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2025-03-21T14:48:31.055792
| 2020-05-24T07:23:35 |
361197
|
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"Tony Huynh",
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|
Stack Exchange
|
Finding minimum weight perfect matchings in sparse bipartite graphs
Question:
What can be recommended for finding optimal perfect matchings in large bipartite graphs with small vertex degree if the edge-weights are positive real values?
I am looking for algorithms of which freely available implementations exist or that don't require heavy machinery for implementation.
Apart from that I am also looking for graph modifications that can speed up the calculation of the optimal matching edges; subtracting vertex potentials may be in that vein.
The fractional matching polytope is equal to the matching polytope for bipartite graphs, so you can just solve a small linear program.
pyMCFSimplex seems to best fit my needs.
"It is a free Python port of a Python Wrapper for MCFSimplex. pyMCFimplex is a Python-Wrapper for the C++ MCFSimplex Solver Class from the Operations Research Group at the University of Pisa. MCFSimplex is a piece of software hat solves big sized Minimum Cost Flow Problems very fast through the (primal or dual) network simplex algorithm."
scipy.sparse.csgraph.min_weight_full_bipartite_matching is an alternative
|
2025-03-21T14:48:31.055906
| 2020-05-24T07:25:45 |
361199
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361199"
}
|
Stack Exchange
|
Perfect Cayley graphs for abelian groups have $\frac{n}{\omega}$ disjoint maximal cliques
Let $G$ be a perfect/ weakly perfect Cayley graph on an abelian group with respect to a symmetric generating set. In addition let the clique number be $\omega$ which divides the order of graph $n$. Then, would $G$ have $\frac{n}{\omega}$ disjoint maximal cliques, that is $\frac{n}{\omega}$ maximal cliques of order $\omega$ which are mutually vertex disjoint.
The result is immediate if $G$ is complete multipartite, or even if $G$ is a unitary cayley graph. From here, we know that, if $\omega|n$, $G$ would be CIS(clique intersection stable), that is every maximal clique intersects every maximal independent set. I guess it should work in this case, and also for other graphs on abelian groups. Will the result hold even for Cayley graphs of non-abelian groups? Thanks beforehand.
This would work for any perfect Cayley graph. Since we have $n$ vertices and the clique number is $\omega$, we have $n$ different $\omega$-cliques (of course several intersecting). Hence, any maximal independent set would contain $\frac{n}{\omega}$ vertices in it. In fact, the vertices would be distributed equitably in $\omega$ independent sets in a $\omega$ coloring. By the paper linked in the question, we have that every maximal clique intersects every maximal independent set. Therefore each disjoint maximal independent set of vertices has one vertex of each maximal clique. Thus, there are $\frac{n}{\omega}$ vertex disjoint maximal cliques.
|
2025-03-21T14:48:31.056025
| 2020-05-24T07:55:13 |
361203
|
{
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"MaoWao",
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|
Stack Exchange
|
Riesz transform of fractional operators
I am interested in Riesz transforms linked to the fractional Laplacian and other fractional operators. I have been hunting down in the literature to find related results but I have not been able to find any.
Any related references would be greatly appreciated.
I do not think it is immediately clear what is the notion of a Riesz transform corresponding to a fractional operator. Do you mean something like $\partial_x L^{-1}$, where, say, $L = (-\Delta)^s$?
I am looking for Riesz transforms of the form $D L^{-1/2}$ with $L=(-\Delta)^{\alpha/2}$ $\alpha \in (0,2)$ and $D$ some (fractional) derivative operator.
Then it looks like you are after fractional powers of the usual Riesz transforms, right? I do not think these have been studied much, but general theory of singular integrals should apply. (By the way, out of curiosity: what is your motivation to study these operators?)
Meyer's inequality with a stable reference measure
Some results on the Riesz transform for the fractional Laplacian with (a kind of) fractional derivative are contained in this paper by Junge, Mei and Parcet: https://arxiv.org/pdf/1407.2475.pdf (see 1.4.A). Is this something of the kind you were looking for?
thanks. I have been aware of this paper published in JEMS. I was wondering if generalizations of the fractional laplacian have been considered in the literature and if the continuity results with dimension free bound would have been obtained by transference method.
You can find an elementary and very detailed approach to fractional operators and Riesz transforms in
http://www.pitt.edu/~hajlasz/Notatki/Harmonic%20Analysis4.pdf
See in particular the last section, pages 109-117. I hope it will be helpful.
Thank you for the reference and your answer. I am looking for something intrinsically fractional for the "first order" derivative operator used to define the Riesz transform.
|
2025-03-21T14:48:31.056184
| 2020-05-24T08:31:48 |
361206
|
{
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"Doug Liu",
"gdb",
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|
Stack Exchange
|
Compare bounded (unbounded) derived categories of D-modules
I am reading [HTT, D-modules, Perverse Sheaf and Representation Theory]. In [HTT, 1.5.7, page 32], it is claimed that an equivalence of categories between bounded derived category with quasi-coherent cohomologies and bounded derived category of quasi-coherent sheaves of $D_X$ modules
$$D_{qc}^b(\mathrm{Mod}(D_X))\sim D^b(\mathrm{Mod}_{qc}(D_X)).$$
It says that "It will not be used in what then follows and the proof is omitted." The refference it cited is [Borel, Algebraic D-modules VI.2.10. page 222], whose proof is really short (almost 1 page). However, it seems there are some problems.
Firstly, some notations do not make sense. In [Borel], $\mathcal{A}$ is a quasi-coherent sheaf of $\mathcal{O}_X$-algebras (Not necessarily be commutative. In our case, $D_X$ is non-commutative). Let us assume $X$ is a "good" scheme. It is denoted by $\mu (\mathcal{A})$ the category of quasi-coherent $\mathcal{A}$-modules, where quasi-coherent means quasi-coherent over $\mathcal{O}_X$. Our goal is to prove the fully faithfulness
$$\mathrm{Hom}_{D(\mu (A))}(F^\cdot, G^\cdot)=\mathrm{Hom}_{D_{qc}(\mathrm{Mod}(\mathcal{A}))}(F^\cdot, G^\cdot),~F^\cdot,G^\cdot\in D^b(\mu (\mathcal{A})),$$
In the second set $\mathrm{Hom}_{D_{qc}(\mathrm{Mod}(\mathcal{A}))}(F^\cdot, G^\cdot)$ a morphism is represented by a third complex $H^\cdot$ which may not in $\mu(\mathcal{A})$, and a quasi-isomorphism. In [Borel], the following notation of $\mathrm{Hom}^\cdot$-complex seems uncorrect
$$\mathrm{Hom}^\cdot_{D(\mu (A))}(F^\cdot, G^\cdot) .$$
I will assume that this is a typo in the following.
Secondly, I do not understand how it was able to reduce to the affine case in [Borel]. I do know the adjointness of pushforward and pullback
$$\mathrm{Hom}_{\mathcal{A}}(F^\cdot,i^*(\tilde{J}))=\mathrm{Hom}_{\mathcal{A}|_U}(F^\cdot|_U,\tilde{J}).$$
But I am not sure if the following is true
$$\mathrm{Hom}_{D(\mathcal{A})}(F^\cdot,i^*(\tilde{J}))=\mathrm{Hom}_{D(\mathcal{A}|_U)}(F^\cdot|_U,\tilde{J})$$
Thirdly, I have another possible explanation using $\mathrm{Hom}^\cdot$-complex and $R\mathrm{Hom}$ by using an injective resolution of $G^\cdot$ in $K^+(\mathrm{Mod}_{qc}(\mathcal{A})))$, say $I^\cdot \in \mathrm{Ob}(K^+(\mathrm{Mod}_{qc}(\mathcal{A})))$. Then
$$H^0R\mathrm{Hom}_{\mu(\mathcal{A})}(F^\cdot,G^\cdot)=H^0\mathrm{Hom}^{\cdot}_{\mu(\mathcal{A})}(F^\cdot,I^\cdot)\cong \mathrm{Hom}_{D^+(\mu(\mathcal{A}))}(F^\cdot,G^\cdot)$$
This approach finally suffers from I do not know if an injective object in $\mu(\mathcal{A})$ is injective in $\mathrm{Mod}(\mathcal{A})$. In [HTT, 1.4.14, page 29] it was particularly added a comment "We do not claim that $I$ is injective in $\mathrm{Mod}(D_X)$". In [Thomason and Trobaugh, Higher Algebraic K-theory of schemes and of derived categories, Appendix B.4, page 410], it is explained that in a general scheme, the right derived funtor in $\mathrm{Mod}(\mathcal{O}_X)$ and the right derived funtor in $\mathrm{Mod}_{qc}(\mathcal{O}_X)$ could have different effects on quasi-coherent objects [SGA 6, II App. I 0.2]. While in Noetherian case, they are the same, essentially because the injective objects in $\mathrm{Mod}(\mathcal{O}_X)$ are also injective in $\mathrm{Mod}_{qc}(\mathcal{O}_X)$ on a Noetherian scheme [Hartshorne, Residues and Duality, 7.18, 7.19, page 133].
I noticed that in [Bökstedt-Neeman, Homotopy limits in triangulated categories, corollary 5.5], it is true for a quasi-compact and separated scheme that the natrual inclusion
$$D(\mathrm{QCoh}(\mathcal{O}_X)) \to D_{qc}(\mathrm{Mod}(\mathcal{O}_X))$$
is an equivalence of category
But their proof is pretty long and I am not familiar with them. I do not know what is essential so far and whether it extends to our case for D-modules.
Also see [Stacks project, 36.3] for the affine case, whose proof is not so short. These kind of results are unbounded.
Bernstein himself in his notes [Bernstein, Algebraic theory of D-moduls, page 14, lemma] claimed a similar without proof. He simply said "properties of coherent D-modules imply the natural morphism
$$D_{coh}(\mathrm{Mod}(D_X))\to D(\mathrm{Mod}_{coh}(D_X))$$
is an equivalence of categories". But it seems [Bökstedt-Neeman]'s long argument implies that this should not be simple.
I am a beginner, so I am not sure if this question is important. However, it seems that we could restrict ourselves in just one of them safely and put this question aside. Bernstein himself in his notes [Bernstein, Algebraic theory of D-moduls, page 22, remark] said that "I do not know whether the natural morphism $D(Hol(D_X))\to D_{hol}(D_X)$ is an equivalence of categories. In a sense, I do not care."
I guess one can adapt the proof from Stacks Project in this situation. The key is to construct functors in both direction. The functor $D(qc-D_X) \to D_{qc}(D_X)$ is just the natural ``inclusion''. To construct the functor in the other direction, one considers the inclusion $qc-D_X-Mod \to D_X-Mod$, this admits a right adjoint $D_X-Mod \to qc-D_X-Mod$by the result analogous to Tag 077P. Then its derived functor gives a map $D(D_X) \to D(qc-D_X)$ its restriction to $D_{qc}(D_X)$ is called $RQ$. These two functors are adjoint by definition, so one only needs to check that ...
that the unit and counit maps are isomorphism. But in order to check this, we can forget about $D_X$-mod structure (one needs to check that $RQ$ is compatible with the functor $RQ: D_{qc}(\mathcal{O}_X) \to D(qc-\mathcal{O}_X)$) and work with just $\mathcal{O}_X$-modules. Then in the affine case, this can be done directly using that $RQ_X=\widetilde{R\Gamma(X, -)}$ and the case of a general sep'd quasi-compact scheme can be reduced to the affine case by means of Tag 09T6.
Dear @gdb, I wanted to express my appreciation for the strategy you outlined, which can be found in Thm. A.0.3 of my draft at https://webusers.imj-prg.fr/~haohao.liu/DFM.pdf. I found your approach to be insightful and valuable for my writing. I hope my comment is not offensive in any way, and I kindly ask you to let me know if you have any concerns. Thank you for your time and expertise.
|
2025-03-21T14:48:31.056558
| 2020-05-24T11:21:11 |
361210
|
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"Carlo Beenakker",
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|
Stack Exchange
|
Are there any extensive treatments on rational zeta series?
I've been trying to find an extensive, in-depth treatment of rational zeta series. Via the Wikipedia article on the topic, I've found two articles on this subject. While they are certainly very informative, I wonder whether there is a more comprehensive treatment that -- among perhaps other things -- provides an introduction on the summation methods, summarizes all the known results to date, shows what the open problems in the domain are, and shows how the summation of zeta terms relates to finding the sum of terms of other (Dirichlet) series.
I've asked a related question on MSE. The question above is a slightly modified and somewhat more elaborate version of it.
A lot of analytic number theory textbooks does what you are asking for. Perhaps the best one available currently is “Analytic Number Theory” by Iwaniec and Kowalski.
@StanleyYaoXiao Hi Stanley, thank you for your suggestion. I've looked through the table of contents of the book you mentioned, but I haven't been able to find a (sub)chapter on rational zeta series. Could you please tell me in which section(s) of the book they write about rational zeta series?
There is a book by Srivastava, Hari M., Junesang Choi, "Series Associated with the Zeta and Related Functions" that contains many series. Springer 2001. But is is at the Faculty at this time of coronavirus and it is only what I have in mind,
@juan Thank you for your suggestion. I've checked out the table of contents and some of the other pages, and it does seem to include a number of rational zeta series
two articles by Derek Orr: Generalized rational zeta series for $\zeta(2n)$ and $\zeta(2n+1)$ and
Approximations for Apery's constant $\zeta(3)$ and rational series representations involving $\zeta(2n)$
@CarloBeenakker Yes, those are nice articles, thank you. They're not entirely what I'm looking for though. I seek for an article that provides an overview of the research done on rational zeta series so far
A book that features a vast amount of rational zeta series is "Zeta and q-Zeta Functions and Associated Series and Integrals" by H. M. Srivastava and J. Choi (2012). It is a revised and expanded version of the book suggested by juan in the comments.
It includes the following series:
$$\sum_{k=1}^{\infty} \frac{\zeta(2k)-1}{k\cdot2^{2k}} = \log\Big{(} \frac{3 \pi}{8} \Big{)}$$
$$\sum_{k=2}^{\infty} (-1)^{k} \frac{\zeta(k)-1}{3^{k}} = \frac{11}{12} - \frac{\pi}{18}\sqrt{3} - \frac{1}{2}\log(3) $$
$$\sum_{k=1}^{\infty} \{ \zeta(k)-1 \} \Big{(} \frac{4}{3} \Big{)}^{2k} = \frac{39}{14} - \frac{2}{9}\sqrt{3} $$
$$\sum_{k=1}^{\infty} \frac{2^{2k+1}-1}{5^{2k}}\zeta(2k+1) = \frac{5}{4}\log(5) - \frac{\sqrt{5}}{2}\log(2+\sqrt{5})$$
and a great deal more.
|
2025-03-21T14:48:31.057230
| 2020-05-24T11:56:39 |
361212
|
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|
Stack Exchange
|
Lorentzian manifolds of negative spacelike sectional curvature
Suppose $(M,g)$ is a simply connected Lorentzian manifold of signature $(-,+,\ldots,+)$ and such that the sectional curvature of any space-like surface is non-positive. Is it true that there are no conjugate points or cut points along space-like geodesics?
I think you may want to add the hypothesis that $M$ be simply connected, because there are simple examples in which there are cut points: If you let $M = \mathbb{R}\times S$ with metric $g = -\mathrm{d}t^2 + g_0$ where $(S,g_0)$ is a compact Riemannian manifold with negative sectional curvature, this satisfies your curvature hypothesis, but there are clearly cut points for the geodesics in ${0}\times S$. Also, certain discrete quotients of $\mathrm{SL}(2,\mathbb{R})$ endowed with its usual bi-invariant Lorentzian metric (of type (-,+,+)$ give such examples.
Yes that’s a great suggestion. I was thinking of adding this assumption as well.
|
2025-03-21T14:48:31.057342
| 2020-05-24T12:49:35 |
361213
|
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"Piotr Achinger",
"Will Sawin",
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"k.j."
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|
Stack Exchange
|
Drinfeld basis of the universal formal deformation of a supersingular elliptic curve
Now I'm trying (5.3.2) of Katz-Mazur's Arithmetic moduli of elliptic curves.
Let $k$ be an algebraically closed field of characteristic $p > 0$, $E_0$ a supersingular elliptic curve over $k$, and $E/W[[T]]$ the universal deformation of $E$.
(Where $W$ is the Witt vector ring with the residue field $k$.)
Let $A$ be a finite $W[[T]]$-algebra which is local, $\mathfrak{E} = E \times_{W[[T]]} A$, $P, Q \in \mathfrak{E}(A)$ be a Drinfeld $p^n$-basis
(i.e., a pair of points such that $\mathfrak{E}[p^n] = \sum_{(a,b) \in (\mathbb{Z}/p^n)^2} [aP + bQ]$ as Cartier divisors), and $X$ be a parameter of the formal group of $\mathfrak{E}$.
Then $P, Q$ both lie in the formal group of $\mathfrak{E}$.
So let $f = X(P), g = X(Q) \in \mathfrak{m}_A$.
Then for an artinian local ring $R$ with the residue field $k$ and for a map $A \to R$ with $f, g \mapsto 0$, we have $P \times_A R = Q \times_A R = 0$.
Does the word "$P$ lies in the formal group" mean that $P : \operatorname{Spec}A \to \mathfrak{E}$ induces $\operatorname{Spf}A \to \hat{\mathfrak{E}}$?
And if so, how can I show it?
I know the Grothendieck existence theorems.
(In particular $\operatorname{Hom}(X, Y) \cong \operatorname{Hom}(\hat{X}, \hat{Y})$ over a complete noetherian local ring.)
But $\hat{\mathfrak{E}}$ is the formal completion along the $0$-section, not along $\mathfrak{m}_A \mathscr{O}_{\mathfrak{E}}$.
So we cannot apply it.
And I know that since $E_0$ is supersingular, $P \times k = Q \times k = 0 \in E_0(k)$.
Where is the quote taken from? What is a Drinfeld $p^n$-basis?
The formal completion on the zero section should equal the formal completion on the maximal ideal since the base ring $A$ is already complete.
@WillSawin Thank you very much! Would you explain it in a little more detail? I think that in order to show that the completion along the 0 section and one along $\mathfrak{m}_A$ are the same, we need these two closed subsets are the same.
And now this fails.
It's not true that to show two completions are the same, you need the two closed subsets to be the same.
@WillSawin For a nice scheme $X$ and its closed subset $Z$, the fomal completion along $Z$ is the space $Z$ with some sheaf of rings.
So I think we need it.
No, in the affine / local case, we're just talking about a ring, not a space with a sheaf of rings.
@WillSawin Sorry, I don't understand. What should I show?
The completion along 0 is the topological space $\operatorname{Spec}A$ with the sheaf $\lim \mathscr{O}/\mathscr{I}^n$, where $\mathscr{I}$ is the sheaf of ideals corresponding to the 0 section.
One along $\mathfrak{m}_A$ is the topological space $E_0$ with the sheaf of rings $\lim \mathscr{O}/\mathfrak{m}^n \mathscr{O}$.
These are completely different.
(These underlying spaces have different dimensions. And so it seems to be nonsense to say that these sheaves are isomorphic or not.)
Just show the global sections are isomorphic (at least to start).
|
2025-03-21T14:48:31.057582
| 2020-05-24T12:55:06 |
361214
|
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|
Stack Exchange
|
An arrangement of entangled squares
Is there an arrangement of finitely many axes-parallel squares in the plane, of $k$ different colors, such that:
The squares of each color are pairwise-disjoint;
Each square overlaps at least $4$ squares of some other color (e.g. each red square overlaps $4$ green or $4$ blue squares; each blue square overlaps $4$ red or $4$ green squares; etc.)?
For $k=2$ colors, Barry Cipra proved that the answer is no. Can his argument (or another one) be extended to $k>2$?
EDIT:
The beautiful constructions by Manfred Weis show that, for $k\geq 4$, there exists an arrangement in which each square overlaps at least 4 other squares that may be of different colors (e.g. a red square may overlap two green and two blue squares). But my intention (which was probably not clear enough) was that each square should overlap at least 4 other squares of the same color (e.g. a red square should overlap 4 green squares, or 4 blue squares).
as I have demonstrated, that a solution exists for k>3, the question boils down to as to whether a solution with k=3 is possible.
I assume you are asking whether there is a $k$ for which such arrangement exists. If yes, this is equivalent to the following uncolored version: Does there exist an arrangement of finitely many squares where each square overlaps with four pairwise disjoint squares?
There is such arrangement, for some large $k$. In the text below, I did not try to optimize $k$.
1. First, we reduce the problem to the following uncolored version:
There is an arrangement $\mathcal A$ of finitely many squares such that each $a\in\mathcal A$ overlaps with four pairwise disjoint squares in $\mathcal A$.
Indeed, given such arangement $\mathcal A$, for each square $a\in \mathcal A$ we introduce a color $C_a$, find four pairwise disjoint squares overlapping with $a$, and put into an arrangement $\mathcal B$ their copies having color $C_a$. Then $\mathcal B$ satisfies the requirements: each square in it is a copy of some $a\in\mathcal A$, and hence the color $C_a$ fits for it.
2. Now we construct a required $\mathcal A$. We introduce the colors to distinguish the squares; those colors have a few in common with the colors in the initial problem.
On each step of the process, we say that a square is good if it satisfies the required property, and bad otherwise. The degree of a square is the maximal number of its pairwise disjoint overlappers.
Start with a $10\times 10$ grid of just-non-overlapping red squares, and put into any interior vertex of the grid a blue square. All squares are good, except for the boundary red ones (see left figure).
A convenend property is that for any point $x$ in the upper-eft quarter of the figure (within the grid), we may take a large square with $x$ as an either upper-left, or upper-right, or lower-left corner; this square will be automatically good, as it overlaps with many red squares. We attach such squares (not shown in the figure) to each side of the grid. Now, still boundary red squares are bad, but each of them already has degree 3. [In fact, this is not needed, so it is made just for visibility reasons.]
Now take any two adjacent boundary red squares in the upper-left quarter; the right figure shows them magnified. We do the following chain of operations:
$\bullet$ add a green square; to two its dark-green vertices attach large non-overlapping squares, as described above. The red squares become good, the green one has degree 3 (due to the upper red square and two large oens);
$\bullet$ add a magenta square, and attach two large squares to its dark-magenta vertices --- its degree is 3 (due to the green square and two large ones), the green one becomes good;
$\bullet$ add a yellow square, attach three large squares. All squares in this small area are good (the yellow --- due to the upper red square and three large ones).
Applying such operations to every pair of adjacent boundary red squares (or even splitting them in pairs), we reach the goal.
(The construction is superfluous, but, as I told, I did not try to optimize.)
I do not understand the reduction: "for each square $a\in A$ we introduce a color $C_a$, find four pairwise disjoint squares overlapping with $a$, and put into an arrangement $B$ their copies having color $C_a$". But do you copy the square $a$ itself to $B$? If not, then how will $B$ satisfy the requirements?
I do not copy $a$ at this stage. But, when $a$ participates in some disjoint 4-tuple for some other square, a copy of $a$ appears in $\mathbb B$. And for any such copy, color $C_a$ serves.
This is the construction in the upper-left quarter; for the other quarters you will need a slightly different construction, right?
Sure; just reflect this construction accordingly.
Beautiful construction. Thanks!
A solution for $k=4$ colors can be constructed from the arrangements of squares that square the square in the followig way: if an square has a corner in common with the bounding square and is neighbor to only 3 inner squares, then split the neighboring square whose corners are strictly inside the bounding square into four equal parts.
That is illustrated with A. J. W. Duijvestijn's smallest squared square.
the squares with sidelengths 33 and 27 are only neighbors to 3 other squares; splitting inner squares with sidelengths 4 and 8 makes all squares neighbors of at least 4 other squares; increasing the size of each square by $\epsilon$ while preserving the location of their centers makes each square overlap with at least four others.
If we take as vertices of a planar graph the squares and as edges pairs of overlapping squares, then a vertex coloring of that graph yields a coloring of the squares such that no overlapping pair has the same color.
one of the possible 4-colorings of the squares that remains valid after a slight center-preserving enlarging of the squares
The smallest satisfying arrangement of squares from which a satisfying example can be obtain by slightly enlarging the squares of the arrangment:
Consider the "42" at the bottom-right. It overlaps 24, 18, 16, 37, but these cannot be of the same color - they must be of at least two different colors. So it does not overlap 4 squares of the same color.
It is possible for infinitely many squares:
tile the plane with a shape made up of 5 squares with centers at $(-1,0),(0,0),(+1,0),(0,-1),(0,+1)$ each of a different color and sidelengths equal to e.g. $0.6$.
The centers of squares of the same color are the corners of a rotated square grid of knight moves.
The depicted arrangement is also utilized in the printing business, however with different colors.
Looks interesting, but is this an arrangement of finitely many squares?
@ErelSegal-Halevi sorry, I missed the restriction to finitely many squares; Iwilledit my answer accordingly.
Also, the squares in the picture do not overlap
@ErelSegal-Halevi yes, but that is easily fixed by increasing the sidelengths of the squares; I just found is easier for illustrating the pattern.
|
2025-03-21T14:48:31.058053
| 2020-05-24T13:04:42 |
361215
|
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|
Stack Exchange
|
Generalized partitions
Let $\kappa>0$ be a cardinal and $X$ be a set. We set $[X]^\kappa = \{A \in {\cal P}(X): |A| = \kappa\}$.
If ${\cal A}\subseteq {\cal P}(X)$ we say that ${\cal B} \subseteq {\cal P}(X)$ is an ${\cal A}$-partition if for every $X \in {\cal A}$ there is a unique $Y\in {\cal B}$ with $X\subseteq Y$. Note that ${\cal A}$-partitions may not exist, for instance if ${\cal A}$ has 2 distinct members such that one is a subset of the other. Trivially, if ${\cal A} = \{\{x\}: x\in X\}$ then an ${\cal A}$-partition is just a partition in the conventional sense.
If $a \in b \in \omega$ with $a > 1$, is there ${\cal B}\subseteq [\omega]^b$ such that ${\cal B}$ is an $[\omega]^a$-partition?
Can you just construct $\mathcal{B}$ greedily by at every step, chosing the minimal $a$ element subset $S$ that is not already a subset of an element in $\mathcal{B}$ and adding $S \cup T$ to $\mathcal{B}$ where $T$ is a set of cardinality $b-a$ containing only large numbers that are not in any subset already in $\mathcal{B}$?
|
2025-03-21T14:48:31.058153
| 2020-05-24T13:25:21 |
361217
|
{
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"authors": [
"Carlo Beenakker",
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}
|
Stack Exchange
|
Balanced Gray codes for powers of 2
All of the binary 4-bit cyclic balanced Gray code sequences can be formed from simple reversals, bit-permutations, and circular shifts of the one Wikipedia example:
toggle bit number: 0 1 0 2 0 3 2 1 2 3 0 3 1 2 1 3
Is this also true for 8 bits? If not, what is the multiplicity?
I'd like to know the multiplicity for 256 bits also if that's easy.
crossposted -- https://math.stackexchange.com/questions/3688990/balanced-gray-codes-for-powers-of-2 -- please don't, to avoid wasteful duplication of efforts.
Ok, I'll delete the question from the "easy site", I copied it here because no one could answer it.
By the way, I don't actually need to include reversals above (since a 1-for-2 permutation gives the reversal). I'm curious about all symmetries, including this "reverse symmetry", for higher powers.
Please define the terms: simple reversals, bit-permutations, and multiplicity (of what?).
|
2025-03-21T14:48:31.058245
| 2020-05-24T13:26:53 |
361218
|
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|
Stack Exchange
|
Degree $6$ curves in cubic surface
What are the degree $6$ curves in a cubic surface $X$ other than complete intersection of $X$ with a quadric?
Is there any such degree $6$ curve in X? If yes, then can it contain an elliptic curve of degree $4$?
Yes, a lot. For every $g$ with $0\leq g\leq 4$, there are many families of curves of degree $6$ and genus $g$. And of course you can have the union of an elliptic curve of degree 4 and of a conic, if this is the meaning of your last question. Please look at "cubic surfaces" on any textbook of algebraic geometry.
why all such curves contained in a cubic surface ? why the last example in not in any quadric ?
A cubic surface is obtained by blowing up 6 points in the plane. From this description it is easy to construct all examples you want.
|
2025-03-21T14:48:31.058330
| 2020-05-24T14:35:02 |
361221
|
{
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361221"
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|
Stack Exchange
|
Degree $6$ curve in $\mathbb{P}^3$
Let a degree $6$ curve $C$ is given as $(Q_1 \cap Q_2) \cup l^2$ where $Q_i's$ are qudrics and $l$ is a line intersecting $Q_1 \cap Q_2$. Is it true that $C$ always lies on a quadric ?
No. The quadric must belong to the pencil spanned by $Q_1$ and $Q_2$; if a line is contained in such a quadric, it intersects $E=Q_1\cap Q_2$ in 2 points. Just take for $l$ a general line through a point of $E$.
@abx: "The quadric must belong to the pencil spanned by Q1 and Q2" - can you please explain why? Is it true that the space of quadrics containing a given curve is at most two dimensional?
Certainly not in general! But the space of quadrics containing $Q_1\cap Q_2$ is indeed the pencil spanned by $Q_1$ and $Q_2$ — this follows from an easy Koszul complex argument.
Yes. I realized that. Sorry for silly question. However if $Q_1 \cap Q_2$ is contained in a cubic which is singular along the line $l$, then there is a quadric containing $Q_1 \cap Q_2$ and $l$.
|
2025-03-21T14:48:31.058434
| 2020-05-24T15:28:36 |
361225
|
{
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"Jochen Wengenroth",
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"santker heboln"
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|
Stack Exchange
|
Are nuclear operators closed under extensions?
Given $X_i, Y_i$ Banach spaces, $f_j, g_j, T_i$ bounded linear operators for $i=1,2,3$ and $j=1,2$. We have the following diagram
$\require{AMScd}$
\begin{CD}
0 @>>> X_1 @>f_1>> X_2 @>f_2>> X_3 @>>> 0\\
@V VV @V T_1 VV @V T_2 VV @V T_3 VV @V VV \\
0 @>>> Y_1 @>>g_1> Y_2 @>>g_2> Y_3 @>>> 0
\end{CD}
with two horizontal topologically short exact sequences. If $T_1$ and $T_3$ are nuclear operators, does it imply that $T_2$ is nuclear as well? A reference to problems of this general form, would be most welcome.
Just to get the terminology straight in my head: by topologically short exact do you mean the morphisms are bounded linear maps but you have short exactness in the category of vector spaces and linear maps? In other words, $f_1$ has closed range, and not just "$f_1(X_1)$ is dense in $\ker(f_2)$"?
I mean the first, i.e. $f_1$ has closed range. In that sense what I mean by "topological exactness" is algebraic exactness.
The answer is no: you can even have $T_1=T_3=0$ and $T_2$ equal to the identity $id$ on an infinite dimensional Banach space.
Indeed, consider the following commutative diagram with exact rows:
$$\begin{CD}
0@>>> 0 @>0>> X @>id>> X @>>> 0\\
&&@V0VV @VV{id}V @VV0V\\
0@>>>X @>>id> X @>>0> 0 @>>> 0
\end{CD}
$$
See this paper for related results.
That's the case $A=0$ of my example, I believe :) But yours is a cleaner and simpler approach
In the generality stated, the question has a negative answer (which took me embarrassingly long to spot); the point is that when $T_1$ and $T_3$ are not assumed injective or surjective they give us very little traction on what $T_2$ is, in contrast to the intuition one might have from the Five Lemma. The "silly" counterexample might be useful for other settings so here are the details; it should work in any category with zero object and an appropriate notion of "binary sum".
Take objects (=Banach spaces if you wish to be concrete) $A$ and $B$ and let $\iota_L: A\to A\oplus B$, $\pi_L: A\oplus B \to A$, $\iota_R:B\to A\oplus B$ and $\pi_R: A\oplus B \to B$ be the usual embedding and projection operators.
Take as your top row the short exact sequence $A \stackrel{\iota_L}{\to} A\oplus B \stackrel{\pi_R}{\to} B$ and as your bottom row the short exact sequence $B \stackrel{\iota_R}{\to} A\oplus B \stackrel{\pi_L}{\to} A$. The vertical arrow on the left is the zero map $A\to B$, the vertical arrow on the right is the zero map $B\to A$, and the middle arrow is $(0,{\rm id_B}) : A\oplus B \to A\oplus B$. Then everything commutes.
To turn this into a counterexample for the original question, just take $B$ to be your favourite infinite-dimensional Banach space and $A$ to be an arbitrary Banach space.
On the other hand, I think I can prove the following: suppose I am given everything in your initial diagram except the middle arrow $T_2$, so that we have (strict) exactness on the top row and bottom row and nuclear operators $T_1:X_1\to Y_1$, $T_3:X_3\to Y_3$. Then $T_1$ has a nuclear extension $R:X_2\to Y_1$, $T_3$ has a nuclear lift $S:X_3\to Y_2$, and defining $\theta= g_1R+Sf_2$ gives a "middle arrow" which is nuclear and does make all the squares commute. So depending on your intended applications, this might be of some use; it says we can manufacture an "extension" of $T_1$ and $T_3$ which is nuclear. Furthermore, even if one wants to show that a given $T_2$ is nuclear, this construction might help since under certain extra conditions one may be able to prove that $\theta=T_2$.
This was a quick response! In fact I just wanted to edit my question, to say that $T_1, T_2$ and $T_3$ should be surjective. Your answer is interesting nonetheless.
@santkerheboln before you edit the question, I should point out that surjective operators on Banach spaces are nuclear if and only if they are finite-rank. So your modified question will have a positive answer for trivial reasons
Indeed, I guess a more sensible assumption (to get a positive result and also in view of intended applications) would be that the squares commute. In any case, the question as stated has been answered exhaustively. Thank you (all)!
I now think the conditions the $T_i$ having dense ranges for $i = 1,2,3$ and the squares commuting should give the desired equality $\theta = T_2$ in your approach.
Addendum: Actually, Jochen's example shows that even if $T_i$'s are assumed to be surjective, $\theta$ may not equal $T_2$. However, if $T_i$ have dense ranges, I think $\theta$ could be "close" to $T_2$, it seems unitarily equivalent.
Of course, Yemon was faster than me. But I want to emphasize that
the point is very elementary linear algebra: The simple commutative diagram
$\begin{CD}
0 @>>> \mathbb R @>f_1>> \mathbb R^2 @>f_2>> \mathbb R @>>> 0\\
@V VV @V T_1 VV @V T_2 VV @V T_3 VV @V VV \\
0 @>>> \mathbb R @>>f_1> \mathbb R^2 @>>f_2> \mathbb R @>>> 0
\end{CD}$
with the natural inclusion $f_1(x)=(x,0)$ and projection $f_2(x,y)=y$ in the rows shows shows that $T_2$ is not at all determined by $T_1$ and $T_3$. If $T_2$ is given by a matrix $\begin{bmatrix} a&b\\ 0&c\end{bmatrix}$ you get nothing for $b$.
Actually I spent an embarrassingly long time trying to turn the observations mentioned at the end of my answer into a proof that the original question had a positive answer. It wasn't until I tried to work out why I was getting stuck that the basic point observed by you and M.Gonzalez hit me
It also took me far too long...
|
2025-03-21T14:48:31.058772
| 2020-05-24T16:00:23 |
361229
|
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|
Stack Exchange
|
Finitely generated commutative rings with the same profinite completion
Let $R_1$ and $R_2$ be two finitely generated commutative rings. Assume that their profinite completions are isomorphic: $\widehat{R_1}\cong \widehat{R_2}$.
Suppose that $R_1$ is a domain. Does it imply that $R_2$ is a domain as well?
The isomorphism between $\widehat{R_1}$ and $\widehat{R_2}$ is equivalent to the existence of a bijection $\phi: \operatorname{maxSpec}(R_1)\to \operatorname{maxSpec}(R_2)$ between the sets of maximal ideals, such that for all $M\in \operatorname{maxSpec}(R_1)$ the corresponding completions $(R_1)_{(M)}$ and $(R_2)_{(\phi(M))}$ are isomorphic.
Since this took me time to figure out: here are two non-isomorphic f.g. scalar rings with isomorphic profinite completions: $\mathbf{Z}[1/6]\times\mathbf{Z}$ and $\mathbf{Z}[1/2]\times\mathbf{Z}[1/3]$.
How about $R_2 = \mathbf{F}_p[x,y]/(xy)$ (two lines intersecting at a point) and $R_1 = \mathbf{F}_p[x,y]/(y^2 - x^2(x-1))$ (irreducible nodal cubic)?
@PiotrAchinger: From the profinite completion you can read off the number of $\mathbb{F}_p$-points. This is $2p-1$ for $R_2$ and $p-1$ for $R_1$, so the completions are not isomorphic. In general, I guess zeta functions could be an approach to the problem.
|
2025-03-21T14:48:31.058886
| 2020-05-24T16:24:01 |
361231
|
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|
Stack Exchange
|
Examples of Heyting categories that are not toposes?
When explaining how Heyting categories can model first order logic it would be nice to be able to give some small example and contrast it with Set-semantics. I realized however that I don't know of any Heyting category which is not also a topos. It would be nice to have more concrete examples to give.
A search netted me a master thesis that discussed the Heyting category structure on some categories of graphs. Do you have other examples of Heyting categories that are not toposes?
EDIT: Syntactic categories are Heyting categories which are not necessarily toposes as pointed out by godelian. I am however looking for examples not arising out of logic (:
Any first-order (intuitionistic) theory gives rise to a small Heyting category: its syntactic category. See e.g. Johnstone
Yes, that's true, I had something more concrete in mind though, that I could show someone who is less familiar with categorical logic.
In the category of sets, take the full subcategory whose objects are the countable sets. This looks to me like a Heyting category, also satisfying classical logic.
A similar sort of example is the category of classes (relative to some model of set theory).
In case someone else stumbles upon this question: there are several examples of "natural" Heyting categories arising in algebraic logic. Ghilardi and Zawadowski (2002) have some results concerning this, which imply that the dual category to finitely presented Heyting algebras is a Heyting category. With duality theory this can be given a more concrete description, for instance with Esakia duality. There are other examples of this kind, intimately related to the question of when a variety of a model completion, which have been explored in that literature. Moreover, such examples are not toposes, as also shown by Ghilardi and Zawadowski.
|
2025-03-21T14:48:31.059055
| 2020-05-24T17:06:07 |
361233
|
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|
Stack Exchange
|
An inequality for the Hessian of eigenfunctions of the Laplacian on compact manifolds
Let $(M,g)$ be a compact Riemannian manifold, and let $\Delta$ be the Laplace-Beltrami operator. Let $\lambda_1 >0$ be the first positive eigenvalue. That is, there exists a non-trivial function $\phi$ satisfying $\Delta \phi + \lambda_1\phi = 0$, and that $\lambda_1$ is the smallest such real number. Then one can always normalize the eigenfunction such that $\max_M\phi = 1$ and $\min_M = -k\geq -1$. I have two questions:
Can one further choose an eigenfunction such that the maximum is attained at a unique (or at worse an isolated) point? For example, for a sphere $\mathbb{S}^2$ with the round metric, the eigenfunctions are given by $\cos d$, where $d$ is the distance to a fixed point (say the north pole N). The north pole in this case is the unique maximum. On the other hand if we take the product $\mathbb{S}^2\times\mathbb{S}^2$ with the product of the round metrics, and if we let $d_1$ and $d_2$ be the distance functions (from the respective north poles) on each factor, then each $u_i = \cos d_i$ is an eigenfunction, but the point where they attain maximums are clearly is unique (for instance $u_1$ takes maximum value on $\{N\}\times\mathbb{S}^2$). But in this case one can take $u = (\cos d_1 + \cos d_2)/2$ to be a noramlized eigenfunction with a unique maxima. So my question is whether such a normalized eigenfunction exist on all compact Riemannian manifolds $(M,g)$.
