant finite-dimensional.
Oh, yes, of course you're tensoring over $k$.
Here's one way of constructing counterexamples for finite groups.
Suppose $M$ is a periodic $kG$-module with period $p$: i.e., the $p$th syzygy $\Omega^pM$ is isomorphic to $M$, but $\Omega M\not\cong M$. Then
$$(\Omega M)^{\otimes p}\cong \Omega^pM\otimes M^{\otimes (p-1)}\cong M^{\otimes p},$$
up to projective direct summands.
If $G$ is a $p$-group, so that all projective $kG$-modules are free, then taking the direct sum of $|G|$ copies of $M$ and of $\Omega M$, and adding a free direct summand to whichever is smaller, we can get two modules $X$ and $Y$ of the same dimension, and then $X^{\otimes p}\cong Y^{\otimes p}$ (since they're isomorphic up to free direct summands and have the same dimension).
For example, take $p=2$ and $G$ the quaternion group $Q_8$. The trivial module $k$ has period $4$, so $M=k\oplus\Omega^2k$ has period $2$. $M$ has dimension $10$ and $\Omega M$ has dimension $14$, so in this case we can take the direct sum of two copies of each module (rather than $|G|=8$ copies), to get modules $X=k\oplus k\oplus\Omega^2k\oplus\Omega^2k\oplus kG$ and $Y=\Omega k\oplus\Omega k\oplus \Omega^3k\oplus\Omega^3k$ with $X^{\otimes 2}\cong Y^{\otimes 2}$.
Examples of suitable periodic modules for odd $p$ can be found in
Carlson, Jon F., Periodic modules with large periods, Proc. Am. Math. Soc. 76, 209-215 (1979). ZBL0419.20011.
On the other hand, I think that $X$ and $Y$ must have the same Brauer character if $X^{\otimes p} = Y^{\otimes p}$ so that they are isomorphic if one adds additional assumptions such as simplicity.
Very helpful! Now I wonder about the situation for connected reductive groups...
@user125639 You should probably accept this answer and ask a separate question for the connected case (with a link to this question).
|
2025-03-21T14:48:30.055463
| 2020-03-12T00:29:24 |
354749
|
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"authors": [
"Brendan McKay",
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|
Stack Exchange
|
Ramsey graph for C4 graph
Does there exist a ramsey graph of C4 such that an induced subgraph has a monochromatic C4, no matter how the edges are colored?
You need to write your question more carefully, with all the conditions.
If $m\gt n$ and $p\gt n\binom m2$ then the complete bipartite graph $K_{m,p}$ has the property that, in any edge coloring with at most $n$ colors, it has a monochromatic induced subgraph $K_{2,2}=C_4$.
I.e., if $n$ is finite, we can take $m=n+1$ and $p=n\binom{n+1}2+1$, while if $n=\aleph_\alpha$ we can take $m=\aleph_{\alpha+1}$ and $p=\aleph_{\alpha+2}$.
|
2025-03-21T14:48:30.055537
| 2020-03-12T02:43:53 |
354755
|
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|
Stack Exchange
|
Is the Jordan decomposition for reductive groups algebraic?
Let $G$ be a connected affine algebraic group over $\mathbb{C}$. It's a known fact that elements of $G$ admit a decomposition into semisimple and unipotent elements. Namely, choose a faithful representation $G \subseteq GL(n,\mathbb{C})$. Then given $g \in G$, the Jordan decomposition states that $g = su$, where $s$ is semisimple, $u$ is unipotent, and $su = us$. The crucial fact is then that $G$ contains both $s$ and $u$, and this decomposition is independent of the representation. Therefore, the decomposition $g = su$, as well as the notions of being semisimple and unipotent, are intrinsic to $G$. Furthermore, the decomposition is preserved by morphisms of algebraic groups.
However, if we allow holomorphic maps of Lie groups, the decomposition is not respected. For example, the exponential map $exp : \mathbb{C} \to \mathbb{C}^{*}$ sends unipotent elements to semisimple elements!
So this makes me wonder whether the decomposition depends on the algebraic structure of $G$.
For definiteness, here's my question: Let $G_{1}$ and $G_{2}$ be connected complex reductive groups over $\mathbb{C}$, and let $\phi: G_{1} \to G_{2}$ be a holomorphic group isomorphism. Does $\phi$ preserve the Jordan decomposition?
My feeling is that the answer is yes, and that it will come down to something like $G_{i}$ admitting a unique algebraic structure compatible with the group structure, and any holomorphic isomorphism being automatically algebraic.
Yes, it's easy to check that any biholomorphic isomorphism $G_1\to G_2$ between reductive complex groups is algebraic. One can suppose $G_1,G_2$ connected. Then one reduces to the case of semisimple groups on the one hand (which is essentially immediate) and the case of tori. Since all $n$-dim tori are isomorphic, we can consider biholomorphic automorphisms of the $n$-tori. At the Lie algebra level we have $GL_n(\mathbf{C})$. To pass to the group it has to preserve the lattice kernel, so we get $GL_n(\mathbf{Z})$ as biholomorphism group and this is algebraic.
Thanks for the comment. Could you give a bit more detail?
See Proposition D.2.1 in these notes of Brian Conrad for a proof: http://math.stanford.edu/~conrad/papers/luminysga3.pdf
I upvoted the two answers in the comments but the fact that you are asking also follows easily from algebraicity of holomorphic representations. Four possible proofs are outlined in the link above. The difference between reductive and semisimple is immaterial: just use the fact that a connected reductive group is a product of a semisimple group and a torus to conclude that it acts algebraically.
Now realize $G_2$ as a Zariski-closed subgroup of $GL_n({\mathbb C})$. This gives you a holomorphic representation $\phi: G_1 \rightarrow GL_n({\mathbb C})$. Its algebraicity explained above seals the deal.
I don't think it's true that a connected reductive group is a product of a semisimple group and a torus, although it is certainly a finite quotient of such a product. For example, I think that $\operatorname{GL}_n$ is already not such a product, although it is the quotient $(\operatorname{SL}_n \times \operatorname{GL}_1)/\mu_n$.
Nonetheless, I think that the spirit of the argument is easy enough to carry save: just replace $\phi : G_1 \to G_2$ by $G_1' \times \operatorname Z(G_1)^\circ \to G_1 \xrightarrow\phi G_2 \to \operatorname{GL}n \to \operatorname{SL}{n + 1}$, where $G_1'$ is the derived group of $G_1$ (which is semisimple).
|
2025-03-21T14:48:30.055777
| 2020-03-12T05:37:36 |
354759
|
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|
Stack Exchange
|
Lie groupoid $G$ extensions and principal $\text{Out}(G)$ bundles over Lie groupoids
I am reading the paper Non abelian differentiable gerbes by C. Laurent-Gengoux et.al.
The definition $3.4$ of the paper goes as follows:
Definition : Let $X_1\xrightarrow{\phi} Y_1\rightrightarrows M$ be a Lie groupoid $G$-extension with kernel $\mathcal{K}$. Then $\text{Out}(\mathcal{K},G)\rightarrow M$, considered as an $\text{Out}(G)$-torsor (principal bundle) over the Lie groupoid $[Y\rightrightarrows M]$, is called the band of the $G$-extension $\phi$.
Question : Fix a Lie group $G$ and a Lie groupoid $[Y_1\rightrightarrows M]$. Suppose $P\rightarrow M$ is a principal $\text{Out}(G)$-bundle over the Lie groupoid $[Y_1\rightrightarrows M]$, does there exist a Lie groupoid $G$-extension $X_1\rightarrow Y_1\rightrightarrows M$, such that, the associated band is isomorphic, as principal $\text{Out}(G)$ bundles, to the principal bundle $P\rightarrow M$?
|
2025-03-21T14:48:30.056116
| 2020-03-12T06:59:33 |
354761
|
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|
Stack Exchange
|
Subgradient chain rule
Suppose $$F:\mathbb{R}^n \to \mathbb{R},\; F(x)=\mathrm{max}_\mathrm{eig}(C-\mbox{diag}(x)).$$
I am trying to find a subgradient of $F$ at $x_0$. A subgradient of $\mathrm{max}_\mathrm{eig}$ is given by $xx^T$ where $x$ is an eigenvector with the largest eigenvalue. However, when I try to find a subgradient of the diagonal map $\mbox{diag}$ I realize the subgradient, just like its gradient, is not a matrix but a tensor. How should I proceed with this derivation?
This tensor times the matrix $xx^T$ gives a vector and this is a subgradient, if I am not mistaken.
@Dirk Yes I would suspect so, from the multidimensional chain rule, is there a nice way to carry out the derivation then?
|
2025-03-21T14:48:30.056201
| 2020-03-12T07:29:28 |
354763
|
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|
Stack Exchange
|
How many numbers $\le x$ can be factorized into three numbers which form the sides of a triangle?
Note: Posting in MO since it was unanswered in MSE
Definition: We say that a natural number $n$ has triangular divisors if it has at least one triplet of divisors $n = d_1d_2d_3, 1 \le d_1 \le d_2 \le d_3$, such that $d_1,d_2$ and $d_3$ form the sides of a non-degenerate triangle.
Eg: $60$ has triangular divisors because $60 = 3.4.5$ and $3,4,5$ form a triangle. Note that another triplet of divisors of $60 = 1.4.15$ does not form a triangle but because of the triplet $3,4,5$ the number $60$ qualifies a number with triangular divisors. On the other the number $10$ does not have any triplet of triangular divisors.
The first few numbers in this sequence are
$$
1,4,8,9,12,16,18,24,25,27,32,36,40,45,48,\ldots
$$
I am interested in the density of these numbers. In the linked questions users commented that the almost all integers are expected this property since because most numbers will have a several small prime factors so the natural density was initially thought to be $1$.
However, quite counter intuitively, in one of the long comment which was posted as an answer in the linked question, it was proved that if $n$ has triangular divisors then the largest prime factor of $n$ is less than $\sqrt n$ which immediately implies that the natural density of numbers with triangular divisors is $ < 1 - \log 2 \approx 0.3069$. Experimentally, the data shows that the natural density approaches $0$. Let $f(x)$ be the number of integers $\le x$ with triangular divisors. We have
$$
f(46732002) = 3630678
$$
The graph of $\frac{f(x)}{x}$ vs $x$ given below shows that the density decreases as $x$ increases.
Experimental data: Let $f(x)$ be the number of integers $\le x$ with this property. The graph of $\frac{f(x)}{x}$ vs. $x$ is shown below.
A simple curve fitting gives $\frac{a}{\log x}$ as a good fit with $R^2 = 0.9977$ which suggests that $f(x)$ growth rate somewhere close to a constant times the prime counting function $\pi(x)$. This is rather counter intuitive as mentioned in the comments that we expect almost all integers to have this property.
Higher density of even numbers: A curious observation is the there are significantly more even numbers with triangular divisors as compared to odd numbers. Let $f_o(x)$ be the number of odd numbers $\le x$ with triangular divisors. The graph of $\frac{f_o(x)}{f(x)}$ is shown below.
Question: How many numbers $\le x$ have triangular divisors and why are even numbers more dense than odd numbers?
Related question: Reshaping an object into two integer sided cuboids without changing the total volume.
A fun fact is that for the first values of your sequence (call it $S$), $n\in S$ implies $nr_{0}(n)\in S$, where $r_{0}(n)=\inf{r>0,(n-r,n+r)\in\mathbb{P}^{2}}$. If it's true, it may shed light on your numerical observations.
This has the feel of a project euler problem (which isn't necessarily a bad thing)
Your sequence 1,4,8,9,12,16,18,24,25,27,32,36,40,45,48,… is not in the Online Encyclopedia of Integer Sequences https://oeis.org/ You should include it!
This problem is related to the higher dimensional multiplication table problem (specifically, the three dimensional version). From the work of Tenenbaum, Ford, and Koukoulopoulos this is now well understood in many cases. From their work, one can show that the number of $n$ up to $x$ with triangular divisors is
$$
\ll \frac{x}{(\log x)^{\alpha}}(\log \log x)^{\frac 12}
$$
where
$$
\alpha=1- \frac{2}{\log 3} +\frac{2}{\log 3} \log \frac{2}{\log 3} = 0.27016\ldots.
$$
The power of $\log \log x$, which is the thrust of the work of Ford and Koukoulopoulos, is probably not correct here. But there should be a corresponding lower bound of size $x/(\log x)^{\alpha +o(1)}$; it should be possible to establish this from the techniques in Koukoulopoulos.
Let me now indicate why an upper bound of this magnitude holds. We want to count $n \le x$ that may be written as $n=abc$ with $a\le b\le c$ and $c<a+b$. Clearly we must have $a\le x^{\frac 13}$. Consider first the terms where $a\le x^{\frac 13}/\log x$. Given $a$, note that we must have $b\le \sqrt{x/a}$ and then $c$ lies in the range $b \le c<a+b$, so that given $a$ and $b$ there are at most $a$ choices for $c$. Thus the number of such $n$ with $a\le x^{\frac 13}/\log x$ is
$$
\le \sum_{a\le x^{\frac 13}/\log x} \sum_{a \le b\le\sqrt{x/a}} a \ll
\sum_{a\le x^{\frac 13}/\log x} \sqrt{xa} \ll \frac{x}{(\log x)^{\frac 32}},
$$
which is acceptably small.
It remains to count solutions where $x^{\frac 13} \ge a>x^{\frac 13}/(\log x)$. Note that $b$ (which is $\ge a$ and $\le \sqrt{x/a}$) then lies in the interval $x^{\frac 13}/\log x$ to $x^{\frac 13} (\log x)^{\frac 12}$. Break the possible values for $a$ and $b$ into dyadic intervals $A < a\le 2A$ and $B< b\le 2B$. There are about $(\log \log x)^2$ such dyadic intervals, and both $A$ and $B$ are about $x^{\frac 13}$, and we can assume that $B \le \sqrt{x/A}$. Now we apply Theorem 2.6 from Koukoulopoulos's thesis which gives
$$
\#\{ n\le 4x: \ ab |n, \ \ A< a\le 2A, \ \ B< b\le 2B\} \asymp \frac{x}{(\log x)^{\alpha} (\log \log x)^{\frac 32}}.
$$
Since there are $\ll (\log \log x)^2$ such dyadic blocks, we conclude that the number of $n=abc$ with $a>x^{\frac 13}/(\log x)$ is
$$
\ll \frac{x}{(\log x)^{\alpha}} (\log \log x)^{\frac 12}.
$$
This gives the claimed upper bound.
|
2025-03-21T14:48:30.056551
| 2020-03-12T07:51:14 |
354764
|
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|
Stack Exchange
|
singular support in the singular case
For any constructible sheaf (or D-module) $\mathcal{F}$ over a smooth variety $X$ over $\mathbb{C}$, there is a notion of singular support $SS(\mathcal{F})$ that lives in the cotangent bundle $T^{*}X$ of $X$.
Now, suppose that $X$ is singular (quasi-projective for simplicity), can we define in the same way some singular support that will have the same functoriality property as in the smooth case.
Of course, we can embed $X$ into a smooth $X'$, but can we make it independent of the choice and would $SS(\mathcal{F})$ will still live in $T^{*}X$?
Won't any two embeddings of the same dimension be locally equivalent in the analytic or etale topology? Independence should follow from that.
the drawback of such a definition is for instance if one would have functoriality w.r.t to proper morphisms, because there is no reason that it extends to the embedding.
For $f: X \to Y$ proper, embed $X$ into $A$, embed $Y$ into $B$, and then embed $X$ into $A \times B$ diagonally via $f$. This works fine if $X$ and $Y$ are quasiprojective, I think. I agree it's beneficial to have a general theory though.
@WillSawin, do you know if the containment $SS(F) \subset T^*X$ holds in the singular case?
@user125639 If we embed $X \subset A$ then $T^* X$ is not a subspace of $T^* A $, but rather a quotient, so the inclusion $SS(F) \subset T^* X$ would not be well-defined.
Fair enough. But is there some "a priori" estimate in terms of X in which SS(F) lives?
|
2025-03-21T14:48:30.056684
| 2020-03-12T09:00:36 |
354767
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/354767"
}
|
Stack Exchange
|
Existence of subsequences convergence with weak topology
Let $\left\{ {{\varphi _n}} \right\}$ is the sequence bounded in ${L^\infty }\left( {0,\infty ;H_0^1\left( {0,1} \right)} \right)$. Is there exists $\varphi \in {L^\infty }\left( {0,\infty ;H_0^1\left( {0,1} \right)} \right)$ and subsequence $\left\{ {{\varphi _{{n_k}}}} \right\}$ such that ${\varphi _{{n_k}}} \to \varphi $ in weak star topology of ${L^\infty }\left( {0,\infty ;H_0^1\left( {0,1} \right)} \right)$ and for almost every $t \in \left( {0,\infty } \right)$ we have ${\varphi _{{n_k}}}\left( t \right) \to \varphi \left( t \right)$ weakly in ${H_0^1}\left(0,1 \right)$.
There will certainly be a weak-* convergent subsequence because $L^\infty(0, \infty; H)$ is the dual of the separable Banach space $L^1(0, \infty; H)$ and so its bounded sets are weak-* metrizable and relatively compact.
You can't expect an a.e. convergent subsequence though; this is not even true for $L^\infty(0, \infty; \mathbb{R})$ (consider $f_n(x) = e^{inx}$).
|
2025-03-21T14:48:30.056780
| 2020-03-12T10:33:02 |
354769
|
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"Benjamin Steinberg",
"Ville Salo",
"YCor",
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|
Stack Exchange
|
Is Thompson's group definably orderable?
Is Thompson's group $F$ definably left-orderable? definably bi-orderable?
Orderability definitions: Recall that a group $G$ is left-orderable (resp. bi-orderable) if it admits a left-invariant (resp. bi-invariant) total order. If $S$ is a submonoid of $G$ such that $S\cup S^{-1}=G$ and $S\cap S^{-1}=\{1_G\}$ (call this a cone in $G$), then $\le_S$ defined by: $g\le_S h\Leftrightarrow g^{-1}h\in S$ is a left-invariant total order, and $S$ is conjugation-invariant iff $\le_S$ is bi-invariant. Conversely if $\le$ is a left-invariant total order then $S_\le=\{g:g\ge 1\}$ is a cone (which is conjugacy-invariant iff $\le$ is bi-invariant. We have bi-orderable $\Rightarrow$ left-orderable $\Rightarrow$ torsion-free.
Definability: a subset of a group $G$ is definable if it can be described using logical Boolean operators, quantifiers on group elements, group operations, and parameters in the group: for instance for $a,b\in G$, one can consider the definable subset of $G$: $\{x\in G:\forall y\in G,\exists z\in G:[a,z][b,z^3]=[x,y^5] \vee (x^2=y^2=1)\}$ (this example is totally random). Similarly one can define a definable subset of $G^n$ for every $n$.
A group $G$ is definably (left/bi)-orderable if it admits a (left/bi)-invariant total order that is a definable subset of $G^2$, or equivalently whose positive cone is a definable subset of $G$.
One motivation is that such a group satisfies a single sentence $\Phi$ such that every group satisfying $\Phi$ is also (left/bi)-orderable (and hence torsion-free). Another is whether one can interpret an infinite total order in Thompson's group.
Examples:
(a) The trivial group is definably bi-orderable (this is the only obvious example).
(b) The cyclic group $\mathbf{Z}$ is not definably left-orderable, and more generally any group with a nontrivial abelian direct factor is not definably left-orderable. (Indeed, for such torsion-free $G$, the theory of $G\times\mathbf{Z}/p\mathbf{Z}$ tends to the theory of $G$ when the prime $p$ tends to infinity, so no $\Phi$ as above can exist).
(c) The Heisenberg group over $\mathbf{Z}$ (or $\mathbf{R}$) is definably bi-orderable. This is not hard but a bit tricky.
(d) The free group $F_n$, $n\ge 2$, which is bi-orderable, is not definably left-orderable (according to an expert I asked, every definable submonoid of a free group is a subgroup).
Concerning the question: actually, the bi-invariant total orders on Thompson's group are classified (by Navas and Rivas, arXiv link; GGD 2010) and this is very explicit: I don't know if any of those is definable.
Did you check that your formula doesn't define a biorder on it? (j/k)
Do you allow constants from G
@BenjaminSteinberg yes (this is explicit in the definition I gave of definable: "and parameters in the group").
Sorry i missed that
(About $F_{n\ge 2}$: this group is model-theoretically stable by Sela, so can't interpret an infinite total order, so there's no definable total order, even without invariance assumption.)
Yes, Thompson's group $F$ is definably bi-orderable.
Let $a$ be some element of $F$ with the support of $a$ equal to $(0,1/2)$. Let $b$ be some element of $F$ with the support of $b$ equal to $(1/2,1)$.
We will rely upon the following facts
If $g$ and $h$ are in $F$ then $[a^g,b^h] = 1_F$ if and only if $(1/2)g \leq (1/2)h$.
$F$ acts transitively on the dyadic rationals.
