added
stringdate 2025-03-12 15:57:16
2025-03-21 13:32:23
| created
timestamp[us]date 2008-09-06 22:17:14
2024-12-31 23:58:17
| id
stringlengths 1
7
| metadata
dict | source
stringclasses 1
value | text
stringlengths 59
10.4M
|
|---|---|---|---|---|---|
2025-03-21T14:48:29.676683
| 2020-01-27T12:21:13 |
351241
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dmitri Panov",
"M. Winter",
"Magma",
"https://mathoverflow.net/users/108884",
"https://mathoverflow.net/users/135257",
"https://mathoverflow.net/users/943"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625677",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351241"
}
|
Stack Exchange
|
Can I build infinitely many polytopes from only finitely many prescribed facets?
Given a finite set of convex $d$-dimensional polytopes $\mathcal P$, for some $d\ge 2$.
Question: Is it true that there are only finitely many different convex $(d+1)$-dimensional polytopes whose facets are solely (scaled and rotated versions of) polytopes in $\mathcal P$?
Some clarifications
A face of a polytope is the intersection of the polytope with a touching hyperplane, so subdividing facets does not count here.
In general, two $(d+1)$-polytopes shall be considered as different if they differ not just in scale and orientation.
By the usual rigidity arguments, given the shape of facets and their connections, the metric of the polytope is uniquely determined. Hence, if we can built different polytopes, they will be combinatorially different as well.
Example
There are only finitely many polyhedra that can be built from any finite set of regular polygons, but as far as I know, this result is by enumeration (see, e.g. Johnson solids).
I think this is indeed true. This should follow from the rigidity of convex spherical polytopes and a few more lemmas. I'll explain how to deal with cases $d\ge 3$.
Lemma 1, rigidity. Two convex spherical cobminatorially equivalent polytopes with isomertric faces are isometric.
Lemma 2, bounded volume. The boundary of any convex subset of round $\mathbb S^n$ has volume bounded from above by the volume of round $\mathbb S^{n-1}$.
Lemma 3, gap. Suppose we have a spherical convex polytope in $S^n$ of diameter less than $\pi-\varepsilon$ then the volume of the dual polytope is bounded from below by $c(n,\varepsilon)>0$.
Lemma 4, sum of dual solid angles. Let $P$ be a convex $d+1$-polytope. For each of its vertices consider the dual cone. Then the sum of solid angles of such dual cones is equal to the volume of the round $\mathbb S^d$.
Proof of finitness. So, suppose we have a finite number of convex $d$-polytopes $P_1,\ldots P_k$. Denote by $S_1,\ldots, S_m$ the spherical $d-1$ polytopes corresponding to vertices of $P_1,\ldots, P_k$.
Suppose we have a $d+1$-polytope $P$ with faces homothetic to $P_1,\ldots P_k$. Then for each vertex of $P$ we get a convex spherical $d$-polytope whose faces are composed from $S_1,\ldots, S_m$. Note the the number of such spherical polytopes is finite up to isometry, by Lemmas 1 and 2. Moreover, all of them have diameter less than $\pi-\varepsilon$ for some $\varepsilon$ . So by Lemma 3 the volumes of spherical polytopes dual to those (coming from vertices) are at least $c(d,\varepsilon)$. Hence by Lemma 4 the total number of vertices of $P$ is at most $vol(S^d)/c(d,\varepsilon)$. Hence we have only finite number of combinatorial types of $P$.
Thank your for your answer! I need to better understand your definitions: why do you explicitly speak of spherical polytopes, and why are they subsets of $\Bbb S^n$ and not $\Bbb R^n$?
Yes, spherical polytopes are subsets in $\mathbb S^n$ bounded by geodesic $n-1$-spheres. To each vertex $p$ of an Euclidean polytope $P $ in $\mathbb R^n$ we associate a spherical polytope in $\mathbb S^{n-1}$: take a small radius $r$ sphere $S_r^{n-1}$ in $\mathbb R^n$ centred at $p$ and take the intersection $S_r^{n-1}\cap P$. This is the spherical polytope, just scale it by $1/r$ so that it lies in $\mathbb S^{n-1}$ (i.e. in a sphere of radius $1$).
