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2025-03-21T14:48:32.074977
| 2020-09-20T14:51:32 |
372155
|
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|
Stack Exchange
|
Reference for Grothendieck's theorem on representation of unramified functors
In the Exposé 294 of the Bourbaki Seminar of the year 1964-1965, Murre gives an outline of proof of a theorem of Grothendieck giving necessary and sufficient conditions of representability by a scheme for a functor which is unramified and separated. This is very interesting because it is probably the first general representability theorem that did not use projective methods: a precursor to Artin's representability theorems.
I am wondering if there exists a more complete and cleaned-up treatment somewhere in the literature, with full proofs. This would be nice for at least two reasons:
The outline of Murre is very sketchy at times, and some details would be very welcome.
There are some complements and generalizations of the theorem whose proof is lacking. For instance, after the statement of the main theorem there is a Note saying « it is possible to prove a statement similar to theorem 1 with "unramified" replaced by "locally quasi-finite" over S. »
Does anyone know such a treatment ?
I agree that this result would benefit from being reformulated using algebraic spaces. Eg show representability by algebraic spaces using Artin's criterion, and then use the fact that a quasi-finite separated morphism of aglebraic spaces is representable by schemes. I feel the same about Raynaud's, Un critère d’effectivité de descente, Th 4.1
Right! On the other hand, the specific situation of unramified functors allows to simplify a lot the deformation-theoretic arguments, which is what Grothendieck does (and is the reason why he was able to find an "Artin criteria" theorem before algebraic spaces where discovered). This makes this result interesting.
Well formal versality is rather easy to check using Schlessinger's criterion. Then you need algebraicity (for effectivity). Artin's approximation is a big theorem, but can be used as a black box. In practice the hard part to check is openness of versality. But for a smooth alg space (over the base), this is a lot easier to check (no obstruction to deformations). In particular this apply in the étale case. Now an unramified morphism is étale locally a closed immersion. So essentially you just need the étale case of Artin's axioms, and then use representability of loc. qf separated morphisms.
So I think we mostly agree that unramified functors simplify the deformation-theoretic arguments, but my point of view is that more generally the smooth case simplify deformation-theoretic arguments. To get the unramified case we just need the fact that a sheaf which is étale locally an algebraic space is algebraic.
|
2025-03-21T14:48:32.075439
| 2020-09-20T15:50:46 |
372158
|
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|
Stack Exchange
|
Do we have classification (upto Morita equivalence) of Lie groupoids?
Vague question is the following:
Is there a classifcation of Lie groupoids?
Slightly less vague question is the following:
Is there a (short?) list of "types" of Lie groupoids such that any arbitrary Lie groupoid is Morita equivalent to one from the list?
Some Lie groupoids I know are the following:
Lie groupoids coming from manifolds $[M\rightrightarrows M]$,
Lie groupoids coming from Lie group $[G\rightrightarrows *]$,
Lie groupoids coming from action of a Lie group on a manifold $[G\times M\rightrightarrows M]$,
Lie groupoids coming from foliation on a manifold $\mathcal{F}(M)$,
Lie groupoids coming from a submersion, the submersion groupoid $[M\times_NM\rightrightarrows M]$
Lie groupoids coming from a principal $G$-bundle, the Gauge groupoid,
Lie groupoids coming from a vector bundle over $M$, the general linear Lie groupoid $[GL(E)\rightrightarrows M]$.
Do we know any "interesting" Lie groupoids that do not belong (not Morita equivalent) to the above list?
Only classification I know is if a Lie groupoid is proper and étale, then it is locally looks like action Lie groupoid.
I might have forgot one or two types, but these are the only types of Lie groupoids I have come across till now.
Edit: It turns out my question was not conveyed correctly. I do know that there is no hope of classifying Lie groups. I do not want to classify Lie groups. For me, all Lie groupoids that comes from Lie groups are of "same type". So, the classification I am looking for is only based on types. I think I conveyed it correctly now. If this is not clear, please do ask for more clarification.
Lie groups sit inside Lie groupoids via the delooping construction, and there is no hope of classifying Lie groups.
@DmitriPavlov It looks like I have not mentioned it clearly.. Yes, I do know that there is no hope of classifying Lie groups... I do not want to classify Lie groups :D For me Lie groupoids coming from Lie groups are of "same type"... Does it now make it clear what I was trying to understand?
What is a “type” of a Lie groupoid? Your list of 7 items looks like a fairly ordinary list of examples of Lie groupoids that one would give in a typical expository text after a definition. Clearly, there are many more examples, and I do not understand why one would expect to exhaust all Lie groupoids by adding other examples to this list.
@DmitriPavlov "type" of a Lie groupoid is not a standard terminology, I just invented now... I call Lie groupoids coming from Lie groups as one type.. Lie groupoids coming from manifolds as one type.. Lie groupoids coming from foliations as one type and so on... Let me rephrase this in another way. Are there any "interesting" examples of Lie groupoids which are not Morita equivalent to any of the above 7 types? Here, interesting does not have a unique meaning and I leave it to you to interpret in a fair manner :)
So it appears that the actual question being asked here is “What are some interesting examples or constructions of Lie groupoids?”. This is radically different from classifying Lie groupoids in any sense, since a typical example of a classification in this area is the Cartan–Killing classification of simple Lie group, a precise and complete result. This newly revealed question, on the other hand, is clearly open-ended and does not have (or allow for) a complete answer.
You don't need étaleness to get locally like an action groupoid: proper is enough.
Given an appropriate Čech cocycle, one can construct a bundle gerbe from it à la Hitchin–Chatterjee. It will look like none of these, in general.
@DavidRoberts "It will look like none of these, in general".. not even locally? Moerdijk assumes the Lie groupoid is proper and etale to say it is locally an action Lie groupoid.. I saw this in https://arxiv.org/abs/math/0203100 (page 8).. May be it is true with out etale assumption..
@DmitriPavlov I am not asking for interesting examples or constructions of Lie groupoids.. I do not know why it is conveying in that way :O I do not know what other term to use other than classification for what I have mentioned before... Let me rephrase it again... I take all proper etale Lie groupoids to be of "same type" as all of them are locally action Lie groupoids...
Well, up to Morita equivalence, every proper Lie groupoid is glued together out of orthogonal actions of compact Lie groups on open unit balls. You don't need étaleness. I was thinking of gerbes not up to stable iso/Morita equivalence, so I guess I take that one back. Every U(1)-gerbe is, up to stable iso, the pullback of the universal gerbe which can be written down as a lifting gerbe (which looks like a proper action of an infinite-dimensional Lie group).
@DavidRoberts oh, ok. Do you think the question is not well phrased? If anything is not clear, I would like to make it clearer..
I don't know. The gauge groupoid one can be replaced by the same construction for an arbitrary fibre bundle, and there is the fundamental groupoid of a manifold (also has a Lie groupoid structure), but I think this might be covered by the foliation groupoid example, I can't remember precisely. I mean, all of the examples you give are essentially functorial constructions of Lie groupoids from other data. But who knows what an arbitrary Lie groupoid looks like? There is also the example of bundles of Lie groups, that's not on your list. Really you can only ask for more examples...
@PraphullaKoushik: It appears that I am not alone in being confused as to the actual meaning of your question; David Roberts voiced similar concerns. Maybe you could give us some existing example of “classification into types” that you find acceptable.
@DavidRoberts "But who knows what an arbitrary Lie groupoid looks like?" :) I would like to know about this :D.. I will try to make the question clearer..
@DmitriPavlov I did not doubt when you said the question is confusing.. :) I do think it is slightly unclear.. I will make it clearer..
@DmitriPavlov I could not make it any better than "what an arbitrary (good enough) Lie groupoid looks like?"... Do you have any "comment" for a special case or variant of this question? As it is, it looks open ended... :(
@DavidRoberts I could not make it any better than "what an arbitrary (good enough) Lie groupoid looks like?"... Do you have any "comment" for a special case or variant of this question? As it is, it looks open ended... :(
@PraphullaKoushik yes, it's open ended. Maybe the best you can ask for is for more functors that produce Lie groupoids from other data that end up being not Morita equivalent to the ones on the list. As it is, your list has reasonable overlap (1. and 2. are both special cases of 3., for example; and there's a relation between 4. and 5., since a submersion defines a foliation of its domain, 6. and 7. are special cases, as I mentioned, of the groupoid arising from a rather general fibre bundle...)
@DavidRoberts Yes. I think the same. I will leave it as it is and edit if I can get some interesting special case as a question.
|
2025-03-21T14:48:32.075853
| 2020-09-20T16:52:54 |
372161
|
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|
Stack Exchange
|
Hitchin map and vector bundles
I've been learning a bit about automorphisms of moduli spaces of vector bundles and the Hitchin map.
I'm reading this paper of Indranil Biswas, Tomas L. Gomez, V. Munoz, and I have a problem about prop $5.1$
Let $E$ be a semistable vector bundle over a smooth projective curve $X$ over $\mathbb{C}$ and $K_X$ be the canonical bundle of $X$.
$H$ is the Hitchin map, $H:H^0(End_0E\otimes K_X)\rightarrow W$ $(W:=\bigoplus H^0(X, K_X^{\otimes i}))$
and $h_r$ is the composition of $H$ and the projection $W\rightarrow W_r$ $(W_r:=H^0(X,K_X^{\otimes r}))$
$h_r:H^0(End_0E\otimes K_X)\rightarrow W_r$
Then prop $5.1$ says that
$H^0(End_0E\otimes K_X(-x))$$=\{\psi\in H^0(End_0E\otimes K_X) \mid h_r(\psi+\phi)\in H^0(K_X^{\otimes r }(-x)),\forall\phi \in h_r^{-1}(H^0(K_X^{\otimes r}{(-x)}))\}$
By using prop$5.1$, we can get the vector bundle
$\mathcal{E}\rightarrow X$
whose fiber over $x \in X$ is $\mathcal{E}_x=H^0(End_0E\otimes K_X(-x))$
But I don't understand why prop$5.1$ allows us to construct such vector bundle.It shows that $dim_{\mathbb{C}}\mathcal{E}_{x}$ is constant?
This may be a stupid question.
Thank in advance.
Let $F=End_0 E \otimes K_X$, let $p,q:X\times X\to X$ be the projections, and let $\Delta\subseteq X\times X$ be the diagonal. Isn’t $p_(\mathcal{O}(-\Delta)\otimes q^ F)$ the bundle you want?
The “constant dimension” condition + Grauert’s theorem (cohomology and base change) will tell you that it is locally free with formation compatible with base change (so the fibers are what you want).
Thanks for comments. I think what you are saying is right and it is very helpful. But how does your comment relate to the proposition 5.1 ?
|
2025-03-21T14:48:32.075989
| 2020-09-20T17:32:15 |
372162
|
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"Benoît Kloeckner",
"Iosif Pinelis",
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|
Stack Exchange
|
Taking points uniformly inside a general finite geometric domain
It is well known that if we want to take $n$ uniformly and randomly points inside a circle of radius $r$ and centered at the origin the following apparently correct approach for generating $x$ and $y$
$$ x= U \cos(\theta), \;y= U \sin(\theta)$$ where $U$ is a uniform variate in $(0,1)$ and $\theta$ is uniform variate in $(0, 2 \pi)$, does not work. Rather the correct way to generate $x$ and $y$ coordinates is to use:
$$ x= \sqrt{U}\cos(\theta),\; y= \sqrt{U} \sin(\theta) $$ with $U$ and $\theta $ as mentioned above. I wonder is there a some way of generating a of $n$ points inside the bounded region defined by
$$\mathbf{D}=\{(x,y):f(x,y)=0\},$$
given that $ \mathbf{D}$ forms a closed region with smooth boundary.
For instance let's say
$$\mathbf{D}=\{ (x,y):\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\},$$
how can we create $n$ points that are uniformly distributed inside the ellipse. Thanks for any hints/responses in advance!
For an ellipse, one can rescale the coordinates so that the region becomes a disk and then sample in the way you mentioned.
However, in general sampling efficiently from irregular regions (or distributions) is a really hard problem. If the dimension is low and the region is not too crazy, you can find some hypercube $C$ which contains $D$ and sample uniformly from $C$ while rejecting draws that don't belong to $D$. However, if the dimension is high or the shape is too irregular, you end up throwing away the vast majority of your draws, which really limits the effectiveness of this brute force approach.
Instead, the common thing to use in practice is some variant of Markov chain Monte-Carlo, which attempts to walk around the space in a random way. This is much more efficient to implement and by various ergodic theorems should converge to uniform sampling in the limit. Unfortunately, in practice it's very difficult to determine whether you've let it run long enough for the convergence to actually occur. Statisticians have developed a whole slew of heuristics for this problem, but it's very hard to say things rigorously.
Edit: in fact, it is possible to sample uniformly from an ellipsoid in any dimension by taking an affine change of coordinates so that the region is a ball. You then sample the radius $\sim r^{n-1}$. To determine the angle, you take $n$ independent draws from the standard univariate normal distribution. It turns out that the angle of the resulting vector will be equidistributed in $\mathbb{S}^n$, which you can use to choose an angle quickly. This example is a bit magical, and not at all what you should expect for more general regions.
The approach (already alluded to in one of the answers) of sampling uniformly from a larger set and then throwing out the samples that you don't want is known as rejection sampling. You'll find all sorts of useful ideas in the linked Wikipedia article and references thereof.
A quick way to generate a random point uniformly distributed in a bounded region $D$ is to generate a random point $P$ uniformly distributed in a rectangle $R$ containing $D$ and, if $P\notin R$, then discard $P$ and continue until you have as many random points as you want.
For instance, here is the generation (in Mathematica) of $3000$ random points uniformly distributed inside the ellipse centered at the origin with half-axes $a=2$ and $b=1$:
Here the "waste" fraction is $1-\frac{\pi ab}{4ab}\approx0.21$, about $21\%$, no problem at all.
Alternatively, one may generate a random point uniformly distributed in an arbitrary measurable plane region $D$ of positive area without any waste, as follows. For real $x$, let
$$F(x):=F_D(x):=\frac{A(x)}{A(\infty)},$$
where $A(x)$ is the area of the region $\{(s,t)\in D\colon s\le x\}$. So, $F$ is a cumulative probability distribution function (cdf), which is actually the cdf of the abscissa of a random point uniformly distributed in $D$. For any $u\in(0,1)$, let
$$F^{-1}(u):=\min\{x\in\mathbb R\colon F(x)\ge u\},$$
the quantile function corresponding to $F$. So, if $U$ is a random variable (r.v.) uniformly distributed on the interval $(0,1)$, then the distribution of the r.v.
$$X:=F^{-1}(U)$$
will coincide with the distribution of the abscissa of a random point uniformly distributed in $D$.
If now the conditional distribution of a r.v. $Y$ given $X=x$ is the uniform distribution on the one-dimensional set
$$D_x:=\{y\in\mathbb R\colon(x,y)\in D\},$$
then the random point $(X,Y)$ will be uniformly distributed in $D$.
So, the generation of a $(X,Y)$ uniformly distributed in $D$ is reduced to the generation of two random points on the real line.
For instance, here is the generation (in Mathematica) of $2000$ random points uniformly distributed in the ellipse $E:=\{(x,y)\in\mathbb R^2\colon100 (x - y)^2 + (x + y)^2 \le4\}$:
Here there is no "waste" at all, but the volume of calculations is much greater than in the previous example.
In this particular case, it would be more economical to rotate the ellipse appropriately to make its axes horizontal and vertical and then use the approach of the previous example (without rotation, there could be too much waste).
In response to the comment by Timothy Budd, who wrote: "RandomPoint[Disk[{0, 0}, {a, b}], n] achieves the same result but is over a hundred times faster (for $n=3000$). Of course, this may be just due to low-level optimization."
I think the advantage of Mathematica's command RandomPoint[] over the function QQ[] defined above is mainly due to two things: (i) QQ[] produces (pseudo-)random points one-by-one, whereas RandomPoint[] apparently works with entire lists/arrays and (ii) RandomPoint[Disk[{0, 0}, {a, b}], n] takes into account the knowledge that the region is a (stretched) disk.
The image below of a Mathematica notebook shows that, when QQ[] is modified to a command QQQ[] operating on entire lists, RandomPoint[Disk[{0, 0}, {2, 1}], 3000] is only $0.0051967/0.0028878<2$ times faster than QQQ[2, 1, 3000]. Moreover, the command RandomPoint[ImplicitRegion[x^2/4 + y^2 <= 1, {x, y}], 3000], which does not let Mathematica know that the region is a (stretched) disk, is $0.157363/0.0051967>30$ times slower than QQQ[2, 1, 3000]:
Note that is quickly only in low dimension. In high dimension, your domain will represent an exponentially small part of the rectangle, and this approach is infeasible (mentioned in Gabe K's answer).
@BenoîtKloeckner : The question was about plane domains. Also, I have now described an alternative, waste-free approach.
Although not very relevant to the question as we do not know Mathematica's implementation, QQ[a_, b_, n_] := RandomPoint[Disk[{0, 0}, {a, b}], n] achieves the same result but is over a hundred times faster (for n=3000). Of course, this may be just due to low-level optimization.
Since @Timothy and Iosif both mention Mathematica's RandomPoint[], I will note that the method internally used, very roughly, corresponds to discretizing the given region into simplices, and then sampling uniformly inside those simplices, with the choice of simplex weighted by their content (e.g. area for triangles, volume for tetrahedra, etc.). See e.g. this thread on mathematica.SE .
@TimothyBudd : Thank you for your comment. I have added a response to it, at the end of the answer.
@J.M.isn'tamathematician : Thank you for this information. I did think about triangulation -- seems reasonable, especially maybe for high dimensions.
|
2025-03-21T14:48:32.076472
| 2020-09-20T18:35:38 |
372166
|
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|
Stack Exchange
|
Set-theoretic geology III: inside the core
Thanks to Jonas, Asaf, and Gabe I understand a little more of grounds and the mantle (or mantles, because it looks like there may be more than one).
But, set-theoretic geology, or so it seems to me, should not be exclusively about grounds: the Earth has a core, and in fact perhaps several strata of cores, till it reaches the mantle (or some intermediate zone).
So, how about doing the opposite? Rather than penetrating a model $M$ by erosion, what if we know that there is a solid center ( say $L^M$, assuming that the model is not $L[G]$ for some generic set), and grow that minimal core as far as we can go?
In other words, let us define a class of M-non grounds,
$$\mathrm{CORE}= \{ N\ \nsubseteqq M, N \vDash ZFC \land \nexists G M=N[G] \}$$
and determine its structure. In some cases this class is empty, but suppose it is not.
Question:
what can be said of the non grounds of $M$? Is this class a upward directed partial order? When do this class reach the mantle (in the sense that I cannot find anything below the mantle that is not in CORE)?
Are there any layers for some models which lie between the outer core and the intersection of all mantles?
Sounds like Jules Verne's Voyage au Centre de la Terre $\dots$
Models of ZF, of course. Many of which are in fact grounds, but sometimes not all of them.
Assuming that in the definition of CORE, $G$ is a generic filter for set-forcing, I think that you need some non-triviality condition on M, but I'm not sure what exactly. For example, in $M=L[0^{#}]$ you can construct two $L$-Cohen generics $r_0, r_1 \in {}^\omega 2$ such that their bit-wise xor is $0^{#}$. Since $0^{#}$ cannot be introduced by any set forcing, both $L[r_0], L[r_1]$ are in CORE, but there is no model $N$ of ZF between $L[r_0] \cup L[r_1]$ and $M$.
Thanks Yair! Yes, G is supposed to be in M, and a generic filter. Not sure what you mean for non-triviality: I did not ask for CORE to be upward directed, I only ask when does happen. For instance, if $M=L[0^{#}]$, CORE is made of all the $L[r}]$. , and it has a quite interesting structure: in some cases, as in your example, two non-grounds do not have a common refinement, in some cases they do.
Also, do you require the ordinals of $N$ to be the ordinals of $M$? Or do you allow $N$ to be shorter?
I was thinking of the standard scenario, ie the ordinals are On^M, but of course it is also interesting to know what happens in case N is shorter.
Here are a few observations about CORE.
Claim: It is consistent that CORE is not pairwise upwards directed.
Proof:
Let $\mathbb{P}_0$ be the class forcing for the Easton product of the Cohen forcing $\mathrm{Add}(\alpha^+, 1)$, over all cardinals $\alpha$, in $L$. Let $G$ be $L$-generic for $\mathbb{P}_0$. Then, we can pick a partition $A \cup B$ of the class of cardinals, with both $A, B$ being proper classes. Then $N_0 = L[G \restriction A], N_1 = L[G \restriction B]$ are both in CORE, and have no common upper bound in CORE.
On the other hand:
Claim: It is consistent that CORE is upwards directed:
Proof: Let $\mathbb{P}_1$ be the backwards Easton support iteration of $\mathrm{Add}(\alpha^+, 1)$ for $\alpha$ successor of a singular cardinal in $L$. Let $G$ be an $L$-generic and let $M=L[G]$.
Sub Claim: For every $N \in CORE^M$, there is an ordinal $\alpha$ such that $N \subseteq L[G \restriction \alpha]$. In particular, CORE is upwards directed.
Sketch of Proof: Let $\alpha$ be minimal (successor of singular) such that $L[G\restriction \alpha] \not\subseteq N$ and let $x\in N$ be a set of ordinals of minimal rank such that $x \notin L[G \restriction \alpha]$. Then $x$ has to be a fresh set over $L[G \restriction \alpha]$, and by minimality if has to be a subset of $\alpha^+$. By gap-type arguments, $x$ together with its name (which is in $L$) codes the set $G \restriction \alpha$, and thus $L[G \restriction \alpha] \subseteq N$.
Finally, large cardinals seems to have negative affect on the directedness of CORE:
Claim: Let $\kappa \in M$ measurable and $2^\kappa = \kappa^{+}$. Then, there are $N_0, N_1 \in CORE^M$, and $x \in N_1$, $y \in N_0$, such that $N_0[x] = N_1[y] = M$.
Proof: Let $\mathcal{U}$ be a normal ultrafilter on $\kappa$ and let $N$ the ultrapower by $\mathcal{U}$. Let us construct inside $M$ two $N$-generic filters $G_0, G_1 \subseteq \mathrm{Add}(\kappa^{+}, 1)$, such that their bitwise xor codes $\mathcal{U}$. This is possible, since $2^\kappa = \kappa^{+}$ (both in $M$ and in $N$). Let $N_0 = N[G_0], N_1 = N[G_1]$ and note that $N_0[G_1] = N_1[G_0] \supseteq N[\mathcal{U}] = M$. QED
On the other hand, if $A$ is a set of ordinals and $A^{\#}$ exists, then $L[A] \in CORE$, since no set forcing in $L[A]$ can introduce a sharp for $A$.
So under the large cardinal axiom "every set has a sharp" (which follows from the existence of class of measurable cardinals, for example), $\bigcup CORE = V$, so CORE can certainly contain sets which are not in the mantle.
By the way, the name CORE is a bit confusing. There is no clear connection between the Core Model ($K$) and the models in CORE.
I am happy to go with whichever name you come up with, but remember: it must be in line with set geology
|
2025-03-21T14:48:32.076850
| 2020-09-20T19:58:01 |
372169
|
{
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|
Stack Exchange
|
Maximal order of an order-preserving map
Let $X$ be a finite partially ordered set, let $f\colon X\to X$ be an order-preserving map [edit: meaning $x\le y\implies f(x)\le f(y)$], and let $x_0$ be an initial point. Define $x_n = f(x_{n-1})$ for all $n$; then the sequence $(x_n)$ is ultimately periodic. What is its worst-case period? I.e. what are the minimal $k<\ell$ such that $x_k=x_\ell$?
With no order assumption, one could have $\ell=\#X$ since the map $f$ could cycle through all of $X$. I'm particularly interested in the case when $X$ is the family of subsets of a set $Y$, and $x_0$ is a small subset. Does there then exist a non-trivial bound, say polynomial in $\#Y$ and exponential in $\#x_0$?
Nice question. In the case you care about, when $X=2^{Y}$ is a Boolean lattice, maybe you can argue that the worst case would be if $f$ is a permutation on $Y$.
I can never remember whether "order-preserving" means $x\lt y\implies f(x)\lt f(y)$ or $x\le y\implies f(x)\le f(y)$, please remind me.
@SamHopkins So the answer would be related somehow to Landau's function?
@bof: Regarding Landau's function- well, sort of, except that if we want to keep track of $#x_0$ we should only take the lcm of $#x_0$ cycle sizes. Regarding $<$ vs. $\leq$, I think $\leq$ is the usual one for defining morphisms of posets.
@bof I would rather state it with $\le$ (for the application I have in mind), I'll update the question for clarity.
@SamHopkins I have trouble understanding what kind of structure there is on order-preserving maps. As an example, map all subsets of size $<k$ to the empty set, all subsets of size $>k$ to $Y$, and permute arbitrarily the sets of size $k$. Here we're saved because either $#x_0$ is large or it hits immediately $\emptyset$.
You can have the poset with no relations except things like $x\le x$ and $f$ maps everything around in a cycle.
@BrendanMcKay: I believe the OP understood this ("With no order assumption, one could have $\ell=#X$ since the map $f$ could cycle through all of $X$."); thus the focus on the Boolean lattice case.
This is a comment about a special case that I think gives a negative answer to the last question. Let $\varOmega$ be a set of size $n$ and let $B$ be the lattice of its subsets with the inclusion order.
Now let $\phi:\varOmega\to\varOmega$ be any function. For $X\subseteq\varOmega$, define $X^\phi=\lbrace \phi(x) \mid x\in X\rbrace$ (duplicates removed of course).
Now it is elementary that $f:B\to B$ by $X\mapsto X^\phi$ is order-preserving. The question is: what can be said about the eventual cycle length of the trajectory of an element of $B$?
Consider the case that $\phi$ is a permutation. Take $x_0\in B$ to have one element from each of the cycles of $\phi$. Then the trajectory of $x_0$ is a cycle of length equal to the order of $\phi$ (as an element of the symmetric group).
For example, if $n=2 + 3 + \cdots + p_k$ (sum of first $k$ primes) and $\phi$ has one cycle of each prime size, then $\#x_0=k$ and the length of the cycle is $N = 2\times3\times\cdots\times p_k$.
I understand that
$N = \exp\bigl((1 + o(1)) k \log k\bigr)$.
So $N$ is larger than exponential in $\# x_0$.
Also, $n\sim \frac12 k^2\log k$. If I got the inversion right, $\# x_0=\Theta(\sqrt{n/\log n})$ and $N=\exp(\Theta(n^{1/2}\log n))$.
Nice, so this is basically using Landau's function as mentioned in the comments.
Thanks -- it shows that not much can be hoped for, even in that (presumably nicest) case. I'll accept it as a (negative) answer!
|
2025-03-21T14:48:32.077249
| 2020-09-20T19:58:36 |
372170
|
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|
Stack Exchange
|
Operator norm and spectrum
I am wondering about when an operator norm coincides with the maximum eigenvalue of an operator and there is one particular aspect that confuses me quite a lot.
Let's say we have a symmetric positive continuous linear operator
$$
T : L^2(\Omega) \rightarrow L^2(\Omega)
$$
with maximum eigenvalue $\lambda>0$ so that $T u = \lambda u$ for some eigenfunction $u\in L^2(\Omega)$. Then (if I understand it correctly), it should hold
$$
\lambda = \sup_{v \in L^2(\Omega) \setminus \{ 0 \}} \frac{\| Tv \|_{L^2(\Omega)}}{\| v \|_{L^2(\Omega)}}.
$$
Next, let's assume that the operator has a smoothing effect such that $\mbox{image}(T) \subset H^1_0(\Omega)$ and that it is also $H^1$-continuous (I am thinking of $T$ as the inverse of an elliptic differential operator). In this case we can interpret the operator as
$$
T : H^1_0(\Omega) \rightarrow H^1_0(\Omega)
$$
The spectrum should remain unchanged, so that I would think that
$$
\lambda = \sup_{v \in H^1_0(\Omega) \setminus \{ 0 \}} \frac{\| Tv \|_{H^1_0(\Omega)}}{\| v \|_{H^1_0(\Omega)}}.
$$
However, the statement
$$
\sup_{v \in H^1_0(\Omega)} \frac{\| Tv \|_{H^1_0(\Omega)}}{\| v \|_{H^1_0(\Omega)}} = \lambda = \sup_{v \in L^2(\Omega)} \frac{\| Tv \|_{L^2(\Omega)}}{\| v \|_{L^2(\Omega)}}.
$$
looks wrong to me. Is it? If so, where is the mistake in my arguments? I feel like I have a very basic misunderstanding here.
If the operator is not self-adjoint/symmetric, then the formula is for singular value, not eigenvalue.
Thanks! I missed the point with the singular values. Let's say the operator is also symmetric, is the statement true then?
A brief remark (though not essential for the question): your assumption that $T$ be continuous from $L^2$ to $H^1_0$ is redundant; this follows automatically from the closed graph theorem.
If we have the identity $Tu = \lambda u$ for $u\not = 0$ (strongly in $H^1(\Omega)$), then $(T - \lambda I) u = 0$, which means that $(T - \lambda I)$ cannot be invertible. This should not change depending on if we interpret $T$ as an operator on $L^2(\Omega)$ or on $H^1_0(\Omega)$. What do I miss here?
Good point, i take that back.
I don't think there's anything very mysterious about this. Under the stated assumptions, we secretly know that $|Tv|/|v|$ (with both norms) is maximized by the maximum eigenvector, and then of course it doesn't matter which norm we use.
@ChristianRemling I think the concern by OP here is that one supremum is taken over a smaller space (in the sense that one embeds continuously (even compactly - Rellich Kondrachov!)) into the other, but the suprema are the same, when it seems like the supremum over the smaller space should be smaller (possibly strictly) than that taken over the largest space.
Do we know that $T$ is self-adjoint as an operator on $H_0^1(\Omega)$?
@MateuszKwaśnicki: No, the operator need not be self-adjoint on $H_0^1$, and this is apparently the main problem here..
@ChristianRemling: In fact, the quotient of the $H^1_0$-norms need not be maximized by the leading eigenvector (as follows, for instance, from my answer below).
The major point here is that, for an operator $S$ on a Banach space (or Hilbert space) $X$, the number $\sup_{x \in X \setminus\{0\}} \frac{\|Sx\|}{\|x\|}$ is not the spectral radius of $S$ but the operator norm. The operator norm is always $\ge$ the spectral radius, but we cannot expect equality in general.
On a Hilbert space, one sufficient condition for equality of the operator norm and the spectral radius is that the operator be self-adjoint or, more generally, normal.
But as pointed out in a comment by Mateusz Kwaśnicki, if $T$ is self-adjoint on $L^2$, this does not imply that $T$ is self-adjoint on $H^1_0$ since the inner product there is different from the inner product on $L^2$.
Here is a concrete counterexample:
Let $\Omega = (0,2\pi)$ (endowed with the non-normalised Lebesgue measure) and define $z,v \in H^1_0 := H^1_0(\Omega)$ by
\begin{align*}
z(x) & = \frac{|\sin(x)|}{\sqrt{\pi}}, \\
v(x) & = \sin(\frac{1}{2}x)
\end{align*}
for all $x \in (0,2\pi)$.
We define the operator $T$ on $L^2$ by
$$
Tf = \langle f, z\rangle_{L^2} \cdot z
$$
for all $f \in L^2$. Then $T$ is a self-adjoint rank-$1$ projection on $L^2$ whose norm and spectral radius are thus equal to $1$. Clearly, the range of $T$ is a subset of $H^1_0$.
The restriction of $T$ to $H^1_0$ is again a non-zero projection and thus still has spectral radius $1$. But the operator norm of $T$ on $H^1_0$ is strictly larger than $1$. Indeed, we have
$$
\|T\|_{H^1_0 \to H^1_0} \ge \frac{\|Tv\|_{H^1_0}}{\|v\|_{H^1_0}} = \frac{\sqrt{512}}{\sqrt{45}\pi} > 1
$$
(we need to compute a few integrals to obtain the equality in the middle, but the computations are rather straightforward).
This proves that the operator norms of $T$ on $L^2$ and on $H^1_0$ are distinct, although the spectral radius on both spaces is $1$. In particular, $T$ cannot be self-adjoint (and not even normal) on $H^1_0$.
EDIT: An additional observation. While, in the example above, equality of the spectral radii on both spaces follows from the fact that $T$ acts as a projection on both spaces, I thought it might be worthwhile to point out that the equality of the spectral radii is actual a general fact:
Proposition. (Equality of spectral radii) Let $V,X$ be complex Banach spaces such that $V$ is continuously embedded in $X$. Let $T: X \to X$ be a bounded linear operator such that $TX \subseteq V$. Then the spectral radius of the operator $T: X \to X$ coincides with the spectral radius of the restriction $T|_V: V \to V$.
Proof. We use the spectral radius formula
$$
(*) \qquad r(S) = \lim_{n \to \infty} \|S^n\|^{1/n}
$$
which holds for the spectral radius $r(S)$ of each bounded linear operator $S$ on a complex Banach space.
For each $n \in \mathbb{N}$ the operator $(T|_V)^n = (T^n)|_V: V \to V$ factorizes as
$$
V \hookrightarrow X \overset{T^{n-1}}{\longrightarrow} X \overset{T}{\longrightarrow} V,
$$
so the spectral radius formula $(*)$ implies that $r(T|_V) \le r(T)$. But conversely, the operator $T^n: X \to X$ factorizes as
$$
X \overset{T}{\longrightarrow} V \overset{(T|_V)^{n-1}}{\longrightarrow} V \hookrightarrow X,
$$
so the spectral radius formula $(*)$ also implies that $r(T) \le r(T|_V)$. This proves the proposition.
Remark. What is quite nice about the proposition above is that it does not rely on eigenvalues, so no compactness assumption on the operator $T$ is needed.
|
2025-03-21T14:48:32.077684
| 2020-09-20T20:08:02 |
372173
|
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|
Stack Exchange
|
Characterization of random variables whose tensor powers have subexponential "small-ball" probabilities
Is there a succinct characterization of all random variables $\zeta$ on $\mathbb R$ with the following properties
1. Symmetry: $\zeta \overset{d}{=} - \zeta$.
2. Small-ball probability: there exists a constants $\alpha > 0$ and $u_0 \in (0,\infty]$ such that $P(\|\zeta^\otimes\| \le u\sqrt{n}) \le (\alpha u)^n$, for all $u \in [0,u_0)$ and positive integer $n$. Note that $\zeta^\otimes$ is a random vector on $\mathbb R^n$ with iid coordinates having the same distribution as $\zeta$.
Of course, $\zeta \sim N(0,1)$ fits the bill. I'm looking for general characterization.
In case (and only in case) it turns out that the space of all such distributions is too "vast and unstructured", consider the following 3rd axiom
3. Sub-Gaussianity: There exists $\sigma>0$ and $u_1 \ge 0$ such that $P(|\zeta| \ge u) \le 2e^{-u^2/(2\sigma^2)}$, for all $u \ge u_1$.
Let $X:=(X_1,\dots,X_n)$, where the $X_j$'s are iid copies of $\zeta$. Then the problem is about conditions for
$$P(\|X\|\le u\sqrt n)\le C^n u^n\tag{1}$$
for some real $C>0$, all natural $n$, and all small enough real $u>0$ (and then for all real $u>0$, possibly with a different $C>0$).
If the distribution of $X_1$ has a density bounded by some real $c>0$, then the distribution of $X$ has a density bounded by $c^n$, so that
$$P(\|X\|\le u\sqrt n)\le c^n (u\sqrt n)^n|B_n|\le C^n u^n$$
for some real $C>0$, all real $u>0$, and all natural $n$, where $|B_n|$ is the volume of the unit ball $B_n$ in $\mathbb R^n$; so, (1) holds.
On the other hand, (1) cannot hold (for $C,u,n$ specified above) if the distribution of $X_1$ has an atom at $0$. Moreover, (1) cannot hold for any real $C>0$, any natural $n$, and all small enough real $u>0$ if the distribution of $X_1$ has a density $p$ such that $p(x)\to\infty$ as $x\to0$. Indeed, then for each natural $n$ we have
$$P(\|X\|\le u\sqrt n)/u^n=\int_{u\sqrt n\,B_n}\prod_{j=1}^n (p(x_j)\,dx_j)/u^n\to\infty$$
as $u\downarrow0$.
Thanks for the input. I was hoping there would be broad classes distributions strictly between atomic and bounded density, which satisfy the conditions.
@dohmatob : In your question, you mentioned symmetry and subggaussianity as prospective conditions for (1). Did you also mention there (non-)atomicity of the distribution and boundedness of the density? Anyhow, I have now also added that the boundedness of the density cannot be relaxed by allowing the density to explode to $\infty$ near $0$. So, the relevant conditions are shown to be in terms of boundedness of the density.
Thanks for the update. As a side note, I didn't mention those because I thought the were examples and non-examples (Indeed, in my question, I used boundedness of density to prove that $N(0,1)$ was a member of the sought-for family).
|
2025-03-21T14:48:32.077893
| 2020-09-20T20:49:17 |
372175
|
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|
Stack Exchange
|
Characterizing 'very homogeneous' finitely valued stochastic processes
Fix a positive integer $n$. Let $X = \{X_i\}_{i \in \mathbb{N}}$ be a discrete time stochastic process such that each $X_i$ is a $\{0,\dots,n-1\}$-valued random variable. Suppose that the joint probability distributions of any finite sequence of $X_i$'s only depends on the order of their indices, or to be more precise suppose that $X$ satisfies the following:
For any $k\in \mathbb{N}$, any two increasing sequences of indices $i_0<i_1 < \cdots i_{k-1}$ and $j_0 < j_1<\cdots<j_{k-1}$, and any function $f : \{0,\dots,k-1\} \to \{0,\dots,n-1\}$, $$\mathbb{P}(X_{i_0} = f(0) \wedge X_{i_1} = f(1) \wedge \cdots \wedge X_{i_{k-1}} = f(k-1)) = \mathbb{P}(X_{j_0} = f(0) \wedge X_{j_1} = f(1) \wedge \cdots \wedge X_{j_{k-1}} = f(k-1)).$$
Call such a stochastic process 'strongly homogeneous.' I'm trying to understand what the set of strongly homogeneous stochastic processes looks like. This is my approach so far:
The set of $\{0,\dots,n-1\}$-valued discrete time stochastic processes can be understood as the set of Borel probability measures on the (compact) space $A = \{0,\dots,n-1\}^{\mathbb{N}}$. This is a subset of Banach space of (regular Borel) signed measures on $A$. Let $S$ be the set of such measures corresponding to a strongly homogeneous stochastic process. It's not hard to check that $S$ is a convex, weak* closed set, and therefore that the Krein-Milman theorem applies to it. This gives us that every element of $S$ is in the weak* closure of the convex hull of the set of extreme points of $S$ (where a point is extreme if it is not the convex combination of any other elements of $S$). This leads to my precise question.
Question: What are the extreme points of $S$?
Note that the extreme points of the set of all probability measures on $A$ is precisely the set of Dirac measures on $A$, but a similar statement here is not sufficient. For instance, if $n=2$, then the only strongly homogeneous Dirac measures are those concentrated on the constant $0$ sequence or the constant $1$ sequence, but convex combinations of these do not have the measure corresponding to a sequence of i.i.d. fair coin flips in their weak* closure.
My suspicion is that the measures corresponding to i.i.d. sequences are the extreme points, but I haven't proved either that they are all extreme points or that all extreme points are of that form. (Proving that they are all extreme points should be easy, however.)
This only answers the first part of your question:
The strongly homogenous processes are , in particular, stationary. Extreme points of stationary processes are exactly the stationary ergodic processes. The i.i.d. measures are ergodic, hence they are extreme among stationary processes, which implies they are extreme among strongly homogenous processes.
Strongly homogenous processes that are also strong mixing are necessarily i.i.d., but this still requires dealing with ergodic strongly homogenous processes that are not strong mixing.
Thank you very much. Do you have any good references for some of these statements (such as ergodic being extreme among stationary processes)?
Cornfeld and Sinai https://www.springer.com/gp/book/9781461569299 should have it. This is really immediate from the definition of ergodicity. In particular, if an ergodic measure $\mu$ is the average of two stationary measures $\mu_1$ and $\mu_2$, then they must be absolutely continuous to $\mu$. The Radon Nikodym derivatives $\frac{d\mu_i}{d\mu}$ are invariant functions, hence a.s. constant.
|
2025-03-21T14:48:32.078148
| 2020-09-20T21:57:30 |
372178
|
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|
Stack Exchange
|
Alternative definition of Kolmogorov complexity
In Kikuchi's paper Kolmogorov complexity and the second incompleteness theorem the Kolmogorov Complexity (KC) of $x$ is defined as
$$ K(x) = \mu e (\varphi_e(0) \simeq x) \, . $$
This seems to give exponentially larger outcomes then the more common (rough) definition of $K(x)$ as "the length of the smallest computer program running on some fixed universal TM that returns $x$". How does Kikuchi's definition match up with the usual KC definition? Is this a common alternative? Does any major result in KC change/not work under this new definition, or can we just move everything by an exponent and rely on its monotonicity?
Can you explain the notation in the definition? I think it's "smallest $e$ such that the $e$th program (in some enumeration) outputs $x$ on input $0$."
@usul That's how I read it as well, with the extra condition that this $e$th program should terminate, of course. Kikuchi doesn't say any more about it.
It does seem to me that Kikuchi is using an exponentiated version of regular Kolmogorov complexity. For example, he says things like "we say a number $x$ is random if $x \le K(x)$", which makes sense if we think of an encoding of a number as needing about $\log_2 n$ bits. I have not seen this version in other places, but since exponentiation is monotonic, I don't see how it would change things much.
Can you specify which “usual theorems” you have in mind? This would make the question more precise and concrete.
@user76284 I don't know that much about the general theory of KC, so my question is open: does any major result in KC change/not work under this new definition? One thing that is quite different is that no two strings can have the same complexity under Kikuchi's definition. But I'm not sure whether this impacts anything significant. Like Artemy says exponentiation is monotonic, so it seems we're all good. But the definition is quite unusual, hence maybe someone with more experience in KC can tell whether it really boils down to the same thing or not (but with everything exponentiated).
If you don’t get a response here, it might be worth asking on the Theoretical CS SE.
Kikuchi mentions Odifreddi's Classical Recursion Theory (North-Holland, 1989) as the source of that particular definition, and that's also the earliest occurrence I know of. Another paper using the same definition, also citing Odifreddi's book, is P. Raatikainen's On interpreting Chaitin's incompleteness theorem (JPL 27:569-586, 1998).
@user103227 Where does Kikuchi mention that?
Last paragraph of the introduction: "we [...] define Kolmogorov complexity following Odifreddi".
@user103227 Is that In Odifreddi volume I or II? Does he motivate the definition? Thanks, btw for your help :). I'll also try shooting Raatikainen an email.
@Jori It's in volume I. I don't have a physical copy available, but in the pdf available on Odifreddi's website, the definition is on page 151.
|
2025-03-21T14:48:32.078407
| 2020-09-20T21:59:20 |
372179
|
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|
Stack Exchange
|
Separating spheres in $3$-manifolds of positive scalar curvature and mean convex boundary
Recently, A. Carlotto and C. Li proved a complete topological classification of those compact, connected and orientable $3$-manifolds with boundary which support Riemannian metrics of positive scalar curvature and mean-convex boundary. Namely, if $M^3$ is such a manifold, then there exist integers $A, B, C, D \geq 0$ such that $M$ is diffeomorphic to a connected sum of the form
\begin{align*}
P_{\gamma_1} \# \cdots \# P_{\gamma_A} \# \mathbb{S}^3/ {\Gamma_1} \# \cdots \# \mathbb{S}^3 / {\Gamma_B} \# \left( \#_{i=1}^C \mathbb{S}^2 \times \mathbb{S}^1 \right) \setminus \left( \sqcup_{i=1}^D B_i^3 \right),
\end{align*}
where $P_{\gamma_i}$, $i \leq A$, are genus $\gamma_i$ handebodies; $\Gamma_i$, $i \leq B$, are finite subgroups of $SO(4)$ acting freely on $\mathbb{S}^3$, $B_i^3$, $i \leq D$, are disjoint $3$-balls in the interior.
My question: Can we classify, in terms of $A,B, C, D$, the $3$-manifolds $M$ of the form above in which any smoothly embedded $2$-sphere in the interior separates $M$?
For instance, if $(A,B,C,D) = (1, 0, 0, 0)$, then this holds. Indeed, if $M = P_{\gamma_1}$, then $H_2(M) = 0$, so the connecting homomorphism $H_2(M, \partial M;\mathbb{Z}) \to H_1(\partial M; \mathbb{Z})$ is injective. Since an embedded $2$-sphere has no boundary, it lies in the kernel of this map, and thus equals to $0$ in $H_2(M, \partial M; \mathbb{Z})$. This means it separates $M$.
Every embedded 2-sphere will separate M if and only if C=0, since the handlebody summands and spherical summands are irreducible.
@JoshHowie deleting balls and making connected sums preserve irreducibility?
No it doesn't preserve irreducibility, but removing 3-balls which are disjoint from any 2-sphere we are interested in, will not change whether the 2-sphere will separate. I have written an answer with more details.
