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2025-03-21T14:48:30.114624
| 2020-03-24T18:49:02 |
355626
|
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|
Stack Exchange
|
What should I call a "differential" which cubes, rather than squares, to zero?
If I had a vector space with a linear endomorphism $D$ satisfying $D^2 = 0$, I might call it a differential and study its (co)homology $\operatorname{ker}(D) / \operatorname{im}(D)$. I might say that $D$ is exact if this (co)homology vanishes. I would especially do this if $D$ increased by 1 some grading on my vector space.
But I don't have this structure. Instead, I have a vector space with an endomorphism $D$, which increases by 1 some grading, satisfying $D^3 = 0$ but $D^2 \neq 0$. Then there are two possible "(co)homologies": $\operatorname{ker}(D) / \operatorname{im}(D^2)$ and $ \operatorname{ker}(D^2)/\operatorname{im}(D) $. It is an amusing exercise that if one of these groups vanishes, then so does the other, so that it makes sense to talk about $D$ being "exact".
Surely this type of structure has appeared before. Does it have a standard name? Where can I read some elementary discussion?
How about nilpotent endomorphism of order 3 and degree 1? It's a bit long winded I guess.
A fractional differential, or a divisor of a differential?
A diffferential?
I know nothing, but it seems like one place to start searching the literature is here https://arxiv.org/abs/1509.00438
Maybe I'm being obtuse, but since $(D^2)^2 = 0$, why not just call $D^2$ a differential and call $D$ a square root of the differential?
@SteveHuntsman : That doesn't seem to reflect the requirement that $D^3=0$.
A cube root of zero?
In algebraic topology, this was considered by Moore and Peterson in their 1972 paper "Modules over the Steenrod algebra", later by Miller and Wilkerson in 1981, "Vanishing lines for modules over the Steenrod algebra", and others. They did not give it any special name.
Not entirely serious but... trifferential?
@AlexArvanitakis Hah!
Many years ago, when I was a graduate student, I remember seeing a couple of papers on the homologies of operators satisfying $\partial^p=0$, generalizing the case $p=2$. I seem to remember that they were by somebody like Steenrod, and it might evan have been in the Annals, sometime in the 40s or 50s.
Unfortunately, I'm at home now and not able to access MathSciNet to look it up. However, I do remember that there was something like a set of axioms, generalizing the Steenrod axioms, for the various '$p$-homologies' that could be associated to topological spaces using such operators.
I forget now why I was interested in them. When I'm back in my office (maybe tomorrow), I'll try to find it on MathSciNet.
Here is one by Kapranov https://arxiv.org/abs/q-alg/9611005
The paper you're thinking of is probably by Mayer, "A new homology theory", Annals 1942. There was a resurgence of interest in these objects after the paper of Kapranov that Phil mentioned and if you follow the citation trail from Kapranov's paper on Google scholar you'll find plenty of references.
@DanPetersen: Thanks (and thanks to Phil, too)! Yes, Mayer, that's who it was, and now that you mention the title, that's the paper I had in mind. I hadn't thought about that stuff in more than 40 years. You saved me the trouble of rooting around in MathSciNet.
I recall looking through Kapronovs paper before I actually understood anything about homology or cohomology.
I would just call it a module over the truncated polynomial algebra $k[D]/D^3$. Your two flavors of homology appear as positive odd-degree and even-degree groups in $\operatorname{Ext}^*_{k[D]/D^3}(k,M)$. (This is $2$-periodic in positive degrees.) The "exactness" should hold precisely if $M$ is free as such a module.
The same interpretation works also for usual chain complexes, by the way. The algebra appearing there is the exterior algebra $k[D]/D^2$, and the $\operatorname{Ext}$-groups are actually $1$-periodic in positive degrees and recover the usual notion of homology.
If you want to be fancy, you can localize the derived category of $k[D]/D^3$ (or $k[D]/D^2$) by killing free modules. If you do this correctly, you obtain the so-called "stable module category" (a stable $\infty$ or dg-category), in which the mapping complex from $k$ to $M$ is actually a fully periodic version of the above $\operatorname{Ext}$, so in some sense is precisely described by your two different "homologies" of $M$.
That interpretation of homology as $\mathrm{Ext}_{k[D]/D^2}$ is very neat! Do you know any treatment that develops this interpretation?
@PeterLeFanuLumsdaine The Ext interpretation is not terribly surprising. Note that the minimal resolution of $\mathbb k$ over $\Lambda=\mathbb k[t]/(t^2)$ is periodic given by multiplication by $t$ on $\Lambda$ everywhere. Hence, if $M$ is a chain complex, then $\mathsf{Ext}\Lambda(\mathbb k, M)$ identifies naturally with the complex where we have $M$ everywhere and the map is $d :M\to M$, so that its homology is just $H*(M)$ in each degree. It is also true that this is given by $\mathsf{Tor}^\Lambda(\mathbb k,M) $.
Such objects, for more general values of $3$, or at least the graded version (i.e., $\mathbb{Z}$-graded objects where $D$ is a degree one map with $D^N=0$) have attracted some interest in the representation theory of finite dimensional algebras in recent years, under the name of "$N$-complexes".
The fairly recent paper
Iyama, Osamu; Kato, Kiriko; Miyachi, Jun-Ichi, Derived categories of $N$-complexes, J. Lond. Math. Soc., II. Ser. 96, No. 3, 687-716 (2017). ZBL1409.18013.
may be of interest to you for the fairly lengthy list of relevant references in the introduction.
You could also call this a "curved differential", or a "curved complex", motivated by the differential induced by a connection with non-trivial curvature:
consider a vector bundle $E \to M$ with connection on a manifold $M$. This induces a differential $d_E \colon \Omega^p(M;E) \to \Omega^{p+1}(M;E)$ on forms on $M$ with coefficients in $E$. The differential satisfies $d_E^2 = F_E \wedge (-)$, where $F_E$ is the curvature 2-form of the connection. By the Bianchi identity it always follows that $d_E^3 = 0$. I think this would also be in agreement with the notion of curved $L_\infty$-algebras.
don't we have $d_E^3= F_E\wedge d_E$ instead?
The situation similar to what you are describing happens when people talk about the so-called N-Koszul algebras, originally defined by R. Berger in his paper "Koszulity for nonquadratic algebras" (J. Algebra 239 (2001), 705-734). Basically, for an algebra with homogeneous relations of degree N, the way one constructs a complex that determines whether the algebra is "homologically well behaved" (an analogue of the Koszul complex of a quadratic algebra) is obtained from a $D^N=0$ situation by considering the chain complex where the differential is equal to $D$ in odd places and to $D^{N-1}$ in even places. Apart from the paper of Kapranov mentioned in comments, this is the most common source of this kind of phenomena I am aware of.
What about converting this trifferential to an ordinary differential?
If $D$ acts on a space $X$, let $X_1$ and $X_2$ be two copies of $X$. Then define a differential $Q$ on the direct sum $X_1\oplus X_2$ by
$$
Q X_1 \subseteq X_2\,,\quad QX_2 \subseteq X_1\,,\qquad Qx_1=Dx_1\in X_2 \forall x_1\in X_1,\quad Qx_2=D^2 x_2 \in X_1\forall x_2\in X_2\,.$$
Clearly $D^3=0\iff Q^2=0$, no?
EDIT: Assuming $D$ is degree $1$ on $X$, then if you shift the degree on $X_2$ by $+1$ relative to $X$ and assuming $X_1$ has the same degree assignment as $X$ does then I think $Q$ is also of degree 1. Nevermind that.
$\text{ker}(Q) = \text{ker}(D) \oplus \text{ker}(D^2)$. $\text{Im}(Q) = \text{Im}(D^2) \oplus \text{Im}(D)$. So your homology groups are (the direct sum of) those of Kapranov, which are in the case of $D^p = 0$, given as $_k H(X) = \text{ker}(D^k)/\text{Im}(D^{p-k})$.
@MikeMiller: thanks, that's good to know. The obvious generalisation to $D^p=0$ seems to have the same homology groups so your formula for $_kH(X)$ is a nice check
|
2025-03-21T14:48:30.115295
| 2020-03-24T19:10:30 |
355627
|
{
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"Johnny T.",
"Laurent Moret-Bailly",
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|
Stack Exchange
|
Question about the statement of the going-down theorem of Cohen-Seidenberg in Mumford
In Mumford's red book the statement of the Going-Down Theorem (Chapter II Section 8) is as follows.
Let $f: X \to Y$ be a finite morphism. Assume that $Y$ is an irreducible normal scheme. Assume that for all $x \in X$, no non-zero element of $O_{Y, f(x)}$ is a $0$-divisor in $O_{X,x}^{*}$. Then for every pair of points $x_1 \in X$, $y_0 \in Y$ such that
$f(x_1) \in \overline{\{ y_0 \}}$,
there is a point
$x_0 \in f^{-1}(y_0)$
such that
$x_1 \in \overline{ \{ x_0 \} }$.
By "no non-zero element of $O_{Y, f(x)}$ is a $0$-divisor in $O_{X,x}^{*}$", does this mean that it is ok if a non-zero element of $O_{Y, f(x)}$ gets mapped to $0$ in $O_{X,x}$? or does it mean every non-zero element of $O_{Y, f(x)}$ gets mapped to a non-zero element which is not a $0$-divisor in $O_{X,x}^*$? Thank you.
PS Instead of editing this question, I decided to ask separately On an application of the going-down theorem of Cohen-Seidenberg in Mumford, which was the reason why I was trying to understand the statement in the first place.
To me, a "zero divisor" in a ring $R$ is an element $a$ such that multiplication by $a$ in $R$ is not injective. This includes $0$ if $R$ is not the zero ring.
@LaurentMoret-Bailly That is what I first thought as well, but then he does put the $$, as in a $0$-divisor in $O_{X,x}^{}$... Do you think that changes the meaning?
$R^*$ usually denotes non zero elements in a ring $R$..
@PraphullaKoushik Right, so I was wondering because he writes a $0$-divisor in $R^{*}$, is he excluding the zero element?
If you exclude 0 as you say, the result is obviously false: take $X={x}$ (a closed point of $Y$), $f$ the inclusion, and $y_0\neq x$.
@LaurentMoret-Bailly I don't quite follows what you mean. To apply the result one needs to specify a pair of points $x_1 \in X$ and $y_0 \in Y$ such that $f(x_1) \in \overline{{ y_0 }}$. But with your example one can not get such a pair to begin with?
Yes, of course: $x_1=x$, $y_0$=any generization of $x$ in $Y$, assuming of course $Y\neq{x}$.
@LaurentMoret-Bailly I see your point now thank you. And if I understand you correctly, then for example you are considering the morphism $f: \operatorname{Spec} \kappa(x) \to Y$, where the unique point is mapped to $x$ and $\kappa(x)$ is the residue field of $x$. In this case the map of the stalks is $O_{Y, x} \to O_{Y, x}/\mathfrak{m}x \hookrightarrow \kappa(x)$. This means some of the non-zero elements of $O{Y,x}$ gets mapped to $0$, so by the $0$-divisor he is probably including $0$.. Is there an easy example where this morphism is finite?
It is always finite if $x$ is a closed point.
Ok, thank you very much!
That mark * is not a star on the ring but an asterisk referring the reader to a footnote at the bottom of the page. This the one instance where the Springer edition is clearer than the original.
@roysmith O right...
|
2025-03-21T14:48:30.115527
| 2020-03-24T19:11:16 |
355629
|
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"Bazin",
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|
Stack Exchange
|
Stationary phase in spherical integral
I'm trying to find asymptotics for an oscillatory integral on $\mathbb{S}^{n-1}$, for which my advisor said I should use stationary phase arguments. The particular, he claims that:
If $\lambda\gg 1$, then
$$I(\lambda,x) = \int_{\mathbb{S}^{n-1}}(x\cdot y)e^{i\lambda(x\cdot y)}\,d\sigma(y),$$
is $\mathcal{O}(|x|(\lambda |x|)^{-\frac{n-1}{2}})$ when $|x|\geq \lambda^{-1}$.
When I hear stationary phase, I think of working with operators of the form
$$
L = \frac{\nabla_y(x\cdot y)}{i\lambda |\nabla_y (x\cdot y)|^2} \cdot \nabla_y,
$$
since then $L^N[e^{i\lambda (x\cdot y)}] = e^{i\lambda (x\cdot y)}$, for any $N \geq 1$, and I can use integration by parts to move these operators over to the $(x\cdot y)$ term. However, wouldn't that require that the integral be defined over an $n$-dimensional region, rather than an $(n-1)$-dimensional surface?
I wouldn't be struggling so much if the gradient $\nabla_y$ could be taken in Cartesian coordinates. But we have a surface integral in $d\sigma(y)$, meaning that we would need to parametrize our surface with $n-1$ parameters, say $\omega = (\omega_1, \ldots, \omega_{n-1}) \in \Omega \subset \mathbb{R}^{n-1}$, with any derivates now being taken in these new variables. Specifically $\nabla_y$ would generate an $n$-vector, while $\nabla_{\omega}$ would generate an $n-1$-vector. So how should I go about applying the same kind of stationary phase arguments to this new integral?
$$
I(\lambda, x) = \int_{\Omega} (x \cdot y(\omega)) e^{i\lambda (x\cdot y(\omega))} \,dV(\omega)
$$
Now that $y$ depends on $\omega$, the gradients $\nabla_\omega (x\cdot y(\omega))$ become much trickier to get a grasp on. I'm particularly struggling trying to argue how we can find regions where $|\nabla_\omega (x\cdot y(\omega))| > 0$, so that our $L$ type operators are properly defined.
Am I going about this all wrong? If I can get the big-oh asymptotics I mentioned above in some other way, it doesn't really matter. I just need to prove these results as a lemma to something bigger. Any help is much appreciated!
You have
$
I(\lambda, x)=x\cdot\int_{\mathbb S^{n-1}} ye^{i \lambda x\cdot y} d\sigma(y)=x\cdot J(x,\lambda)
$
and you claim that for $\vert x\vert \lambda \ge 1$, you have
$$
J(x,\lambda)=O((\vert x\vert \lambda)^{-\frac{n-1}{2}}).
$$
Indeed, using coordinate charts and a finite partition of unity, you are reduced to the case where
$$
J(x,\lambda)=\int_{\mathbb R^{n-1}} a(z) e^{i\lambda (x'\cdot z+ x_n\sqrt{1- \vert z\vert^2})}dz, \quad\text{$a\in C^\infty_0(\mathbb R^{n-1})$ supported near $0$, $x=(x', x_n)\in \mathbb R^{n-1}\times \mathbb R$}.
$$
Let us set $\phi(x,z)=x'\cdot z+ x_n\sqrt{1- \vert z\vert^2}$. We have
$$
\frac{\partial \phi}{\partial z}= x'-(1- \vert z\vert^2)^{-1/2} z x_n,
$$
which vanishes at $z=0$ when $x'=0$. Then you calculate the Hessian of $\phi$ at $z=0$
and get
$$
\phi''_{zz}=-x_n.
$$
The stationary phase in $n-1$ dimensions gives the sought estimate with $O((\vert x_n\vert \lambda)^{-\frac{n-1}{2}})$ (note that you know that $\vert x_n\vert \lambda \ge 1$, since you are near $x'=0$).
I'm still reading through this carefully, but thank you! I have a couple small clarification questions, though, so if you don't mind I'll just enumerate them here in a series of separate comments:
By bringing the integral inside of the dot product, you are effectively making $\mathbf{J}(x,\lambda)$ a vector-valued function, correct? But isn't your cutoff function, $a(z)$ a scalar-valued function? Should I just be interpreting $a(z)$ as one scalar component of $\mathbf{J}(x,\lambda)$? If not, then wouldn't this seem to contradict the definition for $\mathbf{J}(x,\lambda)$ you had earlier?
Assuming we have clarified the issue with $\mathbf{J}(x,\lambda)$ above, lets call $J_k(x, \lambda)$ the $k$-th component of the vector. Then I can see how stationary phase around $(x,z) = (0,0)$ gives us the $(|x_n| \lambda)^{(n-1)/2}$ behavior, but this would only be asymptotic behavior in each coordinate. Moreover, every $J_k$ would satisfy the same big-oh bounds involving the absolute value $|x_n|$ only; so how do I put these all together to get a uniform bound involving the norm, $|x|$?
@Patch (1) $J$ is indeed vector-valued, but you can handle each coordinate separately. (2) You start with $x$ such that $\vert x\vert \lambda\ge 1$; then you look at the stationary phase method with an integral near a given point $y_0$ of the sphere. You check if you have a stationary point: if not you get a better estimate with $O((\vert x\vert \lambda)^{-N})$ for all $N$. If you land on a stationary point, you check the Hessian. Note that the choice of $y_0$ in the answer above is $e_n$.
Thanks, again!!
|
2025-03-21T14:48:30.115837
| 2020-03-24T19:29:12 |
355630
|
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|
Stack Exchange
|
Ternary error correction codes
Let`s define ternary ECC as a code that its codewords can be defined by $ \{ xyz f(y,z) f(x,z) f(x,y) | x,y,z \in \{0,1\}^m \} $ for some function $f$. $f$ returns bitstring of constant length.
Are there any known good error correction codes that are ternary?
Such a family of LDPC codes would be best.
Is there a reason it won't be good(in terms of distance, rate)?
It might be useful in a construction I have. I just wanted to make sure it is not known already before I dive in.
Thanks
Your question is not clear. Please give more details if you want appropriate answers.
I improved this, but let me know what is unclear, if it is still unclear.
Does $xy$ means the concatenation of $x$ and $y$? Please give an example for such codes and the function $f$
x, y, z are bitstrings of length m. Concatenation makes sense.
This is an interesting question, but the choice of "ternary" seems unfortunate-- there's a already a whole literature on ternary ECCs, where "ternary" means "base-3". Your codes are binary with special structure.
@Mark, I don't know what exactly OP has in mind for $f$, but nothing OP has written leads me to think its output is meant to be of length $m$. Maybe it's some kind of parity check, and its output is a single bit.
Sorry, I wanted to update my comment saying it was GF field not a long time ago by mistake. Had a timeout. The function f returns bitstring. Sorry for that.
OK, but the length of that bitstring seems to be important. That is, the answer to your question seems to depend on just how long that bitstring is.
Well, if the function $f$ has range $GF(2)^m$, represented by $GF(2^m)$ if convenient, it has rate 1/2. Such a function can really control symbol ($GF(2^m)$ ) not bit errors so it is a code over $GF(2^m)$ of length $n=6$ and rate 1/2 (dimension 3). If the code is MDS [best possible] it has symbol distance at most $n-k+1=6-3+1,$ 4, so could correct double symbol errors.
A Reed-Solomon code would achieve this, and is the optimal such code. But we do need $2^m+1\geq n=6,$ (since Reed Solomon codes are essentially evaluation codes) so $m$ would have to be at least 3.
Edit: If $f$ maps into $GF(2)$ as suggested by Gerry Myerson, then this is a single error correcting code with $n=3m+3$ and $3$ parity checks. If $3m+3=2^n-1,$ then a Hamming code will do, and no fancy $f$ could do better.
|
2025-03-21T14:48:30.116025
| 2020-03-24T21:58:44 |
355639
|
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|
Stack Exchange
|
Integration on an non-orientable manifold
Suppose $M_n$ is a $n$ dimensional non-orientable manifold.
I am interesting in knowing whether the following statements are true:
A characteristic class $w_{n}^{(p)} \in H^{n}(M_n, \mathbb{Z}_p)$ can be integrated on $M_n$ only when $p=2$.
Suppose $n_1+n_2=n$, and $n_i, (i=1,2)$ are positive integers. $w_{n_i}^{(p)} \in H^{n_i}(M_n, \mathbb{Z}_p), i=1,2$. Then $w_{n_1}^{(p)}\cup w_{n_2}^{(p)}$ can be integrated on $M_n$ only when $p=2$.
If not true, is there a counter example?
see also https://mathoverflow.net/q/16632/27004
What do you mean by (1)? It follows from Poincare duality that for a closed non-orientable manifold of dimension $n$, we have $H^n(M;R) \cong R/2$. If $R = \Bbb Z/p$ for $p$ odd, then this group is just zero, so I suppose under any reasonable interpretation the answer to both (1) and (2) is (for $p$ prime) "yes, only when $p=2$". If you didn't just mean $p$ prime, and $p$ is even, then $(\Bbb Z/p)/2 = \Bbb Z/2$, so the answer to (1) and (2) is "the integral is an element of $\Bbb Z/2$".
|
2025-03-21T14:48:30.116124
| 2020-03-24T22:17:17 |
355640
|
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|
Stack Exchange
|
Spectrum of "classical" operators
Lately, I've been reading a couple of papers from different one-dimensional PDE contexts on which operators like $\mathcal{L}:=-\partial_x^2+c_*+\Phi$ repeatedly appear. Usually, on these contexts $\Phi$ is a smooth exponentially decaying function and $c_*\in\mathbb{R}$ is a positive constant.
I am very surprised that in all of these papers the authors, just by knowing these facts, they immediately conclude that the continuous spectrum of $\mathcal{L}$ is exactly $[c_*,+\infty)$ and the rest of the spectrum consists on a finite number of eigenvalues. My question is, how do they know that the continuous spectrum starts exactly at $c_*$? I've seen this at least five times for different values of $c_*$ and different functions $\Phi$ (all of them smooth with exponential decay). Does anyone have an explanation for this? Or any reference?
A second question (I know that the difficulty of the question can exponentially grow now so I am actually very happy just by understanding the previous part!): What if I consider a non-smooth but still exponentially decaying $\Phi$? For instance $\Phi=e^{-\vert x\vert}$? The previous "result" still holds?
The 3D case is treated here around page 88. I can't remember whether they did the 1D case, or how it was proved.
Let $A$ be a self-adjoint operator with domain $D(A)\subset\mathcal H$ ($\mathcal H$ is some Hilbert space). An operator $C$ with $D(A)\subset D(C)$ is called relatively compact with respect to $C$ if $C(A-zI)^{-1}$ is compact for some (hence all) $z\notin\sigma(A)$. Paraphrasing Corollary 2, page 113 Section XIII.4, in [1], we have
If $C$ is relatively compact with respect to $A$, then $B:=A+C$ is closed with domain $D(B)=D(A)$, and $$\sigma_{ess}(B)=\sigma_{ess}(A).$$
In fact, this is elementary once one realises that $C(A-zI)^{-1}:\mathcal H\to \mathcal H$ is compact if and only if $C:D(A)\to\mathcal H$ is compact.
In your case, setting $A = -\partial_x^2 + c_*$ and $C$ the multiplication operator by $\Phi$, is it not hard to prove that $C:H^2(\mathbb R)\to L^2(\mathbb R)$ is compact. We then obtain that
$$
\sigma_{\mathrm{ess}}(\mathcal{L}) = \sigma_{\mathrm{ess}}(-\partial_x^2 + c_*).
$$
Since it is well know that $\sigma_{\mathrm{ess}}(-\partial_x^2 + c_*) = [c_*,+\infty)$, the result follows.
For your second question, this answer is yes the result hold for $\Phi(x) = e^{-|x|}$. In fact it will hold in dimension $d\leq 3$ for any $\Phi\in L^2(\mathbb R^d)$. As explained earlier, it is enough to show that multiplication by $\Phi$ is (well-defined and) compact from $H^2(\mathbb R^d)\to L^2(\mathbb R^d)$; let us prove this result.
Let $(f_n)_{n\geq0}$ be a bounded sequence in $H^2(\mathbb R^d)$, $d\leq 3$. The core of the argument is the following fact.
$(f_n)_{n\geq0}$ is bounded in $L^\infty$ and, up to extraction, converges $L^\infty$-locally to some function $f$.
Note that $f$ then has to be bounded as well. Once this is established, we see that
$$ \begin{align*}
\limsup |\Phi f_n-\Phi f|_{L^2}^2
& = \limsup \int|\Phi|^2\cdot|f_n-f|^2 \\\\
& \leq \limsup |f_n-f|_{L^\infty([-k,k]^d)}^2\cdot\int|\Phi|^2{\mathbf 1}_{[-k,k]^d} \\\\
& \quad + \limsup (|f_n|_{L^\infty}+|f|_{L^\infty})^2\cdot\int|\Phi|^2{\mathbf 1}_{([-k,k]^d)^\complement} \\\\
& \leq |\Phi|_{L^2}^2\cdot0
+ 4M^2\cdot\limsup \int |\Phi|^2{\mathbf 1}_{([-k,k]^d)^\complement}
\end{align*} $$
for $M$ a bound on the norms $|f_n|_{L^\infty}$, and $\limsup$ the limit superior along a convergent subsequence. Because $\Phi$ is in $L^2$, Lebesgue's dominated convergence theorem guarantees that the limit is zero, and $\Phi f_n$ converges to $\Phi f$ as expected.
Let us turn to the proof of the fact. The boundedness of $(f_n)_{n\geq0}$ on compact sets follows from the continuous embeddings from $H^2(\ell+[0,1]^d)$ to $\mathcal C^0(\ell+[0,1]^d)$ for all $\ell\in\mathbb Z^d$ (because $d\leq3$). Since the norm of these injections does not depend on $\ell$, $(f_n)_{n\geq0}$ is in fact uniformly bounded over $\mathbb R^d$
As for the convergence up to extraction, according to the usual Sobolev embeddings/inequalities, the sequence of restrictions $\big((f_n)_{|[-k,k]^d}\big)_{n\geq0}$ is relatively compact in $\mathcal C^0([-k,k]^d)$ for all $k\geq1$ (we use again that $d\leq3$). We can then extract diagonally a subsequence $(f_{\sigma(m)})_{m\geq0}$ that converges to some continuous function $f$ uniformly over the compact sets, concluding the proof of the fact.
[1] Reed, M., Simon, B. (1978). Methods of Modern Mathematical Physics. IV Analysis of Operators. New York: Academic Press.
|
2025-03-21T14:48:30.116416
| 2020-03-25T01:39:44 |
355645
|
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"Giorgio Metafune",
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|
Stack Exchange
|
Singular integral operator
I am working on a problem involving the Biot-Savart law in fluid dynamics. I found a theorem of singular integral which is intimately related to my research.
Assume that $K(x)$ is a classical Calderon-Zygmund kernel in $\mathbb{R}^{3}$ and $f$ is a smooth function satisfying
\begin{equation}
|f(x)|\leq\frac{1}{(1+r)^{a}} \quad \text{for} \quad 0<a<2,
\end{equation}
and
\begin{equation}
|\nabla f(x)|\leq\frac{1}{(1+r)^{b}} \quad \text{for} \quad 0<b<2,
\end{equation}
where $r=\sqrt{x_{1}^{2}+x_{2}^{2}}$. Decays only appear in horizontal directions. No information on the vertical direction.
Let $Tf(x)=\int K(x-y)f(y)dy$. Then there exists a constant $c_{0}$ such that
\begin{equation}
|Tf(x)|\leq\frac{c_{0}}{(1+r)^{d}},
\end{equation}
where $d$ can be any number strictly less than $min\{a,b\}$.
If you can show me that this theorem is sharp, there is no more problems. Otherwise, my questions:
Someone suggested me that this theorem might be able to get improved. Since it is a pointwise or $L^{\infty}$ estimate for the singular integral which was not covered by any existing theory. Is that possible to prove something like $d=\frac{a+b}{2}$ which will bring great help to my project.
One may change the second condition i.e. the horizontal decay of $\nabla f$ or add something new like the decay of $\nabla^{2} f$.
I’m not an expert on harmonic analysis. I am very interested in my question 2. Any suggestions are welcome. Thanks!
The estimate looks rather fishy, as written: consider the case when $a=b$ and $f$ that is essentially $1/r^a$ times the characteristic function of the cylinder $x_1^2+x_2^2\le r^2$ with large $r$ (you need just length $1$ to descend to $0$ with your gradient restriction). Then for the points slightly outside the cylinder, you have all function on one side, so the oscillation in the kernel does not help and you get at least $r^{-a}\log r$ integrating the distance from the point to the power $-3$. What am I missing?
@fedja It is not clear to me. Are you taking one function or a family? And if $r=\sqrt {x_1^2+x_2^2}$, which is the cylinder? Thank you.
@GiorgioMetafune Yeah, my using the same $r$ notation as in the OP is confusing, indeed. Replace it by capital $R$ and view it as a family of functions. The point is that the declared estimate cannot hold without logarithmic correction though the example I gave is almost the worst case scenario.
@fedja Thanks. But replace $r$ by $R$ only in the definition of the cylinder or everywhere or...? By the way, even if the estimate were true, I do not believe it can be improved to the average of $a,b$. This does not happen when the function decays in all directions.
@fedja Yes, you are correct. There is a log factor on the numerator. I have edited my question. More comments are welcome!
Could you give a reference for the estimate you quote?
@GiorgioMetafune Please read Lemma 3.2 of the paper arxiv 1801.07420. https://arxiv.org/pdf/1801.07420.pdf
|
2025-03-21T14:48:30.116646
| 2020-03-25T02:19:12 |
355646
|
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|
Stack Exchange
|
Maximize sum of supermodular functions over nested sets
Let $R$ be a function that maps a set and a positive integer to a real positive number. We have that for any positive integer $t$ and $S \subseteq \{1, \ldots, t\}$, $R(S, t)$ satisfies:
For all $t < t'$, $R(S, t) < R(S, t')$. For all $A \subset S$, $R(A, t) > R(S, t)$.
For all $i < S_{[1]}$, $R(S, t) = R(S-i, t-i)$ where $S_{[1]}$ denotes the smallest element of $S$ and $S - i = \{s - i: s \in S\}$ .
For all $A \subseteq B \subseteq S$, $i \in S \setminus B$,
$$R(A \cup \{i\}, t) - R(A, t) \leq R(B \cup \{i\}, t) - R(B,t).$$
(Note that the above supermodularity is only valid for a given $t$, but not for different t. If it helps, one can think of $t$ as time and the set $S$ as some kind of consumption time stamps.)
Before going to the optimization objective, for notation simplicity, I will use $[T]$ to denote the set $\{1, \ldots, T\}$, $S_{[i]}$ to denote the $i-$th smallest element in $S$ and $S_{[i]:[j]}$ to denote the subset $\{s \in S: S_{[i]} \leq s \leq S_{[j]}\}$. Now, the goal is to maximize the below objective: for a given $T$,
$$\max_{S \subseteq [T]} \sum_{i=2}^{|S|} R(S_{[1]:[i-1]}, S_{[i]}) + \sum_{j=2}^{T - |S|} R(S^c_{[1]:[j-1]}, S^c_{[j]}),$$ where $S^c$ is the complement of $S$, i.e., $S^c = [T] \setminus S$.
I am quite confused about where to start to approach this problem. One suggestion I got is to try to first write the objective as a difference between two supermodular functions, but I am unsure how to come up with such decompositions. I would really appreciate any suggestions on where to start, how to think of this problem, relevant literature or good keywords to search for existing work on similar problems. Thanks in advance!
Nitpick: for rule 2, should $i \leq S_{[1]}$ be $i < S_{[1]}$ (or alternatively, should $S$ stat at 0)? Less of a nitpick: in rule 3, $i \in {1, \dots, t} \backslash S$ doesn't make sense, as $S = {1, \dots, t}$; do you mean $i \in S \backslash B$?
Thanks for the clarification questions! 1)Yes, you are right! 2) Yes, you are right :). I will edit the question accordingly!
|
2025-03-21T14:48:30.116810
| 2020-03-25T04:51:12 |
355650
|
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|
Stack Exchange
|
Where did this presentation of Gödel's theorem appear?
This question was asked and bountied at MSE, with no response.
Many years ago I ran into the following proof of Gödel's first incompleteness theorem
(here $T$ is an "appropriate" theory of arithmetic; see Jones and Shepherdson - Variants of Robinson's essentially undecidable theory $R$):
First, we observe that Tarski's undefinability theorem can be slightly tweaked to say that for all $M\models T$, the theory of $M$ can't be the standard part of an $M$-(parameter-freely-)definable set. Next, by the usual representability arguments we have that every computable set is invariantly definable: if $X\subseteq\mathbb{N}$ is computable and $M\models T$ then $X$ is the standard part of a definable set in $M$. Now if $S$ were a computable satisfiable completion of $T$ we would get a contradiction by looking at $M\models S$. If we want, we can then replace "satisfiable" with "consistent" via the completeness theorem.
My question is:
Where did this argument appear?
I'm not asking whether this is a "genuinely different" argument (although that said, see here). Rather, I'm interested in its presentation in terms of invariant definability. This was a notion first introduced by Kreisel following Godel/Herbrand and subsequently studied by others (see e.g. Kreisel - Model-theoretic invariants: Applications to recursive and hyperarithmetic operations, Moschovakis - Abstract Computability and Invariant Definability, Apt - Inductive definitions, models of comprehension and invariant definability). My vivid recollection is that the argument above appeared as a footnote in the first paper mentioned above, but it turns out it's not there after all.
EDIT: I think I finally figured it out! At around the same time that I was reading Kreisel's invariant definability paper, I was also looking at Kikuchi/Tanaka, On formalization of model-theoretic proofs of Godel's theorems for completely different reasons. This paper goes into careful detail on internal model-constructions, and in particular Corollary 5.6 has a very "modal" feel to it (if $\mathfrak{M}_0$ can "see"/build according to a specific type of procedure $\mathfrak{M}_1$, then we get $\Sigma_1$-persistence from $\mathfrak{M}_0$ to $\mathfrak{M}_1$). I think while trying to understand the details of the K/T paper I noticed the (to me anyways) affinity with invariant definability ideas.
Based on the lack of reference provided here and at MSE, as well as my own lack of success in finding a reference, I'm going to tentatively say that this particular approach has not appeared as such in print. I'll delete this answer of course if someone supplies a reference in which it appears, but I think at present it's appropriate to move this off the unanswered queue (I've made my answer CW since the line between answer and non-answer here is a bit blurry, so this doesn't feel quite right as a reputation-producing answer).
Very belatedly, in response to Joel's answer: I see a meaningful difference between Fitting's argument and the one in the OP - which I'll call "Kreiselian" - due to the former's reliance on representability of provability. (For simplicity, below I'll conflate sentences/formulas with their Godel numbers and leave some things unoptimized.)
Consider the following (basically the "right" conclusion of the Kreiselian argument, while staying in the realm of arithmetic for simplicity):
$(\star)\quad$ Suppose $\mathfrak{C}$ is a class of models of Robinson's $\mathsf{R}$ with $\mathbb{N}\in\mathfrak{C}$. Then there is no complete theory $T$ in the language of arithmetic such that
$T$ is "$\mathfrak{C}$-satisfiable" (= some $\mathcal{M}\in\mathfrak{C}$ satisfies $T$) and
$T$ is "$\mathfrak{C}$-invariantly-definable" (= there is a formula $\varphi$ such that $\varphi^\mathcal{N}\cap\mathbb{N}=T$ for every $\mathcal{N}\in\mathfrak{C}$ - here I'm conflating sentences with their Godel numbers for simplicity).
Fitting's argument can be summarized as "provability is representable but truth is not." In the system-of-structures framework provability corresponds to the common theory of the class $\mathfrak{C}$ under consideration, and so Fitting's argument proves $(*)$ in the special case that $\mathfrak{C}$'s common theory is not too complicated. But no such assumption is needed for the Kreiselian argument.
Of course, the generality of $(\star)$ doesn't seem to have any use - I don't know of any interesting applications where the system $\mathfrak{C}$ is not reasonably nice, with better arguments thus applying. But this is at least a mathematically meaningful statement which Fitting's argument, unless I'm missing something, doesn't yield.
This argument is essentially similar to the argument of Mel Fitting in his article, "Russell's paradox, Gödel's theorem" Chapter in book: Raymond Smullyan on self reference, 47–66, Outstanding Contributions to Logic, 14, Springer, Cham, 2017.
We had used this article as one of the readings in my Graduate Phil Logic seminar last term.
Fitting argues via Russell's paradox using Tarski's theorem on the non-definability of truth, in a manner that I find very similar to the argument you describe (although he doesn't use your invariant terminology, and perhaps you may regard his argument as not the same as what you intend). Namely, he argues that although provability is representable, truth is not, because of Russell's theorem, and so the incompleteness theorem follows.
Part of his point in that article is that this way of arguing avoids the need for explicit self-reference.
At a glance I think there are meaningful differences between the two arguments, although I haven't had time for a close reading yet; once I've got a chance, I'll (accept your answer after changing my mind or) add a response to it on my answer.
Sure, no problem, it might be a bit more different than what you had in mind.
Very belatedly, I've edited my answer to include a response. Let me know what you think!
|
2025-03-21T14:48:30.117334
| 2020-03-25T05:27:37 |
355652
|
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|
Stack Exchange
|
Movement of repelled particles in a ball
EDIT:
Given a system of $N\geq 3$ charged point particles in $\mathbb{R}^3$ of the same charge which interact according to Coulomb law (thus they repell one from each other). Is it possible that the system remains in a fixed ball all the time? (For $N=2$ this is impossible, and this is what I expect in general.)
More precisely, denote $m_1,\dots, m_N>0$ the masses of the particles. Assume that the $i$th particle acts on $j$th one with the force
$$\vec F_{ij}=\frac{k_ee_ie_j}{|\vec x_j-\vec x_i|^3}\cdot (\vec x_j-\vec x_i), $$
where $k_e>0$ is a constant, $e_i$ is a charge of $i$th particle such that $e_ie_j>0$, $\vec x_i$ is the location of the $i$th particle. The equations of motions are
$$m_j\frac{d^2 x_j}{dt^2}=\sum_{i\ne j}\vec F_{ij}, \mbox{ where } j=1,\dots,N.\,\,\,(1)$$
The question is whether there is a solution such that for some $R$ one has
$$||\vec x_i(t)||<R \mbox{ for all } t>0, \, i=1,\dots, N.$$
ADDED: I expect that this is impossible. In fact I expect that not only for Coulomb law, but still in greater generality. Assume that the equations (1) are satisfied when the force $\vec F_{ij}=\vec F_{ij}(x_i,x_j)$ has the same direction as the vector $\vec x_j-\vec x_i$. Assume moreover that if all points are in a fixed ball of the radius $R$ then for some constant $\varepsilon >0$ such that
$$||\vec F_{ij}||>\varepsilon.$$
Is there a solution of (1) such that all the point are in the ball of radius $R$ for all $t>0$?
Why should one expect this to be possible? At least for $N=2$ (Coulomb scattering) it looks impossible.
@gmvh : I expect that this is impossible.
Nitpick: For $N=1$, it's possible.
@MichaelEngelhardt: Ok, corrected.
https://en.wikipedia.org/wiki/N-body_choreography
@SteveHuntsman Those are, unfortunately, for an attractive force; the repulsive case is very different.
An idle thought, I don't know if it can pan out but it's a possible angle: consider particles by their distance from the origin. Then if at any time $t_0$ the particle furthest from the origin is outward-moving (i.e., its velocity has positive dot product with the vector from the origin to its position), that will be true for all times $t\gt t_0$. It may be possible to show that in that case velocity of the most distant body is $\Omega(1/r)$, in which case the distance from the origin would have to grow as at least $\Omega(\log r)$; then all that's left is showing that it's true at some point.
(To clarify a bit: it won't necessarily always be the case that that specific particle is outward-moving, but that whatever particle is currently furthest from the origin will be outward-moving.)
Let $r(t)$ be the distance of the farthest particle from the origin. It is piecewise analytic. When $r$ is not analytic, a particle has just passed another particle and it is easy to see that $r'$ doesn't decrease. When $r$ is analytic, the net force on an extremal particle is directed outwards and bounded below by some fixed $\epsilon$ (thanks to a short calculation with the $1/r^2$-law and the assumption that $r<R$), so $r''>\epsilon$. These two conditions show $r\to\infty$. This argument breaks down when the force is weaker than a $1/r$-law or isn't analytic.
If all the particles remained in a bounded domain, the virial theorem would apply. In the case of a radial inverse square power law, it states that twice the asymptotic time average of kinetic energy of the system equals minus the asymptotic time average of its potential energy. However, while the kinetic energy is always nonnegative, the potential energy for repulsive coulomb forces is positive, contradiction.
Very nice. We can also check how the $N=1$ case (cf. nitpick to OP) escapes this: Then, the virial theorem reduces to $0=0$ (the particle can, and must, remain at rest) and there is no contradiction. As soon as $N>1$, the potential energy is strictly positive, and the contradiction arises.
Let $B$ be the smallest ball such that all $N$ particles remain inside $B$ for all $t\geq0$.
Either the trajectory of one of the particles intersects $\partial B$ at some finite time $t_0$, or there is one particle and a sequence $(t_n)_{n\in\mathbb{N}}$ with $\lim \limits_{n \to \infty} t_n ~=~\infty$ such that
the particle position at $t_n$ has distance $<1/n$ from $\partial B$, and no other particle at $t_n$ is closer to $\partial B$.
In the first case the radial velocity of the particle at $t_0$ is zero, and therefore the radial component of its acceleration must be $\leq 0$, in contradiction to the fact that the radial component of all forces is positive.
In the second case for each $\epsilon>0$ one can find a time $t_n$ such that the radial acceleration of the particle is less than $\epsilon$. But the radial component of the force from the other particles has a global lower bound because they cannot get arbitrarily close to the particle due to global energy conservation, but have to stay inside the sphere. Choosing a sufficiently small $\epsilon>0$ therefore leads to a contradiction.
I am not sure why in the second case that at time $t_n$ the radial acceleration is less than $\epsilon$? May I get more details?
One can even choose points $t_n$ with radial acceleration $\leq 0$.
The particle is either oscillating towards $\partial B$,
then $(t_n)$ can be chosen to be a subsequence of the locally closest points,
in which accelerations are $\leq 0$, or the particle is creeping towards $\partial B$
with oscillating non-negative radial velocity, then the $t_n$ can be chosen to be points
of locally maximal radial velocity, again with accelerations $\leq 0$,
or the radial velocity is creeping towards zero, and again from some time on radial acceleration is $\leq 0$.
|
2025-03-21T14:48:30.117744
| 2020-03-25T05:48:10 |
355653
|
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|
Stack Exchange
|
Is a normal covering of the total space of a principal bundle also a principal bundle?
Let $G\to E \to B$ be a principal $G$-bundle over $B$. Take a normal covering $\bar{E}$ of $E$. Does $\bar{E}$ admit a principal bundle structure? Namely, $\bar{G}\to \bar{E}\to \bar{B}$, such that $\bar{G}$ and $\bar{B}$ are normal coverings of $G,B$ respectively.
Do you mean a principal bundle over the same base, or...?
@abx I have edited my question just now.
Without any loss of generality, we may assume that all spaces $B$, $E$ and $\bar{E}$ are connected. Let us first consider that case of $\bar{E}$ being the universal cover $\tilde{E}$ of $E$ (of course I am assuming that spaces are nice enough to admit universal covers). We want to establish the existence of a regular cover $\bar{B}\rightarrow B$ fitting in the commutative diagram
$\require{AMScd}$
\begin{CD}
\tilde{E} @>>> E\\
@V V V @VV V\\
\bar{B} @>>> B
\end{CD}
in which the columns are principal bundles. Denote the transformation of $E$ which
$g\in G$ induces by $\theta_g:E\rightarrow E$. Let $G'$ be the set of all self-homeomorphisms of $\tilde{E}$ that fit into the a commutative diagram of the form
\begin{CD}
\tilde{E} @>>> \tilde{E}\\
@V V V @VV V\\
E @>>\theta_g> E
\end{CD}
for some $g\in G$. It is not hard to show that $G'$ is a group of self-homeomorphisms of $\tilde{E}$ acting freely on $\tilde{E}$, and $\tilde{E}/G'$ could be identified with $E/G$; see the related posts:
https://math.stackexchange.com/questions/1667068/on-lifting-an-action-of-g-on-x-to-an-action-of-g-on-tildex
https://math.stackexchange.com/questions/1672763/is-the-quotient-x-g-homeomorphic-to-tildex-g
The group $G'$ is an extension of the discrete group ${\rm{Deck}}(\tilde{E}/E)\cong\pi_1(E)$ by $G$: The group of deck transformations of the universal cover $\tilde{E}\rightarrow E$ is precisely the normal subgroup formed by those elements of $G'$ that lie above $\theta_{{\rm{id}}_G}={\rm{id}}_E$. Any two different lifts of a transformation $\theta_g$ in the previous diagram differ by a deck transformation. Hence the quotient $G'\big/{\rm{Deck}}(\tilde{E}/E)$ may be identified with $G$, and $G'\rightarrow G$ is therefore a normal covering with ${\rm{Deck}}(\tilde{E}/E)$ as its fiber.
In this setting where $\bar{E}=\tilde{E}$, one may take $\bar{B}$ to be $B$ itself, and $\bar{G}$ to be $G'$. In that case, we have a fibration
$$\bar{E}=\tilde{E}\rightarrow \tilde{E}/G'\cong E/G\cong B=\bar{B}$$
which is the quotient map for the free action on $\tilde{E}$ of the normal cover $\bar{G}=G'$ of $G$.
Now let us take an arbitrary normal connected cover $\bar{E}\rightarrow E$. This may be regarded to be an intermediate cover of
$\tilde{E}\rightarrow E=\tilde{E}\big/{\rm{Deck}}(\tilde{E}/E)$; that is, to be in the form of $\bar{E}=\tilde{E}\big/H\rightarrow E=\tilde{E}\big/{\rm{Deck}}(\tilde{E}/E)$ where $H\unlhd{\rm{Deck}}(\tilde{E}/E)$. A lift $\bar{E}\rightarrow\bar{B}$ must be in the form of
\begin{CD}
\bar{E}=\tilde{E}\big/H @>>> E=\tilde{E}\big/{\rm{Deck}}(\tilde{E}/E)\\
@V V V @VV V\\
\bar{B}=\tilde{E}\big/N @>>> B=\tilde{E}\big/G'
\end{CD}
where $N$ is a normal subgroup of $G'$ containing $H$ as a normal subgroup with $G'/N$ discret. In that situation the columns of the diagram above are principal bundles with structure groups $N/H$ and $G'\big/{\rm{Deck}}(\tilde{E}/E)=G$. So the problem seems to group-theoretic: If $H\unlhd{\rm{Deck}}(\tilde{E}/E)\unlhd G'$, does there exist a subgroup $N$ with $H\unlhd N\unlhd G'$ and $G'/N$ discrete? The key point is that $H\unlhd{\rm{Deck}}(\tilde{E}/E)\unlhd G'$ does not imply $H\unlhd G'$; otherwise, one could take $N$ to be $G'$ itself. Recall a basic fact from group theory: If $H$ happens to be a characteristic subgroup of
${\rm{Deck}}(\tilde{E}/E)$, then we do know that $H\unlhd G'$. So there are definitely cases that the answer is positive; but in general, more information on the extension
$$
1\rightarrow{\rm{Deck}}(\tilde{E}/E)\rightarrow G'\rightarrow G\rightarrow 1
$$
is required.
|
2025-03-21T14:48:30.118003
| 2020-03-25T06:02:21 |
355654
|
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|
Stack Exchange
|
Finding the conjugate of a function
I know that the Fenchel conjugate of a function is
$$f^*(x^*) = \sup_x\{\langle x, x^*\rangle - f(x)\}.$$
However, how do I find the Fenchel conjugate of the function
$$f(x) = \frac{1}{p}\sum\limits_{i=1}^n |x_i|^p$$ where $1 < p < \infty$.
I have tried differentiating the equation and taking it to be $= 0$ but I cannot seem to reach any answer.
This survey of the Legendre Fenchel transform discusses this example (see example 4). I do not fully trust their computations (they seem to use $|x|$ to denote $|x|_2$, $|x|_p$, and $|x|_q$ for $(1/p)+(1/q) = 1$, and "cancel" certain inequivalent norms). Still, it may be a useful resource if you are stuck --- in particular I see no issue with the computation of the maximizer of the supremum being $x = \frac{x^}{|x^|_p^{\frac{p-2}{p-1}}}$, which sounds like information you do not yet know.
Trying to be careful tracking the relevant norms throughout, I end up getting to $$f^(x^) = |x^|_p^q\left(\frac{|x^|_2^2}{|x^|_p^2} - \frac{1}{p}\right)$$ where $q$ is the Holder conjugate of $p$. Upon identifying $|\cdot|_q, |\cdot|_p, |\cdot|_2$ this recovers $f^(x^) = \frac{|x^|^q}{q}$, the result the survey claims (as a sanity check to my computation).
Thanks for the answer! It seems like I have not learnt this material, just wondering, for the example in the linked pdf, how did they perform the steps (61-62) and (63-64). How did they convert to and fro a norm?
In 61 -> 62 I believe the expressions should be $z = |y|^{p-2} y$, so an absolute value, not a norm (that distinction won't matter for what I'm about to say, as both absolute values and norms are homogenous). Taking the absolute value of both sides, and using that for any constant $c$ $|cx| = c|x|$, we recover their step. 63 -> 64 simply takes the expression for $|y|$ (line 63) and substitutes it into equation 61.