The set $S_1:= \left\{ f \in F \mid [a,b^f] = 1_F \right\}$ is the set of elements $f$ of $F$ for which $(1/2)f \geq 1/2$.
The set $S_2:= \left\{ f \in F \mid [a^f,b] \neq 1_F \right\}$ is the set of elements $f$ of $F$ for which $(1/2)f > 1/2$.
The set $S_3:= \left\{ f \in F \mid \exists g \in F \text{ with } [a^{gf},b^g] \neq 1_F \right\}$ is the set of elements $f$ of $F$ for which there is some dyadic rational $d \in (0,1)$ with $(d)f>d$ (here $d = (1/2)g$).
The set $S_4:= \left\{ f \in F \mid \exists g \in F \text{ with } [a^{gf},b^g] \neq 1_F \text{ and } \forall h \in F \text{ we have } [a^h,b^{hf}] = 1_F \vee [a^g,b^h] = 1_F\right\}$ is the set of elements $f$ of $F$ for which there is some dyadic rational $d \in (0,1)$ with $(d)f>d$ and for all dyadic rationals $e \in (0,1)$ either $(e)f \geq e$ or $e \geq d$ (here $d = (1/2)g$ and $e = (1/2)h$.
Equivalently $S_4$ is the set of non-identity elements $f$ of $F$ for which the right gradient at the infimum of the support of $f$ is strictly greater than $1$.
The union $\{1_F\} \cup S_4$ forms the cone of a bi-order on $F$. Specifically $\{1_F\} \cup S_4$ is the cone of $\preceq^+_{x^-}$ from the article of Navas and Rivas linked to in the question.
EDIT: Since people seem to be interested in the case of $[F,F]$ and Chehata's group I have added below a slightly stronger argument that applies to them.
Let $G$ satisfy the following
Every element of $G$ has only finitely many components of support.
For any $0 < u < v < 1$ and $0 < w < x < y < z < 1$ there is $g \in G$ with $w < (u)g < x < y < (v)g < z$.
There is some element $a$ (that we now fix) of $G$ with a single component of support $(p,q)$ bounded away from $0$ and $1$.
$G$ could be either of $[F,F]$ and Chehata's group.
Fix a non-identity elements $b$ of $G$ with the support of $b$ a proper subset of the support of $a$.
Let $S_5$ be the set of $g$ in $G$ such that $[g,a] = 1_G = [g,b]$.
For $h \in G$ write $\bar{h}$ for the boundary of the support of $h$. If $h$ is in $S_5$ then $\mathrm{supt}(h) \cap \bar{b} = \varnothing = \bar{h} \cap \mathrm{supt}(a)$. It follows that $\mathrm{supt}(h) \cap \mathrm{supt}(a) = \varnothing$. An element of $G$ whose support is disjoint from the support of $a$ is easily in $S_5$ so $S_5$ is the set of elements of $G$ whose supports do not intersect the support of $a$.
Fix a non-identity element $c$ of $G$ with the support of $c$ a subset of $(q,1)$.
Let $S_6$ be the set of $g$ in $G$ such that there exists $h \in G$ with $[a^h,a] = 1_G = [a^h,b]$ and $[a^h,c] \neq 1_G \neq [a^h,a^g]$.
We will now argue that $S_6$ is the set of those $g$ in $G$ with $q < (q)g$.
Let $g$ be in $S_6$. There exists a conjugate $a^h$ of $a$ whose support does not intersect $\mathrm{supt}(a)$ but does intersect $\mathrm{supt}(c)$ and does intersect $\mathrm{supt}(g)$.
Any conjugate of $a$ must have a single component of support so either $\mathrm{supt}(a^h) \subseteq (0,p)$ or $\mathrm{supt}(a^h) \subseteq (q,1)$. Since the support of $a^h$ intersects the support of $c$ we must have $\mathrm{supt}(a^h) \subseteq (q,1)$. Since the support of $a^g$ intersects the support of $a^h$ it follows that the support of $a^g$ intersects $(q,1)$. Now it follows that $q < (q)g$.
If $g$ is in $G$ and $q < (q)g$ then $g$ is in $S_6$ by condition 2. above.
$S_6$ now corresponds to $S_1$ from the original proof and the rest of the construction works similarly.
For the non-trivial direction of 1. (I'm sure it's simpler than this though):
If $(1/2)g = t$ and $(1/2)h < t$ then $(t)b^h \neq t$. If $(t)b^h < t$ then for small $\epsilon > 0$ we have $(t+\epsilon) a^g b^h = (t+\epsilon) b^h = x < t$ so $(t+\epsilon) b^h a^g = x a^g \neq x$ because $x$ is in the support $[0,t]$ of $a^g$. If $(t)b^h > t$, then let $t'$ be such that $(t')b^h = t$ and we have
$(t'+\epsilon) b^h a^g = (t'+\epsilon) b^h$ because $(t'+\epsilon) b^h > t$, while
$(t'+\epsilon) a^g b^h = (t'+\epsilon) b^h$ would imply $(t'+\epsilon) a^g = t'+\epsilon$, contradicting $t' < t$.
Great! It seems that this definition works under the bare assumption of a subgroup $G$ of $\mathrm{Homeo}^+([0,1])$ with a dense orbit $Gx$ on the open interval $]0,1[$, such that there exists $a,b$ with support $]0,x[$ and $]x,1[$. This defines $S_4$ as the (conjugation-invariant) set of $g\in G$ such that for some $y$ with $yg>y$ and $zg\ge z$ for all $z\le y$. That this defines a (strict) total order requires something on $G$. It seems enough that $G$ acts piecewise analytically.
@VilleSalo just use that if they commute then the support of each one is invariant by the other one.
@YCor So if I've followed correctly, this argument implies that Chehata's group $G(I)$ (which contains his example of a simple bi-orderable group) is definably bi-orderable.
@shane.orourke yes, this works for a wealth of known "rich" subgroups of $\mathrm{Homeo}^+([0,1])$. Actually, with only assumption existence of $a,b$ and density, one already gets a definable partial bi-ordering (which is not trivial, since $a,b>1$).
@shane.orourke I've actually had a look at Chehata's 1952 paper. His group is the set of piecewise affine increasing homeomorphisms of an ordered field, that are identity near $\pm\infty$. Thus the argument does not apply, since we don't have $a,b$ at disposal. For the same reason, the argument does not apply to the derived subgroup $[F,F]$ (which is simple and consists of elements of $F$ that are identity outside $[1/n,1-1/n]$ for some $n$).
@YCor But I think the group $G(I)$ (where $I$ is a compact interval) does satisfy your condition, and is thus definably bi-orderable, although the simple subgroup $G(I')$ (where $I'$ is the interior of $I$) does not.
@shane.orourke OK, I agree. Actually $F$ is similar to $G=\mathrm{PL}(I)$: just defined by restricting to a segment and adding dyadic constraints on slopes and breakpoints, and similarly, $[F,F]$ is simple as well as $[G,G]$. It seems that $F$ and $[F,F]$ are best-known now, but indeed Chehata's work (1952) is anterior to R. Thompson's groups.
|
2025-03-21T14:48:30.057456
| 2020-03-12T11:22:59 |
354772
|
{
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"authors": [
"Emil Jeřábek",
"Erik D",
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|
Stack Exchange
|
Markov's principle from constant domain logic
I am looking for a proof of and/or a reference for the result that Markov's principle can be proved in the framework of constant domain logic. By constant domain logic, I mean intuitionistic logic plus the axiom
$\forall x(P(x) \,\vee\, Q) \to \forall xP(x) \,\vee\,Q \quad$ where x is not free in $Q$.
This result is alluded to, without proof, in the entry about intuitionistic logic in the Stanford encyclopedia of philosophy:
https://stanford.library.sydney.edu.au/archives/fall2008/entries/logic-intuitionistic/#KriSemForIntLog (I have found no other reference.)
$\forall x,(A(x)\lor\neg A(x))\to\forall y,(\neg A(y)\lor\exists x,A(x))\to\forall y,\neg A(y)\lor\exists x,A(x)$. So, that’s even stronger than MP: decidability is preserved by existential quantification.
Thank you. If you re-post that as an answer I will mark this as solved.
Also, if you happen to have a reference where this (or similar) was first noticed, I'd be happy to know.
Emil Jeřábek answered in a comment:
$\forall x(A(x) \vee \neg A(x)) \:\to\: \forall x (\exists y A(y) \vee \neg A(x))
\:\to\: \exists y A(y) \vee \forall x(\neg A(x))$
So, that’s even stronger than MP: decidability is preserved by existential quantification.
|
2025-03-21T14:48:30.057565
| 2020-03-12T11:38:37 |
354775
|
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"Denis Serre",
"Gustave",
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|
Stack Exchange
|
Common eigenvalues for two Sturm-Liouville problem
Does exist in literature any results concerning the common eigenvalues for the two eigenvalue problems of the form
$$y''(x)=\lambda^2 y(x)+\lambda a(x)y(x), \ x\in(0,1), $$$$z''(x)=\lambda^2 z(x)-\lambda a(x)z(x), \ x\in(0,1),$$
with boundary conditions $$y(0)=y(1)=z(0)=z(1)=0.$$
Things are trivial if $a$ is constant function, but for $a=a(x)$ I couldnt find any way to handle this problem. Any suggestions?. Thank you.
Do you mean the Dirichlet boundary condition (everything $=0$) ?
@DenisSerre Yes thank you sir. I have edited the post.
It appears that what you need is the tensor Bezoutian for operator polynomials. Its definition and relation to the counting of the common eigenvalues is briefly reviewed in the following article (Theorem 9), where also references are given for further details:
Lancaster, Peter, Common eigenvalues, divisors, and multiples of matrix polynomials: A review, Linear Algebra Appl. 84, 139-160 (1986). ZBL0627.15004.
Your operator polynomials (in $\lambda^{-1}$, to make the constant coefficient exactly equal to $1$, as in Theorem 9 above) are $P_x^\pm(\lambda^{-1}) = 1 \pm \lambda^{-1} a(x) - \lambda^{-2}\partial_x^2$. Their Bezoutian works out to be the oprator matrix $B = [B_{ij}]$, whose operator entries are defined by the identity
$$
\frac{P^+_{x_1}(\lambda^{-1}) P^-_{x_2}(\mu^{-1}) - P^+_{x_1}(\mu^{-1}) P^-_{x_2}(\lambda^{-1})}{\lambda^{-1} - \mu^{-1}}
= \sum_{i=0}^1 \sum_{j=0}^1 \lambda^{-i} \mu^{-j} B_{ij} .
$$
In this particular case, we get
$$
B = \begin{bmatrix}
a(x_1) + a(x_2) & \partial_{x_1}^2 - \partial_{x_2}^2 \\
\partial_{x_1}^2 - \partial_{x_2}^2 & a(x_2) \partial_{x_1}^2 + a(x_1) \partial_{x_2}^2
\end{bmatrix} ,
$$
acting on $[\begin{smallmatrix} u(x_1,x_2) \\ v(x_1,x_2) \end{smallmatrix}]$, with $u$ and $v$ satisfying Dirichlet boundary conditions on the square $(x_1,x_2) \in [0,1]^2$.
According to the theorem in Lancaster's review, the dimension of the kernel of $B$ counts the number of common eigenvalues of $P^\pm(\lambda)$ (excluding $\lambda=0$, I think, but that value is never an eigenvalue under Dirichlet boundary conditions).
Thank you @Igor Khavkine for this great answer. I've taken a look at this paper, I found that it deals with bounded operator case. I 'm not sure that this can be generalized to unbounded operator case. Thank you in advance to clarify more things.
@Gustave, just a remark. If you multiply both of your operator polynomials by $(\partial_x^2)^{-1}$ (with the inverse defined by Dirichlet boundary conditions), then the operator coefficients of the powers of $\lambda$ become bounded. It's not hard to work out the new corresponding $B$ operator matrix.
Thank you sir for the answer. Best regards.
|
2025-03-21T14:48:30.057762
| 2020-03-12T13:28:31 |
354778
|
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|
Stack Exchange
|
Finite concatenation-free languages
Suppose, $A$ is a finite alphabet. $L \subset A^*$ is a language. Let's call $L$ concatenation-free iff $\forall u, v \in L$ we have $uv \notin L$.
Does there exist some function $c: \mathbb{N} \to (0; 1)$, such that for any finite language $L \subset A^*$, there exists a concatenation-free sublanguage $L_0 \subset L$, such that $|L_0| \geq c(|A|)|L|$?
The only thing I currently know about this problem, is that we can take $c(1) = \frac{1}{3}$. That is a direct consequence of Erdos-Sidon theorem, that states:
$\forall A \subset \mathbb{Z}$ $\exists$ a sum-free $A_0 \subset A$, such that $|A_0| \geq \frac{|A|}{3}$
However, I do not know how to deal with $|A| \geq 2$.
This question on MSE
|
2025-03-21T14:48:30.057845
| 2020-03-12T13:30:38 |
354779
|
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|
Stack Exchange
|
Generalization of positive definiteness for a non-smooth quadratic form
I have a "quadratic form" defined as follows:
$$f(\vec x) = \sum_{i<j} c_{ij} x_i x_j + \sum_i a_i(x_i) x_i^2$$
where the coefficients $c_{ij}$ are fixed constants, but the coefficients $a_i(x_i)$ can depend on the sign of $x_i$:
$$a_i(x_i) = \cases{
a_i^+ & $x_i \ge 0$ \\
a_i^- & $x_i < 0$}$$
I assume that both $a_i^+, a_i^-$ are positive. I also assume all the quantities here are real for simplicity.
I need to determine whether $f(\vec x)$ is "positive definite", that is, $f(\vec x)\ge 0$ for all $\vec x$.
If $a_i^+ = a_i^-$ for all $i$, then this question reduces to computing the eigenvalues of a matrix with entries $(a_i; c_{ij})$ and checking whether they are all positive.
In my case, where $a_i^+$ can be different from $a_i^-$, it seems that I have to compute an eigenvalue decomposition for each possible matrix (one for each quadrant, so exponentially many), and if the eigenvector points in the direction of the quadrant, check the sign of the corresponding eigenvalue. But this approach is combinatorially intractable.
Is there a better way of characterizing when $f(\vec x) \ge 0$ for all $\vec x$?
Two obvious observations:
If you replace $a_i=min(a_i^_,a_i^-)$ and the corresponding quadratic form is positive definite then f is positive definite.
If you check positive definiteness for predefined sign's of the $x_i's$ and you have a term $c_{ij}x_ix_j$ which is always positive it suffices to show positive definitess with a $c_{ij}$ changed towards 0.
You could replace $x_i=u_i-v_i$ with $u_i, v_i\geq 0$, $u_i v_i=0$. Then we have $a_i(x_i)x_i^2= a_i^+ u_i^2+a_i^- v_i^2$. Hence you will get a standard quadratic form and you need to prove that for nonnegative $u_i, v_i$ it is positive. You then do not have exponentially many cases. The system is however twice as large.
|
2025-03-21T14:48:30.057996
| 2020-03-12T15:09:51 |
354786
|
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"Gerry Myerson",
"Noah Schweber",
"Shahrooz",
"Taras Banakh",
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"https://mathoverflow.net/users/19885",
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"https://mathoverflow.net/users/3684",
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"https://mathoverflow.net/users/8133"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/354786"
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|
Stack Exchange
|
Topology and infinite number of primes
One of the strange proofs (among the other beautiful proof) in the book "Proofs from the book" is the fifth one, which uses a special topology on the set of integer numbers, to prove there are infinite prime numbers.
My question is:
Is this method special just for this case, or is there anything deeper and this technique (or its generalization) can be used for some other kinds of problems?
A related question (by @ToddTrimble comment):
Is Fürstenberg's topology useful?
When unwound straightforwardly, the "topological" proof is really just the usual proof in disguise (and it doesn't use anything more than the language of topology in the first place). That said, I am under the impression that the topology introduced there is actually interesting, just not really for that reason.
If anybody could mention which topology it is, it would help answering the question. (If it's the profinite one, it definitely has a huge number of uses, say $p$-adic numbers and so on.)
Certainly it's the profinite topology, i.e., the topology inherited from the profinite completion $\hat{\mathbb{Z}}$ that appears in the adeles, as pointed out by Chandan Singh Dalawat here: https://mathoverflow.net/q/42589/2926
@Todd Trimble It seems that I must delete this question, yes?
It's your call. I'm not suggesting you should. Maybe someone would like to say something even more enlightening than the hints given so far in comments; dunno.
It is generous!
PS the "Furstenberg topology" is the terminology given by some subcommunity to the profinite topology.
There is a modfication of the Furstenberg topology, called the Golomb topology (see e.g https://mathoverflow.net/q/285557). It is generated by the base consisting of the arithmetic progressions $\mathbb N\cap(a+b\mathbb Z)$ with coprime $a,b$. By the famous Dirichlet Theorem, the set of prime numbers is dense in the Golomb topology (but not in the Furstenberg one). Golomb popularized this topology expecting to find a topological proof of the famous Dirichet Theorem on density of primes. But till now such a topological proof has not been found.
@Taras I am aware of Dirichlet's Theorem on primes in arithmetic progressions, but that's not a density result. What density result do you have in mind, please?
@GerryMyerson I had in mind the density in topological sense (as the density of the set of primes in the Golomb topology).
|
2025-03-21T14:48:30.058205
| 2020-03-12T15:26:53 |
354787
|
{
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"authors": [
"Jeremy Rickard",
"R. van Dobben de Bruyn",
"https://mathoverflow.net/users/105652",
"https://mathoverflow.net/users/22989",
"https://mathoverflow.net/users/82179",
"nikola karabatic"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/354787"
}
|
Stack Exchange
|
Projective module which splits off sequence of submodules, but not the sum
Does there exist an example of a module $X$ over some ring $R$ together with submodules $T_i$ such that:
$X$ is projective,
$X$ splits as an internal direct sum $X\cong T_1\oplus T_2\oplus \ldots \oplus T_n\oplus S_n$ (with some $S_n$) for every $n$,
$X$ does not split off the infinite direct sum $\bigoplus_{i=1}^\infty T_i$,
$R$ is hereditary.
Remark: I also don't know the answer without the last condition, so this would already be interesting, though for my specific application I definitely need all conditions.
In the second condition, do you just want an abstract isomorphism for each $n$, or do you want the sum of the submodules actually to be a direct summand?
I meant an internal direct sum, as in the answer below, added it to the question -- thanks
Here is an example if we interpret all direct sums as internal direct sums.
Example. Let $R$ be a discrete valuation ring with uniformiser $\pi$ and fraction field $K$. Let $X = R^{(\mathbf N)}$, and let $T_i$ be the free rank $1$ submodule with basis $\pi e_{i+1}-e_i$. Then the natural map
$$\bigoplus_{i=1}^n T_i \to X$$
is injective with image $T_{\leq n} = \operatorname{span}(\pi e_2 - e_1, \ldots, \pi e_{n+1} - e_n)$, because the latter clearly has rank $n$. Moreover,
$$S_n = \bigoplus_{i > n} Re_i \subseteq X$$
is a complement of $T_{\leq n}$: one easily sees that $S_n \cap T_{\leq n} = 0$, and they span $X$ because $e_n = \pi \cdot e_{n+1} - (\pi e_{n+1} - e_n)$, etcetera. But if $T = \bigoplus_{i \in \mathbf N} T_i = \sum_i T_i \subseteq X$, then
\begin{align*}
X/T &\stackrel\sim\to K\\
e_i &\mapsto \pi^{-i}.
\end{align*}
This surjection does not split because $X$ has no infinitely divisible elements. $\square$
What's going on is that we wrote $K$ as a filtered colimit of surjections $S_n \twoheadrightarrow S_{n+1}$ of free modules:
$$K = \underset{\substack{\longrightarrow \\ n}}{\operatorname{colim}}\ S_n.$$
Each $X \twoheadrightarrow S_n$ has a splitting $S_n \hookrightarrow X$, but $X \twoheadrightarrow K$ does not.
Just a comment that this nice construction also works for $R=\mathbb{Z}$. Let $0\to T\to X\to\mathbb{Q}\to0$ be a free resolution of $\mathbb{Q}$ and write $T$ as a direct sum $T=\bigoplus_iT_i$ of copies of $\mathbb{Z}$.
@JeremyRickard absolutely! That's actually the example I started with, but the DVR case was easier.
|
2025-03-21T14:48:30.058393
| 2020-03-12T16:30:11 |
354788
|
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"url": "https://mathoverflow.net/questions/354788"
}
|
Stack Exchange
|
Connected graphs $G$ with $\delta(G) > 1$ and long minimum size roundtrips
Let $G = (V,E)$ be a finite, connected, simple, undirected graph. By a roundtrip of $G$ we mean a map $r:\{0,\ldots,n\} \to V$ for some $n\in\mathbb{N}$ with the following properties:
$r$ is surjective,
$\{r(k), r(k+1)\} \in E$ for all $k \in \{0, \ldots, n-1\}$, and
$r(0) = r(n)$.