If further the Ricci curvature is bounded below, say $\mathrm{Ric}_g\geq n-1$, where $n$ is the dimension, it is well known that $\lambda_1\geq n$. then does there exist an eigenfunction $u$ such that $$\mathrm{Hess}(u) \geq -\frac{\lambda_1}{n}ug?$$ If the answer to this is affirmative, then such an eigenfunction will clearly have an isolated maximum (since the Hessian will be negative definite at the maximum), hence answering the first question also in affirmative. It is of course well known (Obata) that $\lambda_1 = n$ if and only if $(M,g)$ is isometric to the round sphere. Moreover, in this case we necessarily have that for any eigenfunction $u$, both sides of the above inequality must in fact be equal. This follows from a simple integration by parts argument. On the other hand, suppose $\lambda_1>n$, then once again taking the example of $\mathbb{S}^2\times\mathbb{S}^2$ it is clear that the above inequality of the Hessian will not hold for all eigenfunctions. My guess is that the Hessian inequality is too strong, and is likely wrong. But a counter example would be nice.
By Uhlenbeck's theorem, generically the first nonzero eigenfunction is unique up to scaling. So there's not really any choice in your first question. (Uhlenbeck also shows generically eigenfunctions are Morse, which I think answers the question about isolated maxima.)
@Neal - Thanks for the reference on Uhlenbeck's theorem. In my case of course the Riemannian metric is given, and I dont seem to have any room to perturb it to make use of Uhlenbeck's theorem. But in any case, as a side note, if the Riemannian metric is Kahler, then would Uhlenbeck's theorem provide a nearby Kahler metric that has a one dimensional eigenspace?
The answer to question 1 is no. On a two-dimensional torus $\mathbb{S}^1 \times \mathbb{S}^1$, the first nontrivial eigenspace consists of $\mathrm{Span}\{\cos(x),\sin(x),\cos(y),\sin(y)\}$. Each eigenfunction in this eigenspace is separable, i.e. it is the product of a lower-dimensional function and a constant function.
I've asked a related question about whether separable eigenfunctions are the only example of this: https://mathoverflow.net/questions/384367/non-separable-laplace-beltrami-eigenfunctions-have-isolated-critical-points-ref.
|
2025-03-21T14:48:31.059422
| 2020-05-24T17:19:28 |
361234
|
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|
Stack Exchange
|
Solving recurrence of a three variable function
I am fairly new to generating functions and have been trying to solve the following recurrence for a computer science problem.
$$ f(k,d,n) = \sum_{i=1}^{n-1} \binom{n-2}{i-1} \left(\frac{1}{2}\right)^{n-2} \left(\sum_{j=0}^{k} f(k-j,d-1,n-i)\cdot f(j,d-1,i)\right)$$
Note that
$$
f:\mathbb{N}^3 \rightarrow [0,1]
$$
with the following base cases:
$$
\begin{split}
f(k,d,1)& = \begin{cases}
1 & (k=0 \wedge d!=0) \vee (k=1 \wedge d=0)\\
0 & \text{otherwise}
\end{cases} \\
f(k,d,2)&= \begin{cases}
1 & ((k=0 \wedge d!=1) \vee (k=2 \wedge d=1))\\
0 & \text{otherwise}
\end{cases} \\
\end{split}
$$
and the following domains:
$$
\begin{split}
n &\in \{1,2,3,4,\ldots\} \\
k &\in \{0,1,2,\ldots,n-1,n\} \\
d &\in \{-\infty,\ldots,-2,-1,0,1,2,\ldots,\infty\} \\
\end{split}
$$
From f(k,d,n) and base cases we can derive the following for d<0:
$$ f(k,d,n)= \begin{cases}
1 & k=0 \\
0 & k>0
\end{cases} \\
$$
I would like to mention that $f(k,d,n)$ can also be expressed as the following:
$$ f(k,d,n) = \sum_{i=1}^{n-1} \binom{n-2}{i-1} \left(\frac{1}{2}\right )^{n-2} \left(\sum_{j=0}^{k} f(j,d-1,n-i)\cdot f(k-j,d-1,i)\right)$$
Thank you for your time and help in advance.
Define
$$F_d(x,y) := \sum_{k\geq 0}\sum_{n\geq 1} f(k,d,n) x^k \frac{y^{n-1}}{(n-1)!}.$$
Then the recurrence is equivalent to
$$\frac{\partial F_d}{\partial y}(x,y) = F_{d-1}(x,y/2)^2,$$
while the initial conditions imply
$$F_0(x,0)=x,\quad F_d(x,0)=1\ \text{for}\ d\ne 0$$
and
$$\frac{\partial F_1}{\partial y}(x,0)=x^2,\quad \frac{\partial F_1}{\partial y}(x,0)=1\ \text{for}\ d\ne 1.$$
This can be seen as a generalization of your previous question.
Well, I was trying to look at the same problem from different perspective and was hoping to be able to find an "easier" formula for it, since the formula in my previous question did not have non recursive expression.
@KokoNanahji: The g.f. for $\sum_{k=0}^n kf(k,d,n)$ is $\frac{\partial F_d}{\partial x}(1,y)$. If we denote it as $H_d(y)$, then $H_d'(y) = 2 F_{d-1}(1,y/2) H_{d-1}(y/2)$.
thank you very much for your quick response. I deleted my question because I figured out a way to simplify $\sum_{k=0}^{n} k \cdot f(k,d,n)$ using the constraint that I forgot to mention which is $\sum_{k=0}^{n} f(k,d,n)=1$. Let $f(d,n)=\sum_{k=0}^{n} k \cdot f(k,d,n)$ then $f(d,n)=\sum_{i=0}^{n-2} \binom{n-2}{i} \left(\frac{1}{2}\right)^{n-3} \cdot f(d-1,i+1) $
|
2025-03-21T14:48:31.059574
| 2020-05-24T17:41:18 |
361239
|
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|
Stack Exchange
|
Ensemble averaging in a random graph (or network) in the large $N$ limit
I have a random graph/network described by the adjacency matrix $(a_{ij})_{N\times N}$ where $a_{ij}=1$ with probability $p$. Each node in the graph is associated with a continuous quantity $\eta_i=\sum_j a_{ij}\eta_j$ in $\mathbb{R}$. In this framework the ensemble average of $\eta_i$, denoted by $\bar{\eta}$ is just $\mathrm{E}[\eta_i]$. Similarly, $\overline{\eta^2} = \mathrm{E}[\eta_i^2]$. I want to understand the derivation for
$\overline{\sum_j a_{ij}\eta_j} = Np\bar{\eta}$
and
$\overline{\sum_j a_{ij}\eta_j-Np\bar{\eta})^2} = Np(\overline{\eta^2} - p\bar{\eta})\quad\quad\quad\quad\quad$ (*)
The first formula (correct me if I am wrong, please) should come from the reasoning that
$\mathrm{E[\sum_j a_{ij}\eta_j]} = \sum_j \mathrm{E[a_{ij}]}\mathrm{E}[\eta_j] = \bar{\eta}\sum_jp = Np\bar{\eta}$
But for the second formula I cannot find a correct derivation. I actually obtain $\overline{\sum_j a_{ij}\eta_j-Np\bar{\eta})^2} = Np(\overline{\eta^2} - \bar{\eta})$, and I cannot figure out where the $p$ factor inside the parenthesis in (*) comes from. Should I opt for a typo (*). Or is anyone that can provide a correct derivation?
Instead of downvoting, comment please and be more constructive
Ok, it seems that this section of StackExchange is really made for you to figure out your own answer... After some lengthy frustrating trial and error, there is no typo in the given solution. The mistake I was doing was in the calculation of $\mathrm{E}[(\sum_ja_{ij}\eta_j)^2]$. In the development of this multinomial square, you need to be careful when you compute the expectation of the mixed terms. In fact:
$\mathrm{Var}[\sum_ja_{ij}\eta_j] = \mathrm{E}[(\sum_ja_{ij}\eta_j - Np\bar{\eta})^2] = \mathrm{E}[(\sum_ja_{ij}\eta_j)^2] - N^2p^2\bar{\eta}^2$
$=\mathrm{E}[\sum_ja_{ij}^2\eta_j^2] + \mathrm{E}[\sum_ja_{ij}\eta_j\sum_{k,k\ne i}a_{ik}\eta_k] - N^2p^2\bar{\eta}^2 = Np\cdot \overline{\eta^2} + Np\cdot(N-1)p\cdot \bar{\eta}^2 - N^2p^2\bar{\eta}^2$
$=pN(\overline{\eta^2}-p\bar{\eta}^2)$
I hope that it can help others. FYI this type of calculation occurs very often in the study of neural networks.
|
2025-03-21T14:48:31.059724
| 2020-05-24T18:04:56 |
361242
|
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"Bedovlat",
"Yemon Choi",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361242"
}
|
Stack Exchange
|
Self-intersecting irreducible real projective elliptic surface
I will say that a homogeneous polynomial $P(X)\in\mathbb{R}[X]$ ($X=(X_1,\ldots,X_n)$) is elliptic if its zero locus in $\mathbb{R}^n$ is $\{0\}$.
I will say that the zero locus of a homogeneous polynomial $P(X)\in\mathbb{R}[X]$ in $\mathbb{C}^n$ is self-intersecting (on a semi-real line) if for some $x,y\in\mathbb{R}^n$ the polynomial $P(x+\lambda y)$ of complex variable $\lambda\in\mathbb{C}$ has a root of multiplicity higher than one.
Question: What is an example of a real elliptic homogeneous polynomial $P(X_1,\ldots,X_n)\in\mathbb{R}[X_1,\ldots,X_n]$, irreducible over $\mathbb{C}$, such that its zero locus in $\mathbb{C}^n$ is self-intersecting?
Discussion: If I remove the condition of ellipticity then $P(X,Y,Z)=Y^2Z-X^2Z-X^3$ gives a self-intersection along the real line $(0,0,1)+\lambda(0,1,0)$.
I have zero background in algebraic geometry, so please, be as specific as possible. Thank you.
This question is posted on MSE here.
As a user for several years, you should already know that cross-posting to MO within 24 hours of posting to MSE is probably not long enough to give the MSE users a chance to answer. FWIW, as I am not an algebraic geometer, if the question had only been posted on MO I would not necessarily have voted to move it to MSE; but since it started on MSE, it seems more appropriate to wait a while there before opening a duplicate on MO
As a user for several years, I can bet that no-one is going to answer any of my questions in MSE. I am posting there first only as a tribute to the formal notion that the question may be indeed very elementary for a specialist in the subject, and they may prefer to answer it in MSE rather than in MO. But that is only a theoretical possibility and has never happened. If you answer this question on MSE, you are welcome to close it on MO. But as long as there is no answer in any of them, I see no reasoning in closing a question.
Fair enough. As it happens, I haven't downvoted or voted to close; my comment was prompted just by seeing the question come up in the "review votes to close" queue
|
2025-03-21T14:48:31.059888
| 2020-05-24T18:17:57 |
361244
|
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"Josiah Park",
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|
Stack Exchange
|
Pseudo-differential operators and differential operator
I am totally new to pseudo-differential operators and I’m wondering if a differential operator is a pseudo-differential operator.
So, I want to show , using the definition of the symbol given by Hörmander definition of symbols (see Pic), that if $p(x,\xi) =\sum_{|\alpha|\le m} a_{\alpha}(x) \xi^{\alpha}$ is a symbol of differential operator, it’s also a symbol in $S^m$ but I could not bounded the coefficients in x...
Thanks
This is all satisfactorily explained here: https://ncatlab.org/nlab/show/pseudodifferential+operator#every_differential_operator_is_a_pseudodifferential_operator.
In my opinion the question is ill-defined. What should be the definition of a differential operator? The class of objects which are called differential operator in the ncatlab is certainly much more narrow than what a researcher in differential operators would it expect to be.
Hörmander's definition of a pseudo-differential operator on an open subset $\Omega$
of $\mathbb R^n$ in the class $\text{Op}S^m$ ($m\in \mathbb R$) is the following: take a symbol $a$, that is a smooth function on $\Omega\times\mathbb R^n$ such that
\begin{multline}
\forall K \text{ compact }\subset \Omega, \forall \alpha, \beta\in \mathbb N^n,\ \exists C_{K\alpha\beta},\ \forall (x,\xi)\in K\times\mathbb R^n,\\ \quad
\vert(\partial_x^\alpha\partial_\xi^\beta a)(x,\xi)\vert\le C_{K\alpha\beta} (1+\vert \xi\vert)^{m-\vert \beta\vert},
\tag{1}\end{multline}
and consider the operator $\text{Op}(a)$ defined on $C_c^\infty(\Omega)$ by the formula
$$
(\text{Op}(a) u)(x)=\int e^{2iπ x\cdot \xi} a(x,\xi)\hat u(\xi) d\xi.
$$
Then if you consider a differential operator
$P=\sum_{\vert \gamma\vert\le m} p_\gamma(x) D_x^\gamma
$
with $p_\gamma\in C^\infty(\Omega)$, you find that it is a pseudo-differential operator of order $m$ on $\Omega$ since if $u\in C^\infty_c(\Omega)$, you have
$$
(P u)(x)=\sum_{\vert \gamma\vert\le m}p_\gamma(x)\int e^{2iπ x\cdot \xi}\xi^\gamma \hat u(\xi) d\xi
=\int e^{2iπ x\cdot \xi}p(x,\xi) \hat u(\xi) d\xi,
$$
with $p(x,\xi)=\sum_{\vert \gamma\vert\le m}p_\gamma(x)\xi^\gamma$; the symbol $p$ obviously verifies (1).
|
2025-03-21T14:48:31.060040
| 2020-05-24T18:50:40 |
361245
|
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"Aleksandar Milivojević",
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|
Stack Exchange
|
Finite CDGA model for a compact manifold
Is it true that a compact smooth manifold always has a finite-dimension commutative dg algebra model?
Same question can be asked about compact CW complexes.
More generally, is it true that a CW complex of finite type always has a CDGA model of finite type?
PS. I always assume the coefficients are rationals or any other field of characteristic zero.
The main problem here is when H^1 is non-trivial.
For me it is the same thing. I mean graded commutative if this is the question, which is the same as commutative in the symmetric monoidal category of dg spaces.
A positive answer to the third question implies a positive answer to the second, because we can always truncate a CDGA at the top degree where cohomology is non-trivial. A positive answer to the second implies a positive answer to the first, because any compact smooth manifold admits a handle decomposition.
Do you work with rational coefficients? If not, then the quotient $M$ of $S^3$ by the standard $\mathbb{Z}/3$-action with $\mathbb{F}_3$-coefficients is a counterexample. Here the Massey-product $\langle x,x,x\rangle \in H_2$ is nonzero, but for commutative DGAs such a Massey-product would always be zero (choose the same chain bounding $x*x$ on the left and on the right).
HenrikRuping, I do assume the coefficients are rational numbers. It is a question in Rational Homotopy Theory. I updated the question.
The answer to the first question is yes in the simply-connected case, by Lambrechts-Stanley: https://arxiv.org/abs/math/0701309. But judging by your first comment, you seem to be aware of this.
@MarkGrant an alternative argument for a simply connected $n$-complex (with no assumption on Poincare duality): take a minimal model for the space, and set all elements strictly above degree $n$ to 0; we get an induced map to the space that is an iso on cohomology everywhere except in deg $n$, where we have introduced some kernel by getting rid of the deg $n+1$ elements. Introduce elements in deg $n-1$ to kill this kernel, and map them to 0. Then again set all elements in degree $>n$ to 0. This gives a finite model; note that in the presence of $H^1$ we might have to repeat the last step.
|
2025-03-21T14:48:31.060209
| 2020-05-24T18:57:55 |
361246
|
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"Yemon Choi",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361246"
}
|
Stack Exchange
|
Is $F: B(H)\to B(H\otimes K)$, $X\mapsto X\otimes \mathbb{1}_K$ completely contractive?
Take Hilbert spaces $H$ and $K$. Consider a linear map $F: B(H)\to B(H\otimes K)$, $X\mapsto X\otimes \mathbb{1}_K$.
Is it true that $F$ is completely contractive? If it is, I would be very grateful for references.
It is a complete isometry, with respect to the standard operator space structures on both sides. The particular statement you give is a special case of "injective $*$-homomomorphisms are isometries" applied to each matrix level
So I am not sure if there is an explicit reference as such; the statement seems to follow from the fact I stated, and the way that the standard matrix norms are defined for Cstar algebras
|
2025-03-21T14:48:31.060282
| 2020-05-24T19:07:43 |
361249
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361249"
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|
Stack Exchange
|
General construction of enveloping C*-algebra, left/right-regular representation, etc
In a number of contexts (e.g. groups, crossed products, groupoids, Fell bundles) there are similar constructions of enveloping C*-algebras and left/right-regular representations that incorporate intertwining operators, subrepresentations, cyclicity, matrix coefficient functions, etc. Is there a generalized version of this construction?
|
2025-03-21T14:48:31.060336
| 2020-05-24T19:37:58 |
361251
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361251"
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|
Stack Exchange
|
Discriminant mod 8
Let $f,g\in\mathbb{Z}[X]$ be monic polynomials of degree $n$, with discriminants $\Delta_{f}$ and $\Delta_{g}$, such that $f=g$ in $\mathbb{F}_{2}[X]$ and $\Delta_{f}$ is odd.
Is there an elementary way to see that $\Delta_{f}\equiv\Delta_{g}$ mod 8?
(For even $\Delta_{f}$ this fails, as $f=X^{2}$ and $g=X^{2}+2X$ show.)
By a theorem of Stickelberger, if $r$ is the number of irreducible factors of $f$ in $\mathbb{F}_{2}[X]$, then $\Delta_{f}\equiv1$ mod 8 iff $n-r$ is even, and $\Delta_{f}\equiv5$ mod 8 otherwise. But this is a non-trivial result in algebraic number theory...
|
2025-03-21T14:48:31.060400
| 2020-05-24T19:46:23 |
361253
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Jeff Yelton",
"https://mathoverflow.net/users/24757"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361253"
}
|
Stack Exchange
|
primes of multiplicative reduction for elliptic curves with rational $\ell$-torsion
Very recently, I made an observation from scanning lists of elliptic curves on the LMFDB that leads me to the following (unproven) statement:
Fix $\ell \in \{5, 7\}$. Let $E$ be an elliptic curve over $\mathbb{Q}$ with a rational $\ell$-torsion generator. Then if $p \neq \ell$ is a prime at which $E$ has multiplicative reduction, and if $p$ appears in the prime factorization of the denominator of $j(E)$ with exponent not divisible by $\ell$, we have $p \equiv \pm 1$ (mod $\ell$).
I don't know how to get a spreadsheet of $j$-invariants of elliptic curves satisfying the hypothesis given above so I couldn't check this efficiently, but this was the case for each of the dozens of elliptic curves I checked. I noticed other strong trends as well, such as the fact that primes $p \neq \ell$ appearing in the factorization of the \textit{numerator} of $j(E)$ with exponent not a multiple of $\ell$ tend to satisfy $p \equiv \pm 1$ (mod $\ell$), but this is not true all of the time -- the exceptions are always smaller primes and always appear with exponent $3$ in the data I checked. Also, such a statement fails badly when I loosen the assumption about a rational $\ell$-torsion point to requiring only that the mod-$\ell$ Galois image be a Borel subgroup.
Is there any obvious proof or reason to expect that the main assertion above is true? Right now I don't see any way to prove it using rational points on modular curves, but I'm really not an expert on modular curves. No strategy using Neron models is jumping out at me either.
Let $E/\mathbb{Q}$ be an elliptic curve with a rational $\ell$-torsion point $P$ for $\ell\in\{5,7\}$.
Let $p$ be a prime different from $\ell$. Then the formal group $\hat{E}(p\mathbb{Z}_p)$ has no $\ell$-torsion, which shows that the reduction of $P$ is not the point at infinity.
Suppose $E$ has bad reduction at $p$. I claim that it follows that the reduction is multiplicative. This is because the group of connected components for additive reduction has order coprime to $\ell$ and hence $P$ would have to be a non-singular point. But then the reduction would have order $p$.
Now if $E$ has multiplicative reduction, there are two cases. Either the point $P$ reduces to the singular point or to a non-singular non-zero point.
In the frist case the group of components must now be a cyclic group of order divisible by $\ell$. Hence the reduction is split multiplicative and the Tamagawa number and the $p$-adic valuation of $j(E)$ are both divisible by $\ell$.
In the second case (that is what you observe), the point $P$ is a non-zero point in the group of non-singular points in $E(\mathbb{F}_p)$. This group is either cyclic of order $p-1$ or cyclic of order $p+1$ depending on whether the reduction is split or non-split. In the first case $\ell\mid p-1$ and in the second $\ell\mid p+1$.
Thanks! It seems our proofs are essentially equivalent; I was delayed half an hour posting mine and reading yours due to a power outage in my area.
Never mind -- I think I can answer this. The answer occurred to me as soon as I posted the question; I guess sometimes the process of writing it out for other people helps trigger the realization that there is a pretty straightforward answer.
Suppose that $E$ has multiplicative reduction at $p$ and $p$ appears in the factorization of the denominator of $j(E)$ with exponent $e$ with $(e, \ell) = 1$. If $E$ has non-split multiplicative reduction at $p$, we extend the base field $\mathbb{Q}$ to an appropriate quadratic extension, so we may assume that $E$ is defined over a number field $K$ and has split multiplicative reduction at a prime $\mathfrak{p}$ whose residue field $k$ has order either $p$ or $p^2$. Now the special fiber of the Neron model of $E$ at the prime $\mathfrak{p}$ has cyclic component group of order $e$ with the identity component being a copy of $\mathbb{G}_m / k$, so that the order of the special fiber as an algebraic group is $e(p-1)$ or $e(p^2-1)$. Since $\ell \neq p$, the hypothesized rational point of order $\ell$ reduces to a $k$-point of order $\ell$ in the special fiber of the Neron model, so we get $\ell | e(p-1)$ or $\ell | e(p^2-1)$. Since by hypothesis $\ell \not| e$, this implies that $p \equiv \pm 1$ (mod $\ell$).
This doesn't seem to explain the trend towards primes in the numerator of $j(E)$ being $\equiv \pm 1$ (mod $\ell$), but it answers my main question.
|
2025-03-21T14:48:31.060688
| 2020-05-24T20:41:17 |
361259
|
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|
Stack Exchange
|
Reference for "every 5-dimensional polytope has a 3-gonal or 4-gonal face"
It seems to be folklore that every 5-dimensional convex polytope has a 3-gonal or 4-gonal face of dimension two. I was not able to track down a source for that claim.
Alternatively, I would be interested in a nice and short proof of the statement.
Not an answer, but a possible lead. I found in
Grünbaum, Branko. Convex Polytopes. Vol. 221. Springer Science & Business Media, 2013. (MSN)
the following passage:
Grünbaum, p.224b.
Unfortunately I cannot access the reference [f], which is On low-dimensional faces that high-dimensional polytopes must have, by Gil Kalai. But perhaps this will help: @GilKalai.
The doi of the cited article is 10.1007/BF02122781 The result is in Theorem 1.
Thanks to @EFinat-S, the entertainingly titled Kalai - On low-dimensional faces that high-dimensional polytopes must have (MSN). Sadly for this situation, @-ing GilKalai won't work, since they haven't yet been involved in the conversation.
@LSpice: Thanks re @-ing. I did not understand that. I guess to protect users from spamming.
The general reason for why anything with SO software is the way it is is that it is intended to fit only the exact needs and preferences of its creators. (For example, you can't @ from questions at all, so it wouldn't work even if GilKalai were here; you can only @ one user per comment, and it mustn't be the author of the parent post; etc.) Lots of such quirks are documented at https://meta.stackexchange.com/questions/43019/how-do-comment-replies-work . MO has some special treatment grandfathered in, but, as far as I know, nothing affecting @-replies.
@LSpice I changed the link to a doi link, rather than a Springer one, since they are prone to breaking. And I included author and title in the plain text of the answer, for better information content/reference.
|
2025-03-21T14:48:31.060851
| 2020-05-24T21:38:20 |
361263
|
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"url": "https://mathoverflow.net/questions/361263"
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|
Stack Exchange
|
Inclusion $H^1_0(\Omega)\cap C^k(\Omega)\subset C^0(\overline\Omega)$?
Let $\Omega\subset {\bf R}^d$ be a bounded domain with a Lipschitz boundary. Assume that a function $u$ is $C^k$ inside $\Omega$ and that $u$ also belongs to $H^1_0(\Omega)$. Can one conclude that $u$ is continuous on the closure $\overline{\Omega}$? (If necessary, one can assume $k=\infty$.)
The answer is no. If I will find time I will write more details tomorrow.
$u=z/r$ (multiplied by a cut-off), $r=\sqrt{x^2+y^2+z^2}$ for $z>0$ is a counterexample since continuity at 0 fails.
|
2025-03-21T14:48:31.060923
| 2020-05-24T22:05:05 |
361266
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361266"
}
|
Stack Exchange
|
Are all classes Stiefel-Whitney classes?
When I thought of this question, I was sure it must have been asked before on this site, but I could't find anything. Maybe my search skills are lacking, or maybe the question is obvious and it's my math skills that are lacking. Anyway, here it goes.
For a $CW$-complex $X$ let $sw^*X$ be the subring of $H^*(X,\mathbb{F}_2)$ generated by all classes which are Stiefel-Whitney classes of some vector bundle over $X$. It is not hard to see that $sw$ is a proper subfunctor of mod 2 homology. For example (and this might be overkill) if you take the right dimensional sphere $S^n$, then by Bott periodicity, $KO(S^n)=0$, so $sw^*S^n=0$.
Now let $SW^*X$ be the subring generated by all classes which are either Stiefel-Whitney classes of some vector bundle over $X$, or suspensions or desuspensions of such classes.
$\textbf{Edit}$: Perhaps it wasn't clear from context, but I want $SW^*$ to be a functor, so I force it to be closed under pullbacks. For that reason I am puzzled by Nicholas Kuhn's suggested answer below. Also, we know in retrospect that $H\mathbb{F}_2^*X$ is a summand in $MO^*X$, and that thing is sort of tautologically built out of characteristic classes...
Is $SW^*X=H^*(X,\mathbb{F}_2)$?
I suppose the question is equivalent to something like: does the identity map of $K(\mathbb{F}_2,n)$ factor, stably, through some $BO(m)$?
A 1968 paper in Topology by Anderson and Hodgkin shows that
$KO^*(K(\mathbb F_2, n)) = 0$ if $n \geq 2$. This implies that if $n \geq 2$, then no nonzero classes in $H^*(K(\mathbb F_2,n);\mathbb F_2)$ are SW classes. (And of course, $BO(1) = K(\mathbb F_2,1)$.)
Sorry, why does this show that no nonzero classes can be pulled back from some suspension of some $BO(m)$?
John, correct me if i'm wrong, but this follows from (i) F_2-cohomology of K(F_2,n) is polynomial algebra and (ii) multiplication in deg>0 cohomology of any suspension is trivial. But I don't see by now any trivial reason for the desuspension statement
@John, upd: for desuspension of the form \Sigma^N K(n)\to BO(\infty) it follows from Nicholas answer because after adjunction we obtain a map and as a class in KO^*(K(n)) he is equal to zero. Is there are a way to see why suspension on both sides is zero on cohomology?
|
2025-03-21T14:48:31.061194
| 2020-05-22T02:07:15 |
361012
|
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"sort": "votes",
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|
Stack Exchange
|
Whitney sum formula for topological Pontryagin classes
Is there a Whitney sum formula for topological rational Pontryagin classes? I thought the answer is yes, but now I cannot find a reference. Is it even true? The PL case would also be of interest.
Let p(E) denote the total (integral) Pontryagin class of a real bundle E. It's not too hard to show that p(E (+) F) = p(E) * p(F) mod 2-torsion. The 2-torsion term is stated in Theorem 1.6 of Brown's "The Cohomology of BSO_n and BO_n with Integer Coefficients".
@sdk: Brown's paper is about vector bundles, while my question is about locally trivial bundles with fiber homeomorphic to $\mathbb R^n$.
Ah, yes. Sorry, didn't read carefully enough.
Yes. A simple argument is that $BO \to BTOP$ is a rational equivalence and an H-space map (in fact even an infinite loop map), so it follows from the Whitney sum formula for vector bundles.
Edit: The argument for this is as follows. Let $\mu : BTOP \times BTOP \to BTOP$ be the map corresponding to Whitney sum of (stable, topological) bundles. The question is whether the identity
$$\mu^* p_n = \sum_{ a + b = n} p_a \otimes p_b$$
holds. As the map $BO \to BTOP$ is a rational equivalence and an H-space map, it is equivalent to verify this equation in the cohomology of $BO$ instead, where it indeed holds.
Thank you! For my record here is a reference: on p.215 of Boardman-Vogt's book "Homotopy invariant algebraic structures on topological spaces" the authors say (roughtly) that since $O\to Top\to F$ are topological monoid homomorphisms under the Whitney sum, these maps are infinite loop space maps, and hence so are the corresponding maps of classifying spaces $BO\to BTop\to BF$. Actually, instead of this last sequence they write $BO\to Top\to F$ which I assume is a misprint.
Is it true more generally that for any topological monoid (or at least group) homomorphism $G\to H$ the corresponding map of classifying spaces $BG\to BH$ is an infinite loop map?
@ConnorMalin: thank you, I see now that Boardman-Vogt's result has quite a few technical assumptions; they speak of symmetrical monoidal functors. I gather, the point is that Whitney sum is commutative.
Sorry, I deleted my comment in fear that I was missing something. For posterity, I said that usually a group will not be an infinite loop space.
@Oscar: actually, I am confused. It seems your argument only proves that topological Pontryagin classes satisfy the Whitney sum formula on vector bundles. I need it for topological bundles. Basically, there are two H-maps which are rational equivalences from $BSO$: one to $BSTOP$ and the other one to the product of $K(\mathbb Z, 4i)$ spaces where $i$ varies over positive integers. The latter map is the integral Pontryagin class. One needs to show that the topological Pontryagin class completes the square rationally. None of the maps from $BSO$ can be reversed, I think.
@Igor Belegradek: stably and rationally, every topological bundle has a unique vector bundle structure; this is just rephrasing that the map BSO->BSTOP is a rational equivalence. The fact that this map is one of H-spaces says exactly that the induced vector bundle structure is compatible with sums. Note that it suffices to know that BSO->BSTOP is a rational equivalence; you do not need to use that the map to the product of the K(Z,4i)s is one (however, to prove the former, one needs the latter).
@archipelago and@Oscal: The edit does not address my confusion. I do not understand the claim "rationally, every topological bundle has a unique vector bundle structure". If say $X$ is a finite complex, the cokernel $[X,BO]\to [X,BTOP]$ is finite. The inclusion could be like $2\mathbb Z\subset\mathbb Z$. If $x\in [X,BTOP]$, which vector bundle is $x$ equivalent to rationally? If $t\in [X, BTOP]$ has order $n$, i.e. $nt=0$, how to we know that $p(t)=1$? Why is $p(x+t)=p(x)p(t)$? Here $p=1+p_1+p_2+\dots$, the total Pontryagin class.
If $K_Q$ is the product of $K(4i,\mathbb Q)$ spaces over all integers $i>0$, then all we have is an isomorphism $[BTOP, K_Q]\to [BO, K_Q]$ which sends the topological Pontryagin class to the usual one.
$[X,BSO_\mathbb{Q}]\rightarrow [X,BSTOP_\mathbb{Q}]$ is an isomorphism. The rational Pontryagin classes of $x\in [X,BSTOP]$ only depend on its image in $[X,BSTOP_\mathbb{Q}]$.
|
2025-03-21T14:48:31.061488
| 2020-05-22T03:48:17 |
361015
|
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|
Stack Exchange
|
Extension of outer unit normal vector to interior
Suppose we have a bounded smooth domain $\Omega$ in $\mathbb{R}^n$, so there exists an outer unit normal vector field $\eta$ everywhere on the boundary. Can we extend it to the interior satisfying some conditions? To be more specific, for example, can we construct a vector field $ X$ on $\Omega$ satisfying
(1) $X|_{\partial \Omega}=\eta;$
(2) $X$ is compactly supported in $\delta$-neighborhood of boundary $\Omega_\delta=\{x\in\Omega: d(x,\partial\Omega)\leq \delta\}$ for very small $\delta$.
(3) For example,the vector field $X$ is bounded in C^1: $\|X\|_{C^1(\bar{\Omega})}\leq C/\delta$ where C is a universal constant.
As it is pointed out, I should let the boundary be at least $C^2$ such that (3) makes sense. So for simplicity, let's just consider domain with smooth boundary. My thought was to think $\eta=(\eta_1,\cdots,\eta_n)$, so $\eta_i$ are smooth functions on the boundary, we can use the partition of unity to extend these functions to functions $X_i$ in some neighborhood of boundary. Then certainly we will get bounds in $C^k$ for $X_i$. I am just wondering if we can extend it to $\delta$-neighborhood and make those bounds (such as $C^1$ norm) depending on $\delta$, for example, bounds like $C\delta; C/\delta; C\delta^2;C\log\delta...$ where $C$ is a universal constant. How do we make this work exactly?
Edit: Maybe those bounds such as $C\delta,C\delta^2,..$ don't look reasonable since we expect that the derivative should become greater when $\delta$ is smaller. So it only makes sense to expect bounds like $C/\delta;C/\log\delta..$
@PiotrHajlasz, Thanks, I will think about it more.
I think this question would be more appropriate for the Mathematics Stack Exchange as it is a standard exercise.
If $\partial\Omega\in C^2$, then the answer is yes. We need $C^2$ as this condition implies that the vector normal to the boundary is $C^1$ (since the normal vector id defined through derivatives).
According to the collar neighborhood theorem, there is $d>0$ and a diffeomorphism of class $C^1$:
$$
\Phi:\partial\Omega\times(-1,1)\to (\partial\Omega)_d=\{x:\operatorname{dist}(x,\partial\Omega)<d\}
$$
Let
$$
\pi_1:\partial\Omega\times(-1,1)\to\partial\Omega,\ \ \pi_1(x,t)=x,
$$
and
$$\pi_2:\partial\Omega\times(-1,1)\to\partial\Omega,\ \ \pi_2(x,t)=t,
$$
be projections on components.
Then $\tilde{X}=\eta\circ\pi_1\circ\Phi^{-1}$
is a $C^1$ extension of the vector field $\eta$ to a neighborhood of the boundary.
Let
$$
\phi\in C_0^\infty(-1,1),
\quad
\text{with}
\quad
\text{$\phi(t)=1$ for $t\in [-1/2,2]$}
\quad
\text{and $\phi_\epsilon(t)=\phi(t/\epsilon)$.}
$$
Then $|(\phi_\epsilon)'(t)|\leq C/\epsilon$ and hence
$$
X=\tilde{X}\cdot (\phi_\epsilon\circ\pi_2\circ\Phi^{-1})
$$
satisfies the required condition (with $\epsilon$ in place of $\delta$).
could you recommend a reference for just an extension $X\in C^2(\overline{\Omega}): , |X(x)|\le 1$ in $\Omega$? Thanks in advance.
|
2025-03-21T14:48:31.061701
| 2020-05-22T05:35:44 |
361017
|
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"Anton B",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361017"
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|
Stack Exchange
|
Is the Singer cycle preserved by field automorphisms and graph automorphisms?
Let $T=\operatorname{PSL}_n(q)$ with $n$ a prime number. Then the $\mathscr{C}_3$ subgroup $M=\langle x\rangle{:}\langle\sigma\rangle$ of $T$ is isomorphic to $\mathbb{Z}_{\frac{q^n-1}{(q-1)(n,q-1)}}{:}\mathbb{Z}_n$, where $x$ comes from the Singer cycle.
Note that $\sigma$ has a matrix which is a permutation matrix corresponding to $(1,2,\dots,n)$ in $\operatorname{SL}_n(q)$. It follows that $\langle\sigma\rangle$ is preserved by any outer automorphism of $T$.
My question is: is $\langle x\rangle$ preserved by any outer automorphism of $T$? It is true for the diagonal automorphism and the $\mathscr{C}_3$ subgroup of $\operatorname{PGL}_n(q)$ is isomorphic to $\mathbb{Z}_{\frac{q^n-1}{q-1}}{:}\mathbb{Z}_n$. How about the field automorphism and the graph automorphism? How can we find a matrix form of $x$ in this case?
That is to say, if $o\le\operatorname{Out}(T)$, then is the $\mathscr{C}_3$ subgroup of $T.o$ just $M.o$? And how about the case when $T=\operatorname{PSU}_n(q)$ and $M$ is isomorphic to $\mathbb{Z}_{\frac{q^n+1}{(q+1)(n,q+1)}}{:}\mathbb{Z}_n$, where $n$ is still a prime number? In this case we only need to consider the field automorphism.
There is no uniform shape for $x$. But there is the way to construct it in a particular situation (for given $n$ and $q$), see M. W. Short, "The primitive soluble permutation groups of degree less than 256," page 15. But it is going to be the matrix in not the same basis in which the matrix of $\sigma$ is a permutational matrix most probably.
For other questions see "The subgroup Structure of the Finite Classical Group" by P. Kleidman and M. Liebeck, it looks like $\S 4.3$ has some answers. Equation (4.3.8) shows that there is a field automorphism of $T$ normalising $\langle x \rangle$.
So the answer to the question is yes, this maximal subgroup extends to a maximal subgroup of ${\rm Aut}(T)$.
@DerekHolt I can now see that a field automorphism normalises $\langle x\rangle$ by Anton's comment. But how about the graph automorphism (even though the answer seems to be yes)?
This is true by Proposition 4.3.6.(I) of Kleidman and Liebeck's book "The Subgroup Structure of the Finite Classical Groups", which says that, in all cases for the linear and unitary groups, there is a unique conjugacy colass of maximal ${\mathscr C}_3$-subgroups.
In fact in your situation it is easy to prove it directly using Zsigmondy's theorem: with a few exceptions, if $q$ is a prime power and $n>1$, then there is a prime $r$ dividing $q^n-1$ that does not divide $q^m-1$ for any $m<n$. So your maximal subgroup is in fact the normalizer of a (cyclic) Sylow $r$-subgroup of ${\rm PSL}(n,q)$, and then the result is clear.
The only exceptions to Zsigmondy's theorem that apply here are with $n=2$ and $q$ a Fermat prime, in which case the groups are the normalizers of Sylow $2$-subgroups.
|
2025-03-21T14:48:31.061907
| 2020-05-22T07:38:16 |
361019
|
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"A.Skutin",
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"url": "https://mathoverflow.net/questions/361019"
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|
Stack Exchange
|
Question about polynomials in $\mathbb{C}[x, y, z]$
What can be said about pairs of polynomials $P, Q\in\mathbb{C}[x, y, z]$, such that $\frac{\partial P}{\partial y}\frac{\partial Q}{\partial z} - \frac{\partial P}{\partial z}\frac{\partial Q}{\partial y}\in\mathbb{C}[x]$?