That result seems obvious, given results in the literature. From a technique of Bray (see also Miao), the double of the manifold admits a positive scalar curvature metric, which is symmetric with respect to reflection. Then Coda-Marques' classification of 3-manifolds with positive scalar curvature and the understanding of fixed points of involutions on them (which comes from the orbifold theorem) completes the picture. See: https://mathoverflow.net/a/368937/1345
Every embedded 2-sphere will separate $M$ if and only if $C=0$.
Proof: Suppose $C=0$, and let $\Sigma$ be 2-sphere embedded in $M$. Let $\{S_j\}$ be a collection of 2-spheres which decompose $M$ into prime summands. Look at the intersection of $\Sigma$ with $\{S_j\}$. Let $\Delta$ be an innermost disk on some $S_k$. We can surger $\Sigma$ along $\Delta$, which will decompose $\Sigma$ into two 2-spheres. Repeating this process until there are no intersections with $\{S_j\}$, $\Sigma$ is decomposed into a collection of embedded $2$-spheres which we call $\Sigma'$. Then $\Sigma$ will be non-separating in $M$ if and only if some component of $\Sigma'$ is non-separating in $M$.
Each component of $\Sigma'$ is contained within a single prime summand of $M$. We then cut $M$ along $\{S_j\}$, and glue $3$-balls onto each $2$-sphere boundary component of the resulting $3$-manifolds. It is well-known that handlebodies and closed spherical $3$-manifolds are irreducible, which means that every embedded $2$-sphere bounds a $3$-ball. Thus each component of $\Sigma'$ is separating in its respective prime summand and hence in $M$. Removing the $D$ $3$-balls from $M$ which are disjoint from $\Sigma$ and $\Sigma'$ does not affect whether $\Sigma$ is separating. Therefore $\Sigma$ is separating in $M$.
Conversely, if $C\neq 0$, then we can find a non-separating 2-sphere $\Sigma''$ in some $S^2\times S^1$ component which is disjoint from each $B_i$ and each $S_j$. Furthermore the dual $1$-sphere to $\Sigma''$ is disjoint from each $B_i$ and each $S_j$. Therefore $\Sigma''$ is non-separating in $M$.
What do you mean by dual 1-sphere?
A surface in a 3-manifold is non-separating if there exists a circle which intersects that surface transversely exactly once.
Yes, this is true. I just didn’t understand how do you conclude that such a curve exists for $\Sigma’’$.
Such a circle exists in $S^2\times S^1$, and can be isotoped to be disjoint from any collection of balls, so the circle still exists in $M$.
Perfect, thank you for the detailed answer.
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2025-03-21T14:48:32.078715
| 2020-09-20T23:01:52 |
372181
|
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|
Stack Exchange
|
Decidability of a first-order theory of hyperreals
The theory of real closed fields is decidable. The hyperreals satisfy that theory, so we can interpret statements in the theory of real closed fields as being about hyperreals.
If we add a unary predicate for "is a standard real number" to the language, is the theory still decidable?
Any reason for the anonymous downvote?
Do you mean adding a predicate "standard real"? Because the predicate "is a real number" in the theory of real-closed fields would, in principle, be interpreted as "everything".
@AsafKaragila standard real, yes.
What happens if you take an ultrapower of that structure, then your "new" hyperreals will have the standard reals predicate cover the "old" hyperreals.
@MattF. That would be equivalent to $y-x$ is a positive infinitesimal. Is there a way to express being an infinitesimal without quantifiers?
@MattF. sorry, meant that $y-x$ is positive and not infinitesimal. And how do you express division without quantifiers.
Related
Yes, the theory is decidable.
If $F$ is an ordered field and $R\subseteq F$ a non-cofinal subfield, then
$$O=\{x\in F:\exists u\in R\:(-u\le x\le u)\}$$
is a convex valuation ring of $F$, with maximal ideal
$$I=\{x\in F:\forall u\in R_{>0}\:(-u\le x\le u)\}.$$
$R$ embeds as a cofinal subfield in the residue field $O/I$; in general, the embedding may be proper, but if $R=\mathbb R$, then $R=O/I$, as $\mathbb R$ is complete.
Thus, let $T$ be the theory of structures $(F,R,+,\cdot,<)$ such that
$F$ is a real-closed field,
$R$ is a non-cofinal subfield of $F$, and
the canonical embedding of $R$ into the residue field $O/I$ as defined above is surjective (and therefore an isomorphism).
Then $T$ is a recursively axiomatized theory, it is valid in the hyperreal structures $({}^*\mathbb R,\mathbb R)$, and it is complete (see below), hence it is decidable, and axiomatizes the first-order theory of $({}^*\mathbb R,\mathbb R)$.
As pointed out in a comment by Erik Walsberg (thanks!), the completeness of $T$ is a special case of a more general result on tame elementary extensions of o-minimal structures due to Van den Dries and Lewenberg [1]. (Here, tameness is basically the axiom 3 above.) Their results also show that $T$ is model-complete, and in fact, that it has quantifier elimination in a language expanded with function symbols for roots of polynomials (which make the theory of real-closed fields universally axiomatized) and for the “standard part” map $\mathrm{st}\colon O\to R$ such that $x-\mathrm{st}(x)\in I$.
Let me indicate how to prove a weaker result: the theory $T_0$ of structures $(F,O)$ such that
$F$ is a real-closed field,
$O$ is a proper convex subring of $F$,
is complete and decidable. This follows from the Ax–Kochen–Ershov principle, which states that two henselian valued fields of residue characteristic $0$ with elementarily equivalent residue fields and value groups are elementarily equivalent.
First, it is an easy consequence of basic facts about valued fields that if $F$ is a real-closed field with a convex valuation ring $O$, then the valued field $(F,O)$ is henselian, the residue field $O/I$ is real-closed, and the value group $F^\times/O^\times$ is divisible.
Thus, if $(F,O)$ and $(F',O')$ are two models of $T_0$, their residue fields are elementarily equivalent by completeness of the theory of real-closed fields, and their value groups are elementarily equivalent by completeness of the theory of divisible totally ordered abelian groups, hence the valued fields $(F,O)$ and $(F',O')$ are elementarily equivalent by the AKE principle.
As shown by Cherlin and Dickman [2], $T_0$ has quantifier elimination in the language of ordered rings expanded with the predicate
$$x\mid y\iff y\in xO.$$
References:
[1] Lou van den Dries, Adam H. Lewenberg: $T$-convexity and tame extensions, Journal of Symbolic Logic 60 (1995), no. 1, pp. 74–102, doi: 10.2307/2275510. On JSTOR.
[2] Gregory Cherlin, Max A. Dickmann: Real closed rings II. Model theory, Annals of Pure and Applied Logic 25 (1983), no. 3, pp. 213–231, doi: 10.1016/0168-0072(83)90019-2.
Neat! You wouldn't happen to know if quantifier elimination would work, like Matt F. speculated in a comment under the question?
The language of real closed fields has the $>$ relation. The language that the Ax–Kochen–Ershov principle talks about wouldn't have that, right?
I think it is worth pointing out that completeness of this theory is a special case of the work of van den Dries and Lewenberg on tame pairs. They also get a quantifier elimination in a reasonable language. Their results cover the analogous structure when one replaces the field of real numbers by some o-minimal expansion of the field of real numbers. https://projecteuclid.org/euclid.jsl/1183744679
@ErikWalsberg Excelent, thank you. I will look at it when I get to the office.
For $T$, (3) is just saying that every hyperreal that is bounded by reals is infinitesimally close to a real, right?
Yes. ~~~~~~~~~~
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2025-03-21T14:48:32.079110
| 2020-09-20T23:30:33 |
372182
|
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|
Stack Exchange
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Does Hahn-Banach for $\ell^\infty$ imply the existence of a non-measurable set?
Working over ZF but without the Axiom of Choice (AC), assume that the Hahn–Banach Theorem holds for $\ell^\infty$. Does it follow that there exists a set of real numbers that is not Lebesgue measurable?
The motivation for this question is the old question of what would be involved in replacing ZFC by ZF+LM plus some weakened version of AC (here LM denotes the axiom "all sets of reals are Lebesgue measurable"). Now, analysts are accustomed to freely invoking theorems such as Hahn–Banach and Krein–Milman, and are going to dislike having to check some picky logical detail every time they do so. In light of Solovay's model, one naïve hope would be to define some notion of "tame" function space, and prove, using DC (the axiom of dependent choice), that all the standard theorems hold for tame function spaces, and that any function space that arises "in practice" is tame.
Unfortunately, $\ell^\infty$ is not tame. We can see this even without a precise definition of "tame" because by applying the Hahn–Banach theorem to $\ell^\infty$, one can show that $(\ell^\infty)^* \ne \ell^1$. On the other hand, it is known that ZF+DC+BP (Baire property) implies that $(\ell^\infty)^* = \ell^1$.
These inconvenient facts limit our options. One option would be to regard $(\ell^\infty)^* = \ell^1$ as a feature and not a bug; students coming to the subject for the first time might find it to be a pleasant fact. But it might be a hard sell for analysts who have grown up using Hahn–Banach freely and who "know" that $(\ell^\infty)^* \ne \ell^1$.
Another option would be to ignore AC for the moment and just examine the extent to which Hahn–Banach is compatible with LM. That brings me to the stated question.
Whether or not it's a feature or a bug, working in settings where $(\ell^\infty)^=\ell^1$ breaks some nice features of the category of Banach spaces, unless one redefines the dual space functor in some way (I'm guessing something like $X^$ not a Banach space but some vector space with some kind of localic structure on its "unit ball"). For instance, it is a nice feature of the category Ban that subobjects of reflexive objects are reflexive, and that theorem is lost in the world where $(\ell^\infty)^*=\ell^1$
Have you checked Howard and Rubin?
@Nate: That can't be right. We know that free ultrafilters imply the existence of non-measurable sets. But the others seem to be open.
@AsafKaragila: You're right, I shouldn't have said "ultrafilter".
According to this answer, it is open whether the existence of a non-principal finitely additive probability measure on $\mathbb{N}$, or a Banach limit, implies the existence of a non-measurable set. Since those are the most obvious objects that one constructs with Hahn-Banach on $\ell^\infty$, this may perhaps be open as well.
Does anyone even use Hahn-Banach on $l^\infty$? I can't think of a single time I've needed it for any non-separable space. Of course, no choice is needed for Hahn-Banach for separable spaces.
@NikWeaver, I guess HB for $\ell^\infty$ is crucial for the existence of the Dixmier trace which in turn has been used by Connes in applications to non-commutative geometry.
@NikWeaver: Does anyone even use $\ell^\infty$? :P
@AsafKaragila: Sure, it's the vector space from hell in constructive mathematics, so it gets "used" to scare students: "So you think $\ell^\infty$ has a norm? The devil has stolen your soul!"
@Andrej: What soul? I thought constructivists don't have a soul, because one cannot construct it, and instead it needs to be axiomatised (and then you either have or you don't, but it's hard to prove uniformly which one).
@AsafKaragila: Touché!
@NateEldredge: I finally got Howard and Rubin's book out of the library. The existence of a non-measurable set is Form 93 but, as one might suspect, Hahn-Banach for $\ell^\infty$ isn't a form per se, I didn't see any other implications that obviously settled my question. As for the website, it appears to be totally broken. I couldn't get any of the features to work.
|
2025-03-21T14:48:32.079554
| 2020-09-21T01:22:23 |
372187
|
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|
Stack Exchange
|
Do discrete embeddings of surface groups not necessarily carry an embedding of SL_2?
We can get arithmetic lattices isomorphic to free groups in $\mathrm{SL}_2\mathbb{R}$, so in general we can’t expect homomorphisms of lattices into semisimple Lie groups to say much about $\mathrm{SL}_2\mathbb{R}$, even if we require the image to be discrete. In particular, unlike in higher-rank cases, such homomorphisms won’t necessarily extend to homomorphisms of $\mathrm{SL}_2\mathbb{R}$.
However, if $\Gamma<\mathrm{SL}_2\mathbb{R}$ is cocompact and we have an discrete embedding (by which I mean a map which is both a homeomorphism and an isomorphism onto its image) of $\Gamma$ into some semisimple Lie group $G$, will we not necessarily get a copy of $\mathrm{SL}_2\mathbb{R}$ in $G$ containing (the image of) $\Gamma$?
By the considerations above, the image of $\Gamma$ in such a copy of $\mathrm{SL}_2\mathbb{R}$ might not look like it originally did (meaning the embedding of $\Gamma$ hasn’t necessarily extended to all of $\mathrm{SL}_2\mathbb{R}$), but part of me imagines that the image of $\Gamma$ should still determine something. Is this naïve?
Homomorphisms of a surface group into a lie group $G$ form a nice space, the character variety (with $2g$ coordinates in $G$ and one equation). The ones whose image lands in a copy of $SL_2(\mathbb R)$ are a high codimension subset (their closure should just consists of themselves plus maps that factor through a degneration of $SL_2(\mathbb R)$). Your condition that the embedding be homeomorphic to its image is not quite an open condition but it seems more generic than not.
There are vast number of counterexamples. For instance, see the following paper for surface subgroups of $SL(3,\mathbb{R})$ that are Zariski dense in the entire group: http://web.math.ucsb.edu/~long/pubpdf/CoCompact_IJM.pdf
Up to conjugation, there only finitely many homomorphisms $SL(2,R)\to G$ (where $G$ is, say, connected and semisimple). At the same time, the dimension of the space of homomorphisms $\Gamma\to G$, up to conjugation, is $(2p-2)dim(G)$, where $p$ is the genus, provided $p\ge 2$.
@Tara That’s exactly the kind of thing I was looking for. If you’d like to post that as an answer, then I’ll accept it.
Homomorphisms from surface groups are very flexible, so there are indeed such examples.
For instance, use: Breuillard–Gelander–Souto–Storm (Dense embeddings of surface groups.
Geom. Topol. 10 (2006), 1373–1389) (DOI link)
They proved, for a surface group $\Gamma$ ($\pi_1$ of oriented closed surface of genus $\ge 2$) that, among others, $\Gamma$ embeds densely into arbitrary nontrivial connected semisimple Lie groups.
To deduce a discrete embedding, argue as follows: choose a dense embedding into $G$ (with $G$ pretty arbitrary, even $G=\mathrm{SL}_2(\mathbf{R})$ is fine) and consider a standard discrete embedding into $\mathrm{PSL}_2(\mathbf{R})$. Then the diagonal embedding into $G\times\mathrm{SL}_2(\mathbf{R})$ has a Zariski-dense image, so its image is not contained in any copy of any subgroup locally isomorphic to $\mathrm{SL}_2(\mathbf{R})$.
Variant, not using the above reference [BGSS]:
Choose two Fuchsian embeddings $i_1,i_2:\Gamma\to H$ where $H=\mathrm{PSL}_2(\mathbf{R})$, such that the images of $i_1$ and $i_2$ are not conjugate under $\mathrm{PGL}_2(\mathbf{R})$. Then the diagonal embedding $i_1\times i_2:\Gamma\to H\times H$ works (its image is Zariski-dense in $H^2$).
Sadly, this does not answer the posed question since it is about discrete and faithful representation (an embedding in question is required to be a homeomorphism to its image).
@MoisheKohan indeed, right... the word "discrete" is indeed useful to emphasize. I'm leaving the (cw) answer anyway; hopefully the discrete example in the comment will be posted too. I tend not to read (or read too quickly) a definition of a term if I'm already familiar with the term, so can miss an unusual additional assumption.
@MoisheKohan oh, it's straightforward to deduce discrete embeddings too: I added the missing argument.
Yes, it is correct now. The stronger result however is that every semisimple Lie group without compact factors and of dimension $>3$ contains a discrete subgroup isomorphic to a surface group and not contained in any subgroup locally isomorphic to $SL(2,R)$.
@MoisheKohan sure it's not the best result (probably it's even true with "Zariski dense"). I was rather looking at the easiest approach. Note that the very elementary argument I give at the end carries over all $G$ containing a subgroup locally isomorphic to $\mathrm{SL}_2\times\mathrm{SL}_2$ (this covers $SL_4$ and $Sp_4$ but not $SL_3$ or rank one groups). Also one of the most important and studied examples is that of quasi-Fuchsian groups.
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2025-03-21T14:48:32.079901
| 2020-09-21T02:30:35 |
372191
|
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|
Stack Exchange
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A rather non-$F_\sigma$ Borel set
I asked this question at MSE a week ago, but received no answer, so I cross-post it here.
I obtained a negative answer to this MSE question provided each metric space $X$ such that $|X|=\frak c$ and density $d(X)<\frak c$, contains a Borel set $B$ such that $|B\setminus C|=\frak c$ for each $F_\sigma$-subset $C$ of $X$ with $C\subset B$. My question is whether the latter claim holds. I guess this is known (and true), but it is hard to find a reference. Thanks.
My try. I guess using Theorem 22.4 from [Kech] I can show the claim when $X$ is Polish. To prove the claim for a separable $X$, by Proposition 12.1 from [Kech], it suffices to prove it for subspaces of an arbitrary fixed Polish space.
References
[Kech] A. Kechris, Classical Descriptive Set Theory, Springer, 1995.
Isn't this true of every Borel set B which is not $F_\sigma$, since, by the perfect set property for Borel sets, the alternative to $B \setminus C$ being of cardinality continuum is that it is countable?
@PaulLarson This alternative holds even for Borel sets of a non-Polish space $X$?
I see. I wasn't reading very carefully. I guess not, since your space could just be (0,1) along with a subset of (1,2) of size aleph_1?
@PaulLarson Right.
Here's an argument that the statement is false if the Continuum Hypothesis fails and the covering number for the null ideal is the same as the continuum. Wellorder the Borel sets of reals as $\langle B_{\alpha} : \alpha < \mathfrak{c} \rangle$. Choose for each $\alpha < \mathfrak{c}$ an $F_{\sigma}$ set $C_{\alpha} \subseteq B_{\alpha}$ such that $B_{\alpha} \setminus C_{\alpha}$ is null and a real $x_{\alpha}$ not in $B_{\beta} \setminus C_{\beta}$ for any $\beta < \alpha$. Let $X = \{ x_{\alpha} : \alpha < \mathfrak{c}\}$. Then any Borel subset of $X$ is $B_{\alpha} \cap X$ for some $\alpha$. Furthermore, $C_{\alpha} \cap X$ is an $F_{\sigma}$ subset of $B_{\alpha} \cap X$, and $(B_{\alpha} \cap X) \setminus (C_{\alpha} \cap X)$ is contained in $\{ x_{\beta} : \beta \leq \alpha\}$ which has cardinality less than $\mathfrak{c}$.
As for the consistency of the statement that CH is false and $\mathrm{cov}(\mathcal{N}) = \mathfrak{c}$, this is a standard consequence of MA + not-CH (see Theorem 26.39 of the 2003 edition of Jech's Set Theory). The consistency of MA and not-CH is Theorem 16.13 of Jech.
I have to revise my earlier claim that the statement in question follows from CH. This appears to be true for spaces $X$ having a Borel subset which is not $F_{\sigma}$, by the idea in my original comment : if $B$ has an $F_{\sigma}$ subset $C$ such that $B \setminus C$ is countable, then $B$ is $F_{\sigma}$. On the other hand, one can run the proof in the first paragraph of this answer under CH to produce a set of reals of cardinality $\mathfrak{c}$ such that every Borel set is $F_{\sigma}$. So the statement would fail for such a space.
Your construction works and so a negative answer to my question is consistent. This is a surprising and bad news for me, because it shows that my idea to obtain an answer for an original MSE question will not work and so I accepted your answer. Fill free to refer to it in an answer to my original MSE question in order to mark it as answered.
I don’t see how my statement follows from the Continuum Hypothesis. Moreover, I don’t see a need in your construction that both $\frak c$ and the covering number $\operatorname{cov}(\mathcal N)$ or the null ideal $\mathcal N$ are both $\frak c$ (and a consistency reference of this is still needed). I think that your construction works provided $\operatorname{cov}(\mathcal N)=\frak c$, and this is consistent (for instance, it holds under CH).
You are right that $\mathfrak{c} = \aleph_{2}$ can be replaced with $\mathfrak{c} > \aleph_{1}$, and also about that typo. I've made those changes. I'll try to add something soon about the claim that CH is sufficient.
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2025-03-21T14:48:32.080201
| 2020-09-21T05:55:35 |
372196
|
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|
Stack Exchange
|
Reference request: the theory of currents
I am a graduate student and want to study the theory of currents. What is a good reference for a beginner?
I should be familiar with the theory of distributions or generalized functions on $\mathbb R^n$.
de Rham's book "Differentiable manifolds - forms, currents, harmonic forms"
Federer's book "Geometric measure theory" contains everything from the beginning :p
@HassanJolany Really? An introduction for a beginner?
https://webusers.imj-prg.fr/~tien-cuong.dinh/Cours2005/Master/cours.pdf
@ChrisGerig de Rham's book is certainly well written and I usually end up recommending it to everyone interested in currents. But that being said, it is not really a good introduction to what is now understood to be the theory of currents, as Federer and Fleming kind of totally transformed de Rham's ideas just a few years later.
The theory of currents is a part of the geometric measure theory. Unfortunately, Federer made the subject completely inaccessible after he wrote his famous monograph:
H. Federer, Geometric measure theory. Die Grundlehren der mathematischen Wissenschaften, Band 153 Springer-Verlag New York Inc., New York 1969.
The problem is that the book contains `everything' (well, almost) and it is unreadable. After this book was published, people did not dare to write other books on the topic and only the bravest hearts dared to read Federer's Bible.
In my opinion the first accessible book on the subject is
L. Simon, Lectures on geometric measure theory. Proceedings of the Centre for Mathematical Analysis, Australian National University, 3. Australian National University, Centre for Mathematical Analysis, Canberra, 1983.
You can find it as a pdf file in the internet. Note that this book was written 14 years after Federer's book and there was nothing in between.
I would also suggest:
F. Lin, X. Yang, Geometric measure theory—an introduction. Advanced Mathematics (Beijing/Boston), 1. Science Press Beijing, Beijing; International Press, Boston, MA, 2002.
I haven't read it, but it looks relatively elementary (relatively, because by no means the subject is elementary).
The last, but not least is
F. Morgan, Geometric measure theory. A beginner's guide. Fifth edition. Illustrated by James F. Bredt. Elsevier/Academic Press, Amsterdam, 2016.
You will not learn anything form that book as it does not have detailed proofs, but you can read it rather quickly and after that you will have an idea about what it is all about.
Wikipedia also lists Fomenko, Anatoly T. (1990), Variational Principles in Topology (Multidimensional Minimal Surface Theory), Mathematics and its Applications (Book 42), Springer, Kluwer Academic Publishers, ISBN<PHONE_NUMBER>308; and Mattila, Pertti (1999), Geometry of Sets and Measures in Euclidean Spaces, London: Cambridge University Press, с. 356, ISBN 978-0-521-65595-8; but I do not know if they are good.
@OlegLobachev I don't know Fomenko's book, but Mattila's book does not discuss currents.
One important note about Morgan's book is that at least its first half is meant as a kind of companion to Federer's magnum opus, i.e. they share the same notation and for many theorems Morgan will only give the important ideas and then point to the precise corresponding section of Federer's book, where all the technical details are carried out. Also, without wanting to spoil too much, the book is worth at least a look for the illustrations alone.
A beginner friendly introduction can be found in chapter 7 of the book "Geometric Integration Theory " by Krantz and Parks. It is from 2008 and written in a modern and clear style and it starts nearly from "zero".
You are right. That is a good reference. I forgot about it.
http://www.math.wustl.edu/~sk/books/root.pdf
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2025-03-21T14:48:32.080537
| 2020-09-21T07:04:35 |
372197
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"Zach Teitler",
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"https://mathoverflow.net/users/158000",
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"https://mathoverflow.net/users/44191",
"https://mathoverflow.net/users/45250",
"https://mathoverflow.net/users/45581",
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|
Stack Exchange
|
Tools from other disciplines useful to mathematics research?
Obviously, mathematics provides essential tools for physicists, biologists, economists, engineers and many others to use in their research. Equally obviously, physics, biology, economy and engineering provide the inspiration for research topics and directions in mathematics. But this isn't what this question is about. Instead, I'm wondering about tools provided to mathematicians by other disciplines.
Are there any examples of tools (excluding software) from another discipline that you personally have used in your mathematical research?
E.g., I find dimensional analysis (as in assigning dimensionfull units to variables) a very useful tool from physics, which can greatly help in finding errors during algebraic manipulation of equations.
Probably better clarify that you want to exclude "software" as an answer, or at least lump all computers and computer programs into a single answer, so that you don't get a zillion answers that just list out a whole bunch of programs that are useful for math.
@ZachTeitler good point! I've edited the question.
I was thinking more in terms of intellectual tools, so I guess dictionaries do count to a certain extent.
... but coffee-making equipment doesn't count?
@shane.orourke ... but coffee-making equipment is so universally useful in research (in any discipline) that it falls in the same essential category as chairs, desks, pens, paper and wastebaskets.
"An old joke tells that mathemtics is the second-cheapest department at any college because all you need is a pencil, paper and a wastebasket. Philosophy is the cheapest because you don’t need the wastebasket."
This is more math teaching than research, so I don't think it belongs as an answer, but I've found the concept of "scope" from programming useful when teaching things where care with quantifiers is necessary. Not extremely useful - students who know one usually understand the other without the connection being made explicit - but it occasionally helped.
The William G. Pritchard Fluids Lab conducts experiments–actual, physical experiments with messy, wet fluids–in the Penn State Math Department. Perhaps this counts as drawing on laboratory techniques of physics, engineering, etc.
I haven't personally used this in my research, though. So I'm marking this answer as community wiki.
Good point - there is also a fluids lab at DAMTP in Cambridge, http://www.damtp.cam.ac.uk/research#Fluid%20and%20Solid%20Mechanics.
That there is a wet lab in a US math department is probably a historical accident..but the world of fluid dynamics is full of applied math folks collaborating with physicists and engineers...so in itself it isn't unusual.
John Bush, at MIT, runs the Applied Mathematics Laboratory (which is a fluid mechanics lab): https://math.mit.edu/~bush/ . He did postdoctoral work at Cambridge.
Two books in my library (but, sadly, as yet unread) claim to use physical methods (i.e., methods of physics, specifically, mechanics) to solve - or suggest a way of solving - mathematical problems:
The Mathematical Mechanic: Using Physical Reasoning to Solve Problems (Levi, 2009)
INTRODUCTION
...the book does exact revenge—or maybe just administers a pinprick—agsinst the view that mathematics is a servant of physics. In this book physics is put to work for mathematics, proving to be a very efficient servant (with apologies to physicists). Physical ideas can be real eye-openers and can suggest a strikingly simplified solution to a mathematical problem. ...
Some Applications of Mechanics to Mathematics (Uspenskii, 1961) - a slim volume, reproducing a lecture to (Russian) high school students:
FOREWORD
The applications of mathematics to physics (in particular, to mechanics) are well-known. We need only open a school text-book to find examples. The higher branches of mechanics demand a complex and refined mathematical apparatus.
There are, however, mathematical problems for whose solutions we can succeessfully use the ideas and laws of physics. A number of problems of this kind, soluble by methods drawn from mechanics (namely, using the laws of equilibrium) were given by the author in his lecture ...
Somewhat related: https://mathoverflow.net/questions/347134/gadgets-as-primality-tests
Likewise somewhat related: https://mathoverflow.net/questions/172109/optical-methods-for-number-theory
I hope I correctly got what you mean by "tools". Google maps is a great tool for some applications of graph theory, optimization and even topology. I have seen several papers that use the map tools for developing optimal or sub-optimal algorithms on finding the shortest routes or statistical analysis etc.
For example, a brief search leads to this paper and a handful of others.
Since this is now CW, I can just as well add some answers.
As Matt F. mentioned in comments, language skills are often useful in mathematical research (e.g. for accessing literature written in French or Russian, or indeed in English in the case of second-language speakers), and those can certainly be considered tools from another discipline.
Loosely related: https://mathoverflow.net/questions/247314/learning-german-and-russian-for-reading-old-mathematical-papers-in-these-languag
Since you mentioned dimensional analysis, you might enjoy the book Street-Fighting Mathematics: The Art of Educated Guessing and Opportunistic Problem Solving
by Sanjoy Mahajan. From the book description: "Sanjoy Mahajan builds, sharpens, and demonstrates tools for educated guessing and down-and-dirty, opportunistic problem solving across diverse fields of knowledge—from mathematics to management. Mahajan describes six tools: dimensional analysis, easy cases, lumping, picture proofs, successive approximation, and reasoning by analogy."
I have built physical models (wood, paper, brass, plastic) of geometric objects (polyhedra, dissections, tensegrities) that have helped me see and prove theorems.
Steinhaus's Mathematical Snapshots is an inspiration. Dover offers the ebook. Better: find a used hard copy.
@MattF. Fair comment. The "other discipline" might be art, or carpentry.
or, these days, 3D printing.
@MichaelLugo I do that too, but left if out since computer applications were discouraged). You can find stuff here https://www.cs.umb.edu/~eb/slicecubes/ and here http://gallery.bridgesmathart.org/exhibitions/2018-joint-mathematics-meetings/ebolker .
BTW, this is the easiest possible away to understand Desargues theorem (in 3 dimensions) ... - Also - the most astonishing and unexpected thing I have ever seen: I moved to a new neighborhood about 15 years ago and while driving around saw in someone's front yard a bunch of large stick objects made of bamboo sticks (he had lots of bamboo plants). Each was about 4' tall. All of the platonic solids, of course. Quite a few Archimedean solids. And a proof of Desargues' theorem!!! 6' high made of bamboo in his front yard!
I find that thinking about types (as in object-oriented programming) is really helpful. This can help clarify things (and in particular, help students).
For example, one cannot usually add a matrix-type and a partition-type.
I suppose this is closely connected to category theory and/or type theory,
but one does not need any background in those fields, in order to think about types; perhaps this is similar to the "units"-connection mentioned in a different answer.
Even though OOP is the first place where one encounters the concept of types conciously (say, as a beginning CS student), and so this is totally fine for the scope of the question, I still want to point out that having types does not require being object-oriented. In fact, I would argue that the programming languages which most cleanly embrace the concept of types are typically not object-oriented (Haskell is probably the best example). In a way, the object-oriented paradigm forces you to lump together the types (i.e. what objects are) with all the things you can do with them.
Sorry to post yet another self-answer, but this is completely separate from my other answer.
Polymath-style projects or sites like MathOverflow are tools for mathematical research that could be considered applications of social science to mathematics research. This is especially true if incentive systems like reputation scores or badges are being used, which are ultimately based on insights from psychology and/or economics (which can to some extent be understood to be the study of incentives).
He lists lots of works by Edward Tufte, who (while his background is in statistics) worked a lot on political economy.
The book "Peopleware", which he lists, while written by programmers, starts from the statement "The major problems of our work are not so much technological as sociological in nature".
And of course UI design is ultimately based on perceptual psychology, even if most people writing books about it are computer scientists. It wouldn't work if it weren't.
It's not clear how useful this would actually be, but the idea of using "citizen science" (a somewhat sociology-based notion which has been developed into a useful tool for — primarily observation-based — scientific research by astronomers, botanists and zoologists, among others) for mathematics research has been floated here on MO: Can pure mathematics harness citizen science?. It would be nice to know whether any of the projects proposed in answers to that question ever came to fruition.
Citizen science is expanding, see https://www.zooniverse.org/
|
2025-03-21T14:48:32.081595
| 2020-09-21T07:49:06 |
372200
|
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|
Stack Exchange
|
Enumerating subspaces of $\mathbb{F}_q^n$ in terms of words and inversions
When $q$ is a prime power, then on the one hand the $q$-binomial coefficient $\binom{n}{k}_q$ equals the number of $k$-dimensional subspaces of $\mathbb{F}_q^n$, and on the other hand it is the generating function of the sequence which sends $r$ to the number of words in two letters $X,Y$ of length $n$ with $k$ occurances of $X$ and with $r$ inversions (i.e. places where $Y$ comes before $X$). Therefore, there must a bijection
$$\{k\text{-dimensional subspaces of } \mathbb{F}_q^n\} \\ \downarrow{\small\cong}\\ \coprod_{r=0}^{k(n-k)} q^r \cdot \{\text{ words in X,Y of length } n \text{ with } k \text{ X's and } r \text{ inversions}\}$$
My question is if you can write down a nice, explicit bijection. I require that its description is independent of the theory of $q$-binomial coefficients, thus doesn't use the calculation of the cardinalities, and that it is also without recursion (since you can easily transform the inductive proof of the equality of cardinalities into a recursive bijection; this doesn't count here).
For $k=1$ the bijection looks as follows: It sends $\langle a_1,\dotsc,a_n \rangle$ to $(a_1/a_{r+1},\dotsc,a_r/a_{r+1}) \cdot Y^r X Y^{n-r-1}$, where $r$ is maximal with $a_{r+1} \neq 0$.
You may act similarly as follows.
Let $V$ be a $k$-dimensional subspace of $\mathbb F_q^n$. Take any its base, put its elements into the rows of some matrix, and make it to the reduced row echelon form $B$ (which is unique). The rows of $B$ still form a base of $V$.
Let $r_1$, $r_2$, $\dots$, $r_k$ be the indices of leading elements in rows $1,2,\dots,k$. Then you put into correspondence to $V$ the tuple of all elements of $B$ which are non-fixed (i.e., those not in the leading columns, and not to the left of leading elements), along with the word $Y^{r_1-1}XY^{r_2-r_1-1}XY^{r_3-r_2-1}X\dots XY^{n-r_k}$ read backwards. The freedom of the coefficients provides the exponent of $q$ being
$$
(n-r_1-(k-1))+(n-r_2-(k-2))+\dots+(n-r_k),
$$
and the number of inversions (in the reversed word) is clearly
$$
(n-r_k)+(n-r_{k-1}-1)+(n-r_{k-2}-2)+\dots+(n-r_1-(k-1)),
$$
so they trivially coincide.
Thank you! I still need to check the details. For some reason I have never heard that the reduced row echelon form only depends on the subspace (which you seem to say), do you have a reference for that?
The reduced row echelon form of a matrix is unique —- see, e.g., https://en.wikipedia.org/wiki/Row_echelon_form ;) You may prove this just by choosing it from bottom to top. On the other hand, the matrices which can be obtained from a certain matrix via elementary row transforms are those with the same linear span of rows.
Ok. I was aware that it only depends on the matrix, but your 2nd sentence explains why it then also only depends on the subspace generated by the rows.
Can you explain a bit more what $n-r_1-(k-1)$ is counting? Is it a column of non-fixed elements (where?), or a row?
Say, a second row contains $r_2-1$ leading zeros, a $1$, and $k-2$ zeros on the subsequent leading columns. All other elements are ''free'', and they are $n-r_2-(k-2)$ many.
This is very standard. It is the Bruhat decomposition (a.k.a. Schubert cell decomposition) of the Grassmannian, in the case of finite fields. See the proof of Proposition 1.7.3 of Stanley's "Enumerative Combinatorics," Vol. 1, 2nd ed: http://math.mit.edu/~rstan/ec/ec1.pdf
This bijection is described very clearly (but without reference to Schubert cells etc.) in D. E. Knuth, Subspaces, subsets, and partitions, J. Combin. Theory Ser. A 10 (1971), 178–180, https://www.sciencedirect.com/science/article/pii/0097316571900227
|
2025-03-21T14:48:32.082003
| 2020-09-21T07:58:07 |
372201
|
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"url": "https://mathoverflow.net/questions/372201"
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|
Stack Exchange
|
Principal bundles over marked Riemann surface $\mathcal{M}_{g,h,r,s}$
Let $\mathcal{M}_{g,h,r,s}$ be a Riemann surface with genus $g$, $h$ boundary components, r interior marked points, and $s$ marked points on the boundary $\partial \mathcal{M} = \Sigma$. In the case where $s=0$, the isomorphism classes of Principal G-bundles over $\mathcal{M}_{g,h,r}$ are in bijection to $\dfrac{Hom( \pi_1(\mathcal{M}_{g,h,r}), G)}{G}$ where those homomorphisms have to have match the conjugacy classes specified on the boundaries. My question is how this generalizes to the case $s>0$. Do we need to specify conjugacy classes on each segment (i.e. in between the marked points) of the marked boundaries? Is there a bijection to something resembling $\dfrac{Hom( \pi_1(\mathcal{M}_{g,h,r}), G)}{G}$? I guess the simplest context could be for a principal bundle over a disk with 2 marked points on its boundary.
I would greatly appreciate any good references that explain the construction of principal bundles over $\mathcal{M}_{g,h,r,s}$.
|
2025-03-21T14:48:32.082099
| 2020-09-21T08:08:51 |
372202
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/372202"
}
|
Stack Exchange
|
Concentration inequality for the supremum of $L_2$ norm of a vector-valued Gaussian process with iid components
Let $\Omega$ be a compact subset of $\mathbb R^p$ and let $f_1,\ldots,f_k$ be zero mean identically distrubuted Gaussian processes on $\Omega$ such that $f_1(x),\ldots,f_k(x)$ are independent $x \in \Omega$. Thus $f:=(f_1,\ldots,f_k)$ can be seen as a vector-valued Gaussian process with iid components at each point in space. Let $\lambda:= \mathbb E[f_1(0)^2]$. Now, given $x \in \Omega$, define $\nu(x) := \|f(x)\| := (\sum_{j=1}^k f_j(x)^2)^{1/2}$. I'm interested in concentration inequalities for $\|\nu\|_\infty := \sup_{x \in \Omega}\nu(x)$.
In the special case where $k=1$, the Borell-TIS inequality kicks-in to give
$$
\begin{split}
&\forall u\ge 0,\; \mathbb P(\|\nu\|_\infty > \mathbb E[\|\nu\|_\infty] + u) \le e^{-u^2/(2\sigma^2)},\\
&\text{ where }\sigma^2 := \sup_{x \in \Omega}\mathbb E[\nu(x)^2] = \mathbb E[\nu(0)^2] = \lambda
\end{split}
$$
Question. How to get concentration ienqualities for $\|\nu\|_\infty$ when $k \ge 2$ ?
Some notes on the case $k \to \infty$
Define a random field $Z_k$ on $\Omega$ by setting $Z_k(x):= \dfrac{1}{\sqrt{2\lambda k}}(\|f(x)\|^2-\lambda k) = \dfrac{1}{\sqrt{2\lambda k}}(\sum_{j=1}^kf_j(x)^2-\lambda k) $, for all $x \in \Omega$.
I don't know if there is an appropriate notion of CLT which could kick-in here, to give the limiting distribution of the random field $Z_k$.
Consider the real-valued centered Gaussian process $(X_{t,a}\colon(t,a)\in T\times B_k)$, where
$$X_{t,a}:=\sum_{j\in[k]}a_j f_j(t),$$
$T:=\Omega$, $B_k$ is the unit ball in $\mathbb R^k$, and $[k]:=\{1,\dots,k\}$. Then
$$\|\nu\|_\infty=\|X\|_\infty:=\sup\{|X_{t,a}|\colon t\in T, a\in B_k\}$$
and
$$EX_{t,a}^2=\sum_{j\in[k]}a_j^2 Ef_j(t)^2\le\sup_{t\in T,j\in[k]}Ef_j(t)^2=:\sigma^2.$$
So, by the Borell--TIS inequality for the process $(X_{t,a})$, for all real $u\ge0$
$$
P(\|\nu\|_\infty>E\|\nu\|_\infty+u)=P(\|X\|_\infty>E\|X\|_\infty+u)\le e^{-u^2/(2\sigma^2)}.
$$
|
2025-03-21T14:48:32.082230
| 2020-09-21T08:21:30 |
372203
|
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|
Stack Exchange
|
On Ext-duals of injective modules for commutative rings
Let $R$ be a commutative noetherian ring and $I=E(R/p)$ the injective hull of the module $R/p$ for a prime ideal $p$.
Question: Is there a (more) explicit description of the $R$-modules $Ext_R^i(I,R)$ for $i \geq 0$ (at least in special cases such as $R$ being Gorenstein)? When are those modules finitely generated?
|
2025-03-21T14:48:32.082291
| 2020-09-21T09:35:45 |
372206
|
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"Carlo Beenakker",
"fs l",
"https://mathoverflow.net/users/11260",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/372206"
}
|
Stack Exchange
|
A problem of matrix polynomial expansion
The problem is
$b = (1, -1)^\top, c = (1, 1)^\top, A \in \mathbb{R}^{2 \times 2}$, suppose the sum of reverse diagonal elements of $A$ is zero (i.e., $A_{12} + A_{21} = 0$), prove that the sum of reverse diagonal elements of $\sum\limits_{r=0}^{n-1} A^r c b^\top A^{n-1-r} $ is zero for any $n \in \mathbb{N}^{+}$.
In fact, this is my conjecture and I have tested many examples in my computer. For diagonal case, it is easy to prove it, but for the general case I do not know how to do it. One idea come to my mind is to write $A$ as the sum of a diagonal matrix and an anti-diagonal matrix, then expand $A^r$ by binomial expansion, but unfortunately they do not commute.
Could someone give me some hints?
Thanks!
Someone told me a simple method, I decide to post it here.
Note that for any $A \in \mathbb{R}^{2 \times 2}$, $A_{12} + A_{21} = 0$ if and only if
\begin{equation*}
A^\top = \sigma^{-1} A \sigma
\end{equation*}
with
\begin{equation*}
\sigma = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
\end{equation*}
Let
\begin{equation*}
J = \sum\limits_{r=0}^{n-1} A^r c b^\top A^{n-1-r}
\end{equation*}
then when $A_{12} + A_{21} = 0$ we have
\begin{align*}
J^\top & = \sum\limits_{r=0}^{n-1} \left(A^\top\right)^{n-1-r} b c^\top \left(A^\top\right)^{r} \\
& = \sigma^{-1} \sum\limits_{r=0}^{n-1} A^{n-1-r} c b^\top A^{r} \sigma \\
& = \sigma^{-1} J \sigma
\end{align*}
Therefore
\begin{equation*}
J_{12} + J_{21} = 0
\end{equation*}
This is a bit of a brute force approach, but it's effective. Note that the sum of the reverse diagonal elements of a $2\times 2$ matrix $M$ equals ${\rm tr}\,\sigma M$ with
$$\sigma=\begin{pmatrix}0&1\\1&0\end{pmatrix}.$$
For the most general form of the matrix
$$A=\begin{pmatrix}a&b\\ -b&c\end{pmatrix},\;\;\text{and for}\;\;D=\mathbf c\mathbf b^{\rm T}=\begin{pmatrix}1&-1\\1&-1\end{pmatrix},$$
I calculate
$$J(r,n)={\rm tr}\,\sigma A^r DA^{n-1-r}=$$
$$=\frac{2^{-n-1} (a+c-z)^{-r} (a+c+z)^{-r}}{(a-2 b-c) \left(a c+b^2\right)} \left[\left(z (a+c)-(a-c)^2+4 b^2\right) (a+c+z)^n (a+c-z)^{2 r}-\left(z (a+c)+(a-c)^2-4 b^2\right) (a+c-z)^n (a+c+z)^{2 r}\right],$$
with the definition $z=\sqrt{(a-c)^2-4 b^2}$.
Then I evaluate for $n\geq 1$ the sum
$$\sum_{r=0}^{n-1}J(r,n)=\frac{2^{-n-1} (a+c) \left((a-c)^2-4 b^2-z^2\right) \left((a+c-z)^n-(a+c+z)^n\right)}{z (a-2 b-c) \left(a c+b^2\right)}.$$
Substitution of the definition of $z$ finally gives the desired result
$$\sum_{r=0}^{n-1}J(r,n)=0.$$
Details of the calculation: I may assume $b\neq 0$ (otherwise $A$ is diagonal and the identity follows trivially). Then the matrix $A$ is diagonalizable when $b\neq \tfrac{1}{2}|a-c|$, in the form
$A=U\Lambda U^{-1}$ with $$U=\left(
\begin{array}{cc}
z-a+c & -z-a+c \\
2 b & 2 b \\
\end{array}
\right),\;\;\Lambda={\rm diag}\,\left(\tfrac{1}{2} \left(-z+a+c\right),\tfrac{1}{2} \left(z+a+c\right)\right)$$
With this decomposition we can readily evaluate $A^r=U\Lambda^r U^{-1}$.
If $b=\tfrac{1}{2}(a-c)\neq 0$ we instead use the Jordan decomposition $A=VJV^{-1}$ with
$$V=\left(
\begin{array}{cc}
-1 & -\frac{2}{a-c} \\
1 & 0 \\
\end{array}
\right),\;\;J=\left(
\begin{array}{cc}
\frac{a+c}{2} & 1 \\
0 & \frac{a+c}{2} \\
\end{array}
\right).$$
Then $A^r=VJ^r V^{-1}$, with $J^r=2^{-r} (a+c)^r\left(
\begin{array}{cc}
1 & 2 r (a+c)^{-1} \\
0 & 1 \\
\end{array}
\right)$.
Thanks for providing this method by using trace trick! But I am a bit confused how to derive the trace equality $(tr(\sigma A^r D A^{n-1-r}))$? Could you give me some hints? Thanks!
you diagonalize the matrix $A=U\Lambda U^{-1}$, with diagonal $\Lambda$, and then substitute $A^p=U\Lambda^p U^{-1}$, hence ${\rm tr},\sigma A^r DA^{n-1-r}={\rm tr},\tilde\sigma\Lambda^r \tilde{D}\Lambda^{n-1-r}$, with $\tilde\sigma=U^{-1}\sigma U$ and $\tilde{D}=U^{-1}DU$.
