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2025-03-21T14:48:29.752086
| 2020-01-31T18:54:52 |
351649
|
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|
Stack Exchange
|
Can this number be interpreted as a fractal dimension?
Under Goldbach's conjecture, let's denote for a large enough integer $n$ by $r_{0}(n)$ the quantity $\inf\{r>0,(n-r,n+r)\in\mathbb{P}^2\}$ and by $k_{0}(n):=\pi(n+r_{0}(n))-\pi(n+r_{0}(n))$.
Let's now consider the density of primes around $n$ at two different scales: first at large scale, hence a global density, $\delta_{0}(n):=\frac{\pi(2n)}{2n}$ and then at small scale, hence a local density, $\delta_{1}(n):=\frac{k_{0}(n)}{2r_{0}(n)}$.
My idea is that, writing $\delta_{1}(n):=\delta_{0}(n)^{\mathcal{D}_{n}}$, the number $\mathcal{D}_{n}$ can be interpreted as the fractal dimension of a "fractal prime curve" in the plane, and thus should be a real number less than $2$. This would imply that $\frac{2r_{0}(n)}{k_{0}(n)}\ll\log^{2} n$ and possibly that $r_{0}(n)\ll\log^{C}n$ for some positive constant $C$.
So, can this interpretation of $\mathcal{D}_{n}$ as a fractal dimension be made rigorous and thus be proven to be less than $2$?
|
2025-03-21T14:48:29.752296
| 2020-01-31T19:00:22 |
351650
|
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|
Stack Exchange
|
A question on the problem of Dirichlet 2
Let $U$ be an open set in $\mathbb{R}^{n}$ with $n\geq2$ and $V$ an open set containing the boundary $\partial U$ of $U$. Suppose $u$ is subharmonic on $V$. We know that the generalized solution of Dirichlet problem $H^{U}_{u}(x)$ exists, i.e. is a harmonic function on $U$ that $\to u(y)$ as $x\to y$, for all regular point $y\in\partial U$. My question is: can we say that
$$u(x)\leq H^{U}_{u}(x)$$ for all $x\in V\cap U$?
Notice that the answer would be yes, if $u$ was subharmonic on $U$, and not on a neighborhood of the boundary of $U$ only.
The answer is no. E.g., let $U:=\{x\in\mathbb R^n\colon|x|<1\}$, $V:=\{x\in\mathbb R^n\colon1/2<|x|<2\}$ (or $V:=\mathbb R^n\setminus\{0\}$), $u(x):=|x|^{2-n}-1$ for $x\in V$ if $n\ge3$, and $u(x):=-\ln|x|$ for $x\in V$ if $n=2$. Then $u$ is harmonic and hence subharmonic on $V$. However, $u>0=H_u^U$ on $V\cap U$.
|
2025-03-21T14:48:29.752389
| 2020-01-31T19:30:42 |
351653
|
{
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"David Roberts",
"J Cameron",
"Jiří Rosický",
"Kevin Carlson",
"Simon Henry",
"Tim Campion",
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"url": "https://mathoverflow.net/questions/351653"
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|
Stack Exchange
|
Does the homotopy category of spaces admit a weak generating set?
As a follow-up to this question, let $\mathcal C$ be a category and $\mathcal S \subseteq \mathcal C$ a class of objects. Say that $\mathcal S$ is weakly generating if the functors $Hom_{\mathcal C}(S,-)$ are jointly conservative, for $S \in \mathcal S$. That is, a map $X \to Y$ in $\mathcal C$ is an isomorphism if and only if it induces a bijection $Hom_{\mathcal C}(S,X) \to Hom_{\mathcal C}(S,Y)$ for each $S \in \mathcal S$.
Question 1: Does the homotopy category of spaces admit a small generating set? (For example, as Simon Henry asks, do finite CW complexes work? How about the spheres?)
Of course, by Whitehead's theorem, the homotopy category of pointed connected spaces admits a generating set given by the spheres. But I'm not sure about unpointed spaces.
Note that the singleton set comprising the contractible space $\ast$ is a generator in the $\infty$-category of spaces, since $X \to Y$ is an equivalence if and only if $Map(\ast, X) \to Map(\ast,Y)$ is an equivalence. But passage to the the homotopy category discards the higher homotopy of the mapping spaces.
Question 2: More generally, if $\mathcal C$ is an accessible $\infty$-category, then does the homotopy category $h\mathcal C$ admit a small generating set? What if we assume that $\mathcal C$ is presentable?
Again, if $\mathcal C$ is $\kappa$-accessible, then the class $\mathcal C_\kappa$ of $\kappa$-compact objects forms a generating set in $\mathcal C$, but it's not clear if it forms a generating set in $h\mathcal C$. In fact, I think that Question 2 (in the "presentable" case) is equivalent to Question 1: if the answer to Question 1 is affirmative, so that $\mathcal S$ is a generating set for the homotopy category of spaces and $\mathcal T$ is a generating set for $\mathcal C$, then the set of spaces $S \ast T$ for $S \in \mathcal S, T \in \mathcal T$ forms a generating set for $h\mathcal C$. Here $\ast$ denotes copowering.
One result in this direction is Rosicky's Theorem, which says (in model-independent language) that if $\mathcal C$ is a presentable $\infty$-category, then the canonical functor $h\mathcal C \to Ind_\kappa(h\mathcal C_\kappa)$ is essentially surjective and full for some $\kappa$. For my purposes, it would suffice to know that this functor is conservative for some $\kappa$.
Naive question: what the status of the class of finite CW complexes regarding question 1 ? Is it known they are not enough, or you don't know ?
@SimonHenry I don't know -- but I imagine that somebody does! If finite CW complexes don't work, it's hard to imagine that anything will! Maybe I should add that as an additional question.
By the way, Rosicky’s paper had an error invalidating the main claims, which remain open as far as I know. The Arxiv version has been updated: https://arxiv.org/pdf/math/0506168.pdf
@KevinCarlson Ah, thanks! I was about to use that result for something!
@TimCampion Yes, it would be a very useful theorem-Neeman shows it implies the dual of a well generated triangulated satisfies Brown representability, for instance.
A counter-example to my "theorem" is in https://arxiv.org/pdf/1102.3240.pdf.
This paper by Kevin Carlson and Dan Christensen says that the answer to question one is no: No set of spaces detects isomorphisms in the homotopy category, arXiv:1910.04141.
Ah, perfect! (Though rather disappointing.) Thanks!
@TimCampion it's worth pointing out that in another paper Kevin Carlson shows that there is a 2 categorical generating set. Maybe he'll chime in to say more. https://arxiv.org/pdf/1802.04439.pdf
@TimCampion It all comes down to the existence of noninvertible group homomorphisms which, according to groups of bounded size, appear to be conjugate to the identity. In the 2-morphisms of the homotopy 2-category we get our hands on homotopies with endpoints fixed, repairing the mod-conjugacy problem and allowing the spheres to weakly generate. The homotopy 2-category of any presentable infinity category also gets a weak generator. I find it mysterious, spaces not being an $(\infty,2)$-category, that their homotopy 2-category should be the nice thing, but so it goes.
Carlson's paper is called On Whitehead’s theorem beyond pointed connected spaces (arXiv:1802.04439). @Kevin that result is remarkable to me. Why tori? Is there a good conceptual reason?
@DavidRoberts There’s nothing very special about the tori here. It’s just that the family of maps $\pi_1 L^n f$, $L$ the free loop space functor, see how $f$ acts on $\pi_n$ at each base point, and can be seen in the homotopy 2-category. I find it less intuitive to work with unbased mapping spaces from spheres, but Raptis convinced me that in fact the spheres are a generator as well.
|
2025-03-21T14:48:29.752725
| 2020-01-31T19:47:54 |
351655
|
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|
Stack Exchange
|
Approximation of adapted continuous random function by adapted polynomials
Given filtration $\mathcal{F}_t$. Let $f(t,x)$ is adapted with respect to $\mathcal{F}_t$ continuous in $t$ and $x$ function. Question: Is there sequence of adapted functions $f_n(t,x)$ such that $f_n(t,x)$ is polynomial in $x$ and $f_n(t,x)$ converges to $f(t,x)$ uniformly on each compact $x \in[a,b]$? It seems the answer is yes, but how to prove it?
This statement is used in the proof of Ito formula, when $f(t,x)$ if itself random adapted function.
|
2025-03-21T14:48:29.752800
| 2020-01-31T20:23:38 |
351657
|
{
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"authors": [
"Piotr Hajlasz",
"Wlod AA",
"https://mathoverflow.net/users/110389",
"https://mathoverflow.net/users/121665"
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|
Stack Exchange
|
Generalized Nikodym sets
Let me quote en.wikipedia about the original Nikodym's example (the definitions above I've written myself just for MO):
a Nikodym set is a subset of the unit square in $\ \mathbb R ^2\ $ with the complement of Lebesgue measure zero, such that, given any point in the set, there is a straight line that only intersects the set at that point. The existence of a Nikodym set was first proved by Otto Nikodym in 1927.
End of quote. See https://en.wikipedia.org/wiki/Nikodym_set.
In the above example, we can talk about the open square (or else, we may remove the border points of the square from the set and the things would still work). The $\ L\ $ would be the intersections of the straight lines with the open square. This would define a little bit nicer Nikodym set.
I am surprised that the original Nikodym set was provided in $\ (0;1)^2\ $ instead of $\ \mathbb R^2,\ $hence
Question 1: Let $\ L\ $ be the set of all straight lines in $\ \mathbb R^2\ $ Does there exist a respective Nikodym set, i.e. $\ A\subseteq \mathbb R^2\ $ such that the complement of $\ A\ $ has measure $0\,$ and
$$ \forall_{x\in A}\exists_{\ell\in L}\quad \ell\cap A=\{x\} $$
?
Question 2: Does there exist $\ A\ $ as in Question 1 and such that $\ A = \{2\cdot x: x\in A\}\ $ ?
I'll stop now while additional natural questions come to one's mind; some of them after reading the whole quoted en.wikipedia article.
The answer to Question 1 is yes. This is a result of Falconer, Corollary 6.6 in
K. J. Falconer,
Sets with prescribed projections and Nikodým sets.
Proc. London Math. Soc. (3) 53 (1986), no. 1, 48–64.
The result states as follows:
Theorem.
Let $1\leq m<n$. Then there exists a set $K\subset\mathbb{R}^n$ with
$\mathcal{L}^n(K)=0$, such that for each $x\in\mathbb{R}^n\setminus K$, there is a $m$-plane $P$ such that $P\cap (\mathbb{R}^n\setminus K)=\{ x\}$.
This answers positively Question 1 with $A=\mathbb{R}^n\setminus K$.
Thank you, Piotr, for an extra general answer. My Q2 was initially a bit different -- I asked about a Nykodym set in $\ A\subseteq(-1;1)^2\ $ such that $\ (-1;1)^2\cap B = A,\ $ where $\ B:={2\cdot x:x\in A}.\ $ Answer Yes to this modified Q2 could provide Yes to Q1.
Nikodym (sorry for my typo).
(It seems, Piotr, that you have 2 typos -- the last two $\ \mathbb R^n\setminus K\ $ should be simply $\ K,\ $ I guess).
@WlodAA No typo. I followed Falconer's notation. The set $\mathbb{R}^n\setminus K$ has full measure so in your notation $A=K\setminus\mathbb{R}^n$.
Nie kijem, to pałką -- then not the last two but the first $\ \mathbb R^n\setminus K\ $ should be simply $\ K;\ $ am I wrong again?
@WlodAA You are right. There was a typo in Falconer's paper. Dziękuję!
Piotr, u'r welcome! :)
|
2025-03-21T14:48:29.753006
| 2020-01-31T20:48:50 |
351659
|
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"Frode Alfson Bjørdal",
"Noah Schweber",
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|
Stack Exchange
|
Set of definable real numbers?
Is there a set theory at least as strong as $KP\omega$ which has as a theorem that there is a set $\mathbb{D}$ of precisely the definable real numbers?
How do you define "definable real number"? The obvious approach is via truth, which isn't definable.
Yes, I now recall Joel David Hamkins had some on this which entailed indefinability in ZF.
I had in mind definable by a first order condition as in the subset axiom.
Again, how do you define "definable by a first order condition"?
The subset axiom already uses the notion of a first order definable condition, which is sometimes called a Skolem condition. Basically it is a condition expressed using only the set theoretic language. Weyl was onto the same thing earlier, and both were motivated by making Zermi’s theory more precise.
"The subset axiom already uses the notion of a first order definable condition" Not really - it's a scheme of axioms, one for each formula. That's not helpful for what you're trying to do here, since you need a single axiom treating all possible definitions at once.
@Noah Schweber You’re right! We would indeed need a truth predicate for such a construction, and cannot have one I guess for strong theories. This is what Hamkins was onto, I think. But perhaps we may have one for weaker theories?
Well, Tarski applies to any reasonably strong theory whatsoever, so the only hope would be something like the overkill situation I indicated - where the objects you want to define all happen to have simple definitions.
You are right again. The Diagonal Lemma makes it impossible. So such constructions cannot be had in classical style set theories. Thanks!
This is only a partial answer to the question.
The immediate problem here is phrasing the question properly. The predicate "is a definable real number" is not definable in any natural way, so it doesn't even make sense in general to ask whether "Every real is definable" (or "For every set, the set of definable reals in that set is a set") is a theorem of a given system. (And note that if it did, the question would be trivial: ZFC proves that $\mathbb{R}$ exists and has the full Separation scheme, so if we could refer to definability then ZFC would prove "there is a set of all definable reals.")
Contrast this with e.g. the construction of the $L$-hierarchy: definability in a "small structure" is perfectly definable - e.g. the predicate "$x$ is definable in $M$" is definable - but definability in the universe poses a problem. Also contrast this with ordinal definability, which surprisingly is definable after all.
The only way around this problem I see is to go for an overkill solution. While definability itself is not definability, $\Sigma_n$-definability is definable for each fixed $n$. We would therefore be quite happy if "definable" and "$\Sigma_n$-definable" were to coincide for reals. This can happen - for example, in $L_{\omega_1^{CK}}$ the $L$-ordering itself is definable by a single formula and has ordertype $\omega_1^{CK}$, so each real $r$ can be defined as "The unique real in the $\alpha$th place of the $L$-ordering of the universe" for an appropriate computable (hence definable) well-ordering $\alpha$. (In fact, note that this gives more generally that every element of $L_{\omega_1^{CK}}$ is definable by a $\Sigma_n$ formula for some fixed small $n$ - at a glance, $n=3$ should be enough.)
Of course as you observe this by itself doesn't solve the problem: we want the class of $\Sigma_n$-definable reals to be a set (and this fails in $L_{\omega_1^{CK}}$, in particular). I think the natural question at this point is:
Is there some natural number $n$ such that the theory $KP\omega$ + "$\mathbb{R}$ exists" + "Every real is $\Sigma_n$-definable" is consistent?
I believe this is the "right" precisiation of your question (although unfortunately I don't have an answer at the moment).
I don’t see that this settles whether there is a set of the definable real numbers.
@FrodeAlfsonBjørdal You're right, here the definable reals form a definable class (namely all of $\mathbb{R}$) but not a set. Let me see if I can fix that ...
I like your precisiation and hope it can work.
So may I interpret the last two as $\exists x(\forall y(y\in x\leftrightarrow y \ is \ [ a \ \Sigma_n-definable\ real \ number] ))?$
|
2025-03-21T14:48:29.753336
| 2020-01-31T21:26:58 |
351663
|
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"Asaf Karagila",
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|
Stack Exchange
|
Undetermined Banach-Mazur games in ZF?
This question was previously asked and bountied on MSE, with no response. This MO question is related, but is also unanswered and the comments do not appear to address this question.
Given a topological space $\mathcal{X}=(X,\tau)$, the Banach-Mazur game on $\mathcal{X}$ is the (two-player, perfect information, length-$\omega$) game played as follows:
Players $1$ and $2$ alternately play decreasing nonempty open sets $A_1\supseteq B_1\supseteq A_2\supseteq B_2\supseteq ...$.
Player $1$ wins iff $\bigcap_{i\in\mathbb{N}} A_i=\emptyset$.
ZFC implies that there is a subspace of $\mathbb{R}$ with the usual topology whose Banach-Mazur game is undetermined; on the other hand, it's consistent with ZF+DC (and indeed adds no consistency strength!) that no subspace of $\mathbb{R}$ does this ("every set of reals has the Baire property").
However, when we leave $\mathbb{R}$ things get much weirder. My question is:
Does ZF alone prove that there is some space $\mathcal{X}$ whose Banach-Mazur game is undetermined?
Controlling the behavior of all possible topological spaces in a model of ZF is extremely hard for me, and I suspect the answer to the question is in fact yes. In fact, I recall seeing a pretty simple proof of this; however, I can't track it down or whip up a ZF-construction on my own (specifically, everything I try ultimately winds up being a recursive construction killed by having too many requirements to meet in the given number of steps).
I removed my answer since I'm not sure how to fix the issues you guys brought up. I'm not sure that it's unfixable, just I can't do it now.
This is only a partial answer. ZF + DC + 'every Banach-Mazur game is determined' is inconsistent.
Let $X$ be the set of all functions of the form $f: \alpha \rightarrow \{0,1\}$, with $\alpha$ an ordinal such that for any ordinals $\beta < \gamma$ with $\omega \cdot \gamma + \omega \leq \alpha$, the sets $(n \in \mathbb{N} : f(\omega \cdot \beta + n) = 1)$ and $(n \in \mathbb{N} : f(\omega \cdot \gamma + n) = 1)$ are distinct.
Let $\tau$ be the topology on $X$ generated by sets of the form $U_f = (g \in X : g \supseteq f)$, and let $\mathcal{X} = (X,\tau)$.
Claim: Player $1$ does not have a winning strategy for the Banach-Mazur game on $\mathcal{X}$.
Proof of claim: For any strategy $S$ for player $1$. Let $T$ be the tree of all initial segments of plays against $S$ such that each play by player $2$ is of the form $U_f$ for some $f \in X$ of length $\omega \cdot \alpha$ for some $\alpha$. If this tree fails to be pruned, then player $1$ has in some play of the game played a set $V$ such that for any $U_f \subseteq V$, $f$ enumerates every element of $2^{\mathbb{N}}$ extending $\sigma$ for some $\sigma \in 2^{<\omega}$. This would imply that $2^{\mathbb{N}}$ can be well-ordered, allowing us to construct a Bernstein set, contradicting our assumptions. Hence this must be a pruned tree of height $\omega$, so by dependent choice it has a path. Let $g$ be the union of all $f$ such that $U_f$ is on that path somewhere. By construction, $g \in X$, so we have that the strategy where player $2$ blindly plays the moves in this path wins against $S$.
So, it must be the case that player $2$ has a winning strategy $S$. For any $f \in X$ with length $\omega \cdot \alpha$ for some $\alpha$, let $T_f$ be the strategy for player $1$ that plays $U_f$ on player $1$'s zeroth move and on player $1$'s $n+1$st move, if player $2$ played $V$, then player $1$ plays $V \cap \bigcup_{\sigma \in 2^n} U_{f\frown \sigma\frown 0}$, if this is non-empty, and otherwise player $1$ plays $V \cap \bigcup_{\sigma \in 2^n} U_{f\frown \sigma\frown 1}$. (Note that since the union of these two is $V$, one of these must be non-empty.) Since $S$ is a winning strategy, for any $f\in X$ of length $\omega \cdot \alpha$ for some $\alpha$, the play of $T_f$ against $S$ must result in a nonempty intersection. By construction, for any $g$ and $h$ in that intersection, $g(\omega \cdot \alpha + n) = h(\omega \cdot \alpha +n)$ for all $n<\omega$, and the set $(n \in \mathbb{N} : h(\omega \cdot \alpha + n) = 1)$ must be distinct from $(n \in \mathbb{N} : f(\omega \cdot \beta + n) = 1)$ for any $\beta < \alpha$.
Therefore $S$ gives us a uniform procedure for picking a real not on a given well-ordered list of reals. By iterating this gives us a well-ordering of the reals. Therefore we can construct a Bernstein set from $S$ and we have that the Banach-Mazur game on that set is not determined, which contradicts our assumption. Therefore ZF + DC proves that there is an undetermined Banach-Mazur game.
Right now I don't see how to use a failure of DC to build an undetermined game.
Bernstein, not Berenstein... :-)
Berenstain, not Berenstein...
@Gabe Not all games are bears. You can define an analogous notion to a mouse, and then finally have Bernstein bears. But that has yet to happen.
@GabeGoldberg So would this question be about Berenstain Baires? (I'll put that on the list of math books for children, along with If you give a mouse a measure.)
When I was looking around at papers related to this I found out that 'barely Baire spaces' are a thing.
Yes, but if we have bears, we can have a "barely Baire bear", or "a Bernstein bear is barely a Baire bear". Or some other pun like that.
@JamesHanson Surely those should be called "Bairely spaces." :P (In some languishing project of mine I looked at sets of reals with Banach-Mazur winning strategies satisfying extremely mild computational tameness properties - I called these sets "Bairely acceptable." I should restart that project just so I can use that term ...)
|
2025-03-21T14:48:29.753704
| 2020-01-31T21:38:30 |
351666
|
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|
Stack Exchange
|
Sobolev spaces complement of Hausdorff codimension 2, restriction theorem
Let $X$ be an open domain in $R^n$. Let $E$ be a subspace of $X$ with Hausdorff dimension $m$. Fix $k$ and $p$. What are the optimal assumptions on $m$ and $n$ so that the trivial map $W^k_p(X) \to W^k_p(X \setminus E)$ becomes an isomorphism?
I am mostly interested in the case $k = 1$ and $p = 2$, and in that situation, it seems to me that the optimal bound is $m \leq n - 2$...
Here the Sobolev space $W^k_p(X)$ is defined as completion of smooth functions on $X$ with respect to the Sobolev norm (and not by the restriction from $R^n$).
If $E$ is a closed set such that Hausdorff measure $H^{n-p}(E)$ is $\sigma$-finite, then its capacity satisfies $Cap_p(E)=0$ and it follows that $W^{1,p}(X)=W^{1,p}(X\setminus E)$. Sobolev $W^{1,p}$ functions simply `do not see' sets of capacity zero.
In particular if $E$ is a linear subspace of dimension $m\leq n-p$, then $W^{1,p}(X)=W^{1,p}(X\setminus E)$. You can find more about capacity in the following books:
L. C. Evans, R. F. Gariepy, Measure theory and fine properties of functions. Revised edition. Textbooks in Mathematics. CRC Press, Boca Raton, FL, 2015.
W. P. Ziemer,
Weakly differentiable functions.
Sobolev spaces and functions of bounded variation. Graduate Texts in Mathematics, 120. Springer-Verlag, New York, 1989.
J. Malý, W. P. Ziemer, Fine regularity of solutions of elliptic partial differential equations. Mathematical Surveys and Monographs, 51. American Mathematical Society, Providence, RI, 1997.
|
2025-03-21T14:48:29.753829
| 2020-01-31T22:36:55 |
351672
|
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|
Stack Exchange
|
Volume ratio of complete flat manifolds
Let $(M^n,g)$ be a complete flat Riemannian manifold. Suppose there exists a number $s \in (n-1,n]$ such that for some point $p \in M$
$$
\limsup_{r \to +\infty} \frac{\text{Vol}\,B(p,r)}{r^s}>0.
$$
Can we prove that $(M^n,g)$ is isometric the Euclidean space?
Yes, it is true.
Note that $M$ is isometric to a quotient of the Euclidean space $\mathbb{E}^n$ by a totally discontinuous free isometric action of a group $\Gamma$.
Your condition implies that the soul of $M$ is a single point.
It follows that $\Gamma$ fixes a point in $\mathbb{E}^n$.
Since the action is free, the group $\Gamma$ is trivial.
|
2025-03-21T14:48:29.753913
| 2020-01-31T23:40:28 |
351678
|
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|
Stack Exchange
|
Gauss-Manin and Hilbert modular forms
There is a geometric formulation of Hilbert modular forms (HMFs) that parallels that for classical modular forms (sections of a line bundle over the moduli space of Hilbert-Blumenthal Abelian varieties). However, it doesn't give the automorphic forms on $G = \text{Res}_{\mathcal{O}_F/\mathbb{Z}}\text{GL}_2$, but rather on the subgroup $G^*$, where $G^*(R) = \{g \in \text{GL}_2(\mathcal{O}_F\otimes_\mathbb{Z}R) \mid \text{det}(g) \in R^\times \subset (\mathcal{O}_F \otimes_\mathbb{Z}R)^\times \}$. By restricting automorphic forms to this subgroup, we may view the space of HMFs for $G$ as a subspace (which is in fact a summand) of the space of HMFs for $G^*$, and there is an explicit projector $\mathfrak{e} \colon$ {HMFs for $G^*$} $\to$ {HMFs for $G$} given by averaging over some group action.
There's also a Gauss-Manin connection $\nabla$ that sends the nearly algebraic HMFs of weight $k$ and type $r$ for $G^*$ to the space of nearly algebraic HMFs of weight $k+2$ and type $r+1$ for $G^*$. Viewing an HMF for $G$ as an HMF for $G^*$, we can apply $\nabla$ to it, but a priori it won't yield an HMF for $G$ (i.e., an element of that subspace).
I'm looking for a reference on how this projector $\mathfrak{e}$ and the connection $\nabla$ interact. I've tried to argue that they commute, but I think anything in this direction would require knowing what $\nabla$ looks like. (Honestly, I'd love an early reference involving $\mathfrak{e}$ at all. I can't find one easily.)
I imagine the discrepancy here is solely the matter of missing some connected components and the inclusion of $G$-HMFs into $G^{*}$-HMFS and the projector $\mathfrak{e}$ are extending sections to other components respecting group action on $\pi_{0}$ and restricting section to certain components. In this way everything is geometric ($H_{dR}^{1}$, $\nabla$, etc.) so I could only imagine $\nabla$ and $\mathfrak{e}$ commute.
I don't think that's the entire story based on e.g. how $G^* \subset G$ looks $p$-adically for various $p$. But I've only seen $\mathfrak{e}$ in a recent paper (which doesn't mention $\nabla$), and it's not in some of the older references I've looked through.
Actually, if we add the special linear group into the story, it feels like that is what's going on, at least kind of. Can we use the "sections of a bundle on a moduli space" interpretation of modular forms (and HMFs) when we really mean automorphic forms on SL$_2$, not GL$_2$?
I don't think p-adic or any arithmetic consideration really matters here because after all everything a priori is complex algebraic ("classical automorphic forms are adelic automorphic forms"). I can certainly believe that $\mathfrak{e}$ or the inclusion does not interact obviously with rational structures.
What is the main reference you are looking for $\mathfrak{e}$?
My main reference is about $p$-adic properties, it's "On overconvergent Hilbert modular cusp forms" by Andreatta, Iovita, and Pilloni. The projector is discussed in the introduction and in section 4.1. It's available here: http://perso.ens-lyon.fr/vincent.pilloni/AIP2.pdf
There is really nothing scary happening here. The map $\nabla$ will preserve the image of $G$-forms inside $G^$-forms; this is completely obvious if you write down the $q$-expansions (both $\nabla$, and the condition for being in the image of $G$, are completely explicit in terms of Fourier coefficients). The only real subtlety to keep track of is that the naive restriction map from $G$-forms to $G^$-forms is not injective in general (so one has to bulk up the target of the map by summing over spaces of $G^*$ forms for several different levels); this is what is done in AIP.
|
2025-03-21T14:48:29.754199
| 2020-02-01T00:06:39 |
351679
|
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|
Stack Exchange
|
Deligne's "Notes sur Euler-Poincaré: brouillon project"
Is the note "Notes sur Euler-Poincaré: brouillon project" by Deligne available somewhere online? If not, is there a way that I could get a copy of it?
I have a copy of "Notes sur Euler-Poincaré: brouillon project" (which I received from professors Illusie and Saito). I don't feel at liberty to post it on a public web site, but I can email the notes (scan of 14 handwritten pages) to whomever is interested. Just let me know at<EMAIL_ADDRESS>
It might be better that somebody writes an email to Deligne, given that he uploaded some letters on his homepage. I also don't know why he wrote "project" instead of "projet" in French.
@Z.M --- since Desargues' "Brouillon Project" from the sixteenth century, this spelling is not uncommon
@CarloBeenakker: Strange, because the natural topic in French would have been "proje(c)t brouillon", i.e. "draft project" ("brouillon" playing the role of an adjective here, rather than a noun - like "draft").
@AlexM. --- certainly, I think both the order of the words and the spelling points to Desargue; I have by the way written to Deligne, no response so far, so I'll just keep on distributing the pdf by email "for personal use"
Oh christ... I had to wait in order to award the bounty to you. So, when I saw that I could finally give it to you, I just clicked on the first answer that I saw and gave the bounty to the other one... I'm really sorry...
@Gabriel, you might flag this post, or the question, for moderator attention and ask if the mis-click can be corrected.
@Gabriel --- none of this is particularly important, but you might consider repairing the anomaly by issuing a second bounty...
That project is mentioned in Illusie's article "From Pierre Deligne’s secret garden: looking back at some of his letters" (Japan. J. Math. 10, 237–248 (2015) DOI: 10.1007/s11537-015-1514-9). It isn't the full project, but at least some information about it is available there.
|
2025-03-21T14:48:29.754493
| 2020-02-01T00:06:52 |
351680
|
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|
Stack Exchange
|
Do there exist any variational principles on the space of braids (or knots)?
This is very speculative question and I do not know where to start looking up the literature, or if what I am looking for is even mathematically possible/meaningful.
Q: I am interested in finding out whether any variational or action principles have been derived anywhere in the physics or mathematical literature (e.g. in study of geometry of configuration spaces etc.) on the space of braids (or knots).
Explicitly, some example where a Lagrangian is a function of a braid representation, and it is minimized/extremized under some constraints. The resulting equation would be differential equation whose solution gives a trajectory in the space of braids.
https://en.wikipedia.org/wiki/Knot_energy ?
Energy functionals are also used for braids. It goes back to Graham Segal's work, but if you are interested a more modern manifestation see Paolo Salvatore's recent paper "A cell decomposition of the Fulton Mac Pherson operad".
@RyanBudney Thanks. Do you happen to have a reference for Segal's work ?
Configuration-spaces and iterated loop-spaces. Graeme Segal.
Inventiones mathematicae volume 21, pages 213–221(1973)
My Phd supervisor developed a floer theory on spaces of braids: see https://www.few.vu.nl/~vdvorst/NEWPAPERS/IM1f.pdf and later work.
O'Hara introduced knot energies, and a Möbius invariant case was studied by Freedman-He-Wang. For prime knots, Zheng-Xu He subsequently showed that there exists a smooth minimizer (up to Möbius transformation), and now it is known that critical points are analytic.
For braids, one can look at various variational principles by looking at metrics on Teichmuller space of the punctured sphere. The Teichmuller metric is a Finsler metric, with unique minimizing geodesics for pseudo-Anosov braids. The Weil-Petersson metric is a Riemannian metric which is incomplete, but for which pseudo-Anosovs also have unique minimizing representatives.
The minimizing geodesic then gives a canonical braid representative up to conjugacy (by choosing the representatives to have fixed center of mass and constant moment). One can also find unique representatives for braids up to choosing basepoint surfaces (e.g. $n$ points lying on the $x$-axis in $\mathbb{R}^2$).
For the 3-punctured sphere, the Teichmuller metric is the hyperbolic plane, and one can easily compute the geodesic.
One can also look at solutions to the planar $n$-body problem to get braid realizations. Richard Montgomery first proved the existence of solutions to the 3-body problem.
|
2025-03-21T14:48:29.754697
| 2020-02-01T00:41:27 |
351681
|
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|
Stack Exchange
|
Does $\sqrt{n}\sup_x |\Phi(\dfrac{x-\overline{X}}{S})-\Phi(x)|\to0$ in probability?
Let $X_1,...,X_n$ be iid observations from $N(0,1)$. Let $\overline{X}=\dfrac{1}{n}\sum_{i=1}^n X_i$ and $S^2=\dfrac{1}{n}\sum_{i=1}^n (X_i-\overline{X})^2$. Then is it true that $\sqrt{n}\sup_x |\Phi(\dfrac{x-\overline{X}}{S})-\Phi(x)|\stackrel{p}{\to}
0$? Here $\Phi(.)$ is standard normal cdf.
Seems something like Berry Esseen bound but I think I am getting it is stochastically bounded and not really going to 0.
No, it is not true. First fix any $x$ and consider $\overline{X^2}=\frac1n\sum_{i=1}^nX_i^2$, $S^2=\overline{X^2}-\bigl(\overline X\bigr)^2$ and $h(s,t)=\Phi\left(\dfrac{x-s}{\sqrt{t-s^2}}\right)$.
Then use multivariate Delta method to prove that
$$
\sqrt{n} \left(\Phi\biggl(\dfrac{x-\overline{X}}{S}\biggr)-\Phi(x)\right)=\sqrt{n}\biggl(h\biggl(\overline X, \overline{X^2}\biggr)-\underbrace{h(\mathbb EX_1,\mathbb EX_1^2)}_{h(0,1)}\biggr)\xrightarrow{\mathcal D}\mathcal N(0,\sigma_x^2),
$$
where $\sigma^2_x = \nabla h(0,1)^T\cdot \Sigma \cdot \nabla h(0,1)$ and $\Sigma$ is a covariance matrix of $(X_1,X_1^2)$.
Here $\text{Var}(X_1)=1$, $\text{Var}(X_1^2)=2$, $\text{Cov}(X_1, X_1^2)=0$, so
$$
\Sigma=\begin{pmatrix}1 & 0 \cr 0 & 2 \end{pmatrix}
$$
$$
\frac{\partial h}{\partial s}\bigg|_{s=0, t=1} = -\varphi(x) = -\frac{1}{\sqrt{2\pi}}e^{-x^2/2}
$$
and
$$
\frac{\partial h}{\partial t}\bigg|_{s=0, t=1} = -\frac12\varphi(x) = -\frac{1}{2\sqrt{2\pi}}e^{-x^2/2}
$$
Please check the derivatives, although their values do not matter much. And
$$
\sigma^2_x = \varphi^2(x)\cdot 1+\left(\frac{\varphi(x)}{2}\right)^2\cdot 2 = \frac32\varphi^2(x).
$$
Say, for $x=0$,
$$
\sqrt{n} \left(\Phi\biggl(\dfrac{x-\overline{X}}{S}\biggr)-\Phi(x)\right) \xrightarrow{\mathcal D} \mathcal N\biggl(0,\frac{3}{4\pi}\biggr)
$$
For any $\varepsilon>0$
$$
\liminf_{n\to\infty}\mathbb P\left(\sqrt{n}\sup_x \biggl|\Phi\biggl(\dfrac{x-\overline{X}}{S}\biggr)-\Phi(x)\biggr|>\varepsilon\right) \geq \liminf_{n\to\infty}\mathbb P\left(\sqrt{n} \biggl|\Phi\biggl(\dfrac{0-\overline{X}}{S}\biggr)-\Phi(0)\biggr|>\varepsilon\right)
$$
$$
=\mathbb P(|Y|>\varepsilon)=2\Phi\left(-\frac{2\sqrt{\pi}\varepsilon}{\sqrt3}\right)>0
$$
where $Y\sim N\bigl(0,\frac{3}{4\pi}\bigr)$.
Yup I realised that, essentially by using delta method after posting this question. Thanks anyway!
|
2025-03-21T14:48:29.755089
| 2020-02-01T02:16:12 |
351686
|
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|
Stack Exchange
|
Extension of subharmonic functions at infinity
Let $W$ be the complement of a compact set $K$ in $\mathbb{R}^{n}$, and $u$ a subharmonic function on $W$. Can we find, under some conditions, a function $\tilde{u}$ that is subharmonic on $W\cup\{\infty\}$ and coincides with $u$ on $W$?
(Too long for a comment.)
What exactly is your definition of a (sub)harmonic function on $W \cup \{\infty\}$?
You can always project your function to the unit $n$-sphere using the stereographic projection: $$\mathbb{R}^n \ni x \mapsto \phi(x) = p + 2 |x' - p|^{-2} (x'-p) \in \mathbb{S}^n ,$$
where $x' = (x_1, \ldots, x_n, 0) \in \mathbb{R}^{n+1}$, $p = (0, \ldots, 0, 1) \in \mathbb{R}^{n+1}$ and $\mathbb{S}^n = \{x \in \mathbb{R}^{n+1} : |x| = 1\}$. More precisely, the function $$v(x) = |x' - p|^{2 - n} u(\phi^{-1}(x))$$
is subharmonic on $\phi(W) \subseteq \mathbb{S}^n$. (This is a close relative of the Kelvin transformation.)
With this identification, you can treat $\infty$ just as every other point of $\mathbb{R}^n$ (or $\mathbb{S}^n$). The problem is that the value of (the extension of) $v$ at $p$ is not the same as the hypothetical value of $u$ at infinity; in fact, $v(p) = \lim_{|x| \to \infty} |x|^{n-2} u(x)$.
I do not see any other reasonable notion of (sub)harmonicity on $W \cup \{\infty\}$. One could naively try to require that $u(\infty)$ is equal to (or does not exceed) the average of $u$ over an arbitrary sphere which contains $\mathbb{R}^n \setminus W$ in its interior. But this definition is not consistent: for example, the only harmonic functions in $W \cup \{\infty\}$ would be constants.
Edited: After reading another question by M. Rahmat, I realised that the last paragraph of my answer was wrong. What I called a "naive" approach actually works in dimensions $n \geqslant 2$, and it was an idea developed by Brelot in 1940s; see [M. Brelot. Sur le rôle du point à l'infini dans la théorie des fonctions harmoniques. Ann. Sci. ́Ecole Norm. Sup., 61:301–332, 1944].
The situation is special when $n=2$. Then $\infty$ can be treated as any other point, and removable singularity theorem is applicable.
|
2025-03-21T14:48:29.755270
| 2020-02-01T03:59:43 |
351688
|
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|
Stack Exchange
|
Conjectured primality test for specific class of $N=4kp^n+1$
Can you provide a proof or counterexample for the following claim?
Let $P_m(x)=2^{-m}\cdot((x-\sqrt{x^2-4})^m+(x+\sqrt{x^2-4})^m)$ .
Let $N= 4kp^{n}+1 $ where $k$ is a positive natural number , $ 4k<2^n$ , $p$ is a prime number and $n\ge3$ . Let $a$ be a natural number greater than two such that $\left(\frac{a-2}{N}\right)=-1$ and $\left(\frac{a+2}{N}\right)=-1$ where $\left(\frac{}{}\right)$ denotes Jacobi symbol. Let $S_i=P_p(S_{i-1})$ with $S_0$ equal to the modular $P_{kp^2}(a)\phantom{5} \text{mod} \phantom{5} N$. Then $N$ is prime if and only if $S_{n-2} \equiv 0 \pmod{N}$ .
You can run this test here.
I have verified this claim for $k \in [1,500]$ with $p \leq 97$ and $n \in [3,50]$ .
Further generalization of the claim
A
Let $P_m(x)=2^{-m}\cdot((x-\sqrt{x^2-4})^m+(x+\sqrt{x^2-4})^m)$ .
Let $N= 2kp^{n} + 1 $ where $k$ is a positive natural number , $ 2k<2^n$ , $p$ is a prime number and $n\ge3$ . Let $a$ be a natural number greater than two such that $\left(\frac{a-2}{N}\right)=-1$ and $\left(\frac{a+2}{N}\right)=-1$ where $\left(\frac{}{}\right)$ denotes Jacobi symbol. Let $S_i=P_p(S_{i-1})$ with $S_0$ equal to the modular $P_{kp^2}(a)\phantom{5} \text{mod} \phantom{5} N$. Then $N$ is prime if and only if $S_{n-2} \equiv -2 \pmod{N}$ .
You can run this test here.
B
Let $P_m(x)=2^{-m}\cdot((x-\sqrt{x^2-4})^m+(x+\sqrt{x^2-4})^m)$ .
Let $N= 2kp^{n} - 1 $ where $k$ is a positive natural number , $ 2k<2^n$ , $p$ is a prime number and $n\ge3$ . Let $a$ be a natural number greater than two such that $\left(\frac{a-2}{N}\right)=1$ and $\left(\frac{a+2}{N}\right)=-1$ where $\left(\frac{}{}\right)$ denotes Jacobi symbol. Let $S_i=P_p(S_{i-1})$ with $S_0$ equal to the modular $P_{kp^2}(a)\phantom{5} \text{mod} \phantom{5} N$. Then $N$ is prime if and only if $S_{n-2} \equiv -2 \pmod{N}$ .
You can run this test here.
The "if and only if" statement fails for
$$[p,n,k,a] \in \{ [3, 4, 1, 100], [3, 4, 1, 225], [3, 6, 13, 2901] \}$$ and many others. In these cases, $N$ is not prime, but the congruence $S_{n-2}\equiv 0\pmod{N}$ still holds.
|
2025-03-21T14:48:29.755378
| 2020-02-01T04:13:58 |
351689
|
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|
Stack Exchange
|
Groebner basis of a toric ideal
I know about toric ideals that it is a sort of binomial ideal i.e. generated by $x^u - x^v$, where $Au = Av $ ( A is the associated matrix). So by finding all integer solutions of $AX = 0$, can we somehow find the generators of the toric ideal? More specifically, can we determine the grobner basis? With the help of the integer solutions?
I think the term you’ll want to lookup is Markov basis. I think the Markov basis will always be a Gröbner basis, but I’m not sure.
Can you provide/suggest any meterial for this topic?
I believe "Markov basis" refers to any generating set for the toric ideal -- so typically not a Groebner basis. You also don't generate the ideal from only considering $u-v$ in some lattice basis for the kernel of $A$ --- you need to saturate out various coordinate hyperplanes. There is pretty good software for computing these sorts of things -- normaliz and 4ti2 come to mind.
|
2025-03-21T14:48:29.755481
| 2020-02-01T06:43:53 |
351692
|
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|
Stack Exchange
|
Is it true, the space of embeddings segments is homotopy equivalent to the subspace of all line segments?
Consider the space of smooth embeddings of the segment in the plane with the compact-open topology. Denote by X the quotient space obtained from the equivalence relation $a \sim b$ if and only if $\mathrm{Im}~ a = \mathrm{Im}~ b$
Consider the retraction $f: X \to X$ such that $f(s)$ is the line segment connecting the ends of s (it is obvious that the function is continuous and idempotent). Is it true, that $f$ is homotopic to the $id$? if not, can it be otherwise proved that the subspace of all segments is a deformation retract?
Most of the techniques that I've tried to address this question founder on issues with the point set topology on X. The function "arc length" $X \to \Bbb R$ is not continuous - in fact, it is not even bounded above by a continuous function - because, eg, one can converge to a straight line in the compact-open topology with a sequence of very tightly packed helix-shaped curves.
@TylerLawson: It might help to get some clarification from the author. It looks like you are thinking of the $C^0$ version of the compact-open topology. There are versions adapted to the smooth mapping space as well, and it's unclear if this is what the author means. More often it would just be called the weak topology, or the Whitney topology, in that context.
Here are some remarks which may be relevant.
First of all it seems to me that the correct topology to use is the Whitney $C^\infty$-topology on the embedding space.
Let $M$ be an closed manifold. There is a free action of the group of diffeomorphisms $\text{Diff}(M)$ on the space of embeddings $E(M,\Bbb R^n)$. The orbit space of the action, call it $S(M,\Bbb R^n)$, is identified with subspace of all submanifolds of $\Bbb R^n$ that are diffeomorphic to $M$. I remember hearing as a graduate student that $E(M,\Bbb R^n)\to S(M,\Bbb R^n)$ is a Serre fibration. This result may actually be due to Cerf.
If a version of the result in (2) holds for compact manifolds $M$ with boundary $\partial M$, then the map $E(M,\Bbb R^n)\to S(M,\Bbb R^n)$ is a fibration where the group acting in this case is the diffeomorphisms of $M$ which restrict to the identity on $\partial M$. (However, I do not know if this is true.)