It will be, somewhat expectedly, $$f^*(x^*) = \frac{\|x^*\|_q^q}{q}, \quad \frac1q + \frac1p = 1$$ as indicated in the comments to the OP. The paper linked there was already rightfully identified as sketchy, the derivation there is wrong on several levels, which is why I am including this answer.
I will use $q$ for the conjugated exponent to $p$ in the following without mentioning this explicitly. Let $$G(x) := \langle x^*,x \rangle - f(x).$$ First note that $\frac1q \|x^*\|_q^q$ is an upper bound on $G$ since $$\langle x^*,x\rangle \leq \frac{\|x^*\|_q^q}{q} + \frac{\|x\|_p^p}{p}.$$ Thus, if a stationary point $\bar x$ of $G$ realizes this upper bound, we are done.
Now, $$\partial_j G(x) = x^*_j - |x_j|^{p-2}x_j,$$ hence $\nabla G(x) = 0$ if and only if $x^*_j = |x_j|^{p-2}x_j$ for each $j$. (The linked paper gets this wrong and then proceeds to mix up all norms.) It is easy to see that $\alpha(t) = |t|^{p-2}t$ and $\omega(s) := |s|^{q-2}s$ are bijective functions on $\mathbb{R}$ which are inverse to each other. Hence $x^*_j = |x_j|^{p-2}x_j$ if and only if $x_j = |x^*_j|^{q-2}x_j^*$. So this is the designated maximizer which we call $\bar x$. There holds $\|\bar x\|_p^{q-1} = \|x^*\|_q$ and thus indeed
$$G(\bar x) = \|x^*\|_q^q - \frac1p \|x^*\|_q^q = \frac{\|x^*\|_q^q}q.$$
To clarify, is the above answer correct for my question? Since the linked paper is solving for f(x) = norm(x)^p, whereas my question is solving for f(x) = abs(x)^p or are they the same?
@JustAPerson the $f$ you give in the question is $f(x) = \frac1p |x|_p^p$.
|
2025-03-21T14:48:30.118261
| 2020-03-25T07:27:19 |
355655
|
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|
Stack Exchange
|
On the statement of Noether-Lefschetz theorem
Noether-Lefschetz theorem says that, If $d \geq 4$, then a surface $S \subset \mathbb P^3$ of degree $d$ having general moduli has Picard group $\text{Pic}(S) \cong \mathbb Z.\mathcal O(1)$.
That means if the surface $S$ is of particular kind(of general moduli) then it's Picard group is generated by $\mathcal O_S(1)$ and hence is isomorphic to $\mathbb Z$. Alternatively one can say the restriction map $\text{Pic}(\mathbb P^3) \to \text{Pic}(S)$ is an isomorphism.
At this point my confusion is regarding the first part of the theorem:" $S$ having general moduli".
In the literature it's mentioned that "of general moduli" means that there is a countable union $V$ of subvatrieties of the projective space $\mathbb P^N=|\mathcal O_{\mathbb P^3}(d)|$ of smooth surfaces of degree $d$ in $\mathbb P^3$ such that the property $\text{Pic}(X) \cong \mathbb Z$ holds for $X \in \mathbb P^N - V$.
More precisley, does this mean that the first part of the theorem says assume existence of such $V$ with the said property and choose $S \in \mathbb P^N - V$. But then the way we have chosen $S$ we automatically have $\text{Pic}(S) \cong \mathbb Z$(Surely I am missing the point here). Then is the phrase "general moduli" independent of the surface $S$?
Any help from anyone is welcome
The theorem means: There exists $V \subset \mathbb{P}^N$ which is a countable union of subvarieties, such that for all $S \in \mathbb{P}^N - V$, it holds that $\operatorname{Pic}(\mathbb{P}^3) \to \operatorname{Pic}(S)$ is an isomorphism. In other words, it asserts (not assumes) existence of such a $V$. The surface $S$ is chosen after $V$ is established.
@ZachTeitler, Thanks a lot for the clarification. Is the existence of such a $V$ for $d \leq 3$ known?
There is no such V for $d=2$ or 3. Smooth complex surfaces of these degrees have Picard groups of rank 2 and 7 respectively.
|
2025-03-21T14:48:30.118433
| 2020-03-25T08:47:12 |
355658
|
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"Daniel Loughran",
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|
Stack Exchange
|
Computing the class group of a quadratic function field
I am asking for a reference in which I can find tools to answer questions like the following: Let $K=\mathbb{F}_q(X)$ be a rational function field over the finite field with $q$ elements. Let $E/K$ be a quadratic extension. How can I compute the class groups of $K$?
I'm not interested in the general case, but in the special case where $E = K(\sqrt{D})$, where $D$ is an irreducible polynomial of even degree $\deg D=2g+2$ and $g,D$ may vary (that is, statistical results are also welcomed).
Edit. I've been asked to be more specific in my question: Given a finite field $\mathbb{F}_q$ and a prime $\ell$ (say $\ell=3$ to be concrete), can is there always an hyperelliptic curve $y^2=D(x)$ with $D$ irreducible of degree $2g+2$ such that the class group has a order divisible by $\ell$?
Isn't the class group just the group of $\mathbb{F}_q$-rational points on the jacobian of the smooth projective curve $C$ with function field $E$? Your curve here is a hyperelliptic curve, and I think tmagma can computer the points on the jacobian.
Thanks for you comment Daniel. It indeed is. But I am trying to have something more conceptual. For example, can I find for every finite field, say of characteristic neq 2, an elliptic curve y^2 = D(x), with D of degree 4 and irreducible such that the number of points is even, or divisible by three?
I see. Well the question you have written above seems very different to the question in your comments. Can you ask a more specific question which has a well-defined answer? Or do you just want a list of all papers which consider class groups of function fields?
About the question in the comments: Honda-Tate theory tells you exactly which abelian varieties can exist over any finite field and how to calculate their number of rational points via the Weil conjectures. Now they may not all be jacobians. But in the elliptic curve case they are, so I would guess that this would allow you to answer your question.
The answer to the precise question is yes. See Theorem 1.2 of:
"Homological stability for Hurwitz spaces and the Cohen-Lenstra conjecture over function fields," https://arxiv.org/pdf/0912.0325.pdf.
This implies that a positive proportion of imaginary quadratic extensions of $\mathbb{F}_q(x)$ has class number divisible by $\ell$, at least for $q$ large and $q \not \equiv 1 \bmod \ell$.
The corresponding infinitude is easier and was known previously in more generality (see the discussions following Theorem 1.2).
This is more of a too-long comment than an answer.
The specific question in the comments, "can I find for every finite field, say of characteristic neq 2, an elliptic curve $y^2 = D(x)$, with $D$ of degree 4 and irreducible such that the number of points is even" has a nice positive answer: yes, for any irreducible $D$. And the same trick applies to produce hyperelliptic examples in every odd genus.
Let $g \ge 1$ be odd, and let $F$ be an irreducible polynomial of degree $2g+2$ over $\mathbb{F}_q$ with $q$ odd (I will use $F$, rather than $D$, for the polynomial, so that I can use $D$ for divisors). We define a genus-$g$ hyperelliptic curve
$
H: y^2 = F(x)
$
over $\mathbb{F}_q$. Let $D_\infty$ be the (degree 2) divisor at infinity on $H$.
The polynomial $F$ factors over $\mathbb{F}_{q^2}$ as $F = GG'$ where $G$ and $G'$ are irreducible polynomials of degree $g+1$ over $\mathbb{F}_{q^2}$, conjugate over $\mathbb{F}_q$. Let $\{\alpha_1,\ldots,\alpha_{g+1}\}$ be the roots of $G$ and $\{\alpha_1',\ldots,\alpha_{g+1}'\}$ be the roots of $G'$. Since $g$ is odd, we can define divisors $D := \sum_{i=1}^{g+1}(\alpha_i,0) - ((g+1)/2)D_\infty$ and $D' := \sum_{i=1}^{g+1}(\alpha_i',0) - ((g+1)/2)D_\infty$ on $H$, both of degree $0$ and both defined over $\mathbb{F}_{q^2}$. Now $D$ and $D'$ are not principal (they correspond to the ideals $(G(x),y)$ and $G'(x),y)$, but by construction $2D = (G)$ and $2D' = (G')$, so $[D]$ and $[D']$ are 2-torsion divisor classes exchanged by Galois. But in fact they are the same class, because $D + D' = (y)$; so $[D] = [D']$ is an $\mathbb{F}_q$-rational divisor class of order 2, and therefore the divisor class group of $H$ has even order.
|
2025-03-21T14:48:30.118718
| 2020-03-25T10:38:51 |
355662
|
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|
Stack Exchange
|
Isomorphism of Archimedean real closed fields with infinite transcendence degree
Are all countable Archimedean real closed fields with infinite transcendence degree isomorphic?
No, just take a proper inclusion between two such subfields of $\mathbf{R}$. (The only possible isomorphism would be the identity.)
Thank you very much!
|
2025-03-21T14:48:30.118767
| 2020-03-25T11:04:51 |
355665
|
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"url": "https://mathoverflow.net/questions/355665"
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|
Stack Exchange
|
Effaceability conditions in the derived category
In abelian categories effaceability of functors is often an interesting property.
Is there any general equivalent condition on derived functors in the derived category?
For example, for a functor in the category on constructible sheaves, can effaceability be detected in the derived category of constructible sheaves?
|
2025-03-21T14:48:30.118838
| 2020-03-25T11:38:02 |
355666
|
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"Bazin",
"Carlo Beenakker",
"HenrikRüping",
"Willie Wong",
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"wlad"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/355666"
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|
Stack Exchange
|
Does the generalised directional derivative satisfy any version of the chain rule?
Is there a chain rule of any kind for the generalised directional derivative (of the Clarke type)? There is certainly a chain rule for the generalised gradient.
The generalised directional derivative is: $$f^\circ(x;v)=\limsup_{y \to x, t \downarrow 0} \frac{f(y+tv) - f(y)}{t},$$
where $x,v \in \mathbb R^n$ for some $n$, and $f:\mathbb R^n \to \mathbb R^m$. Albeit, the definition is valid over any Banach space (but I'd like to keep things to finite dimensions for simplicity's sake).
Some information about it can be found here: https://www.encyclopediaofmath.org/index.php/Clarke_generalized_derivative
[edit]
The naive version of the chain rule is false: Consider $f(x) = |x|, g(x) = -x$. We have that $(f\circ g)^\circ(0;1) = 1$ while $f^\circ(g(0); g^\circ(0;1)) = f^\circ(0;-1) = -1$. What I'm looking for must therefore be an inequality.
Please provide some context. Are you working in an arbitrary Banach space?
@YCor $\mathbb R^n$
What is the point of the variable $y$ in the definition?
a typo, $x$ in the numerator on the right-hand-side should read $y$
So if $m=n=1$ and $f$ is a function whose slope alters between $+1$ and $-1$ faster and faster as $x$ goes to zero, we get that $f^\circ(0)=1$ since close to one, there is always a point with slope $+1$ and we get $(-f)^\circ(0)=1$, since the same is true for $-f$. Doesnt that contradict the chai rule
@CarloBeenakker thank you
@HenrikRüping It could be an inequality instead. Basically, the generalised gradient already satisfies only a weak version of the chain rule
It may be helpful if you can quote in your question the known weak version of the chain rule satisfied by the generalized gradient.
According to theorem 8.14 of https://arxiv.org/pdf/1708.04180.pdf, we have that for locally Lipschitz $g:Y\to \mathbb R$ and Frechet-differentiable $f: X \to Y$, that
$$(g\circ f)^\circ(x;v) \leq g^\circ(f(x);f'(x)v).$$
Equality holds if $g$ is regular.
We can furthermore say that:
$$(g\circ f)^\circ(x;v) \geq -g^\circ(f(x);-f'(x)v)$$
under the same conditions.
A nice and non-trivial extension of the chain-rule occurs from the DiPerna-Lions theory of rough vector fields: take on an open subset $\Omega$ of $\mathbb R^n$ a vector field $X$ with $L^\infty_{loc}(\Omega)$ coefficients and null divergence such that $X\in W^{1,1}_{loc}(\Omega)$. Let $u$ be an $L^\infty_{loc}(\Omega)$ function such that $Xu\in L^1_{loc}(\Omega)$. Then
$$
X(u^2)=2u Xu.
$$
The previous chain-rule formula is the main point to prove uniqueness of weak solutions for these vector fields. There are generalizations in several directions: instead of looking at $u^2$, you may check
$$
X(F(u))=F'(u) Xu, \quad F\in C^1.
$$
Also, you can relax the regularity $W^{1,1}$ to $BV$, play a bit with regularity $W^{1,p}$ and relax as well the condition on the divergence by requiring only absolute continuity wrt Lebesgue measure.
I fail to see how this addresses the question in the OP. Can you give more details?
Well the chain rule is about finding the derivative of $F\circ u$. My remark above is highlighting the fact that when you have a limited amount of regularity, say when the function $u$ is not better than $L^\infty$, if you have the weak information that $Xu$ belongs to $L^1$, you can still apply the chain rule and this is the core argument for the DiPerna-Lions theory for first-order linear equations with $W^{1,1}$ coefficients.
Even for finite-dimension, you have only an inclusion instead of equality in the chain rule. Basically, naively performing the chain rule by multiplying Clarke subdifferentials can result in a set that strictly contains the Clarke subdifferential of the composition. This is made clear in the book of Clarke.
If you want something like the Clarke subdifferential which admits a chain rule by design, I invite you to read about conservative fields.
The straightforward generalization of the usual chain rule would give
$$(f\circ g)^\circ(x,v)=v\cdot\bigl(Dg(x)\bigr)^{\rm T}\cdot\bigl(\nabla f(y)\bigr),$$
with $Dg$ the Jacobian matrix and $g(x)=y$.
except I don't think that's true...
consider $f(x) = |x|$, $g(x) = -|x|$, $h(x) = f(x) + g(x)$. We have $f^\circ(0;1) = 1$, $g^\circ(0;1) = 1$, while $h^\circ(0;1) = 0$. This contradicts the addition rule, and therefore your naive version of the chain rule
I may edit the question to include these examples
I've included a simple example in the question
|
2025-03-21T14:48:30.119415
| 2020-03-25T12:01:29 |
355668
|
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|
Stack Exchange
|
Restriction of real irreducible 2-Brauer characters to subnormal subgroups
Question: Find a finite group $G$, a subnormal subgroup $H$ of $G$, a real-valued irreducible $2$-Brauer character $\chi$ of $G$ and a real-valued irreducible $2$-Brauer character $\mu$ of $H$ such that $\mu$ occurs with odd multiplicity $\geq3$ in $\chi_H$.
Context: Let $N$ be a normal subgroup of a finite group $G$, let $p$ be a rational prime and let $\chi$ be an irreducible ($p$)-Brauer character of $G$. Then according to Clifford theory $\chi_N=e(\theta_1+\dots+\theta_t)$, where $e,t$ are positive integers and $\theta_1,\dots,\theta_t$ is a $G$-orbit on the irreducible Brauer characters of $N$. We say that $\chi$ lies over each $\theta_i$, and that $\theta_i$ has multiplicity $e$ in the restricted character $\chi_N$.
From now on $p=2$. Let $\theta$ be a real-valued irreducible Brauer character of $N$. Then by a recent result of Rod Gow and the author, there is a unique real-valued irreducible Brauer character $\chi$ of $G$ such that $\theta$ has odd multiplicity in $\chi_N$. Moreover this multiplicity is $1$ (so $\chi_N$ is the sum of the $G$-orbit of $\theta$). Equivalently, $\theta$ has a unique real-valued extension to its stabilizer in $G$.
For fans of solvable groups, this result follows from well-known properties of Isaac's canonical set B$_{2'}(G)\subseteq{\rm Irr}(G)$, and a result of I. M. Richards: if $|G:N|$ is odd then each real-valued ordinary irreducible character of $N$ has a unique real-valued extension to its stabilizer in $G$. Our proof, for all finite groups, uses a straightforward cohomological argument over a perfect field of characteristic $2$.
Suppose now that $H$ is a subnormal subgroup of $G$, and $\mu$ is real-valued irreducible Brauer character of $H$. Iterating the above result, there is a unique real-valued irreducible Brauer character $\chi$ of $G$ such that $\mu$ has odd multiplicity in $\chi_H$. Of course the irreducible constituents of $\chi_H$ may no longer be all $G$-conjugate. It is easy to see that the multiplicity must be $1$ if there exists $N$ with $H\unlhd N\unlhd G$.
Jürgen Müller has provided me with two families of examples, where the restriction multiplicities are arbitrarily large and odd. We describe these below, with his kind permission.
First let $p$ be a prime such that $p\equiv1$ (mod $4$), let $L :=\operatorname{GL}_2(p)$, let $S$ be a Sylow $2$-subgroup of $L$, and set $N := N_L(S)$. Then $S =C_s\wr C_2$, where $s$ is the largest $2$-power dividing $p-1$ and the outer $2$ can be chosen as the involution $t:=\begin{bmatrix}0&1\\1&0\end{bmatrix}$. Moreover, $N = \operatorname{Z}(L) S = C_{(p-1)/s}\times S$ has order $2s(p-1)$.
Next, let $E = Z_p + Z_p$ be the natural $L$-module, and let $G := E\rtimes N$ be the natural semidirect product. Now let $u \in E$ be an eigenvector of $t$ w.r.t. the eigenvalue $-1$. Then $H := \langle u,t\rangle$ is dihedral of order $2p$. Further, $H$ is subnormal in $G$ as it is normal in $E\rtimes(\operatorname{Z}(L)\times\langle t\rangle)$ and $G/(E\rtimes\operatorname{Z}(L))$ is nilpotent.
Now all irreducible $2$-modular Brauer characters of $H$ are self-dual, and, next to the
trivial character $1$, there are $(p-1)/2$ characters $\phi_i$ of degree $2$. Finally, $G$ has a unique, hence self-dual, irreducible $2$-modular Brauer character $\phi$ of degree $s(p-1)$, whose restriction to $H$ decomposes as $(p-1)$ copies of $1_H$ and $(s-1)$ copies of $\sum_i \phi_i$. Here $s-1 \geq 3$ is odd, and can be made arbitrarily large by chosing $p$ appropriately.
The case $p\equiv -1$ (mod $4$) can be treated along very similar lines, only the description of the Sylow $2$-subgroup $S$ of $L :=\operatorname{GL}_2(p)$ and the Clifford theory became slightly more tricky. Letting $s$ be the largest $2$-power dividing $p+1$, we have $S = \operatorname{QD}_{4s}$, a quasi-dihedral group, and $\operatorname{N}_L(S) = C_{(p-1)/2}\times S$.
Choosing $G$ and $H = D_{2p}$ as before, utilising a suitable non-central involution of $S$, again there is a (unique) self-dual, irreducible $2$-modular Brauer character $\phi$ of $G$ of degree $s(p-1)$, whose restriction to $H$ decomposes as $(p-1)$ copies of $1_H$ and $(s-1)$ copies of $\sum_i \phi_i$.
For the whole series, the Brauer character $\phi$ has defect $1$, hence belongs to a block of defect $1$ and is the only irreducible Brauer character of this block.
|
2025-03-21T14:48:30.119803
| 2020-03-25T12:23:12 |
355669
|
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|
Stack Exchange
|
Sobolev topology on essentially compactly supported Sobolev-"functions"
The locally convex space of essentially compactly-supported $p$-integrable "functions" $\operatorname{L}_{\mathrm{comp}}^p(\mathbb{R}^d,\mathbb{R})$ is defined as the set
$$
\bigcup_{n \in \mathbb{N}} \left\{
f \in L^p(\mathbb{R}^d,\mathbb{R}):\, \operatorname{ess-supp}(f)\subseteq [-n,n]^d
\right\},
$$
topologized with the injective limit topology in the category of LCSs with continuous linear maps as morphisms, where the colimit is taken over the injective system $$\left\{L^p(\mathbb{R}^d,\mathbb{R}):\, \operatorname{ess-supp}(f)\subseteq [-n,n]^d\right\}_{n \in \mathbb{N}}.$$
Fix a $k \in \mathbb{N}$, $1\leq p<\infty$ and let $W^{k,p}_{\mathrm{comp}}(\mathbb{R}^d):=W^{k,p}(\mathbb{R}^d) \cap \operatorname{L}_{\mathrm{comp}}^p$. How are the subspace topologies on $W^{k,p}_{\mathrm{comp}}(\mathbb{R}^d)$ comparable? Is the subspace topology induced by restricting the subspace topology on $L^p_{\mathrm{comp}}(\mathbb{R}^d,\mathbb{R})$ to $W^{k,p}_{\mathrm{comp}}(\mathbb{R}^d)$ at-least as fine than the one obtained by restricting the Sobolev topology $W^{k,p}$ to $W^{k,p}_{\mathrm{comp}}(\mathbb{R}^d)$?
Why should this be true? The $L^p_{comp}$-topology gives nothing for the distributional derivatives! You probably know that because of the stricness of the inductive system the relative topology of $L^p_{comp}$ on the steps is just the $L^p$-topology.
But in this post:
https://mathoverflow.net/questions/347318/can-l1-loc-be-represented-as-colimit?noredirect=1&lq=1
don't we find that the topology on $L^p_{comp}$ is strictly finer?
Indeed, the $L^p_{comp}$ topology is stricly finer than the $L^p$-topology, but on the steps they coincide. There is no contradiction.
Oh my brain filtered the word "steps" sorry. I've never seen this terminology, you mean the finite "sub-colimits?". In that case definitely.
To make my comment an answer: No. To see this, fix a cube $K$ and look at $L^p_K=\{f\in L^p: \text{ ess-supp}(f) \subseteq K\}$. On this space, the $L^p_{comp}$-topology coincides with the $L^p$-topology (this follows from the stricness of the inductive limit) and the $W^{k,p}$-topology is strictly finer on this subspace.
Ah, you're right this is very clear.
|
2025-03-21T14:48:30.119968
| 2020-03-25T13:17:30 |
355674
|
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"Ari Shnidman",
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"url": "https://mathoverflow.net/questions/355674"
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|
Stack Exchange
|
Can each ideal class contain an ideal with norm equal to $1$?
Let $K$ be an imaginary quadratic number field. Let $\mathcal O$ be an order in $K$. Can it happen, that there are $h(\mathcal O)>1$ fractional proper $\mathcal O$-ideals, representing the ideal classes of $\mathcal O$, each with norm equal to $1$? Does this always happen?
(Note: I have edited the answer to cover all cases.)
Let's call this "property 1". I claim that $\mathcal{O}$ has property 1 if and only if $h(\mathcal{O}_K)$ is odd.
First let me prove it when $\mathcal{O} = \mathcal{O}_K$. Then for a fractional ideal to have norm 1, it must be of the form $I/\bar I$, where $\bar I$ is the Galois-conjugate ideal (here I am tacitly using that ideals in $\mathcal{O}_K$ have unique factorization into prime ideals). Since conjugation is inversion in the class group, this means that the ideal class is a square. If all ideal classes are squares then the squaring map is surjective, hence injective, and so the 2-torsion subgroup is trivial. So the group has odd order.
Now consider the general case where $\mathcal{O}$ is an order of index $f$ in $\mathcal{O}_K$. I'll show that $\mathcal{O}$ has property 1 iff $\mathcal{O}_K$ has property 1. If $I$ is a proper fractional $\mathcal{O}$-ideal of norm 1, then $\mathcal{O}_KI$ (the product lattice) has norm 1, since norm is multiplicative. Since the map $\pi_f \colon \mathrm{Pic}(\mathcal{O}) \to \mathrm{Pic}(\mathcal{O}_K)$ sending the class of $I$ to the class of $\mathcal{O}_KI$ is surjective, this shows that if $\mathcal{O}$ has property 1, then so does $\mathcal{O}_K$.
Conversely, suppose $\mathcal{O}_K$ has property 1. Then for any ideal class $[I]$ in $\mathrm{Pic}(\mathcal{O}_K)$, I may choose $I$ to have norm 1 and also to be prime to $f$. Now, one can check that the ideals classes in the pre-image $\pi_f^{-1}([I])$ all have representatives $J$ which are simply index $f$ sublattices in $I$. [Let me not prove this here, but this is exactly what Siegel is doing explicitly in the paper you link to in the comments below. Note, in general not all index $f$ sublattices are proper $\mathcal{O}$-ideals. In fact, the number of such is precisely the size of the kernel of the map $\pi_f$, at least assuming there are no extra units in $\mathcal{O}_K^\times$.] Note that each these $\mathcal{O}$-ideals $J$ has norm 1 as well. So property 1 holds for $\mathcal{O}$.
As for your question of how often this happens, the 2-part of the class group is related to the number of primes dividing the discriminant. If, for example, there are at least 2 odd primes $p, q$ dividing the discriminant of $K$, then the 2-part is non-trivial. Indeed, the unique ideal above $p$ is 2-torsion in the class group of the maximal order, but not principal. But if $K$ has prime discriminant then indeed the class number is odd.
On the other hand, in this question we have even class number and at the same time each class is represented by a fractional ideal of norm equal to $1$. How is this possible?
Ah, you're right! My proof does not actually show the "only if" in the case of non-maximal orders (because it's conceivable that the ideal in question is not prime to the conductor, and hence my tacit use of unique factorization of ideals is not allowed). I will edit in the correct statement and proof.
Why and how is then Siegel computing the character $\left(\frac{f}{N(\mathfrak c)}\right)$? This character depends on norm modulo $f$, but if in each class is a fractional ideal of norm $1$ should not be this always the trivial character?
No idea, I don't have time to stare at his paper. Maybe it has to do with the definition of the norm though. In the "proper" definition (the one I was using) the norm of a proper O-ideal is computed relative to O itself (not O_K). Maybe Siegel is computing the norm as the index relative to O_K.
|
2025-03-21T14:48:30.120250
| 2020-03-25T13:37:51 |
355675
|
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|
Stack Exchange
|
Reference request: a survey of (linear) Krein-Rutman theory
I'm looking for a survey article or book chapter where a rather exhaustive treatment of the Krein-Rutman theory of positive linear operators an ordered Banach spaces is given.
Motivation. Some explanation is probably required. Classicly, one finds two types of "Krein-Rutman theorems", and many versions thereof, in the literature:
Let $X$ be a real Banach space that is ordered by a closed cone $X_+$ (i.e., $\alpha X_+ + \beta X_+ \subseteq X$ for all $\alpha, \beta \in [0,\infty)$, and $X_+ \cap (-X_+) = \{0\}$). Let $T: X \to X$ be a bounded linear operator which is positive in the sense that $TX_+ \subseteq X_+$.
Theorem 1. ("Weak version") Assume that $X_+$ is total (i.e. its span is dense in $X$). If $T$ is compact, then:
The spectral radius $r(T)$ is in the spectrum of $T$.
If $r(T) > 0$, then $r(T)$ is an eigenvalue of $T$ with an eigenvector in $X_+$, and an eigenvalue of the dual operator $T'$ with an eigenvector in the dual cone $X_+'$.
Theorem 2. ("Strong version") Assume that $X_+$ has non-empty topological interior $\operatorname{int}(X_+)$ in $X$ and that $T$ is compact and strongly positive in the sense that $Tx \in \operatorname{int}(X_+)$ for every $0 \not= x \in X_+$. Then:
$r(T) > 0$.
The eigenvalue $r(T)$ is algebraically simple, and the corresponding eigenspace is spanned by a vector in $\operatorname{int}(X_+)$.
$r(T)$ is the only spectral value of $T$ with modulus $r(T)$.
$r(T)$ is the only eigenvalue of $T$ that admits an eigenvector in $X_+$.
These results are easy to find in numerous places in the literature, and so are various improvements thereof (for instance, "compact" is obviously too strong an assumption; it suffices if the essential spectral radius $r_e(T)$ is strictly smaller than $r(T)$).
However, I find many treatments of this circle of ideas in the literature rather unsatisfactory, for the following reasons:
These results are typically used as auxiliary results for other purposes (for instance, in PDE theory), so they are often presented in an ad-hoc manner rather in a structure theoretic way.
The underlying ideas are typically not exploited in a comprehensive way. For instance, it is easy to prove a version of Theorem 1 if the compactness assumption is replaced with the assumption that $T/r(T)$ is weakly almost periodic and that $T$ has an eigenvalue of modulus $r(T)$; but I have not seen this anywhere, yet.
In the setting of general ordered Banach spaces, most versions of Theorem 2 that I have encountered (with some exceptions, though) use the rather strong - and, imho artificial - assumption that the cone has non-empty interior. On the other hand, though, in the setting of Banach lattices, an entire Krein-Rutman theory (which is, in the context of Banach lattices more often called Perron-Frobenius theory) is available without this assumption (the assumption "strong positivity" then has to be replaced with some kind of irreducibility assumption; see, for instance, Section~V.5 in H. H. Schaefer's book "Banach lattices and positive operators (1974)"). I find this to be a strange state of affairs since it seems desirable to view the corresponding results on Banach lattices and Theorem 2 above as special cases of the same general theory.
The question. So my question is, in more precise terms: Is there a survey article or book chapter about linear Krein-Rutman theory in ordered Banach spaces that satisfies the following requirements:
(i) It uses a structure theoretic approach, i.e. it does not focus on ad-hoc methods but rather on theory building.
(ii) It is comprehensive in the sense that it (a) presents very general versions of the theorems rather than special cases (for instance, it assumes $r_e(T) < r(T)$ rather than compactness of $T$) and (b) tries to exhaust the available methods and to provide counterexamples which demonstrate optimality of results.
(iii) It is unifying in the sense that it does not assume the interior of the positive cone to be non-empty unless really necessary - such that the known results on Banach lattices fit into the general framework. It particular, it should also contain a theory a irreducible operators on ordered Banach spaces.
Remarks.
If you know a survey which meets only a proper (and non-empty) subset of the requirements (i)-(iii), I will still be happy to learn about it (although it does not completely answer the question).
The only reference that I know which comes somewhat close is the Appendix of H. H. Schaefer's book "Topological vector spaces (1971)". It satisfies requirement (i), but then, again, it falls in some respects short of (ii) and, to a large extent, also of (iii).
What this question is not about. Please note:
I am not asking for single generalizations of the above theorems. I know that there are plenty of them, and I know where to find many of them. I am looking for a survey where theses results and generalizations are discussed in a single place.
Non-linear Krein-Rutman theory is interesting and important, but it is not part of this question. (But, of course, a survey which covers both the linear and the non-linear theory would qualify as an answer, as long as the linear part is also treated in detail.)
I am not asking for literature that focusses merely on Banach lattices (or on ordered Banach space with the Riesz decomposition property).
I am not asking for ideas or methods how one could generalize, for instance, Theorem 2 to a theory that fits requirement (iii). I am aware of such methods, but I would like to have a citable reference where they are presented.
|
2025-03-21T14:48:30.120628
| 2020-03-25T14:34:23 |
355678
|
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|
Stack Exchange
|
Matching generating function of 2-lifts
Let $P_G$ denote the matching generating function of a finite simple bipartite graph $G$.
Let now $H$ be a $2$-lift of $G$. We know (see for example Proposition 5.3.3 in Barvinok's book Combinatorics and Complexity of Partition Functions) that
$$ P_H(t)\le P_G(t)^2\quad \forall t\ge 0. $$
I was wondering if anything interesting could be said about the relation between the roots of $P_H(t)$ and the roots of $P_G(t)$.
More generally, what can be said about $P_H(t)$ in terms of $P_G(t)$ other than the above statement?
|
2025-03-21T14:48:30.120696
| 2020-03-21T19:40:35 |
355395
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/355395"
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|
Stack Exchange
|
Absoluteness and the scale property for $Π^2_2$ or $Σ^2_2$
Under the diamond principle $◊$ and large cardinal axioms, which of the two pointclasses $Π^2_2$ or $Σ^2_2$ is expected to have the scale property?
Because conditional $Σ^2_2$ absoluteness under $◊$ and large cardinal axioms remains conjectural, I am fine with the answers using additional assumptions. Some assumption like $◊$ is necessary since one can use a generic extension without the Continuum Hypothesis (CH) to 'scramble' $Δ^2_1$ sets, and with CH but without $◊$, one can still scramble $Δ^2_2$ sets.
The scale property is the most important uniformization-like property in descriptive set theory; it implies prewellordering and uniformization. For important non-self-dual pointclasses in descriptive set theory, commonly either the class or its dual has the scale property, though there are exceptions. (Under large cardinal axioms, the main exception up to $(Σ^2_1)^\text{uB}$ is if there is a gap in complexity levels during which new $(Σ^2_1)^\text{uB}$ statements do not appear; uB means universally Baire.)
Under CH and existence of an iterable model with a measurable Woodin cardinal, $Π^2_1$ has the scale property (The length-$ω_1$ open game quantifier propagates scales). There, $Π^2_1$ corresponds to the game quantifier for open analytical games of length $ω_1$. Presumably, $Π^2_1$ in $V$ also corresponds to $Σ^2_1$ in the minimal iterable inner model with a proper class of indiscernible Woodin cardinals, with that model satisfying $(Σ^2_1)^\text{uB} = Σ^2_1$.
Fine-structural models relying on the conventional comparison using $ω_1+1$ iterability (for countable elementary substructures) are limited to reals that are $Δ^2_2$ in a countable ordinal. However, perhaps there is a maximal such model, with the model able to compute $Σ^2_2$ truth (without being $Σ^2_2$ correct). For comparison, $L$ is $Σ^1_2$ correct and only has reals that are $Δ^1_2$ in a countable ordinal.
Hugh Woodin has conjectured that the transition beyond $Δ^2_2$ occurs around supercompact up to a measurable. He identified an obstruction that prevents certain models (that add extenders without adding their iteration strategies) from reaching supercompact up to a measurable, and conversely, if the models he is developing (In Search of Ultimate-L) work out and there is no failure of iterability, the obstruction does not occur much earlier. I agree that the transition beyond $Δ^2_2$ is unlikely below long extenders.
Woodin also connected conditional $Σ^2_2$ absoluteness with determinacy of certain games (Beyond $Σ^2_1$ Absoluteness). This connection might be helpful for getting the scale property. Two caveats:
- The impossibility $Σ^2_2(\mathcal{I}_\text{NS})$ absoluteness under the $Ω$-conjecture has since been retracted.
- The connection might be inapplicable if say $(Σ^2_1)^\text{uB} = Π^2_2$ (in which case $Π^2_2$ would have the scale property, with $Σ^2_2$ absoluteness under $◊$ violating/transcending the $Ω$-conjecture).
$Σ^2_2$ maximality
Assuming a proper class of measurable Woodin cardinals:
- $Σ^2_1$ statements are absolute between all generic extensions of $V$ satisfying CH. Moreover, under CH, for universally Baire $A$, $Σ^2_1(A)$ is universally Baire.
- $Σ^2_2$ maximality: there is a generic extension of $V$ that satisfies all $Σ^2_2$ statements that hold in some generic extension of $V$ satisfying CH (Regular embeddings of the stationary tower and Woodin’s $Σ^2_2$ maximality theorem).
I think that the $Σ^2_2$ maximality gives a good approximation to the $Σ^2_2$ absoluteness with respect to the structure of $Π^2_2$ and $Σ^2_2$ sets, and in any case, despite uncertainty about what lies beyond, I will accept the scale property in such a model as an answer.
My guess is that if you start with a fine-structural iterable model with a proper class of measurable Woodin cardinals (caveat: iterability remains an open problem) below long extenders, then in the $Σ^2_2$ maximality extension, $Π^2_2$ equals $(Σ^2_1)^\text{uB}$ and hence has the scale property.
The fact that a $Π^2_2$ statement holds in all generic extensions of $V$ satisfying CH suggests that there is some structure to enforce it. And a countable mouse (a fine-structural model) $M$ with a measurable Woodin cardinal $δ$ and a universally Baire iteration strategy can use a generic extension over the extender algebra $W_{δ,δ}$ to essentially conjure up an arbitrary subset $A$ of $ω_1$ in every generic extension of $V$. $M$ can then quantify over $A$ and use "for all $A$ ..." to certify a $Π^2_2$ statement. The stratification of $Π^2_2$ statements according to the minimal complexity of $M$ in turn likely leads to the scale property.
A good explanation of the extender algebra can be found in The extender algebra and $Σ^2_1$-absoluteness, but briefly, we have the following. In every $V[G]$, every $A⊂ω_1^{V[G]}$ is $W_{δ,δ}^{M'}$-generic over some iterate $M'$ of $M$ with $δ^{M'}=ω_1^{V[G]}$ and $M'[A]∈V[G]$, hence the $Π^2_2$ certification is valid. (Informally, we get every relevant $A$, and $M'[A]∈V[G]$ ensures that $M$ cannot cheat about $A$.) Conversely, for every forcing condition $p$ for $W_{δ,δ}$ with $p⊢δ=ω_1$, a generic set $A$ below $p^{M'}$ already exists in $V$ for an iterate $M'$ of $M$ with $δ^{M'}=ω_1$ (and we can also insert any $ω_1$ reals (or using CH, $ℝ$) into the extension), which should ensure that the quantification over $A$ is not too broad (or at least, this suffices for the $Σ^2_1$ absoluteness under CH).
|
2025-03-21T14:48:30.121074
| 2020-03-21T19:50:32 |
355396
|
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|
Stack Exchange
|
Regularity theory for parabolic PDEs in fractional Sobolev spaces
I am trying studying on my own parabolic PDEs throughout the book "Linear and Quasi-linear Equations of Parabolic Type" by Vsevolod A. Solonnikov, Nina Uraltseva, Olga Ladyzhenskaya and the existence and uniqueness of solutions for linear parabolic PDEs is proved for a fractional Sobolev space on page $320$:
$\textbf{Theorem 5.1.}$ Suppose $l > 0$ is a nonintegral number and the coefficients of the operator $\mathcal{L}$ belongs to the class $H^{l,l/2} \left( \overline{D^{(T)}_{n+1}} \right)$, then for any $f \in H^{l,l/2} \left( \overline{D^{(T)}_{n+1}} \right), \ \phi \in H^{l + 2} \left( \overline{E}_n \right)$ problem $(5.2)$ has a unique solution from the class $H^{l/2,l/2+1} \left( \overline{D^{(T)}_{n+1}} \right)$. It satisfies the inequality
$$|u|_{D^{(T)}_{n+1}}^{(l + 2)} \leq c \left( |f|_{D^{(T)}_{n+1}}^{(l)} + |\phi|_{E_n}^{(l + 2)} \right)$$
with the constant not depending on $f$ and $\phi$.
I did not find that the solution in $H^{l,l/2} \left( \overline{D^{(T)}_{n+1}} \right)$ imply some regularity in the classical sense in the book.
I was looking for this in other sources I only find that the solution remains in $C^{0,\alpha}$ under the assumptions of the theorem $8.2$ on page $55$ of this lecture notes and remains in $C^{0,\beta}$ for $\beta < \alpha$ according the following corollary of the theorem $8.2$:
$\textbf{Corollary 1.9}$ Let $s \in (0,1)$ and $p \in [1,+\infty)$ such that $sp > n$. Let $\Omega$ be a $C^{0,1}$ bounded domain of $\mathbb{R}^n$. Then the embedding
$$W^{s,p}(\Omega) \hookrightarrow C^{0,\beta}(\Omega)$$
is compact for every $\beta < \alpha$ with $\alpha := \frac{sp - n}{p}$.
This corollary is on page $9$ of "Variational Methods for Nonlocal Fractional Problems" by Giovanni Molica Bisci, Vicentiu D. Radulescu, Raffaella Servadei.
I would like to some references to regularity theory for parabolic PDEs, i.e., results that prove that solutions in $H^{l,l/2} \left( \overline{D^{(T)}_{n+1}} \right)$ imply that the solution is classical. I also would like to know if there is some Schauder Theory for parabolic PDEs as in the item $(ii)$ of the theorem $4.5$ of this lecture.
Thanks in advance!
|
2025-03-21T14:48:30.121237
| 2020-03-21T20:03:35 |
355397
|
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|
Stack Exchange
|
Understanding the definition of stacks
First of all I should apologies if this question does not count as a research level one. I asked the same question on MathUnderflow and didn't receive any answer. Let me cross post (copy and paste) it here.
I am trying to understand the notion of sheaves and stacks. Intuitively, sheaves are bit easy to understand as a gluing of compatible families of sets assign to opens sets of a topological space. In other words, it is a contravariant functor $\mathcal{F}:\mathbf{Open}(X)^{\operatorname{op}}\to\mathbf{Set}$ such that
$$\mathcal{F}(U)=\lim\left(\prod_{i\in I}\mathcal{F}(U_i) \rightrightarrows \prod_{j,k\in I^2}\mathcal{F}(U_j\cap U_k)\right)$$
where $X$ is a topological space, $U$ is an open set of $X$ and $\{U_i\}_{i\in I}$ is any open cover of $U.$ Further, it is easy to imagine a sheaf as a étalé space over $X.$
Then I started reading about stacks using this notes and nlab as my primary sources. I learned that a stack is a contravariant functor $\mathcal{F}:\mathcal{C}^{\operatorname{op}}\to\mathbf{Grpd}$ satisfying a descent property and, categories fibered in groupoids over $\mathcal{C}$ is an intuitive way to think about stacks, where $\mathcal{C}$ is a site (category equipped with a coverage). Now I have following questions:
How can I understand the descent property for stack? To be more specific, how can triple fiber products (intersections) appear in the equalizer fork diagram?
What categories fibered in groupoids over $\mathcal{C}$ corresponds to stacks? Is this the correct analogue of étalé space of a stack?
https://mathoverflow.net/questions/271987/how-is-a-stack-the-generalisation-of-a-sheaf-from-a-2-category-point-of-view?rq=1 might be of some use.
@user170039: That is quite an old nickname, actually.
I found this video is quite helpful.
A canonical example of a sheaf of sets on a topological space $X$
is the sheaf that sends an open subset $U$ of $X$ to the set of continuous real-valued functions on $U$.
The gluing property then says that a continuous function on a union of open subsets $U_i$ of $X$ is the same thing as a collection of continuous functions
$f_i: U_i \to \mathbb{R}$ such that $f_i$ and $f_j$ coincide on $U_i∩U_j$.
A canonical example of an ∞-sheaf (alias stack) of groupoids on a topological space $X$
is the ∞-sheaf that sends an open subset $U$ of $X$
to the groupoid of finite-dimensional continuous real vector bundles on $U$.
(Isomorphisms in this groupoid are continuous fiberwise linear isomorphisms
of vector bundles on $U$.)
The gluing property then says that a vector bundle on a union
of open subsets $U_i$ of $X$ is the same thing as a collection of vector bundles $V_i$ on $U_i$,
together with isomorphisms $t_{i,j}: V_i→V_j$ of vector bundles restricted to $U_i∩U_j$,
and such that $t_{j,k}t_{i,j}=t_{i,k}$ on $U_i∩U_j∩U_k$.
This last condition is known as the cocycle condition
and in some textbooks vector bundles are defined in this manner.
So the point of triple intersections is that isomorphisms
of vector bundles over pairwise intersections must themselves
satisfy a higher coherence identity.
This condition is trivial for sheaves of sets because two functions
can be equal in exactly one way, unlike vector bundles,
which can be isomorphic in many different ways.
To answer the second question: the analog of the etale space of a sheaf of sets
is the etale stack of an ∞-sheaf of groupoids.
This stack is no longer an ∞-sheaf on the original topological space,
but rather on the site of all topological spaces.
(Some care must be taken when dealing with size issues here,
since topological spaces do not form a small category,
but I suppress these issues here for simplicity.)
This etale stack can be constructed in many different ways.
For example, there is a unique homotopy cocontinuous functor
from ∞-sheaves of groupoids on a topological space $X$
to ∞-sheaves of groupoids on all topological spaces
that sends a representable sheaf given by an open subset $U$ of $X$
to the representable sheaf of $U$ as an object in the site of all topological spaces.
The image of a given ∞-sheaf $F$ of groupoids under this functor $E$
is the etale stack $E(F)$ of $F$, which is equipped with a canonical morphism
(in the category of ∞-sheaves of groupoids on topological spaces)
to the representable sheaf of $X$.
If we now take the ∞-sheaf of sections of the resulting map $E(F)→X$
of stacks, we recover the original ∞-sheaf $F$.
(There are many other constructions, of course.
For example, one could work instead with topological groupoids,
or rather, localic groupoids, instead of sheaves of groupoids on topological spaces.)
What categories fibered in groupoids over $\mathcal{C}$ corresponds to stacks?
A category fibered in groupoids over $\mathcal{C}$ is given by a functor $p_{\mathcal{F}}:\mathcal{F}\rightarrow \mathcal{C}$ satisfying certain conditions (I am not writing the definition as I assume you already know what is a category fibered in groupoids); look at Definition 4.2 in the paper Orbifolds as stacks?
Put a Grothendieck topology on the category $\mathcal{C}$; considering it as a site.
Given an object $U$ of the category $\mathcal{C}$, we consider its fiber; a category, denoted by $\mathcal{F}(U)$, defined as
$$\text{Obj}(\mathcal{F}(U))=\{V\in \text{Obj}(\mathcal{F}):\pi_{\mathcal{F}}(V)=U\},$$
$$\text{Mor}_{\mathcal{F}(U)}(V_1,V_2)=\{(f:V_1\rightarrow V_2)\in \text{Mor}_{\mathcal{F}}(V_1,V_2):\pi_{\mathcal{F}}(f)=1_U\}.$$
Given a cover $\{U_\alpha\rightarrow U\}$ of the object $U$ (remember that we fixed a Grothendieck topology), we consider its descent category, denoted by $\mathcal{F}(\{U_\alpha\rightarrow U\})$.
An object of the category $\mathcal{F}(\{U_\alpha\rightarrow U\})$ is given by the following data:
for each index $i\in \Lambda$, an object $a_i$ in the category $\mathcal{F}(U_i)$,
for each pair of indices $i,j\in \Lambda$, an isomorphism $\phi_{ij}:pr_2^*(a_j)\rightarrow pr_1^*(a_i)$ in the category $\mathcal{F}(U_i\times_{U}U_j)$
satisfying appropriate cocycle condition.
Now, given a categroy fibered in groupoids $p_{\mathcal{F}}:\mathcal{F}\rightarrow \mathcal{C}$, an object $U$ of $\mathcal{C}$ and a cover $\mathcal{U}(U)=\{U_\alpha\rightarrow U\}$ of $U$ in $\mathcal{C}$, there is an obvious functor $$p_{\mathcal{U}(U)}:\mathcal{F}(U)\rightarrow \mathcal{F}(\{U_\alpha\rightarrow U\})$$
(which you might have guessed already but let me say that), at the level of objects $$a\mapsto ((a|_{U_\alpha}),(\phi_{ij}))$$
A category fibered in groupoids
$p_{\mathcal{F}}:\mathcal{F}\rightarrow \mathcal{C}$ is said to be a
stack if, for each object $U$ of $\mathcal{C}$ and for each cover
$\mathcal{U}(U)=\{U_\alpha\rightarrow U\}$, the functor
$$p_{\mathcal{U}(U)}:\mathcal{F}(U)\rightarrow \mathcal{F}(\{U_\alpha\rightarrow U\})$$ is an equivalence of categories.
Now, you can ask what does equivalence of categories has anything to do with "sheaf like" properties? For a functor $\mathcal{D}\rightarrow \mathcal{C}$ to be an equivalence of categories, along other things, for each object $d\in \mathcal{D}$ we need an object $c\in \mathcal{C}$ such that, there is an isomorphism $F(c)\rightarrow d$.
Let $((a_\alpha),\{\phi_{\alpha\beta}\})$ be an object of $\mathcal{F}(\{U_\alpha\rightarrow U\})$. For this, by equivalence of categories, gives an element $a\in \mathcal{F}(U)$ that maps to $((a_\alpha),\{\phi_{\alpha\beta}\})$. That is, given an object $U$ of $\mathcal{C}$, an open cover $\{U_\alpha\rightarrow U\}$, for each collection of objects $\{a_\alpha\in \mathcal{F}(U_\alpha)\}$ that are compatible in some sense, there exists an object $a\in \mathcal{F}(U)$, such that, under appropriate restriction of $a$, you get the objects $a_{\alpha}$. This should remind the notion of sheaf on a topological space. This is how a stack is seen as a generalization of sheaf.
References :
Notes on Grothendieck topologies, fibered categories and descent theory.
Orbifolds as stacks?
How is a Stack the generalisation of a sheaf from a 2-category point of view?
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2025-03-21T14:48:30.121815
| 2020-03-21T20:15:38 |
355398
|
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|
Stack Exchange
|
separable support of Borel measure, with tau-additive measure and full support
I have a problem with Proposition 7.2.10 in Bogachev's Measure Theory Volume II book on page 77 (I have link to my drive with that book https://drive.google.com/file/d/1CgzgWhEiNPU1vy0YkyiVjGLH3Qiq1aCL/view). Supposedly, from that proof one can deduced that $\text{supp}\,\mu$ is separable if $\mu$ is $\tau$-additive Borel measure on metric space $(X, d)$ and $\mu$ has full support ($\mu(X \setminus \text{supp}\,\mu) = 0$). I don't see it, and also I don't know what is $\Gamma$ in this proof. I think, it is not specified. Maybe anyone knows answers for my questions or at least one of them? I'd appreciate.