Ok, I see. But this case is for diagonalizable matrix right? What if $A$ is not diagonalizable?
I added some details, including the special case $b=|a-c|/2$ when $A$ is not diagonalizable.
Yes, I see it, thanks for your reply!
|
2025-03-21T14:48:32.082522
| 2020-09-21T09:55:49 |
372208
|
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|
Stack Exchange
|
Central limit theorem for chi-squared random field on $\mathbb R^p$
Let $X:x \mapsto X(x)$ be a centered stationary Gaussian process on the $\Omega:=\mathbb R^p$, such that $X(x) \overset{d}{=}X(x')$ for all $x,x' \in \Omega$. Set $\sigma^2 := \mbox{Var}(X(0)) = \mathbb E[X(0)^2]$. Let the random fields $X_1,\ldots,X_N$ be iid copies of $X$, and define a random process $Z_N$ on $\Omega$ by
$$
Z_N(x) := \frac{1}{\sqrt{2N\sigma^2}} \left(\sum_{i=1}^N X_i(x)^2-N\sigma^2 \right),\;\forall x \in \Omega.
$$
Question. Is there a central limit theorem (perhaps under further conditions on the base field $X$) for the limiting distribution of the random field $Z_N$ when $N \to \infty$ ?
$\newcommand\si\sigma\newcommand\Om\Omega$To determine the limit distribution of the process $Z_N$ (and even the distribution of the process $X$), it is not enough to know only $\si^2$; one also has to know the covariances $r_{x,y}:=Cov(X(x),X(y))$ for $x,y$ in $\Om$.
Then, using e.g. the joint moment generating function of $(X(x),X(y))$, one can see that $$Cov(X(x)^2,X(y)^2)=R(x,y):=2r_{x,y}^2.$$
So, by the multivariate central limit theorem, for each $(x_1,\dots,x_k)\in\Om^k$, the corresponding finite-dimensional distribution of the process $Z_N$ (that is, the joint distribution of $(Z_N(x_1),\dots,Z_N(x_k))$) will converge to the corresponding finite-dimensional distribution of a centered Gaussian process $Z$ (that is, the joint distribution of $(Z(x_1),\dots,Z(x_k))$), which will be the $k$-variate normal distribution with zero mean and covariance matrix $(R(x_i,x_j)/(2\si^2))_{i,j=1}^k$.
To address such questions as into what functional vector space $V$ can the realizations of the limit process $Z$ and even of the original process $X$ be placed and whether one has the convergence of the full distributions of the processes $Z_N$ to the full distribution of the process $Z$ in $V$ require specific information on the covariances $r_{x,y}$. See e.g. Talagrand's book.
Thanks for the answer and the reference to Talagrand's book.
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2025-03-21T14:48:32.082680
| 2020-09-21T11:13:42 |
372212
|
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|
Stack Exchange
|
4-dimensional cohomology $\mathbb{CP}^2$'s
Let $M$ be a closed, smooth $4$-manifold with integral cohomology ring isomorphic to that of $\mathbb{CP}^2$, is it diffeomorphic to it?
Do you have any reason to believe that this would be true?
If the manifold has a Morse function with three critical points, then by Eells-Kuiper it is homotopy equivalent to $CP^2$. In particular, it is simply connected and then by Freedman homeomorphic to $CP^2$. The question of exotic smooth structures on $CP^2$ is open, albeit a conjecture of Fintushel and Stern says there are infinitely many.
No. If $\Sigma$ is any homology 4-sphere with non-trivial fundamental group, $\mathbb{CP}^2 \# \Sigma$ is a homology $\mathbb{CP}^2$ with non-trivial fundamental group. (Here $\#$ denotes connected sum.) There are many examples of homology 4-spheres: for instance, Kervaire produced many examples in his paper Smooth homology spheres and their fundamental groups.**
Whether there is any simply connected 4-manifold with the same homology as $\mathbb{CP}^2$ but that is not diffeomorphic to it, is an important open question, which is somewhat related to the 4-dimensional smooth Poincaré conjecture.
** The main statement is about homology $n$-spheres for $n$ strictly larger than 4, but the remark at the end of the introduction mentions that the constructive part of the proof works for $n = 4$ as well.
Sorry, i do not understand your example since I required the space to be a 4-dimensional manifold.
Maybe you mean connect sum?
Indeed, I meant connected sum! That was quite a typo...
@MarcoGolla Do you know if in addition we say X is irreducible and co-homology ring isomorphc to CP^2 then is it homeomorphic to CP^2?
I don't know off hand. I think it's a good question, but I also think it's hard to check irreducibility. (Note that irreducibility must be defined with more care than in the 3-d case, since there might be non-trivial homotopy 4-spheres that are invertible.)
@MarcoGolla The reason I am curious because Donaldson proves that for even intersection form with no 2-torsion in $H_1$, $b_2^+=1$ implies $S^2\times S^2$. So similar thing should be true for CP^2. And here when I am saying irreducible, that means topologically there is no non-trivial 3-sphere. If that makes sense?
Sorry, I automatically read "diffeomorphic" instead of "homeomorphic" in your previous comment. In any case, I still don't know the answer to your question, but I'm inclined to believe that the answer is "no" (via some surgery construction).
Is it known if a simply connected homology-$CP^2$ is homeomorphic to $CP^2$? That avoids the smooth Poincaré difficulties.
@KevinCasto: almost. Freedman's theorem implies that if X is a simply-connected smooth 4-manifold with the same cohomology as $\mathbb{CP}^2$, then it is homeomorphic to it. (However, there is a simply-connected, non-smoothable topological 4-manifold with the same cohomology as $\mathbb{CP}^2$.)
@MarcoGolla I have found an example, see below.
This was originally going to be a comment to Marco Gallo's answer, but got too long. I figured since this provides an alternative solution to the problem, I'll post it as an answer.
An alternative answer to this question are the fake projective planes, these are algebraic surfaces which have the same cohomology as $\mathbb{CP}^2$, but nontrivial fundamental groups. They have universal cover the complex unit ball which topologically is just $\mathbb{R}^4$.
I offer this additional answer since they provide examples very different to Marco Galla's answer and answer a question of Anubhav Mukherjee's in the comment. Either they are irreducible or have an irreducible summand that is a cohomology $\mathbb{CP}^2$.
Suppose $X$ is one of the fake projective planes. If $X$ is irreducible, we're done. If $X$ is reducible, then it admits a connect sum decomposition $X = Y \# \Sigma$ where $Y$ is a cohomology $\mathbb{CP}^2$ and $\Sigma$ is a homology $S^4$ that is not a homotopy $S^4$. Again if $Y$ is irreducible and not homeomorphic to $\mathbb{CP}^2$, we're done. If $Y$ is reducible, repeat this process again until you get either an irreducible cohomology $\mathbb{CP}^2$. This process must terminate by compactness.
This gives a connect sum decomposition which we'll write as $X = Y \# \Sigma$ (slightly abusing notation here) where $Y$ is an irreducible cohomology $\mathbb{CP}^2$ and $\Sigma$ is a homology $S^4$ which is the connect sum of all of the homology $S^4$'s we found. If $Y$ is not $\mathbb{CP}^2$, we're done.
Now we see that $Y$ can not be $\mathbb{CP}^2$. Suppose $Y = \mathbb{CP}^2$ and consider the universal cover $f: \tilde{\Sigma} \rightarrow \Sigma$. Take a point $p \in \Sigma$ and it's inverse image $f^{-1}(p)$. We can now construct a cover $f^*: Y \# |f^{-1}(p)|\mathbb{CP}^2 \rightarrow \Sigma \# \mathbb{CP}^2 = X$ by connect summing a $\mathbb{CP}^2$ at each point in $f^{-1}(p)$. This covering space is simply connected since it's summands are simply connected and so it is the universal cover. This universal covering space has 2nd homology coming from the $\mathbb{CP}^2$ summands. This contradicts $X$ having universal cover $\mathbb{R}^4$. Thus $Y$ is not $\mathbb{CP}^2$ and is an irreducible cohomology $\mathbb{CP}^2$.
Can you please tell me the description of such algebraic surfaces which are fake projective spaces? Or any references?
Kai, the fake projective planes only have the rational cohomology of $CP^2$, they have torsion in $H_1$. See Theorem 9.1 https://arxiv.org/abs/math/0512115v2
Fake projective planes are irreducible. More generally, aspherical manifolds of dimension at least three are always irreducible.
|
2025-03-21T14:48:32.083046
| 2020-09-21T11:49:42 |
372215
|
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|
Stack Exchange
|
non constant regular function derivative is zero
Let $R$ be a Noetherian regular $k$-algebra (where $k$ any field of char = 0) of dimension greater than 0. Is it true that $H^{0}_{dR}(R \lvert k) = k$?.
More generally we could ask, Is it true that $H^{0}_{dR}(R \lvert k)$ is finite $k$-algebra?
In other words I am asking suppose a regular function $P \in R$ such that its derivative is zero i.e. $d(P) = 0$ in $\Omega^{1}_{R}$, then does $P$ necessarily belongs to $k$? where, $d : R \rightarrow \Omega^{1}_{R}$ is the Kahler derivation.
I know this holds for polynomial algebra and I thought it should be consistent with the fact in manifold that a smooth function which derivative is zero must be constant.
Probably you want to assume $Spec(R)$ is connected?
Sure we can assume that as this is also needed in manifold case as well.
Take $k=\mathbb{Q}$ and $R$ be any finite extension .
Ok well you are right. But in some sense these are kind of trivial counter example which I don't want. May be I should be precise about my question and should say dimension of R is greater than 0.
OK, building on Mohan's comment, take $R=\mathbb{Q}(i)[x]$. A bit more precision perhaps?
I think the correct statement should be that $H^0_{\rm dR}(R/k)$ is a finite $k$-algebra (and hence a product of finite field extensions).
Yes I agree with @Achinger. May be I should edit the question.
Answer to your edit: no. Take $k=\mathbb{Q}$, $R=\overline{\mathbb{Q}}[x]$.
It seems like only possible counter example to the above claim are of the same flavour i. e. Take an algebaric extension of rational so that it's module of differential is zero and attach finite many variable. Is it true these are the only possible way the above claim could fail.
|
2025-03-21T14:48:32.083195
| 2020-09-21T12:06:19 |
372216
|
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|
Stack Exchange
|
Divergence as infinitesimal volume change on a Finsler manifold
Let $M$ be a smooth manifold and $Z$ a smooth vector field on it.
It generates a family of diffeomorphisms $\phi_t:M\to M$ by demanding that $\phi_0=\operatorname{id}$ and $\partial_t\phi_t(x)=Z(\phi_t(x))$.
(Local properties are enough for me, but you can make things work globally if $M$ is, say, closed.)
I want to understand how these diffeomorphisms change a volume measure or form for small $t$.
The Riemannian case was simple enough for me.
If $dVol_g$ is the volume measure of a Riemannian metric $g$, there are smooth positive functions $f_t(x)$ so that $dVol_{\phi^*_tg}(x)=f_t(x)dVol_g(x)$.
A calculation shows that
$$
\partial_tf_t(x)|_{t=0}
=
\operatorname{div}(Z)(x),
\tag{1}
$$
so the divergence of $Z$ describes the infinitesimal volume change of displacement by the vector field $Z$.
What happens when I have a volume measure arising from a Finsler metric $F$ on $M$?
There are different volume forms; I have heard of Busemann–Hausdorff and Holmes–Thompson, but there may be others.
What kinds of divergences do the different volume forms correspond to in this sense?
Or is there a volume form corresponding to a natural Finsler divergence so that (1) holds?
In short:
Is there a Finsler setting (measure and divergence) where (1) works?
Once a smooth volume is given (in the form of a n-form $\mu$) the divergence is well defined through the Lie derivative $L_Z\mu = div(Z) \mu$, and it describes the infinitesimal change of volume through the flow of Z as in the Riemannian case. No matter where $\mu$ comes from, it is only a differential concept.
@Raziel That sounds promising! Every volume form gives rise to a concept of divergence. But then the question is: Are both $\mu$ and $\operatorname{div}(Z)$ some canonical Finsler objects? (In the Riemannian case the canonical choice for both objects satisfies that condition.) Or if one starts with any of the Finsler volume forms, what is the resulting concept of divergence?
As I said before, once a volume is fixed, the divergence is uninquely defined by the formula $L_Z \mu = div_\mu(Z) \mu$, provided that $\mu$ is regular enough (C^1 is enough). If \mu is a "canonical Finsler objetct", then also $div_mu$ will be.
|
2025-03-21T14:48:32.083368
| 2020-09-21T12:32:12 |
372219
|
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|
Stack Exchange
|
About the definition of lineal convexity
I have been trying to understand the definition of lineal convexity. I am reading the article Duality of functions defined in lineally convex sets by Christer O. Kiselman. For a set $A\subset \mathbb{C}^n\setminus \{0\}$ they have defined the dual complement and lineal convexity as follows.
I have understood the definitions and also understood that $A$ is lineally convex iff $A=A^{**}$. But I haven’t used $A$ being a homogeneous set anywhere. Can anyone tell why in the definition is $A$ being homogeneous mentioned?
Also another reference to lineal convexity can be found in the book Notions of convexity by Lars Hormander, page 290, is defined as
A non-homogeneous set cannot be linearly convex.
@NajibIdrissi But the symmetrized bidisc is non homogeneous, but it is linearly convex(also called hypoconvexity). See Theorem 2.3 here https://link.springer.com/content/pdf/10.1007/BF02922097.pdf
Maybe I am missing something, but suppose $A$ is not homogeneous and let $z \in A$ such that $tz \not \in A$ for some $t \neq 0$. Then how can there be a hyperplane passing through $tz$ not intersecting $A$? Since hyperplanes are homogeneous themselves, any hyperplane containing $tz$ must contain $z = \frac{1}{t} \cdot tz$, right?
@NajibIdrissi Yes. You are correct. Well does it make a difference if we consider an affine Hyperplane instead of an hyperplane. Because that seems to be the difference between the two definitions. And they have used the affine Hyperplane definition in the Symmetrized bidisc paper.
Well, one book says "lineally convex" while the other says "linearly convex".
|
2025-03-21T14:48:32.083516
| 2020-09-21T12:41:13 |
372222
|
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|
Stack Exchange
|
Examples of "natural" finitely generated groups with an undecidable conjugacy problem
I am looking for natural groups with undecidable conjugacy problem. By natural, I mean that the word problem should be decidable, and the group should be given by some natural action. I know that $\mathbb{Z}^d \rtimes F_m$ (with a suitable action of $F_m$) has undecidable conjugacy problem. That's very nice, but I'd like to know other examples. I do not care about finite presentation, and I'm also fine with the group being a f.g. subgroup of something natural and geometric, which maybe simplifies things. A concrete case I was not able to resolve is whether all f.g. subgroups of right-angled Artin groups have decidable conjugacy problem.
Šunić, Zoran; Ventura, Enric, The conjugacy problem in automaton groups is not solvable., J. Algebra 364, 148-154 (2012). ZBL1261.20034.
It's trivial that all f.g. subgroups of RAA groups have decidable word problem. If you meant conjugacy, it's known that some f.g. subgroup of $F_2\times F_2$ has unsolvable conjugacy problem.
It was indeed a typo. Your latter sentence then solves my question completely. Could you write it out? Apologies if this was too easy.
You can find a reference to the $F_2 \times F_2$ result in C.F.'s Miller's book/thesis from 1971. Miller also showed that if $S_1$ and $S_2$ are recursively enumerable subsets of $\mathbb{N}$, then $S_1$ is Turing reducible to $S_2$ if and only if there exists a f.g. recusrively presented group whose word problem has the Turing degree $S_1$ and whose conjugacy problem has the Turing degree of $S_2$. You might be interested in this (and the construction is rather straightforward modulo the naturality of a group with undecidable conjugacy problem).
Thanks, I was not aware of the reference! The latter result I've heard of, can you do that in a RAAG as well, or is there some HNN magic or the like?
References to Miller are [19] and [20] here: https://arxiv.org/abs/0708.4331
My own interpretation of the fact that there are f.g. subgroups of $F_2\times F_2$ with undecidable conjugacy problem, is not that they are "natural" instances of f.g. groups with undecidable conjugacy problem, but rather that they are "non-natural" f.g. subgroups of $F_2\times F_2$... of course "natural" is subjective!
@YCor: We agree, it's just that this question is a compromise between a good question of general interest and me actually just needing a very specific thing for boring reasons.
If $N$ is a finitely generated torsion-free normal subgroup of a hyperbolic group $H$ such that $H/N$ has undecidable word problem, then $N$ has undecidable conjugacy problem. Therefore, Rips' construction gives many examples of f.g. groups with decidable word problem but undecidable conjugacy problem. (However, the groups are not f.p., and their "natural"-ness is debatable!). Reference is: Theorem 1.2 of A. Martino, and A. Minasyan. "Conjugacy in normal subgroups of hyperbolic groups." Forum Mathematicum. Vol. 24. No. 5. De Gruyter, 2012 (doi).
(This is probably related to the $F_2\times F_2$ example, although Martino and Minasyan do use properties of hyperbolic groups in their proof. However, their proof is pleasantly short, and the properties used are pretty basic (e.g. an element has finite index in its centraliser, etc.) so it may generalise.)
Chuck Miller in [Miller, Charles F., III On group-theoretic decision problems and their classification. Annals of Mathematics Studies, No. 68. Princeton University Press, Princeton, N.J.; University of Tokyo Press, Tokyo, 1971] proves the following two rather nice and natural examples.
Theorem III.10. The free product of two free groups with finitely generated amalgamation can have unsolvable conjugacy problem. Further, the finitely presented HNN extension of a free group can have unsolvable conjugacy problem.
(Note that Miller calls HNN extensions 'Strong Britton extensions').
Now by Bass-Serre theory, there is a natural action of an amalgamated free product/HNN on the associated Bass-Serre tree, which should satisfy your "natural action" criterion.
Edit: The result mentioned by YCor can also be found in Miller's book.
Theorem III.23 The group $F_2 \times F_2$ has a finitely generated subgroup with undecidable conjugacy problem.
An important side remark, however, is that $F_2 \times F_2$ itself has decidable conjugacy problem, as do all RAAGs, in linear time. See [Crisp, John; Godelle, Eddy; Wiest, Bert; The conjugacy problem in subgroups of right-angled Artin groups. J. Topol. 2 (2009), no. 3, 442–460.].
This is nice, but not quite as nice as I understood from the comments, i.e. literal subgroup of RAAG.
(I guess this is, nicer in a different direction.)
This is a different example. There are also f.g. subgroups of $F_2 \times F_2$ which have undecidable conjugacy problem; I wanted to give YCor some time to write that answer up.
Gotcha! I did realize this has nicer generators (presumably).
@Carl-FredrikNybergBrodda no, feel free to expand your own post, I don't need to answer just to provide a reference. Actually I don't know these examples (unlike the membership issue in f.g. subgroups of $F_2^2$ which I understand).
@VilleSalo Miller gives explicit generators and relations (modulo the generators and relations for a f.p. group with undecidable word problem). It should not be too horrible to work through to get an explicit presentation (and also Bass-Serre tree), though I have not done this.
@YCor I don't understand the examples well; they are quite artificial, and tie together several constructions, but the overall strategy is not too different from the membership problem, afaik.
I think these F2xF2 examples are also related to the free abelian group semidirect product free group examples. I vaguely remember a talk by Ventura and he said something to that effect maybe using that F_2xF_2 ends in GL_4(Z). This may be nonsense. It was long ago
@BenjaminSteinberg: Certainly $F_2 \times F_2$ at least embeds in GL$_6(\mathbb{Z})$. But not the other way around, and I was precisely hoping for something slightly simpler than the semidirect product example (which I learned about from an MO post of yours).
(On a second reading, I guess you were suggesting something more specific with your comment which I didn't understand, and were just stating the embeddability in GL$_4(\mathbb{Z})$ as a fact.)
You can embed F_2 in GL_2 and hence F_2xF_2 into GL_4 by direct sum. They my belief is they embed the undecidable issues of F_2xF_2 in the orbit problem but maybe I am wrong
@BenjaminSteinberg: The undecidability of the conjugacy problem in semidirect products $\mathbb{Z}^d \rtimes F_m$ by Sunic-Ventura uses orbit-undecidability, which comes from a result of https://ddd.uab.cat/pub/prepub/2007/hdl_2072_9160/Pr766.pdf which in turn uses $F_2 \times F_2$ and Mihajlova's trick.
OK. I remembered a reduction to F_2xF_2 but didn't remember the details
|
2025-03-21T14:48:32.084055
| 2020-09-21T13:30:01 |
372226
|
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|
Stack Exchange
|
stability of two-sided sectional curvature bounds in Lorentzian geometry
Suppose that $(M,g)$ is a Lorentzian manifold of signature $(-,+,\ldots,+)$. Given a two plane $\Pi=\textrm{Span}\{X,Y\}$ with $X,Y \in T_pM$, we say that $\Pi$ is non-degenerate if
$$ g(X,X)g(Y,Y)-g(X,Y)^2 \neq 0.$$
Moreover, given a non-degenerate two-plane we say that it is timelike or spacelike if the above quantity is negative or positive respectively. Finally, the sectional curvature $\textrm{Sec}(\Pi)$ for a non-degenerate two plane $\Pi$ as above is defined by
$$\textrm{Sec}(\Pi)=\frac{g(R(X,Y)X,Y)}{ g(X,X)g(Y,Y)-g(X,Y)^2 }.$$
Question. Suppose that given any non-degenerate space-like two plane in $(M,g)$ its sectional curvature is bounded from above by a fixed $K_1<0$ and that given any non-degenerate time-like two plane in $(M,g)$ its sectional curvature is bounded from below by a fixed $K_2>K_1$. Suppose that $\widetilde{g}$ is another Lorentzian metric on $M$ that is obtained by adding a sufficiently small smooth compactly supported tensor $h$ to $g$. Is it true that there exists constants $K_1'$ and $K_2'$ such that similar sectional curvature bounds hold for the perturbed metric $\widetilde g$ with the new constants $K_1'$ and $K_2'$ in place of $K_1$ and $K_2$?
Non-degerate space-like two plane has positive curvature by definition. How can it be bounded above by $K_1<0$?
To add on to Ettore Minguzzi's answer:
If $g(R(X,Y)X,Y) < 0$ for all pairs $X,Y$ as he defined, then since this is an open condition stability is automatic for small smooth compactly supported perturbations.
If $g(R(X,Y)X,Y) = 0$ for some pair, then stability can be violated.
This can be seen from taking $\tilde{g} = e^{2\phi} g$ a small conformal change, where $\phi\in C^\infty_0$ is small. We will choose further that $\phi$ to have a critical point at the base of $X,Y$
Since $\tilde{g}$ is conformal, $X,Y$ are still a pair with the same properties. Our goal is to try to make the corresponding curvature tensor
$ \tilde{g}(\tilde{R}(X,Y)X,Y) > 0$. From well-known formulas since we chose $\nabla\phi = 0$ at the point of interest, we have
$$ \tilde{g}(\tilde{R}(X,Y)X,Y) = (-\tilde{g}\odot \nabla^2\phi)(X,Y,X,Y)$$
where $\odot$ is the Kulkarni-Nomizu product.
The orthogonality properties of $X$ and $Y$ further implies (I'm too lazy to figure out the currect sign)
$$ \tilde{g}(\tilde{R}(X,Y)X,Y) = \pm\tilde{g}(X,X) \nabla^2_{Y,Y}\phi $$
So choosing $\phi$ suitably convex/concave in the direction of $Y$ you can make the quantity positive.
In particular, applying this construction to Minkowski space gives you a counterexample to stability (of Minguzzi's condition).
Returning to your actual question: it turns out that the case $g(R(X,Y)X,Y)=0$ is automatically ruled out for smooth manifolds.
Using Minguzzi's analysis, consider
$$ \rho(t) = g(R(X, Y+tU)X, Y+tU)$$
and
$$ \sigma(t) = g(X,X)g(Y+tU,Y+tU) - g(X, Y+tU)^2$$
Observe that $\sigma(t)$ vanishes at $t = 0$ to only first order.
The requirement that $\rho(t) / \sigma(t) > K_2$ when $t > 0$ and $\rho(t) / \sigma(t) < -K_1$ for $t < 0$ requires that $\rho$ is strictly negative when $t \neq 0$. So if $\rho(0) = 0$ this would make it a critical point and hence vanish to second order.
But this would mean that $\lim_{t\to 0} \rho(t)/\sigma(t) = 0$ which is a contradiction.
Therefore in the end, we find that you must be in the situation where stability holds.
Your condition is somewhat restrictive, and might not be what you want. Let $Y$ be a future directed lightlike vector and let $X$ be a spacelike vector orthogonal to it. Consider $Y'=Y+tU$ where $U$ is a unit future directed timelike vector. Then for $t<0$ with $\vert t\vert $ sufficiently small, $\textrm{Span}(X,Y')$ is spacelike, that is, $g(X,X)g(Y',Y')-g(X,Y')^2$ is positive actually going to zero for $t\to 0$. For $t>0$, $\textrm{Span}(X,Y')$ is timelike and $g(X,X)g(Y',Y')-g(X,Y')^2$ is negative actually going to zero for $t\to 0$. Thus if $g(R(X,Y)X,Y)> 0$ the 'spacelike' sectional curvature for $t\to 0-$ goes to plus infinity while in the 'timelike' sectional curvature for $t\to 0+$ goes to minus infinity, so the two bounds cannot be satisfied. We conclude that $g(R(X,Y)X,Y)\le 0$ for a generic pair as above. Maybe from here one can obtain further constraints on the Riemann tensor.
|
2025-03-21T14:48:32.084324
| 2020-09-21T13:51:45 |
372229
|
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|
Stack Exchange
|
2 K3s and cubic fourfolds containing a plane
Two K3 surfaces show up when talking about cubic fourfolds containing a plane. Let $P\subset X\subset \mathbb{P}^5$ be the plane inside the cubic. Since $P$ is cut out by 3 linear equations then $X$ has equation $L_1Q_1+L_2Q_2+L_3Q_3=0$, with $L_i$ and $Q_i$ respectively of degree 1 and 2. Hence $X$ contains the octic K3 complete intersection of the 3 quadrics.
On the other hand by projecting off $P$ one gets a quadric fibration $\pi: Bl_P(X) \to \mathbb{P}^2$, and the relative Hilbert scheme of lines in the fibers is a Brauer Severi variety over a degree 2 K3.
Is there any known relation between these two K3s? Are they FM partners or have similar cohomology?
Is the octic K3 uniquely determined up to isomorphism? I mean we can choose the L_i differently, which changes the Q_i, which conceivably -- to me -- changes the K3.
@RP: good point, I nedd to think about it but probably you are right. But still the question makes (some) sense - I guess.
Indeed what is well-defined is a net of 2-dimensional quadrics in $X$ (Weil divisors). This does not define a K3 in $\Bbb{P}^5$.
FM partners of a given K3 form a discrete set in the moduli space, so a general member of a non-trivial family of K3 (like intersections of three quadrics from your question) cannot be a partner of the K3 associated with the quadric fibration.
Still that degree 8 (and sectional genus 5) K3 must have something to do with the cubic fourfold. In the $\mathcal{C}_d$ cases where there is an associated K3, the surface has degree $d$ and genus $2d-2$. This is why I suspect that when such a cubic 4fold has an associated K3 (for example if it is in $\mathcal{C}8\cap \mathcal{C}{14}$ or 26), then this K3 may be related to the octic.
@IMeasy: Do you mean degree $d$ and genus $d/2 + 1$?
@Sasha: sure it was a typo! $(d+2)/2$, thanks. But still the question stays open.
|
2025-03-21T14:48:32.084492
| 2020-09-21T14:40:53 |
372235
|
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|
Stack Exchange
|
Kan replacement of finite $\mathbb{Q}$-type simplicial set
Sorry if this question is maybe a bit basic, but it is on a rather specialized topic so I think it is more appropriate for MO than SE.
Suppose that $X$ is a simplicial set that has finitely many
non-degenerate simplices in every simplicial dimension. Then of course
$X$ has finite $\mathbb{Q}$-type (i.e. $H_n(X;\mathbb{Q})$ is finite
dimensional as $\mathbb{Q}$-vector space for all $n\ge1$).
Let $Y$ be a fibrant resolution of $X$, that is, $Y$ is a Kan complex
and we have a weak equivalence $X\stackrel{\sim}{\to}Y$. I would like
to say that $Y$ also has finite $\mathbb{Q}$-type. Is this true?
This is where I get stuck: we know that being of finite $\mathbb{Q}$-type is equivalent to $\pi_n(Y;\mathbb{Q})$ is finite dimensional for all $n\ge2$ and $H_1(Y;\mathbb{Q})$ is finite dimensional. The first part of the condition holds because it holds for $X$ and $Y$ is weakly equivalent to $X$. However, I am unsure how to show that $H_1(Y;\mathbb{Q})$ is finite dimensional.
Any help and/or references would be greatly appreciated! I am also iterested in the case where we only assume $X$ to be of finite type (which should work with the same proof, hopefully).
Isn't it true that a weak equivalence is always a homology equivalence, and therefore the map $X\to Y$ induces an isomorphism in homology, with any coefficients?
Weak equivalences of simplicial sets induce isomorphisms on simplicial homology groups.
In particular, the property of having finite-dimensional rational
homology groups is preserved under weak equivalences.
The easiest way to prove this from scratch
is the observe that the free simplicial module functor
(in this case, we use modules over rational numbers)
is a left Quillen functor from the Kan–Quillen model structure
on simplicial sets to the projective model structure
on simplicial modules
because the right adjoint preserves fibrations and acyclic fibrations.
In particular, the free simplicial module functor preserves weak equivalences
because all simplicial sets are cofibrant.
Thus, weak equivalences of simplicial sets
are sent to weak equivalences of simplicial modules,
which after passing to normalized chains become
quasi-isomorphisms of chain complexes,
hence induce isomorphisms on homology groups.
Great, thank you so much!
|
2025-03-21T14:48:32.084649
| 2020-09-21T14:49:37 |
372236
|
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|
Stack Exchange
|
When is the cotangent complex perfect?
Let $X\rightarrow S$ be a proper flat morphism of schemes.
When is the cotangent complex $L_{X/S}$ perfect ?
It is well known, that for local complete intersections the cotangent complex is perfect, but are there examples where perfectness still holds without the lci assumption ?
And if there are examples, is there some general theorem which deals with perfectness in the non-lci case ?
https://mathoverflow.net/questions/114282/cotangent-complex-and-lci-morphisms
I would guess you just need locally of finite presentation
Avramov's theorem already answers the question: for flat maps, we have lci <=> perfect cotangent complex (over a noetherian base). Relatedly, being locally finite presentation is way too weak (take any non-lci variety).
|
2025-03-21T14:48:32.084741
| 2020-09-21T15:27:36 |
372240
|
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|
Stack Exchange
|
Can all (inverse) trigonometric functions with periodic iterates be characterized?
I wonder whether all (composites of) trigonometric and inverse trigonometric functions with periodic functional iterations can be found. In order to specify what I mean by that, let's introduce some notation first.
Let $f^{[n]} (x)$ denote the function $f$ that has been iterated $n$ times with itself. So initially, we have $f(x) := f^{[1]}(x)$. Then we have $f^{[2]} (x)= (f \circ f)(x)$, $f^{[3]}(x) = (f \circ f \circ f)(x)$, etc. We call a function periodic if there is some positive integer $k$ such that $f^{[1+k]}(x) = f^{[1]}(x)$. If that is the case, the period of of the function $f(\cdot)$ is equal to $k$.
Now we turn to some examples of periodic (inverse) trigonometric functions.
Examples
Consider the function $f(x) := f^{[1]} (x)= \sin(\arccos(x)) = \sqrt{1-x^{2}}$. Then we obtain $f^{[2]} (x)= x$, and $f^{[3]} (x)= \sqrt{1-x^{2}} = f(x)$. Thus, the function $f(\cdot)$ is periodic, and has a period of $3 - 1 = 2$. The same applies to the function $v(x) = \cos(\arcsin(x))$.
Define the function $g(x) := g^{[1]}(x) = \tan( \cot^{-1}(x)) = \frac{1}{x}$. Then, $g^{[3]}(x) = g(x)$, so it's also periodic and has a period of $2$, too.
Let $h(x) := h^{[1]}(x) = \sin(\arctan(\cos(\arcsin(x)))) = \sqrt{1-x^{2}}$. Like the first example, it's periodic and has a period of two.
Counterexamples
Let $p(x) := p^{[1]}(x) = \cot(\arccos(x)) = \frac{x}{\sqrt{1-x^{2}}}$. Then, $g^{[n]}(x) = \frac{x}{\sqrt{1-nx^{2}}}$ for $n \geq 1$, so it is not periodic.
Consider $m(x) := m^{[1]}(x) = \cot(\arcsin(x)) = \frac{\sqrt{1-x^{2}}}{x}$. Then, $m^{[n]}(x) = \sqrt{ \frac{F_{n} - F_{n+1}x^{2}}{F_{n}x^{2} - F_{n-1}} \cdot (-1)^{n+1} } $ for $n >1$, where $F_{n}$ is the $n$'th Fibonacci number. Thus, it is not periodic.
Define $l(x) := l^{[1]}(x) = \cos(\arctan(x)) = \frac{1}{\sqrt{1+x^{2}}} $. In this case, $l^{[n]} (x) = \sqrt{\frac{F_{n} + F_{n-1}x^{2}}{F_{n+1} + F_{n}x^{2}}}$ for $n>1$. Again, not a periodic function.
This is what I've found so far. The list is not exhaustive.
Questions
Can all periodic composites of trigonometric and inverse trigonometric functions be found?
What are their periods? Are there any examples of functions with periods greater than two?
Are there any periodic trigonometric functions that do not involve inverse trigonometric functions?
Note: let's discard trivial examples like $q(x) = \cos(\arccos(x)) = x$, which have a period of zero.
Is there a reason to require periodic functions to return to the first iterate, rather than some other iterate? I don't know if it makes a difference. (For that matter, I'd prefer the base case of the iteration to be $f^{[0]} = \operatorname{id}$, but that's just a matter of taste.)
Also, what branches do you choose for the inverse functions? How do you ensure that your functions are iterable—e.g., do you allow a function that is periodic only on the subset of its domain where the iterate is defined, or do you insist that your functions be globally defined?
@LSpice w.r.t. your first comment: no, I did not have a specific reason in mind. It just seemed like the most natural thing to require. As for your second comment: the function does not have to be globally defined, it can be iterable only on the subset of the domain where its iterate is defined.
|
2025-03-21T14:48:32.084953
| 2020-09-21T16:11:11 |
372244
|
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|
Stack Exchange
|
An identity for Ext for rings
Let $A$ be a two-sided noetherian ring (which we should assume to be Gorenstein first so that everything is well defined, otherwise it is only well defined up to a conjecture, which states that every non-zero module has finite grade). For simplicity we can also assume first that $A$ is a finite dimensional algebra and modules are finitely-generated (but non-finitely generated examples or non-Gorenstein examples are also welcome but Im mainly interested in finite dimensional algebras).
For a module $M$, define the grade of $M$ as: $g_M:= \inf \{ i \geq 0 | Ext_A^i(M,A) \neq 0 \}$.
Define the Ext-dual of $M$ to be $U(M):=Ext_A^{g_M}(M,A)$ and the double Ext-dual of $M$ as
$G(M):=U(U(M))=Ext_{A^{op}}^{g_{Ext_A^{g_M}(M,A)}}(Ext_A^{g_M}(M,A),A)$.
Note that $G(M)$ is always non-zero.
Question: Do we have always that $G^l(M) \cong G^{l-1}(M)$ (at least in the stable category of $A$) for some $l \geq 1$ and indecomposable modules $M$, so that the sequence of the $G^l(M)$ becomse stationary?
This is true for $A$ selfinjective or hereditary (in those cases we have $G(M) \cong M$ for all $M$). In all examples this was even true for $l \leq 2$ so I wonder whether we have $G^2(M) \cong G(M)$ (at least in the stable category).
It seems that any module $M$ whose double $A$-dual $M^{**}$ is not reflexive gives a counterexample. In this case $G(M) = M^{**}$ is a summand of $G^2(M) = (M^{**})^{**}$ with non-trivial complement, and $G^2(M)$ can't be reflexive since this property is inherited by summands. Repeating this argument with $G(M)$ instead of $M$, we see that $G^l(M)$ is a summand of $G^{l+1}(M)$ with non-trivial complement for each $l \geq 0$.
Here's an example over a non-Gorenstein ring:
Take $S$ to be the unique simple module over $A = k[x,y]/(x,y)^2$. Then $g_S = 0$ and its $A$-dual is $S^* \cong S^{\oplus 2}$ with $g_{S^*} = 0$, so that $G(S) = S^{**} \cong S^{\oplus 4}$ and in general $G^{l}(S) = S^{\oplus 4l}$.
Thanks, do you also know an example where $add(G^l(M))$ does not become stationary?
I don't. That sort of stability would be very interesting if it did occur.
I will do some more experiments, especially with local algebras to see whether this might be true.
If you're intent on Gorenstein rings, then I'd suggest taking a look at Auslander-Gorenstein rings (in particular any commutative Gorenstein ring). Over such rings the grade of $Ext^n(M, A)$ is always $n$, and there are various sequences relating $M$ to its double Ext-dual, for example in the work of Levasseur (http://www.math.univ-brest.fr/perso/thierry.levasseur/files/regular_rings.pdf). My guess is modules over these rings might satisfy the stability $G(M) = G^2(M) = G^3(M) = \cdots$ that you wanted.
(Small correction: The grade of $Ext^n(M, A)$ is always at least $n$, and if $n = g_M$ then the grade of $Ext^{g_M}(M, A)$ is also $g_M$.)
|
2025-03-21T14:48:32.085158
| 2020-09-21T17:30:57 |
372248
|
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"Dieter Kadelka",
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|
Stack Exchange
|
Intersection of a Poisson bridge and a Brownian bridge
Take a Poisson process $N_t$, a Brownian motion $W_t$ and constants $T > 0$ and $a > 0$. Suppose $N$ and $W$ are independent. I'm interested in the probability that $W$ does not cross over $a + N_t$ during the interval $[0,T]$. To this end, I want to first calculate the above probability conditioned on $N_T = n$ and $W_T = x$, $(x < n)$.
Thus I want to calculate
$$
P\bigl(W_t < a + N_t , \forall t \leq T \; \bigl| \; W_T = x, N_T = n \bigl).
$$
Given that the conditioned on $N_T$, the jump times of $N$ are uniformly distributed over the interval $[0,T]$, and that the average path of the Poisson bridge is simply the straight line joining the points $(0,a)$ and $(T,n)$, my intuition is that this conditional probability will end up being the same as the probability that the Brownian bridge between $(0,0)$ and $(T,x)$ remains below the straight line joining $(0,a)$ and $(T,n)$.
Can someone suggest a way to approach this problem?
Is it correct that you assume that $N$ and $W$ are independent processes? Otherwise nothing can be concluded.
@DieterKadelka yes, I forgot to mention that. I'll put that in now. Thank you.
I found a result by Perry, Stadje and Zacks that calculates an integral equation satisfied by the distribution of first meeting time of a Brownian motion and an independent compound Poisson process (not bridges). I can't easily deduce my conjecture from there.
I guess I could adapt their argument to come up with a new integral equation, but I foresee difficulties.
(1/2) What you are after is the probability $q(a,t)$ that the Lévy process $X_t := W_t - N_t$ does not exceed level $a$ up to time $t$. This is the fundamental question studied in fluctuation theory of Lévy processes. Usually, things get pretty complicated here, and unless you are extremely lucky, explicit expressions are not available. A general approach is as follows. The bi-variate Laplace transform of $q(a,t)$ is given by the Pecherskii–Rogozin identity in terms of Wiener–Hopf factors. These are usually given in an integral form only. Nevertheless, it is sometimes possible...
(2/2) ...to invert the Laplace transform and get a reasonable expression for $q(t,a)$. If you are interested, I can provide some entry points to literature, and, time permitting, have a look and see if there is any hope for workable expressions for this particular process.
Hi Mateusz. Yes that probability can be derived easily from the Kendall's identity. In fact, writing $\tau_x$ as first passage time of $X$ over level $x$, $f(t,x)dx = P(X_t \in dx)$ and $g(t,x)dt = P(\tau_x \in dt)$, we have $g(t,x) = x f(t,x)/t$.
Developing this, in our case when the Levy process has a non-zero diffusion component, $f$ is a convolution of normal density $\phi$ and Poisson or some other infinitely divisible distribution $H$: $f(t,x) = \int \phi(t,x-y) H(t,dy)$, so that by Kendall, $g(t,x) = \int x \phi(t,x-y) / t H(t,dy)$. The integrand $x \phi(t, x-y )/t$ is the first passage density of Brownian motion over the STRAIGHT LINE through points $(0,x)$ and $(t, x-y)$. This is where the straight line intuition in my question came from. I was hoping to find an independent proof.
@MateuszKwaśnicki and of course, I forgot to mention, Kendall's identity only holds when level crossing occurs continuously, i.e., without overshoots (with probability 1), which is indeed the case in my setting. More generally, Kendall holds for spectrally negative (positive) processes with non-lattice distributions crossing a level boundary from below (above).
Classical result - I know of at least 4 proofs.
@zab: Ah, I thought you meant a two-sided Poisson process. For the one-sided, of course Kendall's identity applies. I'll think about the "alternative proof" question and will get back if I have anything insightful to add.
@MateuszKwaśnicki Thank you. I took the liberty to send you an email.
|
2025-03-21T14:48:32.085450
| 2020-09-21T18:21:36 |
372251
|
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|
Stack Exchange
|
Dependence of the complexity of solving polynomial sytems on the multidegree
Let $f_1,\ldots,f_n\in \mathbb{Q}[X_1,\ldots,X_n]$ be a system of $n$ polynomials in $n$ indeterminant, which only has finitely many solutions. Supose that the each of the variables $X_i$ appears at most with degree $d_i$ in the polynomials. What can be said about the complexity of finding the solutions in this case?
It will also depend (rather badly, I guess) on the coefficients.
What do you mean by "finding" here? What are your inputs and outputs? Inputs seem like they should be easy - the coefficients of the $f_i$ - and I have one idea for outputs, but it isn't "nice" (specifically, it gives a set of single-variable polynomials $P_i$ such that if $(x_1, \dots)$ is a common zero of $(f_1, \dots)$, then $x_i$ is a solution of $P_i$; this gives "too many" answers because it "solves" each coordinate separately, meaning it doesn't eliminate "crossover" between two solutions).
@FelipeVoloch At least for my solution, the complexity shouldn't depend on coefficients - if we let the coefficients of the $f_i$ be indeterminate, then the coefficients of the $P_i$ can still be determined in terms of the $f$ coefficients. They're polynomials, even. There may be cases where the complexity drops from that, but there is an algorithm that works in the generic case.
|
2025-03-21T14:48:32.085571
| 2020-09-21T18:40:50 |
372253
|
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|
Stack Exchange
|
Dual of $n$-cosheaf is an $n$-sheaf?
For a site $\mathcal{C}$, an $n$-cosheaf is a functor $\mathcal{F}:\mathcal{C}\rightarrow $n$\text{-}\mathfrak{Cat}$ such that for any cover $(U_i\rightarrow U)_i$, we have
$$\prod_{i,j}\mathcal{F}(U_i\times_U U_j)\rightrightarrows \prod_i \mathcal{F}(U_i)\rightarrow \mathcal{F}(U)$$
is an $n$-pushout. Does it then follow that the contravariant functor
$$H_{\mathcal{F}}:\mathcal{C}^{op}\rightarrow n\text{-}\mathfrak{Cat},\quad X\mapsto \text{Hom}_{n\text{-}\mathfrak{Cat}}(\mathcal{F}(X),\mathfrak{C})$$
is a $n$-sheaf, where $\mathfrak{C}$ is a $n$-category? Basically what I'm asking is if the "$n$-Hom" functor sends pushouts to pullbacks.
The case I'm most interested in is that of cosheaves of $\infty$-groupoids and groupoids.
Yes, Homs send colimits to limits always.
I agree with Qiaochu, modulo the fact that your definition of an $n$-cosheaf is wrong. The right definition of a sheaf valued in an $\infty$-category $C$ says that for any cover $U_\bullet \to U$, we have that $\mathcal F(U)$ is canonically equivalent to the $\infty$-categorical limit of the functor $\Delta \to C$, $[n] \mapsto \prod_{i_0,\dots,i_n} \mathcal F(U_{i_0} \times_U \cdots \times_U U_{i_n})$. If $C$ is an $n+1$-category like $Cat_n$, then I believe you can truncate this simplicial object at level $n+1$ and still get things right. Cosheaves should be defined by the analogous colimit.
|
2025-03-21T14:48:32.085694
| 2020-09-21T19:06:27 |
372254
|
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|
Stack Exchange
|
What groups should I test my conjecture on?
I have a conjecture that a certain criterion is enough for two groups to be isomorphic. I tested it on all pairs of groups up to size 12, and it worked like a charm. I know, however, that groups are strange and that it is very likely that my conjecture will break for larger groups. I have made myself a python library that handles making groups and seeing they have certain properties/are isomorphic, but I do not know what pairs of groups to test my conjecture on.