Assuming the statement in (3) is true, then take $M = I = [0,1]$. The diffeomorphism group $\text{Diff}(I \text{ rel } \partial I)$ in this case is weakly contractible. It follows that the map $E(I,\Bbb R^n) \to S(I,\Bbb R^n)$ is a weak homotopy equivalence.
It seems to me that the OP's question is: "Is the restriction map $S(I, \Bbb R^2) \to S(S^0,\Bbb R^2)$ an equivalence?" This seems harder to me.
$E(I,\Bbb R^2)$ is weakly equivalent to $S^1$. Assuming the fibration assertion, it would follow that $S(I,\Bbb R^2)$ is also weakly equivalent to $S^1$. But $S(S^0,\Bbb R^2)$ is identified with $S^1$ as well (it's a configuration space of unordered subsets of $\Bbb R^2$ of cardinality two). So I think the OP's question will follow from what I remarked.
|
2025-03-21T14:48:29.755698
| 2020-02-01T11:48:06 |
351698
|
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|
Stack Exchange
|
Sampling algorithms on convex polytopes
Let $f=\mathbf{c}\cdot\mathbf{x}$ be the optimization objective function whose parameter vector $\mathbf{x}\in\mathbb{R}^n$ is subject to the following constraints in the very well-known linear-programming problem
$$
\mathbf{A}\mathbf{x}=\mathbf{b}\in\mathbb{R}^m\;,\quad \mathbf{0}\leq\mathbf{x}\leq\mathbf{v}
$$
Suppose that the space of feasible solutions of the above linear system is a convex polytope; my questions are the following:
How are the vertices of said polytope found in linear programming? Kindly reference the theoretical argument for that.
What are the MCMC algorithms to sample points from the polytope s.t. it eventually finds the optimal point for the object function?
Perhaps the vertex enumeration problem is of interest to you.
So does that mean it is an open problem and I can't find the vertices of a convex polytope in general?
For one thing, there can be exponentially many vertices, e.g. the hypercube. So more needs to be said about what it means to "find" the vertices.
Here is a link for software and references: https://rdrr.io/cran/volesti/man/sample_points.html
|
2025-03-21T14:48:29.755833
| 2020-02-01T13:01:37 |
351702
|
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|
Stack Exchange
|
Notation for "the" left adjoint functor
As far as I know, there is no "official" notation for the left adjoint of a functor $F : \mathcal{C} \to \mathcal{D}$ if it exists. I have seen the notation $F^*$ sometimes, but this looks only nice when $F$ is already written as $F_*$, which is not practical. (This notation is then motivated by direct and inverse image functors. And it seems to be quite common for adjunctions between preorders aka Galois connections.) Similarly, I have seen the notation $F_*$ for the right adjoint of $F$, which only looks nice when $F$ is already written as $F^*$. I have also seen the notation $F^{\dagger}$ for the right adjoint, which looks nice, but then how would you denote the left adjoint if it exists? Perhaps ${}^{\dagger} F$? I don't want to start a debate here what is a good notation or not, since this is subjective anyway and is not suited for mathoverflow. I would like to know:
Are there any textbooks, influential papers or monographs on category theory which have introduced a notation for the left adjoint of $F$? Is there any notation which has been used by multiple authors?
Just to avoid any misunderstanding: Of course there is the official notation $F \dashv G$ when $F$ is left adjoint to $G$, but $\dashv$ is a relation symbol. I am interested in a function symbol (which makes sense since left and right adjoints are unique up to canonical isomorphism if they exist).
I've no idea how widely used it is, but I like $F^L$ for the left adjoint and $F^R$ for the right one.
@Adrien Thanks. Can you give an example of a book where it is used?
You could write it as a formula using limits: $d\mapsto \lim(\pi:d/F\to C)$.
I think the use of $F$ for a start is problematic. Often $F$ is the left adjoint as it is often used for 'free' with $U$ or similar standing for 'underlying'. Different contexts require different notation but I have used $L$ for the left adjoint and $R$ for the right when discussing adjoint functors when teaching category theory. It might help if you gave more idea of the context in which you are wanting this.
@TimPorter I completely agree. But sometimes the letters $U,L,R$ are already taken, or when no other functor is in the context, it is very common to name the functor $F$. By the way, Borceux calls $F$ the right and $G$ the left adjoint, yikes!
@Adrien, how about ${^{\text L}F}$ and $F^{\text R}$? (EDIT: Ah, this is basically @FredRohrer's answer.)
In EGA <IP_ADDRESS>-3 (from the 1971 Springer edition) the right adjoint and the left adjoint of a functor $F$ are denoted by $F^{\rm ad}$ and ${}^{\rm ad}\!F$, respectively.
I am very happy about this answer, and I accept it right away (even though there might be other books with other notations) because EGA I has/had such a huge impact. I didn't read all of the preliminaries, so I wasn't aware that this notation is introduced there. Well spotted!
Yet, this leads to the question why this notation is not common, even not in EGA itself (right?).
Let me also remark that the same section introduces a notation for the bijection $$\hom(F(x),y) \xrightarrow{\sim} \hom(x,F^{\text{ad}}(y)),$$ namely $f \mapsto f^{\flat}$, and $g \mapsto g^{\sharp}$ for the inverse. At least, this notation can be seen a lot in EGA.
I am quite sure that the "ad"-notation does not occur in EGA outside the cited section.
I've seen the notation $\lceil f \rceil$ and $\lfloor f \rfloor$ being used for the isomorphism, mainly in a few comp sci papers
@WorldSEnder I think I've used this notation myself, and it comes by analogy to the fact that floor and ceiling are adjoints to the inclusion of posets $\mathbb{Z} \hookrightarrow \mathbb{R}$.
The floor is very nice and was always my prefered one. Yet (depending on the context of course) you may be willing to use the ad notation.
I always use the $\flat$ and $\sharp$ notation as well, but in the exactly opposite way. It seems to make more sense to me.
P. Gabriel uses $F_\lambda$ for the left adjoint and $F_\rho$ for the right adjoint. See for instance p. 344 of the article "Covering spaces in representation theory" by K. Bongartz and P. Gabriel (Invent. Math. 65, 1982) (EuDML).
|
2025-03-21T14:48:29.756151
| 2020-02-01T13:58:50 |
351703
|
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|
Stack Exchange
|
Generate algorithmically an elliptic curve with its exact class group structure?
Is it possible to generate an elliptic curve $E$ (randomly), together with knowing its class group $\mathrm{Cl}(\mathcal{O})$ structure? where $\mathcal{O}$ is its endomorphism rings $\mathsf{End}(E)$ or $\mathbb{F}_q$ rational endomorphism ring, $\mathsf{End}_{{\mathbb F}_q}(E)$, i.e. those endomorphism with $\mathbb{F}_q$ coefficients. Below, I give some context on why I am interested in it.
Any reference is appreciated.
Context: The ideal class groups $\mathrm{Cl}(\mathcal{O})$ are of great interest in cryptographical scenario e.g. the Cl encryption scheme in those homomorphic encryption literatures. It also relates nearly to Couveignes-Rostovtsev-Stolbunov (CRS) based cryptography, such as CSIDH, in which we have class group randomly acts on a class of isogeneous elliptic curves. However, instead of computing the exact class group structure in the first place, to my knowledge most of them chose to indirectly compute these group actions, (e.g. using "elliptic curve tricks" such as modular polynomials, division polynomials or finding torsion points). If one could construct an elliptic curve (especially supersingular ones) together with its class group structure, then we could directly benefit from those class group technique such as (low-level) algorithmic improvements from binary quadratic form etc.
|
2025-03-21T14:48:29.756260
| 2020-02-01T14:13:16 |
351704
|
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|
Stack Exchange
|
Which finite solvable groups have solvable automorphism groups?
Is it possible to give a reasonable description of those finite solvable groups $G$ such that $A = {\rm Aut}(G)$ is also solvable?
The central case to deal with is that in which $G$ is a $p$-group of nilpotence class $2$ (and $G$ may be assumed to be of exponent $p$ if $p$ is odd and exponent dividing $4$ if $p = 2$). All other cases can be reduced to this one: here is a sketch of the argument (via fairly standard group theory):
Firstly, let $G$ be a general finite solvable group, and $A$ be its automorphism group. Let $F = F(G)$ be the Fitting subgroup of $G$, and let $C = C_{A}(F)$. Then $[G,F,C] = [F,C,G] = 1$, so $[C,G,F] = 1$. Hence $[G,C] \leq Z(F)$ and $[G,C,C] = [C,G,C] = 1$. Then $[C,C,G] = 1$ and so $C$ is Abelian. Hence $A$ is solvable if and only if $A/C$ is solvable. But $A/C$ embeds in ${\rm Aut}(F)$. Hence if ${\rm Aut}(F)$ is solvable, then so is $A$.
We are now effectively reduced to the case that $G$ is a $p$-group. Now known results (for example of J.G. Thompson and H. Bender) show that there is a characteristic subgroup $H$ of $G$ with $C_{G}(H) \leq H$ such that $H$ has nilpotence class at most $2$ (and exponent as stated above depending on whether $p$ is odd or $p = 2$) and with $C_{A}(H)$ a $p$-group. Hence if ${\rm Aut}(H)$ is solvable, then so is $A$. This reduces us to the stated case.
Later edit: It might be worth remarking that when $G$ is a solvable group with $\Phi(G) = 1,$ then a priori we can only be sure that ${\rm Aut}(G)$ is solvable when $F(G)$ has cyclic Sylow $p$-subgroups for all primes $p \geq 5$ and $F(G)$ has elementary Abelian Sylow $2$-subgroups of rank at most $2$ and elementary Abelian Sylow $3$-subgroups of rank at most $2.$
Aren't these groups which have a filtration by characteristic subgroups where the quotient of each group by the next is the additive group of a finite field and the automorphism action on each is via multiplication by elements of that finite field?
Doesn't the reduction to $p$-groups only go in one direction? $S_3^4$ has solvable automorphism group, but $(\mathbb Z/3)^4$ and $(\mathbb Z/2)^4$ do not.
@WillSawin : For the first question, you may be right, though I don't see the whole argument myself. As for the second question, yes, I didn't say that it was equivalent to asking the same question for the reduced $p$-groups, the questions are indeed not equivalent. But the conclusion of the outlined argument is that unless $G$ has a class (at most) $2$-characteristic $p$-subgroup $P$ (of exponent $p$ or $4$) with non-solvable automorphism group (and $P$ containing its centralizer in $O_{p}(G)$) , then $G$ has a solvable automorphism group.
@MarkSapir: Tht's a nice question, somewhat related to this oone, whose answer I don't know, but others might.
Just a remark: If $G$ and $Aut(G)$ are solvable, and $G$ acts faithfully and irreducibly on an elementary abelian $p$-group $A$, then the semidirect product $AG$ has solvable automorphism group.
In the specific case of groups of prime exponent $p$ and nilpotency class $\le p-1$, we boil down to a similar question about nilpotent Lie algebras over $\mathbf{F}_p$, which has no definite answer (even in nilpotency class $2$). In a sense, it is generically the case that the automorphism group is solvable, and the contrary means "lots of symmetry".
@YCor : Thanks. It is also generically the case that most $p$-groups have outer automorphism group a $p$- group by a reasonable measure, but those which can occur as $O_{p}(G)$ for a reasonably complicated solvable group $G$ must already have fairly complex automorphism groups, so there may still be something to be said in that case.
I wouldn't say this. For instance, I wouldn't say that a $p$-group of exponent $p$ has generically nilpotency class $\ge 3$, and when the nilp. class is $2$ there is an aut. of order 2. I'm saying generic not necessarily in the sense of counting, but having in mind complex nilpotent Lie algebras (which thus descent on $\mathbf{F}_p$ in some way, unique or not). The variety of nilpotent $d$-dimensional Lie algebras has several connected components, so there is no "genericity" in some sense. Nevertheless I expect no whole component consists of algebras with non-solvable derivation Lie algebra.
|
2025-03-21T14:48:29.756654
| 2020-02-01T15:38:05 |
351706
|
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|
Stack Exchange
|
Understanding Frobenius's Theorem
Let $\Gamma_g$ be the fundamental group of a closed Riemann surface of genus $g$. Let $G$ be a finite group. Then a theorem of Frobenius states that
$$
|\mathrm{Hom}(\Gamma_g, G)|/|G|= \sum_{\chi} \big(|G|/\dim(\chi)\big)^{2g-2}
$$
where $\chi$ ranges over irreducible complex representations of $G$.
Question: What is the correct way to understand this theorem? Can one give a canonical bijection between two sides?
Proofs of this result which I have seen (e.g. the one in the paper of Hausel-Rodriguez Villegas) were not very illuminating for me.
Minor comment: I thought that people refer to the $g=1$ case as Frobenius's theorem, and the general case as Mednykh's formula.
One thing is that this formula can be interpreted in terms of $2d$ TQFT. For instances see Teleman's discussion here https://math.berkeley.edu/~teleman/math/barclect.pdf.
Proposition I of Aizenbud, Avraham; Avni, Nir_, Representation growth and rational singularities of the moduli space of local systems, (2016) may be somewhat related to this.
|
2025-03-21T14:48:29.756762
| 2020-02-01T15:40:44 |
351707
|
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"Joel David Hamkins",
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|
Stack Exchange
|
Is Proper Class Choice equivalent to Global Choice?
Working in "MK-Regularity-Limitation of Size + Replacement for sets", call it the Base theory, let's coin the following axiom:
Axiom of Super-Choice:$$\forall \ relation \ R \ \exists F \subset R \ (F: dom(R) \to rng(R))$$ Where: $$R \text { is a relation } \equiv_{df} \forall r \in R \ \exists x,y \ (r=\langle x,y \rangle)\\dom(R)=\{x| \exists y (\langle x,y \rangle \in R)\} \\rng(R)=\{y|\exists x (\langle x,y \rangle \in R)\}$$
I think this would be equivalent to Global Choice over the Base theory.
Now if we weaken the above to:
Axiom of Proper Class Choice:$$\forall \ relation \ R \ \text{ with proper class rows } \\\exists F \subset R \ (F: dom(R) \to rng(R))$$
where: $$ z \text{ is a row of }R \iff
\\ \exists x \in dom(R)[z=\{y| \langle x,y \rangle \in R\}]$$
A relation $R$ is with proper class rows, if every row of it is a proper class.
Would that still be equivalent to Global Choice over the Base theory?
If not, then if we also add Choice over sets [i.e., every set has a choice function], would that result in proving Global Choice?
You might find this useful: http://jdh.hamkins.org/the-global-choice-principle-in-godel-bernays-set-theory/. The theorem there gives several equivalent formulations of global choice over Gödel-Bernays set theory, which would be a natural base theory.
@JoelDavidHamkins, Thanks for the reference!
Yes, this is still equivalent: given any relation $R$, consider the new relation $$R^{bigrows}=\{\langle x,y\rangle: \exists a,b(y=\langle a,b\rangle\wedge \langle x,a\rangle\in R)\}.$$ Basically, $R^{bigrows}$ just "pads out" the rows of $R$ with a dummy coordinate.
But from a choice function $F$ for $R^{bigrows}$ we can extract one for $R$ itself: consider $$x\mapsto \pi_0(F(x))$$ (where $\pi_0$ denotes projection onto the left coordinate).
Basically, restricting attention to proper classes is never going to make things substantively easier - the idea here is the same, although the implementation is different, as in my answer to your earlier question.
|
2025-03-21T14:48:29.756915
| 2020-02-01T16:39:04 |
351709
|
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|
Stack Exchange
|
Graph connectivity after deleting an f-factor
Let $G(V,E)$ be an undirekted $k$-vertex-connected, $k$-regular graph
and let $F$ be an $f$-factor of $G$ consisting of a set of $f$-vertex-connected components, $f<k$.
Question:
what is the vertex-connectivity of $G\setminus F$, is it $k-f$, resp., what is the highest lower bound on the resulting vertex-connectivity?
Make $G$ from two highly-connected pieces joined by a matching of $k$ edges. It has connectivity $k$. Now take $F$ to be a perfect matching that includes the edges of the cut. $G\setminus F$ is then disconnected. So there is no general lower bound except 0.
|
2025-03-21T14:48:29.757001
| 2020-02-01T17:51:51 |
351712
|
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|
Stack Exchange
|
Can we characterize the set of neoclassical production functions?
INTRODUCTION
The neoclassical production function is the main building block in neoclassical growth theory, and consequently the main building block of modern macroeconomic theory. Mathematically, the neoclassical production function $F$ is a function that satisfies Assumption 1 and 2 below. In economics, the neoclassical production function is a production function that satisfies Assumption 1 and 2 below and that defines the output $Y$ by a given amount of capital $K$, labor $L$, and technology $A$: $$Y=F(K,L,A).$$
My questions given below are only about the mathematical definition of a neoclassical production function, leaving philosophical and empirical issues aside. Assumption 1 and 2 below are taken from pages 29 and 33 in the popular graduate-level textbook in economics by Acemoglu (2009).
Typically, only the constant returns to scale (CRS) Cobb-Douglas production function is given as an example of a neoclassical production function. It can be symbolically defined by $$F(K,L,A)=K^{\alpha}(AL)^{1-\alpha}$$ for some $\alpha\in(0,1)$. (This is the so-called Harrod-neutral representation of the Cobb-Douglas function and is related to Uzawa's theorem. There are two other equivalent representations.) In all graduate-level textbooks in macroeconomics that I know of, no other type of neoclassical production functions is defined! With less restrictive assumptions on the production function than Assumption 1 and 2, one can show that the Inada conditions imply that the production function must be asymptotically Cobb-Douglas (Barrelli et al., 2003; Litina et al., 2008).
QUESTIONS
Define the set of neoclassical production functions as the set of functions that satisfy Assumption 1 and 2 below.
Can we characterize the set of neoclassical production functions?
What are some examples of neoclassical production functions that do
not fall in the class of Cobb-Douglas production functions?
ASSUMPTION 1 (Continuity, Differentiability, Positive and Diminishing Marginal Products, and Constant Returns to Scale)
The production function $F:\mathbb{R}_+^3\to\mathbb{R}_+$, where $\mathbb{R}_+$ is the set of nonnegative real numbers, is twice differentiable in $K$ and $L$, and satisfies over $(K,L,A)\in\mathbb{R}^2_{++}\times\mathbb{R}_+$, where $\mathbb{R}_{++}$ is the set of positive real numbers, $$F_K(K,L,A)\equiv\frac{\partial F(K,L,A)}{\partial K}>0,\quad F_L(K,L,A)\equiv\frac{\partial F(K,L,A)}{\partial L}>0,$$ and $$F_{KK}(K,L,A)\equiv\frac{\partial^2 F(K,L,A)}{\partial K^2}<0,\quad F_{LL}(K,L,A)\equiv\frac{\partial^2 F(K,L,A)}{\partial L^2}<0.$$ Moreover, $F$ exhibits constant returns to scale in $K$ and $L$. That is, $F(K,L,A)$ is homogenous of degree 1 in $(K,L)$ so that for any $\lambda\in\mathbb{R}_+$ and $(K,L,A)\in\mathbb{R}^3_+$ we have that $$F(\lambda K,\lambda L, A)=\lambda F(K,L,A).$$
ASSUMPTION 2 (Inada Conditions)
$F$ satisfies the Inada conditions $$\lim_{K\to0^+}F_K(K,L,A)=\infty\text{ and }\lim_{K\to\infty}F_K(K,L,A)=0\text{ for all }L>0\text{ and all }A\geq0,$$ and $$\lim_{L\to0^+}F_L(K,L,A)=\infty\text{ and }\lim_{L\to\infty}F_L(K,L,A)=0\text{ for all }K>0\text{ and all }A\geq0.$$ Moreover, $$F(0,L,A)=0$$ for all $(L,A)\in\mathbb{R}^2_+$.
REFERENCES
Acemoglu, D. (2009). Introduction to Modern Economic Growth. Princeton University Press.
Barelli, Paulo; Pessôa, Samuel de Abreu (2003). "Inada conditions imply that production function must be asymptotically Cobb–Douglas". Economics Letters. 81 (3): 361–363.
Litina, Anastasia; Palivos, Theodore (2008). "Do Inada conditions imply that production function must be asymptotically Cobb–Douglas? A comment". Economics Letters. 99 (3): 498–499.
Do you want $\mathbb{R}^+$ to be the set of (strictly) positive numbers instead? $K^{\alpha}$ is not differentiable at 0 for $\alpha \in (0,1)$.
@KevinCasto (+1) Thanks for your comment. No. Assumption 2 requires that $F(0,K,L)$ is defined. However, I want to make the restriction that the neoclassical production function is only differentiable over open sets in its domain. The Cobb-Douglas production function should satisfy this requirement. I revised the description of Assumption 1 to account for your example. My previous description of Assumption 1 was taken directly from page 29 in Acemoglu (2009).
That's what's called a "one sector" model. You occasionally see two-sector models, etc. The main reason why you see one-sector Cobb-Douglas models is that developed economies seem to satisfy the "balanced growth" condition, and it's easy to get that condition out of one-sector Cobb-Douglas. This blog post gives the condition, and talks other kinds of models that can satisfy it: https://growthecon.com/blog/BGP/
@arsmath (+1) Thanks for your comment. I agree that one reason for why we often see the Cobb-Douglas production function in textbooks is that it is consistent with Uzawa's theorem and with the empirically observed constant growth rate regime during the period after the industrial revolution in countries that are developed today. However, Uzawa's theorem does not imply a Cobb-Douglas production function. It implies a Harrod-neutral production function, and the Cobb-Dogulas production function is one example. See Theorem 2.6 on page 60 in Acemoglu (2009). The link does not answer my questions.
Something seems to be wrong here. Your Cobb-Douglas production function has a negative partial derivative in $K$.
@IosifPinelis That was a typo. It is corrected now. Thanks for noticing.
That's why I left a comment. I wanted to draw your attention to the two-sector models mentioned in that blog post.
Concerning your Question 1: "Can we characterize the set of neoclassical production functions?" -- This set is already characterized, tautologically, by its definition, as the set of all functions $F$ satisfying Assumptions 1 and 2. There seems to be no reason/way for there to exist a better characterization.
Concerning your Question 2: "What are some examples of neoclassical production functions that do not fall in the class of Cobb-Douglas production functions?"
First here, note that Assumptions 1 and 2 impose almost no restrictions on the dependence of $F(K,L,A)$ on $A$. So, e.g., any function $F$ given by the formula
$$F(K,L,A)=K^a L^{1-a}h(A),$$
where $a\in(0,1)$ and $h$ is any positive function, will do.
Second, note that your set of neoclassical production functions is a cone. So, any linear combination of functions in this set with positive coefficients, as well as appropriate limits of such linear combinations, will do. E.g., all the functions $F$ given by the formula
$$F(K,L,A)=Lf(K/L)h(A)\tag{1}$$
will do, where $h$ is any positive function and
$$f(z)=\int_{(0,1)}z^a\mu(da)$$
for all all real $z\ge0$, where in turn $\mu$ is any positive measure on $(0,1)$. In particular, here we can take
$$f(z)=\int_{(0,1)}z^a\,da=\frac{z-1}{\ln z}$$
for $z\in(0,1)\cup(1,\infty)$, with $f(0)$ and $f(1)$ defined by continuity,
$$f(z)=\int_{(0,1)}z^a\,a\,da=\frac{1-z+z \ln z}{\ln^2 z}$$
for $z\in(0,1)\cup(1,\infty)$, with $f(0)$ and $f(1)$ defined by continuity, or, more generally,
$$f(z)=\int_{(0,1)}z^a\,a^n\,da= \frac{\Gamma (n+1)-\Gamma (n+1,-\ln z)}
{(-\ln z)^{n+1}}$$
for real $n\ge0$ and $z\in(0,1)\cup(1,\infty)$, with $f(0)$ and $f(1)$ defined by continuity, where $\Gamma (m,u)$ represents the incomplete Gamma function.
Of course, here one can also use (possibly infinite) linear combinations with different functions $h$.
Third, one can certainly construct neoclassical production functions $F$ as in (1) with $f$ other than (possibly infinite) linear combinations of the functions $z\mapsto z^a$ with $a\in(0,1)$ (as was done in the second item above). For instance, one can glue pieces of such linear combinations over non-overlapping subintervals of $[0,\infty)$ in any manner just so that to preserve the continuity and differentiability.
|
2025-03-21T14:48:29.757438
| 2020-02-01T18:53:02 |
351716
|
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|
Stack Exchange
|
Almost transferred model structures
Let $F : \mathcal{C} \leftrightarrows \mathcal{D} : U$ be a Quillen adjunction between cofibrantly generated model categories. The model structure on $\mathcal{D}$ is called transferred if $U$ preserves and reflects weak equivalences and fibrations. It follows that cofibrations of $\mathcal{D}$ are generated by $F(\mathrm{I})$, where $\mathrm{I}$ is a set of generating cofibrations of $\mathcal{C}$.
Now, assume that cofibrations of $\mathcal{D}$ are generated by $F(\mathrm{I})$ and that $U$ reflects fibrant objects (or even all fibrations). Is the model structure on $\mathcal{D}$ necessarily transferred in this case? If these conditions hold and the transferred model structure exists, then it necessarily coincides with the given one. Thus, the question can be reformulated as follows. Is there a model category $\mathcal{D}$ satisfying conditions given above such that the transferred model structure on $\mathcal{D}$ does not exist?
This is too long for a comment, so I'm putting it here. I can think of three places to look for such an example. First, you could check out Example 2.8 in this paper I wrote with Michael Batanin: Bousfield Localization and Eilenberg-Moore Categories, arXiv:1606.01537.
In that example, we proved that the transferred model structure doesn't exist, so you could check if there is a different model structure on the category of operads valued in Ch($\mathbb{F}_2)$ that has the property you want.
Second, it is well-known that the transferred model structure on commutative differential graded algebras does not exist in characteristic $p>0$. So, CDGA($\mathbb{F}_p)$ doesn't have a transferred model structure. It does have a model structure discovered by Donald Stanley: Determining Closed Model Category Structures, (1998) (link),
and you could check if Stanley's model structure has the property you want. Seems plausible.
Third, you could try to construct such an example using the techniques of Reid Barton at this answer:
Counter-example to the existence of left Bousfield localization of combinatorial model category
You'd want to pick $C$ and $D$ to be very small, like Reid did. I would try to work this out, but I just don't have time (and probably won't all semester).
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2025-03-21T14:48:29.757620
| 2020-02-01T20:12:39 |
351718
|
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"Martin Rubey",
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|
Stack Exchange
|
Cyclic action on Kreweras walks
A Kreweras walk of length $3n$ is a word consisting of $n$ $A$'s, $n$ $B$'s, and $n$ $C$'s such that in any prefix there are at least as many $A$'s as $B$'s, and at least as many $A$'s as $C$'s. For example, with $n = 3$, one Kreweras walk is: $w = AABBCACCB$. These are the same as walks in $\mathbb{Z}^2$ from the origin to itself consisting of steps $(1,1)$, $(-1,0)$, and $(0,-1)$ which always remain in the nonnegative orthant (treat $A$'s as $(1,1)$ steps, $B$'s as $(-1,0)$ steps, and $C$'s as $(0,-1)$ steps). Kreweras in 1965 proved that the number of Kreweras walks is $\frac{ 4^n(3n)!}{(n+1)!(2n+1)!}$ (OEIS sequence A006335). Many years later, in the 2000's, the Kreweras walks became a motivating/foundational example in the theory of "walks with small steps in the quarter plane" as developed by Mireille Bousquet-Mélou and her school. They also are related to decorated planar maps and in particular are a key ingredient in recent breakthrough work relating random planar maps to Liouville quantum gravity.
I discovered a very interesting cyclic action on Kreweras walks, which apparently had not been noticed previously. Let me refer to this action as rotation. To perform rotation on a Kreweras walk $w$, first we literally rotate the word $w =(w_1,w_2,...,w_{3n})$ to $w' = (w_2,w_3,...,w_{3n},w_1)$. With the above example of $w$, we get $w' = ABBCACCBA$. This is, however, no longer a valid Kreweras walk. So there will be a smallest index $i$ such that $(w'_1,...,w'_i)$ has either more $B$'s than $A$'s, or more $C$'s than $A$'s. Then we create another word $w''$ by swapping $w'_i$ and $w'_{3n}$ (which is always an $A$). For example, with the previous example, we have $i = 3$ (corresponding to the second $B$ in the word), and we get $w'' = ABACACCBB$. It's not hard to see that the result is a Kreweras walk, which we call the rotation of the initial Kreweras walk. The sequence of iterated rotations of our initial $w = AABBCACCB$ example looks like
$$ 00 \; AABBCACCB \\
01 \; ABACACCBB \\
02 \; AACACCBBB \\
03 \; ACACABBBC \\
04 \; AACABBBCC \\
05 \; ACABBACCB \\
06 \; AABBACCBC \\
07 \; ABAACCBCB \\
08 \; AAACCBCBB \\
09 \; AACCBABBC \\
...$$
In particular, $3n = 9$ applications of rotation yields the Kreweras walk which is the same as our initial $w$ except that the $B$'s and $C$'s have swapped places. If we did another $9$ applications we would get back to our initial $w$.
Conjecture: For a Kreweras walk of length $3n$, $3n$ applications of rotation always yields the Kreweras walk which is the same as the initial walk except with the $B$'s and $C$'s swapped (so $6n$ applications of rotation is the identity).
(So my question is obviously: is my conjecture right?) I've thought about this conjecture a fair amount with little concrete progress. I've done a fair amount of computational verification of this conjecture: for all $n \leq 6$, and for many thousands more walks with various $n \leq 30$.
Where this action comes from: The Kreweras walks of length $3n$ are in obvious bijection with the linear extensions of a poset $P$, namely, $P=[n] \times V$, the direct product of the chain on $n$ elements and the 3-element ``$V$'' poset with relations $A < B$, $A < C$. I became aware of this poset thanks to this MO answer of Ira Gessel to a previous question of mine, which cited this paper of Kreweras and Niederhausen in which the authors prove not just a product formula for the number of linear extension of $P$, but for the entire order polynomial of $P$. The rotation of Kreweras walks as I just defined it is nothing other than the famous (Schützenberger) promotion operation on linear extension of a poset (see this survey of Stanley's for background on promotion). There are few non-trivial classes of posets for which the behavior of promotion is understood (see section 4 of that survey of Stanley's), so it is very interesting to discover a new example. In particular, all known examples are connected to tableaux and symmetric functions, etc.; whereas this Kreweras walks example has quite a different flavor.
Some thoughts: The analogous rotation action on words with only $A$'s and $B$'s (i.e., Dyck words) is well-studied; as explained in section 8 of this survey of Sagan's on the cyclic sieving phenomenon, it corresponds to promotion on $[2]\times[n]$, and in turn to rotation of noncrossing matchings of $[2n]$. There is a way to view a Kreweras walk as a pair of noncrossing partial matchings on $[3n]$ (basically we form the matching corresponding to the $A$'s and $C$'s, and to the $A$'s and $B$'s). But this visualization does not seem to illuminate anything about the rotation action (in particular, when we rotate a walk, one of the noncrossing partial matchings simply rotates, but something complicated happens to the other one).
As mentioned earlier, there is a bijection due to Bernardi between Kreweras walks and decorated cubic maps, but I am not able to see any simple way that this bijection interacts with rotation.
On a positive note, it seems useful to write the $3n$ rotations of a Kreweras walk in a cylindrical array where we indent by one each row, as follows:
$$
\begin{array} \,
A & A & B & b & C & A & C & C & B \\
& A & b & A & C & A & C & C & B & B \\
& & A & A & C & A & C & c & B & B & B \\
& & & A & c & A & C & A & B & B & B & C \\
& & & & A & A & C & A & B & B & b & C & C \\
& & & & & A & c & A & B & B & A & C & C & B \\
& & & & & & A & A & B & b & A & C & C & B & C \\
& & & & & & & A & b & A & A & C & C & B & C & B \\
& & & & & & & & A & A & A & C & C & B & c & B & B \\
& & & & & & & & & A & A & C & C & B & A & B & B & C
\end{array}
$$
In each row I've made lowercase the $B$ or $C$ that the initial $A$ swaps with. We can extract from this array a permutation which records the columns in which these matches occur (where we cylindrically identify column $3n+i$ with column $i$). In this example, the permutations we get is $p = [4,3,8,5,11,7,10,9,15] = [4,3,8,5,2,7,1,9,6]$. The fact that this list of columns is actually a permutation (which I don't know how to show) is equivalent to the assertion that the position of the $A$'s after $3n$ rotations is the same as in the initial Kreweras walk. Furthermore, this permutation $p$ determines position of $A$'s. Namely, the $A$'s are exactly the $p(i)$ for which $p(i) < i$. In our example, these are $2$, $1$, and $6$, corresponding to $i = 5,7,9$. Also, you can see how the $3n$ rotations "permute" the position of the $A$'s from $p$ as well. To do that, write down a new permutation $q$ from $p$: $q$ is the product of transpositions $q = (3n, p(3n)) \cdots (2, p(2)) \cdot (1, p(1))$. Then $q$ exactly tells us how the $A$'s are permuted. In our example, as we process the transpositions of $q = (9,6)(8,9)(7,1)(6,7)(5,2)(4,5)(3,8)(2,3)(1,4)$ right-to-left on the positions $\{1,2,6\}$ of $A$'s we see $1 \to 4 \to 5 \to 2$; $2 \to 3 \to 8 \to 9 \to 6$; and $6 \to 7 \to 1$. Note that the $A$'s end up changing places, and that they each are involved in a different number of swaps. Another thing worth noting is that the permutation $p$ does not determine the Kreweras walk (even after accounting for the $B \leftrightarrow C$ symmetry).
In spite of these observations, the lack of any connection to algebra (e.g., the representation theory of Lie algebras), and the lack of any good "model" for these words, makes it really hard to reason about how they behave under rotation.
EDIT:
Let me add one example which may indicate some subtlety. Let's define a $k$-letter Kreweras word of length $kn$ to be a word consisting of $n$ A's, $n$ B's, $n$ C's, $n$ D's, etc. for $k$ different letters such that in any prefix there are at least as many $A$'s as $B$'s, at least as many $A$'s as $C$'s, at least as many $A$'s as $D$'s, etc. So $3$-letter Kreweras words are the Kreweras walks discussed above, and $2$-letter Kreweras words are the Dyck words. We can define rotation for $k$-letter Kreweras words in exactly the same manner: literally rotate the word, find the first place the inequalities are violated, swap this place with the final $A$ to obtain a valid word (and this corresponds to promotion on a certain poset).
For the case $k=2$, note that $kn$ applications of rotation to a $k$-letter Kreweras word of length $kn$ results in a word with the $A$'s in the same position (because this is just rotation of noncrossing matchings). For the case $k=3$, apparently $kn$ applications of rotation results in a word with the $A$'s in the same position (because apparently the $B$'s and $C$'s switch). But for $k > 3$, it is not true necessarily that $kn$ applications of rotation results in a word with the $A$'s in the same position. For instance, with $k=4$ and $n=3$, starting from the word $w=AADACCDCBDBB$, 12 rotations gives us:
$$ 00 \; AADACCDCBDBB \\
01 \; ADACCDABDBBC \\
02 \; AACCDABDBBCD \\
03 \; ACADABDBBCDC \\
04 \; AADABDBBCDCC \\
05 \; ADABDBACDCCB \\
06 \; AABDBACDCCBD \\
07 \; ABDAACDCCBDB \\
08 \; ADAACDCCBDBB \\
09 \; AAACDCABDBBD \\
10 \; AACDCABDBBDC \\
11 \; ACDAABDBBDCC \\
12 \; ADAABDBBDCCC $$
where the $A$'s do not end up in the same positions they started in. So something kind of subtle has to be happening in the case $k=3$ to explain why they do.
Ah, I have a student working on very similar-sounding questions!
I was inspired a bit by the paper https://arxiv.org/abs/1805.01254 which does not seem to have anything to do with walks at a glance, but it is there...
@PerAlexandersson: for other instances of "rotation of lattice paths restricted to a cone" see this paper of Fontaine-Kamnitzer: https://arxiv.org/abs/1212.1314.
I really like this! Doing a few examples by hand, it seems like for $k=4$, after $4n$ steps, the former position of Bs and the new position of the Bs have zero overlap, and same for C and D. Your conjecture obviously implies a stronger statement for $k=3$.
@HughThomassupportsMonica: Very nice observation! Doing some large random examples it looks like this could be the case!
In fact my claim, for k=3, already implies your conjecture. Thinking in terms of promotion where we don't decrement the entries (so the each number just moves through the poset from maximum to minimum and is replaced by itself $+3n$ in a maximum position again) every number $t$ starts life as a maximum, and my conjecture then implies that when $t+3n$ appears in the filling, as a maximum again, it must be the other one. The rest follows. If my conjecture is true, that could be the explanation for $k>3$ being different from $k=3$.
@HughThomassupportsMonica: I don't completely understand this last comment. But, I believe that the claim "all the B's become non-B's and all the C's become non-C's" is sketched in Max Alekseyev's answer below: his argument about how the "type" of A changes when it swaps, and the "type" of other As stays the same (this part needs to be explained more), shows that the B's become C's and vice-versa- it seems to leave open the possibility that these later become A's, but that would still show they end different than they started.
@HughThomassupportsMonica: The main thing I am not understanding in your last comment is how it shows the difference between $k=3$ and $k>3$.
Sorry, I retract what I said in my last comment. I no longer think that what I claimed allows you to prove your statement for $k=3$. I find it plausible that there could be something a bit stronger than my claim which does allow you to prove your claim for $k=3$ but which still leaves the door open to chaos when $k=4$, but that's just a feeling. (I should probably delete that comment and this one.)
@HughThomassupportsMonica: I think it's okay to leave the comment as it's a reasonable idea.
Martin Rubey and I solved my conjecture.
The basic idea of the proof is as follows. First to a Kreweras word $w$ we associate what we call its bump diagram, which is just the union of the two noncrossing partial matchings of $\{1,2,...,3n\}$ associated to $w$ (the one for the A's and B's and the one for the A's and C's), drawn as a graph in the obvious way. For example, with $w=AABBCACCB$ its bump diagram is
We also think of this diagram as a set of ordered pairs ('arcs'); in this example that set is
$$\{ (1,4),(1,8),(2,3),(2,5),(6,7),(6,9)\} $$
We extract a permutation $\sigma_w$ of $\{1,2,...,3n\}$ from the bump diagram as follows.
For $i=1,2,...,3n$, we take a trip from position $i$. We start traveling from position $i$ along the unique arc ending at $i$ (if $w_i=B$ or $C$), or the "shorter arc" beginning at $i$ (if $w_i=A$), and we continue until we reach a "crossing" of arcs. When we hit a crossing of arcs $(i,k)$ and $(j,\ell)$ with $i \leq j < k < \ell$ (note that we allow the case $i=j$), we follow the following "rules of the road":
if we were heading towards $i$, then we turn right and head towards $\ell$;
if we were heading towards $\ell$, then we turn left and head towards $i$;
otherwise we continue straight to where we were heading.
When we finish our trip at position $j$ then we define $\sigma_w(i) := j$.
For example, to compute $\sigma_w(3)$: we start traveling along the arc $(2,3)$ heading towards $2$; then we come to the crossing of $(2,3)$ and $(2,5)$ and we turn right, heading towards $5$; then we come to crossing of $(1,4)$ and $(2,5)$ and we turn left, heading towards $1$; then we come to the crossing of $(1,4)$ and $(1,8)$ and we turn right, heading towards $8$; then we come to the crossing of $(1,8)$ and $(6,9)$, but we just continue straight to $8$; and so we finish our trip at $8$. So $\sigma_w(3)=8$.
Or to compute $\sigma_w(7)$: we start traveling along the arc $(6,7)$ heading towards $6$; then we come to the crossing of $(6,7)$ and $(6,9)$ and we turn right, heading towards $9$; then we come to the crossing of $(1,8)$ and $(6,9)$ and we turn left, heading towards $1$; and then we come to the crossing of $(1,4)$ and $(1,8)$, but we just continue straight to $1$; and so we finish our trip at $1$. So $\sigma_w(7)=1$.
We could compute the whole permutation is $\sigma_w = [4,3,8,5,2,7,1,9,6]$.
You might notice that this example $w$ is the same as the original post and that this permutation $\sigma_w$ is the same as the "permutation" $p$ defined in terms of the cylindrical rotation array.
Indeed, this always happens (that the trip permutation is the same as the permutation from the cylindrical rotation array). It follows from the main lemma behind the overall proof, which is
Lemma. If $w'$ is the rotation of $w$, then $\sigma_{w'} = c^{-1} \sigma_w c$ where $c= (1,2,3,...,3n)$ is the "long cycle."
As a remark, these trip permutations come from the theory of plabic graph (cf. Section 13 of Postnikov's paper https://arxiv.org/abs/math/0609764).
Since $\sigma_w$ does not completely determine $w$, to finish the proof we need to keep track of a little more data. For that purpose, we define $\varepsilon_w=(\varepsilon_w(1),...,\varepsilon_w(3n))$, a sequence of $3n$ letters which are B's or C's, defined by
$$ \varepsilon_w(i) := \begin{cases} w_{\sigma(i)} &\textrm{if $w_{\sigma(i)}\neq A$} \\ w_{\sigma(\sigma(i))} &\textrm{if $w_{\sigma(i)}= A$}. \end{cases} $$
Similarly to the previous lemma, we can show
Lemma. If $w'$ is the rotation of $w$, then $\varepsilon_{w'} = (\varepsilon_w(2),\varepsilon_w(3),...,\varepsilon_w(3n),-\varepsilon_w(1))$ with the convention that $-B=C$ and vice-versa.
The above lemmas easily imply that the $3n$th rotation of $w$ is obtained from $w$ by swapping B's and C's.
Martin and I will post a preprint to the arXiv with all the details soon.
EDIT: The paper is now on the arXiv at https://arxiv.org/abs/2005.14031.
This is not an answer, but too long for a comment.
This promotion operator is (apparently) governed by local rules, similar to https://arxiv.org/abs/1804.06736, as follows:
regard each path as a sequence of coordinates, that is, $A$ adds $(1,1)$, $B$ adds $(-1,0)$ and $C$ adds $(0,-1)$ to the current coordinate
create a cylindrical array from each promotion orbit, for example, for the path $AABBCACCB$
${\scriptstyle\begin{array}{llllllllllllllllllll}
0,0 & 1,1 & 2,2 & 1,2 & 0,2 & 0,1 & 1,2 & 1,1 & 1,0 & 0,0 \\
&0,0 & 1,1 & 0,1 & 1,2 & 1,1 & 2,2 & 2,1 & 2,0 & 1,0 & 0,0 \\
&&0,0 & 1,1 & 2,2 & 2,1 & 3,2 & 3,1 & 3,0 & 2,0 & 1,0 & 0,0 \\
&&&0,0 & 1,1 & 1,0 & 2,1 & 2,0 & 3,1 & 2,1 & 1,1 & 0,1 & 0,0 \\
&&&&0,0 & 1,1 & 2,2 & 2,1 & 3,2 & 2,2 & 1,2 & 0,2 & 0,1 & 0,0 \\
&&&&&0,0 & 1,1 & 1,0 & 2,1 & 1,1 & 0,1 & 1,2 & 1,1 & 1,0 & 0,0 \\
&&&&&&0,0 & 1,1 & 2,2 & 1,2 & 0,2 & 1,3 & 1,2 & 1,1 & 0,1 & 0,0 \\
&&&&&&&0,0 & 1,1 & 0,1 & 1,2 & 2,3 & 2,2 & 2,1 & 1,1 & 1,0 & 0,0 \\
&&&&&&&&0,0 & 1,1 & 2,2 & 3,3 & 3,2 & 3,1 & 2,1 & 2,0 & 1,0 & 0,0 \\
&&&&&&&&&0,0 & 1,1 & 2,2 & 2,1 & 2,0 & 1,0 & 2,1 & 1,1 & 0,1 & 0,0 \\
&&&&&&&&&&0,0 & 1,1 & 1,0 & 2,1 & 1,1 & 2,2 & 1,2 & 0,2 & 0,1 & 0,0 \\
&&&&&&&&&&&0,0 & 1,1 & 2,2 & 1,2 & 2,3 & 1,3 & 0,3 & 0,2 & 0,1 & 0,0 \\
&&&&&&&&&&&&0,0 & 1,1 & 0,1 & 1,2 & 0,2 & 1,3 & 1,2 & 1,1 & 1,0 & 0,0 \\
&&&&&&&&&&&&&0,0 & 1,1 & 2,2 & 1,2 & 2,3 & 2,2 & 2,1 & 2,0 & 1,0 & 0,0 \\
&&&&&&&&&&&&&&0,0 & 1,1 & 0,1 & 1,2 & 1,1 & 1,0 & 2,1 & 1,1 & 0,1 & 0,0 \\
&&&&&&&&&&&&&&&0,0 & 1,1 & 2,2 & 2,1 & 2,0 & 3,1 & 2,1 & 1,1 & 1,0 & 0,0 \\
&&&&&&&&&&&&&&&&0,0 & 1,1 & 1,0 & 2,1 & 3,2 & 2,2 & 1,2 & 1,1 & 0,1 & 0,0 \\
&&&&&&&&&&&&&&&&&0,0 & 1,1 & 2,2 & 3,3 & 2,3 & 1,3 & 1,2 & 0,2 & 0,1 & 0,0
\end{array}}$
consider any square of four coordinate in this array
\begin{array}{ll}
\lambda & \nu\\
\kappa & \mu
\end{array}
and let $\tilde\mu = \kappa + \nu - \lambda$.