"Therefore, we obtain a family $\Gamma$ of open sets of $\mu$-measure zero such that their union has a positive $\mu$-measure."
Yeah, but I can's see why $\Gamma$ exist. From what we know it ?
Following Bogachev's notation and terminology, let $S_\mu$ be the intersection of all closed sets of full $\mu$-measure, and say $\mu$ "has support" if $S_\mu$ itself has full measure. A useful fact is that $x \in S_\mu$ if and only if every open neighborhood of $x$ has positive measure.
Now suppose $\mu$ is a finite measure, as in the proof. If $S_\mu$ were not separable, then as Bogachev notes, it would be possible to find an uncountable collection of disjoint open balls centered at points of $S_\mu$. By the useful fact above, they must all have positive measure, and in particular, for some $n$, there are infinitely many with measure at least $1/n$, contradicting the finiteness of $\mu$.
As for $\Gamma$, at this point we are supposing that $\mu$ is a measure without support, so that $\mu(S_\mu^c) > 0$. Now by the fact above, every $x \in S_\mu^c$ has an open neighborhood $U_x$ of measure zero; necessarily $U_x \subset S_\mu^c$ and without loss of generality $U_x$ is a ball. Take $\Gamma = \{ U_x : x \in S_\mu^c\}$. Their union is $S_\mu^c$ which has positive measure.
Thanks! You made it understable. Author's reasoning was too fast for me.
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2025-03-21T14:48:30.121990
| 2020-03-21T21:18:55 |
355405
|
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|
Stack Exchange
|
Diophantine equation $10^n-a^3-b^3=c^2$
Consider the Diophantine equation:
$10^n-a^3-b^3=c^2$, for $a$, $b$, $c$, and $n$ positive.
Has this equation infinitely many solutions?
I think you really want to provide a motivation for the question and show your own efforts in solving this problem --- to improve the reception here at MO.
Feels like an olympiad problem.
Variation now posted as https://mathoverflow.net/questions/355415/has-this-diophantine-equation-infinitely-many-solutions-for-a-b-not-multiple
Yes, it has.
Note that using (the truly genius observation!) $10^n = 8\cdot 10^{n-1}+10^{n-1}+10^{n-1}$; it suffices to choose $n\equiv 1\pmod{3}$, and $n\equiv 1\pmod{2}$. That is, take $n=6\ell+1$, $\ell\in\mathbb{N}$. Then,
$$
10^n=10^{6\ell+1} = (2\cdot 10^{2\ell})^3 + (10^{2\ell})^3 + (10^{3\ell})^2.
$$
Thus,
$$
(a,b,c,n)=(2\cdot 10^{2\ell},10^{2\ell},10^{3\ell},6\ell+1),\quad \ell\in\mathbb{N}
$$
is a parametric family along which you get infinitely many solutions.
are there solutions with a, b c >1 and not multiple of 10?
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2025-03-21T14:48:30.122085
| 2020-03-21T22:34:55 |
355411
|
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|
Stack Exchange
|
Closed form of coprime numbers
Denoting primes $p_n$, I'm interested in the order of the set of natural numbers below a given limit that are divisible by the nth prime, but not by any smaller prime. Explicitly, with $A(M,n) = \{ x\leq M : p_n | x$ and $p_k\nmid x, \forall k<n \}$, what is $|A(M,n)|$?
For small n some case work gives a formula, but no promise of generalizing to all n. Also, if take $B_n = \lim_{M\to \infty} |A(M,n)|/M$, we have a nice closed form $$ B_n = \frac{1}{p_n} \prod_{k=1}^{n-1} \frac{p_{k}-1 }{p_{k}}. $$ However, this is also not quite sufficient. I've been unable to find a name for $A(M,n)$ in terms of known objects, or bounds on its order. Any help would be appreciated.
A "$k$-rough number" is a positive integer whose prime factors are all greater than or equal to $k$. The Buchstab function is used to count rough numbers. You want the number of $p_n$-rough numbers, minus the number of $p_{n+1}$-rough numbers.
Or, the $p_n$-rough numbers up to $M/p_n$.
|
2025-03-21T14:48:30.122290
| 2020-03-21T23:27:49 |
355416
|
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"Igor Rivin",
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|
Stack Exchange
|
signatures of quasi-gram matrices
Suppose I have a finite subset $\mathcal{M}$ of a Banach space $B$ $\mathcal{M}=p_1, \dots, p_n,$ and I create the following ``Gram'' matrix $G_{\mathcal{M}}:$
$$g_{ij} = \frac{\|p_i\|^2 + \|p_j\|^2 - \|p_i - p_j\|^2}{2}.$$ Notice that if $B$ is a Hilbert space, then this is just the usual Gram matrix, and, in particular, positive (semi)-definite. The the question is: what is known about the signature of $G_{\mathcal{M}}?$ Experiment reveals that almost all the eigenvalues are positive when $B$ is $L^1,$ , but it is not obvious to me why that should be so. A secondary (presumably much easier) question is what is the relationship between the signature of this matrix and the Cayley-Menger matrix?
You probably know this, but when $B$ is $L^1$ or subspace of $L^1$, it is hypermetric and thus the distance metric $d(x_i,x_j)$ of any $n$ points satisfies a negative-type inequality (Assuad, sur les inégalités valides dans $L^1$). Perhaps these inequalities imply your condition for the "Gram" matrices (?). Nice question.
@alvarezpaiva I did NOT know that. In fact in the (real life) example this came from exactly one eigenvalue is negative, and, bizarrely, the modulus is almost exactly the same as that of the largest positive eigenvalue.
|
2025-03-21T14:48:30.122400
| 2020-03-21T23:32:08 |
355417
|
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|
Stack Exchange
|
Defining the cospecialization in topology
Below is an excerpt from part V of Deligne's Étale cohomology - starting points.
Let $X$ be a complex analytic variety and $f:X\to D$ a morphism from $X$ to the disk. We denote by $[0,t]$ the closed line segment with endpoints $0$ and $t$ in $D$ and by $(0,t]$ the semi-open segment. If $f$ is smooth, then the inclusion $$j:f^{-1}(0,t]\hookrightarrow f^{-1}[0,t]$$ is a homotopy equivalence: we can push the special fiber $f^{-1}(0)$ into $f^{-1}[0,t]$. [I think this should be $f^{-1}(0,t]$.]
In practice, for $t$ small enough, $f^{-1}(0,t]$ will be a fiber bundle on $(0,t]$ so that the inclusion $$f^{-1}(t)\hookrightarrow f^{-1}(0,t]$$ will also be a homotopy equivalence. We then define the cospecialization morphism to be the homotopy class of maps $$\mathrm{cosp}:f^{-1}(0)\hookrightarrow f^{-1}[0,t]\overset{\simeq}{\leftarrow}f^{-1}(0,t]\overset{\simeq}{\hookleftarrow}f^{-1}(t).$$ This construction can be expressed in pictorial terms by saying that for a smooth morphism, the general fiber swallows the special fiber.
Let us assume no longer that $f$ is necessarily smooth (but assume that $f^{-1}(0,t]$ is a fiber bundle over $(0,t]$. We can still define a morphism $\mathrm{cosp}^\bullet$ on cohomology provided $j_\ast\mathbf Z=\mathbf Z$ and $\mathrm R^qj_\ast \mathbf Z=0$ for $q>0$. Under these assumptions, the Leray spectral sequence for $j$ shows that we have $$\mathrm H^\bullet(f^{-1}[0,t],\mathbf Z)\overset{\cong}{\to}\mathrm H^\bullet(f^{-1}(0,t],\mathbf Z)$$ and $\mathrm{cosp}^\bullet$ is the composite morphism $$\mathrm H^\bullet(f^{-1}(0),\mathbf Z)\leftarrow \mathrm H^\bullet(f^{-1}[0,t],\mathbf Z)\overset{\cong}{\to}\mathrm H^\bullet(f^{-1}(0,t],\mathbf Z)\overset{\cong}{\to}\mathrm H^\bullet(f^{-1}(t),\mathbf Z).$$
Questions.
How to use smoothness of $f$ to get a deformation retraction of $j$ (the inclusion)? If I understand correctly, smoothness implies the underlying $C^\infty$ map is a submersion and therefore admits an Ehresmann connection. This allows to use parallel transport to obtain a horizontal flow but without properness it may not be defined on entire fibers, so but I don't see how to get a deformation retraction...
What's a "canonical" example of a non-smooth $f:X\to D$ which is well behaved (say, a fiber bundle over the punctured disk) but with the inclusion $j$ not a homotopy equivalence?
What is the geometric interpretation of the conditions in the definition of the cohomological cospecialization map? What is an example where they fail?
|
2025-03-21T14:48:30.122585
| 2020-03-22T02:08:21 |
355421
|
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|
Stack Exchange
|
Cohomology of elementary abelian $p$-groups, i.e. $H(G,{\mathbb F}_p)$ with $G\cong{\mathbb F}_p^r$
I have two questions.
$\bf 1.$ First, a reference request. Let $G\cong{\mathbb F}_p^r$ for some integer $r\geq 0$ and let $V=G^*={\rm Hom}(G,{\mathbb F}_p)$. Then $(H(G,{\mathbb F}_p),+,\cup )$ is a ring,
$$H(G,{\mathbb F}_p)\cong\begin{cases}S(V)&p=2\\
\Lambda (V)\otimes S(V)&p>2\end{cases}.$$
Moreover, if $p=2$ then $V={\rm Hom}(G,{\mathbb F}_p)$ identifies as $H^1(G,{\mathbb F}_p)$; if $p>2$ then $V$ from $\Lambda (V)$ identifies as $H^1(G,{\mathbb F}_p)$, while $V$ from $S(V)$ identifies with the image of $V=H^1(G,{\mathbb F}_p)$ via the Bockstein boundary map $\beta :H^1(G,{\mathbb F}_p)\to H^2(G,{\mathbb F}_p)$, which happens to be injective.
An alternative description is
$$H(G,{\mathbb F}_p)\cong\begin{cases}{\mathbb F}_2[x_1,\ldots,x_r]&p=2\\
\Lambda (x_1,\ldots,x_r)\otimes{\mathbb F}_p[y_1,\ldots,y_r]&p>2\end{cases},$$
where $x_1,\ldots,x_r$ are a basis of $V$ and $y_i=\beta (x_i)$.
These results are proved via Kunneth formula.
My question is where I can find these results so that I can quote them. I saw them in a paper and in a book, but with no reference given. It seems that people regard them as "common knowledge". In the paper I mentioned the authors simply said "Recall that...", as if everybody knows this, but some need to be reminded in case they forgot.
$\bf 2.$ The second question is whether there are explicit formulas for these isomorphisms in the literature.
If $p=2$, then the isomorphism ${\mathbb F}_2[x_1,\ldots,x_r]\to H(G,{\mathbb F}_2)$ is given by $x_{i_1}\cdots x_{i_n}\mapsto x_{i_1}\cup\cdots\cup x_{i_n}\in H^n(G,{\mathbb F}_2)$.
If $p>2$, then the isomorphism $\Lambda (x_1,\ldots,x_r)\otimes{\mathbb F}_p[y_1,\ldots,y_r]\to H(G,{\mathbb F}_p)$ is given by $x_{i_1}\wedge\cdots\wedge x_{i_s}\otimes y_{j_1}\cdots y_{j_t}\mapsto x_{i_1}\cup\cdots\cup x_{i_s}\cup y_{j_1}\cup\cdots\cup y_{j_t}\in H^{s+2t}(G,{\mathbb F}_p)$.
How about the reverse isomorphisms? Did anybody see anything published regarding this problem?
I did obtained explicit formulas for the reverse isomorphisms, where the elements of $H(G,{\mathbb F}_p)$ are written in terms of normalized cocycles. However, I don't know wether these results are new. Somebody might have thought about them before.
They're essentially exercises, compute it for $r=1$ and then invoke Künneth, and I'd expect every book to include it: Adem--Milgram's classic book for example (Corollary II.4.3 and Theorem II.4.4).
You can look at the identical homology case in Brown's classic book (Theorem V.6.6) and in particular, his description in Section V.5.3 shows how to build your reverse isomorphism (starting from a free resolution of the group module $\mathbb{F}_pG$).
Thank you for the answer. Corollary II.4.3 and Theorem II.4.4 in Adem--Milgram is exactly what I needed. Funny thing, this is the book I was mentioning for having the result without a reference. I saw the result for $p>3$ at the beginning of section III.3.
It does answer the first question.
However, for the second question, Section V.5.3 in Brown doesn't deal with the case $G\cong{\mathbb F}_p$, but with the case when $G$ is cyclic. So I guess I should go ahead with writing a small paper on the explicit formula for the isomorphism. (Provided I don't find out that the result already exists. Maybe somebody saw it somewhere.)
I'm not, by far, an expert in cohomology. I only took a one semester reading course from Brown's book about 20 years ago and then forgot everything. Only recently I needed some cohomology in my work.
Sorry, I meant ${\mathbb F}_p^r$. Forgot to add the exponent $r$. I'll try to read more carefully that section in Brown to see if it leads to explicit closed formulas for the isomorphisms. I'm using a different approach.
I'm almost done with writing the paper. Here is the result.
If someone saw anything similar, then please let me know.
We have a basis $s_1,\ldots,s_r$ of $G$ over ${\mathbb F}_p$ and a basis $x_1,\ldots,x_r$ of $V=G^*=H^1(G,{\mathbb F}_p)$, which is dual to $s_1,\ldots,s_r$. Recall that $y_i=\beta (x_i)$ where $\beta :H^1(G,{\mathbb F}_p)\to H^2(G,{\mathbb F}_p)$ is the Bockstein boundary map.
The easier to state is the result in the case $p=2$.
$\bf Theorem~1$ If $p=2$ the isomorphism $H^*(G,{\mathbb F}_2)\to{\mathbb F}_2[x_1,\ldots,x_r]$ is given by
$$[a]\mapsto\sum_{1\leq i_1,\ldots,i_n\leq r}a(s_{i_1},\ldots,s_{i_n})X_{i_1}\cdots X_{i_n}$$
for every $a\in Z^n(G,{\mathbb F}_2)$.
The case $p>2$ is more complicated and it requires some extra definitions.
First note that $\Lambda (x_1,\ldots,x_r)\otimes{\mathbb F}_p[y_1,\ldots,y_r]$ has a structure of graded algebra which makes the isomorphism $\Lambda (x_1,\ldots,x_r)\otimes{\mathbb F}_p[y_1,\ldots,y_r]\to H^*(G,{\mathbb F}_p)$ an isomorphism of graded algebras. For every $n\geq 0$ the homogeneous component of degree $n$ is
$$(\Lambda (x_1,\ldots,x_r)\otimes{\mathbb F}_p[y_1,\ldots,y_r])^n=\bigoplus_{2k+l=n}\Lambda^l(x_1,\ldots,x_r)\otimes{\mathbb F}_p[y_1,\ldots,y_r]^k.$$
(Here ${\mathbb F}_p[y_1,\ldots,y_r]^k$ denotes the homogeneous polynomials of degree $k$.)
For every $1\leq i\leq r$ and $m\geq 0$ we define $x_i^{(m)}\in (\Lambda (x_1,\ldots,x_r)\otimes{\mathbb F}_p[y_1,\ldots,y_r])^m$ by
$$x_i^{(m)}=\begin{cases}1\otimes y_i^k&m=2k\\
x_i\otimes y_i^k&m=2k+1\end{cases}.$$
We have a basis of $\Lambda (x_1,\ldots,x_r)\otimes{\mathbb F}_p[y_1,\ldots,y_r]$ made of all products $x_{i_1}\wedge\cdots\wedge x_{i_l}\otimes y_1^{k_1}\cdots y_r^{k_r}$ with $1\leq i_1<\cdots <i_l<r$ and $k_1,\ldots,k_r\geq 0$.
For $1\leq i\leq r$ we put $l_i=1$ if $i\in\{ i_1,\ldots,i_l\}$ and $l_i=0$ otherwise. Then
$$x_{i_1}\wedge\cdots\wedge x_{i_l}\otimes y_1^{k_1}\cdots y_r^{k_r}=x_1^{(n_1)}\cdots x_r^{(n_r)},$$
where $n_i=2k_i+l_i$.
It follows that $x_1^{(n_1)}\cdots x_r^{(n_r)}$, with $n_1,\ldots,n_r\geq 0$, are a basis of $\Lambda (x_1,\ldots,x_r)\otimes{\mathbb F}_p[y_1,\ldots,y_r]$.
We consider the action of the symmetric group $S_n$ on $C^n(G,{\mathbb F}_p)$ given by
$$\sigma a(u_1,\ldots,u_n)=sgn (\sigma)a(u_{\sigma^{-1}(1)},\ldots,u_{\sigma^{-1}(n)})~\forall u_1,\ldots,u_n\in G.$$
If $n_1,\ldots,n_r\geq 0$ with $n_1+\cdots +n_r=n$ then we denote by $Sh(n_1,\ldots,n_r)$ the set of all $(n_1,\ldots,n_r)$-shuffles.
$$Sh(n_1,\ldots,n_r)=\{\sigma\in S_n\, :\,\sigma (h)<\sigma (h+1)\,\forall h,\, h\neq n_1+\cdots +n_i\,\forall 1\leq i\leq r-1\}.$$
The condition from the defintion of $Sh(n_1,\ldots,n_r)$ also writes as $\sigma (n_1+\cdots +n_{i-1}+1)<\cdots <\sigma (n_1+\cdots +n_r)$ $\forall 1\leq i\leq r$.
If $1\leq i\leq r$, $m\geq 0$, $k=[m/2]$
and $q_1,\ldots,q_k$ are nonnegative integers, then we define
$s_{i,m,q_1,\ldots,q_k}\in G^m$ by
$$s_{i,m,q_1,\ldots,q_k}=\begin{cases}(s_i^{q_1},s_i,\ldots
s_i^{q_k},s_i)&m=2k\\
(s_i,s_i^{q_1},s_i,\ldots s_i^{q_k},s_i)&m=2k+1\end{cases}.$$
$\bf Theorem~2$ If $p>2$ the isomorphism $\Lambda
(x_1,\ldots,x_r)\otimes{\mathbb F}_p[y_1,\ldots,y_r]\to H^*(G,{\mathbb F}_p)$ is given by
$$[a]\mapsto\sum_{n_1+\cdots
+n_r=n}c_{n_1,\ldots,n_r}x_1^{(n_1)}\cdots x_r^{(n_r)}$$
for every $a\in Z^n(G,{\mathbb F}_p)$, where
$$\begin{aligned}
c_{n_1,\ldots,n_r}=(-1)^{\frac{l(l-1)}2}\sum_{\sigma\in
Sh(n_1,\ldots,n_r)}\sum_{1\leq q_{i,j}\leq p-1}\sigma
a(s_{1,n_1,q_{1,1},\ldots,q_{1,k_1}},\ldots,s_{r,n_r,q_{r,1},\ldots,q_{r,k_r}}),
\end{aligned}$$
with $l=|\{ i\,\mid\, 1\leq i\leq r,\, n_i\text{ is odd}\}|$ and
$k_i=[n_i/2]$.
Here by the sum $\sum_{1\leq q_{i,j}\leq p-1}$ we mean that every
variable $q_{i,j}$, with $1\leq i\leq r$ and $1\leq j\leq k_i$, takes
values between $1$ and $p-1$.
Also $(s_{1,n_1,q_{1,1},\ldots,q_{1,k_1}},\ldots,s_{r,n_r,q_{r,1},\ldots,q_{r,k_r}})\in G^n$
is the concatenation of the sequences
$s_{i,n_i,q_{i,1},\ldots,q_{i,k_i}}$ for $1\leq i\leq r$, of lengths
$n_1,\ldots,n_r$.
I posted an article with this result on arXiv:
https://arxiv.org/abs/2005.11868
Before sending it to be published, I want to make sure it is original. If anybody saw a similar result somewhere, then please let me know.
Also please let me know if you saw somewhere the so-called ${\mathcal I}$-cochains I introduced in the first section. (Perhaps with other name, other notation.) I have already asked about them on mathoverflow, but didn't get any answer.
An alternative description of normalized cochains in terms of tensor powers of the augmented ideal
I am a little confused by your paper. As Chris Gerig pointed out in his answer, this isomorphism is well-known and can be found in many classic references. So I don't understand your question about originality...
Maybe he was a little hasty. Later he wrote a comment, but it disappeared, I guess he deleted it. In Brown V.5.3, which he suggested I should look at, $G$ is cyclic. I need $G={\mathbb F}_p^r$. Even if $r=1$, when $G$ is cyclic, V.5.3 only produces a reverse isomorphism if we describe the elements of $H^*(G,{\mathbb F}_p)$ in terms of that special resolution of period 2 for the cyclic groups. I need it in terms of the normalized bar resolution. To go from one to the other one needs an explicit homotopy, which is technically difficult. Even if you mange to do this, how about the case $r>1$?
I'm not commenting on originality, but $r>1$ follows from $r=1$ as I mentioned, and even Brown writes (for the proof of his Theorem V.6.6): "If G is cyclic, this follows easily from our earlier computations [V.5.3].The general case can now be deduced by using the Künneth formula and direct limits as in the proof of [Theorem] 6.4".
@Constantin-NicolaeBeli Ah, I understand a little better what you're trying to do. The computation of $H_*$ relies on the shuffle product, which on the cohomology level presumably translates to some Hopf algebra structure. This is implicitly present in your results as well, but shouldn't there be a direct way to deduce this from Brown Thm. V.6.6? (To be clear, what Chris Gerig is saying is that dualising the isomorphism on homology of V.6.6 gives a map in the direction you want on cohomology.)
Unfortunately, you are talking about things I'm not familiar with. I never worked with $H_$ and at this moment I can't even state its definition. I too noticed the same action of the symmetric group and the shuffle permutations, which appear in my formulas. Very likely it's not a mere coincidence. If there is some duality between $H^$ and $H_$, then this might be used to determine my coefficients $c_{n_1,\ldots,n_r}$. In my paper I only use the basic properties of $H^$ and the existence of the isomorphism whose inverse I want to find. (Corollary II.4.3 and Theorem II.4.4 in Adem--Milgram.)
Group homology is not that different from group cohomology. Use the same standard resolution of the $\mathbf Z[G]$-module $\mathbf Z$, but instead of applying $\operatorname{Hom}{\mathbf Z[G]}(-,\mathbf F_p)$ you apply $-\otimes{\mathbf Z[G]}\mathbf F_p$. The duality between $H_$ and $H^$ follows immediately from the definition (use that $\mathbf F_p$ is a field). The reason homology is more convenient for this problem is that mapping out of a tensor algebra is easier than into one (because the former has a universal property), and also because algebras are more familiar than coalgebras.
OK, I finally get it. (Or I think so.) We have a pairing $\rho :H^n(G,{\mathbb F}_p)\times H_n(G,{\mathbb F}_p)\to {\mathbb F}_p$. Both $H^n(G,{\mathbb F}_p)$ and $H^n(G,{\mathbb F}_p)$ have dimension $N=\binom{n+r-1}r$. I want to write $[a]\in H^n(G,{\mathbb F}_p)$ in the basis $\alpha_1,\ldots,\alpha_N$ of $\in H^n(G,{\mathbb F}_p)$, which is made of $\tau (x_1^{(n_1)}\cdots x_r^{(n_r)})$ with $n_1+\cdots +n_r=n$. If $\rho$ is non-degenerate and $\beta_1,\ldots,\beta_N$ is the dual basis of $H_n(G,{\mathbb F}_p)$ then $[a]=\sum_kc_k\alpha_k$, with $c_k=\rho ([a],\beta_k)$. This might work.
|
2025-03-21T14:48:30.123209
| 2020-03-22T02:42:57 |
355424
|
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"Rahman. M",
"Todd Eisworth",
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|
Stack Exchange
|
Strong Chains in $^{\omega_1}\omega_1$ of length $\omega_3$
In a previous question, I asked about the impact of strong chains in $^{\omega_1}\omega_1$ (e.g., sequences of functions $\langle f_\alpha:\alpha<\kappa\rangle$ in $^{\omega_1}\omega_1$ that are strictly increasing modulo the ideal of finite sets) on cardinal characteristics of the continuum.
Koszmider [1] proved it is consistent that there is a strong chain in $^{\omega_1}{\omega_1}$ of length $\omega_2$, and Veličković and Venturi showed that Neeman's techniques (finite approximations with side conditions consisting of elementary submodels of two types) can be adapted to obtain the same result. Certainly this latter proof does not adapt in a straightforward manner to yield chains of even longer length. I am less familiar with Koszmider's original proof, but that seems to work only for $\omega_2$ as well.
Question:
Is it consistent that there is a sequence $\langle
f_\alpha:\alpha<\omega_3\rangle$ in $^{\omega_1}\omega_1$ that is increasing modulo the ideal of finite sets? What is known about this?
[1] Koszmider, Piotr, On strong chains of uncountable functions, Isr. J. Math. 118, 289-315 (2000). ZBL0961.03039.
[2] Veličković, Boban; Venturi, Giorgio, Proper forcing remastered, Cummings, James (ed.) et al., Appalachian set theory 2006–2012. Based on the Appalachian set theory workshop series during the period 2006–2012. Cambridge: Cambridge University Press (ISBN 978-1-107-60850-4/pbk). London Mathematical Society Lecture Note Series 406, 331-362 (2013). ZBL1367.03094.
I had asked Boban about this question and I think it is still open. I like this problem very much and it is really interesting. It can serve as a test question for finding the method of size conditions with finite conditions which can add objects of size $\aleph_3$. I don't remember if Boban has mentioned any result around this problem to me. I will ask him in the next online meeting!
@Rahman.M I'm still trying to map out what is known and unknown, so I should have more questions on here soon. Shelah has some ZFC limits on what can happen larger cardinals.
|
2025-03-21T14:48:30.123473
| 2020-03-22T05:26:29 |
355429
|
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|
Stack Exchange
|
An upper bound on $\mathbb{E}\bigg[\bigg(\sum_{i=1}^{k}(X^{\top}A_{i}X)^{2}\bigg)^{q}\bigg]$
Let $X\in\mathbb{R}^{d}$ have independent, mean zero subgaussian entries, and $A_{1},\ldots,A_{k}$ be fixed $d\times d$ matrices that have zeros on the diagonal. I would like to upper bound the quantity
\begin{equation}
\mathbb{E}\bigg[\bigg(\sum_{i=1}^{k}(X^{\top}A_{i}X)^{2}\bigg)^{q}\bigg],
\end{equation}
for $q\in\mathbb{N}$.
Without the square on the quadratic form, this computation is easy as one can pull the summation inside and use results for the moments of subexponential random variables ($X^{\top}BX$ is subexponential.) With the square, however, it seems difficult. My idea is to use a decoupling trick to replace $X^{\top}A_{i}X$ with $X^{\top}A_{i}X'$, condition on $X'$, and then pull the summation in ($X'$ is an independent copy of $X$).
Vershynin's textbook on High Dimensional Probability (Theorem 6.1.1) gives
\begin{equation}
\mathbb{E}[f(X^{\top}AX)]\le \mathbb{E}[f(4X^{\top}AX')].
\end{equation}
for $f:\mathbb{R}\rightarrow\mathbb{R}$ convex and $A$ diagonal-free.
A multivariate version of this result might be helpful. Any hints?
|
2025-03-21T14:48:30.123578
| 2020-03-22T09:05:55 |
355434
|
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"url": "https://mathoverflow.net/questions/355434"
}
|
Stack Exchange
|
Chart in $1$-parameter family of Lagrangians in a Kähler manifold
Let $(X,\omega,J)$ be a complex $n$-dimensional Kähler manifold ($\omega$ Kähler form, $J$ complex structure) and $L \subset X$ be a closed real-analytic Lagrangian submanifold. Furthermore, let $L_{t}$ be a analytically varying $1$-parameter family of real-analytic Lagrangian submanifolds of $X$ such that $L_{0} = L$ (here one should think of the parameter $t$ as sufficiently small) and not intersection each other, i.e. $L_{t} \cap L_{t'} = \emptyset$, $\forall t,t'$. Furthermore, let $p\in L$ be a point.
Question: Does there exists a holomorphic chart $\varphi : U \rightarrow V$ around $p$, where $U \subset X$, $V \subset \mathbb{C}^{n}$ and $p \in U$ such that:
$V \cap (\mathbb{R}^{n} \times \{0\}) \not= \emptyset$ (here we view $\mathbb{C}^{n} = \mathbb{R}^{n} \oplus i\mathbb{R}^{n}$);
$\forall t$ we have $\varphi(U\cap L_{t}) = V\cap (\mathbb{R}^{n}\times \{t\} \times \underbrace{\{0\}}_{\in \mathbb{R}^{2n-1}})$?
If this is not true (or you think that it might not be true) do you have any idea if there exists something similar (in any sense) for a non-intersection $1$-parameter family of Lagrangian submanifolds?
What do you mean by "zero velocity"?
How should one apply here implicite function therem to obtain such a holomorphic chart?
Can you provide more details, please?
In any holomorphic chart, real analytic submanifolds remain real analytic. If your Lagrangian manifolds are not real analytic, they cannot become real analytic in holomorphic coordinates. In fact, you cannot even arrange that they lie in a real analytic hypersurface, let alone become these particular real analytic submanifolds.
Besides real analyticity, there is an invariantly defined section of the normal bundle of each Lagrangian manifold $L_t$, which, at each point of $L_t$, gives the normal component of the velocity of that point as $t$ varies. The tangential component is not defined, as $L_t$ is only defined as a submanifold, unparameterized. If this normal velocity field vanishes at some point of some $L_t$, then no such holomorphic coordinates can exist.
I think that if that normal velocity field is nonzero, and the family $L_t$ is real analytic, then you can get such holomorphic coordinates, but I have no tried to work the details out yet.
ok I will add real-analytic.
I see now what you mean by velocity. Yes, I assume that the velocity in non-zero at each point. How can one apply here the implicite function theorem?
|
2025-03-21T14:48:30.123737
| 2020-03-22T09:50:13 |
355437
|
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|
Stack Exchange
|
Moments of the prime counting function given the moments of the second Chebyshev function
I have read this article (Montgomery and Soundararajan: Primes in short intervals. http://arxiv.org/abs/math/0409258 ). In the second page of the article, it is stated that the mean and variance of $\psi(n+h)-\psi(n)$, in a given range of h and n, tends to $h$ and $h \log\frac{h}{n}$ respectively, being $\psi(n)$ the second Chebyshev function. On the other hand, I have read this question (Are the primes normally distributed? Or is this the Riemann hypothesis?) in which it is stated that the mean and variance of $\pi(n+h)-\pi(n)$ tends to $\frac{h}{\log n}$ and $\frac{h}{(\log n)^{2}} \log\frac{h}{n}$ in that range. My question is, ¿ How can I obtain the moments of the prime counting function from the moments of the second Chebyshev function. ?
Should be $\log n/h$ throughout. In a short interval $[n,n+h]$ the weighting by $\Lambda$ is just weighting by approximately $\log n$; don't worry about prime powers since there are so few of those.
|
2025-03-21T14:48:30.123835
| 2020-03-22T13:30:41 |
355447
|
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"Alexandre Eremenko",
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|
Stack Exchange
|
Square-integrable unbounded function
In R.D. Richtmyer, Principles of Advanced Mathematical Physics, p.85 an example is given of a continuous and square-integrable on $\bf{R}$ function, which is not bounded at infinity:
$$f(x)=x^2\exp{(−x^8\sin^2{x})}.$$
Intuitively, it can be expected that $f(x)$ is square-integrable as its peaks become more and more narrow. But how can one rigorously prove that $f(x)$ is indeed square-integrable?
By looking at the measure of the sets ${ x:|f(x)|>c}$.
For $k=1,2,\dots$, let
$$I_k:=\int_{|x-k\pi|<1/k}f(x)^2\,dx
=\int_{|x-k\pi|<1/k}x^4\exp(-2x^8\sin^2 x)\,dx.$$
Then, as $k\to\infty$,
$$I_k\asymp k^4\int_{|x-k\pi|<1/k}\exp\{-(2+o(1))(k\pi)^8\sin^2 x\}\,dx \\
=k^4\int_{|x|<1/k}\exp\{-(2+o(1))(k\pi)^8\sin^2 x\}\,dx \\
=k^4\int_{|x|<1/k}\exp\{-(2+o(1))(k\pi)^8 x^2\}\,dx \asymp1,$$
by substitution $u=(k\pi)^4 x$.
(Here we use the usual notation: $A\asymp B$ meaning $A=O(B)$ and $B=O(A)$.)
Hence,
$$\int_{-\infty}^\infty f(x)^2\,dx\ge\sum_{k=1}^\infty I_k=\infty.$$
So, $f$ is actually not square-integrable.
Reasoning similarly (but using, say, $h_k:=1/k^{b/3}$ instead of $1/k$ in $|x-k\pi|<1/k$), one can see that for any real $a,b>0$, letting
$$f(x):=|x|^a\exp(-|x|^b\sin^2x),$$
we have the following:
$f$ is continuous, but unbounded at $\infty$.
$f$ is square-integrable iff $2a-b/2<-1$.
To get this result, we also note that for $k=1,2,\dots$
$$\int_{h_k\le|x-k\pi|\le\pi}x^{2a}\exp(-2x^b\sin^2 x)\,dx \\
=O(k^{2a}\exp\{-(2+o(1))(k\pi)^b h_k^2\})=O(1/k^c)$$
for any real $c$.
In your example, we have $a=2$ and $b=8$, so that $2a-b/2=0\not<-1$, and so, your $f$ is not square-integrable.
Just another way of seeing it: instead of cutting the integral with respect to $h_k$, one can first change $\sin(x)^2$ to $Cx^2$ (or $\varepsilon x^2$, depending on the direction of the estimate), then perform the change of variable $h_k\cdot y=x-k\pi$. Then the integral between ± something of order $1/h_k$ is less than the integral over $\mathbb R$, and more than the integral between ± something of order 1.
Richtmyer says it is square-integrable, but it seems you are right (I have done some numerical calculations). The same statement is reproduced in https://arxiv.org/abs/quant-ph/9907069. Maybe there was a typo in Richtmyer.
@ZurabSilagadze : Perhaps, Richtmyer meant $a=1$ instead of $a=2$ (indeed, there is no reason to take $a=2$ to get an unbounded $f$, when $a=1$ would suffice). Then, still with $b=8$, we would have $2a-b/2=-2<-1$, just what is needed.
@ Iosif Pinelis I think prefactor in your solution should be $(k\pi)^4$, not $k^4$. What is $c$ in the general case and why you cannot use $|x-k\pi|<1/k$ in this case also?
@ZurabSilagadze : (i) With the (standard, I think, and now explained) meaning of $\asymp$, positive real factors don't matter. (ii) As was stated, $c$ is any (constant) real number. (iii) I cannot use $h_k=1/k$ if $b\le2$, because I need $k^b h_k^2$ to be much greater than $\ln k$ for large $k$ -- in particular, to get the bound $O(1/k^c)$ for all real $c$.
|
2025-03-21T14:48:30.124039
| 2020-03-22T14:18:29 |
355449
|
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"Dominic van der Zypen",
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|
Stack Exchange
|
Increasing the "shuffling distance" by iterating a permutation $\varphi: \omega \to \omega$
Motivation. I was wondering about the following when playing a card-shuffling game with my elder son.
If $\varphi: \omega \to \omega$ is a bijection, we define the shuffling distance of $\varphi$ by $$sh(\varphi) = \min\{|n-\varphi(n)| : n \in \omega\}.$$
Let $\varphi^{(1)}=\varphi$ and $\varphi^{(k+1)} = \varphi\circ\varphi^{(k)}$ for all integers $k \geq 1$.
What is an example of a bijection $\varphi: \omega \to \omega$ such that $sh(\varphi^{(k+1)}) > sh(\varphi^{(k)})$ for all integers $k\geq 1$?
There's indeed an example of such a function.
Given a an infinite subset $I\subset\omega$, call indexation of $I$ the unique increasing bijection $\omega\to I$. If $(x_n)$ is its indexation, define $f_I$ the permutation of $I$, mapping $x_1\mapsto x_0$, $x_{2n+1}\mapsto x_{2n-1}$ ($n\ge 1$), $x_{2n}\mapsto x_{2n+1}$ ($n\ge 0$). Hence $f_I^k$ maps $f_{2n+1}\mapsto f_{2n+1-2k}$ for $n\ge k$, maps $f_{2n+1}\mapsto f_{2k-2n-2}$ for $0\le n\le k-1$, and maps $f_{2n}\mapsto f_{2n+2k}$.
Say that a subset $I$ of $\omega$ is "good" if it is infinite, and, $(x_n)$ being its indexation, it satisfies $x_n>2x_{n-1}$ for all $n\ge 1$. If $I$ is good, then $\min_n|x_n-f_I^k(x_n)|=x_{k+1}-x_k$, and $(x_{k+1}-x_k)$ is strictly increasing.
Given a partition $\mathcal{P}$ of $\omega$ into good subsets, let $f$ be the permutation of $\omega$ whose restriction to $I$ is $f_I$ for every $I\in\mathcal{P}$. Then $(\min_n|x_n-f^{k}(x_n)|)_k$ is strictly increasing. The existence of such a partition is obtained by a straightforward induction.
Maybe the answer is no if one requires in addition that $f$ belongs to the wobbling group, namely $\sup_n|f(n)-n|<\infty$.
Wonderful proof - and a nice conjecture in the comment, thanks @YCor!
It's an expectation, but I don't call it conjecture (at least on my behalf).
|
2025-03-21T14:48:30.124181
| 2020-03-22T15:59:42 |
355453
|
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|
Stack Exchange
|
How to prove the relationship between Stern's diatomic series and Lucas sequence $U_n(x,1)$ over the field GF(2)?
I found the bit count of Lucas sequence $U_n(x,1)$ over the field GF(2) is Stern's diatomic series, I want to know the reason?
https://oeis.org/A002487 : Stern's diatomic series
https://oeis.org/A168081 : Lucas sequence $U_n(x,1)$ over the field GF(2)
I always work mod 2. Let $f(n)$ be the bit count (number of nonzero
coefficients) of $U_n(x,1)$. $U_n(x,1)$ satisfies
$$ U_{2n}(x,1) = xU_n(x,1)^2 $$
$$ U_{2n+1}(x,1) = (U_n(x,1)+U_{n+1}(x,1))^2. $$
From the first equation $f(2n)=f(n)$, and only odd powers of $x$
can have nonzero coefficients. From the second equation, only even
powers of $U_{2n+1}$ can have nonzero coefficients. Hence no
cancellation occurs when we add $U_n(x,1)$ and $U_{n+1}(x,1)$, so
$f(2n+1)=f(n)+f(2n+1)$. Thus $f(n)$ satisfies the same recurrence and
same initial conditions $f(0)=0$ and $f(1)=1$ as Stern's diatomic sequence.
Addendum. The two formulas above can be proved by induction on
$n$. Namely, writing just $U_n$ for $U_n(x,1)$,
$$ U_{2n}=xU_{2n-1}+U_{2n-2} = x(U_{n-1}+U_n)^2+xU_{n-1}^2
=xU_n^2 $$
and
$$ U_{2n+1} = xU_{2n}+U_{2n-1} = x^2U_n^2+(U_{n-1}+U_n)^2
=U_n^2+(xU_n+U_{n-1})^2=U_n^2+U_{n+1}^2. $$
|
2025-03-21T14:48:30.124278
| 2020-03-22T16:03:00 |
355454
|
{
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"Drew Heard",
"IJL",
"Nicholas Kuhn",
"Theo Johnson-Freyd",
"Todd Leason ",
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}
|
Stack Exchange
|
How does the Steenrod algebra act on $\mathrm{H}^\bullet(p^{1+2}_+, \mathbb{F}_p)$?
Let $p$ be an odd prime. The $\mathbb F_p$ cohomology of the cyclic group of order $p$ is well-known: $\mathrm{H}^\bullet(C_p, \mathbb F_p) = \mathbb F_p[\xi,x]$ where $\xi$ has degree 1, $x$ has degree 2, and the Koszul signs are imposed (so that in particular $\xi^2 = 0$). As a module over the Steenrod algebra, the only interesting fact is that $x = \beta \xi$, where $\beta$ denotes the Bockstein. The rest of the Steenrod powers can be worked out by hand.
There are two groups of order $p^2$. The $\mathbb F_p$ cohomology of $C_p \times C_p$, including its Steenrod powers, is computable from the Kunneth formula. For the cyclic group $C_{p^2}$, you have to think slightly more, because there is a ring isomorphism $\mathrm{H}^\bullet(C_p, \mathbb F_p) \cong \mathrm{H}^\bullet(C_{p^2}, \mathbb F_p)$, but the Bockstein vanishes on $\mathrm{H}^\bullet(C_{p^2}, \mathbb F_p)$. Still I think the Steenrod algebra action is straightforward to write down.
I want to know about the groups of order $p^3$. The abelian ones are not too hard, I think, and there are two nonabelian groups. The one with exponent $p^2$ is traditionally denoted "$p^{1+2}_-$", and the one with exponent $p$ is traditionally denoted "$p^{1+2}_+$". I care more about the latter one, but I'm happy to hear answers about both. And right now I care most about the prime $p=3$.
The cohomology of these groups was computed in 1968 by Lewis in The Integral Cohomology Rings of Groups of Order $p^3$. Actually, as is clear from the title, Lewis computes the integral cohomology, from which the $\mathbb F_p$-cohomology can be read off using the universal coefficient theorem. For the case I care more about, Lewis finds that $\mathrm{H}^\bullet(p^{1+2}_+, \mathbb Z)$ has the following presentation. (I am quoting from Green, On the cohomology of the sporadic simple group $J_4$, 1993.) The generators are:
$$ \begin{matrix}
\text{name} & \text{degree} & \text{additive order} \\
\alpha_1, \alpha_2 & 2 & p \\
\nu_1, \nu_2 & 3 & p \\
\theta_j, 2 \leq j \leq p-2 & 2j & p \\
\kappa & 2p-2 & p \\
\zeta & 2p & p^2
\end{matrix}$$
(For the $p=3$ case that I care most about, there are no $\theta$s, since $2 \not\leq 3-2$.) A complete (possibly redundant) list of relations is:
$$ \nu_i^2 = 0, \qquad \theta_i^2 = 0, \qquad \alpha_i \theta_j = \nu_i \theta_j = \theta_k \theta_j = \kappa \theta_j = 0$$
$$\alpha_1 \nu_2 = \alpha_2 \nu_1, \qquad \alpha_1 \alpha_2^p = \alpha_2 \alpha_1^p, \qquad \nu_1\alpha_2^p = \nu_2 \alpha_1^p,$$
$$ \alpha_i\kappa = -\alpha_i^p, \qquad \nu_i\kappa = -\alpha_i^{p-1}\nu_i,$$
$$ \kappa^2 = \alpha_1^{2p-2} - \alpha_1^{p-1}\alpha_2^{p-1} + \alpha_2^{2p-2}, $$
$$ \nu_1 \nu_2 = \begin{cases} \theta_3, & p > 3, \\ 3\zeta, & p = 3. \end{cases}$$
From this Green (ibid.), for example, writes down a PBW-type basis.
Question: What is the action of the Steenrod algebra been on $\mathrm{H}^\bullet(p^{1+2}_+, \mathbb F_p)$?
I'm not very good at Steenrod algebras. Does the ring structure on the $\mathbb Z$-cohomology suffice to determine the action? For instance, the additive structure of $\mathrm{H}^\bullet(G, \mathbb Z)$ already determines the Bockstein action on $\mathrm{H}^\bullet(G, \mathbb F_p)$. If there is a systematic way to do it, where can I learn to do the computations?
If I am very lucky, someone will answer my question by an example: work out the case $p=3$, and explain the steps used. Of course, references will also be accepted.
There are two excellent answer. I want to accept both.
If $P$ is the group of order $p^3$ and exponent $p$, its mod $p$ cohomology ring is known to have its depth = its Krull dimension = rank of a maximal elementary abelian subgroup = 2. A theorem of Jon Carlson then implies that the product of restriction maps
$$ H^*(P;\mathbb F_p) \rightarrow \prod_E H^*(E;\mathbb F_p)$$
is monic, where the product is over (conjugacy classes of) subgroups $E \simeq \mathbb Z/p \times \mathbb Z/p$.
Thus the ring you care about, viewed as an algebra equipped with Steenrod operations (I would call this an unstable $A_p$--algebra) embeds in a known unstable algebra. Thus if you know how algebra generators of $H^*(P;\mathbb F_p)$ restrict to the various $H^*(E;\mathbb F_p)$'s, it shouldn't be hard to calculate Steenrod operations.
For example, David Green and Simon King's group cohomology website tells you cohomology ring generators and relations, and these restrictions for the group of order 27. (See https://users.fmi.uni-jena.de/cohomology/27web/27gp3.html) I'll let you take it from here.
[By the way, you began your question with a remark that there isn't so much to the cohomology ring $H^*(C_p;\mathbb F_p)$ as a module over the Steenrod algebra. Yes, it is trivial to compute, but it is a deep theorem, with huge unexpected consequences, that this is a injective object in the category of unstable $A_p$--modules. See the book by Lionel Schwartz on the Sullivan conjecture for more detail.]
Edit the next day: Thanks to Leason for pointing out that my map doesn't detect for $p>3$. To get the detection result, one needs that the depth = Krull dimension, and this turns out to only happen in the one case that I looked up carefully: $p=3$. So in the other cases, the depth will be 1 = rank of the center, and one needs a more general detection theorem, pioneered by Henn-Lannes-Schwartz in the mid 1990's, and then explored by me in various papers about a decade ago. (Totaro later wrote about this in his cohomology book: this is the result mentioned by Heard.) In the case in hand, the range of the detection map for $H^*(P;\mathbb F_p)$ will need one more term in the product: Let $C < P$ be the center: a group of order $p$. The multiplication homomorphism $C \times P \rightarrow P$ induces a map of unstable algebras
$$ H^*(P;\mathbb F_p) \rightarrow H^*(C;\mathbb F_p) \otimes H^{*\leq 2p}(P;\mathbb F_p)$$
where the last term means truncate above degree $2p$. That the number $2p$ works to detect all remaining nilpotence is an application of my general result: it can be determined by understanding the restriction to the center.
At any rate, for that original group of order 27 and exponent 3, this isn't needed. (The group of order 27 and exponent 9 will also need that extra factor in the detection map range.)
In general, one only knows that the kernel of the product of restrictions to maximal elementary abelian subgroups has nilpotent kernel. Do you have a reference for the stated injectivity?
In fact (up to p=3) it isn't injective! For, since all maximal subgroups (in the exponent p case) are elementary abelian of rank 2, the kernel of the product map is the essential cohomology. Now a theorem of Minh ("Essential cohomology and extraspecial p-groups") says that the essential cohomology is non trivial, except in case p=3.
@ToddLeason: Do I understand that Nicholas's answer solves the problem I care most about --- namely the group $3^{1+2}_+$ --- but misses some cohomology when $p>3$?
The discrepancy occurs because 'depth = its Krull dimension' is not true in general (e.g., when p = 5, the ring has depth 1, see https://users.fmi.uni-jena.de/cohomology/125web/125gp3.html). When p = 3, the stated injectivity is true (for a reference, see Corollary 12.5.3 of Carlson--Townsley--Valeri-Elizondo--Zhang). There is actually a slightly more complicated description for all primes that allows you to find an unstable algebra for which the cohomology injects into - this is Theorem 13.21 in 'Group Cohomology and Algebraic Cycles' by Burt Totaro.
@Theo Johnson-Freyd: Yes.
I think the answer you want can be found in the following article:
AUTHOR = {Leary, I. J.},
TITLE = {The mod-{$p$} cohomology rings of some {$p$}-groups},
JOURNAL = {Math. Proc. Cambridge Philos. Soc.},
FJOURNAL = {Mathematical Proceedings of the Cambridge Philosophical
Society},
VOLUME = {112},
YEAR = {1992},
NUMBER = {1},
PAGES = {63--75},
ISSN = {0305-0041},
Ian's thesis, complete with Steenrod operations! Folks interested in this area should note that techniques are available, and have been for quite a while, but the details take both work and organization.
I was itching to answer this question myself, but you beat me to it. My article also gives examples of elements (in degrees $6,8,\ldots 2(p-1)$ that are not detected on proper subgroups. Since all the generators are in degrees at most $2p$, there is really only the Bockstein and $P^1$ to worry about when computing the Steenrod algebra action.
|
2025-03-21T14:48:30.125056
| 2020-03-22T16:03:46 |
355455
|
{
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"Antoine Balan",
"Chris Gerig",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/355455"
}
|
Stack Exchange
|
The Seiberg-Witten equations for forms
I define equations like the Seiberg-Witten equations for forms of a riemannian four-manifold $(M,g)$. $\alpha \in \Lambda^2 (TM)$ and $\theta \in \Lambda^1(TM)$.
$$
d\alpha+\theta \wedge \alpha=0
$$
$$
d\theta_+=\frac{\alpha_+}{||\alpha ||}
$$
The gauge group acts:
$$
f.(\alpha,\theta)=(f\alpha,\theta- df/f)
$$
Can we define invariants?