The integer $n$ is called the length of the roundtrip $r$. An easy inductive argument shows that we can select $n$ such that $n \leq 2|V(G)|$. Let $\rho(G)$ denote the minimum length of any roundtrip in $G$.
For $k \geq 1$ let ${\frak r}(k)$ the maximum $\rho(G)$ where $G = (V,E)$ is a connected graph with $|V| =k$ and $G$ has no vertex of degree $1$. We have ${\frak r}(k) \leq 2k$ for all positive integers $k$.
Question. What is $$\lim\sup_{k\to\infty}\frac{{\frak r}(k)}{k}?$$
(Note: I ask for $\lim\sup$ instead of $\lim$ because while I am quite certain that the limit exists, I haven't been able to prove it.)
The answer is $2$. To see this, let $G$ be the graph which consists of two triangles connected by a path with $k-4$ vertices. Then $G$ has $k$ vertices and the length of a shortest roundtrip is $2k-4$. Since $\lim_{k \to \infty} \frac{2k-4}{k}=2$, the answer is at least $2$. On the other hand, you have already noted ${\frak r}(k) \leq 2k$ for all positive integers $k$, so the answer is at most $2$.
|
2025-03-21T14:48:30.058517
| 2020-03-12T18:10:53 |
354796
|
{
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"authors": [
"Brendan McKay",
"https://mathoverflow.net/users/9025"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/354796"
}
|
Stack Exchange
|
Number of subgraphs with matching of size $n$ for a complete bipartite graph
Say we have a $K_{n,n}$ bipartite graph (i.e. a complete bipartite graph with $n$ nodes on each side). We induce a subgraph by deleting some subset of edges. There are $2^{n^2}$ possible subgraphs. How many of these subgraphs still contain a matching of size $n$. Is there a known non-trivial upper bound?
Furthermore, how can you compute the number of subgraphs which contain a matching of at least size $x$ for $x\leq n$?
It looks like the number of such subgraphs is known only up to $n=7$. See OEIS A227414 and Example 10.4 in Permutohedra, Associahedra, and Beyond, by Alexander Postnikov.
$n=8$ would be a fairly easy calculation. One day...
If a bound down to a factor of $1+o(1)$ will do: Almost all bipartite graphs $G$ with one side $X$ and the other side $Y$ still have a perfect matching.
Here is a sketch of the proof: Form a subgraph $G$ of $K_{n,n}$ as follows: For each edge $e$ in $K_{n,n}$, flip a fair coin as to whether to keep $e$ in $G$ or not. Then every subgraph of $K_{n,n}$ has the same probability of resulting from this. However, one can show via a counting argument that with high probability, $G$ is such that every set $S$ in one side has at least $|S|$ neighbours on the other side. [Indeed, check out proofs that say that random bipartite $k$-regular graphs for $k \ge 3$ are expanders, similar principle at work here.]
|
2025-03-21T14:48:30.058633
| 2020-03-12T19:18:17 |
354800
|
{
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"authors": [
"Alessio Di Lorenzo",
"https://mathoverflow.net/users/90127"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/354800"
}
|
Stack Exchange
|
Estimates on divergence-type operator for the matrix
Is there any result (Schauder-like estimates, $L^2$ estimates or similar) to equations of the form
$$
{\rm div}(Av)=f
$$
where $A$ is the "unknown" (i.e. I would like estimates on $A$ depending on $f$,$v$)?
Thanks!
Of course not, because this equation is far from being elliptic. Actually, it is even under-determined, in the sense that you have only one equation, for $n^2$ unknowns (where the matrix is $n\times n$).
Let me however give you a result in this direction, that I discovered two years ago, which has important consequences in various domains.
Let $A$ be symmetric, with entries in the space $\cal M$ of bounded measures. Suppose that $A$ is positive semi-definite. Suppose at last that ${\rm div}(A\vec e_i)\in\cal M$ for every $i\in[1,n]$. Then $(\det A)^\frac1n$, which is a priori a bounded measure, is actually an element of $L^{\frac n{n-1}}$. This qualitative result is associated with a functional inequality.
Thanks! It is obvious indeed that one needs less generality, and I was looking for results on these lines
|
2025-03-21T14:48:30.058733
| 2020-03-12T19:30:05 |
354801
|
{
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"authors": [
"Neil Strickland",
"https://mathoverflow.net/users/10366"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/354801"
}
|
Stack Exchange
|
Homological and homotopical equivalence of complex analytic varieties
Consider a map between two complex analytic varieties of finite type $f:X\to Y$. Suppose that $f$ induces isomorphisms on cohomology with (constant) integral coefficients. Under what reasonable hypotheses can we conclude that it induces an equivalence on homotopy groups?
I refer in particular to this article https://arxiv.org/pdf/1710.05366.pdf, Proof of proposition 3.17 on page 14, where the Authors use a result of this kind. I cannot understand why classical hypotheses like nilpotency of the spaces are verified.
Thank you in advance.
I don't think that the proof you cite can be considered complete. There is no obvious reason why homology equivalence should be sufficient in that context.
|
2025-03-21T14:48:30.058827
| 2020-03-12T20:04:01 |
354803
|
{
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"authors": [
"LSpice",
"Nate Eldredge",
"Siddharth Bhat",
"https://mathoverflow.net/users/123769",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/4832",
"https://mathoverflow.net/users/84768",
"reuns"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/354803"
}
|
Stack Exchange
|
Logarithm of the Fourier transform?
I've found this paper on the logarithm of the discrete fourier transform which proves that
$$
log F = 1/4 i \pi (I - (1 +i)F + F^2 - (1 - i)F^3)
$$
where $F$ is the unitary discrete Fourier transform operator.
Is there a similar analogous version known for the infinite dimensional Fourier transform?
I am asking this from the point of view of trying to better understand cumulants, whose generating function is the logarithm of the characteristic function. The characteristic function is the fourier transform of the PDF of a random variable, thus motivating this question.
EDIT: Please take as many "niceness" assumptions as needed to answer the question regarding convergence.
First one has to worry about convergence of the logarithm in the infinite-dimensional setting ….
The spectrum of the Fourier transform on $L^2(\mathbb{R})$ is ${\pm 1, \pm i}$, so it should suffice to choose any polynomial which agrees with (your favorite branch of) the complex logarithm on those four points.
@NateEldredge I'm really unfamiliar with the "proper" fourier transform. I've only used it from a "physics" standpoint. Could you please expand your comment into an answer, as I do not really understand what you've said? That would be very helpful for me :)
Reading your question again, though, I think you may be mixing up two different things. The cumulant is the log of the Fourier transform of a specific function. But the paper you link, and my comment, are about taking the log of the Fourier transform operator in the sense of spectral theory. It's the difference between $\log(F[f])$ and $(\log F)[f]$. It's like the fact that if $A$ is a matrix and $v$ is a vector, $A^2 v$ is not the same as taking $Av$ and squaring each entry.
Let us start with the identity $F^4=I$. As a result, we have formally
\begin{align}
\ln F&=\ln(I+F-I)=\sum_{k\ge 1}\frac{(-1)^{k-1}(F-I)^k}{k}
\\&=
\sum_{1\le k\le 3}\frac{(-1)^{k-1}(F-I)^k}{k}+
\sum_{k\ge 4}\frac{(-1)^{k-1}}{k}\sum_{0\le l\le 3}\binom{k}{l} F^l(-1)^{k-l}
\\&\hskip94pt+
\sum_{k\ge 4}\frac{(-1)^{k-1}}{k}\sum_{k\ge l\ge 4}\binom{k}{l} F^l(-1)^{k-l},
\end{align}
and since in the last sum $F^l=F^j$ with $j\equiv l\mod 4$, we get a polynomial in $F$ with degree 3, that is
$$
\ln F= a_0+a_1 F+a_2 F^2
+a_3 F^3.$$
I guess that it is straightforward to get the expressions of the $a_j$ from the above identity. Also you can use the fonction $g(x)=e^{-π x^2}$ which is such that $(F-I) g=0$ implying that
$$
0=(\ln F)(g)=(a_0+a_1 +a_2
+a_3)(g)\Longrightarrow \sum_{0\le j\le 3}a_j=0,
$$
and in fine you will get the same coefficients as yours.
$$F=e^{\frac{1}{4} \pi i \left(D^2-x^2+1\right)}$$
This is for unitary case.
Your operator doesn't converge for most kind of functions.
For $F$ any linear map such that $F^4=Id$ then $F= \sum_{k=0}^3 i^k T_k$ where $T_k =\frac14 \sum_{m=0}^3 i^{-mk} F^m $, $T_k^2=T_k,T_k T_l = 0$ which gives that $(\sum_{k=0}^3 c_k T_k)^n = \sum_{k=0}^3 c_k^n T_k$.
With $L=\sum_{k=0}^3 \frac{ki\pi}{2} T_k$ we get that $\exp(L)=\sum_{n=0}^\infty \sum_{k=0}^3 \frac{(\frac{ki\pi}{2})^n}{n!} T_k=\sum_{k=0}^3 \exp(\frac{ki\pi}{2}) T_k= F$.
|
2025-03-21T14:48:30.059056
| 2020-03-12T20:36:52 |
354804
|
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"Alexander Woo",
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|
Stack Exchange
|
Parametrization of Plucker Equations for Grassmannian $Gr(2,n)$ by ordered quadruples
As David Speyer pointed out in a comment to this MO question, Number of Plücker relations for a Grassmannian,
'many people mean specific lists of relations when they say "the Plucker relations"'.
I'm looking for a reference to the parametrization of the set of Plucker equations for the Grassmannian $Gr(2,n)$ by ordered 4-sets of indexes. Here is a precise theorem:
Point $x$ belongs to $Gr(2,n)$ iif its Plucker coordinates $\Delta_{a,b}$ satisfy the Plucker equation
$$ \Delta_{i,k} \Delta_{j,l} = \Delta_{i,j} \Delta_{k,l} + \Delta_{i,l} \Delta_{j,k} $$
for every ordered quadruple $1 \leq i < j < k < l \leq n$.
Is there a paper that explicitly mentions such a parametrization? If not, what might be an illuminating passage from the standard parametrizations to the one above?
This theorem is precisely stated (for all $Gr(k,n)$ or $G/B$) and proved in several textbooks, for example Fulton's Young Tableaux, or Miller and Sturmfels's Combinatorial Commutative Algebra.
|
2025-03-21T14:48:30.059282
| 2020-03-12T21:01:35 |
354808
|
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"Nate Eldredge",
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|
Stack Exchange
|
Prove that fractional Brownian motion is not a semimartingale using the p-variation
What follows, up to the horizontal line, is taken from Rogers "Arbitrage with fractional Brownian motion".
Consider an interval $[0,T]$ on which is defined the fractional Brownian motion $B$, and consider its partitions $\pi_n = \{t^n_k = \frac{kT}{n} : 0\le k\le n\},\ n\in\mathbb N$.
Let $p\ge1$, the $p$-variation of $B$ is
$$
V_p(B) = \lim_{n\to\infty} \sum_{k=0}^{n-1} |B(t^n_{k+1})-B(t^n_k)|^p =
\begin{cases}
\infty, & \text{if }\ pH < 1, \\
0, & \text{if }\ pH > 1.
\end{cases}
$$
If $H>1/2$ we can choose $p\in(1,\frac1H)$ so that $pH<1$, then the $p$-variation is infinite, hence the quadratic variation of $B$ is infinite too.
If $H<1/2$ we can choose $p>2$ so that $pH<1$, then again we obtain that the $p$-variation and the quadratic variation of $B$ are infinite.
In both cases the quadratic variation of $B$ is not finite, hence the fBm is not a semimartingale for $H\ne1/2$.
Could somebody further explain the above reasoning? I don't fully get what has to be proved, is it related to the fact that a semimartingale has to have finite variation? But which variation: quadratic, p-variation or another one?
Moreover, I don't understand how to deduce what the quadratic variation is, given that we know the p-variation. Is it related to the fact that given $p_1<p_2$ then $V_{p_2}\le V_{p_1}$?
Fianlly, what about the case $H=1/2$, in which $B$ is the usual Brownian motion? If we take $p\in(1,2)$ then we are still in the case $pH<1$ and so the $p$-variation is infinite hence the quadratic variation of $B$ is infinite too, contradicting the fact that B is a martingale.
I think you've misstated Rogers' argument for $H > 1/2$. Read page 98 of the paper again.
The key is that a semimartingale always has finite quadratic variation, and if its quadratic variation is zero, then it is of bounded (1-)variation.
It is true that if $p_1 < p_2$ then $V_{p_2} \le V_{p_1}$. More is true: if $V_{p_1} < \infty$ then $V_{p_2} = 0$.
And in the $H=1/2$ case, it is absolutely true that Brownian motion has infinite $p$-variation for every $p < 2$, but that does not imply its quadratic variation is infinite, and indeed it is not.
@NateEldredge Oh ok I got it thank you very much!
@NateEldredge Even stronger, Brownian motion has infinite $p$-variation for every $p\leq 2$. The $2$ variation of Brownian motion is infinite a.s.
@user341290: You mean in the sense where we take the supremum over partitions, instead of the limit in probability as mesh goes to zero?
@NateEldredge Yes. Small point, but it is good to clarify in questions like this.
Assume $B$ is a semimartingale, then it has finite quadratic variation.
Recall that if $s < b$ then $V_b \le V_s$.
If $H<1/2$ we can choose $p>2$ s.t. $pH<1 \implies V_p = \infty \implies \infty\le V_2 \implies V_2 = \infty$, i.e. the quadratic variation ($p=2$) is infinite too: contradiction.
If $H>1/2$ we can choose $p\in(\frac1H,2)$ s.t. $pH>1 \implies V_p = 0 \implies V_2 \le 0 \implies V_2 = 0 \implies B$ must have finite variation. But on the other hand, for $p\in(1,\frac1H)$ we have $V_p = \infty$, hence $B$ cannot have finite variation: contradiction.
Either way, if $H\ne\frac12$, the fBm is not a semimartingale.
Let $B^{H}_{\cdot}$ denote the fBM with Hurst parameter $H\in (0,1)$. This is a centered Gaussian process with Covariance function (i.e.: $\Sigma(t,s) =\mathbb{E}[B_t^HB_s^H] $):
$$
\Sigma(t,s)= 2^{-1}\left(
t^{2H} - s^{2H} - |t-s|^{2H}
\right)\qquad \boldsymbol{(1)}.
$$
We may therefore apply this Komologorov-Chestov argument (i.e.: Theorem 1 in these lecture notes) to (1) deduce that $B_{\cdot}^H$ has a version with $H$-Hölder-continuous paths. Therefore, it must have a version with finite $H^{-1}$-variation.
Since we know that every semi-martingale has finite quadratic variation then if $H\neq 2^{-1}$ $B_{\cdot}^H$ is not a semi-martingale.
Note: This doesn't show that it is a (semi-)martingale for $H=2^{-1}$.
|
2025-03-21T14:48:30.059538
| 2020-03-12T23:30:48 |
354811
|
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"Augusto Santi",
"GH from MO",
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|
Stack Exchange
|
Is it possible to determine whether the sequence $\,a_0=p,\;a_{n+1}=(a_n-2)\cdot a_n+2\,$ will reach another prime number?
Given a prime $\,p\,$ let's consider the following sequence:
$a_0=p$
$a_{n+1}=(a_n-2)\cdot a_n+2$
Is it possible to determine whether the sequence $\,a_n\,$ will reach, sooner or later, another prime number?
Some examples:
for $\,p=2$, $\;\;a_1=2\;\;$ (prime)
for $\,p=3$, $\;\;a_1=5\;\;$ (prime)
for $\,p=29$, $\;\;a_2=614657\;\;$ (prime)
for $\,p=31$, $\;\;a_5=185302018885184100000000000000000000000000000001\;\;$ (prime)
Many thanks.
[ Added ]
Experimental evidence shows that, up to 100, only for the following prime numbers $\,p\,$ the sequence $\,a_n\,$ should never reach another prime:
$$13,19,23,43,53,59,61,71,73,79$$
In order to find out the possible divisors of $\,a_n=(p-1)^{2^n}+1\,$ (see the answer of GH from MO) the following result can be exploited:
the only prime divisors of $\,a_n\,$ are of the form $\,k\cdot2^{n+1}+1$.
Example:
$$(13-1)^8+1=17\cdot97\cdot260753=(1\cdot2^4+1)\cdot(6\cdot2^4+1)\cdot(16297\cdot2^4+1)$$
Notice that we can't even establish the infinitude of primes of the form $n^2+1$, let alone with the added condition of $n$ being a power of $p-1$.
@GjergjiZaimi: I think this question is slightly different. I suspect that there exists a prime $p$ such that $a_n$ is never prime, but probably we will never be able to decide this.
Your question can be reformulated as follows.
Question. If $p$ is a prime, does there always exist a positive integer $n$ such that $(p-1)^{2^n}+1$ is also a prime?
I believe that this question is out of reach at present (my guess is that the answer is "no", but we will never know). Similar to the well-known questions on Fermat numbers and its generalizations.
Maybe did you want to write $(p-1)^{2n}+1$?
@AugustoSanti: For $p=31$, your $a_5$ is $30^{32}+1$, not $30^{10}+1$. In general, your $a_n$ is $(p-1)^{2^n}+1$, as I wrote. The exponents are $2$-powers.
Sorry ... you're absolutely right
@MarkSapir: Excellent point!
|
2025-03-21T14:48:30.059685
| 2020-03-13T02:52:59 |
354819
|
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"url": "https://mathoverflow.net/questions/354819"
}
|
Stack Exchange
|
Coefficents of these partition-based polyomials are $0, \pm1$
This is a follow up on my earlier MO question.
Given an integer partition $\lambda=(\lambda_1,\dots,\lambda_{\ell(\lambda)})$ of $n$ where $\ell(\lambda)$ is the length of $\lambda$, associate $\tilde\lambda'=\lambda,0$ found by appending one extra zero at the right end of $\lambda$. Further, define the following numerics $a(\lambda)_j=\tilde\lambda_j-\tilde\lambda_{j+1}$ for $j=1,2,\dots,\ell(\lambda)$.
For example, if $\lambda=(3,2,1,1)$ and $\tilde\lambda=(3,2,1,1,0)$ and $a(\lambda)=(1,1,0,1)$.
QUESTION. Is it true that the coefficients of the polynomial $A_n(q)$ are all in $\{-1,0,1,2\}$?
$$A_n(q):=\sum_{\lambda\vdash n}q^{n-\lambda_1}
\prod_{a(\lambda)_j\geq1}\frac{(q^{2a(\lambda)_j}-1)(q-1)}{q+1}.$$
REMARK. In fact, it appears that only coefficient of the middle-term can possibly be equal to $2$.
Since you're summing over all partitions, any reason why we need to take the conjugates first? The only reference to $\lambda$ itself seems to be the exponent of $q$ in the outer sum, and that can be replaced by $\lambda'_1$, right?
@user44191: edited accordingly.
|
2025-03-21T14:48:30.059801
| 2020-03-13T03:34:21 |
354821
|
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"Carlo Beenakker",
"Emil Jeřábek",
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"url": "https://mathoverflow.net/questions/354821"
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|
Stack Exchange
|
If $x^x=2$ then is $x$ expressible using elementary functions?
I have a curious question. Let $x∈\mathbb{R}^+$ such that $x^x=2$. I am aware that the Gelfond–Schneider theorem implies that $x$ cannot be algebraic. However, is it still possible that $x$ can be expressed in terms of elementary functions applied to rationals? I saw this post but that is concerning the whole Lambert-W function and not a single point. Is anything known about this $x$? If not, is anything known about the cardinality of $\{ x : x∈E ∧ x^x=c ∧ c∈\mathbb{Q}^+ ∖ \{ k^k : k∈\mathbb{Z}^+ \} \}$, where $E$ is defined as the set of all reals that can be expressed using elementary functions applied to rationals?
By "elementary function applied to rationals" I mean any of the following:
Rational functions with rational coefficients
Exponential function
The inverse of any of the above
The composition of any of the above
Constants $0,1,i$
Since some people want a precise definition of "inverse", here is one: For any meromorphic function $f$, let $f^{-1}(z)$ be defined as $w = re^{it}$ where $r∈\mathbb{R}_{≥0}$ and $0≤t<2π$ iff $f(w) = z$ and $r$ is minimum and then $t$ is minimum. If no such $w$ exists then $f^{-1}(z)$ is undefined.
As usual for questions of this sort, you need to define what an elementary function is. The most obvious definition allows constant functions, but then all reals are so expressible! Chow's "What is a closed-form number?" (MSN) provides one possible way of formulating the question.
@LSpice: I see I have to be more specific since for some reason constant functions are elementary...
Isn't $E$ countable? That would give your set cardinality at most $\aleph_0$. And it's trivially at least $\aleph_0$ because of $1^1, 2^2, 3^3, \dots$.
@NateEldredge: Argh of course. I wanted something other than those trivial solutions. Sorry about that.
Downvoters should explain why they think this question is not fit for MO.