Then, apparently, we have
$
\mu = \begin{cases}
\tilde\mu &\text{if $\tilde\mu$ has positive coordinates}\\
\tilde\mu + (2,1) &\text{if the first coordinate of $\tilde\mu$ is negative}\\
\tilde\mu + (1,2) &\text{if the second coordinate of $\tilde\mu$ is negative}
\end{cases}
$
Possibly we can get a proof that the occurrences of $\tilde\mu$ with a negative coordinate yield a permutation, assuming that the local rules are correct.
First we paste the triangular region to the right of the first $3n$ rows into the empty region below the diagonal (and removing the final $3n-1$ rows). The pasting must be done in a way such that there is precisely one $(0,0)$ coordinate in each row and column:
${\scriptstyle\begin{array}{llllllllllllllll}
0,0 & 1,1 & 2,2 & 1,2 & 0,2 & 0,1 & 1,2 & 1,1 & 1,0 & 0,0 \\
1,0 & 0,0 & 1,1 & 0,1 & 1,2 & 1,1 & 2,2 & 2,1 & 2,0 & 1,0 \\
2,0 & 1,0 & 0,0 & 1,1 & 2,2 & 2,1 & 3,2 & 3,1 & 3,0 & 2,0 \\
2,1 & 1,1 & 0,1 & 0,0 & 1,1 & 1,0 & 2,1 & 2,0 & 3,1 & 2,1 \\
2,2 & 1,2 & 0,2 & 0,1 & 0,0 & 1,1 & 2,2 & 2,1 & 3,2 & 2,2 \\
1,1 & 0,1 & 1,2 & 1,1 & 1,0 & 0,0 & 1,1 & 1,0 & 2,1 & 1,1 \\
1,2 & 0,2 & 1,3 & 1,2 & 1,1 & 0,1 & 0,0 & 1,1 & 2,2 & 1,2 \\
0,1 & 1,2 & 2,3 & 2,2 & 2,1 & 1,1 & 1,0 & 0,0 & 1,1 & 0,1 \\
1,1 & 2,2 & 3,3 & 3,2 & 3,1 & 2,1 & 2,0 & 1,0 & 0,0 & 1,1 \\
0,0 & 1,1 & 2,2 & 2,1 & 2,0 & 1,0 & 2,1 & 1,1 & 0,1 & 0,0
\end{array}}$
(This array does not satisfy the local rules along the diagonal.)
Now we consider one square of four, but instead of its corners label the four edges with $\lambda-\kappa$, $\nu-\lambda$, $\mu-\kappa$ and $\nu-\mu$. There are 11 different squares occurring, two of which correspond to a $b$ or $c$ respectively. For these two, the labels on parallel edges are different, for the others, they are same. Put a bullet in the squares whose parallel edges have distinct labels.
In the case at hand, we obtain
${\def\x{\huge\bullet}
\scriptstyle\begin{array}{llllllllllllllllll}
& A & & A & & B & & B & & C & & A & & C & & C & & B &\\
B & & A & & A & & A & \x & B & & B & & B & & B & & B & & B\\
& B & & A & & B & & A & & C & & A & & C & & C & & B &\\
B & & B & & A & \x & B & & B & & B & & B & & B & & B & & B\\
& B & & B & & A & & A & & C & & A & & C & & C & & B &\\
C & & C & & C & & A & & A & & A & & A & & C & \x & C & & C\\
& B & & B & & C & & A & & C & & A & & C & & A & & B &\\
C & & C & & C & & C & & A & \x & C & & C & & C & & C & & C\\
& B & & B & & C & & C & & A & & A & & C & & A & & B &\\
A & & A & \x & B & & B & & B & & A & & A & & A & & A & & A\\
& B & & A & & C & & C & & B & & A & & C & & A & & B &\\
C & & C & & C & & C & & C & & C & & A & \x & C & & C & & C\\
& B & & A & & C & & C & & B & & C & & A & & A & & B &\\
A & \x & B & & B & & B & & B & & B & & A & & A & & A & & A\\
& A & & A & & C & & C & & B & & C & & B & & A & & B &\\
B & & B & & B & & B & & B & & B & & B & & B & & A & \x & B\\
& A & & A & & C & & C & & B & & C & & B & & B & & A &\\
A & & A & & A & & A & & A & & A & \x & C & & C & & C & & A\\
& A & & A & & C & & C & & B & & A & & B & & B & & C &
\end{array}}$
It remains to show that in each row of "vertical" labels only $A$ and one other letter occurs, and in each column of "horizontal" labels, only $A$ and one other letter occurs, except that for the "horizontal" labels below the diagonal, we have to swap $B$ and $C$.
I believe this follows from the local rules.
Thanks for this, it is really nice. I think it should be straightforward to show that these local rules hold. Do you think that it is possible that this promotion operation fits into the tensor invariants picture, but for a rank 2 affine or other Kac-Moody algebra?
I have no idea. One further observation: the occurrences of $\tilde\mu$ with one coordinate negative are (apparently) precisely the entries of the permutation.
Right: when we remove the initial $A$ (1,1) step, the resulting path will go out of the 1st orthant exactly once, and precisely at the first $B$ or $C$ step to which this $A$ is matched.
I am still struggling with the final (crucial) step: that there is precisely one bullet in every column.
Indeed, this seems like the crux. Maybe thinking about how this case is different from the $k>3$ cases (for which the 'one bullet in every column' property fails) discussed in my edit to the original question will be illuminating?
I checked the case $k=4$. Indeed, the local rules for this case are the most straightforward generalisation of the rules for $k=3$. Thus it is very unlikely that the local rules alone explain the order of promotion.
Again not (quite) an answer, but too long for a comment:
Here is another way to obtain the (conjectural) permutation.
As running example, let $p = A B A A C C B C A B B C$.
The algorithm to obtain the permutation can be reformulated as follows:
to determine the exceedences, proceed using promotion.
So we have, in one line notation and using $x$ for unknowns,
$\pi = [2, 8, 6, 5, x, 7, x, 11, 10, x, 12, x]$.
to determine the deficiencies,
let "openers" be the positions of the $A$'s in the path: $\{1, 3, 4, 9\}$.
let "closers" be the positions of the deficiencies (i.e., the
indices of the unknowns) $\{5, 7, 10, 12\}$
match each closer $c$ (beginning with the smallest) with the nearest
opener $o$, such that $p_{\pi(o)}$ differs from $p_c$.
$\pi(5) = 1$ because $p_{5}=C$ and $p_{\pi(4)}=p_5$ and $p_{\pi(3)}=p_6$ equal $C$
$\pi(7) = 4$ because $p_7=B$ and $p_{\pi(4)}=p_5=C$
$\pi(10) = 3$ because $p_{10}=B$ and $p_{\pi(3)}=p_6=C$
$\pi(12) = 9$ because $p_{12}=C$ and $p_{\pi(9)}=p_{10}=B$
Although this looks much more complicated, it might be easier to prove
that this algorithm works: we "only" have to show that there is a
matching from closers to openers such that each closer is matched to a
smaller opener with the correct label.
I think this goes as follows: consider the square diagram obtained from the cyclindrical growth diagram by appropriate use of scissors and glue:
\begin{array}{lllllllll}
A& b& A& A& C& C& B& C& A& B& B& C\\
B& A& A& A& C& C& B& c& A& B& B& C\\
B& C& A& A& C& c& B& A& A& B& B& C\\
B& C& C& A& c& A& B& A& A& B& B& C\\
b& C& C& C& A& A& B& A& A& B& B& C\\
A& C& C& C& B& A& b& A& A& B& B& C\\
A& C& C& c& B& B& A& A& A& B& B& C\\
A& C& C& A& B& B& C& A& A& B& b& C\\
A& C& C& A& B& B& C& B& A& b& A& C\\
A& C& c& A& B& B& C& B& B& A& A& C\\
A& C& A& A& B& B& C& B& B& C& A& c\\
A& C& A& A& B& B& C& B& b& C& C& A\\
\end{array}
It is an immediate consequence of the local rules (and probably also of the definition of promotion) that in every column above the main diagonal the letter in the first line repeats, until it is eventually replaced by an $A$, and the same holds for the letters below the diagonal. (At this point we do not know that the non-$A$ letter below the diagonal is different from the non-$A$ letter above the diagonal.)
Observe, however, that the non-$A$ letter in each column below the diagonal equals the replaced letter (indicated as lower-case $b$ or $c$ in the example) in the corresponding row. This is the case because promotion appends the replaced letter to the Kreweras walk.
It remains to show that promotion of the path intertwines with rotation of this permutation (regarded as a chord diagram).
|
2025-03-21T14:48:29.759139
| 2020-02-01T20:52:39 |
351721
|
{
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"Carlo Beenakker",
"Descartes Before the Horse",
"Wolfgang",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351721"
}
|
Stack Exchange
|
Evaluation of $\int \prod_{j=1}^u \frac{x+j}{j-x}~dx$
Let $I_{n,k} = \frac{(n+k)!}{k!(k-1)!(n-k)!}$. This is a sort of generalization of the Apéry's numbers, with $I_{n,n} =$ the $n$-th Apéry number. I am studying integrals of the form:
$$f_u(x)=\int \prod_{j=1}^u \frac{x+j}{j-x}~dx.$$
Where $u$ is a natural number. For $u>2$, I have shown that
$$\tag{1}f_u(x) = C+ (-1)^ux+\sum_{w=1}^{u}(-1)^w \log(x-w)I_{u,w}.$$
For example,
$$f_3(x)=-x-60\log(x-3)+60\log(x-2)-12\log(x-1).$$
I am looking for clarification on two things:
My eq. $(1)$ does not work for $u=1,2$. It generates a function that is almost what the integral evaluates to; whereas $f_1(x) = -x-2\log(1-x)$, eq. $(1)$ gives me $-x+2\log(x-1)$. Is there any amendment I can make to $(1)$ to ensure that it holds for all natural $u$?
My eq. $(1)$ surely appears like it should be written as one sum, but I have not been able to manipulate the summand to include the $(-1)^u x$ term. Is there a way to rewrite $(1)$ as a single sum, barring $C~$?
What if the arguments of your logs are negative? Shouldn't you have log|▪︎| everywhere?
notice that $\log(x-w)$ and $\log(w-x)$ only differ by an (imaginary) constant, so this can be absorbed in the $C$ in $f_u(x)$; the only difference between $u=1,2$ and $u>2$ that matters is that the sign in front of the logarithm is different from $(-1)^w$; if I am allowed to change the sign of the definition of $I_{u,w}$ for $u=1,2$, I'm done.
are you sure your equation (1) is correct? I think $-\sum_{w=1}^u$ should be $+\sum_{w=1}^u$
@CarloBeenakker you are correct, fixed.
I may be mistaken, but I get
$$f_u(x)=\int \prod_{j=1}^u \frac{x+j}{j-x}~dx= C+ (-1)^ux+\sum_{w=1}^{u}(-1)^w \log(x-w)I_{u,w}$$
and this is correct for all $u=1,2,3,...$, so it seems issue 1 is resolved.
Thank you. If no answer addressing the second part of my question is posted in some time, I will accept this.
|
2025-03-21T14:48:29.759303
| 2020-02-01T23:36:33 |
351731
|
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"authors": [
"Tereza Tizkova",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/351731"
}
|
Stack Exchange
|
A classification of $G_{\delta\sigma}$ zero-dimensional spaces?
Among separable metrizable spaces:
Cantor set is the unique compact zero-dimensional space without isolated points.
$\mathbb Q$ is the unique countable space without isolated points
$\mathbb R \setminus \mathbb Q$ is the unique zero-dimensional, $G_\delta$-space with no compact neighborhood.
$\mathbb Q ^\omega$ is the unique zero- dimensional, first category $F_{\sigma\delta}$-space with the property that no nonempty clopen subset is a $G_{\delta\sigma}$-space.
Question. Is there a simple classification of zero-dimensional $G_{\delta\sigma}$-spaces which have no compact neighborhoods?
The simplest examples would be $\mathbb Q$, $\mathbb R\setminus \mathbb Q$, and $\mathbb Q\times (\mathbb R\setminus \mathbb Q)$. Are there many others?
Is there a nice characterization of $\mathbb Q\times (\mathbb R\setminus \mathbb Q)$?
Could you please provide a source or proof for the $\mathbb{Q}^{\omega}$ part? I am interested mainly in proof that no nonempty clopen subset is a $G_{\delta\sigma}$-space.
This paper by Van Mill from 1981 gives a characterisation of $\Bbb Q \times \Bbb P$ (where $\Bbb P$ is a common notation for the irrationals) in Thm 5.3:
If $X$ is separable metrisable and zero-dimensional, $\sigma$-complete and nowhere complete and nowhere $\sigma$-compact then $X \simeq \Bbb Q \times \Bbb P$.
Where by complete I mean topologically complete (i.e. in this context: completely metrisable) and a nowhere-$P$ space is one where no non-empty open subset has property $P$, so $\Bbb P$ and $\Bbb Q$ are nowhere locally compact, e.g.)
I think $\Bbb Q \times C$ (with $C$ the Cantor set) is another example for your list.
A lot of information can be found in van Engelen's PhD-thesis from 1985: homogeneous zero-dimensional absolute Borel sets, where he shows there are are $\omega_1$ many homeomorphism types of subsets of $C$ that are both $F_{\sigma \delta}$ and $G_{\delta\sigma}$, also separately written up here. In his thesis he also gave the first characterisation of $\Bbb Q^\omega$ (now relegated to the appendix of it). Look up Van Engelen's work from around that time for related results.
|
2025-03-21T14:48:29.759452
| 2020-02-02T01:07:31 |
351734
|
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"Joe Silverman",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/351734"
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|
Stack Exchange
|
Bounds on degrees of minimal polynomials of infinite degree algebraic extension
If $E/F$ is algebraic extension of finite degree $n$, then if $\alpha \in E$ is an element, then the degree of minial polynomial $m_\alpha$ for $\alpha$ is at most $n$. Even better, $\deg m_\alpha$ divides $n$. Are there some interesting bounds (with possibly some more conditions on $E,F,\alpha...$) on degree of $m_\alpha$ when the extension $E/F$ is of infinite degree?
In characteristic $0$, every finite extension has a primitive element. So if $[E:F]=\infty$, then you can find a sequence of $E_n\subset E$ with $[E_n:F]$ finite and going to $\infty$. Now take $E_n=F(\alpha_n)$, and the degree $m_{\alpha_n}\to\infty$. There are infinite extensions with interesting divisiblity properties, for example extensions such that for every $\alpha\in E$, the degree $m_\alpha$ is a power of some fixed prime $p$. But the power will grow.
There are purely inseparable extensions $E/F$ of infinite degree
for which there exists a prime power $q$ such that
every $x \in E$ satisfies an equation $x^q = a$ for some $a \in F$.
For an example with $q=2$, take
$F = {\bf F}_2(x_1,x_2,x_3,\ldots)$
with the $x_n$ algebraically independent, and
$E = F(x_1^{1/2}, x_2^{1/2}, x_3^{1/2}, \ldots)$.
|
2025-03-21T14:48:29.759565
| 2020-02-02T05:28:52 |
351741
|
{
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"François Brunault",
"Yemon Choi",
"https://mathoverflow.net/users/6506",
"https://mathoverflow.net/users/763"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351741"
}
|
Stack Exchange
|
A deduction in paper of Tanguy Rivoal and Wadim Zudilin which I am unable to think about
This question is related to Paper of Wadim Zudilin and Tanguy Rivoal "A note on odd zeta values".
Some users are saying that I should not post question related to papers in MO but site mentions questions from graduate text books and papers could be asked. Please give some hint and a question easy to you might not necessarily be easy to me .
Adding images
[] 3
I am unable to prove in last line of 4 th image how $(\phi_{n})^{-3} {d_n}^{A+2} q_{j,n}$ is an integer for odd j $\geq$ 5 but first part for $j=0$ I could prove.
I tried 2nd part 3 times but could not prove it. Can someone please help.
You have a mathematical question, but in my humble opinion it would be much better if you can phrase it differently from just extracting and copy-pasting from the article. Making this effort may also help you understand which theory you need. This probably belongs to MSE (unless you have a specific reason to believe there is some argument missing or some explanation needed (at research level), but you don't mention such a reason here).
I see. General advice I can give you is that when you face a question you have no idea how to solve, try to reformulate it in your own words and notation, detaching it from the context, and transforming it into a natural (possibly more general) statement with no or little notations from the original article. If you still cannot solve it, this will make a fine question for MSE, or MO if it happens to be more difficult than expected. Of course, other people can tell you the solution of the first question, but you will learn and benefit much less than with the "reformatting" process.
Of course I also understand things are not so simple when you don't have easy access to references from your location.
This question already had an accepted answer. Why have you bumped it with an edit?
Expression (6) from the paper is
$$ q_{j,n} = \sum_{m=0}^n (j-2)(j-1)p_{j-2,m} $$
for odd $j\geq 5$. It follows that
$$ \Phi_n^{-3}d_n^{A+2}q_{j,n} = \sum_{m=0}^n (j-2)(j-1)\Phi_n^{-3}d_n^{A+2}p_{j-2,m} \in \mathbb{Z}.$$
The argument before the part you quote shows that
$$ \Phi_n^{-3}d_n^{A-j}p_{j,m} \in \mathbb{Z}$$
for any $1\leq j\leq A$ and any $0\leq m\leq n$. Since $d_n=\mathrm{lcm}(1,2,\ldots,n)$ is an integer we know that
$$(j-2)(j-1)\Phi_n^{-3}d_n^{A+2}p_{j-2,m}\in\mathbb{Z}.$$
As the sum of integers $\Phi_n^{-3}d_n^{A+2}q_{j,n}$ is itself an integer. Although not stated at this point in the proof, this deduction is only valid when $1\leq j-2\leq A$, but the range of $j$ we are looking at for $q_{j,n}$ is odd $j$ with $5\leq j\leq A+1$, so this is valid where needed.
|
2025-03-21T14:48:29.759753
| 2020-02-02T06:17:48 |
351742
|
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"Yemon Choi",
"erz",
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"url": "https://mathoverflow.net/questions/351742"
}
|
Stack Exchange
|
Criterion for weak convergence of sequences
Let $E$ be a normed space and let $F\subset E^{*}$. It is known that $F$ is dense if and only if the restriction of $\sigma(E,F)$ on $B_E$ coincides with the weak topology.
Hence, if $F$ is dense and we have a bounded net $e_\alpha$ and $e\in E$ such that $\left<e_\alpha,f\right>\to \left<e,f\right>$, for every $f\in F$, then $e_\alpha\to e$ weakly. In particular, that works for sequences. However, density of $F$ seems to be too strong for sequences.
Is there a nice condition weaker than density of $F$ that would imply that every bounded sequence $e_n\in E$ converge weakly to $e\in E$ as long as $\left<e_n,f\right>\to \left<e,f\right>$, for every $f\in F$?
If $F$ is not dense then clearly testing on $F$ will not determine convergence, since you can always add to your $e_n$ any $h_n \in F^\perp$. So I am not sure what you are hoping for here
@YemonChoi $F$ still can be weak* dense, and so we have all information about $e_n$ by testing on $F$
I realised my error/misreading as soon as I'd finished typing and I thought I'd deleted my comment, but evidently not. Anyway I agree that my initial comment is insufficient reasoning (but see Matt's answer below)
In a nutshell, no, at least in the separable case. Let $F\subseteq E^*$ be not norm dense, and with $F$ (norm-) separable. By Hahn-Banach there is $M\in E^{**}$ which is non-zero and annihilates $F$. Let $f_0\in E^*$ with $\langle M,f_0 \rangle=1$.
I shall use Helly's Lemma (which I have failed to find an online reference for; it follows from e.g. the principle of local reflexivity) which says that if $N\subseteq E^*$ is finite-dimensional and $M\in E^{**}$ then for $\epsilon>0$ we can find $x\in E$ with $\|x\|\leq \|M\|+\epsilon$ and $\langle M,f\rangle = \langle f,x\rangle$ for $f\in N$.
Let $(f_n)$ be a norm-dense sequence in $F$. For each $n$ there is hence $x_n\in E$ with $\|x_n\| \leq \|M\|+1/n$, with $\langle M,f_0\rangle = \langle f_0,x_n\rangle$ and with $\langle M,f_k\rangle = \langle f_k,x_n\rangle$ for $k\leq n$. Thus $(x_n)$ is not norm-null. I shall show that $\langle f,x_n \rangle \rightarrow 0$ for each $f\in F$.
(This follows as $(x_n)$ is bounded and $(f_n)$ is dense in $F$. To give the details, for $f\in F$ and $\delta>0$ there is $k$ with $\|f-f_k\|<\delta$ and so if $n\geq k$ then $|\langle f,x_n\rangle| $ $\leq |\langle f-f_k,x_n\rangle| + |\langle f_k,x_n\rangle| $ $= |\langle f-f_k,x_n\rangle| + |\langle M, f_k\rangle| $ $= |\langle f-f_k,x_n\rangle| $ $\leq \delta (\|M\|+1/k)$.)
|
2025-03-21T14:48:29.759948
| 2020-02-02T06:31:52 |
351743
|
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"Nandakumar R",
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"მამუკა ჯიბლაძე"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/351743"
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|
Stack Exchange
|
Polyhedra that can pack 3-space only in a non-vertex-to-vertex fashion
Question: Are there polyhedral units (convex or otherwise) that can pack 3D space without gaps only such that the arrangement is not vertex-to-vertex?
Same question can be asked with 'edge to edge'.
Note 1: In a vertex to vertex arrangement, any vertex of any unit touches any other unit only at a vertex - and not at an intermediate point of an edge or a face. In an edge to edge arrangement, any edge of any unit cannot touch another unit at an interior point of a face.
It appears unlikely that there are convex polyhedral units that pack 3-space only such that some vertices touch another unit at an interior point of a face.
Note 2: In 2D, at least some of the pentagons that tile the plane (discovered by Marjorie Rice) appear to tile only in non-vertex-to-vertex manner.
Do rectangular prisms of the tilings alluded to in Note 2 work?
Thank you very much. Yes, they should work as examples of polyhedral which pack perfectly only when an edge of a unit touches the interior of faces of another unit. But are there non-prismatic polyhedrons that pack perfectly and have this property (Prisms are 2D in some sense, aren't they?)? And one still suspects that there probably are no convex polyhedrons that are perfect packers only when some vertices touch an interior point of a face of another unit.
It seems that there are convex polyhedra which tile 3D space, for which any such tiling must have a vertex of some tile lying on the interior of a face of another tile.
Consider the polyhedron which is the convex hull of the following points:
$$(0,0,4), (2,0,2), (2,0,-2), (0,0,-4), (-2,0,-2), (-2,0,2), (3,\sqrt{3},0), (-3,\sqrt{3},0), (0,2\sqrt{3},2), (0,2\sqrt{3},-2)$$
This polyhedron can tile an unbounded hexagonal prism: Place three copies of the tile by rotating the shape defined above by 0, 120, and 240 degrees around the edge through $(0,2\sqrt{3},2)$ and $(0,2\sqrt{3},-2)$. Then take a copy of those three tiles, rotate them by 60 degrees around the same line and shift by 4 in the $z$ direction. These six tiles can then be repeated indefinitely by shifting by 8 in the $z$ direction.
So far, everything is vertex-to-vertex. But notice that there are regularly spaced vertices lying in the middle of each side of each hexagonal prism, and the location of these are offset between adjacent sides of the prism. We can put two hexagonal prisms together aligning these points vertex-to-vertex, but then the prism which touches those two cannot be placed to have all the vertices line up.
Of course, I have not proven that there is not some other way to tile 3D space with this tile. But I suspect that the only tilings with this tile involve forming hexagonal prisms, in which case the result holds. Note also that the polyhedron I described can be stretched by an arbitrary factor in the $z$ direction to form a family of examples of this form, in case the parameters I chose above happen to admit some surprising alternate tiling.
Assume "polyhedron" is being understood simply as a (numerically) finite intersection of halfspaces.
Then you well could tile 3D space by the "polyhedron" 1, which is just the upper halfspace $(0,0,1)\cdot (x,y,z)\ge 0$, and the "polyhedron" 2, which is just the lower halfspace $(0,0,-1)\cdot (x,y,z)\ge 0$.
This tesselation then obviously isn't vertex to vertex, as neither of those "polyhedra" have any vertices at all. Moreover both used "polyhedra" are clearly convex each. (It's just that those are not compact.)
Btw., by the same argument that very tesselation is not edge to edge either, as neither of those "polyhedra" have any edges at all.
--- rk
Are not these strictly speaking vertex-to-vertex and edge-to-edge? I mean, since there are no instances of the condition to check, the condition is trivially satisfied.
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2025-03-21T14:48:29.760210
| 2020-02-02T07:21:23 |
351744
|
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|
Stack Exchange
|
Over derivations with an inusual property
Let $A$ be a ring finitely generated over $\mathbb{Z}$. Let $D$ be a derivation on $A$. Let $D^p$ be the composition of $D$ with itself, $p$ times. We suppose that $D^p(x)$ belongs to the ideal $pA$, for every prime number $p$. Moreover, we suppose that there exist $T$ in $A$ such that $D(T)=1$.
Let $B$ be a subring of $A$ such that $B$ contains $T$. Let $K(B)$ be the field of fractions of $B$. We suppose that $K(A)|K(B)$ is algebraic. My question is if $D$ becomes an operator over $K(B)$ for a arbitrary $B$ with this property. It is $D(K(B))\subset K(B)$. Is posible that there exist a derivation whit the property of my question?
I do that question because if $K$ is a field of function of some component of $A/pA$. Let $L$ be the space of solutions of $D(x)=0$ in characteristic $p$. Then the condition over $D$ imply that $K|L$ is a purelly inseparable and $K=L(T)$. So if $K_1$ is such that $K|K_1$ is separable then $K_1=(L\cap K_1)(T)$. Therefore $D(K_1)$ is in $K_1$. Using the Nakayama's lemma $D(K_1)\subset K_1$ for every field such that $K(A)|K_1$ be algebraic and $T$ belongs to $K_1$. Therefore I have a derivation with that property.
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2025-03-21T14:48:29.760316
| 2020-02-02T09:24:16 |
351747
|
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|
Stack Exchange
|
Real number which is different from all rationals
By diagonalization, it is possible to construct a real number $r \in [0,1]$ such that for every rational $q \in [0,1]$, there exists an index $i \in \mathbb{N}$ such that $r_i \neq q_i$ (where $x_i$ is the $i$'ith digit in $x$).
Can we make a stronger claim, and construct a real number $r \in [0,1]$ such that for every rational $q \in [0,1]$ there exists an index $i \in \mathbb{N}$ such that for every $j \geq i$, $r_j \neq q_j$?
This is not a research-level question, and as such it is off-topic for this site. It might be more appropriate for https://math.stackexchange.com . Anyway, a hint: think about $q=k/9$ (assuming decimal digits).
Sorry! I didn't know this website is for research level questions. Should I delete the question?
No. Consider the rational numbers 0, 0.111..., 0.222..., 0.333.., ..., 0.888...
Any real number either shares infinitely many digits with one of these, or it has only finitely many digits which are not 9. But then it is after some point only 9s, so is equal to a terminating rational and has infinitely many 0s.
If you prefer not to allow this equivalence, add 0.999... to the above list.
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2025-03-21T14:48:29.760886
| 2020-02-02T09:46:25 |
351750
|
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|
Stack Exchange
|
Representing a nonlinear elliptic PDE as an energy minimization problem
I need to solve a PDE in 2D representing a (time-independent) nonlinear diffusion process. The unknown function is $\phi(x,y)$ and its gradients create fluxes $\vec J$ through a nonlinear relation:
$$\vec J = (\phi+h)^\alpha\frac{\vec\nabla\phi}{\sqrt{|\vec \nabla\phi|}} \ .$$
The PDE to be solved (representing steady-state flow) is
$$
\operatorname{div}\vec J = \partial_x J_x+\partial_y J_y =0 \ .
$$
$h(x,y)$ is a given function and the values of $\phi$ are known on the boundary of the domain (Dirichlet problem).
For various reasons, both numerical and analytic, I want to represent the problem as an energy-minimization problem. That is, to find a functional $\mathcal{L}(\phi, \vec\nabla\phi)$ whose minimizer is a solution to the PDE (or yet in other words: a functional whose Euler-Lagrange equation is the PDE).
Is there a general way to do this, or to prove that it is impossible? Without the $(\phi+h)^\alpha$ term I can easily guess the functional, but I couldn't make it work with it.
As already said by Kosh, you are trying to solve an inverse problem in the calculus of variation: in its classical formulation, given a system of PDE, the problem consists in finding a functional whose Euler-Lagrange equations are exactly the given PDE(s).
To my knowledge, a necessary and sufficient condition for a system of PDE (jointly with its boundary/Cauchy conditions) to be the Euler-Lagrange equation(s) of a Lagrangian functional (meaning functional which is the integral of a function whose arguments, apart from a "independent" variable $\mathbf{x}$, are the unknown function $\phi$ and its lower order derivatives) is still unknown.
However, it is possible to approach the problem from a more general yet concrete point of view: consider a general operator
$$
\phi\mapsto\mathsf{N}(\phi)(\mathbf x)-f(\mathbf{x}),\label{op}\tag{OP}
$$
where
$\mathsf N$ is a linear or nonlinear operator (possibly your differential operator, an integral operator, a system of PDEs etc.),
$\phi\in\operatorname{Dom}(\mathsf N)$, the domain of $\mathsf N$ in an appropriate function space, and
$f\in\operatorname{Range}(\mathsf N)$ the range of $\mathsf N$ again in an appropriate function space.
The kernel of this operator is characterized by the following equation
$$
\mathsf N(\phi)(\mathbf x)=f(\mathbf{x}).\label{1}\tag{1}
$$
If there exists a general functional $F$ such that its functional derivative vanish on the set of solution(s) of \eqref{1}, i.e.
$$
\left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}F(\phi+\varepsilon\psi)\right|_{\varepsilon =0}\!\!\!=0\;\;\,\forall\psi \in \operatorname{Dom}(\mathsf N)\iff \phi(\mathbf x)\text{ is a solution of \eqref{1},}\label{gvf}\tag{GVF}
$$
the problem \eqref{1} is said to admit an extended variational formulation and the operator \eqref{op} is said to be a potential operator (or conservative operator according to [2] §3 p. 152).
Enzo Tonti (see [1], chapter III §11.1 p. 94 and chapter IV §§17.1-17.4, pp. 168-162 and the recent paper [2] or the Author's web site, where a brief survey of the result is given) proved that, under fairly general conditions, a such a functional $F$ always exists. Precisely, Tonti's theorem says that every linear / nonlinear problem \eqref{1}, provided some natural hypotheses are assumed, admits an extended variational formulation. The proof is constructive in that it shows that it is possible to explicitly construct a compact self-adjoint invertible linear operator $\mathsf K$ such that the functional
$$
F(\phi)=\int\limits_\Omega \mathsf N(\phi)\mathsf K\big(\mathsf N(\phi)-f\big)\mathrm{d}\mathbf{x}\label{2}\tag{2}
$$
satisfies \eqref{gvf}: moreover, if $\mathsf{K}$ is also a positive definite operator (i.e. $\langle v, \mathsf{K}v\rangle>0$ for all functions $v\in \operatorname{Dom}(\mathsf K)$ such that $v\not\equiv 0$), the the solutions of \eqref{1} are the minimum points of the functional \eqref{2} ([2] §4 theorem 2 pp. 155-156). Thus applying the techniques described in [1] and [2] you can surely find variational formulation for your problem, even if it cannot be expressed as the Euler-Lagrange equation(s) of an appropriate energy functional.
Notes
When $\phi, f$ belong to function spaces on sufficiently regular domains $\Omega$, then the Green's operator associated to the Green's function for a any symmetric linear PDE on $\Omega$ can be chosen as the needed $\mathsf{K}$ operator. When $\Omega$ is not regular, there are nevertheless other possible choices: some of them are described in §5, pp. 156-161 of [2].
The analytical approach to the inverse problem of the calculus of variations is perhaps comprehensively described in the wonderful monograph [1], which is however up to date up to 1989: on the other hand [2] surveys nicely the theory of potential operator up to the more recent development due to the Author.
In remark 17.1 of [1] chapter IV, §17.2 pp. 171-172, Filippov gives a brief but very interesting survey on the conditions of potentiality for a nonlinear differential operator (ordinary or partial): perhaps, if something more tailored to the form of a given equation respect to \eqref{2} is needed, it would be worth to have a look at the references cited there.
References
[1] Filippov, Vladimir Mikhailovich, Variational principles for nonpotential operators. With an appendix by the author and V. M. Savchin, Transl. from the Russian by J. R. Schulenberger. Transl. ed. by Ben Silver, Translations of Mathematical Monographs, 77. Providence, RI: American Mathematical Society (AMS). pp. xiii+239 (1989), ISBN: 0-8218-4529-2. MR1013998, ZBL0682.35006.
[2] Tonti, Enzo, "Extended variational formulation", Vestnik Rossiĭskogo Universiteta Druzhby Narodov, Seriya Matematika 2, No. 2, 148-162 (1995). ZBL0965.35036.
In your equation (2), the reduction of differentiability on $\phi$ that would emerge in classical calculus of variation through integration by parts is not there (say $N(\phi)\equiv \phi_{xx}$ then $I=1/2\int (\phi_x)^2$). What are the consequences in terms of constructing approximate solutions?
@pluton I'm really not so able in the direct methods in the Calculus of variation to give you a definitive answer. Nevertheless I can guess that possibly this does imply not much in terms of the smoothness of the approximating functions (which are usually local splines in a finite elements approach) while this implies that you get smoother solutions respect to the cases where the operator is symmetric or can be made so by an integrating factor and you can reduce the order of differentiability. Thus possibly you have a higher regularity.
@pluton but again I can be wrong. Moreover much depends on the structure of $\mathsf K$: it can assume the structure of the Green's function of an elliptic operator, and thus its interactions with $\mathsf N$ can be tricky and cause a reduction in the order of differentiability similar to the one happening when integration by part is possible.
In general, the problem can be complicated because some nonlinearities arise as Lagrange multipliers. Therefore, an energy functional (whose Euler-Lagrange equations coincide with the nonlinear PDE) can exist provided that some constraints to the energy functional are imposed.
Anyhow, the general problem is called The Inverse Problem of the Calculus of Variations.
There is a quite recent book on that. You can take a look at it.
https://www.springer.com/gp/book/9789462391086
|
2025-03-21T14:48:29.761391
| 2020-02-02T09:54:48 |
351751
|
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|
Stack Exchange
|
Differential forms on standard simplices via Whitney extension vs diffeological structure
The standard simplices $\Delta^n \subset \{\mathbf{x}\in\mathbb{R}^{n+1}\mid x_0 + \ldots + x_n =1 \} =: \mathbb{A}^n$ carry two natural sorts of smooth differential forms:
Those differential forms on the interior of $\Delta^n$ that extend smoothly to a neighbourhood of $\Delta^n$ in $\mathbb{A}^n$ (this definition is used for instance by Dupont in his book Curvature and Characteristic Classes). This is implicitly using Whitney's extension theorem, I think.
Consider $\Delta^n$ as a diffeological subspace of $\mathbb{A}^n$, where a function $\phi\colon \mathbb{R}^k \to \Delta^n$ is a plot if and only if the composite $\mathbb{R}^k \to \Delta^n \hookrightarrow \mathbb{A}^n$ is smooth. A differential form $\omega$ on $\Delta^n$ then consists of the data of a differential form $\phi^*\omega$ for each plot $\phi$ with a compatibility condition when a plot factors through another via a smooth map between Euclidean spaces.
The first of these is more of a "maps out" viewpoint, and probably corresponds to a natural smooth space structure defined via smooth real-valued functions. The second is a "maps in" viewpoint. Note that the $D$-topology arising from the diffeology in 2. above is the standard topology on the simplex. We then get a cochain complex of differential forms of each type, as exterior differentiation can be defined in the more-or-less obvious way in each case.
My question is: how do these relate? Is one a subcomplex of the other? Or are they quasi-isomorphic, via a third cochain complex?
The motivation is that Dupont's simplicial differential forms on semisimplicial manifolds $X_\bullet$ look like they should be differential forms on the fat geometric realisation considered as a diffeological space, since a simplicial differential form is more or less descent data for the sheaf of differential forms and the 'cover' $\coprod_{n\geq 0} \Delta^n\times X_n \to ||X_\bullet||$, assuming the first definition above. However, if his differential forms on $\Delta^n$ (or more precisely on $\Delta^n\times X_n$) aren't diffeological differential forms, they don't give a form on $||X_\bullet||$ as a diffeological space. I guess all we really need is a map of cochain complexes from the first to the second given above.
The two chain complexes are isomorphic for any $n≥0$.
Fix some $n≥0$, $k≥0$ and consider $k$-forms on the $n$-simplex.
I will use the notations $Ω_e^k$ and $Ω_d^k$ for forms of type 1 and 2 respectively.
I also use the notation $Δ_d^n$ for the diffeological $n$-simplex
so that $\def\Hom{\mathop{\rm Hom}} \def\R{{\bf R}} \def\d{\,{\rm d}} Ω_d^k=\Hom(Δ_d^n,Ω^k)$.
First, we have a canonical map $ι\colon Ω_e^k→Ω_d^k$ that
sends $ω∈Ω_e^n$ to the map $Δ_d^n→Ω^k$
that sends a smooth map $φ\colon S→Δ_d^n$ to the $k$-form $φ^*ω∈Ω^k(S)$.
Secondly, the map $ι$ is injective,
which follows from the following two observations.
By the Yoneda lemma the restriction of $ι(ω)\colon Δ_d^n→Ω^k$ to the sheaf represented by the open interior of $Δ_d^n$ equals the restriction of $ω$ to the open interior of $Δ^n$.
Furthermore, any two forms in $Ω_e^k$ whose restrictions to the open interior of $Δ^n$ coincide must be equal (by continuity).
Thirdly, the map $ι$ is surjective.
This can be shown using a two-step construction:
in the first step we construct a certain element $ω$ of $Ω_e^k$
starting from some given element $ψ$ of $Ω_d^k$
and in the second step we show that $ι(ω)=ψ$.
First, let's briefly examine the simplest nontrivial case $d=n=1$.
We have a form $ψ∈Ω_d^1(Δ_d^1)=Ω_d^1([0,1])$,
and we already know its restriction $f(x) \d x$ to the open interval $(0,1)$.
In particular, $f$ is a smooth function on $(0,1)$.
Pulling back $ψ$ along the plot $\R→\R$ ($x↦x^2$)
yields some 1-form $g(x) \d x$, where for any $x≠0$ we have $g(x)=2xf(x^2)$
and $g$ is a smooth function on $(-1,1)$.
Since $g$ is odd, we have $g(0)=0$,
so the even function $h$ given by $h(x)=g(x)/(2x)$ is smooth on $(-1,1)$
and we have $h(x)=f(x^2)$ for all $x≠0$.
Set $f(0)=h(0)$.
We claim $f$ is smooth on $(-1,1)$.
Indeed, $h'(x)=2xf'(x^2)$ for $x≠0$,
and since $h'$ is odd, we have $h'(0)=0$
and $h'(x)/(2x)$ is a smooth function on $(-1,1)$
whose value at $x=0$ is $f'(0)$.
We repeat this step by induction, proving $f$ has derivatives of all orders at 0.
The general case is nothing else than a multivariable paremetrized
version of the above argument.
More precisely, we argue as follows.
For the first step, suppose we are given some $ψ∈Ω_d^k$.
Given some point $x∈Δ^n$, we would like to define $ω(x)$.
Denote by $d≥0$ the codimension of the stratum of $Δ^n$ that contains $x$.
Parametrize the simplex $Δ^n$ using $n$ coordinates
in such a way that $x_1=⋯=x_d=0$
and the other $n-d$ coordinates of $x$ are strictly positive,
so that the points $y$ of the stratum containing $x$ are defined by the relations $y_i≥0$ for $1≤i≤d$.
Decompose $$ω=∑_I g_I ∏_{i∈I} \d x_i$$ into its individual components with respect to this coordinate system.
Pick one such component $g_I ∏_{i∈I} \d x_i$; we would like to define $g_I$ as a smooth function on some open neighborhood $U$ of the given stratum of $Δ^n$, which is parametrized by the remaining $n-d$ coordinates $x_{d+1}$, …, $x_n$.
Consider the smooth map $U→Δ^n$ that (in the coordinates introduced above)
sends $x_i↦x_i^2$ for $1≤i≤d$ and $x_i↦x_i$ for $i>d$.
This defines a morphism $τ\colon U→Δ_d^n$, so we have a form $τ^*ψ∈Ω^k(U)$.
Now take the coefficient $h$ of $τ^*ψ$ before $∏_{i∈I} dx_i$.
Take the partial derivative of $h$ with respect to all coordinates $x_i$
such that $i∈I$ and $i≤d$.
Divide the resulting function by $2^{\#(I∩\{1,…,d\})}$.
This is the function $g_I$.
This argument also proves that the resulting $k$-form $ω$ is smooth.
For the second step, we have to show that $ι(ω)=ψ$.
Reusing the notation of the previous paragraph,
consider some arbitrary plot $φ\colon S→Δ_d^n$ such that $φ(s)=x$ for some $s∈S$.
The first $d$ coordinates of $φ$ must be nonnegative in a neighborhood $U$ of $s$.
Since $φ$ is differentiable, the first derivatives of these $d$ coordinates must vanish at the point $s$.
Thus, taking the square root of each of the first $d$ coordinates produces a smooth map $ε\colon U→\R^n$ such that $φ=τε$, where the map $τ$ was defined in the previous paragraph.
Now $φ^*ι(ω) = ε^* τ^* ι(ω) = ε^* τ^*ψ = φ^*ψ$,
where $τ^*ι(ω)=τ^*ψ$ by definition of $ω$.
Cool, thanks! I'm glad this works out so smoothly (heh). So Dupont's construction really gives a connection on the universal bundle considered as a diffeological object, nice.
|
2025-03-21T14:48:29.761879
| 2020-02-02T10:31:36 |
351753
|
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|
Stack Exchange
|
What do you call an object constructed as part of a proof?
I find myself wanting to talk about parts of a proof, e.g. the role played by mathematical expressions within a proof.
When proving a theorem it is common to construct some kind of object and then prove this has certain properties. E.g. in the standard proof that there are infinitely many primes we assume the primes are finite $p_1,\cdots, p_k$ and the consider $n=p_1\times \cdots \times p_k +1$. We then establish properties of $n$ which lead to a contradiction. What do you call the act of constructing $n$, or $n$ itself?