You're asking a big question (almost homework-like), like "show me how to build a cohomology theory", without demonstrating any attempt yourself, which is why I voted to close. I recommend understanding how the ordinary Seiberg-Witten invariants are defined, and check it for yourself (do you have an elliptic equation, transversality, appropriate index, compactness, etc.).
But there is a real analogy between these equations and those of Seiberg-Witten. The question deserves to be posed.
trivial remark: if we think of $\theta$ as associated to a connection, then the first equation just says that $\alpha$ is a parallel $2$-form with respect to that connection. The second equation seems to say something about the curvature of this connection.
|
2025-03-21T14:48:30.125167
| 2020-03-22T16:07:47 |
355456
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/355456"
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|
Stack Exchange
|
Representable functors and symplectic co-tangent bundles
I've been banging my head against something that I feel should follow from abstract non-sense, and I hope someone here can set me straight.
Let $\mathcal{M}$ be the category of smooth manifolds, with morphisms given by smooth maps. Suppose I have a functor
$F: \mathcal{M}\rightarrow Sets$
so that there exists a smooth manifold $M_{F}$ and a natural isomophism
$F(-)\simeq C^{\infty}(-, M_{F}).$
As with any smooth manifold, there exists a canonical symplectic structure $\omega$ on $T^{\star}M.$
Question: Is there a notion of cotangent space/bundle/functor (sorry for the vagueness) to $F$ which recovers the canonical symplectic structure $\omega$ on $T^{\star}M?$
If it makes things any easier, you are free to assume that $F$ is a lax groupoid valued functor and is a representable stack (for your favorite Grothendieck topology on $\mathcal{Man}$), though I don't know if this makes much of a difference for this question.
I've heard of the cotangent complex to a stack, and this is likely the kind of thing I'm looking for, but wading through the technicalities is proving difficult.
Ideally, I'm interested if something like the following is true. Let $(x, \alpha)\in T^{\star}M_{F}.$ Then look at $T_{(x, \alpha)}T^{\star}M_{F}.$ The symplectic form on $T^{\star}M_{F}$ induces a linear isomorphism
$T_{(x, \alpha)}T^{\star}M_{F}\simeq (T_{(x, \alpha)}T^{\star}M_{F})^{\star}\simeq Hom_{\mathbb{R}}(T_{(x, \alpha)}T^{\star}M_{F}, \mathbb{R}).$
Does this isomorphism arise in some natural way from some cotangent type object attached to the functor F?
It seems this question is already interesting for functors $F$ of the form
$F(U)=C^{\infty}(U,M)$
for some smooth manifold $M.$ But sadly, already here, I don't understand the obvious thing that I should do.
Thank you for any help you can provide.
I am a little skeptical about this because taking a cotangent bundle is not functorial.
|
2025-03-21T14:48:30.125315
| 2020-03-22T17:17:52 |
355459
|
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"Shreya",
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"url": "https://mathoverflow.net/questions/355459"
}
|
Stack Exchange
|
Clarification: Using Hensel's Lemma to determine $K_v$-rational points on a curve
From Silverman's AEC page 332:
I need to understand why the determination of the following local kernel
$$
ker \Big( H^1(G_v, E[\phi]) \rightarrow WC(E/K_v)[\phi] \Big)
$$
is straightforward. The book says that it is the same as answering the questions whether a curve has a point over a complete local field (which I understand).
This is further reduced by Hensel's Lemma to checking whether the curve has a point in some finite ring $R_v/ \mathcal{M}_v^e$ for some easily computable integer $e$.
Now, as far as I know and understand, Hensel's Lemma says that for a polynomial $f(x) \in R_v[x]$, if it has a root in $R_v/\mathcal{M}$, then it lifts to a unique root in $R_v$ and hence $K_v$.
However, this version does not seem to be directly used here. I suspect that maybe some several variable version is being used? And where did that $e$ come from?
I would really appreciate it if someone could explain it to me. Thank you.
you really shouldn't crosspost. Anyway, you've slightly misstated Hensel's lemma, you left out the assumption that $f(x)$ has a simple root in $R_v/\mathcal{M}_v$. That's where the $e$ is coming from. In general, if $f(x)$ has a root of higher multiplicity in $R_v/\mathcal{M}_v$, then you need to work in $R_v/\mathcal{M}_v^e$. So if the curve is non-singular modulo $v$, then you can take $e=1$, and it really does reduce to the 1-variable Hensel lemma, since the curve has dimension 1. However, if the curve is singular and it has a singular point defined over $R_v/\mathcal{M}_v$, that's not enough to conclude that the point can be lifted, you need to work with a larger value of $e$.
Love it when the author replies himself. Thank you!
|
2025-03-21T14:48:30.125464
| 2020-03-22T17:18:08 |
355460
|
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|
Stack Exchange
|
A generalization of discrete Hilbert's transform (Montgomery's inequality)
In the paper "Hilbert's inequality", Montgomery and Vaughan proved that a generalization of the discrete Hilbert transform is bounded in $\ell^2$. The inequality reads as follows
$$ \Big| \sum_{k\neq n}\frac{a_k \overline{b_n}}{\lambda_k-\lambda_n} \Big| \leq \frac{\pi}{\delta} \Big(\sum_{k=1}^{\infty} |a_k |^2 \Big)^{1/2}\Big( \sum_{n=1}^{\infty} |b_n |^2 \Big)^{1/2}, $$
where $\{a_k\}, \{ b_n \}\in \ell^2 $, $ \lambda_n $ is an increasing sequence of real numbers such that
$$ \delta:= \inf_{k n}| \lambda_k-\lambda_{k+1}|. $$
Of course $\delta$ is assumed to be strictly positive. Also the constant appearing in the inequality $\pi/\delta$ is optimal. Quite surprisingly all proofs I managed to find use strongly the Hilbert space structure of $\ell^2$.
Therefore I would like to ask if anything is known for the this inequality when considered on $\ell^p, p\neq 2$. Namely, is it true
$$\Big| \sum_{k\neq n}\frac{a_k \overline{b_n}}{\lambda_k-\lambda_n} \Big| \leq C(p,\delta) \Big(\sum_{k=1}^{\infty} |a_k |^p \Big)^{1/p}\Big( \sum_{n=1}^{\infty} |b_n |^q \Big)^{1/q}, $$
where $1<p<\infty$, $q$ is the conjugate exponent of $p$ and $C(p,\delta)>0 ?$
(I wouldn't venture so far as to ask for an optimal constant in this case, given the difficulty of the problem for the classical discrete Hilbert transform.)
What kind of inequality are you envisioning? If it involves $\ell^p$ and $\ell^q$ norms, you might as well use $\ell^2$. And I don't know how to imagine an inequality without the dual $\ell^q$ norm.
@Lucia The inequality involves $p$ and $q$ norms as you said (I edited the question so its more clear) but I don't really understand what you mean by saying you might as well use the $\ell^2$ norm.
One can transfer the continuous $L^p$ theory to this discrete setting without difficulty.
Let's normalise $\sum_k |a_k|^p = \sum_n |b_n|^q = 1$. Consider the two quantities
$$ X_1 := \sum_{k \neq n} \frac{a_k \overline{b_n}}{\lambda_k - \lambda_n}$$
$$ X_2 := \sum_{k, n} p.v. \int_{{\bf R}^2} \varphi(s) \varphi(t) \frac{a_k \overline{b_n}}{(\lambda_k+s) - (\lambda_n+t)}\ ds dt$$
where $\varphi$ is a bump function of total mass 1. It is not difficult to show that
$$ p.v. \int_{\bf R} p.v. \int_{\bf R} \varphi(s) \varphi(t) \frac{1}{(\lambda_k+s) - (\lambda_n+t)}\ dt$$
is equal to $\frac{1}{\lambda_k - \lambda_n} + O_\delta( |k-n|^{-2} )$ when $k \neq n$ and $O_\delta(1)$ when $k=n$, so we have $X_1-X_2 = O_{p,\delta}(1)$ by Schur's test. One can also write $X_2$ as
$$ p.v. \int_{\bf R} \int_{\bf R} \frac{f(x) g(y)}{x-y}\ dx dy$$
where
$$ f(x) := \sum_k a_k \varphi(x-\lambda_k)$$
and
$$ g(y) := \sum_n b_n \varphi(x-\lambda_n)$$
so from the $L^p$ boundedness of the continuous Hilbert transform we have $X_2 = O_{p,\delta}(1)$, and the claim follows.
Let me deal with a continuous situation. Let $\lambda:\mathbb R\rightarrow\mathbb R$ be an increasing $C^1$ diffeomorphism and let $u,v$ be in $L^2(\mathbb R)$. We have with $\phi=\lambda^{-1}$,
$$
A=\iint \frac{u(y)\overline{u(x)}}{iπ(\lambda (x)-\lambda(y))} dx dy=
\iint \frac{u(\phi(t))\overline{u(\phi(s))}}{iπ(s-t)}\phi'(t)\phi'(s) ds dt,
$$
so that with $U(t)=u(\phi(t))\phi'(t)^{1/2}$, we find
$$
A=\iint \frac{U(t)\phi'(t)^{1/2}\overline{U(s)\phi'(s)^{1/2}}}{iπ(s-t)}ds dt,
$$
and thus assuming
$
0<m\le \phi'(t)\le M<+\infty
$
we get the $L^p$ boundedness properties from those of the Hilbert transform.
Thank you very much for the answer. I think a similar reasoning can be found also here link
In the same idea a paper "SHARP NORM INEQUALITIES FOR THE TRUNCATED HILBERT TRANSFORM" by Enrico Laeng .
Thank you for your contribution, but which part of the paper you think it is related to the Montgomery inequality ?
I read this paper a long time ago, but maybe you can generalize
|
2025-03-21T14:48:30.125829
| 2020-03-22T17:51:45 |
355462
|
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"Narutaka OZAWA",
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|
Stack Exchange
|
Representations of Banach algebras
If $A$ is a Banach algebra and $L$ a left ideal of $A$, consider the representation $T_{L}$ of $A$ into the algebra $B(A/L)$ of bounded linear operators on the quotient space $A/L$ defined by $T_{L}(a)(b+L)=ab+L$. When is $T_{L}(A)$ closed in the norm of $B(A/L)$? Certainly this happens if $A/L$ is finite-dimensional. Are there other cases? Thanks, Paul.
How did this question arise?
I am currently studying the invariant subspaces of strictly cyclic Banach algebras and their reflexivity (as operator algebras).
Well, it would be easy to give examples. In "reasonable" cases the kernel of $T_L$ is $L$ and the norm of $T_L(a)$ equals the norm of $a + L$ in $A/L$. So $T_L(A)$ is isometric to $A$ and therefore complete. There will be degenerate examples where that isn't true, though. A good question would be: is $T_L(A)$ always closed?
Thanks, but I do not know when this happens in general. It may happen even if the norms are not equal.
The kernel is a two sided ideal, cannot equal L (a left ideal),
In case $A$ is a C*-algebra and $L$ is a closed left ideal, $T_L(A)$ is isometrically isomorphic to the C*-algebra $A/\ker T_L$. I guess it's not that simple for general Banach algebras.
Thanks Narutaka. I need these facts for more general Banach algebras. I could assume a little more: what if L is a maximal left ideal?
T_L (A) is not in general a C*-algebra even if A is. Am I missing something? Thanks Narutaka.
|
2025-03-21T14:48:30.125966
| 2020-03-22T18:03:27 |
355463
|
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"მამუკა ჯიბლაძე"
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|
Stack Exchange
|
Elementary questions about vanishing cycles and emerging cycles
Let $X\to D$ be a proper $C^\infty$ map with $D$ an open disk about the origin in some Euclidean space. Suppose $0\in D$ is the only singular value, i.e that over $D^\times=D\setminus \left\{ 0 \right\}$ the map is a submersion. A choice of Riemannian metric on $X$ furnishes a notion of gradient. Flowing along $-\nabla\|f\|^2_2$ furnishes a map $X\to f^{-1}(0)$ pushing the generic fibers radially inwards into the special fiber (this tacitly requires properness). Precomposition with inclusion of a generic fiber defines the specialization map for any $t\in B^\times$ from a generic fiber to the special fiber.$$\psi_t:f^{-1}(t)\to f^{-1}(0)$$
Question 1. It is geometrically intuitive that specialization maps with respect to different Riemannian metrics are homotopic. Is this true? How to prove it?
Remark. It's been asked here whether the specialization map is surjective. Nobody's answered yet!
According to MO and many other sources,
A vanishing cycle is a homology class of the generic fibre that shrinks to zero in the special fibre.
"Shrinking in the special fiber" seems to suggest following the specialization map. It seems an elementary interpretation of a vanishing $n$-cycle is an element in the kernel of $\pi_n(\psi_t)$ or $\mathrm H_n(\psi_t)$, i.e an $n$-cycle that dies upon specialization. Similarly, an emerging $n$-cycle would be an element of the associated cokernel (or at least complement of the image).
Question 2. In this paper by Meigniez, the author gives a direct definition of (representative of) vanishing and emerging cycles, but seems to have the reverse convention for direction. He writes on page 2 "when moving from a given fiber to
immediately neighboring ones over some path in the base", which is opposite from the direction of specialization. Is this convention used elsewhere?
Question 3. Where are emerging cycles in the literature? Alternatively, what's the "role played by vanishing cycles" as opposed to emerging cycles?
Is the map always defined? Cannot the gradient flow sort of smear out when approaching the special fiber? In principle its trajectories may fail to have single limit in the special fiber...
@მამუკაჯიბლაძე I think the properness hypothesis precludes smearing out, but maybe I am wrong.
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2025-03-21T14:48:30.126141
| 2020-03-22T18:06:32 |
355464
|
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|
Stack Exchange
|
Finite-dimensional argument for Morse-Smale pairs?
Using the Sard-Smale theorem, it is relatively easy to show that Morse-Smale pairs on a manifold $M$ (i.e. pairs $(f,g)$ where $g$ is a metric on $M$, $f$ is a Morse function on $M$, and the stable/unstable disks corresponding to th flow of $-\nabla f$ intersect transversely) are dense in the appropriate topology. However, this argument requires the use of Banach manifolds. For the sake of a more accessible exposition, I was hoping to find an argument that uses only finite-dimensional methods, such as Thom transversality.
My thought was the following: Perhaps if we fix the metric $g$, we can show that functions $f$ such that $(f,g)$ is Morse-Smale are dense. (Just looking for a function seems to be more in the realm of classical applications of Thom transversality than also looking for a metric.) The Sard-Smale result certainly guarantees that this will be true for a generic $g$, but will it hold for any $g$? It would certainly seem very strange to me if there were a metric $g$ not admitting any such function $f$.
The Sard-Smale result certainly guarantees that this will be true for a generic , but will it hold for any ?
If $g$ is fixed, you can certainly use the Sard-Smale theorem to prove the existence of a Morse function $f$ so that the pair $(f,g)$ is Morse-Smale.
And yes, there's also a proof with less heavy machinery, see for instance Theorem 6.6 in the book "Lectures on Morse Homology" by Banyaga and Hurtubise.
|
2025-03-21T14:48:30.126261
| 2020-03-22T18:08:20 |
355465
|
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|
Stack Exchange
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Hyperbolic manifolds with infinite cyclic fundamental group
It is a well known fact that there is a correspondence between complete hyperbolic $n$-manifolds up to isometry and discrete subgroups of isometries of the hyperbolic space $\mathbb{H}^n$ that act freely on $\mathbb{H}^n$ up to conjugation.
The correspondece is given by $\Gamma < Isom(\mathbb{H}^n)\mapsto \mathbb{H}^n/\Gamma$ and the inverse is given by the map $M\mapsto \pi_1(M)\hookrightarrow Isom(\tilde{M})$ where $\tilde{M}\simeq \mathbb{H}^n$ is the fundamental cover of $M$.
The requirement that $\Gamma$ acts freely on $\mathbb{H}^n$ is equivalent to requiring that there are no elliptic isometries in $\Gamma$ or equivalently if every element in $\Gamma$ has infinite order.
In particular any parabolic or hyperbolic element in $Isom(\mathbb{H}^n)$ generates an infinite cyclic subgroup. This will correspond to manifolds with $\pi_1 M \simeq \mathbb{Z}$.
What are the complete hyperbolic manifolds with fundamental group $\mathbb{Z}$? Can we say something at least in the case of $3$-manifolds?
Yes, in dim 3, it consists in classifying conjugacy classes of non-elliptic elements (modulo inversion) in the isometry group. The isometry group can be viewed as those $z\mapsto\frac{az+b}{cz+b}$ and $z\mapsto\frac{a\bar{z}+b}{c\bar{z}+d}$. If I'm correct, up to conjugacy and inversion, we get the loxodromics: $z\mapsto az$, with $|a|>1$ of nonnegative imaginary part, and $z\mapsto a\bar{z}$ with $a$ positive real. And the horocyclic (aka parabolic): $z\mapsto z+1$ and $z\mapsto\bar{z}+1$.
In dimension 2: identifying the isometry group with $\mathrm{PGL}_2(\mathbf{R})$, we get $x\mapsto ax$ for $|a|>1$, $x\mapsto x+1$. Converted into an action on the upper half-plane, this yields $z\mapsto az$ and $z\mapsto -a\bar{z}$ for $a>1$, $z\mapsto z+1$.
This consists in classifying non-elliptic elements of the Lie group $\mathrm{Isom}(\mathbf{H}^n)\simeq\mathrm{PO}(n,1)$ up to conjugacy and inversion.
One can do separately loxodromics and horocyclics ("parabolics").
Loxodromics: they have two invariants: the translation length (a positive real number), and the transverse isometry, namely an isometry of $\mathbf{H}^{n-1}$ fixing a point (up to conjugation fixing this point), and this is classified by an element of $\mathrm{O}(n-1)$ up to conjugation [and inversion].
Horocyclics: they are classified by their action on the horosphere (which is a non-geodesic copy of the Euclidean space $\mathbf{R}^{n-1}$, modulo conjugation [and inversion] by the whole group of similarities. Hence, by a non-elliptic isometry of $\mathbf{R}^{n-1}$, modulo conjugation by similarities. In general, the horocyclic is orthogonal direct sum of a nontrivial translation and an element of $\mathrm{O}(n-2)$. Hence horocyclics are classified by conjugacy classes of $\mathrm{O}(n-2)$.
The horocyclic case corresponds to the existence of a cusp in the quotient manifold. If one sticks to orientable manifolds, one should restrict to $\mathrm{SO}(n-1)$ in the loxodromic case and $\mathrm{SO}(n-2)$ in the horocyclic case.
["And inversion" will not play any role since it follows that all isometries are conjugate to their inverse, since this holds in $\mathrm{O}(k)$ for every $k$.]
Let's specify in small dimension:
$n=2$: loxodromics are classified by a positive real number, and a sign (preserving or not the orientation). There's a single horocyclic (orientation-preserving).
$n=3$: loxodromics are classified by a positive real number, and by an element of $\mathrm{O}(2)$ up to conjugation (hence, either a rotation of angle in $[0,\pi]$, or a reflection). Horocyclics: it can be a translation, or a glide reflection.
$n=4$: loxodromics are classified by a positive real number, and by an element of $\mathrm{O}(3)$ up to conjugation (hence a rotation or antirotation of angle in $[0,\pi]$). Horocyclics: classified by some conjugacy class of $\mathrm{O}(2)$.
Topological classification:
actually, in the orientable case, the quotient manifold is analytically diffeomorphic to $\mathbf{R}^{n-1}\times (\mathbf{R}/\mathbf{Z})$, and in the non-orientable case, it is analytically diffeomorphic to $\mathbf{R}^{n-2}\times (\text{Möbius})$.
Indeed, in both case one sees that the isometry is analytically conjugate to a non-elliptic isometry of the Euclidean space $\mathbf{R}^n$. Such an isometry can be conjugated to have the form $f:(t,y)\mapsto (t+1,Sy)$ with $t\in\mathbf{R}$, $y\in\mathbf{R}^{n-1}$ and $S\in\mathrm{O}(n-1)$. If $S\in\mathrm{SO}(n-1)$, there is a 1-parameter subgroup $(S^t)$ with $S^1=S$, and conjugating $f$ by the analytic self-diffeomorphism $(t,y)\mapsto (t,S^ty)$ yields a translation. If $S\notin\mathrm{SO}(n-1)$, write $f$ as $(t,u,z)\mapsto (t+1,-u,Tz)$ with $z\in\mathbf{R}^{n-2}$ and $T\in\mathrm{SO}(n-2)$. Then conjugating as above only on the last variable conjugates to $(t,u,z)\mapsto t+1,-u,z)$ and this yields the requested description.
Thank you YCor. I'm interested in the topological type of these object.
For example, in 3D do we get solid tori and solid klein bottles or what else? It's not immediate to understand the quotient by these actions.
@WarlockofFiretopMountain It somewhat trivializes (the homeomorphism type is determined by the dimension and orientability). I added a paragraph.
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2025-03-21T14:48:30.126581
| 2020-03-22T18:16:48 |
355466
|
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|
Stack Exchange
|
Metrizability of a topological vector space where every sequence can be made to converge to zero
This is a follow-up to this answer.
If $E$ is a (real or complex) topological vector space, we say that a sequence $\{x_n\}_{n=1}^\infty$ in $E$ can be made to converge to zero if there exists a sequence $\{\alpha_n\}_{n=1}^\infty$ of strictly positive real scalars such that $\lim_{n\to\infty} \alpha_n x_n = 0$.
In the aforementioned answer, I compared this with two related notions, and I gave a generic example of a space where every sequence can be made to converge to zero: take a metrizable topological vector space and pass to a weaker (i.e. coarser) topology, possibly with a different dual. My question is whether every example is of this form. In other words:
Question. If $E$ is a topological vector space where every sequence can be made to converge to zero, is there a finer linear topology on $E$ that is metrizable?
Maybe it is false in general, but true in certain special cases? (E.g. if $E$ is locally convex/complete/separable, or if $E$ the strong dual of a Fréchet space?)
For the counterexample below we need a little lemma about barrelled spaces (i.e., every barrel = closed, absolutely convex, and absorbing set is a $0$-neighbourhood):
If $(E,\mathcal T)$ is a barrelled locally convex space which has a finer metrizable vector space topology $\mathcal S$ then $(E,\mathcal T)$ is metrizable.
Indeed, let $(V_n)_{n\in\mathbb N}$ be a basis of the $0$-neighbourhoods in $(E,\mathcal S)$ und $U_n =\overline{\Gamma(V_n)}^{\mathcal T}$ the closed absolutely convex hulls of $V_n$. These are barrels in in $(E,\mathcal T)$ and hence $\mathcal T$-neighbourhoods of $0$. On the other hand, every $\mathcal T$-neighbourhood of $0$ contains a closed absolutely convex one which itself contains some $V_n$ (because $\mathcal S$ is finer) and hence some $U_n$. This means that $(U_n)_{n\in\mathbb N}$ is a countable basis of the $0$-neighbourhoods of $(E,\mathcal T)$ which is thus metrizable.
Now the counterexample: Let $I$ be an uncountable index set and
$$E=\{(x_i)_{i\in I}\in \mathbb R^I: \{i\in I: x_i\neq 0\} \text{ is countable}\}$$
endowed with the relative topology of the product topology on $\mathbb R^I$. According to proposition 4.2.5(ii) in the book Barrelled Locally Convex Spaces of Bonet and Perez-Carreras, $E$ is barrelled. Since every $0$-neighbourhood only gives conditions on finitely many coordinates there is no countable basis of all $0$-neighbourhoods, i.e., $E$ is not metrizable and hence, by the lemma, it does not admit a finer metrizable vector space topology. On the other hand, for every sequence $x^n\in E$ the union $J$ of the supports of the $x^n$ is countable, so that $\{x^n:n\in\mathbb N\}$ is contained in the subspace $F=\mathbb R^J\times \{0\}^{I\setminus J}$ which is metrizable. Therefore, there is a sequence of $t_n>0$ such that $t_nx_n$ converges to $0$ in $F$ and hence also in $E$.
As you also asked for the particular case of strong duals of Frechet spaces $X$: This is not only covered by the answer of Tomasz Kania but it is rather trivial: If $X$ is not isomorphic to a Banach space there is a fundamental sequence of semi-norms $p_n$ which are pairwise non-equivalent. Then $X'=\bigcup_{n\in\mathbb N} X_n'$ where $X_n'$ is the space of functionals which are continuous with respect to $p_n$. Choosing $\phi_n\in X_n'\setminus X_{n-1}'$ there is no sequence of $t_n>0$ such that $t_n\phi_n\to 0$ because every bounded subset of $X'$ is contained in some $X_n'$.
EDIT. Another counterexample is the space of absolutely summable families $$\ell^1(I)=\{(x_i)_{i\in I}: \sum_{i\in I} |x_i|<\infty\}.$$
Every absolutely summable family has only countably many non-zero terms and hence a somehow natural topology (of course, besides the Banach space topology of the $\ell^1$-norm) is that of the inductive limit $\ell^1(I)=\lim\limits_\to \ell^1(J)$ for all countable subsets $J\subseteq I$. This finer topology is barrelled but not metrizable (hence it does not admit a finer metrizable vector space topology) but every sequence can be made bounded.
It's true, at least for (DF)-spaces as proved by Kąkol and Saxon.
Let $X$ be a (DF)-space. Then the following are equivalent:
$X$ admits a finer normed topology;
$X$ has property $C_4$, that is, for any sequence $(x_n)_{n=1}^\infty$ in $X$, there is a sequence of positive reals $(t_n)_{n=1}^\infty$ s.t. $0\in \overline{\{x_n t_n\colon n\in \mathbb N\}}$.
It is not inconceivable that the requirement of having a finer normed topology can be replaced with having a weaker metric topology at the expense of dropping the hypothesis of being a (DF)-space.
For details, see Proposition 15.5 in
Jerzy Kąkol, Wiesław Kubiś, and Manuel López-Pellicer, Descriptive Topology in Selected Topics of Functional Analysis, Springer US, 2011.
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2025-03-21T14:48:30.126886
| 2020-03-22T19:14:24 |
355468
|
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|
Stack Exchange
|
Existence of the inverse Fourier transform, Carr Madan
I have a function $C_T(k)$ that is not $L_1$, because its limit in negative infinity is a constant.
So I dampened it by $ e^{\alpha k} $. Let's call the transformed function (of the dampened function) $\psi$.
My aim was to compute a few things on $\psi$, and then take back the function to the time domain, and multiply by $ e^{-\alpha k} $ to get back my original function. However, I am not sure why $\psi$ would be integrable thus Fourier transformable.
My question would be, is the function $\psi$ invertible ?
If the answer to the question is "it depends", does it help to know that $\psi$ can be writen as a bell behaved function of a characteristic function ? Like a constant times a characteristic function ?
I know that if the PDF related to that characteristic function had jump, thus it would mean that $\psi$ is not integrable since a Fourier transform is always continuous. But here, the PDF has no jump. Thus I don't know if this much information would allow me to say anything about the integrability of $\psi$.
In the worst case scenario that the information I am giving is not enough, what should I look at to know if $\psi$ is integrable without having an analytical expression for it ?
Thank you.
Post Scriptum :
I am trying to be as precise as I can, without giving you the full details of the domain and the research paper I am dealing with. However, by experience, I know people will complain about my non capacity of being precise enough, so here is, in advance, the precise paper and the precise moment that confuses me. It is here in Carr Madan's paper, page 3 :
http://citeseerx.ist.psu.edu/viewdoc/download?doi=<IP_ADDRESS>8.4044&rep=rep1&type=pdf
Can you talk slightly about the domains of your function? Multiplying by $e^{ak}$ will make it explode in one of the directions for instance. I think what you're after anyway is called the laplace transform, and terry tao has good notes (and the inversion) of it. Anyway if you're talking about fourier as $L^2(R) \to L^2(R)$ it is invertible (Parsevals formula shows the $L^2$ norm of the fourier is like that of the original function you started with).
@Andy thanks for the comment. Right I was thinking about Parseval's formula, but Fourier is taking an $L^1$ function to an $L^1$ function right ? Or have I missed something ? In the paper, they also talk about square integrability, and I don't get what s the link between that and having a Fourier transform.
No, consider the indicator of [0,1], for many $t$ the fourier is $1/t$. However if your function is smoother, it will have rapidly decreasing fourier transform, which will put it back in $L^1$. For instance if you have a function in $C^k$ with the derivative in $L^1$, then integration by parts shows it goes down at least like $1/t^k$.
Your function $\psi$ is certainly a tempered distribution, then you have a valid inversion formula.
@Andy right sorry. Imeant it goes from integable functions to continuous functions, but we don't know anything about the final space. However, what you're saying, is that the Fourier transform of a L1 inter L2 function is going to be in L1 inter L2?
@MarineGalantin No, my indicator example is in $L1 \cap L2$. You are correct that integrable functions go to continous ones
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2025-03-21T14:48:30.127127
| 2020-03-22T19:53:04 |
355471
|
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|
Stack Exchange
|
Finite $p$-groups of co-class $3$, class at least $4$ and some controlled generator growth
I am trying to prove the following comment (Ref. https://link.springer.com/article/10.1007/s00605-016-0938-5 Page-684, Rmk3.2):
Let $G$ be a finite $p$-group of co-class $3$, class $\geq 4$. Then $G$ satisfies $d(Z_2(G)/{Z(G)}) = d(G) d(Z(G))$ if and only if $d(G) = 2$. Here $d$ denote the rank of the corresponding group and co-class is $n-c$ if $|G|=p^n$ and $G$ has class $c$.
My attempt is to connect the question to the discussion in the thread: Groups in which lower central series and upper central series coincide
"$\Rightarrow$" Assuming the condition, we have $d(Z_2(G)/{Z(G)}) \geq 2$. Suppose if possible $|Z_2(G)/{Z(G)}| \geq p^3$. Then we can directly check that $G/Z_{c-1}(G) \cong C_p \times C_p, Z_2(G)/{Z(G)} \cong C^3_p, Z(G) \cong C_p$ and $Z_i(G)/{Z_{i-1}(G)} \cong C_p$ for the remaining upper central quotients. Now since $G$ and $G/{Z(G)}$ has same co-class, using Lem.4.1.13 (Ref. https://books.google.co.in/books/about/The_Structure_of_Groups_of_Prime_Power_O.html?id=34khoLiyP_QC&redir_esc=y) we have $Z(G) = \gamma_c(G)$. At this point proving that $G$ is a UL-group (i.e., the lower and upper central series of $G$ coincide) would imply this side. However, we know that $\gamma_{c+1-i}(G) \subseteq Z_i(G)$ (i.e., gamma-groups are smaller).
I am stuck here. Any ideas?
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2025-03-21T14:48:30.127245
| 2020-03-22T20:46:40 |
355472
|
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|
Stack Exchange
|
The Lie algebra of the subgroup of $GL(n)$ preserving a given variety
Let $V=k^n$ for an algebraically closed field $k$ of characteristic 0, and let $W \subseteq V$ a subspace. Let $G_W\subseteq GL(V)$ be the set of invertible linear maps that preserve $W$, i.e.
$$
G_W=\{x \in GL(V): x(W)=W\},
$$
and let
$$
\mathfrak{g}_W=\{X \in \mathfrak{gl}(V) : X(W) \subseteq W\},
$$
where $\mathfrak{gl}(V)$ is the Lie algebra of $GL(V)$, identified with the set of linear maps on $V$ under the commutator bracket. It is known (see e.g. Humphrey's Linear Algebraic Groups section 13.8) that $\mathfrak{g}_W$ is the Lie algebra of $G_W$.
My question is: If $Y \subseteq V$ is an arbitrary irreducible affine variety, what is the Lie algebra of its preserver $G_Y$? I have a guess that would be consistent with the above result:
Guess: Suppose $Y \subseteq V$ is a homogeneous irreducible affine variety (i.e. $Y$ is a cone: $v \in Y \iff \alpha v \in Y$ for all $\alpha \in k$). Let
$$
\mathfrak{g}_Y=\{X \in \mathfrak{gl}(V) : X(\mathscr{L}(Y)) \subseteq \mathscr{L}(Y)\},
$$
where $\mathscr{L}(Y) \subseteq V$ is the tangent space to $Y$ at $0$, i.e., if $Y=V(f_1,\dots, f_m)$, then $\mathscr{L}(Y)=V(d_0f_1,\dots, d_0 f_m)$, where $d_0 f (x)= \sum_{j=1}^n \frac{\delta f}{\delta x_i} (0) x_i$. Then $\mathfrak{g}_Y$ is the Lie algebra of $G_Y$.
Is this guess correct? If so, I would appreciate a proof or a reference to a proof.
A quick proof of the above result for when $Y=W$ is a subspace is provided by ShinyaSakai in the comments to this question
EDIT
@abx quickly disproved my guess, so my new question is simply: What is the Lie algebra of $G_Y$? I would appreciate any relevant references in this vein.
Wrong. If $Y$ is not contained in a hyperplane, the $f_i$ are homogeneous of degree $>1$, hence their derivatives vanish at $0$, and $\mathfrak{g}_Y=\mathfrak{gl}(V)$.
@abx Thanks! Do you have any idea what the Lie algebra of $G_Y$ is then (or references to results in this vein)?
See https://mathoverflow.net/questions/165900/reference-for-an-algebraic-group-preserving-a-cubic-form for the cubic case. I believe if the degree is > 2 the group is usually finite, so you won't get interesting information from the Lie algebra. (Of course for degree 2 you get the orthogonal group)
Doc, you are right but only amorally. You need to replace the tangent vectors with jets to capture the behavior of your cone.
Let $I(Y)$ be the ideal of zeroes of your $Y$. Then
$$
Lie (G_Y) = \{ X \in {\mathfrak{gl}}(V) | X(I(Y))\subseteq I(Y)\}.
$$
Now you know that $I(Y)$ is homogeneous. Pick a finite set of its generators. Let $n$ be the highest degree of a generator from your set. Consider $n$-cojets
$$
J^\ast(Y) := I(Y)/(I(Y)\cap I(0)^{n+1}) \subseteq J^\ast := I(0)/I(0)^{n+1}
$$
and $n$-jets
$$
J(Y) := J^{\ast}(Y)^\perp \subseteq J := (J^\ast)^\ast
$$
where $I(0)$ is the principal ideal.
This yields a desired "finite-dimensional" condition
$$
Lie (G_Y) = \{ X \in {\mathfrak{gl}}(V) | X(J(Y))\subseteq J(Y)\}.
$$
I'm having trouble verifying your first line $Lie(G_Y)=...$ How does $X \in \mathfrak{g}\mathfrak{l}(V)$ act on $f \in I(Y)$?
$f\in k[V]$ on which $X$ naturally acts
Okay, we can think of the action of $GL(V)$ on $V$ as an algebraic group homomorphism $GL(V) \rightarrow GL(k[V])$, the differential of which is a Lie algebra homomorphism $\mathfrak{g}\mathfrak{l}(V) \rightarrow \mathfrak{g}\mathfrak{l}(k[V])$, which gives our action of $\mathfrak{g}\mathfrak{l}(V)$ on $k[V]$. Is there a formula for this action? I would greatly appreciate a reference.
Just come from an algebraic cloud down to the analytic Earth. The elementary matrix $E_{i,j}$ acts as a vector field $x_i\frac{\partial}{\partial x_j}$. I am not sure about a reference. Probably, Theorie der Transformationsgruppen by Lie :-))
Thanks. I think I see it up here in the clouds. Any algebraic group $G$ action on an affine variety $X$ induces a natural action of $\mathfrak{g}$ on $k[X]$. A standard example is when $X=G$ and $g \cdot h= hg^{-1}$, in which case $\mathfrak{g}$ acts on $k[G]$ by right convolution (i.e., as a left-invariant derivation). A good reference is Humphreys Linear Algebraic Groups. It's not hard to extend this argument to get a formula for an arbitrary variety $X$.
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2025-03-21T14:48:30.127639
| 2020-03-22T21:52:57 |
355476
|
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|
Stack Exchange
|
Diverging solution to a SDE
I'm considering the SDE, with $B$ the brownian motion and $\beta$ a scalar (it can be negative)
$$ X_t = x_0 + \int_0^t (\beta + X_s^2) ds + B_t $$
and I would like to show that $X_t$ almost surely diverges to infinity in finite time.
I really don't know how to do it since usually I have to control $X_t$, does anyone have a tip ?
Thanks :)
Let us begin by showing that $X$ diverges to $+\infty$, possibly as time goes to infinity. Set
$$f : x\mapsto \int_0^x\exp\left(-2\beta y-\frac 23y^3\right)\mathrm dy. $$
The point of $f$ is that $f(X)$ is a local martingale, possibly up to the explosion time $\tau$ of $X$ ($f$ is solution to $(\beta+x^2)\partial_xf+\frac12\Delta f=0$, in fact all such solutions can be written as $af+b$).
Notice that $f$ is increasing. Moreover, it is bounded above, since the exponential term goes to zero fast enough. Since the explosion of $X$ can only occur if $X$ diverges to $+\infty$ ($X_t$ is at least $x_0+\beta t+B_t$), $f(X)$ is in fact a local martingale for all times, setting by convention $f(X_t)=\sup f$ for all $t\geq\tau$.
As a local martingale bounded above, $f(X)$ converges almost surely, and given that $f$ is increasing and isn't bounded below, $X$ converges in $\mathbb R\cup\{+\infty\}$. We need to show that $\lim X_t$ cannot be finite with positive probability.
Notice that
$$ B_{t+1}-B_t = (X_{t+1}-X_t) - \int_t^{t+1}(\beta+X_s^2)\mathrm ds. $$
In particular, if $X$ converges to a finite limit $\ell\in\mathbb R$, then $B_{t+1}-B_t$ converges to $ -\beta-\ell^2$. Hence the event that $X$ converges to a finite limit is included in the event that $B_{t+1}-B_t$ converges, which obviously has measure zero.
So $X$ diverges to $+\infty$. I will now try to use deterministic arguments. Let us assume for now the following
Fact.
Let $Y$ be a process such that $$ Y_t \geq -C + \int_0^t \left(Y_s^2-\alpha^2\right)\mathrm ds $$
for some constants $C,\alpha>0$, and all $t$ possibly up to some explosion time $\tau$ (explosion means “leaves all compact subsets”).
Then either $\liminf_{t\to\tau} Y_t\leq\alpha$ or $Y$ diverges to $+\infty$ in finite time.
Note that for any $\alpha>0$ such that $\alpha^2>-\beta$, almost surely there exists a (random) constant $C>0$ such that $B_t\geq -C-(\alpha^2+\beta)t$ for all $t>0$. In particular,
$$ X_t \geq x_0 - C + \int_0^t\left(X_s^2 - \alpha^2\right)\mathrm ds. $$
Since $X$ diverges to $+\infty$, obviously its limit inferior is not bounded above, so the fact implies that $X$ must undergo explosion in finite time.
Now onto the proof of the fact. Suppose that the limit inferior of $Y$ is larger than $\alpha$. Then for all $t$ large enough (which here means close enough to $\tau$), $Y_t^2>\alpha^2+\varepsilon$ for some $\varepsilon>0$. According to the inequality, $Y$ must then diverge, possibly in infinite time.
Setting
$$ I_t = \int_0^t \left(Y_s^2-\alpha^2\right)\mathrm ds, $$
we see that
$$ I'_t = Y_t^2 - \alpha^2 \geq \frac12(Y_t+C)^2 \geq \frac12I_t^2 $$
for all $t$ large enough. In particular (note that $I_t>0$ for $t$ large enough),
$$ \frac{\mathrm d}{\mathrm dt}\left(-\frac1{I_t}\right)
= \frac{I'_t}{I_t^2}
\geq \frac12 $$
for all $t$ large enough, hence
$$ Y_t\geq -C + I_t\geq-C+\frac 2{T-t} $$
for some $T>0$ and all $t$ large enough, so that $Y$ explodes in finite time.
Great, thanks for the trick ! But if I'm not missing something, I think that we have just showed that $X_t$ diverges almost surely to infinity. We do not have showed that it's in finite time. But I should be able to conclude by myself though.
You are absolutely right! I forgot to add the end of the argument, I'll edit in a bit.
It should now be complete. I feel like the proofs of the fact are somewhat inefficient, but I cannot find something more convincing. Tell me which one you prefer and I will delete the other.
Yes thank you very much ! I achieved the proof by myself yesterday, and I used the comparison with the deterministic situation as well. I have previously solved the deterministic equation in all possible case so the fact is implied by this previous work. Your second proof is quite clear though :)
|
2025-03-21T14:48:30.127905
| 2020-03-22T22:41:50 |
355477
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/355477"
}
|
Stack Exchange
|
Jacobian fibration of an abelian fibration
Let $f \colon S \rightarrow C$ be a minimal elliptic surface and let $g \colon J \rightarrow C$ be its jacobian fibration. In this case, we know that the fibers of $g$ are better behaved that the ones of $f$. There is an explicit classification of what fiber one gets after taking the jacobian fibration (see, for instance, the chapter on elliptic surfaces in Algebraic Surfaces, by Shafarevich, Proceedings of the Steklov Institute of Mathematics. American Mathematical Society, 75).
This information can be rendered numerically by means of the log canonical threshold: for every $o \in C$, $\mathrm{lct}(J,g^*o) \geq \mathrm{lct}(S,f^*o)$ (see Lemma 1.6 here for the inequality between boundary divisors in the canonical bundle formula, https://arxiv.org/pdf/alg-geom/9305002.pdf, and section 3 here for the relation between lct and the coefficients of the boundary divisors https://arxiv.org/pdf/1210.5052.pdf).
My question is the following. Do we have a similar description for fibrations of higher-dimensional abelian varieties? We can assume we are in the following situation: C is a smooth curve, $f\colon X\rightarrow C$ is a relatively minimal fibration whose geometric generic fiber is an abelian variety, and $g\colon J\rightarrow C$ is a relatively minimal model of the corresponding jacobian fibration.
1) Do we still have the inequality between the log canonical thresholds?
2) Can we get an explicit description of what the new fiber is (maybe just in some cases)?
|
2025-03-21T14:48:30.128024
| 2020-03-23T01:19:17 |
355481
|
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"D.W.",
"https://mathoverflow.net/users/37212"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/355481"
}
|
Stack Exchange
|
Best known bound on feedback arcset in high-girth directed graphs?
Let $G$ be a directed graph with $n$ vertices and $m$ edges such that every directed cycle in $G$ has length at least $m/k$. An arcset of $G$ is defined as a set of edges $X$ whose removal from $G$ leaves an acyclic graph. What is the best known bound on the size of the smallest such $X$? The best I have been able to find in the literature is a bound $|X| \leq O(k \log k \log \log k)$ due to Seymour (see 1.4 here: https://link.springer.com/article/10.1007/BF01200760 ).
Meanwhile, the best known worst-case lower bound is $|X| \geq O(k \log k)$, which can be obtained by taking an undirected expander graph and orienting the edges appropriately. Has this gap of $\log \log k$ been closed?
Cross-posted: https://cstheory.stackexchange.com/q/46518/5038, https://mathoverflow.net/q/355481/37212. Please don't cross-post.
|
2025-03-21T14:48:30.128103
| 2020-03-23T01:50:14 |
355482
|
{
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"authors": [
"Tim Campion",
"https://mathoverflow.net/users/2362"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/355482"
}
|
Stack Exchange
|
Retracts in the bicategory of spans
I would like to show that the category of sets and spans between them, seen as a $(2,1)$-category, is Cauchy complete, i.e. has splitting of (homotopicaly coherent) idempotent.
Ideally I would also like to prove that the retract of a set $X$ in this bi-category are all subsets of $X$, and that all this also works if we replace set by a category with finite limits.
I think I have a poof of this (If I'm not mistaken), but it is very long and technical and I wanted to know if there is literature about this, or if there is a simpler or more conceptual way to obtain these results.
I think I worked something out here but ... erm ... I definitely skipped some bits that seemed technical :)
|
2025-03-21T14:48:30.128196
| 2020-03-23T01:52:59 |
355483
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Alex Kruckman",
"Asaf Karagila",
"P. Grabowski",
"https://mathoverflow.net/users/127260",
"https://mathoverflow.net/users/2126",
"https://mathoverflow.net/users/7206"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/355483"
}
|
Stack Exchange
|
The theory of a model of a theory that knows all formulas true in almost all its models
We are in first order logic world.
Let $\sigma$ be a finite signature and $T$ a consistent theory of $\sigma$.
Due to Löwenheim–Skolem theorem, we can consider the $\underline{set}$ of all at most countable models of $T$ up to elementary equivalance(does not change what formulas hold).
Let $\mu$ be a nonprincipial ultrafilter of it, denote by $\Omega^T_\mu$ the ultraproduct of
all those models with respect to $\mu$.
Clearly, the theory of $\Omega^T_\mu$ are formulas that are true for almost all with respect to $\mu$ models of $T$.
In particular, this theory, call it $T(\Omega^T_\mu)$, has $T$ as a subset.
My questions are:
Does the equality $T(\Omega^T_\mu)=T$ hold?
Does it depend on $\mu$? Is the intersection over all $\mu$ equal to $T$?
If they are not equal, what the difference means? Furthemore, we could consider a sequance of theories $T\subset T(\Omega^T_\mu) \subset T(\Omega^{T(\Omega^T_\mu)}_{\mu'})\subset\ldots$. Does it stabilize?
Does $T(\Omega^T_\mu)$ complete $T$ in any nice sense?
I suspect that all those questions are classical and already answered, but I do not know where
to look for them. Just before pandemia I have finished my first course on mathematical logic and I have been wondering about this object since then.
Well, "almost all" can wildly vary depending on the choice of the ultrafilter.
What is a $\underline{set}$? $T$ has a proper class of at most countable models - but only a set up to isomorphism (for example, you can restrict to just those models whose domains are subsets of $\omega$).
@AsafKaragila Exacly! There I use "almost all", but in fact I mean almost all with respect to $\mu$. I have edited it.
@AlexKruckman You have a good point. I mean the set of models up to elementary equivalance, since that notion doesn't change what formulas hold.
In this answer, I will basically repeat things that zeb said in their nice answer, but arranged differently (in a way that might or might not be more clear).
Let $\text{Mod}_{\leq \aleph_0}(T)$ be the "set" of all at most countable models of $T$. Your question is a bit ambiguous about what this means - I'll come back to that later. Morally, picking an ultrafilter on $\text{Mod}_{\leq \aleph_0}(T)$ is the same as picking a completion of $T$.
In one direction, if you pick an ultrafilter $\mu$ on $\text{Mod}_{\leq \aleph_0}(T)$, you can take the complete theory of the ultraproduct, $T(\Omega^T_\mu)$ in your notation. This is a complete theory, and it contains $T$, so it's a completion of $T$.
In the other direction, given a completion $T\subseteq T'$, for each sentence $\varphi\in T'$, consider the set of models $M_\varphi = \{M\in \text{Mod}_{\leq \aleph_0}(T)\mid M\models \varphi\}$. Then $F_{T'} = \{M_\varphi\mid \varphi\in T'\}$ has the finite intersection property, so we can extend it to an ultrafilter. And moreover for any ultrafilter $\mu$ extending $F_{T'}$, we have $T(\Omega^T_\mu) = T'$, by Łoś's Theorem.
But now there's a subtlety here: You want $\mu$ to be a non-principal ultrafilter.
If by $\text{Mod}_{\leq \aleph_0}(T)$, you mean a set of isomorphism representatives for the models of $T$, so this set contains just one copy of each model of $T$ up to isomorphism, then if $T'$ has only one (at most countable) model (i.e. if it's the theory of a finite structure, or if it's $\aleph_0$-categorical), then the ultrafilter $\mu$ you get by the above construction will be principal. That is, picking a non-principal ultrafilter on $\text{Mod}_{\leq \aleph_0}(T)$ picks a completion of $T$, but we never get a completion that is $\aleph_0$-categorical or the theory of a finite structure. Worse, if $T$ already has only finitely many (at most countable) models up to isomorphism, then there is no non-principal ultrafilter on $\text{Mod}_{\leq \aleph_0}(T)$ - this corresponds to the fact that $T$ has no completion with more than one (at most countable) model.
These issues go away if $\text{Mod}_{\leq \aleph_0}(T)$ contains infinitely many copies of each model up to isomorphism (e.g. if it's more like the proper class of all at most countable models of $T$). Then when we extend $F_{T'}$ to an ultrafilter $\mu$, we can always make sure $\mu$ is non-principal.
So the answer to your questions are:
(1) $T(\Omega_\mu^T) = T$ if and only if $T$ is already complete.