Do we even know, for example, that $x$ is not a rational multiple of $e$? As far as I know it is not known if $e^e$ is rational.
isn't this basically asking whether the Lambert W-function can be expressed in terms of elementary functions? this MSE posting seems relevant --- https://math.stackexchange.com/a/2161098/87355
@CarloBeenakker OP linked this exact thread in the question. As they remark, the linked post is about the function, but this question is (essentially) about specific values of Lambert W. Even nonelementary functions can take elementary values.
Won't every $x\in E$ belong to the set you query about? Since if $x$ is elementary, so is $x^x$. The less trivial question is for $x^x\in E$ but $x\not\in E$; presumably "most" (in some sense) $x$ with $c=x^x\in E$ will not belong to $E$, like the example with $c=2$ (which I would bet many of my marbles on not being elementary).
@Wojowu: I fixed the mistake. I was thinking about the function mapping $c>0$ to $x$ such that $x^x=c$, and thinking about what happens when it is applied to rationals, but mistakenly put "$c∈E$" instead.
@MarkSapir I don’t think that’s any easier, as we lack tools to separate linear time from slightly super-linear time. (Note that the number is computable in time $O(M(n)\log n)$, see e.g. Wikipedia; that’s $O(n(\log n)^2)$ using the best known multiplication algorithm.) Also, I’m not sure how is it relevant, as the vast majority of constants defined by elementary functions are also likely not computable in linear time.
@MarkSapir They are all computable in time $O(M(n)\log n)$ (i.e., $O(n(\log n)^2)$), as that is the complexity of elementary functions. And as witnessed here, this also holds if you throw in Lambert’s function, and more generally the inverse of any (sufficiently smooth) function that you can already construct. I stress that here, computing $x$ means, given $n$, compute an $n$-digit approximation of $x$, i.e., something within distance (1/2 of the value of the last digit position) from $x$, but not necessarily smaller than $x$. If you really need to compute the $n$th digit of the infinite ...
... decimal expansion of $x$, the $O(M(n)\log n)$ time algorithm only works if you know a bound on the irrationality measure of $x$.
This is really an extended comment.
As Dan Richardson explains in his paper The elementary constant problem, there are different classes of numbers that you might be interested in. Your number certainly belongs to what is now usually called the class of elementary numbers, because it is a solution to the system of equations
$$\eqalign{e^z &= 2 \cr xy&=z \cr e^y&= x\cr}$$
Now one could ask whether your number is a Liouvillian number, meaning that it's obtained by a finite
sequence of algebraic, exponential, or logarithmic extensions of the rationals. But this allows for a more generous definition of "inverse" than you seem to want. As LSpice noted in a comment, you could also ask whether your number is a closed-form number in the sense of my paper, but this may be a more restrictive class than you want.
In any case, I think that the answer to your question is likely to be unknown since these types of questions tend to be very difficult.
Thank you for your answer! Yes, this was just a question out of curiosity, and I just want to know if any result of this sort is known, whether with more or less restrictive classes of 'nice' numbers. Ordinarily, I expect such questions to be beyond known mathematics, but I have seen surprising results (to me) before.
|
2025-03-21T14:48:30.060165
| 2020-03-13T04:43:21 |
354822
|
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|
Stack Exchange
|
Metaplectic group $Mp(2n)(\mathbb{A}_F)$ splits over $Sp(2n)(F)$?
My question is the title.
In some literature, authors seem to use this without assumption.
Is it ture in general?
It would be helpful if you could please cite some sources where you have seen this used.
@Uriya, I saw this in the hypotheses of the covering morphism $pr$ in I.1.3 of the book 'Spectral decomposition and Eisenstein series' by Moeglin and Waldspurger. I guess that the authors made this hypotheses keeping the double covering map from metaplectic group to symplectic group in their mind.
Yes. This is true in general. In fact, the definition of automorphic forms on Mp depends on this splitting. If my memory is correct, Moeglin and Waldspurger dealt with more general case than the metaplectic double cover Mp.
@Qing, Thank you very much! I referee the book ‘Spectral decomposition and Eisenstein series' by Moeglin and Waldspurger but couldn’t find proof of the generalization of this. Would you please let me know the page of the book?
I don't think Moegline-Waldspurger has a proof. The proof is probably in one of the two Weil's classical papers. The point is, MW has to assume there is such a splitting because they considered more general covers. For those more general covers, the existence of such splittings is not guaranteed. Since you only ask the metaplectic cover, the answer should be yes.
@Qing, I found some reference which says that Weil’s proof on the existence of theta functional guarantees $Sp(2n)(F)$ splits in $Mp(2n)(\mathbb{A})$. I don’t understand this well. Do you know this?
|
2025-03-21T14:48:30.060305
| 2020-03-13T06:21:38 |
354823
|
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"A. Bailleul",
"François G. Dorais",
"GH from MO",
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|
Stack Exchange
|
Infinitely many primes don't split?
Milne's note says when $L,K$ are number fields and $L/K$ is an abelian extension, infinitely many primes don't split. I want to ask if it's right in general case and how to prove it?
Thank you!
Overkill: Chebotarev's density theorem - https://en.wikipedia.org/wiki/Chebotarev%27s_density_theorem
I believe existence of non-split primes in abelian extensions was historically the main stumbling block for providing fully algebraic proofs of class field theory (but don't quote me on that historical account). An algebraic proof now exists, I believe due to Chevalley, but is quite involved, see Proposition VII.4.6 of Milne's CFT. There is a simpler analytic proof too (which works in all Galois extensions), see Theoren VI.4.6 ibidem.
You can use Chebotarev's density theorem as François said, or mimick the proof of Dirichlet's theorem in this particular case. All you need to know is that $L$ is the class field of some congruence class group $I_{\mathfrak m}/H$ of $K$, and that a prime of $K$ splits in $L$ if and only if it is in $H$.
Assume that all but finitely many primes split in $L/K$. Then the Dedekind zeta function $\zeta_L(s)$ agrees with $\zeta_K(s)^{(L:K)}$ apart from finitely many Euler factors. Comparing the order of the pole $s=1$, we conclude that $(L:K)=1$, that is, $L=K$.
Added. I understood split as "completely split". If split is understood as "has at least one degree one prime over it", then the above still works with some modifications. I am grateful to Levent Alpoge for this observation (see his comment below).
Why do you get a power of $\zeta_K$? Doesn't that need the stronger assumption that all but finitely many primes split completely in $L$? (I read "don't split" as "inert", but maybe I misread the question.)
@R.vanDobbendeBruyn: I understood split as "completely split".
@R.vanDobbendeBruyn If you interpret "don't split" as "stay inert", then this is false if $L/K$ is not cyclic. Indeed, if $\mathfrak p$ stays inert (in particular is unramified), then $Gal(L/K)$ injects into the Galois group of the residue fields, which is cyclic.
Why is it considerably harder? In the Galois closure L’/K the expected density (= 1/[L’:K]) of primes of K split completely in L’ (same Dedekind zeta argument). Hence the density of primes of K that split completely in L is at least 1/[L’:K]. The rest (forgetting a finite set) have at least one degree one prime over them. That’s too big by the Dedekind zeta argument (1 + ([L:K]-1)/[L’:K]\leq 1 implies L = K).
@alpoge: Thanks, that's a nice observation! I updated my post accordingly.
see https://mathoverflow.net/a/325267/23291
|
2025-03-21T14:48:30.060527
| 2020-03-13T07:16:55 |
354824
|
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|
Stack Exchange
|
A permutation statistic and determinantal identity
I'm trying to read this paper, Total positivity, Grassmannians and networks by Postnikov (https://arxiv.org/abs/math/0609764) and I'm stuck on Lemma 5.1, which is essentially an identity about maximal minors of a matrix.
Lemma 5.1. Let $I$ and $J$ be two $k$-element subsets in $[n]$, $K = I\setminus J$ and $L = J \setminus I$. Also let $r = |K| = |L|$. Then the following identity holds for the Plücker coordinates in $Gr_{kn}$
$$\Delta_J \cdot \Delta_I^{r-1} = \sum_{\pi: K \to L}(-1)^{\operatorname{xing}(\pi)}\prod_{i\in K}\Delta_{(I\setminus\{i\})\cup\{\pi(i)\}}.$$
Please see the paper for the relevant definitions. The crossing number is (I believe) not the usual one by his definition. I can't see why the following claims in the proof are true:
$\operatorname{xing}(\pi) = \binom{r}{2}-\operatorname{inv}(\pi)$
This identity, which he reduces the above to, is true:
$$\det B = \sum_{\pi: K \to L}(-1)^{\binom{r}{2}-\operatorname{inv}(\pi)}\prod_{i \in [r]}(-1)^{r-j}b_{i,\pi(i)}.$$ I'm happy that $$(-1)^{\binom{r}{2}-\operatorname{inv}(\pi)}\prod_{i \in [r]}(-1)^{r-\pi(i)}b_{i,\pi(i)} = (-1)^{\binom{r}{2}-\operatorname{inv}(\pi) + \binom{r}{2}}\prod_{i \in [r]}b_{i,\pi(i)} = (-1)^{-\operatorname{inv}(\pi)}b_{i,\pi(i)}.$$ This seems really close to the definition of the determinant, but off by a minus sign. Maybe I made a mistake somewhere.
Thank you!
On the second question: $\left(-1\right)^{-a} = \left(-1\right)^a$.
As for the first question, keep in mind that Postnikov has WLOG assumed that $I $ and $J $ are rather special subsets.
Incidentally, you seem to be reading an outdated version of Postnikov's preprint. See https://math.mit.edu/~apost/papers/tpgrass.pdf for a newer one.
Oh oh oh oh. Ok that makes so much more sense now. Thank you!!
|
2025-03-21T14:48:30.060664
| 2020-03-13T07:31:25 |
354825
|
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|
Stack Exchange
|
For which G is BLG weak homotopy equivalent to LBG?
Let $G$ be a (Edit: path-)connected topological group. Under what additional hypotheses on $G$ is it true that $LBG$ is a classifying space for $LG$? (or, I guess equivalently, when is $LBG \sim BLG$?) Here I'm taking the free loop space, and the compact-open topology on it. I know it is true for strong hypotheses, such as $G$ a compact Lie group. If one can find a model of $BG$ that is locally contractible and paracompact, then by Atiyah and Bott's The Yang-Mills Equations over Riemann Surfaces (doi:10.1098/rsta.1983.0017), Proposition 2.4, I believe it is possible. So, alternatively, for what $G$ is it true that $BG$ can be thus chosen?
[UPDATE: There were some mistakes in the first version. Here is a more careful account.]
I'll work everywhere with CGWH spaces, so I have a Cartesian closed category.
Note that $BLG$ is always path-connected, but $\pi_0(LBG)=\pi_0(G)/\text{conjugacy}$, so we need to assume that $G$ is path-connected. (The question says connected, but this may be a bit stronger; I do not know whether there are connected topological groups that are not path-connected.)
For any space $X$ and any $t\in S^1$ we have an evaluation map $\epsilon_{X,t}\colon LX\to X$. For the special case of the basepoint $1\in S^1$ we write $p_X=\epsilon_{X,1}\colon LX\to X$. This is always a Hurewicz fibration. If $X$ is based we have a fibre sequence $\Omega X\xrightarrow{i_X}LX\xrightarrow{p_X}X$.
Now let $G$ be a topological group. We write $EG$ and $BG$ for the usual simplicial constructions, so $EG$ is contractible and has a free $G$-action with orbit space $BG$. Simplicial methods also give a natural commutative diagram
$\require{AMScd}$
\begin{CD}
G @>k_{G}>> EG @>r_{G}>> BG\\
@Vj_GVV @VV l_G V @VV 1 V\\
\Omega BG @>>> PBG @>>> BG
\end{CD}
Both $EG$ and $PBG$ are contractible, and the bottom row is a Hurewicz fibration. If the top row is also a Hurewicz fibration, we can conclude that $j_G\colon G\to\Omega BG$ is a homotopy equivalence. If the top row is merely a Serre fibration or quasifibration, we can still conclude that $j_G$ is a weak equivalence. I do not know what are the minimal conditions for the top row to be a quasifibration.
Next, given $u\in BLG$ and $t\in S^1$ we have a homomorphism $\epsilon_{G,t}\colon LG\to G$ and thus a map $B\epsilon_{G,t}\colon BLG\to BG$ and thus an element $(B\epsilon_{G,t})(u)\in BG$. We would like to define $f_G\colon BLG\to LBG$ by $(f_G(u))(t)=(B\epsilon_{G,t})(u)$. To justify this we need to check continuity in $t$ and then in $u$. This in turn needs some continuity properties of the functor $B$, which can be proved using some abstract nonsense with CGWH spaces and Cartesian closure. (The initial version of this answer referred to a natural map in the opposite direction, but I think that does not actually exist.) We now want to construct a diagram as follows:
\begin{CD}
B\Omega G @>Bi_G>> BLG @>Bp_G>> BG \\
@V f'_G VV @V f_G VV @VV 1 V \\
\Omega BG @>>i_{BG}> LBG @>>p_{BG}> BG
\end{CD}
Constructions that we have already discussed provide all spaces and maps except for $f'_G$. On the top row we note that $(Bp_G)\circ (Bi_G)$ is trivial, and on the bottom row we know that $i_{BG}$ is the fibre of $p_{BG}$, so there is a unique way to fill in $f'_G$. The bottom row is always a Hurewicz fibration. The top row is obtained by applying $B$ to a Hurewicz fibration of topological groups, but it is not clear exactly what we get from that. If the top row is at least a Serre fibration, we see that $f_G$ is a weak equivalence iff $f'_G$ is a weak equivalence.
Finally, define $\tau\colon S^1\wedge S^1\to S^1\wedge S^1$ by $\tau(s\wedge t)=t\wedge s$. From the definitions one can check that the following diagram commutes:
\begin{CD}
\Omega G @>j_{\Omega G}>> \Omega B\Omega G \\
@V \Omega j_G VV @VV \Omega f'_G V \\
\Omega^2 BG @>>\tau^*> \Omega^2 BG
\end{CD}
If $j_G$ and $j_{\Omega G}$ are weak equivalences, we conclude that $f'_G$ is also a weak equivalence.
All this assumes that we start with the simplicial definition of $BG$. One could instead consider an axiomatic characterisation of $BG$, which might include the condition that the map $EG\to BG$ is a Serre fibration. The idea should be that $[X,BG]$ should biject with the set of isomorphism classes of principal $G$-bundles over $X$, but one would need to restrict attention to paracompact $X$, or to principal bundles over arbitrary $X$ that admit a numerable trivialising cover. I do not know how the technicalities would work out.
Why "connected" is not enough ?
@GSM Because $\pi _0(X)=[S^0,X]$ is the set of homotopy class of maps from $S^0$ to $X$, in other words, the set of path-connected components, and not the set of connected components.
@Neil re connected vs path-connected. Yes, was being a bit sloppy there. Path connected is fine by me, I'm not considering topological groups with bad local connectedness properties.
Hmm, maybe I accepted this too quickly. How do we know the left and right maps are weak equivalences? Certainly the spaces involved are weakly equivalent, but it's not obvious to me the the right map even is a weak homotopy equivalence.
My understanding is that $G\to \Omega BG$ is always a weak homotopy equivalence (essentially in Proposition 4.66 of Hatcher, but also spelled out at https://math.stackexchange.com/a/442866/). That relies on $EG \to BG$ being a fibration, but if it is a locally trivial bundle it is a Serre fibration, and if it is a numerable bundle, it is a Hurewicz fibration. (The latter case holds if ${e} \to G$ is a closed cofibration.) So this deals with your first and last points.
The extension $\Omega G\to LG \to G$ is continuously split, giving a semidirect product, so the usual simplicial construction of $ELG$ (which is a topological group) is a semidirect product of the topological groups $EG$ and $E\Omega G$. This might be leveraged to show that $BLG \simeq (EG\ltimes E\Omega G)/(G\ltimes \Omega G) \to BG$ is a fibration.
And it turns out that $(EG\ltimes E\Omega G)/(G\ltimes \Omega G) \sim EG\times_G B\Omega G$, where $G$ acts on $B\Omega G$ by conjugation, as it does so on the simplicial space whose geometric realisation is $B\Omega G$. Thus if $EG \to BG$ is a locally trivial bundle, so is the associated bundle $EG\times_G B\Omega G \to BG$, hence is a fibration. So to me it seems that if we only require that $EG \to BG$ is a locally trivial bundle, then the desired map is a whe.
Incidentally, a proof that $LBG$ is fibrewise homotopy equivalent over $BG$ to $EG \times_G G$, when $G$ is of the homotopy type of a CW-complex is given in the appendix to https://arxiv.org/abs/0811.0771
$\Omega LX\cong L\Omega X$. Therefore if $X$ is a delooping of $G$ then $LX$ is a delooping of $LG$.
|
2025-03-21T14:48:30.061178
| 2020-03-13T12:39:40 |
354830
|
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|
Stack Exchange
|
Fixed points of recursive functions with finite range
Let $\phi$ be a programming system satisfying the UTM Theorem (i.e., $\phi$ is a $2$-ary partial recursive function such that the list $\phi_0,\phi_1,\ldots$ includes all $1$-ary partial recursive functions, where $\phi_i=\lambda x.\phi(i,x)$ for all $i\in\mathbb{N}$). Suppose that for all recursive functions $f$ with finite range, there exits an $n\in\mathbb{N}$ such that $\phi_n=\phi_{f(n)}$. If $g$ is any recursive function (with finite or infinite range), is there an $n\in\mathbb{N}$ such that $\phi_n=\phi_{g(n)}$.
Do I understand correctly that your effective programming system satisfies the u-t-m theorem (by definition), but you did not require that it satisfy the s-m-n theorem?
I suppose I am asking in what way your numbering differs from an admissible numbering.
I'd guess you could build a counterexample using a variation of the Friedberg numbering construction. Let $g(n)=n+1$, and mostly make the $\phi_i$ distinct, but occasionally choose a large $i$ and small $j$ and have $\phi_i$ repeat $\phi_j$ to defeat a given $f$.
Expanding on my comment above. There is a Friedberg listing of the partial recursive functions, i.e. a 2-ary p.r. function $\phi$ such that $\phi_i = \lambda x.\phi(i,x)$ lists every p.r. function without repetition. We may assume that $\phi_0$ is the empty function.
Now we'll define a new list $\psi$ with $\psi_{2i} = \phi_i$ for all $i$. We also promise that for each $2i+1$, there will be a $j < 2i+1$ with $\psi_{2i+1} = \psi_{j}$. By induction, we may assume that $j$ is even. Now let $g$ be any total computable function such that $g(n)$ is always an even number strictly greater than $n$. Then $g$ has no fixed points in this system. (Because a fixed point of $g$ would give us $j < i$ with $\phi_j = \phi_i$.)
It remains only to define $\psi_{2i+1}$. If $i = \langle e, k\rangle$, where this is the standard pairing function, then we let $\psi_{2i+1}$ be the empty function until $\phi_e(2i+1)\!\downarrow = j < 2i+1$. Once this occurs, we make $\psi_{2i+1}$ copy $\psi_j$. Then $2i+1$ is a fixed point of $\phi_e$.
If $f = \phi_e$ is total and has a bounded range, then for a sufficiently large $k$, $f(2\langle e,k\rangle+1) < 2\langle e, k\rangle+1$, so $f$ will have a fixed point.
|
2025-03-21T14:48:30.061386
| 2020-03-13T15:32:29 |
354836
|
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|
Stack Exchange
|
Show that $\big(s(. |C_n)\big)_n$ is equicontinuous on $X^*$
Let $(X,\|.\|)$ be a separable Banach space with dual $X^*$. $\mathcal{P}_{wkc}(X)$ be the class of nonempty, weakly-compact and convex subsets of $X$. For any $C\in\mathcal{P}_{wkc}(X)$ we define its support function $s(.|C):X^*\to \overline{\mathbb{R}}$ by:
$$
s(x^*|C):=\sup_{x\in C} \langle x^*,x \rangle\qquad \forall x^*\in X^*
$$
and its radius by :
$$
\|C\|:=\sup_{x\in C}\|x\|
$$
Let $(C_n)_n\subset \mathcal{P}_{wkc}(X)$ such that :
$$
\exists M>0~,~\forall n\in\mathbb{N}~,~\|C_n\|\leq M
$$
Show that $\big(s(. |C_n)\big)_n$ is equicontinuous on $X^*$.
An idea to make this proof please.
What is the range of the mapping $x^*\mapsto (s(.|C_n))_n$?
I do not know sir. can you help me please
If one doesn't know the question, how can one give an answer? Could it be that you are actually talking about the map $x^\mapsto s(x^|C)$?
Ahh, but i have already defined $s (.|C)$, sir. it is: $$x^\to s (x^|C)=\sup_{x\in C} \langle x^*,x \rangle $$
Sorry, I think I misunderstood your question; please ignore my remarks!