I think that having names for things is important, especially when talking about them. For example, many simple proofs by contradiction could be a contrapositive instead. Having a word "contrapositive" is very helpful indeed in discussing such proofs, and explaining to students the difference between contradicting the hypothesis, and a general external contradiction such as $1=0$.
The word "ansatz" is widely used "is an educated guess or an additional assumption made to help solve a problem, and which is later verified to be part of the solution by its results" (Wikipedia). I find having a word for an ansatz is very helpful.
I don't have a word for "a particular object constructed as a device within a proof, built to establish certain conditions must hold".
I have lots of other examples of such objects, but no name for them.
My favourite proposal at the moment is "gadget", or "proof-gadget" for emphasis. This is a relatively positive, utilitarian word (not used elsewhere in mathematics as far as i can tell). Gadgets are something which a built for a particular purpose, and are often igneous.
I can, of course, use words as I see fit (The Humpty Dumpty defence) but I'd like to ask if anyone knows of words used routinely for this purpose?
Thanks,
Chris
One can state the infinitude of primes as: given the primes $p_1,\ldots, p_n$, the prime divisors of $p_1\cdots p_n +1$ are not among the given primes. And these new primes are what you have constructed. There are many theorems in the literature that are given as mere existence statements, but which could be restated as a construction together with a property of the constructed object. One word for this is 'proof relevant mathematics', where the content of the proof is meaningful and relevant outside the proof environment.
I think the word scholium might be partially relevant here. I seem to recall Peter Johnstone uses it to denote something like a corollary, but which follows from something in a previous proof, not from the statement of the result. (Edit: found it: footnote 7 on page xiv, in the Preface to Sketches of an Elephant)
Proclus' commentary on Euclid's Elements split every theorem and proof into six parts (see e.g. https://www.jstor.org/stable/639502): enunciation, setting out, definition of goal, construction, proof, conclusion. Typically the construction involves many auxiliary lines and figures. So, if pretentiousness is not a problem, how about κατασκευή? See for instance the fourth sense 'a device, a trick' here: https://en.wiktionary.org/wiki/κατασκευή.
"gadget" has a specific meaning in computational complexity proofs.
Interesting, thanks Brendan. Could you post a DOI to a computational complexity proof using the term?
@DavidRoberts: I don't think I would refer to the "consideration" of the number $p_{1} \cdots p_{n} +1$ in the proof of Euclid IX-20 as a scholium: https://mathoverflow.net/a/261056/1593
@JoséHdz.Stgo. yes, but one might record as a scholium the (trivial) observation that the procedure $(n_1,\ldots,n_k) \mapsto 1+\prod_{j=1}^k n_j$ results in a number that is never $0$ modulo each $n_j$.
@ChrisSangwin: When the construction is not well motivated, people resort to the term "deus ex machina" sometimes...
It is unfortunate that the "standard" proof of the infinitude of primes assumes only finitely many exist and deduces a contradiction. The way Euclid did it is better: Assume $P$ is any finite set of primes (e.g. ${,5,7,}$) and show that the prime factors of $1+\prod P$ are not in $P$ (e.g. $1 + (5\times7) = 2\times2\times3\times3,$ so that there are more primes than those in $P.$ Making it a proof by contradiction adds an extra complication that serves no purpose and confuses some students, as follows:$,\ldots\qquad$
$\ldots,$A confused student may think that if $p_1,\ldots,p_n$ are the smallest $n$ primes, then $p_1\times\cdots\times p_n+1$ has been shown to be prime in every instance. That is false: $(2\times3\times5\times7\times11\times13)+1 = 59\times509.$ Then when a student learns of such counterexamples, they think that that shows Euclid's proof is erroneous. But Euclid's proof is in fact sound.
I suspect that the erroneous belief that Euclid's proof is by contradiction originated with Dirichlet's book on number theory, published in 1859.
@MichaelHardy your comments are not relevant to this question and getting into an argument about Euclid's proof here is just a distraction.
@MichaelHardy I'm not sure I understand, although I read this opinion multiple times: if you're not arguing by contradiction, how do you conclude that there are infinitely many primes from the fact that you can generate, from any given finite set of primes, a number having divisors outside the set?
@AlessandroDellaCorte It’s a proof by negation, not by contradiction. I agree with Sam that a discussion about Euclid’s proof here is off-topic.
Igneous?? Ingenious, surely.
@SamHopkins : Chris Sangwin gave great prominence to Euclid's proof in his question. And in so doing, he misled people (in just the way so many mathematicians do about this topic).
@AlessandroDellaCorte : I don't understand your comment. If every finite set of prime numbers can be extended to a larger finite set of prime numbers, then the list of prime numbers keeps on going.
@MichaelHardy …contradicting the statement that that list contained every prime number, no? It seems to me that your correct remark about the fact that $n$ is not necessarily prime but just has a prime factor other than $p_1,\dots,p_k$ has little to do with the proof being by contradiction or not.
@AlessandroDellaCorte : Only the assumption that the primes one starts with are the only primes that exist can lead authors to write that $1+\prod P$ is not divisible by any primes and "is therefore itself prime" (quoting G. H. Hardy in A Course of Pure Mathematics, Cambridge University Press, 1908, pages 122–3). Thus that error results from that assumption, and that assumption is absent when one doesn't assume at the outset that there are only finitely many primes.
@MichaelHardy I took a closer look at Euclid's original statement and proof. It seems to me that whether the proof is by contradiction or not depends on how you translate the claim in modern mathematical language. If you say "no finite set of primes contains every prime" then it is by contradiction indeed. If you say "given any finite set of primes, there exists a prime which is not in the set", then it's not. Probably the latter makes more sense, so you're right. But in any case this has little to do with the fact that the "new" prime is $n$ itself (wrong) or one of its factors (correct).
@AlessandroDellaCorte : I don't actually understand the difference in meaning between the statement that "no finite set of primes contains every prime" and the statement that "given any finite set of primes, there exists a prime which is not in the set". $$\begin{align} & \lnot\exists\text{ finite set } S ,,, \forall\text{ prime } n ,,, n\in S. \ {} \ & ,,,\forall\text{ finite set } S,,, \exists\text{ prime }n,,, n\notin S. \end{align}$$ One translation says "Prime numbers are more than any proposed multitude of prime numbers" or something close to that.
Two kinds of "objects" popping up in proofs come to my mind:
One defines object X and then proves that X has some required properties. In this case I'd call X a "candidate" when introducing it, and I think this is not uncommon.
One assumes something and then defines the object Y (as the number $n$ in your example) to disprove it. In this case I'd call Y just a "counterexample", no? In your case, $n$ is a counterexample to the claim that the prime divisors of any number are among $p_1,\dots,p_k$.
I'm not saying that those are the only cases of "objects in proofs", but perhaps they're the most important ones, and I can't immediately point at cases that are really so different from 1. or 2.
This question was asked four years ago, but never got an answer (though, it had some discussion in the comments). This question could have also been asked on academia.SE but it's not clear it's really on topic there either. Let me try to answer so that this doesn't linger forever on the unanswered queue.
With any question about writing, in addition to the guiding principle of not inventing new terminology for something that already has standard terminology, another good guiding principle is do what's best for your reader. Having prepared thousands of pages of writings in my research and teaching, I've found that what's best for the reader is usually whatever is clearest. Hence, when I make a construction in one of my papers and want to refer to it elsewhere in the paper, I give it a numbered environment, just like an equation, definition, lemma, proposition, theorem, conjecture, etc.
Construction 2.1: Under the assumption that there are only finitely many primes $p_1, \dots, p_k$, let $P = p_1p_2\cdots p_k + 1$.
Later in the paper, maybe I need to do the same trick again, and can say "Using the same technique as Construction 2.1, we now ..."
I've also had things like:
Standing Hypothesis 1.1: We assume all model categories are cofibrantly generated in this paper.
or
Agreement 2.1: Let us agree that by "operad" we mean "reduced operad" in what follows.
I think this is MUCH clearer than words like Scholium and Ansatz, that have the potential to throw off readers who have not seen words like that before. I teach so many students from Asia for whom English is already a second language, and words that draw from yet a third language tend to throw them off quite a lot.
Now suppose you're teaching a course on proof-writing, or writing a paper about how we write mathematics, and want a way to refer to the general idea of a construction done inside a proof. In that case, I think "gadget" is a good choice, even though it has a technical meaning elsewhere, because most students/readers know what a gadget is, especially if I mention the real-world meaning of the term and why I chose that word for this concept. Based on the comments, it sounds like there is no standard term. If you want to avoid "gadget" because of the connection to computational complexity, you can use "gizmo" instead.
Lastly, I want to point out that one way to avoid gizmos in proofs is to create lemmas, following Terry Tao's advice. You could imagine doing all your writing in such a way that, wherever you were constructing something inside a proof, you deliberately pulled that out into a lemma like:
Lemma 3.1: If $L = \{p_1, \dots, p_k\}$ is a finite list of prime numbers, then $P = p_1\cdots p_k + 1$ is divisible by a prime not in $L$.
This technique of writing is extremely clear and helps the reader focus on one proof at a time, instead of a proof within a proof. Incidentally, this gives me another idea. You could use "inception" to refer to the phenomenon of proofs/constructions within existing proofs of other statements.
I hope you wouldn't put Lemma 3.1 in a paper. I guess it's true, since there is no such list, but I'd call that a misleading kind of truth since it tempts so much to the analogous false statement where "finite list containing all prime numbers" is replaced by "finite list containing only prime numbers". By contrast, "… then $p_1\dotsm p_k + 1$ is divisible by a prime not on the list" leads to no such confusion.
It is unfortunate that the "standard" proof of the infinitude of primes assumes only finitely many exist and deduces a contradiction. The way Euclid did it is better: Assume $P$ is any finite set of primes (e.g. ${,5,7,}$) and show that the prime factors of $1+\prod P$ are not in $P$ (e.g. $1+(5×7)=2×2×3×3,$ so that there are more primes than those in $P.$ Making it a proof by contradiction adds an extra complication that serves no purpose and confuses some students, as follows:$,\ldots\qquad$
$\ldots,$A confused student may think that if $p_1,\ldots,p_n$ are the smallest $n$ primes, then $p_1\times\cdots\times p_n+1$ has been shown to be prime in every instance. That is false: $(2\times3\times5\times7\times11\times13)+1 = 59\times509.$ Then when a student learns of such counterexamples, they think that that shows Euclid's proof is erroneous. But Euclid's proof is in fact sound.
David, Might Lemma 3.1 be clearer as: If $p_1,\ldots,p_k$ is a finite list of prime numbers, then $P=p_1\cdots p_k+1$ is divisible by a prime that is not in the list. From there, most students would agree that there can't be a finite list that contains every prime number.
The purpose of this answer was to focus on how to write in general, not on this specific lemma, that I only chose because it was in the original post. Since the choice seems to have generated heat, I rephrased it as suggested by @JoeSilverman. Thanks. I know that some people have very strong feelings about this proof regarding the infinitude of primes, but I encourage readers to focus on the main point of the answer, not on the example. As a topologist, I would have preferred an example of proof writing in topology, but trying to align with the OP here.
@MichaelHardy Let's not get off topic. The question is about proofs within proofs, not about the infinitude of primes. The OP only used this proof because it's well known. Let's avoid digressing into opinions about the proof itself, and instead focus on the question, which involves math writing.
I agree that "gadget" is a fine word, and the fact that it has a technical meaning in one subfield isn't such a big deal: the list of words that have technical meanings is very long. I also agree that the discussions of Euclid's proof in the comments are off topic here.
Along the lines of "gizmo", there's "whatsis", "thingamabob", "whatchamacallit", and "doohickey". Can't wait to see this in a math paper: "Thingamabob 3.1 follows immediately from Doohickey 2.7".
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2025-03-21T14:48:29.762977
| 2020-02-02T11:20:38 |
351756
|
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|
Stack Exchange
|
Identifying the edges that are essential for biconnectivity
Question:
If $G(V,E)$ is a biconnected symmetric graph, is it possible to identify the edges, whose deletion destroys biconnectivity, in the following way:
determine the union $B:= ST\cup F_1$ of a spanning tree and a maximal forest, that is edge-disjoint with $ST$
determine a second maximal forest $F_2$ that has no edges in common with $B$.
take as the critical edges those edges of $B$, that are adjacent to a vertex of degree 2 or connect different trees of $F_2$.
I found several counter-examples, but found the flaw in each before posting! So maybe the answer is yes. But why do you suspect this?
@AaronMeyerowitz the idea is essentially based on the observation, that removing any of the essential edges generates cut vertices and that a "pure" cut vertex doesn't harm the connectivity of a tree whereas the removed edges is essential for biconnectivity and separates a pair of trees in $F_2$; that is essentially the basis of the naive idea.
I suspect that the sketched reasons also generalize to detecting the edges that are essential for $k$ connectivity, i.e. of any edge, whose deletion destroys $k$ connectivity.
This seems like a counterexample: take 3 cliques $C_1$, $C_2$ and $C_3$, each on say $5$ vertices. Connect $C_1$ and $C_2$ with three vertex-disjoint edges, and connect $C_2$ and $C_3$ with 2 edges. Now it is possible to choose $ST$ to be a spanning tree going from $C_3$ to $C_2$ to $C_1$ and back to $C_2$, and $F_1$ a disjoint spanning tree going from $C_1$ to $C_2$ to $C_3$, so that the following holds: the remaining graph has three connected components, $F_2$ has a tree in every component, and $B$ connects the trees in $C_1$ and $C_2$. Hence the procedure will return an edge between $C_1$ and $C_2$, which is not critical.
That looks good. But $C_3$ is irrelevant and I am not sure $5$ is enough.
Without $C_3$ the graph would be 3-connected. And it seemed to work with 5-cliques, but you might as well take larger ones.
Given a bi-connected graph $G$, say an edge $e$ is destructive if $G-e$ has a cut-vertex. Say that $e$ is $(T,F_1,F_2)$-critical if there is a spanning tree $T$ and forests $F_1,F_2$ as in you description with respect to which $e$ critical.
Must a $(T,F_1,F_2)$-critical edge be destructive? The answer is no. Here is a example based on the answer given by @smapers.
Here is a $14$ point graph with $38$ edges. If desired, add another $4$ edges on the left and $3$ on the right to make each a $K_7$. Either way, there are no destructive edges. The tree $T$ is the green path, The forest $F_1$ is the red tree And $F_2$ consists of two blue paths. Then the central red edge is critical but not destructive.
If one wants $G$ to have some destructive edges and be exactly bi-connected then, as suggested, add some new vertices and edges making a $K_5$, connect it to the $K_7$ on the left two new edges which will become part of $T$ (which will lose an edge elsewhere to avoid a cycle) and then add enough edges to make $F_1$ and $F_2$ valid.
This illustrates:
Do the $(T,F_1,F_2)$-critical edges include all the destructive edges? Again, no. One can certainly arrange to have some of the destructive edges in $T.$
If $e$ is destructive is there some way to choose the tree and forests so that it is critical? Yes. Assign $e$ to $F_1$ with the rest to be determined. Now $G-e$ is connected with a cut-vertex $v$. Start with all the edges on $v$ and extend that to a spanning tree $T$. Now $G-e-T$ is disconnected with the vertices of $e$ in distinct connected components. So $F_2$ will have those two vertices in distinct distinct trees. How ever we form $F_1$ (using $e$ ) and $F_2$, The destructive edge $e$ is critical.
Note that some of the trees in $F_2$ might be isolated vertices.
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2025-03-21T14:48:29.763284
| 2020-02-02T11:35:42 |
351757
|
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|
Stack Exchange
|
Smooth integers with lower bound on $\omega(n)$
Define $(b,c)$-smooth integers to be integers having all prime factors bigger than $c$ and smaller than $b$.
Probability a number is $(b,1)$-smooth is governed by the Dickman function while probability of number of prime divisors being at least $d$ is governed by the Erdos-Kac formulation.
Is there a known distribution for $(b,c)$-smooth integers?
If we want the joint distribution of $(b,c)$-smooth integers with at least $d$ prime divisors then is there a formulation?
The reason I am asking is it seems we cannot have independence of smoothness and number of prime divisors since it quite does not make sense if the number of divisors is between $\log\log n+0.1\sqrt{\log\log n}$ and $\log\log n+0.2\sqrt{\log\log n}$ then the number cannot be $(b,1)$-smooth where $b=o(f(n))$ at a suitable function $f(n)$ with probability $1-o(1)$ and I do not know to capture this and identify $f(n)$ with independence assumption and identify involved probabilities.
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2025-03-21T14:48:29.763388
| 2020-02-02T12:09:16 |
351758
|
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|
Stack Exchange
|
Matching book embedding of Cartesian products of graphs
In the book embedding of a graph $G$ , each vertex of $G$ is placed on the spine and each edge is placed in the pages without crossing each other edge. If vertices have degree at most one in each page, the book embedding is matching . The minimum number of pages in which a graph can be matching book embedded is called matching book thickness. For Convenience, we denote the matching book thickness of a graph $G$ by $\mathrm{mbt}(G)$.
For the Cartesian product of a complete graph $K_n$ and a path $P_m$, I want to know $\mathrm{mbt}(K_n\Box P_m)$. Since $\mathrm{mbt}(K_n)=n$, it is not hard to see that $\mathrm{mbt}(K_n\Box P_2)=n+1,$ where $P_2=K_2$. For the case $ K_n\Box P_3$, I guess $\mathrm{mbt}(K_n\Box P_3)=n+2$. But I have no idea about the proof.
I will appreciate it if someone could give any suggestions.
The general problem of matching book thickness for the Cartesian
product of a cycle and a complete graph is addressed in a preprint
which just popped up on RGate from Feb 2, 2020 by Z. Shao, Y. Liu and Z. Li [1] (arXiv link)
It appears that they've answered your question. In fact, however,
there is some additional information which I can provide.
[1] uses the theorem of Shannon Overbay, that dispersibility
implies bipartiteness for a regular graph to obtain the lower
bound which is then achieved by construction. The statements
in the argument suggest that generalizations should be possible.
For Overbay's Theorem, see her thesis [2] at
this link
The matching book thickness terminology is used in these slides "circLayouts.pdf" [3] of mine,
where 'mbt' is determined for a class of circulant graphs.
The reference to "On book embeddings with degree-1 pages"
in the arXiv paper should be replaced by "circLayouts.pdf".
Ref.[3] was done using an older version of Mathematica
by Wolfram Research, Inc. Unfortunately, the code is not
currently functional and so the paper is "legacy" and has
some key misprints: "Regular" was omitted from the conjecture
on p. 5. A follow-up conjecture on p. 9 omitted "Bipartite."
Finally, the figure is for n = 7, not n = 10.
@Kainen Thank you for the answer. I know the theorem of Shannon Overbay, that a regular and dispersalbe graph is bipartitite. It is valid for $K_n\Box C_m$ to obtain the lower bound $\Delta +1$ because their regularity and non-bipartiteness. It is easy to get $mbt(K_n\Box P_3)\leq n+2.$ But for the lower bound, i have no idea how to use Shannon's Theorem here for $K_n\Box P_3$ is not regular. Could you please explain it to me in detail?
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2025-03-21T14:48:29.763592
| 2020-02-02T12:42:38 |
351761
|
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|
Stack Exchange
|
CBB inequality and two comparison triangles / 4-point condition for CBB spaces
Assume $(X,d)$ is a CBB($\kappa$) space with $\kappa > 0$. (That is we can find comparison triangles in the model space $(M_\kappa^2, \bar{d})$ and the reverse of the CAT inequality holds; Definition 5.1.12 in Lipschitz Functions - Ştefan Cobzaş, Radu Miculescu, Adriana Nicolae).
Now my question is the following. Assume there are $4$ points $x,y, z_1, z_2$ in $X$ and we take comparison triangles $\Delta(\bar{x},\bar{y},\bar{z_1})$ and $\Delta(\bar{x},\bar{y},\bar{z_2})$ in $M^2_\kappa$ such that the distance between $\bar{z_1}$ and $\bar{z_2}$ becomes minimal. Under what conditions does it hold that $d(z_1, z_2) \geq \bar{d}(\bar{z_1}, \bar{z_2})$?
I think this follows directly from Kirszbraun theorem in Alexandrov spaces...
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2025-03-21T14:48:29.763703
| 2020-02-02T13:32:06 |
351764
|
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"Bugs Bunny",
"Matthew Daws",
"Nate Eldredge",
"YCor",
"Yemon Choi",
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|
Stack Exchange
|
Empty interior of union of cosets?
The following question arises from trying to understand Lemma 1.3(ii) of arXiv:math/0405063. I believe a particular case of the proof (and in fact I think the proof is essentially equivalent to this claim) is:
Let $G$ be a locally compact group. Let $C,D$ be cosets (not assumed open, closed etc.) each of which has empty interior. Then $C\cup D$ also has empty interior.
This is not try in general topology, of course: let $C,D$ be the rational, respectively, irrationals, in $\mathbb R$. However, I cannot decide if being a coset rules out this sort of example. Is the claim true, and if so, what is a proof?
@NateEldredge I think Matt means Lemma 1.3(ii), which -- FWIW -- the preceding text claims is following/adapting a similar argument given in Rudin's book Fourier Analysis on Groups
For a coset (=right coset, =left coset) in a topological group, having empty interior is equivalent to being non-open.
BTW, Matt, in that bit of the proof, are they not taking cosets whose closures have empty interior? (Not a rhetorical question, I find that bit of the paper a bit hard to follow.)
Corrected. Yes, sorry, that was a typo. I must admit to not having checked Rudin's book; I'll do this tomorrow when I'm in the office... (But my question still stands, I think, unless I am very much mistaken. I am trying to understand how to get the displayed equation at the top of page 10).
This is false. Take $G=(\mathbf{Z}/2\mathbf{Z})^\mathbf{N}$ and let $H$ be a dense subgroup of index 2 (there are many, since $G$ has only countably many closed subgroups of index 2 but has $2^c$ subgroups of index 2). Then $G=H\cup (G-H)$ and both $H,G-H$ have empty interior.
@YemonChoi: I don't think so. As I said, top of page 10. I think you could have $K_i = G$ already, and $N_{1,1}$ and $N_{1,2}$ my $C,D$. The proof then claims that the closure of $G \setminus (C\cup D)$ is all of $G$, which is equivalent to my question.
@YCor: Great! That would do it. Okay, I guess I need to go back to Rudin and try to reconstruct the non-abelian version of his proof...
Matt: a word of warning, if you read M &N's comments about Lemma 1.3(i) they are suggesting that the original arguments in Rudin may be incomplete, see https://mathoverflow.net/questions/232351/a-possible-mistake-in-walter-rudin-fourier-analysis-on-groups
@YemonChoi: Indeed, I think Rudin is also wrong on page 86, for the same reason Ilie and Spronk appear to be wrong.
@YCor Since the question is aimed at clarifying/correcting a gap in an argument in research-level harmonic analysis (and may even be detecting errors in a standard reference, namely Rudin's book) maybe you can post your comment as an answer. As I am somewhat familiar with the Ilie-Spronk paper and the literature that cites it, I should point out that Matt's question seems to be the result of digging rather deeper into the proofs than most people have done during the last 15 years
This is false. Take the (compact abelian) group $G=(\mathbf{Z}/2\mathbf{Z})^\mathbf{N}$ and let $H$ be a dense subgroup of index 2 (there are many, since $G$ has only countably many closed subgroups of index 2 but has $2^c$ subgroups of index $2$, and clearly a subgroup of index 2 is either closed or dense). Then $G=H\cup (G\smallsetminus H)$ and both $H$ and its coset $G\smallsetminus H$ have empty interior.
With an additional assumption that there exists an interior point $x\in C\cup D$ such that $x\in C\cap D$, one can prove this fact for any topological group. It doesn't require local compactness.
Note that $e$ is in the interior of $x^{-1}C\cup x^{-1}D$ and both sets are subgroups. Thus, WLOG, $x=e$ and $C$, $D$ are subgroups.
Now pick an open neighbhourhood of $e$ such that $U\subseteq C\cup D$. Pick another open neighbhourhood of $e$ such that $V^2\subseteq U$.
Now an easy algebraic argument shows that $V$ must by a subset of $C$ or $D$. Otherwise, pick $y\in V\setminus D$ and $z\in V\setminus C$. The product $yz$ must be in $U$, so in $C$ or $D$. Both lead to a contradiction.
A basis of compact open subgroups?? I must be misunderstanding, because that doesn't seem to be true at all for say, $\mathbb{R}$, not even if you meant "relatively compact".
Yes, you are right. This is for totally disconnected only...
I will fix it now...
Fixed. Sorry for the earlier misunderstanding.
So, why are $x^{-1} C$ and $x^{-1} D$ both subgroups? This would hold if $x \in C \cap D$ but I don't see why that would have to be true.
Because they are cosets containing $e$
Setting aside the WLOG for a moment, since I think it's causing confusion: you seem to be claiming that the original cosets $C,D$ must have nonempty intersection. I don't see how that follows from the assumptions.
@NateEldredge seems to have identified a gap in the argument near the start. You start with x that lies in the union of $C \cup D$, and then you claim that $x^{-1}C$ and $x^{-1}D$ are subgroups. This only works if $x$ belongs to $C$ and $x$ belongs to $D$
The edited version doesn't address the original question, and the extra assumption seems much stronger than the setting in which @MatthewDaws originally asked his question.
|
2025-03-21T14:48:29.764073
| 2020-02-02T14:39:00 |
351768
|
{
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"authors": [
"Jochen Wengenroth",
"https://mathoverflow.net/users/21051"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351768"
}
|
Stack Exchange
|
Topology on function spaces for pointwise convergence
Suppose I have some collection of maps $T_\lambda: C^\infty(\Omega)\rightarrow C^\infty(\Omega)$ which are linear and parameterized by a parameter $\lambda >0$. (Perhaps more generally take $C^\infty(\Omega)$ as a Hilbert Space or Banach Space of functions.)
If we have the property that for every $x\in \Omega$ and every $f \in C^\infty(\Omega)$, then
$$
\lim_{\lambda\rightarrow 0} T_\lambda f(x) = Tf(x).
$$
Can I express this in terms of a well-known operator norm on $C^\infty(\Omega)?$ If so, what is the strongest topology in which this is the case?
You can endow $C^\infty(\Omega)$ with the (very weak) locally convex topology of pointwise convergence having the seminorms $p_E(f) \max{|f(x)|: x\in E}$ with finite subsets $E$ of $\Omega$ and the space of continuous linear operators $L(C^\infty(\Omega))$ with the locally convex topology of "simple convergence" having the seminorms $q_{F,E}(T)= \max{p_E(T(f)): f\in F}$ with finite subsets $E$ of $\Omega$ and $F$ of $C^\infty(\Omega)$. I don't think that this is very useful.
|
2025-03-21T14:48:29.764174
| 2020-02-02T15:02:06 |
351769
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/351769"
}
|
Stack Exchange
|
Scattering theory for Coulomb potential
Both physical and mathematical theories of quantum scattering seem to be well developed in the case when the potential (or a more general perturbation of the Laplacian) decays fast enough at infinity and satsifies some regularity assumptions. For example under such rather general assumptions existence of $S$-matrix is proved in Ch. XIV of Hormander's "The Analysis of Linear Partial Differential Operators II".
However the Coulomb potential does not satisfy these assumptions of fast decay at infinity. Moreover my impression is that this is not just a technical issue, but the whole theory should look different for the Coulomb potential.
I am wondering if there is any notion of $S$-matrix for the Coulomb potential and how it looks like.
A reference would be helpful. Apologies if the question is not advanced enough- I am not an expert.
One way to define wave operators in the long range case is to modify the free motion appropriate, see e.g. Lars Hörmander, "The Analysis of Linear Partial Differential Operators IV pp 276-331" , https://link.springer.com/chapter/10.1007/978-3-642-00136-9_7 and J. Dollard, Quantum-mechanical scattering theory for short-range
and Coulomb interactions, Rocky Mountain Journal of Mathematics
1, 5-88 (1971) .
Another standard reference that discusses scattering theory of long range potentials at length is the third volume of Reed-Simon.
The $1/r$ Coulomb potential needs to be regularized, typically this is done by studying the Yukawa potential $e^{-\alpha r}/r$ and taking the limit $\alpha\rightarrow 0$ at the end. A recent critical examination of this procedure can be found in Regularization of the Coulomb scattering problem (2004).
|
2025-03-21T14:48:29.764327
| 2020-02-02T15:26:06 |
351771
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/351771"
}
|
Stack Exchange
|
How to regard negative PSH function with neat analytic singularities as a generalization of Green-type function?
I am reading this paper:A SIMPLIFIED PROOF OF OPTIMAL L2-EXTENSION THEOREM AND EXTENSIONS FROM NON-REDUCED SUBVARIETIES by Hosono. https://arxiv.org/pdf/1910.05782.pdf.
The setting is as follows.Let $\Omega\subset\subset\mathbb C^n$ be a bounded pseudoconvex domain and $V\subset \Omega$ be a closed submanifold.The Green-type function $G$ on $\Omega$ with poles along $V$ is defined by :$G\in PSH(\Omega)$,$G<0$ on $\Omega$,and for some continuous functions $A$ and $B$ on $\Omega$,
$$\log d^2(z,V)+A(z)\leq G(z)\leq \log d^2(z,V)-B(z).$$
In the same time,$\psi$ is a negative PSH function on $\Omega$ with neat analytic singularities,i.e. for each point $x\in\Omega$,there exist a neighborhood $U$ of $x$, a positive number $c>0$,a finite number of holomorphic functions $g_1,...,g_N\in\mathcal O(U)$,and a smooth function $u\in C^\infty (U)$ s.t.
$$\psi(z)=c\log (|g_1(z)|^2+...+|g_N(z)|^2)+u(z)$$
for $z\in U$.
Then he says,we will regard $\psi$ as a generalition of Green-type functions $G$ in page 5.I mean，how to comprehend this sentence？For $\psi$,can I find explicit $A(z),B(z)$?
|
2025-03-21T14:48:29.764437
| 2020-02-02T15:29:34 |
351772
|
{
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"Gerry Myerson",
"Josiah Park",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/351772"
}
|
Stack Exchange
|
Maximizing $\iiint|(x-z)\times(y-z)|d\mu d\mu d\mu$ over probability measures on the unit circle
What probability measure(s) maximize the quantity $\iiint_{\mathbb{S}^1}|(x-z)\times(y-z)|d\mu(x)d\mu(y)d\mu(z)$?
The answer appears to be uniform measure, since informally it appears better to have more triangles in the support of $\mu$ which the function $|(x-z)\times(y-z)|$ computes the area of.
Is there an argument that shows this is true generally for $\mathbb{S}^{d-1}$?
Edit: Thanks to @fedja for the answer to the main question and the request for clarification. While it could be possible to replace the area of triangles with volumes of simplices (and so on), what was meant by generalization was most immediately the question of whether uniform measure maximizes $$\iiint_{\mathbb{S}^2}|(x-z)\times(y-z)|d\mu(x)d\mu(y)d\mu(z).$$
Probably not. Let's take a small disk of radius $r$ on the sphere and replace the area measure in it by the delta-measure at the center. Then it looks like you'll gain $cr^2$ for a typical height of the triangle with one vertex in the disk, so the total gain will be $cr^4$ while the loss will come from the triangles with 2 vertices in the disk (order $r^5$) and three vertices in the disk (order $r^8$). It is not a 2-dimensional equator either because in high dimensions for the uniform measure most triangles are about equilateral with side $\sqrt 2$ and $\frac{\sqrt 3}2>\frac 3{2\pi}$.
Ok. There is a semidefinite programming argument which shows that surface measure is optimal (among other optimizers) for $|(x-z)\times(y-z)|^2$ and $\mathbb{S}^2$. It is curious for $|(x-z)\times(y-z)|$ what should be the optimizer in the case uniform measure is not optimal.
Interesting. To be honest, I have no idea what to suggest for the optimizer since it looks like we have a break of symmetry here (an asymmetric solution of a symmetric problem) and such things are pretty hard to guess in general...
Arghhh... I made a (very) stupid mistake in my computations. Just forget everything I said about the sphere for now.
Follow-up question, https://mathoverflow.net/questions/352236/maximizing-iiintx-z-timesy-zd-mu-d-mu-d-mu-over-probability-measures-on
Yes, it is true for the circle (though the reason is not quite the one you suggested). We shall consider the discrete version of the problem, which is to put some odd number $n\ge 3$ of points (think of them as of being assigned the probability of $1/n$ each) on the circle so that the sum of triangle areas is maximized. Then an optimal configuration exists by compactness. We just want to show that it should be an equispaced distribution. Since you can obtain any measure on the circle as a weak limit of such discrete measures (and the integrand is continuous, so a small displacement of the measure does not change the integral much), you'll get the required optimality of the uniform measure as a limiting statement.
Take any of the points $e$ and enumerate all other points by the numbers $1,\dots,n-1$ clockwise starting from $e$ (so the whole system is $e,e_1,\dots,e_{n-1}$. Fix all points except $e$ and think a bit of how the total double area depends on the position of $e$. You'll realize quite soon that it is just some constant plus $v_e\times e$ where
$$v_e=\sum_{1\le k< m \le n-1}(e_m-e_k).$$
If you now consider for fixed $\ell<\frac n2$ all vectors $e_m-e_k$ with [$(\ell\le k< m\le n-\ell$) and ($k=\ell$ or $m=n-\ell$)], you'll see that $v_e=\sum_{\ell<\frac n2}A_\ell(e_{n-\ell}-e_\ell)$ with some $A_\ell>0$. You will be able to increase the sum of (even oriented) areas if $e$ is not perpendicular to $v_e$, so in the optimal configuration we must always have $\langle e,v_e\rangle=0$ for every choice of $e$.
Now recall the famous "drive around the circle without running out of gas" problem (more precisely, its particular case when all gas stations have the same amount of fuel sufficient for driving $\frac 1n$ of the entire circle). The conclusion of it is that you can choose $e$ so that the arc distance to $e_k$ is at most $\frac k n$ of the whole circle for every $k=1,\dots,n-1$, which makes all scalar products $\langle e,e_{n-\ell}-e_\ell\rangle$ with $\ell<\frac n2$ non-positive with the only chance of the equality $\langle e,v_e\rangle=0$ when the arc distance from $e$ to $e_k$ is exactly $\frac kn$ for all $k=1,\dots,n-1$.
As to higher dimensions, can you, please, specify first what you mean by "this"? (there are several ways to try to generalize the setup, so the word seems rather ambiguous to me :-) )
|
2025-03-21T14:48:29.764739
| 2020-02-02T18:05:24 |
351777
|
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"url": "https://mathoverflow.net/questions/351777"
}
|
Stack Exchange
|
Finding divisors with canonical singularities in a moving linear system
I apologize if the question is too naive or trivial:
We know that any reduced divisor in a smooth variety has Gorenstein singularities. However, I don't know if there's a cone theorem for Gorenstein varieties.
Let $|L|$ be a moving linear system on a smooth projective variety $X$. Can we find some $D \in |L|$ such that $D$ has canonical singularities? (Ideally I want something to which Cone Theorem applies). Thanks in advance.
If "moving" means "movable" (i.e. no fixed component), the answer is no: take the pencil $\lambda (X^3+Y^3)+\mu Z^3=0$ in $\mathbb{P}^3$.
|
2025-03-21T14:48:29.764823
| 2020-02-02T18:23:38 |
351778
|
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"Greg Martin",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/351778"
}
|
Stack Exchange
|
Is it possible to multiply two series to get as a result all composite numbers?
I was toying with the following problem:
Is it possible to find two infinite integer sequences $(a_n), (b_n)>0$ such that $\sum_{n=1}^{\infty}\frac{1}{(a_n)^s}\cdot \sum_{n=1}^{\infty}\frac{1}{(b_n)^s}=\sum_{n=1}^{\infty}\frac{1}{(c_n)^s}$ for every $s>1$? Here $c_n$ denotes the $n$-th composite number.
I can show that without loss of generality, $1\in a_n$ and for every $x$ with $\Omega(x)=2, x\in (b_n)$ but this did not help much.
Can someone provide an answer to this problem?
Can you point me towards a formula which defines what $c_n$ in the multiplied series is given arbitrary $a_n$ and $b_n$? Usually, I see the multiplication of series as a power series thing.
@Zemyla: $a_i^sb_j^s=(a_ib_j)^s$, so $c_n=a_ib_j$ for some $i,j$. The question is equivalent to asking: is it possible to find $(a_n)$, $(b_n)$ such that every composite number $c$ has exactly one factorization of the form $c=a_ib_j$, while prime numbers (and $1$) have no factorizations of that form?
Using @GregMartin 's characterization of the problem, we can rewrite further: write $X = \mathbb{N}^\infty$ to denote the set of all eventually-zero sequences of natural numbers (where we include $0$). Can we find $A, B \subset X$ such that $A + B = X \backslash ({e_i} \cup {(0)})$, where $e_i$ is the sequence with $1$ in the $i$th position and $0$ everywhere else, and where we think of the sum as a multiset (in other words, so there is no repetition)?
This is the answer of Greg Martin, with the correction of Mark Sapir, and details added.
Write $\Omega(n)$ for the number of prime factors of $n$ (counted with multiplicity), and $\Omega_{\operatorname{odd}}(n)$ for the number of odd prime factors of $n$ (counted with multiplicity), so $\Omega(n) = \Omega_{\operatorname{odd}}(n) + v_2(n)$ (where $v_2$ is the valuation at $2$).
Example. Let $A = \{1,2,4,8,\ldots\} = 2^{\mathbf Z_{\geq 0}}$, and let
$$B = \{n\ |\ \Omega(n) = 2\} \cup \{n \text{ odd}\ |\ \Omega(n) \geq 3\}.$$
For $n \in \mathbf Z_{>0}$, the number of representations $n = a \cdot b$ with $a \in A$ and $b \in B$ is $1$ if $n$ is composite, and $0$ otherwise.
Proof. Given $n \in \mathbf{Z}_{>0}$ composite (i.e. $\Omega(n) \geq 2$), define $k \in \mathbf Z_{\geq 0}$ as follows:
If $\Omega_{\operatorname{odd}}(n) \geq 2$, set $k = v_2(n)$.
If $\Omega_{\operatorname{odd}}(n) = 1$, set $k = v_2(n) - 1$.
If $\Omega_{\operatorname{odd}}(n) = 0$, set $k = v_2(n) - 2$.
In cases 2 and 3, note that $k \geq 0$ since $\Omega(n) \geq 2$. Then set $a = 2^k$ and $b = \tfrac{n}{a}$. Then $n = a \cdot b$, and clearly $a \in A$. We also have $b \in B$:
In case 1 above, $b$ is odd with $\Omega(b) \geq 2$;
In case 2 above, $b$ is even with $\Omega(b) = 2$;
In case 3 above, $b = 4$.
This shows existence of the desired decomposition. For uniqueness, assume $n = a \cdot b$ with $a \in A$ and $b \in B$. Let $m = v_2(n)$. Then $\Omega_{\operatorname{odd}}(b) = \Omega_{\operatorname{odd}}(n)$, so
If $\Omega_{\operatorname{odd}}(n) \geq 2$, then $\Omega_{\operatorname{odd}}(b) \geq 2$, which by definition of $B$ forces $b$ odd, hence $a = 2^m$.
If $\Omega_{\operatorname{odd}}(n) = 1$, then $\Omega_{\operatorname{odd}}(b) = 1$, which by definition of $B$ forces $b$ even and $\Omega(b) = 2$, hence $a = 2^{m-1}$.
If $\Omega_{\operatorname{odd}}(n) = 0$, then $\Omega_{\operatorname{odd}}(b) = 0$, which by definition of $B$ forces $b = 4$, hence $a = 2^{m-2}$.
This shows that $(a,b)$ must be as constructed above. Finally, since all elements of $B$ are composite, any integer of the form $n = a \cdot b$ with $a \in A$ and $b \in B$ is composite. $\square$
One way of doing this is by taking $(a_n)=(1,2,4,8,16,\dots)$ and by taking $(b_n)=(4, 9, 15, 21, 25, 27, 33, 35, 39, 45, 49, 51, \dots)$ to consist of $4$ together with the sequence of odd composite numbers. EDIT: as Wlod AA points out, one should also include $2p$ in the $b$ sequences for all odd primes $p$.
This solution can be modified by replacing the special prime $2$ with any other prime; it might well be possible to replace $\{2\}$ with a larger set of primes and generalize the construction.
What about 6? Am I missing something?
@KonstantinosGaitanas, actually, $\ 2^n\cdot 4,=,2^{n+2}.$
$\newcommand\N{\mathbb N}$
Define the nondecreasing sequences $(A_n)_{n\in\N}$ and $(B_n)_{n\in\N}$ of subsets of $\N=\{1,2,\dots\}$ recursively as follows:
$$A_1:=\{1\},\quad B_1:=\{4\};$$
for $n\ge2$,
$$
(A_n,B_n):=\left\{
\begin{aligned}
(A_{n-1},B_{n-1})&\text{ if }c_n\in A_{n-1}B_{n-1},\\
(A_{n-1},B_{n-1}\cup\{c_n\})&\text{ if }c_n\notin \N B_{n-1},\\
(A_{n-1}\cup\{a^*_n\},B_{n-1})&\text{ if }c_n\in\N B_{n-1}\setminus A_{n-1}B_{n-1}\\
&\text{ and }a^*_n>\max A_{n-1}, \\
(A_{n-1},B_{n-1}\cup\{c_n\})&\text{ if }c_n\in\N B_{n-1}\setminus A_{n-1}B_{n-1}\\
&\text{ and }a^*_n\le\max A_{n-1},
\end{aligned}
\right.
$$
where
$$a^*_n:=\min(\N\cap(c_n/B_{n-1})).$$
Let now
$$A:=\{a_1,a_2,\dots\}:=\bigcup_{n\in\N}A_n,\quad B:=\{b_1,b_2,\dots\}:=\bigcup_{n\in\N}B_n,$$
where $a_1<a_2<\cdots$ and $b_1<b_2<\cdots$.
Then the product of $A$ and $B$ equals $C:=\{c_1,c_2,\dots\}$, where $A,B,C$ are considered multisets. That is, for each $c\in C$ there is a unique pair $(a,b)\in A\times B$ such that $c=ab$.
The identity in question now follows.
For an illustration, note that, in particular,
$$A_{50}=\{1, 2, 4, 8, 16\},$$
$$B_{50}=\{4, 6, 9, 10, 14, 15, 21, 22, 25, 26, 27, 33, 34, 35, 38, 39, 45, 46, \
49, 51, 55, 57, 58, 62, 63, 65, 69\}.$$
The illustration strongly suggests $A = {2^i}, B = {4} \cup {2p} \cup C$, where $C$ is the set of odd numbers with at least 2 (not necessarily distinct) prime factors.
|
2025-03-21T14:48:29.765267
| 2020-02-02T18:56:25 |
351780
|
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|
Stack Exchange
|
Reference for discrete Laplacian on $\mathbb{Z}$
For $x\in \mathbb{R}^\mathbb{Z}$, let the discrete Laplacian be defined as
\begin{align*}
(\Delta x)_k = 2x_k-x_{k+1}-x_{k-1}.
\end{align*}
I am looking for good references about its spectrum (or eigen-structure), compactness, and properties of the semigroup it generates.
The spaces to consider include the usual $l^p(\mathbb{Z})$, for my purpose, it is also interesting to work on weighted space $l^p_\rho(\mathbb{Z})$ in which, provided a summable weight $\rho$ with $\sum_k \rho_k <\infty$, the norm is given by $$\|x\|^p_{p,\rho}= \sum_{k\in\mathbb{Z}} |x_k|^p\rho_k.$$
There won't be any references that discuss this operator at length since it's very easy to analyze: just take Fourier transforms $Fx = \sum x_n e^{int}$ to represent $\Delta$ as multiplication by $2(1-\cos t)$ in $L^2(-\pi,\pi)$. This immediately answers all your questions (for example, the spectrum equals $[0,4]$, is purely ac of multiplicitly $2$).