In the case $d=2$ it is not possible if you exclude coplanar faces.
Proof:
Consider the set $S$ of all angles of the faces in $\mathcal P$. Since $\mathcal P$ is finite, there's a minimal such angle, let's say it is $\varepsilon > 0$. Now consider the set $V$ of all angles below $2\pi$ you can produce by summing elements of $S$. $V$ is finite, since each element of $V$ is a sum of at most $2\pi/\varepsilon$ elements of $S$. So there is a maximal element of $V$, let's say it's $2\pi-\delta$ where $\delta > 0$.
Now let's choose a polyhedron made from polygons in $\mathcal P$. At each vertex, this polyhedron must have nonzero angular defect, since otherwise the polyhedron would not be convex, or it would have coplanar faces. So the angular defect is $2\pi-v$ for some $v \in V$, so it is at least $\delta$. Now by Descartes' theorem on total angular defect the total angular defect is equal to $4\pi$, so the number of vertices of the polyhedron is bounded by at most $4\pi/\delta$.
Because of this, the total sum of angles at vertices can be at most $(4\pi/\delta)(2\pi-\delta)$. Since each face has an angle sum of at least $\pi$, the amount of faces is bounded by $n := (4/\delta)(2\pi-\delta)$, and this value depends only on $\mathcal P$.
Now there are only finitely many ways to glue together at most $n$ shapes of $\mathcal P$, and each such way describes a unique polyhedron (if any) by Cauchy's theorem. Therefore the amount of polyhedra you can form using faces in $\mathcal P$ is finite.
I'm fairly certain it is possible to generalize this proof to other values of $d$.
Thank you. My argument for $d=2$ was exactly along these lines and I hoped to generalize. But I do not know about a higher dimensional analogue of Descartes' theorem.
There's Chern's theorem which could discretize to the desired statement, but it only applies in the case of even $d$.
Not an answer, but an "almost counterexample" for $d=2$ that came to my mind when I saw that question. Starting with a regular dodecahedron, we can successively insert circular "belts" of hexagons, in a way that the whole polyhedron is convex at each stage. The illustrations should give an idea. The shapes of the hexagonal facets can come very close to each other (in the appropriate sense), though it is clear that there can only be so many of a precise shape.
That's a great illustration of what I was afraid would happen (at least in higher dimensions).
Perhaps you should add a condition to exclude coplanar facets, i.e., to insist upon
strict convexity. Otherwise the deltahedra lead to an infinite number of combinatorially (but not geometrically) distinct polyhedra:
Wikipedia image.
For me, a face of a convex polytope is the intersection of it with a touching hyperplane, so your example would still have only four facets. I will edit this into the question.
|
2025-03-21T14:48:29.677065
| 2020-01-27T12:54:46 |
351243
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"KConrad",
"Rupert",
"YCor",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/15482",
"https://mathoverflow.net/users/3272"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625678",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351243"
}
|
Stack Exchange
|
is there a unique measure on a local field?
Suppose we consider a local field and forget about the topology for a moment and consider the set of all measures on some non-trivial $\sigma$-algebra over the field that makes the field operations measurable. Has there been any research done into the nature of this set?
Given any measure, you can always multiply it by a positive number to get another measure, so without normalizing in some way there is no uniqueness (such as with Haar measure). Also, there is a Dirac (point) measure at each point of the local field. Is there a purpose behind your question or is it random curiosity?
This is with regard to a project I am doing where we have formulated a notion of "building in the category of measure spaces" and I am trying to classify such buildings.
Being measurable does not depend on the measure...
|
2025-03-21T14:48:29.677164
| 2020-01-27T13:01:19 |
351244
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"DamienC",
"Daniil Rudenko",
"https://mathoverflow.net/users/21620",
"https://mathoverflow.net/users/7031"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625679",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351244"
}
|
Stack Exchange
|
What is the meaning of the coefficients of the Alekseev-Torossian associator
Drinfeld associators became a central object in mathematics and mathematical physics. They appear in deformation quantization, quantum groups, in the proof of the formality of the little disks operad, Lie theory (through the Kashiwara--Vergne conjecture), in works related to multiple zêta values, etc...