(2) Yes, it definitely depends on $\mu$ (unless $T$ is complete). If you remove the restriction that $\mu$ be non-principal, the intersection of all theories of the form $T(\Omega^T_\mu)$ is $T$. If you require $\mu$ to be non-principal, this intersection is the set $T_{\text{cof}}$ of all sentences which are in all but finitely many models in $\text{Mod}_{\leq \aleph_0}(T)$. If $\text{Mod}_{\leq \aleph_0}(T)$ contains infinitely many copies of each model up to isomorphism (e.g. if this "set" is actually the proper class of all at most countable models of $T$), then again $T_{\text{cof}} = T$.
(3) The "difference" consists of moving from $T$ to a completion. The chain stabilizes after the first step, i.e. $T(\Omega_\mu^T) = T(\Omega_{\mu'}^{T(\Omega_\mu^T)})$, since $T(\Omega_\mu^T)$ is already complete.
(4) Yes, $T(\Omega_\mu^T)$ is a completion of $T$.
The theory of $\Omega_{\mu}^T$ will automatically be a complete theory which extends the original theory $T$, just because it is the theory of a particular model, and every statement is either true or false in any particular model. It will usually contain more than just the statements that are true in all but finitely many of the models of $T$.
If your original theory $T$ has a statement $\phi$ which is true in infinitely many countable models and false in infinitely many other countable models, then it is easy to construct an ultrafilter $\mu$ which contains the set of models where $\phi$ holds, and to construct another ultrafilter $\mu'$ which contains the set of models where $\phi$ fails (since any filter is contained in an ultrafilter). So $\phi$ will hold in $\Omega_{\mu}^T$ but not in $\Omega_{\mu'}^T$.
For example, the first order theory $N$ of the natural numbers (Peano arithmetic) is known to have infinitely many non-isomorphic models which contain (non-standard) numbers encoding proofs of contradictions in Peano arithmetic, as well as infinitely many non-isomorphic models which do not contain numbers encoding proofs of contradictions in Peano arithmetic, so you can construct ultrafilters $\mu, \mu'$ such that $\Omega_{\mu}^N$ is a model of Peano arithmetic which asserts the inconsistency of Peano arithmetic, while $\Omega_{\mu'}^N$ is a model of Peano arithmetic which asserts the consistency of Peano arithmetic.
Edit: A stronger observation than the above is this. Suppose $T'$ is any extension of the theory $T$ such that $T'$ has infinitely many nonisomorphic countable models. Then there is a non-principal ultrafilter $\mu$ such that $\Omega_{\mu}^T$ is a model of $T'$. Again, to prove this you just start with the filter generated by complements of singletons and by the set of models of $T'$, and complete it to an ultrafilter.
|
2025-03-21T14:48:30.128644
| 2020-03-23T02:40:55 |
355485
|
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|
Stack Exchange
|
Ergodic automorphism is mixing of all orders
I'm trying to solve an exercise in "Ergodic Theory with a view towards Number Theory".
The exercise is suppose $T$ is an ergodic automorphism on an compact abelian group, show that it is mixing of all orders.
(The idea is to use the dual basis of characters and ergodicity being equivalent to no finite periods).
I am unable to show the mixing of all orders.
Thus we need to show for character $\chi_i$ that unless they're all trivial, then as $n_i \to \infty$, we can't have $\chi_0 \times \chi_1(T^{n_1}) \times \ldots \times \chi_k(T^{n_1 + \ldots +n_k}) = 1$.
I can't even refute $\chi_0 (T^{2^n}) \chi_1(T^n) = \chi_2$.
Basically I would love to have a repitition of some sort so that I can cancel and induct.
Please give me a hint.
Begin by thinking about the dual automorphism to a hyperbolic toral automorphism. You can project the dual group onto an "expansive" direction. Any relation of the sort you are looking at can be shifted by the automorphism so that all powers are negative except for one large positive one. In the projection, this would mean that a huge number is the sum of very small numbers, a contradiction. The kind of argument works whenever there is some sort of expansiveness. Then you need to show all ergodic group automorphisms do have some sort of expansiveness.
Much more is true, see my paper Ergodic group automorphisms are exponentially recurrent, Israel J. Math. 41 (1982), 313-320.
What is more interesting is that this fails for commuting group automorphisms (or algebraic ${\Bbb Z}^d$-actions), the classic example due to Ledrappier. Amazingly, it does remain true in the case of commuting toral automorphisms, however the proof uses some very deep number theory about additive relations in fields (see Cor. 27.6 in Dynamical Systems of Algebraic Origin by Klaus Schmidt).
Thanks for the hint! I'm trying to implement it for the hyperbolic toral case. For simplicity I started by assuming the matrix diagonalizable- your argument gives that the only problem can appear on eigenspaces of eigenvalues of modulus 1. Since they are difficult to handle, I hoped I can avoid handling them by using Kronkers theorem to get a nonnorm 1 eigenvalue, and hope no integer vector is orthogonal to it. Are there any properties of the dual group in general I should be aware of ? I would love for it to be finitely generated for example
All you need is an eigenvalue outside the unit circle, and you're right that by Kronecker there must be one since otherwise you violate ergodicity. But there are situations in which there is no geometric hyperbolicity at all, for automorphisms of solenoids rather than tori. For these, there is an arithmetic hyperbolicity that serves as a replacement. This is part of a beautiful "global" view, based on treating all completions of $\Bbb Q$ equally, both $p$-adic and real, and each plays a cooperative role.
My problem is even if I have an eigenvalue $\lambda$ outside the unit circle with eigenvector $e$, my vectors $v_i$ can be orthogonal to it, thus not letting me project along it. I think there is no choice but to try to understand what is going for nonroots of unity on the circle. Regarding your second point that sounds fascinating, do you have a good source?
Use the eigenspace decomposition, where "project" doesn't mean orthogonal projection but rather projection along the complementary eigenspace. There's a very short paper with Schmidt, Bernoullicity of solenoidal automorphisms and global fields, Israel J. Math 87 (1994), 33-35, which has this point of view. Also a paper with Ward Automorphisms of solenoids and p-adic entropy, Ergod. Theory & Dynam. Syst. 8 (1988), 411-419, where the p-adic expansion and contraction using p-adic eigenvalues is essential to getting the complete picture.
A more thorough understanding of this happens when considering the joint action of $d$ commuting automorphisms. See my paper with Einsiedler, Algebraic $\BbbZ ^d$-actions of entropy rank one, Trans. AMS 356 (2004), 1799-1832. It's really a beautiful "global" picture of what's going on.
|
2025-03-21T14:48:30.128935
| 2020-03-23T03:13:43 |
355487
|
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"F Zaldivar",
"Felipe Voloch",
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|
Stack Exchange
|
Modern reference for Andre Weil's 'Sur les courbes algébriques et les variétés qui s'en déduisent'
I'm currently interested in the cardinality of the set of values of a polynomial over a finite field.
I found a paper
Saburo Uchiyama, Sur le Nombre des Valeurs Distinctes d'un Polynome a Coefficients dans un Corps Fini, Proceedings of the Japan Academy 30, Issue 10 (1954) p. 930–933, doi:10.3792/pja/1195525873, Project Euclid.
In the paper, the author uses some theorem of Weil on an asymptotic formula of the cardinality of the zero set of an absolutely irreducible polynomial $f^{\ast}(u,v)=\frac{f(u)-f(v)}{u-v}$.
So I want to study 'quickly' Weil's theorem on this occasion. But the 1948 book of Weil 'Sur les courbes algébriques et les variétés qui s'en déduisent' (WorldCat) seems to be out of print and might not be modern.
I have a little familiarity with the first 3 chapters of Hartshorne.
I hope someone can recommend a reference. Thank you.
Hartshorne Exercise 1.10 of Chapter V, Section 1.
@FelipeVoloch Thank you very much for your comment. I think the exercise gives a roadmap. May I ask one more reference that might be more comprehensive please?
You may find a sketch, with some details, of the proof of Hasse-Weil bound for the number of ${\mathbb F}_q$-rational points on a smooth projective curve defined over the finite field ${\mathbb F}_q$ in the notes posted on the following link: https://www.ocf.berkeley.edu/~rohanjoshi/2019/11/25/the-riemann-hypothesis-for-curves-over-finite-fields/
|
2025-03-21T14:48:30.129073
| 2020-03-23T03:37:59 |
355488
|
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"paul garrett"
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|
Stack Exchange
|
Sobolev convergence of Fourier series
Consider $f\in H^{\sigma}(S^1)=W^{\sigma, 2}$ (the usual Sobolev space on the circle) and let $S_Nf$ be its truncated Fourier series $S_Nf = \sum_{|n|\leq N} \hat{f}(n)e^{2\pi i n x}$. I am looking for the theory for inequalities of the following type:
$$\|f-S_Nf\|_{H^s} \leq CN^{c(s,\sigma)} \|f\|_{H^{\sigma}} \, ,$$
What values of $s>0$ work for any given $\sigma >0$? What is known of the constant $C$ and of the rate $c(s,\sigma )$?
Motivation: I am familiar with the equivalent theory for orthogonal polynomial expansions (see Canuto and Quarteroni, Math. Comp. 1982), but figured out that the Fourier theory is older, and maybe simpler.
See Lemma A. 43 in [A. Kirsch: An Introduction to the Mathematical Theory of Inverse Problems. Springer, New York, 1996].
Thanks! Is this tight, in terms of spaces, rates, and constants?
@YidongLuo my comment above
The rates is tight for the periodic Sobolev space $ H^r, r \in \mathbb{R} $ in terms of the regular estimate proof.
Let us start with pointing out that $f\in H^\sigma$ is equivalent to
$$
(\langle n\rangle^\sigma\hat f(n))_{n\in \mathbb Z}\in \ell^2(\mathbb Z),
\quad \text{with $\langle n\rangle=\sqrt{1+n^2}$.}
$$
Then you have for $s<\sigma$
$$
\Vert f-S_N(f)\Vert_{H^s}^2=\sum_{\vert n\vert> N}\langle n\rangle^{2s}\vert\hat f(n)\vert^2=
\sum_{\vert n\vert> N}\langle n\rangle^{2\sigma}\vert\hat f(n)\vert^2\langle n\rangle^{2s-2\sigma}\le\langle N\rangle^{2s-2\sigma}\Vert f\Vert^2_{H^\sigma}.
$$
Yes, in my opinion, this is the most sensible answer possible here. Also, I think it is essentially sharp, apart from being the natural argument.
|
2025-03-21T14:48:30.129241
| 2020-03-23T03:48:28 |
355489
|
{
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"authors": [
"Gil Kalai",
"JCK",
"Lucia",
"Sextus Empiricus",
"Shahrooz",
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Stack Exchange
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Suggestions for reducing the transmission rate?
What are suggestions for reducing the transmission rate of the current epidemics?
In summary, my best one so far is (once we are down to the stay home rule) to discretize time, i.e., to introduce the following rule for the general populace not directly involved in necessary services:
If members of your household go to public services on a certain day, the whole household should not use any public service for 2 weeks after. That way you still can get infected but cannot infect without knowing it.
Do you have some suggestions? Good models to look at? No predictions please, just advices what to do.
Edit: In more detail:
Model (the simplest version to make things as clear as possible): There are several categories of people $C_i$ that constitute portion $p_i$ of the population and have a certain matrix $A$ of interactions per day. Then, if $x_i(t)$ is the number of ever infected people in category $C_i$ by the time $t$, the driving ODE is
$$
\dot x(t)=\alpha A[x(t)-x(t-\tau)]
$$
where $\tau$ is the ("typical") time after which the sick person is removed from the population and $\alpha$ is the transmission probability. In this model the exponential growth is unsustainable if $\alpha\lambda(A)\tau<1$ where $\lambda$ is the largest eigenvalue of $A$. We do not know $\alpha$ (though we can try to make suggestions how to reduce it, most such suggestions are already made by the government). The government can modify $A$ by issuing orders. Some orders merely reduce $a_{ij}$ to $0$, but the government cannot shut essential public services completely this way.
Questions: What is $A$, which entries $a_{ij}$ are most important to reduce, how to issue a sensible order that will modify them, and by how much they will reduce the eigenvalue?
First suggestion for these 4 answers: There are two categories of people: ordinary population that only goes to public services and
public servants that both provide services and go to them. There is only one ("averaged") type of service involving a dangerous client-server interaction and all infection goes there. The portion of public servants in the population is $p$. The server sees $M$ clients a day. Then the current social interaction matrix (say, for the grocery store I've seen yesterday) is $A=\begin{bmatrix}0 & M(1-p)\\ Mp & 2Mp\end{bmatrix}$ (ordinary population does not transmit to ordinary population, servers transmit to clients who can be both servers and clients, clients transmits to servers. The largest eigenvalue is $M(p+\sqrt p)$. The lion's share comes from $\sqrt p$, which is driven by the off-diagonal entries, The order should be issued as above, the effect that ordinary people never come to the service infected, which will remove the left bottom corner and drop the largest eigenvalue to $2Mp$. Assuming $p=1/9$ (not too unrealistic), the drop will be two-fold even if you leave the service organization as it is.
That ends the solution I propose in mathematical language.
In layman terms, the public will completely do this part (you cannot ask for more) and still have some life, and we can concentrate on the models of how servers should be organized.
Edit:
Time to remove the non-relevant part and add some relevant thoughts about what else we can help with plus the response to JCK.
First of all, It is very hard to formulate the orders correctly. The stay home rule really means "avoid all close contacts outside your household except the necessary interactions with public servants providing vital services to you" (and even that version is, probably flawed). It is not about dogs, etc., as the Ohio version reads now. If everybody understood and implemented that meaning, my suggestion could be formulated as I said. However the intended meaning really is
When going to public services, minimize the probability that you can infect others as much as feasible and consider it to be $0$ if in the last two weeks nobody from your household had a contact with a stranger and nobody in the household had any symptoms.
Now it is more to the point, but also more complicated. And if a professional mathematician like myself is so inept, imagine the difficulties of other people.
So, within that model, what would be the best formulation of the order to give?
Second, the set of questions I asked is clearly incomplete.
One has to add for instance "What assumption can be wrong and what effect that will have on the outcome under the condition that the order is given in the currently stated form. I have never seen a book that teaches the influence of the order formulation on the possible model behavior and that may be a crucial thing now. The interaction between the formal logic and differential equations within a given scenario is a non-existing science (or am I just ignorant of something? That reference would really be useful).
Third, if we have a particular question (say, how much to reduce and how organize the public transportation, which is NY and Tel Aviv headache now), what would be a good mathematical model for just that and what would be the corresponding order statement under this model?
The questions like that are endless and if there were ready answers in textbooks, the governments would just implement them already instead of having 7-hour meetings. So I can fairly safely conclude that they are not there.
What I tried with my model example was, in particular, to show that there may be some non-trivial moves in even seemingly optimal situations (strict stay home order and running only the absolutely vital services at the minimal rate that still allows to serve the population) that also make common sense and can be used by everyone right now and right here. Finding such moves can really help now. The main real life question now is "What can I (as government, business, or individual) do to reduce the largest eigenvalue of the social interaction matrix?" Now show me the textbook that teaches that and I'll stop the "ballspitting" and apologize for the wasted time of the people reading all this.
Hmm much as I understand the urge to ask this question, I would have preferred MO to be a "safe space" that I could come to in order to get away from the Coronavirus. And, unlike Gil Kalai's question, I don't see any mathematical connection here.
@Lucia I addressed your doubts and concerns the best I could at 5AM, now back to sleep. You are right, I should have said more in the question formulation. My only excuse is that I don't ask questions too often :-)
Could you add existing mathematical tags? (applied-mathematics? mathematical-biology? graph-theory? pr.probability?) Also creating "epidemics" would have been more useful than "coronavirus".
@MattF Criticism accepted. Will reformat and add details today. Just give me some time.
@MattF Better now? Thanks a lot for constructive criticism. I need it.
I don't think we should artificially create so-called "coronavirus safe spaces" for ourselves, because there is not such a thing! COVID-19 is deeply influencing our life and work in all ways, and it would be absurd to pretend that research mathematics is "safe" from it. Moreover, I think we should all agree that as research scientists we have a social responsibility to put our knowledge to solve real, imminent problems affecting the entirety of mankind.
That said, I agree with @MattF. that the question should be research mathematics focused. Since MathOverflow is a math forum, it would be totally appropriate (and morally correct) to discuss mathematically-oriented methods and solutions to this global crisis.
My immediate instinct as an economist is that any useful model is going to have to incorporate maximizing behavior: People will voluntarily change their behavior in response to the transmission probabilities (to protect themselves, and also, perhaps, because they care about protecting others). So the parameter $M$, for example, should be determined within the model, not taken as a constant.
@StevenLandsburg M is currently defined as the number of people a public servant sees per day, which is an observable for every particular service. You are right, voluntary action is better than an order, but for mathematics the difference between the two manifests only if the interests are opposite and here they coincide. In my suggestion every person implementing the rule would reduce the off-diagonal product and thus the $\sqrt$ part, so it is not all or nothing. What do you think of the analysis itself?
@fedja: I think you've misunderstood. My point has nothing to do with the merits of voluntary action. Instead my point is that voluntary actions, whether desirable or not, are going to happen, and your model should account for them. A public servant who currently sees $M$ clients per day is going to choose to see fewer clients if he believes those clients might infect him, so a higher infection rate should lead to a lower $M$. To take $M$ as given is to deny that workers have ways of slowing themselves down so that they take fewer risks. I think that is not realistic.
@StevenLandsburg You are right: I've been to the grocery store 3 days ago, and cashiers looked nervous. The lower M is, the better, of course. But if the rule is implemented for 2 weeks in a row, the general public will become safe for servants and one would need only to care about servants being served. Just let them wear some badges when going as clients. Anyway, that starts looking like a meaningful discussion. Want to open a chat?
@fedja: I'd love to open a chat, but not today, as I am ten minutes away from teaching my first online class. This should be an adventure!
@StevenLandsburg OK, I'll wait :-)
With respect, there's an enormous literature out there on mathematical epidemiology, and I think digging into that is probably a better course of action than spitballing on MO. For example, "Mathematical Epidemiology", Fred Brauer et al, Springer Lecture Notes vol 1945. Or "An Introduction to Mathematical Epidemiology", Martchava.
@JCK That is totally correct in the sense that there are plenty of experts who are doing exactly that now. However sometimes a non-expert can come with an idea the experts may miss. And that is what I'm trying to do. The question about the literature has been asked by Gil Kalai already and got good answers. It may be a good read when the infection is over. What is needed now is some simple ideas that can help to answer some urgent questions quickly. If the general problem I stated has been considered and solved, I'll sleep much better tonight :-) Upvoting your comment, but continuing.
@xuq01 I am all for online discussion and collaboration to better understand and model the spread of infection etc, but it is a long-standing tenet that MO is not a labspace for conducting research. It could provide a jump-off point for interested parties to make contact and collaborate, but this site is not a forum and it is not Wikipedia
@xuq01 "Moreover, I think we should all agree that as research scientists we have a social responsibility to put our knowledge to solve real, imminent problems affecting the entirety of mankind." I respect your passion but I do not think it is shared universally, and should you succeed in landing an academic career you may feel less than happy to be told you should not work on the academic questions you wish to because other problems are deemed more important to society.
@xuq01 Moreover, there is a certain amount of hubris in seeking to take over from the epidemiologists (as opposed to, say, working with them). Personally I think that right now the problems are less to do with modelling and more to do with data, with politics, and with resources.
@YemonChoi I totally agree with your point and this sentiment. But "you should not work on the academic questions you wish to because other problems are deemed more important to society" - that's not what I mean. If importance to society is the only criterion, I would have become a doctor or an environmental scientist, rather than a mathematician or computer scientist. However, I don't think that devoting a little bit of this site to relevant discussions is too much. Of course I don't mean "everyone should stop doing their current research to work on epidemiology"!
@YemonChoi Also, "taking over from the epidemiologists": no, I totally don't mean that. I think we have a misunderstanding here. I know that MO is not "the right place" for such discussions, but it is totally "a useful place", and we can also make fruitful discussions while keeping ourselves within the scope and the rules of the site. In other words, I support doing "whatever we can" within the scope and rules of MO.
Just for clarity: the row and column indices for the matrix A correspond to categories of people?
"What are suggestions for reducing the transmission rate of the current epidemics?" This is a very broad (but interesting) question. Could you make it more clear what the actual question is regarding the mathematical content. Is it some sort of a question about matrices?
Maybe by good setting in $SIR$ model, we can found the best conditions for public transfer! Anyway, I think your question is interesting.
This is just a slight expansion of my comment.
When the environment changes, behavioral parameters (that is, parameters like $M$ that describe people's behavior --- in this case the behavior of public servants deciding how many clients to serve each day) are going to change. Therefore such parameters should not be taken as constants; they should be determined within the model.
This means we need to be able to predict how $M$ will change in an unprecedented circumstance. Fortunately, we have a lot of relevant data. For example, consider the function $f(p)$ that tells you how much income a person is willing to forgo in order to avoid a probability $p$ of death. At least in the United States, we know (somewhat roughly and with various caveats) that when $p$ is small, $f(p)\approx \$10,000,000\times p$. (Theory predicts, and evidence seems to confirm, that $f(p)$ is linear for small $p$.) We infer this, for example, from the premiums you have to pay people in order to get them to take on dangerous jobs, or from the amount people are willing to pay for safety devices. I'm not sure whether $p\approx 1\%$ counts as small for this purpose, but there are data available that will help decide that.
(This, incidentally, is precisely what economists mean when they say that "In the United States, the value of a life is about $10,000,000".)
The best way to account for all this is to assume that people are maximizing some functionn $U$ which takes as arguments things like income, social interaction, time spent being sick, and probability of death. Try to estimate the function $U$ by observing the choices people make in a great variety of ordinary circumstances. In other words, observe their behavioral parameters in ordinary times, assume that those behavioral parameters are the solutions to some maximization problem, and try to infer what's being maximized.
Now when the pandemic comes along, you've got to assume that something is unchanged; otherwise you have no basis to make any predictions whatsoever. The idea is to assume that what's unchanged is the maximand $U$, and that the pandemic represents a change in the constraints subject to which people are trying to maximize. Having estimated $U$, having assumed it's fixed, and writing down the new constraints, you can calculate the new behavioral parameters that result from the new maximization problem.
Now your model is essentially a fixed point problem: Behavioral parameters (that is, the solutions to the maximization problem) cause changes in behavior, which cause changes in the way the pandemic spreads (that is, the constraints on the maximization problem), which cause changes in behavioral parameters. The solution to the model is a fixed point of that process.
You can also estimate how the fixed point will change if you add additional constraints, such as penalties for going outside, prohibitions on meetings, etc.
This sort of modeling is what economists try to do all the time. I expect, but do not know, that one could say the same of epidemiologists. No model is perfect, but economists have learned the painful lesson that some models are a lot less perfect than others, and that fixed behavioral parameters are generally a hallmark of such models.
OK, but as long as the infection curve goes up, $M$ won't be voluntarily increased by anyone, and since all entries of $A$ are nonnegative, the current $M$ whatever it is is the worst case scenario, so reducing $\lambda(A)$ is the first priority. But you made an excellent point that when we reach a descending part, the people may want to increase $M$ and ruin the game. One solution is to issue a decree to not increase the operation rate (which is what $M$ is) for any business, but if you can solve this with incentives rather than regulations, that would be an excellent thing.
I expanded my comment to you in a separate thread :-)
Here is another model from Gil Kalai last idea:
We have the moving compressible fluid in which the particles can teleportate, diffuse, and organize the motion, while contamination can only diffuse. Each particle can be contaminated or not. You see contamination levels but not individual contamination. How to move?
The answer is that unorganized teleportation should be excluded (uncontrolled private cars), diffusion should be reduced (Israeli 100 meters from home rule for walking out), the movement of all particles should be towards the higher contamination density until the maximum allowed density of the fluid is reached there (meaning sending goods only from warehouses in less contaminated areas to the shops in more contaminated areas, but never the other way around, sending the hospital cases to the hospitals in the highest contamination areas until full capacity is reached there, the doctors working in the nearest to home hospitals whenever the switch is possible, etc, etc. Has this been implemented anywhere?
Now the next question: how to let China return to the normal mode and provide goods for the rest of the world? That would require more than hemotaxis in the model: we need a flow. It cannot be maintained forever, of course, but it is possible for quite a long time. What would that mean in practical terms? First, no one can enter China (they believe their cases now are just "imported", but why do we need any import? BTW, no question is rhetorical now). No shipments of goods should be sent to China; all hospital cases should be moved to the closest hospital to where they get most cases at the moment (by military helicopters or planes if it is not 911 and they can afford now to test every symptomatic case and move it), and the areas that should be opened to the relaxed life should be furthest from the areas with cases. In other words, if the cases are imported, they should stay at the border, the doctors treating them should stay at the borders, and somewhere along the borderline a deserted area should be created, so China will split itself into productive inland and defensive outskirts. In the fluid terminology, in addition to hemotaxis on the patch boundary towards the contamination, there should be hemotaxis away from the boundary inside until the patch separates into the outer layer and the inner bulk with minimal density in between (transit transport only, no real inhabitance).
How much of that is feasible is another question. But does it make sense?
Response to Steven Landsburg:
The main point Steven made was that M is at least as important as $A$.
How to reduce M? (the interaction between public servants and ordinary people)? That, as Steven correctly pointed out, is in the hands of people. The servants just serve what you need. If you reduce what you need, you can fill your supplies to the household capacity less frequently and thus reduce $M$ locally in the matrix $A$ (at last I have enough preliminary explanations to speak in the language in which I think). Example: I eat mainly grain now (bought a few bags at Costco some time ago). That produces zero garbage emission, so my garbage bin fills slowly and I can put it out, probably, 5 times less frequently, so most of the time the truck bypasses my house and I am not afraid of contaminated garbage been 5 times longer than an average person, so I have an egoistic incentive to do it as well. Besides, I just have no supply of required garbage bags :-) ). That potentially reduces M five times for the garbage pickup part of $A$. And that is where the virus teleports: the garbage trucks go everywhere. But it teleports at the speed of car and we can teleport ideas at the speed of light (Internet). The proposed modus operandi for propagation of thought is exactly what I posted in the beginning: share your thoughts on what to do in the widest network available to you where you have enough reputation to be listened to. We'll fight the car teleportaion of virus with the light speed teleportation of ideas. That is not even restricted to country: explain to your friends everywhere what is going on, what's the global strategy, why you are doing what you are doing to support it, and talk to them in the common language between you and them trying to keep it as simple as possible, but not simpler than that. Everyone is free to implement the rules that are not weaker than the current regulations in his household (I follow max(Israel,Ohio)) and think how to advance them more. That's why the government does not impose the draconian regulations itself. They hope that graduate reduction will be met by gradual self-restriction and we'll bargain not at the strict line but on a curve allowing freedoms where they are needed and using freedoms only if they are needed. That's the advantage of democracy, used properly. The advantage of dictatorship is the speed of bargaining. What we'll get depends on how we act.
If you get lost, please don't downwote, just ask questions :-)
Here is a suggestion that may get downvoted. I make it not because it is a good idea, but because some modification may lead to a good idea.
Get everyone sick to get better.
If you hunt down my WordPress blog (grpaseman) you will see a mild expansion of this idea. The crux is to interrupt the replication of SARS-CoV-2 by introducing a different virus whose systemic effects are known , are not fatal, but use the same resources that SARS-CoV-2 would use to replicate. This might delay replication inside a host long enough for the immune system to manufacture a response.
The idea has problems and needs thinking through. If it suggests a better path to interrupt replication at a different level, it will be worth all the down votes.
Gerhard "Getting Downvotes To Inspire Others" Paseman, 2020.03.26.
Right now it is exactly what is going on but in a different setting: the humanity, as a whole, uses the internet to spread the information about the existence of virus and enforcing the long-distance behavior of communities (cells in biological language). The main problem is that the body (community) reacts only if it experiences pain (anxiety) in the immediate physical (geographical) vicinity. If we could make pain (anxiety) felt everywhere at once but at the safe level, the body (society) and pass the blueprint, the body (society) would react way faster.
So, if we could just transfer the pain (generically understood) and the virus blueprint from lungs to other organs as well and start the production of antibodies everywhere at the same time that might help. Just my crazy 2 cents.
The modification is exactly that: you do not need to or should use the same resources. All you really need is to get the antibodies that fight both. If you use the same resources, you will just kill more cells in the lungs and there will be zero advantage. Am I making sense now?
The exact reformulation of the OP idea that corresponds to what I wrote: Get everybody who is better to feel sick. Alas, that is exactly the idea of the vaccination but vaccination uses the same blood network as the virus. Can we use the neural network somehow in the body to magnify the pain in the lungs? That would not lead to cure but it will lead to quick and sure detection and the result should be different from cough, of course. Some symptom like you have for no common illness. Now the suggestion is complete and I wrote the whole train of thought. We can just inject a self-test this way.
This post and especially the comments is firmly in "not even wrong" territory.
Mathematically speaking it looks that if we split the categories $C_i$ into subcategories with zero (or small) cross transmission edges then this suppress the exponential growth at some scale. (And this is also the case for other geometric limitation on the graph of possible transmission, e.g if there are no edges for people who arge geographically remote). In technical terms one need to make the transmission graph (and every large subgraph) a very "bad" expander.
You can also separate in time and turn the blocks on and off one after another. If you could do it with 16 blocks (i.e., only some particular 1/16th of the population is out running the services and the rest stays at homes (especially strictly on their days 2-5 from their day to be out when they can be stealth virus carriers) .on every given day and can only call for medical emergency response), we would be done. I mean, make $A(t)$ depending on $t$ turning on one block a time on any given day and forcing the rest to be identity. That would sort of raise $A$ to the power $1/16$.
Moreover, the time splitting may work no matter what the actual dynamics is. Misha Sodin told me that the Israeli army is already doing something like that, apparently, letting people in group $C_i$ out to do the work on day $i$ and educating the rest how to run necessary services when they are locked in (if I understood him right). In principle, this model can be applied to the whole population.
The time splitting is also better because we have no guarantee that each particular block $C_i$ has a smaller largest eigenvalue than the whole matrix, and every geographical block is infected by now. That would be "the" solution, if only it were feasible... Thanks for the response, BTW :-)
Dear fedja, this may mean that not allowing (non essential) people to use cars is essential. (In some cases public transformation is stopped but there is no limitation on private cars.)
Dear Gil, I tried to bring your idea about cars to its logical extreme. See if it makes sense.
I have one rather practical answer, which does not answer the question in the body, but possibly in the title.
I was very surprised to learn that a (small) hospital in my home town was doing their scheduling ("rostering" might be the correct term, but I don't know) by hand.
They were clever enough to ask for help just in time. Again surprisingly: informally via a mathematician they knew privately. Thus, they now have a professional university team developing a plan that fits their new needs - minimal service, minimizing the probability that a large portion of their staff simultaneously drops out.
I guess that such a strategy might also make sense for other community services, who might not even be aware of this fact.
(Disclaimer: I am not helping with the implementation, but I was lucky to know someone who knew someone who was a professional in this field, who luckily agreed to do it.)
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2025-03-21T14:48:30.131730
| 2020-03-23T08:47:11 |
355496
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Stack Exchange
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What is the geometric shape of the Monster sporadic group?
Conway made the comment that the Monster group represents the symmetries of a shape in 196,883 dimensions, something like a "star you hang on a Christmas tree."
My question is, What do we know (or conjecture) about the enigmatic shape whose symmetries are captured by the Monster? Is the shape analogous to a star (pointy) or convex like a polygon / polytope, or is it more likely to be something with a different topology, analogous to a high-dimensional torus?
Conversely, what do we know it is not like, what have we ruled out?
P.S. Also, if anyone knows Conway, please wish him well, and maybe they might ask him this question?
Update May 2020 RIP Mr Conway
I don't know what Conway meant, but when you have a faithful representation, then you can built an orbit polytope in that dimension with the monster group as its symmetry group. If its a real representation, then its a convex polytope, otherwise its more like a set of point in a complex vector space.
Here is described a "Monster graph" whose automorphism group is the Monster group (described by Griess). Maybe this graph could be the 1-skeleton of a polytope in 196883 dimensions? http://www.neverendingbooks.org/the-monster-graph-and-mckays-observation
It is possible that Conway was referring to the generic construction that works for all finite groups equipped with faithful representations, given in the other answers. However, I think it is more likely that Conway was referring to a construction that is specific to the monster, hinted at in Ian Agol's comment.
In section 14 of Conway's 1985 Inventiones paper, "A simple construction for the Fischer-Griess monster group", Conway points out that in the 196883 dimensional representation, there is a collection of distinguished lines fixed by centralizers of elements in conjugacy class 2A (in fact, 196883 decomposes as 1+4371+96255+96256 under the action of such a centralizer). Elements in this conjugacy class are known as Fischer involutions or transpositions, and the lines are called axes of transpositions. The centralizer of a transposition is a double cover of Fischer's Baby Monster sporadic group, so there are $\frac{|\mathbb{M}|}{2|\mathbb{B}|} \sim 9 \times 10^{19}$ of these axes. Choosing a nonzero point on an axis, and taking its orbit yields an arrangement of points (or we may consider the convex hull). I claim that this arrangement has Monster symmetry.
To prove this, we use the fact that for a pair $(x,y)$ of 2A elements, their product lies in 2A if and only if the angle between the axes is a particular value. In fact, the conjugacy class of the product is uniquely determined by the angle, except for classes 3C and 4B, as we can see from the "McKay E8 diagram" in Conway's paper:
We therefore have a natural construction of a graph from this polytope, whose vertices correspond to axes (or orbit elements on the axes), and whose edges correspond to those pairs whose product lies in class 2A. This graph is the "monster graph" mentioned in Ian Agol's comment, and Griess showed that its automorphism group is precisely the monster.
These axes appear in the theory of vertex algebras, in the following way. The 196883 dimensional representation is naturally embedded as the space of Virasoro primary vectors in the weight 2 subspace of the "Moonshine Module" vertex operator algebra $V^\natural$. On each axis, there is a distinguished "Ising vector" that generates a vertex algebra isomorphic to the $L(1/2,0)$ minimal model. Miyamoto showed that any Ising vector in a vertex algebra yields a "Miyamoto involution", and for $V^\natural$, these are precisely the 2A elements.
Ooh, I much prefer this construction! The minimal permutation representation of the Monster has the same number of elements<PHONE_NUMBER>2009186000) as there are vertices in your polytope, so your polytope is (in a certain sense) the 'simplest' polytope with Monster symmetry. Elegant!
This sounds fascinating, but, to a physics-educated layman, what is the nature of the shape of this "monster graph" / polytope? Conway referred to it as like a "star" you hang on your Christmas tree. Is there anyway of describing it in non-jargon? Is it a simple polytope in many dimensions, like a high-D tetrahedron? Or more intricate?
He said, why is it there? He had something SPECIAL in mind.
I can't visualise the shape from the jargon, I'm afraid, even though I do enjoy the jargon :)
@JamesEadon Fundamentally, it is a collection of about $10^{20}$ points in 196883-dimensional space. If you had five points symmetrically arranged in the plane, you could choose to connect them like a pentagon, or like a star (or perhaps try some more exotic thing). Similarly, there are options here, but I do not have the vocabulary to describe them. Most people would choose the analog of "pentagon", i.e., the convex hull, because it is conceptually simpler.
That makes a lot of sense, thanks. Are the points regularly spaced, e.g. like a regular hexagon, or a dodecahedron, etc?
Also, you mention a "pentagon". Does the number 5 turn up as a symmetry of the shape, or was "pentagon" just a pure analogy?
@JamesEadon The pentagon was an analogy that was useful because of the link to star shapes. The points are very regularly spaced, in the sense that the shape has lots of symmetry.
@S.Carnahan: Do these points form $196883-1$ dimensional simplexes?
@Rudi_Birnbaum Assuming you are asking whether the boundary of the convex hull decomposes into flat 196882-simplices joined to their neighbors at nonzero angles, I do not have an answer to that question.
@S.Carnahan, thanks for precision of the question. I just learned its called facets. So maybe then like that: Are the facets of the convex hull regular simplices? Maybe its worthwhile finding out.
Can we give the Schläfli symbol for this polytope?
@Rudi_Birnbaum If such a symbol could be given, it would require 196882 elements.
@S.Carnahan, yes exactly!
I have found an interesting quote: "The Mathieu groups M${11}$ and M${22}$ are not automorphism groups of any polytopes.". Is this in contradiction to what has been written above?
@Rudi_Birnbaum I cannot read the article because of the paywall. However, given the name of the article, I suspect the author is restricting his view to regular polytopes, and didn't bother to include the word "regular" in that particular sentence.
@Rudi_Birnbaum I couldn't find that sentence in the linked article, but it does appear in this one. S. Carnahan is correct that the author is discussing regular polytopes, but they are also "abstract", which refers to a very different concept. Of course, many important lattices such as the Leech lattice include $\mathrm M_{11} ! $ as a subgroup of their automorphism group.
@Rudi_Birnbaum I was inspired by your comment to construct a polytope with an automorphism group of $\mathrm M_{11} \hspace {-1pt}$. It consists of $66$ vectors in $\mathbb Z^{11} ! $. I found some truly interesting things along the way. I've posted a lengthy write-up here.
@Rudi_Birnbaum Unfortunately, the answer I provided was incorrect, but the feat is in fact possible. I have turned the post linked in my previous comment into an open forum for constructing simple polytopes with an automorphism group of $\mathrm M_{11} \hspace {-0.5pt} $, where "simple" can be interpreted in number of equally valid ways.
@S.Carnahan how do you know that the convex hull doesn’t have any additional edges?
@DanielSebald I'm pretty sure the convex hull has many additional edges. To construct the graph, I am restricting my attention to those edges that come from certain central angles.
The bit.ly link in a comment above points to a PDF, uploaded on ResearchGate, of the article Polytopes Derives from Sporadic Simple Groups by Michael I. Hartley and Alexander Hulpke (January 2010, Contributions to Discrete Mathematics, 5:106-118). Just posting this in case the URL shortener ends up breaking in the future.
In the penultimate chapter of Sphere Packings, Lattices and Groups, the authors define a $196884$-dimensional real vector space and a faithful representation of the Monster group on that space.
Now, because we know the degrees of the irreducible representations of the Monster, this representation must necessarily decompose as the direct sum of a trivial 1-dimensional representation and a faithful real $196883$-dimensional representation.
Choose an vector $v$ in that $196883$-dimensional subspace in general position, and normalise it to have unit length. Let $X$ be the orbit of $v$ under the action of the Monster; it follows that $X$ has the same number of elements as the Monster. Moreover, $X$ is a subset of the unit sphere.
Let $P$ be the convex hull of $X$. Then $P$ is a $196883$-dimensional sharply vertex-transitive convex polytope with symmetry group isomorphic to the Monster group.
EDIT: S. Carnahan's answer provides a more elegant construction, taking $v$ to be a point on a line fixed by a Fischer involution instead of a point in general position. The resulting polytope has<PHONE_NUMBER>2009186000 vertices, which is minimal (as any permutation representation of the Monster has at least this many vertices).
I just want to point out that the claim $\mathrm{Aut}(P)\cong M$ (where $M$ is the monster group) is far from obvious. It can be proven though.
Hi Adam, thanks for answering. You say, "sharply vertex-transitive convex polytope" - I understand convex polytope as a higher-dimensional hexagon, say. But what does "sharply vertex-transitive" mean, in language a university MSc. physics / computer science grad can understand, in the context of the Monster?
Basically, I'm trying to visualise the shape of the monster in terms of a 2D or 3D analogue. Is it like a high-dimensional regular hexagon or similar? It's not a completely regular polytope, is it?
@JamesEadon If you did the same thing with the alternating group A5 and its 3-dimensional irreducible representation, you'd get a [not necessarily uniform] snub dodecahedron (see https://en.wikipedia.org/wiki/Snub_dodecahedron for a picture). 'Sharply vertex-transitive' means that, given any ordered pair $(u, v)$ of vertices, there's a unique element $g$ of the symmetry group such that $g(u) = v$.
@JamesEadon to me it seems it cant be a $196883$ dimensional simplex, because that would have the symmetry of the fully symmetric group of the $97239461142009186000$ vertices ($S_{97239461142009186000}$), I suppose.
@Rudi_Birnbaum I was thiking myself about the symmetry. The monster group has a fair bit of structure, I believe, implying (I'm guessing) that the shape might not be as simple as some kind of a high-D simplex.
It is not too surprising that the Monster group $M$ is the symmetry group of something geometrical.
E.g. every group is the group of symmetries of some convex polytope.
You can even make it a vertex-regular polytope $P$, which means that $\mathrm{Aut}(P)\cong M$ acts regularly on the vertices of $P$. So the vertices of $P$ can be set in one-to-one correspondence with the elements of $M$.
A convex polytope is not much more than its set of vertices, and if you are okay with that point of view than it is easy to construct from any faithful representation a discrete set of $>10^{20}$ points in $>$ 196883 (complex) dimensions for which the (unitary) symmetry group is isomorphic to $M$.
Here are the details
Let $\rho:M\to \mathrm O(\Bbb R^d)$ be some irreducible faithful orthogonal representation of the monster group $M$.
Then $\Gamma:=\mathrm{im}(\rho)\subset\mathrm O(\smash{\Bbb R^d})$ is a matrix group isomorphic to $M$.
For every point $p\in\smash{\Bbb R^d}\setminus\{0\}$ we obtain the so called orbit polytope
$$P:=\mathrm{Orb}(\Gamma,p):=\mathrm{conv}(\Gamma p)\subset\Bbb R^d,$$
that is, the convex hull of the orbit of $p$ under $\Gamma$.
This is a $d$-dimensional convex polytope and clearly has $\Gamma\subseteq\mathrm{Aut}(P)$.
One can show [1] that for an appropriate choice of $p$ this gives equality, that is $\mathrm{Aut}(P)\cong M$.
Now, your representation might not be real, but complex valued. But from that we can get a real representation of twice the dimension (and therefore a convex polytope of twice the dimension).
Alternatively, you can let $\rho:M\to\smash{\mathrm U(d)}$ be a unitary representation, $\Gamma:=\mathrm{im}(\rho)\subset \mathrm U(n)$ as well as $p\in\smash{\Bbb C^d\setminus\{0\}}$.
In that case the orbit $\Gamma p\subset\smash{\Bbb C^d}$ does not yield a convex polytope but just a discrete set of points in a complex vector space.
Again, it is possible [2] to choose $p$ appropriately so that
$$M\cong \mathrm{Aut}_{\mathrm U(n)}(\Gamma p),$$
a set of points in (complex) dimension 196883 with the monster group $M$ as its (unitary) symmetry group.
[1] Babai, László. "Symmetry groups of vertex-transitive polytopes."
[2] Friese, Erik. "Unitary Groups as Stabilizers of Orbits"
Which kind of "set of points" is $X$ in the unitary case? In which sense it is "196883-dimensional"?
@Qfwfq It is a set of points in $\Bbb C^d$, so 196883 complex dimensions.
(Ah, yes, it's just a discrete set. For some reason I didn't realize you don't make an analogous operation to the convex envelope in the unitary case.)
Dear M. Winter. The jargon does give me a feel for the amazing and beautiful mathematics involved. It's mind-blowing actually. If I may, please can you translate your answer into something understandable by a MSc Physics / Computer Science grad? Is this thing a high-dimensional hexagon or something? I'm just trying to visualise it, in 3D or 2D analogue terms. A weirdly special shape for such symmetry to only exist at such a specific high dimension?
@JamesEadon In essence: yes, there is a high dimensional shape whose symmetries are exactly described by the Monster group. In fact, this shape is a polytope (a 196883-dimensional version of a hexagon, cube, ...). But it is the nature of the Monster group that this polytope exists only in such a high dimension, and all smaller dimensional examples will miss some of the complexity of that group. This polytope is nothing special though, as something like this exists for almost all groups. It is also not regular, or otherwise special besides its relation to the Monster group (as far as I know).
|
2025-03-21T14:48:30.132945
| 2020-03-23T09:13:19 |
355498
|
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|
Stack Exchange
|
Simple proof that the arithmetic genus is non-negative
I take an irreducible and reduced closed curve $C\subseteq \mathbb{P}^n$, defined over an algebraically closed field $k$ and define the arithmetic genus $p_a(C)$ as the integer such that the Hilbert polynomial of $C$ (or of its ideal of polynomials vanishing on it) is
$$h_C(X)=\deg(C) X+1-p_a(C).$$
I would then like to have a simple proof that $p_a(C)\ge 0$. Of course, one may say that it is because it is the dimension of a vector space obtain using cohomology, but I would like a proof not involving cohomology, as it is for a master work and in my course of algebraic geometry I (unfortunately you will say) did not define the cohomology. It would then take a lot of time for the student to understand the cohomology just to prove this.
In the course, we defined the Hilbert Polynomial of any ideal, computed local intersections, Bézout theorem in $\mathbb{P}^n$ and studied blow-ups of surfaces. We also proved that $p_a(C)=g(C)$ when $C$ is smooth, where $g(C)$ is given by Riemann-Roch (the smallest integer such that $\ell(D)\ge \deg(D)+1-g$ for each divisor $D$ on $C$). I would like the student to use the fact that $p_a(C)\ge 0$ to bound the type of singularities of a plane curve and to show that one can have a resolution by blowing-up the singular points and repeating this process finitely many times.
Thanks for your help.
Mumford (Complex Projective Varieties, section 7) has the following, reasonably simple proof.
Let $d$ be the degree of $\mathrm{C}$, $m$ big enough such that $h_{\mathrm{C}}(m)=\mathrm{dim}_{\mathbf{C}} (\mathbf{C}[\mathrm{T}_0,\ldots,\mathrm{T}_n]/\mathrm{I}(\mathrm{C}))_m$ and $md/2>p_a$. Embed $\mathrm{C}$ into $\mathbf{P}^N$ by the degree $m$ Veronese embedding; let $\mathrm{L}\subset\mathbf{P}^N$ be a linear space containing $\nu_m(\mathrm{C})$ such that $\nu_m(\mathrm{C})$ is nondegenerate in $\mathrm{L}$. Then $\dim(\mathrm{L})=h_\mathrm{C}(m)=md+1-p_a$, in other words $\mathrm{L}\simeq\mathbf{P}^{md-p_a}$, and the degree of $\nu_m(\mathrm{C})$ is $$md=\mathrm{deg}(\nu_m(\mathrm{C}))\geqslant\mathrm{codim}(\nu_m(\mathrm{C}))+1\geqslant md-p_a$$
since $\nu_m(\mathrm{C})$ is nondegenerate in $\mathrm{L}$.
Thanks. It is a nice proof that I did not know.
Maybe what follows should be modified depending on what is exactly assumed in the question; nevertheless, let me try.
We fix $m$ positive integer large enough so that $h_C(m)=dim H^0(C,\mathcal{O}_C(m))$. We have to show that $dim H^0(C,\mathcal{O}_C(m)) \leq deg(D)m +1$.
Let $H$ be a degree $m$ hypersurface of $\mathbb{P}^n$ intersecting $C$ in finitely many points of $C$, all smooth (such $H$ exists as $C$ is reduced and so has finitely many singular points). One can view the intersection $D:=H \cap C$ as an effective Weil divisor on the smooth part of $C$, we have $D=\sum_i a_i p_i$ with $p_i$ smooth point of $C$ and $a_i$ positive integer multiplicity. We have $\sum_i a_i =deg(C)m$.
The equation defining $H$ restricted to $C$ defines a rational trivialization of the line bundle $\mathcal{O}_C(m)$, singular along $D$. Therefore, sections of $C$ can be identified with rational functions $f$ on $C$ such that $div(f)+D \geq 0$ (remark that this makes sense as $D$ is made of smooth points of $C$). Concretely, $f$ is allowed to have a pole of order at most $a_i$ at the point $p_i$. Near each point $p_i$, fix a local coordinate and consider the polar part of the Laurent expansion of $p_i$ (terms with negative power of the local coordinate): it consists of $a_i$ coefficients. If $f$ and $g$ have the same polar parts at every $p_i$, then $f-g$ is a regular function on $C$ and so a constant as $C$ is reduced and irreducible. Therefore the kernel of the map from $H^0(C,\mathcal{O}_C(m))$ to the space of polar parts is one-dimensional and so $dim H^0(C,\mathcal{O}_C(m)) \leq (\sum_i a_i)+1=deg(C)m+1$.
Thanks for the proof. I have to try to see if I can use it with what I did in the course.
|
2025-03-21T14:48:30.133246
| 2020-03-23T09:48:58 |
355499
|
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|
Stack Exchange
|
Bounds on Fourier coefficients for $GL(3)$
I am referring for instance to this question about coefficients of automorphic forms on $GL(3)$. I know that the Ramanujan on average bound is known and gives
$$\sum_{n^2 m < x} |\lambda(n,m)|^2 \ll x^{1+\varepsilon}.$$
Is there anything known (in terms of upper bounds, with explicit dependence in the fixed $m$ or $n$) about the partial sums (except trivially bounding by the above) :
$$\sum_{n^2< x} |\lambda(n,m)|^2 \qquad \text{and} \qquad \sum_{m < x} |\lambda(n,m)|^2 ?$$
On the Ramanujan conjecture $\lambda(m,n) \ll 1$. As there is no cancellation in the second sum, essentially (upto $x^\epsilon$) the best upper bound which one may expect for that is $x^{1+\epsilon}$.