Isn't $s(.|C)$ Lipschitz continuous with Lipschitz constant $|C|$ and hence your family uniformly Lipschitz continuous with constant $M$?
|
2025-03-21T14:48:30.061486
| 2020-03-13T15:36:23 |
354837
|
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|
Stack Exchange
|
Renewal functions inequalities
I came along the statement that for $x \geq z$, if $U(x)$ is a renewal function, there exists a constant $K$ such that
\begin{align}
U(x) - U(x-z) \leq U(z) \leq K (z+1).
\end{align}
This is not clear to me. For instance, Blackwell's renewal theorem gives $U(x) - U(x-z) = O(1)$, but not the above... It seems to be a rather basic statement. References are also welcome.
Renewal functions are subadditive. For a reference, this article (in Example 1) points to [Dal, Section 4], [Fell, Ch XI]:
[Dal] D. J. Daley, Upper bounds for the renewal function via Fourier methods, Annals of Probability 6 (1987), 876-884. MR0494547
[Fel] W. Feller, An introduction to probability theory and its applications, Volume II, 2nd ed., Wiley, 1971. MR0270403
This gives the first inequality by the very definition, and the other one by a simple estimate $$U(z) = U(1 + 1 + \ldots + 1 + (z - \lfloor z\rfloor)) \leqslant \lfloor z\rfloor U(1) + \sup\limits_{[0,1]} U.$$
|
2025-03-21T14:48:30.061594
| 2020-03-13T16:51:54 |
354843
|
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|
Stack Exchange
|
Preservation theorem for intersection of relations on a fixed set
It is a well known fact that any intersection of equivalence relations on a fix set is itself an equivalence relation. The same holds for a few other common relational theories, such as partial orders. I'm curious about a syntactic characterization of theories with this property. All of the examples I can think of are axiomatized by universal Horn sentences (i.e. sentences of the form $\forall \bar{x} \varphi_0 \wedge \dots \wedge \varphi_{n-1} \rightarrow \varphi_n$, with $\varphi_0,\dots,\varphi_n$ atomic formulas), but I'm not sure if this is a precise characterization.
Fix a language $\mathcal{L}$ with a single relational symbol $R$. We'll say that an $\mathcal{L}$-sentence or theory $\Sigma$ is preserved under relational intersections if whenever $(X,A)$ and $(X,B)$ are models of $\Sigma$ on the same underlying set $X$, then $(X,A\cap B) \models \Sigma$, and $(X,X^n) \models \Sigma$, where $n$ is the arity of $R$ (this being the empty intersection).
Question 1: Is there a syntactic characterization up to logical equivalence of sentences that are preserved under relational intersections?
All of the examples I know of also have the property that they are preserved under arbitrary relational intersections. In a similar situation--sentences that are preserved under intersections of substructures--preservation under finite intersection gives preservation under arbitrary intersections, but the idea of that proof doesn't really work in this context.
Question 2: If $\Sigma$ is preserved under relational intersections, does it follow that for any (possibly empty) family $(X,A_i)\models \Sigma$ indexed by $i \in I$, we have $(X,\bigcap_{i \in I} A_i)\models \Sigma$?
This feels like something that should be classical model theory, but I can't find any references.
It’s not just universal Horn sentences: all sentences that do not use $R$ at all, such as there exist at least $k$ objects, also trivially have this property, and in general are not equivalent to universal Horn sentences.
Very good point.
@EmilJeřábek3.0 Perhaps more intricately, "If $R$ holds of at most $n$ tuples then the universe has at least $k$ objects."
This is only a very partial answer, but it may be useful. I can characterize those theories which are preserved under substructure and arbitrary relational intersections.
Let's say a sentence is universal $(R,\neq)$-Horn if it has the following form: $$\forall x\, \left(\left(\bigwedge_{i=1}^n \varphi_i\right)\rightarrow \psi\right)$$
where $\psi$ is an instance of $R$ in context $x$, and each $\varphi$ is either an instance of $R$ in context $x$ or $x_i \neq x_j$, where $x_i$ and $x_j$ are distinct variables in $x$.
It is easy to see that any universal $(R,\neq)$-Horn theory is preserved under substructure and arbitrary relational intersections.
Let $\Sigma$ be a theory which is preserved under substructure and arbitrary relational intersections. Let $\Sigma'$ be the set of all universal $(R,\neq)$-Horn consequences of $\Sigma$. Then every model of $\Sigma$ is a model of $\Sigma'$, and we want to show the converse.
Let $(M,R)\models \Sigma'$, and let $$\mathcal{R} = \{R'\subseteq M^n\mid R\subseteq R' \text{ and } (M,R')\models \Sigma\}.$$ If we can show that $\bigcap_{R'\in \mathcal{R}} R' = R$, then we're done: $\Sigma$ is preserved under arbitrary relational intersections, so $(M,R)\models \Sigma$.
Clearly $R\subseteq \bigcap_{R'\in \mathcal{R}} R'$, so it suffices to show that for every tuple $b\in M^n$ such that $b\notin R$, there is some $R'\in \mathcal{R}$ such that $b\notin R'$.
So fix such a $b$ and consider the $L(M)$-theory $$\Sigma\cup \{a\neq a'\mid a\neq a'\in M\} \cup \{R(a)\mid a\in M^n\text{ and }M\models R(a)\}\cup \{\lnot R(b)\}.$$ This is consistent by compactness, using the fact that $M$ satisfies all universal $(R,\neq)$-Horn consequences of $\Sigma$. Let $N$ be a model, and let $(M',R')$ be the induced substructure of $N$ with domain the interpretation of the constant symbols in $L(M)$. We can identify $M'$ with $M$, since distinct constant symbols have distinct interpretations. Since $\Sigma$ is preserved under substructure, $(M,R')\models \Sigma$, and by construction $R\subseteq R'$ and $b\notin R'$, as desired.
It's now an easy application of compactness to show that a sentence is preserved under substructure and arbitrary relational intersections if and only if it is equivalent to a finite conjunction of universal $(R,\neq)$-Horn sentences.
A few comments:
There's a slight ambiguity in your question about whether "arbitrary intersection" includes the empty intersection. It seems to me to be more natural to include the empty intersection, i.e. if $\Sigma$ is preserved under arbitrary intersections, then $(M,R)\models \Sigma$ when $R = M^n$. The argument above uses this interpretation (since $\mathcal{R}$ might be empty!). But then when we talk about being closed under (finite) relational intersections, we should include the empty intersection in addition to binary intersections.
If you want to "arbitrary intersection" to mean "arbitrary nonempty intersection", you just need to adjust the definition of universal $(R,\neq)$-Horn to allow $\psi$ to be $\bot$. Then in the argument, you can show that $\mathcal{R}$ is non-empty, since $$\Sigma\cup \{a\neq a'\mid a\neq a'\in M\} \cup \{R(a)\mid a\in M^n\text{ and }M\models R(a)\}$$ is consistent.
To adjust this argument to remove the assumption of preservation under substructure seems tricky: in the compactness argument to find the model $N$, we'd need to ensure that the domain doesn't grow, which forces us to think about omitting the partial type $\{x\neq a\mid a\in M\}$. So the syntactic characterization has to be strong enough to ensure not just consistency but also that this partial type is omittable. It's not immediately clear to me how to do this.
Adjusting arbitrary intersections to binary intersections seems much harder. My guess is that there is no nice characterization unless your Question 2 has a positive answer.
I'm always for the empty intersection. Am I understanding correctly that we could in principle allow the $\varphi_i$'s to be of the form $x=y$, but this can be removed by renaming variables?
Also I believe you can get preservation under arbitrary intersections from preservation under binary intersections and passing to substructures by a proof analogous to the proof for intersections of substructures.
Given $(X,A_i)$, take an ultraproduct over the family of finite intersections of the $A_i$'s using an ultrafilter extending the filter generated by sets of the form ${\bar{j} \in I : i \in \bar{j}}$, where $A_{\bar{j}} = \bigcap_{i \in \bar{j}}A_i$, then pass to the substructure whose underlying set is the diagonal embedding of $X$. This structure will be $(X, \bigcap_{i \in I}A_i)$.
@JamesHanson Comment 1: Yes, exactly. On the other hand, you can't take $\psi$ to be of the form $x = y$ if you allow the empty intersection. If you don't allow the empty intersection, you can take $\psi$ to be of the form $x = y$, but this isn't necessary, since you can add $x\neq y$ into the conjunction and replace $\psi$ with $\bot$. Comments 2 and 3: Ah, nice, I'm glad Question 2 has a positive answer in this case. Of course, you still need to assume $\Sigma$ is preserved by binary relational intersections and the empty relational intersection - you don't get it for free.
@JamesHanson And I think you should unaccept my answer for now - hopefully someone will be able to answer the real question (without preservation under substructure tacked on).
Can you think of an example of a theory that is preserved under relational intersections and only has infinite models but isn't equivalent to a universal $(R,\neq)$-Horn theory plus axioms asserting the existence of infinitely many elements?
Yes, I think this works: $(\forall x\forall y R(x,y))\lor (\exists x\forall y \lnot R(x,y))$, plus the axioms saying the domain is infinite.
|
2025-03-21T14:48:30.062431
| 2020-03-13T17:15:50 |
354845
|
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|
Stack Exchange
|
Connection between Volkenborn integral and Haar measure on $\mathbb{Q}_p$
This may be a rather elementary question, but I haven't been able to figure it out on my own, and the literature appears to be eerily silent on the topic.
Since $\mathbb{Q}_p$ is a locally compact group with $\mathbb{Z}_p$ as a compact subset, there exists a unique Haar measure $\mu$ on $\mathbb{Q}_p$ with $\mu(\mathbb{Z}_p)=1$.
The Volkenborn integral for functions $\mathbb{Z}_p\to\mathbb{Q}_p$ is defined by
$$
\int_{\mathbb{Z}_p} f = \lim_{n\to\infty}\frac{1}{p^n}\sum_{x=0}^{p^n-1}f(x),
$$
takes values in $\mathbb{Q}_p$, and is not translation invariant.
But now suppose I define an integral over $\mathbb{Z}_p$ for functions $\mathbb{Z}_p\to\mathbb{R}$ using the same expression as the Volkenborn integral. Then it seems that I can integrate the functions $\mathbb{1}_{a+p^n\mathbb{Z}_p}$ to get $\frac{1}{p^n}=\mu(a+p^n\mathbb{Z}_p)$, and because the integral so defined is clearly ($\mathbb{R}$-)linear in $f$, it would appear that it agrees with the integral over the Haar measure (at least when both exist).
Since such a Volkenborn-style integral for $\mathbb{R}$-valued functions is never discussed in the literature that I could find, I wonder if I am missing some obvious problem with the construction here, or if this is just a well-known rarely-mentioned fact.
This is a well knowm fact, and is not rarely mentioned. In articles and books on $p$-adic distributions, this is usually the first example given, and authors usually call it the ($p$-adic valued) Haar measure (which is not a $p$-adic measure). It is the same as the Volkenborn integral. You may look at Koblitz's book, Mazur's Bourbaki report, Colmez's Asterisque survey, etc.
Thanks for the pointers, but to be quite clear: my question is about functions $\mathbb{Q}_p\to\mathbb{R}$, so I'm not sure how this is a $p$-adic valued measure. Is this just nomenclature?
Probably I'm not understanding you correctly. The Haar measure is always $\mathbb{R}$ valued, by definition. Now, in this case it is $\mathbb{Q}$ valued, hence may be interpreted as $\mathbb{Q}_p$ valued. Hence the name "$p$-adic Haar measure" is just nomenclature (as far as I know). Actually, there is no translation invariant, $p$-adic valued, bounded measure on $\mathbb{Z}_p$.
OK, now I understand you: the Haar measure is interpreted as $\mathbb{Q}_p$ valued, which leads to statements such as it being unbounded (cf. Jon's answer), i.e. relative to $|\cdot|_p$. When integrating $\mathbb{Q}_p$-valued functions, that makes sense, but when integrating $\mathbb{R}$-valued functions, I think the real value is the only one that fits; I never saw the latter case being discussed, but that may be because number theorists just aren't terribly interested in it.
Perhaps you can find some discussion on this in Schikhof's book on ultrametric calculus and/or in Van Rooij's book on ultrametric functional analysis. I don't have my copies at hand, but I'm pretty sure they are in libgen.
Thanks, I'll try to find those and have a look.
This does not just vary with translation, it's non-canonical. The idea is that knowing the size of the compact open subsets $p^n\mathbb{Z}_p$ should give you a measure (i.e., an integral) by the following, if it exists:
$$ \int_{\mathbb{Z}_p} f d\mu = \lim_{n \to \infty} \sum_{a=0}^{p^n-1} f(\text{any representative of }a+p^n\mathbb{Z}_p) \mu(a+p^n\mathbb{Z}_p). $$
You should compare this with the usual Riemann integral on $\mathbb{R}$ to see that it makes sense. It seems that your Volkenborn integral just chooses a specific representative, $a$, for $a+p^n\mathbb{Z}_p$.
To see the issue, try to integrate $f(x)=x$ where you choose $\{0,1,\dots,p^n-1\}$ as your representatives, and then do it where you choose $\{p^n,1,\dots,p^n-1\}$. You should get different answers! (I believe they are $\pm\frac{1}{2}$.)
In general, algebraic number theorists say that the "Haar measure" is actually only a $\textit{distribution}$, since $|\mu(p^n\mathbb{Z}_p)|_p = p^n$ is unbounded. If $\mu$ tells us the size of each compact open $|\mu(p^n\mathbb{Z}_p)|_p$ is bounded, we call it a measure. Standard analysis techniques can show that integrals of continuous functions are well-defined with respect to such a measure. EDITED TO ADD: The same proof should work regardless of the metric on your target space, as long as you have that the sizes are bounded. So since $|\mu_{\text{Haar}}(a+p^n\mathbb{Z}_p)|_\infty = \frac{1}{p^n} \leq 1$ is bounded when the target is $\mathbb{R}$, this shows that your Volkenborn integral agrees with the more canonical definition for functions $f \colon \mathbb{Z}_p \to \mathbb{R}$.
(On the other hand, I'm told that analytic number theorists call distributions measures, and call measures "strict measures" -- with these names, integration continuous functions is well defined with respect to strict measures, but not necessarily over measures).
Thanks, but I'm not sure how your example relates to the question: I specifically asked about $\mathbb{R}$-valued functions, which $f(x)=x$ isn't. EDITED TO ADD: With your addition in edit, it sounds like you agree that this works in the case where the target is $\mathbb{R}$?
This is an addendum to Jon's answer, just to clarify some points on the Volkenborn integral. This was studied by Volkenborn, in a slightly more general form, because of its relation to Bernoulli numbers and polyomials, hence to zeta functions. Namely,
$$B_n(x)=\int_{\mathbb{Z}_p}(x+t)^n dt$$
is the $n$-th Bernoulli polynomial (I think this identity is due to Witt). This was probably inspired by the sums that appeared in the work of Kubota and Leopoldt on the $p$-adic zeta function. Hence, in practice it is an extremely useful way to handle Bernoulli polynomials, and actually a lot of their properties have trivial proofs using the Volkenborn integral.
But, the definition of this "integral" by means of "Riemann sums" is kind of artificial once you notice that this is part of a more general theory, and because this integral depends on the choice of the representatives as Joel point out. I think that the theoretically correct way of thinking the Volkenborn integral of a (say, strictly differentiable) function $f$ is as a "Riemann sum" which happens to give the correct value of the $p$-adic distribution $\mu(a+p^n\mathbb{Z}_p)=p^{-n}$ (which modulo normalization, happens to "be" the Haar measure) applied to $f$. This is done in detail in Colmez's Fonctions d'une variable $p$-adic, section II.3.3 in page 32.
|
2025-03-21T14:48:30.062859
| 2020-03-13T18:20:57 |
354848
|
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|
Stack Exchange
|
Thinnest rigid packings of the plane
A packing of the plane with copies of any shape is called rigid (or "stable") if every unit is fixed by its neighbors, i.e., no unit can be translated without disturbing others in the packing. We are interested in thin rigid packings of the plane, rigid packings that that leave the largest possible fraction of the plane uncovered.
https://mathworld.wolfram.com/RigidCirclePacking.html shows a way to achieve the thinnest rigid packing of the plane with unit circles - with density zero.
What is the thinnest rigid packing of the plane with unit squares? Is it a non-lattice arrangement? I do not know any references.
Which is the convex shape for which the densest and thinnest rigid packings of the plane differ the least in terms of coverage of the plane?
Note: For example, with thin rectangles or thin triangles, we observe (https://nandacumar.blogspot.com/2020/03/thinnest-rigid-packings-contd.html) that difference in coverage can be arbitrarily high: indeed, the densest packing of the plane is a perfect tiling and the thinnest rigid pack can leave an arbitrarily large fraction of the plane uncovered.
See this question and answers for packings with zero density.
Why do you write "probably" the thinnest? The packing by Böröczky cited on that page and in the linked question has density zero. Are you suggesting there may be ones with negative density?
thanks... made a corrective edit.
The question now is: is there a shape which does not admit density zero rigid packings?
For unit squares I believe you should be able to use the structure in the mathworld illustration, replacing the circles with alternating coordinate-aligned squares and 45-degree-rotated squares so that each square of one orientation is wedged between three squares of the other orientation.
Thanks TonyK. the link you gave shows the construction given in https://mathworld.wolfram.com/RigidCirclePacking.html is not quite optimal (zero density) - The construction drawn by Prof. O'Rourke does not look a straight enhancement of the picture given in the above Mathworld page. And yes, after some experimentation, I am unable to see how the kind of construction beginning with a hexagonal lattice of disks can be naturally done with unit squares.
And Wojowu, I too would like an answer to the question you raised. Thanks
The construction given on https://mathoverflow.net/questions/145314/are-there-locally-jammed-arrangements-of-spheres-of-zero-density probably works for zero density arrangements with half-disks (semidisks) too - this appears to follow from all boundaries between a region with disks and the empty region being concave polylines as noted there.
|
2025-03-21T14:48:30.063080
| 2020-03-13T22:25:06 |
354858
|
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"Rahul Sarkar",
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|
Stack Exchange
|
Existence of a Euler-like formula for the continuous image of $S^1$ in an orientable surface
Let $\mathcal{M}$ be a compact 2-manifold, and let $\gamma: S^1 \rightarrow \mathcal{M}$ be a continuous map (you can assume piecewise smooth if it is convenient), with the property that the set $A = \{p \in S^1 : \exists \; q \in S^1, \; p \neq q, \text{ s.t. } \gamma(p) = \gamma(q) \}$ is finite. Define an equivalence relation on $\mathcal{M} \setminus \gamma(S^1)$ that identifies points as follows: $x \thicksim y$ iff $x,y$ are path-connected in $\mathcal{M} \setminus \gamma(S^1)$. The equivalence classes are the ``interiors of the faces'' of $\mathcal{M}$ induced by this construction. Notice that each equivalence class is open (this can be done by choosing any Riemannian metric on $\mathcal{M}$, and then each point in $\mathcal{M} \setminus \gamma(S^1)$ is contained in an open ball that does not intersect $\gamma(S^1)$), and so $\mathcal{M} \setminus \gamma(S^1)$ is a disconnected set, which is the union of its connected components (the equivalence classes).
First, can someone help me prove that the number of equivalence classes is also finite?
Now onto the actual research question: One can interpret this construction as a drawing of a graph on $\mathcal{M}$, whose vertices are the finite set of points $\gamma(A)$, and the edges of the graph are the segments of the curves joining two points. So now you have all the ingredients for an "Euler-like" formula: vertices, faces, edges. I want to know if any such formula is known or not, both in the orientable and non-orientable cases. I'd also be interested to know if any special cases are known.
The particular case I'm interested in is when $\mathcal{M} = S^1 \times S^1$, and additionally $\gamma$ has the property that if $\gamma(p) = \gamma(q)$ for distinct $p,q \in S^1$, then $\not \exists \; r \in S^1$ s.t. $\gamma(r) = \gamma(p)$.
I do not think there is such Euler-like formula without further assumptions. Such a formula would have to be independent of the topology of the surface, because you could always connect-sum on a handle and increase the (orientable or non-orientable) genus, so it would have to depend only on $V, E, F$ and hold for all graphs. This seems unlikely. The trouble is that you are not making any assumptions about the topology of the faces. If you assume that the faces are all homeomorphic to discs then you can probably get somewhere.
@dvitek how about an inequality instead of an equality?
|
2025-03-21T14:48:30.063261
| 2020-03-13T22:56:13 |
354859
|
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"url": "https://mathoverflow.net/questions/354859"
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|
Stack Exchange
|
Decomposition of the space of Radon measures with respect fractional harmonic capacity?
It is well know that there is a generalization of Lebesgue decomposition theorem in the following way:
Any non negative Radon measure can be decomposed uniquely into the sum of an absolutely continuous measure and a singular term with respect to the harmonic capacity. The absolute continuous term itself can be decomposed (not uniquely) to a function in $L^1$ and a function in $H^{-1}$ (dual of $H_0^1$).