I am interested in examples of such $P, Q$.
The similar question can be asked if we replace $\mathbb{C}[x]$ with any field $F$ of characteristic zero and $\mathbb{C}[x, y, z]$ with $F[y, z]$.
The pair of such polynomials forms a set? I don't know what this question aims at.
These pairs form a $\mathbb{C}[x]$-submodule in $\mathbb{C}[x, y, z]^2$. For example, it obviously holds if $P$ or $Q$ in $\mathbb{C}[x]$.
Is that set of pairs closed under addition? That’s a nonlinear differential operator...
To start with you could ask the question in $L[y,z]$ for $L$ a field of characteristic zero, then specify to $L=\mathbf{C}(x)$, and then intersect with $A[y,z]$ with $A=\mathbf{C}[x]$.
@ZachTeitler is right. Actually, for every $(P,Q)$ we can write $(P,Q)=(P,P)+(0,Q-P)$, and both pairs $(P,P)$ and $(0,Q-P)$ satisfy the equation, but not $(P,Q)$ in general.
It is not additive. Right now I am also interested in any not trivial examples. By trivial I mean $P, Q\in\mathbb{C}[x, y]$, or $P, Q\in\mathbb{C}[x, z]$, also $(P, Q) = (f(x)\cdot y, g(x)\cdot z)$.
For any scalar ring, on the $L$-algebra $L[y,z]$, the usual product along with ${P,Q}=\partial_yP\partial_zQ-\partial_zP\partial_yQ$ form a commutative Poisson algebra. That is, ${\cdot,\cdot}$ is a Lie bracket, and is a derivation (for the multiplication) with respect to each variable. I don't know to which extent this can answer the question, but at least it's a very well-studied kind of structure. See Wikipedia: Poisson algebra.
For the last question (for simplicity, take $F=\mathbb{C}$), I suppose you know that the these can be split into two cases, one in which you get that Jacobian to be zero and the other non-zero. The latter leads you to the notorious Jacobian conjecture and the former can be completely characterized.
So if I understand it correctly, from Jacobian conjecture for $F = \mathbb{C}(x)$ there exist polynomials $R, T\in\mathbb{C}^{[2]}$ with $0\not= R(P, Q) = T(P, Q)\in\mathbb{C}[x]$.
|
2025-03-21T14:48:31.062084
| 2020-05-22T07:44:04 |
361021
|
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|
Stack Exchange
|
Asymptotic development of Integral of $e^xx^r$
Let $\alpha \in (0,1)$ and $\delta \in (0,1/2)$ be fixed, and consider the following integrals for each integer $j \geq 0$:
$$I_j(u):= \frac{e^u}{u^{j+\alpha}} \int_{-u\delta}^0 e^t t^{j-1+\alpha}\left(1+\frac{t}{u}\right)^{-1}dt, \hspace{2mm} u>0$$
Show that for any integer $k \geq 0$ and for any integer $0 \leq j \leq k$, each of these integrals as an asymptotic development of the form
$$I_j(u) = \frac{e^u}{u^{\alpha+j}} \left( d_{0,j} + \frac{d_{1,j} }{u} + \cdots + \frac{d_{k,j}}{u^{k}} + O\left(\frac{1}{u^{k+1}}\right) \right) \text{ as }u \rightarrow \infty$$
Attempts: For $1 \leq j \leq k$, I expanded the $(1+t/u)^{-1}$ as a geometric series upto $k$ terms and got
$$I_j(u) = \frac{e^u}{u^{j+\alpha}} \left(\sum_{n=0}^k \frac{(-1)^n}{u^n} \int_{-u \delta}^0 e^t t^{j-1+\alpha+n} dt + O\left(\frac{1}{u^{k+1}}\right) \right)$$
So, I'm down to showing there exist constants $c_{0, j}, \cdots , c_{k, j}$ such that for each $0 \leq n \leq k$,
$$\int_{-u \delta}^0 e^t t^{j-1+\alpha+n} dt = c_{n, j} + O \left( \frac{1}{u^{k+1}} \right)$$
But I'm not sure how to proceed from here (I tried integrating by parts but got too many terms...does that look promising?) or if this will work. Also will something similar work for $j=0$? Thanks.
P.S.: This is related to this unanswered question I asked on MSE a long time ago: https://math.stackexchange.com/questions/3667949/integrating-an-asymptotic-development
With notation there, what I had done was separate the integral of $\gamma$ into one over $\gamma_1$ and one over $\gamma_2$, and tried obtaining an asymptotic development of the desired form for $\gamma_1$ (with that along $\gamma_2$ being analogous). For $\gamma_1$, I substituted $x:=e^u$ (the original integral there is in terms of $x$) and with $x>1$, we have $u>0$. The integral so obtained is equal to the real integral
$$\int_{1-\delta}^1 e^{uv} (v-1)^{j-1+\alpha} v^{-1} dv$$
wherein I made the substitution $v := 1 + t/u$ to get the above collection of integrals. Perhaps those integrals could be directly/better estimated by complex analytic methods, in which case I would also appreciate a direct estimation of the integral I started with in the link.
You want to show that for some complex $b$ we have
$$\int_{-v}^0 e^t t^a\,dt=b+O(v^{-c})$$
where $v:=u\delta\to\infty$, $a:=j-1+\alpha+n$, $c:=k+1$.
This is true. Indeed, let $b:=\int_{-\infty}^0 e^t t^a\,dt$. Then, by l'Hospital's rule, for any real $c$
$$\Big|b-\int_{-v}^0 e^t t^a\,dt\Big|\le\int_{-\infty}^{-v} e^t |t|^a\,dt
\sim e^{-v} v^a=O(v^{-c})$$
as $v\to\infty$.
|
2025-03-21T14:48:31.062247
| 2020-05-22T07:44:50 |
361022
|
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"url": "https://mathoverflow.net/questions/361022"
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|
Stack Exchange
|
Derivative of a polynomial $P(z)$ and the derivative of the conjugate reciprocal of $P(z)$
Let $P(z)=\sum_{n=0}^na_nz^n$ be a polynomial of degree $n$ having no zeros in $|z|<1.$ Let $Q(z)=z^n\overline{P(1/\overline{z})}.$ Then it is an easy exercise to show that $\Re\left(zP'(z)/P(z)\right)= \sum_{k=1}^n\Re\frac{z}{z-z_k}\leq n/2$ on $|z|=1,$ since each $\Re\frac{z}{z-z_k}\leq 1/2$ on $|z|=1.$
But using the property $\Re (z)\leq 1/2$ iff $|z|\leq |1-z|$ with little simplification we will have then on $|z|=1,$
$$|P'(z)|\leq |nP(z)-zP'(z)|=|Q'(z)|.$$
Suppose we keep reducing the radius of the disc containing no zeros inside, what may happen to this inequality? Is there any closed form relation between $|P'(z)|$ and $|Q'(z)|$ on $|z|=1?$
For example $P(z)=z^n+K^n, K\leq 1,$ we have the equality relation $K^n|P'(z)|= |Q'(z)|$ on $|z|=1.$ This makes us feel that for a polynomial $P(z)$ having no zeros in $|z|<K, K\leq 1$there may be a closed expression of the kind $f(K)|P'(z)|\leq |Q'(z)|$ on $|z|=1,$ where $f(K)$ is positive and $f(K)\rightarrow 0$ as $K\rightarrow 0.$ What may be that
$f(K)?$
If the polynomial has all its zeros on $|z|=K,$ then $|P'(z)|K^n\leq |Q'(z)|.$
I think it should be of something like $f(K)=\frac{K^n}{1+((|a_0|/|a_n|)-K^n)(1-K^n)}.$ It is purely by intution, I may be incorrect also!
|
2025-03-21T14:48:31.062379
| 2020-05-22T07:50:21 |
361023
|
{
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Stack Exchange
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Rings $R$ such that every [regular] square matrix with entries in $R$ is equivalent to an upper triangular matrix
Let $\text{M}_n(R)$ be the ring of $n$-by-$n$ matrices with entries in a commutative unital ring $R$. Theorem III in
C.R. Yohe, Triangular and Diagonal Forms for Matrices over Commutative Noetherian Rings, J. Algebra 6 (1967), 335-368
provides a characterization of the Noetherian rings $R$ with the property that every matrix in $\text{M}_n(R)$ is equivalent to a diagonal matrix: It turns out that this is the case if and only if $R$ is a direct sum of PIDs and completely primary PIRs, where ``completely primary'' means a local ring with a nilpotent maximal ideal. With this in mind, here are my questions:
(1) Is there any analogous characterization for the rings $R$ with the property that every matrix in $\text{M}_n(R)$ is equivalent to an upper triangular matrix? (2) Does the property hold for any choice of $R$? (3) And if the answer to the previous question is no, what if we restrict attention to the regular matrices in $\text{M}_n(R)$?
[EDIT] The answer to questions (2) and (3) is in the negative, as shown by Mohan in their answer, and that's good to know. On the other hand, I'm really hoping for someone to come up with a reference where it's actually proved that, if $R$ is taken from a class of commutative rings that is sufficiently interesting and sufficiently larger than direct sums of PIDs and completely primary PIRs, then every regular square matrix with entries in is equivalent to an upper triangular matrix. This would not be a characterization in the vein of Yohe's theorem, but still... [END OF EDIT]
Every matrix $A \in \text{M}_n(R)$ can be brought to upper triangular form by elementary row transformations; that is, there exist elementary matrices $E_1, \ldots, E_k \in \text{M}_n(R)$ such that $E_1 \cdots E_k A$ is an upper triangular matrix. But the elementary matrix corresponding to a row-multiplying transformation need not be invertible in $\text{M}_n(R)$; although it is definitely invertible in $\text{M}_n(\mathcal Q(R))$ when $A$ is regular, with $\mathcal Q(R)$ being the total ring of fractions of $R$. Unfortunatley, I don't see how this helps answering any of my questions (I'm especially interested in the last one).
Glossary. By a ``regular matrix'', I mean a regular element in the multiplicative monoid of $\text{M}_n(R)$; or equivalently, a matrix $A \in \text{M}_n(R)$ whose determinant is a regular element of $R$. An element $a$ in a (multiplicatively written) monoid $H$ is regular (or cancellable) if the functions $H \to H: x \mapsto ax$ and $H \to H: x \mapsto xa$ are both injective.
When you say equivalent, do you mean by conjugating with a non-singular matrix? What is your definition of regular matrix?
@Mohan By "equivalent to", I mean "associated to". Two elements $x$ and $y$ in a monoid $H$ — in the present case, the multiplicative monoid of $\text{M}_n(R)$ — are associated if $x=uyv$ for some units $u, v \in H$. AFAIK, "equivalent" is still a standard (although unfortunate) term used in linear algebra in place of "associated"; and it's also used by Yohe in the paper cited in the OP. The answer to your other question is along the same lines: By a "regular matrix", I mean a matrix that is a regular element in the multiplicative monoid of $\text{M}_n(R)$; I've added a definition to the OP.
Your definition of regular element of a monoid is not what any semigroup theorist world use. Regular means von Neumann regular $xax=x$ for some $a$
@BenjaminSteinberg I see. I borrowed it from ring theory (https://en.wikipedia.org/wiki/Zero_divisor). And I'm not the only one doing so!
If you are interested in non-Noetherian rings, the Hermite rings in the sense of Kaplansky, or K-Hermite rings in T. Y. Lam's terminology ("Serre's Problem on Projective Modules") have the triangular reduction property for every $n$. As K-Hermite rings are Bézout rings, Noetherian K-Hermite rings are just principal ideal rings (PIRs), which can be characterized as direct sums of finitely many PIDs and special PIRs (Yohe's complete primary PIRs). Note that K-Hermite = Hermite (f.g. stably free modules are free) + Bézout.
@LucGuyot On this side of the world, non-Noetherian rings are very welcome. I didn't know about Hermite rings (in the sense of Kaplansky), thanks for the pointer. Do you have a precise reference to a proof that (commutative) Hermite rings have the triangular reduction property for every $n$?
@SalvoTringali Check the Appendix of Section I.4 in "Serre's Problem on Projective Modules" of T. Y. Lam. You will find even more results on K-Hermite rings there. If you don't have access to it, you can always resort to "Elementary divisors and modules", I. Kaplansky, 1949. See also "Diagonal reduction of matrices over rings" (2012) by Bogdan Zabavsky.
On the subject of Hermite rings, this is also why you won't be able to get an analog of the Noetherian result. Hermite rings are precisely the rings in which matrices admit triangular reduction, equivalently in which 1x2 matrices admit diagonal reduction (theorem 3.5 in Kaplansky). Being Hermite is a very Bezout-like condition, in fact as soon as you sufficiently restrict zero divisor behavior Bezout=Hermite. See the paper Elementary Divisor Rings and Finitely Presented Modules by Larsen, Lewis, and Shores
Actually I take it back, I think you'll get a fine analog of the Noetherian result if you restrict the minimal prime spectrum of your ring. To get a direct sum decomposition you'd probably want to assume finitely many minimal primes. Basically this would amount a decomposition theorem for Bezout rings with finitely many minimal primes
Small note: there is no such thing as a 'direct sum of rings'. The category of rings does not have biproducts, and the coproduct is given by tensor product. I think you mean 'finite product of rings'.
@R.vanDobbendeBruyn The statement is copied from Theorem III in Yohe's paper (cited in the OP). And, the last lines of the proof of Lemma 2.4 in the paper read, ``Hence R is a commutative noetherian ring with the property that every localization at a maximal ideal is an integral domain. Such a ring is well known to be a direct sum of integral domains." I don't think this refers to a finite product of rings. Should I?
@SalvoTringalo For any ring $R$ with finitely many minimal primes $P_i$ that is locally a domain, we have $R \cong \prod R/P_i$. It's easy to check this locally. So yes Yohe is referring to a finite product of rings.
@R.vanDobbendeBruyn I thought maybe the author used "direct sum" because the decomposition naturally arises from an ideal decomposition of $R$. That said, I didn't read the paper, and also maybe it's just an instance of sloppiness.
It is a well-known artefact mostly carried over from the pre-category theory days, so it's no surprise that it appears in a 60s paper. But we know better now!
@BadamBaplan I see. I was interpreting "direct sum'' in Yohe's paper as to mean "direct decomposition'' in the sense of Definition 7 from Sect. I.8.11 in Bourbaki's Algebra I (this is the same thing as the ideal decomposition you refer to in your last message). However, I had missed that there were only finitely many summands in the decomposition. Thanks!
To clarify, that isn't what I meant when I said ideal decomposition. I meant that $R$ is isomorphic as a module to a direct sum of ideals (in which case the summand ideals have a natural ring structure and identity).
As Luc Guyot mentioned, check out Kaplansky's paper Elementary Divisors and Modules from 1949.
Kaplansky calls a ring Hermite when every $1 \times 2$ matrix is equivalent to a diagonal matrix, and shows that equivalently a ring is Hermite iff for every matrix $M$ there exists an invertible matrix $U$ such that $MU$ is upper triangular.
To get a finite product decomposition result like Yohe's, we obviously need to assume that $R$ has finitely many minimal primes.
On the other hand, we have the following
Theorem If $R$ is a Bézout ring with finitely many minimal primes then there is a finite set of idempotents $e_i$ such that the ring $e_iR$ is a Hermite ring with a unique minimal prime and $R \cong \prod e_iR$. Hence $R$ is Hermite.
This is Theorem 2.2 in Elementary Divisor Rings and Finitely Presented Modules by Larsen, Lewis, and Shores.
Maybe we can say more about the structure of these summands. For example, it is easy to show that in any Bézout ring with a unique minimal prime $P$, the ideal $P$ is essential unless $R$ is a domain. Indeed, if $I \cap P = 0$, then $\operatorname{Ann}(a) = P$ for any $a \in I$ and clearly any $a \in I$ is nonzero in every localization of $R$. Since $R$ locally has totally ordered ideals, this implies that $P$ is locally $0$, hence $0$, i.e. $R$ is a domain.
So to sum up
Conclusion Let $R$ be a ring with finitely many minimal prime ideals. Then the following are equivalent:
$\ \ (1)$ $R$ is a finite direct product of Bézout rings each of which is either a domain or has a unique minimal prime ideal which is essential.
$\ \ (2)$ Matrices over $R$ are equivalent to triangular matrices.
I'm not sure if you can say very much in general about the structure of Bézout rings with essential unique minimal prime ideal. But, here's one observation: A Bézout ring with a unique minimal prime ideal $P$ has the property that every non-nilpotent element divides every nilpotent element.
Indeed, let $b$ be nilpotent and let $a$ be not nilpotent, i.e. $b \in P$ and $a \notin P$. By Bézoutness, pick $c,d,u,v,r$ such that $ac + bd = r$, $ru = a, rv = b$. Deduce that $r \notin P, v \in P$. It follows that $cv + du -1 \in P$, and we deduce that $du$ is a unit. Therefore $a \mid b$.
Note that if $R$ is additionally Noetherian, we can easily deduce that $P$ is the unique prime ideal of $R$, which exactly recovers Yohe's result. To see this, note first that it suffices to assume $R$ is local with maximal ideal $M$, in which case it has totally ordered ideals and $\bigcap_n M^n = 0$, which implies $M^n \subseteq P$ for some $n$ and thus $M \subseteq P$. Without the Noetherian hypothesis, $R$ can have infinite Krull dimension.
As for your 3rd question, I don't have much to say off the top of my head except that it seems really hard. From your own observations, any ring which is its own total ring of fractions (i.e. regular elements are units) will have this property, and that is quite a broad class of rings, members of which sometimes have ostensibly very little in common.
How can a $1\times 2$ matrix (i.e., non-square) matrix be equivalent to a diagonal (thus square) one?
@YCor A diagonal matrix is just defined as a matrix that is zero off of the main diagonal. That's a reasonable question though.
There is still one thing that I find very unsatisfactory with K-Hermite rings: I still ignore whether trigonal reduction of every $2 \times 2$ matrices (the topic of this question) implies trigonal reduction of every $1 \times 2$. It might not be the case. Determining the ring with trigonal reduction of $2 \times 2$ matrices would be an even more on-topic answer. (Mohan's answer is a step forward in this direction.)
@Luc Guyot I agree, what I wrote is really a long comment that answers a question somewhere between (1) and the later edit, which asks for sufficiently general conditions. I'd also be interested to know what trigonal reduction of $2 \times 2$ entails for a ring. I will think on it. Even for domains, intuitively feels like a much weaker property than f.g. ideals.
I will answer only the second and third question in your list, the first is too general and open ended.
Take $R=K[x,y,z,w]$, polynomial ring in four variables over a field $K$. Take the $2\times 2$ matrix $M$ consisting of the four variables as entries. It is regular, but it is not equivalent to an upper triangular matrix.
If it is, then by determinant considerations, one of the diagonal entries have to be a non-zero constant since $\det M$ is irreducible in $R$. We have $uMv=N$ with $u,v$ invertible, $N$ has a non-zero constant entry. But, put all variables equal to zero and then $M(0)=0$ and then $N(0)$ must be zero, which is impossible, since one entry is a non-zero constant.
Nice! It's good to see a counterexample to the 2nd and 3rd questions. On the other hand, I'm really hoping for someone to come up with a reference to something like "If $R$ is taken from a class of commutative rings that is sufficiently interesting and sufficiently larger than direct sums of PIDs and completely primary PIRs, then every regular square matrix with entries in $R$ is equivalent to an upper triangular matrix." This would not be a characterization in the vein of Yohe's theorem, but it seems plausible that something along these lines is out there in the literature.
@SalvoTringali I had suspected that and if you had stressed that, I wouldn't have bothered to answer. For the first, do you have some specific class of rings in mind?
No, I don't. I'd be happy with any pointer to the literature where the question has been tackled.
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2025-03-21T14:48:31.063513
| 2020-05-22T08:00:10 |
361024
|
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Stack Exchange
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Reduce PDE to ODE by dilation symmetry
I am reading Rodrigues, Henrion, and Cantwell - Symmetries and analytical solutions of the Hamilton–Jacobi–Bellman equation for a class of optimal control problems, p.753.
Consider the following PDE: $$q_1x_2^2+q_2x_2^2+V_{x_1}x_2-\frac{V^2_{x_2}b^2}{4r}=0.$$
This PDE has the following dilation symmetry:
$$\tilde{x}_1=e^sx_1,\, \tilde{x}_2=e^sx_2, \, \tilde{V}=e^{2s}V.$$
Note that $\tilde{V}_{\tilde{x}_1}=e^sV_{x_1}.$ So the above PDE in the tilde variables becomes
$$e^{2s}\bigg(q_1x_2^2+q_2x_2^2+V_{x_1}x_2-\frac{V^2_{x_2}b^2}{4r}\bigg)=0.$$
So we can form the following characteristic equations $$ \frac{d x_1}{x_1}=\frac{d x_2}{x_2}=\frac{dV}{2V}.$$
my question is from the following statement,
Integrating and rearranging terms, the PDE is invariant under the change of variables: $\alpha= \frac{x_2}{x_1}$, $V=x_1^2 G(\alpha).$
How to understand the above statement? Does the dilation symmetry implies the above? How to see "being invariant"?
I cannot find the quoted statement in the cited paper; instead, I read on page 753 that $\alpha=x_2/x_1$ and $V=x_1^2 F(\alpha)$ are integrals of the characteristic equations, which indeed they are. Are we talking about the same paper??
@CarloBeenakker Yes, that is what I say. The "invariant" statement is from its extension paper. https://ieeexplore.ieee.org/document/8443024 (p.206, same example)
What is G here?
@PiyushGrover $G$ is a function of $\alpha$ to be calculated (and is calculated in tha paper).
Here's an answer, at least as I understand your question.
Geometrically, you can understand the solutions to your PDE as graphs of surfaces in $\mathbb{R}^3$ given by $(x_1,x_2,V(x_1,x_2))$ (at least locally). From this viewpoint, to say that a PDE has a symmetry, is to say that a solution surface "moved" in the direction of the symmetry will also be a solution surface to your PDE. This may mean that a solution surface moves along itself.
The dilation symmetry you give is precisely the 1-parameter family of diffeomorphisms (or flow) generated by the vector field $X=x_1\partial_{x_1}+x_2\partial_{x_2}+2V\partial_V$. The change of variables you mention come from the invariant surfaces of this flow. That is, the surfaces in $\mathbb{R}^3$ that flow along themselves. These correspond to level sets of functions $f=f(x_1,x_2,V)$ such that $X(f)=0$ (these are the invariant functions, or first integrals of $X$). In this case, all invariant functions are generated by the two independent invariant functions $\alpha=x_2/x_1$ and $G=V/x_1^2$ (I have note yet prescribed $G$ as a function of $\alpha$ here).
As it seems we are concerned with solutions to the PDE that are invariant under the symmetry, we want to understand the PDE purely in terms of the invariants $(\alpha,G)$. This means we want a differential equation involving the two invariants. Since $V=V(x_1,x_2)$, then using the second invariant function, we conclude that $G=G(\alpha)$, so that $V(x_1,x_2)=x_1^2 G(\alpha)$. Throwing this into the PDE produces an ODE which is the reduction.
I've thrown details and rigor under the bus here (for example, there are issues concerning group actions and quotient manifolds). For a rigorous treatment of this approach I suggest looking at Peter Olver's book "Applications of Lie Groups to Differential Equations." Specifically, you can learn about symmetry invariant solutions in Chapter 3.
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2025-03-21T14:48:31.063755
| 2020-05-22T08:55:40 |
361027
|
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is the Lyapunov exponent a continuous function of the invariant measure w.r.t weak-* topology?
I hope for some relerant results for the following question:
Is the lyapunov exponent continuous with respect to the measure?
Assume $M$ is a manifold, $f$ is a diffeomorphism on $M$, $m$ is an invariant measure. Then we have the Lyapunov exponent. Fix $M$ and $f$,is the largest Lyapunov exponent continuous with respect to the weak-* topolpgy on the measure? Do we have any result regarding the linear cocycle?
En effet! now fiexd...
Have a look at the recent survey by Viana
thank you very much
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2025-03-21T14:48:31.063833
| 2020-05-22T09:02:19 |
361028
|
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Mean curvature flow and knot theory
I am wondering if the mean curvature flow of one-dimensional submanifolds of $\mathbb{R}^3$ is understood well enough to give some perspective on (and hopefully a proof of) something like the Fary-Milnor theorem.
For reference, the Fenchel theorem (1929) says that if $c:S^1\to\mathbb{R}^3$ is a smooth embedding, then the total curvature is at least $2\pi$. The Fary-Milnor theorem (1949/50) says that if $c$ forms a nontrivial knot, then the total curvature is at least $4\pi$.
Steven Altschuler ("Singularities of the curve shrinking flow for space curves", JDG 1991) showed that if $c_t:S^1\to\mathbb{R}^3$ is a one-parameter family of smooth embeddings satisfying the mean curvature flow, then
$$\frac{d}{dt}\int_{S^1}|\kappa_t|\leq -\int_{S^1}\tau_t^2|\kappa_t|$$
where $\tau_t$ is the torsion of $c_t$. So if $\int_{S^1}|\kappa_0|<4\pi$ then certainly $\int_{S^1}|\kappa_t|<4\pi$ for all positive $t$. So it seems like one could hope for a proof of the Fary-Milnor theorem which is more or less directly analogous to Hamilton-Perelman's proof of the Poincaré conjecture or of the topological classification of closed 3-manifolds with nonnegative and positive scalar curvature.
The problem would rest on understanding the singularities of the mean curvature flow. Altschuler seems to have shown that a singularity of the mean curvature flow in this setting is characterized by blowup of the curvature (just like Ricci flow), and that a nontrivial tangent flow is given either by an Abresch-Langer solution or the grim reaper solution. This seems to be directly parallel to the Perelman-Brendle theorem which says that the analogous blowup limit of a finite-time singularity of a Ricci flow on a compact 3-manifold is either a quotient of shrinking round cylinders or the Bryant soliton, which was (in a weaker version) Perelman's breakthrough result.
So it seems like the key ingredients are there. Can they be put together? It seems like the basic problem is that I don't know what the "surgery" analogue might look like or how it might be relevant.
So, more generally, forgetting about Fary-Milnor theorem in particular, could one hope to use mean curvature flow for any sort of application in knot theory? Perhaps the proper analogue of the Hamilton-Perelman approach would decompose a given knot as a connected sum of prime knots, and would give some canonical representation thereof? This seems to be comparable to the geometrization conjecture, although the existence and uniqueness of such a decomposition seems to be already known in knot theory.
In order that I have a reasonably concrete question:
Ricci flow with surgery on 3D compact manifolds is closely analogous to mean curvature flow of mean-convex surfaces in $\mathbb{R}^3$ (Brendle-Huisken). Is there an analogy, or a conjectural analogy, to mean curvature flow of curves in $\mathbb{R}^3$, or in 3-manifolds? Are there (conjectural) applications in knot theory?
I couldn't find any references on the web. I am aware of Perelman's use of mean curvature of curves in a Ricci flow background to show finite extinction time of Ricci flow on simply-connected 3-manifolds, but the only detailed version which I know to exist of this, Morgan-Tian's book, seems to have some basic errors (cf. Bahri "Five gaps in mathematics" and some followups on the arxiv)
One obstacle for this is that in $\mathbb{R}^3$, curve shortening flow does not preserve the embedding of curves. Unlike in the plane, curves can cross along the flow which affects the knottedness.
It's also worth noting that the Abresch-Langer solutions have a saddle point property, in that small deformations of them generally do not flow to them but either several copies of a circle or develop a local kink (i.e. a grim reaper curve). Also, all the singularities are planar, so aren't knotted.
Thanks, I should have remembered your first point... so any good application would have to come with some precise understanding of the extrinsic distance function? The second point is certainly interesting, but is it so crucial? As far as I know the Hamilton-Perelman analysis didn't need to know anything about stability of singularities
This question reminds me of a result of Brian White who used mean curvature flow of surfaces surfaces bounding a curve with curvature $<3\pi$. https://mathscinet.ams.org/mathscinet-getitem?mr=4527825
This is really a comment that's too long to fit in 500 characters, but let me try to explain the obstacles I mentioned in the comments. Sorry for how large the pictures are. I'll edit this if I figure out how to resize them.
Curve shortening flow three dimensions can self- intersect, which is the main roadblock. However, there are a few other issues that seem challenging, even if we can overcome this first problem. I'll try to give some pictures (all of which are from Wikipedia) to demonstrate the challenge of understand the limits of spatial CSF from initial data.
Suppose we start with a knot and we somehow know a priori that it does not self intersect under CSF. The question is whether the limiting curve can be used as some sort of model for that knot. The issue we immediately run into is the result of Altschuler you mentioned; the limiting curves are planar, so cannot be knotted. To get around this, one might imagine that the solution is to consider the limiting curve along with some sort of crossing diagram. In some cases, this might work, but I suspect that most of the time it does not.
For instance, if we start with a trefoil knot, it might be the case that the limit is what happens on the left side of the following picture by Au [1], where the circle is covered twice. This is a trefoil with minimal total curvature, but it might not be what we hoped for. It's not the Abresch-Langer solution that genuinely looks like a flattened trefoil, but this is pretty much the best case scenario otherwise. Even here, I don't think this is what happens generically.
For a less ideal example, suppose we start with a figure-8 knot with rotational symmetry, as shown below.
![Taken from Wikipedia]
(https://upload.wikimedia.org/wikipedia/commons/0/05/Blue_Figure-Eight_Knot.png)
This also curvature at least curvature at least $2 \pi$. However, here the "minimal diagram" looks instead like this (rotated by 90 degrees).
(source: wikimedia.org)
This is not going to be the limit of CSF, which tells us that for this initial condition, either CSF has a local singularity or else becomes unknotted. It seems to be the latter, but a proof would take some careful analysis.
Somewhat related to all of this is that we can't easily rule out Type 2 singularities, which are essentially local kinks. In two dimensions, surgery might be useful for local singularities because they only happen when the curve crosses itself. If you cut around an intersection to make two curves, you expect to be able to continue the flow. For spatial curves, it's much less clear when Type 2 singularities emerge. Perhaps there is a version of Grayson's theorem that gives sufficient conditions for an initially embedded curve to develop a Type 1 singularity, but I'm not aware of it. There is a version if the curve is contained on the surface of a sphere, but that's not directly relevant here.
Au, Thomas Kwok-Keung, On the saddle point property of Abresch-Langer curves under the curve shortening flow, Commun. Anal. Geom. 18, No. 1, 1-21 (2010). ZBL1217.53067.
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2025-03-21T14:48:31.064331
| 2020-05-22T09:11:37 |
361029
|
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|
Stack Exchange
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Invariant theory of the indefinite orthogonal groups
I believe the following statements are true:
Let $V$ be a finite-dimensional real vector space with a positive-definite inner product $g$. Let $g_{\otimes n}$ denote the natural extension of $g$ to $V^{\otimes n}=V\otimes\cdots\otimes V.$ Let $\rho:\text{O}(V,g)\to\text{O}(V^{\otimes n},g_{\otimes n})$ be the standard group homomorphism. Then
The $g_{\otimes n}$-invariant linear maps $V^{\otimes n}\to\mathbb{R}$ exist only when $n$ is even, and they are spanned by the complete contractions
$$x_{\sigma(1)}\otimes\cdots\otimes x_{\sigma(n)}\mapsto g(x_{\tau(1)},x_{\tau(2)})\cdots g(x_{\tau(n-1)},x_{\tau(n)})$$
where $\sigma$ and $\tau$ range over all permutations of $\{1,\ldots,n\}$.
If $W\subset V^{\otimes n}$ is a linear subspace such that $(\rho(T))(W)\subset W$ for all $T\in\text{O}(V,g)$ and such that the dimension of the vector space
$$\Big\{\text{bilinear symmetric }S:W\times W\to\mathbb{R}\text{ s.t. }(\rho(T))^\ast S=S\quad\forall T\in\text{O}(V,g)\Big\}$$
is one, then there is no nontrivial proper linear subspace $U\subset V$ with $(\rho(T))(U)\subset U$ for all $T\in\text{O}(V,g).$
My reference is chapter 9 in Besse's book on four-dimensional geometry. My questions are the following:
Does the above remain true if $g$ is only non-degenerate? It seems like the argument for #2 may break down since it relies on orthogonal projection. I couldn't make a guess on #1 since I'm not familiar with Weyl's book on the classical groups.
Where can I find a reference with such a statement? Weyl's book seems to only suggest that the indefinite case is not so different from the definite case, but it's hard for me to tell.
It seems like the article by Peter B. Gilkey "Local invariants of a pseudo-Riemannian manifold", Math. Scand. 36 (1975), 109-130 is relevant, but it's very hard for me to read.
(I asked this question on math.stackexchange with no response.)
You might look at Claudio Procesi's book Lie Groups, where a lot of classical invariant theory is explained in a convenient modern language.
Assertion 1 is about linear maps, so it holds over $\Bbb R$ if and only if it holds over $\Bbb C$, and hence it has nothing to do with positive definiteness of $g$.
What do you mean by $g_{\otimes n}$-invariant linear maps? Invariant with respect to what group?
I mean the group $\text{O}(V^{\otimes n},g_{\otimes n})$
|
2025-03-21T14:48:31.064507
| 2020-05-22T09:41:22 |
361031
|
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|
Stack Exchange
|
On the intersection numbers of the generators of $\text{Pic}(X)$ of a smooth quintic surface
There exists the following result in the literature: There exists a polarized $K3$ surface $(X, H)$ of genus $3$ and a smooth irreducible curve $C$ on $X$ satisfying $C^2 =4$, $C.H=6$ such that $\text{Pic}(X) \cong \mathbb Z[H] \oplus Z[C]$. The theorem follows from [https://arxiv.org/pdf/math/9805140.pdf] theorem $1.1(iv)$.
Now Let's consider $X$ to be a smooth quintic hypersurface in $\mathbb P^3$ with Picard number $2$. Then I think it can be shown that $\text{Pic}(X) \cong \mathbb Z[H] \oplus Z[H']$, where $H$ is the hyperplane class and $H'$ is some divisor. Now in order to locate the ample line bundles in $\text{Pic}(X)$ using the Nakai-Moishezon criterion, we must know the intersection numbers $H.H'$ and $H'^2$.
In this context my question is the following: Does there exist in the literature an analogous existence result as the first-mentioned theorem for smooth quintic hypersurface with Picard number $2$?
To be more precise: Does there exist a polarized smooth quintic hypersurface $(X, H)$ in $\mathbb P^3$ and a smooth irreducible curve $C$ for which the intersection numbers $C^2$ and $C.H$ are known and $\text{Pic}(X) \cong \mathbb Z[H] \oplus Z[C]$?
Can someone give me any reference which could be even remotely useful in the context of finding out such $(X,H)$ and $C$
Any help from anyone is welcome.
A natural source of surfaces in $\mathbb{P}^{3}$ with Picard number $> 1$ is given by linear determinantal surfaces, i.e. zero sets of square matrices of linear forms. In what follows, let $X \subseteq \mathbb{P}^{3}$ be a smooth linear determinantal surface of degree $d \geq 2$ (smoothness can arranged by taking the $d \times d$ matrix to be sufficiently general) and let $H \in {\rm Pic}(X)$ be the class of a hyperplane section.
It can be checked that $X$ contains a curve $C$ which is the degeneracy locus of a $d \times (d-1)$ matrix of linear forms. For degree/genus reasons, this $C$ cannot be a complete intersection of $X$ and another surface in $\mathbb{P}^{3}$.
There is a Noether-Lefschetz-type result which says that if $X$ is a generic linear determinantal surface of degree $d,$ the Picard group ${\rm Pic}(X)$ is isomorphic to $\mathbb{Z}H \oplus \mathbb{Z}C.$ Here is a reference for this result, as well as others which go further than determinantal surfaces:
Lopez, Angelo Felice; Noether-Lefschetz theory and the Picard group of projective surfaces, Mem. Amer. Math. Soc. 89 (1991), no. 438, x+100 pp.
EDIT: The aforementioned result on generic linear determinantal varieties is a corollary of the following result of Lopez; if $C \subseteq \mathbb{P}^{3}$ is a smooth connected curve whose homogeneous ideal is generated in degree at most $d-1,$ then the general surface of $S$ of degree $d$ containing $C$ is smooth and ${\rm Pic}(S) \cong \mathbb{Z}H \oplus \mathbb{Z}C.$ This holds in particular when $d \geq 2$ and $C$ is a line in $\mathbb{P}^{3}.$
thanks for the answer. Is the curve $C$ mentioned in the answer irreducible? Are the intersection numbers $C.C$ and $C.H$ immediate from the discussion? (apology in advance, if this is trivial to compute). Finally, Is theorem $(II)3.1$ mentioned there relevant in this context?
also as the general surface of degree $d \geq4$ is not linear determinantal, but linear pfaffian if $d \leq 15$, so in this case are we talking about smooth quintic surfaces which is not general?
The curve $C$ is cut out by the maximal minors of a $d \times (d-1)$-submatrix of the matrix of linear forms cutting out $X,$ so by general results on determinantal varieties we can ensure $C$ is smooth and connected, and therefore irreducible. The degree $C \cdot H$ and genus can of course be read off the Hilbert polynomial, which in turn can be read off the Eagon-Northcott resolution of the associated determinantal ideal. You can then determine $C \cdot C$ from the adjunction formula on $X.$ I can't access the reference at the moment, so I can't speak to the relevance of Theorem (II)3.
As to your second comment, you are correct that the smooth surfaces in under discussion are not general when $d \geq 4.$
In this context I have one further question : Does it make sense to expect the existence of a smooth quintic hypersurface $X$ in $\mathbb P^3$ containing a line $L$ (even if this is rare) such that $\text{Pic}(X)$ is generated by the hyperplane section $H$ and the line $L$?
See the edit above.
thank you very much for the answer. Can you give me a reference of the general result you just mentioned?
You're welcome! This general result is contained in the monograph of Lopez cited in the answer.
|
2025-03-21T14:48:31.064828
| 2020-05-22T10:06:02 |
361032
|
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|
Stack Exchange
|
Locating the typed version of Hoàng Xuân Sính's thesis on Gr-categories
Hoàng Xuân Sính was a Vietnamese student of Grothendieck who defended her thesis on Gr-categories (now called weak 2-groups). The thesis, handwritten and in French, can be found at
https://pnp.mathematik.uni-stuttgart.de/lexmath/kuenzer/sinh.html
According to the nLab, there exists also a typed version (perhaps in English?):
https://ncatlab.org/nlab/show/Ho%C3%A0ng+Xu%C3%A2n+S%C3%ADnh
Can anyone provide a link to that version?
a summary in English (handwritten) is at https://pnp.mathematik.uni-stuttgart.de/lexmath/kuenzer/thesis_sinh_summary_sinh.pdf
The TeXed version can be found on
https://webusers.imj-prg.fr/~leila.schneps/grothendieckcircle/SinhThesis.pdf
Cristian David Gonzalez Avilés is the author of the TeX version. For reasons unknown to me, the TeX version numbers the theorems/propositions etc. differently than in the handwritten original.