This question concerns number theoretic or motivic aspects of the coefficients of the Alekseev-Torossian associator. Even though there are plenty of associators (they form a torsor on the Grothendieck--Teichmuller group), there are essentially three explicitely known ones:
the KZ associator, which is a generating series for multiple zêta values.
the Deligne associator, which is a generating series for single-valued multiple zêta values.
the Alekseev-Torossian (AT) associator, whose coefficients have been described by Furusho in https://arxiv.org/pdf/1708.03231.pdf (Theorem 3.3).
I find it hard to understand the number theoretic and/or motivic meaning of the coefficients of the AT associator. In particular, there is a formal series in one variable $x$ associated with every associator (the log of its Gamma function), whose coefficients are:
zêta values for the KZ associator.
singled valued zêta values for the Deligne associator. In particular the coefficient of $x^2$ is $0$.
Bernoulli numbers for the AT associators. In particular, odd powers are $0$.
Is there any realization (in the sense of motives) that would send the odd motivic zêta values to $0$?
I am not sure that I can give a fully coherent answer, but to the best of my knowledge coefficients of the AT associator should be closely related to Hodge Correlators, introduced by Goncharov.
Thank you Daniil Rudenko. This is actually a very interesting suggestion.
|
2025-03-21T14:48:29.677312
| 2020-01-27T13:09:46 |
351246
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"IJL",
"M. Winter",
"Ruy",
"https://mathoverflow.net/users/108884",
"https://mathoverflow.net/users/124004",
"https://mathoverflow.net/users/41291",
"https://mathoverflow.net/users/97532",
"მამუკა ჯიბლაძე"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625680",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351246"
}
|
Stack Exchange
|
What is the convex hull of the quaternionic symmetries of the 3 dimensional cube?
It is well known that there are exactly five 3-dimensional regular convex polyhedra, known as the Platonic solids.
In 1852 the Swiss mathematician Ludwig Schlafli found that there are exactly six regular convex 4-polytopes (the generalization of
polyhedra to 4 dimensions) and that, for dimensions 5 and above, there are only three!
The six regular 4-polytopes are:
NAME VERTEXES EDGES FACES CELLS
Hypertetrahedron 5 10 10 5
Hypercube 16 32 24 8
Hyperoctahedron 8 24 32 16
24-cell 24 96 96 24
Hyperdodecahedron 600 1200 720 120
Hypericosahedron 120 720 1200 600
The easiest ones to be described are the first two:
a model for the hypertetrahedron may be obtained as the convex hull of the canonical basis in $\mathbb R^5$ (hence a
4-dimensional object), while a model for the hypercube is the Cartesian product $[0, 1]\times[0, 1]\times[0, 1]\times[0,
1]$.
As in the case of 3 dimensions, the dual of a regular 4-polytope is also a regular 4-polytope and it turns out that the six regular
4-polytopes found by Schlafli are related to each other via duality as follows.
Hypetetrahedron <-> Itself
24-cell <-> Itself
Hypercube <-> Hyperoctahedron
Hyperdodecahedron <-> Hypericosahedron
This means that one needs only describe the 24-cell and the hypericosahedron for all of them to be known. In other words:
Hypertetrahedron = convex hull of the canonical basis in 5 dimensions
Hypercube = [0,1]x[0,1]x[0,1]x[0,1]
Hyperoctahedron = dual of the hypercube
Hyperdodecahedron = dual of the hypericosahedron
24-cell ???
Hypericosahedron ???
The description of the last two 4-polytopes above may be obtained by considering the quaternions $\mathbb H$.
Viewing $\mathbb R^3$ within $\mathbb H$ via the map
$$(x,y,z)\mapsto xi+yj+zk, $$
it is well known that every quaternion
$q$, with $\Vert q\Vert=1$, gives a rotation $R_q$ on $\mathbb R^3$ via the formula
$$
R_q(v) = qvq^{-1}, \quad \forall v \in \mathbb R^3.
$$
In fact the correspondence $q\mapsto R_q$ is a two-fold covering of $SO(3)$ by the unit sphere in $\mathbb H$.