For the first sum for the same reason the best possible bound would be $x^{1/2+\epsilon}$. To prove that note that $\lambda(n,m)=\overline{\lambda(m,n)}$. So
$$\sum_{n^2<x}|\lambda(n,m)|^2=\sum_{n<\sqrt{x}}|\lambda(m,n)|^2\ll x^{1/2+\epsilon},$$
using the bound from the second sum.
But how do you get the bound on the second sum? (for now it seems a bit circular)
I thought that $n$ in the second sum is fixed, say $n_0$. Then the second sum is bounded by $\sum_{n^2m<n_0^2x} |\lambda(n,m)|^2\ll_{n_0} x^{1+\epsilon}$ from the first estimate in your question.
But this is wasteful since we have a rough $n^2$ bound on the right then, which is a loss compared to Ramanujan conjecture. Is there any way to do better?
If you are asking for Ramanujan on average uniformly in $n$, to prove that it will be as hard as proving Ramanujan: Let $n$ be any large prime bigger than $x$ so that $(m,n)=1$ for all $m<x$ so that $$\lambda(n,m)=\lambda(n,1)\lambda(1.m).$$ If we have $$\sum_{m<x} |\lambda(n,m)|^2\ll x^{1+\epsilon}$$ uniformly in $n$ then using the fact that $$\sum_{m<x}|\lambda(1,m)|^2\gg x^{1-\epsilon}$$ we obtain $$\lambda(n,1)\ll x^\epsilon < n^\epsilon.,$$ i.e. Ramanujan.
|
2025-03-21T14:48:30.133411
| 2020-03-23T10:01:50 |
355500
|
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|
Stack Exchange
|
Notation of $P^+$-families - bibliography searching
have you ever met with notation of $P^+$-families in other papers than Iian B. Smythe "A local Ramsey theory for block sequences" and his phd?
Thank you in advance
$P^+$-filters are quite common in the literature. Can you provide the definition of a $P^+$-family?
Thank you for the answer. Up to my knowledge, a family $A$ is "standard" $P^+$ - family if each countable subset of $A$ has pseudointersection in $A^+$. But I'm looking for a non-standard understending of this notation. The reason is following: I've written a paper with the definition (in short): let us call {\it contur} first iteration of Fr'{e}chet filter (by Frolik sum = tensor produkt =..) The family $A$ with fip {\it is not quasi finer than a contur} if for each countable family $B$, such that $A \cup B$ fas fip, there is no contur $C$ such that $\langle A \cup B \rangle \supset C$.
I've send a manuscript to IJM and the referee write, that this definition is well known as a $P^+$-family. Since obviousely those definition describes different classes of families, thus I'm looking "non-standard" definition of the $P^+$-family. By the way, if you would like to look at the paper an old version is on Arxive https://arxiv.org/pdf/1803.03862v3.pdf (with one incorectness and a different notation: for "is not quasi finer than a contur" is used "is not EQ-subbase of the contur"), a new wersion I'll add to Arxive soon.
Best regards
|
2025-03-21T14:48:30.133566
| 2020-03-23T10:56:21 |
355505
|
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"sound wave"
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|
Stack Exchange
|
Solve a 2-dimensional optimal control problem via Riccati nonlinear equation
Consider the 2-dimensional optimal control problem of the LQR kind
$$
\min_u \int_0^\infty (x^T Q x + u^TRu) \, dt \quad\text{such that}\quad \begin{cases}\dot x(t) = Ax(t)+Bu(t) \\ x(0) = \begin{pmatrix}1\\-1\end{pmatrix}\end{cases}
$$
with $x=\begin{pmatrix}x_1 \\ x_2\end{pmatrix}$, $u=\begin{pmatrix}u_1\\u_2\end{pmatrix}$, $Q=B=I$ (identity), $R=\gamma I$, $A=\begin{pmatrix}-\alpha & \alpha\\\beta & -\beta\end{pmatrix}$, $\alpha>\beta>0$ and $\gamma>0$.
Solve it using the stationary Riccati equation
$$
0 = Q+A^TS+SA-SBR^{-1}BS,\quad \text{with}\quad S=\begin{pmatrix}s_1 & s_2 \\ s_2 & s_3\end{pmatrix}.
$$
In order to find $x$ and $u$ we have to combine the equation for $\dot x$ with the equation for the optimal control $u = -R^{-1}B^TSx = -\frac1\gamma Sx$. So, firstly, we have to find the expression for $S$.
The Riccati equation is simplified as
$$
0 = I+A^TS+SA-\frac1\gamma S^2
$$
which is equivalent to the following nonlinear system
$$
\begin{cases}
-\dfrac{s_1^2}{\gamma}-2\alpha s_1 -\dfrac{s_2^2}{\gamma}+2\beta s_2+1=0\\
\alpha s_1 - \alpha s_2 - \beta s_2 + \beta s_3 - \dfrac{s_1s_2}{\gamma}-\dfrac{s_2s_3}{\gamma}=0\\
-\dfrac{s_2^2}{\gamma}+2\alpha s_2 - \dfrac{s_3^2}{\gamma} - 2\beta s_3+1 = 0
\end{cases}
$$
How to solve such a non-linear system to find expressions for $s_1, s_2$ and $s_3$? I think there is a fast way to solve either the system or directly the Riccati equation in matrix form, but I don't know how.
The solutions provided by Matlab, using the code below, are so long that it let me think is not the correct way to solve the problem
syms x y z a b g
eqn1 = 0 == -x^2/g-2*a*x-y^2/g+2*b*y+1;
eqn2 = 0 == a*x-a*y-b*y+b*z-x*y/g-y*z/g;
eqn3 = 0 == -y^2/g+2*a*y-z^2/g-2*b*z+1;
[x,y,z] = solve([eqn1, eqn2, eqn3], [x, y, z])
The solution of a Riccati equation can be found by determining the eigenvectors of the Hamiltonian; see e.g. here on Wikipedia. This is your best hope for a closed-form symbolic solution, in my view. In your case, the Hamiltonian is
[
\begin{bmatrix}
-\alpha & \alpha & -\gamma^{-1} & 0\\
\beta & -\beta & 0 & -\gamma^{-1}\\
-1 & 0 & \alpha & -\beta\\
0 & -1 & -\alpha & \beta
\end{bmatrix}.
]
I don't see immediately how to diagonalize it, but maybe someone here has a more trained eye than me.
Thanks, I did not know about this method. Calling H the Hamiltonian matrix, the diagonalization can be performed in matlab with [V,D] = eig(H), where V is the matrix of eigenvectors and D the diagonal matrix of eigenvalues. However, also in this case the values computed are very long.
Using V and D, how can I construct the matrices $U_1$ and $U_2$ cited in the page you linked? In particular, see https://i.imgur.com/WbfKw7I.png
$\begin{bmatrix}U_1 \ U_2\end{bmatrix}\in\mathbb{C}^{2n\times n}$ ($n=2$ in your case) is obtained by concatenating horizontally the $n$ eigenvectors that are associated to eigenvalues in the left half-plane. In (succinct) Matlab, U = V(:, real(diag(D)) < 0). If you have eigenvalues with real part zero, then it is trickier (but that shouldn't happen in your case).
Thank you very much for your help. Problem is that the command real(diag(D)) doesn't work if D contains symbolic variables, does it? Anyway, I computed the eigenvalues for several values of the parameters, respecting the constraints $\alpha>\beta>0$ and $\gamma>0$. If parameters are smaller than 0.5, then eigenvalues are likely complex numbers, otherwise they are real numbers, but in all cases two eigenvalues have negative real parts and the other two have non-negative real parts. In one case two eigenvalues were 0. If all eigenvalues are real, it is easy to compute U and S=U2*inv(U1).
I don't understand why $\begin{bmatrix}U_1 \ U_2\end{bmatrix}$ is a complex valued matrix. On wiki page it is written "Since $H$ is Hamiltonian, if it does not have any eigenvalues on the imaginary axis, then exactly half of its eigenvalues have a negative real part.". Doesn't this mean that if $H$ does not have any complex eigenvalue then half of its eigenvalues are negative? They are assuming that eigenvalues are not complex. so why they talk about "negative real part"?
If the eigenvalues are complex, $U$ (as we constructed it) is complex. If the eigenvalues are real, $U$ is real. That wiki page is assuming that $H$ may have complex eigenvalues, but not imaginary eigenvalues (that's two different things).
Ah ok thanks, so $H$ may have complex eigenvalues, but their real parts have to be different from 0, otherwise they are imaginary.
Assume we have found the expression for $S$, in order to find $x(t)$ I think we have to plug the expression for $S$ inside the optimal control $u(t) = -R^{-1}B^TSx(t)$, and then plug it inside $\dot x(t) = Ax(t)+u(t)$, compute the integral and apply the initial condition $x(0)$, right?
Depending what you mean by "solve the problem", yes, you might have to do it.
I mean to solve the minimization problem, i.e. find $x(t)$ and $u(t)$ such that the initial integral attains the minimum value in such a way that initial condition and eq for $\dot x$ hold
A user found a way (https://mathoverflow.net/questions/355910/spectral-decomposition-of-a-4-times4-real-nonsymmetric-matrix-with-unknown-ele/) to diagonalize the Hamiltonian matrix, do you think it is possible to solve the problem (i.e. find the expression for $x(t)$) using it? I'm trying to compute the inverse of $U_1$ but I cannot find a handy way to do it. Using the user's notation $U_1 = \begin{bmatrix}
-\left(B^{\rm{T}}-\lambda_1{\rm{I}2}\right)\mathbf{w}{1r} &
-\left(B^{\rm{T}}-\lambda_2{\rm{I}2}\right)\mathbf{w}{2r}
\end{bmatrix}$
|
2025-03-21T14:48:30.133926
| 2020-03-23T11:38:59 |
355509
|
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|
Stack Exchange
|
Vanishing theorems on a non-compact manifold
In complex geometry, various vanishing theorems for cohomology
groups of a hermitian line bundle E over a compact complex manifold X have been found.
My question is
Is there some vanishing theorems over a general noncompact complex manifold exist? (Except shose on Stein manifolds)
A complex manifold of dimension $n$ is non-compact if and only if $H^n(X,{\mathcal F})=0$ for any coherent sheaf ${\mathcal F}$ on $X$. This is the only general vanishing result that I know of on non-compact manifolds.
But other than that, there are some other vanishing theorems on non-compact manifolds. For instance, if $X$ is $q$-complete, then $H^r(X, {\mathcal F})=0,\forall r\geq q$.
Or if $X$ is weakly $1$-complete, and $L$ is a positive line bundle on $X$, then $H^{n,q}(X,L)=0$, $\forall q\geq 1$.
In general, for a non-compact manifold, you know nothing about the cohmology groups, they can even be non-Hausdorff.
$H^{q, n}(X, L)$ or $H^{n, q}(X, L)$ in para3?
You're right, thank you.
|
2025-03-21T14:48:30.134039
| 2020-03-23T11:54:35 |
355512
|
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|
Stack Exchange
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The $S$-unit equation for functions on curves
Let $X$ be a smooth projective connected curve over a number field $k$, and let $S \neq \emptyset$ be a finite set of closed points of $X$. The curve $Y = X \setminus S$ is affine, and we denote by $R$ the $k$-algebra of regular functions on $Y$.
The $S$-unit equation for $k(X)$ is the equation $f+g =1$, with $f,g \in R^\times \setminus k^\times$; in other words $f$ and $g$ are two non-constant rational functions on $X$ whose zeros and poles are contained in $S$.
For example, in the case $Y = \mathbb{P}^1 \setminus \{0,1,\infty\}$, the pair of functions $(f,g) = (t,1-t)$ is a solution of the $S$-unit equation. In fact, if $f$ is an homography preserving $\{0,1,\infty\}$ then $1-f$ has the same property, and $(f,1-f)$ is a solution. So there are at least 6 solutions.
Mason proved that there exists an effecitve bound (depending on the cardinality of $S$ and the genus of $X$) on the degrees of the possible solutions $f,g$; see e.g. Zannier, Some remarks on the $S$-unit equation in function fields, Acta Arith. 64 (1993) no. 1, 87--98.
Is it expected that the number of solutions $(f,g)$ is actually finite? Are there methods or algorithms to find these solutions in practice?
I am interested in the following particular cases:
$X=E$ is an elliptic curve and $S$ is a finite subgroup of $E$;
$X$ is a modular curve and $S$ is the set of cusps of $X$.
The following paper might be relevant: https://arxiv.org/pdf/1610.08377.pdf
The set of solutions to the $S$-unit equation for $k(X)$ is finite. Let me explain why. (You can "theoretically" find all solutions, as the finiteness eventually boils down to the "effective" finiteness result of de Franchis-Severi on maps of curves.)
Let $k$ be a number field, let $X$ be a smooth projective geometrically connected curve over $k$, let $S$ be a finite set of closed points of $X$, and let $Y := X \setminus S$. Let $R = \mathcal{O}(Y)$.
Claim. The set of solutions $(f,g)$ of the $S$-unit equation $f+g =1$ for $X$ (with $f$ and $g$ thus in $R^\times \setminus k^\times $ ) is in bijection with the set of non-constant morphisms $Y\to \mathbb{P}^1_k \setminus \{0,1,\infty\}$.
Proof of Claim. Let $(f,g)$ be a solution of the $S$-unit equation in $k(X)$. Then $f:Y\to \mathbb{G}_{m,k}$ is a non-constant morphism such that $1-f$ also defines a morphism to $\mathbb{G}_{m,k}$. Thus $f(Y) \subset \mathbb{G}_{m,k} \setminus \{1\}$. Conversely, if $f$ is a non-constant morphism from $Y$ to $\mathbb{P}^1_{k}\setminus \{0,1,\infty\}$, then $1-f$ is also such a morphism. This concludes the proof. QED
Let $K$ be an algebraic closure of $k$. Note that $Hom_k(Y,C) \subset Hom_K(Y_K,C_K)$. Thus,
to answer your question, we can work over an algebraically closed field $K$ of characteristic zero. (That is, you can as well let $k$ be any field of characteristic zero.)
The finiteness of the set of solutions will boil down to finiteness results for hyperbolic curves.
Let me recall what a hyperbolic curve is. From now on, let $K$ be an algebraically closed field of characteristic zero.
Hyperbolic curves. Let $C$ be a smooth quasi-projective connected curve over $K$. We say that $C$ is hyperbolic if $2g(\overline{C}) - 2 + \#( \overline{C}\setminus C )>0$. Equivalently, $C$ is non-hyperbolic if and only if $C$ is isomorphic to $\mathbb{P}^1_K$, $\mathbb{A}^1_K, \mathbb{A}^1_{K}\setminus \{0\}$, or a smooth proper connected genus one curve over $K$.
We will need the following topological lemma on hyperbolic curves. (For your purposes we really just need that $\mathbb{P}^1_k\setminus \{0,1,\infty\}$ has a finite etale cover of genus at least two. This can be proven by considering $\mathbb{P}^1_k\setminus \{0,1,\infty\}$ as an (open) modular curve and taking a modular curve of high enough (even) level.
Topological Lemma. If $C$ is a hyperbolic curve over $K$, then there is a finite etale morphism $D\to C$ with $D$ a smooth quasi-projective connected curve over $D$ such that the genus of $\overline{D}$ is at least two. (This is obvious if $\overline{C}$ itself is of genus at least two. Thus, we reduce to the case that $C = \mathbb{P}^1_K\setminus \{0,1,\infty\}$ or that $C $ is $E\setminus \{0\}$ with $0$ the origin on an $E$ an elliptic curve over $K$. In these two cases, one can explicitly construct $D$.
Hyperbolic curves satisfy many finiteness properties. One of them is the following version of the theorem of De Franchis-Severi. An integral quasi-projective curve is of log-general type if its normalization is of log-general type, i.e., hyperbolic.
Theorem. [De Franchis-Severi] Let $C$ be an integral quasi-projective curve over $K$ whose normalization is of log-general type. Then, for every integral quasi-projective curve $Y$ over $K$, the set of non-constant morphisms $Y\to C$ is finite.
Proof of Theorem. Note that the normalization $\widetilde{Y}\to Y$ is surjective. Therefore, replacing $Y$ by its normalization if necessary, we may and do assume that $Y$ is smooth. Now, every non-constant morphism $Y\to C$ is dominant and will factor uniquely over the normalization of $C$. Thus, we may and do assume that $C$ is smooth.
Now, we use the Topological Lemma. Thus, let $D\to C$ be a finite etale morphism with $D$ of genus at least two. Let $d:=\deg(D/C)$. If $Y\to C$ is a morphism, then the pull-back $Y':=Y\times_C D$ is finite etale of degree $d$ over $Y$. Since $K$ is algebraically closed of characteristic zero, the set of $Y$-isomorphism classes of finite etale covers $Y'\to Y$ of degree $d$ is finite. Thus, we may and do assume that $C=D$. Now, note that every non-constant morphism $Y\to C$ extends to a non-constant morphism $\overline{Y}\to \overline{C}$. However, there are only finitely many such maps as $\overline{C}$ is of genus at least two. QED
Remark. In the last paragraph of the previous proof we use the finiteness theorem of de Franchis-Severi for compact connected Riemann surfaces of genus at least two. (It just happens to be that this "compact" version implies the analogous "affine" version. This is no longer true in higher dimensions.) The "compact" finiteness result also holds in higher dimensions: if $C$ is a proper variety of general type and $Y$ is a proper variety, then the set of dominant rational maps $Y\dashrightarrow C$ is finite. This was proven by Kobayashi-Ochiai. (You can use this to show that, for every integral quasi-projective variety $Y$ over $K$, the set of non-constant morphisms $Y\to \mathbb{P}^1_K\setminus\{0,1,\infty\}$ is finite.)
Thank you Ariyan! This not only answer my question but also gives interesting information and directions. I must look at de Franchis-Severi for compact R.S. of genus at least 2 (how does one prove that the set of non-constant maps $\overline{Y} \to \overline{C}$ is finite? I only know the degree and ramification is bounded). Also, when you say "effective" finiteness, do you mean bounding the cardinality of the solutions, or the degrees/heights of the solutions?
You are welcome François. Bounding the degree is one part of the proof (it tells you that the Hom-scheme parametrizing morphisms from $\bar{Y}$ to $\bar{C}$ is of finite type). However, one can show that a finite morphism $f:C\to D$ of higher genus curves is rigid. I can recommend p. 227 of Mazur's Arithmetic on Curves https://projecteuclid.org/download/pdf_1/euclid.bams/1183553167 . If you need a reference for the finiteness statement I proved above: https://projecteuclid.org/download/pdf_1/euclid.dmj/1077303201
Concerning effectivity: one can bound the number of such maps as well their degrees, and there should be papers on this . One can even "theoretically" find all of them. One can namely write down equations for the Hom-schemes that play a role.
Bounding the number can even be done in higher dimensions; see Theorem 2.3 of https://arxiv.org/pdf/math/0311086.pdf In Section 3 the author of that paper discusses effective versions of de Franchis-Severi.
Thank you very much for the references and additional information. I see, if the Hom-scheme is finite then it is possible in theory to find all of its points.
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2025-03-21T14:48:30.134619
| 2020-03-23T11:55:46 |
355513
|
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|
Stack Exchange
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What is a geometric construction corresponding to elliptic curve addition for Poncelet's Porism?
Background
At least since Griffiths and Harris [1] we know that the geometric construction "draw the next tangent" appearing in Poncelet's Porism corresponds to addition of a constant in the elliptic curve group. See here for some animations.
The question
A natural follow-up question seems to be that of the question title. More precisely, as highlighted here, because of the dependence of choice of identity, we should probably study the natural ternary operation. Specifically, given a general pair conics $C, D \subset \mathbb{P}^2$, together with points $x_1, x_2, x_3$ in:
$$X = \{(p, l) \in C\times D^* ~|~ p \in l\},$$
is there a natural geometric construction corresponding to $x_1 - x_2 + x_3$? I'd even settle for some useful special case, e.g., when $x_2$ corresponds to one of the four points of $C \cap D$.
Further waffle
Thinking about this yesterday I decided that it might be worth looking for natural morphisms from $X$ to $\mathbb{P}^1$ since the divisors of such maps will give identities in the group operation. In a slightly-ridiculous thought experiment, I tried to put myself in the position of somebody with the data of a cubic in $\mathbb{P}^2$ who didn't know the "collinear iff sum to zero" rule for cubics in Weierstrass form but who did know there was a group law, and was searching for a geometric construction. At least in this setting, the idea to look for natural morphisms to $\mathbb{P}^1$ is fruitful since the ratio of two linear forms gives a map to $\mathbb{P}^1$. This gets you most of the way to the "Weierstrass rule", and you just have to spot that you should choose one of the two forms to define a line through an inflection point, which is then a natural identity.
However, try as I might, I couldn't come up with anything useful for $X$ in the Poncelet construction. For a general point $(p, l) \in X$ we have a natural isomorphism $l \simeq \mathbb{P^1}$ since there are three distinguished points on $l$:
$p$,
the other intersection point of $l$ and $C$,
the point of tangency at $D$,
but this seems to be of no use. ¯\_(ツ)_/¯
[1] Griffiths, P., and Harris, J. "On Cayley's explicit solution to Poncelet's Porism", Enseign. Math., 24, 31–40 (1978).
I was wondering this myself while teaching a (somewhat) elementary geometry class. I'm as stumped as you are.
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2025-03-21T14:48:30.134816
| 2020-03-23T13:00:24 |
355515
|
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Stack Exchange
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Fully invariant measures for rational functions
Let $f(z)$ be a rational function of degree $d \geq 2$, with complex coefficients. I am interested in fully invariant measures for the dynamical system $(\mathbb C_\infty,f)$, where $\mathbb C_\infty$ is the Riemann sphere. By a fully invariant measure, I mean a probability measure $\mu$ such that $f^\ast \mu = d \mu$. (Such a measure also satisfies $f_\ast \mu =\mu$).
We can limit ourselves to ergodic ones (those which are not barycenter of two other fully invariant measures).
Those measures which have finite support are easy to describe: from the classical theory of Montel-Fatou-Julia, there exists a larger finite subset if $E$ such that $f^{-1}(E)=E$ and it has at most two elements; every fully invariant measure with finite support has support in $E$, so there are $0$, $1$ or $2$ finite support fully invariant ergodic measures.
From results of Ljubisch and Freyre-Lopez-Mane, there is also the so-called "natural measure" $\mu$, which is defined as the limit of $(\frac{1}{d}f^\ast)^n \nu$ where $\nu$ is any smooth measure on $\mathbb C_\infty$, or any Dirac measure $\delta_x$ for $x \not \in E$. It is ergodic, has the Julia set for support, and many other nice properties.
My question is
Are there always other ergodic fully invariant measures? If so, how to prove it and construct one?
Does the full invariance property $f^*\mu=d\mu$ have any ramification about the metric entropy of $\mu$? There is another paper (https://link.springer.com/content/pdf/10.1007/BF02584743.pdf) by Mañé which proves that the measure of maximal entropy is unique. So if $h_\mu=\log(d)$, the only answer is the construction of Lyubich and Freire-Lopes-Mañé which yields an MME by taking iterated preimages.
What exactly do you mean by the pullback measure $f^*(\mu)$? Measures are usually pushed forward.
Since you mention the measure of maximal entropy as an example, I assume that you mean that, for a set where the function is injective, you take the pushforward for the corresponding inverse branch, scaled by 1/d to get another probability measure. Then I believe that the measure of maximal entropy is the only such "balanced" measure on the Julia set.
@Lasse You are right that measures are usually pushed forward. But in this case, where $f$ is a finite map of degree $d$, that is the fibers of $f$ has always $d$ points (for a finite number of fibers we need to count the points with multiplicity for this to be true, and we do), the we $f^\ast \mu = \mu (f_\ast)$ where $f_\ast(z)=\sum_{y \in f^{-1}(z)} f(y)$, the points $y$ being counted with multiplicity.
The unique measure of maximal entropy $\mu_f$ supported on the Julia set of a rational map $f$ of degree $d \geq 2$ is indeed the unique balanced measure for $f$, i.e., the only probability measure $\mu$ not charging the exceptional set and satisfying $f^*\mu =d \cdot \mu$. As you already noticed in the comments, uniqueness of a measure with this property is explicitly stated in the mentioned paper by Freire, Lopes and Ma\~ne in their Theorem, part (d) (page 46). The proof of this statement is on p. 55 and the argument goes as follows: for any balanced measure $\mu$ it is shown that $\mu$ is absolutely continuous with respect to $\mu_f$ and the ergodicity of $\mu_f$ implies that $\mu=\mu_f$ (existence and ergodicity of $\mu_f$ are proved earlier in the paper). No assumption of non-atomicity, no reference to critical points or classification of Fatou components is employed in the proof of this uniqueness statement.
Another way to prove uniqueness of balanced measure is to use potentials of measures on the Riemann sphere $\mathbb{C}_\infty$ introduced as in
F. Berteloot, V. Mayer, Rudiments de dynamique
holomorphe, Vol. 7 of Cours Sp\'ecialis\'es,
Soci\'et\'e Math\'ematique de France, Paris (2001)
They give a streamlined treatment based on prior results by Fornaess and Sibony, Hubbard and Papadopol, Ueda and others.
Consider the cone $\mathcal{P}$ of functions $U$ plurisubharmonic on $\mathbb{C}^2$ and satisfying $U(tz)=c\log|t|+U(z)$ with a constant $c=c(U) >0$. Each such function defines a positive measure $\mu_U$ on $\mathbb{C}_\infty$ by $\langle \mu_U, \Phi \rangle =\int_{\mathbb{C}_\infty}(U \circ \sigma)\frac{i}{\pi}\partial\bar{\partial}\Phi$ for every smooth test function $\Phi$ with support in the domain of definition of the section $\sigma$ of the natural projection $\Pi: \mathbb{C}^2\setminus \{0\} \to \mathbb{C}_\infty$. Furthermore, every positive measure $\nu$ on $\mathbb{C}_\infty$ is defined by a function $U \in \mathcal{P}$ (unique if required to satisfy $\sup_{\|z\|\leq 1}U(z)=0$), specifically by $U(z)=\int_{\mathbb{C}_\infty}\log\frac{|z_1w_2-z_2w_1|}{\|w\|}d\nu([w])$ (Th\'eor`eme VIII.9 in this reference). This is called the potential of $\nu$.
Now, if a measure $\nu$ is balanced, then its potential $U$ satisfies $F^*U=d\cdot U$ Lemme VIII.12), hence $\frac{1}{d^n}F^{*n}U=U$ for every $n$. Here $F$ denotes a lift of $F$ to $\mathbb{C}^2$. Taking limits in $L^1_{loc}$ as $n \to \infty$ we get $U=G_f$ (Th\'eor`eme VIII.15), the potential of the Lyubich-Freire-Lopes-Ma\~ne measure $\mu_f$. Lifts are not unique, but this does not cause a problem.
If you relax the assumption on a measure supported on Julia set to $f_*\mu = \mu$, then there can be more measures satisfying it, even ergodic ones, besides the measure $\mu_f$. Of course the entropy will be less than $\log d$, sometimes even $0$. For more details on this see
S. P. Lalley, Brownian motion and the equilibrium measure on the Julia
set of a rational mapping, Ann. Probab. 20, 4 (1992),
1932--1967.
I mentioned critical points and non-atomicity because there was a complaint that (depending on the definition of the pullback) the definition of "fully invariant" could be subtly different from "balanced", since some points have different numbers of preimages from $d$. My point was that this is irrelevant due to the absence of exceptional points in the Julia set (of course, once you go into the proof, it is likely irrelevant in any case).
The measure of maximal entropy is the unique measure that is "fully invariant" in your sense. I believe that this already follows from the original proofs - indeed, it is well-known that if you take a point mass at some non-exceptional point, and keep pulling back, you will converge to the measure of maximal entropy. This should be enough to deduce the claim.
In the paper "Conformal and harmonic measures on laminations associated with rational maps" by Lyubich and Kaimanovich, the theorem on the existence of the measure of maximal entropy is stated as follows.
THEOREM. Any rational map f has a unique balanced measure $\kappa$. Moreover, $\operatorname{supp}(\kappa) = J(f)$, and the preimages of any point $z\in J(f)$ (excluding, possibly, two exceptional points) are equidistributed with respect to $\kappa$:
$$ \lim_{n\to\infty} \frac{1}{d^n} \sum_{\zeta\colon f^n(\zeta)=z} \delta_{\zeta} = \kappa,$$
where the limit is taken with respect to the weak topology on the space of probability measures on $J(f)$.
(Here a "balanced" measure is a fully-invariant measure, in your terminology, supported on the Julia set.)
Thank you and +1. The Theorem you quote from Lyubich and Kaimanovich is Theorem 4.2 page (68), but attributed to Lyubish 93 (and Brolin 65 but that's just for polynomial maps). Surely Lyubish 93 is in fact Lyubish 83: I have a paper by Lyubish with the same tithe and in the same journal published with the same page numbers, but published in 1983. But I can't find where the uniqueness of a balanced measure is asserted in Lyubish. Only is stated the uniqueness of a measure of maximal entropy.
So the situation is now that I am sure that there are no other fully invariant measure, but I still do not have a proof of it. You suggested that it may follows from the fact that I recalled that for all non-exceptional $x$, $\lim \frac{1}{d^n} (f^\ast)^n \delta_x = \kappa$. Would you have a proof of that?
@Joël I believe that the existence of MME is established independently by Lyubich and Freire-Lopes-Mañé. If you take a look at the latter, (https://link.springer.com/content/pdf/10.1007/BF02584744.pdf) on the first page you immediately see the construction of $\kappa$ as $\lim\frac{1}{d^n}(f^n)^*\delta_x$. The uniqueness is proven in another paper of Mañé (link.springer.com/content/pdf/10.1007/BF02584743.pdf).
Thank you KhashF. Unfortunately, I do not have access on the paper on Mañé, and it is not reviewed on mathscinet. But from the title, and the world "maximal" in it, I deduce that he proves the uniqueness of a measure of maximal entropy $\log d$. That's not enough to answer my question, since we do not know that a measure of maximal entropy is necessarily fully invariant.
Continuation of the preceding comment. In Freire-Lopes-Mañé, there is a uniqueness statement. The natural measure $\kappa$ is the unique measure satisfying that for every Borel set $A \subset \mathbb C_\infty$ such that $f_{|A}$ is injective, $\kappa(f(A))=d\kappa(A)$. If we knew that every fully invariant measure satis fies this property, we would be done. But unfortunately, this is false. If $x$ is a fixed exceptional point (for example $\infty$ when $f$ is a polynomial), then $\delta_x$ is fully invariant, and $f$ is injective on the Borel set $x$, but we have $\delta_x(f(x))=\delta_x(x)$.
Still, this is perhaps a basis to prove the uniqueness result we need.
@Joël If the measure is supported on the Julia set, then it is non-atomic. So the finitely many critical points have measure zero (unlike for the measure supported on an exceptional point), and any "fully invariant" measure is indeed balanced in the sense you cite.
The way that the paper by Freire-Lopes-Mañé makes sense of $f^*\mu=d\mu$ is the following: ''For any Borel subset $A$ of $\Bbb{C}_\infty$ with $f\restriction_A$ injective, one has $\mu(f(A))=d.\mu(A)$.'' (See p. 46 of this paper.)
One observation is that such an ergodic measure $\mu$ is either supported on the Julia set or is one of those measures with finite support that you mentioned in your question. To see this, notice that if ${\rm{supp}}(\mu)\not\subseteq\mathcal{J}$, then $\mu$, being ergodic, must assign zero mass to the backward invariant closed subset $\mathcal{J}$. So there are two cases:
Suppose ${\rm{supp}}(\mu)\subseteq\mathcal{J}$. It is not hard to see that in this case the support must indeed coincide with the Julia set $\mathcal{J}$: If the open subset $\mathcal{J}-{\rm{supp}}(\mu)$ of the Julia set is non-empty, by Montel's Theorem the union $\bigcup_nf^{-n}\left(\mathcal{J}-{\rm{supp}}(\mu)\right)\cap\mathcal{J}$ coincides with $\mathcal{J}$; but it is of measure zero, a contradiction.
Suppose $\mu(\mathcal{J})=0$, so $\mu$ assigns $1$ to the Fatou set $\mathcal{F}$. Recall two major theorems: The Absence of Wandering Fatou Components and The Classification of Periodic Fatou Components. The measure $\mu$ must assign a positive measure to one of the finitely many periodic Fatou components $U$; such a component must be a member of a cycle -- say of period $p$ -- of either immediate attracting (or super-attracting) basins, immediate parabolic basins, or finally, a cycle of rotation domains (Siegel disks or Herman rings). In the latter case, $f$ is injective on $U$ and hence $f^*\mu=d\mu$ implies $\mu(U)=\mu(f^p(U))=d^p.\mu(U)$ contradicting $\mu(U)>0$. The same idea could be employed to show $\mu(U)=0$ if $U$ is the immediate basin of a parabolic periodic point: The dynamics is injective near such a point (is in the form of $z\mapsto z+1$ in a suitable local coordinate). Finally, let us consider the case where $U$ is the immediate basin of attraction for a periodic point $z_0$ of period $p$. If $\mu(U)>0$, one can consider the system $\left(U,f^p\restriction_U,\frac{1}{\mu(U)}.\mu\restriction_U\right)$. As all orbits converge to $z_0$, the non-wandering set of this system is $\{z_0\}$. Hence the support of $\frac{1}{\mu(U)}.\mu\restriction_U$ is $\{z_0\}$; that is, $\mu(U-\{z_0\})=0$. We conclude that if $\mu(\mathcal{F})=1$, the support of $\mu$ consists of finitely many attracting cycles. So ${\rm{supp}}(\mu)$ is a completely invariant finite subset of the sphere and hence lies in the exceptional set of the complex systems which is of cardinality at most two. Examples are $\mu=\delta_\infty$ when $f$ is a polynomial or a measure of the form $\mu=\frac{1}{2}\left(\delta_0+\delta_\infty\right)$ when $f(z)=\frac{1}{z^d}$.
Definitely, the first case where $\mu$ is a fully invariant measure with the Julia set $\mathcal{J}$ as its support is more interesting. I agree with @Lasse Rempe-Gillen that the only such a measure in this situation is the measure of maximal entropy. This could be verified directly in some of the well known cases of the Julia dynamics. For instance, suppose $f$ is in the shift locus: $(\mathcal{J},f\restriction_{\mathcal{J}})$ is topologically conjugate to the one-sided shift $\left(\{0,\dots,d-1\}^{\Bbb{N}_0},\sigma\right)$ on $d$ symbols. The only fully invariant measure of the shift system is the
$\left(\frac{1}{d},\dots,\frac{1}{d}\right)$-Bernoulli measure (whose pullback is the measure of maximal entropy on $\mathcal{J}$). That is because if $\mu$ is fully invariant, for any choice of symbols $x_0,\dots,x_{k-1}\in\{0,\dots,d-1\}$, the iterates $\sigma,\dots,\sigma^k$ of the left shift $\sigma$ are all injective on the cylinder set $[x_0,\dots,x_{k-1}]$. Hence
$1=\mu\left(\{0,\dots,d-1\}^{\Bbb{N}_0}=\sigma^k([x_0,\dots,x_{k-1}])\right)=d^k.\mu\left([x_0,\dots,x_{k-1}]\right)$; so $\mu$ assigns $\frac{1}{d^k}$ to a cylinder set $[x_0,\dots,x_{k-1}]$.
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2025-03-21T14:48:30.135691
| 2020-03-23T13:04:20 |
355516
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Stack Exchange
|
Progress on a problem list
There is a list of open problems in my sub-field that was published in a journal some time ago and has had an impact on the area.
Many of the problems have been solved, some have partial solutions, and some are still unsolved.
I am considering trying to write a survey of the current status of this list. However, I am concerned about a few publication-related questions:
(1) Can my survey reproduce the statements of the problems? Of course I would not take credit for the problems, but I am concerned that I would be reproducing a large proportion of the content of a published paper.
(2) Is the answer to Question (1) different if I am trying to publish the survey in a journal vs. just post on arxiv vs. just post to my website?
(Regarding (2), it may be that no journal would be interested; that is a secondary concern.)
(3) Can anyone point to examples of surveys like this, in any area of mathematics? (Surveys discussing the current status of a list of problems in a specialized area. Surveys and books about the Clay Millennium Problems abound, but how about more specialized examples?)
Thank you.
I don't know about published papers, but there have been a few MO questions like what you're suggesting: see https://mathoverflow.net/questions/349406 and https://mathoverflow.net/questions/265493; but note they got some pushback on meta: https://meta.mathoverflow.net/questions/4423.
Oh now I see that the question about Thurston's list actually does link to a paper like you're talking about: https://doi.org/10.1365/s13291-014-0079-5.
There are a couple of books surveying the Hilbert problems, but that fails your "specialized area" requirement.
Q3: A classic of this type is Erdős on Graphs : His Legacy of Unsolved Problems
This book is a tribute to Paul Erdős, the wandering mathematician once
described as the "prince of problem solvers and the absolute monarch
of problem posers." It examines the legacy of open problems he left to
the world after his death in 1996.
Q1: of course you will want to reproduce the open problems themselves, to make the survey self-contained.
Q2: such reproduction falls under "fair use" of published material, so it can be included in a publication without copyright infringement.
Thank you. I accepted this answer of Carlo Beenakker because it answered all three parts of my question. I appreciate all the responses.
Q3: For group theory, you will probably not find many more comprehensive surveys like this than the Kourovka Notebook; this has been active since 1965, and is regularly updated with new problems, which problems have been solved, and a quick reference to where the solution appears.
Problems have been proposed by hundreds of mathematicians from all over the world, the difficulty of problems ranges from PhD level to well-known problems that remain open for decades. More than fifty years “Kourovka Notebook” serves as a unique means of communication for researchers in Group Theory and nearby fields of mathematics. Probably the most striking illustration of its success is the fact that more than three-quarters of the problems from the 1st edition of 1965 have now been solved.
Thank you for sharing. This is a great project.
|
2025-03-21T14:48:30.135965
| 2020-03-23T14:03:42 |
355519
|
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|
Stack Exchange
|
simplicial complex of two covers
Given two covers $\{U_a,U_b,\dots\}$ and $\{V_1,V_2,\dots\}$ of a space $X$, what is the appropriate idea of simplicial complex? As far as I see there are two ideas, and I was wondering where these were discussed and what their mapping properties are with respect to the complexes for the single covers. Apologies if this is well known, I am not an expert in this topic, and would grateful appreciate any help or references.
1) The sub complex of the categorical product complex, given by vertices $(a,1)$ etc. and corresponding to the cover $U_a\cap V_1$. The sub complex comes about by deleting empty intersections, so $\{(a,1),(b,2)\}$ would be deleted if $U_a\cap V_1 \cap U_b\cap V_2=\emptyset$.
2) Using multi-indices we have $U_\alpha=U_{\alpha_1}\cap \dots\cap U_{\alpha_k}$ and $V_\pi=U_{\pi_1}\cap \dots\cap U_{\pi_s}$. Now take the complex given by pairs $(\alpha,\pi)$ such that $U_\alpha\cap V_\pi\neq\emptyset$. It is not even obvious to me that this would give a simplical complex however - what would it give? Whatever it is, it is smaller than the previous complex as, for example $\{(a,1),(b,2),(c,1)\}$ and $\{(a,1),(b,1),(c,2)\}$ for the first complex both correspond to the single "simplex" $\big\{\{a,b,c\},\{1,2\}\big\}$ for the second.
This work is motivated by considering simplicial complexes as data structures in computer science.
Given a single cover $\mathcal{W}=\{W_i\}_{i\in I}$ of a space $X$ by nonempty sets, the nerve complex $\mathcal{N}(\mathcal{W})$ has $I$ as its vertex set, and it has $[i_0,\ldots,i_k]$ as a $k$-simplex if $\cap_{j=0}^k W_{i_j}\neq\emptyset$.
To see if I have understand your definitions correctly, is it correct to say that your complex in (1) is the nerve complex $\mathcal{N}(\mathcal{U}\cap\mathcal{V})$, where $\mathcal{U}\cap\mathcal{V}$ is the cover of $X$ given by $\mathcal{U}\cap\mathcal{V}=\{U_\alpha\cap V_j~|~U_\alpha\in\mathcal{U}\text{ and }V_j\in\mathcal{V}\}$? [For this nerve complex, if $U_\alpha\cap V_j=\emptyset$, then it would not be included as a vertex.]
Also, is it correct to say that your complex in (2) is the nerve complex $\mathcal{N}(\mathcal{U}\cup\mathcal{V})$, where $\mathcal{U}\cup\mathcal{V}$ is the cover of $X$ given by $\mathcal{U}\cup\mathcal{V}=\{U_a,U_b,\ldots\}\cup\{V_1,V_2,\ldots\}$?
The comment about (1) is correct. For (2) it is not the same as there $U_a$ itself would be a vertex, but this is not allowed under (2). The vertices are the same as (1) but the higher dimensions are reduced.
I don't yet understand the definition of (2) then. The vertices in (2) are all pairs $(a,\pi)$ where $U_a\cap V_\pi\neq\emptyset$. What do you mean when you say that ${{a,b,c},{1,2}}$ could be a simplex of (2) --- what are the vertices in this simplex? Are the vertices of ${{a,b,c},{1,2}}$ given by $(a,1), (a,2), (b,1), (b,2), (c,1), (c,2)$?
It is because the empty set is not allowed in a multi-index $(\alpha,\pi)$. This likely means that we have a quotient of a simplical complex, not a simplical complex...
|
2025-03-21T14:48:30.136177
| 2020-03-23T14:17:02 |
355521
|
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|
Stack Exchange
|
Explicit lifting characterization of complete lattices among posets?
It's well-known that the complete lattices are characterized among all posets as the regular-injectives. That is, a poset $L$ is a complete lattice if and only if $L$ has the right lifting property with respect to the class $Emb$ of all embeddings of posets. This remarkable characterization would be even more useful if there were some smaller, more explicit class of embeddings $\mathcal E \subset Emb$ sufficient to check injectivity. That is,
Questions:
Is there a good sub-class $\mathcal E \subset Emb$ of all poset embeddings such that a poset $L$ is a complete lattice if and only if $L$ has the right lifting property with respect to all the embeddings in $\mathcal E$?
What if we restrict attention to finite posets? That is: Is there a good subclass $\mathcal E^{fin} \subset Emb^{fin}$ of all embeddings of finite posets such that a finite poset $L$ is a lattice if and only if $L$ has the right lifting property with respect to all the embeddings in $\mathcal E^{fin}$?
What is meant by "good" is a bit subjective, but to start I'd be happy with anything non-tautological. Ultimately, it would be nice to have a class $\mathcal E$ or $\mathcal E^{fin}$ which is "explicit" in some sense, so that it actually becomes easier to check that a poset $L$ is complete via lifting properties. Normally, I'd hope for $\mathcal E$ to be small, but I'm pretty sure this is not possible.
One candidate I have in mind for $\mathcal E$ would be the collection of embeddings $\{S \to S^\triangleright \mid S \text{ discrete}\} \cup \{S \to S^\triangleleft \mid S \text{ discrete}\}$ which add a new top or bottom element to a discrete poset $S$. But I suspect this is too naive, as it would mean that any $\infty$-directed and $\infty$-codirected poset is a complete lattice -- this is probably false.
Does it make sense to try and go back to Baer's criterion for modules? This might be an inspiration for defining the proper $\mathcal{E}$. In Popescu's book on abelian categories one can find a generalized version of Bear's criterion which holds for nice abelian categories with a generator.
You are right that $\mathcal E$ cannot be small because complete lattices are not closed in posets under $\lambda$-filtered colimits for any regular cardinal $\lambda$. A candidate for $\mathcal E$ consists from embeddings of posets to their Mac-Neille completions. Since the Mac-Neille completion of a finite poset is finite, it works in the finite case too. But I think that you would not consider it to be "good".
After a bit of thought, I think I have a pretty good set for $\mathcal E^{fin}$ and a pretty good class for $\mathcal E$.
Proposition: Let $L$ be a finite poset. Then $L$ is a lattice if and only if $L$ has the right lifting property with respect to the following set $\mathcal E^{fin}$ of embeddings:
$\emptyset \to 1$
$2 \to 2^\triangleleft$ and $2 \to 2^\triangleright$ where $2$ is the discrete poset with two elements and these morphisms add a cone and cocone respectively.
$\newcommand{\bbowtie}{\bowtie\mkern-17mu\bullet\mkern17mu}$ $\bowtie \to \bbowtie$ where $\bbowtie$ is the 5-element poset $x,y < a < p,q$ and $\bowtie$ is the full sub-poset on $x,y,p,q$.
Proof:
It suffices to show that $L$ is a meet-semilattice, i.e. that for every finite set $S \subseteq L$, the poset $L \downarrow S$ of elements under $S$ has a top element. It suffices to consider the cases (i) when $S$ is empty and (ii) when $S$ has two elements. Moreover, every finite directed poset has a top element, so it will suffice to show that (i) $L$ is directed and (ii) for every $p, q \in L$, the poset $L \downarrow \{p,q\}$ is directed. (i) follows from (1) and the second part of (2). (ii) follows from the first part of (2) and (3).
Analogously, using $\infty$-directedness in place of directedness, we have
Proposition: Let $L$ be a poset. Then $L$ is a complete lattice if and only if $L$ has the right lifting property with respect to the following class $\mathcal E$ of embeddings:
$\emptyset \to 1$
$S \to S^\triangleleft$ and $S \to S^\triangleright$ for each discrete poset $S$ (i.e. we add a top element and a bottom element, respectively, to $S$)
$\newcommand{\bbowtie}{\bowtie\mkern-17mu\bullet\mkern17mu}$ $\bowtie_{S,S} \to \bbowtie_{S,S}$ for each set $S$, where $\bbowtie_{S,T}$ is the poset whose under lying set is $S \amalg T \amalg \{a\}$, with $S < a < T$, and $\bowtie_{S,T}$ is the full subposet on $S \amalg T$.
|
2025-03-21T14:48:30.136457
| 2020-03-23T15:18:41 |
355526
|
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|
Stack Exchange
|
How can I see the relation between shtukas and the Langlands conjecture?
The following bullet points represent the very peak of my understanding of the resolution of the Langlands program for function fields. Disclaimer: I don't know what I'm writing about.
Drinfeld modules are like the function field analogue of CM elliptic curves. To see this, complexify an elliptic curve $E$ to get a torus $\mathbb{C} / \Lambda$. If $K$ is an imaginary quadratic field, those lattices $\Lambda$ such that $\mathcal{O}_K \Lambda \subseteq \Lambda$ correspond to elliptic curves with CM, meaning that there exists a map $\mathcal{O}_K \to \operatorname{End} E$ whose 'derivative' is the inclusion $\mathcal{O}_K \hookrightarrow \mathbb{C}$. Now pass to function fields. Take $X$ a curve over $\mathbb{F}_q$, with function field $K$, and put $C$ for the algebraic closure of the completion. Then we can define Drinfeld modules as an algebraic structure on a quotient $C / \Lambda$.
Shtukas are a 'generalisation' of Drinfeld modules. According to Wikipedia, they consist roughly of a vector bundle over a curve, together with some extra structure identifying a "Frobenius twist" of the bundle with a "modification" of it. From Goss' "What is..." article, I gather that some analogy with differential operators is also involved in their conception.
Shtukas are used to give a correspondence between automorphic forms on $\operatorname{GL}_n(K)$, with $K$ a function field, and certain representations of absolute Galois groups. For each automorphic form, one somehow considers the $\ell$-adic cohomology of the stack of rank-$n$ shtukas with a certain level structure, and I presume this cohomology has an equivariant structure that gives rise to a representation.
While this gives me a comfortable overview, one thing I cannot put my finger on is why things work the way they work. I fail to get a grasp on the intuition behind a shtuka, and I especially fail to see why it makes sense to study them with an outlook on the Langlands program. This leads to the following questions.
Question 1. What is the intuition behind shtukas? What are they, even roughly speaking? Is there a number field analogue that I might be more comfortable with?
Question 2. How can I 'see' that shtukas should be of aid with the Langlands program? What did Drinfeld see when he started out? Why should I want to take the cohomology of the moduli stack? Were there preceding results pointing in the direction of this approach?
I think shtukas are best understood ahistorically. I would start with the modular curves, but specifically with the (geometric) Eichler-Shimura relation. This says that the Hecke operator at $p$, viewed as a correspondence on $X_0(N) \times X_0(N)$, when reduced at characteristic $p$, is just the graph of Frobenius plus the transpose of the graph of Frobenius. The relevance of this fact to the proof of a Langlands correspondence that relates traces of Frobenius acting on cohomology to eigenvalues of Hecke operators acting on cohomology should be unsurprising, even if completing the argument required much brilliant work by many people.
Now for higher-dimensional Shimura varieties, one cannot always generalize this simple geometric Eichler-Shimura relation, and instead must state and prove a suitable cohomological analogue of it.