Here I'm of course assuming that you defined the operator on $\ell^2(\mathbb Z)$.
If you are looking for earliest references then see Phillips and Wiener, "Nets and Dirichlet problem" (1923).
Could you be more precise about the space on which you want information about this operator. In particular, are you really interested about your questions in the whole topological vector space $\mathbf{R}^\mathbf{\mathbf{Z}}$?
|
2025-03-21T14:48:29.765391
| 2020-02-02T19:28:14 |
351782
|
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|
Stack Exchange
|
Is every limit-closed, accessibly-embedded full subcategory of a presentable $\infty$-category reflective?
Let $C$ be a presentable $\infty$-category and let $D\subseteq C$ be a full subcategory closed under limits and sufficiently-filtered colimits. If $D$ is known to be accessible, then by the adjoint functor theorem it follows that the inclusion $D \subseteq C$ has a left adjoint (and it then also follows that $D$ is presentable). Is it really necessary to check that $D$ is accessible, though?
In ordinary category theory, this shortcut is available: Theorem 2.48 of Adamek and Rosicky states that if $C$ is a locally presentable 1-category and $D \subseteq C$ is a full subcategory closed under limits and $\kappa$-filtered colimits for some $\kappa$, then $D$ is accessible and so the inclusion $D \subseteq C$ has a left adjoint (and hence $D$ is in fact locally presentable).
The proof that $D$ is accessible uses the Adamek and Rosicky's Corollary 2.36, which says that if $C$ is accessible and $D \subseteq C$ is a full subcategory closed under $\kappa$-filtered colimits and $\kappa$-pure subobjects for some $\kappa$, then $D$ is accessible. As far as I know, the theory of pure subobjects in the $\infty$-categorical setting has not appeared in the literature, but I believe that this theorem continues to hold $\infty$-categorically. What I'm not so sure of is the second part of the proof: Adamek and Rosicky's Remark 2.31 which shows that a $\kappa$-pure subobject $c$ of an object $d$ in a locally $\kappa$-presentable category $C$ is contained in the smallest full subcategory $D \subseteq C$ containing $d$ and closed under limits and $\kappa$-filtered colimits.
Or perhaps there is an alternate proof not using the theory of pure subobjects?
A few years later, this is proven, by Ragimov and Schlank.
|
2025-03-21T14:48:29.765522
| 2020-02-02T19:52:55 |
351785
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/351785"
}
|
Stack Exchange
|
Naive generalization of determinant from matrices to higher rank tensors
Recall that using the Levi-Cevita symbol the determinant can be written as
$$\operatorname{det} A=\frac{1}{n!}\epsilon_{i_1\dots i_n}\epsilon_{j_1\dots j_n}A_{i_1j_1}\dots A_{i_nj_n}$$
Some computations that I do recently turned out to produce objects of the type
$$\epsilon_{i_1\dots i_n}\epsilon_{j_1\dots j_n}\epsilon_{k_1\dots k_n}\epsilon_{l_1\dots l_n}T_{i_1j_1k_1l_1}\dots T_{i_nj_nk_nl_n}$$
which look like a freshmen-style generalization of determinant to tensors. In my computations only even-rank tensors appear.
Is there a name for such things? Do they have any interesting properties?
https://math.stackexchange.com/q/1213130
Hey! This might be exactly what you're looking for :)
@MoritzEissler Thanks for the link! The question there is indeed the same, but the answer just states the formula from my post without giving it a name of explaining why is that a good definition. Another suggestion was to check out hyperdeterminant, but as far as I can see it is different from my formula.
One interesting property is that this determinant is $0$ when the tensors have an odd number of indices.
|
2025-03-21T14:48:29.765632
| 2020-02-02T20:27:04 |
351786
|
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|
Stack Exchange
|
Bar notation in Bourbaki’s *Lie groups*, Chap. IX
I am looking at Chapter IX (Compact Real Lie Groups), §4, Exercise 8 (translation). Given a complex subspace $\mathfrak p$ in the complexification $\mathfrak g_{\mathbf C}$ of some $\mathfrak g$, they start talking about
$$
\overline{\mathfrak p}.
$$
While I think I know what that bar means (conjugation of $\mathfrak g_{\mathbf C}$ w.r.t. $\mathfrak g$ as in e.g. Knapp), I have tried and failed to ascertain:
Q: Is this notation out of nowhere, or actually introduced some place in the book (or treatise)?
Your question seems to be specifically about Bourbaki (namely the treatise), so I added the tag. The notation has nothing to do with Lie algebras, since it makes sense in the complexification of an arbitrary real vector space. Given Bourbaki's principles it's certainly introduced somewhere, and I would have guessed somewhere in the Algebra books, but can't locate it there.
Neither could I! (Further re-tagged in agreement with your comment.)
The notion of conjugation with respect to a real form is introduced on p. 16. Searching for 'conjugaison' gives several results, most of the form (translated) "let $\sigma$ be the conjugation" or "let $\tau$ be the conjugation" (and then references to conjugacy classes). I can't figure out how to search for overbars, but all occurrences near the word 'conjugaison' seem to apply, properly, to complex numbers. I think that the answer is probably that it comes out of nowhere, or at least nowhere in Chapter IX.
It is clearly explained in Appendix II of the chapter. Indeed one may argue that a reference to that appendix would have been appropriate — though at this stage it is likely that the reader won't have any doubt about the meaning of that notation.
@abx Is it? That Appendix would seem to define $\overline{\mathfrak p}$ as “$\mathfrak p$ where $\alpha\in\mathbf C$ now acts by $v\mapsto\overline\alpha v$”, whereas the exercise talks about $\mathfrak p+\overline{\mathfrak p}$ and $\mathfrak p\cap\overline{\mathfrak p}$.
@Francois Ziegler: Oops, you are right, I was too hasty. Of course this is the same vector space, but viewed inside $\mathfrak{g}_{\mathbb{C}}$.
|
2025-03-21T14:48:29.765840
| 2020-02-02T20:28:40 |
351787
|
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"Ben McKay",
"GSM",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/351787"
}
|
Stack Exchange
|
Fibration of principal bundles
Let $G$ be a topological group, let $f:X\rightarrow Z$ be a $G$-equivariant map of (left) $G$-spaces such that
$X\rightarrow X/G$ and $Z\rightarrow Z/G$ are principal $G$-bundles.
$f$ is a fibration.
Let $\rho: G\rightarrow H$ be a morphism of topological groups. Is the induced map
$$ H\times_{G}X \rightarrow H\times_{G}Z$$ a fibration ? where the (right) action of $G$ on $H$ is induced by $\rho$
Edit: I was trying the following proof, but I think it is incomplete may be someone could help.
First: $f$ induces a continuous map of topological spaces $\hat{f}: X/G\rightarrow Z/G$.
The pullback of the map $Z\rightarrow Z/G$ along $\hat{f}: X/G\rightarrow Z/G$ is exactly (up to isomorphism) the map $X\rightarrow X/G$.
Second: The pullback of the map $H\times_{G}Z\rightarrow Z/G$ along $\hat{f}$ is exactly (up to isomorphism) the map $H\times_{G}X\rightarrow X/G$.
In the first and second item we use the fact that we have $G$-principal bundles and $H$-principal bundles.
If we can proof that $\hat{f}$ is a fibration then we are done! So my question would be answered if $\hat{f}$ is a fibration. If $f$ is a fibration does it follow that $\hat{f}$ is a fibration ?
What exactly do you mean by "map" of $G$-principal bundles? If it is the usual definition (equivariant and commutes with the projection maps) then it must be an isomorphism...
@JohnGreenwood I have edited my question, I hope it is clear.
Is $f$ $G$-equivariant?
@BenMcKay yes f is G-equivariant.
Can we assume Z/G is paracompact?
@JohnGreenwood does it help if Z/G is paracompact ? I will be happy to see if it works in the case of paracompact spaces...
The previous answer was getting a bit too complicated. Locally $f$ looks like $O_i \times G \rightarrow U_i \times G$ and this is a fibration since it is the restriction of the original fibration $f$. Now $\hat f$ locally looks like $O_i \rightarrow U_i$, and this is a fibration since it is a retract of a fibration. Thus locally $\hat f$ is a fibration, and so, thanks to Dold, $\hat f$ is a fibration.
Ah, It was your first claim I was having trouble convincing myself of! But "Dold" still requires numerability right?
I think the answer is "yes" if $Z/G$ is paracompact and there is a cover by contractible neighborhoods over which $p_{Z}: Z\rightarrow Z/G$ is trivial.
By paracompactness such a cover admits a numerable refinement $\{U_{i}\}$, and by a standard theorem it suffices to check that $\hat{f}$ is a fibration over each $U_{i}$.
Now $p_{Z}^{-1}(U_{i})\simeq U_{i}\times G$ by assumption, and $U_{i}$ is contractible. Let $F$ be the fiber of $f$. Let $g:G\rightarrow Z$ be the orbit of some point in $p_{Z}^{-1}(U_{i})$. Then $f^{-1}(p_{Z}^{-1}(U_{i}))\simeq U_{i}\times g^{*}X$. Note that the second factor is an $F$-fibration over $G$.
Now suppose we're given a homotopy $H:Y\times I\rightarrow U_{i}$ and a lift $h_{0}:Y\times\{0\}$ to $\hat{f}^{-1}(U_{i})$. Clearly $h_{0}$ lifts to $p_{Z}^{-1}(U_{i})$, and by the above it also lifts to $f^{-1}(p_{Z}^{-1}(U_{i})$. Since the composite $X\rightarrow Z\rightarrow Z/G$ is a fibration, we get a lift of the homotopy $H$ to $X$. Composing with the projection to $X/G$ and commutativity of the obvious square involving $X,Z,X/G$ and $Z/G$ provides us with a lift of $H$ to $X/G$ that lands in $\hat{f}^{-1}(U_{i})$ as desired.
Thanks! Where do you use that the spaces Z/G has a cover by contractible neighborhoods? is it a crucial assumption ?
I use it to split off a factor of U_i in the last equality of the second paragraph, I'm not sure how crucial it is to the general problem though!
|
2025-03-21T14:48:29.766091
| 2020-02-02T20:38:07 |
351788
|
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|
Stack Exchange
|
Why does MAGMA claim that the automorphism group of a curve is trivial?
I have been trying to compute the Automorphism group of a curve using MAGMA with no success. This is what I have tried: I have tried to compute the Automorphism group of the curve $y^3=x^4-x$ and no matter what I try -- it produces a trivial AutomorphismsGroup over rationals and I am not able to extend the scalars.
Attempt 1:
K:=Rationals();
P<x,y,z>:=ProjectiveSpace(K,2);
C:=Curve(P,y^3*z-x^4+x*z^3);
G:=AutomorphismGroup(C);
#G;
1
The group is trivial... However, we know that over the complex numbers the group is $C_9$, cyclic group of order 9. But, I am not able to extend the scalars to complex numbers and get this result...
L:=AlgebraicClosure(K);
D:=BaseChange(C,L);
G:=AutomorphismGroup(D);
^
Runtime error in 'AutomorphismGroup': Curve must have a function field
or
L:=ComplexField();
D:=BaseChange(C,L);
^
Runtime error in 'BaseChange': Ring must be exact
I have tried to compute the group via function fields:
Attempt 2:
A<x>:=PolynomialRing(K);
B<y>:=PolynomialRing(A);
F:=FunctionField(y^3-x^4+x);
G:=AutomorphismGroup(F);
#G;
1
and
L:=ComplexField();
BaseChange(F,L);
^
Runtime error in 'BaseChange': Bad argument types
Argument types given: FldFun[FldFunRat[FldRat]], FldCom
First question:
How can one compute the Automorphism group of the curve over the complex numbers using MAGMA?
Second question:
Is it possible to compute (in MAGMA) the Automorphism group of the Jacobian and then, using the Torelli theorem, the automorphism group of a curve?
AutomorphismGroup computes the automorphism group over the base field, and it only works with certain types of base fields - in particular, it won't work over the reals, complexes, and the "algebraic closure" (none of which are really suited to geometric computations). But you can get what you want by working over a number field, and in your example it's not hard to see which one you need:
> K := CyclotomicField(3);
> P<x,y,z> := ProjectiveSpace(K, 2);
> C := Curve(P, y^3*z - x^4 + x*z^3);
> G := AutomorphismGroup(C);
> #G;
3
This doesn't really help if you can't work out the field of definition of the automorphisms you want in advance. If you're working with hyperelliptic curves, though, you can use GeometricAutomorphismGroup to get the automorphism group over the algebraic closure.
|
2025-03-21T14:48:29.766254
| 2020-02-02T20:46:50 |
351789
|
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"Martin Brandenburg",
"Meric",
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}
|
Stack Exchange
|
What is the etale cohomology of G_m on Spec Z?
I apologize if this is straightforward, but I can't seem to find a reference: What is $H^1_{et}(\mathrm{Spec}(\mathbb{Z}), \mathbf{G}_m)$? It shouldn't be as simple as noting that the etale fundamental group of Spec Z is trivial, since Spec Z does admit non-trivial etale covers, just not finite ones.
It's much simpler: $H^1_{et}(X,\mathbf{G}_m)$ is always the Picard group $\mathrm{Pic}(X)$ for any scheme $X$. If $X = \mathrm{Spec}(R)$ for a PID R (e.g., $R = \mathbf{Z}$), then this group is trivial by algebra.
@Meric This is an answer (not a comment).
What is a non-trivial etale cover of Spec(Z)? Asking for a friend.
The map $\mathrm{Spec}, \mathbb{Z}[\sqrt{-1}, \frac{1}{2}] \times \mathbb{Z}[\frac{1 + \sqrt{5}}{2}, \frac{1}{5}] \to \mathrm{Spec} , \mathbb{Z}$ works. Extensions of number rings are ramified along the vanishing of their discriminants, so if we delete that vanishing, we have an etale map. Doing this with a pair of number rings with relatively prime discriminants gives us a surjective etale map.
(My algebraic number theory is rusty, so I'm using the table on https://en.wikipedia.org/wiki/Quadratic_field#Orders_of_quadratic_number_fields_of_small_discriminant to build my specific example.)
You could also just look at Zariski covers, like $\mathbf{Z} \to \mathbf{Z}[1/2] \times \mathbf{Z}[1/3]$. These covers are enough to compute (via Cech cohomology) the relevant cohomology group anyways.
|
2025-03-21T14:48:29.766670
| 2020-02-02T20:51:44 |
351790
|
{
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],
"authors": [
"Francois Ziegler",
"Gerhard Paseman",
"Gerry Myerson",
"LSpice",
"Oscar Lanzi",
"Paul B. Slater",
"Steven Stadnicki",
"Yemon Choi",
"https://mathoverflow.net/users/19276",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/3402",
"https://mathoverflow.net/users/3684",
"https://mathoverflow.net/users/44191",
"https://mathoverflow.net/users/47134",
"https://mathoverflow.net/users/7092",
"https://mathoverflow.net/users/763",
"https://mathoverflow.net/users/86625",
"user44191"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/351790"
}
|
Stack Exchange
|
Are $n \times n$ special orthogonal matrices, all the entries of which have the same absolute value, possible for $n \neq 4$?
As I noted in my preceding question https://math.stackexchange.com/questions/3510189/give-a-general-class-to-which-a-specific-4-times-4-special-orthogonal-matrix
in equation (62) of their recent publication https://arxiv.org/abs/1708.05336, "Separable Decompositions of Bipartite Mixed States", Li and Qiao present the matrix $Q \in \mbox{SO}(4)$,
\begin{equation}
Q=\frac{1}{2}\left(
\begin{array}{cccc}
1 & -1 & -1 & 1 \\
-1 & -1 & 1 & 1 \\
-1 & 1 & -1 & 1 \\
1 & 1 & 1 & 1 \\
\end{array}
\right).
\end{equation}
For $n=3, 5, 6$, I have tried (via direct enumeration) unsuccessfully to construct analogous $n \times n$ special orthogonal matrices, in which all the (equal) entries of the last column and row are (for probabilistic reasons) positive, and the remaining $n^2- 2n +1$ entries are all equal in absolute value.
Such matrices might be helpful in extending the Li-Qiao framework to the construction of separable decompositions of length $n \neq 4$. (It is not clear, however, that matrices must be of the specific requested form to so extend their framework. Perhaps, other than having the last row and columns positive, all remaining entries could be unrestricted, other than for the orthogonality requirement.)
It was observed by Robert Israel in the noted preceding question that Q is proportional to a Hadamard matrix. However, the next larger-sized ($8 \times 8$) Hadamard matrices are not orthogonal in character, so this does not seem to be a productive direction to take. (But as the comments below of others and mine indicate I was in error in making this claim.)
How is an 8 by 8 Hadamard matrix "not orthogonal in character"?
I think you meant to link to https://math.stackexchange.com/questions/3510189/give-a-general-class-to-which-a-specific-4-times-4-special-orthogonal-matrix
Echoing @user44191: Hadamard matrices are precisely the $\pm 1$-valued matrices such that $A^\top A$ is equal to a scalar multiple of the identity, so rescaling one of these always gives an orthogonal matrix whose entries share the same absolute value. The determinant 1 condition should be easy to check in various examples.
Also, there are matrices called complex Hadamard matrices of all orders which can supply such orthogonal matrices with entries from the complex numbers. Gerhard "Going Off In Another Dimension" Paseman, 2020.02.02.
Thanks for the comments! Well, I had taken the Kronecker product of a $2 \times 2$ and a $4 \times 4$ Hadamard matrix to presumably get an $8 \times 8$ one, which proved to be not orthogonal, that is its matrix product with its transpose was not proportional to the identity. But it looks like I should probably recheck my computations/references--in view of the comments---just rechecked. My $H_2$ was miscoded--so mea culpa and the $H_8$ is orthogonal. But what about $n$ not a mulitple of 4.
If you wish to change your question, then you should edit the question itself.
Note that the construction works both ways; the existence of matrices of the form you're looking for is precisely equivalent to the existence of a Hadamard matrix, by scaling of the entries. The Wikipedia page on Hadamard matrices notes that they can only exist in dimensions $1$, $2$, and $4n$, so you won't find any other examples.
Any power of $2$ works, really.
Well, if $Q\in\mathrm{SO}(n)$ satisfies your (original) conditions then so does
$
\frac1{\sqrt2}
\begin{pmatrix}
-{}^tQ&Q\\
\phantom{-}{}^tQ&Q
\end{pmatrix}
\in\mathrm{SO}(2n).
$
Added: As S. Stadnicki since commented, your desired set $\mathrm S$ of possible orders $n$ is contained in and conjecturally equal to $\{1,2\}\cup4\mathbf N$ (Hadamard conjecture);* the above construction ($\cong$ Sylvester’s) just shows $\smash{2^{\mathbf N}\subset\mathrm S}$. (Voting to close, as R. Israel had really said all this at the mis-linked question.)
* Paley (1933, front page) proved $\mathrm S\subset\{1,2\}\cup4\mathbf N$ thus: Assume w.l.o.g. that all entries are $\pm1$ and $Q$ has 3 distinct columns $u,v,w$. Then their orthogonality gives $$
n=\|u\|^2=\langle u+v,u+w\rangle=\sum\nolimits_i(u_i+v_i)(u_i+w_i),
$$
a sum all of whose terms are $0$ or $\pm 4$.
Unfortunately it's only in the comments, but @PaulB.Slater has revised the question to require that $n$ not be a multiple of $4$.
OK--thanks for the answer! So, how about a conjecture on my part that for $m$ not a multiple of 4, one can not have an SO(m) matrix, with the entries of the $m$-th row and column all equal and positive, and all the remaining entries of the same absolute value as the entries of the last row and column. So, Hadamard matrices would be special in this regard.
Hasn't the $n \in 4 \mathbf{Z}$ case been settled by the answer of Ziegler, given that the orginal $4 \times 4$ $Q$ is in SO(n)?
@PaulB.Slater I double the size each time, so I think it solves only powers of 2 — your own arises that way starting from the 1$\times$1 identity. But yes, if you now want to exclude those it actually shouldn’t be by changing the question but by asking another. (Else no one knows who’s been answering what.)
OK, point well-taken, FZ. So, my conjecture should be for integers that are not powers of 2, rather than integers that are not multiples of 4.
In reading the ArXiv paper referenced in the post, the authors are using matrices with complex entries. (Near the beginning they talk about generators of SO(2), one of them being a square root of -I.)
If one is looking for complex matrices in an orthogonal group SO(n), one can choose a scaled version of a complex Hadamard matrix of order n, where n is any positive integer, to get a matrix with entries having the same norm (absolute value), as well as having a row and a column having all entries equal to 1/(scale value, which may be n or square root of n).
(For the section being considered having display (62), it is unclear to me if the authors restrict themselves to matrices with real entries. For the purposes of the question, it seems to me that using matrices with complex entries is appropriate for carrying out their analysis, and that restricting the order to accommodate real Hadamard matrices is unnecessary.)
Gerhard "Complex Numbers Are Numbers Two" Paseman, 2020.02.03.
|
2025-03-21T14:48:29.767113
| 2020-02-03T00:06:19 |
351797
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/351797"
}
|
Stack Exchange
|
Glueing local systems over union of compact Riemann surfaces
Let $X,Y$ be two connected, non-singular compact Riemann surfaces such that $X$ intersects $Y$ transversely at two distinct points. Let $L$ be a $\mathbb{C}$-local system on $X$. Let $L'$ be the trivial local system on $Y$. Suppose that the fibers of $L$ and $L'$ are isomorphic. Does there exist a local system $L''$ on $X \cup Y$ such that its restriction to $X$ (resp. $Y$) is $L$ (resp. $L'$)?
Intuitively, I would think this is not always true i.e., there should be some restriction on $L$. This is roughly because (due to Riemann-Hilbert correspondence) local systems are in 1-1 correspondence with representations of the fundamental group. Since fundamental group is a topological invariant (in the curve case depending only on the genus) and the arithmetic genus of $X$ is different from that of $X \cup Y$, the space of local systems on $X \cup Y$ should vary from that on $X$. This should give rise to extra conditions on $X$.
However, I do not understand why the kernel of the following morphism should not always be a local system:
$$L \oplus L' \to \mathbb{C}^r_{X \cap Y}$$
for $L_{X \cap Y} \cong \mathbb{C}^r_{X \cap Y} \cong L'_{X \cap Y}$
and the morphism is given by a choice of identification of the fibers of $L$ and $L'$ at the points $X \cap Y$. Here $\mathbb{C}^r_{X \cap Y}$
denotes the skyscraper sheaf supported at $X \cap Y$ with fibers $\mathbb{C}^r$.
|
2025-03-21T14:48:29.767237
| 2020-02-03T00:06:22 |
351798
|
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"VS.",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351798"
}
|
Stack Exchange
|
Probabilistic interpretation of square free numbers and other properties
We can use the Lindberg condition to show the distribution of number of prime divisors of an integer approaches Gaussian.
Is there a similar probabilistic formulation for square free numbers? That is, is it reasonable to say the probability of an uniformly random integer being square free is $\frac6{\pi^2}$ with suitable probability distribution?
Are there non-trivial and deceptive situations where such probability distribution interpretations break down?
The standard way to formalize this thought is via the concept of "density".
This is because you of course cannot uniformly randomly select $n\in\mathbb{N}$, but you can instead examine (for a subset $A\subseteq\mathbb{N}$, using the notation $[n] = \{1,\dots,n\}$):
$$d(A) = \lim_{n\to\infty}\frac{|[n]\cap A|}{|[n]|}$$
I believe this can analogously be written as:
$$d(A) = \lim_{n\to\infty}\Pr_{x\leftarrow\mathcal{U}([n])}[x\in A]$$
Highlighting the probabilistic aspect of it.
It is known that the natural density of square free numbers is $6/\pi^2$, precisely as you expect. Natural density has many properties similar to what one would expect for "probabilistic" statements about "uniformly random integers", but has some downfalls (it is not defined for all subsets $A\subseteq\mathbb{N}$).
There are other notions of densities which can be examined in those situations, but I am no expert on the benefits/shortcomings of them.
The link you presented on distribution does not talk about probability distribution in Erdos Kac sense. Perhaps all square free numbers conspire against non square free numbers? I am looking for valid arguments for probability distributions.
|
2025-03-21T14:48:29.767483
| 2020-02-03T00:57:15 |
351800
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/351800"
}
|
Stack Exchange
|
The closure of a locus in $\overline{\mathcal{M}}_{0,n}(\mathbb{P}^1, n)$
Consider the closure $K \subset \overline{\mathcal{M}}_{0,n}(\mathbb{P}^1, n)$ in the stable maps space of the locus $K_0$ of maps $(f : C \to \mathbb{P}^1, p_1, \ldots, p_n)$ where $C \cong \mathbb{P}^1$ is smooth and the set of marked points is exactly the preimage of $\infty \in \mathbb{P}^1$; $\{p_1, \ldots, p_n\} = f^{-1}(\infty)$. I want to understand the boundary strata $K \setminus K_0$.
There is an obvious necessary condition for a pointed stable map $(f : C \to \mathbb{P}^1, p_1, \ldots, p_n)$ to be in $K$, which is that $f(p_i) = \infty$ for each $i$. In particular, $K$ is contained in the intersection of the evaluation loci $ev_i^{-1}(\infty)$ for $i = 1, \ldots, n$ where $ev_i : \overline{\mathcal{M}}_{0,n}(\mathbb{P}^1, n) \to \mathbb{P}^1$ are the usual evaluation maps. Let us call this Condition 0.
However, this condition is not sufficient. For example there is a stratum paramerizing $f : C_0 \cup_p C_1 \to \mathbb{P}^1$ where the restriction $f_0 : C_0 \to \mathbb{P}^1$ is a degree $n$ map from a smooth rational curve with $f_0(p) = \infty$, $f$ constant on $C_1$, and all the marked points lying on $C_1$. This stratum is contained $ev_i^{-1}(\infty)$ for each $i$ but is larger dimensional than $K$.
The reason is that being in the closure of $K_0$ imposes some extra conditions on the map. For degree reasons, a map $f : C \to \mathbb{P}^1$ in $K_0$ must be unramified at all the marked points $p_i$ but the only way marked points can collide is if the map becomes ramified at $\infty$. This gives us a condition on certain nodes lying over $\infty$ which I think can be phrased as follows:
Condition 1: Let $C_0 \subset C$ is a connected union of components lying in $f^{-1}(\infty)$. Suppose $C_0$ is attached to components $C_1, \ldots, C_k$ at points $x_1, \ldots, x_k$ where $f|_{C_i}$ is non-constant. Then the number of marked points lying on $C_0$ is the sum of the ramification of $f|_{C_i}$ at $x_i$.
Condition 2: For each component $C_0 \subset C$ with $f|_{C_0}$ non-constant, each point of $f|_{C_0}^{-1}(\infty)$ is either a marked point or a node of $C$.
These two conditions are related but its not immediately clear to me exactly how. Note that a general point of the stratum described above not contained in $K$ satisfies neither condition $1$ nor $2$.
Question 1: Are conditions $0$, $1$ and $2$ sufficient for a stable map to be contained in the closure of $K$? If not, is there some other description of the boundary strata of $K$?
I think a quick dimension count shows that the dimension of the strata satisfying conditions $0$, $1$ and $2$ have dimension at most the dimension of $K$, and in fact strictly smaller than the dimension of $K$ if you exclude $K_0$ itself. This gives some evidence that Question 1 has a positive answer. For example, if one knew that the expected dimension of the locus satisfying these conditions is the dimension of $K$, then these dimension bounds would be enough. However, I don't know how to show this.
It turns out that the answer is yes and these conditions are sufficient.
This question was answered in greater generality by Gathmann (Absolute and relative Gromov-Witten invariants of very ample hypersurfaces. Duke
Math. J., 2002, Proposition 1.14) building off of earlier work of Vakil (The enumerative geometry of rational and elliptic curves in projective space. J. Reine Angew. Math., 2000, Theorems 4.13 and 6.1).
We wrote a direct proof of the special case considered in the question (namely target $(\mathbb{P}^1, \infty)$) in Section 2 of this preprint.
|
2025-03-21T14:48:29.767733
| 2020-02-03T01:57:12 |
351806
|
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|
Stack Exchange
|
Classifying space BG and contractable space EG
This question is probably not research level that's why I asked it previously on MSE a week ago. Unfortunately it doesn't get much attention there and I thought I would try it here.
Choose a arbitrary discrete group $G$. The classifying space $BG$ of $G$ is classically constructed by forming a certain contractable $\Delta$-complex $EG$ (on concrete construction of $EG$: see below) endowed with an action by $G$. $BG$ is obtained by taking quotient $BG:= EG/G$.
The concrete construction of the $\Delta$-complex $EG$ works as follows: The $n$-simplices of $EG$ are the ordered tuples
$$[g_0,g_1,...,g_n] \cong \Delta_n =\left\{x\in \mathbb {R} ^{n}:x=\sum _{i=0}^{n}t_{i}v_{i}\ {\text{with}}\ 0\leq t_{i}\leq 1\ {\text{and}}\ \sum _{i=0}^{n}t_{i}=1 \right\}$$
with $g_i \in G$. The $v_i$ are spanning $\Delta_n$. We obtain a $\Delta$-complex by attaching $n$-simplices to the $(n − 1)$ simplices $[g_0,g_1,..., \hat{g}_i,...,g_n]$ in standard way as a standard simplex attaches to its faces. Here $\hat{g}_i$ means that this
vertex is deleted.
Question. Does there exist a constructive way to show that $EG$ is contractable. By constructive I mean how to construct an explicite homotopy $h_t: EG \times I \to EG$ which contracts $EG$ to a point.
That is $h_0(EG) =EG, h_1(EG) = \{*\}$. How looks it concretely geometrically?
I know some abstract arguments like Whitehead's theorem that also provide a reason for contrability of $EG$ but this is not what I'm looking for.
I tried to define such homotopy as follows: let $p \in [g_0,...,g_n]$. Then $h_t$ "slides" step by step $p$ along $(n+1)$-simplex $[e,g_0,...,g_n]$ to $[e]$ ($e \in G$ the identity element). What I mean by "step by step along $[e,g_0,...,g_n]$"?
If we use again the identification $[g_0,g_1,...,g_n] \cong \Delta_n$ then as long as we sitting "inside" $\Delta_n$ we can interpret the $g_i$ as spanning vectors $v_i$. Let $p = \sum _{i=0}^{n}t_{i}g_{i}$. As $[g_0,g_1,...,g_n] \subset [e, g_0,g_1,...,g_n]$ we can interpret $[g_0,g_1,...,g_n]$ as $x = t_{-1}e + \sum _{i=0}^{n}t_{i}g_{i} \in \Delta_{n+1}$ with $t_{-1}=0$. That is the "point" $p_e:= 1 \cdot e \in [e, g_0,g_1,...,g_n]$ is not contained in $[g_0,g_1,...,g_n]$ and we can define a unique line $l_pe$ which contains $p_e$ and is perpendicular to $[g_0,g_1,...,g_n]$ in our geometric picture $[g_0,g_1,...,g_n] \subset [e, g_0,g_1,...,g_n]$ corresponding to $\Delta_n \subset \Delta_{n+1}$.
This line uniquely intersects $[g_0,g_1,...,g_n]$ in a unique point $p_l$. Then we say that our homotopy slide $p$ along the unique line through $p $ and $p_l$ up to the first contact with a boundary of $[g_0,g_1,...,g_n]$.
Let this boundary be $[g_0,..., \hat{g}_i,...,g_n]$. Then we play the same game with $[g_0,..., \hat{g}_i,...,g_n]$ and $[e, g_0,..., \hat{g}_i,...,g_n]$.
Does this approach work? And is there known a more conventional "textbook" (that I still haven't found) way for the construction of $EG$?
The ``abstract nonsense'' explanation can be found here: https://ncatlab.org/nlab/show/decalage
@BertramArnold: so intuitively decalage "cleans" the obstuctions (like degenerations) of a simplicial set (more precisely it's geom realization) to be contractable?
The easiest way to construct an explicit contracting homotopy
is to observe that EG is the geometric realization of the nerve of the groupoid G//G,
which has G as its set of objects and exactly one morphism between any pair of objects.
The nerve functor sends equivalences of groupoids
to homotopy equivalences of simplicial sets,
and the geometric realization functor sends homotopy equivalences of simplicial sets
to homotopy equivalences of topological spaces.
Thus, it remains to construct an equivalence between G//G and the trivial groupoid
* with a single object and morphism.
This is easy: there is just one functor G//G→*,
and for the inverse functor take *→G//G that picks the identity element of G.
The composition *→G//G→* is identity, whereas the composition G//G→*→G//G
is isomorphic to the identity via the unique choices of morphisms in G//G.
Could you lose few words on your last argument that the composition G//G→*→G//G is homotopic to identity "via the unique choices of morphisms in G//G". I not fully understand what you mean. By constuction between every $g,h \in G//G$ there exist exactly one map. Why does this imply the desired claim?
@MortyPB: A natural transformation F→G (in our case, F=G//G→*→G//G and G=id) is a collection of morphisms F(X)→G(X) that satisfies a certain commutativity condition. The data of morphisms F(X)→G(X) is uniquely prescribed because we already know F(X) and G(X) and there is exactly one morphism of the form F(X)→G(X). Likewise, any square diagram (in fact, any diagram) in G//G automatically commutes because any pair of morphisms between the same objects coincides.
|
2025-03-21T14:48:29.768053
| 2020-02-03T02:18:49 |
351807
|
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|
Stack Exchange
|
Is the complement of a zero-dimensional subset of the plane path-connected?
Let $X$ be a zero-dimensional subset of the plane $\mathbb R ^2$. Is $\mathbb R ^2\setminus X$ necessarily path-connected? I feel the answer must be yes but I need a reference. If it helps, assume $X$ is nowhere dense.
If the zero-dimensional set $X$ is not closed, then the answer is "no".
To construct a suitable example, take any open bounded neighborhood $U\subset\mathbb R^2$ of zero, whose boundary $\partial U$ does not contain a topological copy of $[0,1]$. For example, for $U$ we can take a bounded connected component of the complement of the union of two suitable pseudoarcs in the plane. Then the set $$X=\mathbb R^2\setminus\{\vec a+\tfrac1n\partial U:\vec a\in\mathbb Q^2,\;n\in\mathbb N\}$$ will have the desired property: it is zero-dimensional and its complement $\mathbb R^2\setminus X$ does not contain a copy of $[0,1]$ (by the Baire Theorem) and hence is not path-connected.
At least if $X$ is compact, the answer is yes. Indeed, by Corollary 2 of Theorem IV 3 in:
W. Hurewicz, H. Wallman, Dimension Theory. Princeton Mathematical Series, v. 4. Princeton University Press, Princeton, N. J., 1941,
compact sets separating points in $\mathbb{R}^{n+1}$ must have topological dimension $n$.
In particular, compact sets separating points in $\mathbb{R}^2$ must have (topological) dimension $1$ or $2$ so sets of dimension $0$ cannot separate points.
You know, reading your answer makes me wonder which specific kind of topological dimension OP had in mind (but then again R^2 is polish so it doesn't really matter...)
|
2025-03-21T14:48:29.768198
| 2020-02-03T02:37:21 |
351808
|
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|
Stack Exchange
|
Find torsion classes using flat bundles
My question refers to a discussion from this older thread on Neron-Severi group of a Kähler manifold. In the comments below Ted Shifrin's answer there arose a discussion when the map $H^2(X,\mathbb{Z}) \to H^2(X,\mathbb{C})$ induced by the canonical inclusion $\mathbb{Z} \subset \mathbb{C}$ may be not injective. As Ted pointed out the main reason that prevents this map being injective is the presence torsion in $H^2(X,\mathbb{Z})$.
Then he continues: "...From the line bundle side, you can see this with flat bundles (whose curvature forms vanish, and so the de Rham representative is the zero form)."
Unfortunately I not understand how his example & argument for line bundles by considering to flat bundles work. Definitely, as the curvature forms of flat bundles vanish, their de Rham representative in $H^2(X,\mathbb{C})$ is zero. That's fine.
Why does it provide an example for torsion elements in $H^2(X,\mathbb{Z})$? Does their de Rham representative not already vanish in $H^2(X,\mathbb{Z})$ as the curvature is zero? But then it not provide an example for a torsion element.
Does anybody understand what Ted had by the quoted remark in mind? I tried to ask but probably that was quite obvious point and I'm just too fool to see it.
The de Rham representative is a differential form so does not give an element of $H^2(X, \mathbb{Z})$.
Lefschetz theorem tells you that any torsion class in $H^2(X,\mathbb{Z})$ is the first Chern class of a line bundle, then necessarily flat. Like the thread you quote, the question would be more appropriate in MSE.
@abx:The Lefschetz theorem (at least which I know) says that the first Chern map $c_1: H^1(X,\mathcal{O}_X^)\rightarrow H^2(X,\mathbb{Z})$ (that associating to a holomorphic line bundle its Chern class $c_1$) induces isomorphism $H^1(X,\mathcal{O}_X^)/ker(c_1) \cong H^{(1,1)}(X,\mathbb{Z})$, where the right object is given by $H^{(1,1)}(X,\mathbb{Z})=Im(H^2(X, \mathbb{Z}) \to H^2(X, \mathbb{Z})) \cap H^{1,1}(X)$. As $H^1(X,\mathcal{O}_X^*)$ classifies holom line bundles $c_1$ associates as you said first Chern classes to line bundles. (reference: D. Huybrecht's Complex Geometry page 133).
One point I not understand: you say that Lefschetz theorem tells that any torsion class $H^2(X,\mathbb{Z})$ that is the first class of a line bundle (in other words is contained in the image of $c_1$,right?), is neccessary flat, right? What I don't understand is that does the asumption imply that a torsion element of $H^2(X,\mathbb{Z})$ which is contained in the image of $c_1$, is already $0$? By def $im(C_1)=H^{(1,1)}(X,\mathbb{Z})=Im(H^2(X, \mathbb{Z}) \to H^2(X, \mathbb{Z})) \cap H^{1,1}(X)$ says that image of $c_1$ has no torsion. Or do I misunderstand your point?
ps: yes, you are right that this question has indeed a MSE-level. I have tried to ask it to weeks ago https://math.stackexchange.com/questions/3519053/understand-torsion-using-flat-bundles but unfortunately it doesn't get much attention.
No, the image of $c_1$ can have torsion — in fact, let me repeat that any torsion class in $H^2(X,\mathbb{Z})$ is $c_1(L)$ for some $L$ — just use the exponential exact sequence.
@abx: Could you give reference for the version of Lefschetz thm you mentioned above that states that all torison classes are hit by certain line bundles via $c_1$? I not see now I can derive it instantly from that one on page 133 from Huybrecht's book. Ok and so the conclusion that these preimages of torsion classes must be flat follows simply by definition as $H^2(X,\mathbb{Z}) \to H^2(X,\mathbb{C})$ kills torsion...
This follows from the exact sequence $H^1(X,\mathscr{O}_X^*)\rightarrow H^2(X,\mathbb{Z})\rightarrow H^2(X,\mathscr{O}_X)$ that you'll find in any book on complex geometry, e.g. Griffiths-Harris.
sorry for digging out this topic but as I recently was reading your comments on statement confuses me: you say "Lefschetz theorem tells you that any torsion class in $H^2(X,\mathbb{Z})$ is the first Chern class of a line bundle, then necessarily flat." I agree with you that using exponential sequence we conclude that any torion class in $H^2(X,\mathbb{Z})$ has a primage by Chern map $c_1$. That's just the exactness. But you say also that any torsion class comes from a flat line bundle.
Why? A flat bundle $L$ is characterized by vanishing curvature $\Omega_L=0$. Of course we can project the curvature to it's class in $H^2(X,\mathbb{C})$ and since it's obviously zero in $H^2(X,\mathbb{C})$, then it's a torsion. That's clear. But why we can conclude also in other direction? That is the that a line bundle that is associated by $c_1$ to a torsion class, is flat?
|
2025-03-21T14:48:29.768489
| 2020-02-03T02:50:46 |
351809
|
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"Iosif Pinelis",
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|
Stack Exchange
|
Is there a bound on the norm of the product of second moment matrix with random vector?
Let $X_1,\dots,X_n$ be vectors in $\mathbb{R^d}$. Assume all of the vectors are inside the unite $\ell_2$ ball, but outside the ball of radius $r$ for some $r \in (0,1)$, i.e. $r \leq \|X_i\| \leq 1$ . Let $P$ be a vector in the probability simplex $\Delta_n$ with $P_i>0$ for all $i$. Consider the second moment matrix $\Sigma(P) = \sum_{i=1}^n P_i X_i X_i^\top$. Assume the $X_i$s are such that $\Sigma(P)$ is full rank. Does the following bound always hold? If not, when does it hold?
$$\|\Sigma(P)^{-1}X_j\| \leq \frac{1}{r P_j} \quad \forall j\in \{1,\dots,n\}$$
For instance, if $n=d$ and $X_i=e_i$ are the canonical basis vectors of $\mathbb{R}^d$, then this bounds holds with equality.
$\newcommand\Si{\Sigma}$
$\newcommand\X{\mathbf X}$
The answer is no. Indeed, let $p_i:=P_i$, $p:=P$, $\X:=(X_1,\dots,X_n)$, and $\Si_\X:=\Si(p)$. At least one of the vectors $\Si^{-1}X_j$ is nonzero, for some $j$, because otherwise the matrix
$$I=\Si_\X^{-1}\Si_\X=\sum_1^n p_i\Si_\X^{-1}X_i X_i^T$$
would be zero. So, there is some $j$ such that $c:=\|\Si_\X^{-1}X_j\|>0$. Replacing now $\X$ by $a\X$ for real $a>0$ and letting $a\to0$, we will have
$$\|\Si_{a\X}^{-1}(aX_j)\|=\frac1a\,\|\Si_\X^{-1} X_j\|=\frac ca\to\infty,$$
so that the inequality
$$\|\Si_{a\X}^{-1}(aX_j)\|\le\frac1{p_j}$$
will fail to hold for small enough $a$.
The OP has edited the question, thus invalidating this answer. However, even after the edit, the answer remains no. E.g., let $n=2$, $p_1=p_2=1/2$, $X_1:=[1,1]^T/\sqrt2$, and $X_2:=[1,1-h]^T/\sqrt2$ with $h\downarrow0$. Then $\|X_1\|=1$, $1\ge\|X_2\|\sim1$, but
$$\|\Si_{\X}^{-1}(X_1)\|=\frac{2 \sqrt{2} \sqrt{h^2-2 h+2}}{h}
\sim\frac4h
\not\lesssim2=\frac1{p_1}.$$
What if the $X_i$ are further restricted to be outside a ball of certain radius? So basically $r \leq|X_i|\leq 1$ for some fixed $r \in (0,1)$. Could such an inequality hold as a function of $r$, maybe something like $\frac{1}{r p_j}$
@SudeepRaja : If you have additional questions, please ask them in separate posts. Edits should not invalidate an answer.
Noted. I'm new to stack overflow. Didn't know the etiquette.
|
2025-03-21T14:48:29.768664
| 2020-02-03T03:40:40 |
351813
|
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|
Stack Exchange
|
valuation of a derivative in a completion
Let $q$ be a power of a prime $p$ and $w$ be an irreducible polynomial of $\mathbb F_q[T]$. Denote by $\mathbb C_w$ the completion of an algebraic closure of $K_w$, the completion of $\mathbb F_q(T)$ for the normalized valuation $v$ ($v(w)=1$).
Denote by $K^{\text{sep}}_w$ the separable closure of $K_w$ in $\mathbb C_w$. One knows that the derivative on $\mathbb F_q(T)$ can be extended uniquely in $K^{\text{sep}}_w$ (one still denotes it by $'$). My question: for every $\alpha\in K^{\text{sep}}_w$, does one have $v(\alpha')=v(\alpha)-1$ or $v(\alpha')\ge v(\alpha)$ if $p\mid v(\alpha)$. Obviously, it is true for $\alpha$ in the completion of $\mathbb F_q(T)$.