Letting $P_{20}$ be the icosahedron in $\mathbb R^3$, consider the quaternionic symmetries of $P_{20}$, which I will write as
$\mathbb {HS}(P_{20})$, defined to be the
set of all unit quaternions $q$ such that $R_q$ leaves $P_{20}$
invariant. In symbols
$$
\mathbb {HS}(P_{20}) =\{q\in \mathbb H: \Vert q\Vert=1,\ R_q(P_{20})=P_{20}\}.
$$
Well, the convex hull of $\mathbb {HS}(P_{20})$ in $\mathbb R^4$ turns out to be a model for the hypericosahedron!
Since the symmetries of a regular polyhedron are the same as the symmetries of its dual, it is clear that the symmetries
of $P_{12}$, the dodecahedron, gives nothing new: the convex hull of $\mathbb {HS}(P_{12})$ is just another model for the
hypericosahedron.
Passing to the (self dual) tetrahedron, call it $P_4$, the convex hull of $\mathbb {HS}(P_{4})$ gives a model for the
remaining 4-polytope, namely the 24-cell, completing the description of the six Schlafli's 4-polytopes.
Question: What is the convex hull of the quaternionic symmetries of the 3 dimensional cube?
If I am not mistaken, this 4-polytope has 48 vertexes and 144 edges, so it is not in Schlafli's list and hence cannot be regular.
EDIT: Yes I was mistaken about the number of edges which is in fact 336 according to M. Winter's answer below!
How you came to the number of 144 edges?
I found the vertices numerically, computed the minimum distance among them, and checked how many pairs share that distance. I guess this is consistent with your answer (below).
This is the disphenoidal 288-cell, which is the dual of the bitruncated 24-cell.
This is also mentioned in the "Geometry" section of the Wikipedia article on the 288-cell.
It has 48 vertices, and 336 edges. However, 144 of these are of the shortest length, and I suppose you have counted these.
The symmetry group of the tetrahedron is "the half" of the symmetry group of the cube (as the tetrahedron is the 3-dimensional demicube).
You already know that the tetrahedral symmetries give you the 24-cell.
In the same way, the 24-cell is "the half" of the disphenoidal 288-cell: the latter is the convex hull of the union of a 24-cell and its dual (which is a 24-cell as well, but differently oriented).
Thanks for your quick and thorough answer! What puzzled me in the first place is that, since the cube is so regular, why wouldn't its quaternionic symmetries also give a regular 4-polytope?
@Ruy To be honest, I am more amazed by the fact that the other two actually give regular polytopes. At the moment, I see no a priori reason for why the resulting polytope should have such an exceptionally high symmetry. On the other hand, in low-dimensional geometry many coincidences happen.
Explicitly, the vertices are: eight vertices with one coordinate $\pm1$ and all others zero; 24 vertices with two coordinates $\pm1/\sqrt2$ and two others zero; and sixteen vertices $(\pm1/2,\pm1/2,\pm1/2,\pm1/2)$
If $G$ is a non-cyclic subgroup of the quaternions of order $n$ containing $-1$, the action of $G\times G$ on the set $G$ has kernel of order 2. The map $g\mapsto g^{-1}$ also permutes the set $G$. These together give $n^2$ symmetries for the polytope defined as the convex hull of $G$. A polytope is regular if and only if it has the same number of full flags as symmetries. In the case when $G$ is the group of symmetries of the cube, there are too many flags for it to be regular despite having $2304=48^2$ symmetries.
|
2025-03-21T14:48:29.677733
| 2020-01-27T15:38:12 |
351258
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Abdelmalek Abdesselam",
"Alexander Schmeding",
"Tom Copeland",
"https://mathoverflow.net/users/12178",
"https://mathoverflow.net/users/46510",
"https://mathoverflow.net/users/7410"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625681",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351258"
}
|
Stack Exchange
|
Update on "Hopf algebras: their status and pervasiveness" by Hazewinkel
Hazewinkel wrote this article in 2005. Perhaps it's time for an update.