When we go to the function field world, on the other hand, it pays to be very naive. We want to define a moduli space of some kind of object where Hecke operators act, and Frobenius acts, and these two actions are related. As David Ben-Zvi described in his answer, we understand what kind of object Hecke operators act on, and how - they act on vector bundles, or more generally $G$-bundles, and they act by modifying the bundle at a particular point in a controlled way. Frobenius also acts on $G$-bundles, by pullback. But these actions have nothing to do with each other.
The solution, then, is to force these actions to have something to do with each other in the simplest possible way - demand that the pullback of a $G$-bundle by Frobenius equal its modification at a particular point, in a certain controlled way. In fact, we can freely do this at more than one point, producing a space on which Frobenius acts like any desired composition of Hecke operators at different points.
The actual way Drinfeld came up with the definition was totally different, and did involve differential operators. He first came up with Drinfeld modules by an inspired analogy with the moduli spaces of elliptic curves. He then realized an analogy with work of Krichever, which he realized should lead to an analogous object defined using sheaves, the resulting definition was a shtuka. Precisely, the relation is that a rank $r$ Drinfeld module is the same thing as a $GL_r$-shtuka with two legs (corresponding to the standard representation of $GL_r$ and its dual under the geometric Satake isomorphism, in the modern language) at two points, where one is allowed to vary and the other is fixed at the point "$\infty$", and furthermore where we require that the induced map of vector bundles at the point $\infty$ is nilpotent.
So moduli spaces of Drinfeld modules will be certain subsets of moduli spaces of shtukas. However, the relationship between Drinfeld modules and shtukas is rarely used to study either - the research on both of them is usually quite separate.
Thank you for this great answer which solves one of my long confusion!
I was hoping someone arithmetically qualified would take this on, but here are some comments from a geometer. One nice perspective I learned from Wei Zhang's ICM address - namely, over function fields of curves over finite fields you have moduli spaces of shtukas with arbitrarily many legs (points in the underlying curve where a modification is taking place) -- these come with a map to a Cartesian power of the curve, where we remember only the location of the modification. Over number fields there are only analogs of moduli of shtukas with no legs (the arithmetic locally symmetric spaces which are the home of automorphic forms) and with one leg (Shimura varieties, with the defining map to $Spec(Z)$ being the analog of the "position of leg" map to the curve.
Which is to say, general moduli of shtukas DON'T have an obvious number field analog, rather they are certain generalizations of Shimura varieties (eg moduli of elliptic curves) that make sense over function fields. [Though over local fields there are now analogs of arbitrary moduli of local shtukas.]
Before saying what they ARE exactly, maybe it's worth saying a reason why it's clear they could be useful in the Langlands program. Namely, they carry the same symmetries (Hecke correspondences) as the [function field versions of] arithmetic locally symmetric spaces. As a result, their etale cohomology carries an action of a Hecke algebra, commuting with its natural Galois action. Once you find this out, and you take the point of view that we are looking for a vector space with commuting actions of Galois groups and Hecke algebras in which we might hope to realize the Langlands correspondence, then etale cohomology of moduli of shtukas is a natural place to look, and I imagine this is close to Drinfeld's reasoning (the same picture is one explanation for the role of Shimura varieties).
Anyway to say what they are start with Weil's realization that the function field analog of an arithmetic locally symmetric space is the set of $F_q$ points of the moduli stack of $G$-bundles on a curve:
$$Bun_G(C)(F_q)= G(F)\backslash G(A_F) / G(O_{A_F})$$
where $F$ is the field of rational functions on a smooth projective curve $C$ over a finite field, $A$ and $O_A$ are the adeles and their ring of integers.
So this is just a discrete set, but it comes from a rich geometric object over the algberaic closure of $F_q$. Next you realize that this set is given as fixed points of Frobenius acting on the stack $Bun_G(C)$ over the algebraic closure
$$Bun_G(C)(F_q)= (Bun_G(C))^{Frob}$$
-- i.e. the moduli stack of $G$-bundles equipped with an isomorphism with their Frobenius twist.
Then you say, ok let's relax this condition. Given two $G$-bundles you have the notion of a modification at a point $x\in C$ (or a finite collection of points) -- namely an isomorphism between the two bundles away from these points, with a fixed "relative position" at these points (relative positions measure the pole of this isomorphism at the points it degenerates -- you want to bound or prescribe this pole in a trivialization-independent way). This is the geometric source of [spherical] Hecke correspondences.
So now you can ask for the following data: a G-bundle, together with an isomorphism with its Frobenius twist away from finitely many points (the "legs") and fixed relative positions. These are shtukas! Also note that at points of the curve away from the legs we've "done nothing", from which it follows that the same Hecke correspondences that act on the original automorphic space act on the moduli of shtukas and their cohomology.
From the "modern" point of view (cf Vincent Lafforgue), we shouldn't just fix some minimal collection of "legs" and relative positions at those legs, but consider the entire tower of moduli of shtukas with arbitrary many legs and relative positions, and in particular pay attention to the algebraic structure ("factorization") we get by letting the positions of the legs collide. Lafforgue showed that this structure is enough to see the Langlands correspondence -- or rather one direction, it explains how spaces of automorphic forms "sheafify" over the space of Galois representations. I recommend Gaitsgory's "How to invent shtukas" and Nick Rozenblyum's recent lectures at the MSRI Intro workshop for the Higher Categories program for a very natural explanation of all of this starting from the geometric Langlands POV.
To flesh this out just a tiny bit, geometric Langlands replaces functions on $Bun_G(C)(F_q)$ by the study of the stack $Bun_G(C)$, where again the fundamental object of study is the action of Hecke correspondences (this time acting on categories of sheaves rather than vector spaces of functions). From this starting point you recover the entire story of shtukas very naturally by thinking categorically about the Grothendieck-Lefschetz trace formula for Frobenius acting on $Bun_G(C)$ -- cohomology of shtukas is just what you get when you apply the trace formula to calculate trace of Frobenius composed with a Hecke correspondences. So from this point of view shtukas are not "something new we've introduced", but really a structural part of thinking abstractly on the Hecke symmetries of $Bun_G$.
${\bf Edit:}$ I can't resist adding an "arithmetic field theory" perspective. From the Kapustin-Witten point of view on the Langlands program as electric-magnetic duality in 4d topological field theory, shtukas have the following interpretation.
First of all a curve over a finite field plays the role of a 3-manifold, so a 4d TQFT attaches to it a vector space (here the [functional field version of] the space of automorphic forms).
Second [spherical] Hecke correspondences are given 't Hooft lines, which are codimension 3 defects (AKA line operators) in the field theory. These are labeled by the data that labels relative positions of bundles on a curve (representations of the Langlands dual group).
So you find that the theory has not just one vector space attached to a "3-manifold" like a curve over a finite field, but one attached to each labeling of a configuration of points of the "3-manifold" by these data. (relative positions). This is how cohomology of moduli of shtukas appear (schematically of course!) in the physics POV.
Shimura varieties are locally symmetric spaces. How are the two analogies (Shimura variety = one leg, locally symmetric space = no leg) consistent with each other?
@GTA Shimura varieties are not locally symmetric spaces. They are schemes over the integers whose fiber over the complex numbers has an underlying manifold which is a locally symmetric space. Shtukas with one leg at a point $p$ are the analogue of the reduction of the Shimura variety mod $p$, and Shtukas with one varying leg are the analogue of the Shimura variety as a scheme over $\mathbb Z$. For questions about automorphic forms which only depend on the manifold, not on the Shimura variety, shtukas with no legs can be used.
@WillSawin I am confused how this could happen in the context of shtukas. Can a fiber of the moduli of shtukas with some number of legs at a point of the base be just Bun_G?
@GTA No, the fibers will all have the same dimension, which will almost never be the dimension of $\operatorname{Bun}_G$. The analogy might not be perfect in all respects - although maybe there is an explanation here.
@GTA From the point of view of Langlands, the key property of modular curves (for concreteness, but the same applies to other Shimura varieties) is that their cohomology splits as a sum over automorphic forms of two-dimensional Galois representations associated to them. On the other hand, the key property of locally symmetric spaces is that (generalized) functions on them are modular curves.
@GTA To relate these two notions, justifying why locally symmetric spaces are the special fiber, you have to specialize at $\infty$, then turn the Galois representations into Hodge structures, then split the Hodge structures into 1-forms and conjugate 1-forms, then view the $1$-forms as generalized functions. This Hodge theory story doesn't work the same way as function fields, so there's no reason for the two kinds of spaces to have such a direct relationship.
|
2025-03-21T14:48:30.137760
| 2020-03-23T15:19:00 |
355527
|
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|
Stack Exchange
|
Structure of non-big divisors in an abelian variety
Let $A$ be an abelian variety over $\mathbb{C}$. If $A$ has an effective non-big divisor, then $A$ is not simple. (In a simple abelian variety, every non-zero effective divisor is ample.)
What can we say about the structure of effective non-big divisors in $A$?
On A we should have effective=>nef. Then the big cone equals the ample cone. And effective non big is the same as strictly nef. Any such divisor (at least if it’s a $\mathbb{Q}$-divisor) should be numerically equivalent to the pull back of an ample divisor from some quotient map of abelian varieties $A\rightarrow B$.
I think when you say "strictly nef" you really mean "nef but not ample". "strictly nef" is actually a term of its own: it means $D \cdot C > 0$ for all curves $C$.
Can you expand on what you would like to know about "the structure of effective non-big divisors"? Yosemite Stan's comment gives one possible answer; is that the kind of thing you were looking for?
I was looking for what @YosemiteStan explained. Basically, from what I understand now, non-big divisors are pull-backs of ample divisors along some quotient map $A\to B$. Can you make this into an answer?
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2025-03-21T14:48:30.137886
| 2020-03-23T15:26:32 |
355528
|
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|
Stack Exchange
|
Group law in universal central extension of Thompson's group T
I'm having some troubles with universal central extensions and associated cocycles; in particular, I want to understand the group law of the universal central extension of the Thompson's group $T$.
Let us consider the universal central extension $E$ of the Thompson's group $T$; I know that it is possible to describe $E$ as the set $H_2(G; \mathbb{Z}) \times T$ with the group law defined in the following way $$(a,t)\cdot (a',t')=(a+_{H_2(G)}a'+_{H_2(G)} \phi(t,t'), t \cdot_Tt'),$$
where $\phi \in C^2(T; H_2(T; \mathbb{Z}))$ is the 2-cocycle associated to the extension $E$ (we denote its cohomology class by $[\phi]$), and is characterized by the following universal property: for any abelian group $A$ and for any $v \in H^2(T;A)$ there is a unique $f:H_2(T; \mathbb{Z}) \rightarrow A$ such that $v=H^2(T,f)[\phi]$ (here we are using that since $T$ is perfect we have a universal coefficient isomorphism $H^2(T;A) \simeq Hom(H_2(T; \mathbb{Z}), A)$).
I know that $H^2(T; \mathbb{Z}) \simeq \mathbb{Z} \oplus \mathbb{Z}$ and is generated by the
Euler class $eu$ and a certain class $\alpha$, which can be described explicitely in terms of the Godbillon-Vey cocycle (see https://eudml.org/doc/140083, Corollary 4.5). Moreover, I know that also $H_2(T; \mathbb{Z})\simeq \mathbb{Z} \oplus \mathbb{Z}$ by the universal coefficient isomorphism above.
I want to understand why the cocycle $\phi$ associated to the universal central extension $E$ is $(t,t') \rightarrow (Eu(t,t'), A(t,t'))$, where $A$ is a 2-cocycle representing the class $\alpha$. Also references are appreciated.
What's a reference for the given description of $H_2(T,\mathbf{Z})$?
@YCor "Sur un grup remarquable de diffeomorphisme di cercle", Ghys and Sergiescu (I've come across the statement studing "The spectrum of simplicial volume", by Heuer and Loeh, https://arxiv.org/abs/1904.04539, Theorem 5.5)
I've looked into the Ghys-Sergiescu paper (available here), the 2-cocycle $2\alpha$ seems to be related to the "Godbillon-Vey class", and more precisely to a discretized version of the "Bott-Thurston 2-cocycle". The 2-cocycle is explicitly computed in Thm E (Cor. 4.5). The interpretation of the corresponding central extension is maybe implicit but (at the moment) beyond my understanding.
Yes, I knew the description of $2 \alpha$ in terms of the Godbillon-Vey cocycle, but I think that in order to compute the cocycle associated to the universal central extension it should be enough knowing that $\alpha$ and the Euler cocycle generate $H^2(T; \mathbb{Z}$ (but still, I don't know how to make it explicit).
This is part of the context which is useful in order to answer the question...!
Ok, I will add it to the question!
At this point I'm not sure I understand the question, and whether the result just follows formally passing from $H_2$ to $H^2$ in some standard way.
Perhaps Section 6.6 of https://arxiv.org/pdf/1105.0559.pdf helps?
|
2025-03-21T14:48:30.138103
| 2020-03-23T15:50:06 |
355530
|
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}
|
Stack Exchange
|
Eigenvalues of an integral operator
Let $K\in L^2((0,1)\times(0,1))$ and consider the operator defined in $L^2(0,1)$ by
$$Lu(x):=u(x)-\int_0^1K(s,x)u(s)ds.$$
What kind of assumption might I impose on $K$ such that $\lambda=1$ will be not an eigenvalue of the operator $L$?. Any ideas?. Thank you.
If $\lambda=1$ is an eigenvalue, then $\int_0^1 K(s,x)u(s) ds=0$ for the corresponding eigenfunction. Hence a condition to rule this out is that $\{K(s,x),x \in (0,1)\}$ spans a dense subspace of $L^2$.
Thank you Mr Renardy for the answer. I'm wondering if there exists such a kernel. For instance, the constant kernel doesn't fulfill this property. Do you have an example in mind. Thank you Mr Renardy again.
|
2025-03-21T14:48:30.138176
| 2020-03-23T16:47:24 |
355536
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/355536"
}
|
Stack Exchange
|
Cumulant of functions of weakly dependent random variables
Suppose $X_1,\dots,X_4$ are Gaussian random variables with mean and variance
$$\mathbf E X_i = 0,\quad \mathbf E X_i^2=1.$$
Furthermore suppose that the random variables have a certain weak correlation
$$\mathbf E X_1 X_2 X_3 X_4=\kappa(X_1,X_2,X_3,X_4)=\epsilon,$$
but that all other joint cumulants $\kappa$ are at least $O(\epsilon^2)$. Note that in particular the random variables cannot be jointly Gaussian.
Now I am wondering whether something general can be said about the joint cumulant of $f(X_1),\dots,f(X_4)$ for well behaved functions $f$? As an example, it seems that for $f(x)=x^{2k-1}$ we have
$$\kappa(X_1^{2k-1},\dots,X_4^{2k-1})=\epsilon (2k-1)!!^4 + O(\epsilon^2)$$
and
$$\kappa(X_1^{2k},\dots,X_4^{2k})=O(\epsilon^2).$$
In light of this it seems natural to conjecture that
$$\kappa(f(X_1),\dots, f(X_4)) = \epsilon\Bigl(\mathbf E_{X\sim N(0,1)} f'(X)\Bigr)^4+O(\epsilon^2)\tag{*}.$$
Question: Is (*) correct? If yes, can it be proved using something other than polynomial approximation?
May I ask for the reason of the down vote? Is this question too elementary for this site?
|
2025-03-21T14:48:30.138269
| 2020-03-25T14:43:35 |
355679
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/355679"
}
|
Stack Exchange
|
How large can a planar triangulation be that embeds bi-Lipschitz in a ball of $\mathbb{R^3}$?
Let $G$ be a finite plane graph all bounded faces of which are triangles. For example, $G$ could be
the 1-skeleton of a triangulation of a topological disc. Let $f: V(G) \to \mathbb{R}^3$ be a bi-Lipschitz map, i.e.
$$d(x,y) ≤ |f(x)−f(y)| ≤ Cd(x,y) (1)$$
for a fixed $C\geq 1$ and every $x,y\in V(G)$, where $d(x,y)$ denotes the graph distance.
Note that (1) implies an upper bound on the degrees of $G$ (depending on $C$).
Let $diam(G)$ denote the diameter of $G$.
Can $G$ have more than $\Theta(diam(G)^2)$ vertices?
What is the supremal $r$ such that there is a family ${G_i}$ of graphs satisfying (1) and $|V(G_i)| = \omega(diam(G_i)^r)$?
The question can also be stated with the graph G replaced by a Riemann surface, and $|V(G_i)|$ replaced by its area.
Embedding a portion of the triangular lattice `flat’ shows that $r\geq 2$, and $r\leq 3$ is easy by a volume argument. Any bounds in (2,3) would be welcome.
|
2025-03-21T14:48:30.138383
| 2020-03-25T15:36:53 |
355682
|
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|
Stack Exchange
|
Orthogonality condition of symmetric matrix pencil
Let $P(\lambda)=\lambda M−L\in \mathbb{R}^{n \times n}$ be a matrix pencil with symmetric nonsingular matrix $M$ and $L$ is a weighted Laplacian matrix of a connected graph. Clearly $(0,1_n)$ is an eigenpair of $P(\lambda)$ and suppose $(\lambda_i,x_i) \in \mathbb{R} \times \mathbb{R}^n,\,i=1:n−1$ be the eigenpairs of $P(\lambda)$ then show that $x_i$ can be chosen such that $x^T_iLx_j=0\, \forall \,1\leq i\neq j \leq n−1$ and $x^T_iLx_i=1$ where $1_n$ denotes the vector with all entries are 1.
X-posted: https://math.stackexchange.com/q/3594612/339790
@RodrigodeAzevedo I have asked this question in math stackexchange too. I am expecting some positive response on this query.
@Saheb Users are normally discouraged from opening two versions of the same question (even if they're on different sites in the SE network)
@Omnomnomnom I got your point.
|
2025-03-21T14:48:30.138490
| 2020-03-25T15:45:37 |
355684
|
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|
Stack Exchange
|
Cobordism invariants: topological v.s. geometric
Some cobordism invariants are not cohomology classes. Such as the $\mathbb{Z}_{16}$-valued eta invariant $\eta$ of $\Omega_4^{Pin^+}$, the $\mathbb{Z}_8$-valued Arf-Brown-Kervaire invariant ABK of $\Omega_2^{Pin^-}$, and the $\mathbb{Z}_4$-valued quadratic enhancement $q(a)$ of $\Omega_2^{Pin^-}(B\mathbb{Z}_2)$ where $a\in H^1(M,\mathbb{Z}_2)$. The last invariant is defined as follows: Any 2-manifold $M$ always admits a $Pin^-$ structure. $Pin^-$ structures are in one-to-one correspondence with quadratic enhancements
$$q: H^1(M,\mathbb{Z}_2)\to\mathbb{Z}_4$$
such that
$$q(x+y)-q(x)-q(y)=2\int_M x\cup y\mod4.$$
In particular,
$$q(x)=\int_M x\cup x\mod2.$$
There are many more such examples, I only mention these.
We say that a cobordism invariant is topological if it can be defined purely using topological data, for example, if it is a cohomology class.
While we say that a cobordism invariant is geometric if it can be defined purely using geometric data like metric, connection, and curvature.
These two definitions have no confliction, a cobordism invariant can be both topological and geometric.
My question: Determine whether the cobordism invariants mentioned above are topological and geometric. In general, are there any examples of cobordism invariants which are geometric but not topological? Are there any examples of cobordism invariants which are topological but not geometric?
For example, the eta invariant $\eta$ is discussed in this question.
Thank you!
Can you give precise definitions to "a cobordism invariant is topological", resp "geometric"? To me some invariant associated to a manifold with extra structure is topological if it depends on little more than the smooth structure, eg an invariant of spin manifolds might be called topological because there are finitely many spin structures on a given smooth manifold. That the $\eta$ invariant of index theory normally depends on structure like the metric/connection just says to me that your $\Bbb Z/16$-valued $\eta$ extracts topological information from a normally geometric object.
@MikeMiller Thank you! I have added the definitions.
$\newcommand{\ko}{\mathit{ko}}
\newcommand{\MTSpin}{\mathit{MTSpin}}
\newcommand{\Z}{\mathbb Z}$
As Mike points out in his comment, it's not obvious what it means for a bordism invariant is
“topological” or “geometric.” The bordism invariants you mention can be described
topologically, and some bordism invariants which you might think of as topological also admit geometric
descriptions.
Pontrjagin numbers are oriented bordism invariants which look topological: take a cohomology class on your manifold
and evaluate it on the fundamental class. For example, there's an injective map $\Omega_4^{\mathrm{SO}}\to\mathbb
Z$ sending a closed, oriented $4$-manifold $X$ to $\langle p_1(X), [X]\rangle$; here $p_1$ is the first Pontrjagin
class of $X$.
However, there is an equivalent, “geometric” definition of this invariant: choose a connection on the
vector bundle $TX\to X$, and let $F$ be its curvature. Then one can make sense of $\mathrm{tr}(F\wedge
F)\in\Omega^4(X)$, and Chern-Weil theory proves that
$$
-\frac{1}{8\pi^2}\int_X \mathrm{tr}(F\wedge F) = \langle p_1(X), [X]\rangle.
$$
Certainly a connection is geometric data, so this invariant is both “topological” and
“geometric.”
The pin$^\pm$ bordism invariants you mention admit geometric interpretations (via $\eta$-invariants), but also
topological ones, though the topology is harder to see. First, let's reframe the above bordism invariant in terms
of Thom spectra: the characteristic class $p_1\in H^4(B\mathrm{SO})$ defines via the Thom isomorphism a cohomology
class in $\tilde H^4(M\mathrm{SO})$, hence a map $M\mathrm{SO}\to\Sigma^4 H\mathbb Z$, and upon taking $\pi_4$, we obtain
the map $\Omega_4^{\mathrm{SO}}\to\mathbb Z$ from above.
The point of this reformulation is that by replacing $H\mathbb Z$ (i.e. ordinary cohomology) with other cohomology
theories, we can describe the invariants you've mentioned above.
As a warm-up, take the Arf invariant of a spin surface, which defines an isomorphism $\Omega_2^{\mathrm{Spin}}\to\mathbb
Z/2$. There are several different ways to define it, but here's one in line with our above description of $p_1$:
we have the Atiyah-Bott-Shapiro map $\mathit{ABS}\colon\MTSpin\to\ko$,1 and upon taking $\pi_2$, this
yields a map $\Omega_2^{\mathrm{Spin}}\to \pi_2\ko\cong\mathbb Z/2$. One can unwind this definition through the
Pontrjagin-Thom isomorphism and obtain a description of the Arf invariant through integration of $\ko$-cohomology
classes.
Next, the Arf-Brown-Kervaire invariant. There is a splitting $\mathit{MTPin}^-\simeq\MTSpin\wedge
\Sigma^{-1}\mathit{MO}_1$, where $\mathit{MO}_1$ is the Thom spectrum of the tautological bundle $\sigma\to
B\mathrm O_1$. Therefore, smashing the Atiyah-Bott-Shapiro map with $\Sigma^{-1}\mathit{MO}_1$, we obtain a map
$$
\mathit{MTPin}^-\simeq\MTSpin\wedge \Sigma^{-1}\mathit{MO}_1\longrightarrow \ko\wedge\Sigma^{-1}\mathit{MO}_1.
$$
Taking $\pi_2$, we get a map
$$\Omega_2^{\mathrm{Pin}^-}\to \pi_2(\ko\wedge \Sigma^{-1}\mathit{MO}_1)\cong
\widetilde{\ko}_3(\mathit{MO}_1)\cong\Z/8,$$
and this is the Arf-Brown-Kervaire invariant.2 This can also be described in terms of a pushforward in
twisted $\ko$-theory.
The same approach works $\Omega_4^{\mathrm{Pin}^+}\to\Z/16$, using this time the splitting
$\mathit{MTPin}^+\simeq\MTSpin\wedge\Sigma\mathit{MTO}_1$; here $\mathit{MTO}_1$ is the Thom spectrum of the
virtual bundle $-\sigma\to B\mathrm O_1$. Smashing the Atiyah-Bott-Shapiro map with $\Sigma\mathit{MTO}_1$ and
taking $\pi_4$ yields a map $\Omega_4^{\mathrm{Pin}^+}\to \widetilde{\ko}_3(\mathit{MTO}_1)\cong\Z/16$.
The same approach works for $\Omega_2^{\mathrm{Pin}^-}(B\Z/2)\to\Z/4$:
$$\Omega_2^{\mathrm{Pin}^-}(B\Z/2)\cong\pi_2(\MTSpin\wedge \Sigma^{-1}\mathit{MO}_1\wedge (B\Z/2)_+),$$
which maps to $$\pi_2(\ko\wedge \Sigma^{-1}\mathit{MO}_1\wedge (B\Z/2)_+) = \widetilde{\ko}_3(\mathit{MO}_1\wedge
(B\Z/2)_+)\cong\Z/4.$$
If you're asking which bordism invariants come from cohomology classes, the answer is that characteristic classes
for $n$-dimensional $G$-bordism live in $H^n(BG;A)$, where $A$ is a coefficient group; a $G$-structure on a
manifold $M$ pulls back this class to $M$, and then we evaluate it on the fundamental class. In general, this won't
capture everything: for example, if two $G$-structures have the same underlying orientation, their values on any
cohomological bordism invariant will agree. Hence, for example, the Arf invariant isn't cohomological, as there are
different spin structures on a torus which induce the same orientation, but have different Arf invariants. This is
why topological descriptions of such bordism invariants use generalized cohomology.
1: Here, given a $G$-structure $X\to B\mathrm O$, $\mathit{MTG}$ means the Thom spectrum whose homotopy groups are
the bordism groups of manifolds with a $G$-structure on the tangent bundle, rather than on the stable normal
bundle. This distinction is irrelevant for spin bordism, but important for pin$^\pm$ bordism.
2: The Arf-Brown-Kervaire invariant depends on a choice of an 8th root of unity; depending on this choice, we might
need to compose with an automorphism of $\Z/8$ to obtain “the” Arf-Brown-Kervaire invariant. The same
caveat applies to the $\Z/16$ and $\Z/4$ invariants.
Thank you for your nice answer! I have added the definitions for "topological" and "geometric" and I ask whether they contain each other.
|
2025-03-21T14:48:30.139057
| 2020-03-25T15:59:22 |
355685
|
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|
Stack Exchange
|
Isometry type of alcoves in affine Coxeter complexes
Let $W$ be an irreducible affine Coxeter group (say of type $\widetilde{X}_n$), and let $\Sigma$ be the associated Coxeter complex. Thus, $\Sigma$ is an $n$-dimensional Euclidean space tesselated by isometric copies of a given simplex $A$ (namely, by the alcoves of $\Sigma$). I would like to know what the isometry type of $A$ is, depending on $\widetilde{X}_n$ (at least for the classical types $\widetilde{A}_n$, $\widetilde{B}_n$, $\widetilde{C}_n$ and $\widetilde{D}_n$).
More specifically, if $1$ is the length of the shortest edge of $A$, what are the other possible lengths for the edges of $A$?
(For instance, in type $\widetilde{A}_2$, all edges have length $1$).
Hint: Look at the automorphism group of the extended Dynkin diagram.
Did you compute them for $n\leq 4$? This should give you a clear picture what it is. All the necessary equations for the calculation are in Bourbaki, Lie Groups and Lie Algebras, ch 4-6...
@BugsBunny Do you know of any reference where these calculations are made, or can you show on an example (eg $\widetilde{D}_4$) what has to be done?
I guess this is no longer a comment. So I post it as an answer.
Note that even in Type A, you do not get a regular simplex: the simplex has cyclic symmetry (as explained by the comment of Moishe Kohan), but not full symmetry.
Go to Bourbaki, Groupes et Algebres de Lie, Ch.4-6 where the fundamental weights $\varpi_1, \ldots , \varpi_n$ of $X_n$ are listed. The vertices of the simplex are
$$0 \ \mbox{ and } \ x_k \varpi_k, \ k=1,\ldots,n$$
where $x_k>0$ is such that $x_k \varpi_k$ lies on the fixed hyperplane of the last Coxeter generator of the affine Weyl group. This hyperplane is given by the equation
$$
\langle z , \alpha_0^{\vee}\rangle =1
$$
where $\alpha_0^\vee$ is highest coroot. The highest roots are also listed in the same book. Don't forget to swap $B_n \leftrightarrow C_n$, when looking for the highest coroot, because the coroot system is Langlands dual to the root system.
No, I don't know where this calculation has been carried out. Please, compute the length of all the edges of this simplex yourself, and put it here for the benefit of the whole humankind:-))
Upvoted for the last sentence. Carrying out routine calculations and putting them somewhere publically accessible is an underestimated service to humanity.
|
2025-03-21T14:48:30.139259
| 2020-03-25T16:01:24 |
355686
|
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|
Stack Exchange
|
Hardness of concave minimization problem
I have an optimization problem $\underset{x}{\min} ~ c(x) - k \cdot x$ where $c(x)$ is a non-decreasing concave function with $c(0) = 0$, $x \in C \subseteq \mathbb{R}^d_{\geq 0}$. By non-decreasing, I mean the partial derivative of $c(x)$ for every dimension of $x$ is positive. Is this problem NP-Hard?
I am aware that harish (https://cstheory.stackexchange.com/users/10385/harish) asked a question, Maximizing a convex function with linear constraints, URL (version: 2012-10-17): https://cstheory.stackexchange.com/q/12310 and that one is NP-Hard. In addition, I also searched online and found that concave minimization problems can be NP-hard in general [Pardalos and Rosen] (https://epubs.siam.org/doi/pdf/10.1137/1028106). I am just wondering for my specific concave minimization problem, is it hard to solve? In addition, are there any survey papers on the hardness of some specific concave minimization problems?
I apologize, but I forgot to mention that the feasible region $C$ is bounded, i.e. $\|C\|_2 \leq \gamma$.
I think I found one paper by Pardalos, Panos M., and Stephen A. Vavasis. "Quadratic programming with one negative eigenvalue is NP-hard." Journal of Global Optimization 1.1 (1991): 15-22. They have the conclusion that
$\min f(x) = \frac{1}{2} x^T Q x + c^T, s.t. Ax \leq b$ is NP-hard when $Q$ is $n\times n$ symmetric negative semidefinite matrix. Does this prove my question? Because $c(x)$ in my question will be concave if and only if its Hessian matrix is negative semidefinite.
What do you mean by non-decreasing function $c:\mathbb R^d \rightarrow \mathbb R$ ?
If your problem has a solution $x^* \ne 0$, then $0$ is also a solution. Indeed, consider the function
$$\varphi(t) = c(t \, x^*) - k\cdot (t \, x^*).$$
Since $x^*$ is a solution, we have
$$\varphi(0) \ge \varphi(1).$$
However, if we would assume
$$\varphi(0) > \varphi(1),$$
concavity implies $\varphi(t) \to -\infty$ as $t \to \infty$. This is a contradiction since $1$ is the minimizer of $\varphi$.
Hence, $\varphi(0) = \varphi(1)$ and this implies that $x^{**} = 0$ is also a solution to your original problem.
The proof makes sense, but the conclusion seems kind of strange to me. Does this mean $\underset{x}{\min} c(x) - k\cdot x$ = 0? However, consider a simple example when $c(x) = \sqrt{x}$ and $k = 1$. Apparently, $\underset{x}{\min} \sqrt{x} - x = - \infty$ instead of 0? So where goes wrong in this problem.
Your example does not have a minimizer. If the problem has a minimizer, then $0$ is a minimizer. Otherwise, the infimal value is $-\infty$.
What if the feasible region of $x$ is bounded? Specifically, $x \in X \subseteq \mathbb{R}^d$ and $|X|_2 \leq \gamma$, does your conclusion still hold? As far as I can understand, the contradiction doesn't exist because 1 is only the minimizer of $\phi$ within the bounded region $X$, but it's not the minimizer for the whole infty range.
If the feasible region is bounded, my conclusion fails. The main argument was that we can look at rays starting in $0$.
|
2025-03-21T14:48:30.139470
| 2020-03-25T16:03:52 |
355687
|
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|
Stack Exchange
|
Riemann-Hilbert correspondence versus Simpson correspondence
I couple of days ago, I asked extensively the same question on Stack-exchange (see https://math.stackexchange.com/questions/3592151/riemann-hilbert-correspondence-versus-simpson-correspondence)and go no answer.
Let us assume that X is a connected, smooth complex algebraic variety. Then the Riemann-Hilbert correspondence tells us that the functor which sends a flat connection with regular singularities on a vector bundle of X to its asocciated monodromy representation is an equivalence of categories.
Furthermore, the Simpson correspondence tells us that there is an equivalence of categories between the category of complex representations of the fundamental group of curves and the category of semi-stable Higgs bundles with trivial Chern class.
Its seems to me that the Simpson correspondence should be a consequence of the Riemann-Hilbert correspondence, especially since the category of Higgs bundles is roughly the tangent category of the category of vector bundles. However, based on the number of papers written on this, and the fact that Simpson wrote a ICM note on this, this is clearly not the case. So I guess I must miss what the additional content of the Simpson correspondence is. Could someone help me?
"a couple of days ago" is actually yesterday, isn't it? best practice would be to wait at least a week for an answer before crossposting.
From my point of view, Riemann-Hilbert and non-abelian Hodge are really two independent statements - though the statement of the latter may be wound up in the former in some sense.
There are three different types of objects at play:
1) Higgs bundles (Dolbeault),
2) flat connections (de Rham),
3) reps of $\pi_1$ (Betti).
Riemann-Hilbert relates 2) and 3) via the operation of taking a connection to its monodromy. Simpson's non-abelian Hodge correspondence relates 1) with either 2) or 3) depending on your perspective (in any case, one can go between them via Riemann-Hilbert).
It might be helpful to keep in mind how abelian Hodge theory works. Here, if $X$ is a compact Kahler manifold, say, there are three vector spaces:
1) $H^1_{Dol}(X) = H^1(X;\mathcal O) \oplus H^0(X;\Omega^1)$
2) $H^1_{dR}(X)$
3) $H^1_{Betti}(X;\mathbb C) = Hom(\pi_1(X),\mathbb C)$
The analogue of Riemann-Hilbert here is the de Rham theorem identifying 2) and 3). On the other hand, the analogue of non-abelian Hodge is, well, abelian Hodge theory, which says that 1) and 2) may be identified (via yet another object: the harmonic forms).
There is no sense in which Hodge theory follows from the de Rham theorem, and I would say that the former involves much more subtle and intricate ideas.
|
2025-03-21T14:48:30.139651
| 2020-03-25T18:11:18 |
355693
|
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|
Stack Exchange
|
Every convex set is of locally finite perimeter
I need to prove that every convex subset of $\mathbb{R}^n$ is of locally finite perimeter.
$E$ is of locally finite perimeter if there exists a vector-valued Radon measure $\mu_E$ s.t. the Gauss Green theorem holds: that is for each compactly supported vector field $T$
$$ \int_{E}div(T)=\int_{\mathbb{R}^n}T\cdot d\mu_E. $$
Moreover the perimeter of $E$ is defined as the total variation of $\mu_E$, that is $P(E;A):=|\mu_E|(A)$.
Let me state the following lemma which the book suggests to use:
Let $H_t=\{x : e\cdot x<t \} $ for $t\in \mathbb{R}$ and $e\in S^{n-1}$ (an half space) and $E$ a set of locally finite perimeter with $|E|<\infty $. Then
$$ \mu_{E\cap H_t}= (\mu_E)_{|_{H_t}}+ eH^{n-1}_{|_{E\cap H_t}}. $$
From this follow that $H^{n-1}(E\cap \partial H_t)\leq P(E; H_t)$, $P(E\cap H_t)\leq P(E)=P(E;\mathbb{R}^n)$.
Let now $C$ be a convex set, this happens if and only if $\bar{C}=\bigcap_n H_n$ where $H_n$ are closed half spaces.
The suggestion from the book is the following: first prove that if $E$ is of finite perimeter and $C$ is convex, then $P(E\cap C)\leq P(E)$ (which is an easy consequence of the second inequality in the last result and of the fact that $C$ is a countable intersection of half spaces). Then refine this argument to prove that every convex set if of locally finite perimeter.
I didn't get the suggestion of "refine" the argument, and so i am asking for help.
Thanks to everyone who will use time to respond me
A. Ninno can you specify the book? In particular I need the formula above for the intersection between $E \cap H_t$. Is Maggi's book? If yes, in the book the author wrote this formula for a set of finite perimeter in $\mathbb{R}^n$ and you for a locally finite perimeter with $|E| < \infty$. Why? I would need your version :)
First assume that $E$ is compact. Then, your inequality says that you can approximate it from above by a sequence $E_n$ of convex polytopes with decreasing perimeters. Then, the sequence $\mu_{E_n}$ is weak*-precompact by Banach$-$Alaoglu theorem, so we can assume by passing to a subsequence that $\mu_{E_n}\to \mu$. Hence
$$
\int_E \mathrm{div}\,T=\lim_{n\to\infty}
\int_{E_n} \mathrm{div}\,T=\lim_{n\to\infty} \int Td\mu_{E_n}=\int Td\mu,
$$ and we are done.
If $E$ is not compact, intersect it with balls of large radii and and pass to a limit.
Thank you very much!
For convenience of future readers i add that the same argument that you presented works with the compactness of the sets of finite perimeter: that is
If ${ E_h}_h$ are of finite perimeter and they satisfies $sup_h P(E_h)<\infty$ and $E_h\subset B_R$ for a certain R
Then there exists $E$ a set of finite perimeter s.t. $E_h\to E$ and $\mu_{E_h}\to^{\star}\mu_E$.
|
2025-03-21T14:48:30.139860
| 2020-03-25T18:51:52 |
355694
|
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"Johnny T.",
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|
Stack Exchange
|
On an application of the going-down theorem of Cohen-Seidenberg in Mumford
There is a following result in Mumford's red book of schemes (Chapter II Section 8). Here $R$ is a valuation ring with algebraically closed fraction field $k$.
Let $Z \subset \mathbb{P}^n_k$ be an irreducible closed subset of dimension $r$ and let $\mathcal{Z} = \overline{ \{ i(Z)\} }$, where $i: \mathbb{P}^n_k \to \mathbb{P}^n_R$. Thee exist $(r+1)$ linear forms defined by $\ell_0 = ... = \ell_r = 0$, which we denote by $\mathcal{L}$, is disjoint from $\mathcal{Z}$. Let $\tau$ be the projection $\mathbb{P}^n_R \backslash \mathcal{L} \to \mathbb{P}^r_R$. Then giving any structure of closed subscheme, $\tau: \mathcal{Z} \to \mathbb{P}^r_R$ is a finite surjective morphism.
Then later on he applies the Going-Down theorem (Question about the statement of the going-down theorem of Cohen-Seidenberg in Mumford) to this morphism $\tau: \mathcal{Z} \to \mathbb{P}^r_R$. One of the conditions is that no non-zero element of $O_{\mathbb{P}_R^r, \tau(x)}$ is a $0$-divisor in $O_{\mathcal{Z},x}$. I have been trying to prove that this condition is satisfied, and I would greatly appreciate any explanation of this. Thank you.
Below is my attempt:
By restricting to affine open $D(\ell_j)$, where $\ell_j \neq 0$, without loss of generality it suffices to consider
$$
\operatorname{Spec}{R[\frac{X_0}{\ell_0}, \ldots, \frac{X_n}{\ell_0}]/I} \to
\operatorname{Spec}{ R[\frac{X_0}{\ell_0}, \ldots, \frac{X_n}{\ell_0}]} \to
\operatorname{Spec}{R[\frac{Y_1}{Y_0}, \ldots, \frac{Y_r}{Y_0}]},
$$
where $D(\ell_0) \cap \mathcal{Z} = \operatorname{Spec}{ R[\frac{X_0}{\ell_0}, \ldots, \frac{X_n}{\ell_0}]/I}$.
Thus it suffices to consider the ring homomoprhisms
$$
\tau^{\#}: R[\frac{Y_1}{Y_0}, \ldots, \frac{Y_r}{Y_0}] \to R[\frac{X_0}{\ell_0}, \ldots, \frac{X_n}{\ell_0}] \to R[\frac{X_0}{\ell_0}, \ldots, \frac{X_n}{\ell_0}]/I,
$$
where the first map is defined by
$$
\frac{Y_j}{Y_0} \mapsto \frac{\ell_j}{\ell_0}.
$$
Let $Q$ be a prime ideal of $R[\frac{X_0}{\ell_0}, \ldots, \frac{X_n}{\ell_0}]/I$ and $Q' = (\tau^{\#})^{-1}(Q)$.
Then we consider the map of stalks by looking at the following commutative diagram
$\require{AMScd}$
\begin{CD}
R[\frac{Y_1}{Y_0}, \ldots, \frac{Y_r}{Y_0}] @>>> R[\frac{X_0}{\ell_0}, \ldots, \frac{X_n}{\ell_0}]/I \\
@VVV @VVV\\
R[\frac{Y_1}{Y_0}, \ldots, \frac{Y_r}{Y_0}]_{Q'} @>>> (R[\frac{X_0}{\ell_0}, \ldots, \frac{X_n}{\ell_0}]/I)_{Q}.
\end{CD}
But then it seems possible that
for some $j$, $Y_j/Y_0 \mapsto \ell_j/\ell_0$ and this $\ell_j/\ell_0$ may be in $I$. And in this case we don't have that no non-zero element of $O_{\mathbb{P}_R^r, \tau(x)}$ is a $0$-divisor in $O_{\mathcal{Z},x}$...
Mumford's footnote to the statement of Going Down, (p.252, original redbook), says the condition needed is that all associated points of OsubZ lie over the generic point of the target. Does that help, if say Z is chosen reduced as well as irreducible?
I am a novice on this sort of thing, but my reading, say of Ravi's course notes, suggests that an irreducible reduced variety has only one associated point, its generic point. Thus a surjection of varieties would seem to always satisfy the condition in Mumford's footnote. The connection between the two versions of the condition seems to be that a zero divisor is exactly a section that vanishes at an associated point. Does this do it? As to your computation, algebra is my weak point, but wouldn't the concern raised in your next to last sentence imply that tau is not surjective?
@roysmith I think it's clear now. Thank you!
You are quite welcome. I am myself currently reading the redbook, and found chap. II.8 particularly challenging. I am up to III.1 now and looking forward to the payoff geometry in sections ≥ III.3 !
|
2025-03-21T14:48:30.140117
| 2020-03-25T19:00:06 |
355695
|
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|
Stack Exchange
|
Classification of absolute 2-limits?
Let $\mathcal V$ be a good enriching category. Recall that an enriched limit weight $\phi: D \to \mathcal V$ is called absolute if $\phi$-weighted limits are preserved by any $\mathcal V$-enriched functor whatsoever. If this is a familiar concept, please jump to my question at the end. But otherwise, I will provide some context. For instance,
Idempotent splitting is absolute for $Set$-enrichment and (hence) for any enriching category;
Finite products are absolute for enrichment in abelian groups, commutative monoids, etc;
Limits of Cauchy sequences are absolute for enrichment in $([0,\infty],\geq,+)$;
$\infty$-categorically, finite limits are absolute for enrichment in spectra;
Eilenberg-Moore objects for idempotent monads are absolute for enrichment in $Cat$.
There are a number of general things to say here:
(Street) A $\mathcal V$-weight $\phi: D \to \mathcal V$ is absolute if and only if $\phi$ has a left adjoint $\bar \phi: D^{op} \to \mathcal V$ in the bicategory of $\mathcal V$-categories and $\mathcal V$-profunctors.
Following Lawvere and the third example above, a $\mathcal V$-category with all absolute $\mathcal V$-limits is called Cauchy complete. The Cauchy completion of a $\mathcal V$-category $C$ is obtained by freely adjoining absolute $\mathcal V$-limits to $C$, and may be constructed as the $\mathcal V$-category of absolute weights on $C$. This construction exhibits the Cauchy complete $\mathcal V$-categories as a reflective subcategory of $\mathcal V\text{-}Cat$.
Thus a $\mathcal V$-weight $\phi: D \to \mathcal V$ is absolute if and only if it lies in the Cauchy completion of $D$.
These general facts are useful in working out explicit characterizations of the absolute weights for particular $\mathcal V$:
An ordinary ($Set$-enriched) category $C$ is Cauchy-complete iff $C$ has all split idempotents; a $Set$-enriched weight $\phi: D \to \mathcal V$ is absolute iff it is a retract of a representable (and in particular an ordinary conical limit is absolute iff the indexing diagram has a cofinal idempotent);
An $Ab$-enriched category $C$ is Cauchy-complete iff $C$ has split idempotents and finite products;
A generalized metric space is Cauchy-complete iff it is Cauchy-complete in the usual sense;
A spectrally-enriched $\infty$-category is Cauchy-complete iff it is stable with split idempotents;
But I don't know an analogous statement for $Cat$-enrichment.
Question: When does a 2-category $C$ have all absolute weighted 2-limits?
Clearly $C$ must have split idempotents and Eilenberg-Moore objects for idempotent monads -- do these suffice? Or are there more absolute 2-limit weights out there which can't be built from these?
I believe this is answered for general weak n-categories in https://arxiv.org/pdf/1905.09566.pdf (I'm not putting this as an answer because I havn't read the paper in enough detail to be really sure it answer your question, also it deals with weak, and not strict n-categories)
@SimonHenry Thank you-- this paper is inspiring, and potentially fully answers my question! It's also frustrating -- they seem to be trying to work with weak $(n,n)$-categories in a "model-independent" way, but with the exception of one section at the end, they are not very clear about precisely what general properties they require their model of weak $n$-categories to satisfy. As a result, many of the definitions, statements, and proofs feel imprecise to me. In fact, I think I'd feel more comfortable if they just claimed that their results hold for strict $n$-categories rather than weak ones!
I see your point, though I guess if you only care about the case of 2-categories that shouldn't really be a problem. For the record, if you plan to study this in more details, I would be interested if you wrote as an answer to your own question the details of what their results means for 2-categories.
I just came across this question. Our paper https://arxiv.org/pdf/1905.09566.pdf mentioned by @SimonHenry claims an answer to Tim's question. I am very sympathetic, though, to Tim's comment about models and model independence. To respond to his comment about strict versus weak: I actually don't believe our results in too strict a world. In particular, if you input a strict 2-category, say, and then write down our Karoubi completion, it will not be strict. You could strictify for 2-categories, but I think for 3-categories you may find non-strictifiable things coming out of our construction.
In my paper Condensations in higher categories joint with Davide Gaiotto, we claim the following answer. It deserves someone to write a careful model-dependent paper confirming it. I believe Martin Szyld has been thinking about how to do this.
Define a 2-idempotent to be an endomorphism $e : X \to X$ together with a retract $\mu: e^2 \Leftrightarrow e : \mu^*$, by which I mean 2-morphisms $\mu$ and $\mu^*$ such that $\mu \cdot \mu^* = \mathrm{id}_e$, which is associative, coassiciative, and Frobenius. [In spite of the name "$\mu^*$", it is just another 2-morphism, and isn't required to be the adjoint or dual in any sense beyond the requirement $\mu \cdot \mu^* = \mathrm{id}_e$. In particular, $\mu$ constrains but does not determine $\mu^*$.]
By Frobenius I mean that the three natural maps $e^2 \Rightarrow e^2$, namely $\mu^* \cdot \mu$, $(\mu \circ \mathrm{id}_e) \cdot \mathrm{assoc} \cdot (\mathrm{id}_e \circ \mu^*)$, and $(\mathrm{id}_e \circ \mu) \cdot \mathrm{assoc}^{-1} \cdot (\mu^* \circ \mathrm{id}_e)$, are all equal. [Actually, the Frobenius axiom together with $\mu \cdot \mu^* = \mathrm{id}_e$ imply the associativity of $\mu$ and coassociativity of $\mu^*$.] Here my notation is that $\circ$ is the composition in the 1-morphism direction, and $\cdot$ for 2-morphisms, and $\mathrm{assoc} : e \circ e^2 \cong e^2 \circ e$ is the associator.
This same notion goes by many names. For instance, it is a nonunital separated monad, or a nonunital special Frobenius monad. It is almost but not quite the same as "separable monad" used by Douglas and Reutter. In our paper, we give it yet another name — "condensation monad". But Reutter and I have started saying "2-idempotent" when we talk to each other, and perhaps it is the best name.
A 2-idempotent $(X,e,\dots)$ splits when there is an object $Y$, 1-morphisms $f : X \leftrightarrows Y: f^*$ [again, in spite of the name, I don't require any duality/adjunction], and some 2-morphisms and equations which I will list. First, I require the data of a retract $\phi : f\circ f^* \Leftrightarrow \mathrm{id}_Y : \phi^*$, i.e. 2-morphisms such that $\phi \cdot \phi^* = \mathrm{id}_{\mathrm{id}_Y}$. Now, using the retract $(\phi,\phi^*)$, I claim that you can give the composition $f^*\circ f$ the data of a 2-idempotent. Specifically, you set $\mu : (f^* \circ f) \circ (f^* \circ f) \Rightarrow (f^* \circ f) \circ (f^* \circ f)$ to be what you get by using an assotiator $(f^* \circ f) \circ (f^* \circ f) \cong f^* \circ (f \circ f^*) \circ f$ and then applying $\mathrm{id}_{f^*} \circ \phi \circ \mathrm{id}_f$ and then applying some unitors; and $\mu^*$ is the reverse. The last datum needed to say that $(X,e,\dots)$ splits is an isomorphism $e \cong f^* \circ f$ of 2-idempotents on $X$.