What groups form a "representative sample" of all groups? What paris of groups would it be absolutely crucial for me to test my hypothesis on?
I don't see the need for any answer to be complete, but any ideas for pairs that would be interesting to test on would be appreciated.
EDIT:
The conjecture is that if two groups have the same count of elements of every order, then they must be isomorphic. I proved this for abelian groups, and am now wondering if it is true in general.
if you can't disclose the conjecture it is unlikely a helpful answer can be given.
@CarloBeenakker thank you for your input, I edited the question.
(You seem to write "count" instead of "number" and "degree" instead of "order"?) There are two non-isomorphic groups of order $p^3$ and exponent $p$, for odd $p$.
"What groups form a "representative sample" of all groups?": the answer depends on the question, but typically: groups of small order, $p$-groups of small order (which here give maybe the easiest examples, order 27, although order 16 might too, yet more complicated by hand). If you learn GAP, you can quickly make such tests for groups, say of order $\le 100$.
Although your specific conjecture has been addressed, and the question of exactly which groups to test will vary, I think it's fair to say the answer of "how should I use my algorithm?" is "don't." As @YCor says, GAP knows many small groups, and is tuned to work with them in an efficient way. With no insult meant to your programming abilities, it is likely that GAP will be far faster, so it'll matter less which groups exactly you test, since you can test more of them.
@NickS Maybe I made an error in my calculations since I did them by hand. This was one of the reasons why I wanted to switch to systematic testing on a computer.
@Nick: I'm not sure where you're getting the groups of $3$ elements of order $4$ from (we only get blocks of $p-1$ elements of order $p$ when $p$ is prime). If I'm not mistaken the order profiles of $\mathbb{Z}_4 \times \mathbb{Z}_2, D_8, Q_8$ are ${ 1, 2^3, 4^4 }, { 1, 2^5, 4^2 }, { 1, 2, 4^6 }$.
@QiaochuYuan My mistake, I forgot that the orbit of an element of order 4 contains an element of order 2 :)
Your conjecture is false. Probably an explicit counterexample is easy to write down but here's an existence proof that counterexamples are plentiful: asymptotically it's known (Higman-Sims) that there are $p^{ \frac{2}{27} n^3 + O(n^{8/3})}$ groups of order $p^n$, for $p$ a prime. The elements of such a group have one of $n+1$ possible orders $1, p, \dots p^n$, and so the number of possible different order profiles of such a group is at most the number of compositions of $p^n$ into at most $n+1$ parts, which is ${p^n + n \choose n}$ which only grows at best like $O(p^{n^2})$ (edit: and see the comments for more on this). So asymptotically there are many more groups of order $p^n$ than there are possible order profiles.
Edit: This MO answer contains the explicit counterexample of the Heisenberg group $H_3(\mathbb{F}_3)$ and $C_3^3$ (order $27$, so $p = 3, n = 3$), which are not isomorphic but have the same order profile (all non-identity elements have order $3$). This MO answer to the same question says the smallest counterexamples have order $16$.
Edit #2: It is maybe worth saying that the fast growth of the number of groups of order $p^n$ kicks in already for pretty small values of $n$. For $n = 10$ there are about $49$ million groups of order $2^{10} = 1024$ and these account for over $99\%$ of the groups of order $\le 2000$. See, for example, this blog post. Most of these groups are $2$-step nilpotent so have elements of orders $1, 2, 4$, and so as YCor says in the comments these groups have at most $(2^{10} - 1) + 1 = 1024$ order profiles, so by pigeonhole we know at least some collection of $\approx 49000$ of these groups have the same order profile. Conjecturally "almost all" finite groups are $2$-step nilpotent $2$-groups so this is in some sense "typical."
I guess this huge number comes from groups of exponent 4, which narrows down the number of "profiles". Actually the analogue for odd $p$ is even simpler, since the huge number of groups of order $p^n$ comes from groups of exponent $p$.
@YCor: yes, of course, thank you. The lower bound comes from $2$-step nilpotent groups which only have elements of orders $1, 2, 4$ so in fact we only need to count compositions of $2^n$ into at most $3$ parts, so ${2^n + 2 \choose 2}$.
Rather the number $(2^n)$ of composition $2^n-1$ into two parts, since there's a single element of order 1.
|
2025-03-21T14:48:32.086159
| 2020-09-21T19:15:59 |
372255
|
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|
Stack Exchange
|
Separable non-flat simple ring extension
Let $R \subseteq S$ be two commutative $\mathbb{C}$-algebras such that:
(1) $R$ and $S$ are integral domains.
(2) $Q(R)=Q(S)$, namely, their fields of fractions are equal.
(3) $S=R[w]$, for some $w \in S$.
(4) $S$ is separable over $R$, namely, $S$ is a projective $S \otimes_R S$-module via $f: S \otimes_R S
\to S$ given by: $f(s_1 \otimes_R s_2)=s_1s_2$.
Should such $S$ be flat over $R$? I guess no, so please it would be nice to see a counterexample.
Is there a fifth condition that would guarantee flatness of $R \subseteq S$?
Perhaps adding a fifth condition (5) $R$ is a UFD (or at least integrally closed) would guarantee
flatness of $R \subseteq S$? (I am not sure).
The above is (almost) question 3 of this question.
Also asked here without comments.
Thank you very much!
Take $R$ to be the coordinate ring of the nodal curve $\mathbb{C}[t^2-1, t(t^2-1)]$ and $S$ to be its normalization $\mathbb{C}[t]$. It satisfies (1), ..., (4): The first three are immediate. For (4), note that since $R$ is noetherian, the projectivity of $S$ as an $S \otimes_R S$-module is equivalent to $S$ being unramified over $R$ (Theorem 2.5 in Auslander-Buchsbaum, On Ramification Theory in Noetherian Rings, American Journal of Mathematics, 1959). It is enough to check that maximal ideals are unramified. Since we are in equi-characteristic zero, it suffices to show that for every maximal ideal $\mathfrak{q}$ of $S$, $(\mathfrak{q} \cap R) S_{\mathfrak{q}} = \mathfrak{q} S_{\mathfrak{q}}$. Let $\mathfrak{q} = (t-\alpha)$, $\alpha \in \mathbb{C}$.
Then $(\mathfrak{q} \cap R) S =
(t^2 - \alpha^2, t(t^2-1)- \alpha(\alpha^2-1))$.
If $\alpha \neq 0$, then $t+\alpha \not \in \mathfrak{q}$
so $t-\alpha \in (\mathfrak{q} \cap R)S_{\mathfrak{q}}$.
If $\alpha = 0$, then $t^2-1 \not \in \mathfrak{q}$
so $t \in (\mathfrak{q} \cap R)S_{\mathfrak{q}}$.
Either way, $(\mathfrak{q} \cap R)S_{\mathfrak{q}} =
\mathfrak{q}S_{\mathfrak{q}}$. However, $S$ is not a flat $R$-module.
Further comments:
If $R$ is noetherian and $R \rightarrow S$ is a finite (as it is in the
example), then $S$ is $R$-flat if and only if it is $R$-projective; since
$\mathrm{Spec} R$ is connected, for every prime $R$-ideal $\mathfrak{p}$,
the fibre $\dim_{\kappa(\mathfrak{p})}\kappa(\mathfrak{p}) \otimes_R S$
does not depend on $\mathfrak{p}$, so $\kappa(\mathfrak{p}) \otimes_R S =
\kappa(\mathfrak{p})$, since $Q(R) = Q(S)$. Hence for maximal $R$-ideals
$\mathfrak{p}$, the map $R/\mathfrak{p} \to S/\mathfrak{p}S$ is an
isomorphism; by the Nakayama lemma, $S/R = 0$, i.e., $S=R$.
If $w = \frac{1}{r}$ for some $r \in R$, then $S$ is flat.
If $R$ is integrally closed, but $S$ is not obtained by inverting one
element of $r$, then I don't know an example where $(4)$ holds. For
example, let $R = \mathbb{C}[x,y]$ and $S = \mathbb{C}[x, \frac{y}{x}]$.
This ring map comes from blowing up $\mathbb{C}^2$ at the origin.
Note that $xS$ defines the exceptional divisor in the affine open set of
the blow-up given by $\mathrm{Spec} S$.
If $\mathfrak{q}$ is a maximal $S$-ideal containing $xS$,
then $\mathfrak{q} \cap R = (x,y)R$ and
$(\mathfrak{q} \cap R)S = xS$. Hence $S$ is not unramified over $R$.
Thank you very much! Interesting. Please, do you think it is possible to find a mild condition that will guarantee flatness of $R \subseteq S$ satisfying (1)-(4)?
I have expanded my earlier answer with some comments.
Thank you very much for expanding your answer!
Please, could you take a look at my recent question, which also deals with separability: https://mathoverflow.net/questions/371948/separability-of-mathbbcx-over-its-mathbbc-subalgebras?noredirect=1&lq=1 Any ideas are welcome! Thank you.
In particular, I do not mind to restrict the above quoted question to the following: Assume that $R=\mathbb{C}+(h) \subseteq \mathbb{C}[x]$ is separable.(hence the four conditions of my current question are satisfied). Notice that $R$ is integrally closed iff $R=\mathbb{C}[x]$, so let's assume that $R$ is not integrally closed (this holds for any $h$ of degree $\geq 2$). Then is it possible to find an exact form of such $h$?
|
2025-03-21T14:48:32.086427
| 2020-09-21T19:33:47 |
372257
|
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|
Stack Exchange
|
Real part of a holomorphic section of a vector bundle
Let $F\to M$ be a holomorphic vector bundle over a complex manifold $M$ and let $s:M\to F$ be a no-zero section. Let $E$ be the complexification of $F$, and suppose that $E$ admits a holomorphic structure (the transition maps need not be the ones arising from $F$). Furthermore, let $\nabla$ be the Chern connection of $E$.
We'll say that $s$ is the real part of a holomorphic section if there is a holomorphic section $v$ and a real section $t$ such that, in a trivialization, we can express $$v = s+it$$
My question: is there a coordinate free way to express this condition? That is, a coordinate free manifestation of the Cauchy-Riemann equations?
In some special situations, it is possible to do so in coordinates. Or if $F=TM$, and $J$ is the almost complex structure on $M$, then one just needs to check if the (1,0)-component $s-iJs$ is holomorphic. I'm not sure about a general vector bundle. I should also remark I'm mostly interested in the case where $M$ is a Riemann surface.
|
2025-03-21T14:48:32.086520
| 2020-09-21T19:44:46 |
372258
|
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|
Stack Exchange
|
Touchard / van der Pol's identity for the sum of divisors and an elliptic curve for perfect numbers
In Touchard (1953) it is mentioned that the sum of divisors $\sigma(n)$, satisifies the following recurrence relation ($n>1$):
$$n^2(n-1) = \frac{6}{\sigma(n)} \sum_{k=1}^{n-1}(3n^2-10k^2)\sigma(k)\sigma(n-k)$$
Substituting in this equation a perfect number $n = \sigma(n)/2$ we find that $P=(n:n^2:1)$ is a point of the elliptic curve:
$$E_n: y^2=x^3+x^2 a_n -b_n$$
where
$$a_n = \sum_{k=1}^{n-1} 9 \sigma(k) \sigma(n-k)$$
and
$$b_n = \sum_{k=1}^{n-1} 30k^2 \sigma(k) \sigma(n-k)$$
This elliptic curve can be defined for every $n>1$.
Computations with SAGEMATH suggest, that the torsion subgroup of this elliptic curve is trivial (for all natural numbers $n$), and that the order of the point $P$ is $+\infty$ ( conjecture: $P$ is a point of $E_n$, if and only if $n$ is perfect).
Question: Can this be proven?
Question: Does this curve have any interesting properties?
Thanks for your help!
Your assertions are for any n or just for perfect n's?
@Xarles: The curve can be defined for any $n>1$ and seem to always have trivial torsion group. $P$ seems to be a point of $E_n$ if and only if $n$ is a perfect number. In this case $P$ seems to have infinite order.
Torsion is trivial for most elliptic curves; probably with a bit of hard work one could determine for which $n$, if any, it isn't trivial. The denominator of $2P$ is even it seems, that would show that it is of infinite order. Unless I am mistaken $P$ is on $E_n$ if and only if $n$ is perfect by the initial formula (and it will not have anything to do with $E_n$). The curve $E_n$ has positive rank for many other $n$. So, I am not sure where this is heading.
|
2025-03-21T14:48:32.086919
| 2020-09-21T20:40:38 |
372260
|
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"Igor Belegradek",
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"url": "https://mathoverflow.net/questions/372260"
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|
Stack Exchange
|
Functions which are periodic along every geodesic
In an effort to understand some geometric rigidity theorems, I am curious about the following: let $(M,g)$ be a complete Riemannian manifold and suppose there is a nonconstant real-valued function on $M$ which has a period of 1 when restricted to any unit-speed geodesic of $M$. Does this place restrictions on $(M,g)$?
My feeling is that the vast majority of manifolds cannot support such a function. The round sphere clearly can.
edit: if I understand correctly- according to the article arxiv.org/abs/1511.07852 of Radeschi and Wilking (Invent. Math. 2017) given by Igor Belegradek in the comments, a theorem of Wadsley (J. Diff. Geom. 1975) shows that, for every Riemannian manifold all of whose geodesics are closed, every function on $M$ satisfies the above condition with some constant period; the theorem is that all lengths of closed geodesics must be multiples of a single number, when $\pi_1(M)$ is finite
Does your assumption imply that all geodesics are periodic? If so there is a book by Besse "Manifolds all of whose Geodesics are Closed" which studies such manifolds.
I'm not sure. Supposing it does, I think my condition should be significantly more restrictive, since I assume the lengths of geodesics in Besse's manifolds aren't usually all multiples of the same number
I think the state of the art is in https://arxiv.org/abs/1511.07852, which is "On the Berger conjecture for manifolds all of whose geodesics are closed" by Radeschi and Wilking.
thanks for the article, I've added a comment about it to the question
If $f\circ \gamma$ has period $1$ for all unit speed geodesics $\gamma:\mathbb{R} \to M$ then in particular $f(x) = f(y)$ whenever $x$ and $y$ are the endpoints of a geodesic segment of length $1$.
This implies that $f(x) = f(y)$ whenever there is a sequence $x_0 = x,x_1,\ldots,x_n = y$ such that $x_i$ and $x_{i+1}$ are the endpoints of a segment of length $1$ for $i = 0,\ldots,n-1$. Following Sunada call such a sequence a $1$-geodesic chain.
If any two points can be joined by a $1$-geodesic chain then $f$ would have to be constant. In a couple of papers of Sunada (e.g. Theorem C from "Mean value theorems and ergodicity of certain random walks" Compositio Mathematica 1983) conditions under which any two points can be joined by a $1$-geodesic chain are given.
His results imply that, in order for a non-constant $f$ to exist, there must be a point $o \in M$ such that for any unit speed geodesic with $\gamma(0) = x$ one has that $\gamma(n)$ is conjugate to $x$ along $\gamma$ for all integer $n$.
This implies in particular that $M$ is compact and its fundamental group is finite (since the condition above passes to the universal covering space as well).
For surfaces this shows (Theorem G in the aforementioned paper by Sunada) that $M$ would have to be either the sphere or the projective plane with a metric such that all geodesics from $o$ have period $2$.
Thanks, very interesting!
If $\Gamma$ is a finite subgroup of $SO(n+1)$, does there always exist a non-constant such $f$ on $S^n/\Gamma$? (Maybe this is obvious)
|
2025-03-21T14:48:32.087180
| 2020-09-21T21:30:53 |
372263
|
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|
Stack Exchange
|
Formula expressing symmetric polynomials of eigenvalues as sum of determinants
The trace of a matrix is the sum of the eigenvalues and the determinant is the product of the eigenvalues. The fundamental theorem of symmetric polynomials says that we can write any symmetric polynomial of the roots of a polynomial as a polynomial of its coefficients. We can apply this to the characteristic polynomial of a matrix $A$ to write any symmetric polynomial of eigenvalues as a polynomial in the entries of $A$.
I stumbled upon an explicit formula for this. Let $A$ be an $n \times n$ matrix and $a_1, \dots, a_n$ be its eigenvalues. Then we have the following identity, provided the left hand side is a symmetric polynomial:
$$
\sum_{i \in \mathbb{N}^n} p_i a_1^{i_1} \cdots a_n^{i_n} = \sum_{i \in \mathbb{N}^n} p_i \det(A_1^{i_1}, \dots, A_n^{i_n})
$$
The determinant $\det(A_1^{i_1}, \dots, A_n^{i_n})$ on the right hand side is the determinant of a matrix with those column vectors, where $A_i^k$ is the $i$-th column of the $k$-th power of $A$. The left hand side is a symmetric polynomial of the eigenvalues of $A$, and the right hand side is a polynomial of the entries of $A$.
Example: if $A$ is a $2\times 2$ matrix, then $$a_1 a_2^2 + a_1^2 a_2 = \det(A_1, A_2^2) + \det(A_1^2, A_2)$$
Proof. Let $p(A) \in End(\bigwedge^n V^*)$ be given by $p(A)f(v_1,\dots,v_n) = \sum_{i\in \mathbb{N}^n}f(A^{i_1}v_1,\dots,A^{i_n}v_n)$. We have $End(\bigwedge^n V^*) \simeq \mathbb{R}$ and $p(A)$ is the right hand side of the identity under this isomorphism. Since $p(A)$ was defined basis independently, the right hand side is basis independent, and we get the left hand side in the eigenbasis. $\Box$
Link to detailed proof and slight generalization to an identity on several commuting matrices. E.g. for commuting $2\times 2$ matrices $A,B$:
$$a_1 b_1 a_2^2 + a_1^2 a_2 b_2 = \det(AB_1, A_2^2) + \det(A_1^2, AB_2)$$
This identity looks like it should be a few hundred years old, especially since the proof is quite simple, but I have not seen this in linear algebra courses. Is this a well known identity? Where should I look to learn more about these types of identities? Or, maybe I am mistaken and the identity is false? (though I have also empirically tested it with a computer program) I apologize if this question is too basic for mathoverflow; I am only doing pure mathematics for fun. I initially asked elsewhere but was advised to ask here. Thanks!
That's nice. I do not know the reference. When $p_i=\prod_{k=1}^n H(i_k)$ for certain function $H$ defined on ${0,1,\ldots}$, both parts factorize and are equal to the determinant of $\sum_m H(m)A^m$. The functions of such type generate all symmetric functions $i=(i_1,\ldots,i_n)\mapsto p_i$, so the result follows.
That sounds cool! What is the name of these functions $H$ and where could I find a proof that they generate all symmetric functions of natural number tuples? (which I think means that any symmetric $p$ can be written as a linear combination of those special $p$'s?)
At first, the linear span of such functions is the same as the algebra generated by them. Next, it suffices to consider $p$ with a finite support $A$. Then for $H(m)=m+\alpha$ for $m\in A$ (hereafter: and 0 otherwise) we get a function of the form $p_i=(i_1+\alpha)...(i_n+\alpha)$. Varying $\alpha$ and taking linear combinations we get any elementary symmetric polynomial in $i_1,..., i_n$. They generate an algebra of all symmetric polynomials, and any symmetric function on $A$ is represented by a symmetric polynomial.
Thanks! I think I understand all the steps except one. I could not understand why the linear span is the same as the algebra. The degree of these special $p_i$ seems to be $\leq n$ so I'd expect that the same holds true for their span, whereas in general the $p_i$ can have any degree. Could you help me understand where my reasoning went wrong? By the way, it is easy to prove the reverse implication, that any symmetric polynomial can be written in elementary symmetric polynomials, by taking $A$ to be the companion matrix of a polynomial in the identity :)
because the product of two such functions is again such a function.
Ah, now I understand. I misunderstood your initial comment and thought there was a particular known family of functions $H$ that you were talking about, but you are talking about the $p_i$ that are of that form for any $H$. Sorry for the confusion and thank you for the explanation.
@FedorPetrov : Perhaps you could collect your comments into an answer even though they don't directly answer the "reference request" part of the question.
@TimothyChow done, and I found an issue preparing this answer.
I put a concise version of my proof in the question.
This is not a reference, but a short proof.
We use the following (probably known, but see later) lemma on representing a symmetric tensor as a linear combination of rank-1 symmetric tensors.
Lemma. Let $A$ be a finite set, $K$ an infinite field. Denote by $\mathcal S$ the set of symmetric functions $p:A^n\to K$. Then $\mathcal S$ is the $K$-span of rank-one functions, that is, the functions of the type $h(x_1)h(x_2)\ldots h(x_n)$, where $h:A\to K$.
Proof. Note that the product of two rank-one functions is a rank-one function. Thus the linear space $\mathcal T$, generated by rank-one functions, coincides with the $K$-algebra generated by them.
We may suppose that $A\subset K$. For $k=0,1,\ldots,n$ denote $e_k(x_1,\ldots,x_n)$ the elementary symmetric polynomial, that is, $\varphi_t(x_1,\ldots,x_n):=\prod(1+tx_i)=\sum_{k=0}^n t^ke_k$. We identify $e_k$ and the corresponding element of $\mathcal S$. Choosing $n+1$ distinct values $t_1,\ldots,t_{n+1}\in K$ and solving the corresponding (Vandermonde's) linear system of equations we represent each $e_k$ as a linear combinations of $\varphi_{t_i}\in \mathcal T$. Thus $e_k\in \mathcal S$ for all $k=0,1,\ldots,n$. It is well known that $e_k$'s generate the algebra of symmetric polynomials (over any field). Thus any symmetric polynomial function belongs to $\mathcal T$. It remains to note that any symmetric function $f\in \mathcal S$ may be represented by a symmetric polynomial. Indeed, a symmetric function $f$ may be represented as $F(e_1,e_2,\ldots,e_n)$ for certain function function $F$ defined on the corresponding finite set (because the values of $e_1,\ldots,e_n$ determine the values of $x_1,\ldots,x_n$ up to permutation). $F$ in turn coincides with a polynomial function on this finite set. $\square$
Now we may prove your theorem for finitely supported function $i\mapsto p_i$. Due to Lemma it may be supposed to have the form $p_i=\prod_{k=1}^n H(i_k)$ for a certain finitely supported function $H$ on $\mathbb{N}$ (as OP, I denote here $\mathbb{N}=\{0,1,\ldots\}$). In this case both parts of your identity are equal to $\det (\sum_m H(m)A^m)$.
Comment. Lemma does not hold for finite fields. For example, if $A=K=\{0,1\}$. Then the function $x+y+z$ is not a linear combination of rank-one functions 1, $xyz$, $(x+1)(y+1)(z+1)$: if $x+y+z=a+bxyz+c(x+1)(y+1)(z+1)$, then for $y=0,z=1,x=a$ we get $0=1$. I must make a warning that in the subject-related paper "Symmetric tensors and symmetric tensor rank" by Pierre Comon, Gene Golub, Lek-Heng Lim, Bernard Mourrain (SIAM Journal on Matrix Analysis and Applications, 2008, 30 (3), pp.1254-1279) this statement, after equation (1.1), is stated for any field, although proved for complex numbers, and the proof uses that a non-zero polynomial has non-zero values.
In any case, you may always enlarge the ground field and safely think that it is infinite.
"which in turn may be symmetrized" why? We aren't in characteristic $0$ here.
Thank you @darijgrinberg, I changed the previously incorrect argument. Is it better now?
The proof of the lemma looks good now! Not sure how you derive the theorem from it. (I'm not currently at the best of my abilities, hwoever.)
My main problem is that $p_i$ is not a function, it's a constant. Are you talking about $p$ instead? But how does the decomposition into coefficients survive replacement by a different polynomial? And what is the finite set $A$ in your application of the lemma?
$p_i$ is a function in $i$
Ah, now it makes sense!
Concerning the reference request:
Several text books [1,2] give the theorem and proof for elementary symmetric polynomials $s_k=$ sum of all $k\times k$ principal minors of the $n\times n$ matrix. This also covers the trace ($s_1$) and the determinant ($s_n$).
Matrix
Analysis and Applied Linear Algebra by Carl D. Meyer (Equation 7.1.6 on page 494, screenshot)
Matrix
Analysis by Roger A. Horn and Charles R. Johnson (Theorem
1.2.12 on page 42, screenshot).
Update: I have searched quite extensively for sources that give the generalized formula for complete homogeneous symmetric polynomials, but without success. The derivation could be analogous to the published derivation for the elementary symmetric polynomials, expanding ${\rm Det}\,(A+xI)^{-1}$ instead of ${\rm Det}\,(A+xI)$, but I have not seen it published.
Thank you! So in terms of the formula, we get $s_k$ on the left hand side if we take $p_i = 1$ if the tuple $i$ has $k$ ones and $n-k$ zeroes, and $p_i = 0$ otherwise. Then some of the $A$'s on the right hand side get raised to the 1st power and some to the 0th power, and after Laplace expansion on the columns that have the 0th power, we get a sum over all $k\times k$ principal minors of $A$. I'll try to get access to those books. Is by any chance the proof by looking at the coefficient of $x^k$ in $\det(A + xI)$? I believe I have seen that proof before. P.S. I loved your lectures at Leiden!
I added links to screenshots of each proof, and yes, that is the approach.
Thanks a lot for the screenshots!
Nice update! $\det(I - xA)^{-1} = \det(I + xA + x^2 A^2 + \cdots)$. Then expanding this by multilinearity of the determinant, and taking the coefficient of $x^k$ one gets the right hand side of the general formula, and taking the Jordan form of $A$ and using the generating function of complete homogeneous symmetric polynomials one gets the left hand side. Power sum symmetric polynomials are easy too, of course, via the trace. The proof for the general case is not much more difficult than these special cases though :)
Check out: https://en.wikipedia.org/wiki/MacMahon_Master_theorem
|
2025-03-21T14:48:32.087845
| 2020-09-21T21:53:54 |
372265
|
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}
|
Stack Exchange
|
A simple clarification on Riesz decomposition theorem
Let $D$ be a domain of $\mathbb{R}^{m}$ and let
$K(x)= \log|x|$ if $m=2$, and $K(x)=|x|^{2-m}$ if $m>2$. According to Riesz decomposition theorem (Hayman and Kennedy, "subharmonic functions", vol. 1, pg 104) if $u$ is subharmonic on $D$, then there is a unique Borel measure $\mu$ such that for all compact $E$ in $D$ we have
$$u(x)=\int_{E}K(x-\zeta)d\mu(\zeta)+h(x)$$
where $h$ is harmonic on the interior
of $E$.
Here is my question: is this equation valid for all $x\in D$ or all $x\in E$?
It holds only for $x\in E$. Since the measure is not restricted to $E$, the function does not have to be harmonic outside $E$.
It depends on the convention. You can also understand the claim as $u$ can be represented in the above form in $D$ with $h$ that is harmonic in the interior of $E$ (and subharmonic in the rest of the domain). The key point is that for every compact subset of $D$, the Riesz measure is finite, so the potential is well-defined but this may be not so in the entire domain $D$.
But why do you say that h is subharmonic in the rest of the domain?
(You need a minus sign in front of $\int_E K(x-\zeta)\,d\mu(\zeta)$.)
The integral $\,-\!\!\int_E K(x-\zeta)\,d\mu(\zeta)$ is subharmonic throughout $D$ and harmonic in $D\setminus E$. If $u$ admits a harmonic majorant in $D$, hence a least such majorant (call it $k$), then $h$ (which depends on $E$) can be expressed as $\,-\!\!\int_{E^c} K(x-\zeta)\,d\mu(\zeta)+k(x)$ for $x\in D$.
|
2025-03-21T14:48:32.087989
| 2020-09-21T21:55:40 |
372266
|
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|
Stack Exchange
|
Length of the arc of a Fourier series
I'm working modeling the behavior of periodic variable stars and I have a question about reducing the expression of a parameter involved in this analysis.
Let $f(t)$ be a Fourier series define as:
$$f(t)= \langle m \rangle + \displaystyle{\sum_{k=1}^{N} A_{k}\sin \left(\frac{2\pi kt}{P} + \phi_{k}\right)}.$$
The variable $t$ is real and defined in the so-called phase space, where $0\leq t \leq 1$. These series describe a periodic process with period $P$, $A_i$ and $\phi_i$ the individual Amplitude and phases for all the $N$ harmonics; and $\langle m \rangle$ the average value of the time-series that contains the process. All the shown variables are known. I want to figure out the length of this Fourier series using the definition:
$${\displaystyle s=\int _{a}^{b}{\sqrt {1+\left(f'(x)\right)^{2}}} dx.}$$
If we define $x_k:= \frac{2\pi k}{P}$, the derivative of the Fourier series $f'(t)$ will be:
$$ f'(t) = \displaystyle{ \sum_{k=1}^{N} A_{k} x_{k} \cos(x_{k} t + \phi_{k}).}$$
Then:
\begin{split}
(f'(t))^{2}=\displaystyle{\left(\sum_{k=1}^{N} A_{k} x_{k} \cos(x_{k} t + \phi_{k})\right)^{2}}\\
= \left(\sum_{i=1}^{N} A_{i} x_{i} \cos(x_{i} t + \phi_{i})\right) \times \left(\sum_{j=1}^{N} A_{j} x_{j} \cos(x_{j} t + \phi_{j})\right) \\
= \sum_{i,j=1}^{N} a_i a_j \cos(x_{i} t + \phi_{i}) \cos(x_{j} t + \phi_{j}),
\end{split}
where we also define $a_k := A_k x_k$. Finally our lenght will be:
$${\displaystyle s=\int _{a}^{b}{\sqrt {1+ \sum_{i,j=1}^{N} a_i a_j \cos(x_{i} t + \phi_{i}) \cos(x_{j} t + \phi_{j}) }}\ \ dt.}\ \ (1)$$
The integration limits are $a=0$ and $b=1$, given we live in this convenient phase space. Now comes the big question:
Did you have some idea or trick to attack and solve this integral? I tried with some trigonometric manipulations, but I got not great progress. I'm trying to obtain the less-expensive function to compute and so far, is the expression (1).
Thanks for reading this and every kind of feedback will be awesome :)
by "solve" you mean a closed form expression? that is not likely to be forthcoming, but you can of course evaluate it numerically.
In the case a Fourier series approximating a jump function (i.e. when Gibbs phenomenon kicks in), there is this paper "Gibbs' phenomenon and arclength" (https://link.springer.com/article/10.1007/BF02511544) by Strichartz but the paper does not seem to contain a technique to calculate these integrals…
@CarloBeenakker By solve I mean I want to obtain the more compact expression to compute numerically.
Thanks @Dirk for the paper, I will look it carefully ;)
|
2025-03-21T14:48:32.088302
| 2020-09-21T22:55:59 |
372270
|
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|
Stack Exchange
|
Bimodule Ext for Dynkin path algebras
Let $A=kQ$ be a path algebra of Dynkin type $Q$ and $B=A^{op} \otimes_k A$ the enveloping algebra of $A$. Note that $mod-B$ is just the category of $A$-bimodule and $A$ is a $B$-module.
For a B-module $M$ with $g_M:=grade(M):=\inf \{i \geq 0 | Ext_B^i(M,B) \neq 0\}$ define $W(M):=DExt_B^{g_M}(M,B)$ , where $D=Hom_K(-,K)$ is the duality. $W(M)$ is agian a $B$-module and we can apply $W(-)$ again to get $W^i(M)$ for $i \geq 0$ in this way.
Question: Is it true that there is always exist $l,m$ with $m \geq 1$ such that $W^l(A) \cong W^{l+m}(A)$? Is it even true that $W^1(A) \cong W^{m}(A)$ for some $m \geq 2$?
Note that for $A$ linear oriented of type $A_n$, we have $W(A) \cong DHom_A(D(A),A)$ as $A$-bimodules and in all other types we have $W(A) \cong DExt_A^1(D(A),A)$ as $A$-bimodules.
For example for linear oriented $A_3$ we have $W^1(A) \cong W^5(A)$.
In this example, $W^1(A)$ and $W^2(A)$ have grade 2 and $W^3(A)$ and $W^4(A)$ have grade 0 ($W^4(A)$ is the unique indecomposable projective-injective $B$-module).
There is probably an easy reason in case this is true for all $Q$.
It is not true for non-Dynkin $Q$ for example for the Kronecker quiver it is false (I would think it is false for every non-Dynkin $Q$ and maybe there is a relation to the preprojective algebra).
This obvervation might be true for more general algebras. I have not yet found a counterexample for representation-finite algebras with finite global dimension.
|
2025-03-21T14:48:32.088438
| 2020-09-21T23:07:25 |
372271
|
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|
Stack Exchange
|
What is the big-O time complexity of computing $1/N$ to $\log_{2}(N)$ bits of precision?
I am considering large integer values of $N$ (100 or more digits in base-$10$).
In my algorithm, I need to be able to compute the reciprocal of $N$ with enough precision that the repetend will have been produced exactly. (I estimate this to be to $\lfloor \log N \rfloor$ digits or $\lfloor \log_{2} N \rfloor$ bits)
If I employ ordinary long division, how may I estimate the big-$\mathcal{O}$ time complexity of calculating $\frac{1}{N}$ with the desired level of precision?
I posted question(s) to this effect over on the Mathematics Stack Exchange, but have yet to garner an answer. I know that the complexity of the division of two $n$-digit numbers is $n^{2}$ using ordinary division, but that says nothing about the degree of precision required for non-terminating decimal expansions.
Thank you for your help.
Off the bat, I'm thinking about Newton's root-finding search on $f(x)=Nx-1$, it converges very quickly, though the exact complexity might be difficult. https://en.wikipedia.org/wiki/Root-finding_algorithms
@DUO Yes. I think we can obtain as much precision as we want with Newton's Method. And if I remember correctly, that approach has Quadratic time complexity. And if I could be certain that that were also the case here, I would be happy. I'm just thinking that with the size of $N$ and the required precision, it could be worse than the complexity that Wikipedia, for example, gives for Newton's Method.
Actually, because the function is linear, it is guaranteed consistent performance (no extrema to get stuck in), as long as you start from pre-defined initial value, like $x_0=1$ or $x_0=0$. Also, you should be more concerned about space-- what are you doing with $N$ with "hundreds of digits", as you say? In addition, this is more suited for the Computer Science Stack Exchange.
The number of digits in the period of $1/N$ can be as big as $N-1$.
See https://en.wikipedia.org/wiki/Computational_complexity_of_mathematical_operations . The complexity is $O(M(n))$, where $n=\log N$, and $M$ is the complexity of $n\times n$ bit multiplication. The best currently known algorithm makes $M(n)=O(n\log n)$.
@GerryMyerson Is there a theorem that for any $n$ there are infinitely many $p$ such that $n$ is a primitive root mod $p$?
Also, note that the $O(M(n))$ complexity bound is asymptotically optimal. For division $X/Y$, this follows easily from $XY=2^m/((2^m/X)/Y)$. For $1/X$, this follows from https://cstheory.stackexchange.com/a/25750 : an algorithm for computing $n$-bit approximations of $1/X$ for integers $X$ yields an algorithm for computing $n$-bit approximations of $1/X$ for dyadic rationals $X\in[1,2]$ of roughly the same complexity, which requires time $\Omega(M(n))$ by the linked answer.
@user no, for the simple reason that there are no primes for which $4$ is a primitive root. Even if you throw out such trivial counterexamples, there is no $n$ for which it is proved that there are infinitely many primes for which $n$ is a primitive root. But it is very strongly believed that every $n$ which isn't disqualified for trivial reasons is a primitive root for infinitely many primes, and it is known that there are at most three exceptions. Look it up!
@GerryMyerson I found that the relevant conjecture is Artin's conjecture, but otherwise have no clue where to start looking it up, as this is outside my usual area; any suggestions?
@user44191 Yes, Artin's conjecture is relevant, and there's probably a Wikipedia page on it to get you started. Hooley proved it's implied by the generalized Riemann Hypothesis, and Heath-Brown proved there are at most two (not three, as I mistakenly wrote before) prime exceptions.
Denote by $M_b$ the complexity of multiplying two $b$-digit integers $z = xy$. One easily sees that this is essentially obtained by convolving the $b$-dimensional vectors of digits $x*y$. The school algorithm is a "slow convolution" algorithm that takes $O(b^2)$, but fast convolution algorithms give rise to $M_b = O(b\log b)$ or $M_b = O(b\log^2 b)$ algorithms, see e.g. the Schönhage–Strassen algorithm.
As described by Brent,
whatever the complexity of your multiplication algorithm, the Newton iteration for the reciprocal also takes $M_b$ pairwise operations on digits, provided that the number $b_k$ of digits at the $k$th Newton iteration grows in the right way (i.e. geometrically). In the first few Newton iterations, the accuracy is poor so you use few bits of accuracy. As you converge to $1/x$, you use more and more bits of accuracy.
As pointed out elsewhere, the period of $1/N$ could be as large as $N$ so you're looking at $O(N\log N)$ or $O(N\log^2 N)$ running time and $O(N)$ space.
Edit: in your title, you also ask about computing $1/N$ to $O(\log(N))$ bits of accuracy, which is possibly not enough to reveal the repeating pattern of $1/N$. As per above, computing $1/N$ with $O(\log(N))$ bits of accuracy, can be achieved in just a hair more than $O(\log(N))$ digitwise operations.
|
2025-03-21T14:48:32.088805
| 2020-09-22T00:39:54 |
372275
|
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|
Stack Exchange
|
Supersymmetry charge $Q$ as anti-linear and anti-unitary operator
We know the supersymmetry (SUSY) charge $Q$ satisfies the following relation respect to fermion parity operator $(-1)^F$:
$$
(-1)^F Q + Q (-1)^F :=\{Q, (-1)^F \} =0
$$
which defines the anti-commutator to be zero.
The requirement of SUSY charge $Q$ includes that
$Q$ is a Hermitian operator.
$[Q,H]=0$, $Q$ commutes with the Hamiltonian $H$ operator. $H$ is also Hermitian.
$Q^2$ is bounded from below. (Usually proportional to the Hamiltonian $H$ operator.)
Usually, in the literature, $Q$ is a linear and unitary operator. But can we have $Q$ to be instead antilinear and antiunitary?
My question is about the following, can we introduce a (new) SUSY charge called $Q'$ satisfy the additional less-common properties (other than satisfying the previous common properties mentioned above):
$Q'$ is an antilinear operator.
$Q'$ is an antiunitary operator.
Note that the (Hermitian) adjoint of the $Q'$ is also an antilinear and antiunitary operator. In fact, the (Hermitian) adjoint of the $Q'$ can be made to be the same $Q'$; thus $Q'$ can be regarded as Hermitian, or $Q'=Q'^\dagger$. See for example: https://physics.stackexchange.com/q/45227/12813.
Also, the product of two antilinear and antiunitary operators $Q'^2$ become a linear and unitary operator. Such as the complex conjugation (antilinear and antiunitary) $K$, whose square $K^2=+1$ is an identity (linear and unitary). Thus obeying conditions 4. and 5., do not seem to conflict with conditions 1.2.3. earlier.
Also, are there existing or previous literature introducing SUSY charge $Q'$ to be also antilinear and antiunitary?
Suppose you are given a super Hilbert space $\mathcal{H} = \mathcal{H}_0 \oplus \mathcal{H}_1$, with bosonic and fermionic subspaces $\mathcal{H}_0$ and $\mathcal{H}_1$ respectively. Define a new super Hilbert space $\mathcal{H}' = \mathcal{H}_0 \oplus \overline{\mathcal{H}_1}$, where you have complex-conjugated the fermionic subspace but left the bosonic subspace intact. The space of even unitary operators on $\mathcal{H}$ is $U_0(\mathcal{H}) = U(\mathcal{H}_0) \times U(\mathcal{H}_1) \cong U(\mathcal{H}_0) \times U(\overline{\mathcal{H}_1}) = U_0(\mathcal{H}')$. But an odd operator on $\mathcal{H}$ takes $\mathcal{H}_0$ to $\mathcal{H}_1$. As a result, the odd linear operators on $\mathcal{H}$ are the same as odd antilinear operators on $\mathcal{H}'$, and vice versa.
As a result, your "antilinear susy algebra" has the same representation theory as the usual linear susy algebra. There is no benefit to making the change that you suggest, no new examples, and only an emotional cost that you have to handle nonlinear operators.
I remark that $Q^2 \propto \hat H$ is never unitary in examples: it is an unbounded self-adjoint operator, with spectrum bounded below (as you say). Similarly, $Q$ is never (anti)unitary. Rather, the supersymmetry operator $Q$ should be required to be an (unbounded) odd self-adjoint operator. There are various equally-valid conventions for the meaning of "odd self-adjoint", and they affect the proportionality constant in the expectation that $Q^2 \propto \hat H$. The issue is the following conflict. We expect that the Jordan product, and in particular the square, of self adjoint operators is self-adjoint. But we also expect that the Lie product of self-adjoint operators is skew-adjoint. Well, is $Q^2$ the Jordan product of $Q$ with itself (as it would be in the even case) or the Lie product (since $[Q,Q] = QQ - (-1)^{|Q||Q|} QQ$)? For further details of the different possible conventions, I recommend Section 23 of Greg Moore's notes on Linear Algebra.
Thanks, I had voted up -- can you clarify the opening question remark you made:
"But we also expect that the Lie product of self-adjoint operators is skew-adjoint. Well, is $Q^2$ the Jordan product of $Q$ with itself (as it would be in the even case) or the Lie product (since $[Q,Q] = QQ - (-1)^{|Q||Q|} QQ$)? " What follows next?
@wonderich I meant the following. If you are a category theorist, then the thing that would seem most natural to you is to think of $Q^2 = \frac12[Q,Q]$ as the Lie product. The Lie product of self-adjoint operators is skew-adjoint, and $Q$ was supposed to be self-adjoint, so we should have an equation like $Q^2 = \sqrt{-1} \hat{H}$. But a physicist might prefer to work with underlying vector spaces of a super vector space. In terms of underlying vector spaces, $Q^2$ is the Jordan product. The result is that the category theorist and the physicist will adopt different conventions about ...
... the meaning of "odd self-adjoint operator". The conventions will be intertwined by multiplying odd operators by an 8th root of unity. Actually, you can see the convention question really early. What is the symmetry law for the Hilbert pairing? I.e. if $v,w \in \mathcal{H}$ are homogeneous odd elements, do we have $\langle w, v\rangle = (\pm1) \overline{\langle v, w\rangle}$? A category theorist would say $(-1)$, but a physicist might say $+1$. In particular, a category theorist would commit to deciding that for $v$ a homogeneous odd element, $|v| = \langle v, v\rangle$ is pure-imaginary!
Again this is no fundamental difference: the two conventions are intertwined by multiplying $\langle v, w\rangle$ by $i^{|v|}$. (But be a bit careful with the Hilbert pairing on a tensor product of super Hilbert spaces!) These decisions propagate along. For example, the "spectrum" of an odd operator is a funny thing, because the eigenvectors are inhomogeneous, and so not something a category theorist can talk about directly. The category theorist chose her sign conventions so that some things would be clean, and the cost is that the spectrum of $Q$ is dirty: it is in $\sqrt[4]{-1}\mathbb{R}$.
Thanks so much. Is your Lie product exactly the same or some special case o the Lie superalgebra's super bracket? https://en.wikipedia.org/wiki/Lie_superalgebra#Definition
@wonderich Yes.
|
2025-03-21T14:48:32.089190
| 2020-09-22T02:57:40 |
372282
|
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|
Stack Exchange
|
Rigidity case of a geometric theorem for $3$-manifolds with boundary
Let $(M^3,g)$ be a compact Riemannian $3$-manifold with boundary. In a paper by L. Ambrozio, he considers the set $\mathcal{F}_M$ of all immersed disks in $M$ whose boundaries are curves in $\partial M$ that are homotopically non-trivial. Two quantities are defined:
$$\mathcal{A}(M,g) = \inf_{D \in \mathcal{F}_M} \operatorname{Area}(D) \quad \text{ and } \quad \mathcal{L}(M,g) = \inf_{D \in \mathcal{F}_M} \operatorname{Length}(\partial D).$$
He then proves the following theorem:
If $\mathcal{F}_M \neq \emptyset$ and $\partial M$ is mean-convex (positive mean curvature), then
$$\frac{1}{2} \mathcal{A}(M,g) \inf_M R_M + \mathcal{L}(M,g) \inf_{\partial M} H^{\partial M} \leq 2 \pi,$$
where $R_M$ is the scalar curvature of $M$ and $H^{\partial M}$ is the mean curvature of $\partial M$. Moreover, if equality holds, then the universal cover of $(M,g)$ is isometric to a cylinder $(\Sigma_0 \times \mathbb{R}, g_0 + dt^2)$, where $(\Sigma_0, g_0)$ is a disk with constant Gaussian curvature $\inf_M R_M/ 2$ and $\partial \Sigma_0$ has constant geodesic curvature $\inf_{\partial M} H^{\partial M}$ in $\Sigma_0$.
My question is whether the converse of the last statement in this theorem holds. Namely: if the universal cover of $(M, g)$ is isometric to a cylinder $(\Sigma_0 \times \mathbb{R}, g_0 + dt^2)$, where $(\Sigma_0, g_0)$ is a disk with constant Gaussian curvature $\inf_M R_M/ 2$ and $\partial \Sigma_0$ has constant geodesic curvature $\inf_{\partial M} H^{\partial M}$ in $\Sigma_0$, does equality hold?