For example, updating item
34: Ordinary differential equations much work has been done on the underlying Hopf algebra (HA) of Lie-Butcher numerical methods for solving autonomous and non-autonomous ODE's (evolution equations) by , e.g., Ebrahimi-Fard, Hans Munthe-Kaas and their colleagues. See, e.g.,
i) "Butcher series: A story of rooted trees and numerical methods for evolution equations" by McLachlan, Modin, Munthe-Kaas, and Verdier
ii) "Lie-Butcher series, Geometry, Algebra and Computation" by Munthe-Kaas and K. Føllesdal
This is closely allied to item
81: Quantum theory
through the HA (combinatorial Faa di Bruno bialgebra, trees, Feynman diagrams) of renormalization to which Alain Connes, Christian Brouder, David Broadhurst, Dirk Kreimer, Kurush Ebrahimi-Fard, Hector Figueroa, Jose Gracia-Bondia, Hans Munthe-Kass, Loic Foissy, Karen Yeats, Paul-Hermann Balduf, et al. have contributed.
52: Convex polytopes (relabeled)
Hopf monoids have been introduced to explain the association of permutohedra and associahedra with compositional and multiplicative inversion and optimization:
"Hopf monoids and generalized permutahedra" by
Marcelo Aguiar and Federico Ardila.
It would be motivational and useful in pursuing research in these topics if others would list some of their favorite interests under appropriate items and give associated authors and/or papers, particularly of an introductory nature.
Please feel free to note your own work (with no false modesty--if you are taking the time and effort to publish, you must feel it could be of interest to others, otherwise ...).
For 81, see https://mathoverflow.net/questions/88184/what-is-the-significance-of-non-commutative-geometry-in-mathematics/350985#350985
60: Probability theory and stochastic processes
Hopf algebras feature in the work of Hairer and collaborators on the applications of the theory of regular structures to the solution of general singular SPDEs. See in particular:
"Algebraic renormalisation of regularity structures", by Bruned, Hairer and Zambotti in Inventiones 2019.
"An analytic BPHZ theorem for regularity structures", by Chandra and Hairer.
ah beat me to it, but good references
Related "Operads of (noncrossing) partitions, interacting bialgebras, and moment-cumulant relations" by Ebrahimi-Fard, Foissy, Joachim Kock, and Frédéric Patras. https://arxiv.org/abs/1907.01190
I think Ebrahimi-Fard, Patras and Speicher are now also investigating Hopf algebraic relations in free probability: arXiv:2001.03808: Wick polynomials in non-commutative probability
@AlexanderSchmeding: +1 Yes. Hazewinkel in his 2004 article already mentioned this type of applications to free probability, combinatorics of cumulants etc. The applications to regularity structures are new, of course, because they were introduced by Hairer in 2013.
60: Probability theory,... There is even more movement in the stochastic PDE community with Hairer,s regularity structures. The groups appearing there are character groups of certain Hopf algebras and there is the massive work by
Bruned, Hairer and Zambotti to uncover the algebraic framework, leading them to certain Hopf algebras:
Algebraic renormalisatiom of regularity structures arxiv:1610.08468.
Also the whole topic of rough paths (Now Msc2020 60Lxx) is quite connected, as rough paths can be viewed as paths again in character groups of certain Hopf algebras. See e.g.
What does the group action of a rough path in a Lie group look like?
This can be found In most modern treatments in the guise of dealing with the tensor algebra and shuffle products. Some modern more algebraic developments are also included in the works of Ebrahimi-Fard, Manchon et al.
22Exx
Infinite-dimensional Lie theory, turns out that character groups of Hopf algebras are often infinite-dimensional Lie groups (this provides Lie group structures for many well-known examples, such as the Butcher group from numerical analysis):
Character groups of Hopf algebras as infinite-dimensional Lie groups, Ann. Inst. Fourier (Grenoble), 66 no. 5 (2016), p. 2101-2155 arXiv:1501.05221
Lie groups of controlled characters of combinatorial Hopf algebras, arXiv:1609.02044
The geometry of characters of Hopf algebras, Computation and Combinatorics in Dynamics, Stochastics and Control. Abelsymposium 2016. Abel Symposia, vol 13 DOI: 10.1007/978-3-030-01593-0_6, arXiv:1704.01099
Just came across this association in the last few days since it's related to noncrossing partitions and thence to compositional and multiplicative inversions and free probability.