I'm a little uncomfortable calling someone who TeXed a previously existing handwritten thesis the "author" of the TeXed version of that thesis.
|
2025-03-21T14:48:31.064940
| 2020-05-22T10:08:11 |
361033
|
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"url": "https://mathoverflow.net/questions/361033"
}
|
Stack Exchange
|
On smooth extensions of functions
Let $f(x) = \left(I - \hat{n}\hat{n}^T \cdot\textbf{1}_{\vec{n}^TAx \geq 0}\right)Ax$, where $I$ is the identity matrix, $A$ is a (symmetric) $d\times d$ positive definite matrix, $\hat{n}$ is an arbitrary unit vector and $\textbf{1}$ is the indicator. Let $S = \{x:\hat{n}^Tx \leq 0\}$ and consider $f:S \to S$. I want to understand whether or not $f$ has a smooth extension $\tilde{f}:\mathbb{R}^d \to \mathbb{R}^d$. Are there any references? The reason I am asking is because I want to use stable-manifold theorem on $f$ but $S$ is a manifold with boundary.
Thank you!
I know there exists Whitney's theorem but i am not sure if it is applicable because of the boundary.
|
2025-03-21T14:48:31.065022
| 2020-05-22T10:30:03 |
361034
|
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|
Stack Exchange
|
is this process a Markov one?
Here is the problem I can't solve.
Let $\xi_n$ $(n=1,2,3,\dots)$ be a sequence of i.i.d. random variables on $\mathbb{R}$ with density $p(x)>0$, let $\eta_n=\sum_{i=1}^{n}\xi_i^2$. Define $$\zeta_t = \eta_{[t]}(t-[t]) \times \eta_{[t]+1}(-t+[t]+1),$$ where $[t]$ denotes the integer part of $t$. Is $\zeta_t$ a Markov process?
I tried to prove that given process is Markov by definition: $$\mathbb{P}(\zeta_{t+s}\in A |\mathcal{F}_{\le t}) =\mathbb{P}(\zeta_{t+s}\in A| \mathcal{F_{=t}}),$$ where $A \in \mathcal{F}_{\ge t}$, $\mathcal{F}_{\le t} = \sigma(\zeta_s: s\le t)$, $\mathcal{F}_{\ge t} = \sigma(\zeta_s: s\ge t)$, $\mathcal{F}_{= t} = \sigma(\zeta_t)$.
Here is my first attempt so far
$$\mathbb{P}(\zeta_{t+s}\in A |\mathcal{F}_{\le t})=\mathbb{P}(\eta_{[t+s]}(t+s-[t+s]) \times \eta_{[t+s]+1}(-t-s+[t+s]+1)\in A|\mathcal{F}_{\le t})=$$
$$=\mathbb{P}\left(\left(\sum_{k=1}^{[t+s]}\xi_k^2(t+s-[t+s])\right)\times\left(\sum_{i=1}^{[t+s]+1}\xi_i^2([t+s]+1-t-s)\right)\in A|\mathcal{F}_{\le t}\right)=$$
$$=\sum_{k=1}^{[t+s]}\sum_{i=1}^{[t+s]+1}\mathbb{E}(\mathbb{1}_A\xi_k^2(t+s-[t+s])\xi_i^2([t+s]+1-t-s)|\mathcal{F}_{\le t}),$$ using that $\mathbb{P}(A|\mathcal{F})=\mathbb{E}(\mathbb{1}_A|\mathcal{F})$.
My second attempt was to make an example of suitable random variables and probability space, such that $\zeta_t$ is not Markov. I began to think it is not Markov as $\zeta_t$ depends on $\eta_{[t]+1}$, so it depends on the future in some sense because $[t]+1>t$.
I understand that it is kind of "not Mathoverflow question", but I did not receive answer on MathStackexchange and I faced this problem during my term paper, so may be this is a reason to post it here.
Do you really mean "$\times$" and not "$+$" in the definition of $\zeta_t$? It looks like $\zeta_t = 0$ whenever $t$ is an integer.
Yes, there really should be $\times$ in the definition of $\zeta_t$. Can you tell, please, how do you know that $\zeta_t=0$ for integer $t$?
I thought $\eta_{[t]} (t - [t])$ is the product of $\eta_{[t]}$ and $t - [t]$, and the latter is zero when $t$ is an integer?
No, it's not markov. If you see the process at times 1.25, 1.5 then you know $\eta_1, \eta_2$
and so you know the process perfectly in the near future, which is not the case if you only see the process at time 1.5
@MateuszKwaśnicki, sorry, I really misread your comment. Sure, you are right
@mike, thanks for your comment. Can you make it little more detailed? I mean, if we see the process $\zeta_t$ at times 1.5, 1.25, then we know that $\zeta_{1.5} = 0.5^2\cdot\eta_1\cdot\eta_2$ and $\zeta_{1.25} = 0.25\cdot0.75\cdot\eta_1\cdot\eta_2$. But how do I know $\eta_1$ and $\eta_2$?
You only know $\eta_1\eta_2$, but this still gives you $\zeta_t$ for all $t\in [1.5, 2)$, showing that the process is not Markov.
The process is not Markov in general. Indeed, let $X_i:=\xi_i$, $S_n:=\eta_n=\sum_1^n X_i^2$, and $$Z_t:=\zeta_t=(t-[t])([t]+1-t)S_{[t]}S_{[t]+1},$$
where $P(X_i=0)=P(X_i=1)=1/2$. Then
$$Z_{3/2}=\tfrac14\,X_1(X_1+X_2),\quad Z_2=0,\quad
Z_{5/2}=\tfrac14\,(X_1+X_2)(X_1+X_2+X_3).$$
So, the conditional distribution of $Z_{5/2}$ given $Z_2$ is the same as the unconditional distribution of $Z_{5/2}$. On the other hand, the conditional distribution of $Z_{5/2}$ given $Z_{3/2},Z_2$ is not the same as the unconditional distribution of $Z_{5/2}$, because $Z_{5/2}$ depends on $Z_{3/2}$:
$P(Z_{3/2}=0)=P(X_1=0)=1/2$ and $P(Z_{5/2}=0)=P(X_1+X_2=0)=1/4$, whereas
$$P(Z_{3/2}=0,Z_{5/2}=0)=P(X_1+X_2=0)=1/4\ne P(Z_{3/2}=0)P(Z_{5/2}=0).$$
So, $(Z_t)$ is not Markov.
If you insist that the distribution of $X_1$ be absolutely continuous with a strictly positive density, then you can appropriately approximate the discrete distribution by such an absolutely continuous one.
Alternatively, you may e.g. assume that $X_1\sim N(0,1)$. Then $EZ_{3/2}=1$, $EZ_{3/2}=5/2$, but $EZ_{3/2}Z_{3/2}=27/2\ne EZ_{3/2}EZ_{5/2}$, so that here too $Z_{5/2}$ depends on $Z_{3/2}$.
This is a good exercise. As far as I understand the OP, the situation is as follows. Given a deterministic non-negative function $\phi$ on the interval $[0,1]$ with $\phi(t)=0\iff t \in\{0,1\}$ and a sequence of random variables $Z_n$, one defines
$$
\zeta(t)=\phi(t-n) \cdot Z_n \;, \qquad n\le t\le n+1 \;.
$$
The question is when $\zeta(t)$ is Markov, and the answer to this question is pretty obvious: if and only if the variables $Z_n$ are independent (look at the Markov condition at integer times).
In the original question
$$
\phi(t)=t(1-t)
$$
and
$$
Z_n=(\xi_1^2+\dots+\xi_n^2)(\xi_1^2+\dots+\xi_{n+1}^2) \;,
$$
which are pretty obviously not independent unless $\xi_i^2$ are a.s. constant.
|
2025-03-21T14:48:31.065429
| 2020-05-22T10:31:34 |
361035
|
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|
Stack Exchange
|
completeness of $\mathcal M(\Omega)$ without any topological assumptions?
Let $(\Omega,\Sigma)$ be a measurable space (no reference measure is chosen!), and $V$ a finite-dimensional normed vector space.
Note carefully that I am not choosing any topology on $\Omega$, so the $\sigma$-algebra $\Sigma$ is a priori not induced by any Borel structure whatsoever.
The total variation $|\mu|$ of a $V$-valued measure is defined as
$$
|\mu|(E)=\sup\limits_{\pi}\sum\limits_{E_i\in E}|\mu(E_i)|,
$$
where the supremum is taken among all possible disjoint partitions $\pi=\cup E_i$ of the measurable set $E\in\Sigma$. (the set-function $|\mu|$ is always a positive measure, see [Rudin, real and complex analysis]). For an arbitrary measure we denote
$$
\|\mu\|:=|\mu|(\Omega).
$$
We denote $\mathcal M(\Omega)$ the set of all measures with finite total variation $\|\mu\|<\infty$, and $\mathcal M(\Omega)$ is therefore a normed vector space with the above total variation norm.
Question: is $(\mathcal M(\Omega), \|\cdot\|)$ automatically a Banach space?
When $\Sigma$ is the Borel algebra then this is true of course, because we can identify $\mathcal M(\Omega)$ with the topological dual $C_b(\Omega;V^\ast)^\ast $ and the dual of a Complete vector space is automatically complete (and in fact $\|\mu\|=\sup\limits_\phi \int \phi(x)\cdot d\mu(x)$ with $\cdot$ denoting the finite-dimensional $V,V^\ast$ pairing). However, I've never seen the statement written in this full generality, so I'm wondering whether this is actually true or not?
Two statements here are in need of correction. You say that the dual of a complete vector space is always complete. If you mean that the dual of a complete normed space (aka Banach space) is complete, then this is true but the completeness condition is superfluous. If you mean a complete topological vector space (even lcs), it is false. Secondly, if $X$ is ,say, a completely regular space (in particular, locally compact or metrisable), then the dual space of $C_b(X,V^\ast)$ is indeed a space of measures on the Baire $\sigma$-algebra, but not the one you are interested in.
Thanks for pointing this out, I meant indeed a Banach space, not a general topotlogical vector space. Regarding the second part of your comment: I was not aware of these subtleties. Out of curiosity: what would be the dual space (space of measures on the Baire $\sigma$-algebra) that you mentioned for completely regular spaces?
Indeed $(\cal{M}(\Omega),\|\cdot\|)$ is a Banach space. For $V = \mathbb{R}$ or $V = \mathbb{C}$ you can find this result in Dunford/Schwartz (1957), Linear Operators I, ch. III.7.4, in particular p. 161. For arbitrary Banach space $V$ this also holds true, but with a sligthly different norm (see p. 160). For finite dimensional $V$ this norm is equivalent to your norm. Proof as in Lemma III.1.5 of Dunford-Schwartz.
Yup, that does the job. (Although I find this book is a bit outdated for my taste and hard to read, nothing beats the classic old-timers). Thank you very much @Dieter.
|
2025-03-21T14:48:31.065666
| 2020-05-22T10:42:53 |
361037
|
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|
Stack Exchange
|
Probability that a Voronoi cell contains exactly k random points
Consider two independent point processes in the unit square $[0,1]^2$. The two point processes are identically independent and typically binomial/Poisson. One, say $\Phi^*$, is used to generate a Voronoi tessellation of $N^*$ cells. The other one, $\Phi$ with $N$ points, is used to build the statistics of the counts of points of $\Phi$ in the Voronoi cells associated with $\Phi^*,N^*$.
Gilbert's argument (Ann. Math. Stat., 1962) for a randomly picked given point in plane to be contained in a Voronoi cell of area $s$ is that, known the p.d.f. of cell areas -- be it $f(s)$ --, then the probability of $f(s|X=1)=sE[s]^{-1}f(s)$ where $X$ is a random variable that expresses the count of points in $s$. How can I generalize the argument to any $X$ value? In principle, if I consider $k$ points randomly drawn and independently, I thought of considering $f^k(s|X=1)$, but it does not match simulations. Any suggestion?
Can you give more details about the reference you quote? Voronoi cells, in the simplest case, are associated to a set of points in the plane (or more generally a metric space). They need not have equal areas. Please give more details. https://en.wikipedia.org/wiki/Voronoi_diagram
@LiviuNicolaescu The reference is to the classic Gilbert's paper in Ann. Math. Stat. 1962 (in particular the statement is at p. 963, bottom, second paragraph of Section 4).
@LiviuNicolaescu I should mention, that in my case I suppose to know $f(s)$ -- the cell area distribution of the random Voronoi tesselation -- the analytical form of this p.d.f. is not known, but can be well approximated by Gamma distributions.
Thanks! I will check it out
Could you explain the model a bit more? By definition of the Voronoi cells, there is one and only one point in each cell. I'm assuming their seeds are given by a Poisson point process? Are you sampling more points once the domains are chosen? And if so, how?
@PierrePC I re-elaborated the question, hopefully clarifying the premises. Thanks for pining in.
@maurizio Thank you for indulging me. In the case where $\Phi$ is Poisson, then $X$ is by definition a Poisson variable, with parameter proportional to the area of the cell. I guess that means $\mathbb P(X=k)=\int f(s)e^{-k}s^k/k!\mathrm ds$, provided the intensity of the process is 1.
If $\Phi$ is binomial, more or less the same thing happens, $X$ is a binomial with parameters $(n,p)$ where $n$ is the intensity of the global process and $p$ is the the area $s$ of the cell. Here and above I assumed that $s$ meant the area within the square $[0,1]^2$ and, as you said, that $\Phi$ consists of points in the same square.
@PierrePC Thanks. However I think I am confused at this point. The solution that you point out agrees with what I figured out myself in this post few days ago when I also posted this question. But my question was referencing to finding $f(s|X=k)$ starting from $f(s|X=1)$, rather than $\mathbb{P}(X=k)$. The only way I see it is by $\partial_s f_{S,X}(x,s)$ where $f_{S,X}$ is the integrand in $\mathbb{P}(X=k)$. I am wondering if I can find $f_S(s|X=k)$ in another way. $f_S$ is here the probability that given a Voronoi cell of area $s$ it contains $k$ points.
@maurizio Tell me where I'm wrong. $f(s|X=k)=\mathbb P(X=k|\mathrm{Area}=s)\cdot f(s)/\mathbb P(X=k)$ by Bayes' law, and we know all these probabilities by the above. For instance in the Poisson case of intensity 1, $P(X=k|\mathrm{Area}=s)=e^{-s}s^k/k!$, $f(s)$ is assumed to be known, and $\mathbb P(X^k)=\int f(s)e^{-s}s^k/k!\mathrm ds$.
@PierrePC Yes. There is nothing wrong with this derivation. But I am wondering if I can find a way to derive $f(s|X=k)$ void of Bayes' law, just by geometry considerations. One way I was thinking of is as following. Gilbert also discusses the probability $\mathbb{P}(b)$ that given two randomly picked points at distance $b$, they belong to the same cell. We know the $f_b(b)$ for two randomly picked points in $[0,1]^2$. Then for $k$ points in the same cell, we can perhaps think of $k-1$ pairs of points, at distance from each other $b_i$...
|
2025-03-21T14:48:31.065994
| 2020-05-20T10:03:19 |
361038
|
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|
Stack Exchange
|
Čech functions and the axiom of choice
A Čech closure function on $\omega$ is a function $\varphi:\mathcal P(\omega)\to\mathcal P(\omega)$ such that (i) $X\subseteq\varphi(X)$ for all $X\subseteq\omega$, (ii) $\varphi(\emptyset)=\emptyset$, and (iii) $\varphi(X\cup Y)=\varphi(X)\cup\varphi(Y)$ for all $X,Y\subseteq\omega$; in other words, it obeys the Kuratowski closure axioms except possibly the idempotent law $\varphi(\varphi(X))=\varphi(X)$.
In 1947 E. Čech asked if there exists such a closure function (on any set, not necessarily $\omega$) which is also surjective and nontrivial (not the identity map). Čech's question has been answered in the affirmative under various set-theoretic assumptions, including ZFC. The ZFC example is rather complicated and makes heavy use of the axiom of choice; it seems unlikely that one could construct such a thing without choice.
Question. Is it consistent with ZF that there is no nontrivial surjective Čech closure function $\varphi:\mathcal P(\omega)\to\mathcal P(\omega)$?
That is, there is no function $\varphi:\mathcal P(\omega)\to\mathcal P(\omega)$ satisfying the conditions (i)-(iii) above and also (iv) for every $Y\subseteq\omega$ there exists $X\subseteq\omega$ such that $\varphi(X)=Y$, and (v) $\varphi(X)\ne X$ for some $X\subseteq\omega$. (Condition (ii) is now redundant, as it follows from (i) and (iv) that $\varphi(X)=X$ if $X$ is finite.)
I could imagine several candidate models: Maybe Cohen's model is a suitable example with the Dedekind-finite set of reals "spoiling things", but more likely would be Solovay's model where there are no MAD families, etc.
Also a candidate: the Feferman–Levy model where the reals are a countable union of countable sets, or the Truss model (although if this holds there, I'd imagine the Solovay model would already capture this).
I think it would be a good point in time to migrate this to MathOverflow. I don't have an obvious solution at hand, and I have quite a lot of things on my hands right now (which is why I might be missing something obvious).
Is it correct to think as "in ZF there's such a function" as "one can construct such a function"? More precisely, is it true (from some general principle) that if "there is such a function" is a theorem of ZF, then there exists $f_0\in P(\omega)^{P(\omega)}$ satisfying the given condition, and a formula of set theory $F(x)$ with only parameter $x$, such that for every $f\in P(\omega)^{P(\omega)}$, $F(f)$ holds iff $f=f_0$?
@bof I'd be happy with a weaker definition of "constructible". I'm just wondering whether the slogan "(exists in ZF)=(can be constructed)" has any foundation.
@YCor: Are you asking for a definable class (the class of functions such that blah blah, defined by formula) whose elements are individually not definable (so that you cannot write down such $F$'s to pick out individual elements)?
@YCor: I got a partial answer to this in this question.
@bof Out of Theorem 1 and Theorem 2 in the Galvin-Simon paper, which one is using AC in its proof?
@Haim Looks to me like they both use choice. My guess is, if you can construct such a function without choice, it will be by a completely different method.
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2025-03-21T14:48:31.066241
| 2020-05-22T11:17:20 |
361041
|
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|
Stack Exchange
|
An extension of symplectomorphism group
$\DeclareMathOperator\GL{GL}\DeclareMathOperator\Sp{Sp}$Let $\omega=\sum dx_i\wedge dy_i$ be the standard symplectic structure of $\mathbb{R}^{2n}=\mathbb{R}^{n}\times \mathbb{R}^n$.
We consider the following two
extensions of $\Sp(2n,\mathbb{R})$, the group of linear isomorphisms of $\mathbb{R}^{2n}$ preserving $\omega$:
1) Let $G$ be the group of all $A\in \GL(2n,\mathbb{R})$ which map all isotropic subspaces to isotropic subspaces. (A closed subgroup of the general linear group containing the symplectic group $\mathrm{Sp}(2n,\mathbb{R})$.)
2) Let $H$ be the group of all elements in $\GL(2n,\mathbb{R})$ which map all symplectic subspaces to symplectic subspaces.
A non-linear construction as above can be introduced on a symplectic manifold $M$: $G(M)$ is the group of all diffeomorphisms of $M$ whose derivatives map isotropic subspaces to isotropic subspace. And similarly $H(M)$ is the group of all diffeomorphisms whose linear parts map symplectic subspaces to symplectic subspaces.
Is there a terminology for Lie groups $G$, $H$, $\bar{H}$ and their Lie algebras and also the corresponding structures on symplectic manifolds? Is there any relation between $G$ and $\bar H$? What can be said about their first fundamental groups, as it is customary to compute the first fundamental groups of classical Lie groups?
Inspired by the concept of symplectic vector fields as Lie algebra of symplectomorphism group, what can be said about the Lie algebra of $G(M)$ and some dynamical interpretations on a symplectic manifold?
Does "symplectic subspace" mean "subspace on which the symplectic form is non-degenerate"? (I originally asked what $\bar H$ was, but deleted it when I realised you meant the closure.)
Yes The symplectic subspace are meant as you mentioned.$H$ is not closed subset of general linear group so we consider its closure $\bar H$
Is there an easy example to show that $H$ isn't closed?
+1 for this comment Note that the limit of a sequence of symplectic subspaces is not necessarilly symplectic subspace (In grassmanian topoloy) but the limit of isotropic subspace is again an isotropic subspace. For example The 2 dimensional space $(x,y,\epsilon y, \epsilon x)$ is a symplectic subspace only when $\epsilon \neq 0$
@LSpice I think that "diffeomorphisms whose linear part/ derivative" was correct and better than with plural. Googling "functions whose image" yields thrice more occurrences than "functions whose images".
@YCor, both your usages sound OK to me (although obviously I prefer the one I used!), but "diffeomorphisms whose derivative maps [A] to [B]" sounds exceedingly strange to me. Of course, it's AliTaghavi's post, so, if they prefer it that way, then I have no objection, and apologise for the unwanted change.
@LSpice I see. Actually probably I'd have phrased with "whose derivative at every point"... as you say OP will decide.
Thanks for the example!
Short answer: $G = H$ is the group of conformal symplectic linear maps. What follows is a proof of this (which I've simplified slightly from what I originally wrote):
1) $G$ is the group of conformal symplectic linear maps
First, some notation. We write $\{e_i\} \cup \{f_j\}$ for the standard basis for $\mathbb{R}^{2n}$, i.e. with $\omega(e_i,e_j) = 0$, $\omega(f_i,f_j) = 0$, and $\omega(e_i,f_j) = \delta_{i,j}$.
Suppose $\phi \in G$. We begin by singling out the isotropic subspaces $A = \mathbb{R}^n \times \{0\}$ and $B = \{0\} \times \mathbb{R}^n$. The following lemma is one way you might think about the group $\mathrm{Sp}(2n,\mathbb{R})$ in the first place: as being defined by its action on an isotropic subspace and a choice of isotropic complement.
Lemma: Given $\phi$, there is a unique element $\psi \in \mathrm{Sp}(2n,\mathbb{R})$ such that the following three properties hold:
$A = (\psi \circ \phi)(A)$, with $(\psi \circ \phi){\big|}_{A} = \mathrm{id} \colon A \rightarrow A$
$B = (\psi \circ \phi)(B)$
Given this lemma, we have that $\psi \circ \phi = \begin{pmatrix}\mathrm{id} & 0 \\ 0 & T\end{pmatrix}$, and it suffices to find conditions on $T \in \mathrm{GL}(n,\mathbb{R})$ so that isotropic subspaces are preserved by $\psi \circ \phi$. It suffices to consider just two cases:
For $i \neq j$, the span of $e_i$ and $f_j$ is isotropic, so its image under $(\psi \circ \phi)$, the span of $e_i$ and $Tf_j$, must also be isotropic. This proves that $T$ is diagonal.
Also for $i \neq j$, we can use the subspace spanned by $e_i + f_j$ and $e_j + f_i$, which will give us the condition that the diagonal entries of $T$ are all equal. Hence $T = c \cdot \mathrm{id}$ for some constant $c \neq 0$.
Finally, we see that with $T = c \cdot \mathrm{id}$, $(\psi \circ \phi)^*\omega = c \cdot \omega$, which clearly preserves isotropic subspaces. So this property completely characterizes $G$.
In terms of nomenclature, $G$ might be called the group of conformally symplectic linear operators, and the nonlinear theory would fall under the umbrella of (locally) conformal symplectic geometry, which is slightly more general than what you ask (i.e. $\omega$ itself does not need to be symplectic, only (locally) conformally symplectic, which is the more natural setting). I know only a little about the theory, and am far from an expert on the history, so I'll say nothing more that there exists literature on the subject.
As for computations of classical invariants, we see that $G \cong \mathbb{R}^* \times \mathrm{Sp}(2n,\mathbb{R})$, so there's nothing interesting to say about computing its standard invariants that can't already be said for $\mathrm{Sp}(2n,\mathbb{R})$.
2) $G=H$
Suppose $\omega(v,w) \neq 0$ and $\omega(v,x) = 0$. Then $\omega(v,w+tx) \neq 0$ for all $t \in \mathbb{R}$, so $v$ and $w+tx$ always span a symplectic subspace. Hence, if $\phi \in H$, since $\phi$ preserves symplectic subspaces, we require $$0 \neq \omega(\phi(v),\phi(w+tx)) = \omega(\phi(v),\phi(w)) + t\omega(\phi(v),\phi(x))$$ for all $t \in \mathbb{R}$. Hence, we must have $\omega(\phi(v),\phi(x)) = 0$ whenever $\omega(v,x) = 0$. In particular, $\omega$ preserves isotropic subspaces, so $H \subseteq G$. Meanwhile, every conformal symplectic linear map certainly preserves symplectic subspaces, so $G \subseteq H$. Hence $G=H$.
Remark: Even though the limit of symplectic subspaces may not be symplectic, as in the comments, this is not enough to show $H$ is closed, since if a sequence of elements of $H$ realizes this collapsing of a symplectic subspace onto a non-symplectic one, it can (and must by what I've just written) limit to a singular matrix.
Thank you very much for your attention to my question abd your very interesting answer
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2025-03-21T14:48:31.066690
| 2020-05-22T11:25:39 |
361043
|
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|
Stack Exchange
|
Computing the fractional Laplacian of power function
Is it possible to compute explicitly the fractional Laplacian (in $\mathbb R^n$) of a power function $|x|^p$?
Yes: $(-\Delta)^{\alpha/2} [|x|^p] = 2^\alpha \Gamma((p+n)/2) \Gamma((\alpha-p)/2) (\Gamma((p+n-\alpha)/2) \Gamma((-p/2))^{-1} |x|^{p-n-\alpha}$ whenever defined.
This follows easily from the composition rule for the Riesz potential kernel. See Theorem 3.1 in my survey [Kwaśnicki, Fractional Laplace Operator and its Properties, in: A. Kochubei, Y. Luchko, Handbook of Fractional Calculus with Applications. Volume 1: Basic Theory, De Gruyter Reference, De Gruyter, Berlin, 2019], DOI:10.1515/9783110571622-007 for a rigorous statement and discussion.
@MateuszKwaśnicki Thanks! Could you clarify the $x$ dependence in your result? I expect it to be something like $-C_{N,\alpha} |x|^q$.
Ah, I noticed a typo in my previous comment: should be $|x|^{p-\alpha}$, not $|x|^{p-n-\alpha}$, sorry. This is exactly as you say, with $q = p - \alpha$, but the constant depends on $n$, $\alpha$ and $p$.
@MateuszKwaśnicki Thanks again. Oh, now I understand; in my browser the formula inside the comment is not properly displayed (I see it divided in two pieces with "whenever defined" in the middle). For the sake of clarity, could you please write it in an answer?
Sure, I will write it as an answer in a minute. Strange: the same rendering error happened here, but it disappeared when I refreshed the tab.
I noticed that Thm 3.1 is a wrong reference; should be Table 1. Once again sorry for all errors in the comments above.
Here it is:
Proposition: Let $\alpha \in (0, \infty)$, $p \in (-n, \alpha)$, and
$$
f(x) = |x|^p .
$$
Then $(-\Delta)^{\alpha/2} f(x)$ is well-defined for $x \ne 0$, and
$$
(-\Delta)^{\alpha/2} f(x) = 2^\alpha \frac{\Gamma(\frac{p+n}{2}) \Gamma(\frac{\alpha-p}{2})}{\Gamma(\frac{p+n-\alpha}{2}) \Gamma(-\tfrac{p}{2})} \, |x|^{p - \alpha} .
$$
Here we understand the right-hand side is zero if $\tfrac{p+n-\alpha}{2}$ or $-\tfrac{p}{2}$ is a non-positive integer.
This is sort of standard, as it follows from early work of M. Riesz on what is now known as the Riesz potential kernel. Sample reference is the first entry in Table 1 in my survey:
M. Kwaśnicki, Fractional Laplace Operator and its Properties, in: A. Kochubei, Y. Luchko, Handbook of Fractional Calculus with Applications. Volume 1: Basic Theory, De Gruyter Reference, De Gruyter, Berlin, 2019, DOI:10.1515/9783110571622-007
but there are obviously older sources.
|
2025-03-21T14:48:31.066872
| 2020-05-22T11:26:40 |
361044
|
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|
Stack Exchange
|
Smoothing continuous functions in metric space
Let $(X,\rho)$ be a metric space.
For any $f:X\to\mathbb{R}$, define the local Lipschitz constant of $f$ at $x$ by
$$ \Lambda_f(x) := \sup_{x'\in X\setminus\{x\}} \frac{|f(x)-f(x')|}{\rho(x,x,')}
.
$$
I conjecture that for every continuous $f:X\to[0,1]$ and $\eta>0$ there is an
$h:X\to[-1,1]$ such that $\overline f:=f+\eta h$ satisfies
$$
\Lambda_{\overline{f}}(x) \le \Lambda_f(x)
$$
for all $x\in X$
and
$$
|\Lambda_{\overline f}(x)-\Lambda_{\overline f}(x')|
\le
\eta^{-1}\rho(x,x')\max\{\Lambda_f(x),\Lambda_f(x')\}^2
$$
whenever $x,x'\in X$ and at least one of
$\Lambda_{\overline f}(x),\Lambda_{\overline f}(x')<\infty$.
Question: Are results of this kind known? Conversely, is there an inherent reason why the above (or some close variant) cannot possibly hold? A brief search seems to indicate that results in this general spirit are obtained here: https://arxiv.org/pdf/1812.11087.pdf ; am I at least correct about the "general spirit" part?
The "continuous" assumption can probably be dropped.
|
2025-03-21T14:48:31.066975
| 2020-05-22T11:37:57 |
361045
|
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|
Stack Exchange
|
Continuous surjection from $X(D_n)$ onto $\operatorname{Homeo}_0(D_n)$
Let $n>1$ and let $\mathfrak{X}(D_n)$ denote the set of continuous vector fields on the closed disc $D_n\subseteq \mathbb{R}^n$. Let $\operatorname{Homeo}_0(D_n)$ be the set of homeomorphism of the disc $D_n$ fixing $0$ which are isotopic to the identity. Equip both these with the compact-open topology.
Is there an explicit surjective continuous map from $\mathfrak{X}(D_n)$ onto $\operatorname{Homeo}_0(D_n)$?
You're asking for a continuous map with no further assumption? it might exist for some "bad" reason (bad meaning quite unrelated to the context, working with $\mathrm{Homeo}_0(D_n)$ replaced with quite general nonempty connected Polish spaces). (This might be explicit too!)
I added the word explicit. Otherwise, it goes well with me if it's for some unrelated reason just as long as it can be expressed explicitly.
But usually these kind of "general type reasons" can also be made explicit. For instance, a continuous surjection $[0,1]\to [0,1]^2$ can be made explicit, and one could imagine finding an explicit continuous surjection $X(D_n)\to \mathcal{H}$ ($\mathcal{H}$ separable infinite-dim Hilbert) as well as an explicit continuous surjection $\mathcal{H}\to \mathrm{Homeo}_0(D_n)$.
The following "one" is straight-forward: $f \to \pi'\circ f\circ \iota$, where $\iota:D_n\to \mathbb{R}^n$ is the inclusion and ${\pi}$ is the identity on $D_n$ and on $\mathbb{R}^n-D_n$ it sends $x$ to $\frac{x}{|x|}$. However, when I try to incorporate the conditions that the image is a homeomorphism then such a map breaks (likewise with the constraint that the origin must be fixed).
By the way your title doesn't match your question (homeo vs continuous surjective)
Thanks. What space did you have in mind when you described $\mathcal{H}\to Homeo_0(D_n)$?
This may be related to YCor's comment: https://www.jstor.org/stable/1971161?seq=1#metadata_info_tab_contents
|
2025-03-21T14:48:31.067132
| 2020-05-22T12:09:45 |
361047
|
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"leo monsaingeon"
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|
Stack Exchange
|
Correct translation of the title from French
It is not a research question. But it is a question to working researchers, who is aware of old and new terminology of calculus of variations. That is why I ask it here.
I want to figure out correct translation of the title of an old article.
The article is about calculus of variations, but unfortunately I am not aware of terminology of the subject.
The article is: "Sur les solutions périodiques et les extrémales fermées du calcul des variations. Math. Ann. 110(1935), 63-96."
In particular I can not decide should "extrémales fermées" be translated as "closed extremal" or it should be rather translated as "closed extremum".
Neither, since it is plural.
@BenMcKay Ok, Do you know what would be a correct translation of the title?
How would you distinguish "extrema" from "extremals"? I think they are synonyms, but maybe you have in mind a technical distinction that would be relevant here. For me, I would translate it as "On the periodic solutions and closed extremals of the calculus of variations".
Definitely extremals ($\ne$ extrema). For me, “On periodic solutions and closed extremals in the calculus of variations”.
Same as @François Ziegler
@BenMcKay I think extremals are in the domain of the functional while extrema are in the range. So extremals are the places where extrema are attained.
From Bliss, "Lectures on the Calculus of Variations", sec. I..7: "An admissible arc defined by functions $y(x), z(x)$ having continuous first and second derivatives, and satisfying the equations (7.1), is called an extremal."
|
2025-03-21T14:48:31.067302
| 2020-05-22T12:16:37 |
361048
|
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"Abdelmalek Abdesselam",
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"Sergei Akbarov",
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|
Stack Exchange
|
On Köthe sequence spaces
I asked this a week ago at math.stackexchange, but without success.
As far as I understand, there are several meanings of the notion of the Köthe sequence space, in particular, Hans Jarchow in his "Locally convex spaces" defines it as the space $\Lambda(P)$ of sequences $\lambda:{\mathbb N}\to {\mathbb C}$ satisfying the condition
$$
\forall \alpha\in P\quad \sum_{n=1}^\infty \alpha_n\cdot|\lambda_n|<\infty,
$$
where $P$ is an arbitrary set of sequences with the properties:
1) $\forall\alpha\in P$ $\forall n\in{\mathbb N}$ $\alpha_n\ge 0$,
2) $\forall\alpha,\beta\in P$ $\exists\gamma\in P$ $\forall n\in{\mathbb N}$ $\max\{\alpha_n,\beta_n\}\le\gamma_n$
3) $\forall n\in{\mathbb N}$ $\exists\alpha\in P$ $\alpha_n>0$.
Jarchow mentions the space $\Lambda(P)$ from time to time in his book to illustrate (sometimes to formulate) different results, but without a summary about $\Lambda(P)$.
I wonder if there is a text where the results on $\Lambda(P)$ are systematized? I think the main properties of $\Lambda(P)$, like barreledeness, nuclearity, reflexivity, Heine-Borel property, completeness in different senses, etc. can be stated on one page (these are properties of $\Lambda(P)$ as a topological vector space, but its properties as just a vector space are interesting as well). Can anybody enlighten me if such a text exists?
Jarchow gives some conditions (for example, on p.497 he explains when $\Lambda(P)$ is nuclear), but the whole picture remains vague, and I even must confess that some elementary properties of $\Lambda(P)$ are not clear for me. For example, is it true, that if a sequence $\omega_n\ge 0$ has the property
$$
\forall\lambda\in \Lambda(P)\quad \sum_{n=1}^\infty \omega_n\cdot|\lambda_n|<\infty
$$
then there are $\alpha\in P$ and $C>0$ such that
$$
\forall n\in{\mathbb N}\quad \omega_n\le C\cdot\alpha_n
$$
?
I can prove this only in the case when $P$ has a countable cofinal subset (excuse me my ignorance).
Pietsch, A.: Nuclear locally convex spaces, vol. 66 in Ergebnisse der Mathematik und ihrer Grenzgebiete. Springer-Verlag, New York, Heidelberg, 1972. Translated from the second German edition. This should contain a lot of info on Koethe spaces.
@StefanWaldmann: that deserves to be posted as an answer I think.
You are very optimistic, Sergei!
In the countable case (or, only slightly more general: if there exists a cofinal countable subset) $\Lambda(P)$ is Fréchet, and you find many results about this case, e.g., in the book Introduction to Functional Analysis of Meise and Vogt, chapter 27. But even in this case, the characterization when $\Lambda(P)$ is reflexive or Montel (=Heine-Borel-Property) is a quite difficult theorem (this is called the Dieudonné-Gomes theorem). Of course, for Fréchet spaces barrelledness is for free, but I don't know of a characterization in terms of $P$ in the uncountable case (this is related to the explicit question at the end of your post -- my guess is that this is not always true: The hypothesis means that $\omega$ defines a linear functional on $\Lambda(P)$ and the conclusion means its continuity).
For the dual case of countable inductive limits of weighted Banach sequence spaces a lot of work has been done (e.g., by Bierstedt and others) to describe the dual again as weighted space and to characterize barrelledness in this situation. Again this is quite subtle, beyond the case of inductive limits of Banach spaces there are results of Vogt as well as Bierstedt and Bonet -- and if you really want to have a counterexample to your explicit question you should study their work.
Other than reflexivity or the Heine-Borel property there are many locally convex properties which are directly defined in terms of the semi-norms (Schwartz or nuclearity) -- for such conditions it is no difference whether $P$ is countable or not.
OK, on request, the following reference as answer:
Pietsch, A.: Nuclear locally convex spaces, vol. 66 in Ergebnisse der Mathematik und ihrer Grenzgebiete. Springer-Verlag, New York, Heidelberg, 1972. Translated from the second German edition.
This should contain a lot of info on Koethe spaces. I liked it in particular because of the nice Grothendieck-Pietsch criterion on nuclearity. This is easily checked for Köthe spaces and, I guess, one of the ways to check nuclearity for many other function spaces: namely, you establish an isomorphism to a suitable Köthe space.
Stefan, I see only the nuclearity criterion in this book (Proposition 6.1.2). Am I missing something?
@Sergei Akbarov maybe you're right. I don't have the book here at home, so I can't check. In any case, I think it was my first encounter with Köthe spaces, that's why I remembered the book. Another source is the one Jochen Wengenroth mentioned: Meise&Vogt is now available even as an english translation. Very nice one...
This is an addendum to the information that you already have. Firstly, there are three sources of substantial results on Köthe spaces—unsurprisingly the first volume of his monograph (which also has useful references), Grothendieck‘s dissertation (and his textbook on topological vector spaces) and the monograph by Valdivia suggested above. Secondly, as already mentioned, they are a useful source of models for spaces of test functions and distributions. There is a useful unified approach to this which is, I hope, worth mentioning. Suppose that we have an unbounded, self-adjoint operator $T$ on Hilbert space. (Typically, one uses differential operators of mathematical physics whose spectral properties are well known—Sturm Liouville operators, the Laplace Beltrami operator on suitable domains or manifolds, possibly with boundary conditions, and Schrödinger operators).
Then the intersection of the domains of definition of its powers has a natural Fréchet space structure and many of the classical spaces of test functions can be obtained in this way. A dual construction leads to the corresponding spaces of distributions. The connection with Köthe spaces is provided by the fact that if the spectrum of $T$ is discrete, i.e., consists of a sequence of eigenvalues, then this space (and its dual) are Köthe spaces which can be described explicitly in terms of this sequence. An advantage of this correspondence is that it can easily be generalised to give, e.g., ultradistributions, Roumieu spaces and Sobolev spaces, even of infinite order.
One example of the benefits is that this approach can be used to give transparent proofs of important results in various contexts—the classical case is the celebrated kernel theorem of Laurent Schwartz.
+1. I also mentioned the proof of the Kernel theorem being easy (in fact trivial) given the sequence space isomorphism. What references do you know which do that explicitly? The only one I know is an article by Barry Simon in J. Math. Phys. for the multilinear form version of the theorem. In my linked answer I consider the version about continuous linear maps from $S$ to $S'$. I would be interested in learning about more works in the same spirit.