Have a look at the Gauss-Bonnet Theorem. https://en.wikipedia.org/wiki/Gauss%E2%80%93Bonnet_theorem
@IanAgol is $\mathcal{A}(M,g)$ realized by a slice?
Yes. The equality of the formula is equivalent to Gauss-Bonnet for a slice.
It is intuitive that $\mathcal{A}$ is realized by a slice, but how to prove it?
Apply the gauss-bonnet
|
2025-03-21T14:48:32.089361
| 2020-09-22T03:18:14 |
372283
|
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|
Stack Exchange
|
Some special subgroups of formal groups
Let $G$ be a 1-dimensional, commutative formal group over a ring $R$. Give $G$ a coordinate $x$ and let $A\subset R$ be the subring generated by the coefficients of the corresponding formal group law $F(x,y)= \sum_{ij}a_{ij}x^iy^j$. So $G$ is really defined over $A$.
Call a finite subgroup $K\subset G$ special if it is the kernel of a homomorphism $T:G\rightarrow \phi^*G$ for some ring map $\phi:A\rightarrow R$. ($\phi^*G$ is the formal group over $R$ with formal group law $\phi^*F(x,y)=\sum_{ij}\phi(a_{ij})x^iy^j$.)
What's an example of a finite subgroup $K\subset G$ that is not (isomorphic to) a special subgroup? (I guess that would be the same as asking that $G/K$ does not inject into any $\phi^*G$.)
In all the cases I've tried (including the additive, multiplicative, and universal formal group laws) I seem to have convinced myself that all subgroups are special. Which leads me to suspect that maybe special subgroups are not so special.
$\textbf{Edit}$: To be explicit, a subgroup $K$ corresponds to a monic polynomial $f_K(x)\in R[x]$ with nilpotent lower order coefficients and such that $f(F(x,y))\equiv 0\ \text{mod}\ (f(x),f(y))$. That subgroup $K$ is special if furthermore there is some invertible power series $u(x)\in R[[x]]$ and a ring map $\phi:A\rightarrow R$ as above such that $u(F(x,y))f(F(x,y))=\phi^*F(u(x)f(x),u(y)f(y)).$
What's a finite subgroup of a formal group?
@QiaochuYuan It's a closed formal subscheme that's finite and free and is also a group. That's equivalent to the data of a monic polynomial $f(x)$ with nilpotent lower order coefficients such that $f(F(x,y))=0$ mod $(f(x),f(y))$.
Your description of specialness in the Edit seems to be demanding that there would be an endomorphism $u\cdot f$ of $F$, vanishing on the kernel of $T$, so that in some sense $G/(\ker T)\cong G$ as formal groups. Have I misconstrued?
@Lubin So $u\cdot f$ "is" $T$ - it is not quite an endomorphism of $F$ but rather sends $F$ to $\phi^*F$. And instead of $G/\text{ker}T\cong G$ all I have is $G/\text{ker} T\hookrightarrow \phi^*G$.
Let me specialize heavily to the case of formal groups (group laws) of dimension one over a $p$-adic ring $\mathfrak o$, i.e. the ring of integers of a finite extension $k$ of $\Bbb Q_p$.
I still am uncertain about what category you’re thinking of. If we restrict further to formal groups of finite height (the endomorphism $[p]$ being of finite degree $p^h$), then these things become $p$-divisible groups, or, if you like, ind-finite objects. For instance the kernel of $[p^n]$ will be a finite $\mathfrak o$-group-scheme, $K_n=\ker([p^n])=\mathrm{Spec}(\mathfrak o[[x]]/([p^n](x))\,)$, and you have natural maps $K_n\hookrightarrow K_{n+1}$, and you see that $\projlim\mathfrak o[[x]]/([p^n](x))\cong\mathfrak o[[x]]$. In this sense, your $G$, if indeed a formal group of finite height over $\mathfrak o$, is the union of its finite subgroups. This is the viewpoint that I tend to work with.
Now, let’s consider just one fairly simple case, where the formal group law has all its coefficients in an unramified extension $A$ of $\Bbb Z_p$, even in $\Bbb Z_p$ itself, and suppose the height is $h=2$ for simplicity. This means that $[p](x)\equiv px+ux^{p^2}\pmod{x^{p^2+1}}$, where $u$ is a unit of $A$, and the congruence ignores all terms in the power series of degree $>p^2$. Look at the Newton polygon and see that all the $z\in\overline k$ with $v_p(z)>0$ and $[p](z)=0$ have $v_p(z)=\frac1{p^2-1}$, plus of course $0$. So $p^2$ in all, and thus they form an elementary $p$-group of order $p^2$.
Now take any of the cyclic subgroups of $\ker[p]$, call it $\Gamma$. One proves that
$$
\pi_\Gamma(x)=\prod_{\gamma\in\Gamma}F(x,\gamma)\,,
$$
which is defined over a totally ramified extension $A'$ of $A$ (actually of degree $p+1$), is a morphism into another formal group, which I will abuse language in calling $G/\Gamma$.
I ask you to believe that I have shown you a formal group $G/\Gamma$ that, as far as I can see, will prove to you that $\Gamma$ is not a special subgroup of $G$, once you see that the formal group law of $G/\Gamma$ is not isomorphic to that of $G$, not even with a morphism $\varphi^*$ of the type you allow. (I think, because I’m not sure what properties you allow $\varphi^*$ to have.)
How do I know that $G/\Gamma$ is nothing like $G$? By Newtonian magic, you see that the Newton polygon of $[p]_{G/\Gamma}$ has vertices at $(1,1)$, $(p,\frac1{p+1})$, and $(p^2,0)$. The important fact is that this polygon is not the same as that of $[p]_G$; and since the shape of the Newton polygon of $[p]$ is an invariant, it follows that there is no way for $G/\Gamma$ to be isomorphic to $G$.
In the appropriate category, the map from $G$ to $G/\Gamma$ is onto. You can show, for instance, that if $v_p(\eta)>0$, there is $\xi$ in a finite extension of $k(\eta)$ such that $v_p(\xi)>0$ and $\pi_\Gamma(\xi)=\eta$.
(All of this is in an old and poorly-written paper of mine, Finite subgroups and isogenies…. EDIT:The “Newtonian magic” involves the “Newton copolygon”, also called the valuation function. I’ll bet a nickel that somebody else has explained it better than I can, but it’s in a later paper of mine, Canonical subgroups of formal groups, and I fear that it’s at most a little better-written than the other. Copolygon talk begins on p. 109.)
Thanks alot for this! Some questions: what is the smallest ring over which $G/\Gamma$ is defined (I think that's $A'$ in the above?) and how does it relate to $A$ and $\mathfrak{o}$? How can I show that $G/\Gamma$ does not even inject into any $\phi^G$? ($\phi^$ is any map $Spf\mathfrak{o}\rightarrow SpfA$? Also, where can I learn Newtonian magic (preferably without selling my soul)? I looked through your paper and couldn't find "Newton" anywhere.
Yes, if $A$ was unramified over $\Bbb Z_p$, $A'$ is the smallest. Without changing the Newton polygon, the only maps are isomorphisms, if you’re talking the category of one-dimensional formal groups. I thought that the paper I quoted has the Newtonian magic explained, but really it’s in a later paper. I’ll edit the answer itself to include the reference.
Consider the case of a formal group $G$ of finite height over a complete local Noetherian ring $R$ of residue characteristic $p>0$. For each $m$ there is a finite $R$-algebra $S$ that classifies finite subgroups of $G$ of order $p^m$ in the sense that $R$-algebra homomorphisms $S\to T$ biject with subgroup-schemes $A<\text{spec}(T)\times_{\text{spec}(R)}G$ such that $\mathcal{O}_A$ is free of rank $p^m$ over $T$. The structure of this classifying ring is described in my paper Finite subgroups of formal groups; there is another version on my home page with additional exposition. In another paper I showed how this ring arises from a calculation in algebraic topology, in the case where $G$ is the universal deformation of a formal group $G_0$ over a finite field $F$. In the case where $F$ is of prime order, every finite subgroup of $G_0$ is the kernel of a power of Frobenius, and we can use the universal deformation property to deduce a kind of specialness property for $G$. Specifically, given $\alpha\colon R\to T$ and a finite subgroup $A<\alpha^*G$ there is another map $\beta\colon R\to T$ with $(\alpha^*G)/A\simeq\beta^*G$. There is a similar but slightly more involved statement in the case where $|F|$ is not prime. In algebraic topology this is all closely bound up with the theory of power operations in $H_\infty$ ring spectra. This is explained in a paper by Charles Rezk. The similar specialness property of Lazard's universal FGL is similarly bound up with the $H_\infty$ structure of the complex cobordism spectrum $MU$, via Quillen's fundamental theorem that the homotopy ring $\pi_*(MU)$ is canonically isomorphic to the Lazard ring.
Thanks, is the claim you're making that any (or many?) formal group coming from a ring spectrum that admits an H-infinity complex orientation has only "special" subgroups? Because the reason I actually posted this question is that I had convinced myself that something like that is true!
That's certainly the right idea; I'm not sure exactly what technical assumptions you need to support it. It should work for $K(n)$-local even periodic $H_\infty$ ring spectra. If $E$ is even periodic and $p$-complete then in many cases it is a retract of $\prod_nL_{K(n)}E$. I don't remember what is the maximum generality in which that is known. Using that fact, one should be able to generalise the statement about special subgroups.
|
2025-03-21T14:48:32.089941
| 2020-09-22T03:26:08 |
372284
|
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|
Stack Exchange
|
Relation between automorphic representations and Laplacian eigenvalues
Let $G=SL(2, \mathbb{R})$. I would like to understand how and why automorphic forms correspond to automorphic representations, and how general is this fact (for other groups).
Let $\Gamma$ be a lattice in $G$, $X = \Gamma\backslash G$ and $Y=X/K \simeq \Gamma \backslash \mathbb{H}$ for a compact subgroup $K$. The group $G$ acts on $X$, hence on the right on functions on $X$ or on $Y$.
Take an automorphic form $\phi$ (i.e. an eigenfunction of the Laplace operator on $Y$). Introduce the space of $G$-translates of $\phi$, that is $V_\phi$ the span of the $g\phi = \phi(g^{-1}\cdot)$ for all $g\in G$. This is a $G$-invariant subspace, and in fact an irreducible unitary representation of $G$.
Take an irreducible unitary representation $(\pi, V)$ of $G$, and assume it has a $K$-fixed vector $v$. Suppose you have an embedding $V \to L^2(X)$ and define $\phi$ to be the image of $v$ by this embedding. Is it necessarily an eigenfunction of $\Delta$? That would mean that the unitary irreducible representations of $G$ (with a $K$-fixed vector and an embedding in $L^2(X)$) exactly correspond to eigenfunctions of $\Delta$. Does it have a meaning in the setting of more general groups?
$G$ does not have any compact open subgroup. Moreover, a Laplace eigenfunction only generates an irreducible unitary representation of $G$ if it is also a Hecke eigenform. Finally, $G$ should act on the right inside $\phi$ in order to maintain left $\Gamma$-invariance. I recommend that you read a standard textbook (e.g. Bump: Automorphic forms and representatiions, or Goldfeld-Hundley: Automorphic Representations and $L$-Functions for the General Linear Group I-II).
|
2025-03-21T14:48:32.090078
| 2020-09-22T04:41:36 |
372287
|
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|
Stack Exchange
|
Can Q(R) embed to Q((R ⊗ S )/ P)
Let $R, S$ be Noetherian $k$-algebra, where $k$ is a field, and $P \otimes S$ is Noetherian.
let $P$ be a prime ideal of $R \otimes S$ such that $P \cap (R \otimes 1) = (0) = P \cap (1 \otimes S)$, then it is obvious $R \hookrightarrow (R \otimes S)/P$
Can $Q(R) \hookrightarrow Q((R \otimes S) / P)$ ? where $Q(R)$ is classical quotient ring of $R$
Have you tried appealing to the universal property of Q(-)?
This problem has been solved~
|
2025-03-21T14:48:32.090276
| 2020-09-22T06:23:13 |
372289
|
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"Gerald Edgar",
"Hirofumi Shiba",
"Piotr Hajlasz",
"YOTAL",
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"juan"
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|
Stack Exchange
|
An example that the sum of two Borel sets which is not a Borel set in n-dimensional Euclidean space
By sum of two sets I mean $A+B := \{x+y:x \in A \quad y \in B\}$, and there is a tip in a book of real analysis by Zhou Minqiang which says:
“If $A,B$ are Borel sets in $\mathbb{R}^{n}$, $A+B$ may not be a Borel set.”
I want to know some specific examples.(Maybe $\mathbb{R}^{1}$ ?)
Any comments will be helpful.
I heard that we can construct a closed set in $\mathbb{R}^{2}$ project to a non-Borel set. Is that true?
A closed set in $R^2$ is a countable union of compacts. So its projection is a countable union of compact sets. Therefore it is a Borel set.
Yes. You are right. Maybe just a Borel set in $\mathbb{R}^{2}$ @juan
Yes, there are Borel sets with projection non Borel. Lebesgue, in one of his paper pretended to proof that the projection of a Borel set is a Borel set. Lusin detected the error. This started the Theory of analytic sets. But the examples, I think, are always difficult.
@juan It was not Lusin who detected the error, but Souslin, a student of Lusin. This is why projections of Borel sets are called Souslin sets. Then Lusin and Sierpinski developed the major part of the theory of Souslin sets.
Is there a simple example? Well, your set $A+B$ would be analytic but not Borel. Is there a "simple" example of that?
This is a result of Erdos and Stone: https://www.ams.org/journals/proc/1970-025-02/S0002-9939-1970-0260958-1/S0002-9939-1970-0260958-1.pdf
Thanks a lot. Is there any easy example?
I'm not sure if it's easy, but this answer provides another example in $\mathbb{R}^3$ and can be generalized into the case $\mathbb{R}^n$ where $n\ge3$.
Short Answer
The core idea is to exploit a non-Borel analytic set (a.k.a. Souslin set) $A'_1\subset[-1,1]$ which is also the image of the first projection of the Borel set $A'\subset[-1,1]^2$. This non-Borel set $A'_1$ is then mapped onto the cylinder
$$
\left\{(x_1,x_2,x_3)\in\mathbb{R}^3\mid x_2^2+x_3^2=1\right\}.
$$
Denoting it $A\subset\mathbb{R}^3$, then we get $(A+B(0,1))\cap(\mathbb{R}\times\{0\}^2)=A'_1$, a scenario that is impossible if $A+B(0,1)\in\mathcal{B}(\mathbb{R}^3)$.
Long Explanation
For every uncountable Polish space, there exists a non-Borel Souslin set, as is explained in Kechris 1995 p.85 Theorem 14.2.
Taking $X=[-1,1]$, we can construct a non-Borel Souslin set $A_1'\in\Sigma^1_1(X)\setminus\mathcal{B}(X)$, and using this $A_1'$ we are going to construct a counterexample for $n\ge3$.
Here, we are in need of the following characterization of Souslin sets:
Theorem (Bogachev 2007 p.24 Theorem 6.7.2, Kechris 1995 p.86 14.3.)
Let $X$ be a Souslin space, a Souslin set which is also Hausdorff, and let $A\subset X$ its subset. The following are equivalent:
$A$ is a Souslin set;
$A$ can be represented as $A=\mathrm{pr}_1(F)$, where $F\overset{\text{closed}}{\subset} X\times\mathbb{N}^\infty$;
$A$ can be represented as $A=\mathrm{pr}_1(B)$, where $B\subset X\times\mathbb{R}$ is Borel measurable.
Here we take $X:=[-1,1]$ and $A:=A_1'$, we can find a Borel measurable subset $A'\subset[-1,1]^2$ such that $A_1'=\mathrm{pr}_1(A')$.
The next step is crucial, where we map the Borel subset $A'$ to a cylinder
$$
C:=\{(x_1,x_2,x_3)\in\mathbb{R}^3\mid x_2^2+x_3^2=1\},
$$
using a homeomorphism $\psi:\mathbb{R}^2\to\mathbb{R}^2$ which satisfies
$$
\psi([-1,1]\times\{0\})\subset\{(x_1,x_2)\in\mathbb{R}^2\mid x_1^2+x_2^2=1\}.
$$
Such a homeomorphism $\psi$ takes the segment $[-1,1]$ on the $x_1$-axis into the unit circumference $S^1$ in the $(x_1,x_2)$-plane.
Using $\psi$ as a building block, we constract a homeomorphism $\Psi:\mathbb{R}^3\to\mathbb{R}^3$ by
$$
\Psi(x_1,x_2,x_3):=(x_1,\psi(x_2,x_3)).
$$
Such a homeomorphism $\Psi$ pastes the set $A'$ onto the surface of the cylinder $C$.
Thus, $A:=\Psi(A')\subset\mathbb{R}^3$ now satisfies the following properties:
$$
\biggl(A+B(0,1)\biggr)\cap\biggl(\mathbb{R}\times\{0\}^2\biggr)=A_1',
$$
where $A_1'\notin\mathcal{B}(X)$ is non-Borel and $B(0,1)\subset\mathbb{R}^3$ is a closed unit ball centered at the origin. This scenario is impossible if $A+B(0,1)$ is Borel measurable, since $\mathbb{R}\times\{0\}^2$ is Borel measurable.
This idea is stimulated from Lurio et. al. 2014, Example 2.4.
True Appreciation for such a specific counterexample.
and if you can write it down as a self-contained answer I'll merge such kind of questions in SE.(If I have time)
@YOTAL Happy if that helped! I've just updated it!
|
2025-03-21T14:48:32.090603
| 2020-09-22T07:37:43 |
372292
|
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|
Stack Exchange
|
Kac-Rice formula and Borell-TIS inequalities for gradient-flow of centered gaussian random field
Let $x\mapsto g(x)$ be a centered gaussian random field on $\mathbb R^m$. Let $x_0 \in \mathbb R^n$, and (assuming regularity conditions) consider the gradient-flow
$$
\dot{x}(t) = -\nabla g(x(t)), \;x(0) = x_0.
$$
Integrating the above system gives
$$
g(x(t)) = g(x_0) + \int_0^t \langle \nabla g(x(t)),\dot{x}(t)\rangle dt = g(x_0)-\int_0^t\|\nabla g(x(t))\|^2 dt.
$$
Question 1. Is there a (Kac-)Rice formula for the zero-crossings of $g(x(t)$, namely $\#\{t \in [0, T] \mid g(x(t)) = 0\}$ ?
One of the things I'm interested in are upper-bounds for $g(x(t))$ as a function of $t$.
Question 2. Can one obtain a Borell-TIS bound for the $\sup_{0 \le t \le T} g(x(t))$.
Note. I'm not necessarily looking for a clear-cut answer (though this would be really cool), but general guidelines on how to go about this.
Examples
As a working examples, one could consider the following "simple" fields
Linear gaussian random field wherein $g(x) = w^Th(x)$, with $w \sim N(0,I_k)$ and deterministic $h \in \mathcal C^1(\mathbb R^n \to \mathbb R^k)$.
Stationary gaussian random fields
Here's a toy model that is truly linear
$$g(x) =\frac{1}{2}\sum_{i=1}^n \Lambda_i x_i^2, $$
where $\Lambda_i$ are i.i.d. $N(0,1)$ then
$$x(t)= \Big(e^{-t\Lambda_1} x_1(0),\dotsc, e^{-t\Lambda_n} x_n(0)\Big)$$
$$U(t):= g(x(t))=\frac{1}{2}\sum_{i=1}^n \Lambda_ie^{-2t\Lambda_i} x_i(0)^2. $$
Denote by $N_T(U)$ the number of zeros of $U(t)$ on the interval $[0,T]$. Denote by $p_{U(t)}(u)$ the probability density of $U(t)$. Then the Kac-Rice formula state that $\newcommand{\bE}{\mathbb{E}}$
$$\bE\big[\; N_T(U)\;\big]=\int_0^T\bE\big[ \; |U'(t)|\;|\; U(t)=0\;\big] p_{U(t)}(0) dt, $$
where $\bE[-|-]$ denotes the conditional expectation.
Alternatively we have $\newcommand{\bP}{\mathbb{P}}$
$$
\bE\big[\; N_T(U)\;\big]=\underbrace{\bE\big[\; N_T(U)\;|\;g(U(T))>0\;\big]}_{=0}\bP\big[ g(U(T))>0\big]+\bE\big[\; N_T(U)\;|\;g(U(T))<0\;\big]\bP\big[ g(U(T))<0\big]
$$
$$
=\bE\big[\; N_T(U)\;|\;g(U(T))<0\;\big]\bP\big[ g(U(T))<0\big]
$$
$$ =\underbrace{\bE\big[\; N_T(U)\;|\;g(U(0))>0,g(U(T))<0\;\big]}_{=1}\bP\big[ g(U(0))>0,g(U(T))<0\big]+ \underbrace{\bE\big[\; N_T(U)\;|\;g(U(0))<0,g(U(T))
<0\;\big]}_{=0}\bP\big[ g(U(0))<0,g(U(T))>0\big]
$$
$$
=\bP\big[ g(U(0))>0,g(U(T))<0\big].
$$
Thanks for the worked example (upvoted). Can this integral be evaluated explicity to get $E[N_T(U)] =$ some explicit function of $T$ ? I've expected that such a "simple" model would produce an explicit answer down the line :)
What about the limiting case when $n \to \infty$, does anything simplify (e.g via some LLNs / CLT-type argment) ?
Concerning last but one comment, if we consider the even simpler example where $g(x):= \Lambda^Tx$, then $U(t) = g(x(t)) = g(x(0) + t\Lambda) = \Lambda^Tx(0) + t|\Lambda|^2$, so that $U'(t) = |\Lambda|^2$. Thus the formula would give $E[N_T(U)] = \int_0^TE[|\Lambda|^2 \mid \Lambda^Tx(0) + t|\Lambda|^2] p_{U(t)}(0)dt$, but I don't quite see how to proceed to get a concrete number.
I don't know how to compute that conditional expectation. Note that $\mathbb{E}[N_T(U)|g(x_0)<0]=0$ and $\lim_{T\to\infty} \mathbb{E}[N_T(U)|g(x_0)>0]=1$ a.s..
Maybe one can conjecture $E[N_T(U) \mid g(x_0) > 0] \ge 1 - \exp(-\alpha(T))$ where $\alpha$ is a nonnegative increasing function of $T$ ? (or something like that)
You need to solve for $t$ the equation $$\sum_{i=1}^n\lambda_i e^{-2\lambda_i} x_i^0=0,$$ and view it as a function of $x_i^0,\lambda_i$. Note that this equation has no solution if $\lambda_i >0$ for any $i$. For example if $n=2$, $a,b>0$ the equation $ae^{-at}x-be^{-2bt} y=0$ so $$1-\frac{b}{a}e^{2(a-b)t}\frac{y}{x}=0$$, $$t=\log\left(\frac{xa}{by}\right)/2(a-b).$$ For higher $n$ analyzing $t(\lambda_i,x_j^0)$ is more involved.
Thanks for the details.
In the posted answer, what do you mean by $g(U(0))$, etc. ? Did intend to say $U(0)$ ? Also, there is a stray dollar-dollar sign in your post.
$U(t)$ is a trajectory of the random gradient flow $\dot{U}(t)=-\nabla g\big(U(t)\big)$ and $g\big(U(0)\big)$ is the value of $g$ at the initial point on the trajectory.
|
2025-03-21T14:48:32.090866
| 2020-09-22T07:57:45 |
372293
|
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|
Stack Exchange
|
Entropy of flow and time-1 map
Let $\Phi=(\phi_t)_{t\in \mathbb{R}}$ be a continuous flow on a compact metric space $X$. Let $\mu$ be a $\phi_1$-invariant measure. Then it is not hard to verify tht $\int_{0}^{1} \phi_t\mu dt$ is $\Phi$-invariant measure. I would like to whether it is possible to compare the value of entropies $h(\mu)$ and $h(\int_{0}^{1} \phi_t\mu dt)$. Are they equal? Thanks.
|
2025-03-21T14:48:32.091365
| 2020-09-22T09:33:02 |
372300
|
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"Aoki",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/372300"
}
|
Stack Exchange
|
Definition of Hitchin map
This may be a dumb question.
$\mathcal{M}(r,d)$ is a coarse moduli scheme for semistable pairs $(E,\phi:E \rightarrow K_X \otimes E)$ of rank $r$, degree $d$ on a smooth projective curve $X$ over $\mathbb{C}$.
Then,the Hitchin morphism from $\mathcal{M}(r,d)$ to $H^0(X,K_X) \times ...\times H^0(X,K_X^{\otimes r})$ is defined a pair $(E,\phi)$ to its characteristic polynomial. Explicitly, $(E,\phi) \mapsto (\operatorname{Tr}(\wedge^1\phi),...,\operatorname{Tr}(\wedge^r\phi))$
But strictly, this is not a scheme morphism in the sense of Hartshorne's algebraic geometry.
My idea:
Let $U_i=Spec(B_i)\subset{\mathcal{M}(r,d)}$ be affine cover. First, define $Sym(V^*)\rightarrow B_i$.
If it is possible to glue them, we get the morphism $\mathcal{M}(r,d) \rightarrow Spec(Sym(V^*))$
$(V:=H^0(X,K_X) \times ...\times H^0(X,K_X^{\otimes r}))$, using $Hom_{sch}(X,Spec(R))\simeq Hom_{ring}(R,\Gamma(X,O_X))$
But can we gule them? Or this is a completely wrong direction?
Yes, you are on a wrong track. All you have to prove is that $(E,\phi)\mapsto \operatorname{Tr}\wedge^{i}\phi $ is a regular function on $\mathscr{M}(r,d)$. This is clear on a fine moduli space (or stack), then use basic descent.
Another way of seeing it (perhaps a rewording of abx's comment?) is: the construction $(E,\phi)\mapsto\mathrm{tr}\wedge^i \phi$ works well in families cause it's done fiberwise. This gives a map from the moduli functor (i.e. the functor associating to $S$ the set of appropriate families on $S$) to (the functor of points of) your $SpecSym(V^)$. By definition of coarse moduli space, this induces a unique map of schemes from the coarse moduli space to $SpecSym(V^)$ which makes the diagram of the previous maps commute hence is the given map on points/objects.
Thanks for comments. But what do you mean by regular function? regular map in the category of algebraic variety?
@Aoki A regular function on $X$ is an element of $\mathcal O_X$, equivalently a map to $\mathbb A^1$. Since you're mapping to a vector space here, you want a tuple of regular functions I think.
@Tabes Bridges I think Hitchin map should be a morphism of the category of scheme. But $(E,\phi)\mapsto \operatorname{Tr}\wedge^{i}\phi$ is only defined at closed points and isn't a morphism between schemes.And, I don't know the usual definition of regular between schemes.
What, for you, is the definition of a coarse moduli space?
It is the scheme M satisfying certain universal property to the functor $\mathscr{M}(r,d): S \mapsto$ ${$ isom class of flat family over $S$ of semistable pairs $(E,\phi) }$
@Aoki: This scheme turns out to be a variety... Before studying the Hitchin map, it might be useful.to study the classical theory of moduli spaces of vector bundles.
@abx Thanks, I'll study it.
Then the main point of Qfwfq's argument is to check that $\operatorname{Tr} \wedge^i \phi$ makes sense as a map from $S$ to a vector space for a flat family over $S$ of (semistable) pairs $(E,\phi)$. This is fairly straightforward.
|
2025-03-21T14:48:32.091573
| 2020-09-22T11:08:55 |
372303
|
{
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"Taras Banakh",
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|
Stack Exchange
|
Integration theory for functions and values with values in topological rings
I am curious whether somebody ever tried to generalize the classical theory of Lebesgue integral to functions and measures with values in Hausdorff topological rings.
The generalization of a measure is straightforward: given a topological ring $R$ and a $\sigma$-algebra $\mathcal A$ on a set $\Omega$, define an $R$-valued measure as a function $\mu:\mathcal A\to R$ such that
$\bullet$ $\mu(A\cup B)=\mu(A)+\mu(B)$ for any disjoint sets $A,B\in\mathcal A$;
$\bullet$ $\mu(\bigcup_{n\in\omega}A_n)=\sum_{n\in\omega}\mu(A_n)$ for any sequence $(A_n)_{n\in\omega}$ consisting of pairwise disjoint sets in the algebra $\mathcal A$.
Given a simple $\mathcal A$-measurable function $f:\Omega\to R$ and an $R$-valued measure $\mu$, define the integral
$\int f d\mu$ as the (finite) sum $\sum_{y\in R}y\cdot\mu(f^{-1}(y))$.
So, the question:
Is anything known about topological rings $R$ for which the $R$-valued integral can be defined for some reasonably wide class of functions and so-generalized integral has all basic properties of the usual Lebesgue integral?
You mean Hausdorff topological ring for assumptions to be meaningful (still topological abelian group is enough to define the measure). Also when you write $\sum_\omega \mu(A_n)$, you implicitly mean that this sum exists, which probably means that $\sum_{k=0}^n\mu(A_k)$ converges... or you assume anything stronger? anyway this bare convergence assumption forces $\sum_{k=0}^n\mu(A_{f(k)})$ to converge to the same limit for every permutation $f$ of $\omega$, which is quite close to "absolute convergence" in spirit.
Yes, I assume that $R$ is a Hausdorff topological ring, if necessary, commutative. And writing $\sum_{n\in\omega}\mu(A_n)=\mu(\bigcup_{n\in\omega}A_n)$ I understand that the series converges to $\mu(\bigcup_{n\in\omega}A_n)$. And the convergence is unconditional (since $\mu(\bigcup_{n\in\omega}A_n)$ does not depend on the order of the summands). In infinite-dimensional Banach spaces the unconditional convergence is strictly weaker than the absolute convergence. But in general topological rings the absolute convergence is undefined (only unconditional can be defined).
I would start with Kaplansky, Irving. "Topological rings." Bulletin of the American Mathematical Society 54.9 (1948): 809-826. and refer to later works that cite it.
@rschwieb Thank you. This paper of Kaplansky has only 46 citations. None of them involves measure. There is however another paper of Kaplansky with the same title in Amer. J.Math. with 289 citations. Maybe this will lead to something interesting.
In general one could consider three Hausdorff topological abelian groups $A,B,C$ with a continuous $\mathbf{Z}$-bilinear map $A\times B\to C$, such a measure valued in a $A$, consider functions $f$ valued in $B$, and define the integral for simple $f$ as $\sum_{y\in B}f(\mu(f^{-1}({y})),y)$.
@YCor It seems that in Functional Analysis they develop something even more general: the integral of operator-valued function over a vector-valued measure, see. e.g. https://www.imath.kiev.ua/~tesko/pdf/tesko_2012%20-%20the%20integration%20of%20operator-valued%20functions%20with%20respect%20to%20vector-valued%20measures.pdf
@YCor But at the moment I do not see why such an integral is well-defined for bounded measurable functions into the topological ring $\ell_2$ (with coordinatewise multiplications). Maybe this is not true?
This is a comment but too long. The leap from the classical case to your general one is, of course, huge but if one takes a more modest one, namely to functions with values in locally convex algebras and their non locally convex analogues, one sees quite clearly what can happen. To be concrete, we consider the rings of continuous, resp. (equivalent classes of) measurable functions say on the reals, functions from the interval with values in these spaces and finally their integrals with respect to Lebesgue measure). (This fits into your scheme since we can consider the reals as the subring of the constant functions). The first case is well-studied and well-behaved, in particular, continuous or even bounded measurable functions are integrable, but this is no longer true in the non-locally convex case. The kindergarten reason for this is that while convex combinations of small things are small in the first situation, this can fail in the second one—the standard way of defining an integral (e.g. the Riemann one) is to take convex combinations of function values and then proceed to take a limit.
The two things you lose in going from the l.c.s. case to the t.v.s. are duality and convexity arguments which play a vital role in vector or algebra valued integration. I suppose that what I am trying to say is that without some substitute for these, problems could arise in the very general situation you are envisaging.
Thank you for your answer. Could you give me a reference where the locally convex case is studied. Observe that duality theory cannot be applied since for a linear functional $x^$ and elements $x,y$ of the ring the value $x^(xy)$ cannnot be expressed via $x^(x)$ and $x^(y)$. So, the vector-valued integral does not project to the field of scalars. This is the case if either you integrate vector-valued functions by scalar measures or integrage scalar functions by vector-valued measures, but not vector-valued functions by vector-valued measures.
I only considered the case of integration with respect to a scalar measure since I wanted to show how thing can go wrong even in this special case and to indicate some of the pitfalls that would have to be avoided in the general situation you described. I have no knowledge about work on vector valued functions being integrated with respect to vector valued measures.
|
2025-03-21T14:48:32.091896
| 2020-09-22T11:22:17 |
372305
|
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|
Stack Exchange
|
Gysin map and $B\mathbf{G}_m$, confusion
Write $\text{Sh}(X)$ for the triangulated/stable $\infty$ category of $\ell$-adic sheaves on $X$, and $k\in\text{Sh}(X)$ fo the unit object. In playing around with $\text{Sh}(B\mathbf{G}_m)$ I've found something disturbing I'd like to resolve.
Following 7.2 of [DG12], we can identify $\text{Sh}(B\mathbf{G}_m)$ with the modules in $\text{Sh}(\text{pt})$ for the algebra $B=H^*(\mathbf{G}_m,k)^\vee\in\text{Sh}(\text{pt})$. Note $B=k[u]/u^2$ where $|u|=-1$. Cohomology is
$$H^*(B\mathbf{G}_m,-)\ :\ M \ \longrightarrow\ \text{Ext}_B(k,M).$$
For instance, $k\mapsto k[t]$ where $|t|=2$ and $B\mapsto k$.
Now let $V$ be a vector bundle over $B\mathbf{G}_m$ (=vector space with $\mathbf{G}_m$ action) of rank $r$. Write $i$ for the zero section and $j$ for the complementary open embedding. Applying $i^*$ to the distinguished triangle $i_!i^!k\to k\to j_*j^*k$ gives the Gysin sequence
$$i^!k \ \stackrel{e}{\longrightarrow} \ i^*k \ \longrightarrow \ i^*j_*j^*k$$
Note that $i^!k\simeq k[-2r]$ and $e$ is multiplication by the Euler class. In particular, if $V$ has nonzero $\mathbf{G}_m$ weights, then on cohomology $e=t^{2r}:k[t]\to k[t]$ so $H^*(i^*j_*j^*k)$ is finite dimensional.
Question: however, what on earth is $i^*j_*j^*k$? Forgetting the $B$-module structure,
$$i^*j_*j^*k\ =\ k\oplus k[-2r+1]$$
by pulling back the Gysin sequence to $\text{pt}$. In the rank $r=1$ case, this means that as a $B$ module it's $k\oplus k[-1]$ or $B$. Only the latter has finite dimensional cohomology it must be that:
$$i^*j_*j^*k\ = \ B \ =\ H^*(\mathbf{G}_m)^\vee.$$
However, when the rank is greater than one there seems to be no $B$-module structure on $k\oplus k[-2r+1]$ which gives finite dimensional cohomology!
What on earth is going on here?
Can't it have generators in degrees $0$ through $-2r+1$, differentials from degree $-2r+2$ to $2r+3$, $-2r+4$ to $-2r+5, \dots, -2$ to $-1$, and $u$ sends $0$ to $-1$, $-2$ to $-3$, $\dots, -2r+2$ to $-2r+1$?
@WillSawin Thanks for the reply. Doesn't $u$ have to commute with the differential? I admit to not having checked 7.2 of [DG12] in detail but I thought that was what was meant by $B$-algebra.
I think $du$ and $ud$ are both $0$ on the complex I wrote down, so they commute. This is because $d$ and $u$ go different ways because $d$ is a cohomological differential and $u$ is in degree $-1$.
I think there's lots of nice points of view here. (i) this is a prototypical case of Koszul duality; you could write down an explicit Koszul complex implementing the equivalence between (finite dimensional) B-modules and (perfect) k[t]-module, and then apply it k[t]/t^{2r}; (ii) if you do this you'll basically see the following play out: t corresponds to the extension with B in the middle and k's on each side, so then t^{2r} corresponds to splicing together 2r-copies of this extension, and this gives Will's answer; (contd)
(iii) maybe the confusion stems from the following line of thought: I want to define an algebra map B-->End(M) for some k-module, but in the example at hand it looks like nothing is in degree -1 (for large r), so how could there possibly be an interesting map here? The key is that we are in derived land! B is not free as an associative algebra: we have to choose a homotopy (or cycle) killing u^2, and then we have to choose something killing the resulting massey product <u,u,u>, etc. In the example at hand, it turns out that choosing an interesting Massey product at the 2rth-ish step
is what builds this module. Said slightly differently: on homotopy groups the action of B looks trivial, but there is a nontrivial massey product like <u,u, ..., u, x>=y which relates the two generators in $k\oplus k[-2r+1]$.
It seems my confusion was just down to confusing which direction $u$ went in! Normally I would delete this question in such a case, but I think Dylan and Will's comments make it worth keeping, thank you.
|
2025-03-21T14:48:32.092518
| 2020-09-22T11:49:47 |
372306
|
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|
Stack Exchange
|
When does a gaussian quadratic form converge (in probability) to a constant?
Let $(h_{ij})_{i,j \in \mathbb N}$ be a sequence of real numbers (deterministic) and let $x_1,\ldots,x_n,\ldots$ be a sequence of iid $N(0,1)$ randm variables. For each positive integer $n$, consider the quadratic form $q_n:=\dfrac{1}{n}\sum_{i=1}^n\sum_{j=1}^nh_{i,j}x_ix_j$.
Question. Under what conditions on the sequence $(h_{ij})$ does there exist $c \ge 0$ sucht aht $q_n \to c$ in probability ? Is there some other kind of convergence that might hold here ?
Note. In the special case $h_{ij} = \delta_{ij}$, we have $q_n = \dfrac{1}{n}\sum_{i=1}^n x_i^2 \overset{p}{\longrightarrow}1$.
I think you should replace $H$ by $H_n$. Otherwise its irritatig.
Indeed. Updated.
First, notice that w.l.o.g. you can assume that the matrix $H_n$ is diagonal (from rotational invariance of the isotropic Gaussian).
You thus are interested in
$$
\frac{1}{n} \sum_{i=1}^n \lambda_i(H_n) X_i^2.
$$
So the condition
$$\sum_{i=1}^n \lambda_i^2(H_n)/n^2 \to 0$$
is the key.
Then, one has that
$$
Var\left(\frac{1}{n} \sum_{i=1}^n \lambda_i(H_n) X_i^2\right)= 2\sum_{i=1}^n \lambda_i^2(H_n)/n^2 \to 0.
$$
and
$$
\mathbb{E}\left(\frac{1}{n} \sum_{i=1}^n \lambda_i(H_n) X_i^2\right)= \sum_{i=1}^n \lambda_i(H_n)/n.
$$
Ah, I should've thought of rotational invariance and SVD. Thanks!
So what kind of convergence (and to what) does that condition imply ? (My holy grail would be convergence to a constant...)
Ok, i guess if $\sum_{i=1}^n \lambda_i(H_n)/n^2 \to 0$, then we have $q_n \overset{L^2}{\longrightarrow} \lim_n \mathbb E[q_n] = \lim_n \sum_{i=1}^n\lambda_i(H_n)/n =: c$, and therefore (by Markov's inequality) $q_n \overset{p}{\longrightarrow} c$
|
2025-03-21T14:48:32.092641
| 2020-09-22T12:29:41 |
372308
|
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|
Stack Exchange
|
Prime ideals and class group equations
Let $K$ be a degree $n \geq 3$ extension over $\mathbb{Q}$, and let $\mathcal{O}_K$ be its ring of integers. We say a rational prime $p$ splits completely in $\mathcal{O}_K$ if the principal ideal $(p) = \mathfrak{p}_1 \cdot \cdots \cdot \mathfrak{p}_n$ for pairwise distinct degree one prime ideals $\mathfrak{p}_i$, $1 \leq i \leq n$ (indeed, we do not want $p$ to ramify in $\mathcal{O}_K$). Let $C_K$ be the ideal class group of $\mathcal{O}_K$. For an ideal $I \subset \mathcal{O}_K$, let $[I]$ denote the corresponding ideal class in the class group.
Let $g_1, \cdots, g_{n-1}$ be arbitrary (not necessarily distinct) elements of $C_K$. Does there necessarily exist a rational prime $p$ which splits completely in $\mathcal{O}_K$ such that the ideal class equations $g_i = [\mathfrak{p}_i]$ holds for $1 \leq i \leq n-1$? Note that there are at most $n-1$ degrees of freedom since the product of the $\mathfrak{p}_i$'s is principal.
No. If $K$ is Galois, since the Galois group act transitively on the $\mathfrak{p_i}$, their class must be the same up to automorphism of the class group. Thus you can take as a counterexample any Galois extension $K$ of degree at least $3$ with nontrivial class group, $g_1$ the identity and $g_2$ any nonidentity element.
|
2025-03-21T14:48:32.092749
| 2020-09-22T12:42:14 |
372309
|
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|
Stack Exchange
|
Asymptotic behaviour of an integral. How should I proceed?
Let us consider the following SDE: $$dY_t=b(Y_t)dt+\sigma(Y_t)dW_t\tag{1}$$ with $b, \sigma: (l, r)\to\mathbb{R}$, $−\infty \leq l < r \leq \infty$ bounded functions on compact intervals of $(l, r)$. In particular, $$b(Y_t)=(u-(u+i)Y_t)$$ $$\sigma(Y_t)=o\sqrt{(Y_t)(1-Y_t)}$$ with $u$, $i$ and $o$ arbitrary parameters. Hence, focus will be on the following SDE: $$dY_t=(u-(u+i)Y_t)dt+o\sqrt{(Y_t)(1-Y_t)}dW_t\tag{2}$$ I must check whether the process $\{X_t\}$ remains within the interval $(l,r)$ or not for each $0\leq t\leq T$.
To this, I use the Feller test for explosions. Such a test requires that the following two integrals must be defined and computed: $$p(x)=\int_c^x \exp\bigg\{-2\int_c^{\xi}\frac{b(\zeta)}{\sigma^2(\zeta)}d\zeta\bigg\}d\xi\tag{3}$$ $$v(x)=\int_c^x\frac{2(p(x)-p(y))}{p\hspace{0.1cm}'(y)\sigma^2(y)}dy\tag{4}$$ with $c\in(l,r)$. According to Feller test, probability that the process at least touches the bounds of interval $I$ equals $1$ or is less than $1$ according to whether $v(l+)=v(r-)=\infty$ or not. Let us fix $(l,r)=(0,1)$ and $c=\frac{1}{2}$.
I would like to study the asymptotic behaviour of the integral $(4)$ with $c=\frac{1}{2}$ at bounds $l=0$ and $r=1$, but I have not any experience with analyses like that. Is there a good standard method or is it just a matter of manipulation? Could you please help me understand how could I study asymptotic behaviour of $(4)$?
Is anything known about the signs of $u$ and $i$?
No, nothing as far as I know @IosifPinelis
In a certain sense, if I properly understood Feller's test spirit, it could help me find values for parameters $i$, $u$, $o$ such that bounds are NOT touched by the process. In the first instance, however, the problem would consist in understanding the asymptotic beahviour of $v(x)$ for $x=0$ and $x=1$
Direct calculations show that for $a:=i/o^2$, $k:=u/o^2$, and $x\in(1/2,1)$
$$p(x)=\int_{1/2}^x(2-2s)^a(2s)^k\,ds,$$
whence
$$p'(x)=(2-2x)^a(2x)^k\asymp(1-x)^a$$
and, for $a>-1$ and $y\in(1/2,x)$,
$$p(x)-p(y)=\int_y^x(2-2s)^a(2s)^k\,ds\asymp\int_y^x(1-s)^a\,ds
\asymp(1-y)^{a+1}-(1-x)^{a+1}\le(1-y)^{a+1}.$$
So,
$$v(x)\ll\int_{1/2}^x\frac{dy}{(1-y)^{a+1}}\,(1-y)^{a+1}\le1.$$
So, the right endpoint, $r=1$, is not touched by the process for any $a>-1$. The consideration of the case $a\le-1$ is similar, with the same conclusion, so that the right endpoint, $r=1$, is not touched by the process for any real $a,k$.
The left endpoint, $l=0$, is considered similarly. Here we can also use the symmetry $x\leftrightarrow1-x$ and $a\leftrightarrow k$, which reduces the consideration of the left endpoint to the already completed consideration of the right endpoint. Thus, we conclude that the left endpoint, $l=0$, is not touched by the process for any real $a,k$.
|
2025-03-21T14:48:32.092922
| 2020-09-22T12:53:51 |
372312
|
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|
Stack Exchange
|
Basic question about convergence of top and penultimate eigenvalues of a sequence of operators
$\newcommand{\norm}[1]{\|#1\|}$ $\newcommand{\ab}[1]{\langle #1\rangle}$ $\newcommand{\C}{\mathbf C}$
Questions
Let $I$ be the unit interval.
Let $H=L^2(I)$ and $T:H\to H$ be a compact self-adjoint operator.
Let $f_n:I\to I$ be a sequence of function in $L^\infty(I)$ such that $f_n\to \mathbf 1$ in $L^\infty$ where $\mathbf 1$ denotes the function $I\to \C$ which takes the value $1$ everywhere.