I will be thankful to any one that will help me answering the following question: do there exist such kind of decomposition involving the fractional harmonic capacity?
|
2025-03-21T14:48:30.063335
| 2020-03-13T23:17:16 |
354861
|
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|
Stack Exchange
|
Inseparable field extensions of degree p and linear independence
Let $F$ be a field of characteristic $p$; let $\alpha \in F$ such that $\alpha \neq \beta^p$ for any $\beta \in F$, and let $K := F(x)$ where $x=\sqrt[p]{\alpha}$.
Is it true that the elements $1,(x-1)^{p-1},(x^2-1)^{p-1},\dots,(x^{p-1}-1)^{p-1}$ are linearly independent over $F$?
Please write the nonstandard notation $F[x : x^p = \alpha]$ more clearly. You didn't even define what $\alpha$ means. Presumably $\alpha$ is a nonzero element of $F$ that is not a $p$th power in $F$. If so, say that. And write the field as $F(\sqrt[p]{\alpha})$ or as "$F(x)$ where $x^p = \alpha$.''
For $k \geq 1$, $(x^k - 1)^{p-1} = (x^k-1)^{p}/(x^k-1) = (\alpha^k-1)/(x^k-1)$ and $\alpha^k-1$ is in $F$ and is not 0 (because $\alpha^k = 1 \Rightarrow \alpha$ is algebraic over $\mathbf F_p$, but that would contradict the assumed inseparability of $K/F$), so the linear independence of that list over $F$ is equivalent to the linear independence of $1, 1/(x-1), 1/(x^2-1), \ldots, 1/(x^{p-1}-1)$. An $F$-linear relation among that list is reminiscent of partial fraction decompositions, and maybe you can find a contradiction from that point of view.
|
2025-03-21T14:48:30.063439
| 2020-03-14T01:38:31 |
354866
|
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|
Stack Exchange
|
$G_{\infty}$ (also known as $E_2$)-operad in terms of trees
It's well known that the $A_{\infty}$ and $L_{\infty}$ operads, being resolutions of the associative and Lie operads, admit descriptions as free operads of certain trees.
The description I am referring to for the $A_{\infty}$ operad is described in Voronov's notes:
http://www-users.math.umn.edu/~voronov/8390/lec9.pdf
Similarly the $L_{\infty}$ operad is described in section 2 of the paper by Voronov, Stasheff and Kimura:
https://arxiv.org/pdf/hep-th/9307114.pdf
Does the $G_{\infty}$-operad, the free resolution of the Gerstenhaber (Poisson) operad admit a similar description in terms of trees? How can one define it explicitly?
Yes. All of your examples fall within the techniques of Koszul duality theory for operads, and for this you can consult the standard reference of J.-L. Loday and B. Vallette. But also see G. Ginot's thesis http://www.numdam.org/item/AMBP_2004__11_1_95_0/ for explicit descriptions of $\mathsf{Gers}_\infty$ and $\mathsf{Pois}_\infty$.
|
2025-03-21T14:48:30.063641
| 2020-03-14T02:02:40 |
354868
|
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"Ali Taghavi",
"Moishe Kohan",
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|
Stack Exchange
|
Non-commutative analogue of a certain fact in differential geometry
In the literature, is there a non-commutative analogue of the fact that every Riemannian manifold whose isometry group has sharp dimension must be a constant curvature manifold?
What do you mean by "sharp dimension"? Do you mean that $M$ is $n$-dimensional and $dim(Isom(M,g))=n(n+1)/2$? Such manifolds need not be flat, they are manifolds of constant curvature.
I mean so. Tes you are right I revise the question
@MoisheKohan thanks for your helpful comment.
|
2025-03-21T14:48:30.063709
| 2020-03-14T07:01:54 |
354872
|
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"sort": "votes",
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|
Stack Exchange
|
Additivity of the upper Banach density
The following notion of upper Banach density was defined (Definition 2.1(c)) by Hindman and Strauss in their paper 'Density in arbitrary semigroups':
Definition: Let $S$ be a semigroup, let $\mathcal{F}=\{F_i\}_{i\in \mathcal{I}}$ be a net in $\mathcal{P}_f(S)$, and let $A\subseteq S$. Then the upper Banach density of $A$ with respect to $\mathcal{F}$ is $$d_{\mathcal{F}}^*(A)=\sup\big\{\alpha: (\forall i\in \mathcal{I})(\exists j>i)(\exists x\in S\cup \{1\})(|A\cap(F_jx)|\geq \alpha |F_j|)\big\}.$$
Note that for the standard Følner sequence $\mathcal{F}=\{F_n\}_{n\in\mathbb{N}}$, where $F_n=\{1,2,...n\}$, the corresponding upper Banach density is never additive. My question is
Question: For a fixed semigroup $S$ (we can assume this to be $\mathbb{N}$), does there exists a net $\mathcal{F}$ in $\mathcal{P}_\mathrm{f}(S)$, so that the corresponding upper Banach density becomes completely additive, i.e. for each $A,B\subseteq S \text{ such that } A\cap B=\emptyset,\,d_{\mathcal{F}}^*(A\sqcup B)=d_{\mathcal{F}}^*(A)+d_{\mathcal{F}}^*(B)$?
For every family $\mathcal F\subset\mathcal P_f(\Bbb N) $ both a set $\{1, 3,4, 7,8,9, 13,14,15,16,\dots\} $ and its complement have upper Banach density $d^*_{\mathcal F}$ equal to $1$.
|
2025-03-21T14:48:30.063812
| 2020-03-14T07:27:29 |
354873
|
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|
Stack Exchange
|
A converse of the Erdős-De Bruijn Theorem?
For the chromatic number $\chi(G)$ of a simple, undirected graph, there is a "compactness" theorem by Erdős and De Bruijn stating that if an infinite graph $G$ has finite chromatic number, then there is a finite subgraph $G_0\subseteq G$ such that $\chi(G_0) = \chi(G)$.
A converse statement would be
$(\text{S})$ Let $k>0$ be an integer. Whenever a graph $G$ has the property that every finite subgraph $G_0$ of $G$ can be colored with $k$ colors, then $G$ can be colored with $k$ colors.
Is $(\text{S})$ true?
Isn't $(S)$ the usual statement of the Erdös-De Bruijn theorem? From the second line in the wikipedia link: "It states that, when all finite subgraphs can be colored with $c$ colors, the same is true for the whole graph"
But even if you take the statement in the question, $(S)$ is an immediate corollary. Suppose $\chi(G_0)=k$ for every finite $G_0\subset G$. Obviously $\chi(G)\geq k$. But we can't have $\chi(G)>k$ because that would be witnessed by a finite subgraph.
This can be proven by applying compactness for 1st order logic, right?
@Monroe that's right
Apparently yes. By googling, I found the assertion in this 1951 paper of de Bruijn and Erdős, and the first page contains several further references.
|
2025-03-21T14:48:30.063935
| 2020-03-14T08:58:03 |
354875
|
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|
Stack Exchange
|
Undecidable definition of mathematical expressions?
I am arguing a bit on Facebook regarding the definition of a mathematical expression. Some argue that equations are not expressions (and there are a few possibly dubious online sources which states this).
Let's say that equations are not expressions, but both sides of an equation are considered expressions.
Now,
$$
\sum_{k=1}^n |\{x \in \mathbb{C} : x^k-1=0\}|
$$
is still an expression, but it contains equations. So, by not allowing equations in the set of expressions, we run into the issue that parts of an expression may not be expressions anymore.
The question I have is: Is it a decidable problem if a finite string of math symbols is an expression or not?
As this is really about the definition of what an expression is, the above (in my opinion wrong) "definition" makes the question more interesting, I think.
To reword, is there a definition of the term mathematical expression which sounds reasonable, but is troublesome in the sense that it is in general undecidable if a finite string IS an expression?
The notion that equations aren't expressions probably comes from programming. In most C-like languages an expression is something that evaluates to a value at runtime, like a + 2, while an assignment like b = a + 2; is a statement. Many people associate assignments with equations, probably because of the symbol =. Somewhat confusingly, the equality test b == a + 2 is an expression since it evaluates to a Boolean value...
True, byt high level languaged such as mathematica allow for assigning equations to variables. This is quite convenient
Part "/2" of an expression "1/2", is very often not an expression.
As you say, this is really a question of the definition of "expression". It sounds to me like you're taking it to mean "constant term". If so, given any reasonable mathematical language it will be decidable whether a given finite string is a constant term.
The concern that "parts of an expression may not be expressions any more" puzzles me --- why would you expect this?
I suppose it wouldn't be too hard to cook up a language in which you could write things like "$\sum_{k=1}^\infty a_n$ where $a_n = 0$ or $1$ depending on whether the $n$th Turing machine halts on null input" or whatever, and declare that things like this only count as constant terms if they converge. That could make termhood undecidable.
|
2025-03-21T14:48:30.064127
| 2020-03-14T11:07:57 |
354882
|
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|
Stack Exchange
|
Can extensionality and separative-reflection interpret all rules of set theory?
Language: first order logic with equality, with additional primitives of membership $\in$ and a constant symbol $V$ standing for the class of all sets.
Extensionality: $\forall z (z \in x \leftrightarrow z \in y) \to x=y$
Reflection: if $\varphi$ is a formula in $\mathcal L(=,\in)$, and if $\psi$ is a formula in which $y$ is not free, then all set closures of: $$ \exists x (\varphi) \to \forall \vec{c} \ \exists x \big[\varphi\wedge \exists y \in V \ y=\{z \in x| \psi\} \big]\;; $$ are axioms.
It's nice to know that just these two axioms can interpret the whole of ZFC, since it straightfowardly prove all axioms of K(W) theory (see page 3).
It's known that ZFC can interpret all set statements of Ackermann's set theory, also this theory does that! Clearly this theory proves all axioms of Ackermann's set theory except those of foundation and class construction! Now the questions are:
Can this theory interpret all the rules of Ackermann's set theory including class comprehension as well?
if we restrict the parameters of $\varphi$ to just two, and change $\vec{c}$ to just $c$, would the resulting theory still interpret ZFC?
What does "set closure" mean? Perhaps adding all possible quantifiers of the form $\exists z\in V$? So $\psi$ has 2 kinds of variables, those that you call $c$ and those that I call $z$ here?
@Goldstern yes those bounded ($\in V$) quantified $z$ variables occur at the beginning of the statement (before the first $\exists x$), and they are what I meant by set closures. $\psi$ might have the $z$'s free in them, or both the $z$ and $c$ variables, or non of them. It important to realized that the $c$ string of variables is unbounded! However the parameters of $\varphi$ are only the $z$ variables, i.e. all of them are bounded in $V$.
|
2025-03-21T14:48:30.064292
| 2020-03-14T11:18:25 |
354885
|
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|
Stack Exchange
|
Convergence of fraction of expectation values
Let $X_1,...,X_n$ be iid normal random variables.
I am looking for a strategy to establish the following limit for fraction of expectation values
$$\lim_{N \rightarrow \infty} \frac{E(\prod_{1\le i < j\le n} \vert X_i-X_j \vert^{1/n})}{E(\prod_{1\le i < j\le n-1} \vert X_i-X_j \vert^{1/n})}=1.$$
Does anybody have any ideas what to use for this limit?
The Mehta integral is
$$M_n(\gamma):=E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}
=\prod_{j=1}^n\frac{\Gamma(1+j\gamma)}{\Gamma(1+\gamma)}.$$
So, your fraction under the limit sign is
$$\frac{M_n(1/(2n))}{M_{n-1}(1/(2n)}=\frac{\Gamma(3/2)}{\Gamma(1+1/(2n))}\to\Gamma(3/2)\approx0.886227.$$
thank you, what a nice way of seeing this. I am just wondering: Imagine we would not have this explicit formula at hand, do you see any way to give upper/lower bounds on the limiting value?
@Sascha : Another way to find the limit is to use (say) Theorem 3.1 of Chen and Shao https://projecteuclid.org/euclid.bj/1179498762 for U-statistics, since $\sum_{1\le i<j \le n}\ln|X_i-X_j|$ is a U-statistic. This will work even without the normality assumption on the $X_i$'s. If you want more details, post this additional question separately.
|
2025-03-21T14:48:30.064397
| 2020-03-14T11:46:18 |
354888
|
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|
Stack Exchange
|
Approximate a given number by factorials
For a pair of integers $a$ and $b$, can we approximate a given number by $\log_{a^{!^n}}b^{!^m}$ where $a^{!^n}$ is iterated factorial operator? Are there theorems to deal with this kind of questions?
|
2025-03-21T14:48:30.064445
| 2020-03-14T12:09:26 |
354889
|
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|
Stack Exchange
|
Quantum dimension in the Drinfeld center
Let $\mathcal{C}$ be a spherical tensor category. It is known that the Drinfeld center of $\mathcal{C}$ is modular (and therefore also spherical), see for example, Corollary 8.20.14 in [1]. Recall the notion of dimension in spherical tensor categories obtained by taking the quantum trace of the identity.
Question: For a given object $Z = (X,\gamma)$ in the Drinfeld center is there a simple way to express the dimension $d(Z)$ in terms of $X$ and $\gamma$?
[1] Etingof, Pavel; Gelaki, Shlomo; Nikshych, Dmitri; Ostrik, Victor, Tensor categories, Mathematical Surveys and Monographs 205. Providence, RI: American Mathematical Society (AMS) (ISBN 978-1-4704-2024-6/hbk). xvi, 343 p. (2015). ZBL1365.18001..
Isn't the spherical structure you put on the Drinfeld center "equal" to that of $\mathcal{C}$? I.e. $d(Z)=d(X)$?
You are quite correct, this can be seen by doing Exercise 7.13.6. in the reference I provided. I feel a bit silly asking just a trivial question!
|
2025-03-21T14:48:30.064538
| 2020-03-14T12:30:25 |
354892
|
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|
Stack Exchange
|
Generalization of minimal selection theorem
Consider a metric space $X$ and a set-valued map $F : X \to \mathbb{R}^{n}$. We define the minimal selection
\begin{equation*}
m(F(x)) := \arg\min \big\{ \lvert u \rvert : u \in F(x) \big\},
\end{equation*}
where $\lvert \, \cdot \, \rvert$ denotes the 2-norm.
The minimal selection theorem in question can be stated as follows [e.g. Corollary 9.3.3 in Aubin and Frankowska]:
Let $F$ be a continuous (upper and lower semicontinuous) set-valued map from a metric space $X$ to $\mathbb{R}^{n}$ with nonempty closed convex images. Then the minimal selection is continuous.
My question is if there exist analogous theorems with $m$ replaced by
\begin{equation*}
m'(F(x)) := \arg\min \big\{ f(x,u) : u \in F(x) \big\},
\end{equation*}
where $f$ is a continuous function such that $f(x, \, \cdot \,)$ is strictly convex on $F(x)$?
I think the Berge Maximum theorem could be applicable?
https://en.wikipedia.org/wiki/Maximum_theorem
See page 116;
[C. Berge, Topological Spaces: including a treatment of multi-valued
functions, vector spaces, and convexity. CC, 1997]
While the provided version of the Berge Maximum theorem is not a generalization of the minimum selection theorem as posed in the question, I was indeed unaware of its existence. This seems like a promising avenue of further research. Thank you!
|
2025-03-21T14:48:30.064654
| 2020-03-14T12:48:05 |
354893
|
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|
Stack Exchange
|
Making area/volume calculations that use SIA rigorous
There are some intriguing "proofs" using Smooth Infinitesimal Analysis of theorems concerning areas and volumes. Some examples:
A proof that $\sin'(0) = 1$.
A proof that the surface area of a cone is $\pi r \sqrt{h^2 + r^2} + \pi r^2$
A proof of the Fundamental Theorem of Calculus.
A proof of the arc-length formula.
These "proofs" can be found in the book A Primer of Infinitesimal Analysis by JL Bell. To what extent can these proofs be made rigorous?
Using the surface area of a cone example, we need some way of defining surface area before carrying out the computation.
[edit]
Concerning where the lack of rigour is: The book appears to treat geometric figures as being made up of infinitely many infinitesimal simplices. It calculates the area/volume/length of an infinitesimal simplex, and then uses integration to calculate the area/volume/length of the whole figure. In calculating the area of an infinitesimal simplex, it is assumed that these simplices satisfy classical rules of Euclidean geometry, like Pythagoras's Theorem, $A=\frac12bh$, and so on. No explicit limiting argument is used.
You can find similar calculations in the freely available article An Introduction to Smooth Infinitesimal Analysis by M O'Connor.
What does 'rigorous' mean? I.e., in what sense are they not already rigorous? (I haven't read the book, so I don't know about these specific proofs, but my understanding is that SIA already is a rigorous discipline. In fact, the book's subtitle is "A rigorous, axiomatically formulated presentation of the 'zero-square', or 'nilpotent' infinitesimal.")
@LSpice See my latest edit
@MattF. Bell doesn't really provide a definition. The article I linked to uses similar sorts of arguments
@MattF. OK, but why can you apply Pythagoras's Theorem to the small triangle? Maybe I'm nitpicking
@CarloBeenakker I thought my question was a bit silly. And I thought your answer was wrong too. So it didn't make sense to keep it. Anyway, if you'd prefer it, I'll undelete the question
|
2025-03-21T14:48:30.064821
| 2020-03-14T13:10:28 |
354895
|
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|
Stack Exchange
|
Positions in the Wythoff array
Suppose that $x$ and $y$ are positive integers. How can the position of $x+y$ in the Wythoff array (A035513) be predicted from the positions of $x$ and $y$?
Background. The Wythoff array begins with
1 2 3 5 8 13 21 34 55 89 144 ...
4 7 11 18 29 47 76 123 199 322 521 ...
6 10 16 26 42 68 110 178 288 466 754 ...
9 15 24 39 63 102 165 267<PHONE_NUMBER> ...
12 20 32 52 84 136 220 356<PHONE_NUMBER> ...
14 23 37 60 97 157 254<PHONE_NUMBER> 1741 ...
17 28 45 73 118 191 309<PHONE_NUMBER> 2118 ...
19 31 50 81 131 212 343<PHONE_NUMBER> 2351 ...
22 36 58 94 152 246<PHONE_NUMBER> 1686 2728 ...
As an example, take $x=11$ and $y=26$, from positions $(2,3)$ and $(3,4)$. The sum $37$ is in position $(6,3)$.
Are you looking for something that starts with 37 and returns that the position is (6,3) or something that starts with (2,3) AND (3,6) and returns that the position is (6,3)... perhaps without ever finding 11,26 and 37?
Aaron, the second option - to start with positions and return the position of the sum, perhaps without ever finding the numbers in the positions. If a closed form isn't available, then bounds on the number of algorithm steps may be of interest.
Because the Wythoff array is an encoding of the Zeckendorf representation (the representation of an integer as a sum of nonconsecutive Fibonacci numbers), one can use algorithms that, given the Zeckendorf representations of two integers $x$ and $y$, find the Zeckendorf representation of their sum $z=x+y$, as presented in Efficient Algorithms for Zeckendorf Arithmetic (2012).
Here is an expansion on Carlo's answer. It is more in the nature of clarifying what is sought as well as a nice parallel.
First another well known permutation of the integers is the the 2-ary array (A054582). For technical reasons is it convenient to index rows and columns starting with $0$
$\left[ \begin {array}{cccccc} 1&2&4&8&16&32\\ 3&6&
12&\mathbf{24}&48&96\\ 5&10&\mathbf{20}&40&80&160\\
7&14&28&56&112&224\\ 9&18&36&72&144&288
\\ 11&22&\mathbf{44}&88&176&352\end {array} \right]$
The analogous question is
Given pairs $(i,j)$ and $(k,\ell)$ find the pair $(p,q)$ giving the position of the sum of the entries in the two given positions.
for example given $(1,3)$ and $(2,2)$ we wish to get back $(5,2)$ since the corresponding entries are $24$ $20$ and $44$
It is not required to find $24,20$ and $44$ nor is it forbidden.
That is one strategy because we easily find
DECODE: The find the entry at position $(i,j),$ multiply $(2i+1)\cdot 2^j$
ENCODE: To find the position of $n$ factor out the twos to get $n=(2i+1)\cdot 2^j$
So then an easy solution to the given problem is decode $(i,j)$, decode $(k,\ell)$, add the results (however one adds) and then encode the sum.
It is relevant to use binary rather than decimal for the array hence
$\left[ \begin {array}{cccccc} 1&10&100&1000&10000&100000
\\ 11&110&1100&11000&110000&1100000
\\ 101&1010&10100&101000&1010000&10100000
\\ 111&1110&11100&111000&1110000&11100000
\\ 1001&10010&100100&1001000&10010000&100100000
\\ 1011&10110&101100&1011000&10110000&101100000 \end{array}\right]$
DECODE: Write $i$ in binary (better yet, use binary) write another $1$ then end with $j$ $0$'s This gives you the entry in binary.
So given $(10,3)$ we write $10$ in binary then a $1$ then $3$ zeros $1010\ 1\ 000$ i.e. $10101000$ Given a second pair we could write a second binary string and then perform binary addition. It is true that $21\cdot 2^3=168=10101000_2$ but in some sense we stayed closer to the given information. Especially if instead of calling it row $10$ we call it row $1010$
ENCODE: Given $n$ convert it to binary (better yet, get it in binary). The number of terminal $0$'s give the column and the head, removing the last $1$ gives the row.