Yes, Simons used it for the case where $T$ is the standard $n$-dimensional Schrödinger operator to get the result for the tempered distributions. The same proof works whenever you have an increasing sequence of positive eigenvalues such that $\frac1 {\lambda_n^\alpha}$ is summable for some positive $\alpha$
Good question. I think these sequence spaces deserve to be better known because they provide a rich bank of concrete examples for things related to the theory of topological vector spaces which can be dauntingly abstract.
Another resource with an extensive discussion of these spaces is the book by Manuel Valdivia "Topics in Locally Convex Spaces". It has a long chapter on sequence spaces including the particular case of echelon spaces which was a key class of examples used in the work of Grothendieck when he discovered the notion of nuclear spaces.
By the way, my previous somewhat related question
The "Spaces of Schwartz distributions are finite dimensional" challenge
was about figuring out nice properties of $P$ which would ensure $\Lambda(P)$ would behave, for all practical purposes, like a finite-dimensional space, i.e., it would be nuclear, (strongly) reflexive,...(fill in the blanks).
Addendum: Following Jochen's comment, I should add that providing examples is not the only motivation for spending time learning what a sequence space is. Spaces that matter are sequence spaces (modulo TVS isomorphism). I in fact would go further in this philosophy, in particular with regards to teaching distributions TVS's etc., not per se but for the needs of mathematical physics, probability,... As can be seen from my other posts listed below. Even in an introductory course, I think it makes sense investing time at the beginning to prove the sequence space isomorphism theorems once and for all, and then proving all the needed theorems like the kernel, Fubini for distributions, Bochner-Minlos, Prokhorov, Lévy continuity,...with sequence spaces.
Can distribution theory be developed Riemann-free?
https://math.stackexchange.com/questions/3512357/understanding-the-proof-of-schwartz-kernel-theorem/3512932#3512932
https://math.stackexchange.com/questions/2623515/schwartz-kernel-theorem-and-dual-topologies/2647815#2647815
I think the view that sequence spaces are mainly interesting because they provide examples is not entirely correct. Many concrete spaces of smooth or holomorphic functions or of distributions are (e.g., by the coefficient funtionals of a Schauder Basis) isomorphic to sequence spaces. That is why Köthe investigated them -- that they often provide counterexamples is a Benefit but not their "raison d'être".
I agree. This is why I need them.
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2025-03-21T14:48:31.068053
| 2020-05-22T13:09:04 |
361055
|
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|
Stack Exchange
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If $(Y,T)$ is a connected minimal system with a symbolic extension of linear word complexity, is $(Y,T)$ equicontinuous?
Let $(Y,S)$ be a minimal topological dynamical system such that $Y$ is connected. A simple example of a system like this is an irrational rotation of the circle, and it is known that Sturmian sequences are symbolic extension of these systems and that have linear word complexity function. More generally, any irrational torus rotation has a symbolic extension of linear complexity (actually, a substitutive subshift).
My question is about the converse: if $\pi:(X,T) \to (Y,T)$ is a factor map such that $(X,T)$ is a minimal linear-complexity subshift and $Y$ connected, then is $(Y,S)$ equicontinuous?
I think that the low complexity of $(X,T)$ and the connectedness hypothesis on $Y$ forces $(Y,T)$ to be of "low complexity" as well; maybe, equicontinuous.
Did you mean to also say $Y$ is connected in the statement of your question?
@DanRust Yes! I edited the question to make it clearer.
By linear complexity, do you mean liminf or limsup?
I was thinking in lim sup:
$$ p(n) \leq Cn,\quad \forall n\in\mathbb{N}$$
But a counterexample with
$$ \liminf_{n\to\infty}p(n)/n < +\infty $$
it is also interesting for me.
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2025-03-21T14:48:31.068173
| 2020-05-22T13:53:18 |
361057
|
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|
Stack Exchange
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Question on the proper sub-representation of induced representation
$\DeclareMathOperator\Ind{Ind}$Let $G$ be a reductive group over a $p$-adic local field $F$, and $P=MN$ a parabolic subgroup.
Let $\sigma$ be an irreducible representation of $M(F)$ and consider its unnormalized induced representation $\Ind_P^G(\sigma)$. Let $\pi$ be a subrepresentation of $\Ind_P^G(\sigma)$. For an arbitrary element $v$ in $\sigma$, we can choose $f \in \pi$ such that $f(e)=v$.
Let $U^-$ be the unipotent radical of the opposite parabolic subgroup $P^-$ of $P$. Then I am wondering whether we can choose $U^-$-invariant $f \in \pi$ such that $f(e)=v$? If $\pi$ is the full induced representation, it is possible because $P \cap U^- = \{e\}$. But I don't know whether it holds for proper sub-representation of $\Ind_P^G(\sigma)$.
Furthermore, I am also wondering whether we can choose $f\in \pi$ so that $f(e)=v$ and $f$ has small support near $e$. This also holds when $\pi$ is the full induced representation. But I don't know whether it is possible for proper sub-representation $\pi$.
Thank you very much!
(PS: I am especially think of the case when $\pi$ is the image of the local intertwining operator $M(s) : \Ind_P^G(\sigma\cdot |\det|^s) \to \Ind_P^G( \sigma\cdot |\det|^{-s})$ defined by $M(s)f(s)(g)= \int_{N}f(wng)dn$, where $w$ is the longest Weyl element. If it is difficult to consider general $\pi$, how about the case when $\pi=M(s)(\Ind_P^G(\sigma\cdot |\det|^s) )$?)
The image of $\operatorname{Res}^G_M \pi \to \sigma$, $f \mapsto f(e)$, is an $M$-submodule, since $\pi(m)f \mapsto f(m) = \sigma(m)f(e)$. So it is all of $V$ unless every vector in $\pi$ vanishes at $e$, which implies that $\pi$ is the $0$ representation, since $\pi(g)f \mapsto f(g)$ for all $f$ in the space of $\pi$ and all $g \in G$. (Also, I encourage you never to use $\nu$ for a vector; I guarantee it will lead to confusion at some point.)
@LSpice, Thank you very much! It helps me a lot!
@LSpice, can we choose such $f$ which is either $U^{-}$-invariant or sufficiently small support near $e$? (here, $U^{-}$ is the unipotent radical of the opposite parabolic subgroup $P^{-}$ of $P$) These are possible if $\pi$ is the full induced representation. But I am wondering if it holds for a proper subrepresentation $\pi$. How do u think of it?
I don't have a specific counterexample, but I suspect that there's too much flexibility with subrepresentations to insist that they all must contain functions of small support. As for $U^-$-invariance, choose a compact open subgroup $K$ such that $G = P K$, and replace $f$ by $\int_{U^- \cap K} \pi(u^-)f,\mathrm du^-$.
@LSpice, Thank you very much for sharing your wisdom. You think that it might be impossible to choose $f$ which has small support. On your answer to my second question, I am wondering why $(\int_{U^- \cap K} \pi(u^-) f du^-)(e)=f(e)$. Why this does holds?
Oh, good point. I was thinking that $f$ is identically $f(e)$ on $U^-$, but actually it's identically $f(e)$ on $U$. Now that I don't mix up the parabolic and its opposite, I also doubt that this one holds for all subrepresentations, though again without an explicit counterexample. (I think that these revised questions are more interesting, and, if you move them into the question, I would remove my close vote.)
@LSpice, Thank you very much for your comments. I revised my question!
@LSpice, how about to restrict $\pi$ to the image of intertwining operator appearing in the proof of analytic continuation and functional equation of Eisenstein series? It looks so difficult to analyze the image of intertwining operator.
What do you mean by ''$f$ has small support near $e$''? By definition, $f(n)=f(e)$, for $n\in N$, which means that $f$ can not have compact support in $G$. If you meant to say ''small support'' in $N\backslash G$ that also seems unlikely. Because $f(nm)=\sigma(m)v$ for $m\in M$. This is nonzero (for $v\neq 0$), for instance, if $M$ is a torus.
@Subhajit Jana, Oh I am sorry for having not given enough explanation. I mean small supports of $e$ in $U^-$ modulo $P$.
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2025-03-21T14:48:31.068423
| 2020-05-22T13:53:24 |
361058
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|
Stack Exchange
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Is this elementary formula for the parabolic segment new?
Recently (May 2020) a formula for the area of the parabolic segment (i.e. the region enclosed by a parabola and a line), in terms of the coefficients of the Cartesian equations, has been published by an Italian high school student in MatematicaMente (n. 269).
Let the parabola and the line have equations $y=ax^2+bx+c$ and $mx+q$, respectively, and assume the line does intersect the parabola. Then the published formula is
$$A=\frac{\sqrt{[(b-m)^2-4a(c-q)]^3}}{6a^2}$$
where $A$ is the area of the corresponding parabolic segment.
If the parabola's equation has normal form $y=ax^2$, then the formula reduces to
$$A=\frac{\sqrt{(m^2+4aq)^3}}{6a^2}\;\;.$$
Is the above formula actually new? By this, I mean: can it be found anywhere in the literature (articles, preprints, books,...) prior to, say, 2019? If yes, was it proven by an elementary proof (using Cartesian coordinates geometry) or did it use integral calculus?
The coordinates proof involves a quite tedious computation but no original idea. So, it is quite surprising if indeed nobody, since the time of Decartes, had ever thought of seeing how the classical Archimede's result related with coordinate geometry.
While this clearly isn't research level, it seems both interesting, and well in line with the type of history questions we get here. I encourage you to leave a comment explaining a vote to close; particularly, if you think it belongs to HSM, I encourage you to keep in mind the resistance of a lot of our most valued members to moving their history questions there from here.
The formula for the area of the parabolic segment is usually written more concisely in terms of the two intersection abscissas $x_1$ and $x_2$,
$$A=\tfrac{1}{6}|a||x_1-x_2|^3.$$
See for example this elementary derivation.
Since $|x_1-x_2|=|a|^{-1}\sqrt{4 a (q-c)+(b-m)^2}$, being the difference of the two solutions of the quadratic equation $ax^2+(b-m)x+(c-q)=0$, the formula in the OP follows immediately.
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2025-03-21T14:48:31.068593
| 2020-05-22T14:38:28 |
361064
|
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|
Stack Exchange
|
Reference for homotopy colimit = total complex
I'm looking for a reference for the following fact:
take a simplicial chain complex $ X:\Delta^{op}\to Ch_{\geq 0}(\mathcal A)$ for $\mathcal A$ a nice abelian category (say, cocomplete with enough projectives, although I'm willing to add more hypotheses, because the $\mathcal A$ I want to use it for is the category of connective modules over some connective dga ); then a model for the homotopy colimit of $X$ is the total complex of the bicomplex associated to $X$.
I know a proof (I haven't checked the details so I'm not sure it works for an arbitrary dga - it works at least for discrete rings), so that's not what I'm looking for (except if you have an especially short and elegant one, then it wouldn't hurt to see it); I'm mostly looking for a reference.
I know the result is mentioned in Dugger's A primer on homotopy colimits (proposition 19.9) but there seems to be no proof in there - so I'll add the criterion that the reference should contain a proof.
This may be related to this question, which relates the total complex and the diagonal - since there is an answer with a reference there, it would also suffice to provide a reference for the fact that the diagonal is a model for the homotopy colimit (actually, this would be enough for other reasons : one can use the diagonal model for simplicial objects that land in $\mathcal A$, and then use homotopy cofinality of $\Delta^{op}\to \Delta^{op}\times \Delta^{op}$ to get the result for an arbitrary simplicial chain complex).
For the latter, I know references for simplicial sets, but not for simplicial $\mathcal A$-objects (and in the case of a discrete ring, one may use this as well via the usual adjunction).
The answers given here seem to be unsatisfactory given the comments below.
Here, the question itself provides a sketch of proof for $\mathbb Z$ which I think can be adapted to the general case, but the adjunction that is mentioned does not seem crystal clear to me (if you could explain it, that would also be great) and it's not a reference.
This is a duplicate of https://mathoverflow.net/questions/194010/reference-for-homotopy-colimits-of-cochain-complexes-via-totalization-of-dou, which is already referenced in the main post.
@DmitriPavlov: you're right but the answers there didn't seem to solve my problem so I wasn't sure what the etiquette was in that situation
See Problem 4.23 and Problem 4.24 (with proofs) of Ulrich Bunke's Differential cohomology.
The underlying abstract machinery for computing homotopy (co)limits
via homotopy (co)ends is presented by
Sergey Arkhipov and Sebastian Ørsted
in Homotopy (co)limits via homotopy (co)ends in general combinatorial model categories.
Thank you for your answer ! However it seems that the first reference only deals with $\mathcal A = \mathbf{Ab}$ (as far as I can tell + the proof 4.24. uses a claim that I would typically prove using the desired result, that is "every element in $Ch^{\Delta^{op}}$ is a homotopy colimit of a diagram of $c(\mathbb Z)$'s " ). For the second one, I think what I'm missing is an understanding of $Tot$ as a coend (or as an end, for the dual version) - do you maybe have references for that ?
@MaximeRamzi: There is nothing special about Z, the proof works in the same way for any locally presentable abelian category. Also, 4.23 works in the opposite category, giving you the desired presentation of a colimit as a coend.
As for your question about presenting totalization as a 1-coend, this is a purely 1-categorical question. It suffices to observe that chains on Δ^n only have a single summand (Z in degree n) that does not come from Δ^k for k<n. This is precisely the copy of Z responsible for totalization via shifting in degree by n. I have never seen this (simple) argument proved explicitly in the literature.
well, the proof can certainly go through in an arbitrary locally presentable abelian category, but as it is written, it does seem to use some stuff from $\mathbb Z$ (specifically the chain complex $C_*(\Delta^n)$ and its equivalence to $\mathbb Z$, that is, you need some monoidal unit or something) ; and since I'm looking for a reference, I'd like one that explicitly states that it works for an arbitrary good enough abelian category (although if I don't find one, I'll probably end up citing that one and saying "it goes through similarly in more generality"). For the coend, yes you're right !
(of course I'm not claiming that you can't adapt $C_*(\Delta^n)$ to an arbitrary abelian category, I'm just saying that as written, it is not that general)
@MaximeRamzi: No, chains on Δ^n have nothing to do with the abelian category A. Chains on Δ^n take values in the enriching category of abelian groups, and any cocomplete abelian category is enriched over abelian groups. The proof of 4.23 makes no use of A being the category of abelian groups, the hom in formula (33) can be replaced by the enriching hom over abelian groups instead, which is also a left Quillen bifunctor.
Ah I see, you mean that you can just define it to be a functor $\underline{\hom} : (Ch(\mathbb Z)^\Delta)^{op} \times Ch(\mathcal A)^\Delta \to Ch(\mathcal A)$ ?
@MaximeRamzi: Yes, that's how the Ab-enrichment works here.
Alright I guess the difference is small enough for it to be good enough. I would have rather liked some place where it's explicitly stated in full generality, but I guess that must be hard to find
@MaximeRamzi: As you can see from the responses to the other question, numerous experts do not know of such a place. In general, the state of literature on homotopy theory for such fairly elementary questions is often pretty bad. Even the paper about coends that I cited supposedly explains "well-known" facts, but was written only in July 2018.
Yup, you seem to be right. I'll accept your answer because I feel like it's the best I'll get
I wrote a note for referential purposes. I hope this helps.
Arakawa, K. (2023). Homotopy Limits and Homotopy Colimits of Chain Complexes. arxiv.2310.00201
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2025-03-21T14:48:31.069254
| 2020-05-22T14:43:55 |
361065
|
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|
Stack Exchange
|
Pairs of integers whose product is one more or less than a prime
Given a positive integer N it is often possible to pair each of the integers 1, 2, 3, ..., N with a different integer between N + 1 and 2 N so that the product of each pair is one less or more than a prime.
For example, if N = 10, such a pairing is (1,12), (2,11), (3,14), (4,13), (5,16), (6,15), (7,18) (8,17), (9, 20), and (10, 19).
Is this possible for infinitely many N? For all, but a finite number of N?
Have you tried to list such $N$ for small values, say $N\le 10000$? it should be easy with any mathematical software.
If my program is correct, it works for all $N$ from 2 to $200$.
Almost certainly. Note that all the products have to be even, and avoid being "in the middle" of large prime gaps, which (for small n) is about a third of the even numbers, and each such rules out something less than N pairings. So (2 13) and (2 17) and pairs multiplying to 50 and 56 are ruled out, etc. This should leave enough for a Hall argument to work. Gerhard "Plan For Many June Nuptials" Paseman, 2020.05.22.
@GerhardPaseman: since primes (and hence ${p\pm1}$) have density zero, I'm doubtful that such an argument is going to work.
@Greg, if you have a heuristic argument that there are N which won't work, I think that would be a useful answer post. I will admit the possibility that such N exist, but I think using a density zero argument will itself fail to indicate failure. I think the factorization of p+-1 numbers are the key, and I suspect one can set up a Hall argument based on this. Gerhard "Working Towards A Happy Ending" Paseman, 2020.05.22.
You can find the number of such prime numbers in special case. Denote $qk\pm 1=p$ with $k\in [N+1,2N]$ and $q\in [1,\log^A N]$. From this we will write the sum $\displaystyle \sum_{\substack{q(N+1)\leq p \leq 2qN \\ p \equiv (\pm 1 \mod q)}}1$ that controls the interval for $k$ and counts the prime numbers in the above form. An estimate for such a sum is made as for the proof of Siegel-Walfisz theorem, through equality $\displaystyle {\frac {1}{\phi (q)}}\sum _{\chi }{\bar {\chi }}(a)\chi (n)={\begin{cases}1,&{\text{ if }}n\equiv a{\pmod {q}}\\0,&{\text{ otherwise, }}\end{cases}}$ with $(a,q)=1$. We obtain $\displaystyle \sum_{\substack{q(N+1)\leq p \leq 2qN \\ p \equiv (\pm 1\mod q)}}1\sim \frac{qN}{\phi (q)\log N}$. (Linnik's theorem)[https://en.m.wikipedia.org/wiki/Linnik%27s_theorem] gives a larger interval for $q$, but not enough. Linnik's constant $L$ equal to $2$ will give the desired result since $q\leq (qN)^\frac{1}{2}$, under GRH $L=2+\epsilon$.
Note that any such pairing gives products greater than $N$ and at most $2N^2$, and thus all the products must be even or the number 3. So with one exception, odd numbers at most N are paired with even numbers greater than N, and for N greater than 2 odd numbers greater than N are paired with even numbers less than N. So what pairings do not lead to products of the desired form?
Let r be even with r not adjacent to a prime. The smallest r start with 26, then 34,50,56,64,76,86,92,94,116,118,120,122,124,134. A paring is a factorization of one of these special even numbers (p,q) with the product being r and one of the factors being odd. Skipping the case that the odd number is 1, we have in most cases these numbers have one odd factor, with the exceptions in the previous list being 50,64, and 120. In this small sample, we have on average little more than one forbidden edge for each candidate. This supports the notion that for small N one can always find a mapping.
Also, not every disallowed edge needs to be considered. If 210 were a disallowed product (it isn't) then (2,105) would be a problem edge only for N in (52,105), and (14,15) only for N=14, so not all seven possibilities (excluding (1,210) are in play for any given N. So the number of bad edges blocking a matching appears to be small. Again ignoring 1, these edges are (5,10),(2,13),(2,17),(2,25),(7,8),(4,19),(4,23)(2,47), on up. This means one problem edge for N=5, 3 for N=7, 2 for N=8, 3 for N=9, and so on. Given that when N=10 there are fifty edges with five coming from each vertex, having three problem edges seems like nothing. (If we include 1, there are more problem edges, but there are also many edges coming from 1 on the order of 2N/ln(2N) many.)
It seems reasonable to estimate (and challenging to prove, which I won't do now) that for every N, that the product graph (which associates to each k less than n those j in (N,2N] for which kj is adjacent to a prime) for N described above has each node with a degree about N/(2ln N). This does not prove there is a matching, but I think it makes it very likely, especially since the (set of) vertices connected to a j less than N will often be quite different to (set of) those connected to a k less than N. Unfortunately, we need something like Linnik's theorem to show no isolated vertices in this graph.
Gerhard "Not Strengthening Linnik's Theorem Yet" Paseman, 2020.05.24.
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2025-03-21T14:48:31.069625
| 2020-05-22T14:52:51 |
361066
|
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|
Stack Exchange
|
Properties of the contraction map of the exceptional divisor of a blow-up
Let $X$ be a smooth variety and $Z$ be a smooth subvariety. Consider the blow-up at $Z$ $$\pi:\mathrm{Bl}_Z(X)\rightarrow X$$ and let $E$ be the exceptional divisor. What are the properties of the restriction map $\pi|_E:E\rightarrow Z$?
Of course it is projective, more explicitly, if $Z$ is given by the coherent ideal $\mathcal{I}$ of $X$ then the map is given by
$$E=\underline{\mathrm{Proj}}_Z(\oplus_{n\geq 0}\mathcal{I}^n/\mathcal{I}^{n+1})\rightarrow Z.$$
That is, it is the projective morphism given by the associated graded ring of the ideal $\mathcal{I}$.
Nevertheless, I think that this map should be much more than just projective in general. For example, I would like to know if this map is flat or smooth or if it has any other property relating it to a projection of the form $W\times Z\rightarrow Z$.
If more generally you know something about the general contraction of the exceptional locus of a birrational proper/projective map between smooth varieties it would be really nice! Thanks in advance.
$E$ is a projective bundle over $Z$; it is the projectivisation of either the normal or conormal bundle of $Z$ in $X$, depending on your choice of convention. In particular it is smooth.
Sorry, I meant to write that $\pi_{|E}: E \rightarrow Z$ is smooth.
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2025-03-21T14:48:31.069866
| 2020-05-22T15:39:36 |
361070
|
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|
Stack Exchange
|
Surjection in compact-open topology
Let $Z$, $X$ and $Y$ be topological spaces and let $f:X\to Y$ be a continuous surjection then is the induced map $g \to f\circ g$ from $C(Z,X)$ to $C(Z,Y)$ is continuous. But is it still a surjection?
My issue is that it's not clear if it has a right-inverse...
In many cases, $f_\ast: C(Z,X)\to C(Z,Y)$, $g\mapsto f\circ g$ is not surjective: Put $Z=Y$ and $h=id_Y \in C(Z,Y)$, then $h$ is in the range of $f_\ast$ only if $f$ has a right inverse.
So the sufficient condition is for $f$ to have a continuous right-inverse?
If $f$ has a right inverse $r$ then $g=f\circ(r\circ g)$...
Yes, of course, Flabby.
Let $X = Y = S^1 = \mathbb{R}/ \mathbb{Z}$ be the circle and let $f \colon X \to Y$ be given by $f(t) = 2t$ which is continuous and surjective.
There exists no $g \in C(S^1,X)$ such that $f \circ g = \mbox{Id}_{S^1}$, as $f$ induces the doubling map on $\pi_1$.
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2025-03-21T14:48:31.069974
| 2020-05-22T15:54:31 |
361073
|
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|
Stack Exchange
|
Observable nearly commuting with a "complete" set of commuting observables
Consider the Hilbert space $H = E^{\otimes n}$ where $E=\mathbb{C}^2$.
On $E$ we have an observable $O$ (i.e. a Hermitian matrix) that is diagonalizable in the standard basis with eigenvalues $1$ and $-1$. By tensoring $O$ with the identity on $E^{\otimes n-1}$, and doing so for each of the $n$ possible positions for the factor $E$, we get a set of commuting observables $O_i$, $i=1\dots n$. An observable that commutes with all these observables must belong to the algebra generated by the $O_i$'s.
Now if I have an observable $M$ on $H$ such that the operator norm of $[M,O_i]$ is at most $1$ for all $i$, how far can $M$ be from the algebra generated by the $O_i$'s, in the operator norm? What are explicit examples that are far from the algebra?
Any pointer or relevant remark for related questions welcome.
The $O_i$ are not a complete set of commuting observables --- there aren't enough. The dimension of $H$ is $2^n$ not $n$.
(Unless you mean something different than I do by a "complete" set.)
@NikWeaver that's correct sorry. I meant the ring they generate is a complete set of commuting observables
@NikWeaver: I think the question is basically how far away $M$ is from the space of polynomials in the $O_i$.
@AbdelmalekAbdesselam that's correct, sorry for the bad phrasing
@alesia: not your fault. Dirac is to blame.
Well, you have
$$
[\frac{1}{idx},x]=1/i
$$
although $\frac{1}{idx}$ is far from the Identity.
Yes, are you suggesting that taking the analog of this commutation relation over the finite field F_2^n instead of the reals would give an interesting example?
What does your fraction mean? Should it be $\frac1 i\frac{\mathrm d}{\mathrm dx}$, or is there really some meaning to $\frac1{\mathrm dx}$?
I think the question is really finite dimensional in nature and this example does not help much.
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2025-03-21T14:48:31.070151
| 2020-05-22T17:04:57 |
361081
|
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"Abdelmalek Abdesselam",
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|
Stack Exchange
|
Exponential decay of voltage potential difference
Consider the following adjacency matrix of a complete graph:
$$A=(e^{-|i-j|})_{1\leq i\neq j\leq n}$$
with 0 on the diagonal. Let $D=diag\{d_1,...,d_n\}$ be the degree matrix where $d_i=\sum_{j\neq i}e^{-|i-j|}$. Then $L=D-A$ is the Laplacian. Let $L^\dagger$ be the Moore-Penrose inverse of the Laplacian. I'm interested in the following quantity
$$a_{ij}=|(e_1-e_2)^TL^\dagger(e_i-e_j)|$$
where $e_i=(0,0...,0,1,0,...0)$ with 1 on the ith coordinate. I conjecture that $a_{ij}$ will decay exponentially when both $i$ and $j$ moves away from 1 and 2. Something like $a_{ij}\leq C _1e^{-C_2\min\{i,j\}}$ where $C_1,C_2$ are some constants. From the physics point of view, $a_{ij}$ is the voltage potential difference between $i$ and $j$. It is intuitive that when they are far away from the source, 1 and 2, they should be very small given the structure of the graph.
In fact, my simulation shows that as long as $i,j\neq1,2$, $a_{ij}$ suddenly becomes extremely close to 0. There seems to be no decay, but an acute cut. This phenomenon holds for slight perturbation of $A$, keeping the decaying property.
Is this conjecture true? How can we prove it? What is the rate of decay?
Another quantity that is also interesting is
$$\sum_{i\neq j}e^{-|i-j|}a_{ij}$$
which is the weighted average of potential differences. How can we bound this? For this quantity, I conjecture it is bounded by some constant instead of growing with $n$. The physical meaning of this quantity is the sum of all currents in each edge.
(Update)
Enlightened by the discussion with @Abdelmalek Abdesselam below. We have the Neuman series representation:
$$a_{ij}=|(e_1-e_2)^TD^{-1/2}\sum_{k\geq0}\left(D^{-1/2}AD^{-1/2}-\alpha D^{1/2}JD^{1/2}\right)^kD^{-1/2}(e_i-e_j)|$$
where $J$ is the matrix of all 1s and $\alpha$ is some constant to be chosen. We want to choose $\alpha$ such that the power of the matrix decays fast. How can we achieve this and bound the entries of $D^{-1/2}AD^{-1/2}-\alpha D^{1/2}JD^{1/2}$? A possible choice is $\alpha=1/tr(D)$.
I guess one should say for more precision that $C_1,C_2$ must be independent of $n$. I would have to think a bit about the use of the Moore-Penrose inverse. But for a related problem where you would add some positive quantities on the diagonal so that $L$ is invertible and you can use $L^{-1}$ instead of $L^{\dagger}$, a simple Neuman series will do the trick. It is also the Green function of a (jumping) random walk with a positive killing rate. Another idea is to use the Matrix-Tree theorem to express $L^{\dagger}$.
Yes. $C_1$ and $C_2$ are independent of $n$ as I posted they are constants.
@AbdelmalekAbdesselam Actually, they may depend on $n$, as long as the bound is sharper than they are constants. It is not clear to me that the optimal bound is independent of $n$.
This is a very good question BTW. I think the result should be true and the continuum limit of it is basically saying the derivative of Brownian motion is white noise. To see this generalize your conjecture to $A$ which are not equal but rather bounded by exponentials. The particular case or nearest neighbor $A$ amout to showing decorrelation of local increments of a discrete walk trying to become a Brownian motion.
If $L$ is replaced by $\lambda I+L$ for $\lambda>0$, the analogous result is proved as Lemma 3, page 10 of https://arxiv.org/abs/0901.4756
@AbdelmalekAbdesselam Could the representation of inverse of symmetrized Laplacian in Lemma 19 of http://jmlr.csail.mit.edu/papers/volume15/vonluxburg14a/vonluxburg14a.pdf be useful? (Page 1778)
That's the same idea as the one I told you which is: use a Neumann series. Here it is adapted for the pseudoinverse. You can do that yourself by following the geometric description in https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse and applying an ordinary Neumann series when it comes to step about inverting the map from the orthogonal of the kernel to the range. This orthogonal contains the $e_1-e_2$ etc. you are rightfully using.
Then to finish the job: getting the bound from the Neumann series, you can follow the Lemma 3 in my paper with Procacci and Scoppola.
@AbdelmalekAbdesselam How do you bound the operator norm in this case?
And I do not think spectral is enough in this case, since $e_i-e_j$ is a sparse vector and spectral concerns the whole matrix.
Spectral consideration are a red herring here. You need the explicit Neumann series/random walk representation at the level of matrix elements. Forget about operator norms and read the proof of Lemma 3 in the paper I referenced. If you insist on using norms, what you need is an $L^{\infty}-L^{1}$ norm.
@ AbdelmalekAbdesselam But your proof used operator norm to bound matrix elements.
@nevernevernever: Sorry I spoke too fast. I think the Neumann series should give the result for more general entries than exponentials, but I don't yet see the way to do that because one has to exploit some cancellations due to the test functions summing to zero, i.e., being linear combination of things like $e_i-e_j$. In any case Mateusz gave a nice answer for the exponential "integrable model".
Edit: This turns out to be quite simple. Observe that $a_{1i} / a_{2i} = q$ does not depend on $i \in \{3, 4, \ldots, n\}$. Thus, if $x_1 = 1$, $x_2 = -q$ and $x_i = 0$ for $i \in \{3, 4, \ldots, n\}$, then we clearly have $L x = c e_1 - c e_2$, where $c = \sum_{i=3}^n a_{1i}$. It follows that $$L^\dagger (e_1 - e_2) = c^{-1} x + \operatorname{const}.$$
I leave the previous version of this answer below, as it provides a way to evaluate $L^\dagger$ explicitly.
I cannot say I understand what is really going on here, but at least I have a proof that $a_{ij} = 0$ when $i, j \ge 3$. (I leave my previous comment/answer, as it contains some related stuff that is not included here.)
Notation: Every sum is a sum over $\{1, 2, \ldots, n\}$. We write $q = e^{-1}$ (and actually any $q \in (0, 1)$ will work). Given a vector $x = (x_i)$ we write
$$ \Delta x_i = x_{i+1} + x_{i-1} - 2 x_i $$
if $1 < i < n$.
Given a vector $(x_i)$, we have
$$ Lx_i = \sum_j q^{|i-j|} (x_i - x_j) = b_i x_i - \sum_j q^{|i-j|} x_j ,$$
where
$$ b_i = \sum_j q^{|i-j|} = \frac{1 + q - q^i - q^{n+1-i}}{1 - q} $$
Therefore, when $1 < i < n$, we have
$$ \begin{aligned}
\Delta Lx_i & = \Delta (b x)_i - \sum_j (q^{|i-j+1|}+q^{|i-j-1|}-2q^{|i-j|}) x_j \\
& = \Delta (b x)_i - (q + q^{-1} - 2) \sum_j q^{|i-j|} x_j + (q^{-1} - q) x_i \\
& = \Delta (b x)_i + (q + q^{-1} - 2) L x_i - (q + q^{-1} - 2) b_i x_i + (q^{-1} - q) x_i \\
& = (q + q^{-1} - 2) L x_i + b_{i+1} x_{i+1} + b_{i-1} x_{i-1} - ((q + q^{-1}) b_i - (q^{-1} - q)) x_i .
\end{aligned} $$
A short calculation shows that
$$ b_{i+1} + b_{i-1} = ((q + q^{-1}) b_i - (q^{-1} - q)) $$
(which looks somewhat miraculous, but there must be some insightful explanation for that). Thus,
$$ \Delta Lx_i = (q + q^{-1} - 2) L x_i + b_{i+1} (x_{i+1} - x_i) + b_{i-1} (x_{i-1} - x_i) . $$
Suppose that $x_i = L^\dagger y_i$ for some vector $(y_i)$ such that $\sum_i y_i = 0$. Then $L x_i = L L^\dagger y_i = y_i$. Write $c = q + q^{-1} - 2$. We then have
$$ \Delta y_i - c y_i = b_{i+1} (x_{i+1} - x_i) + b_{i-1} (x_{i-1} - x_i) . $$
In particular, the following claim follows.
Proposition 1: If $1 < i < n$, $y_{i-1} = y_i = y_{i+1} = 0$ and $x_i = x_{i+1}$, then $x_{i-1} = x_i$.
The above result will serve as an induction step. To initiate the induction, we need to study the $i = n$, which is slightly different. In that case:
$$ \begin{aligned}
Lx_{n-1} - Lx_n & = b_{n-1} x_{n-1} - b_n x_n - \sum_j (q^{|n-j-1|}-q^{|n-j|}) x_j \\
& = b_{n-1} x_{n-1} - b_n x_n - (q^{-1} - 1) \sum_j q^{|n-j|} x_j + (q^{-1} - q) x_n \\
& = b_{n-1} x_{n-1} - b_n x_n + (q^{-1} - 1) L x_n - (q^{-1} - 1) b_n x_n + (q^{-1} - q) x_n \\
& = (q^{-1} - 1) L x_n + b_{n-1} x_{n-1} - (q^{-1} b_n - (q^{-1} - q)) x_n .
\end{aligned} $$
This time we have
$$ q^{-1} b_n - (q^{-1} - q) = b_{n-1} , $$
and hence
$$ Lx_{n-1} - Lx_n = (q^{-1} - 1) L x_n + b_{n-1} (x_{n-1} - x_n) . $$
Again we consider $x_i = L^\dagger y_i$ for some vector $(y_i)$ such that $\sum_i y_i = 0$, and we write $d = q^{-1} - 1$. We then have
$$ (y_{n-1} - y_n) - d y_n = b_{n-1} (x_{n-1} - x_n) . $$
As a consequence, we have the following result.
Proposition 2: If $y_{n-1} = y_n = 0$, then $x_{n-1} = x_n$.
For $y = e_1 - e_2 = (1, -1, 0, 0, \ldots)$, we immediately obtain the desired result.
Corollary: If $y = e_1 - e_2$ and $x = L^\dagger y$, then $x_3 = x_4 = x_5 = \ldots = x_n$. Consequently, $a_{ij} = x_i - x_j = 0$ whenever $i, j \ge 3$.
Another consequence of the above result is that if $L^\dagger = (u_{ij})$, then $$u_{i+1,j+1}-u_{i,j+1}-u_{i+1,j}+u_{i,j} = 0$$ whenever $i + 1 < j$ or $j + 1 < i$. Moreover, it should be relatively easy to use Propositions 1 and 2 to evaluate $u_{ij}$ explicitly, and in particular to prove that
$$ u_{ij} = v_{\max\{i,j\}} + v_{\max\{n+1-i,n+1-j\}} + \tfrac{1}{4} |i - j| $$
when $i \ne j$, where $(v_i)$ is an explicitly given vector (in terms of products/ratios of $b_i$, I guess).
Final remark: there is a corresponding result in continuous variable: the Green function for the operator $$L f(x) = \int_0^1 e^{-q |x - y|} (f(x) - f(y)) dy$$ has zero mixed second-order derivative. The proof follows exactly the same line, and is in fact somewhat less technical.
This is such a miracle! When the entries are not $e^{-|i,j|}$ exactly, for example we only have $e^{-|i-j|}\leq A_{ij}\leq 2e^{-|i-j|}$. Then I think they will not be exactly 0, can we bound $a_{kl}$ using similar arguments?
(1/2) A miracle — indeed, I am completely surprised by this result, and I still do not understand how this is possible. I did once look at the continuous counterpart mentioned in the end of my answer and I noticed some nice explicit expressions, but with a different boundary condition. I am kind of shocked that with this particular definition one can still get explicit expressions.
(2/2) Regarding perturbations: I no longer trust my intuition here, but my wild guess would be "no", I think. The above approach seems to heavily use the structure of $e^{-|i-j|}$.
Your updated solution is very insightful. It basically says that $e_1-e_2$ is in the span of the first two columns of $L$. When the entries of $L$ is not exactly $e^{-|i-j|}$, it seems intuitive that $e_1-e_2$ should also roughly lie within the span of the first several columns. The importance of each column to form the vector $e_1-e_2$ by linear combination seems to decay somehow.
This is not an answer, but too long for a comment.
Consider a doubly infinite matrix $L = (q_{ij})_{i,j \in \mathbb{Z}}$ with entries $q_{ij} = -e^{-|i - j|}$ when $i \ne j$, and $q_{ii} = 2 e / (1 - e)$; here $i, j \in \mathbb{Z}$. The symbol of this matrix (i.e. the Fourier series with coefficients $e^{-|j|}$, except at $j = 0$) is:
$$ \psi(x) = \frac{e^2 - 1}{e^2 - 2 e \cos x + 1} - \frac{e + 1}{e - 1} . $$
The symbol of $L^\dagger$ is thus $1 / \psi(x)$ (in the principal value sense), which has a singularity of type $1 / x^2$ at $x = 0$. It follows that in this case
$$ a_{kl} = \frac{1}{2 \pi} \int_{-\pi}^{\pi} \frac{(e^{i x} - e^{2 i x}) (e^{i k x} - e^{i l x})}{\psi(x)} \, dx . $$
In general, the above expression will only have power-type decay as $k,l \to \infty$.
However, for this particular choice of $L$, things simplify a lot. The pseudo-inverse $L^\dagger = (u_{ij})_{i,j \in \mathbb{Z}}$ can be found explicitly, and its entries are $u_{ij} = C_1 - C_2 |i - j|$ when $i \ne j$ and $u_{ii} = C_3$ for appropriate constants $C_1$, $C_2$, $C_3$. Consequently, $a_{kl} = 0$ when $k, l > 2$.
I do not have a clear intuition about what happens in the one-sided case (that is, if we consider an infinite matrix $L$ with entries indexed by $i, j \in \{1, 2, \ldots\}$), let alone the bounded case (with $i, j \in \{1, 2, \ldots, n\}$). My wild guess would be that the symmetry breaks, and there is no hope for any closed-form formula.
However, a quick numerical experiment suggests strongly that we still have $a_{kl} = 0$! More precisely, the entries $u_{ij}$ of $L^\dagger$ apparently satisfy
$$
u_{ij} = v_{\max\{i,j\}} + v_{\max\{n+1-i,n+1-j\}}, v_{n-i} + v_j\} + \tfrac{1}{4} |i - j| \qquad (i \ne j)
$$
for an appropriate vector $v_i$. I find this extremely surprising!