Define $T_n=f_n T$, that is $T_n:H\to H$ takes $\xi$ to $(T\xi)f _n$, that is, the pointwise product of $T\xi$ and $f_n$.
Question 1.
I am trying to see if
$$
\lambda_{\text{max}}(T_n) \to \lambda_{\text{max}}(T)
$$
where $\lambda_\text{max}(T_n)$ denotes the largest eigenvalue of $T_n$ and similarly for $T$.
(We know that since each $T_n$ is compact and has real spectra as argued in the Appendix).
Progress on the Question 1. If $\rho$ denotes the spectral radius, then by the spectral radius formula, we have
$$
\rho(T_n) = \limsup_{k\to \infty} \norm{(T_n)^k}^{1/k} = \limsup_{k\to \infty} \norm{(f_n T)^k}^{1/k}
$$
If $n$ is large enough then we have $\norm{f_n-\mathbf 1}\leq \varepsilon$, which given from above that
$$
\rho(T_n) \leq \lim\sup_{k\to \infty} \norm{f_n^k T^k}^{1/k}\leq (1+\varepsilon)\limsup_{k\to \infty} \norm{T^k}^{1/k} = (1+\varepsilon) \rho(T)
$$
Similarly we can show that for large enough $n$ we have
$$
(1-\varepsilon)\rho(T) \leq \rho(T_n)
$$
giving that $\rho(T_n)\to \rho(T)$ as $n\to \infty$.
Since the spectrum may have negative values, this does not necessarily give $\lambda_{\text{max}}(T_n)\to \lambda_{\text{max}}(T)$ but is a step in that direction.
Also, if $\lambda_{\text{penmax}}(T)$ denotes the second to largest eigenvalue of $T$, then,
Question 2.
Is it also true that
$$
\lambda_{\text{penmax}}(T_n) \to \lambda_{\text{penmax}}(T)
$$
as $n\to \infty$.
If there is something more general known about convergence of spectra (of compact operators or otherwise) then please feel free to share.
Appendix
Let $\varepsilon>0$ be a small positive number and $f:I\to I$ be such that $\norm{f-\mathbf 1}_\infty<\varepsilon$, where $\mathbf 1$ denotes the constant function $1$.
Let $H=L^2(I)$.
Let $T:H\to H$ be a compact operator and define $fT: H\to H$ as $(fT)\xi = f\ T\xi$ for all $\xi\in H$.
We will show that if $T$ is a compact self-adjoint operator then $fT$ is compact with all its eigenvalues real.
If $T$ is compact then there is a sequence $(T_n)$ of bounded linear operators on $H$ such that each $T_n$ has finite rank and $T_n\to T$ in operator norm.
Clearly, $fT_n\to fT$ in operator norm.
Since each $fT_n$ is also of finite rank, we see that $fT$ is indeed compact.
For any $g\in L^\infty(I)$ close to $\mathbf 1$ in the $L^\infty$ norm, define an inner product $\ab{\cdot, \cdot}_g$ on $L^2(I)$ as $\ab{\xi, \zeta}_g = \int_I \xi\bar \zeta g\ d\mu$.
It is easy to check that this is indeed an inner product.
Let us write $H_g$ to denote $L^2(I)$ equipped with the inner product $\ab{\cdot, \cdot}_f$.
Then we have
$$
\ab{(fT)\xi, \zeta}_{1/f} = \ab{f\ T\xi, \zeta}_{1/f} = \ab{T\xi, \zeta} = \ab{\xi, T \zeta} = \ab{\xi, f\ T\zeta}_{1/f} = \ab{\xi, (fT) \zeta}_{1/f}
$$
This shows that $fT$ is self-adjoint with respect to the inner product $\ab{\cdot, \cdot}_{1/f}$.
Cross Posted on MSE.
The variational formulas for eigenvalues should imply that, for any $k$, the $k$-th largest eigenvalue of $T_n$ converges to the $k$-largest eigenvalue of $T$, and similarly for the smallest, see https://en.wikipedia.org/wiki/Min-max_theorem
More generally, for self-adjoint operators (not necessarily compact), the spectrum behaves very-well with respect to perturbations: if $S,T$ are self-adjoint operators with $|S - T |\leq \varepsilon$, then the spectrum of $T$ is contained in the $\varepsilon$-neighborhoud of the spectrum of $S$. This applies here, as (say if $|f_n -1| \leq 1/2$) $f_n T = \sqrt{f_n} (\sqrt{f_n} T \sqrt{f_n}) \sqrt{f_n}^{-1}$ is similar to $\sqrt{f_n} T \sqrt{f_n}$, and $|\sqrt{f_n} T \sqrt{f_n} - T|\leq 10 |T| |f_n-1|$.
@MikaeldelaSalle Thank you for the comment. Can you please give me a reference for the statement that you made about perturbations (that if $S$ and $T$ are self-adjoint with $|S-T|<\varepsilon$ then the spectra of $S$ and $T$ are within $\varepsilon$ neighborhoods of each other). Thank you.
I am bad at references, but I would be surprised it this did not belong to standard textbooks on spectral theory for self-adjoint operators. Here is a short proof. By symmetry and translation, we have to justify that if the spectrum of $S$ does not intersect $[-\varepsilon,\varepsilon]$, then $T$ is invertible. But in that case $|S^{-1}| < 1/\varepsilon$ (the norm of a self-adjoint operator is its spectral radius), so $S^{-1}(S-T)$ has norm $<1$, so $T = S( 1 - S^{-1}(S-T))$ is invertible as a composition of two invertible operators.
@MikaeldelaSalle Thank you. That helps a lot.
|
2025-03-21T14:48:32.093213
| 2020-09-22T12:54:05 |
372313
|
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|
Stack Exchange
|
Compute limit of $\mathbb P(Y \le X_n)$ using limiting information on the sequence of random variables $X_n$
Let $Y$ be a symmetric random variable, $(X_n)_n$ be a sequence of nonnegative random variables, and set $p_n = \mathbb P(Y \le X_n)$. It is known from Slutsky's theorem that, if $c$ is a constant such that $X_n \to c$ in probability, then $p_n \to F_Y(c)$, where $Y$ is the CDF of $Y$.
Question. Can convergence of $X_n$ in probability to a constant $c$, be replace by some other notion of convergence, say $X_n \to X$ (in some sense), for some random variable $X$, such that we can still compute the limit of $p_n$ only via the data $X$ and $Y$ ?
Of course by convergence in distribution, since the limit r,v. here is constant.
It won't work. $X$ is not a constant. I've updated the text to make this point more explicit.
Even if $X_n\to c$ in probability for some real constant $c$, it is not necessary that $P(Y\le X_n)\to P(Y\le c)$ -- you also need to require that $P(Y=c)=0$.
More generally, if the limit of $X_n$ is not a constant, then you need to assume the convergence, not just of the distribution of $X_n$, but of the joint distribution of $X_n$ and $Y$. In particular, if $(X_n,Y)$ converges to $(X,Y)$ in distribution and $P(Y=X)=0$, then, by the Portmanteau theorem, $P(Y\le X_n)\to P(Y\le X)$.
Makes sense. Thanks!
|
2025-03-21T14:48:32.093338
| 2020-09-22T13:29:47 |
372316
|
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|
Stack Exchange
|
What is the convergence rate of the minimum separation distance?
Let $\Omega\subset \mathbb{R}^m$ be a bounded Lipschitz domain. Let $D$ be a countable dense subset of $\Omega$, denoted as $D = \{p_1,p_2,p_3\ldots \}$. Define the minimum seperation distance among first $n$ points of $D$ as
$$h(n) = \min_{1\le i,j\le n,i\ne j}\|p_i-p_j\|_2 .$$
I'd like to know the convergence rate of $h(n)$ as $n\to\infty$. Surely $h(n)\to 0$, but at what rate. Has it got anything to do with the dimension of the space. Is it $$h(n) = O(n^{-\frac{1}{m}})$$ os it simply $h(n) = O(1/n)?$
I feel there are some notions that I am not aware, which is the reason this question is puzzling me.
I think you are interested in $\max_D \min_{i,j}|p_i-p_j|$? right?
no. $h(n) = \min_{p_i,p_j\in E_n,p_i \ne p_j}|p_i-p_j|_2$
Is it just $h(n) = O(1/n)$?
My question could be really simple or silly, just that I don't know the answer. I am sure I am asking what I intended.
@Dirk : I thought Lipschitz domain means bounded already, but learnt just now that it isn't. So to add, $\Omega$ is a bounded domain with a smooth boundary.
If $\Omega$ is any bounded domain in $\mathbb{R}^m$, Let $B_0$ be a ball which contains it, and let $B$ be a ball with the same center but with its radius increased by $1$. Say $R$ and $V$ are the radius and volume of $B$.
Fix $r < {\rm min}(\frac{1}{2}h(n), 1)$. Then the $r$-balls about the points $p_1$, $\ldots$, $p_n$ are disjoint, and as they are all contained in $B$ we have $nv \leq V$, where $v$ is the volume of a ball of radius $r$. Thus $nr^m \leq R^m$, so $r \leq Rn^{-\frac{1}{m}}$. As this is true of every $r < \frac{1}{2}h(n)$, we get $h(n) \leq 2Rn^{-\frac{1}{m}}$.
To see that $O(n^{-\frac{1}{m}})$ is best possible, let $\Omega$ be the unit cube in $\mathbb{R}^m$ and form $D$ as the union, in order, of the sets $D_k = 2^{-k}L \cap \Omega \setminus (D_1 \cup \cdots \cup D_{k-1})$, where $L$ is the integer lattice in $\mathbb{R}^m$. Then $|D_1 \cup \cdots \cup D_k| = |2^{-k}L \cap \Omega| = 2^{km}$, so $h(2^{km}) \geq 2^{-k}$ (the minimum distance between points in $2^{-k}L \cap \Omega$). That is, for $n = 2^{km}$ we have $h(n) \geq n^{-\frac{1}{m}}$ in this example.
If I define $$r(n) = \sup_{x\in\Omega}\min_{1\le i \le n}|p_i-x|_2 $$, what would be the convergence rate of $r(n)$? Will this be the same as that of $h(n)$ or any faster? If this a very different question that needs a separate post?
|
2025-03-21T14:48:32.093513
| 2020-09-22T13:35:48 |
372317
|
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|
Stack Exchange
|
Expectation of the determinant of the inverse of non-central Wishart matrix
Let $A$ be $(n,n)$ central Wishart matrix with $k$ degrees of freedom.
my question is there is a way to estimate the expectation of:
\begin{align}
E[det(I+(I+A)^{-1})]
\end{align}
The joint distribution of the eigenvalues $\lambda_i$, $i=1,2,\ldots n$ of $A$ is known,
$$P(\lambda_1,\lambda_2,\ldots\lambda_n)=c_{k,n}\prod_{i<j}|\lambda_i-\lambda_j|\prod_m e^{-\lambda_m/2}\lambda_m^{(k-n-1)/2},$$
with $c_{k,n}$ a normalization constant.
The desired expectation value is given by
$$U_{n,k}=\mathbb E[\det(I+(I+A)^{-1})]=\int_0^\infty d\lambda_1\int_0^\infty d\lambda_2\cdots \int_0^\infty d\lambda_n\,\prod_{i<j}|\lambda_i-\lambda_j|\prod_m e^{-\lambda_m/2}\lambda_m^{(k-n-1)/2}\left(1+(1+\lambda_m)^{-1}\right).$$
For small $n$ the integrals can be done by quadrature, but the integrals quickly become unwieldy.
For example, for $n=1$, $k>0$ I find
$$U_{1,k}=2^{-\frac{k}{2}} \left[2^{k/2}+\sqrt{e} \,\Gamma \left(1-\tfrac{1}{2}k,\tfrac{1}{2}\right)\right],$$
with $\Gamma$ the incomplete Gamma function.
If you are satisfied with the expectation value of the logarithm of the determinant, then you can use the Marchenko-Pastur distribution to obtain an accurate result for large $n$.
You did not specify whether you have a relation between $n$ and $k$ and whether you care about asymptotics. I will assume that $k/n\to 1$ and that $n\to\infty$, and that your scaling is such that the eigenvalues are of order $1$. (The case $k/n\to\alpha$ can be handled similarly; also, if you meant that the typical eigenvalues are of order $k$, then the situation is much simpler).
Let $\lambda_i$ denote the eigenvalues of $A$. Let $L_k=k^{-1} \sum \delta_{\lambda_i}$ denote the empirical measure. Then
$\log det(I+(I+A)^{-1})=\sum \log (1+(1+\lambda_i))^{-1}= k \langle L_k,g\rangle$ where
$g(x)=\log(1+(1+x)^{-1})$ is continuous and bounded on $R_+$. Now, $L_k$ satisfies a LDP
at speed $k^2$, which means that the probability that $L_k$ is not in a small neighborhood of the Pastur-Marchenko law $\mu$
is exponentially small at scale $k^2$. You then have that
$$k^{-1}\log E \det(I+(I+A)^{-1}) =k^{-1} \log E(e^{k \langle L_k,g\rangle})\to \langle \mu,g\rangle $$
Do you happen to know (or have a reference) if this LDP is known for empirical covariance matrices of iid Sub-gaussian vectors? (i.e. not necessarily independent entries).
I know this is very specific, so no worries if nothing comes to mind. Thank you!
Update: posted my question in longer form https://mathoverflow.net/questions/417364/
|
2025-03-21T14:48:32.093694
| 2020-09-22T14:12:31 |
372319
|
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|
Stack Exchange
|
Explicit description of the extension generated by the square root of a fundamental unit of a real quadratic field
Suppose $K={\mathbb{Q}}(\sqrt{m})$ is a real quadratic field which contains a totally positive fundamental unit $\epsilon$ (hence its norm is $1$). Suppose that $\epsilon \equiv 1 \pmod 8$, so that the quadratic extension $K(\sqrt{\epsilon})$ is unramified over $K$. In particular, it is contained in the genus field; i.e., there exists $d \mid m$ such that $K(\sqrt{\epsilon}) = K(\sqrt{d})$. Is there a recipe for $d$ in terms of $\epsilon$?
A guess: suppose $m$ is odd; then for each odd prime $p \mid m$ let $\tilde{p}$ denote the unique ideal in $K$ dividing $p$, so in particular $\epsilon \equiv \pm 1 \pmod{\tilde{p}}$. Is it true that $d = \prod p$ where the product is over the primes dividing $m$ where $\epsilon \equiv 1$ (or $-1$ as $d$ and $m/d$ give the same extension)?
This has nothing to do with ramification. Write your unit in the form $\varepsilon = t + u \sqrt{m}$; then $t^2 - mu^{2} = 1$, hence $t^2 - 1 = (t-1)(t+1) = mu^2$. If $t$ is odd, then unique factorization implies $t-1 = 2ar^2$ and $t+1 = 2bs^2$, where $ab = m$, hence $t = ar^2 + bs^2$. It is then easily verified that $(r\sqrt{a} + s \sqrt{b})^2 = t + u \sqrt{m}$, and $k (\sqrt{\varepsilon}) = k(\sqrt{a})$. The case where $t$ is even is taken care of similarly.
|
2025-03-21T14:48:32.093811
| 2020-09-22T14:35:08 |
372323
|
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|
Stack Exchange
|
Is this lattice in the Tate module of an elliptic curve, coming from complex-analytic uniformization, stable under Frobenius?
Let $E$ be an elliptic curve over $\mathbb{Q}$, and let $p$ and $\ell$ be two distinct primes of good reduction. Let $T_\ell = T_\ell(E) = \varprojlim E[\ell^n](\overline{\mathbb{Q}})$ be the $\ell$-adic Tate module, and let $F_p \in \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ be a Frobenius element at $p$. Then $F_p$ acts $\mathbb{Z}_\ell$-linearly on $T_\ell$, and this action depends only up to conjugation on the choice of $F_p$. In particular, its characteristic polynomial is well-defined. A basic result is that the coefficients of this characteristic polynomial are integers.
This last fact is usually proved by considering the reduction of $E$ modulo $p$, which does not change the $\ell$-adic Tate module, and using that we can realize the $F_p$-action in characteristic $p$ as coming from an actual morphism of elliptic curves, namely the Frobenius morphism $E \to E^{[p]}$. But I was wondering if it is possible to give a more direct proof, namely by constructing a $\mathbb{Z}$-lattice $\Lambda \subset T_\ell$ (by which I mean a rank 2 free $\mathbb{Z}$-module such that the map $\Lambda \otimes \mathbb{Z}_\ell \to T_\ell$ is an isomorphism) which is preserved by $F_p$ in the sense that $F_p(\Lambda) \subset \Lambda$ (note that one cannot expect equality here since the determinant of $F_p$ acting on $T_\ell$ is $p$). Certainly, if you already know that $F_p$ has integral characteristic polynomial, then you can easily construct such lattices: take any $t \in T_\ell \setminus \ell T_\ell$ that is not an eigenvector for $F_p$, then $\Lambda = t \mathbb{Z} + F_p(t)\mathbb{Z} \subset T_\ell$ is an $F_p$-invariant lattice. So there should be plenty such lattices. But the goal is to construct an $F_p$-invariant lattice without using that we already know that $F_p$ has integral characteristic polynomial.
One potential lattice can be constructed as follows. We choose a complex-analytic uniformization $E(\mathbb{C}) = \mathbb{C}/\Lambda_0$ for some lattice $\Lambda_0 \subset \mathbb{C}$. Then we define a map $\Lambda_0 \to T_\ell$ by sending $\lambda \in \Lambda_0$ to the sequence $(\ell^{-1} \lambda, \ell^{-2} \lambda, \ell^{-3}\lambda, \ldots) \in T_\ell$, which is well-defined because $\ell^{-n}\lambda \in E(\mathbb{C})[\ell^n] = E(\overline{\mathbb{Q}})[\ell^n]$. Let $\Lambda_\ell \subset T_\ell$ be the image of this map. It is not hard to prove that $\Lambda_\ell$ is free of rank 2 and that $\Lambda_\ell \otimes \mathbb{Z}_\ell \to T_\ell$ is an isomorphism. Also note that $\Lambda_\ell$ does not depend on the choice of the uniformization.
Question: Does $F_p(\Lambda_\ell) \subset \Lambda_\ell$ hold?
P.S. I've tried searching for results in this direction in various places, but did not find much. If someone has suggestions for references or keywords to search for, I would be much obliged.
There is a subtle problem with this idea, that causes serious problems. You observed that $\Lambda_\ell \otimes \mathbb Z_\ell = T_\ell$ but didn't find any other information for it. There is a reason for that.
Let $K$ be the field generated by the coordinates of the $\ell$-power torsion points of $E$. Given an $\ell$-power torsion point defined over $F$, to make Frobenius act on it, we need to know its reduction mod $p$, so we need to embed $F$ into the maximal unramified extension $\mathbb Q_p^{ur}$ of $\mathbb Q_p$.
Given a point in the homology of $E_{\mathbb C}$, to find the corresponding point of $F$, we need to express the coordinates as complex numbers, so we need to embed $F$ into $\mathbb C$.
Are these embeddings canonical? Well, if we define $F$ as the field generated by the complex coordinates of $\ell$-power torsion points, then the second embedding is canonical but the first isn't. If we define $F$ as the field generated by the $p$-adic coefficients of torsion points, then the first embedding is canonical but the second isn't. So regardless, there is some ambiguity - we can translate one of our embeddings by an automorphism of $F$ and get one that looks equally reasonable.
How bad is that ambiguity? Fixing an automorphism $\sigma \in \operatorname{Gal}(F/\mathbb Q)$ of $F$, making this change of embeddings corresponds exactly to translating your lattice by the action of $\sigma$ on $T_\ell(E)$. So the set of lattices we obtain your construction is a $\operatorname{Gal}(F/\mathbb Q)$-orbit in $T_\ell(E)$.
For $E$ generic, we have $\operatorname{Gal}(F/\mathbb Q) \cong GL_2(\mathbb Z_\ell)$, so the orbit is quite large. In fact every single lattice $\Lambda$ with $\Lambda \otimes \mathbb Z_\ell = T_\ell$ lies in this orbit, because we can find a matrix in $GL_2$ taking the basis of one such lattice to another. So there is no more information available about these lattices than your initial observation that $\Lambda \otimes \mathbb Z_\ell = T_\ell$!
Of course, there are examples of such $\Lambda$ stable under $F$ and examples not stable under $F$.
For any $E$ non-CM, the situation is the same, because the Galois group is an open subgroup of $GL_2(\mathbb Z_\ell)$ and these act transitively on the set $GL_2(\mathbb Z_\ell)/GL_2(\mathbb Z)$ of lattices $\Lambda$, since $GL_2(\mathbb Z)$ is dense in $GL_2(\mathbb Z_\ell)$.
For $E$ CM, the situation is different, as the Galois group is much smaller. If $p$ is a supersingular prime, then David Speyer's argument shows $\Lambda_\ell$ is never stable under Frobenius. Conversely, if $p$ is an ordinary prime, then the endomorphism $V =p /F$ lifts to an endomorphism of the curve over the CM field and thus an endomorphism of the curve over $\mathbb Q$, thus always preserves $\Lambda_\ell$, and because its determinant is $p$, $F= p/V$ necessarily preserves $\Lambda_\ell$ as well. So for CM curves, Frobenius preserves this lattice if and only if $p$ is ordinary.
Thanks! You are right, the question is subtly but seriously problematic. I actually mentioned in the question already that the action of $F_p$ is well-determined only up conjugation, but somehow I did not realize that this makes the question of whether $F_p(\Lambda_\ell) \subset \Lambda_\ell$ completely meaningless in most cases.
Any construction along these lines is going to run into an obstruction pointed out by Serre. Consider the elliptic curve $E = \{ y^2 = x^3+x \}$ over $\mathbb{Z}[i]$, and let $p$ be a prime which is $3 \bmod 4$. Let $E/p$ be the reduction of $E$ modulo $p$ (which remains prime in $\mathbb{Z}[i]$). Then $E/p$ has the following endomorphisms:
The $p$-power Frobenius $F(x,y) = (x^p, y^p)$ and
The complex muliplication $J(x,y) = (-x, iy)$.
These maps obey $JF=-FJ$, $J^2 = -1$ and $F^2 = -p$.
There do not exist $2 \times 2$ integer matrices obeying these relations. (Proof below.) So there is no construction which associates a $\mathbb{Z}$-lattice to an elliptic curve and is functorial in charateristic $p$. So it is impossible that $J$ and $F$ both preserve your lattice. I didn't think about this in detail, but it seems much more likely that $J$ does than $F$.
Proof that there are not integer matrices obeying $J^2 = -1$, $JF = -FJ$ and $F^2 = -p$: Suppose otherwise. Using $J^2 = -1$, we can choose bases so that $J = \left[ \begin{smallmatrix} 0&-1 \\ 1&0 \end{smallmatrix} \right]$. The equation $JF=-FJ$ means that $F$ is of the form $F = \left[ \begin{smallmatrix} a&b \\ b&-a \end{smallmatrix} \right]$. Then $F^2 = (a^2+b^2) \mathrm{Id}$. There is no solution to $a^2+b^2 = -p$ in integers.
$J$ certainly preserves the lattice because it extends to an endomorphism over the complex numbers. So indeed $F$ must not.
Thanks! This argument is quite famous and I have actually seen this argument before, so I should have known that what I was trying to do had no chance of ever working...
|
2025-03-21T14:48:32.094392
| 2020-09-22T16:16:11 |
372332
|
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|
Stack Exchange
|
Can quotient space be isomorphically isometric to some closed subspace of original one?
Suppose $\mathcal{B}$ is a Banach space which is not isomorphic to a Hilbert space. What condition should I pose on $\mathcal{B}$ such that for every closed subspace $\mathcal{M}$, the quotient space $\mathcal{B}/\mathcal{M}$ is isomorphically isometric to some closed subspace of $\mathcal{B}$?
Edit---------------
The comment of M.González looks really good to me. I am considering a revised problem:
Suppose $\mathcal{B}$ is a Banach space which is not isomorphic to a Hilbert space. What condition should I pose on $\mathcal{B}$ such that for every finite dimensional subspace $\mathcal{M}$, the quotient space $\mathcal{B}/\mathcal{M}$ is isomorphically isometric to some closed subspace of $\mathcal{B}$?
Thanks a lot!
This is not quite what you've asked, but just for completeness's sake: If $\mathcal{B}$ has the property that any arbitrary closed subspace $\mathcal{M}$ is complemented, then $\mathcal{B}$ is already isomorphic to a Hilbert space. In other words, you're asking for situations in which $\mathcal{B}/\mathcal{M}$ is isometric to a subspace but that subspace isn't (in general) a complement of $\mathcal{M}$.
@JohannesHahn Right. That is called Lindenstrauss-Tzafriri theorem.
Do you want conditions on $\mathcal B$ that work for all closed subspaces $\mathcal M$?
@YemonChoi Yes.
Perhaps, you get an idea from non-examples like $\ell^1$ which has every separable Banach space as a quotient.
$\mathcal{B}$ separable containing isometric copies of all separable spaces, like $C[0,1]$.
There is some ambiguity in the phrasing of your revised question. When you are asking for $B/M$ to be isometrically isomorphic to some closed subspace of $B$, are you in fact asking for an isometry $B/M \to B$ which is a right inverse to the quotient map $B\to B/M$? That is a much stronger condition, because it implies all closed subspaces of B are complemented
I am not sure I completely understand the question, but every separable Banach space is a quotient of $\ell_1$ (I know this as the Banach-Mazur theorem) but it is well known that $\ell_1$ does not contain a copy of $c_0$, say because it has the Schur property. This can all be found in a good textbook on Banach space theory.
@G.Rodrigues Sure, I think Jochen mentioned this as a "non-example" to illustrate some constraints
@YemonChoi In the reviesed one, I suppose M is any finite dimensional subspace of B instead of all of closed subspaces. The isomorphical isometry I wanna construct is between B/M and N in the question.
Per @YemonChoi's comment, I think your question should be "What conditions should I impose on $\mathcal B$ such that, for every finite-dimensional subspace $\mathcal M$, …". That is, the condition is chosen before $\mathcal M$.
@LSpice My bad. Thank you for pointing out this. I have revised it. Hope it is clear now.
|
2025-03-21T14:48:32.094609
| 2020-09-22T16:54:28 |
372334
|
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|
Stack Exchange
|
A generalisation of the Catalan numbers
Let $n$ be a nonnegative integer. It is well-known that the number of lattice paths from $(0, 0)$ to $(n, n)$ with steps $(0, 1)$ and $(1, 0)$ that are never rising above the line $y=x$ is given by the Catalan number $C_n$. There are lots of generalisations of Catalan numbers in the literature; however, I have a different approach to these numbers which I could not manage to find any known results about.
Question: Let $a$ and $b$ be nonnegative integers. What is the number of lattice paths from $(0, 0)$ to $(a, b)$ with steps $(0, 1)$ and $(1, 0)$ that are never rising above the line $bx-ay=0$?
Let $C(a, b)$ be the integer valued function that counts the number of such lattice paths. It is clear that
$$
C(n,n) = C_n = \frac{1}{n+1}\binom{2n}{n}
$$
for every nonnegative integer $n$. We also know that $C(a,b)=C(b,a)$, which can be obtained by setting up a bijection between the two sets of such paths by means of reflecting the paths across the line $y=x$, so we can take $a≤b$ without loss of generality.
In the case of $a \mid b$, there is a well-known solution to the problem that is discussed by Hilton and Pedersen, Sands and possibly many others:
$$
C(a,ma) = \frac{1}{ma+1} \binom{a(m+1)}{a}.
$$
Besides, there is a powerful enumeration tool called The Cycle Lemma (see Dvoretzky and Motzkin) which can be used to prove the equations mentioned above. Furthermore, it is possible to solve the problem thanks to the Cycle Lemma when $a$ and $b$ are coprime integers:
$$
gcd(a,b)=1 \Rightarrow C(a,b) = \frac{1}{a+b} \binom{a+b}{a}.
$$
Unfortunately, none of my findings could prove itself useful in finding a general solution to the problem. Are there any existing papers or known results that shed some light on this question? If not, how can we achieve an explicit formula of $C(a,b)$?
You might want to check the natural refinements of the Catalan numbers https://oeis.org/A125181 and the signed version https://oeis.org/A134264.
The $C(a,ma)=C(a,ma+1)$ thing seems kinda special to me, and not representative of the general one runs into in the non-coprime case (btw, these are often called 'Fuss-Catalan numbers') .
The sequences of Fuss-Catalan #s are mentioned in A134264 with a link to the first sequence, https://oeis.org/A001764. A relation to Dyck paths falls out of the connections among the NC-partitions, Dyck paths, and compositional inversion.
In the non-coprime case the answer is much more complicated. See "Rational Dyck Paths in the Non Relatively Prime Case" by Gorsky, Mazin, and Vazirani. They explain that it is a result of Bizley that
$$ \sum_{d\geq 0} C(da,db)z^d = \mathrm{exp}\left(\sum_{d\geq 1} \frac{1}{d(a+b)}\binom{d(a+b)}{da}z^d \right) $$
for coprime $a, b$, which is the best it seems one can do.
The paper of Bizley is: https://doi.org/10.1017/S002026810005424X.
Interestingly, the Gordita et. al. paper is from 2017 and the paper of Bizley is from 1954.
Gordita!?! More seriously, "rational Catalan combinatorics" is a hot topic of late...
Gorsky...@#$#^ autocorrect. It was the 1954 that I thought was interesting.
The argument from that 1954 is not too difficult; but note that $\frac{1}{d(a+b)}\binom{d(a+b)}{da}$ are not always integers (e.g. $\frac{1}{2(2+5)}\binom{2(2+5)}{2*2}$) so it's a little more complicated than a straightforward exponential generating function argument, which is what I initially thought it was.
|
2025-03-21T14:48:32.094950
| 2020-09-22T18:58:06 |
372341
|
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|
Stack Exchange
|
Probability distribution optimization problem of distances between points in $[0,1]$
Let $\mathcal{D}$ be a probability distribution with support $[0,1]$. Let $X, Y, Z$ three i.i.d. random variables with distribution $\mathcal{D}$, and $T$ a random variable uniformly distributed in $[0,1]$ independent from $X$, $Y$ and $Z$. We define $$\Delta=\mathbb{E}\left(1-|x-y|~\big|~x,y<t<z\right)$$ and $$\Delta'=\mathbb{E}\left(1-\min\left(|x-y|,|z-y|\right)~\big|~x,y<t<z\right)~.$$
Question: What is the minimum value of the ratio $\rho=\frac{\Delta}{\Delta'}$ over all probability distributions $\mathcal{D}$?
(If $\mathcal{D}$ is uniform, then $\rho=\frac{16}{17}$. Is there a distribution $\mathcal{D}$ such that $\rho<\frac{16}{17}$?)
Are $X,Y,Z$ independent (it seems like you're assuming that but you don't say explicitly)? Similarly, $T$ independent from $X,Y,Z$?
Yes, thank you @SamHopkins
Random check: if $\mathcal{D}$ is supported on ${0,3/4,1}$ and takes values $0$ and $3/4$ equally often, then I think I calculated $\rho=\frac{32}{33}$.
Thank you @SamHopkins In this case, the result that I calculated seems to depend on how often $0$ and $3/4$ are taken. I think there is misunderstanding that I would like to clarify. There are are only $5$ cases satisfying the condition $x,y<t<z$, which I represent with triplets of variables taken values in the support ${0,3/4,1}$: ${x,y,z}$, ${y,x,z}$, ${(x,y), z, \cdot}$, ${(x,y),\cdot, z}$, ${\cdot,(x,y), z}$, right? How did you calculate $\rho=\frac{32}{33}$?
Sorry, my computation in the comments was wrong. I think it leads to something with $\rho < \frac{16}{17}$.
Namely, let $\mathcal{D}$ be the distribution with $\mathrm{Pr}(\mathcal{D}=0)=\mathrm{Pr}(\mathcal{D}=3/4)=1/N$, and $\mathrm{Pr}(\mathcal{D}=1)=(N-2)/N$, where $N$ is large.
Then the possibilities for $(x,y,z)$ which fit your conditional probability are:
$0 < t < \frac{3}{4}$: $(0,0,\frac{3}{4})$, $(0,0,1)$
$\frac{3}{4} < t < 1$: $(0,0,1)$, $(0,0,1)$, $(0,\frac{3}{4},1)$, $(\frac{3}{4},0,1)$, $(\frac{3}{4},\frac{3}{4},1)$
Only one of these has $z\neq 1$; if $N$ is very large, then that case will occur much less frequently and we can "ignore" it (so we're really doing the limit $N\to \infty$ computation, for convenience).
Let $\delta=1-|x-y|$ and $\delta'=1-\min(|x-y|,|z-y|)$. Then the events to consider, and their probabilities and values, are
$0 < t < \frac{3}{4}$: $(0,0,1)$ - relative prob. $\frac{3}{7}$, $\delta=\delta'=1$
$\frac{3}{4} < t < 1$: $(0,0,1)$ - relative prob. $\frac{1}{7}$, $\delta=\delta'=1$
$\frac{3}{4} < t < 1$: $(0,\frac{3}{4},1)$ - relative prob. $\frac{1}{7}$, $\delta=\frac{1}{4}$, $\delta'=\frac{3}{4}$
$\frac{3}{4} < t < 1$: $(\frac{3}{4},0,1)$ - relative prob. $\frac{1}{7}$, $\delta=\delta'=\frac{1}{4}$
$\frac{3}{4} < t < 1$: $(\frac{3}{4},\frac{3}{4},1)$ - relative prob. $\frac{1}{7}$, $\delta=\delta'=1$
So we can compute
$$\Delta=\frac{3}{7}+\frac{1}{7}+\frac{1}{7}(\frac{1}{4})+\frac{1}{7}(\frac{1}{4})+\frac{1}{7}=\frac{11}{14}$$
$$\Delta'=\frac{3}{7}+\frac{1}{7}+\frac{1}{7}(\frac{3}{4})+\frac{1}{7}(\frac{1}{4})+\frac{1}{7}=\frac{12}{14}$$
$$\rho=\frac{\Delta}{\Delta'}=\frac{11}{12}< \frac{16}{17}$$
As mentioned, really we took the limit $N\to \infty$; but since we got $\rho< \frac{16}{17}$, that means there should be some finite $N$ we can take with $\rho< \frac{16}{17}$, just the computation will be more annoying.
Suppose $m=3/4$ is allowed to vary as well. Then my calculations suggest that $\rho=1-3/35 < 11/12 = 1-3/36$ when $m=4/5$, and that the minimum over all $m$ is $\sqrt{2}-1/2$ at $m=3/2-1/\sqrt{3}$. I think one can get below $8/9$ if you also vary the relative probability of $0$ and $m$.
@aorq: indeed, I did not try to optimize here.
Thank you very much @SamHopkins and aorq ! This problem originated from its "symmetric" version that I just posted here: https://mathoverflow.net/q/372688/115803. Initially, I posted this problem thinking I could easily extend any answer to its "symmetric" (original) version. However, it seems that it is not possible, and I think that the minimum $\frac{16}{17}$ is attained by the uniform distribution $\mathcal{D}$ in the symmetric version. Do you have any idea about whether it is possible to extend this result to the symmetric (original) problem? Thanks.
|
2025-03-21T14:48:32.095212
| 2020-09-22T19:20:59 |
372343
|
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|
Stack Exchange
|
Do pseudodifferential operators represent all physically meaningful quantities in quantum mechanics?
(Qualifier: I know virtually nothing about quantum mechanics)
In classical physics, Newton's laws guarantee that any physically relevant quantity is a function of the position and momentum of the particles in a system studied. Given a function $a(x,\xi)$ of position and momentum, quantizing gives a psuedodifferential operator $a(X,D)$ which somehow generalizes the function $a$ to operate on the wave functions of particles. Does this idea resulting from Newton's laws generalize to quantum mechanics, i.e. do we expect any physically meaningful quantity in quantum mechanics to be represented by a pseudodifferential operator? Are there principles in quantum mechanics that describe this property?
This is appropriate for Physics StackExchange (and the question needs more clarification), or opening a book on QM looking for the buzzword "observables".
Quantum mechanics is far from a well-defined mathematical theory, where such questions could be answered...
If I were particularly prone to point out irony, I might suppose that the OP used the word "quantize" without realizing its origin. But perhaps a more charitable interpretation of the question is this: Do all quantum observables (physically relevant quantities) arise by quantization? The answer may be both Yes and No, depending on how strict one wants to be with the terminology.
I’m voting to close this question because it belongs on physics.se
@IgorKhavkine I admit I know much more about harmonic analysis than quantum mechanics (I'm coming at the theory more as intuition for understanding psuedodifferential operators) :P. My understanding is that quantization is when one generalizes classical measurements to measurements in quantum mechanics. Where might I find discussion on whether observables arise by quantization?
Unfortunately, the last comment hasn't helped me understand how to interpret your question. In your own question, the sentence "Given a function..." is itself a discussion of how observables arise by quantization. Would pointing out the fact that $a(X,D)$ is in fact a quantum observable be enough to close the loop between your question and the information you already possess? Anything detailed discussion of "measurements" would probably involve too much physics and only be a red herring to your intuition.
The spin degree of freedom is generally not represented by pseudodifferential operators (though such a representation can be constructed a posteriori, as pointed out by Francois Ziegler in comments). What matters ultimately are representation-independent statements, i.e., the operator algebras. I don't know whether any physically relevant algebra could in principle be represented using pseudodifferential operators, but spin is a ready example where such a representation at least does not initially arise from the underlying physics.
This I feel is really misleading. The representations of $\mathrm{SU}(2)$ describing spin arise by geometric quantization in holomorphic sections of a line bundle over the classical 2-sphere (Borel-Weil theory), and there the components of spin are represented by differential operators in a very non-“contrived” way.
@FrancoisZiegler - thank you for pointing out the construction. I'll edit and dispose of "contrived". What I was trying to express is that representations of the spin operators weren't originally arrived at by "quantizing" some classical coordinates using differential operators. The representation through differential operators is constructed a posteriori, and in practice seldom invoked in physics applications.
Historically you are quite right. For the longest time the claim was made that spin had no “classical counterpart” of which it would be the “quantization”. Same for $\mathrm{SU}(n)$ with isospin, hypercharge and other baryonic quantities. The development of symplectic geometry has now put us well past that stage.
|
2025-03-21T14:48:32.095487
| 2020-09-22T19:33:02 |
372345
|
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|
Stack Exchange
|
Covering R2 by rectifiable curves and null sets
We have a family indexed by $t \in \mathbb R$ of rectifiable Jordan curves $(\gamma_t)$, such that $\bigcup_{t\in \mathbb R}\gamma_t = \mathbb R^2 \setminus \{0\}$. Moreover, the family is monotone, in the sense that $\gamma_t$ is contained in the closure of the bounded connected component of $\mathbb R^2 \setminus \gamma_T$, for all $t < T$. But they are not necessarily disjoint.
For each $\gamma_t$, there is a set $E_t$ with $0$ arclength measure on $\gamma_t$ and we know that $E : =\bigcup_{t \in \mathbb R} E_t$ is Lebesgue measurable in $\mathbb R^2$.
Can we conclude that $E$ has zero Lebesgue measure?
In the case where $\gamma_t$ are concentric circles centered at $0$, this is just the Fubini-Tonelli theorem. It is very tempting to believe that it is also true assuming only $\gamma_t$ to be rectifiable...
Thanks for your help!
This is false. Look at the article "Fubini Foiled" by Milnor, Math Intelligencer 19, 1997.
Thanks a lot, this is very valuable.
What would be the minimal assumption of the foliation about the dependence on t that would make E zero measure? I think if we have Lipschitz dependence on t then Fubini applies, but anything weaker than Lipschitz? @PabloLessa
I'm not an expert but as far as I understand it you need the holonomy transformations to be absolutely continuous. I'm not sure what other conditions guarantee this. I think a famous result of Anosov's thesis is that stable foliations of C^2 hyperbolic diffeomorphisms satisfy this in spite of being only Hölder continuous. I would look at "On the ergodicity of partially hyperbolic systems" by Keith Burns and Amie Wilkinson and check out the introduction and references.
Thank you for the additional information!
|
2025-03-21T14:48:32.095649
| 2020-09-22T20:30:54 |
372349
|
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|
Stack Exchange
|
Are there infinitely many L-rigs?
$\DeclareMathOperator{\Q}{\mathbb{Q}}$Call "L-rig" any class $\mathcal{L}$ of L-functions of automorphic representations of $\operatorname{GL}_{n}(\mathbb{A}_{\Q})$ for some $n$ belonging to the Selberg class that be closed under both the usual product (which we'll denote by $\times$) and the Rankin-Selberg convolution (which we'll denote by $\otimes$), containing the respective neutral elements $s\mapsto 1$ and $\zeta$, and fulfilling the abstract algebraic properties making $(\mathcal{L},\times,\otimes,s\mapsto 1,\zeta)$ a rig (ring without negatives).
Does the main result in
automorphy of $m$-fold tensor products of GL(2), Dieulefait 2020
imply the existence of infinitely many non trivial L-rigs?
Moreover, denoting by $\mathcal{M}$ the maximal L-rig under inclusion, can we see it as the analogue for L-rigs of the separable closure of a field? If yes, would it make $\operatorname{Aut}(\mathcal{M})$ isomorphic to some absolute Galois group like, say, $\operatorname{Gal}(\bar{\Q}/\Q)$?
Edit October 25th, 2020: there are at least 3 different L-rigs, namely the trivial one $\mathcal{L}_{0}$ generated by $s\mapsto 1$ and the Riemann Zeta function, $\mathcal{M}$ and its sub-L-rig $\mathcal{D}$ consisting of all self-dual L-functions. Assuming $\operatorname{Aut}(\mathcal{M})$ is isomorphic to some absolute Galois group and the analogue for L-rigs of the fundamental theorem of Galois theory, this absolute Galois group can't be finite (as all such Galois groups are of order at most $2$). It may then be possible to prove that $\operatorname{Aut}(\mathcal{M})$ is profinite.
Edit October 30th, 2020: perhaps a way to show we face a profinite group would be to prove that $\displaystyle{\mathcal{M}}$ is defined by a filtration $(\mathcal{L}_{i}):={(\mathcal{L}(F_{i}))}_{i\in I}$ so that $\mathcal{M}=\varinjlim_{i\in I}\mathcal{L}_{i}$ and $\displaystyle{\operatorname{Aut}(\mathcal{M})\cong\varprojlim_{i\in I}\operatorname{Gal}\left(\frac{\mathcal{L}_{i}}{\mathcal{L}_{0}}\right)}$, where $\mathcal{L}(F)$ is the L-rig generated by $F$, the sequence of intermediate L-rigs $\mathcal{L}_{i}$ being analogues of Galois extensions of $\mathcal{L}_{0}$ defined above.
More exactly the considered Galois group should be $\operatorname{Gal}(\mathcal{K}_{\mathcal{L_{i}}}/\mathcal{K}_{\mathcal{L}_{0}})$ with $\mathcal{K_{L}}$ the field generated by the L-ring $\mathcal{L}$, that we can call an "L-field". Proving $\mathcal{K}_{\mathcal{L}_{0}}\cong\mathbb{Q}$ may imply that $\operatorname {Aut}(\mathcal{M})\cong\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$.
Edit November 1st, 2020: following the answer by nguyen quang do in
https://math.stackexchange.com/questions/2782069/abstract-properties-of-the-absolute-galois-group-over-mathbbq and assuming $\operatorname{Aut}(\mathcal{M})$ has the structure of an absolute Galois group, then it is a profinite group.
Edit November 11th 2020: as $\mathcal{L}_{0}$ is the L-ring generated by the neutral elements, $\mathcal{K}_{\mathcal{L}_{0}}$ is the L-field generated by those neutral elements, and as such is isomorphic to $\mathbb{Q}$. Now, the maximality of $\mathcal{M}$ implies that if the extension of $\mathbb{Q}$ isomorphic to $\mathcal{K}_{\mathcal{M}}$ is algebraic, then it is "its" algebraic closure $\bar{\mathbb{Q}}$.
Edit December 12th 2020: define the "symmetry group" $\operatorname{Sym}(F_{\pi})$ of an element $F_{\pi}:s\mapsto L(\pi,s)=\prod_{v}L_{v}(\pi,s)$ of $\mathcal{M}$ as the stabilizer thereof under the action of $\operatorname{Aut}(\mathcal{M})$ on $\mathcal{M}$. Then any permutation $\sigma$ of the places $v$ leaves $F_{\pi}$ invariant, so that if it induces an automorphism of $\mathcal{M}$, that we'll denote by $\phi_{\sigma}$, the latter induces an isomorphism between $\mathbb{Q}_{v}$ and $\mathbb{Q}_{\sigma(v)}$ as a morphism between fields. But $v\neq v'\Longrightarrow\mathbb{Q}_{v}\not\cong\mathbb{Q}_{v'}$ and in particular, $\sigma$ induces an automorphism of $\mathbb{R}$, (when $v$ is the archimedean place), hence either the identity or the complex conjugation. Hence $\operatorname{Sym}(F_{\pi})\cong\operatorname{Gal}(\mathbb{C}/\mathbb{R})$ if $\pi$ is self-contragredient, and is trivial otherwise.