To connect with things you know: 2-idempotents are a version of monads, and splittings are a version of Eilenberg–Moore objects. The difference is that mine are not unital and are separable, in fact separated.
Then our claim is that a weak 2-category is Cauchy complete when (and only when) it is locally idempotent complete [i.e. all hom-categories are idempotent complete] and also every 2-idempotent splits. If your 2-category is locally idempotent, then a splitting of a 2-idempotent is unique up to unique isomorphism. I forget if this is true without local idempotent completion.
We also claim an explicit construction of the 2- (indeed, $n$-)categorical Karoubi completion. Given any 2-category, the first step is to locally Karoubi complete it. [Easy exercise: write down the 1-morphism composition in the local Karoubi-completion of a 2-category.] Now given a locally Karoubi complete 2-category $\mathcal{C}$, I build a new 2-category whose objects are the 2-idempotents in $\mathcal{C}$. A morphism $(X,e,\dots) \to (Y,f,\dots)$ is a morphism $m : X \to Y$ together with retracts $m\circ e \Leftrightarrow m$ and $f \circ m \Leftrightarrow m$ such that a bunch of equations hold making $m$ into a bimodule and bicomodule and cetera. A "bi-bimodule", perhaps? The 2-morphisms are natural: they are the homomorphisms of these bi-bimodules.
The interesting thing is the composition. Given $(m, \dots) : (X,e,\dots) \to (Y,f,\dots)$ and $(n, \dots) : (Y,f,\dots) \to (Z,g,\dots)$, I can look at $n \circ m : X \to Z$. Now I claim you can write down an idempotent $n \circ m \Rightarrow n\circ m$. The trick is to map $n \circ m \Rightarrow n \circ f \circ m \Rightarrow n \circ m$ where you use the $f$-coaction on $m$, and then the $f$-action on $n$. Well, you could have used the $f$-coaction on $n$ and then the $f$-action on $m$, but it turns out you will get the same thing. Since your 2-category is by assumption locally idempotent complete, you can split this idempotent. Actually, $n \circ m$ was already a bi-bimodule, and the idempotent you write down on it is a moprhism of bi-bimodules. So the splitting is a bi-bimodule. This is the composition in the 2-Karoubi completion.
You can see that to write this down as a bicategory, say, would require making some arbitrary choices: I just told you "split the thing", and not which splitting to take, so even if you started life as a strict 2-category, you won't end up strict, and you should not expect our construction to lead to any statement of the form "this is the Cauchy completion" in the world of strict 2-categories and their functors.
A 2-category is Cauchy complete (in the sense you describe) if and only if idempotents splits, which is, if and only if its underlying ordinary category is Cauchy complete.
This holds more generally in this context: the base of enrichement $\mathcal V=(\mathcal V_0,\otimes,I)$ is symmetryc monoidal closed and locally presentable, and the functor $\mathcal V_0(I,-):\mathcal V_0\to\mathbf{Set}$ is (weakly) cocontinuous and (weakly) strong monoidal. (weakly here means that the induced comparison maps need not be bijections, but just surjections.) In this case a $\mathcal V$-category is Cauchy complete if and only if it has splittings of idempotents. This is proven for example in Corollary 3.16 here.
The idea is that, given an absolute $\mathcal V$-weight $\phi\colon\mathcal D\to\mathcal V$, then the category of elements $\mathcal E$ of $\mathcal V_0(I,\phi_0-)\colon \mathcal D_0\to\mathbf{Set}$ is absolute (in the ordinary sense, i.e. $\mathcal E$-colimits are preserved by any functor) and that $\phi$ can be written as a conical colimit
$$ \phi\cong \textit{colim}(\mathcal{E}_{\mathcal V}\to\mathcal D^{op}\to[\mathcal D,\mathcal V] )$$
indexed on $\mathcal E$.
Examples of such bases are of course $\mathbf{Cat}$ and $\mathbf{Set}$, but also $\mathbf{SSet}$, $\mathbf{Pos}$, and $\mathcal V\textit{-}\mathbf{Cat}$ whenever $\mathcal V$ is locally presentable.
Thanks, this is really neat -- I will have to dig into your preprint! I'm a little surprised, in your last sentence, by the example of $Ab-Cat$ -- I would have thought $Ab-Cat$-enriched categories should have absolute finite coproducts. But thinking about it now, I suppose it's clear that $Ab-Cat-Cat$ doesn't have a zero object, dashing my expectations!
|
2025-03-21T14:48:30.140827
| 2020-03-25T19:52:30 |
355698
|
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|
Stack Exchange
|
On the number of connected functional digraphs recoverable from the preimage set size structure
I am studying the list of inverse images (preimage sets) of some function $f$ at a given inverse depth $j$ -- for each element $x_i$ of a finite domain $X$.
For example,
$P_j=\left[f^{-j}(x_1), f^{-j}(x_2), f^{-j}(x_3),\text{...},f^{-j}( x_n)\right]$
For each of these, we may populate the matrix
$P=\left(\begin{array}{cccc}
f^{-1}(x_{1}) & f^{-2}(x_{1}) & \cdots & f^{-n}(x_{1})\\
f^{-1}(x_{2}) & \ddots\\
\vdots\\
f^{-1}(x_{n}) & & & f^{-n}(x_{n})
\end{array}\right)$
However, of particular interest to me is what happens when we look at the sizes of such matrix elements.
Yielding a matrix $\Sigma$ with entries
$\Sigma=\left(\begin{array}{cccc}
\mid f^{-1}(x_{1})\mid & \cdots & \mid f^{-n}(x_{1})\mid \\
&\ddots\\
\vdots\\
\mid f^{-1}(x_{n})\mid & & \mid f^{-n}(x_{n})\mid
\end{array}\right)$
For example, for $f=(a,b),(b,a),(c,a)$ by calculating the second matrix we get $\Sigma_f=\left(\begin{array}{ccc}
2 & 1 & 2\\
1 & 2 & 1\\
0 & 0 & 0
\end{array}\right)$
Now, each preimage size (Sigma) matrix has to it associated a set of functions which have the matrix as their Sigma matrix. This is the partition of functions $X^X$ by the matrix structure.
I Suspect: 1. In each class $[f]={g:\Sigma g=\Sigma f}$ there is only one single-component functional digraph. 2. There are no two functions with the same size cycle
For example, for $f=(a,b),(b,a),(c,a),(d,b)$ and $g=(a,a),(b,b),(c,a),(d,b)$ we have the same sigma matrix
$\Sigma_{f,g}=\left(\begin{array}{ccc}
2 & 2 & 2 & 2\\
2 & 2 & 2 & 2\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}\right)$
However there is only one connected (maximal) functional digraph $f$
If anyone has some references or a suggestion on how to proceed I'd be grateful... I suspect that holding everything else the same, increasing/decreasing the cycle size in a connected functional digraph changes the preimage structure. Thus, to get cycles of length 1 (in the example above) the graph had to be disconnected. Similarly to increase the cycle size to 3 would require a different preimage structure...
Any help appreciated.
Fournier, Bradford M., "Towards a Theory of Recursive Function
Complexity: Sigma Matrices and Inverse Complexity Measures" (2015).
University of New Orleans Theses and Dissertations. 2072.
https://scholarworks.uno.edu/td/2072
|
2025-03-21T14:48:30.141119
| 2020-03-25T20:22:23 |
355699
|
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|
Stack Exchange
|
Has the equation $p^3-q^2+2=2^3\cdot q$ infinitely many solutions for $p$ and $q$ prime?
Consider the equation:
$p^3-q^2+2=8\cdot q$
Has this equation infinitely many solutions for $p$ and $q$ prime?
No, every Mordell curve has only finitely many integral points.
Actually, there are no prime solutions at all. See my response.
Every elliptic curve over a number field has only finitely many integral points. This was proven by Carl Ludwig Siegel. See chapter IX of Silverman's book "The Arithmetic of Elliptic Curves".
I'm voting to close this question because (a) there was no motivation provided or context for the question (b) the user did not engage with the comments and answers (c) the user has not been active on this account, and seems to instead be creating new burner accounts every few days
According to SAGE, the integral solutions are $(7,-23)$ and $(7,15)$:
sage: EllipticCurve([0,0,8,0,2]).integral_points(both_signs=True)
[(7 : -23 : 1), (7 : 15 : 1)]
Hence there are no (positive) prime solutions.
Actually there is no solution at all.
[8, 27, 125, 343, 1331, 2197, 4913, 6859, 12167, 24389]
[18, 31, 63, 103, 207, 271, 423, 511, 711, 1071]
For the first 10 primes, evaluated at ^3 and ^2 + 8* - 2 respectively. After that the ^3 goes way much faster than the other so no solution at all.
You seem to be assuming that $p$ and $q$ are equal.
|
2025-03-21T14:48:30.141255
| 2020-03-25T20:55:56 |
355701
|
{
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"authors": [
"Monroe Eskew",
"Noah Schoem",
"Otto",
"https://mathoverflow.net/users/11145",
"https://mathoverflow.net/users/119731",
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|
Stack Exchange
|
Can a weakly inaccessible non-weakly-Mahlo cardinal carry a $\kappa$-complete, $\kappa^+$-saturated ideal?
An ideal $I$ on a regular cardinal $\kappa$ is said to be $\mu$-saturated if whenever a family $\langle S_\alpha \mid \alpha<\lambda\rangle$ of subsets of $\kappa$ is such that each $S_\alpha\notin I$ and for $\alpha\neq \beta$, $S_\alpha\cap S_\beta\in I$, then $\lambda<\mu$.
Question: If $\kappa$ is weakly inaccessible and carries a $\kappa$-complete, $\kappa^+$-saturated ideal, must $\kappa$ be weakly Mahlo,
It is known, for instance, that if $\kappa$ carries a $\kappa$-complete, $\kappa$-saturated ideal, then $\kappa$ must be weakly Mahlo. Further, $\kappa$ may carry a $\kappa$-complete, $\kappa^+$-saturated ideal if $\kappa$ is a successor cardinal; these results are summarized in [2]. However, those consistency results for $\kappa$ inaccessible all have that $\kappa$ is Mahlo.
On face value, the argument in [1], Theorem 7.57, doesn't address the question either, as in the final model, there are no inaccessible cardinals.
[1] Foreman, M. (2010). Ideals and Generic Elementary Embeddings. In Handbook of Set Theory (pp. 885–1147). Dordrecht: Springer Netherlands. https://doi.org/10.1007/978-1-4020-5764-9_14
[2] Kunen, K. (1978). Saturated Ideals. Journal of Symbolic Logic, 43(1), 65–76. https://doi.org/10.2307/2271949
Have you checked the following: suppose $\kappa$ is an almost huge cardinal and there exists a $\kappa$-c.c forcing $P$ that forces $\kappa$ to be non-Mahlo (https://mathoverflow.net/questions/168460/destroying-the-mahloness-of-a-cardinal-with-kappa-c-c-forcing), suppose the almost huge-ness is witnessed by $j: V\to M$ such that $\lambda=j(\kappa)$ and $|j(\lambda)|=\lambda$, then do $P*Coll(\kappa, <\lambda)$ (should be analogous to Kunen's analysis).
I haven't but on the face of it this doesn't look unreasonable to me. Is the the almost huge embedding with $|j(\lambda)|=\lambda$ what guarantees the appropriate absorption of $P*Coll(\kappa,<\lambda)$ to $j(P)$?
No but you can always force again. In $V^P$, $j(P)/P$ is $\lambda$-cc, and $Coll(\kappa, <\lambda)$ is $\lambda$-Knaster, hence in $V^{PColl(\kappa, <\lambda)}$, you can force with $(j(P)/P)^{V^P}$ ($\lambda$-c.c), so in $V^{PColl*j(P)/P}$, first lift j to $j^1: V^P\to M^{j(P)}$, then use $|j(\lambda)|=\lambda$ to build a generic over $M$ for $j^1(Coll(\kappa, <\lambda))$ consistent with ${j^1}''G$ ($G\subset Coll(\kappa, <\lambda)$ generic over $V^P$).
@Otto It‘s not that simple or automatic. One has to engineer $P$ so that $j(P)$ absorbs $Col(\kappa,<\lambda)$
In general you can't in this situation. If $j(P)$ were to absorb $Coll(\kappa, <\lambda)$, then by elementarity, $P$ will have to force $\kappa$ to be a successor cardinal, which is not what we want.
|
2025-03-21T14:48:30.141460
| 2020-03-25T21:02:10 |
355703
|
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|
Stack Exchange
|
Equivalence of conditions for torsions of links to be defined
The torsion of a link complement $S^3 \setminus L$ is defined in terms of the twisted chain complex $C_*(S^3 \setminus L; \rho)$, where $\rho : S^3 \setminus L \to \operatorname{GL}_k(\mathbb{k})$ is a representation of the link complement into the group of matrices over a field $\mathbb{k}$. (For the usual Reidemeister torsion/Alexander polynomial, $\rho$ is the representation sending each meridian to a variable $t \in \operatorname{GL}_1(\mathbb{Q}(t))$, but more generally we can consider nonabelian representations.)
For the torsion to be defined, the complex $C_*(S^3 \setminus L; \rho)$ needs to be acyclic.
There are two conditions usually given in the literature. One is that there is some meridian $x$ with $\det(\rho(x) - I_k)$ nonzero. Another is related to the reduced Burau representation. If $L$ is represented as the closure of a braid $\beta$ on $n$ strands, then we have $n$ meridians $x_1, \dots, x_n$ corresponding to the strands of $\beta$. The second condition says that $C_*(S^3 \setminus L; \rho)$ is acyclic if $\det(\rho(x_1 \cdots x_n) - I_k)$ is nonzero.
Either of these conditions is sufficient, but is either necessary? Are they equivalent?
I'm not sure that $\det(\rho(x)-I_k) \neq 0$ is sufficient for $C_*(S^3 \setminus L;\rho)$ to be acyclic. Write $X_L:=S^3 \setminus L$. Let $\omega \in S^1 \setminus \lbrace 1 \rbrace$ and take the one-dimensional representation $\rho \colon \pi_1(X_L) \to \mathbb{C}^\times$ that maps each meridian to $\omega$. This way $\det(\rho(x)-I_1)=\omega-1 \neq 0$. However, I believe that $H_1(X_L;\rho)$ is zero if and only if $\Delta_L(\omega) \neq 0$ (the dimension of $H_1(X_L;\rho)$ is the nullity $\eta_L(\omega$)). Consequently, if $\omega$ is a root of $\Delta_L$, then I think that $C_*(X_L;\rho)$ is not acyclic.
So I think that the condition is necessary but not sufficient: the complex is acyclic if and only if all the $H_i(X_L;\rho)$ vanish. Your condition basically controls $H_0(X_L;\rho)$. If the complex is acyclic, then the $H_0$ must vanish which should imply that $\det(\rho(x)-I_k) \neq 0$.
Now to braids. From now on, $L=\widehat{\beta}$ is a braid closure and $\rho \colon F_n \to GL_k(\mathbb{k})$ is a representation of the free group that extends to a representation of $\pi_1(X_L).$ For the same reason as above, I'm not sure the condition is sufficient. I don't think it is necessary either. Assume $L$ is a knot, that $\beta$ has $n=3$ strands and that $\omega^3=1$. I am taking the same $1$-dimensional representation as above. Now $C_*(X_L;\rho)$ is acyclic ($\det(\rho(x)-I_1)=\omega-1 \neq 0$ and $\Delta_L(\omega) \neq 0$ because for a knot $K$, $\Delta_K(t)$ does not have roots that are prime powers of unity). However, by my choice of $\omega$, I have $\det(\rho(x_1x_2x_3)-I_1)=w^3-1=0$.
In particular, the conditions are not equivalent: for that example $\det(\rho(x)-I_k) \neq 0$ but $\det(\rho(x_1x_2\cdots x_n)-I_k) = 0$.
Finally, a topological remark. Write $D_n$ for the $n$ times punctured disc. The free group $\pi_1(D_n)$ has the $x_i$ as its generators, so morally $\partial D_n$ "is" $x_1x_2\cdots x_n$. Now $X_L$ can be obtained from the exterior of the closure of the braid in the solid torus, by adding on an extra solid torus. Equivalently, $X_L=X_{\widehat{\beta} \cup \partial D_n} \cup (D^2 \times \partial D_n)$. The torsion of $D^2 \times \partial D_n$ is your $\det(\rho(x_1x_2\cdots x_n)-I_k)$. Now the resulting Mayer-Vietoris sequence with twisted coefficients relates the acyclicity of $C_*(X_K;\rho)$ to $\det(\rho(x_1x_2\cdots x_n)-I_k)$.
Thanks for your answer! I think my confusion was caused by the twisted Alexander polynomial literature, where the representations are always tensor products of representations and the abelianization map, viewed as a map to $\mathbb{Z}[t^{\pm1}]$. Then the eigenvalues are multiples of $t$, hence transcendental, so they cannot be roots of the Alexander polynomial.
However, it does seem like you can still extract a well-defined torsion when $\det(\rho(x_i) - I) \ne 0$, even if the complex isn't acyclic. It might be that there's a canonical choice of basis of $H^1$ that is implicitly being chosen.
The context for this question is Proposition 3.6 of this paper, which might need to be modified in light of your answer.
Regarding your first comment. Take $\rho \colon \pi_1(X_L) \to \mathbb{Q}(t)$ mapping each meridian to $t$. If $L$ is an $n$-component boundary link, then its Alexander polynomial is identically zero and the dimension of $H_1(X_L;\mathbb{Q}(t))$ is $n-1$ and yet $\det(\rho(x)-I_1))=t-1 \neq 0$.
|
2025-03-21T14:48:30.141777
| 2020-03-25T21:24:37 |
355704
|
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"Ryan Vaughn",
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|
Stack Exchange
|
Injectivity radius of parallel hypersurfaces
Let $(M,g)$ be a Riemannian manifold and let $N$ be a compact hypersurface isometrically embedded into $M$ and let $\eta$ denote a choice of unit normal vector field on $N$. It is then true that $N$ admits an $\varepsilon$-tubular neighborhood. That is, there exists some $\varepsilon>0$ such that the normal exponential $\exp(t\eta)$ is a diffeomorphism onto its image for $t\in (-\varepsilon, \varepsilon)$. A generalization of the Gauss lemma implies that for each $t$, we have a "parallel" hypersurface $N_t$ which intersects the normal geodesics orthogonally and are distance $t$ from $N$.
I am interested if there has been a result relating the injectivity radius of $N_t$ to the injectivity radius of $N$. More specifically, it seems like it should be the case that the cut locus of $N_t$ varies continuously as one travels along the normal geodesic flow. That is, small change in $t$ should imply small change in the cut time of a point $p(t)$ moving along the normal geodesic. I have not seen a proof of this and to my knowledge the arguments for continuity of the injectivity radius of a manifold do not seem to generalize.
Has this been shown anywhere? The best result that I have found so far is the Riccati equation for tubular neighborhoods in Grays "Tubes" book:
$$
S'(t) = S^2(t)+R(t)
$$
where $S(t)$ is the shape operator of $N_t$ and $R(t)$ is defined by $R[X,Y] = \nabla_{[X,Y]} - [\nabla_X, \nabla_Y]$ on $N_t$. In this way, it seems like one should be able to relate the second fundamental form of $N_t$ to $N$ via integration. However, since the cut locus is not uniquely determined by the shape operator, this does not seem to be sufficient.
Has this result been shown before? Perhaps a weaker result: if we take $t$ to be on a compact interval $[-\varepsilon, \varepsilon]$, can we guarantee that the injectivity radius of $N_t$ does not shrink to zero as $t\rightarrow \varepsilon$?
Any help would be greatly appreciated!
Yes, the two injectivity radii are related.
Let $r>0$ be the injectivity radius from $N$. Then there is a neighborhood $U \subset M$ of $N$ and a Riemannian isometry $\phi : U \to (-r,r) \times N$. Each hypersurface $N_t$ as you describe, for $|t|<r$ will correspond to the slice $\{t\} \times N$ with the identification provided by $\phi$.
Identifying $U$ with $(-r,r)\times N$, the Riemannian metric has the form $dx^2 + h_x$ where, for all $x\in (-r,r)$, $h_x$ denotes a smooth one-parameter family of Riemannian metrics on $N$. This fact is just a generalization of the concept of normal coordinates (this is for sure proved in Gray's book, I think).
Furthermore, for all $p\in N$, the rays $\tau \mapsto (\tau,p)$, for $\tau \in (-r,r)$ are unit speed geodesics that realize the distance to $N$, that is $d_N((\tau,p)) = |\tau|$.
Therefore, if $|t|<r$, we have that the injectivity radius from $N_t$ is equal to $r-|t|$.
EDIT AFTER COMMENT
After Ryan's comment, I realize what the OP was really asking concerns the injectivity radius of the embedded submanifold $N_t$ with the metric induced by $(M,g)$, that is the injectivity radius of $(N,h_t)$, where $h_t$ is a one-parameter family of metrics on a compact manifold $N$ (induced by the ambient metric $g$, in this case, but this is not really important).
Since the function $h \mapsto i_h(N)$, defined on the space of smooth Riemannian structure equipped with the $C^2$ topology, is continous, and since the aforementioned family $h_t$ is smooth in $t$, for small $t$, the result easily follows. The continuity of $h \mapsto i_h(N)$ is proved here:
Ehrlich, Paul E., Continuity properties of the injectivity radius function, Compos. Math. 29, 151-178 (1974). ZBL0289.53034.
and see also the following reference for perhaps simpler statements and a more accessible proof:
Sakai, Takashi, On continuity of injectivity radius function, Math. J. Okayama Univ. 25, 91-97 (1983). ZBL0525.53053.
Since the proof above is not simple, let me show how one can prove your weaker claim. For a given manifold $(N,h)$ there exist a beautiful universal lower bound for its injectivity radius given explicitly in terms of a lower bound $\delta$ for the Ricci curvature, an upper bound $\Delta$ for the sectional curvature, the diameter $D$ and the volume $V$. This is originally due to Cheeger, but the lower bound can be recovered by more "elementary" arguments such as (i) the Klingenberg Lemma and (ii) the Heintze-Karcher inequality (cf. Note <IP_ADDRESS> and references therein in Berger's "Panoramic view of Riemannian geometry). In particular one obtains
$$
i_h(N) \geq \inf\left\{ \frac{\pi}{\sqrt{\Delta}}, c_n \left(\frac{\sqrt{|\delta|}}{\sinh(\sqrt{|\delta|} D)}\right)^{n-1} V \right\}
$$
where $c_n$ is a universal dimensional consant.
Applyting this estimate to your specific case $(N,h_t)$ shows that the injectivity radious cannot become zero for small $t$.
Clearly $t \mapsto i_{h_t}(N)$ can become zero when your family $N_t$ approaches the normal cut locus, as $N_t$ can collapse. Consider for example the unit circle in $\mathbb{R}^2$, for which $N_t$ collapses to a point as $t \to 1$.
Interesting. I don't quite follow the last sentence. It seems you are talking about the "normal cut locus" to $N$, that is, the distance $r$ for which the geodesics intersecting $N$ orthogonally are unique and length minimizing.
I am interested in the injectivity radius of the submanifold $N_t$, i.e. the relationship between the injectivity radius of $N$ endowed with the metric $h_x$ as compared to the metric on $N$.
Thank you for the great edit! This was exactly what I was looking for.
|
2025-03-21T14:48:30.142162
| 2020-03-25T21:26:12 |
355705
|
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|
Stack Exchange
|
Properties of sequences associated to Nakayama algebras
Assume Nakayama algebras are connected and given by quiver and relations. Note that such Nakayama algebras have global dimension at most $2n-2$ in case it is finite and the algebra has $n$ simples.
For $n$ fixed and $1 \leq k \leq 2n-2$ let $a_{n,k}$ be the minimal vector space dimension of a Nakayama algebra with $n$ simples and global dimension $k$.
Question 1: Do we have $a_{n,i} \geq a_{n,i+1}$ for $i=1,...,n-1$ and $a_{n,i} \leq a_{n,i+1}$ for $i=n-1,...,2n-2$?
(What is the name for such sequences by the way? Reverse unimodal?)
Here $a_{n,k}$ for $n=8$:
[ 36, 18, 17, 16, 16, 16, 15, 17, 20, 23, 29, 35, 57, 71 ]
For $n$ fixed and $1 \leq k \leq 2n-2$ let $b_{n,k}$ be the maximal vector space dimension of a Nakayama algebra with $n$ simples and global dimension $k$.
Question 2: Do we have $b_{n,i} < b_{n,i+1}$ for $i$ odd and $b_{n,i}>b_{n,i+1}$ for $i$ even?
Here $b_{n,k}$ for $n=8$:
[ 36, 92, 77, 91, 71, 85, 66, 80, 62, 76, 59, 73, 57, 71 ]
Both questions have a positive answer for $n \leq 9$.
|
2025-03-21T14:48:30.142256
| 2020-03-25T22:52:56 |
355708
|
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|
Stack Exchange
|
Does there exist a rational polynomial $P(x)\in{\mathbb Q}[x]{}$ such that $P(\zeta(s))=\zeta(P(s))$?
let $P(x)\in{\mathbb Q}[x]{}$ be a rational polynomial with $P(1) >1$ and $\zeta $ be the Riemann zeta function , I want to know if there exist a rational polynomial such that $P(\zeta(s))=\zeta(P(s))$ with $s>1$ ?
I have been working on a related problem, and my conclusion is that no injective $P$ fulfilling your requirements exists.
@SylvainJULIEN: Polynomials of degree at least $2$ are not injective on $\mathbb{C}$.
Indeed, but the OP doesn't forbid degree $1$ polynomials.
The question doesn't match with the title.
Extending the argument by GH from MO, $\zeta(P(s))$ has a pole for any $s$ such that $P(s)=1$, while $P(\zeta(s))$ has unique pole for $s=1$. Therefore if $\zeta(P(s))=P(\zeta(s))$, then $P(s)=1$ has unique solution $s=1$ and $P(x)-1=c(x-1)^n$ for some complex $c$ and positive integer $n$, and we get $\zeta(1+c(s-1)^n)=1+c(\zeta(s)-1)^n$. For $s=1+x$ with small $x$ the equality of leading asymptotic terms gives $1/(cx^n)=c/x^n$, $c=\pm 1$. For $s=2$ we get $\zeta(1+c)=1+c(\zeta(2)-1)^n$, i.e., either
(i) $c=1$, $\zeta(2)-1=(\zeta(2)-1)^n$, $n=1$, $P(x)=x$, or
(ii)$c=-1$, $-3/2=-(\zeta(2)-1)^n$, this is impossible (since $0<\zeta(2)-1<1$).
No. If $P$ has a positive leading coefficient, then letting $s$ go to infinity and using continuity of $P$ we get $P(1)=1$. If $P$ has a negative leading coefficient, then $\zeta(P(s))$ has zeros at arbitrarily large $s$, while $P(\zeta(s))$ does not.
Doesn't your argument also apply for polynomials in $\mathbb{R}[x]$?
@SylvainJULIEN: There is a simple proof for all complex polynomials. See my post below.
It is straightforward to see that if $P\in\mathbb{C}[x]$ satisfies $P(\zeta(s))=\zeta(P(s))$, then $P(1)=1$. Note that if $P(\zeta(s))=\zeta(P(s))$ holds for all real $s>1$, then it also holds for all complex $s\neq 1$ by the uniqueness of analytic continuation.
Indeed, $\zeta(s)$ has a pole at $s=1$, hence $P(\zeta(s))=\zeta(P(s))$ also has a pole at $s=1$. This implies that $P(1)=1$.
Added. It seems easy to show that $P(x)=x$ is the only non-constant polynomial satisfying the conditions. (Indeed, this is true, see Fedor Petrov's response.)
Can your argument be used to show that any complex function commuting to $\zeta$ is necessarily continuous?
@SylvainJULIEN, nice idea probably you meant uses of Voronin's universality to show any commuting complex function to R zeta function must be continious
@GH from MO , would be the same simple proof with s lie in the critical strip ?
@zeraoulia rafik: with $s$ in the critical strip, you get that if $z$ is a non trivial zero of $\zeta$, then $P(\zeta(z))=\zeta(P(z))=P(0)$. As this equality doesn't depend on the zero, it means $P$ is a permutation of the multiset of the non trivial zeros of $\zeta$ and thus $P(0)=0$.
@zeraouliarafik: If $P(\zeta(s))=\zeta(P(s))$ holds at infinitely many distinct points $(s_1,s_2,\dots)$ with a limit point in $\mathbb{C}$, then it also holds for all complex $s\neq 1$ by the uniqueness of analytic continuation. Hence we infer $P(1)=1$ as before.
@SylvainJULIEN: My intuition is that there are plenty of non-continuous functions commuting with $\zeta$, but I don't want to think about it.
And as a permutation is bijective, if $P$ is continuous, it is either the identity or the complex conjugation.
@SylvainJULIEN: There are also plenty of continuous functions from $\mathbb{C}$ to $\mathbb{C}$ that permute the zeros of $\zeta(s)$, not just the identity and complex conjugation.
But do these functions commute with $\zeta$?
@SylvainJULIEN: I think there are plenty of continuous and non-continuous functions from $\mathbb{C}$ to $\mathbb{C}$ that commute with $\zeta$. On the other hand, there probably very few holomorphic of meromorphic functions that commute with $\zeta$. At any rate, I stop here a this site is not a discussion board, and also these questions appear to be pretty random (they have little to do with $\zeta$).
Here is a simple argument which applies to polynomials in $C[x]$, and even to most rational functions in $C(x)$. $\zeta$ is a meromorphic function in the plane.
So the Nevanlinna characteristic $T(r,\zeta)$ is defined. It is a positive increasing function, a kind of transcendental analog of the degree of a rational function. Your equation implies
that $(\deg P+o(1))T(r,\zeta)=T(r^d(1+o(1),\zeta).$ Here $d$ is the order of pole
of $P$ at $\infty$, for the case of polynomial, $d=\deg P$, of course.
If $d\leq 0$ then the right hand side is bounded, if $d\geq 1$, we obtain a contradiction since for every transcendental meromorphic function $T(r,f)/\log r\to\infty$. So only the case $d=1$ remains, but for this case the completion of the proof is elementary. Of course I understand that this is too heavy artillery to solve such an elementary question:-)
|
2025-03-21T14:48:30.142842
| 2020-03-25T22:55:04 |
355709
|
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|
Stack Exchange
|
Clarify formula for Steifel-Whitney (Poincaré dual) homology classes in a barycentric subdivision?
Let $X$ be a triangulated manifold of dimension $n$. Let $[W_{n-p}] \in H_{n-p}(X,\mathbb{Z}_2)$, be the homology class that's Poincaré dual to the $p$-th Stiefel-Whitney class $[w_p] \in H^p(X,\mathbb{Z}_2)$. There is a canonical chain-level formula, see Halperin-Toledo, for $W_{n-p} \in Z_{n-p}(X,\mathbb{Z}_2)$ given some barycentric subdivision of a triangulation of $X$: that,
$$W_{n-p} = \sum_{\sigma_p \subset K'} \sigma_p$$
where $K'$ is the set of all simplices in the barycentric subdivision, and $\sigma_p$ runs over the $p$-simplices of K'. So, $W_{n-p}$ is represented by the sum of the duals of all $p$-simplices in $K'$.
I seem to be misunderstanding what this formula is saying, and it doesn't make sense to me, e.g. in 2 dimensions. Let's pick an example of some manifold such that $[w_1]$ and $[w_2]$ don't vanish, like $\mathbb{R}P^2$. I'm seeing two issues with the above formula.
First, in any triangulation of any two manifold, the dual graph is trivalent (since every face in the original 1-skeleton is bounded by a triangle). So, taking the sum of every 1-cell in $K'$ has a nonempty boundary, since at every vertex there are three edges: the boundary would in fact be every single vertex! So, the chain supposedly representing $w_1$ isn't even a cycle.
Second, suppose that the original triangulation had $F$ faces (2-simplices). Then the barycentric subdivision would have $6 F$ faces, which is an even number. So, the sum of all the (dual) 0-cells would be a boundary, since there's an even number of them! And its (mod 2) integral would be 0. This can't represent $[w_2]$ if $[w_2] \neq 0$.
How exactly should one interpret this formula? A picture for 2D may help clarify this.
EDIT: Actually, it seems like the formulas don't pass to the dual lattice at all, they are used to view the chains as living on the triangulation itself. If this is the case, then the formula actually matches up for the case of a triangulation of $\mathbb{R}P^2$, that for its barycentric subdivision there's an odd number of 0-simplices and that all the 1-simplices give a dual of $w_1$
Would you introduce $X$, $n$ and $p$?
I thought this result was due to Whitney? Perhaps I'm mis-remembering.
I guess it is, but I think it first appears in print in Halperin-Toledo (they mention in their paper Whitney didn't print a book in which he proves it). I'll change the wording though.
What does $<$ represent? The subset relation?
@OlivierBégassat, the sum is over all $p$-simplices of K'. I reworded that bit.
Have you looked at the Goldstein-Turner paper, in Proc AMS 58 (1976) on this topic? It appears to be a simplification of Whitney's result.
"since at every vertex there are three edges" seems false to me. K' is not the dual complex, it is the barycentric subdivision of K. The central vertex in the barycentric subdivision of $\Delta^2$ has degree 6, and degree four for the barycenters of 1-simplices in a closed surface. And the comment about faces doesn't seem to make sense. The chain $C_n$ is always just the fundamental class, the sum of each n-simplex. That's dual to $w_0 = 1 \in H^0$. (I now see you caught this in your edit, but maybe the comment about the fundamental class helps others.)
|
2025-03-21T14:48:30.143105
| 2020-03-25T22:59:58 |
355710
|
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"LSpice",
"darij grinberg",
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}
|
Stack Exchange
|
Has vol. 3A of Cullis's "Matrices and Determinoids" been scanned and vol. 3B been archived?
This is a borderline question, but I'm going to risk posing it.
Cuthbert Edmund Cullis (1875?-1955?) was a somewhat obscure British mathematician whose opus magnum was a multi-volume treatise called Matrices and Determinoids. While his notation is somewhat idiosyncratic, it was a fairly systematic exposition of most(?) of the linear algebra known by the 1920s (heavy on determinants and minors, but also including invariant factors and canonical forms, or at least things looking like them), along with lots of new results. Some of his innovations, I believe, would be of interest even to us modern people: in particular, in the first 200 pages of Volume 3A, he seems to algorithmically construct multi-resultants using multisymmetric polynomials (aka MacMahon symmetric polynomials). I have never seen this done anywhere else.
Three volumes of his treatise were published: volume 1, volume 2 and volume 3A. According to his necrologue, he left behind a (plan of?) volume 3B. Thus, I'm wondering:
Question 1. Has Volume 3A ever been digitized?
Question 2. Is the manuscript of Volume 3B available anywhere?
The reason for Question 2 is mostly curiosity, but I have a rather practical reason for Question 1: I'm interested in understanding multi-resultants elementarily and also in applying multisymmetric polynomials. The twist is that I have access to Volume 3A in hardcopy for the rest of this week, but before I leave this library I'd like to know if it's worth scanning the doorstopper (700 pages!) or someone has already done that work.
I had never encountered the term 'necrologue'. Spotlight's search, which I turned to to query the Apple dictionary, helpfully informs me that the word has occurred in my Internet history, on this page.
@LSpice: Wow, I always thought that was a word in English. (Well, almost.)
Q1: Volume 3 part 1 (a.k.a. volume 3A) has been digitized and reissued as a paperback by Cambridge UP, see Amazon. The digital version is online in the HathiTrust digital library, but with limited "search only" access because of copyright restrictions.
With some further search I actually found a library in India that has placed the digital version online (a 54 MB download). Quite possibly not copyright compliant, but in the spirit of the National Emergency Library I guess in these difficult times a download can be justified.
Hmm. I assume HathiTrust's "search only" does not translate into download if you're on the right university network, does it?
Thank you for the download link!
As for copyright -- if it's not compliant now, it will be next year, as it was published in 1925.
|
2025-03-21T14:48:30.143418
| 2020-03-25T23:36:18 |
355714
|
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"url": "https://mathoverflow.net/questions/355714"
}
|
Stack Exchange
|
A non nuclear $C^*$ algebra $A$ for which the algebraic tensor product $A\otimes A$ admits a unique $C^*$ norm
Is there a non nuclear $C^*$ algebra $A$ for which the minimum and maximum $C^*$ norms on $A\otimes A$ coincide?
A somewhat similar question is discussed here.
Pisier https://arxiv.org/abs/1908.02705 very recently constructed a non-nuclear $C^\ast$-algebra $A$ with the weak expectation property (WEP) and the local lifting property (LLP). By a celebrated result of Kirchberg (see Corollary 13.2.5 in Brown and Ozawa's book) it follows that if $B$ has WEP and $C$ has LLP, then $B\otimes_{\max{}} C = B\otimes_{\min{}} C$. Hence $A\otimes_{\max{}} A = A\otimes_{\min{}} A$.
|
2025-03-21T14:48:30.143488
| 2020-03-26T00:23:26 |
355717
|
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|
Stack Exchange
|
Any cobordism invariant made of "characteristic classes", on unorientable manifolds, must be a mod 2 class?
For any cobordism invariant (or simply bordism invariant) quantity $\omega$ that satisfy the conditions:
$\omega$ can be fully decomposed from the cup product of characteristic classes (such as Stiefel Whitney class, Chern class, etc.).
$\omega$ can be defined on an unorientable smooth manifold.
then question:
Could we show that this $\omega$ has to be $\mathbb{Z}/2\mathbb{Z} : = \mathbb{Z}_2$ valued (a mod 2 class)? Is this true or false?
From the recent post: Cobordism invariants: topological v.s. geometric, I recall that:
(1). The 2d $Pin^-$ bordism invariant (defined on unorientable manifolds) from $\Omega_2^{Pin^-}=\mathbb{Z}_8$, where any 2-manifold $M$ always admits a $Pin^-$ structure. The $\mathbb{Z}_8$ is a Arf-Brown-Kervaire (ABK) invariant. $Pin^-$ structures are in one-to-one correspondence with quadratic enhancements
$q: H^1(M,\mathbb{Z}_2)\to\mathbb{Z}_4$ such that $x,y \in \mathbb{Z}_2$, we have
$q(x+y)-q(x)-q(y)=2\int_M x\cup y\mod4.$ In particular,
$q(x)=\int_M x\cup x\mod2.$ The $x,y$ counts as a cohomology class, but the ABK invariant seems not be decomposed from decomposed from a cup product of characteristic classes only? But does this count as a counter example of the above statement?
(2) The 4d $\Omega_4^{Pin^+}(B^2 \mathbb{Z}_2)=\mathbb{Z}_4 \times \mathbb{Z}_{16}$ contains an $\eta$ invariant of $\mathbb{Z}_{16}$, and a quadratic refinement of $\mathbb{Z}_4$ (from $B \in H^2(M,\mathbb{Z}_2)$ to $H^4(M,\mathbb{Z}_4)$. (It is probably similar to Pontryagin square $P(B)$ of $B \in H^2(M,\mathbb{Z}_2)$; but Pontryagin square $P(B)$ is defined on an orientable $SO$ structure, here I am looking at the unorientable $Pin^+$ structure. But does this count as a counter-example of the above statement?
If the above statement is false, can one give counter examples?
Many thanks for comments/answers.
|
2025-03-21T14:48:30.143623
| 2020-03-26T05:55:54 |
355725
|
{
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"site": "mathoverflow.net",
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"url": "https://mathoverflow.net/questions/355725"
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|
Stack Exchange
|
Reference request: Transverse parabolic Schauder estimates
Is there a version of the parabolic Schauder estimates for transversely parabolic linear PDE's on a manifold with a Riemannian foliation for functions that are constant on the leaves of the foliation? If so, could you provide a reference? Many thanks!
|
2025-03-21T14:48:30.143669
| 2020-03-26T08:33:19 |
355729
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Yin Mingzhou",
"https://mathoverflow.net/users/154104"
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"url": "https://mathoverflow.net/questions/355729"
}
|
Stack Exchange
|
Least squares with matrix product constraints
I am encountering a constrained LS problem with the following structure:
$$
\text{min}_Q\ \sum_{i=1}^M ||Q_i X_i-Y_i||_F^2
$$
$$
\text{s.t. }\ Q_M Q_{M-1}\cdots Q_1=I,
$$
where $Q_i,X_i,Y_i\in \mathbb{R}^{n\times n}$ are full rank.
I understand this can be a very hard problem. I just want to know if there is a numerically converging algorithm for this type of problem by any chance.
@FedericoPoloni Sorry, it should be a sum of squared norms, but $Q$ is not orthogonal.
|
2025-03-21T14:48:30.143734
| 2020-03-26T09:01:47 |
355732
|
{
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"Fedor Petrov",
"Samdney",
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"https://mathoverflow.net/users/4312"
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"site": "mathoverflow.net",
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"url": "https://mathoverflow.net/questions/355732"
}
|
Stack Exchange
|
Distinct distances between adjacent equal elements
Let's call a sequence $a_1, \ldots, a_n$ suitable if for any positive integer $d$ there is at most one index $i$ such that $a_i = a_{i + d}$ and all elements $a_{i + 1}, \ldots, a_{i + d - 1}$ are not equal to $a_i$.
For each $k$, I'm interested in longest suitable sequences with all elements in $\{0, \ldots, k - 1\}$. There is a suitable sequence of length $3k - 1$: start with numbers $0, \ldots, k - 1$ in order, followed by first $2k - 1$ elements of A025480. E.g., for $k = 3$ this sequence would look as follows: $0, 1, 2, 0, 0, 1, 0, 2$. It isn't difficult to prove that this pattern works for any $k$.
With brute-force I've discovered a few curious observations:
$3k - 1$ appears to be the maximum length of a suitable sequence with elements in $\{0, \ldots, k - 1\}$;
The number of longest suitable sequences appears to be $k! \times $A002047$[k]$.
How can this be explained?
I think, it is A002047[k-1], not A002047[k]
This can be explained as follows.
Assume that $(a_1,\ldots,a_{n})$ is a suitable sequence.
For every $j\in \{0,\ldots,k-1\}$ denote $A(j)=\{i:a(i)=j\}$ and denote by $m(i)$ and $M(i)$ the minimal and maximal, respectively, elements of $A(j)$. Then $$\sum_{j=0}^{k-1} \left(M(i)-m(i)\right)\geqslant 1+2+\ldots+(n-k),$$
since LHS equals to the sum of distances between adjacent equal elements, and there are at least $n-k$ such distances (exactly $n-k$ if $A(j)\ne \emptyset$ for all $j$). Since all guys $m(j)$ are distinct, their sum is at least $1+2+\ldots+k$, analogously the sum of $M(j)$ is at most $(n-k+1)+\ldots+n$, thus $$
\sum_{j=0}^{k-1} M(i)-m(i)\leqslant (n-k+1)+\ldots+n-(1+\ldots+k)=k(n-k),$$
and we get $k(n-k)\geqslant 1+\ldots+n-k=(n-k)(n-k+1)/2$, or $n\leqslant 3k-1$.
Also we get that if $n=3k-1$ is the length of a suitable sequence $(a_1,\ldots,a_{3k-1})$, then this may be only possible if $a_1,\ldots,a_k$ are all distinct, and so are $a_{2k-1},\ldots,a_{3k-1}$. For $s=1,2,\ldots,2k-1$ denote by $f(i)$ the minimal positive number such that $a_i=a_{i+f(i)}$. Then $f(1),\ldots,f(2k-1)$ are well-defined and form a permutation of $1,\ldots,2k-1$. Also the numbers $i+f(i)$ must form a permutation of $k+1,\ldots,3k-1$. And if this all happens, we get $k!$ corresponding suitable sequences. It remains to note that then $(i-k,f(i)-k,2k-i-f(i))$ for $i=1,2,\ldots,2k-1$ are the columns of a $3\times (2k-1)$ zero-sum array (see definition atA002074), and viceversa.
You can do the following considerations, at first for the Grundy values (A025480) which are given by
$a\left(2n\right) = n \quad \mathrm{and} \quad a\left(2n+1\right) = a\left(n\right)$
At first, we will define $m^{e} := 2n$ (m is even) respectively $m^{o} := 2n + 1$ (m is odd) and hence, we can rewrite this to
$a\left(m^{e}\right) = \frac{m^{e}}{2} \quad \mathrm{and} \quad a\left(m^{o}\right) = a\left(\frac{m^{o}-1}{2}\right)$
If we look at our examples and the definition of Grundy values, we see that starting by any $m^{o}$ the calculation of the final element stops if we reach a $m^{e}$ after an certain number of odd $m^{o}$'s. So, the $m^{e}$ are our termination cases of element computation.
We will determine an equation which connects the starting element $m_{1}^{o}$ with the final termination element $m_{i+1}^{e}$.
For this we get in general for only odd steps:
$m_{i}^{o} = \frac{m_{1}^{o} - \sum_{k=0}^{i-2}2^{k}}{2^{i-1}}\\
= \frac{m_{1}^{o} - 2^{0} - \sum_{k=1}^{i-2}2^{k}}{2^{i-1}}\\
= \frac{m_{1}^{o} - 1 - 2\frac{2^{i-2} - 1}{2 - 1}}{2^{i-1}}\\
= \frac{m_{1}^{o} - 2^{i-1} + 1}{2^{i-1}}$
for all $i \geq 2$, $n \in \mathbb{N}$. To determine the final sequence element, we have to do one even step:
$n_{i+1}^{e} = \frac{m_{1}^{o} - 2^{i-1} + 1}{2^{i-1}} \cdot \frac{1}{2}\\
= \frac{\left(2n_{1}^{o} + 1\right) - 2^{i-1} + 1}{2^{i}}\\
= \frac{2n_{1}^{o} - 2^{i-1} + 2}{2^{i}}\\
= \frac{n_{1}^{o} - 2^{i-2} + 1}{2^{i-1}}$
with $m_{1}^{o} = 2n_{1}^{o} + 1$ and $m_{i}^{e} = 2n_{i+1}^{e}$. We will solve the equation for $n_{1}^{o}$:
$2^{i-1}n_{i+1}^{e} = n_{1}^{o} - 2^{i-2} + 1$
$n_{1}^{o} = 2^{i-1}n_{i+1}^{e} + 2^{i-2} - 1$
Now, we want to use the result from above for some distance examinations.
We want to determine the first appearances of a particular number within the Grundy sequence.
At first at all, we have the first appearance of a number simple given by an even step. So, $n_{i+1}^{e}$ appears for $m_{i+1}^{e} = 2n_{i+1}^{e}$, because of $a\left(m_{i+1}^{e}\right) = a\left(2n_{i+1}^{e}\right) = n_{i+1}^{e}$.
So, to determine when this number $n_{i+1}^{e}$ appears the next, second time, within the Grundy sequence, we simple have take the equation for $i=2$:
$n_{1,1}^{o} = 2^{2-1}n_{i+1}^{e} + 2^{2-2} - 1\\
= 2n_{i+1}^{e}$
Next, we are interested in the positions of sequence elements.
The position $pos$ of a number $n_{i+1}^{e}$ within a simple integer sequence $0,1, \dots, k-2, k-1$ is given by
$n_{i+1,pos}^{e} = n_{i+1}^{e}$
and the position $pos$ of numbers $n_{i}^{u}$ respectively $n_{i}^{e}$ are given by
$n_{i,pos}^{o} = 2n_{i}^{o} + 2 \quad \mathrm{and} \quad n_{i,pos}^{e} = 2n_{i}^{e} + 1$
Now, we want to calculate the position distance for our given problem statement sequence.
$|n_{1,1,pos}^{o} - n_{i+1,pos}^{e}| = 2n_{1}^{o} + 2 - n_{i+1}^{e}\\
= 2\left(2n_{i+1}^{e}\right) + 2 - n_{i+1}^{e}\\
= 4n_{i+1}^{e} + 2 - n_{i+1}^{e}\\
= 3n_{i+1}^{e} + 2$
We start counting the sequence by $1$. Since we want to have a look at the original problem statement with a given pre-sequence $\{0,1,\dots, k-2,k-1\}$, we have to resubstitute the solution by $n_{i+1}^{e} - 1$ to
$|n_{1,1,pos}^{o} - n_{i+1,pos}^{e}| = 3\left(n_{i+1}^{e} - 1\right) + 2\\
= 3n_{i+1}^{e} - 3 + 2\\
= 3n_{i+1}^{e} - 1$
I also wrote it all together in a pdf version with some additional calculations. You can find it here: more detailed pdf Version
"The number of longest suitable sequences appears to be k!×A002047[k]." Can you please explain this more or give an example.
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