Here is the code in Octave, in case anyone is interested. First, we construct $L$ and its pseudo-inverse (denoted U here):
n = 10; # size of the matrix
A = toeplitz(exp(-(0:n-1)));
L = diag(A * ones(n,1)) - A; # matrix L
U = pinv(L); # pseudo-inverse L^\dagger
Next, we verify that the mixed second-order difference of $L^\dagger$ is a tri-diagonal matrix:
D = U(1:n-1, 1:n-1) - U(1:n-1, 2:n) ...
- U(2:n, 1:n-1) + U(2:n, 2:n); # second-order difference of U
This already shows that $L^\dagger$ has the desired structure, but we can verify this directly. First two lines are to extract the vector $v_i$, the other two define the matrix Z with entries
$$
u_{ij} - v_{\max\{i,j\}} - v_{\max\{n+1-i,n+1-j\}}, v_{n-i} + v_j\} - \tfrac{1}{4} |i - j| \qquad (i \ne j)
$$
which should be zero except on the diagonal:
X = U - 0.25 * abs(repmat(1:n, n, 1) - repmat(1:n, n, 1)');
V = X(:, 1) - 0.5 * X(n, 1);
I = repmat(1:n,n,1);
Z = X - V(max(I, I')) - V(max(n + 1 - I, n + 1 - I'));
How did you find the expression of $L^\dagger$ explicitly?
We have $1/\psi(x) = c_1 (e^2 - 2 e \cos x + 1) / (1 - \cos x) = c_2 + c_3 / (1 - \cos x)$; the constant $c_2$ corresponds to $c_2 \delta_{ij}$, and $c_3 / (1 - \cos x)$ corresponds to the Green's function for the simple random walk, $-c_4 |i - j|$.
What if the entries of the matrix is not exactly $e^{-|i-j|}$ and the matrix is not Toeplitz, but the entries decay at the same rate. Can we still obtain explicit expression for $L^\dagger$? Is $a_{kl}$ still 0 for $k,l>2$?
I doubt one can get explicit expressions if $L$ is not exactly $(e^{-|i-j|})$.
I edited my answer/comment. I am now convinced that your $a_{ij}$ is equal to zero, but I have no clue why.
|
2025-03-21T14:48:31.070967
| 2020-05-22T17:20:09 |
361082
|
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|
Stack Exchange
|
Necessary and sufficient condition for a prime to be represented by an arbitrary positive definite binary quadratic form $ax^2+bxy+cy^2$
Given an arbitrary (but fixed) positive definite primitive integral binary quadratic form $g(x, y)=ax^2+bxy+cy^2$, and let $m$ be an arbitrary integer. We will denote the discriminant of $g$ by $D=D_g=b^2-4ac$. The following problems are very interesting to me:
$\color{Red}{\text{A}}$) Which integers can be represented by the quadratic form $g$?
$\color{Red}{\text{B}}$) What is the exact number of representations?
i.e. what is the exact number of integral solutions to the equation $g(x,y)=m$, where $g$ is a fixed quadratic form as $m$ varies?
$\color{Lime}{\text{C) My question is}}$: Is there any necessary and sufficient condition for a prime to be represented by an arbitrary primitive binary quadratic form $ax^2+bxy+cy^2$?
(0) When the form $g$ is the principal form of the form class group $\mathcal{C}(D)$, then we can give a necessary and sufficient condition under which a prime number is representable by $g(x,y)$. (Choose $\epsilon_D \in \{ 0, 1 \}$ such that $D \stackrel{2}{\equiv} \epsilon_D$, then a representative for the principal class would be $P_{D}=x^2+\epsilon_Dxy+\frac{\epsilon_D-D}{4}y^2$)
(To being more precise, it should not be equal, it should just be equivalent)
From now on $D$ would denote an arbitrary negative Discriminant which is fixed, and $P$ will denote the principal quadratic form of discriminant $D$. Also, we will use this notation $h=\vert \mathcal{C}(D) \vert$. Also, by a quadratic form, we will mean a positive definite primitive integral binary quadratic form.
The main theorem of the book «Primes of the form $x^2+ny^2$» says that:
The Main Theorem.
Consider the principal form $P(x, y)$. and let $K:=\mathbb{Q}(\sqrt{D})$,
Then $\mathcal{O}=\mathbb{Z}[\frac{D+\sqrt{D}}{2}]$ would be an order of conductor $f$,
(where $D=f^2d_K$,
and $d_K$ is the discriminant of the imaginary quadratic field $K=\mathbb{Q}(\sqrt{D})$
and $D$ is the discriminant of the quadratic form $g$.)
in the ring of integers $\mathcal{O}_K=\mathbb{Z}[\frac{d_K+\sqrt{d_K}}{2}]$
in the imaginary quadratic field $K=\mathbb{Q}(\sqrt{D})$.
Let $L=K(\alpha)$ be the ring class field of the order $\mathcal{O}=\mathbb{Z}[\frac{D+\sqrt{D}}{2}]$ in the imaginary quadratic field $K=\mathbb{Q}(\sqrt{D})$. And let $f_{D, h}(x)$ be the minimal polynomial of $\alpha$, note that the degree of $f_{D, h}(x)$ is $h$. And let $p$ be a prime such that $p\nmid D \times \text{Disc}(f_{D, h}(x))$, then $p$ can be represented by the principal form $p=P_{D}=x^2+\epsilon_Dxy+\frac{\epsilon_D-D}{4}y^2$ if and only if $(D/p)=1$ and $f_{D, h}(x) \cong 0 \mod p$ has a solution; in other words, it has the form $p=P_{D}=x^2+\epsilon_Dxy+\frac{\epsilon_D-D}{4}y^2$ if and only if both of the equations $x^2-D \cong 0$ and $f_{D, h}(x) \cong 0$ has a solution $\mod p$.
(1) Can we have some generalized results if we replace the principal form with any arbitrary positive definite binary quadratic form? Could we find a similar statement for the necessary and sufficient condition under which a prime number can be represented by an arbitrary positive definite binary quadratic form?
To clarify what I mean:
(2) For instance, if each genus contains only one class, then the answer would be positive, and it is given by congruence classes. Being more precise, we can determine the set of primes represented by $g$, by congruence conditions over $\mathbb{Z}$, if and only if the form class group $\mathcal{C}(D)$, equivalently the proper ideal class group $\mathcal{C}(\mathbb{Z}[\frac{D+\sqrt{D}}{2}])$, is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^r$ (Each genus would contain only one class if and only if these two equivalent conditions are satisfied), Also, in this case we can give a completely satisfying answer to the question $\color{Red}{\text{B}}$. Also, note that there are only finitely many of them, for the case of even discriminant see Euler's Convenient Numbers.
$\color{Green}{\text{Part (3)}}$ Or if $g(x,y)$ is a quadratic form of order $3$, then $\{ P_D, g^{1}, g^{-1} \}$ form a group of order $3$ in the form class group $\mathcal{C}(D)$. Also, we know that the Ideal class group $\mathcal{C}(\mathcal{O})$ is isomorphic to the form class group $\mathcal{C}(D)$, and so there is a corresponding subgroup of order $3$ in $\mathcal{C}(\mathcal{O})$, and let us denote the fixed field of this subgroup by $H$, which would be an intermediate field of $L/K$, of degree $\frac{h}{3}$ over $K=\mathbb{Q}(\sqrt{D})$. Note that the characteristic of $K$ is zero, so there exists a primitive element $\beta$ such that $H=K(\beta)$, and let $f_{D, {\frac{h}{3}}}(x)$ be its minimal polynomial, notice that its degree is $\frac{h}{3}$. Then by The Main Theorem, we can see that a prime number $p$ with $p\nmid D \times \text{Disc}(f_{D, h}(x)) \times \text{Disc}(f_{D, {\frac{h}{3}}}(x))$, is represented by $g(x,y)$ if and only if $(D/p)=1$ and $f_{D, {\frac{h}{3}}}(x) \cong 0 \mod p$ has a solution and $f_{D, h}(x) \cong 0 \mod p$ does not have any solution. Also, we can give similar statements if $g(x,y)$ is a quadratic form of order $n=1, 2, 3, 4, 6$ because these are the only numbers whit $\varphi(n) \leq 2$, and in these cases we can distinguish the element {1} in $\mathbb{Z}/n\mathbb{Z}$ up to multiplication by the subgroup $\{ \pm 1\}$.
(4) Although we can give a necessary and sufficient condition in these special cases, but even this is not satisfactory, and I am not looking for this kind of condition; notice that in the main theorem if $p=P_D(x,y)$, then some equation has a solution $\mod p$. But the necessary and sufficient condition for these cases in $\color{Green}{\text{Part (3)}}$, has a somehow reverse nature: For instance, assume that $h(D)=3$, and also let $g(x,y)$ be a quadratic form of order $3$, then the argument in $\color{Green}{\text{Part (3)}}$ implies that a prime $p$ is represented by this quadratic form, if some equation does not have a solution $\mod p$, so we can not consider the trivial condition for the case $h(D)=3$ as a generalization of The Main Theorem in a reasonable way.
(5) I believe the answer to my question is negative. If we restrict ourselves to the case Discriminant= $-92$ or $-108$, then the class group has $3$ classes in both of the cases, and a complete set of representatives can be given as $P_{23}=x^2+23y^2$ & $g_{23}^{\pm}=3x^2\pm2xy+8y^2$, and $P_{27}=x^2+27y^2$ & $g_{27}^{\pm}=4x^2\pm2xy+7y^2$, and let $f_{-92, 3}=x^3-x-1$ & $f_{-108, 3}=x^3-2$. Let $D=-92$ or $D=-108$, Let $p$ be a prime which does not divide $D$, nor the discriminant of $f_{D, 3}(x)$ and also $(D/p)=1$. Suppose on contrary that the answer to my question is positive, then there exists polynomial $F_{D, 3}(x)$ such that: for every prime $p$ with $(D/p)=1$, exactly one of the polynomials $f_{D, 3}(x)$ or $F_{D, 3}(x)$ has a solution $\mod p$, and the other one does not have a solution, but I can not continue to gain a contradiction.
I am looking for $\color{Green}{\text{any kind of generalization of the main theorem}}$ (I am looking for a generalization from principal form to any arbitrary binary quadratic form, but any other kind of generalizations are welcome, for example, some related facts about quadratic forms in more variables) of that book, and I searched through many works of literature and many papers, but I didn't find anything. I am aware that questions $\color{Red}{\text{A}}$ and $\color{Red}{\text{B}}$ involving too many deep areas in number theory, and maybe giving a complete answer to them is impossible in some sense, and I am not asking for the answers to them. I just asking for the question $\color{Lime}{\text{C}}$; and if the answer is positive, $\color{Green}{\text{to be introduced to other fields involving them}}$, also, anything regarding the questions $\color{Red}{\text{A}}$ and $\color{Red}{\text{B}}$, would be welcome.
$\color{Green}{\text{Any guidance on other areas and directions relating to this problem is welcome.}}$
The answer is really quite straightforward: a prime $p$ is representable by a quadratic form $f(x,y) = ax^2 + bxy + cy^2$ only if $p$ splits in the quadratic field $\mathbb{Q}(\sqrt{\Delta(f)})$, where $\Delta(f) = b^2 - 4ac$ is the discriminant. To give a sufficient condition, the primes $\mathfrak{p}$ above $p$ have to belong to the ideal class corresponding to $f$ via the Gauss composition laws, or its inverse. The exact number of representations by a given form $f$ is not known, only the total number of representations by all classes of the same discriminant.
(a) Dear @StanleyYaoXiao , The necessity part of your answer is very clear to me and we can prove it in many ways, for instance, it is equivalent to the Corollary 2.6 of the first chapter [Page 24] of the second edition of the book Primes of the form $x^2+ny^2$ (that corollary was stated only for even quadratic forms, but it can be reformed to work for all cases. Or it can be deduced by Dedkind Factorization theorem).
(b) Dear @StanleyYaoXiao , For the sufficiency part, by a little work with the Theorem 7.7 of the second chapter [Page 123] of that book, it follows that one of the primes lying above $p$ must belong to the corresponding class to $f$, in the proper ideal class group of the order $\mathcal{O}$, and the other one should belong to the corresponding class to $f^{-1}$.
And both of the proofs of Corollary 2.6 and Theorem 7.7 are done in a quite elementary manner, without any class field theoretic phenomena.
I am not sure what kind of answer you are looking for. The phenomenon you are asking about is controlled by the class group, and in general one cannot reduce it to congruence conditions over the rational integers because the Hilbert class field of most imaginary quadratic fields is not an abelian extension over the rationals.
@StanleyYaoXiao Consider a primitive form $g(x,y)=ax^2+bxy+cy^2$ with discriminant $D=D_g=b^2-4ac < 0$.
We can determine the set of primes represented by $g$, by congruence conditions over $\mathbb{Z}$, if and only if the form class group $\mathcal{C}(D)$ (equivalently the proper ideal class group $\mathcal{C}(\mathbb{Z}[\frac{D+\sqrt{D}}{2}])$) is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^r$. And there are only finitely many of them, for the case of even discriminant see Convenient Numbers. Therefore I am not looking for congruence conditions.
17th version of this question.
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2025-03-21T14:48:31.071502
| 2020-05-22T18:26:45 |
361087
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Stack Exchange
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Status of the $n$ conjecture and, as secondary question or reference request, what about a transfer method for this conjecture $n>3$
The n conjecture is a generalization of the abc conjecture. What is the current status of the $n$ conjecture? See also [1]
Question 1. Can you tell us what about the current status of the $n$ conjecture, we consider $n>3$, of the $n$ conjecture?
One method of creating abc triples, that is triples which are close to being tight with respect to the abc inequality is the transfer method.
Question 2. Is there a version of the transfer method for the $n$ conjecture? (See Section 2.4 of Martin and Miao - $abc$ triples.)
References:
[1] Jerzy Browkin and Juliusz Brzeziński, Some remarks on the abc-conjecture, Math. Comp. 62 (206), (1994), pp. 931–939. (MSN)
[2] Greg Martin and Winnie Miao, abc triples, Funct. Approx. Comment. Math. Vol. 55, N. 2 (2016), pp. 145-176. (MSN)
About the Question 2: A) I've known the transfer method from section 2.4. Transfer method of the corresponding preprint on arXiv with identificator 1409.2974 (I believe thus that this is the corresponding preprint to [2]). B) Also, I don't know if does make sense a LLL method for the $n$ conjecture, or if it is in the literature of the $n$ conjecture in some way.
Many thanks for your edit and help @JoshuaZ
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2025-03-21T14:48:31.071733
| 2020-05-22T18:46:59 |
361090
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Stack Exchange
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Curves on a Kahler manifold
Let $(X,\omega )$ be a compact Kahler manifold. For any $d>0$ are there only finitely many families of curves $C_i$ such that $C_i\cdot \omega <d$? (More precisely, if $C$ is any curve such that $C\cdot \omega <d$, then $C$ belongs to one of the families $C_i$.) I believe the analogous statement for projective varieties follows from well known results for Hilbert or Chow schemes.
I guess you want $X$ to be compact right?
It should be true when $X$ is a projective manifold by considering the map from the Hilbert to Chow schemes. Perhaps the same thing works more generally using Douady and Barlet spaces (assuming compactness)?
Yes, $X$ is compact and the projective case should follow from a Hilbert scheme argument.
From "Bounded sets of sheaves on Kähler manifolds"
By Matei Toma, J. reine angew. Math. 710 (2016), 77–93
Lemma 4.4. Let X be a Kähler manifold, r be an integer and F be a set of compact reduced subspaces of X of bounded degree and all of whose components are of dimension r and contained in a fixed compact subset of X. Then F is bounded.
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2025-03-21T14:48:31.071834
| 2020-05-22T19:27:46 |
361093
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Stack Exchange
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Geometrization/sheafification of elementary formulas
Looking in admiration at Deligne's re-definition of Kloosterman sums as traces of Frobenii acting on stalks of certain complexes of sheaves defined via pull-push from $\mathbb{G}_a$ to $\mathbb{G}_m$ via $\mathbb{G}_m^n$, I realized that this kind of geometrization/sheafification of explicit elementary formulas must be going on under the hood of highly abstract-seeming notions all the time. That's after all how simply stated longstanding classical problems are often solved as well as generalized by current mathematical machinery, but I found the Kloosterman reformulation particularly direct and neat.
I'd love to know other beautiful (in the eyes of the provider, of course) instances of such geometric reformulations of elementary formulas so that one can begin to discern the art of doing so. I'm primarily interested in number theory, but examples from other areas are welcome too.
P.S.1 Can the Kloosterman reformulation somehow be regarded a case of Serre's faisceaux-fonctions dictionary?
P.S.2 I don't think of coding of (non-existent) solutions in Fermat's last theorem as elliptic curves as a good example because while extremely clever, and obviously fruitful, it doesn't seem to me intrinsic or natural (hard-to-define characteristics I admit.) For the sake of delimiting the scope of the question too I'd like to keep it to geometric (or even just sheafy) re-interpretation of simple explicit formulas.
I must disagree (courteously) with your PS2. Both the encoding of Fermat's last theorem into an elliptic curve, and the rncoding of the Mordell conjecture into the Shafarevich conjecture (which is what Faltings used) are examples of encoding the solutions to a Diophantine equation as points on a moduli space, and then using the moduli structure as a fundamental tool in analyzing the solutions to the original problem. Of course, this is somewhat oversimplified, but helps to explain why the proof of FLT through elliptic curves is not simply an isolated trick.
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2025-03-21T14:48:31.071999
| 2020-05-22T20:19:59 |
361098
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A morphism of monads that doesn't preserve thunkability?
Recall that for a monad $(T,\eta,\mu)$ on a category $C$, the Kleisli category $C_T$ has as objects the objects of $C$ and as morphisms $C_T(x,y) = C(x,T y)$. A morphism $f\in C_T(x,y) = C(x,T y)$ is said to be thunkable if $T\eta \circ f = T f \circ \eta$ (in $C$), or equivalently $T \eta \circ f = \eta T \circ f$. (The name comes from viewing Kleisli categories as semantics for call-by-value programming languages, see e.g. the references here.)
Every morphism in the image of the free functor $F:C\to C_T$ is thunkable, and the converse holds if $\eta$ is the equalizer of $\eta T$ and $T \eta$. I am told that every nontrivial monad on $\mathrm{Set}$ satisfies this condition.
If $T_1\to T_2$ is a morphism of monads on $C$, there is an induced functor $C_{T_1} \to C_{T_2}$. My question is: is there a morphism of monads such that this induced functor does not preserve thunkability? The above remark shows that if so, $C$ cannot be $\mathrm{Set}$.
Is it trivial to show that every nontrivial monad on Set satisfies the condition?
@IvanDiLiberti I don't know, I haven't tried.
(BTW there are two "trivial" monads on Set: one is constant at 1, the other is constant at 1 everywhere except it sends 0 to 0.)
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2025-03-21T14:48:31.072126
| 2020-05-22T20:32:01 |
361101
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Stack Exchange
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Laundering PDF files
Is there an easy way to "launder" a PDF file so that it won't appear to have been generated from LaTeX?
(I have a good reason for wanting to do this: I just tried to post an article to the arXiv, but the arXiv software isn't processing my latex source correctly, so I have to circumvent the usual way the arXiv creates a PDF from a source file. I plan to give the arXiv the source file too; I just don't want to get into a long argument with any humans about what I'm doing and why.)
I know this is more of a stackexchange sort of question, but I couldn't find it addressed there, and it strikes me as the sort of issue that mathematicians may face more often than others.
Maybe.https://tex.stackexchange.com/questions/186068/how-to-upload-latex-generated-pdf-paper-to-arxiv-without-latex-sources has some ways, but it seems like arxiv keeps getting better at rejecting such submissions
don't, it's cheating the system, which if you choose to use arXiv is not a best practice.
One solution that will always work is to physically print it and scan it back as a pdf. Alternatively, you can probably open it in Adobe or Preview (on a mac) and then use the print feature to "save as pdf." I did this latter solution once successfully to solve the exact sort of problem you are asking. I can't remember if it was for arxiv or a grant application, but I know it wasn't rendering correctly in their system, but did render correctly on mine, so I just gave them the "laundered" pdf because time was of the essence.
if you print and scan, find/search/select will no longer work...
You could scan with OCR, but that technology is still not great, especially for math papers. In my experience, the Tesseract OCR performs best. Indeed, simply performing OCR on a pdf generated from latex might also launder it. https://github.com/tesseract-ocr/
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2025-03-21T14:48:31.072295
| 2020-05-22T21:01:15 |
361103
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Stack Exchange
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Books on foliations
I am looking for resources (books, notes, lecture video, etc. anything will do although printed material in English is preferable) on foliations which satisfy some or all of the following constraints.
Prerequisites: I am familiar with algebraic topology (in the geometric style, as in Hatcher), differential topology (as in Guillemin-Pollack), and Riemannian geometry (as in do Carmo) along with all other standard undergraduate topics (by which I mean content covered in standard textbooks in real and complex analysis, linear and basic algebra, commutative algebra, classical algebraic geometry, point set topology, curves and surfaces). I am also familiar with characteristic classes (Morita's differential forms book and Madsen-Tornehave), basic symplectic geometry (da Silva), and basic topological/measure theoretic dynamics (earlier chapters of Brin-Stuck).
I am also familiar with physics (general relativity using Caroll, classical mechanics using Goldstein, etc.) if it helps.
Ideally, I am looking for two types of resources. (Notice that the two are not mutually exclusive.)
An exposition of the theory which has a strong geometric taste (much like Hatcher's books) ideally with a lot of pictures and concrete examples. Ideally, the book connects new ideas introduced in the book with older ideas (described in "prerequisites" above).
Collection of problems which allows one to practice applying the theory. I prefer exercises which are not just filling in technical details which the author did not have time for. Instead, I prefer something which allows one to a.) learn key heuristics, and ideally b.) get a sense on why the theory will be important later in one's studies.
So far, I have the following books:
Tamura, Topology of Foliations: An Introduction
Calegari, Foliations and the Geometry of 3–Manifolds
I see in Wikipedia these references (neither of which I know; just references to me): Candel, Alberto; Conlon, Lawrence (2000). Foliations I. Graduate Studies in Mathematics. 23. Providence, Rhode Island: American Mathematical Society. ISBN 0-8218-0809-5.
Candel, Alberto; Conlon, Lawrence (2003). Foliations II. Graduate Studies in Mathematics. 60. Providence, Rhode Island: American Mathematical Society. ISBN 0-8218-0809-5.
There are the two volumes of G. Hector and U. Hirsch, Introduction to the geometry of foliations. (Vieweg, Braunschweig, 1981) which are fairly geometric and explicit. In the first volume, they discuss the classification of foliations on surfaces in some detail.
Geometric Theory of Foliations by César Camacho and Alcides Lins Neto in Portuguese, or in English thanks to Sue Goodman's fantastic translation.
I think it does almost everything you're asking for in terms of pictures, examples, and lovely exercises beyond just letting readers fill in details, though some theorems do leave proofs to the reader.
As is pointed out in the comments the two books by Candel and Conlon and your inclusion in your post of Calegari's book are amazing, but for rather different reasons.
Candel and Conlon could be said to provide the opposite of a geometric experience, in the sense that it provides an opportunity to learn the theory by manipulating the symbols of differential geometry's formalism. The degree of precision and generality is probably unrivaled.
Calegari paints a quite different picture, giving a sophisticated and contextualized presentation. His choice of material to collect in that book seems to have been prescient, in that the book was completed 5 years prior to the publication of, what is now known as, the L-space conjecture in ~2012. The content in Calegari's book is a fantastic preparation for understanding the state of the art techniques in low dimensional foliation theory to attack the L-space conjecture. Worth mentioning is that a proof of the L-space conjecture would provide us with a new, Ricci free, proof of the Poincaré conjecture.
Could you elaborate on your last sentence? Surely it’s probable that any solution to the L-space conjecture will rely on the Poincaré conjecture, or even Geometrisation?
The proof in the graph manifold case does not depend on Poincaré or Geometrisation. To be too brief with Boyer and Clay's work, they use a gluing result for horizontal foliations in Seifert fibered spaces, and when that doesn't quite work, it's a gluing result for contact approximations of foliations. The full graph manifold case is a combination of Boyer-Clay and Sarah Rasmussen. András Juhász writes about a route to the Poincaré through the L-space conjecture in https://arxiv.org/abs/1310.3418. Which to be honest does depend on an additional conjecture, but sort of only open in hyperbolic.
Graph manifolds are geometrisable by definition! Juhasz’s paper looks interesting — thanks!
Haha yes of course they are geometrisable. I see now what you mean, if someone deals with hyperbolic manifolds and then follows the same thread through a gluing result using geometrization tori then yea, it seems eminently reasonable it will depend on Perelman. The Ricci free stuff I was referring to is in the direction of Thurston's universal circles. Thurston's slithering preprints and following work to prove geometrisation by taut foliations. The state of the art efforts I mentioned being lifting universal circle representations to actions on $\mathbb{R}$ to certify left orderability.
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2025-03-21T14:48:31.072697
| 2020-05-22T23:01:51 |
361112
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Stack Exchange
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When is the module of Kahler differentials free?
As the title says, when is the module of Kahler differentials a free module? In particular, are there known conditions or criterions that could be met that ensures that it will be free?
For example, if one has a finitely generated algebra $S=k[x_0,\cdots,x_n]/(f_1,\cdots, f_l)$ over a field $k$, then one could require that the generators induced from the $f_i$ for $\Omega_{S/k}$ be linearly independent. However, this is a very naive approach. I was curious if there something more interesting. For example, if the ring $S=k[x_0,\cdots,x_n]/(f_1,\cdots, f_l)$ (where $l<n$) has the property that the determinant of the matrix $(\frac{\partial f_i}{\partial x_j})_{i,j=1}^l$ is a unit of $S$. I am not entirely sure if that is accurate on the top of my head, but something along those lines.
Another question is, when is the module of differentials reflexive?
Freeness of $\Omega_{S/k}$ is very closely related to smoothness of $S$ as a $k$-algebra (one has to assume also that it has the correct rank, to avoid issues with characteristic $p$); this should be in most commutative algebra resources. If $S$ is not smooth, then it seems a bit subtle exactly when $\Omega_{S/k}$ is reflexive, but in general one expects to have both torsion and cotorsion (see, e.g., https://arxiv.org/pdf/1012.5940.pdf for the failure of reflexivity for some fairly mild singularities).
For $S$ as above, $\Omega^1_X$ is free of rank $=\dim X$ (X=\operatorname{Spec} S$ implies $S=k[x_0,\ldots,x_n, x_{n+1},\ldots x_m]/(g_1,\ldots, g_{l+m})$ where the $g_i$ form a regular sequence, that is $S$ is a complete intersection possibly after adding more variables.
@Mohan Could you explain this a bit further please? Perhaps add some details? This would answer a huge problem I have been facing with these modules!
@DevlinMallory Ah yes, I see what you mean that these modules are closely related to smoothness. I mean this is the essence of smooth varities and when the sheaf of differentials is finite locally free, there is a lot of interaction. Thanks for the reference, that covers part of the question nicely!
To give one more precise reference (for $\Omega ^1_{X/k}$ free of rank $\dim X$ $\Longleftrightarrow \ X$ smooth over $k$): Kunz Kähler differentials, Theorem 8.1.
If $R$ is an $S$-algebra and $\Omega_{R/S}$ is projective of rank $n$, then it is free of rank $n$ if, and only if, it can be generated by $n$ elements. A reference is pg 556 of "Lombardi, Henri, and Claude Quitté. "Commutative algebra: Constructive methods." Traduction anglaise, révisée et augmentée, de l’édition française (Algebre commutative. Méthodes constructives. Calvage et Mounet, 2011). Springer, Berlin 1.2 (2015): 3."
This is just expanding my comment above (which I messed up forgetting dollar signs).
For simplicity, let me assume that $X\subset\mathbb{A}^n$ be a $d$ dimensional smooth variety with $\Omega^1_X$ free of rank $d$ (in characteristic zero, like your situation, it always is smooth, but in positive characteristic, you need to assume smoothness). Then, for a sufficiently large $m$, embed $\mathbb{A}^n\subset \mathbb{A}^{n+m}$ as a linear subspace and then $X\subset\mathbb{A}^{n+m}$ is a complete intersection. Here is a sketch of the proof.
Let $I$ define $X\subset\mathbb{A}^n$. Then one has the Euler sequence,
$$0\to I/I^2\to \Omega^1_{\mathbb{A}^n|X}\to\Omega^1_X\to 0.$$
Thus $I/I^2$ is stably free. So, if we emebed $X\subset\mathbb{A}^{n+r}$, for large $r$, and call $I$ as the defining ideal of $X$ in this larger space, one gets $I/I^2$ to be stably free and large rank. A stably free module of sufficiently large rank is free by Bass's theorem. So, we may assume that $I/I^2$ is free (of rank, the codimension of $X$).
Now adding one more variable, say $y$, one can check that $I+(y)$ is in fact generated by the correct number of elements. For this, first pick a set of elements $f_1,\ldots, f_s\in I$ which generate $I/I^2$. Then by Nakayama, it is easy to see that there exists an element $h\in I$ such that $h(1-h)\in (f_1,\ldots, f_s)$ and $I=(f_1,\ldots, f_s,h)$. Then $I+(y)=(f_1,\ldots, f_s,h+y(1-h))$, proving what you want.
Another interesting source for this could actually be found within the following paper: https://www.jstor.org/stable/2042872?seq=1 "The Conormal Module of an Almost Complete Intersection by Ernst Kunz"
|
2025-03-21T14:48:31.073007
| 2020-05-22T23:46:16 |
361116
|
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|
Stack Exchange
|
Cobordism theory of some weird space
Let $G=SU(3)$ and $N=SO(3)$, then $G/N= SU(3)/SO(3)$ = a 5-dimensional Wu manifold $W$.
The $W$ is a homogeneous space (also a quotient space), but not a group.
Previously, I am aware of the bordism group of $BG$ where $BG$ is the classifying space of a group $G$. The $d$-th bordism group is $\Omega_d^H (BG)$.
Now let us consider the quotient of their classifying spaces $BSU(3)/BSO(3)$ (either the point set quotient or the mapping cone).
I am not aware of the bordism group of
$$\Omega_d^H(BSU(3)/BSO(3))$$
of this space $BSU(3)/BSO(3)$. Are there previous knowledge of computing the bordism group of this type?
Here $H$ can be some structures equipped for the cobordant of manifolds.
As far as I know, nobody has computed these bordism groups. I would guess the standard spectral-sequence approaches that you already know probably work, but I'm not certain.
There's also the long exact sequence for a cofibration...does that not tell you anything useful?
|
2025-03-21T14:48:31.073108
| 2020-05-23T00:18:43 |
361118
|
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|
Stack Exchange
|
Are there analogues of real-valued measurability for larger powersets?
Apologies if the question is somewhat naïve - I'm not a set theorist, just an enthusiast.
One possible intuition we could have about the generalized continuum hypothesis is that power sets are so difficult to well-order that they must be very large on the ordinal scale (so, at the very least, they should be weakly Mahlo). Questions about this intuition have appeared on mathoverflow before, e.g. here, here, and here.
For the continuum, we have the nice result that if the cardinality of the continuum is real-valued measurable, then it is automatically greatly Mahlo, and there is the pleasing result due to Prikry that in this case for all $\aleph_0 \le \kappa < 2^{\aleph_0}$, we have $2^{\kappa} = 2^{\aleph_0}$ (I found these in sections 4 and 5 of Fremlin's Real-valued-measurable cardinals). There is the additional advantage that real-valued measurability is a concept which is easy to explain to a non-specialist, such as myself.
However, these results about real-valued measurability are very specific to the continuum: no larger powerset can be real-valued measurable because measurable cardinals are necessarily strongly inaccessible.
I want to know if there is some analogue of this for larger powersets, such as $\kappa = 2^{2^{\aleph_0}}$. A first guess might be to ask for something like a $\kappa$-additive "surreal-valued" measure, but I don't know of any way to make this idea coherent.
In an attempt to be more specific, what I wish for is some property $X$ such that:
The assumption "Every infinite powerset has property $X$" isn't obviously inconsistent with ZFC,
There is some convincing analogy between property $X$ and real-valued measurability,
Any cardinal $\kappa$ with property $X$ is automatically weakly Mahlo, and
If $2^\kappa$ has property $X$ and $\kappa \le \lambda < 2^\kappa$, then $2^\lambda = 2^\kappa$ (or some variation of this).
Edit: After sleeping on this, I noticed that the first bullet point may be too strong, since it contradicts the Singular Cardinal Hypothesis - perhaps it should be weakened to "For every regular $\kappa$, $2^\kappa$ has property $X$". Also, the second bullet point could be made slightly less vague, by asking for the existence of a cardinal with property $X$ to have equivalent consistency strength to the existence of a measurable cardinal.
Do you want something directly analogous to a measure? Some kind of field-valued additive function on sets?
That would be nice, but I'm willing to settle for a less direct analogy: perhaps an abstraction of the ideal of null sets of such a measure, if the existence of such a thing is sufficiently strong.
|
2025-03-21T14:48:31.073306
| 2020-05-23T03:50:30 |
361126
|
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|
Stack Exchange
|
Lower bounds in the space of compact operators
Let $H$ be a separable Hilbert space, and $K(H)$ the corresponding space of compact operators. Consider the "unit sphere" $S:=\{T\in K(H)|T\geq 0\text{ and }||T||=1\}$. Is it true that, given any pair of operators $T_1,T_2\in S$, there exists another operator $T\in S$ such that $T\leq T_1,T_2$?.
Welcome to MathOverflow! What if $T_1$ and $T_2$ are orthogonal projections with product $0$?
Jochen's counterexample works of course for $\dim H =2$ ...
|
2025-03-21T14:48:31.073379
| 2020-05-23T07:15:37 |
361133
|
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|
Stack Exchange
|
Conjectured primality test for specific class of $N=k \cdot 6^n+1$
Can you provide a proof or a counterexample for the claim given below?
Inspired by Theorem 5 in this paper I have formulated the following claim:
Let $N=k \cdot 6^n+1$ , $k<6^n$ and $\operatorname{gcd}(k,6)=1$. Assume that $a \in \mathbb{Z}$ is a 6-th power non-residue . Let $\Phi_n(x)$ be the n-th cyclotomic polynomial, then:
$$N \text{ is a prime iff } \Phi_2\left(a^{\frac{N-1}{2}}\right)\cdot \Phi_3\left(a^{\frac{N-1}{3}}\right) \equiv 0 \pmod{N} $$
You can run this test here. I have tested this claim for many random values of $k$ and $n$ and there were no counterexamples .
Test implementation in PARI/GP without directly computing cyclotomic polynomials.
EDIT
More generally we can formulate the following claim:
Let $N=k \cdot (p \cdot q)^n+1$ , where $p$ and $q$ are distinct prime numbers, $k<(p \cdot q)^n$ and $\operatorname{gcd}(k,p\cdot q)=1$. Assume that $a \in \mathbb{Z}$ is a $p \cdot q$-th power non-residue . Let $\Phi_n(x)$ be the n-th cyclotomic polynomial, then:
$$N \text{ is a prime iff } \Phi_p\left(a^{\frac{N-1}{p}}\right)\cdot \Phi_q\left(a^{\frac{N-1}{q}}\right) \equiv 0 \pmod{N} $$
You can run this test here.
Test implementation in PARI/GP without directly computing cyclotomic polynomials.
EDIT 2
It seems that this claim can be generalized even further:
Let $N=k \cdot b^n+1$ , $k<b^n$ and $\operatorname{gcd}(k,b)=1$. Let $p_1,p_2,\ldots,p_n$ be a distinct prime factors of $b$. Assume that $a \in \mathbb{Z}$ is a $p_1\cdot p_2\cdot \ldots \cdot p_n$-th power non-residue . Let $\Phi_n(x)$ be the n-th cyclotomic polynomial, then:
$$N \text{ is a prime iff } \Phi_{p_1}\left(a^{\frac{N-1}{p_1}}\right)\cdot \Phi_{p_2}\left(a^{\frac{N-1}{p_2}}\right)\cdot \ldots \cdot \Phi_{p_n}\left(a^{\frac{N-1}{p_n}}\right) \equiv 0 \pmod{N} $$
In one direction (wnen $N$ is prime) the statement is trivial. In the reverse direction, it's false however.
Here is just one counterexample: $n=4$, $k=133$, and $a=11$ with $N=172369=97\cdot 1777$, where we already have
$$\Phi_2(11^{\frac{172369-1}2})\equiv 0\pmod{172369}.$$
My implementation of the test gives correct result. See here
If we use $a=2$ test returns "Composite" for this combination of $k$ and $n$. On the other hand $11$ is 6-th power non-residue so you are right. Maybe, I should add some additional constraints to the claim. Thank you for investigation.
|
2025-03-21T14:48:31.073531
| 2020-05-23T07:18:18 |
361134
|
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|
Stack Exchange
|
Roadmap for Quillen 1
Question
Suppose you grasped and enjoyed reading Quillen's "Higher Algebraic K-theory I". Now, if you could go back in time to when you started studying algebraic topology and create a reading list / roadmap with the above paper as a goal, what would this plan look like?
Here's another version of the question: suppose you're the PhD advisor of a student that's recently finished their undergrad degree and in terms of books background has only read and completed most of the exercises in:
algebraic geometry: all of Q. Liu's book
algebraic topology: first two chapters of W. Massey's intro book
differential geometry: first half of J. Lee's smooth manifolds book
Which books and papers (and in what order) should this student master in order to understand (or at least appreciate) most of Quillen 1?
Outlook
The above questions are likely quite vague wrt to "how do I learn modern algebraic k theory?", but hopefully they're somewhat concrete by stating i) the goal [Quillen 1] and ii) the starting maths background.
If it helps, assume a secondary goal is to eventually focus on studying/appreciating arithmetic problems like Parshin's conjecture.
What do you mean with "has never studied algebraic topology proper"? Has he seen the fundamental group? Homology groups (even deRham)? Some differential topology?
@DenisNardin: great question -- I've attempted to clarify some of the assumed book-level background for the hypothetical new grad student
It's not quite what you're asking but Weibel's K-book is quite nice. The first several chapters are from a more historical perspective so you go through elementary matrices, the whitehead group, K_0, K_1, K_2, Milnor K-theory, and then you start going through various constructions of the algebraic K-theory spectrum.
On a minimal level, I'd say that reading Kan's On c.s.s. complexes is enough. Of course the real answer is that to learn K-theory it would be a very good idea to learn (the basis of) modern homotopy theory first.
For someone who is comfortable with Algebraic Geometry (on the level of Liu's book) but less comfortable with Topology, I'd recommend simply using Srinivas book on Algebraic K-theory.
It's a textbook and it first introduces K-theory "axiomatically", then presents a number of applications in algebraic geometry, and only then starts with the proofs which really need some background in topology. Also, the book pretty much covers a great deal in Quillen I, so if the student works through that book, it could act as a self-contained replacement of (most of) Quillen I.
So, if the student reads Srinivas and Quillen I simultaneously, that should make it waaaay easier.