What if I start with 1, zeta, and the Dirichlet L-function of some quadratic character. Is the rig generated by those three objects just all finite products of those functions (because the Rankin-Selberg of that Dirichlet L-function with itself is the Riemann zeta function)?
Logically, yes.
So there are infinitely many.
Feel free to post this remark as an answer.
I guess an analogous reasoning can be made with Dirichlet characters of any order? In that case there would be infinitely many non isomorphic L-rigs.
The Rankin-Selberg convolution of a quadratic Dirichlet L-function with itself is the Riemann zeta function. Therefore the rig generated by $\{1, \zeta(s), L(s, \chi_d)\}$ consists of all finite products (and powers) of $\zeta(s)$ and $L(s, \chi_d)$. In particular, there are infinitely many L-rigs.
If you start with $\{1, \zeta(s), L(s, \chi)\}$ where $\chi$ is a primitive Dirichlet character, then Rankin-Selberg convolution gives you $L(s, \chi^j)$ for any positive integer $j$. That L-rig is generated by a finite set, depending on the order of $\chi$. So you get infinitely many non-isomorphic L-rigs.
Note that if $\chi^j$ is not primitive, then $L(s, \chi^j)$ should be interpreted as the Dirichlet L-function of the inducing primitive character.
Thank you very much. Would this imply that $\operatorname{Aut}(\mathcal{M})\cong\operatorname{Gal}\left(\bar{\mathbb{Q}}/\mathbb{Q}\right)$?
I think the real issue is whether or not there is anything to be gained by thinking about the collection of all L-rigs. Does it reveal anything beyond what we already believe (namely, that we know the set of L-functions is closed under products and we believe it is closed under R-S convolution)? I don't see it pointing in a useful direction.
Well, I see one. If we consider the torsion subgroup of $\operatorname{Aut}(\mathcal{M})$ that preserve a given L-function $F$, it may be possible to prove this "symmetry group" of $F$ is isomorphic to the group of isometries of the complex plane preserving the multiset of its non trivial zeros as, loosely speaking, preserving an L-function is equivalent to preserving its (non trivial) zeros. And RH can be seen as the "minimality" of this isometry group (the latter being isomorphic to $C_{2}$ rather than to the Klein group as far as zeta is concerned).
To David Farmer: shouldn't the L-rigs generated by Dirichlet L-functions that you consider give rise to L-fields isomorphic to cyclotomic fields?
Those L-rigs may match up with products of cyclic groups, but I don't see there is any deep meaning to that.
This may give rise to L-fields that are finite extensions of $\mathcal{K}{\mathcal{L}{0}}$ of degree $n$ , through the isomorphism $\frac{\mathbb{Z}}{n\mathbb{Z}}\cong\prod_{k}\left(\frac{\mathbb{Z}}{p_{k}\mathbb{Z}}\right)^{\alpha_{k}}$, where the product is to be understood as the cartesian product, where $n=\prod_{k}p_{k}^{\alpha_{k}}$.
I don't see any useful mathematical content in that apparent connection, and I stand by my previous comment that these apparent connections are not pointing in a useful direction.
|
2025-03-21T14:48:32.096224
| 2020-09-22T20:37:45 |
372351
|
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|
Stack Exchange
|
Gcd of linear function
$\DeclareMathOperator\gcd{gcd}$Take $q\in \mathbb N$ and $X>0$ ($q$ not necessarily smaller than $X$). A sum such as
$$\sum_{d\leq X}(q,d)$$
is easily seen to be $\ll q^\epsilon (X+q)$ so that the gcd doesn't make the sum much larger than how it would be without it — the values for which $(q,d)$ are significant are rare.
If I have instead a sum like
$$\sum_{dd'\leq X}(q,d+d')$$
can I still conclude a similar bound, thinking that the $d+d'$ should give just as "random values" to $(q,d+d')$ as did $d$ to $(q,d)$? Or is this completely the wrong way to think about it?
It's of course similar to asking about
$$\sum_{\substack{dd'\leq X\\q\mid d+d'}}1$$
which seems easy enough but I'm still a bit unsure… is this even $\ll (qX)^\epsilon (X/q+1)$?
Which definition of $\ll$ are you using here? I'm used to seeing it to mean something approximating little-o notation, but for $q = 2$, the sum is around $\frac{3}{2} X$, if I'm not mistaken.
If $q = \prod p_i^{n_i}$ is held constant, then the first sum is approximately $\prod (1 + \frac{p_i - 1}{p_i} n_i) X$ for large $X$. There may be few significant large contributors, but if the number of contributors is comparable to $X$, then that can be a large contribution on its own.
$\ll $ means the usual one ($X\ll Y $ if $|X|<CY$, for some positive $C$).
hmm not sure about your second comment (if $q$ is constant then the gcd is irrelevant given the bounds i'm trying for. are you sure that's what you mean or maybe i misunderstood?
I probably should have written $X \prod (1 + \frac{p_i - 1}{p_i} n_i)$ for clarity, but otherwise yes. And that also should disprove the assertion if we instead let $q$ grow: consider $q_i = X_i = 2^i$; then the sum is $(1 + \frac{i}{2}) X$, which is eventually larger than $C (X + q)$ for any $C$.
i don't think i understand this index formula. however, in any case, $iX$ isn't larger than $q^\epsilon (X+q)$
If that's what you meant by "thinking of $q^\epsilon$ bounds as harmless", then I didn't find it clear; I think it's worth rewriting that for clarity. If that is what you meant, then you shouldn't need to add $q$ to $X$ in the first place, I think.
ok, edited. i think i still need the $q$
1. We have
$$\sum_{\substack{dd'\leq X\\q\mid d+d'\\d\leq d'}}1
=\sum_{d\leq\sqrt{X}}\sum_{\substack{d'\leq X/d\\q\mid d+d'\\d\leq d'}}1
\leq\sum_{d\leq\sqrt{X}}\sum_{\substack{c\leq 2X/d\\q\mid c}}1
\leq\sum_{d\leq\sqrt{X}}\frac{2X}{qd}<\frac{2X(1+\log\sqrt{X})}{q}.$$
We get the same bound when the roles of $d$ and $d'$ are interchanged,
hence in the end
$$\sum_{\substack{dd'\leq X\\q\mid d+d'}}1<\frac{2X(2+\log X)}{q}\ll_\epsilon\frac{X^{1+\epsilon}}{q}.$$
2. From the last inequality we infer
$$\sum_{dd'\leq X}(q,d+d')\leq
\sum_{r\mid q}r\sum_{\substack{dd'\leq X\\r\mid d+d'}}1
\ll_\epsilon\sum_{r\mid q}r\frac{X^{1+\epsilon}}{r}\ll_\epsilon (qX)^\epsilon X.$$
top! thanks a lot! ^ ^
|
2025-03-21T14:48:32.096435
| 2020-09-22T21:02:29 |
372353
|
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"url": "https://mathoverflow.net/questions/372353"
}
|
Stack Exchange
|
Estimating the alternating sum $\sum_{j \ge 1} (-1)^j e^{-j^2} j^k$
I have been trying to get a lower bound on the following alternating sum but without much success:
$$
\sum_{j=1}^T (-1)^j e^{-j^2} j^k .
$$
For small values of $k$, this is easy because the first term dominates the remaining terms.
For an upper bound, I can get an estimate using the gamma function by ignoring the sign change. (Basically this reduces to estimating the moments of some Gaussian distribution.)
In general, what I want to estimate are summations of the form
$$
2^{-n} \sum_{j>0} \binom{n}{n/2 + jM} (-1)^j (jM)^k .
$$
(It may be useful to think of $M = C\sqrt{n}$ and $k = 1$ at first, but my goal is to understand the sum for other values of $M$ and $k$ in general.)
Is there any general method for estimating this kind of alternating sums? Thanks.
Since $e^{-j^2}j^k$ is decreasing eventually, we're in the situation of the Leibniz criterion and obtain bounds from this (though perhaps not of the kind you were hoping for).
Thanks for the suggestions! Can you please explain a bit more on how to use de Moivre's formula for $\log n!$ to approximate the sum?
|
2025-03-21T14:48:32.096550
| 2020-09-22T21:17:53 |
372354
|
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"url": "https://mathoverflow.net/questions/372354"
}
|
Stack Exchange
|
2-monads for categories with a class of (co)limits
This question concerns the strictness of (co)completions, at various levels of generality.
In Blackwell–Kelly–Power's Two-dimensional monad theory, the authors state
For instance, the 2-category $\mathbf{Lex}$ of small finitely-complete categories, left-exact functors, and natural transformations is (see Subsection 6.4 below) $T\text{-}\mathbf{Alg}$ for a finitary 2-monad $T$ on Cat.
Where can be found an explicit proof that there is such a 2-monad, as opposed to a pseudomonad? This claim appears in several other papers, but I have only been able to find constructions of a pseudomonad for finite limit completion.
More generally, is there an explicit proof in the literature that, for a suitable class $\phi$ of (co)limits, there is a 2-monad on $\mathbf{Cat}$ for $\phi$-(co)completion, whose algebras and (pseudo)morphisms are $\phi$-(co)complete categories, $\phi$-(co)continuous functors, and natural transformations?
Similarly, is it known whether the small cocompletion of locally-small categories forms a 2-monad?
Most generally, are there any results that establish when a lax idempotent pseudomonad (i.e. KZ-doctrine) on a 2-category $\mathcal K$ may be replaced by a lax idempotent 2-monad on $\mathcal K$ (with isomorphic or equivalent 2-categories of algebras)?
Power–Cattani–Winskel's A Representation Result for Free Cocompletions in particular seems promising, but the characterisation result there still assumes that such 2-monads exist in the first place.
For (1), let $Cat_{\phi,s}$ be the 2-category of categories with specified $\phi$-limits, and 1-cells which strictly preserve these. It's supposed to be obvious that the forgetful functor $Cat_{\phi,s} \to Cat$ is strictly 2-monadic. You can construct the 2-monad via generators and relations just like in universal algebra. That said, I don't know where this has been done explicitly. For (2), you will need to take pseudo-morphisms to "undo" the extra strictness gotten by working with this 2-monad.
Thanks. The construction via a (2-categorical) presentation was the one I had in mind, but that taking pseudomorphisms instead of strict morphisms suffices to obtain exactly the continuous functors seems a little subtle, which is why I would have liked to seen it proven. Additionally, if this construction works for, say, finite completion, I see no reason it would not also work for small completion (with an appropriate notion of large presentation), but several people have expressed to me they would be surprised if the small cocompletion pseudomonad could be strictified into a 2-monad.
Kelly and Lack's paper On the monadicity of categories with chosen colimits answers your questions (1),(2) and (3) affirmatively. The main theorems are Theorem 6.1, 6.2 and 7.1. Their main trick is Lemma 4.1, which allows them to modify a biadjunction (and so a pseudomonad) to a strict 2-adjunction (and so a 2-monad), assuming various hypothesis. I can imagine that something like this lemma might be helpful also in answering you question (4), but I am not aware of seeing any result of that nature.
Regarding your comment on presentations: if you have any presentation for a 2-monad (for instance, for categories with finite limits) and want to check that its pseudomorphisms are the expected ones (in this case, finite limit preserving functors), it is enough to know that those expected morphisms satisfy some natural properties such as doctrinal adjunction (so you don't have to work with the presentation at all). This is described in my paper Two-dimensional monadicity, with your example discussed in Section 7.2.
Thank you, both of those references are perfect!
|
2025-03-21T14:48:32.096807
| 2020-09-22T21:47:09 |
372355
|
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"Aaron Bergman",
"Dmitri Pavlov",
"Donu Arapura",
"Ramiro Hum-Sah",
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|
Stack Exchange
|
Stacks for a string theory student
First, I'm a string theory student hoping to grasp some math involved in some physics developments.
I'm hoping to read the famous Kapustin-Witten Paper "Electric-magnetic duality and the geometric Langlands program" and the related "The Yang-Mills equations over Riemann surfaces".
The following statement serves to explain the origin of my trouble: Let $G$ be a simple complex Lie group and $C$ a Riemann surface. Geometric Langlands is a set of mathematical ideas relating the category of coherent sheaves over the moduli stack of flat $G^{L}$-bundles ($G^{L}$ is the dual Langlands group of $G$) over $C$ with the category of $\mathcal{D}$-modules on the moduli stack of holomorphic $G$-bundles over $C$.
My problem: I have working knowledge of representation theory, but I'm completely ignorant about the theory of mathematical stacks and the possible strategies to begin to learn it. What specifically worries me is how much previous knowledge of $2$-categories is needed to begin.
My background: I've read Hartshorne's book on algebraic geometry in great detail, specifically the chapters on varieties, schemes, sheaf cohomology and curves. My category theory and homological algebra knowledge is exactly that needed to read and solve the problems of the aforementioned book. I'm also familiar with the identification between the topological string $B$-model branes and sheaves at the level of Sharpe's lectures.
My questions: I'm asking for your kind help to find references to initiate me on the theory of stacks given my elementary background and orientation. I'm completely unfamiliar with the literature and the pedagogical routes to begin to learn about stacks.
Any suggestion will be extremely helpful to me.
Update: I have found the paper String Orbifolds and Quotient Stacks
very useful and explicit.
This seems relevant. Personally I got started with Gomez's article and Olsson's book, but nowadays I treat the stacks project as the canonical reference.
Thanks for your kind comment Will Chen.
You don’t need to know what a stack is to understand those papers. This will probably offend all the mathematicians here, but in this context, you can just think of them as quotient spaces where you keep track of what’s going where the action isn’t free.
But if you really want to know, I’d suggest Vistoli’s notes: https://arxiv.org/abs/math/0412512
And, while I’m here, you might find Witten’s comments in section 6 of https://arxiv.org/abs/0802.0999 to understand the role of stacks in the correspondence from the physics side.
Aaron Bergman, I sincerely appreciate the comments and references you share with me. I agree with you, It may not be crucial for a physicist to understand the precise definitions a properties of stacks to understand those papers. The problem is that I'm finding really finding terribly difficult to begin with the papers, and I suspect that the reason is probably my lack of intuition about the underlying concepts.
I was not aware of the Witten's paper you share and looks extremely helpful for my purposes. Vistoli's book also is probably what I was waiting for. My sincere gratitude to you.
@AaronBergman I can't speak for all mathematicians, but I'm not offended, since that's pretty much how I think of a stack.
@AaronBergman: Not only this description cannot offend any mathematician, but in fact, this a perfectly rigorous definition of a stack. The only two subtleties is that the spaces can be more general than manifolds (e.g., they can be infinite-dimensional), which is formally expressed using presheaves of sets, and the group that acts on a space can itself vary as you move around the space, which is formally expressed using the language of groupoids.
I still think that Vistoli's notes on descent, fibred categories, and stacks are the canonical non-infinity-categorical introduction:
Angelo Vistoli, Notes on Grothendieck topologies, fibered categories and descent theory, arXiv:math/0412512.
|
2025-03-21T14:48:32.097120
| 2020-09-22T22:11:41 |
372358
|
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"mme"
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|
Stack Exchange
|
Well-definedness of marking a Riemann surface by diffeomorphisms in the context of Teichmüller spaces
In "An introduction to Teichmüler Theory" of Yoichi Imayoshi and Masahiko Taniguchi the Teichmüller space is defined as follows: fix a compact Riemann surface $R$ of genus $g$, a marking on a Riemann surface $S$ (of genus $g$) is an orientation-preserving diffeomorphism $f:R\rightarrow S$. Two markings $(S,f)$ and $(S',f')$ are declared equivalent if and only if there is a biholomorphic map $h:S\rightarrow S'$ such that the map $g\circ h \circ f^{-1}:R\rightarrow R$ is homotopic to the identity. The equivalence classes give the genus $g$ Teichmuller space $T(R)$.
Is this definition assuming that between any two smooth surfaces there exists an orientation preserving diffeomorphism? Is this not a huge assumption? Would we loose any generality if instead of asking them to be diffeos we asked for homeos? (I don't think it matters, at least in section 1.2 and 1.3 of chapter 1).
EDIT:
I understand that it is assuming the uniqueness of smooth structures mod diffeos. In any case, how does this relate to the notion of defining the Teichmuller space declaring two objects equivalent if they is a biholomorphic map that is is homotopic to the identity? (which seems much more intuitive).
There certainly is an orientation-preserving diffeomorphism between closed surfaces of genus $g$; this is the classification of surfaces. 2) Homemomorphic smooth surfaces are diffeomorphic. This is an old, old theorem, and there is a nice writeup using newer language by Hatcher here.
Hello, thank you very much. 1) I don't get how from the classification of surfaces we can always find an orientation preserving diffeo between any two closed surfaces. We can surely find a homeo and use it so the other surface inherits the smooth structure of the other, but "a priori" without point (2) this would not yield a diffeo for any two smooth structures.
(2) I know of the fact, part of the question comes from that is seems to me that it's quite a big assumption to make. In any case, thank you and I will read the writeup.
"The classification of surfaces" means both the the topological and smooth classification. You can prove it smoothly either by largely the same arguments, taking care with the corners, or by way of Morse theory --- you get a handlebody decomposition and you can do handleslides and handle cancelations to put your decomposition into a standard form.
Okay, thank you very much!!
|
2025-03-21T14:48:32.097316
| 2020-09-22T22:49:48 |
372360
|
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|
Stack Exchange
|
A function with a dense set of periods and two values, except for a countable set, is a constant a.e. with respect to Lebesgue measure
I am reading the book Vector Measures of Diestel and Uhl, especifically example 6 of Sierpinski in Chapter 2, about the construction of a function that is weak$^*$-measurable but not weakly measurable. As part of the construction, the authors find a function $\phi$ on $[0,1]$ that only takes values 0 and 1, except for the countable set of dyadic numbers. They also show that $\phi$ has a dense set of periods, especifically, that $\phi(t+d)=\phi(t)$ for every $t, t+d \in[0,1]$ with $d$ dyadic. If $\phi$ were Lebesgue measurable they conclude that $\phi$ is a constant $k$ almost everywhere with respect to Lebesgue measure, but I cannot see why. For all I see, a constant $k$ repeats at least countable times, but why almost everywhere? Any help explaining this will be very appreciated.
The function is only defined on $[0,1]$, so $t+d$ outside that is excluded.
I have also tried using that the measure of $\phi^{-1}({0})$ is positive or the measure of $\phi^{-1}({1})$ is positive, but I still find no clarity as to why the measure of one of those is 1.
I am also stuck using outer regularity and inner regularity of Lebesgue measure.
This looks suspicious under these assumptions. What about starting with the standard example of a non-measurable set, which has the property that its rational translates partition $\mathbb R$, and then taking $\phi^{-1}(1)$ as its dyadic translates only. This set feels non-measurable also. Maybe $\phi$ has additional properties that you haven't mentioned?
@ChristianRemling I forgot to mention that Diestel and Uhl assume that $\phi$ is Lebesgue measurable (to get a contradiction). From the properties of $\phi$ they also obtain that $\phi(1-t)=1-\phi(t)$ when $t$ is non-dyadic. Now, if $\phi$ were Lebesgue measurable, they would have it is a constant $k$ almost everywhere (this is the part I don't get). Then they would have that $k=1-k$, so $k=1/2$, instead of one or zero. Thus $\phi$ is not Lebesgue measurable.
In this version, this is a quick consequence of Lebesgue's density theorem: let $A=\phi^{-1}(1)$. Almost every point of a measurable set is a point of density. If we had $0<|A|<1$, then both $A$ and $A^c$ have points of density, but this can't be true here because we can take a part of $A$ that almost fills a small interval and use dyadic translates to maneuver this to a point of density of $A^c$.
@ChristianRemling Ok, I get it now. Thank you!
|
2025-03-21T14:48:32.097517
| 2020-09-22T23:27:44 |
372363
|
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|
Stack Exchange
|
Solutions to nonhomogeneous quadratic equation mod $N$
Is there any way to find non-trivial solutions to the equation $x^2 + y^2 - x \equiv 0 \mod{N}$? There are clearly several trivial solutions, for example $(x, y) = (0, 0), (1, 0), (2^{-1}, 2^{-1}), (2^{-1}, -2^{-1})$. By trivial solution, I mean one that holds for all choices of $N$ and hence do not tell us anything about the structure of any specific $N$. In this problem, assume the factorization of $N$ is not known.
The number of different solutions may be proportional to $N$. Are you interested in just one?
Sure, I'd take any solution. Suppose $N = 1234567$. Can you find any nontrivial $(x, y)$?
In order to get solutions of the congruence you are interested in let us consider the equation $x^2+y^2-x-Nz=0$. Using the trivial solution $x=1, y=0, z=0$, we parametrize all rational solutions by taking $x=t+1, y=ut, z=vt$, where $t, u, v$ are rational parameters. Note that for $t=0$ we get our trivial solutions $(x,y,z)=(1,0,0)$. Solving the resulting quadratic equation for $t$, we see that $t = \frac{Nv-1}{u^2 + 1}$, hence
$$
x=\frac{Nv+u^2}{u^2+1},\quad y=\frac{u (Nv-1)}{u^2+1},\quad z=\frac{v (Nv-1)}{u^2+1}.
$$
Since we are interested in integral solutions of the congruence, it is enough to choose integers $u$ in such a way that $u^2+1$ is coprime to $N$. If this condition is satisfied, then we compute $(u^2+1)^{-1}\pmod{N}$ and get the solutions. Note that, to do this we don't need to have a factorization of $N$.
For example, if you take $N=1234567$ it is enough to take $u=2$ and say $v=1$. Then $(u^2+1)^{-1}\pmod{N}$ is equal to $493827$, and the solution (after reduction $\pmod{N}$) is
$$
x=740741, \quad y=246913.
$$
I don't quite follow your answer. I verified that your choice $(x, y)$ is indeed a solution to the equation for $N = 1234567$. Can you explain what you mean by "using the trivial solution $x = 1, y = 0, z = 0$? How did you use this solution? Also, if we take $y = ux$, $z = vx$, we obtain the equation $x^2 + u^2x^2 - x - Nvx = 0$, whose solution is $x = \frac{vN + 1}{u^2 + 1}$, which is not the solution you described... I think your answer might have some small typos. I'm really interested to hear what you figured out, though!
@Gautam Of course you are right. I was to quick. I edited the answer and believe that everything is clear now.
Thanks, this is now clear. Does this capture all solutions? Intuitively, the family of solutions should depend on only one parameter, since there are two unknowns ($x$ and $y$) and one constraint. This leads me to believe that your proposal indeed captures all possible solutions, but I'm not sure how to prove it.
@Gautam My parametrization is invertible over $\mathbb{Q}$ but not necessarily modulo $N$. Indeed, the inverse is given by $u=y/(x-1), v=z/(x-1)$ (here $z=(x^2+y^2-x)/N$) and can be computed modulo $N$ provided that $x-1$ is coprime to $N$.
|
2025-03-21T14:48:32.097985
| 2020-09-22T23:54:36 |
372365
|
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|
Stack Exchange
|
Frobenius reciprocities
An adjunction of the form $\mathrm{Hom}(A \otimes X, Y) \cong \mathrm{Hom}(X, A^* \otimes Y)$ in a rigid monoidal category is sometimes called Frobenius reciprocity. Is there a result that unifies this adjunction with the familiar adjunction between the induction and the restriction of representations? What explains this coincidence in terminology?
Frobenius reciprocity is a tensor-hom adjunction: https://en.wikipedia.org/wiki/Tensor-hom_adjunction
|
2025-03-21T14:48:32.098055
| 2020-09-23T00:07:19 |
372366
|
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|
Stack Exchange
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Equation $u_t - u_{tx} - u_{xx} = 0$
Consider the following heat equation with a perturbation given by a second order mixed derivative:
$$u_t - u_{tx} - u_{xx} = 0$$
Does this equation have a name? How can one prove a wellposedness result for it?
I am not aware of a name for it. How you would prove wellposedness depends on what boundary conditions you want to add to it. If you consider it on the whole real line, Fourier transform in x will give you easy results.
Doing the change of variables $s = 2t - x$, $y = x$, in the $s, y$ coordinates your equation reads $2 u_s + u_{ss} - u_{yy} = 0$. If you take $s$ to be the "time variable" this is just a damped wave equation.
the equation with $u_{tx}$ replaced by $u_{txx}$ appears in many physical problems (search for "heat conduction involving two temperatures"); in what context does your equation arise? It is not invariant for $x\mapsto -x$, which seems strange.
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2025-03-21T14:48:32.098164
| 2020-09-23T05:34:52 |
372377
|
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Stack Exchange
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Subgroups of algebraic groups containing regular unipotent elements
Let G be a simple algebraic group. Let H be a reductive subgroup of G which contains a regular unipotent element of G. Such subgroups were classified by Saxl and Seitz in all good characteristics. I'm actually interested in the characteristic zero version of this result, which apparently goes back to Dynkin. Saxl–Seitz and Dynkin are difficult to read.
I'm wondering if there exists a modern reference for this classification over complex numbers. Or better, could some one please provide the sketch of an argument?
For quick reference, the classification is stated just after Proposition 8 in this paper.
You can read about classification of reductive subgroup of $G$ in Part II of the book "Linear algebraic groups and finite groups of Lie type" by Gunter Malle and Donna Testerman, Cambridge University Press, 2011.
@MikhailBorovoi Thank you for the reference. I did not find the answer to my question in there. They don't seem to be concerned with regular unipotent elements in this book. If you can point to a specific chapter containing relevant material, I would appreciate it.
I do not know a reference, but I have thought about the same question. Here is a sketch using arguments that are in the literature. For some basics about regular unipotent elements, see for example Chapter 4 of [1]. First a reminder:
$(*)$ Let $\Phi^+$ be a system of positive roots on $\Phi$. A unipotent element $\prod_{\alpha \in \Phi^+} x_{\alpha}(c_{\alpha})$ is regular if and only if $c_{\alpha} \neq 0$ for all simple roots $\alpha$.
Let $G$ be simple algebraic group over $\mathbb{C}$ and let $H < G$ be a reductive subgroup containing a regular unipotent element $u \in G$. Denote the root system of $G$ by $\Phi$, with respect to some maximal torus $T$.
Any power of $u$ is also a regular unipotent element and $H/H^\circ$ is finite, so we might as well assume that $H$ is connected. Also $H$ must be semisimple, since $C_G(u)^\circ$ is unipotent and so $u$ is not centralized by any non-trivial torus.
So $H = H_1 \cdots H_t$ is a central product of some simple algebraic groups $H_i$. Write $u = u_1 \cdots u_t$, for some unipotent $u_i \in H_i$. The subgroup generated by $u_i$ lies in the unipotent radical of $H$, so we can assume that the $u_i$ are contained in $U = \prod_{\alpha \in \Phi^+} U_{\alpha}$ with respect to some system of positive roots $\Phi^+$.
Suppose that $t \geq 2$. Then none of the $u_i$ can be regular in $G$, since they are centralized by $H_j$ for $j \neq i$. Since $u$ is regular and commutes with $u_i$, it follows from the lemma below that each $u_i$ is contained in $\prod_{\alpha \in \Phi^+ \setminus \Delta} U_{\alpha}$, where $\Delta$ is the set of simple roots. But then $u \in \prod_{\alpha \in \Phi^+ \setminus \Delta} U_{\alpha}$, which contradicts $(*)$.
Lemma: Let $u, u' \in U$. Suppose that $u$ is regular and $uu' = u'u$. Then either $u'$ is regular or $u' \in \prod_{\alpha \in \Phi^+ \setminus \Delta} U_{\alpha}$.
Proof: Write $u = \prod_{\alpha \in \Phi^+} x_{\alpha}(c_{\alpha})$ and $u' = \prod_{\alpha \in \Phi^+} x_{\alpha}(c_{\alpha}')$. Then by the Chevalley commutator formula $[u,u'] = \prod_{\alpha \in \Phi^+ \setminus \Delta} x_{\alpha}(t_{\alpha})$ for some $t_{\alpha} \in \mathbb{C}$.
Suppose that $[u,u'] = 1$.
We show that if $c_{\alpha}' \neq 0$ for some $\alpha \in \Delta$, then $c_{\beta}' \neq 0$ for any $\beta \in \Delta$ adjacent to $\alpha$ in the Dynkin diagram. To this end, from the Chevalley commutator formula we see that $$t_{\alpha+\beta} = \pm (c_{\alpha}c_{\beta}' + c_{\alpha}'c_{\beta}).$$ So the claim follows since $c_{\alpha},c_{\beta} \neq 0$ by $(*)$.
The Dynkin diagram of $G$ is connected, so the conclusion from this is that either $c_{\alpha}' = 0$ for all $\alpha \in \Delta$, or $c_{\alpha}' \neq 0$ for all $\alpha \in \Delta$. By $(*)$ the latter is same as being regular, so the lemma follows.
The above lemma is Lemma 2.4 in [2].
In any case, $H$ must be simple. Furthermore, we can show that $u$ must be regular in $H$.
Lemma: $u$ is regular in $H$.
Proof: If $u$ is not regular in $H$, then by $(*)$ there exists a non-Borel parabolic subgroup $P_H < H$ such that $u \in R_u(P_H)$. By the Borel-Tits theorem, there exists a parabolic subgroup $P_G$ of $G$ such that $P_H < P_G$ and $R_u(P_H) < R_u(P_G)$. But then $P_G$ is a non-Borel parabolic of $G$ such that $u \in R_u(P_G)$, which is a contradiction by $(*)$.
By results of Jacobson-Morozov and Kostant, in $G$ there is always a simple subgroup $H$ of type $A_1$ which contains a regular unipotent element of $G$. Such a $H$ is unique up to conjugacy in $G$, I will call it a ``regular $A_1$-subgroup''.
Then we still need to consider the case where $H$ is simple of rank $\geq 2$. I won't go through all the details, but at this point we can use some representation theory. Let $X < H$ be regular $A_1$-subgroup of $H$ (hence of $G$). The idea is that we look at the action of $X$ on a small $G$-module $V$, and this will place a heavy restriction on what $H$ and the $H$-module $V \downarrow H$ can be.
For $G$ of classical type you can take $V$ to be the natural module. In this case $V \downarrow X$ is irreducible if $G$ is not of type $D_n$, and $V \downarrow X = V_X(0) \oplus V_X(2n-2)$ in type $D_n$.
For types $G_2$, $F_4$, $E_6$, $E_7$, $E_8$ take $V$ to be an irreducible $G$-module of dimension $7$, $26$, $27$, $56$, $248$ respectively. You can compute the composition factors of $V \downarrow X$ or look at tables in the literature.
Anyway, in all cases $V \downarrow X$ is multiplicity-free with not too many composition factors. The irreducible $H$-modules $W$ such that $W \downarrow X$ is multiplicity-free have been classified in [3]. You don't need this full result since our situation is even more specific, but the basic technique in the beginning of the paper might be helpful for you.
For example, for $c$ the highest weight of $V \downarrow X$, there is no composition factor of highest weight $c-2$ or $c-4$ (so weights $c$, $c-2$, $c-4$ occur with multiplicity one). From this you can already see that in the restriction $V \downarrow H$, the only possible composition factors are $V_H(\varpi_{\alpha})$ (fundamental highest weight corresponding to $\alpha \in \Delta$), where $\alpha \in \Delta$ is an end node of the Dynkin diagram.
Then with similar arguments you can rule out all configurations except those that actually occur (for example $H = F_4$, $G = E_6$, with $V \downarrow H = V_H(\varpi_4) \oplus V_H(0)$).
[1] Humphreys, James E.: Conjugacy classes in semisimple algebraic groups. Mathematical Surveys and Monographs, 43. American Mathematical Society, Providence, RI, 1995.
[2] Testerman, Donna; Zalesski, Alexandre: Irreducibility in algebraic groups and regular unipotent elements. Proc. Amer. Math. Soc. 141 (2013), no. 1, 13–28.
[3] Liebeck, Martin W.; Seitz, Gary M.; Testerman, Donna M.: Distinguished unipotent elements and multiplicity-free subgroups of simple algebraic groups. Pacific J. Math. 279 (2015), no. 1-2, 357–382.
Thanks for the extensive answer Mikko. It appears you assume the you are assuming that H is connected. Is there a classification when H is disconnected as well?
@Dr.Evil: This is what I was hinting at in the beginning of my answer: in characteristic zero if $H < G$ contains a regular unipotent element of $G$, then so does $H^\circ$. So $H^\circ$ is one of these things we have classified, and you can check for them that $N_G(H^\circ) = H^\circ$, so in fact $H = H^\circ$. In positive characteristic you have disconnected examples which are classified by Saxl-Seitz.
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2025-03-21T14:48:32.098728
| 2020-09-23T06:28:05 |
372379
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Stack Exchange
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The evaluation map on twists of a vector bundle and an induced filtration
The following question arose while thinking about a step in the proof of Huybrechts-Lehn, Theorem 1.3.1 (the Harder-Narasimhan filtration for the projective line $\mathbb{P}^{1}$):
Setup: Let $k$ be a field, let $X$ be a projective $k$-scheme, let $\mathcal{O}_{X}(1)$ be a fixed very ample line bundle on $X$, let $E$ be a vector bundle on $X$. For an integer $b \in \mathbb{Z}$, let $\mathrm{ev}_{E(b)} : \mathrm{H}^{0}(X,E(b)) \otimes_{k} \mathcal{O}_{X} \to E(b)$ be the evaluation map and let $E_{b} \subseteq E$ be the image of the twist $\mathrm{ev}_{E(b)}(-b) : \mathrm{H}^{0}(X,E(b)) \otimes_{k} \mathcal{O}_{X}(-b) \to E$.
Question: Why is $E_{b} \subset E_{b+1}$?
(And does this inclusion depend on a choice of global section $s \in \Gamma(X,\mathcal{O}_{X}(1))$?)
Remark: Note that $\mathrm{Hom}_{\mathcal{O}_{X}}(\mathcal{O}_{X}(-b),\mathcal{O}_{X}(-b-1)) = 0$ so there is no map $\varphi : \mathrm{H}^{0}(X,E(b)) \otimes_{k} \mathcal{O}_{X}(-b) \to \mathrm{H}^{0}(X,E(b+1)) \otimes_{k} \mathcal{O}_{X}(-b-1)$ satisfying $\mathrm{ev}_{E(b+1)}(-b-1) \circ \varphi = \mathrm{ev}_{E(b)}(-b)$.
There is a commutative diagram
$$\require{AMScd}
\begin{CD}
H^0(E(b)) \otimes H^0(\mathcal{O}(1)) \otimes \mathcal{O}(-b-1)
@>{\mathrm{ev}_{\mathcal{O}(1)}}>>
H^0(E(b)) \otimes \mathcal{O}(-b)
\\
@VVV @V{\mathrm{ev}_{E(b)}}VV
\\
H^0(E(b+1)) \otimes \mathcal{O}(-b-1)
@>{\mathrm{ev}_{E(b+1)}}>>
E
\end{CD}
$$
where the left vertical arrow is the multiplication of global sections. The top arrow is surjective (because $\mathcal{O}(1)$ is very ample), hence the composition of the top and right arrows has image $E_b$. By commutativitity of the diagram it is contained (canonically) in the image $E_{b+1}$ of the bottom arrow.
Thanks very much. I think your argument shows that we only really need $\mathcal{O}(1)$ to be globally generated: Let $X$ be a scheme, let $E$ be a quasicoherent $\mathcal{O}{X}$-module, let $\mathcal{L}$ be a globally generated line bundle. Then $\operatorname{im} \mathrm{ev}{E} \subseteq \operatorname{im} (\mathrm{ev}_{E \otimes \mathcal{L}} \otimes \mathcal{L}^{-1})$.
Yes, that's correct.
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2025-03-21T14:48:32.098881
| 2020-09-23T06:43:36 |
372380
|
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Stack Exchange
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Using Nelder-Mead to solve system of polynomial equations
I am trying to solve a system of $9$ polynomial equations in $9$ unknowns over the non-negative reals.
Since the equations are quite large and I would like to use VBA, I prefer an algorithm that avoids partial derivatives. Hence, I tried to use the Nelder-Mead (downhill simplex) algorithm. Unfortunately, it doesn't converge. Before investigating my code, I would like to know if Nelder-Mead is even suitable for my task. If not, could you recommend a better algorithm?
P. S. I already solved the problem using Mathematica, so it is solvable but I would like to implement it in MS Excel.
http://scicomp.stackexchange.com
https://www.wolfram.com/products/applications/excel_link/
Minimum-finding routines (wich is what Nelder-Mead/downhill-simplex is) are generally poorly suited to finding zeros of equation systems -- if you add the squares of all equations, you get many spurious local minima in addition to the global minimum corresponding to the zero (assuming your equations have a unique zero).
@gmvh ok that is probably the reason for the poor performance
Maybe just post (a link to) your system of equations?! Are you looking for all solution of just a single soluion?
Minimum-finding routines (which is what Nelder-Mead/downhill-simplex is) are generally poorly suited to finding zeros of equation systems — if you add the squares of all equations, you get many spurious local minima in addition to the global minimum corresponding to the zero (assuming your equations have a unique zero).
If you can't afford to evaluate the Jacobian, your best hope would appear to be Broyden's method.
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2025-03-21T14:48:32.099014
| 2020-09-23T07:02:19 |
372381
|
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Stack Exchange
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What is sequential boundary of a $\delta$-hyperbolic space and how is the Gromov product extended to the boundary?
I have been reading up on $\delta$-hyperbolic spaces. But I am not getting a clear idea of sequential boundary of $\delta$-hyperbolic spaces and how the Gromov product is extended to it. Could somebody please explain it to me?
The generic name is "Gromov-hyperbolic space". $\delta$ is just a quantitative value.
There is the beginnings of a discussion, and many references, here: https://en.wikipedia.org/wiki/Gromov_boundary
The paper by Kapovich and Benakli gives a definition of the "sequential boundary". Usually this is just called the "Gromov boundary". If you have a more precise question feel free to ask!
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2025-03-21T14:48:32.099127
| 2020-09-23T07:47:52 |
372382
|
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"Noah Schweber",
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"Timothy Chow",
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Stack Exchange
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What does T+non-Cons(T) mean?
I am puzzled by the following question, which is about the philosophical meaning of some axiomatic system.
Suppose that a formal axiomatic system (containing arithmetics) $T$ is consistent. Let a statement $S$, expressible within $T$, be indecidable. Then (correct me if I am wrong) both theories $T$+($S$ is true), and $T$+(non-$S$ is true) are consistent.
According to Gödel, the consistency of $T$ is expressible but indecidable within $T$. Thus it seems that the theory $T$+(non-Cons($T$)) is consistent.
What should I understand ? The above argument let me think that if $T$ is consistent, then it is not... Something must be wrong in the few lines above, but I can't see where.
Perhaps it is best to think of $T=$Peano arithmetic. Any model of $S=T+\lnot\mathrm{Con}(T)$ contains nonstandard numbers. For any actual number $n$, $S$ proves that $n$ is not the code of a proof of a contradiction (from the axioms of $T$). However, any model of $S$ contains "fake numbers", objects that do not correspond to actual natural numbers. But $S$ thinks they do. In particular, one such $n$ "codes" a proof of a contradiction...
... but if you try to decode it you will see that this is really gibberish: it may be an infinite proof (there is no such thing), although $S$ thinks it is finite. It may use infinitely many "axioms", some of which are not real axioms, but more gibberish, or finitely many axioms, but some of them are fake. So, this theory has a peculiar interpretation of what it means to be an axiom of $T$, which is probably wrong, and of what it means to be a proof from the axioms of $T$, which is also probably wrong. It sees a contradiction, but from the outside you can tell that this is not really the case
A further remark: one may object that the consistency of, say, $PA$ is expressible in $PA$, because the predicate "being a theorem of $PA$" is not representable in $PA$ (under a recursive godelization) by Gödel's first incompleteness.
I posted an equivalent question on MSE some years ago. There have been some enlightening answeres.
There is indeed an entire book devoted to clearing up ways people get confused about this topic: Torkel Franzén, Gödel's Theorem: An Incomplete Guide to its Use and Abuse.
I highly recommend this article by Raymond Smullyan "Logicians who reason about themselves" https://www.sciencedirect.com/science/article/pii/B9780934613040500284
Link to the Smullyan article that isn't paywalled: http://citeseerx.ist.psu.edu/viewdoc/summary?doi=<IP_ADDRESS>.8548
Would the following be a valid way to restate this issue? The sentence "$S$ is expressible in $T$" isn't accurate, because $T$ gives no semantics, only syntax. Instead, we should say "the metatheoretical statement $S$ can be encoded as a statement $\phi$ which is soundly interpreted in at least one model of $T$". However, it could be (and often is) the case that this sound interpretation occurs only in models satisfying $\phi$ (but never the negation of $\phi$).
For mathematicians of an algebraic bent it may be helpful to think of a nonstandard model of, say, Peano arithmetic as a very funny sort of ring (really a semiring), obtained by starting from $\mathbb{N}$ and adjoining some nonstandard numbers, in the same way we construct $\mathbb{C}$ by starting from $\mathbb{R}$ and adjoining a square root of $-1$. A model of $PA + \text{Con}(\neg PA)$ is like "$\mathbb{N}$ adjoin the Godel code $p$ of a proof of a contradiction in PA."
It's worth saying "Godel code" because $p$ is not a proof that PA is inconsistent. What it is is something stranger: we use the fact that PA can talk about, say, Turing machines to write down a complicated first-order formula $\phi$ which, when applied to standard natural numbers, tells us which standard natural numbers are the Godel codes of a proof of a contradiction in PA. Then we adjoin a "formal solution" $p$ satisfying $\phi$.
$\phi$ has only been constructed to have a particular interpretation when applied to standard natural numbers, and as it turns out the behavior of $\phi$ on nonstandard natural numbers is otherwise pretty unconstrained. There's an analogous but simpler discussion we can have about passing from $\mathbb{R}$ to $\mathbb{C}$ by adjoining a square root of $-1$; you might imagine a 14th century mathematician being very confused about the fact that [the field axioms + the axiom that there exists an element $x$ such that $x^2 = -1$] is consistent, because they're used to a particular interpretation of what $x^2$ means (say that it measures the area of a square of side length $|x|$) that is only valid in $\mathbb{R}$, and the existence of a solution in a larger field involves a different interpretation from the one they're used to.
I am not sure what happens exactly if you attempt to decode $p$ as Andres describes in the comments. I think basically if you ask any question about the proof $p$ is supposed to be encoding which can be expressed as a first-order proposition, $p$ will just respond in whatever way it needs to to avoid a contradiction, but you won't learn anything useful. One possibility (and I haven't thought carefully through how much sense this makes) is that you can only ask questions about, say, the first $n$ lines of the proof, and $p$ will just tell you, for any particular $n$, that the first $n$ lines of the proof are totally fine, no problem here, but the proof may be "infinite" in the sense that $p$ may just sit around proving useless lemmas forever.
There's again an analogous but simpler discussion to have about the first-order properties of $i \in \mathbb{C}$: our 14th century mathematician may attempt to ask questions like "is $i$ contained in the interval $[a, b], a, b \in \mathbb{Q}$?" In $\mathbb{R}$ this question can be asked in the first-order language of fields, because in $\mathbb{R}$ we can encode $x \le y$ as $\exists z : y - x = z^2$. But this encoding only has its intended interpretation in $\mathbb{R}$: in $\mathbb{C}$ every number has a square root so "$x \le y$" is always true in $\mathbb{C}$, so our mathematician will learn that "$i$ is contained in every interval $[a, b]$," except not really, because again the construction we're using to talk about $\le$ only has its intended interpretation in $\mathbb{R}$.
The opposite sort of thing happens if we try to locate $p$: it will happily report that, for any positive integer $n$, it is greater than $n$. It does not follow, of course, that $\forall n : p \ge n$, because PA proves that no such number exists. This is already a little wacky if you're not used to it: if you add a constant symbol $p$ to PA and the axioms $p \ge 0, p \ge 1, p \ge 2, \dots$, this set of axioms is consistent (e.g. by the compactness theorem) and so has a model, which we can take to be an ultrapower of $\mathbb{N}$. It is maybe worth working carefully through why the statement $\neg \forall n : p \ge n$ continues to be true in this ultrapower: it's because we are now quantifying over all nonstandard natural numbers.
Re: the first paragraph, however, it's worth keeping in mind that building a nonstandard model of $\mathsf{PA}$ is much harder than adjoining an element to a (semi)ring: e.g. by Tennenbaum's theorem no nonstandard model of $\mathsf{PA}$ is computable. We can indeed build a nonstandard model of $\mathsf{PA}$ as the free object of a particular kind on one generator, but the relevant kind of object is far more complicated than a semiring.
@Noah: right, there are many more "operations" involved. I think the analogy is helpful as a bridge to more familiar mathematical reasoning about "ordinary" mathematical objects where there's less confusion about what various things mean.
While I don't disagree with the substance of what Qiaochu Yuan and Andrés Caicedo have said, I'm not happy with the terms "gibberish" or "useless."
It's important to bear in mind that when we say "consistency," we have a particular syntactic notion in mind, and that there is no canonical way of expressing it in terms of the semiring operations $+$ and $\times$. It is easy to glibly write down "Con" and think that it means "consistent," but in fact "Con" is an enormously complicated formula. We construct it by noting that syntactic operations on strings can be faithfully mimicked by arithmetic operations on natural numbers. Strings—sequences of symbols—can be faithfully encoded as natural numbers, and proofs—sequences of strings—can also be faithfully encoded as natural numbers. Formally, "Con(PA)" is just an assertion that a certain $x$ satisfying certain properties does not exist.