Recall that there are two common ways to get the binary expansion of $i$: go right to left putting a 1 for odd $n$ and replacing it by $\frac{n-1}2$ or $0$ for even $n$ then replacing with $\frac{n}2.$ OR we can go left to right subtracting off the largest possible power of $2$ , say $2^i$ and putting a $1$ then continue considering $2^j$ for $j=i-1,i-2,\cdots$ each time either subtracting and putting a $1$ or just putting a $0$ according as the current value is or is not at least $2^j.$
The second method applied using the Fibonacci numbers $\cdots 13,8,5,3,2,1$ gives the unique Zeckendorf representation of $n$ as a binary vector with no two consecutive $1$'s. (I don't know if there is an easy right to left method.)
Using this representation turns the array of the question
$\left[ \begin {array}{cccccc} 1&2&3&5&8&13\\ 4&7&11
&18&29&47\\ 6&10&16&26&42&68\\ 9&
15&24&39&63&102\\ 12&20&32&52&84&136
\\ 14&23&37&60&97&157\end {array} \right] $
into
$\left[ \begin {array}{cccccc} 1&10&100&1000&10000&100000
\\ 101&1010&10100&101000&1010000&10100000
\\ 1001&10010&100100&1001000&10010000&100100000
\\ 10001&100010&1000100&10001000&100010000&
1000100000\\ 10101&101010&1010100&10101000&101010000
&1010100000\\ 100001&1000010&10000100&100001000&
1000010000&100001000000\end{array} \right]$
If one has the first column then one has the Zeckendorf representation of the entry in position $(i,j)$ and can do the appropriate addition, which is not quite as easy as binary addition but not too bad. And there are formulas such as $\lfloor i \tau^2 \rfloor-1=\lfloor i \tau \rfloor+i-1.$ Here $\tau$ could be replaced by a sufficiently good ratio of Fibonacci numbers. This still leaves some loose ends such as how to figure out the row from the representation.
|
2025-03-21T14:48:30.065164
| 2020-03-14T14:56:11 |
354898
|
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|
Stack Exchange
|
Mean squared absolute value of inner product of unit vectors
Given a nonempty finite subset $S$ of the unit sphere of $d$-dimensional complex Hilbert space, let $\lambda(S) = \frac{1}{\lvert S \rvert^2} \sum_{x,y \in S} \lvert \langle x,y \rangle \rvert^2$ be the mean squared absolute value of the inner product of two vectors chosen from $S$.
A basic observation from experiment is that $\lambda(S) \geq \frac1d$. This looks simple enough, but I don't know how to prove it, and as I've had no answers on math.stackexchange I am asking the question here.
A second experimental observation is that if $S_1$ and $S_2$ are disjoint then $\lambda(S_1 \cup S_2) \leq \max(\lambda(S_1), \lambda(S_2))$.
If these conjectures are true, it follows that $\mathcal{E}_d = \{S: \lambda(S) = \frac1d\}$ is closed under disjoint unions, and it would be interesting to characterize its "basic" sets, i.e. those sets in $\mathcal{E}_d$ that are not a disjoint union of other sets in $\mathcal{E}_d$ (or alternatively, those with no proper subset in $\mathcal{E}_d$). Clearly these include the orthonormal bases, but there are plenty of others, e.g. $S = \{(1, 0), (\frac12, \frac{\sqrt3}{2}), (\frac12, -\frac{\sqrt3}{2})\}$ for $d = 2$.
The Welch bound gives
$$\lambda(S) = \frac{1}{\lvert S \rvert^2}
\sum_{x,y \in S} \lvert \langle x,y \rangle \rvert^2 \geq
\frac{(\sum_{x \in S} \lvert \langle x,x \rangle \rvert)^2}{d \lvert S \rvert^2}=\frac{1}{d}$$
which is what you want.
There are Welch bound equality sets (do a google search) but achieving equality for general set size $|S|$ is difficult.
Thank you, that was the reference I needed!
|
2025-03-21T14:48:30.065301
| 2020-03-14T15:02:16 |
354899
|
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|
Stack Exchange
|
When is a manifold boundary a deformation retract of its open neighborhood?
For this question a manifold is a locally upper-Euclidean Hausdorff space; paracompactness or second-countability is not assumed, and boundary may be present.
Let $M$ be a manifold. What are sufficient conditions for there to exist open $U \subset M$ such that $\partial M \subset U$ and $U$ (strongly?) deformation retracts onto $\partial M$? (Could this be true for every manifold?)
Clearly $\partial M$ being collared in $M$ is a fairly general sufficient condition (this includes paracompact manifolds). However, I'd be interested in whether there is a more general sufficient condition than this.
I can't find a reference to " locally upper-Euclidean"; what does this mean? Do you mean locally homeomorphic to the upper-half space?
Yes. For each point there exists an open neighborhood homeomorphic to an open subset of $\mathbb{R}^{n - 1} \times \mathbb{R}+$, where $\mathbb{R}+ = {x \in \mathbb{R} : x \geq 0}$, i.e. $M$ is a manifold-with-boundary but not necessarily metrizable.
Thanks kaba, just wanted to be sure I wasn't missing something important :)
I do not understand the question: Boundary of every topological manifold is collared. What else would you need?
@MoisheKohan This is not true if one does not assume metrizability. See for instance this question (and the discussion in the comments): https://mathoverflow.net/questions/344104/collared-boundary-of-a-non-metrizable-manifold
The following theorem and example give (partial) answers. By an $n$-manifold we understand a Hausdorff space whose any point has a neighborhood, homeomorphic to an open subspace of $\mathbb R^n_+=\{(x_1,\dots,x_n)\in\mathbb R^n:x_n\ge 0\}$.
Theorem. For each 1-manifold $M$ the boundary $\partial M$ is a deformation retract of some open set $U\subseteq M$ that contains $\partial M$.
Proof. By the definition of a 1-manifold, for every $x\in\partial M$ there exists a neighborhood $U_x\subseteq M$, homeomorphic to $[0,1)$. Fix a homeomorphism $h_x:U_x\to[0,1)$ such that $h_x(x)=0$. It can be shown that for distinct $x,y\in\partial M$ the sets $U_x,U_y$ are disjoint. Then $U=\bigcup_{x\in \partial M}U_x$ is an open neighborhood of $\partial M$ and the function $H:U\times [0,1]\to U$, $H:(u,t)=h_x^{-1}(t\cdot h_x(u))$ where $u\in U_x$, is a deformation retraction of $U$ onto $\partial M$.
Example. There exists a separable 2-manifold $M$ whose boundary $\partial M$ is homeomorphic to $\mathbb R\times\mathfrak c$. Being non-separable, the boundary $\partial M$ cannot be a retract of a (necessarily separable) neighborhood $U$ of $\partial M$ in $M$.
This example is described by Peter Nyikos as Example 3.6 of his survey paper "The theory of nonmetrizable manifolds" in the Handbook of Set-Theoretic Topology.
Thanks! This goes a long way.
Unless there will be other answers, I'll probably accept this as an answer; the question has been open for almost two years now. The collar provides a reasonable sufficient condition, and the example here shows that a retract is not always possible.
|
2025-03-21T14:48:30.065623
| 2020-03-14T15:37:15 |
354901
|
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"paul garrett"
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|
Stack Exchange
|
Iwasawa decomposition on unitary group of anisotropic kernel
Let $E/F$ be a quadratic extension of number fields. If $V$ is a hermitian space over $E$, let $V=X+V_0+Y$ be its Witt decomposition, where $X,Y$ are maximal totally isotropic subspaces and $V_0$ is anisotropic kernel of $V$ respectively.
Let $P(X)$ be the parabolic subgroup of $U(V)$ stabilizing $X$.
Then we can decompose $U(V)=P(X)K$ where $K$ is a maximal compact subgroup of $U(V)$.
I am wondering what is the Iwasawa decomposition of $U(V_0)$. How can we decompose it?
I also want to know what is the parabolic subgroup of $U(V_0)$.
Any comments will be highly appreciated!
Unless I misunderstand something here... $U(V_o)$ is compact, no? And has no proper ($F$-rational) parabolics.
@Paul, You are right! I think there is no proper parabolic and so no proper Iwasawa decomposition. Thank you very much for comment.
@Paul, from your comment, I think that $U(1)$ has no proper parabolic subgroup. Right?
Right: with a non-degenerate $E$-one-dimensional space, the corresponding unitary group has no proper parabolics... because it has no non-trivial isotropic subspaces.
@Paul, thank you for clear explanation. I always learn much from your comment and answers!:)
|
2025-03-21T14:48:30.065730
| 2020-03-14T15:56:50 |
354902
|
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|
Stack Exchange
|
Learning a new field under publishing pressure?
This question is about real situation and I do not say anything about the field of research or any name. Also, I do not have any special opinion. I just want to say a real story and know the experts idea about this subject. Also, maybe someone experienced same situation and did some valuable work. I am so eager to hear the experts story.
Suppose you are a faculty member of a math department in a university. In this university we have the rule "publish or perish"; I know at least one such university, so the supposed set is not empty.
You are working on a special branch in a special field of mathematics which publishing paper is so difficult. Also, working on this topic is long term project which may have significant implications and results.
Parallel to this scenario, you are going to fire since you did not have sufficient number of publications.
In the other group in this department the situation is heavenly. They publish many papers since in their field publishing is so easy and there are a lot of Q1 journals for submitting and getting accept.
You can spend several months with this group and learn this branch. Also, you can publish enough (as do not fire) with the team work.
But, the problem is that this field is not in your interest and your contribution to published paper is so weak.
What shall you do? Does The end justifies the means?
Comment: Finding another job (to working on your problem independently)is so difficult!
Any ideas and guidance will be appreciated.
Don't see why people are downvoting this question. Maybe the tag general-mathematics is not suited, but the question seems as relevant as https://mathoverflow.net/questions/116734/publishing-a-bad-paper?rq=1 which was highly upvoted.
While this is a good question for another forum, most likely Academia, in this form I think it is actually harmful for MathOverflow. If it could be rewritten so it sounds less like "to research or not to research", I might find a way to fit it in to the mission here. Gerhard "We Are Pro Research Here" Paseman, 2020.03.14.
Different fields of math publish at different rates. However, one important skill that I look for when evaluating hiring and tenure cases is knowing their field (and self) well enough to identify problems they can solve in a reasonable amount of time. To phrase it a different way: it is easy to identify problems that are important but totally out of reach. The real skill is identifying things that are just barely accessible, and then doing them.
@Gerhardpaseman I asked real question in a research level. I can not change it since it describe real situation. Anyway, your comment has a lot of followers since the question is closed now! Feel free.
A first improvement to the question would be to find a more informative (and less poetic) title
@Andy putman you say your ideas but I asked special question. It is obvious that we must know our field to find good question. But sometimes you need to develop special branch. For example, why Sir Wiles traveled to USA?
@YCor feel free to change the title. I do not know new things. But mathematicians always say logical poet!
The title should reflect, or better summarize, what you want to ask. Maybe "Learning a new field under publishing pressure?" reflects the question.
@YCor Thanks for the title.
|
2025-03-21T14:48:30.065988
| 2020-03-14T16:10:07 |
354903
|
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|
Stack Exchange
|
Source of a quote by Ferdinand Rudio
I am looking for the source and context of this quote, found e.g. at St Andrews:
Only with the greatest difficulty is one able to follow the writings of any author preceding Euler, because it was not yet known how to let the formulas speak for themselves. This art Euler was the first to teach.
— F. Rudio
(My emphasis. Slight variants suggest that it could originally have been in another language.)
Does close-mindedness also speak for itself? ;-)
The quote is from a speech Rudio gave at the Town Hall in Zürich on the 6th December 1883; The German original is published in Felix Stähelin, Reden und Vorträge (1956, I have not found it online).
An English translation of the full speech is here. The translated quote reads as follows:
But I cannot move on from reviewing Euler's mathematical work without
having considered an important factor. I have said that mathematics is
a language in which natural phenomena can be described in the simplest
and most comprehensive manner. With this in mind, you will understand
how important it is to express mathematical thoughts themselves as
concisely and clearly as possible. In this respect, Euler's work was
epoch-making. We can be safe to say that the whole form of modern
mathematical thinking has been created by Euler. If you read any
author immediately before Euler, it is very difficult indeed to
understand his terminology, as he has not yet learned how to let the
formulas speak for themselves. This art was not taught until Euler
came along.
Oh! Same site. How did you find it so fast? Google doesn’t.
|
2025-03-21T14:48:30.066138
| 2020-03-14T17:28:32 |
354905
|
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|
Stack Exchange
|
The fastest growing function of given complexity
Let $f$ be a computable function $\mathbb{N} \to \mathbb{N}$
be a computable function. Since a program of a computable function is a finite object we can define plain Kolmogorov complexity of $f$ (we can identify programs as Turing machines, for example).
Now I will talk only about total computable functions.
1) Is there a function $f$ with complexity not greater than $d + O(1)$ such that for every $g \in \mathcal{F}_d$---the set of function with complexity atmost $d$---and for every $x \in \mathbb{N}$ it holds that $f(x) \ge g(x)$?
More precisely: Is there $C$ such that for every $d$ there exists $f$ with Kolmogorov complexity at most $d + C$ such that for every $g$ with Kolmogorov complexity at most $d$ and for every $x \in \mathbb{N}$ it holds that $f(x) \ge g(x)$?
2) Is there a function $f$ with complexity not greater than $d + O(1)$ that growing at $\infty$ faster than any function in $\mathcal{F}_d$, i.e. there exists $C$ such that for every $g \in \mathcal{F}_d$ and for every $x >C$ it holds that $f(x) \ge g(x)$?
3) Is there a rather small subset $F$ (say, $|F| = \text{poly}(d)$) of functions with complexity not greater than $d + O(1)$ such that for every $g\in \mathcal{F}_d$ there exists $f \in F$ that grows faster than $g$?
This could depend heavily on the exact definition of Kolmogorov complexity. Do you have anything precise in mind?
@Wojowu, I think you mean that it can depend on the way of the representation of functions and the choice of optimal decompressor? I agree with it. It is better to restrict the complexity of $f$ by $d + O(1)$. I will fix it, thank you
To me the question is even more confusing after changing to $d+O(1)$. What does a function $f$ with complexity not greater than $d+O(1)$ exactly mean? It's as ambiguous as the phrase a number not greater than $d+O(1)$. I suggest that you rephrase your question in standard first order logic statements instead of notions such as $O(1)$. For the ambiguity of Kolmogorov complexity, it's perhaps better to fix a language first and then argue the effect of changing the language.
Is $d$ meant to be a constant, or a function of $x$? If the idea is that $d$ is constant, then it seems like you could 'run' all the $g\in\mathcal{F}d$ and take the largest of them (plus 1, if you want). This will give you a larger constant $d'$ (dependent on $d$, of course) and a function $f\in\mathcal{F}{d'}$ that satisfies 1)...
@WhatsUp I have wrote more precisely. Is it clear now?
@StevenStadnicki the constant in $O(1)$ should not depend on $d$. I have wrote more precisely.
Yes, part 1 is now a clear statement to me.
Is this plain complexity or prefix-free? With plain, the answer to 2 is yes, since d + O(1) bits is enough to specify how many functions of complexity at most d are total. So a function can wait until that many have converged up to a given value, then output the max.
@DanTuretsky But what if for given $x$, the computation of $g(x)$ converges even though $g$ is not total?
@Wojowu You wait until you see the appropriate number converging on all $y \le x$. For sufficiently large $x$, all non-total $g$ of complexity $\le d$ will fail to converge on some $y \le x$. This is why it works for 2 but not 1.
@DanTuretsky I see, clever!
Alexander Shen gave the full answer: https://arxiv.org/abs/2004.02844
|
2025-03-21T14:48:30.066374
| 2020-03-14T17:51:33 |
354906
|
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|
Stack Exchange
|
On bounding a certain discrepancy between probability distributions on the symmetric group
Disclaimer. This is a follow up to a question I asked and answered on SE https://math.stackexchange.com/q/3579311/168758. The question was about upper-bounds. Here I'm interested in lower bounds, and they seem harder to get...
Question
So, let $\mathfrak S_n$ be the symmetric group of permutations on $n$ objects and let $P$ and $Q$ be a probability distributions on $\mathfrak S_n$ (i.e $P$ and $Q$ are points on the $n!$-simplex). Let $p_{ij}$ be the probability that a random permutation $\sigma$ drawn from $P$ ranks $j$ ahead of $i$, i.e satisfies $\sigma(i) < \sigma(j)$. Note that $p_{ii} = 0$ and $p_{ji} = 1-p_{ij}$ for every $i,j \in \{1,\ldots,n\}$ with $i \ne j$. Consider the quantity $\Delta(P,Q) := \sum_{1 \le i < j \le n}|p_{ij}-q_{ij}|$.
What is a reasonable lower-bound for $\Delta(P,Q)$ in terms of some standard measure of distance (e.g total variation) between $P$ and $Q$ ?
Partial answer: an upper bound via total variation
Claim. $\Delta(P,Q) \le n(n-1)TV(P,Q)$. Moreover, the $n^2$ factor in the bound is tight.
Proof. let $E_{ij} := \{\sigma \in \mathfrak S_n \mid \sigma(i) < \sigma(j)\}$. This is the set of permutations which rank $j$ ahead of $i$. One can then rewrite
$p_{ij} = \mathbb P_{\sigma \sim P}[\sigma \in E_{ij}] = \sum_{\sigma \in \mathfrak S_n}P(\sigma)1_{\sigma \in E_{ij}}$. Thus
$$
\begin{split}
\Delta(P,Q) &:= \sum_{i < j}|p_{ij}-q_{ij}| = \sum_{i < j}\left|\sum_{\sigma \in E_{ij}}(P(\sigma)-Q(\sigma))\right| \le \sum_{i < j}\sum_{\sigma \in E_{ij}}\left|P(\sigma)-Q(\sigma)\right|\
\\
&\le \sum_{i < j}\sum_{\sigma \in \mathfrak S_n}\left|P(\sigma)-Q(\sigma)\right| = \frac{n(n-1)}{2}\sum_{\sigma \in \mathfrak S_n}\left|P(\sigma)-Q(\sigma)\right|\\
&= n(n-1)TV(P,Q),
\end{split}
$$
where the first inequality is Cauchy-Schwarz.
We now show that the $n^2$ factor in the bound is optimal. Indeed if $P$ is a dirac and $Q$ is uniform, then $p_{ij} \in \{0,1\}$ and $q_{ij}=1/2$ if $i \ne j$. Thus, $\Delta(P,Q) = {n\choose 2}(1/2) = n(n-1)/4 = o(n^2)$. $\quad\quad\quad\Box$
There is no non-trivial lower bound as it is well possible that $P\neq Q$, whereas $p_{ij}=q_{ij}$ for all pairs of $i$ and $j$. The reason is that (as you point out), these numbers are nothing but the values of the measure $P$ (resp., $Q$) on the sets $E_{ij}$, and the collection of the sets $\{E_{ij}\}$ is just not big enough to separate measures on the symmetric group. For instance, let $P$ be the uniform measure, and let $Q$ be the uniform measure on the union of the cycle $(1,2,\dots,n)$ and of its product with the involution $\sigma(i)=n-i$. Then $p_{ij}=q_{ij}=1/2$ for all $i,j$.
By the way, your upper bound immediately follows from the inequality
$$|P(A)-Q(A)| \le \|P-Q\|/2$$
(the RHS of which is what probabilists erroneously like to call the total variation) satisfied for any subset of the symmetric group (in particular, for $A=E_{ij}$).
Thanks for the feedback; makes sense. As for the last comment, indeed $TV(P,Q) = \sup_A |P(A)-Q(A)|$ where sup is over measurables sets, and so my bound is loose enough (and not very informative, by the same token) to apply for even more general subsets $E_{ij}$.
Unfortunately, someone thought it was the greatest idea to downvote the question...
It might still be quite interesting to decribe the collections of measures determined by presribed values of $p_{ij}$ and to look at the distances (in an appropriate sense) between these collections.
Thanks again for the input. Let me try to cut-off some dead branches from my problem. So, for $h \in [0, 1/2]$, let us say $P$ verifies the "low noise condition" $\mathbf N(h)$ if $\min_{i < j}|p_{ij}-1/2| \ge h$. Then, one might be interested in bounding $\Delta(P,Q)$ under the following scenarios:
(A) $P$ verifies $\mathbf N(h)$; (B) Both $P$ and $Q$ verify $\mathbf N(h)$; (C) $P$ verifies $\mathbf N(h)$ and $Q=\hat{P}_N$ the empirical version of $P$ based on $N$ iid samples; etc.
|
2025-03-21T14:48:30.066633
| 2020-03-14T18:16:25 |
354908
|
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|
Stack Exchange
|
Morphisms on fibre products
Let $X$ and $Y$ be two quasi compact, separated schemes over $k$, and consider the fibre product $X \times Y$. If we call $p_1$ and $p_2$ the two projections, and we take perfect complexes $F_1, F_2 \in \mathfrak{Perf}(X)$, $G_1, G_2 \in \mathfrak{Perf}(Y)$, then by flat base change we have an isomorphism
$$
\text{RHom}_{X}(p_1^{\ast}F_1, p_1^{\ast}F_2) \otimes \text{RHom}_{Y}(p_2^{\ast}G_1, p_2^{\ast}G_2) \simeq \text{RHom}_{X \times Y}(p_1^{\ast}F_1 \otimes p_2^{\ast}G_1, p_1^{\ast}F_2 \otimes p_2^{\ast}G_2).
$$
I am trying to understand whether this is true even when we take all the complexes to be in $D_{qc}(X)$ and $D_{qc}(Y)$.