As an aside: I don't think any Differential Geometry/Differential Toplogy is needed for any of this whatsoever. I think when people use Serre-Swan stuff or Bott periodicity to motivate why one might find Algebraic K-theory cool, they usually do this to address an audience that likes these things already. So for somebody who doesn't know these things in the first place, it's not needed to teach them that first. Rather, try to get them curious by relying on their background. For an Algebraic Geometry student, this might be a question like: How does the localization sequence for Chow groups (or Pic, for somebody on the level of Liu) continue to the left? (leading ultimately to Bloch's higher Chow groups and all that stuff)
|
2025-03-21T14:48:31.073904
| 2020-05-23T07:57:08 |
361135
|
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|
Stack Exchange
|
Cayley graphs do not have isolated maximal cliques
Let a Cayley graph $G$ of a group $H$ with respect to the generating set $\{s_i\}$ have a clique of order $> 2$. In addition assume the graph $G$ is non-complete. If the clique size is less than half the order of $G$, then is it possible for some group $H$ that $G$ has a unique "disjoint maximal clique". By "disjoint maximal clique", I mean a clique equal to the clique size of the graph, and such that any other clique of same order would not be vertex disjoint with the prior clique.
I don't think so. For, if $(e),(s_1),(s_1\cdot s_2),(s_1\cdot s_2\cdot s_3),\ldots,(s_1\cdot s_2\cdots s_n)$ be the sequence of vertices in a maximal clique, then I think even $(s_1^2),(s_1^3),(s_1^2\cdot s_2),\ldots,(s_1^2\cdot s_2\cdots s_n)$ would also be a sequence of vertices in a maximal clique, where $e$ denotes the identity element. But, what if $s_1$ is an order $2$ or $3$ element. How do we ensure that there always exist a disjoint clique apart from the clique $(e),(s_1),(s_1\cdot s_2),(s_1\cdot s_2\cdot s_3),\ldots,(s_1\cdot s_2\cdots s_n)$? Will this be true at least for the case when $H$ is an abelian/cyclic group? Any hints? Thanks beforehand.
Your question is not clear. "Disjoint" is a property of two things, not a property of one thing.
@BrendanMcKay edited the post.
Your terminology looks confusing. "Unique maximal clique" usually means that it is a maximal clique and there is no other maximal clique.
@FedorPetrov no, my meaning is that there is a maximal clique and no other "vertex disjoint" maximal clique
Rephrased: "Do all maximal cliques in $G$ pairwise intersect"? (Also I don't see why you need to assume $G$ is non-complete: then your condition is trivially satisfied.)
@SamHopkins yes, that is my question.( and yes, complete graphs trivially satisfy the condition and hence would not be a problem!-the problem starts when clique number of $G$ is less than half the order of $G$)
Your definition still seems a bit strange. A disjoint maximal clique is one that is NOT disjoint from any other clique? Is that really what you mean?
@verret yes, my meaning is that a disjoint maximal clique is one that is NOT disjoint from any other MAXIMAL clique.
@verret I think this is related to your paper here
Let $G$ be the linegraph of the complete graph $K_n$ for $n\geq 5$. For some but not all $n$, $G$ is a Cayley graph, see Chris Godsil's answer to another question.
$G$ has $\binom n2$ vertices and degree $2n-4$. The maximum cliques of $G$ correspond to the edges incident with one vertex and so they have size $n-1$. Moreover, the cliques corresponding to two different vertices of $K_n$ have one point in common, namely the edge between those two vertices.
Therefore, $G$ is an example of a Cayley graph for which any two maximum cliques intersect, even though the maximum cliques only have size about the square root of the number of vertices.
I wonder if this example is optimal in some sense.
ADDED: Here is an exposition of Ilya's argument from the comments.
Theorem. If a vertex-transitive graph with $N$ vertices has cliques of size $k$ such that $k^2<N$, then there are two such cliques which are disjoint.
Proof. Take a fixed $k$-clique $C$ and apply a random automorphism $\gamma$. The expected number of elements of $C$ that map to an element of $C$ is $k^2/N$, so $k^2<N$ implies that $C$ must sometimes map to a clique disjoint from itself.
In the case of a Cayley graph of a group $\varGamma$, we can use a random non-identity element of $\varGamma$ to improve the inequality to $k(k-1)<N-1$.
There is a clique size gap of about $\sqrt 2$ between these bounds and the linegraph of a complete graph. So the problem is still missing a complete solution.
It is optimal in some sense: if a Cayley graph contains a clique of size $<\sqrt n$, then it contains its isomorphic copy, from probabilistic reasons
@IlyaBogdanov you mean if a cayley graph has clique size $< \sqrt n$, then it has disjoint cliques of maximal size? Any reference for this?
If a clique size is $k$, there are $n$ copies of one clique (starting from any vertex), and only $k(k-1)+1$ of them intersects the initial one (any vertex of a copy may coincide with any vertex of the origin).
will the property hold for cayley graphs on cyclic or abelian groups at least?
@IlyaBogdanov in saying that there are $n$ cliques of size $k$, you are using the vertex transitivity of the graph, isnt it? So this should hold for vertex transitive graphs also, right?
Yes, I’m using transitivity. Not exactly this argument, but as slight modification works for vertex transitive graphs as well: exactly a proportion of $k/n$ of all copies contains a fixed vertex.
@IlyaBogdanov you mean a vertex transitive graph with clique number $k$ need not have $n$ $k$-cliques?
It needs (with repetitions!), but you may be unable to find $n$ cliques covering the vertices uniformly. Nevertheless, the orbit of one clique covers the vertices uniformly, and I use that fact.
|
2025-03-21T14:48:31.074256
| 2020-05-23T08:25:57 |
361136
|
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|
Stack Exchange
|
Is my ansatz for finding $n$-periodic-points of the exponential-function exhaustive?
The following is about getting help for a proof on existence and indexability of periodic points of the exponential-function, here with base $e:=\exp(1)$.
Update The question is a complete rewriting of the previous formulation of my question which I hope is much better focused and straightforward.
Let us define $f(z):=\exp(z)$ for $z \in \mathbb C$. Iteration may be denoted by $f°^1(z)=f(z)$ and $f°^{h+1}(z)=f°^{h}(f°^1(z))$.
Fixpoints: it is known (for instance W. Bergweiler1, pg 16),
that $f$ has infinitely many fixpoints (=$1$-periodic points) $p_1$,
that all of them are non-real, and
that all of them are repelling.
They may be indexed by the branchindex $k \in \mathbb Z$ used in the Lambert-$W(k;z)$-function like $p_{1:k}$.
Periodic points: it is further known that for all $n$ the sets of $n$-periodic points are as well infinite1(pg.16). Let us denote such a set from now $\mathbb P_n$. So, in generalization of the indexing of $1$-periodic points, one might say, that the set $\mathbb P$ of all $\mathbb P_n , \text{with } n=1 \ldots \infty $ can be indexed by $\mathbb Z^\infty$.
But I think this is not precise enough; I assume, and would like to prove,
(Conj. 1) that actually $\mathbb P_n$ can exactly be indexed by $\mathbb Z^n$ $\qquad \Leftarrow \mathbb {\text{problem to be proved}}$
For my approach arguing for (Conj. 1) I refer to the property, that a fixed- or periodic points, which is repelling for iteration over function $f()$, is attracting for iteration over its inverse, which means is attracting for the iteration over the $\log()$-function. From Bergweiler, pg. 17, I take, that all periodic points are repelling on iteration on $f()$ and thus are all attracting for its inverse.
For convenience of further notation let us define $\log()$ as $g(x):=\log(x)$ and as well the iteration $g°^1(z)=g(z)$ and $g°^{h+1}(z)=g°^{h}(g°^1(z))$.
To make $g()$ a true inverse of $f()$, we'll need the branch index(es) explicite, so let us simply extend the notation
$$g(x,k):=\log(x) + k \cdot C \qquad \text{where } C=2 \pi î$$
This allows to make precisely for some fixed $z$
$$ g(f(z),0)=z $$
but for the reversion of some $z'=z + k\cdot C$ we need
$$ g(f(z'),k)=g(f(z+k\cdot C),k) =g(f(z),k)=z+k \cdot C=z' $$
For adressing periodic points of period-length $n$ we expand the notation further
$$ \begin{array}{} g(z,[k_1])&:= g(z,k_1) \\
g(z,[k_1,k_2])&:= g(g(z,k_1),k_2) \\
g(z,[k_1,k_2,...,k_n])&:= g(...g(g(z,k_1),k_2)...,k_n) \\
&\small \text{where all $k_j \in \mathbb Z$}\\
\end{array}$$
Finally I use $K_n:=[k_1,k_2,...,k_n]$ for the vector of branch-indexes. With this I conjecture now the following:
iterations of each expression $z_{i+1}=g(z_i,K_n)$ are attracting.
we can approximate any periodic point $p_{n,K}$ by simple fixed-point iteration of the previous with some suitable initial value $z_0 \ne 0$ according to $$p_{n,K} = \lim_{i \to \infty} z_{i+1}=g(z_i,K_n)$$ (of course we can increase speed of approximation when Newton-iteration on $g()$ follows).
the iteration for a given $K_n$ is attracting over the whole complex plane except for the initial values $z_0 \in \{0,1,e,e^e,...\}$.
Non uniqueness occurs only for $K=[0]$ (and its non-primitive repetitions $K=[0,0]$,... $K=[0,0,...,0]$) in that the initial value $z_0$ is relevant for to converge towards the $1$-periodic point either in the upper or in the lower half plane.
Main conjecture to be proved: All $n$-periodic points with the exception of the conjugated primary fixed points $p_{1:[0]}$ and $\overline {p_{1:[0]}}$ (which have the same branch index-vector $K=[0]$) are in bijection to the indexes $K_n$ and can be approximated by simple fixed-point iteration over $g(z,K_n)$ (if desired followed by Newton-iteration on $g(z,K_n)$ to speed up convergence).
Remark: I have seen, that with exponential bases different from $e:=exp(1)$ spuriously non-uniquenesses and non-existences of $n$-periodic points occur, which I cannot yet nail down except by giving a couple of heuristic examples. However, large surveys on the exponential with base $e$ -as discussed here- seem to have only that one exception as mentioned in (3.).
An illustration of periodic points of periods $n=1..5$ . Those were found by screening the square $-4-4î...4+4î$ on the complex plane in steps by $1/40$ with the newton-iteration applied. The list has then been checked whether they all agree with the $K_n$-indexing scheme; all found periodic points have a valid $K$-index.
A long & wide discussion (using other bases than $e$, and using another ansatz for partial solutions) can be found at MSE
A handful of used literature: I've found some resources on fixed points and their properties for the exponential function base $e$, but less about $n$-periodic points. The most fruitful so far was the habilitation of Walter Bergweiler, 1991. If there are more comprehensive texts (optimally online available), please leave a comment.
1Bergweiler, Walter, Periodische Punkte bei der Iteration ganzer Funktionen, Aachen: Rheinisch-Westfälische Techn. Hochsch., Math.-Naturwiss. Fak., Habil.-Schr. 51 S. (1991). ZBL0728.30021.
Pg.16:
"Dieses Ergebnis wurde im Jahre 1948 duch Rosenbloom verallgemeinert, der zeigte, daß für jedes $n \gt 2$ unendlich viele periodische Punkte der Periode $n$ existieren"
"Baker im Jahre 1960 (...) bewies (...), daß höchstens eine (von $f$ abhängige) natürliche Zahl $n$ existiert mit der Eigenschaft, daß $f$ nur endlich viele periodische Punkte der primitive Periode $n$ hat."
Pg. 17:
"Satz 2: Es sei $f$ eine ganze transendente Funktion und es sei $n \ge 2$. Dann hat $f$ unendlich viele abstoßende periodische Punkte der primitiven Periode $n$."
"Wir bemerken noch, daß ganze Funktionen keine anziehenden periodischen Punkte zu haben brauchen. Ein Beispiel, (...) ist durch $f(z)=e^z$ gegeben."
Additional readings:
Hellmuth Kneser, [Real analytic solutions of the equation $φ(φ(x))=e^x$ and related functional equations. (Reelle analytische Lösungen der Gleichung $φ(φ(x))=e^x$ und verwandter Funktionalgleichungen.)], J. Reine Angew. Math. 187, 56-67 (1949)(German).Zbl0035.04801.4
Shen, Zhaiming; Rempe-Gillen, Lasse, The exponential map is chaotic: an invitation to transcendental dynamics, Am. Math. Mon. 122, No. 10, 919-940 (2015). ZBL1361.37002.5
Here general aspects of the set of $n$-periodic points are presented in existence-theorems. Even the concept of infinite non-periodic, but not diverging-to-infinity, orbits -as part of the general chaotic behaviour- is covered by the list of theorems.(G.H.)
An introductory article which deals with the question of $\mathbb P_1$ (fixed-) points on the branches of the $\log()$-function by Stanislav Sykora (2016) at his web-space.
P.s. I ask here after I couldn't get much resonance in my questions in MSE here https://math.stackexchange.com/questions/3674391 and here
https://math.stackexchange.com/questions/3677280 . With the question here at MO I tried to put more clarity (and also I focused it to the problem of exhaustivity). I hope this problem is not felt inappropriate for the MO...
|
2025-03-21T14:48:31.074966
| 2020-05-23T08:36:40 |
361137
|
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|
Stack Exchange
|
What is the analog of the symmetrized Jacobi matrix for delay equations?
For a linear system of ODEs in $\mathbb{R}^{n}$ (with the usual inner product), say $\dot{V}(t) = A(t) V(t)$, we know that if $\xi_{1},\ldots,\xi_{k} \in \mathbb{R}^{n}$ and $V_{j}(t)=V_{j}(t,\xi_{j})$ are the corresponding solutions such that $V_{j}(0)=\xi_{j}$, the evolution of the $k$-dimensional volume $|V_{1}(t) \wedge\ldots \wedge V_{k}(t)|_{\bigwedge^{k}\mathbb{R}^{n}}$ satisfies the Liouville trace formula
$$|V_{1}(t) \wedge\ldots \wedge V_{k}(t)|_{\bigwedge^{k}\mathbb{R}^{n}} = |\xi_{1} \wedge \ldots \wedge \xi_{k}|_{\bigwedge^{k}\mathbb{R}^{n}} \operatorname{exp}\left( \int\limits_{0}^{t} \operatorname{Tr} (A(s) \circ \Pi(s)) ds \right),$$
where $\Pi(s)=\Pi(s,\xi_{1},\ldots,\xi_{k})$ is the orthogonal projector onto the space $L(s)$ spanned by $V_{1}(s),\ldots,V_{k}(s)$. To provide effective estimates for the volume decay one usually considers the orthogonal basis $e_{1}(s),\ldots,e_{n}(s)$ of $\mathbb{R}^{n}$ such that $e_{1}(s),\ldots,e_{k}(s)$ span $L(s)$ and proceeds as
$$\operatorname{Tr}(A(s) \circ \Pi(s)) = \sum\limits_{j=1}^{k}(A(s)e_{j},e_{j}) = \sum\limits_{j=1}^{k}\left(\frac{1}{2}(A(s)+A(s)^{*})e_{j},e_{j}\right) \leq \alpha_{1}(s) + \ldots + \alpha_{k}(s),$$
where $\alpha_{i}(s)$ is the $i$-th eigenvalue of $(A(s)+A(s)^{*})/2$ such that $\alpha_{1}(s) \geq \alpha_{2}(s) \geq \ldots \geq \alpha_{n}(s)$. If the linear system comes from the linearization along a trajectory of some nonlinear system, the matrix $(A(s)+A(s)^{*})/2$ is called the symmetrized Jacobi matrix of $A(s)$.
I am interested in such an analog and corresponding estimates for delay equations. Below I explain my attempts and several problems arising on this way.
Let us consider the scalar equation with delay: $\dot{x} = x(t) - \alpha x(t-\tau)$. To make it possible to talk about volumes we consider this equation in the Hilbert space $\mathbb{H} = \mathbb{R} \times L_{2}(-\tau,0;\mathbb{R})$ with the usual inner product. Define the operator $A \colon \mathcal{D}(A) \subset \mathbb{H} \to \mathbb{H}$ as
$$(x,\phi) \overset{A}{\mapsto} \left(\phi(0) - \alpha \phi(-\tau), \frac{\partial}{\partial\theta} \phi\right)$$
for $(x,\phi) \in \mathcal{D}(A) := \{ (x,\phi) \in \mathbb{H} \ | \ \phi \in W^{1,2}(-\tau,0;\mathbb{R}), \phi(0)=x \}$. The evolution equation in $\mathbb{H}$
$$\dot{V}(t)=AV(t)$$
is well-posed and for $\xi \in \mathcal{D}(A)$ we have classical solutions $V(t)=V(t,\xi)$ (see, for example, Bátkai A., and Piazzera S. Semigroups for Delay Equations). It is not hard to prove that for $\xi_{1},\ldots,\xi_{k} \in \mathcal{D}(A)$ we have an analog of the Liouville trace formula
$$|V_{1}(t) \wedge\ldots \wedge V_{k}(t)|_{\bigwedge^{k}\mathbb{H}} = |\xi_{1} \wedge \ldots \wedge \xi_{k}|_{\bigwedge^{k}\mathbb{H}} \operatorname{exp}\left( \int\limits_{0}^{t} \operatorname{Tr} (A \circ \Pi(s)) ds \right)$$
The spectrum of $A$ is determined by the roots of
$$1-\alpha e^{-\tau \lambda} - \lambda = 0$$
If $\alpha \in (0,1)$ there are two real roots $\lambda_{1} > 0$ and $\lambda_{2} < 0$ and the others are located to the left from $\lambda_{2}$. If $\lambda_{1} + \lambda_{2} < 0$ then using the dichotomy we have the exponential decay of $2$-volumes. How this fact can be obtained from the trace formula? It is not hard to see that the adjoint of $A$ is given by the formula
$$ (y,\psi) \overset{A^{*}}{\mapsto} \left(y + \psi(0), -\frac{\partial }{\partial \theta} \psi\right),$$
where $(y,\psi) \in \mathcal{D}(A^{*}) = \{ (y,\psi) \in \mathbb{H} \ | \ \psi \in W^{1,2}(-\tau,0), \psi(-\tau) = \alpha y \}$. We see that for $(x,\phi) \in \mathcal{D}(A) \cap \mathcal{D}(A^{*})$ we have
$$A+A^{*} = (\phi(0)-\alpha\phi(-\tau) + x + \phi(0),0) = (x-\alpha^{2}x + x + x,0).$$
Thus $A+A^{*}$ is a bounded self-adjoint operator in $\mathbb{H}$ with the kernel of codimension one. Clearly, its spectrum has nothing to do with the volumes decay. This is not surprising since to proceed in the second equality from
$$\operatorname{Tr}(A \circ \Pi(s)) = \sum\limits_{j=1}^{k}(Ae_{j},e_{j}) = \sum\limits_{j=1}^{k}\left(\frac{1}{2}(A+A^{*})e_{j},e_{j}\right)$$
we must have $e_{j} \in \mathcal{D}(A) \cap \mathcal{D}(A^{*})$, but we usually have only $e_{j} \in \mathcal{D}(A)$. Moreover, if $e_{j}(s) = (x(s),\phi(s))$ then we have
$$\sum\limits_{j=1}^{k}(Ae_{j},e_{j}) = \sum\limits_{j=1}^{k}\left(\frac{3}{2} |\phi_{j}(0)|^{2} - \alpha \phi_{j}(0) \phi_{j}(-\tau) - \frac{1}{2} |\phi_{j}(-\tau)|^{2}\right)$$
It is not obvious how one may obtain the exponential decay from this expression.
So what's the right approach for the symmetrization of $A$ and the corresponding trace estimates in the case of delay equations?
This isn't my area, so I do apologise if I'm speaking out of turn, but... the construction you introduce To make it possible to talk about volumes doesn't seem at all to me like a natural step from what you're talking about at the start of your question. Is your main interest in evolution of volumes in $\mathbb{R}^d$ or are you more interested in abstract analogs of the Liouville trace formula?
@DCM, delay equations generate infinite-dimensional dynamical systems. There are two approaches for the well-posedness of delay equations and the construction of the corresponding dynamical systems. The first one treats such equations in the space of continuous functions $C([-\tau,0];\mathbb{R})$ and the second one considers the Hilbert space setting. The latter approach allows to talk about volumes and other things concerned with the Hilbert space structure. I'm interested in the evolution of $k$-dimensional volumes in the introduced Hilbert space $\mathbb{H}$.
@DCM, the problem is in that it is easy to obtain the trace formula, but it is not obvious how to derive effective estimates from it (as we could do in the case of ODEs using the eigenvalues of the symmetrized matrix).
|
2025-03-21T14:48:31.075279
| 2020-05-24T23:59:20 |
361272
|
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|
Stack Exchange
|
Characterization of "PSD-Squared" Matrices
$\DeclareMathOperator\DNN{DNN}\DeclareMathOperator\CP{CP}$This question can be thought of as an offshoot of this MO question from a few months ago. Let $M_n(\mathbb{C})$ denote the set of $n \times n$ complex matrices. Whenever I refer to positive semidefinite (PSD) matrices, I assume they are Hermitian. I use "$\odot$" to refer to the entrywise/Hadamard product of matrices, and I will shortly take the (entrywise) complex conjugate of a matrix.
Vague question: What can be said about the set $S$ of matrices
$$
S \mathrel{:=} \{ X \odot \overline{X} : \text{$X \in M_n(\mathbb{C})$ is positive semidefinite}\}?
$$
Some observations: Every member of $S$ is clearly real and entrywise non-negative, and the solution to the MO question linked above shows that every member of $S$ is positive semidefinite. We thus have $S \subseteq \DNN$, where $\DNN$ is the set of doubly nonnegative matrices.
Precise question #1: Does $S = \DNN$?
If the answer to "precise question #1" is "no", several natural follow-up questions arise. For example, the most well-known set of matrices that is contained within $\DNN$ is the set $\CP$ of completely positive matrices. This leads to …
Precise question #2: Does $S = \CP$?
In fact, it's not clear to me that $S \subseteq \CP$ or $\CP \subseteq S$, so either of those inclusions (or refutations of those inclusions) would be interesting as well.
|
2025-03-21T14:48:31.075404
| 2020-05-25T00:26:38 |
361273
|
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|
Stack Exchange
|
Obstruction to the existence of an invariant symplectic connection
Let $M$ be a symplectic manifold with a symplectic action of a Lie algebra $\mathfrak{g}$. I am interested whether there exists a $\mathfrak{g}$-invariant symplectic connection on $M$. Where does the obstruction live and how to construct it?
Let $\nabla$ be any symplectic connection. It is easy to check that the map $\mathfrak{g} \to \Omega^1 (M, \operatorname{End}_{\omega} (TM))$ given by $X \to \mathcal{L}_X (\nabla)$ is a Chevalley-Eilenberg 1-cocycle. Thus, we have an element in $H^1 (\mathfrak{g}, \Omega^1 (M, \operatorname{End}_{\omega} (TM))$. The claim is that this cohomological class is the obstruction to the existence of an invariant symplectic connection.
Suppose that it vanishes. It means that there exists $A \in \Omega^1 (M, \operatorname{End}_{\omega} (TM))$ s.t. for any $X \in \mathfrak{g}$ we have $\mathcal{L}_X (\nabla) = \mathcal{L}_X (A)$, then it is easy to see that $\nabla - A$ is the desirable connection.
Simiraly, if $\nabla$ is any symplectic invariant connection, then $\mathcal{L}_X (\nabla) = 0$ for any $X \in \mathfrak{g}$, then the class is, of course, zero.
|
2025-03-21T14:48:31.075497
| 2020-05-25T02:21:15 |
361275
|
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|
Stack Exchange
|
A question about finitely additive integration
Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space ($\mathbb P$ is countably additive). Let $\{p_\omega: \omega \in \Omega\}$ be a family of (countably additive) probability measures on $(\Omega, \mathcal F)$, and assume that the mapping $\omega \mapsto p_\omega$ is $\mathcal F$-measurable. Let $\mu$ be a finitely additive probability measure on $(\Omega, \mathcal F)$. Suppose the following holds for all $A \in \mathcal F$:
$$\mathbb P(A) = \int p_\omega(A) \mu(d\omega) \tag{1}$$
Can (1) be extended to
$$\int f d\mathbb P = \int\int f(\omega')p_{\omega}(d\omega')\mu(d\omega) \tag{2}$$
for all bounded $\mathcal F$-measurable $f$?
The definition of the finitely additive integral I'm working with is exactly like the usual definition of the Lebesgue integral. In particular, the integral is linear on the space of boudned $\mathcal F$-measurable functions and continuous in the sup-norm. Thus, by linearity, (2) holds for simple functions.
I'm a bit confused about the general case, though. If $f$ is bounded, it can be uniformly approximated by a sequence $f_n$ of $\mathcal F$-simple functions. Then,
$$\int f d\mathbb P = \lim_n \int f_n d\mathbb P = \lim_n \int\int f_n(\omega') p_\omega(d\omega')\mu(d\omega).$$
In order conclude, I think I need that $\int f_n dp_\omega$ converges to $\int f dp_\omega$ uniformly in $\omega$. But I am unable to see that this is the case.
A measurable function $f$ taking values in $[-M,M]$ can be approximated above and below by simple functions: $f_n \le f \le f_n+1/n$ where $f_n(x)=\lfloor nf(x) \rfloor/n$.
Note that $f_n$ takes on less than $3Mn$ values. Now from the case of simple functions it follows that the LHS and RHS of (2) differ by at most $1/n$.
|
2025-03-21T14:48:31.075624
| 2020-05-25T03:09:54 |
361276
|
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|
Stack Exchange
|
Solutions of the differential equation $f'=(f^{-1})^{[n]}$
For clarity, we use the notations $f^{[n]}=f\circ \dots\circ f$ for nesting and $f^{(n)}=\frac d{dx}\left(\frac{d}{dx}\dots\right)f$ for differentiation.
After reading these two posts (here and here) on MO, I couldn't help but wonder what would happen if we were to iterate $f^{-1}$ on that DE. I first started by trying to merge arguments from both pages to see if it could work, to no avail: to my understanding, one has to simply further or adapt the techniques presented on the first page, as the presence of the inverse makes the problem fundamentally different (and more difficult). I haven't managed to go further than the trivial analysis:
Pick an ansatz of the form $ax^b$ and substitute. Equate the powers and the coefficients:
$\begin{cases}
b-1=\frac{1}{b^n} \iff b^{n+1}-b^{n}-1=0 \\
a=b^{\frac{-b^2}{b^2+b+1}}
\end{cases}$
Fortunately, the polynomial in $b$ always has a positive root so there is always this solution. What other solutions are there and how to find them?
I have willingly not specified the domain for $x$, pick whatever suits your answers, or even better, discuss the several possibilities.
One can push this one step forward and explore DEs of the form:
$$F(x,f',...,f^{(n)},f^{-1},(f^{-1})^{[2]},...,(f^{-1})^{[n]})=0$$
as presented similarly by this article.
Any and all ideas are appreciated!
|
2025-03-21T14:48:31.075747
| 2020-05-25T03:10:14 |
361277
|
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"Federico Poloni",
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|
Stack Exchange
|
Product of matrices has real eigenvalues?
Let $A$ be a (symmetric) positive definite matrix and $\hat{n}$ be an arbitrary unit vector. Consider $b,c,d,k$ arbitrary positive integers. I would like to know if the following matrix has real eigenvalues (I would like the more general answer, for more than product of four matrices but even for three I don't know).
$$(I - \hat{n}\hat{n}^T)\cdot A^b\cdot(I - \hat{n}\hat{n}^T)\cdot A^c\cdot (I - \hat{n}\hat{n}^T)\cdot A^d\cdot (I-\hat{n}\hat{n}^T)\cdot A^k\cdot (I-\hat{n}\hat{n}^T).$$
I would like to note that I have asked the same question at mathstackexchange.
It is easy to test 10.000 randomly-generated 20x20 cases, and see if they all have real eigenvalues. Have you tried it? This would give you at least a conjecture.
|
2025-03-21T14:48:31.075821
| 2020-05-25T03:52:36 |
361279
|
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|
Stack Exchange
|
A problem on polynomials
Let $P(z)$ be a polynomial of degree $n$ with $|P(z)|\leq 1$ on $|z|=1$ and $P_m(z)$ be a partial sum of $P(z).$ How large $P_m(z)$ can be on $|z|=1?$
Trivial upper bound is $|1+e^{i \theta}+...+e^{im \theta}|_{L^{1}(\mathbb{T})}$.
What here is fixed and what is allowed to vary?
The trivial upper bound $\max_{|w|=1}|P_{m}(w)|\leq \|1+z+\cdots+z^{m}\|_{L^{1}(\mathbb{T})} \asymp C \log(m)$ that I wrote in the comment is actually sharp in the regime $m=n/2$. Here is the proof.
Notice that $P_{m}(z)$ is convolution of $P(z)$ with $D_{m}(z) = 1+z+\cdots+z^{m}$ on the unit circle, therefore, by the triangle inequality $\max_{|z|=1}|P_{m}(z)| \leq \frac{1}{2\pi}\int_{-\pi}^{\pi}|1+e^{i\theta}+\cdots+e^{i m \theta}| d\theta \asymp C \log(m)$
Next, let us show that the upper bound is sharp in the regime $m=\frac{n}{2}$ where $n$ is large.
Indeed, consider the polynomial
$$
P(z) = z^{n} \overline{\left( 1+\frac{z}{1}+\cdots+\frac{z^{n}}{n}\right)} - z^{n}\left(1+\frac{z}{1}+\cdots+\frac{z^{n}}{n} \right) = z^{n}+\frac{z^{n-1}}{1}+\cdots+\frac{1}{n}-z^{n}-\frac{z^{n+1}}{1}-\cdots-\frac{z^{2n}}{n}
$$
it is of degree $2n$, and $\max_{|z|=1}|P_{n-1}(z)|\geq |P_{n-1}(1)| \asymp \log(n)$. On the other hand let us show that $\max_{|z|=1}|P(z)|\asymp 1$. Indeed, for $z=e^{ix}$ we have
$$
|P(z)| = 2\left| \sum_{k=1}^{n} \frac{\sin(kx)}{k}\right|\leq 2 \int_{0}^{\pi} \frac{\sin(s)}{s}ds \asymp 1 \quad \text{for all} \quad n \geq 1, \; x \in [0, 2\pi)
$$
The last inequality I guess is known "since Democritus".
@user159888 I do not see typos. One small thing is that instead of P_{n-1} I should be writing P_{n}. But both are fine it just gives example of degree 2n with m=n-1
I haven't looked closely, but it feels like this argument might symmetrize to get an upper bound of $O(\log(\min(m,n-m)))$?
Yes, this can be done by writing $P_{m}(z) = P(z)-(P(z)-P_{m}(z))$ the first term is estimated by $1$, and the term $(P(z)-P_{m}(z))$ is upper bounded by $|z^{m+1}+...+z^{n}|_{L^{1}(\mathbb{T})}\asymp \log(n-m)$.
|
2025-03-21T14:48:31.076092
| 2020-05-25T04:24:48 |
361280
|
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|
Stack Exchange
|
The operator equation $AB = \lambda BA$ for self-adjoint operators
Suppose that $A$ and $B$ are self-adjoint bounded linear operators on a Hilbert space and $\lambda \in \mathbb{C}$. It turns out that if $\lambda \notin \{-1, 1\}$ then $AB=\lambda BA \implies AB = BA = 0$.
Does anyone know of any applications of this result?
In the physics context, with $A$ and $B$ creation operators of two identical particles, the fact that only $AB=+BA$ and $AB=-BA$ are nontrivially allowed implies that the particles must be either bosons (even under exchange) or fermions (odd under exchange).
$\require{enclose} \enclose{horizontalstrike}{\style{font-family:inherit;}{\scriptsize\text{Creation operators are not self-adjoint, but still if $AB=\lambda BA$ then $AB=\lambda^2 AB$ hence either $\lambda\in\{-1,1\}$ or $AB=0$.}}}$
For any $q\in (-1,1)$ one can construct operators $A$ and $B$ such that $AB = qBA$, so your last statement is not correct. In the self-adjoint case the condition $AB=\lambda BA$ indeed implies that also $BA = \lambda AB$, hence $AB = \lambda^{2} AB$, but that's not the case in general.
Sooo ... one can't define creation operators for anyons?
local creation operators are fermionic or bosonic, when you exchange the order only the initial and final position need to be specified, so their statistics is governed by the permutation group; for the exchange of anyons the path by which they are exchanged matters, governed by the braid group.
Certainly, but it seems to me that your argument as stated would therefore need some qualification as to what types of creation operators $A$ and $B$ you're considering - roughly speaking, "local", as you say. If the operators involve some Wilson lines connecting them to some reference point, one gets into trouble. Evidently, the OP's observation extends beyond self-adjoint operators, but there are restrictions - it would be interesting to understand those.
|
2025-03-21T14:48:31.076269
| 2020-05-25T05:10:23 |
361282
|
{
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"Ivan Di Liberti",
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|
Stack Exchange
|
Are compact objects in presheaf categories finite colimits of representables?
An object $x$ in a category $\mathsf{C}$ is called compact or finitely presentable if $$\mathrm{hom}(x,-) : \mathsf{C} \to \mathsf{Set}$$ preserves filtered colimits. This concept behaves best when $\mathsf{C}$ has all filtered colimits, e.g. when it is the category of presheaves on some small category $\mathsf{X}$:
$$ \mathsf{C} = \mathsf{Set}^{\mathsf{X}^{\mathrm{op}}} $$
Every representable presheaf is compact. In general, any finite colimit of compact objects is compact. Thus, any finite colimit of representables is compact.
My question is about the converse: in the category of presheaves on a small category, is every compact object a finite colimit of representables?
If you don't want to assume that X is idempotent-complete, then I think you can get a counterexample by looking at retracts of representables that don't exist in your base category, since these will be compact and I don't think you can write them as finite colimits of representables in general. But a particular example with a proof that it isn't a finite colimit of representables isn't coming to me right now.
@WilliamBalderrama: Retracts are colimits over a category with a single object and a single nonidentity morphism, which is idempotent.
@DmitriPavlov Ah, so they are. For some reason I had in mind the sequential colimit along the given idempotent morphism as the way of splitting it.
Maybe a relevant motivational analogy is that compact objects in the poset $2^X$ are precisely finite subsets of $X$.
FWIW this is no longer true in the $\infty$-categorical setting, where "finite" has a slightly different meaning, and taking retracts is no longer a finite colimit. (It fails already when $C$ is a point, by the Wall finiteness obstruction.)
Yes, it is. The reason is:
every object of your presheaf category is a colimit of representables;
so, every object is a filtered colimit of objects which are finite colimits of representables;
so, applying the definition of a compact object, you get a split monomorphism from your compact object $X$ to a finite colimit $T$ of representables. To conclude, write $X$ as the coequaliser of $Id_T$ and the idempotent of $T$ given by your split mono.
Hang on -- this shows that $X$ is a finite colimit of (finite colimits of representables) -- a "2-fold" finite colimit of representables. But how does one turn this into an actual finite colimit of representables? I believe that any retract of $\kappa$-small colimits of representables is a $\kappa$-small colimit of representables when $\kappa$ is an uncountable regular cardinal, but I'm not sure about the case $\kappa = \aleph_0$...
The class of presheaves which are finite colimit of representables is stable under finite colimits:
The class of presheaves which are finite colimit of representables is stable under finite colimits: the stability under finite coproducts is obvious and the stability under coequalisers is easily seen by hand using the universal property of colimits and the fact that Hom(x,-) commutes with colimits for x representable.
I think I see how this works for finite coproducts -- but could you spell out the coequalizers for me?
@TimCampion What am I missing here? By the Fubini theorem for categorical limits and colimits, any finite colimit of finite colimits is equivalent to a colimit out of the product indexing category - which, as a product of finite categories, is finite - so we see a finite colimit of representables pop out at the end. I’m not sure why we need more complicated arguments
@FShrike There's no functor out of any product category in general. In this case, you would need to. lift an idempotent on $T = \varinjlim_i T_i$ to an idempotent on each $T_i$, natural in $i$. There's no reason for such a lift to exist in general.
I think that Aurelien Djament's answer is essentially correct, but I want to nitpick a bit.
If $\mathcal A$ is any locally finitely presentable category and $\mathcal C \subseteq \mathcal A$ is any strong generator of finitely-presentable objects, then every finitely-presentable object $X \in \mathcal A$ lies in the closure of $\mathcal C$ under finite colimits. So $X$ is a finite colimit of finite colimits of ... of finite colimits of objects of $\mathcal C$ -- an "$n$-fold" finite colimit of objects of $\mathcal C$. But $X$ need not be a "1-step" finite colimit of objects of $\mathcal C$. For example, I don't think every finitely-presented group is a finite colimit of copies of $\mathbb Z$.
One might strengthen the hypotheses and ask: if $\mathcal A$ is a locally finitely presentable category and $\mathcal C \subseteq \mathcal A$ is a dense generator, then is every finitely-presentable object $X \in \mathcal A$ a finite colimit of objects of $\mathcal C$? I don't know the answer to this.
But let's focus on the question at hand, i.e. the case where $\mathcal A = \hat {\mathcal C}$ is a presheaf category and $\mathcal C$ is the representables. Let $\tilde {\mathcal C}$ comprise the finite colimits of representables. Then indeed, $\tilde {\mathcal C}$ is closed under finite colimits. This is clear for finite coproducts -- just take the coproduct of the indexing diagrams for the colimits. Now let $A\rightrightarrows B \to C$ be a coequalizer where $A,B \in \tilde {\mathcal C}$. Then there is an epimorphism $\amalg_i X_i \to A$ and a coequalizer diagram $\amalg_j Y_j \rightrightarrows \amalg_k Z_k \to B$ where $X_i,Y_j,Z_k \in \mathcal C$ and the coproducts are finite. The composite maps $\amalg_i X_i \to A \rightrightarrows B$ lift to maps $\amalg_i X_i \rightrightarrows \amalg_k Z_k$. Then we have that $C$ is the coequalizer of the two induced maps $(\amalg_i X_i) \amalg (\amalg_j Y_j) \rightrightarrows \amalg_k Z_k$.
Now I claim that if $f,g \amalg_{i \in I} X_i \rightrightarrows \amalg_{k \in K} Z_k$ are two maps with coequalizer $C$, and if the $X_i$ are representable, then $C$ is the colimit of the following diagram. Indeed, for each $i \in I$, there is a unique $k = k_0(i) \in K$ such that $X_i \to \amalg_{i \in I} X_i \xrightarrow f \amalg_{k \in K} Z_k$ factors through $Z_k$, and similarly a $k_1(i)$ for $g$. The indexing set for our diagram has object set $I \amalg K$, and the nonidentity morphisms are a map $i \to k_0(i)$ and a map $i \to k_1(i)$ for each $i \in I$. Then $C$ is the colimit of the obvious diagram sending $i \mapsto X_i$ and $k \mapsto Z_k$. This diagram is finite if $I$ and $K$ are.
Thus in our case, $C \in \tilde{\mathcal C}$ as desired.
I want to emphasize that here we heavily used the fact that we're in a presheaf category.
I agree that any category which has finite colimits and filtered colimits has all colimits. But Aurelien's second bullet seems to suggest something stronger -- that if $X$ is a colimit of objects of $\mathcal C$, then $X$ is a filtered colimit of finite colimits of objects of $\mathcal C$. I don't have a counterexample, but I'm not sure this is true. The closest I can convince myself of is that $X$ is a coequalizer of coproducts of objects of $\mathcal C$, and therefore a coequalizer of filtered colimits of finite coproducts of objects of $\mathcal C$ -- but this only ensures that $X$ is a finite colimit of filtered colimits of finite colimits of objects of $\mathcal C$.