When all the dust settles, we can examine this monstrous expression "Con(PA)" and confirm that if the quantifiers are interpreted as quantifying over the natural numbers and the symbols $+$ and $\times$ are interpreted as addition and multiplication, then the resulting statement will be true if and only if PA is consistent. But it's very important to notice that "Con(PA)" does not directly "mean" that PA is consistent. It's a formal string that is getting interpreted as a statement about the natural numbers, and the statement about the natural numbers is something that we can see, because of the way we carefully mimicked syntactic operations by arithmetic operations at every step of the construction, will be true if and only if PA is consistent.
If we now interpret "Con(PA)" in terms of the elements of some other semiring, then it is certainly true that the resulting statement will no longer have the property that it will be true if and only if PA is consistent. But I think it is unfair, and more importantly misleading, to say that the statement is "gibberish." "Con(PA)" has a perfectly meaningful interpretation in any semiring satisfying the axioms of PA. What moral obligation does this other semiring have to mimic our intuitions about syntax? None, of course. If we call certain types of semiring elements "proofs" and if that choice of terminology causes us to mistakenly assume that "proofs" will always have the same properties in other semirings that they do in the natural numbers, then that's our fault for choosing terminology that confuses ourselves. Similarly, to use the word "useless" seems to presuppose that the only purpose in life a semiring might have is to satisfy our desire to understand syntax, but who are we to play God to semirings?
It's entirely conceivable to me that the theory of "generalized proofs" in "generalized natural numbers" will one day yield important mathematical insights. Admittedly they don't seem to have done so yet, but that's not a reason to dismiss it all as useless gibberish.
I think you are twisting the meaning. I clearly meant "this is not a proof", "this is not a formula", "this is not even a sequence of 'formulas' " and so on. Anyway, calling gibberish "gibberish" is not unfair to abstract concepts, that's ridiculous.
@AndrésE.Caicedo : Yes, I understand your use of "gibberish" and my use of "unfair" is in the same vein. I'm suggesting that for the purposes of disentangling the stated confusion, it is psychologically helpful to place all models on an equal footing (to the extent possible), rather than unduly privilege one of them.
I believe that is more confusing (requires more logical sophistication, to the point that the question would be unnecessary).
Anyway, the models are not in equal footing, as one is an initial segment of the other. One can imagine a hierarchy of models where each is ill-founded with respect to another one. This would be more "fair", I suppose. Still, there is no equal footing as we can identify the ordering of the models.
@AndrésE.Caicedo yes and no. Yes, if we consider the category of models of PA, then N is the initial object (there is an initial segment which is a copy of N in each of them. No, from the INTERNAL point of view of each of them. Living in a PA model one thinks that it is N (in other words, no model thinks that is is made of standard and non standard numbers) , and, what is more important, the initial segment (the "true" arithmetic) is NOT M-definable. So in this respect Tim's point of view is extremely interesting (I may write my own answer along the same line, but even more radically)
@AndrésE.Caicedo : Certainly the models are not on an equal footing in that sense. But they are equal in the sense that none of them is explicitly a syntactical entity. They're all algebraic entities. I really do believe that this is what confuses non-experts. They see "Con" and think that the formula itself expresses something syntactical. The first thing for them to grasp is that it expresses a perfectly meaningful algebraic fact in any model, and that getting it to line up with something syntactical in the standard model requires careful control.
@Mirco The point is that the claim that there is an N is a metamathematical commitment of some strength. Many of us make it implicitly, but justifying it is harder mathematically. You can imagine a setting where there is no such initial object (any N is illfounded from the point of view of a 'prior' setting). Anyway, look forward to your answer.
@AndrésE.Caicedo holy words! Should be posted on some giant billboard.. Yes, N is a HUGE metamathematical commitment. Nothing wrong with that, of course. What is wrong is sweeping it under the rug. Anyway, shall post my answer, but do not hold your breath in the meantime: right now I am busy cooking my italian risotto, and that is a sacred operation, requiring all my unflinching attention...
@Mirco :-) Enjoy!
The claim that the syntactic concept of "consistency" makes sense is also a metamathematical commitment of some strength.
Agree Tim, in fact I will tackle the bull by the horn exactly on that point: time to "contextualize" consistency!
My answer is more of an extended comment on what Tim has already said. I guess what is at stake here is precisely this question: what does it mean the expression $CON(PA)$?
The common consensus is that it is a metamathematical statement, and indeed it is.
But what is metamathematics? Answer: other mathematics. In principle there is no distinction between Proof Theory of Peano Arithmetics, and , say, the Theory of Sobolev Spaces. In proof theory one studies structures, such as proof trees, which, in their intended meaning, talk about formal proofs of an underlying math theory, in this case Peano Arithmetics. But here is the deal: thanks to a special encoding, one can express certain statements of the "meta"-theory in the base theory, and so, in the case of PA, models of such a theory have some arithmetical statements which code some metamathematical facts on the theory. One of these is the infamous CON(PA).
Now, let us imagine the category of countable models of PA, with corresponding maps. That is our "Arithmetical Multiverse". The chief point to unravel the seeming paradox raised by the PO is that PA does not know anything about actual infinity.
All models, bar none, "think" that they are made of standard numbers.
It so happens that, "from outside" (ie from the perspective of an underlying set theoretical universe), the category above has a distinguished object, namely an initial object. We call that initial object $N$. The metamathematics encoded in $N$ is the "true" metamathematics, and everything else is (again, that is common folklore), gibberish.
But let us take a slightly different look at the story:
let us assume for a moment that all models of PA are in some sense equal. Each has its own coded meta-theory. In some of them, NON-CON(PA) is true. If this statement happened to be true in $N$ it would be an earthquake, because in that case there would be a finite term witnessing a proof of inconsistency. But (assuming the consistency of PA), that is not the case. Note that, although N is an initial segment of all of the arithmetical universes, none has any idea about it, they have no capability to define it with a first-order formula. Everything looks to them just some standard arithmetics.
Now, let us play this game: to every model of PA let us assign a radius, the Radius of Consistency: it is measured by an element in the model, the minimal element (when it is there!) that proves the inconsistency of PA.
If that element does not exist, we say that the radius for that model is unbounded. Note en passant: as all models have the same order type, namely ω + (ω* + ω) ⋅ η, we can actually define the radius of consistency by its position in this order. I shall not pursue this topic further here, for the sake of brevity.
Armed with the Radius of Consistency, we can say that the conventional consistency of PA is the meta-statement that N 's radius is unbounded.
Similarly, there are models of PA who have a much larger radius. Those are exactly the models which are conjured up by the PO.
You may say: all good and well, but these statements are still gibberish to me. Perhaps they encode garbage, useless information, as far as metamathematics goes (obviously, their are legitimate statements from the point of view of algebra).
Perhaps, but not so fast. For instance, sub-theories of PA such as $I\Sigma_1$ are finitely axiomatizable, so for those arithmetical multiverses the axioms involved in the "gibberish "are in fact real axioms. And things may get even odder. I think it is fair to say that so far nobody has done a detailed analysis of these statements of inconsistencies in complete details. There may be many surprises there. But here I stop...
|
2025-03-21T14:48:32.100324
| 2020-09-23T08:30:39 |
372384
|
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|
Stack Exchange
|
Numbers that don't start with (p-1) in base p for any p
Say that an integer $n$ is $p$-leading if its expansion in base $p$ starts with the digit $p-1$. My postdoc, Lifan Guan, asks: are there infinitely many positive integers $n$ that are not $p$-leading for any odd prime $p$? That is: is the set
$$S = \mathbb{N}\setminus \bigcup_{p\geq 3} \bigcup_{d\geq 1} [p^d - p^{d-1},p^d)$$
infinite?
[My estimation is this is interesting, doable and non-trivial. I also suspect it must have been studied before!]
Have you/ has he implemented tests? I'm counting 39422 such numbers in ${0,99999}$ if I did it correctly (with Sage). Also it doesn't seem to be in OEIS, which doesn't recognize '38,39,40,41,49,50,51,52,53,81'
Yes, I recall he got something like that in the same range. The set is most likely infinite (heuristics also suggest that).
Do heuristics say anything on expected lower/upper natural density?
I'd expect the lower natural density to be positive.
... well, at least once you pass to an arithmetic progression modulo a slowly growing product of very small primes.
That's upper natural density, right? By lower and upper natural density of $A$ I mean liminf and limsup of $\frac1n#(A\cap{1,\dots,n})$.
Why doesn't lower natural density work? Effects of very small primes?
I don't claim that, it's just that I don't understand "once you pass to an arithmetic progression modulo a slowly growing product of very small primes".
Well, let's say that I'm guessing that the lower natural density is positive once you take care of small-prime issues (if needed; I haven't thought about it enough to be sure).
As far as I see, the equation (or rather the inequality) is $\lceil \log_p(n)\rceil>\log_p(n)+\frac1{p\ln p}$: does it hold for all prime $p\ge 3$?
Other than that, I could comment that the problem is clearly related to an equidistribution question, but I don't want to influence the discussion too much.
@Seva Well, strictly speaking, the inequality should be $\lceil \log_p(n)\rceil > \log_p(n) - \frac{\log(1-1/p)}{\log p}$, but what you say is a sufficient condition. Since $\sum_p \frac{1}{p \log p}$ converges, it is reasonable to conjecture that the lower natural density will be positive.
... but a simple argument involving adding upper densities does not work, since $\sum_p 1/p$ diverges.
Sorry, I forgot to remove those $n$ such that $n+1$ is prime: given this, I found 35637 such numbers in ${0,\dots,99999}$ (the exact count has little interest, but just in case somebody also computes and compares).
Following some discussion with Harald (offline)... It appears that the answer is YES, the set is indeed infinite.
First ingredient:
We take the idea of adding upper densities, but use logarithmic densities instead of natural densities (logarithmic density meaning one sums reciprocals to $x$, divides by $\log{x}$, and sends $x$ to infinity). A short computation bounds the upper logarithmic density of integers which start with $p-1$ in some base $p > C$ by $\sum_{p > C} \frac{1}{(p-1)\log{p}}$. Unfortunately, this sum is larger than $1$ when $C=2$, so this does not immediately imply a positive answer to the question. So we have to work harder.
2nd component: One shows that if $g_1,\dots,g_m \ge 2$ are integers with $1/\log{g_1}, \dots, 1/\log{g_m}$ being $\mathbb{Q}$-linearly independent, then the fractional parts $\{\log_{g_1} n\}, \dots, \{\log_{g_m} n\}$ are distributed like independent uniform random variables on $[0,1)$, with respect again to logarithmic density. This uses a Weyl-type criterion; the key point is that if $\gamma \ne 0$, then $e^{2\pi i \gamma \log{n}} = n^{2\pi i \gamma}$ has logarithmic mean value $0$. (The condition that $\gamma \ne 0$ explains the linear independence condition required on the numbers $1/\log g_i$.)
Now $1/\log{3}$ and $1/\log{5}$ are easily seen to be $\mathbb{Q}$-linearly independent, and so one gets from the second component that the integers which do not begin with $2$ in base $3$, and do not begin with $4$ in base $5$, comprise a set with logarithmic density
$$ (1-\log_3(3/2)) (1-\log_5(4/5)) \approx 0.543.$$
Then one uses the first component with $C=5$. The sum of $\frac{1}{(p-1)\log{p}}$ for $p>5$ can be proved to be $< 0.53$ (say). It follows that the set in question is indeed infinite, with lower logarithmic density $> 1/100$.
If $\{1/\log{p}\}_{p\ge 3}$ is a linearly independent set (over $\mathbb{Q}$), modifying the above argument would show that the logarithmic density is exactly $\prod_{p \ge 3} (1-\log_p(p/(p-1))$. But I think this linear independence is open.
Doesn't the desired linear independence assumption follow from the fundamental theorem extended to $\mathbb{Q}^{\times}_{+}$?
Does it? That certainly gives the linear independence of the numbers $\log{p}$, but we want $1/\log{p}$.
Yes, I realized that after my comment. I proved years ago that the sum of the reciprocals of the first primes was never an integer, but I'm not sure it can help.
|
2025-03-21T14:48:32.100793
| 2020-09-23T08:42:42 |
372385
|
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|
Stack Exchange
|
Properties preserved in addition of ideals
If I and J are prime (radical) ideals then what are the conditions under which we can define a prime (radical) ideal from I+J?
You mean when is $I+J$ still radical? Cause otherwise take $\sqrt{I+J}$...
Yes when I+J is still radical.
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2025-03-21T14:48:32.100857
| 2020-09-23T09:20:11 |
372388
|
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|
Stack Exchange
|
Reference for matrices with all eigenvalues 1 or -1
In a homological algebra problem I am in the situation that I have an invertible (over $\mathbb{Z}$) integer matrix $X$ and a permutation matrix $Y$ such that $N:=XY$ is a matrix with all eigenvalues equal to 1 or -1.
Question 1 : Does such a situation appear already in other situations/fields in algebra/combinatorics? Is there an interpretation for this or does it have a combinatorial meaning?
Question 2: Do (invertible integer) matrices with all eigenvalues equal to 1 or -1 have a name, or do matrices $X$ as above have a name? Are they studied in the literature?
A square matrix with only 1 as eigenvalue (in every field extension) is called unipotent. So $M$ has all eigenvalues $\pm 1$ iff $M^2$ is unipotent. I don't know it this has a name.
@YCor Thanks, this is already helpful for me.
Hermitian unitary matrices have all eigenvalues equal to $\pm 1$; in the physics context, these are studied as scattering matrices of systems with a chiral symmetry (the trace of this matrix then counts the number of topologically protected "zero-modes").
@CarloBeenakker OP's matrix has integer entries, if I understand correctly, so it cannot be unitary apart from trivial cases (signed permutation matrices).
Your matrix is conjugate (via an invertible integral matrix) to an upper triangular matrix with each main diagonal entry $\pm 1$. I'm not sure that much more can be said, since any matrix with that property has all eigenvalues $\pm 1$.
It's not hard to construct two projections preserving the generalized eigenspaces (by adding/subtracting $I_n$ and dividing by $2$/$-2$ if your matrix is diagonalizable), so projections may be a place to start.
Hi @GeoffRobinson, would it be okay for you to provide a reference for your result? (I found it a bit difficult to prove, as such a matrix is not similar to its Jordan normal form via an invertible integral matrix in general)
I meant that the matrix $N$ is conjugate in the above way to an upper triangular matrix with all diagonal entries $\pm1.$
@GeoffRobinson Thank you, could you tell me where I can find the proof for this result: an invertible integral matrix with all eigenvalue ±1 is conjugate to an upper triangular matrix with all diagonal entries ±1 via an integral matrix
You work by induction on $n$. Identify the $n$-long integerr column vectors with $\mathbb{Z}^{n}$. We may choose a non zero integral column vector $v$ with $Nv = \pm v$, and we may arrange so that the entries of $v$ have gcd 1, which means that $v$ may be extended to a $\mathbb{Z}$-basis for the integer column vectors. This reduces the problem to rank $n-1$ with a little thought.
|
2025-03-21T14:48:32.101093
| 2020-09-23T10:12:30 |
372391
|
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|
Stack Exchange
|
Object classifiers in 1-toposes
In a Grothendieck $\infty$-topos, it is known that, for arbitrarily large regular cardinals $\kappa$, there is a classifier for the class of relatively $\kappa$-compact morphisms. It is also easy to show that this is not the case in 1-toposes, because we might have isomorphic, but not equal, such morphisms classified by the same map. However, we should be able to recover at least part of the definition of an object classifier. Namely, I need to know that in a Grothendieck 1-topos, for arbitrarily large regular cardinals $\kappa$, there is a map $t: U' \to U$ such that for every relatively $\kappa$-compact morphism $f: X \to Y$ there exists a pullback square
$\require{AMScd}$
\begin{CD}
X @>>> U'\\
@VfVV @VVtV\\
Y @>>> U
\end{CD}
(not necessarily unique and such that the map $Y \to U$ doesn't necessarily only classify $f$). I feel that this should definitely be true, but I can't find it anywhere in the literature. It would be very nice to have a reference for it, or a confutation in the unfortunate case I'm wrong. Thanks!
There is a standard strategy that always works. Say that your topos is $\mathsf{Sh}(C)$ for a site $C$. Now, if such an object $U$ exists, it means that
$$U(c) \stackrel{\text{Yon}}{\cong} \text{Nat}(y(c), U) \stackrel{\text{U.P.}}{\cong} {\kappa\text{-compact morphism over } c}.$$
This observation gives you the only possible definition for $U$ and is the standard strategy to build the sub. classifier. Whether this is a sheaf, I don't know.
Your second step is not applicable to my case, because there might be more than one relatively $\kappa$-compact morphism over $y(c)$ corresponding to a map $y(c) \to U$. Of course all the candidates will only differ by an isomorphism, but still it doesn't give a bijection of sets. That's the main issue in moving from $\infty$- to 1-toposes.
Is there any chance that we define an equivalence relation on U so that the quotient is the object that you desire?
I would need an equivalence relation on $\text{Nat}(y(c), U)$ and one on ${ \text{relatively}\ \kappa \text{-compact morphisms over}\ c }$ in such a way that the two quotients are naturally isomorphic, but then I could no longer use Yoneda.
I feel like something like this should work. Let $\mathcal{E}$ be a Grothendieck topos and let $\mathbf{s} \mathcal{E}$ have the model structure where the cofibrations are the monomorphisms and the weak equivalences are the internal weak homotopy equivalences. $\mathbf{s} \mathcal{E}$ has a classifier for small 0-truncated objects, say $\tilde{U} \to U$, which we may take to be a fibration with $U$ fibrant. Pull back along $U_0 \hookrightarrow U$ to get a fibration over a "discrete" object. If the fibres are also "discrete" then we are done, but I don't know if this happens / can be forced.
A weak classifier of k-compact families can be obtained in a 1 topos in a standard way due to hofmann and streicher. For presheaves, there is a strict enough way to give a correct definition of a presheaf in the spirit of Ivan’s attempt (which sadly did not have a functorial action). Streicher observed that by sheafifying this weakly classifying family you get a weakly classifying family for sheaves too! this follows in essence from the left exactness of the localization . If people want me to put the details into an answer, i can try to do so.
@JonathanSterling I worked out a proof myself in the meantime, which is indeed in the spirit of Ivan's comment. On presheaves, it is based on choosing a set of representatives for equivalence classes of relatively $\kappa$-compact morphisms over $y(c)$ as elements of $U(c)$. This can be transferred through a left exact localization, provided that we are not asking for uniqueness of the classifying maps. Is the strategy you have in mind similar to this?
@GiulioLoMonaco the approach i had in mind was the classic one, where you use the equivalence between “slice of presheaf category over a representable” with “presheaves on a slice”. then, one gets a strict functorial action by composition.
@GiulioLoMonaco This is the standard reference for the construction I referred to: http://www2.mathematik.tu-darmstadt.de/~streicher/NOTES/UniTop.pdf
If yours turns out to be different, it might be a nice result that type theorists would appreciate. I encourage you to write it up in an answer.
If $\mathcal{X}$ is a topos, then there is an adjunction $\mathcal{X} \leftrightarrows \mathcal{P(C)}$ with a category of presheaves of sets, where the right adjoint is fully faithful and accessible and the left is left exact.
The idea is very similar to the one spelled out in https://www2.mathematik.tu-darmstadt.de/~streicher/NOTES/lift.pdf, except that where its authors first find a suitable class of relatively $\kappa$-compact morphisms in $\mathcal{P(C)}$ and then consider the class of all morphisms in $\mathcal{X}$ that become relatively $\kappa$-compact in $\mathcal{P(C)}$, my analysis allows to find some $\kappa$ such that the desired class is that of all relatively $\kappa$-compact morphisms in $\mathcal{X}$. A complete proof can be found in Appendix A of http://www.tac.mta.ca/tac/volumes/37/5/37-05.pdf
I'd like to point out that neither strategy is more general than the other. The reason of this is that a class of relatively $\kappa$-compact morphisms in $\mathcal{X}$ need not be sent precisely to a class of relatively $\lambda$-compct morphisms in $\mathcal{P(C)}$ and, conversely, the preimage of a class of relatively $\lambda$-compact morphisms in $\mathcal{P(C)}$ need not be precisely a class of relatively $\kappa$-compact morphisms in $\mathcal{X}$.
However, all relatively $\kappa$-compact morphisms found in the latter proof are indeed sent to relatively $\kappa$-compact morphisms in $\mathcal{P(C)}$.
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2025-03-21T14:48:32.101510
| 2020-09-23T10:30:18 |
372392
|
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|
Stack Exchange
|
Absolute and relative tilings of the hyperbolic plane
In Conway's Symmetries of Things on p. 265 I found these two tilings of the hyperbolic plane with the same vertex configuration $(<IP_ADDRESS>.3)$ (resp. vertex figure, as Conway calls it).
The difference lies in some permutation (of the vertices of one of the pentagons?) but I cannot see the effect of this permutation in the diagrams. I would have expected to see some crossing edges, but all I see are differently coloured pentagons – otherwise the two graphs look the same, i.e. isomorphic (what they presumably are not). What's going on here?
Daniel Huson sent me this plot in much better quality. It was made with Tegula. Thanks to Daniel.
Let $T$ be the pictured tiling, where we forget about the colouring. Then its symmetry group is the maximal discrete subgroup of isometries of $\mathbb{H}^2$ which fixes $T$. Call that symmetry group $\Gamma$. This is the symmetry group of the picture on the right. It acts transitively on the vertices of $T$, and also acts transitively on the centres of the pentagons.
Now let $\Gamma^+$ be the index 2 subgroup of $\Gamma$ which contains the orientation-preserving isometries. Then $\Gamma^+$ also acts transitively on the vertices of $T$, but there are now two orbits for the centres of the pentagons. To see that, consider a pentagon and the 10 triangles nearest to it. On the orange pentagons, pairs of triangles point outwards in a anti-clockwise manner, while they point outwards in a clockwise manner on the yellow pentagons. Hence there is no orientation-preserving isometry taking an orange pentagon to a yellow one.
By absolute, Conway, Burgiel, and Goodman-Strauss mean the full symmetry group $\Gamma$ which acts transitively on the vertices of $T$. By relative, they refer to a proper subgroup of $\Gamma$ which still acts transitively on the vertices of $T$.
Let $e$ be an edge of $T$ which is shared by two adjacent triangles. Then there is a geodesic $m$ in $\mathbb{H}^2$ which contains $e$. A reflection in $m$ maps an orange pentagon onto a yellow one, and vice versa.
In the picture on the left, there are rotations of order 5 about the centres of the pentagons, and also rotations of order 2 about the intersection of any two mirror lines. These intersections are the midpoints of the edge $e$ and its translates by $\Gamma^+$. Hence the orbifold $\mathbb{H}^2/\Gamma^+$ is a $2$-sphere with cone points of orders 5, 5, and 2, and $\chi(\mathbb{H}^2/\Gamma^+)=\frac{-1}{10}$.
In the picture on the right, there are the same rotations, but now the rotation of order $2$ arises as the product of reflections in the two mirror lines passing through that point. So the orbifold $\mathbb{H}^2/\Gamma$ is a disk with one cone point of order 5 and a corner reflector of order 2. Thus $\chi(\mathbb{H}^2/\Gamma)=\frac{-1}{20}$. Then we can see that $\mathbb{H}^2/\Gamma^+$ double covers $\mathbb{H}^2/\Gamma$, which is to be expected since $\Gamma^+$ was an index 2 subgroup of $\Gamma$. This can also be seen by looking at the fundamental regions for the respective group actions.
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2025-03-21T14:48:32.101759
| 2020-09-23T10:44:48 |
372394
|
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|
Stack Exchange
|
How to measure points scattering in a circle
I have circles containing points (x,y). I would like to measure the scattering of the points within the circle.
For example, in the following picture, circle A will have a higher value since the points are much more scattered across the circle.
Notice that the circles have varying value - so we can have a circles with different radiuses.
For example in the following picture, although the points are the same - circle C will have a higher value because the points are scattered across the whole circle.
Do you know a measurement which I can use for such purpose?
Thanks!
Wouldn't mean-square distance from the points' center of mass, divided by the radius of the circle, do the trick?
There exists a variety of measures of uniformity of a point set. See, for example, On assessing spatial uniformity of particle distributions... for an overview, and a critical comparison when applied to real-world data.
There are two distinct classes of uniformity measures: Quadrat-based measures divide the region into a number of small grids, called quadrats, and count the number of points falling into each grid. Distance-based methods focus on the distances between points, such as those between nearest neighbors or between randomly selected locations.
Judging from your pictures, it should be sufficient to consider the root-mean-square distance $\rho$ of the points $\vec{x}_k$ from their center of mass $\vec{\mu}$, divided by the radius $R$ of the circle:
\begin{align}
\vec{\mu} &= \frac{1}{N}\sum_{k=1}^N\vec{x}_k, \\
\rho &= \sqrt{\left(\frac{1}{N}\sum_{k=1}^N\left\lVert\vec{x}_k-\vec{\mu}\right\rVert^2\right)}, \\
S &=\frac{\rho}{R},
\end{align}
where $S$ is your measure of scattering within the circle.
Star discrepancy
The star discrepancy is usually used when thinking about random numbers and low discrepancy sequences, and seems to fit the bill for your task.
|
2025-03-21T14:48:32.101942
| 2020-09-23T11:00:14 |
372395
|
{
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}
|
Stack Exchange
|
Chromatic number of a family of graphs
It is well-known that if a graph has maximum degree $d$, then it is $d+1$ colorable. Say we have $d+1$ graphs $G_1,\ldots, G_{d+1}$ on the same vertex set $V$, and say each $G_i$ has maximum degree at most $d$.
A coloring of $\textbf{G}:=\{G_1,\ldots, G_{d+1}\}$ is just a labelling of the common vertex set of $\textbf{G}$ with $\{1,\ldots, k\}$, for some $k\in\mathbb{N}$. This coloring is proper if for any $i\in\{1,\ldots, k\}$, no edge of $G_i$ can be found between two vertices colored $i$.
My question is whether $\textbf{G}$ admits a proper coloring with just $d+1$ labels.
Another way to formulate this is that we are looking for a partition of the common vertex set $V=V_1\cup \ldots \cup V_{d+1}$, where $V_i$ is an independent set in $G_i$.
The simplest case is when $d=1$. In this case, $\textbf{G}=\{G_1, G_2\}$, and both $G_1$ and $G_2$ are matchings. I claim that $\textbf{G}$ can be properly $2$-colored. Proceed with induction - base case is clear. If there exists a vertex that is isolated in $G_1$, we can remove that vertex, apply induction, and label that vertex $1$. Thus, we can assume every vertex is non-isolated in both $G_1$ and $G_2$, which is only possible if $\textbf{G}$ is an even cycle, its edges alternating between $G_1$ and $G_2$. It follows that $\textbf{G}$ can be properly $2$-colored.
"no edge of $G_i$": do you instead mean "no edge of any $G_j$"?
Correct me if I am wrong, but is your problem equivalent to the following: for each $G_i$ you want to find an independent set $V_i\subseteq V$ ($V$ being the common vertex set), so that $V_1\cup\cdots\cup V_{d+1}=V$. To see the equivalence: interpret the $V_i$ as vertices that are colored with color $i$. We can assume that the $V_i$ are disjoint: if they are not, remove the doubled vertex from one of the problematic sets.
@RobPratt No, I mean no edge of $G_i$ as written. The vertices of the $i^{th}$ label should be an independent set in $G_i$.
@M.Winter That is correct, that's an equivalent formulation.
The concept you introduce is called a cooperative coloring. Check out, e.g., this paper. Theorem 1 (with a reference to another paper) claims a negative answer to your question; but there is other information you may find relevant.
|
2025-03-21T14:48:32.102147
| 2020-09-23T11:17:23 |
372398
|
{
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"Derek Holt",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/372398"
}
|
Stack Exchange
|
Can $1\ne H\cap H^g\lhd H$ happen if $G$ is a primitive permutation group with stabiliser $H$?
Assume everything is finite.
Let $G$ be a primitive permutation group with point stabiliser $G_\alpha$ for some $\alpha$. For $\beta\ne\alpha$, by an arc stabiliser we mean $G_{\alpha\beta}=G_\alpha\cap G_\beta$ and an edge stabiliser we mean $G_{\{\alpha,\beta\}}$, the stabiliser of the set $\{\alpha,\beta\}$. Note that $G_{\{\alpha,\beta\}}\ge G_{\alpha\beta}$.
My question is: can an arc stabiliser $G_{\alpha\beta}\ne 1$ be normal in $G_\alpha$?
This is impossible if $G_{\{\alpha,\beta\}}> G_{\alpha\beta}$. In this case, there exists $g\in G$ such that $\alpha^g=\beta$ and $\beta^g=\alpha$. It follows that $g\in N_G(G_{\alpha\beta})$. Note that $G_\alpha$ is maximal in $G$ and $G_{\alpha\beta}$ cannot be normal in $G$ (otherwise $G_{\alpha\beta}$ will be in the kernel of this action), the normaliser $N_G(G_{\alpha\beta})=G_\alpha$. This gives a contradiction: $g\in G_\alpha$ but $\alpha^g=\beta\ne \alpha$.
I think there might exist an example in the case when $G_{\{\alpha,\beta\}}=G_{\alpha\beta}$. However, by a quick check with Magma there is no such primitive permutation group $G$ with $|G|\le 300$.
An equivalent statement is: Let $G$ be a group and $H$ a maximal and core-free subgroup of $G$. Is it possible that $1\ne H\cap H^g\lhd H$ for some $g\notin H$?
As shown in the comments there by Verret and Holt there is no such example for degree $\le 4095$. I also think the proof or an example, if any, will not be elementary (for example, applying O'Nan-Scott Theorem).
This was initially a MSE question I asked.
@GeoffRobinson I think that remark is essentially equivalent to $G_{{\alpha,\beta} }> G_{\alpha\beta}$. In the language of permutation groups, this is saying that the orbit of $\alpha^g$ under $G_\alpha$ is not self-paired. I conjecture that the answer to the question is no, but that the proof will require CFSG. (The question reminds me of the Sims Conjecture that, for a primitive group $G$, $|G_\alpha|$ is bounded as a function of $|\beta^{G_\alpha}|$ for any $\beta \ne \alpha$, which was evenutally proved using CFSG.)
@DerekHolt : Thanks. I agree that it is likely to be true that there is no such example (but I am not totally sure- for almost simple groups it seems very likely, certainly).
I am reminded of the following generalization by Wielandt of a famous theorem of Frobenius. The Theorem of Wielandt asserts that if a finite group $G$ has a subgroup $H$, such that there is a normal subgroup $H_{0} \lhd H$ such that $H \cap H^{g} \leq H_{0}$ for all $g \in G \backslash H$, then there is $K \lhd G$ such that $G = KH$ and $K \cap H = H_{0}$.
An example was constructed by Pablo Spiga:
https://arxiv.org/abs/2102.13614
"A generalization of Sims conjecture for finite primitive groups and two point stabilizers in primitive groups"
Yes, I have also noticed that paper. Thanks for your comment. We then think this statement holds true for "most" primitive groups. For example, a series of papers by Konygin (references of Pablo Spiga's paper) handle almost all the cases when $G$ is of AS or PA type.
Sorry, this is not an answer, but rather an application of standard techniques of local analysis to obtain substantial structural information about $H\cap H^g$, in case an example exists. The techniques are inventions of John Thompson, George Glauberman, and Helmut Bender.
Proposition: Let $G$ be a primitive permutation group on a set $\Omega$, and let $\alpha$ and $\beta$ be distinct points in $\Omega$. If $1\ne G_{\alpha\beta}\triangleleft G_\alpha$, then $G_{\alpha\beta}$ has the following properties:
(a) $F^*(G_{\alpha\beta})$ is either a $p$-group for some prime $p$, or the direct product of nonabelian simple groups.
(b) If $G_{\alpha\beta}$ is solvable, then $F^*(G_{\alpha\beta})$ is a $2$-group or a $3$-group.
(c) If $P$ is a Sylow subgroup of $G_{\alpha\beta}$, then no nontrivial characteristic subgroup of $P$ is normal in $G_{\alpha\beta}$.
(d) If $F^*(G_{\alpha\beta})$ is a $p$-group, $p>2$, then $G_{\alpha\beta}$ is not $p$-stable.
(e) (Uses Odd Order Theorem) $G_{\alpha\beta}$ has even order.
Proof: First, part (c). Suppose that $P$ is Sylow in $G_{\alpha\beta}$, $P_0\ne 1$ is a characteristic subgroup of $P$, and $P_0\triangleleft G_{\alpha\beta}$. By the Frattini argument $G_\alpha=G_{\alpha\beta}N_{G_\alpha}(P)\le N_{G_\alpha}(P_0)$ so $G_\alpha=N_G(P_0)$ by maximality of $G_\alpha$. Hence $N_{G_\beta}(P)\le N_{G_\beta}(P_0)=G_{\alpha\beta}$. Since $P$ is Sylow in $G_{\alpha\beta}$, it is Sylow in $G_\beta$ and hence also in $G_\alpha\cong G_\beta$. Now $G_\beta=G_\alpha^g$ for some $g\in G$, so $P^{gh}=P$ for some $h\in G_\beta$, by Sylow's theorem. Then $gh\in N_G(P)\le N_G(P_0)$ so $P_0\triangleleft G_\alpha^{gh}=G_\beta$. Then $P_0\triangleleft\langle G_\alpha,G_\beta\rangle=G$, so $P_0=1$ as $G_\alpha$ contains no nontrivial normal subgroup of $G$.
But $P_0\ne1$, contradicttion.
Next, part (a). If $F(G_{\alpha\beta})=1$, then $F^*(G_{\alpha\beta})=E(G_{\alpha\beta})$ and it is the direct product of nonabelian simple groups. So suppose $B=O_p(G_{\alpha\beta})\ne 1$ for some prime $p$. Then $N_G(B)=G_\alpha$. Let $A=O^p(F^*(G_{\alpha\beta}))$, the subgroup of $F^*(G_{\alpha\beta})$ generated by all its $p'$-elements. By the structure of generalized Fitting subgroups, $[A,B]=1$. If $A=1$, then $F^*(G_{\alpha\beta})=B$ and we are done, so assume that $A\ne 1$. Then $N_G(A)=G_\alpha$. Let $a\in A$ be an arbitrary $p'$-element and consider the action of $\langle a\rangle\times B$ on $C=O_p(G_\beta)\triangleleft G_\beta$.
We have $[\langle a\rangle,C_C(B)]\le [A,G_\alpha]\cap[G_\beta,C]\le A\cap C\le O_p(A)$. But $A$ is the elementwise-commuting product of $q$-groups for various primes $q\ne p$, and quasisimple groups. Therefore $[\langle a\rangle, C_C(B)]\le Z(A)$ so $[\langle a\rangle,C_C(B),\langle a\rangle]=1$. As $a$ is a $p'$-element normalizing the $p$-group $C_C(B)$, a standard coprime action lemma implies that $[\langle a\rangle,C_C(B)]=1$.Then the Thompson $A\times B$ Lemma implies that $[\langle a\rangle,C]=1$. But $a$ was an arbitrary generator of $A$, so $[A,C]=1$. This implies in turn that $C\le N_G(A)=G_\alpha$. Since $C$ is a normal $p$-subgroup of $G_\beta$, $C\le O_p(G_{\alpha\beta})=B$. But $B\le O_p(G_\alpha)\cong O_p(G_\beta)=C$, so $B=C$. Since $B\triangleleft G_\alpha$ and $C\triangleleft G_\beta$, $B\triangleleft G$, a contradiction as $B\le G_{\alpha}$.
Now (d). If $F^*(G_{\alpha\beta})$ is a $p$-group, then a Sylow $p$-subgroup $P$ of $G_{\alpha\beta}$ is nontrivial. If also $p>2$ and $G_{\alpha\beta}$ is $p$-stable, then
by Glauberman's $Z(J)$-Theorem, $Z(J(P))\triangleleft G_{\alpha\beta}$ for any Sylow $p$-subgroup $P$ of $G_{\alpha\beta}$. But $Z(J(P))$ is a nontrivial characteristic subgroup of $P$. This contradicts (c).
Now (b). By (a) and solvability, $F^*(G_{\alpha\beta})$ is a $p$-group for some prime $p$. Suppose $p>2$. By (d), $G_{\alpha\beta}$ is not $p$-stable, whence $G_{\alpha\beta}$ has a subquotient isomorphic to $SL_2(p)$. So $SL_2(p)$ is solvable and $p=3$.
Finally, (e). Suppose that $G_{\alpha\beta}$ has odd order, so it is solvable. By (b) and (d), $F^*(G_{\alpha\beta})$ is a $3$-group and $G_{\alpha\beta}$ is not $3$-stable. Hence $G_{\alpha\beta}$ has an $SL_2(3)$ subquotient, which is of even order, contradiction.
These are very good criteriums on such $G_{\alpha\beta}$ (of course, if any). Thank you for your observations.
If $G_{\alpha \beta}$ is solvable, I think you can get that either $G$ involves ${\rm Qd}(3)$ or $G$ involves $S_{4}$. In particular, a Hall ${2,3}$-subgroup of $G$ is neither $2$-closed nor $2$-nilpotent. This needs a Theorem of Stellmacher, as well as Glauberman's $ZJ$-theorem.( using condition c)).
In the above comment, $G$ means $G_{\alpha \beta}.$
|
2025-03-21T14:48:32.102650
| 2020-09-23T11:25:34 |
372401
|
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|
Stack Exchange
|
A weakly sequentially continuous operator which is not weakly continuous
I'm reading some papers where the condition of weak sequential continuity is crucial instead of the weak continuity.
So, let
$T$ an operator between a Banach space $X$ and itself.
$T$ is weakly sequentially continuous if for any $(x_n)_{n\in \mathbb N}$, $x_n \stackrel{w}{\rightharpoonup}x$ in $X$ $\Rightarrow$ $T(x_n) \stackrel{w}{\rightharpoonup } T( x )$ in $X$)*.
$T$ is weakly continuous iff $T$ is continuous with respect to the weak topologies on $X$
While it is clear that 2 implies 1, I want to find a counter-example of an operator which verifies 1 but not 2.
$T$ is not linear
Both assertions are equivalent, and they are also equivalent to "3. $T$ is norm continuous". Proof: "3. => 2." This follows from the existence of the dual operator. "2. => 1." As mentioned in the question, this is obvious. "1. => 3": Let $(x_n)$ be a sequence in $X$ which converges to $0$ in norm. Then $(x_n)$ converges also weakly to $0$. Hence, $(Tx_n)$ converges weakly to $0$ by 1. Thus, $(Tx_n)$ is bounded by the uniform boundedness principle. So $T$ maps sequencs that norm converge to $0$ to bounded sequences; this is sufficient for norm continuity of $T$.
I added the "Banach spaces" tag. By the way, what does the symbol "$)^*$" mean at the end of assertion 1.?
$T$ is not linear
Then just use some fancy space $X$ like $\ell^1$ in which the weak convergence (of sequences) is equivalent to the norm convergence but the weak topology is substantially different from the norm one. For the operator, just take $Tx=|x|e$ where $e$ is any non-zero vector in $X$.
|
2025-03-21T14:48:32.102898
| 2020-09-23T11:33:08 |
372403
|
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"authors": [
"Piotr Hajlasz",
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"url": "https://mathoverflow.net/questions/372403"
}
|
Stack Exchange
|
Does there exist an injective Lipschitz map on the disk whose gradient switches between two given matrices?
While solving a problem in calculus of variations, I came to the following question:
Let $A,B$ be two real $2 \times 2$ matrices with positive determinants, and suppose that $\operatorname{rank}(A-B)=1$.
Let $D \subseteq \mathbb{R}^2$ be the closed unit disk.
Does there exist a Lipschitz map $u:D
\to D$ such that:
$\nabla u \in \{A,B\}$ a.e. in $D$
$u$ is surjective.
$u^{-1}(y)$ is a singleton for almost every $y \in D$.
By a theorem by Ball and James, any $u$ satisfying condition $(1)$ is of the following form:
Write $B-A=an^T$ for some $a,n \in\mathbb{R}^2$, $n$ a unit vector. Then
$$
u(x)=Ax+h(x\cdot n)a+v_0,
$$
for some $v_0 \in \mathbb{R}^2$, and a Lipschitz function $h:\mathbb{R} \to \mathbb{R}$ with $h' \in \{0,1\}$ a.e..
So, it "remains" to examine maps $u$ of the form above.
The assumptions $(1)$ and $(2)$ imply that the average integral of the Jacobian $Ju$ is $1$, so the measure of the set where $\nabla u=A$ is predetermined by $\det A,\det B$.
If it matters, then in my case of application, $A \in \alpha \text{SO}(2)$, where $\alpha>1$, and the singular values of $B$ satisfy:
$\sigma_1+\sigma_2>\beta>1$ and $\sigma_1 \sigma_2 <<1$ very close to zero. Here $\alpha, \beta$ are some fixed parameters.
I deleted my answer since I do not have time to complete details. I still believe my answer is correct. If I can complete details I will undelete it.
|
2025-03-21T14:48:32.103032
| 2020-09-23T12:07:34 |
372405
|
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"Hacon",
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"url": "https://mathoverflow.net/questions/372405"
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|
Stack Exchange
|
Does base point free linear system of polynomials generate higher degree polynomials
Let $k$ be an algebraically closed field. Let $S=k[x_0,\ldots, x_4]$ be the ring of polynomials. We set $S^i$ to the graded piece of degree $i$ polynomials. Let $H$ be a hyperplane of $S^5$ with no base point. Then how can I prove that $H$ generates $S^6$? namely $S^1\cdot H=S^6$. Moreover, is the conclusion true in general (for arbitrary variables and degrees)?
This should be Example 1.8.15. (Green’s theorem) in Lazarsfeld's Positivity I (my e-copy...it may be slightly different in the refs) and there are several far reaching generaizations.
@Hacon: Thank you for pointing out the reference.
|
2025-03-21T14:48:32.103116
| 2020-09-23T12:10:50 |
372406
|
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"Ben McKay",
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"Marco Golla",
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"url": "https://mathoverflow.net/questions/372406"
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|
Stack Exchange
|
For a fiber bundle $f:M\to S^1$, is there a point whose horizontal lift is a circle?
Let $f: M\to S^1$ be a Riemannian submersion, and also a fiber bundle. $M$ may be non-compact, possibly with boundary. For $x\in S^1$, consider the fiber $f^{-1}(x)\subset M$. Is there a point $p\in f^{-1}(x)$, such that the horizontal lift of $S^1$ beginning at p also end at p? If $M$ is the Möbius band, there exist such points.
Every such bundle is a mapping torus of a self-diffeomorphisms of $f^{-1}(x)$. The horizontal lift you ask for exists if and only if this diffeomorphism has a fixed point. For $f^{-1}(x)=(-1,1)$ the Mobius bundle is induced by $t\mapsto -t$ which indeed has a fixed point.
@archipelago: For Mobius band, this surely holds. I ask the general case.
@mathmetricgeometry: archipelago has explained the general case: a mapping torus.
@archipelago: Thank you! Just consider the cyliner $f:S^1\times (-\infty,+\infty)\to S^1$, the diffeomorphisms of $f^{-1}(x)$ is $(0,t)\to (1,t+1)$. Then the Horizontal lifting lines are spiral rising lines.
@archipelago: I think the above cylinder is not a counter example. Their Horizontal lifting are circles. Klein bottle to $S^1$ is a counter example, the fibers are $S^1$, the diff is reflexion, no fixed point. Need Horizontal lifting of base $S^1$ twice, then it's a closed curve.
The 2-sheeted cover of the circle by itself is a counterexample.
By "the" horizontal lift, do you mean "a" horiontal lift? If so,
and if the fibres are connected, then the answer is positive.
The second assertion doesn't seem to be true: the Klein bottle (viewed as the non-orientable $S^1$-bundle over $S^1$) does not have a section.
I cannot agree, Marco. It does have a section (which is weaker than being a product).
Oops. Non-orientable things are confusing. You're right.
|
2025-03-21T14:48:32.103282
| 2020-09-23T12:32:13 |
372407
|
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|
Stack Exchange
|
Extensions of oligomorphic groups
Recall that a permutation group $G\le\mathfrak{S}(\mathbb N)$ is oligomorphic if, for any positive integer $k$, the induced $G$-action on $k$-tuples of distinct elements of $\mathbb N$ has only finitely many orbits.
Let's say that a countable group is oligomorphic if it is isomorphic (as an abstract group) to some oligomorphic permutation group.
Given a short exact sequence of (abstract) countable groups $1\to K \to G \to Q \to 1$, with $Q$ oligomorphic (or even $K$ and $Q$ oligomorphic), are there conditions on the sequence which guarantee that $G$ is oligomorphic?
EDIT: I would even be interested by the case where $K$ is finite. One motivation for me is the following. If $G$ is oligomorphic whenever there is a short exact sequence of (abstract) groups $1\to K \to G \to Q \to 1$, with $K$ finite and $Q$ countable oligomorphic, then every acylindrically hyperbolic group is oligomorphic. Indeed Hull and Osin proved in this article (ArXiv version) that every acylindrically hyperbolic group admits a highly transitive action with finite kernel.
|
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