The algebraic closure of $\mathbf F_q(T)$ does not admit a unique extension of the valuation $v_P$ on $\mathbf F_q(T)$, just as ${\rm ord}_p$ on $\mathbf Q$ does not have a unique extension to the algebraic closure of $\mathbf Q$ (but ${\rm ord}_p$ does extend uniquely from $\mathbf Q_p$ to its algebraic closure). I would expect the notation $\mathbf C_P$ to mean the completion of the algebraic closure of the completion $(\mathbf F_q(T))_P$. Please reconsider what it is you are asking.
It might also be a bit easier to read the question if you picked your notation to avoid using both $P$ and $p$, as the relation $p \mid v_P(\alpha)$ could require extra attention to understand what it is saying (distinguishing a small capital $P$ in $v_P$ from $p$)
You write "for every $\alpha \in K_P$ algebraic over $\mathbf F_q(T)$," but that is redundant since you define $K_P$ to be the separable closure of $\mathbf F_q(T)$ (in a certain field), so automatically all elements of $K_P$ are algebraic over $\mathbf F_q(T)$.
Thanks for the feedback. I edited the question..
Your new notation is confusing: $K_w$ is a completion of $\mathbf F_q(T)$, so the notation $K_w^{\rm sep}$ looks like it should be the separable closure of $K_w$, not the separable closure of $\mathbf F_q(T)$! Would you ever write the separable (= algebraic) closure of $\mathbf Q$ inside $\mathbf C_p$ as $\mathbf Q_p^{\rm sep}$? I hope not.
If $y$ is a root of $y^p-y = 1/T$, then $v(y)=-1/p, v(y')=-2$.
|
2025-03-21T14:48:29.768828
| 2020-02-03T06:20:33 |
351822
|
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|
Stack Exchange
|
induced maps between group $C^*$-algebras
Suppose $G,H$ are two locally compact groups, if there is a injective homomorphism $\phi:G\to H$, can $\phi$ induce the $*$-homomorphism between group $C^*$-algebras $C^*(G)$ and $C^*(H)$? If it exists, is it injective?
No, and that’s much the point of Rieffel (1974): “Actually, if H is a closed but not open subgroup of the locally compact group G, then the group C*-algebra, C*(H), of H is not a subalgebra of C*(G), but rather acts as an algebra of right centralizers [29] on C*(G).”
I saw an example: consider $\Bbb Z\subset \Bbb R$, I cannoe see why there does not exist $$-hommomorphisms between $C^(\Bbb Z)$ and $C^*(\Bbb R)$.
More simply ${0}\subset\mathbf R$. What subalgebra of $C^(\mathbf R)$ would $C^({0})$ be?
But $C^(0)$ is 0, we can get the zero $$ homomorphism from 0 to $C^*(R)$.
The $C^*$-algebra of a one-element group is $\mathbf C$, not ${0}$.
|
2025-03-21T14:48:29.768919
| 2020-02-03T07:20:43 |
351824
|
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|
Stack Exchange
|
Partitionability and colorability of hypergraphs
Motivation. If $\kappa\neq\emptyset$ is a cardinal, then a simple, undirected graph $G=(V,E)$ is $\kappa$-colorable if and only if there is a partition of $V$ into at most $\kappa$ blocks such that every edge $e\in E$ intersects $2$ blocks.
$\kappa$-colorability of a hypergraph. Let $H=(V,E)$ be a hypergraph, that is $V$ is a set and $E\subseteq {\cal P}(V)$. If $\kappa \neq \emptyset$ is a cardinal, we say that a map $c:V\to \kappa$ is a (hypergraph) coloring if for every $e\in E$ with $|e|>1$ the restriction $c|_e$ is non-constant.
$\kappa$-partitionability of a hypergraph. If $H=(V,E)$ is a hypergraph and $\kappa\neq\emptyset$ is a cardinal, we say that $H$ is $\kappa$-partitionable if there is a partition of $V$ into at most $\kappa$ blocks such that for every $e\in E$ with $|e|>1$ we have that $e$ is not a subset of any block of the partition.
It is easy to see that if $H$ is $\kappa$-partitionable, then it is $\kappa$-colorable (just give every block a different color). Does the converse hold?
If the answer is "Yes", then this could be the motivation for the fact that the requirement that a coloring be non-constant on non-singleton edges (instead of injective).
How is a partition functionally different from a coloring? Gerhard "If You Label, It Isn't" Paseman, 2020.02.03.
Oh right, via pre-images, I suppose...
|
2025-03-21T14:48:29.769040
| 2020-02-03T09:05:23 |
351827
|
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|
Stack Exchange
|
A new cardinal characteristic (related to partitions)?
In this post I will discuss some cardinal characteristic of the continuum, related to partitions of $\omega$ and would like to know if it is equal to some known cardinal characteristic.
By a partition of $\omega$ I understand a cover of $\omega$ by pairwise disjoint nonempty subsets. A partition $\mathcal P$ is called finitary if $\sup_{P\in\mathcal P}|P|$ is finite.
A family $\mathfrak P$ of partitions of $\omega$ is called directed if for any two partitions $\mathcal A,\mathcal B\in\mathfrak P$ there exists a partition $\mathcal C\in\mathfrak P$ such that each set $S\in\mathcal A\cup\mathcal B$ is contained in some set $C\in\mathcal C$.
Let $\mathfrak P$ is a family of partitions of $\omega$. An infinite subset $D\subset\omega$ is called $\mathfrak P$-discrete if for any partition $\mathcal P\in\mathfrak P$ there exists a finite set $F\subset D$ such that for any $P\in\mathcal P$ the intersection $P\cap (D\setminus F)$ contains at most one point.
Let $\kappa$ be the smallest cardinality of a directed family $\mathfrak P$ of finitary partitions of $\omega$ admitting no infinite $\mathfrak P$-discrete set $D\subset\omega$.
It can be shown that $\mathfrak b\le\kappa\le\mathfrak c$ (the upper bound follows from the observation that any maximal directed family of finitary partitions has no infinite discrete set, see Proposition 6.5 in this preprint).
Problem 1. Is $\kappa$ equal to some known cardinal characteristic of the continuum?
Problem 2. Is $\kappa=\mathfrak c$ in ZFC?
Problem 3. Find lower and upper bounds on $\kappa$ (which are better than $\mathfrak b\le\kappa\le\mathfrak c$).
Added in Edit. The lower bound $\sup_{U\in\beta\omega}\pi(U)\le\kappa$, suggested by Todd Eisworth can be improved to $\mathfrak s\le \kappa$. One can also prove that $\max\{\mathfrak b,\mathfrak s,\}\le\mathfrak j\le\kappa\le\mathrm{non}(\mathcal M)$ and hence $\kappa$ is not equal to $\mathfrak c$. The cardinal $\mathfrak j$ is discussed in this MO-post.
Possibly a stupid sub-question. For any finitary partition ${\cal P}$ let $\mu({\cal P})$ be the smallest $k\in \omega$ such that for all $P\in{\cal P}$ we have $|P|\leq k$. Let ${\frak P}$ be a directed family of finitary partitions of $\omega$ with the property that there is $N\in \omega$ with $\mu({\cal P}) \leq N$ for all ${\cal P}\in {\frak P}$. (In other words, there is a global "block size bound" for every partition in ${\frak P}$.) Does this imply that ${\frak P}$ admint a ${\frak P}$-discrete set $D\in[\omega]^\omega$?
@DominicvanderZypen Very good question! The answer is "yes". If $\mathfrak P$ is directed and the cardinality of cells is bounded from above, then there exists an increasing sequence of partitions $(\mathcal P_n){n\in\omega}$ in $\mathfrak P$ that converges (in a pointwise sense) to some partition $\mathcal P\infty$. Take a ${\mathcal P_\infty}$-discrete set and notice that it is also $\mathfrak P$-discrete.
Thanks for your answer, Taras!
This is not an answer, but hopefully it's a helpful observation:
(1) If $U$ is an ultrafilter on $\omega$ and $\mathcal{P}$ is a finitary partition of $\omega$, then there is $A\in U$ such that $A\cap P$ contains at most one element for each $P\in\mathcal{P}$.
(As if each piece of the partition has cardinality at most $n$, then there is a $k\leq n$ such that the union of pieces with size exactly $k$ is in $U$. Now split this union up into $k$ pieces in the obvious way, and one of these is in $U$.)
(2) Given an ultrafilter $U$, let $\tau(U)$ be the least cardinal $\tau$ such that some subfamily of $U$ of cardinality $\tau$ fails to have an infinite pseudo-intersection. (We do not require the pseudo-intersection to be in $U$, so $\aleph_1\leq\tau(U)\leq\mathfrak{c}$.)
Observation:
If $U$ is an ultrafilter on $\omega$, then $\tau(U)\leq\kappa$.
Proof. Given a family $\mathfrak{P}$ of finitary partitions of $\omega$ (directed or not), we fix for each $P\in\mathfrak{P}$ a set $A_P\in U$ meeting each element of $P$ in at most one point. If $|\mathfrak{P}|<\tau(U)$ then we can find an infinite pseudo-intersection $X$ for the collection $\{A_P:P\in\mathfrak{P}\}$, and $X$ is $\mathfrak{P}$-discrete.$_\square$
I don't know anything about the cardinals $\tau(U)$. I note that at one point Blass and Shelah claimed to have model containing both simple $P_{\aleph_1}$ and simple $P_{\aleph_2}$ points, but Alan Dow discovered an error in the paper, and I'm not sure if it has ever been repaired. (The existence of a simple $P_{\aleph_1}$-point implies $\mathfrak{b}=\mathfrak{u}=\aleph_1$, while the simple $P_{\aleph_2}$ point is an ultrafilter $U$ with $\tau(U)=\aleph_2$. In such a model, $\kappa$ would be strictly greater than $\mathfrak{b}$.)
Clearly this is all tied up with the topology of $\beta\omega$, so I suspect much more is known by the experts.
Indeed, very interesting! Especially the cardinal $\sup_{U\in\omega^}\tau(U)$. Maybe @Andreas Blass could comment on the consistency of $\mathfrak b<\sup_{U\in\omega^}\tau(U)$?
And what about the consistency of $\sup_{U\in\omega^*}\tau(U)<\mathfrak b$? Is there any model of ZFC where this strict inequality is true?
As far as I know, the error about simple P-points for two different cardinals has not yet been repaired. Shelah and Mildenberger each had arguments to repair it, but neither argument seems to have survived.
I think the second comment by @TarasBanakh can be answered by starting with a model of MA + $\neg$CH and then adjoining $\aleph_1$ random reals. The random reals don't affect $\mathfrak b$, which stays large. But any ultrafilter will have to contain each of the random reals or its complement, and I don't see any chance for a pseudo-intersection (which would have to depend on only countably many of the random reals, by ccc). (Unfortunately, I have no answer yet for the first comment, the one relevant to the actual question.)
Hechler forcing (standard ccc iteration) probably accomplishes the same, no? If I remember, the forcing doesn't add pseudo-intersections to towers in the ground model.
In fact, don't we know $\tau(U)\leq\mathfrak{s}$ for every ultrafilter $U$ on $\omega$ by the same argument?
@AndreasBlass: could you point out where specifically are the errors? thanks
It is consistent that $\mathfrak{b}<\kappa$. See the answer to https://mathoverflow.net/questions/352034/a-question-on-simple-p-aleph-2-points/352661#352661
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2025-03-21T14:48:29.769575
| 2020-02-03T09:32:14 |
351829
|
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|
Stack Exchange
|
Positive definiteness of matrix
This question is about the positive definiteness of a (non-random) matrix that is defined using random variables as follows:
We fix the vector $v=(1,1)$ (yet, it seems the final result does not depend on it) and choose a positive definite covariance matrix $\Sigma \in \mathbb R^{2 \times 2}$.
Let $X$ be a $\mathbb R^2$-valued random variable such that $X$ is distributed according to
$$d\mu(x) \propto e^{-\langle (x-v), \Sigma^{-1} (x-v) \rangle-\vert x_1 \vert^4-\vert x_2 \vert^4} \ dx.$$
We then define the following deterministic matrix using centred random variables $Y=X-\mathbb E(X)$
$$ \langle x,Ay\rangle = -\mathbb E\left(\langle Y,x \rangle \langle Y,\Sigma^{-1} v \rangle \langle Y, y\rangle\right).$$
Fedja proved already in this thread that for any $v$ and $\Sigma$ diagonal(!), the matrix $A$ is positive definite. He understood that in this case the matrix is essentially diagonal, as components $x_1,x_2$ in the measure
$$d\mu(x) \propto e^{-\langle (x-v), \Sigma^{-1} (x-v) \rangle-\vert x_1 \vert^4-\vert x_2 \vert^4} \ dx.$$
factorize.
His proof also shows that at least for the special choice $x=\Sigma^{-1}v$, we have also for non-diagonal $\Sigma $, that $\langle x,Ax\rangle \ge 0.$
We then discussed in the comments whether one can extend the proof to non-diagonal $\Sigma$, but did not succeed so far.
Numerically it seems that for all choices of $\Sigma$ and $v$ I made so far, the matrix $A$ is positive definite.
Here you can find the Mathematica file I was using to verify this Click me
Another way of expressing this matrix $A$ is to write
$$A = \int_{\langle x,\Sigma^{-1} v \rangle>0} (x \otimes x) \langle x,\Sigma^{-1} v \rangle \ (d\mu(x+\mathbb E(X))-d\mu(x-\mathbb E(X))).$$
We observe that for $\langle x,\Sigma^{-1} v \rangle>0$ the integrand $(x \otimes x) \langle x,\Sigma^{-1} v \rangle $ is always positive. So if we could show that "most of the time"
$$\mu(x+\mathbb E(X))\ge \mu(x-\mathbb E(X))$$
this would imply the result.
My question therefore is: How can I show the eigenvalues of $A$ are non-negative?
Numerically, I made the following observations:
The positive definiteness of $A$ holds independent of the choice of $v \neq 0$ and $\Sigma.$
If we write $$d\mu(x) \propto e^{-\langle (x-v), \Sigma^{-1} (x-v) \rangle-\vert x_1 \vert^p-\vert x_2 \vert^p} \ dx.$$
then for $p<2$ the eigenvalues of $A$ become negative, zero for $p=2$ and greater than zero for $p>4.$
This seems to be consistent with what Fedja proved in the one-dimensional case.
|
2025-03-21T14:48:29.769756
| 2020-02-03T11:57:13 |
351833
|
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Stack Exchange
|
How does this orthogonality follow from the map being an isometry?
This is a step of a proof in the book Variational Problems in Geometry by Seiki Nishikawa. I will ignore the background and change some of the statements and notations for simplicity.
Let $(M,g)$ be a smooth Riemannian manifold. Suppose $w:M\to\mathbb R^q$ is an isometric embedding. Let $N$ be a tubular neighborhood of $w(M)$, and $\pi:N\to w(M)$ the canonical projection. Let $\tau(w)$ denote the tension field of $w$. Then since $w$ is an isometric embedding, $\tau(w)$ is orthogonal to $w(M)$ and hence $d\pi(\tau(w))=0$.
What I don't understand is the emphrasized sentence. Why does $w$ being isometric imply $\tau(w)\perp w(M)$?
Edit:
The tension field $\tau(w)$ is an $\mathbb R^q$-valued vector field on $M$, or more precisely, a smooth section of the bundle $w^{-1}(T\mathbb R^q)$. It is defined in coordinates ($x^i$ on $M$ and the standard coordinates on $\mathbb R^q$) by $\tau(w)^r=\Delta w^r+g^{ij}\hat\Gamma_{kl}^rw^k_iw_j^l$, where $g$ is the metric on $M$ and $\hat\Gamma_{ij}^r$ are the Christoffel symbols of $\mathbb R^q$. But since w.r.t. the standard coordinates of the Euclidean space all coefficients of its metric are $0$, $\hat\Gamma_{ij}^r=0$ and $\tau(w)=\Delta w$.
There are a number of ways to see this. One way is to take the covariant derivative of the isometric embedding equation $\partial_iu\cdot\partial_ju = g_{ij}$ and "differentiate by parts". The calculation below is with respect to local coordinates, and $u$ is treated as a $q$-tuple of scalar real-valued functions. Therefore, $\nabla^2_{ij}u = \nabla^2_{ji}u$.
\begin{align*}
0 &= \nabla_kg_{ij}\\
&= \nabla_k(\partial_iu\cdot\partial_ju)\\
&= \nabla^2_{ik}u\cdot \partial_ju + \partial_iu\cdot\nabla^2_{jk}u\\
&= \nabla_i(\partial_ku\cdot\partial_ju) + \nabla_j(\partial_iu\cdot\partial_ku) - 2\partial_ku\cdot\nabla^2_{ij}u\\
&= \nabla_ig_{kj} + \nabla_jg_{ik} - 2\partial_ku\cdot\nabla^2_{ij}u\\
&= -2\partial_ku\cdot\nabla^2_{ij}u
\end{align*}
Since this holds for any $1 \le i, j, k \le \dim M$, it follows that $\nabla^2u(p)$ is normal to $T_pM \subset \mathbb{R}^q$, for every $p \in M$. The tension field is simply $g^{ij}\nabla^2_{ij}u$ and therefore is also normal to $M$.
It's useful to write out the calculation above using Christoffel symbols and the identity
$$ g_{kl}\Gamma^l_{ij} = \partial_ku\cdot\partial^2_{ij}u. $$
Here, $\partial^2_{ij}u$ denotes the second partials of $u$ with respect to local coordinates and not its Hessian with respect to $g$.
|
2025-03-21T14:48:29.769945
| 2020-02-03T12:28:57 |
351834
|
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|
Stack Exchange
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Interior point of a convex polytope
Suppose the convex polytope is the set of feasible solutions $\mathbf{x}\in\mathbb{R}^n$ for the linear system $\mathbf{A}\mathbf{x}=\mathbf{b}\,,\; \mathbf{A}\in\mathbb{R}^{m\times n}$ subject to some (linear) constraints $\mathbf{0}\leq\mathbf{x}\leq\mathbf{v}$; how is it possible to select an interior point of said polytope, i.e. it doesn't lie on an edge or vertex of the polytope?
Does the algorithm requires the knowledge of what the vertices of the polytope are?
Do you mean that it doesn't lie on a proper face, or on an edge or vertex? That wouldn't necessarily be an interior point. Anyway, an idea would be by induction on the dimension of the faces, using the fact that these are aso convex polytopes. Also, you can triangulate.
If you know the vertices, you may take a minimal set and use the definition as the convex hull.
Finding all vertices of the polytope would have the same complexity as the Vertex Enumeration Problem. I do not think that's a practical approach.
The polytope is the intersection of the box $0\leq x \leq v$, and the hyperplanes $Ax=b$. A point in the interior of the box, which also lies on the hyperplanes is what is desired. Consider the following convex optimization problem:
$$
\max_{x\in R^n} \left\{\min \{x_1, v_1-x_1,\cdots, x_n, v_n-x_n\} \right\} ~\mbox{subject to}~ Ax=b ~~\&~~ 0\leq x\leq v.
$$
The rationale is that the numbers within the $\min$ brackets are the distances from the $2n$ hyperplanes defining the box. Thus, maximizing the minimum of those forces the optimal point to be in the interior. The optimization problem can be written as an LP with $n+1$ variables and $m+4n$ constraints. Hence this gives a polynomial time algorithm for determining a point in the interior. Also note that the optimal value is positive if and only if such an interior point exists.
Max min optimization is in P with linear objective over a compact polyhedra?
@VS, thanks for the comment. I have mentioned in my answer that the optimization can be converted to an LP with $n+1$ variables and $m+4n$ constraints. That is, replace the cost function by $t$ (yet another variable), and include constraints $x_i \geq t$, $v_i-x_i\geq t$, $\forall i$. This LP can then be solved using an Interior Point Method.
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2025-03-21T14:48:29.770148
| 2020-02-03T12:45:58 |
351835
|
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Stack Exchange
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Relativistic scattering theory vs non-relativistic one
In relativistic scattering theory (e.g. in quantum electrodynamics) the existence of the $S$-matrix as well as of Moller operators is postulated as far as I understand (although at some stage it has to be modified due to existence of infrared divergences). On the other hand if one considers the non-relativistic scattering problem of several particles with Coulomb potential, it seems to be very different from the usual scattering theory with rapidly decaying potential and much of it does not work (again, this is my impression, and I am not an expert).
I am wondering why one should expect existence of $S$-matrix in quantum electrodynamics with properties similar to $S$-matrix for non-relativistic scattering theory with rapidly decaying potential (rather than Coulomb potential). Do I miss something on the non-relativistic Coulomb scattering?
I guess that this question might not be appropriate for this site as QFT is not yet a mathematical theory. But this question is a direct continuation of this one on this site where I did get some helpful comments. Moreover I imagine that an answer might be known to some mathematical physicists.
From a physical standpoint, higher order corrections in a perturbative solution in QED grants that the Coulomb potential gets screened. This implies that what normally is done to assume a mass term in the Coulomb potential and taking it to go to zero at the end of computation, at the leading order (Born approximation), is justified.
@Jon: Thanks, although your comment is quite concise for me. I will try to guess what you said. If one computes scattering of two charged particles in the tree approximation (thus no screening) in non-relativistic limit then one recovers the Coulomb low like it would be the Born approximation to non-relativistic scattering (the next order of approximation implies Ueling effect, but let's ignore this case). While one gets this compatibility, I do not understand why it makes sense: does the non-relativistic $S$-matrix makes sense for the Coulomb potential? Does Born approximation makes sense?
also at https://physics.stackexchange.com/questions/527229/s-matrix-in-tree-approximation-in-qed-vs-scattering-theory-with-coulomb-potent
@CarloBeenakker: Thanks, that was me. As you see, that did not help...
A rigorous and rather complete description of non-relativistic scattering, including a discussion of long-range potentials, can be found in the following book:
Dereziński, Jan; Gérard, Christian, Scattering theory of classical and quantum $N$-particle systems, Texts and Monographs in Physics. Berlin: Springer. xii,444 p. (1997). ZBL0899.47007.
To handle long-range potentials, the definition of the wave/Møller operators and the corresponding S-matrix must be modified by using different "free dynamics". This idea was introduced by Dollard, as has been mentioned in comments on your previous question. In the quantum 2-body case, the short-range and long-range cases are treated and contrasted respectively in Secs.4.6 and 4.7 of the book. The quantum N-body case with long-range potentials is treated in the last several sections of Ch.6. You may want to read the Introduction section for the whole book, as well as the Introductions of the relevant chapters for a lot of context and for hints which earlier pats of the book to read if you want to quickly jump into chapters 4 or 6.
In relativistic QFT, the scattering problem has a quite different nature, due to the fact that particle number is not conserved. This creates an obstacle to the naive scattering theory in the sense that a state with few incoming particles may give rise to an infinite number of outgoing particles (this is intimately tied in with infrared divergences). These problems appear already at low orders in perturbation theory (a mathematically well-posed setting), so they can be discussed at that level, without the need to go into non-perturbative construction of interacting QFT (which is basically an open problem anyway). Still, there are standard ideas for how to define an S-matrix in this context, in particular for QED, again going back to Dollard's idea of suitably modifying free asymptotic dynamics.
The following paper is a modern mathematical synthesis of these ideas (including historical references), extending them to a larger class of QFTs, which contains QED:
Duch, Paweł, Infrared problem in perturbative quantum field theory, [arXiv:1906.00940].
Trying to understand scattering for non-relativistic toy models of QED, and in particular in the massless case (i.e. long range interactions) is an active area of research in mathematical physics. You can for instance look at the article
"Towards a Construction of Inclusive Collision Cross-Sections in the Massless Nelson Model" in AHP 2012 by Wojciech Dybalski who is an expert in the field.
|
2025-03-21T14:48:29.770499
| 2020-02-03T12:46:45 |
351836
|
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|
Stack Exchange
|
Euler-Poincaré characteristic as a function from a Hilbert scheme of surfaces
Let $H$ be an irreducible component of a Hilbert scheme of surfaces in $\mathbb{P}^n_{\mathbb{C}}$
whose general point corresponds to a smooth irreducible surface.
Consider the function $\chi:U\to\mathbb{Z}$ defined over the subset $U\subseteq H$ parametrizing smooth irreducible surfaces
which sends a surface $S$ to its topological Euler-Poincaré characteristic $\chi(S)$.
What property does the function $\chi$ have? For example, is it constant?
Because $H$ is irreducible and $U$ is open, $U$ is connected, so yes, $\chi$ is constant. This has nothing to do with irreducibility or two-dimensionality of the surfaces - it's a general topological fact about smooth proper maps.
Thanks for the quick answer. Can you give me a reference for the fact that $\chi$ is a smooth proper map?
If you're fine reducing to the analytic setting, https://en.wikipedia.org/wiki/Ehresmann%27s_lemma
|
2025-03-21T14:48:29.770624
| 2020-02-03T13:53:06 |
351838
|
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Stack Exchange
|
Cutting the unit square into pieces with rational length sides
The following questions seem related to the still open question whether there is a point(s) whose distances from the 4 corners of a unit square are all rational.
To cut a unit square into n (a finite number) triangles with all sides of rational length. For which values of n can it be done if at all?
To cut a unit square into n right triangles with all sides of rational length. For which values of n can it be done, if at all?
Remark: If one can find a finite set of 'Pythagorean rectangles' (rectangles whose sides and diagonal are all integers) that together tile some square (of integer side), that would answer this question.
3.To cut a unit square into n isosceles triangles with all sides of rational length. For which values of n can it be done, if at all?
Now, one can add the requirement of mutual non-congruence of all pieces to all these questions. Further, one can demand rationality of area of pieces or replace the unit square with other shapes (including asking for a triangulation of the entire plane into mutually non-congruent triangles all with finite length rational length sides)...
Note: From what has been shown by Yaakov Baruch in the discussion below, cutting the unit square into mutually non-congruent rational sided-right triangles can be done for all n>=4. Indeed, he has shown n=4 explicitly; for higher n, one can go from m non-congruent pieces to m+1 pieces by recursively cutting any of the m right triangular pieces n by joining its right angle to the hypotenuse to cut it into two smaller and mutually similar but non-congruent pieces. That basically settles questions 1 and 2 - the non-congruent pieces case. However, if we need all pieces to be non-congruent and non-similar, the n=4 answer has no obvious generalization to higher n.
References:
1. https://nandacumar.blogspot.com/2016/06/non-congruent-tiling-ongoing-story.html?m=1
On dissecting a triangle into another triangle
Why stop there, and not ask for all ways to cut arbitrary shapes into an arbitrary number of other arbitrary shapes with arbitrary conditions on arbitrary numerical invariants of the shapes?
There are questions on point sets where all or maximally many distances have to be rational. Here the attempt was to think of a family of questions that apply rationality of distances only locally - only sides of each single piece need to be rational; the rest of the distances are free.
You can cut a square into 12 $1/3\times1/4$ rectangles, each of which can be bisected by a diagonal into 2 rational triangles; $n=24$ here. You can then merge some adjacent triangles to form larger triangles and reduce $n$ for both 1. and 2.
Thus for 1. it's easy to achieve $n=5$, and from there any $n\ge 5$. $n=3$ would solve the famous open problem, so the only remaining question for 1. is whether $n=4$ is possible.
As for $n=4$, one can split a $360\times 360$ square into 3 right triangles along the perimeter (360-224-424, 360-105-375, 255-136-289) and a middle triangle (375-289-424).
@YaakovBaruch's beautiful construction for $n=4$:
Edit by Yaakov Baruch. A partition with 8 isoceles triangles (I'm sure not minimal):
Drawing heights in right triangles splits them to produce solutions for $n\ge 5$ too.
Beautiful example! Thanks very much for the illustration too! And the nice thing is all right triangles are mutually-noncongruent. Perhaps such examples can be found for other values of n as well. And yes, the isosceles pieces case...
(It's always amazing what mathematicians find beautiful...)
Searching around the internet I see here that the rational distance problem is known to not have solutions on the square's boundary. So that rules out dissections for $n=3$.
Notice that the middle triangle has rational area, since the other 3 do and the tota areal is 1; any height of the middle triangle therefore splits it into rational ones (as the sum of 2 square roots is only rational if both roots are).
Thanks... for right triangles, if sides are rational, area is automatically rational. So, requiring area too to be rational could be of interest only in the cases such as isosceles triangle pieces.
@NandakumarR. I don't follow your last comment. But what I meant earlier is that if any triangle has rational sides and area, then its heights are rational and they split the bases at rational point. I think though that Gerry Myerson early comment about the proliferation and arbitrariness of possible questions here has merit.
Thanks for the isosceles example. It would be a nice curiosity if there exists a cutting of the unit square into a finite number of mutually noncongruent isosceles triangles. And yes, my last comment only mentioned that if for a right triangle, all sides are of rational length, its area is automatically rational and this property does not hold in general for isosceles triangles.
|
2025-03-21T14:48:29.770985
| 2020-02-03T14:12:20 |
351840
|
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|
Stack Exchange
|
Looking for examples of not injective maps and not surjective maps of the form $ A_{k} (X) \to H_{2k} ( X , \mathbb{Z} ) $
Here: https://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2002/alggeom-2002-c9.pdf, on pages: $ 1 $ and $ 2 $, we find the following paragraph:
For any scheme of finite type over a ground field and any integer $ k>0 $, we will define the so-called Chow groups $ A_k (X) $ whose elements are formal linear combination of $ k $ -dimentional closed subvarieties of $ X $, modulo ''boundaries'' in a suitable sense. The formal properties of this groups $ A_k (X) $ will be similar to those of homology groups. Il the ground field is $ \mathbb{C} $, you might even thought of the $ A_k (X) $ as being ''something like'' $ H_{2k} (X , \mathbb{Z} ) $, although these groups are usually different. But, there is a map $ A_{k} (X) \to H_{2k} ( X , \mathbb{Z} ) $, so you can think of elements in the Chow groups as something that determines a homology class, but this map is in general neither injective nor surjective.
Questions,
After reading this block, and since it is said in this block that, in general, the morphism $ A_k (X) \to H_{2k} (X, \mathbb {Z}) $ is neither injective, nor surjective, can you give me some examples of $ A_k (X) \to H_{2k} (X, \mathbb{Z}) $ maps that are not injective, or that they are not surjective where : $ 4 \leq 2k \leq n-4 $ and $ n - 4 > 0 $, and $ n $ is the dimension of $ X $ ?
Thanks in advance for you help.
Well, the image of the cycle class map has to be contained in $H^{(k,k)}(X,\mathbb{Z}).
For obstructions to infectivity, look up the generalized Bloch conjecture. For obstructions to surjectivity @Qfwfq's comment should be the only rational obstruction; there are other failures of surjectivity known (look up papers giving counterexamples to the integral Hodge conjecture). At the end of the day, assuming Hodge and generalized Bloch, there is a complete characterization of when injectivity/surjectivity fails rationally.
Thank you Qfwfq and dhy. It's clear now. :-)
If $C$ is a smooth projective curve of genus $g \geq 3$ and $J(C)$ is the Jacobian of $C,$ then an Abel curve $C \subset J(C)$ is not algebraically equivalent to its image $-C$ under the negation automorphism, even though $C$ is homologically equivalent to $C.$ This was proved by Ceresa in the paper
https://www.jstor.org/stable/2007078
EDIT: Ceresa also shows in this paper that if $C$ is generic and $1 \leq k \leq g-2$, the cycles $W_k$ and $-W_k$ in $J(C)$ (recall that $W_k$ is the cycle parametrizing effective line bundles of degree $k$ on $C$) are algebraically independent (although they are homologically equivalent). This gives examples of non-injectivity for higher-dimensional cycles.
It should be $g\ge 3$.
@AGlearner Fixed, thanks!
Thank you @Yusuf. :-)
|
2025-03-21T14:48:29.771233
| 2020-02-03T14:19:14 |
351842
|
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|
Stack Exchange
|
Minimality properties of James' space
I am interested in the following question about James' quasi-reflexive Banach space $\mathcal{J}$:
Does there exists a non-Hilbertian subspace $X$ of $\mathcal{J}$ such that $X$ isomorphically embeds into every non-Hilbertian subspace of itself?
Here, by "subspace" I mean "closed, infinite-dimensional vector subspace", and by "Hilbertian" I mean "isomorphic to $\ell_2$".
I vaguely recall having found, one year ago, a paper proving that the answer to this question was no, or at least giving a similar/partial result suggesting that the answer should be no. Problem is, I don't manage to find this paper again, I don't even remember who were the authors and what was the exact result they proved. Do some of you recall having seen something like that?
It's a pitty that the ask-johnson-tag does not exist anymore.
A weaker `block version' is true for the conditional spreading basis (the summing basis) of $\mathcal{J}$: Every seminormalized block basis of the spreading basis has a subsequence either equivalent to an unconditional basis ($\ell_2$) or a convex block sequence equivalent to the basis itself. The result holds in general in spaces with a convex block homogeneous conditional spreading basis. See the section 5 of the following paper
A study of conditional spreading sequences
Spiros A. Argyros, Pavlos Motakis, Bünyamin Sari
@Jochen Wengenroth: As you can see, the ask-johnson-tag is no longer needed on MO. Good riddance!
@Bunyamin Sari: In this paper: https://www.jstor.org/stable/2041285?seq=1#metadata_info_tab_contents, the authors prove that James' space has uncountably many pairwise nonequivalent unconditional basic sequences. This contradicts your claim that the only one is $\ell_2$.
Actually, $\mathcal{J}$ itself cannot have the property I'm asking about: the same paper shows it has non-Hilbertian reflexive subspaces, hence it cannot isomorphically embed into these subspaces.
I agree that the result I quoted doesn't answer your question, sorry I was too quick. I edited the answer. This doesn't contradict the results you quoted either because it involves block subspaces of the conditional basis.
We clearly very much still need the ask-johnson-tag!
Let $X:=\left(\sum_n \oplus E_n \right)_2$ with $d_n(X):=\sup d(E,\ell_2^n)\to\infty$. Is there $Y:=\left(\sum_n \oplus F_n \right)_2\subset X$ so that $d_n(Y)\to \infty$ in much slower pace (so that $X$ doesn't embed into $Y$). This is surely true, and would answer the OP's question in negative.
You don't need that much, Bunyamin, to give a negative to answer the OP's question. Given $a_n \uparrow \infty$, build a space $Y$ of the form $(\sum_n \ell_{p_n}^{m_n})2$ so that $m_n^{|1/p_n -1/2|}\to \infty$ but $a_n/(m_n^{|1/p_n -1/2|})\to \infty$ with $p_n\downarrow 2$. Do this by induction so that a $m_1 + \dots m_n$ is much smaller than $m{n+1}$ and so that $m_n$ dimensional subspaces of $L_{p_{n+1}}$ are $2$-isomorphic to Hilbert spaces (which is the same as $m_n^{|1/p_{n+1} -1/2|} \le 2$). This is easy even if it is a pain to write down the details.
Bill, how do you make sure such spaces $\left(\sum_n \ell_{p_n}^{m_n}\right)_2$ embed into an arbitrary non-Hilbertian subspace of $\mathcal J$.
Given any non Hilbertian subspace of James' space, you construct a space of that form into which the subspace does not embed, which gives a negative answer to the OP's question. Not every non Hilbertian subspace of James' space contains such a space. Conceivably every non Hilbertian subspace of James' space contains an a non Hilbertian $\ell_2$ sum of finite dimensional $\ell_{p_n}$ spaces (where the $p_n$ can be on either side of $2$), but I doubt it.
I think the question asks something harder: if every non-Hibertian subspace $X$ embeds into a non-Hilbertian subspace of $X$ itself.
Oh, I misread the question repeatedly. Thanks.
|
2025-03-21T14:48:29.771609
| 2020-02-03T15:31:41 |
351845
|
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|
Stack Exchange
|
Which plane curves can be harmonically parametrized?
In this question, a “(closed oriented plane) curve” $\Gamma$ will mean a continuous map $f \colon \mathbb{U} \to \mathbb{C}$ where $\mathbb{U} := \{z\in\mathbb{C} : |z|=1\}$ is the unit circle, modulo right-composition by orientation-preserving homeomorphisms $\varphi\colon\mathbb{U}\to\mathbb{U}$. Any such $f$ (i.e. any of the $f\circ\varphi$ defining the curve) will be called a “parametrization” of the curve $\Gamma$.
I'm willing to add any reasonable regularity conditions on $\Gamma$ if they help in answering the question, e.g., piecewise $C^1$ or even $C^\infty$.
Say that a curve $\Gamma$ is harmonically parametrizable iff it admits a parametrization $f \colon \mathbb{U} \to \mathbb{C}$ (see above) such that the Fourier coefficients $(c_k(f))_{k\in\mathbb{Z}}$ of $f$ are zero for all $k<0$: this then allows us to see $f$ as the restriction to $\mathbb{U} = \partial\Delta$ of a continuous function $F$ on the closed unit disk $\overline{\Delta}$ that is holomorphic on the open unit disk $\Delta$ (namely, $F$ is the Poisson integral of $f$, or equivalently $F(z) = \sum_{k=0}^{+\infty} c_k z^k$; see also this question).
More generally, if $k_0\in\mathbb{Z}$ say that $\Gamma$ is $k_0$-harmonically parametrizable iff it admits a parametrization $f \colon \mathbb{U} \to \mathbb{C}$ that $c_k(f)=0$ for all $k<k_0$. (So “harmonically parametrizable” means “$0$-harmonically parametrizable”, and the larger $k_0$ the stronger the condition.)
For example, if $\Gamma$ is a positively-oriented, sufficiently regular (I think “rectifiable” suffices), Jordan curve, then the Riemann conformal mapping theorem guarantees that we can find $F \colon \Delta \to V$ bijective and conformal, where $V$ is the bounded component of the complement of $\Gamma$, extending continuously to $\partial\Delta = \mathbb{U}$ (assuming $\Gamma$ is sufficiently regular), which shows that $\Gamma$ is harmonically parametrizable.
So, two related questions:
Can we find a necessary and sufficient condition on a curve $\Gamma$ (again, assumed sufficiently regular if this is useful) for it to be harmonically parametrizable?
What about $k_0$-harmonically parametrizable for $k_0<0$? Can we find, at least, a reasonable sufficient condition analogous to the “Jordan curve” condition for being harmonically parametrizable?
(For $k_0>0$, being $k_0$-harmonically parametrizable seems to say something about the moments of $\Gamma$ for the harmonic measure vanishing, and this is probably less interesting.)
PS: This question seems related to a kind of converse to the one I'm asking (I want to know if there exists a parametrization change which becomes harmonic, that other question is about what happens upon such a change).
First of all, this is a purely topological problem. Let $\gamma:U\to C$ be a curve.
Your question is when this curve can be reparameterized so that the new parameterization is by boundary values of
an analytic function in the unit disk.
It is necessary that $\gamma$ extends to
a topologically holomorphic map of the unit disk to $C$ (topologically holomorphic map, a. k. a. polymersion, is a map which is topologically equivalent to $z\mapsto z^n$ near every point.
This condition is also sufficient. Indeed, once we have a
topologically holomorphic map, we can pull back the complex analytic structure to the disk, and then use the Uniformization theorem. So for every topologically
holomorphic map $f$ there is a homeomorphism $\phi$ of the disk such that $f\circ\phi$ is holomorphic.
Second, this topological problem is solved in the paper (under some smoothness conditions on the curve):
C. Curley and D. Wolitzer, Btranched immersions of surfaces, Michigan Math. J. 33 (1986) 131-144.
Unfortunately the answer is somewhat complicated and I do not reproduce it here.
|
2025-03-21T14:48:29.771875
| 2020-02-03T17:30:13 |
351854
|
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|
Stack Exchange
|
Quantum Hamiltonian reduction and tensor products
Let $k$ be a field of characteristic zero, $\mathfrak{g}$ a finite-dimensional Lie algebra over $k$, and let $A,B$ associative $k$-algebras.
Suppose that $\mathfrak{g}$ acts on $A$ and $B$, and that each of these actions admit a quantum moment map. Then the associative algebra $A\otimes_kB$ also has a $\mathfrak{g}$ action and admits a quantum moment map.
In this case, is it true that
$$(A\otimes_kB)//\mathfrak{g}\cong (A//\mathfrak{g})\otimes_k(B//\mathfrak{g})$$
where "$//\mathfrak{g}$" is the operation of quantum Hamiltonian reduction?
What is a quantum moment map?
For e.g. $A$, a quantum moment map is an algebra morphism $\mu:\mathcal{U}\mathfrak{g}\to A$ such that for $a\in\mathfrak{g}$, $b\in A$, we have $[\mu(a),b]=a\cdot b$ (where $a\cdot b$ is the action of $a$ on $b$).
This is not true.
It is analogous to asking if $(X \times Y)/G \cong (X/G) \times (Y/G)$ for $G$ a group acting on spaces $X$ and $Y$, which is almost never the case. For example, take $X=Y=G$ with the action by left translation. Then the left hand side is isomorphic to $G$, but the right hand side is a point.
The corresponding counterexample in your setting is the following. Suppose $G$ is a connected complex algebraic group acting on $X=G$ by left translations, then set $A=B=\mathcal D_X$. Then the left hand side is isomorphic to $\mathcal D_X$ but the right hand side is isomorphic to $\mathbb C$. To be even more concrete you could set $G$ to be the additive group $\mathbb C$ acting on itself by left translation.
|
2025-03-21T14:48:29.772016
| 2020-02-03T17:37:25 |
351855
|
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|
Stack Exchange
|
RSK correspondence
Up to now, what are the difference ways we know to define RSK correspondence? I already know:
By insertion and recording tableau.
Ball construction or Viennot's geometric construction.
Growth diagram proposed by Sergey Fomin.
Do you know other models?
There is a very enlightening description of RSK in terms of piecewise-linear maps that goes back to Igor Pak: https://eudml.org/doc/121696
See also the presentation of Robin Sulzgruber: https://dx.doi.org/10.4310/JOC.2020.v11.n2.a3
There is also a direct connection of RSK to important topics in representation theory like Hecke algebras, Kazhdan-Lusztig theory, Springer fibers, etc. and that perspective is probably the most 'canonical'. But it sounds like you are interested in combinatorial constructions.
Dear Sam, thank you so much for your references. it is very good if you can tell me more about the connection of RSK correspondence to important topic in representation theory. I just know a litter bit about RSK. Another connection will provide new point of view.
Do you know about the relation between KL cells and Knuth equivalence? See e.g. arxiv.org/abs/math/9910117 for the basics along these lines.
Dear Sam, it is new for me. Thank you so much!
Greene's theorem describes the shape of the insertion tableau in terms of unions of increasing subsequences of the permutation $w=a_1\cdots a_n$. Thus we can build up the insertion tableau $P(w)$ one step at a time by applying Greene's theorem to the subpermutations (subsequences) of $w$ consisting of the numbers $1,2,\dots,i$. We can similarly compute the recording tableau $Q(w)$ since $Q(w)=P(w^{-1})$.
Dear Prof. Stanley, thank you so much for your comment. This is a very new point of view. Thank you so much!
Dear Prof. Stanley, I am reading
Your book: "Enumerative combinatorics volume 2" and
The article by Prof. Sergey Fomin "Generalized Robinson-Schensted-Knuth correspondence"
to understand the RSK correspondence by growth graph for two-rowed array in lexicographic order. But it seems difficult for me to understand the article of Prof. Fomin since there is no example. I find an example in your book, but it just for the only permutation.
For example, to compute pair $(P,Q)$ of $w=(1,1,2,3,3,3,4,5// 2,3,1,1,3,4,5,1)$.
Could you give me some examples or better references!
@KhanhNguyen: we can reduce the case of two-rowed arrays to permutations by standardization. See the end of Section 7.11 of my book. Growth diagrams are "compatible" with standardization, so growth diagrams for two-rowed arrays reduce to the case of permutations.
Dear Prof. Stanley, I see it, page 321. Thank you so much!
Here, slightly edited, is the first paragraph of Steinberg's paper, An occurrence of the Robinson–Schensted correspondence.