I am fairly positive, because by a result of Ben-Zvi, Francis, Nadler - arXiv:0805.0157 - we know that if consider DG enhancements of these categories (which I will denote by italic letters), then there is a quasi equivalence
$$
\mathcal{D}_X \otimes \mathcal{D}_Y \simeq \mathcal{D}_{X \times Y}.
$$
However, while the right hand side has homotopy category $D_{qc}(X \times Y)$, the tensor product on the left hand side might mess things up at the level of homotopy categories.
My idea was to try to apply the indization functor to the fully faithful functor
$$
\mathfrak{Perf}(X) \otimes \mathfrak{Perf}(Y) \rightarrow \mathfrak{Perf}(X \times Y),
$$
but the indization functor is symmetric monodical only when we consider the tensor product of stale $\infty$ categories (which in this case should be considered as the derived tensor product of DG categories). It is fine to assume that the derived categories are compactly generated, e.g. when the schemes are Noetherian.
Thanks!
This is not true. For example take $F_1 = \bigoplus_{n \in \mathbf{N}} \mathcal{O}_X$, $F_2 = \mathcal{O}_X$, $G_1 = \bigoplus_{m \in \mathbf{N}} \mathcal{O}_Y$ and $G_2 = \mathcal{O}_Y$. Moreover, assume $X = \text{Spec}(k)$ and $Y = \text{Spec}(k)$. Then we see that the left hand side is
$$
(\prod\nolimits_{n \in \mathbf{N}}\ k) \otimes_k (\prod\nolimits_{m \in \mathbf{N}}\ k)
$$
and the right hand side is
$$
\prod\nolimits_{(n, m) \in \mathbf{N} \times \mathbf{N}}\ k
$$
which aren't the same.
|
2025-03-21T14:48:30.066789
| 2020-03-14T19:23:43 |
354914
|
{
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"authors": [
"Alexander",
"Taras Banakh",
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}
|
Stack Exchange
|
A characterization for a topological property
Let $X$ be a compact non-Hausdorff topological space. I am looking for a characterization for the following property on $X$:
Property: For each non-empty closed subset $C$ of $X$ there exists a partition $C=C_1\bigsqcup\cdots\bigsqcup C_n$ of $C$ into closed subsets $C_i$ of $X$ and there exists a non-empty open set $U_C$ such that for each $i$, $U_C\cap C_i$ is contained in a connected component of $C_i$.
$U_C$ should be nonempty, right?
Yes, $U_C$ should be non-empty. Thank you for your comment.
Perhaps some motivation for this question would be helpful? I cannot quite see what this property is about..
For compact Hausdorff spaces this property is equivalent to the scatteredness (= each non-empty closed subspace has an isolated point). For compact (not necessarily Hausdorff) spaces, the scatteredness implies the property in OP. An example of compact non-Hausdorff non-scattered space with the property is the real line endowed with the cofinite topology.
|
2025-03-21T14:48:30.066889
| 2020-03-14T19:37:16 |
354916
|
{
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"authors": [
"Overflowian",
"Sebastian Goette",
"Steve D",
"Tyler Lawson",
"https://mathoverflow.net/users/1446",
"https://mathoverflow.net/users/360",
"https://mathoverflow.net/users/70808",
"https://mathoverflow.net/users/99042"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/354916"
}
|
Stack Exchange
|
$G$ uncountable implies $K(G,1)$ is not a finite CW complex
I have read that $H^i(K(\mathbb{R},1)$) has rank $2^\omega$ for any $i\in \mathbb{N}$ (see Thurston's comment here Nontrivial finite group with trivial group homologies?) therefore $K(\mathbb{R},1)$ is not a finite CW complex and even the $n$-skeleton is not finite for any $n$.
On the other hand there are infinite (discrete) Lie groups $\pi$, such that $K(\pi,1)$ is finite. For example consider $\pi = \pi_1 (E)$ where $E\subset \mathbb{S}^3$ is the knot exterior of a knot.
This seems to be a rather lucky case as we know that $K(F,1)$ is non finite for any discrete finite group $F$.
This made me wonder if the following is true:
Is $K(G,1)$ an infinite CW complex for any $G$ Lie group of dimension greater than 1?
What are other examples of $G$ such that $K(G,1)$ is finite?
Note: infinite CW complex is the same as being non compact.
I think you're looking for groups of type F.
For any finite CW-complex $X$ and any basepoint $x \in X$, the fundamental group $\pi_1(X,x)$ is finitely presented. (This is a consequence of the Seifert-van Kampen theorem.) In particular, the group itself is a quotient of a finitely generated free group, and hence must be a countable set.
However, if $G$ is a Lie group of positive dimension, then the underlying set of $G$ is uncountable. Therefore, no $K(G,1)$ can have the homotopy type of a finite CW-complex.
Thank you Tyler, do you know of other interesting classes of groups (a part from knot exteriors) that have finite $K(\pi,1)$?
@WarlockofFiretopMountain I'm afraid that the examples in the page that Steve D linked to above (esp. torsion-free, finitely-generated nilpotent or hyperbolic groups) cover most of the examples that I know.
I think you should add that you consider $K(G^\delta,1)$, that is, $G$ is equipped with the discrete topology (which it is typically not if called a Lie group of positive dimension). Otherwise, I would not even know what $K(G,1)$ was supposed to mean.
@SebastianGoette sorry maybe I am missing something, doesn't the definition of $K(G,1)$ depend just on the group structure of $G$?
The definition says $\pi_1(K(G,1))\cong G$ and $\pi_k(K(G,1))=0$ otherwise. Now everything hinges on your understanding of "$\cong$". If you say "Lie group of positive dimension", I think that you want the group with its topology and differentiable structure. On the other, $\pi_1(X)$ is classically just a discrete group. But there might be a context where the functor $\pi_1$ can take values in groups with additional structure (in which case $K(G,1)$ should be related to the classifying space $BG$ in that category). I just wanted to avoid any misunderstandings.
|
2025-03-21T14:48:30.067093
| 2020-03-14T19:46:34 |
354918
|
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|
Stack Exchange
|
Degree of automorphic forms, SL(3,Z), and the elliptic Gamma function
In this article, the authors interpret a certain special function, the elliptic Gamma function, defined as
$$
\Gamma(z,\tau,\sigma)=\prod_{j,k=0}^\infty\frac{1-e^{2\pi i((j+1)\tau+(k+1)\sigma-z)}}{1-e^{2\pi i(j\tau+k\sigma+z)}}
$$
with $z,\tau,\sigma\in\mathbb{C}$ and $\mathrm{Im}\,\tau,\mathrm{Im}\,\tau>0$, as the (values of a) generator of an "automorphic form of degree 1", obeying the equation
$$
\Gamma(z/\sigma,\tau/\sigma,-1/\sigma)=e^{i\pi Q(z;\tau,\sigma)}\Gamma((z-\sigma)/\tau,-1/\tau,-\sigma/\tau)\Gamma(z,\tau,\sigma),
$$
for some polynomial $Q(z;\tau,\sigma)$, where the arguments of the Gamma functions are related by the action of some elements of $\mathrm{SL}\mathbb{(3,Z)}$. They call the Jacobi theta function an automorphic form of degree 0, and I am confused by this. I have two questions:
What is the definition of the degree of an automorphic function in this context? I could not find anything in the literature that matches the authors' usage of the term.
I thought that automorphic forms satisy equations of the form $f(g\cdot X)=j_g(X)f(X)$, $g$ being the element of some group $G$ acting on a complex manifold $X$, and $j_g(X)$ being the "factor of automorphy", a nowhere zero function. The above equation for the elliptic Gamma function is not of this form, containing three factors of the function instead of just two, and once again I could not find any other examples in the literature. How does the elliptic Gamma function match the definition of an automorphic form and are there other known examples of this type?
|
2025-03-21T14:48:30.067230
| 2020-03-14T19:54:59 |
354919
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Daniele Tampieri",
"Yongj Tang",
"https://mathoverflow.net/users/113756",
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"url": "https://mathoverflow.net/questions/354919"
}
|
Stack Exchange
|
The derivative of a filter with respect to a output signal
I have two signals, $d(t)$ and $p(t)$, respectively the input and the output of the matching filter $w(t)$, i.e.
$$
d(t)*w(t)=p(t)
$$
where $*$ denotes convolution.The impulse response $w(t)$ may be calculated by going into the frequency domain:
$$
w(t)=F^{-1}\left[\frac{F[p(t)]\overline{F[d(t)]}}{F[d(t)]\overline{F[d(t)]}+\epsilon}\right]
$$
How can I get the derivative of the the filter $w(t)$ with respect to $p(t)$, i.e.
$$\frac{\partial{w}}{\partial{p}}=?$$
I don't understand how do you get your expression for the impulse response $w(t)$: however I'd solve the problem considering $w(t)$ as a functional $w(t)=\mathfrak{w}[p](t)$ of $p(t)$ and then calculating the functional derivative, i.e.
$$
\frac{\partial w}{\partial p}=\frac{\delta w}{\delta p}=\frac{\delta\mathfrak{w}[p]}{\delta p}\triangleq\left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}{\mathfrak{w}[p+\varepsilon v]}\right|_{\varepsilon=0}
$$
where $\delta p=\varepsilon v$ is a variation of the output singnal when the input $d(t)$ is kept constant (from the engineering point of view, possibly due to a parametric variation of the impulse response function).
Precisely, by applying formally the Fourier transform to the Input/Output relation above we have
we have
$$
\hat{d}(\omega)\cdot\hat{w}(\omega)=\hat{p}(\omega)\iff w(t)=\mathscr{F}^{-1}\left[\frac{\hat{p}(\omega)}{\hat{d}(\omega)}\right](t)\label{1}\tag{1}
$$
and thus
$$
\begin{split}
\left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}{\mathfrak{w}[p+\varepsilon v]}\right|_{\varepsilon=0}&=\mathscr{F}^{-1}\left[\frac{\hat{v}(\omega)}{\hat{d}(\omega)}\right](t)\\
&=\mathscr{F}^{-1}\left[\big({\hat{d}(\omega)\big)^{\!-1}}\right]\ast v(t)
\end{split}
$$
Finally (and formally) we can say that
$$
\begin{split}
\frac{\partial w}{\partial p}(\:\cdot\:)& \triangleq\mathscr{F}^{-1}\left[\big({\hat{d}(\omega)\big)^{\!-1}}\right]\ast (\:\cdot\:)\\
&=\int\limits_{-\infty}^{+\infty}\!\!\! \mathscr{F}^{\!-1}\left[\big({\hat{d}(\omega)\big)^{-1}}\right](t-s)(\:\cdot\:)\,\mathrm{d}s
\end{split}\label{2}\tag{2}
$$
i.e. the functional derivative $\frac{\delta w}{\delta p}$ is a convolution operator that maps the variations $v$ of the output signal $p$ to the linear space of (first order) variations of the impulse response $w$ allowing for, from the engineering point of view, an estimation of the magnitude of this quantity.
Notes
All the above analysis is formal since without any assumption on $d(t), p(t)$ the Fourier transforms have no meaning. However, simply assuming $d\in\mathscr{E}^\prime$ (i.e., from the engineering point view, a finite duration of the input signal) and $w, p\in \mathscr{S}^\prime$ (a mild condition which allows to consider even non causal systems) the formal steps done above become perfectly rigorous as an application of the Hörmander/Łojasiewicz solution of the division problem (see this Q&A for more info on this point).
Edit: the meaning of the variation $v$ of the output signal. As the development above shows and as it is explicitly said in the last sentence of the answer, $\frac{\delta w}{\delta p}$ is a convolution operator, not a function. Its argument $v(t)$ is therefore a datum: you cannot deduce it from the input $d$ and the output $p$ of the ideal system. Thus the methodology you may follow in order to use formula \eqref{2} is the following one
Take a real system whose input signal is $d(t)$ and whose expected output signal is $p(t)$: the ideal impulse response of this system is $w(t)$ as expressed by \eqref{1}.
Apply to the system the input signal $d(t)$ and measure, or more generally evaluate, the effective output signal $p_e(t)$.
Put $v(t)=p_e(t)-p(t)$: now you can evaluate $\frac{\delta w}{\delta p}[v](t)$ by using a convolution calculation algorithm, and then you can estimate the effective impulse response
$$
w_e(t)\simeq w(t)+\frac{\delta w}{\delta p}[v](t)
$$
From an engineering point of view, the $\varepsilon$ parameter could be though as the magnitude of the cause that makes $w$ vary, i.e. it could be the working temperature of the system ($\varepsilon=T$), the relative humidity ($\varepsilon=\theta\%$), the parameter variation ($\varepsilon=\text{tol.}\%$), etc. Thus functional derivative allows for an a posteriori (after a performing a set of measurements, for example) determination of the influence the change of those parameter has on the system.
For more information on functional derivatives it is possible to have a look at this Q&A.
Hi, Daniele, I add a small positive number to avoid the denominator of $\hat{d}(\omega)$ to be 0. Could you please explain what the $v(t)$ represents? If I want to calculate the derivative programmatically using the last formula, how to understand the $*$ in the last formula? or how can I do it?
@YongjTang I am posting a revision just now. Please wait a moment.
@YongjTang now the answer is fairly complete. However, feel free to ask for further explanation: probably I'll not answer to you immediately, but I'll do my best to help you.
Daniele, Thank you very much for your kind help. I have read your answer carefully. Maybe I major not in math, and I am not familiar with some mathematical representations. I mainly want to implement it programmatically. I will try my best to read some you suggested and to understand it. Finally, I want to implement it with anumerical solution by using program language. If I have some confusion, I will ask you. Thank you very much.
@YongjTang you’re welcome. And if you really like my answer, please consider accepting it.
@YongjTang it’s a typo. I’ll correct it
@Daneile, I am confused with that you give
$$
w(t)=\mathscr{F}^{-1}\left\frac{\hat{p}(\omega)}{\hat{d}(\omega)}\right
$$
and
$$
\frac{\mathrm{d}}{\mathrm{d}\varepsilon}{\mathfrak{w}[p+\varepsilon v]}_{\varepsilon=0}
= \mathscr{F}^{-1}\left\frac{\hat{p}(\omega)}{\hat{d}(\omega)}\right
$$
They are the same? Could you explain the last formular that how to get the derivative ? e.g. 1 calculate the FT with input data $d(t)$, 2 .... 3 inverse FT 4 convolute what?
@YongjTang For the other comments, I’ll answer later.
@ Daniele, Thanks for your kind help, now I am confused with $v(t)$, it represents what? and how can I calculate it?
@YongjTang, I added a new note emphasized in bold in order to explain the meaning of $v$: in doing so, I hope to have dissipated a few of your doubts. However, despite not being able to answer immediately to your question, I am here.
@ Daniele, detailed explanation. I understand it, and really appreciate your kind help.
|
2025-03-21T14:48:30.067989
| 2020-03-14T20:28:07 |
354920
|
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"Paul Taylor",
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|
Stack Exchange
|
Why is the theory of small categories not algebraic?
In "Partial Horn logic and cartesian categories", E. Palmgren and S. J. Vickers state without proof that "The theory of categories is not algebraic." Is there a reference, or an elementary argument, for this fact? In particular, I'm interested in the setting where "algebraic" refers to the multisorted, potentially infinitary setting.
To be precise, this would entail showing that there is no monadic functor from $\mathbf{Cat} \to \mathbf{Set}/S$ for any set $S$.
One way to see it is that the category of categories is not a regular category, while model of algebraic theory (even multisorted and infinitary) are regular, even exact categories.
This follows from two Facts:
1) A category monadic over Set/S is always an exact category. That is it has quotient by equivalence relation that are effective and universal. It is in particular a regular category. This is showed for example here.
2) The category of categories is not a regular category. An explicit example of regularity (and hence exactness) failing in the category of categories (or posets, or topological spaces) can be found in the example section of the nLab page.
This is strictly speaking the correct answer, but to be more informative one should add that the theory of small categories is an essentially algebraic theory. https://ncatlab.org/nlab/show/essentially+algebraic+theory
of course, but given that the question quote Palmgren & Vickers' paper, I assume this was considered as known.
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2025-03-21T14:48:30.068131
| 2020-03-14T20:39:03 |
354922
|
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|
Stack Exchange
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Quotient of a normal quasi-projective variety by a finite group
I'm a topologist and not an algebraic geometer, but the following question arose in my work.
Let $X$ be a quasiprojective algebraic variety over $\mathbb{C}$ and let $G$ be a finite group acting on $X$. Since $X$ is quasiprojective, we have the quotient variety $X/G$.
Question: if $X$ is smooth, must $X/G$ be normal? Even better, does this hold if $X$ is assumed not to be smooth, but only normal?
I've googled and searched through various books, but I can't find this anywhere (though as I said, I am not a specialist in algebraic geometry, so it could be the case that something far more general is true and I just don't know the how to translate the general statement into the above rather simple-minded one).
If $X$ is normal, so is $X/G$.
Just an algebraic interpretation of @Simpleton's answer in the case of finite group actions. Let $B$ an integrally closed domain with the field of fractions $L$. Let $G$ be a finite subgroup of ${\rm{Aut}}(L)$. Then the extension $L/L^G$ is Galois and $B^G=B\cap L^G$ is a domain (the superscript denotes the invariant subfield/subring) inside $L^G$. It is easy to see that $L^G$ is the field of fractions of $B^G$. An element of $L^G$ integral over $B^G$ is integral over $B$ and hence lies in $B\cap L^G=B^G$. So $B^G$ is integrally closed. Now recall that normality is a local property which for an affine and irreducible variety $X$ translates to the integral closedness of $\mathcal{O}(X)$. Applying the algebraic result above, if $\mathcal{O}(X)$ is integrally closed, then so is $\mathcal{O}(X/G)=\mathcal{O}(X)^G$.
@user31480 But in your action, where $x\mapsto\zeta.x$ and $y\mapsto\zeta.y$, elements like $x^{n},y^{n},x^{n-1}y,...$ are preserved as well. So $B^G$ is much larger than $\Bbb{C}$.
Good point. Sorry I delete my example now.
This is true because $\mathrm{X}/\mathrm{G}$ is a 'categorical quotient' - the projection $\pi:\mathrm{X}\rightarrow \mathrm{X}/\mathrm{G}$ is $\mathrm{G}$-invariant (in the sense that $\pi(gx)=\pi(x)$ for all $x\in \mathrm{X}$ and $g\in \mathrm{G}$) and in fact universal with this property. If $\mathrm{X}$ is normal, consider the normalisation $\widetilde{\mathrm{X}/\mathrm{G}}\rightarrow \mathrm{X}/\mathrm{G}$ of $\mathrm{X}/\mathrm{G}$. Then $\pi$ lifts to a map $\mathrm{X}\rightarrow \widetilde{\mathrm{X}/\mathrm{G}}$, which is $\mathrm{G}$-invariant, and in fact is also a categorical quotient of $\mathrm{X}$ - hence $\widetilde{\mathrm{X}/\mathrm{G}}\rightarrow \mathrm{X}/\mathrm{G}$ must be an isomorphism, and $\mathrm{X}/\mathrm{G}$ is normal.
Thanks! I accepted the other answer since I found it a little more enlightening (though that is probably because I am a low-tech person, so I am more comfortable with arguments that are less abstract).
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2025-03-21T14:48:30.068337
| 2020-03-15T00:21:25 |
354928
|
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|
Stack Exchange
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Left and right topological K-theory of Banach algebras
Let us consider the topological $K$-functor on the category of Banach algebras as described in page $18$ of "Introduction to the Baum–Connes conjecture" by Alain Valette.
The definition is based on invertible groups $\mathrm{GL}_n(A)$where $A$ is a Banach algebra.
What would we obtain if we replace these invertible groups with the space (semigroup) of right or left invertible matrices? Is the resulting structures somehow the same as standard $k$ groups defined with full invertible group $\mathrm{GL}_n(A)$?
The motivation comes from the following situation:
One can find in the literature the concepts left and right topological stable rank which are based on left or right invertibility of a certain element $\sum_{i=1}^n a_i a_i^*$ associated to an $n$-tuple $(a_1,a_2,\ldots,a_n)$ in a $C^*$-algebra. With this motivation we try to consider the left and right K-theories on category of Banach algebras. Does this trivially leads us to the standard K-theory?
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