But using (3), Aurelien's third bullet goes through with some modification. As in any locally finitely presentable category $\mathcal A$ with strong generator $\mathcal C$, any finitely-presentable object is in the closure of the $\mathcal C$ under finite colimits. By (3), in the case $\mathcal A = \hat{\mathcal C}$, the closure of $\mathcal C$ under finite colimits consists exactly of $\tilde{\mathcal C}$, the objects which are "1-step" finite colimits of representables. Here, (3) is actually used in 2 places: first to ensure that the category $\tilde C \downarrow X$ is filtered (this being the diagram which indexes the canonical colimit for $X$), and second to ensure that $\tilde{\mathcal C}$ is closed under retracts.
Thanks for this answer that really works out what's going on here!
Here is another perspective on the problem, using some big guns (Gabriel-Ulmer duality).
Given a small category $C$, let $K$ be its free finite cocompletion. This means $K^{op}$ is the free finite completion of $C^{op}$, which means in turn that for any functor $F: C^{op} \to \mathbf{Set}$, there is a finitely continuous (or left exact) functor $\tilde{F}: K^{op} \to \mathbf{Set}$ that extends $F$ along the canonical inclusion $i: C^{op} \to K^{op}$, and this extension is unique up to unique isomorphism. Put differently, restriction along $i$ induces an equivalence
$$\mathrm{Lex}(K^{op}, \mathbf{Set}) \to \mathrm{Cat}(C^{op}, \mathbf{Set}).$$
In particular, the presheaf category $\mathrm{Cat}(C^{op}, \mathbf{Set})$ is locally finitely presentable. By the way, it's well known that the free finite cocompletion $K$ of a small category $C$ is simply the category of finite colimits of representables: see section 5.9 of Kelly's Basic Concepts of Enriched Category Theory.
On the other hand, Gabriel-Ulmer duality assures us that given a locally finitely presentable category $A$, there is up to equivalence only one finitely complete category $L$ for which $A \simeq \mathrm{Lex}(L, \mathbf{Set})$. Even better, Gabriel-Ulmer duality gives a recipe for obtaining $L$: it is the dual of the category of compact objects in $A$, meaning objects $a$ such that $A(a, -): A \to \mathbf{Set}$ preserves filtered colimits.
Putting all this together, this shows that the category of compact objects in the category of presheaves over $C$ is equivalent to the free finite cocompletion of $C$, or to the category of finite colimits of representable presheaves.
|
2025-03-21T14:48:31.076837
| 2020-05-25T06:08:10 |
361284
|
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|
Stack Exchange
|
Functional inequalities involving the condition $\left(\int_0^t f(x)dx\right)^2 \ge \int_0^t f(x)^3dx$
I was reading the solution to a functional inequality in an article when the author made the following remark without giving any proof: let $f(x): [0, \infty]\to[0, \infty]$ be locally integrable and such that
$$\left(\int_0^t f(x)dx\right)^2 \ge \int_0^t f(x)^3dx$$
for all $t>0$. Then, the following statement is true:
$\int_0^t f(x)^\gamma dx \le \frac{1}{\gamma +1}\left(2\int_0^t f(x)dx\right)^{(\gamma + 1)/2}$ for all positive $t$ and $\gamma \in [1,3]$.
Again, there is no proof in the article, so I don't know if this is fairly easy to prove or very involved. One thing that might be worth mentioning is that the inequalities above become exact when $f(x)=x$. I am wondering if anyone has an idea or have seen these before.
Could you post the paper with the improved constant?
I don't know about the precise multiplicative prefactor $\frac{2^{\frac{\gamma+1}2}}{\gamma+1}$, but I can tell you how to get
$$
\int _0^t f^\gamma\leq \left(\int_0^t f\right)^{\frac{\gamma+1}{2}}.
$$
This is a simple interpolation inequality: for any fixed $\gamma\in (1,3)$ (the boudnary cases $\gamma=1,3$ are immediate) there is a unique $\theta=\theta(\gamma)\in(0,1)$ such that
$$
\frac 1\gamma=\theta\frac{1}{1}+(1-\theta)\frac{1}{3}.
$$
By standard interpolation you get
$$
\|f\|_{\gamma}\leq \|f\|_{1}^{\theta} \|f\|_{3}^{1-\theta}.
$$
Your assumption is equivalent to the "$L^3$ by $L^1$ control" $\|f\|_3\leq \|f\|_1^{\frac{2}{3}}$, which then immediately implies
$$
\|f\|_{\gamma}\leq \|f\|_{1}^{\theta+\frac 23(1-\theta)}.
$$
Leveraging the explicit value of $\theta=\theta(\gamma)$, some straightforward algebra then finally gives the exact exponent $\theta+\frac 23(1-\theta)=\frac{\gamma+1}{2}$.
Could you post or link the paper with the assertion on the improved constant? Thank you
@Giorgio_Metafune: I guess this is a comment for the OP's question, rather than for my answer?
@leo_monsaingeon Yes, true! Sorry, I wrote in the wrong place
@leomonsaingeon, Yes, the general integral relationship is straightforward (directly applying Holder to the condition on $f(x)$). However, the constant is critical.
Perhaps you should make this clear by editing your question? Indeed your prefactor is $<1$ for $\gamma\in(1,3)$ so just Hölder-ing the problem is suboptimal, and I have no clue about how to get there... Is the paper focussed on small $t\to 0$, on large $t\to\infty$, or on any $t$?
|
2025-03-21T14:48:31.077052
| 2020-05-25T06:40:07 |
361286
|
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|
Stack Exchange
|
Path integral derivation of extended TQFT
I know this isn't exactly a math question, but I am asking it here anyway. We define an extended TQFT to be a functor (preserving tensor products) from the $\left(\infty,n\right)$-category of cobordisms to a suitable $\left(\infty,n\right)$-category of vector spaces.
The original Atiyah-Witten definition was a functor for the category of $n$ dimensional cobordisms to $\mathrm{Vect}_{\mathbb C}$. This definition was justified from the path integral in physics.
Can we similarly get an physicsy intuition of extended TQFT from a path integral-like formulation? Any references to a general construction from physics that gives rise to such a functor, starting from the path integral?
Note: I don't want specific example relating to Chern-Simons theory or any other TQFT, but a general construction deriving the extended TQFT axioms from the path integral, or something similar.
It's been a while since I've read Freed 1992, but I thought he did argue for (once) extended TQFT from a path integral perspective.
@Theo, I noticed that you attended Kevin Walkers 2011 minicourse on TQFTs at UC Berkeley. In your notes in lecture 2, it is mentioned that Walker's fields and local relations description and Freed's fully extended Atiyah axioms are equivalent formulations. I haven't seen the lectures yet, but can you tell me the construction? I could send you my e-maid id if communication is easier there.
Wild. I thought I knew where all of my old notes are, but I cannot find my own notes on Kevin's 2011 minicourse.
To answer your question, I think "equivalent" might be a heavy word.
Kevin has developed a very sophisticated theory of the extended operators in certain types of topological field theories. But he tends not to care about some "finiteness" conditions at the very top dimensions.
By now it is pretty well understood that Kevin's TQFTs extend "down" to points. There are some still missing theorems matching Kevin's axiomatization with Lurie's. In low spacetime dimensions, most of this has been proved by Cooke in her thesis. In full generality, the matching will eventually follow from the work by Ayala--Francis on what they have been calling "beta factorization algebras".
I think that https://arxiv.org/pdf/1409.6051.pdf performs exactly what you're asking for : A path integral derivation of (once) extended TQFT. I conjecture that the construction performed at the top of page 4 could be re-done starting with other Lagrangians, to produce the values of the corresponding QFT on codimension 2 surfaces...
The physics motivation for extended QFTs (and not just TQFTs) comes from the locality principle (no spooky action at a distance).
The mathematical expression of locality is the descent property for extended QFTs.
For an early source, see, for instance, Higher Algebraic Structures and Quantization by Daniel S. Freed
and Triangulations, Categories and Extended Topological Field Theories by Ruth J. Lawrence.
The descent property is proved in full generality in arXiv:2011.01208.
Specifically, the assignment to X of the symmetric monoidal (∞,n)-category of extended QFTs with bordisms equipped with a map to X is a stack of symmetric
monoidal (∞,n)-categories with respect to X.
This claim fails unless we work with QFTs extended all the way down to points,
since proving the descent property requires cutting bordisms all the way
down to points.
An informal path integral construction produces an extended QFT
starting from another functorial field theory (quantum or classical) on Y
and performing pushforward along a map Y→X.
The pushforward simply integrates over spaces of n-dimensional manifolds
mapping to Y that have the same image in X.
@Dmiti, what exactly is this 'descent property'? You're talking about evaluating the TQFT by gluing local pieces? I don't know about extended TQFTs with maps to spaces. Can you give me a reference to this stuff? And also the stack you're talking about?
@ChetanVuppulury: The descent property is explained in the third paragraph. I added a reference to a classical paper by Freed that explains how the locality property is connected to extended field theories.
@ChetanVuppulury: And now I added a reference to a new paper that explains and proves the descent property, giving complete details.
The original motivation for extended TQFTs (as introduced by Freed, Lawrence, Baez-Dolan) is indeed giving a finer form of locality, as explained by Dmitri Pavlov. However I think there are two quicker, and arguably more physical, ways to see n-categorical structure in n-dimensional QFTs.
The first is not really about the states of a QFT (as axiomatized in the Atiyah-Segal formalism) but their algebras of observables (extending the distinction between geometric and deformation quantization in the context of quantum mechanics). Namely the theory of factorization algebras as developed in the book of Costello-Gwilliam extracts from the same data as the path integral an n-dimensional factorization algebra of observables. In the topological context such a factorization algebra is the same as an $E_n$ algebra, which is the same as a very connected $(\infty,n)$-category (one with one object, one 1-morphism, ...all the way down).
The second comes from thinking of what ELSE there is in a QFT beyond the path integral -- the most important being the structure of defects of various dimensions. Of these the richest is the notion of a boundary theory (or "boundary condition") for a QFT, which is very loosely "things we can put on the boundary and couple to our theory" -- something like a QFT of one dimension lower that lives on the boundary of manifolds where the bulk carries our given QFT.
In any case, boundary theories in an $n$-dimensional QFT naturally form something like an $(n-1)$-category, which in the cobordism hypothesis formalism for extended TQFT is closely related to what you'd attach to a point. Namely, as morphisms between any two boundary theories you can consider codimension 2 defects that are interfaces between the two theories (think of dividing the boundary of a half-space in $R^3$ into upper and lower halves with a 1-dim interface on the intersection). As 2-morphisms you can consider interfaces between interfaces, and so on and so forth.
To me this is the most compelling way to see that higher categorical structure is physically natural/meaningful. A (somewhat criminal) paraphrase of the cobordism hypothesis says that a fully extended TQFT is determined by its collection of boundary conditions. [Really boundary theories are morphisms between the unit and the object attached by the TQFT to a point, which in general needn't determine this object, but it's a decent ansatz.]
Could you elaborate on the relation between defects and boundary theories (by which I understand holographically related theories, if not please do clarify)?
I don't know the physics but my impression was that holography is not the correct terminology since there's no gravity involved, just a bulk QFT and a theory coupled to it living on the boundary. This is a special case of a domain wall, a (codimension one) interface between two QFTs, where we let the theory on the other side of the wall be trivial. Domain walls can be considered codimension one defects (except for higher codimension defects there is only one theory involved, since space doesn't get disconnected by the defect).
|
2025-03-21T14:48:31.077566
| 2020-05-25T06:56:38 |
361288
|
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|
Stack Exchange
|
Levi-Civita connection from idempotents
Let $(M,g)$ be a closed Riemannian manifold. Let $V$ be a smooth complex vector bundle over $M$. We can write $V$ as the range of an idempotent $E$ in a matrix algebra $M_n(C^\infty(M))$ acting on a trivial bundle $M\times\mathbb{C}^n$. Here $C^\infty(M)$ denotes the complex-valued functions on $M$.
Let $D$ be the trivial connection on $M\times\mathbb{C}^n$ defined by applying the de Rham differential to each entry separately. Then the expression $EDE$ defines a connection on $V$.
Some books call the connection $EDE$ the "Levi-Civita connection" on $V$, "by analogy to the classical situation", and I am trying to understand how this analogy works. In particular, when $V$ is the tangent bundle $TM$ (ignoring the fact that this is a real vector bundle), the Levi-Civita connection depends on the metric $g$, while the expression $EDE$ does not seem to depend on $g$.
Question: Is there a precise sense in which $EDE$ produces the "classical" Levi-Civita connection on $TM$?
Certainly not "a precise sense": your connection depends on the choice of $E$, which is not uniquely defined.
You use the orthogonal projection $E$ for some isometric embedding into Euclidean space, and I am pretty sure that Dirac in his little book on General Relativity shows that you get the Levi-Civita connection.
Thanks, that makes complete sense.
$\newcommand{\R}{\mathbb{R}}$
Yes, it uniquely defines a connection in the case where $V$ is the tangent bundle(the connection won't depend on the choice of orthogonal idempotent) and its the Levi Civita connection. But in general different idempotents from the same metric will define different connections(I think, I'm not sure on this one though).
Let $ V^l \to M^k $ be the vector bundle of rank $l$ over a $k$ manifold and take an isometric embedding of the vector bundle into $\R^{2n}$ with the standard metric. Let $s$ be a section of the vector bundle. Let $e: M \to Mat_{n\times n}$ be the orthogonal projection onto the tangent bundle of $M$.
If we differentiate $s$ in a given direction using the ambient $\R^{2n}$, (we can just take all the directions into account by taking the exterior derivative $ds$), then the result will not be in the vector bundle any more. But $eds$ will be. Hence we define the derivative of $s$ in the $v$ direction to be $\nabla_v s =e\frac{ds}{dv}$
Let us consider the case where $E$ is the tangent bundle of the manifold and $M^k \subset \R^n$. It is possible to write down the expression $e\frac{ds}{dv}$ just using the metric on the manifold.
To show this we show that $eds$ satisfies the axioms of the Levi Civita connection:
If $X,Y,Z$ are sections of our vector bundle, then
\[ X \langle Y,Z \rangle = \langle \nabla_X Y, Z \rangle + \langle
Y, \nabla_X Z \rangle \]
Proof:
\[ \nabla_X Y-\nabla_Y X=[X,Y] \]
If we expand the rhs, \[ \langle \nabla_X Y, Z \rangle + \langle Y, \nabla_X Z \rangle = \langle e \frac{dY}{dX}, Z \rangle + \langle Y , e \frac{dZ}{dX} \rangle = \]
\[\langle \frac{dY}{dX}, e^T Z \rangle + \langle e^TY , \frac{dZ}{dX} \rangle \]
Now $e^T=e$ because $e$ is an orthogonal projection. So this is
\[=\langle \frac{dY}{dX}, eZ \rangle + \langle eY , \frac{dZ}{dX} \rangle=\langle \frac{dY}{dX}, Z \rangle + \langle Y , \frac{dZ}{dX} \rangle=X \langle Y,Z \rangle \]
\[ \nabla_X Y-\nabla_Y X= e \frac{dY}{dX} -e \frac{dX}{dY}=e[X,Y]=[X,Y] \]
Exercise:
Using the above identities, and cyclically permuting $X,Y,Z$ write down an expression for $\nabla_X Y$ in terms of the inner product.
Using the above expression, write down a matrix $A$ of 1-forms in terms of the inner product such that
\[\nabla s=ds-As \]
If you do this exercise, the matrix $A$ you get will be the matrix of christoffel symbols.
Another good exercise is to write out what $A$ is in terms of the idempotent. (You should get $(1-e)de$)
Note that you can't cyclically permute the indices $X,Y,Z$ for an arbitrary vector bundle. I don't think that you get a canonical connection on a vector bundle just given the metric(I think this was one of your questions), and at the very least, the trick above doesn't work.
Aside: I came up with this answer after reading through the comment above about it being in Dirac's General relativity book. I've been trying for the past few months and I can't quite tell what he is doing. I did see though that his argument factors through proving that the axioms of the Levi Civita connection are satisfied without saying that that is what he's doing.
|
2025-03-21T14:48:31.077891
| 2020-05-25T07:28:48 |
361291
|
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|
Stack Exchange
|
Isomorphism between two groups
Let $\mathbf{F}_q$ be finite field of order $q$, where $q$ is an odd-prime (or power of an odd prime). Is there an isomorphism between the following subgroups of unipotent $3\times 3$ matrices over $\mathbf{F}_q$:
$$\left\{ \begin{pmatrix}1 & x& y\\0&1&0\\0&0&1\end{pmatrix} \mid x,y\in \mathbf{F}_q\right\}, \text{ and } \left\{ \begin{pmatrix}1 & x& y\\0&1&x\\0&0&1\end{pmatrix} \mid x,y\in \mathbf{F}_q\right\}?$$
One way is to map generators to generators, and extend it to products. Is there any other isomorphism possible between the above mentioned subgroups?
They are both elementary groups of the same order.
Once the groups are isomorphic, the set of isomorphisms is a principal homogeneous space for the automorphism group (acting by composition on one of the sides).
So here the set of isomorphisms is in bijection with $\mathrm{GL}_{2e}(\mathbb{F}_p)$ where $q=p^e$ (since the group is the additive group of a $2e$-dimensional $\mathbb{F}_p$-vectorspace).
You're right, and there are even more automorphisms. The group is actually $C_p^{2e}$ if $q=p^e$.
|
2025-03-21T14:48:31.078121
| 2020-05-25T08:45:28 |
361294
|
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|
Stack Exchange
|
Reference request: continuity of Cholesky factor
It most books dealing with Cholesky decomposition, or it is variants, one finds a statement of the form if $A$ is symmetric $k\times k$ positive semi-definite (non-negative definite) then the $k\times k$ matrix $L$ solving
$$
A=RR^{\top}.
$$
Note: I do not require that $A$ is positive definite, so $A^{-1}$ may not exist. However, I do require that it is symmetric.
Following his post, we see that under additional constraints there is a unique choice
Theorem 10.9. Let $A\in\mathbb R^{n\times n}$ be positive semidefinite of rank $r$. (a) There exists at least one upper triangular $R\in\mathbb R^{n\times n}$ with nonnegative diagonal elements such that $A = R^TR$. (b) There is a permutation $\Pi$ such that $\Pi^TA\Pi$ has a unique Cholesky factorization, which takes the form
$$
\Pi^TA\Pi=R^TR,\quad R=\left(\begin{matrix} R_{11} & R_{12} \\ 0 & 0\end{matrix}\right),
$$
where $R_{11}$ is $r \times r$ upper triangular with positive diagonal elements.
However, I cannot find the source of book or paper saying map $A \to R$ is continuous.
I have edited the title. The matrix square root is another thing, and proving that it is continuous is nontrivial (for non-symmetric matrices) in my view.
@FedericoPoloni What do you mean? In the case of semi-definite, ut symmetric matrices, would it be easier?
There is a thing called "the (principal) matrix square root", which is defined for all matrices (possibly nonsymmetric) with no real negative eigenvalues and no nontrivial Jordan blocks in zero. It is continuous, but it is nontrivial to prove it. However, it is not what you are asking about here, so I have changed the title. Calling a Cholesky factor "square root" is slightly improper, although I have already heard it in various contexts.
No worries, I worked out what I needed from the answer as you pointed out. However, now I'm interested (purely out of curiousity) do you have a reference to this principle matrix squre-root s, purely our of scientific interest.
Higham's book Functions of matrices. It has a definition on how to extend an arbitrary scalar function to matrices (like you probably already studied with the exponential of a nonsymmetric matrix) in Chapter 1, including some remarks on branches and the principal square root, and then a chapter devoted to the properties of matrix square roots.
Thanks. This will be very interesting!
A subtle issue is that $\Pi$ is not unique here. For instance, if
$$
A = \begin{bmatrix}
1 & 0 & 0\\\\
0 & 0 & 0\\\\
0 & 0 & 0
\end{bmatrix}
$$
then you can take both the identity and $(23)$ as the permutation. Similarly, if $A=I$, then any $\Pi$ will work (and $R=I$).
I don't think you can speak about continuity until you resolve this ambiguity.
How can it be resolved?
It's up to you to define which decomposition exactly you need!
But are there reasonable known criteria on a decomposition, or choice of $\Pi$ as a function of $A$, making the map $A\to \mbox{ Cholesky factor of A}$ continuous? For example with the requirement that $|A -R|_F$ be minimal be such a requirement? (I've never had this type of issue before, so I hope my question is not too misguided).
I am not sure if one can come up with a deterministic criterion for $\Pi$ that solves this problem. For instance, if $A_{2n} = diag(1,1/n,0)$, $A_{2n+1}=diag(1,0,1/n)$, then it seems that one needs to choose the identity permutation on even indices and (23) on odd indices. So the sequence $A_{n}$ converges to the matrix $A$ in my answer, but $\Pi$ is not eventually constant. (The factors $R_n$ converge to $R$, though, so maybe there is a way out.) I don't have a ready answer, sorry.
How do you plan to use this inverse? That is not clear to me. What is "this space"?
Let us continue this discussion in chat.
What is P_n in your reply in chat? Symmetric matrices? Then I guess your map is $A \mapsto A^T A$, not $A \mapsto A^2$.
Numerical Analysis: A Mathematical Introduction page 295.
That only takes care of the positive definite case. I think the real issue that the question raises is about allowing zero eigenvalues.
|
2025-03-21T14:48:31.078433
| 2020-05-25T09:15:20 |
361296
|
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|
Stack Exchange
|
Topological proof that a Vitali set is not Borel
This question is purely out of curiosity, and well outside my field — apologies if there is a trivial answer. Recall that a Vitali set is a subset $V$ of $[0,1]$ such that the restriction to $V$ of the quotient map $\mathbb{R}\rightarrow \mathbb{R}/\mathbb{Q}$ is bijective. It follows easily from the definition that $V$ is not Lebesgue measurable. Now suppose you don't know the Lebesgue measure (which, after all, is not that easy to construct). Is there a topological proof that $V$ is not a Borel subset of $\mathbb{R}$?
The family of sets satisfying the property of Baire is a sigma-algebra, in particular all Borel sets have this property. On the other hand, Vitali set does not have this property. This is a nice instance of Baire property being a convenient substitute of Lebesgue measurability in topological contexts.
Very good, thanks! If you feel like writing this as an answer I'll accept it.
You may be interested in Oxtoby's book Measure and Category that goes through several parallel examples of theorems about measure and Baire category. Wojowu's answer explains the "category analogue" of Vitali's theorem.
Sometimes a convenient substitute for Lebesgue measurability is the property of Baire. Just like Lebesgue measurability, the class of sets with this property is a $\sigma$-algebra containing the open subsets - indeed, open sets clearly have property of Baire, this class is closed under countable unions since meager sets are closed under countable unions, and it's closed under complements essentially since the boundary of an open set is nowhere dense.
It of course follows that every Borel set has the property of Baire. It remains to show the Vitali set doesn't have a property of Baire. Indeed, assume it was, then either $V$ is meager, or it is comeager in some open interval. The former cannot hold, since $[0,1]$ is contained in countable union of translates of $V$, so would be meager itself. In the other hand, if it were comeager in some open interval $I$, then for any rational $q$ the intersection $I\cap(V+q)$, since it's contained in $I\setminus V$, is meager, from which we would conclude $V$ is meager.
This might be a slightly overly detailed answer, but I did try to make it reasonably self-contained - it indeed is quite easier than if we had to build up the theory of Lebesgue measure from scratch!
Just for completeness: a subset $Y$ of a topological space $X$ has the Baire property, or is almost open in $X$ if it differs from an open subset by a meager subset, i.e., can be written as $U\triangle M$ with $U$ open and $M$ meager. And meager means, contained in a countable union of closed subsets with empty interior.
Also I guess that "$V$ is meager in some open interval" should be understood as "there exists a nonempty open interval $I$ such that $V\cap I$ is comeager in $I$". (It took me a while to unravel so I'm adding this as a comment.)
@YCor Thank you for these remarks! Yes, this is precisely what I meant with "comeager in an open interval".
(yes, and sorry for the typo you corrected: I should have quoted "$V$ is co-meager in some open interval)
This might be a slightly overly detailed answer, Not at all. I would expect all answers to be detailed as this. Most of the answers on MathOverflow lack details making them basically useless. Thank you. I just added OP's question to the homework in my measure theory course and I included hints that follow your answer.
|
2025-03-21T14:48:31.078701
| 2020-05-25T10:13:31 |
361302
|
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|
Stack Exchange
|
Conditions for absolute continuity in the Bochner-Schwartz theorem
Suppose that $f$ is a positive-definite Schwartz distribution, that is,
$$\langle\phi,f*\phi\rangle\geq0\qquad\text{for every }\phi\in C_0^\infty(\mathbb R^n).$$
By the Bochner-Schwartz theorem, there exists a unique tempered measure $\mu$ such that $\hat f=\mu$, where $\hat\cdot$ denotes the Fourier transform. I'm interested in the following:
Question. Are there simple conditions that guarantee that $\mu$ is absolutely continuous, i.e., $d\mu(x)=g(x)dx$ for a measurable $g$? What can we say about $g$'s regularity?
If $f$ were continuous, then we could use for example the fact that decay of $f$ translates into regularity for $\hat f$, but since $f$ is only a Schwartz distribution, this seems difficult.
For example, using the formal correspondence
$$\sup_{x\in K}f(x)"="\sup_{\|\phi\|_1=1,~\mathrm{supp}(\phi)\subset K}\langle f,\phi\rangle$$
for any $K\subset\mathbb R^n$, I suppose one could somehow access the "decay" of a distribution by looking at quantities like
$$\sup_{\|\phi\|_1=1,~\mathrm{supp}(\phi)\subset B(0,t)^c}\langle f,\phi\rangle$$
for large $t>0$, where $B(0,t)^c=\{x\in\mathbb R^d:|x|>t\}$ is the complement of the unit ball in $t>0$. It seems to me like results regarding the absolute continuity of $f$'s Fourier transform should be classical, but I've not been able to find any.
The regularity of $g$ is easier It is related to the decay properties of the Fourier transform. If $f$ is in $L^2$ then $\mu$ is absolutely continuous with $g\in L^2(\mathbb{R}^n, dx)$.
@LiviuNicolaescu Indeed, but here I'm hoping to capture a more general phenomenon. For example, both the delta Dirac distribution and the Riesz kernel $f(x)=|x|^{n-\alpha}$ have absolutely continuous Fourier transforms, but neither are in $L^2$. Both of these distributions happen to have some form of decay (much weaker than what would be imposed by continuity+$L^2$, for example), which is what I suspect has something to do with absolute continuity.
|
2025-03-21T14:48:31.078858
| 2020-05-25T10:33:57 |
361304
|
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|
Stack Exchange
|
Stokes's Theorem with singularities on projective line
Let $X$ be a complex manifold and $\omega\in \Omega(X\times \mathbb{P}^1)$ a form. I met the following identity:
$$\int_{\mathbb{P}^1}(\partial_z\bar{\partial}_z\omega) \log|z|^2=\int_{\mathbb{P}^1}\omega\partial_z\bar{\partial}_z\log|z|^2 $$
where $\partial_z,\bar{\partial}_z$ denote the differential operators on $\mathbb{P}^1$.
My question is how can we use Stokes's theorem (in the integration by parts) here given that the function $\log|z|^2$ has singularities at $0$ and $\infty$.
References: http://www.math.stonybrook.edu/~leontak/On%20Bott-Chern%20forms%20and%20their%20applications.pdf page 24, Soulé's "Lecture on Arakelov Geometry" page 81 (top).
Remove tiny neighborhoods of the singularities, use Stokes in the complement of these neighborhoods, and then let the neighborhoods shrink towards the singularities. A good example to have in mind is the proof that $\frac{1}{|x|^{n-2}}$ is a fundamental solution for the Laplacian on $\mathbb{R}^n$
|
2025-03-21T14:48:31.078956
| 2020-05-25T10:54:59 |
361305
|
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|
Stack Exchange
|
Does the compact-open topology retain topological groups?
Let $X$ be a topological space and $Y$ a topological group. Then $C(X,Y)$ is a group, and can also be endowed with the compact-open topology.
Is $C(X,Y)$ in the compact-open topology necessarily a topological group? If not, is there some property of $X$ which will guarantee it?
When $X$ is compactly-generated and Hausdorff, then $C(X,-)$ is right adjoint to $-\times X$, so preserves limits (possibly you need to restrict to $Y$ Hausdorff as well). Any functor that preserves limits sends group objects to group objects. If you are happy with not quite the compact-open topology, then you can just ask that $X$ is core-compact. See discussion at https://ncatlab.org/nlab/show/exponential+law+for+spaces
check out section 2.6 in Narici, Beckenstein - Topological vector spaces
|
2025-03-21T14:48:31.079042
| 2020-05-25T11:07:17 |
361306
|
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|
Stack Exchange
|
Posets which extend centered sets to filters
(Post cross-posted from math.se.)
Suppose $(\mathcal O, \leq)$ is an arbitrary poset. Let us say that $\mathcal O$ is compact if every $\mathcal C\subseteq\mathcal O$ which is centered (any finite subset of $\mathcal C$ has a lower bound in $\mathcal O$) can be extended to a filter in $\mathcal O$.
This is equivalent to the condition that $\mathcal C$ is contained in a directed set (one containing lower bounds for all finite subsets).
For example, if $\mathcal O$ is a meet-semilattice, then it is compact --- it suffices to close $\mathcal C$ under meets.
Not every poset is compact: for instance, if we let $\mathcal O$ consist of the nonempty finite sets of integers and put $A\leq B$ when $A=B$ or $B\subseteq A$ and $B$ is a singleton, then the set $\mathcal C$ of singletons in $\mathcal O$ is centered, but it does not have any centered proper supersets (and it is not a filter).
I was wondering if there is some simpler criterion for this "compactness", or whether it has an established name?
|
2025-03-21T14:48:31.079144
| 2020-05-25T11:36:48 |
361311
|
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|
Stack Exchange
|
G-abelian systems
Let $(\mathfrak{A},\alpha,\phi)$ be a $C^*$-dynamical system made of a unital $C^*$-algebra, a $*$-automorphism and an extremal invariant (i.e. ergodic) state.
Consider the covariant GNS representation $(H_\phi,\pi_\phi, V_{\phi,\alpha},\xi_\phi)$, together with the selfadjoint projection $E_{\phi,\alpha}$ onto the invariant vectors of $H_\phi$ under the unitary $V_{\phi,\alpha}$.
Are there concrete examples for which ${\rm dim}(E_{\phi,\alpha}H_\phi)>1$?
Doesn't the fact that $\dim(E[H]) \geq 2$ contradict the ergodicity of $\phi$? It does in the commutative case.
No Adrian, ergodicity means extremality among invariant states. ${\rm dim}(E[H])=1$ implies ergodicity, but the converse is true under the additional assumption of $G$-abelianess (in our situation $Z$-abelianess), see Prop. 3.1.12 in Sakai's book. The question is that, at my best knowledge, conterexamples don't exist in literature
In commutative case, it does hold true: ${\rm dim}(E[H])=1\iff \phi$ is ergodic. Maybe, it is true also if the support of $\phi$ in the bidual is central.
What is the relation of the title to your question? I.e., would an example of the sort you want be called a G-abelian (or is it $G$-abelian?) system?
The relation is explained in one of the comments above.
|
2025-03-21T14:48:31.079258
| 2020-05-25T12:02:23 |
361313
|
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|
Stack Exchange
|
Non-compact dynamical systems
In topological dynamics, most of the time, we consider the continuous action of a (semi)group $G$ on a compact Hausdorff space $X$. In this context, we can envelop the group in a compact left topological semigroup $E(G,X)$ (the pointwise closure of $G$ in $X^X$) which gives us e.g. the existence of minimal ideals and idempotents within them.
I was wondering, is there some useful theory like this when $X$ is not necessarily compact? I suppose in any case, $E(G,X)$ is a left topological semigroup and not much more can be shown if $G$ and $X$ are arbitrary, but maybe something along the lines of the following hypotheses would help:
$X$ is a complete, separable metric space and $G$ consists of uniform homeomorphisms (the inverses are also uniformly continuous).
Like 1., but $X$ is an arbitrary Hausdorff complete uniform space.
In addition to 1. or 2., $X$ is locally compact.
I suppose you could show quite a lot if $X$ is a disjoint union of compact $G$-spaces, but I am mostly interested in the case when the action is point-transitive.
|
2025-03-21T14:48:31.079364
| 2020-05-25T12:12:12 |
361315
|
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|
Stack Exchange
|
Image of Frobenius element under irreducible representation is diagonalizable
Let $K/ \mathbb Q$ be a Galois extension, and $\rho$ be an irreducible representation of the Galois group $Gal(K/ \mathbb Q)$. Consider an integer prime $p$ which doesn't ramify in $K$, and let $\sigma_p$ denote the conjugacy class of the Frobenius element. Is the linear transformation $\rho(\sigma_p)$ diagonalizable in general? If so, I'd really appreciate a proof or reference (preferably, as elementary as possible). If not, what other conditions are required and what is the best result we can say in that direction?
You have almost no hypotheses, so the answer is clearly no. Just take $K/\mathbb{Q}$ to be finite Galois, e.g. with Galois group $S_3$, and take $\rho$ to be irreducible over a finite field such that some element of $S_3$ is sent to a non-diagonalisable matrix. For $S_3$, the standard representation arising from the isomorphism $S_3\cong {\rm GL}_2(\mathbb{F}_2)$ will do, where the involutions are not diagonalisable. By Chebotarev, every element of your Galois group will be a Frobenius at some $p$.
There is no "best result in that direction", since you have too few hypotheses. For example if $K/\mathbb{Q}$ is finite and the representation is over a field of characteristic $0$, then everything is diagonalisable, but that has nothing to do with Galois representations. In general, (Weil-Deligne) representations for which Frobenii are diagonalisable are called "Frobenius semisimple". You can google that term for lots of literature, but almost none of it will be "elementary".
|
2025-03-21T14:48:31.079489
| 2020-05-25T13:16:27 |
361318
|
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"Mikael de la Salle",
"Moishe Kohan",
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|
Stack Exchange
|
hyperbolic quotient of hyperbolic group
I have a memory of hearing about a result (or perhaps a conjecture), possibly due to Gromov, that, if $G$ is a hyperbolic group and $g \in G$ has infinite order, then the quotient group $G/\langle (g^n)^G \rangle$ is hyperbolic for all sufficiently large $n > 0$.
I have been searching for references, but without success. Can anyone help?.
$\mathbf{Edit}$: After looking at the references in the answer by Mikael de la Salle, I see that I did not state this result correctly. Rather than the statement being for all sufficiently large $n>0$, it should be that there exists and $N>0$ such that $G/\langle (g^{nN})^G \rangle$ is hyperbolic for all $n > 0$. The result stated applies only to non-elementary hyperbolic groups, but for an elementary hyperbolic group this quotient is finite, and so it remains correct.
This is contained in at least Delzant's paper Sous-groupes distingués et quotients des groupes hyperboliques. [Distinguished subgroups and quotients of hyperbolic groups] Duke Mathematical Journal, vol. 83 (1996), no. 3, pp. 661–682, and also in Ol'shanskii's paper SQ-universality of hyperbolic groups, Mat. Sb. 186 (1995), no. 8, 119–132.
I am not an expert, but the first lines of the papers seem to indicate that this result was announced by Gromov, but that the proofs were not all convicing.
You should add a reference to Olshanskiy’s paper from 1995, containing the same result.
I would be happy to do so, but what paper are you refering to ?
SQ-universality of hyperbolic groups, Math. sbornik (1995).
Thanks, that is exactly what I was looking for!
The stronger result is also proved in a paper of Delzant and Gromov:
https://doi.org/10.1112/jtopol/jtn023
The statement you make appeared in Gromov's original paper. See Theorem 5.5.D.
https://link.springer.com/chapter/10.1007/978-1-4613-9586-7_3
Incidentally, this result also follows from group theoretical Dehn Filling. https://doi.org/10.1007/s00222-006-0012-3
@NWMT I'd rather view this as a generalization than an alternative approach.
I believe that the correct reference to Ol'shanskii's work should be A.Yu. Ol'shanskii, On residualing homomorphisms and G-subgroups of hyperbolic groups. Internat. J. Algebra and Comput. 3 (1993), no. 4, 365--409. link
|
2025-03-21T14:48:31.079687
| 2020-05-25T13:44:42 |
361322
|
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Stack Exchange
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Union of Schubert cells being affine
Let $k$ be a field of characteristic zero, $G$ be a reductive group with a Borel $B$ and $\mathcal{F}:=G/B$ the associated flag variety. Let $W$ be the Weyl-group of G.
Then let $S \subset W$ and $Z=\bigcup_{w \in S} C(w) \subset \mathcal{F}$ where $C(w)=BwB/B$ is the Schubert cell associated to $w$.
I'm interested to know when $Z$ is an affine scheme. This is for example the case if all $w \in S$ have the same length. Is this the only case?
It's certainly not the only case; you can take $S = W$, for example.
@LSpice: huh? Wouldn't $S=W$ give the whole flag variety?
Sorry, yes. I missed that CJS was taking the union of the cells in $\mathcal F$, not in $G$.
Do you have a reference for the union of cells of equal dimension being affine? (Or is it obvious and I'm just not seeing it?)
@imakhlin: The union of cells of equal dimension is a disjoint union of affine varieties, hence affine.
@Sasha, thanks. By "disjoint" do you mean that each cell in $Z$ has a neighborhood in $\mathcal F$ that is pointwise disjoint from every other cell in $Z$ (equivalently: the topology on $Z$ induced from $\mathcal F$ is the disjoint union topology)? This property obviously holds if and only if the $w\in S$ are pairwise Bruhat-incomparable. What stops your argument from working in this more general case?
This is essentially an extension of my comment, just to answer the actual "is this the only case?" question. It is not, $Z$ will be affine whenever $S$ is an antichain in the Bruhat order. Indeed, this condition means that no $C(w)$ with $w\in S$ intersects the closure $\overline{C(w')}$ for any other $w'\in S$ which shows that $C(w)$ is open in $Z$. Hence the $C(w)$ are the irreducible components of $Z$ and are also affine, this renders $Z$ affine itself (Hartshorne, Exercise 3.3.2).
Of course, the more interesting underlying question is whether this condition is necessary, I might update this answer if I come up with a proof. (Any algebraic geometers here? Is it at all possible for an affine space to be embedded into an affine variety as a proper open subset? If not, this would give us the answer.)
(To answer your parenthetical question at the end: It is indeed possible, already for surfaces. I don't remember the construction though - maybe it's something like remove a divisor from a Hirzebruch surface? In any case, these examples shouldn't work here, because if $Z$ is open in its closure, then it will contain a $\mathbb{P}^1$ and so can't be affine. On the other hand, if $Z$ is not open in its closure, then the question is a bit ambiguous because there is no natural scheme structure on $Z$.)
@dhy Not sure if I understand the "if Z is open in its closure, then it will contain a P1" part. It would seem that a $Z$ consisting of a single cell is open in its closure but contains no $\mathbb P^1$?
If it is open in its closure and is not an antichain. The point being that then it contains cells $w, w'$ with $l(w)=l(w')+1$, and the union of two such cells is a product of an affine space and a $\mathbb{P}^1.$
Oh, okay, I see now. Yes, if we leave out the cases where $S$ is not convex in the Bruhat order as ambiguous, the rest should be clear.
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