Let $V$ be an $n$-dimensional vector space over an infinite field, $\mathscr F$ the flag manifold of $V$, $u$ a unipotent transformation of $V$, and $\lambda$ the type of $u$, a partition of $n$ whose parts are the sizes of the Jordan blocks for $u$. … The components of $\mathscr F_u$, the variety of flags fixed by $u$, correspond naturally to the standard tableaux of shape $\lambda$. The purpose of this note is to show that the "relative position" of any two components of $\mathscr F_u$ (in general an element of the Weyl group, in the present case an element of $S_n$) is given, in terms of the corresponding tableaux, by the Robinson–Schensted correspondence.
Thanks Timothy!
|
2025-03-21T14:48:29.772281
| 2020-02-03T18:24:01 |
351856
|
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|
Stack Exchange
|
Writing path-dependent conditional expectation in terms of distribution
Let $x$ be an $\mathbb{R}$-valued random variable, then for any bounded and continuous function $f:\mathbb{R}\rightarrow \mathbb{R}$ one may write
$$
\mathbb{E}[f(X)] = \int_{x \in \mathbb{R}} f(x)\pi(x)dx,
$$
where $\pi$ is the density of the law of $X$ (granted that it exists).
In general, if $f:C([0,T];\mathbb{R})$ is a continuous map to $\mathbb{R}$; then given an $\mathbb{R}$-valued process $X_t$ one can view $f(X_{\cdot})(\omega)$ as a random-variable (defined path-by-path). Is there way to write
$$
E[f(X_{\cdot})] = \int_{t=0}^T \int_{x \in \mathbb{R}} f(\cdot) ...\pi_t(x)dxdt?
$$
Or something similar (without integrating over the path-space directly)?
The answer is no. In general, to find $Ef(X_\cdot)$, you need to know the distribution of the entire path $X_\cdot\,$. So, as a minimum, you need to know the so-called finite-dimensional distributions of the process $X_\cdot$, that is, the joint distributions of the random variables $X_{t_1},\dots,X_{t_k}$ for any $t_1,\dots,t_k$ in $[0,T]$ -- whereas the densities $\pi_t$ determine only the individual distributions of the $X_t$'s for $t\in[0,T]$.
|
2025-03-21T14:48:29.772381
| 2020-02-03T18:27:30 |
351858
|
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|
Stack Exchange
|
Circle representations and regular representations
Write $\mathbb T$ for the circle group, $C_n$ for the cyclic group of order $n$, and fix embeddings $C_n \hookrightarrow \mathbb T$. Let $\lambda^i$ be the $\mathbb T$-representation with underlying vector space $\mathbb C$, such that
$$\mathbb T\times \mathbb C \ni (z,w) \mapsto z^i w\in\mathbb C$$
In particular $\lambda^0$ is the trivial representation, and $\lambda=\lambda^1$ is the standard representation. I am interested in the representation
$$[n]_\lambda = \lambda^0 \oplus \lambda^1 \oplus \dotsb \oplus \lambda^{n-1} $$
This has the property that its restriction to $C_n$ is the complex regular representation:
\begin{equation}\tag{$\ast$}
\mathrm{Res}^{\mathbb T}_{C_n} [n]_\lambda = \mathbb C[C_n] = \mathrm{Ind}^{C_n}_e \mathbb C
\end{equation}
Is there a description of $[n]_\lambda$ as a $\mathbb T$-representation—ideally as an induced or coinduced representation—which formally implies $(*)$ for all $n$?
|
2025-03-21T14:48:29.772464
| 2020-02-03T18:29:04 |
351859
|
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|
Stack Exchange
|
Extension of semi-flows on a locally compact metric space
Theorem 2.3, p. 26 from W Shen, Y Yi "Almost automorphic and almost periodic dynamics in skew-product semiflows" states that if there is a semi-flow $\varphi^{t}$, $t \geq 0$, on a locally compact metric space $\mathcal{S}$ such that every point $x \in \mathcal{S}$ has a unique backward orbit then the semi-flow $\varphi^{t}$ can be extended to a flow on $\mathcal{S}$. However, the proof given in the book is not satisfactory. In the most interesting part, where one has to show the continuity of $\varphi^{-t}(x) = (\varphi^{t})^{-1}(x)$ for $t > 0$, it refers to the "inverse function theorem" from a book on topology, in which I cannot find any such statement.
I know that the continuity of the inverse will follow if $S$ is compact (this is a well-known statement) or $S$ is a topological manifold (from the invariance of domain theorem). But I don't know any statement of automatic continuity for the inverse in the case of locally compact spaces.
This is not a complete answer, but an attempt. There exists a continuous bijective map $f \colon \mathcal{S} \to \mathcal{S}$ of a locally compact space to itself that is not a homeomorphism. This map can be contruscted from the standard conitunous bijective map of an half-open interval $[0,1)$ to the unit circle $\mathcal{S}^{1}$ that of course cannot be a homeomorphism. We just take a locally compact space $\mathcal{S} \subset \mathbb{R}^{2}$ as a disjoint union of inifite sequence of copies of $[0,1)$, say $\mathcal{I}_{k}$, and an infinite sequence of copies of $\mathcal{S}^{1}$, say $\mathcal{S}_{k}$, where $k=1,2,\ldots$. Let $f$ map $\mathcal{I}_{k+1}$ into $\mathcal{I}_{k}$, $\mathcal{I}_{1}$ into $\mathcal{S}_{1}$ and $\mathcal{S}_{k}$ into $\mathcal{S}_{k+1}$. Clearly, $f$ is a continuous bijection, but not a homeomorphism. This shows that the reasoning in the book is, indeed, at least unsatisfactory.
However, if we try to construct from this map a continuous analog with $\varphi^{1}=f$, we will have a problem with locally compactness (as for example, in the red point on the figure), which probably occurs since the dimension of the space is increased from one to two.
|
2025-03-21T14:48:29.772895
| 2020-02-03T19:07:06 |
351863
|
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"Abdelmalek Abdesselam",
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|
Stack Exchange
|
Topological spaces containing paths
Let $C(\mathbb{R}^n;\mathbb{R}^d)$ be the space of continuous functions with the uniform-convergence on compacts topology. What are function spaces $X$ building on $C(\mathbb{R}^n;\mathbb{R}^d)$?
$X$ properly contains $C(\mathbb{R}^n;\mathbb{R}^d)$,
The subspace topology on $C(\mathbb{R}^n;\mathbb{R}^d)$ (with respect to $X$) agrees with the topology of uniform convergence on compact subsets,
The elements of $X$ are functions from $\mathbb{R}^n$ to $\mathbb{R}^d$
$C(\mathbb{R}^n;\mathbb{R}^d)$ is dense in $X$.
This doesn't seem very restrictive: you can take any space of functions containing $C(\mathbb{R}^n; \mathbb{R}^d)$ and equip it with the topology of uniform convergence on compacts. What are you actually hoping to achieve?
I out to mention, that I would need $C(\mathbb{R}^n,\mathbb{R}^d)$ to be dense in $X$. This makes the problem much more difficult...
@AIM_BLB: I would say it is a strategic mistake to impose your second condition. Typically useful spaces $X$ would have to have a weaker topology. More precisely, the subspace topology on $C(\mathbb{R}^n,\mathbb{R}^d)$ coming from the topology of $X$ has to be weaker than the intrinsic/natural topology of $C(\mathbb{R}^n,\mathbb{R}^d)$ which is the one you gave. Otherwise you run into the kind of trouble explained in Nate's answer.
An example of useful $X$ would be the space of $\mathbb{R}^d$-valued Schwartz distributions.
Since your space is complete, there is no way you can embed it densely into a a uniform space (in particular, a tvs or even a topological group) in a non-trivial way. Perhaps if you gave some hint of what you want (which new functions you are after, whether the function space should be a vector space, a tvs, a lcs, whether you can weaken the condition on its topology), we could offer some more positive information.
@user131781: in "embed densely", careful about what "embed" means (which is a cultural issue). Folks who say read Munkres and studied point set topology would use the word in the strict sense of topological embedding which imposes equality with the subspace/induced topology. In analysis (think Sobolev embedding theorem), people don't do that and allow the small space to have a stronger topology than the one induced from the big space. This is the point I was trying to make above. This granted, there are lots of $LCS$'s where the given space "embeds densely".
Danke für die Belehrung. See the second bullet point in the query.
@user131781: Bitte schön. I saw the bullet point. In "This is the point I was trying to make above", I was referring to my earlier comment which advised dropping the condition in the second bullet point.
Any such topology will be fairly unpleasant. For instance, the topology of $X$ cannot be induced by any translation-invariant metric $d$.
Lemma. Let $Y_1, Y_2$ be two topological vector spaces whose topologies are induced by translation-invariant metrics $d_1, d_2$, and let $T : Y_1 \to Y_2$ be a continuous linear map. Then $T$ is uniformly continuous.
Proof. Since $T$ is continuous at 0, for any $\epsilon > 0$ there exists $\delta > 0$ such that if $d_1(x, 0) < \delta$ then $d_2(Tx, 0) < \epsilon$. Now if $d_1(x,y) < \delta$, then $d_1(x-y, 0) = d_1(x,y) < \delta$ and we have
$d_2(Tx, Ty) = d_2(Tx-Ty, 0) = d_2(T(x-y), 0) < \epsilon$.
Now recall that $C(\mathbb{R}^n; \mathbb{R}^d)$ is a Fréchet space, so its usual topology is induced by a complete translation-invariant metric $d_0$. By assumption, the identity map $id$ from $(C(\mathbb{R}^n; \mathbb{R}^d), d_0)$ to $(C(\mathbb{R}^n; \mathbb{R}^d), d)$ is a homeomorphism, and so by our lemma, $id$ and $id^{-1}$ are uniformly continuous. In particular, $C(\mathbb{R}^n; \mathbb{R}^d)$ is complete with respect to $d$, and therefore closed in $X$.
Edit. Indeed, $X$ cannot even be a sequential Hausdorff topological vector space. In particular, assuming it is a TVS, its topology cannot be induced by any metric, translation-invariant or not.
In the following, let for brevity $Y = C(\mathbb{R}^n; \mathbb{R}^d)$; the same argument works for any Fréchet space.
Suppose that $Y \subset X$ and that the subspace topology on $Y$ equals the usual topology induced by the complete translation-invariant metric $d_0$ on $Y$. I claim $Y$ is closed in $X$.
Suppose $x$ is in the $X$-closure of $Y$, so that there is a sequence $y_n \in Y$ converging to $x$ in the topology of $X$. Let $\epsilon > 0$ and let $B$ be the open $\epsilon$-ball of the metric $d_0$ centered at $0$. By assumption $B$ is open in the subspace topology of $Y$ inherited from $X$, so there is an $X$-open set $U$ such that $B = U \cap Y$. In particular, $0 \in U$. Now since subtraction is jointly continuous in $X$, there is another $X$-open neighborhood $V$ of $0$ such that for all $a,b \in V$ we have $a-b \in U$.
Since $y_n - x \to 0$ in $X$, there exists $N$ so large that for all $n \ge N$ we have $y_n - x \in V$ (using again the fact that $X$ is a topological vector space). Now if $n,m \ge N$, we have $y_n - x, y_m - x \in V$, so that $y_n - y_m = (y_n - x) - (y_m - x) \in U$. Moreover, $y_n - y_m \in Y$ because $Y$ is a vector space. So $y_n - y_m \in U \cap Y = B$, meaning that $d_0(y_n, y_m) = d_0(y_n - y_m, 0) < \epsilon$, using the fact that $d$ is translation invariant.
Hence $y_n$ is Cauchy in the complete metric $d_0$, so converges in $d_0$-metric to some $y \in Y$. Thus we also have $y_n \to y$ in the topology of $X$. Since the latter is Hausdorff, $x=y$ and thus $x \in Y$.
But it can still be metric?
@AIM_BLB: No. See edit.
Concerning the edit, I think that it is not necessary to assume that $X$ is sequential. The point is that a vector space topology induces a unique uniformity and complete subspaces of Hausdorff topological vector spaces are closed. There are of course different uniformities (even metrics) which may induce the same topology -- but this does not affect the argument because we are speaking about THE uniformity induced by a vector space topology having the entourages (in Köthe's book they are called vincinities) ${(x,y): x-y \in U}$ for $0$-neighbourhoods $U$.
|
2025-03-21T14:48:29.773314
| 2020-02-03T21:15:25 |
351870
|
{
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"Iosif Pinelis",
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|
Stack Exchange
|
Does convergence in law to absolutely continuous limit imply convergence in convex distance?
Let $(X_n)$ be a sequence of $\mathbb{R}^d$-valued random variables converging in distribution to some limiting random variable $X$ whose CDF is absolutely continuous with respect to the Lebesgue measure.
Does it follow that $X_n$ converges to $X$ in convex distance, i.e. that
$$\sup_{h} \lvert \operatorname{E}(h(X)) - \operatorname{E}(h(X_n)) \rvert \to 0,$$
where the supremum is taken over all indicator functions of measurable convex subsets of $\mathbb{R}^d$, if necessary assuming (absolute) continuity of the CDFs of the $X_n$ as well?
Remark 1:
For $d=1$, the implication is true and can be proven by Polya's theorem (convergence in law of real valued random variables towards a limit with continuous CDF implies uniform convergence of the CDF). Is it still true for $d \geq 2$?
Remark 2:
If absolute continuity is replaced by continuity the conclusion is false, see here
What is essential here is that the distribution of $X$ assigns little mass to sets which are essentially $(d-1)$-dimensional.
The standard approach to problems of this kind is to estimate
$$ \operatorname{P}(X_n \in K) - \operatorname{P}(X \in K) $$
from above by
$$ \operatorname{E}(g(X_n)) - \operatorname{E}(f(X)) , $$
and from below by
$$ \operatorname{E}(f(X_n)) - \operatorname{E}(g(X)) , $$
where $0 \leqslant f \leqslant \mathbb{1}_K \leqslant g \leqslant 1$, and $f$ and $g$ are continuous. In $k$-th step we choose $f$ and $g$ in such a way that $\operatorname{E}(g(x) - f(x)) < \tfrac{1}{2 k}$. Convergence of $\operatorname{E}(f(X_n))$ to $\operatorname{E}(f(X))$ and convergence of $\operatorname{E}(g(X_n))$ to $\operatorname{E}(g(X))$ imply that
$$ -\tfrac{1}{k} \leqslant \operatorname{P}(X_n \in K) - \operatorname{P}(X \in K) \leqslant \tfrac{1}{k} $$
for all $n$ large enough.
This works as expected, i.e. leads to convergence of $\operatorname{P}(X_n \in K)$ to $\operatorname{P}(X \in K)$, if (and only if) $\operatorname{P}(X \in \partial K) = 0$: then (and only then) it is possible to choose $f$ and $g$ with the desired property.
Now in order to get uniform convergence for a class of sets $K$ — here the class of convex subsets of $\mathbb{R}^d$ — in every step $k$ we should choose $f$ and $g$ from a fixed finite-dimensional set of functions (which, of course, can well depend on $k$). One solution for the class of convex $K$ is as follows.
By tightness, there is $R > 1$ such that $\operatorname{P}(|X_n| > R) < \tfrac{1}{4 k}$ uniformly in $n$. We will choose (small) $\delta \in (0, 1)$ at a later stage. We cover the ball $\overline{B}(0, R)$ using $d$-dimensional cubes $Q_j$ with edge length $\delta$ and vertices at lattice points $(\delta \mathbb{Z})^d$. To be specific, suppose that the cubes $Q_j$ are open sets. Let $h_j = \mathbb{1}_{Q_j}$ be the indicator of $Q_j$. Then
$$h_1 + \ldots + h_J = 1 \quad \text{a.e. on $\overline{B}(0, R)$}$$
(more precisely: everywhere on $\overline{B}(0, R)$, except possibly on the faces of cubes $Q_j$). We add $h_0 = 1 - \sum_{j = 1}^J h_j$ to this collection. Observe that $h_0 = 0$ a.e. on the complement of $\overline{B}(0, R)$, and hence $\operatorname{E}(h_0(X)) \leqslant \operatorname{P}(|X_n| > R) < \tfrac{1}{4 k}$. Furthermore, by the first part of this answer, we already know that $\operatorname{E}(h_j(X_n)) = \operatorname{P}(X_n \in Q_j)$ converges to $\operatorname{E}(h_j(X)) = \operatorname{P}(X \in Q_j)$ as $n \to \infty$ for every $j = 0, 1, \ldots, J$ (because the distribution of $X$ does not charge the boundaries of $Q_j$).
Given a convex set $K$, we define $f$ to be the sum of all $h_j$ corresponding to cubes $Q_j$ contained in $K$, and $g$ to be the sum of all $h_j$ corresponding to cubes $Q_j$ which intersect $K$. Clearly, $$0 \leqslant f \leqslant \mathbb{1}_K \leqslant g \leqslant 1 \qquad \text{a.e.}$$ As in the first part of the proof,
$$ \operatorname{E}(f(X_n)) - \operatorname{E}(g(X)) \leqslant \operatorname{P}(X_n \in K) - \operatorname{P}(X \in K) \leqslant \operatorname{E}(g(X_n)) - \operatorname{E}(f(X)) . $$
For $n$ large enough we have
$$ \sum_{j = 0}^J |\operatorname{E}(h_j(X_n)) - \operatorname{E}(h_j(X))| \leqslant \tfrac{1}{2 k} , $$
and so
$$ |\operatorname{E}(f(X_n)) - \operatorname{E}(f(X))| \leqslant \tfrac{1}{2 k} , \qquad |\operatorname{E}(g(X_n)) - \operatorname{E}(g(X))| \leqslant \tfrac{1}{2 k} .$$
Thus,
$$ -\tfrac{1}{2k} - \operatorname{E}(g(X) - f(X)) \leqslant \operatorname{P}(X_n \in K) - \operatorname{P}(X \in K) \leqslant \tfrac{1}{2k} + \operatorname{E}(g(X) - f(X)) $$
for $n$ large enough, uniformly with respect to $K$. It remains to choose $\delta > 0$ such that $\operatorname{E}(g(X) - f(X)) < \tfrac{1}{2k}$ uniformly with respect to $K$; once this is proved, we have
$$ -\tfrac{1}{k} \leqslant \operatorname{P}(X_n \in K) - \operatorname{P}(X \in K) \leqslant \tfrac{1}{k} $$
for $n$ large enough, uniformly with respect to $K$, as desired.
By definition, $g - f$ is the sum of some number of functions $h_j$ with $j \geqslant 1$ — say, $m$ of them — and possibly $h_0$. Recall that $\operatorname{E}(h_0(X)) \leqslant \operatorname{P}(|X_n| > R) < \tfrac{1}{4 k}$. It follows that
$$ \operatorname{E}(g(X) - f(X)) \leqslant \tfrac{1}{4 k} + \sup \{ \operatorname{P}(X \in A) : \text{$A$ is a sum of $m$ cubes $Q_j$} \} . \tag{$\heartsuit$} $$
We now estimate the size of $m$.
Lemma. For a convex $K$, the number $m$ defined above is bounded by a constant times $(R / \delta)^{d - 1}$.
Proof:
Suppose that $Q_j$ intersects $K$, but it is not contained in $K$. Consider any point $z$ of $K \cap Q_j$, and the supporting hyperplane $$\pi = \{x : \langle x - z, \vec{u} \rangle = 0\}$$ of $K$ at that point. We choose $\vec{u}$ in such a way that $K$ is contained in $\pi^- = \{x : \langle x - z, \vec{u} \rangle \leqslant 0\}$. If the boundary of $K$ is smooth at $z$, then $\vec{u}$ is simply the outward normal vector to the boundary of $K$ at $z$.
To simplify notation, assume that $\vec{u}$ has all coordinates non-negative. Choose two opposite vertices $x_1, x_2$ of $Q_j$ in such a way that $\vec{v} = x_2 - x_1 = (\delta, \ldots, \delta)$. Then the coordinates of $x_2 - z$ are all positive. It follows that for every $n = 1, 2, \ldots$, all coordinates of $(x_1 + n \vec{v}) - z = (x_2 - z) + (n - 1) \vec{v}$ are non-negative, and therefore the translated cubes $Q_j + n \vec{v}$ all lie in $\pi^+ = \{x : \langle x - z, \vec{u} \rangle \geqslant 0\}$. In particular, all these cubes are disjoint with $K$.
In the general case, when the coordinates of $\vec{u}$ have arbitrary signs, we obtain a similar result, but with $\vec{v} = (\pm \delta, \ldots, \pm \delta)$ for some choice of signs. It follows that with each $Q_j$ intersecting $K$ but not contained in $K$ we can associate the directed line $x_2 + \mathbb{R} \vec{v}$, and this this line uniquely determines $Q_j$: it is the last cube $Q$ with two vertices on this line that intersects $K$ (with "last" referring to the direction of the line).
It remains to observe that the number of lines with the above property is bounded by $2^d$ (the number of possible vectors $\vec{v}$) times the number of points in the projection of $(\delta \mathbb{Z})^d \cap \overline{B}(0, R)$ onto the hyperplane perpendicular to $\vec{v}$. The latter is bounded by a constant times $(R / \delta)^{d - 1}$, and the proof is complete. $\square$
(The above proof includes simplification due to Iosif Pinelis.)
Since the Lebesgue measure of $Q_j$ is equal to $\delta^d$, the measure of $A$ in ($\heartsuit$) is bounded by $m \delta^d \leqslant C R^{d - 1} \delta$ for some constant $C$. Furthermore, since the distribution of $X$ is absolutely continuous, we can find $\delta > 0$ small enough, so that $\operatorname{P}(X \in A) < \tfrac{1}{4 k}$ for every set $A$ with measure at most $C R^{d - 1} \delta$ (recall that $R$ was chosen before we fixed $\delta$). By ($\heartsuit$), we find that
$$ \operatorname{E}(g(X) - f(X)) \leqslant \tfrac{1}{4 k} + \sup \{ \operatorname{P}(X \in A) : |A| \le C R^{d - 1} \delta\} \leqslant \tfrac{1}{2 k} , $$
uniformly with respect to $K$.
To me, the main difficulty here was to prove something like your "$m$ is bounded by a constant times $(R / \delta)^{d-1}$", which of course seems intuitively obvious. Is there is a quick way to show this? Or a reference?
@IosifPinelis: $K' := K \cap B(0, R)$ is convex. Pick $x_0$ in the interior of $K'$. Then $K' = {x : |x - x_0| \le \phi((x - x_0)/|x - x_0|)$ for a Lipschitz function $\phi$ with some Lipschitz constant $L$. Cover the unit sphere with $C (L+1) (R/\delta)^{d-1}$ balls of radius $\delta/R$, centered at $u_j$. Then $B_j := B(x_0+\phi(u_j)u_j, \delta)$ cover $\partial K'$. Each $B_j$ intersects with at most a constant number of supports of $h_j$, and the desired bound follows.
@IosifPinelis: I just realised that in the above comment, $m$ depends on the Lipshitz constant, which in turn depends on the radius of the largest ball $B(x_0, r)$ contained in $K'$. I guess there must be a smarter argument; I'll try to come back to this tomorrow.
@IosifPinelis: I have just edited in a simpler argument into the answer. Also, I extended the argument to arbitrary absolutely continuous distributions, and fixed other errors. Thank you for your comment, and please let me know if you still find any issues.
@MateuszKwaśnicki Thank you VERY much for this very nice and elegant answer! I can follow everything and think your arguments are correct. See my other comment below for some suggestion on how to improve readability of your answer.
@MateuszKwaśnicki For future readers, it would be nice if you could streamline your answer a little bit: 1. With your general estimate $\heartsuit$ which works for any absolutely continuous distribution, $\clubsuit$ and $\spadesuit$ are no longer needed and can be removed, correct? 2. If I'm not mistaken, absolute continuity only enters through the condition $P(X \in \partial K)=0$. If so, this fact could be added. 3. When writing about the bound on $m$: An explanation why you have to do your construction when $\delta \geq R$ at the start would improve readability.
@MateuszKwaśnicki Another quick remark: I think your proof shows in particular that there is a version of the Polya theorem on $\mathhbb{R}^d$: If $\mathbb{R}^d$-valued rv converge in distribution to a limit with absolutely continuous distribution, then their CDF converge uniformly (i.e. they also converge in Kolmogorov distance).
@MateuszKwaśnicki : Nice Lipschitz idea to bound $m$; that was the bottleneck for me. However, with your definition of a partition of unity, I think the claim "Each of these balls intersects at most a constant number of supports of $h_j$" is incorrect -- if you mean here that the constant cannot depend on $J$. I also think that, instead of the $h_j$'s, you can simply use the indicators of cubes $C_j$ of small edge length $\delta$, since $P(X\in \partial C_j)=0$. These small (say left-open) cubes will form a partition of a big cube of edge length $R$.
@r_faszanatas: Done, thanks. If you see any further issues, feel free to edit the answer.
@IosifPinelis: Thanks. The argument was still not entirely correct, though. I replaced it with a still modified version.
@MateuszKwaśnicki : I liked your Lipschitz idea to bound $m$ (except for this idea being combined with a partition of unity). Was anything wrong about that idea? As for the new proof of the bound on $m$, I cannot follow it. In particular, I don't know in what specific way "[...] correspond to at most two cubes". Nor do I understand the jump to the conclusion of "at most $C(R/\delta)^{d-1}$". Also, I think convolutions "with a fixed bump function" are not at all needed, and for those convolutions the decomposition $h_1+\dots+h_J=1$ will not hold.
@IosifPinelis: (1) The "Lipschitz" proof had a gap: the domains of the "Lipschitz functions" could fail to be convex sets, and so the bound on the derivative only proved local Lipschitz property. This is minor, I suppose, but I fail to see a simple workaround. (2) I will add some details to the proof. (3) Although the sum of $h_j$ is still $1$ on the smaller ball $B(0, R)$, convolutions are indeed not needed. I will update the proof. Once again: thanks!
@MateuszKwaśnicki : I think the Lipschitz property was global: If you go from any one point $x$ of the domain of such a Lipschitz function to any other point $y$ of the domain, the function cannot increase by more than $K|x-y|$, where $K$ is the largest slope coefficient of all support hyperplanes at the boundary points corresponding to the points in the domain, just by the definition of a support hyperplane. You don't need derivatives here or any other local notions; support hyperplanes are global.
@IosifPinelis: Ah, that's right. Then, if you prefer the Lipschitz argument, feel free to edit the answer.
$\newcommand{\R}{\mathbb{R}}
\newcommand{\ep}{\varepsilon}
\newcommand{\p}{\partial}
\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}$
This is to try to provide a simplification and detalization of the answer by Mateusz Kwaśnicki.
Suppose that the distribution of $X$ is absolutely continuous (with respect to the Lebesgue measure) and $X_n\to X$ in distribution. We are going to show that then $X_n\to X$ in in convex distance, that is,
$$\sup_K|\mu_n(K)-\mu(K)|\overset{\text{(?)}}\to0$$
(as $n\to\infty$), where $\mu_n$ and $\mu$ are the distributions of $X_n$ and $X$, respectively, and $\sup_K$ is taken over all measurable convex sets in $\R^d$.
Take any real $\ep>0$. Then there is some real $R>0$ such that $P(X\notin Q_R)\le\ep$, where $Q_R:=(-R/2,R/2]^d$, a left-open $d$-cube. Since $X_n\to X$ in distribution and $P(X\in\p Q_R)=0$, there is some natural $n_\ep$ such that for all natural $n\ge n_\ep$ we have
$$P(X_n\notin Q_R)\le2\ep.$$
Take a natural $N$ and partition the left-open $d$-cube $Q_R$ naturally into $N^d$ left-open $d$-cubes $q_j$ each with edge length $\de:=R/N$, where $j\in J:=[N^d]:=\{1,\dots,N^d\}$.
Using again the conditions that $X_n\to X$ in distribution and $\mu$ is absolutely continuous (so that $\mu(\p q_j)=0$ for all $j\in J$), and increasing $n_\ep$ is needed, we may assume that for all natural $n\ge n_\ep$
$$\De:=\sum_{j\in J}|\mu_n(q_j)-\mu(q_j)|\le\ep.$$
Take now any measurable convex set $K$ in $\R^d$. Then
$$|\mu_n(K)-\mu(K)|\le|\mu_n(K\cap Q_R)-\mu(K\cap Q_R)| \\
+|\mu_n(K\setminus Q_R)-\mu(K\setminus Q_R)|
$$
and
$$|\mu_n(K\setminus Q_R)-\mu(K\setminus Q_R)|\le
P(X_n\notin Q_R)+P(X\notin Q_R)\le3\ep.
$$
So, without loss of generality (wlog) $K\subseteq Q_R$. Let
$$J_<:=J_{<,K}:=\{j\in J\colon q_j\subseteq K^\circ\},$$
$$J_\le:=J_{\le,K}:=\{j\in J\colon q_j\cap \bar K\ne\emptyset\},$$
$$J_=:=J_{=,K}:=\{j\in J\colon q_j\cap\p K\ne\emptyset\},$$
where $K^\circ$ is the interior of $K$ and $\bar K$ is the closure of $K$.
The key to the whole thing is
Lemma. $|\bigcup_{j\in J_=}q_j|\le2d(d+2)R^{d-1}\de$, where $|\cdot|$ is the Lebesgue measure.
This lemma will be proved at the end of this answer. Using the absolute continuity of the distribution of $X$, we can take $N$ so large that for any Borel subset $B$ of $\R^d$ we have the implication
$$|B|\le2d(d+2)R^{d-1}\de\implies \mu(B)\le\ep.$$
Using now the lemma, for $n\ge n_\ep$ we have
$$\mu_n(K)-\mu(K)
\le\sum_{j\in J_\le}\mu_n(q_j)-\sum_{j\in J_<}\mu(q_j) \\
\le\sum_{j\in J_\le}|\mu_n(q_j)-\mu(q_j)|
+\mu
\Big(\bigcup_{j\in J_=}q_j\Big)\le\De+\ep\le2\ep.
$$
Similarly,
$$\mu(K)-\mu_n(K)
\le\sum_{j\in J_\le}\mu(q_j)-\sum_{j\in J_<}\mu_n(q_j) \\
\le\sum_{j\in J<}|\mu(q_j)-\mu_n(q_j)|
+\mu
\Big(\bigcup_{j\in J_=}q_j\Big)\le\De+\ep\le2\ep.
$$
So, $|\mu_n(K)-\mu(K)|\le2\ep$. That is, the desired result is proved modulo the lemma.
Proof of the lemma. Since $K$ is convex, for any $x\in\p K$ there is some unit vector $\nu(x)$ such that $\nu(x)\cdot(y-x)\le0$ for all $y\in K$ (the support half-space thing), where $\cdot$ denotes the dot product. For each $j\in[d]$, let
$$S_j^+:=\{x\in\p K\colon\nu(x)_j\ge1/\sqrt d\},\quad
S_j^-:=\{x\in\p K\colon\nu(x)_j\le-1/\sqrt d\},$$
$$J_{=,j}^+:=\{j\in J\colon q_j\cap S_j^+\ne\emptyset\},\quad
J_{=,j}^-:=\{j\in J\colon q_j\cap S_j^-\ne\emptyset\},$$
where $v_j$ is the $j$th coordinate of a vector $v\in\R^d$.
Note that
$\bigcup_{j\in[d]}(S_j^+\cup S_j^-)=\p K$ and hence
$\bigcup_{j\in[d]}(J_{=,j}^+\cup J_{=,j}^-)=J_=$, so that
$$\Big|\bigcup_{j\in J_=}q_j\Big|
\le
\sum_{j\in[d]}\Big(\Big|\bigcup_{j\in J_{=,j}^+}q_j\Big|+\Big|\bigcup_{j\in J_{=,j}^-}q_j\Big|\Big)
\le\de^d
\sum_{j\in[d]}(|J_{=,j}^+|+|J_{=,j}^-|),\tag{*}
$$
where now $|J_{=,j}^\pm|$ denotes the cardinality of $J_{=,j}^\pm$.
Now comes the key step in the proof of the lemma:
Take any $x$ and $y$ in $S_d^+$ such that $x_d\le y_d$. We have the support "inequality" $\nu(x)\cdot(y-x)\le0$, which implies
$$\frac{y_d-x_d}{\sqrt d}\le\nu(x)_d(y_d-x_d)\le\sum_{j=1}^{d-1}\nu(x)_j(x_j-y_j)
\le|P_{d-1}x-P_{d-1}y|,
$$
where $P_{d-1}x:=(x_1,\dots,x_{d-1})$. So, we get the crucial Lipschitz condition
$$|y_d-x_d|\le\sqrt d\,|P_{d-1}x-P_{d-1}y| \tag{**}$$
for all $x$ and $y$ in $S_d^+$.
Partition the left-open $(d-1)$-cube $P_{d-1}Q_R$ naturally into $N^{d-1}$ left-open $(d-1)$-cubes $c_i$ each with edge length $\de=R/N$, where $i\in I:=[N^{d-1}]$.
For each $i\in I$, let
$$ J_{=,d,i}^+:=\{j\in J_{=,d}^+\colon P_{d-1}q_j=c_i\},\quad
s_i:=\bigcup_{j\in J_{=,d,i}}q_j,
$$
so that $s_i$ is the "stack" of all the $d$-cubes $q_j$ with $j\in J_{=,d}^+$ that $P_{d-1}$ projects onto the same $(d-1)$-cube $c_i$. Let $r_i$ be the cardinality of the set $J_{=,d,i}$, that is,
the number of the $d$-cubes $q_j$ in the stack $s_i$. Then for some two points $x$ and $y$ in $s_i\cap S_d^+$ we have $|y_d-x_d|\ge(r_i-2)\de$, whence, in view of (**),
$$\sqrt d\,\sqrt{d-1}\,\de\ge\sqrt d\,|P_{d-1}x-P_{d-1}y|\ge|y_d-x_d|\ge(r_i-2)\de,$$
so that $r_i\le d+2$. So,
$$|J_{=,d}^+|=\sum_{i\in I}r_i\le\sum_{i\in I}(d+2)=(d+2)N^{d-1}=(d+2)(R/\de)^{d-1}.$$
Similarly, $|J_{=,j}^\pm|\le(d+2)(R/\de)^{d-1}$ for all $j\in[d]$. Now the lemma follows from (*).
|
2025-03-21T14:48:29.774442
| 2020-02-03T22:34:10 |
351875
|
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"Craig Franze",
"GH from MO",
"Lucia",
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"https://mathoverflow.net/users/11919",
"https://mathoverflow.net/users/38624"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/351875"
}
|
Stack Exchange
|
Asymptotic behavior of a certain sum of ratios of consecutives primes
I am looking for the asymptotic growth of the following sum
$$\sum_{k=1}^{n}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}$$
where $p_k$ stands for the prime of index $k$.
Manual computations show, for small values of n, a behavior quite similar to that of the sum over naturals
$$\sum_{k=0}^{n-1}(2k+1)=n^2$$
But more accurate simulations with Python suggest that
$\sum_{k=1}^{n}\frac{p_{k+1}+\,p_k}{p_{k+1}-\,p_k}$ ~ $\frac{2}{e}\,n^2\log\log n$
Is there anybody who can confirm this asymptotic behavior and, if it is correct, give a sketch of a proof?
You can probably use $p_n \sim n\log n$. Just blindly putting that in leads to a ratio of $$\frac{(2k+1)\log k+(k+1)\log\left(1+\frac{1}{k}\right)}{\log k +(k+1)\log\left(1+\frac{1}{k}\right)},$$ which explains the $2k+1$ behavior.
It is elementary to prove that the sum grows at least as fast as $n^2$, and at most as fast as $n^2\log n$. The precise asymptotic behavior depends on the distribution of prime gaps $p_{k+1}-p_k$, on which we only have conjectures (see also my Added section below).
It is clear that
$$\#\{k\leq n: p_{k+1}-p_k>\log n\}<\frac{p_{n+1}}{\log n},$$
hence the contribution of $p_{k+1}-p_k>\log n$ is
$$\sum_{\substack{1\leq k\leq n\\p_{k+1}-p_k>\log n}}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}<\frac{p_{n+1}}{\log n}\cdot\frac{p_{n+1}+p_n}{\log n}=O(n^2).$$
Now, for a fixed even $h\leq\log n$, Hardy and Littlewood conjectured that
$$\#\{k\leq n: p_{k+1}-p_k=h\}\sim\frac{n}{\log n}\cdot 2C_2\cdot D_h,\tag{$\ast$}$$
where
$$C_2:=\prod_{p>2}\left(1-\frac{1}{(p-1)^2}\right)=0.66016\dots\qquad\text{and}\qquad D_h:=\prod_{\substack{p|h\\{p>2}}}\frac{p-1}{p-2}.$$
If we believe in this, then integration by parts gives that the contribution of $p_{k+1}-p_k=h$ is asymptotically $n^2 C_2 D_h/h$. Based on this heuristic, it is reasonable to conjecture that
$$\sum_{k=1}^{n}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}\sim C_2\, n^2\sum_{h\leq\log n}\frac{D_h}{h}.$$
It is straightforward that the Dirichlet series of $D_h$ factors as
$$\sum_{h=1}^\infty\frac{D_h}{h^s}=\zeta(s)F(s),$$
where $F(s)$ is an explicit Euler product converging uniformly in $\Re(s)>3/4$, say. Therefore, heuristically,
$$\sum_{k=1}^{n}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}\sim C\, n^2\log\log n,$$
where $C:=C_2F(1)$. Most likely, the constant $C$ is not equal to $2/e$ as suggested by the original post, but I have not checked this.
Added. It think that the known upper bounds for the left hand side of $(\ast)$ allow one to show, unconditionally, that the sum in question is $O(n^2\log\log n)$. As Lucia kindly pointed out, a result of Gallagher's implies that $C=1$.
Hi GH: The singular series constants are $1$ on average -- Hardy-Littlewood probabilities are approximately the same as Cramer probabilities on average. Thus one should have $C=1$ in your final answer.
@Lucia: Thanks for your comment! My $2C_2D_h$ is $\mathfrak{S}({0,h})$, and this also has average $1$ because of translation invariance of $\mathfrak{S}({h_1,h_2})$, right?
Yes, that's right. A slightly more precise approximation is that ${\frak S}({0, h})$ behaves like $1-1/h$ on average, and the negative secondary contribution shows up in some biases.
@Lucia: Thanks again! The negative bias might explain why the OP is seeing $2/e$ instead of $1$. I updated my "Added" section.
|
2025-03-21T14:48:29.774679
| 2020-02-03T23:36:47 |
351878
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351878"
}
|
Stack Exchange
|
Endomorphism ring isomorphic to regular representation
Given the regular representation, $\rho : G \rightarrow GL(K[G])$ define $End_{G}(K[G]) = \{ \phi : K[G] \rightarrow K[G] \text{ s.t. } \rho_{g} \circ \phi = \phi \circ \rho_{g} , \forall g \in G \}$. I want to show that $End_{G}(K[G]) \cong K[G]$. To do so, I observed that $\phi$ is determined by $\phi(1)$ because $\phi(\sum_{g \in G} a_{g} g) = \sum_{g \in G} a_{g} g \cdot \phi(1)$. I was trying to construct an isomorphism $\psi : End_{G}(K[G]) \rightarrow K[G]$ by $\psi(\phi) = \phi(1)$. This is a bijective map, but it is not an isomorphism. Suppose that $\phi_{1}(1) = \sum_{g \in G} a_{g} g$ and $\phi_{2}(1) = \sum_{b \in G} b_{g} g$. Then $\psi(\phi_{1} \phi_{2}) = \phi_{1} \circ \phi_{2} (1) = \phi_{1}(1) \cdot \phi_{2}(1)$, but I am not sure this is true because:
$$\phi_{1}(1) \cdot \phi_{2}(1) = \sum_{g_{1}, g_{2} \in G} a_{g_{1}} b_{g_{2}} g_{1} g_{2}$$.
But
$$\phi_{1} \circ \phi_{2}(1) = \sum_{g_{1} , g_{2} \in G} b_{g_{1}} a_{g_{2}} g_{1} g_{2}$$.
Are these quantities actually equal? I don't think so.
|
2025-03-21T14:48:29.774770
| 2020-02-03T23:40:47 |
351880
|
{
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"authors": [
"Martin Brandenburg",
"https://mathoverflow.net/users/2841",
"https://mathoverflow.net/users/72288",
"user237522"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351880"
}
|
Stack Exchange
|
A certain property of integral domains $A \subseteq B$ with $Q(A) \cap B= A$
I have asked the following question in MSE:
Let $k$ be a field of characteristic zero. Let $A \subseteq B$ be $k$-algebras which are also (commutative) integral domains with fields of fractions $Q(A) \subseteq Q(B)$.
Assume that $Q(A) \cap B=A$.
Define property $P$ as follows:
If $w:=\sum_{i=0}^{n} \tilde{a_i} b^i \in B$, where $\tilde{a_i} \in Q(A), b \in B$,
then $b \in A$ or $\tilde{a_i} \in B$ for all $0 \leq i \leq n$.
By the assumption that $Q(A) \cap B=A$, property $P$ becomes:
If $w:=\sum_{i=0}^{n} \tilde{a_i} b^i \in B$, where $\tilde{a_i} \in Q(A), b \in B$,
then $b \in A$ or $\tilde{a_i} \in A$ for all $0 \leq i \leq n$.
Question: When such $A \subseteq B$ satisfies property $P$? Namely, could one find an additional 'mild' condition which guarantees that
$A \subseteq B$ satisfies property $P$?
Non-answer:
The following four conditions do not guarantee that
$A \subseteq B$ satisfies property $P$:
(i) $B$ is simple over $A$, namely, $B=A[\hat{b}]$ for some $\hat{b} \in B$.
(ii) $B$ is integral over $A$, namely, for every element $b \in B$ there exist
$a_{n-1},\ldots,a_1,a_0 \in A$ such that
$b^n+a_{n-1}b^{n-1}+\cdots+a_1b+a_0=0$.
(iii) The invertible elements of $A$ and $B$ are $k^{\times}$ only.
(iv) $A \subseteq B$ is flat.
(v) Both $A$ and $B$ are UFD's.
Indeed, for example, $A=k[y^2], B=k[y]$, $\tilde{a_1}=\frac{1}{y^2}$, $b=y^3, w=\tilde{a_1}b$; this example appears as an answer here.
Plausible answers:
(1.) $A \subseteq B$ is a separable ring extension? (or separable and flat? Flatness alone is not enough).
(2.) $Q(A)=Q(B)$. Indeed, in this case we get
$B=Q(B) \cap B=Q(A) \cap B=A$. Then $b \in B=A$.
Remarks:
(1) Notice that if $n=0$ then property $P$ is satisfied; indeed,
if $n=0$ then $B \ni w=\tilde{a_0}b^0=\tilde{a_0} \in Q(A)$,
so $\tilde{a_0}= w \in Q(A) \cap B =A$.
(2) There is a nice theorem by H. Bass that appears here (Remark after Corollary 1.3, page 74); it presents conditions on two integral domains $A \subseteq B$ implying that $Q(A) \cap B=A$, but here I am assuming that this equality is satisfied.
Thank you very much for any hints and comments!
Do you want to say $a_i \in B$ for all $i$?
@MartinBrandenburg, thank you. Yes, I want to say that $\tilde{a_i} \in B$ for all $i$ (or $b \in A$). I now edited my question.
I have now found a partial answer to a special case of my question, namely the case where $n \leq 1$ (I suspect that a similar argument holds for any $n \in \mathbb{N}$): If $A,B$ are both UFD's and every irreducible element of $A$ remains irreducible in $B$, then property $P$ is satisfied. Perhaps I will later add the argument.
|
2025-03-21T14:48:29.774969
| 2020-02-04T00:34:32 |
351883
|
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"Timothy Budd",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351883"
}
|
Stack Exchange
|
Solving general two-dimensional recurrence relation
Any techniques for deriving a closed form solution for the following recurrence relation? Or bounds on asymptotic behavior for large $n$?
$$a_{n+1,k} = \sum_{0 \le i \le n} \frac{n!}{i!} a_{i,k-1}$$
where $n,k \ge 0$ and $a_{0,0} = 0$
You will need more boundary conditions; I'm guessing you'll want to set $a_{n, 0}$.
For fixed $r > 1$, if you set $a_{r-1,0} = 1$ and $a_{n,0} =0$ for $n\neq r$ then I believe the solution is given by the $r$-Stirling numbers $a_{n,k}=\left[\begin{array}{c} n \ k+1 \end{array}\right]_r$. See Broder, "The r-Stirling numbers" (1984).
|
2025-03-21T14:48:29.775044
| 2020-02-04T02:47:47 |
351884
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Anton Petrunin",
"Jeremy Brazas",
"LSpice",
"Nick Addington",
"Noah Schweber",
"https://mathoverflow.net/users/1441",
"https://mathoverflow.net/users/16914",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351884"
}
|
Stack Exchange
|
Are open sets determined by paths?
Let $X$ be a topological space, let $U \subset X$, and suppose that for every path $\gamma\colon [0,1] \to X$ the preimage $\gamma^{-1}(U)$ is open. Is it true that $U$ is open? Presumably not in general, but are there reasonable requirements we can put on $X$ to make it true?
To put it another way, the subsets $U$ as above give a topology on $X$, which is the same as or finer than the original topology. When is it the same?
To put it a third way, consider the evaluation map $I \times C(I,X) \to X$, where $I = [0,1]$ and $C(I,X)$ is the set of continuous maps, with (unusually) the discrete topology. When is this is a quotient map?
The $p$-adics (or any non-discrete, totally disconnected space) admit no non-constant paths.
Nice. And the topologist's sine curve will become disconnected in this topology, so maybe we want to say locally path-connected at least.
Any topology on $[0,1]$ strictly coarser than the usual one gives a counterexample - and e.g. the cofinite topology on $[0,1]$ is path connected. I think the right context for this question, at least at first, is (locally) path-connected Hausdorff spaces.
The spaces satisfying this condition are often called “delta generated” spaces.
What about locally path connected spaces?
A space $X$ is called $\Delta$-generated if $U$ is open in $X$ if and only if $\alpha^{-1}(U)$ is open in $[0,1]$ for every path $\alpha:[0,1]\to X$.
It's easy to see that a space $X$ is $\Delta$-generated if and only if $X$ is a quotient space of a disjoint union of copies of $[0,1]$. It follows that every $\Delta$-generated space is a locally path-connected $k$-space. It's also not too hard to show that every first countable locally path-connected space is $\Delta$-generated. Hence,
$$\text{locally path connected & first countable}\Rightarrow \Delta\text{-generated}\Rightarrow\text{locally path connected & }k\text{-space}$$
The category of $\Delta$-generated spaces is very well-behaved, namely, it is a Cartesian closed coreflective subcategory of $\mathbf{Top}$.
The following ``directed hedgehog space" is a locally path-connected space that is not $\Delta$-generated: Let $(A_{\lambda},a_{\lambda})$ be a copy of the based unit interval $([0,1],0)$ for all ordinals $\lambda<\omega_1$. Consider the one-point union $X=\bigvee_{\lambda<\omega_1}A_{\lambda}$ with based point $x_0$ and with the directed wedge topology. This means that $U$ is open in $X$ if $U\cap A_{\lambda}$ is open in $A_{\lambda}$ for all $\lambda$ and if when $x_0\in U$, there exists $\kappa<\omega_1$ such that $A_{\lambda}\subseteq U$ for all $\lambda\geq \kappa$.
Clearly, $X$ is locally path-connected and as the quotient of a compact Hausdorff space, $X$ is a $k$-space. Let $b_{\lambda}$ be the image of $1$ in $A_{\lambda}$, i.e. the end of the $\lambda$-th arc, and set $C=\{b_{\lambda}\mid \lambda<\omega_1\}$. Since $[0,1]$ is sequentially compact and no sequence of countable ordinals converges to $\omega_1$, one can show that every path in $X$ intersects $C$ in at most finitely many points. Even though $C$ is not closed in $X$, $\alpha^{-1}(C)$ is closed in $X$ for every path $\alpha:[0,1]\to X$.
By similar reasoning, the quotient space $\displaystyle X'\cong \frac{(\omega_1+1)\times [0,1]}{\{\omega_1\}\times [0,1]\cup (\omega_1+1)\times\{0\}}$ will be a non-$\Delta$-generated $k$-space but it is not locally path-connected. I imagine that you can build a non-$\Delta$-generated space that is both locally path-connected and a $k$-space, but I don't know one off the top of my head. The $k$-space coreflection of $X$ and locally-path connected coreflection of $X'$ don't seem to work.
|
2025-03-21T14:48:29.775295
| 2020-02-04T03:54:04 |
351885
|
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"Charlotte",
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"https://mathoverflow.net/users/141429",
"vidyarthi"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/351885"
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|
Stack Exchange
|
Algebraic structures on graphs
There are many algebraic structures linked to graphs.
For example one can find zero divisor graphs $[1]$, $[2]$ and many other graphs.
Does there exist any survey paper which characterizes all the graphs obtained till now from algebraic structures?
If someone could provide any survey article which can give some light, I will be grateful
$[1]$ Anderson, D.F. and Livingston, P.S., 1999. The zero-divisor graph of a commutative ring. Journal of Algebra, 217(2), pp.434-447.
$[2]$ Babai, L., 1979. Spectra of Cayley graphs. Journal of Combinatorial Theory, Series B, 27(2), pp.180-189.
you could see the papers by Prof. Ayman Badawi at this link. He has worked on Total graph of a ring, annihilator graph of a ring, etc.
@vidyarthi; can you provide a more detailed answer if possibe
|
2025-03-21T14:48:29.775381
| 2020-02-04T04:02:25 |
351886
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/351886"
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|
Stack Exchange
|
Shifting the group homology of a topological group?
Let $G$ be a topological group. It has a classifying space $BG$, which has homology groups $H_{*}BG$. Changing the topology of $G$ affects the space $BG$ and hence its homology groups.
For example the group $\mathbb{R}$ with its usual topology has $H_{*}B\mathbb{R}\simeq H_{*}pt$. Changing the topology to be much finer, namely the discrete topology, results in a topological group $\mathbb{R}^{\delta}$ that has nontrivial $H_{1}$.
Question: Is it possible to go the other way? I.e. does there exist a topological group $G$, which has a second topology, coarser than the first, that makes the group homology larger?
There are many ways to make "larger" precise--I'm most interested in the case where initial topological group is acyclic. Then the meaning of "larger" is clear.
Motivation: It would help answer this question.
Have you tried the translation-invariant topology on $\Bbb R$ (a set is open iff it's the inverse image of an open set of $S^1$)?
I hadn't, but doesn't that act freely and properly on $\mathbb{R}$?
I don't think that action is continuous.
I was hoping that because this topology gives the subgroup Z of integers the indiscrete topology (making them contractible), the group homology would be closer to that of $S^1$.
Let $p$ be the projection map $\Bbb R \to S^1$. Let $T$ be $\Bbb R$, where we say that a subset $U$ is open if and only if $U = p^{-1} V$ for some open set $V$ of $S^1$. This makes $T$ a topological group, and the maps $\Bbb R \to T \to S^1$ are homomorphisms of topological groups. The space $T$ also has the property that a map $f: X \to T$ is continuous if and only if the composite $p \circ f$ is continuous.
The homotopy type of the space $T$ can also be understood: the projection $p: T \to S^1$ is a homotopy equivalence. To show this, we define $i(x)$ to be the unique element of $[0,1)$ such that $p(i(x)) = x$; then $p \circ i = id$ by definition, so $i$ is continuous. The map $i \circ p$ is also homotopic to the identity: we need to construct a homotopy $H: T \times [0,1] \to T$ which is continuous with various properties, but to verify continuity it suffices for $p \circ H$ to be continuous, and so defining
$$
H(t,s) = \begin{cases} t &\text{if }s = 0,\\i(p(t)) &\text{if }s > 0\end{cases}
$$
works perfectly well.
Therefore, $T$ is a coarser topology on $\Bbb R$ with the same homotopy and homology groups as $S^1$, and (for definitions of $B$ in terms of principal bundles with contractible total space) $BT$ has the same homotopy and homology groups as $BS^1$, larger than the group homology of $\Bbb R$.
|
2025-03-21T14:48:29.775569
| 2020-02-04T05:33:26 |
351890
|
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|
Stack Exchange
|
Highest weight vector as a global section of an affine scheme
Let $G$ be a connected, reductive quasi-split group over a field $k$, acting on an afffine $k$-variety $X$. Let $B = TU$ be a Borel subgroup of $G$ with maximal torus $T$ and unipotent radical $U$. Then $G$ acts on the $k$-algebra of global sections $\mathcal O_X(X)$ of $X$. If necessary, we can assume $k$ has characteristic zero, $k$ is algebraically closed, or $G$ is split.
I recently asked a question on here about what we can say about a fundamental domain for the quotient space of $X$ under the action of $U$ in the case where $X = \mathbb A_k^n$ and $G$ acts by linear automorphisms. Friedrich Knop gave a very helpful answer, and in order to better understand it and the references he provided, I wanted to ask about some basic terminology and results assumed there.
A highest weight vector for this action is a nonzero global section $f \in \mathcal O_X(X)$ with the property that $b.f = \chi(b)f$ for some $k$-rational character $\chi$ of $B$. The stabilizer of the line $kf$ through $f$ is then a parabolic $k$-subgroup of $G$ which contains $B$.
I haven't been able to find much about highest weight vectors defined in this sense. Searching online mostly returns references for the more traditional notion of highest weight vector, which is an element in the one dimensional weight space of a dominant integral weight of a finite dimensional representation of $G$.
What some good references to learn about actions of reductive groups on affine varieties and highest weight vectors in the above sense? Do highest weight vectors always exist? Is there any sense in which they are unique or have a finite multiplicity property?
They are the same as the usual highest weight vectors but the $G$-representation is realized in the coordinate ring of some affine variety $X$. The usual argument goes if $\alpha:G \times X \to X$ is an algebraic action and $f \in \mathcal{O}_{X}$ then $\alpha^*(f) \in \mathcal{O}_G \otimes \mathcal{O}_X$ satisfies $\alpha^{*}(f)(g,x) = \sum_{i=1}^{n} \phi_i(g)f_i(x)$ hence $f$ generates a finite dimensional $G$-stable module spanned by the $f_i$. Thus the coordinate ring can be decomposed as a direct sum (using reductivity + char 0) of the irred. finite dimensional $G$-modules that appear in $\mathcal{O}_X$. By the usual highest weight theory these correspond to dominant weights $\lambda$ and as such there is an $f_{\lambda}$ generating a $G$-module $\cong V(\lambda)$. $f_{\lambda}$ spans the unique line stabilized by $B$ which acts on it through $\lambda$, hence it is a semi-invariant.
If $X$ is affine, normal and has an open Borel orbit (e.g. is spherical) then you can show multiplicity $\leq 1$ for the $\lambda$'s that appear basically imitating the argument for $G$ acting on itself. You can look up "algebraic Peter Weyl" or look at the end of Humphrey's LAG book and Timashev's Homogeneous Spaces and Equivariant Embeddings chapsters 1 and 5.
Thanks. If $X = \mathbb A_k^n$ and $G$ acts irreducibly by linear isomorphisms, is there any connection between highest weight vectors in $X$ and those in $\mathcal O_X(X)$?
|
2025-03-21T14:48:29.775899
| 2020-02-04T07:15:42 |
351891
|
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|
Stack Exchange
|
Asymptotic behavior of $\sum_{k=1}^{n}\frac{p_{k+1}}{p_{k+1}-p_k}$
I refer to my previous question Asymptotic behavior of a certain sum of ratios of consecutives primes.
We can split the sum
$$\sum_{k=1}^{n}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}$$
where $p_k$ stands for the prime of index $k$, into the following two
$\sum_{k=1}^{n}\frac{p_{k+1}}{p_{k+1}-\,p_k}$ ~ $\frac{n\,(n+1)}{e}\,\log\log n$
$\sum_{k=1}^{n}\frac{p_{k}}{p_{k+1}-\,p_k}$ ~ $\frac{(n-1)\,n}{e}\,\log\log n$
Is there anybody who can confirm this asymptotic behavior and, if it is correct, give a sketch of a proof?
My response to your earlier question applies almost verbatim. The heuristic reasoning there gives that
\begin{align*}
\sum_{k=1}^{n}\frac{p_k}{p_{k+1}-p_k}&\sim\frac{C}{2}\, n^2\log\log n,\\
\sum_{k=1}^{n}\frac{p_{k+1}}{p_{k+1}-p_k}&\sim\frac{C}{2}\, n^2\log\log n,
\end{align*}
where the constant $C>0$ is the same as in that post. As I wrote there, this constant is almost surely different from $2/e$. In fact, as Lucia kindly pointed out in a comment, $C=1$.
The difficulty in estimating these sums lies in the erratic behaviour of the denominator $p_{k+1}-p_k$. The numerator $p_k$ (resp. $p_{k+1}$ or $p_k+p_{k+1}$) is easy to handle as it is asymptotically $k\log k$ (resp. $k\log k$ or $2k\log k$).
|
2025-03-21T14:48:29.776005
| 2020-02-04T07:50:49 |
351894
|
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|
Stack Exchange
|
$\lt_{ip}$ is a well-defined well-ordering of iterable set premice
I am cross-posting this question from MSE, where I asked it about $3$ months ago and I decided to ask it here as well.
This question of mine arises from Kanamori's the higher infinite, where he tries to prove the result attributed to Silver and Solovay, that if $\omega_1^{L[U]} = \omega_1$, then there is a $\Pi_2^1$ set without the perfect set property.
Here we are dealing with ZFC$^-$(ZFC minus the Powerset Axiom) premice.
To be more precise, assume that $M$ is a transitive model of ZFC$^-$ and that $U$ is some set in $V$. We say that $\langle M, \in, U\rangle$ is a ZFC$^-$ premouse (at $\kappa$) iff $U$ is an $M$-ultrafilter over $\kappa$ and that for some $\zeta$, $M = L_\zeta[U].$
Also we say that two premice $\langle M, \in, U\rangle$ and $\langle N, \in, W\rangle$ are comparable iff $\exists F \exists \zeta \exists \eta$ such that
$M = L_\zeta[F]$ and that $N = L_\eta[F]$.
Now there is a lemma (called the Comparison lemma) which states that:
If $\langle M, \in, U\rangle$ and $\langle N, \in, W\rangle$ are iterable premice, then they have iterates which are comparable.
Now let $\langle M, \in, U\rangle$ and $\langle N, \in, W\rangle$ be iterable premice, define $\lt_{ip}$ in the following manner:
$\langle M, \in, U\rangle \lt_{ip}\langle N, \in, W\rangle$ iff there exists some $F$ and some $\zeta$ and $\eta$ such that $\langle L_\zeta[F], \in, F\cap L_\zeta[F]\rangle$ is an iterate of $\langle M, \in, U\rangle$ and $\langle L_\eta[F], \in, F\cap L_\eta[F]\rangle$ is an iterate of $\langle N, \in, W\rangle$, such that $\zeta \lt \eta$.
Now Kanamori says that this ordering is a well-defined well-ordering of iterable set premice. And he says that this is straightforward to check, using the comparison lemma.
I am still stuck on showing that this is well-defined. The best idea I had was to show that: if $\alpha$ and $\beta$ are the first indices of the iterations of $M$ and $N$ which are comparable, then all the higher iterates should be comparable in a "coherent" fashion.
So what I did was this: Let $\langle M_\alpha, \in, U_\alpha, \kappa_\alpha, i_{\alpha\beta} \rangle_{\alpha\le\beta\in\text{On}}$ and $\langle N_\alpha, \in, W_\alpha, \lambda_\alpha, j_{\alpha\beta} \rangle_{\alpha\le\beta\in\text{On}}$ be the iterations of $M$ and $N$, respectively. Then let $\alpha$ and $\beta$ be the first ordinals where $M_\alpha$ and $N_\beta$ are comparable. Let $F, \zeta, \eta$ be such that:
$M_\alpha = L_\zeta[F]$ and $N_\beta = L_\eta[F]$. There are $2$ cases:
$(1)$ $\zeta = \eta$: At this point I know that the rest of their iterations should be the same. But I think for totality's sake I have to show that $M = N$. Which is not obvious to me at the moment. (*)
$(2)$ $\zeta \lt \eta$: We can see that $M_\alpha,U_\alpha \in N_\beta$ so that the iteration of $N$ can witness the iteration of $M$ inside it. So for $\delta \in \text{On}$, we can see that $M_{\alpha + \delta} \in N_{\beta + \delta}$, but I can't generalize this argument for all $\delta \ge \alpha$ and $\xi \ge \beta$, i.e. that if $M_\delta$ and $N_\xi$ are comparable via some $G$, then $M_\delta$ falls below $N_\xi$.(**)
(*) and (**) are the two points where I can't finish this argument. Now my question is that, can the above argument be completed? Or is there some other way to prove that $\lt_{ip}$ is well-defined?
Also I would really appreciate any hints or remarks concerning the well-order part.
EDIT I:
The material here can be found in Kanamori's "The Higher Infinite", page $273$, $2$nd edition.
EDIT II:
As Yair Hayut kindly pointed out in the comments below, my definition of $\lt_{ip}$ was flawed. It is now fixed.
Also it was pointed out that it is reasonable to identify each premouse with it's iterates. In this light $(1)$ becomes:
$(1)^*$ In the case $\zeta = \eta$ we should find some premouse $\langle B, \in, O\rangle$ such that both $M$ and $N$ are iterates of $B$. In this case we insure totality.
EDIT III:
Also $(1)^*$ is evidently equivalent to:
$(1)^+$ We have to show that one of $M$ or $N$ is an iterate of the other one.
Note that case (1) can certainly happen for $M \neq N$. Just take $M$ to be any iterated ultrapower of $N$. Your order seems to be a pre-well-order in which a mouse is identified with its iterations. One can well-order each equivalence class by the critical point.
@YairHayut, I actually gave this point some thought before, but I couldn't come up with any concrete examples, since these mice seem to jump to different relative constructive hierarchies in each step. Are there such examples?
I think that your citation of Kanamori is inaccurate - you might need to iterate $M$ and $N$ in order to get to the same $F$ in the definition of $<{ip}$. Now, take $M = L{\alpha}[U]$. Let $N = \mathrm{Ult}(M, U)$. $N$ is also a mouse (of the form $L_{\beta}[U']$ where $U'$ is the image of $U$ under the ultrapower map). In the comparison process of $M$ and $N$, we are going to take the ultrapower of $M$ once and halt.
@YairHayut, Oh, you are absolutlely right. I made a mistake when writing the definition and confused myself. :) Thanks for pointing that out. So the main idea is to identify all premice with their iterations and then look at the order in that way. But even then I still have doubts about (1). I can imagine two different mice being iterated and getting merged at some point. Unless we can prove any pair of such mice are exactly the iterations of some mouse strictly below both?
While it is (sometimes) possible to reconstruct a (fine structural) mouse from its iterates, I think that the arguments for this are more sophisticated than what you need here. Note that in order to compare two mice $M$ and $N$ as in the question, you can simply hit the measures in both of them $\mu = (\max |M|, |N|)^+$ times. Now, it is clear that if there is an iteration that makes $M$ to be an initial segment of $N$ than also the $\mu$-steps iteration will do the same, which is what you need for (2).
@YairHayut, I'm afraid I can't quite follow your reasoning. Can you please explain why doing the iteration for $\mu$ steps is enough? Also it is again not clear to me how we can prove that the iterations behave coherently in the first $\mu$ steps.
Note that in those types of mice, the comparison process is simple:
Lemma: Let $M = L_{\alpha}[U]$, $N = L_{\beta}[W]$ be two mice. Let $\mu$ be a regular cardinal $\mu \geq (\max(\alpha,\beta))^+$. Then $M_{\mu} = L_{\alpha'}[F], N_{\mu} = L_{\beta'}[F]$ where $F$ is the club filter on $\mu$, $M_{\mu}$ is the iteration of $M$ for $\mu$ many steps and $N_\mu$ is the iteration of $N$ for $\mu$ many steps.
Proof: Let $C_M = \{\kappa_{\alpha} \mid \alpha < \mu\}$ be the critical points of the iteration on $M$. Then $C_M$ is a club at $\mu$, and $j_{\mu}(\kappa) = \mu$.
Moreover for all $X \in M_{\mu}$, $$X \in j_{\mu}(U)\iff \exists \zeta < \mu,\ X = j_{\zeta, \mu}(X'), \text{ and } X' \in j_{\zeta}(U)$$
In turn, this is equivalent to $$\forall\zeta' \geq \zeta,\ \kappa_{\zeta'} \in X.$$
On the other hand, if $X \notin j_{\mu}(X)$ then $j_{\mu}(\kappa) \setminus X \in j_{\mu}(U)$ and thus $X$ is disjoint from a club. We conclude that $M_\mu = L_{\alpha'}[F]$ for some ordinal $\alpha'$. The same argument works for $N$.
Lemma: Let $M, N, \alpha', \beta'$ be as above. $M<_{ip} N$ iff $\alpha' < \beta'$.
Proof: First, the order of $\alpha', \beta'$ does not depend on the choice of $\mu$. Moreover, if there is some iteration $M_{\xi}, N_{\eta}$ of different lengths such that $M_{\xi}$ is an initial segment of $N_{\eta}$ than continue by iterating both $M_{\xi}, N_{\eta}$ by $\mu$ many steps where $\mu$ is regular cardinal above the size of both $M_{\xi}, N_{\eta}$. Then by the previous lemma, $M_{\xi + \mu} = M_{\mu}$ is either an initial segment of $N_{\eta + \mu}$ or the other way. The second case cannot occur, since $M_{\xi}$ is an initial segment of $N_{\eta}$ and this is preserved by the longer iteration.
Many thanks for the answer. I think I got it, but for the sake of reassuring myself, I have a question. In the last step we probably should assume towards a contradiction that $N_\eta$ is an intial segment of $M_\xi$ and iterate them $\mu$ times, for a large enough $\mu$, to get a contradiction, right? Because assuming the other way around, we have nothing to prove, no?
Also, I still have some doubts about $(1)^+$. Is it even true? If not, what conditions should we put on our mice for that to happen?(conditions like countability, etc... that don't break the mentioned theorems proof).
|
2025-03-21T14:48:29.776539
| 2020-02-04T09:49:10 |
351897
|
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|
Stack Exchange
|
Signature Map From $p$-Geometric Rough Paths to $T(\mathbb{R})$
Let $f:[0,T]\rightarrow \mathbb{R}^d$ be a p-geometric rough path and let $\mathcal{G}_p^d$ be the collection of all such paths. Does the Lyons signature map define a continuous bijection between
$\mathcal{G}_p^d$ and $T(\mathbb{R}^d)$?
The signature is continuous on the space of $p$-geometric rough paths, but it is not injective since it is parametrisation-independent and invariant under concatenation with "tree-like" pieces. Boedihardjo, Geng, Lyons and Yang showed in this article that these are the only constraints, so that the signature can be inverted if we consider reduced paths modulo reparametrisation. However, even if we identify paths in this way, one does not expect the inverse map to be continuous in the topology of $T(\mathbb{R}^d)$.
Would there be an intuition for the final topology on $T(\mathbb{R}^d)$ induced by the (quotient of) the signature map (defined by the identification you mentioned)?
This would guarantee that the signature map is continuous...
Unfortunately, there isn't even an alternative characterisation of the image of the signature map...
Then could we not define a new topology on $T(\mathbb{R}^d)$ by pushing-forward the topology on the (quotient/identified) space of $p$-geometric rough paths via the (now bijective) "signature" map?
|
2025-03-21T14:48:29.776675
| 2020-02-04T10:51:30 |
351900
|
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|
Stack Exchange
|
Importance of textbooks in health of a sub-discipline
I am interested in published articles, and also more informal writing (blog posts, talk slides etc.) which discuss the importance of textbooks (where this word encompasses research monographs etc.) in the long-term health of a sub-discipline in Mathematics.
Motivation:
I have been thinking of late about how large Mathematics is getting (compared to, say, 50-60 years ago) with many more mathematicians and many more published papers. It also seems to me that at least some areas are becoming increasingly technical.
Furthermore, it can be very hard to follow a field by just reading the original papers: there can be false steps, or incomplete results, which only reach final form after some attempts. Often the pressures of space mean that motivation, or background material, is omitted in articles.
A good textbook can solve all of these problems. It seems to me that especially graduate students, or more established mathematicians seeking to move field, or use results of a different field to their own, face these sorts of problems in the extreme. By contrast, people working in the field probably carry around a lot of the "missing content" in their heads. This then makes me wonder if the lack of textbooks might lead to an ever increasing barrier to entry, and perhaps to sub-disciplines dying out as younger/newer mathematicians do not take up the study.
Hence my question of whether these thoughts have been stated in a longer, more thought out way before.
I hope this is suitable here: I guess also it might belong at academia.stackexchange but I felt that the issues involved are really quite specific to Mathematics (a slowly changing, technically deep field which it's not unusual to look at papers published many, many years ago, for example) where "curating knowledge" in the form of textbooks etc. might be rather different to other subjects. I have asked for this to be made community wiki.
much of the literature on math text books addresses elementary mathematics, as taught in high school; for some thoughts on the use of text books in advanced mathematics, see https://mathoverflow.net/q/13089/11260
Wish there were a textbook for complexification. I'm fairly certain for each correct claim I make here or here, at least 1 mathematician has already proved it.
Matthew Daws and @CarloBeenakker do you know of any textbook that says anything like in the ones I linked in previous comment?
I have rolled back to Matt's original question. The use of CW status on MO usually serves a different role from on other sites, and I think Matt's original wording should be left to stand.
Another interesting issue is: When does new mathematics filter down to the college (undergraduate) level. For example, discrete math vs. calculus.
@JohnSmithKyon FWIW: my impression from my own conversations with Matthew is that the level of mathematics which he has in mind is much higher than the level of your questions about complexification, where it seems that most of what you have been working out and pasting on this website is routine calculation that follows logically from the definitions. (I seem to recall that some of this is outlined in a section of the Kostrikin--Manin book that you refer to.)
thanks @YemonChoi . but still. it's just some though. i'm not saying any of the correct statements are hard to prove. i just found it hard as a beginner to think of this correct statements. it's like the answers are easy, but the questions were hard. for example, do you know any textbook or reference that has joppy's intuition?
@JohnSmithKyon I'm afraid I do not have any interest in further discussion on this matter, or on the particular details that you have been writing out on MO and MSE. Good luck with your own studies, but please be aware that just because people have not written out details that does not mean they do not know how to do so; moreover, part of maturing as a mathematician is to learn to leave out routine calculations in order to focus on the salient details.
The following is purely my opinion, but the specific question that you posed seems to invite such answers.
Strictly speaking, I'd say that the answer to your question is "Yes", a lack of textbooks in a new area is a barrier to entry. However, my experience over the past 30+ years is that areas don't stay "new" for long, and as things become more solidified, people write introductory (graduate level) textbooks. This may be to promote their vision, or it may simply be because they find the subject beautiful and want to share that beauty with others. To take an older example, Grothendieck's revolution in algebraic geometry created a large barrier for entry, but Hartshorne's book appeared when I was in graduate school, and it provided a way in. Was it perfect? No. Have other books, possibly better introductions, appeared since. Sure. But it was there, and I think it's fair to say that it helped train a generation (or more) of algebraic geometers and those in allied fields. (I'm in the latter group.)
So "yes", lack of graduate level texts in a area is a barrier to entry. But is it a long-term problem. I'd suggest that the answer is "No", because a new area of mathematics that's thriving tends to acquire such textbooks.
|
2025-03-21T14:48:29.777070
| 2020-02-04T11:04:00 |
351902
|
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|
Stack Exchange
|
A conjecture about an inequality that involve Ramanujan primes
In this post we denote for integers $n\geq 1$ the $n$-th Ramanujan prime as $R_n$ (thus the sequence A104272 from the On-Line Encyclopedia of Integer Sequences), I add a conjecture that I think can be potentially interesting due that Ramanujan primes are related to the prime-counting function $\pi(x)$ in their definition. It is about if does possible to get a similar expression than the known as Andrica's conjecture. There are some known facts in the literature about the sequence or Ramanujan primes, for example the Wikipedia Ramanujan prime.
Conjecture 1. We've that $\sqrt{R_{n+1}}-\sqrt{R_{n}}<1$ for all integer $n>42$.
Question. Is it possible to prove or refute previous conjecture? Add feedback for previous inequality, even in the case that the problem is very difficut to solve. Many thanks.
My opinion is that this conjecture is interesting and is at research level (I believe that it is interesting in the context of prime gaps, and that will be difficult to elucidate their veracity). I wrote programs in Pari/GP to check the first few hundred of the terms of this sequence to know what about of the mentioned inequality.
All users, on February I decided to remove one of the conjectures to improve the post (more concise and focused to ask a good question for the site MathOverflow), since I was asking about what work can be done about the conjecture I accept the suitable answer, continuining my interest about this. Many thanks for all users.
Not a complete answer, but a bit too long for a comment: Conjecture 1 is very likely to be very difficult if true. The corresponding conjecture for general primes is open. Let $p_n$ be the $n$th prime and $g_n$ be the $n$th prime gap. If one has $\sqrt{p_{n+1}}-\sqrt{p_n} <1 $ for sufficiently large primes then one would have that $g_n = O(p^{1/2})$. The best current result (as far as I'm aware) of that form we can actually prove is that $g_n = O(p^{\frac{3}{4} + \epsilon})$ for any $\epsilon>0$ (due to Nikolai Chudakov)(Edit: see GH's comment below that we have a bound that is $O(p^{\frac{21}{40}})$ due to Baker-Harman-Pintz). Your conjecture would imply that $g_n = O(p^{1/2})$ and is seems to be substantially stronger. Since we have that $R_n$ is in general very close to $p_{2n}$ your conjecture isn't implausible, but likely to be stronger than what we can currently prove.
Many thanks for your nice answer. I'am going to study it, and as soon I can I'm going to upvote it. If you want feel free to add feeback about the second of the conjectures.
Chudakov's bound was improved several times. Currently we know that $g_n=O(p_n^{21/40})$. This was proved by Baker-Harman-Pintz (2000).
@user142929: I have no time for MO now. At any rate, I suggest that you ask this as a separate question.
Possibly unrelated to my question, is I think that a similar Firoozbakht's conjecture for Ramanujan primes maybe is feasible. I don't know if this proposal that I evoke have the best mathematical content and maybe it is in the literature, since I know that Zhi-Wei Sun studied in articles many similar inequalities involving arithmetic functions (I know his preprint in arXiv arXiv:1208.2683 of one of his papers, and I don't know if he, or other mathematician, studied the inequality that I evoke). Isn't required a response, just I add it if you want to think about it.
In past week I've removed the mentioned comment (as outdated). I'm sorry for my insistence in previous weeks and many thanks for your patience @GHfromMO
|
2025-03-21T14:48:29.777349
| 2020-02-04T11:46:46 |
351904
|
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"url": "https://mathoverflow.net/questions/351904"
}
|
Stack Exchange
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Why is this identity about commutators of Lie derivatives true?
I am reading the paper "On splitting methods for Schrödinger-Poisson and cubic nonlinear Schrödinger equations" by Lubich. On page 2147 the author claims
$$[T,V](\psi) = T'(\psi) V(\psi) - V'(\psi) T(\psi) $$
where $T(\psi) = i\Delta \psi$ and $V(\psi) = -i\tilde{V}[\psi]\psi$ with $\tilde{V}([\psi])= \Delta^{-1} |\psi|^2$ where $\psi$ is a complex-valued $H^2$-function on $\mathbb{R}^3$ (so that $\Delta\psi$ is well-defined). It is then claimed that
$$
\begin{split}
T'(\psi) V(\psi) - V'(\psi) T(\psi) = &\; i\Delta(-i\Delta^{-1}(\psi\bar{\psi}) \psi) + i\Delta^{-1}(-i\Delta \psi \bar{\psi})\psi \\
&\;+ i\Delta^{-1}(\psi \overline{i\Delta\psi}) + i\Delta^{-1}(\psi\bar{\psi})i\Delta \psi
\end{split}\label{1}\tag{1}
$$
So the derivative in $T'(\psi)$ is probably just some sort of formal derivative with respect to $\psi$ but what would $(i\Delta \psi)'$ be then? The first term in $(1)$ is cleary coming from $T'(\psi)V(\psi)$ but it doesn't make sense to me from the definition of $T'(\psi)$. It would make sense as $T(V(\psi))$, though. So is $T'(\psi)=T(\psi)$ then?
The other terms are confusing as well. Naively, we would have
$$(\tilde{V}[\psi]\psi)' = \tilde{V}[\psi]'\psi + \tilde{V}[\psi]$$
which, again, doesn't fit with what we have in \eqref{1}.
More formal than above, $T$ and $V$ are vector fields and their and the corresponding differential equation is
$$\partial_t \psi = T(\psi) + V(\psi)$$
In that sense the commutator is the Lie bracket of vector fields.
Maybe someone can give me some clarification?
$T$ is a linear operator so it makes sense that its derivative is itself. I haven't sorted through the other terms yet.
what is $\Delta$? (minus) Laplacian?
Yes, the standard Laplacian
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2025-03-21T14:48:29.777496
| 2020-02-04T12:17:56 |
351906
|
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}
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Stack Exchange
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The existence of $T$-ultrafilters in ZFC
Looking at this MO-problem, my collegue Igor Protasov suggested to ask on Mathoverflow his old question on $T$-ultrafilters hoping that somebody on MO can solve it.
First I recall the necessary definitions.
By a partition of $\omega$ we understand a cover of $\omega$ by pairwise disjoint nonempty subsets. A partition $\mathcal P$ is called finitary if $\sup_{P\in\mathcal P}|P|$ is finite.
Let $\mathfrak P$ is a family of partitions of $\omega$. An infinite subset $T\subset\omega$ is called $\mathfrak P$-thin if for any partition $\mathcal P\in\mathfrak P$ there exists a finite set $F\subset T$ such that for any $P\in\mathcal P$ the intersection $P\cap (T\setminus F)$ contains at most one point.
A free ultrafilter $\mathcal U$ on $\omega$ is called a $T$-point (or else a $T$-ultrafilter) if for any countable family $\mathfrak P$ of finitary partitions of $\omega$ there exists a $\mathfrak P$-thin set $T\in\mathcal U$.
Petrenko and Protasov proved that a free ultrafilter is a $T$-point whenever it is a $P$-point or a $Q$-point.
It is well-known that there are models of ZFC containing no $P$-points and there are models of ZFC containing no $Q$-points. But it is not known yet if there exists a model of ZFC containing neither $P$-points nor $Q$-points. That is why the following question of Protasov is interesting (and non-trivial).
Problem 1. Can the existence of $T$-points be proved in ZFC?
Problem 2. Are Kunen's OK-points $T$-points?
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2025-03-21T14:48:29.777645
| 2020-02-04T12:25:39 |
351908
|
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"authors": [
"LSpice",
"Matthé van der Lee",
"R. Matveev",
"Sergei Akbarov",
"Todd Trimble",
"https://mathoverflow.net/users/148443",
"https://mathoverflow.net/users/18943",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/2926",
"https://mathoverflow.net/users/31923"
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|
Stack Exchange
|
Logical completeness of Hilbert system of axioms
This is really a question about references. The entry in Russian Wikipedia about Hilbert's axioms states, in particular, that completeness of Hilbert's system was proven by Tarski in 1951. The reference is to the Russian encyclopedia of elementary mathematics, to which I don't have access, and I somehow am not able to find any references to this statement in the literature.
I have two questions:
The continuity axioms are not first-order-logic statements. What
does then completeness mean in this situation?
Does anybody know a reference to the original proof by Tarski or any
other proof of this statement, for that matter?
You should not trust Russian Wikipedia. It is written by uneducated people who accidentally seized power. Almost everything written there on mathematics contains a substantial amount of absurdity. Here is an example of a debate on what was written about the foundations of mathematics: https://ru.wikipedia.org/wiki/%D0%9E%D0%B1%D1%81%D1%83%D0%B6%D0%B4%D0%B5%D0%BD%D0%B8%D0%B5:%D0%9E%D1%81%D0%BD%D0%BE%D0%B2%D0%B0%D0%BD%D0%B8%D1%8F_%D0%BC%D0%B0%D1%82%D0%B5%D0%BC%D0%B0%D1%82%D0%B8%D0%BA%D0%B8/%D0%90%D1%80%D1%85%D0%B8%D0%B2/1
@Matt F. The question is though about the statement and the proof by Tarski, not Hilbert.
@Sergei Akbarov. I did not express any opinion about the content of any Wikipedia page.
@MattF. True. I was puzzled by the Wikipedia entry literally saying that completeness of Hilbert's system was proven by Tarski. With the hindsight I would formulate the question differently.
The original is Alfred Tarski's book "The completeness of elementary algebra and geometry", which was due to appear in 1940 but never made it into print because of the outbreak of WW2. An edition appeared after all in 1967 (Institut Blaise Pascal, Paris), but is not easy to come by.
Essentially the same argument is presented in Tarski's 1948/51 "A decision method for elementary algebra and geometry", available here.
Tarski uses an axiomatic setup of elementary geometry different (but equivalent) to Hilbert's, using only points (not lines or planes) as primitive terms, and two relation symbols, $B$ and $D$ (ternary and quaternary, respectively). $Bxyz$ signifies that the point $y$ is on the "line" $xz$ between the points $x$ and $z$ ("betweenness"), while $Dxyzw$ means that the distance between $x$ and $y$ equals that between $z$ and $w$ (the "equidistance" relation). Here, $x$, $y$, $z$ (and $w$) are allowed to coincide.
Full continuity is the second order axiom $\forall_{X}\forall_{Y}((\exists_{a}\forall_{x\in X}\forall_{y\in Y}Baxy)\rightarrow(\exists_{b}\forall_{x\in X}\forall_{y\in Y}Bxby))$, where $X$ and $Y$ are variables ranging over sets of points.
First order continuity is weaker, and is expressed as an axiom schema where $X$ and $Y$ are given as $\lbrace x\mid \phi(x)\rbrace$ and $\lbrace y\mid \psi(y)\rbrace$, respectively, for arbitrary first order formulas $\phi(x)=\phi(x,p_{1},\cdots,p_{n})$ and $\psi(y)=\psi(y,p_{1},\cdots,p_{n})$ that are allowed to contain parameters $p_{1},\cdots,p_{n}$.
A special case of first order continuity is the Circle Axiom, by which a line that contains an interior point of a circle (in the same plane) must meet that circle.
Completeness is the statement that any model $\mathfrak A$ of the axioms of elementary $n$-dimensional geometry without continuity is isomorphic to $K^{n}$ (with the obvious interpretations for the $B$ and $D$ relations) for a Pythagorean ordered field $K$ (that is $K\models\forall_{a}\forall_{b}\exists_{c}(a^{2}+b^{2}=c^{2})$), uniquely determined by $\mathfrak A$ up to isomorphism.
Under full continuity, $K$ must be $\mathbb{R}$, under first order continuity $K$ must be real closed, and for the Circle Axiom $K$ merely needs to be Euclidean (i.e., $K\models\forall_{a}\exists_{b}(a=b^{2}\vee -a=b^{2})$).
This is the content of the Representation Theorem, Th. I, (16.15) in "Metamathematische Methoden in der Geometrie" by W. Schwabhäuser, W. Szmielew and A. Tarski, Springer Hochschultext, 1983, an excellent reference for the metamathematics of elementary geometry (in German).
Edit: Let me add a few comments.
The statement above that Tarki's setup is equivalent to Hilbert's is rather imprecise, as noted by Matt F. and others. Tarski works in first order logic, while a formalization of Hilbert's system is at least unclear. (Still, the axioms in Hilbert's axiom groups I-IV can be derived from Tarski's axioms, as shown in the Schwabhäuser, Szmielew, Tarski text).
For the same reason, it is not clear what it would mean for Hilbert's system to be complete (in the modern sense), and I do not claim that "completeness" of Hilbert's system follows from that of Tarski. Hilbert includes a "completeness axiom", to the effect that his "model" of the axioms in groups I-V (where V is archimedeanity) cannot be extended to a "model" with a larger universe.
To add to the confusion, my use of the word Completeness above (in the body of the answer, in reference to the Representation Theorem) was also unfortunate. Tarski has shown that the first order theory of real closed fields is complete (in the modern sense), and that, as a result, the same goes for the theory of $n$-dimensional elementary geometry (based on Tarski's axioms, with the first order continuity axiom schema included).
@Matt F. Hilbert's axioms, as set forth in his Grundlagen, are, although informal, no less precise. Formalizations of his principles can be deduced in Tarski's system.
@MatthévanderLee How is your comment responsive to Matt's point that Hilbert's system is not first-order? The archimedean and completeness axioms in Hilbert's system fall outside Tarski's system (and so cannot be deduced therein).
@ToddTrimble I agree. Hilbert's archimedean and completeness axioms are not first order statements. Rather, they would be requirements on models of a first order formalization of geometry. So, as suggested by Matt, it would indeed be better to speak of Tarski's formalization as "based on Hilbert's system" rather than "equivalent to it".
Hilbert's completeness axiom ("the model cannot be extended") contradicts the upward Löwenheim-Skolem theorem of first order model theory. (The second order axiom of full continuity ensures that $K=\mathbb{R}$, though.) And archimedeanity comes down to the field $K$ mentioned above being an Archimedean field.
After agreeing that it shouldn't remain, your edit left the assertion that Tarski's system is "different (but equivalent) to Hilbert's".
I do not understand this discussion. Is it not possible to say that the Hilbert system is an extension by definitions of some axiomatic set theory, say, ZFC, and therefore can be considered as a first-order theory? The only vagueness, as far as I understand, lies in the meaning of the expression "equivalent first-order theories". Do people use this term?
@SergeiAkbarov. Hilbert's system is not an "extension" of ZFC, whatever "extension" means. As we all know, there is a model in ZFC for the Hilbert's system. Hilbert's system is not first order, because two of the axioms, Archimedean and completeness use the wrong language (the range of quantification is wrong, in the first case it is natural numbers, which are not inherently present, in the second -- models)
@R.Matveev here is the meaning of the term: https://en.wikipedia.org/wiki/Extension_by_definitions I would say, Hilbert axioms can easily be presented as an extension by definition of ZFC (or NBG, MK, etc.). As, actually, everything in mathematics described in terms of set theory.
@SergeiAkbarov. Any extension of ZFC or any other set theory capable of modelling $\mathbb{N}$ will be incomplete by Gödel's Theorem, while Tarski's system, which is based on Hilbert's as explained above, is complete.
@R.Matveev this is another puzzle for me. If one theory is complete while another incomplete, then what is meant by "equivalence"? What are you arguing about?
@SergeiAkbarov. Hilbert system is equivalent to Tarski, as explained above. This is somewhat informal, because Hilbert's system is slopy by modern standards and contains two statements which are not in the first order logic language. This is explained in the answer by Matthé van der Lee, above. By Tarski's theorem his and Hilberts systems are complete. ZFC is not complete and has nothing to do with this discussion.
@R.Matveev, if there is no definition (of equivalence) is this rigor in declarations apt?
@SergeiAkbarov. I am not sure what your question/concern is. Let me summarize the answer/discussion. 1. Hilbert's system is not a formal system in a modern sense of the word, consequently no mathematically/logically rigorous statement can be made about it. 2. Tarski devised another system of axioms based on (guided by, resembling, similar to, -- not a mathematical notion) Hilbert's. 3. Tarski proved that his system is complete. For more details, see the answer and the discussion.
Let us continue this discussion in chat.
I don't want to continue this discussion in chat. What you explain here, is clear without your explanations. What is not clear (and sounds strange) are declarations like this:
$$
\text{> Hilbert system is equivalent to Tarski, as explained above.}
$$
You should give a definition of this equivalence before writing this, @R.Matveev. And giving advices like this:
$$
\text{> For more details, see the answer and the discussion.}
$$
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