Since compositional inversion is related to vector fields and, therefore, to Lie algebras and groups, solns. to diff eqs, and trees, I always suspect there is a HA with its antipode and dual lurking around somewhere.
5. Combinatorics
Tree hook length formulas can be specified via a Hopf algebra.
A Unifying Approach for Proving Hook-Length Formulas for Weighted Tree Families, Graphs and Combinatorics, 29 no. 6, (2013) p 1839–1865.
Tree hook length formulae, Feynman rules and B-series, Annales de l’Institut Henri Poincaré D, 2 no. 4 (2015), p. 413–430. arXiv:1412.6053.
Though I admit hook length formulas are not that interesting and may already fit under Hazewinkel's description of 5. Also the first paper doesn't actually use Hopf algebras and the second alludes to them.
|
2025-03-21T14:48:29.678226
| 2020-01-27T18:01:54 |
351268
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ashot Minasyan",
"Benjamin Steinberg",
"Joshua Grochow",
"Luc Guyot",
"https://mathoverflow.net/users/15934",
"https://mathoverflow.net/users/38434",
"https://mathoverflow.net/users/7644",
"https://mathoverflow.net/users/84349"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625682",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351268"
}
|
Stack Exchange
|
Does every f.g. group have a minimal presentation?
Call a group presentation $\langle X \,\|\,R \rangle$ minimal if no relator from $R$ is a consequence of the remaining relators, i.e., no $r \in R$ belongs to the normal closure of $R\setminus \{r\}$ in the free group $F(X)$.
Question: Does every finitely generated group have a minimal presentation (with $X$ finite)?
Remark: evidently every finite presentation $\langle X \,\|\,R \rangle$ has a minimal sub-presentation $\langle X \,\|\,R' \rangle$ (where $R' \subseteq R$ has the same normal closure in $F(X)$ as $R$), so the question really concerns infinitely presented groups.
It is possible to construct infinite presentations which do not have minimal sub-presentations. Indeed, let $F=F(a,b)$ be the free group on $\{a,b\}$. One can choose "sufficiently independent" elements $w_1,w_2,\dots \in F$ so that the presentation $$\langle a,b \,\|\, w_i^2,w^2_{i+1}w_i, ~i \in \mathbb{N} \rangle$$ has no minimal sub-presentation. Here $R$ consists of the elements $w_1^2,w_2^2,\dots$, and $w_2^2w_1,w_3^2w_2,\dots$.
However, the above group also has the presentation $\langle a,b \,\|\,w_1,w_2,\dots \rangle$, which may be minimal.
Look at https://arxiv.org/abs/1010.0271. I'm sure @YCor can say more.
Dear Ashot, the answer is negative: combine Theorem. 3.5 and Remark 5.3 of the paper cited by Benjamin Steinberg. A counter-example is given by the nilpotent-by-Abelian group $B$ of Equation (3.2).
Dear Benjamin: many thanks for providing the reference, it is spot on!
Dear Luc: nice paper! I should have read it before asking!
@AshotMinasyan, I believe that paper was one of the first talks I heard by YCor, and so I remembered it right away.
@LucGuyot Why not make that an answer?
@JoshuaGrochow This is an answer now.
The answer is no.
In order to see this, you may combine Theorem 3.9 and Remark 5.3 of [1]. A counter-example is given by the nilpotent-by-Abelian group $B$ of Equation (3.2). Further examples are provided by Remark 5.15.
[1] R. Bieri, Y. de Cornulier, L. Guyot and R. Strebel, "Infinite presentability of groups and condensation", 2014.
|
2025-03-21T14:48:29.678394
| 2020-01-27T18:58:00 |
351271
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ali Taghavi",
"Noam D. Elkies",
"https://mathoverflow.net/users/14830",
"https://mathoverflow.net/users/36688"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625683",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351271"
}
|
Stack Exchange
|
A generalization of polynomial algebra on a Riemann surface
Let $M$ be a $1$-dimensional complex manifold. Let $A$ be the space of all holomorphic functions $f:M\to \mathbb{C}$ such that either $f$ is a constant function or every level set $f^{-1}(c)$ is a finite (probably empty) set.
Is $A$ an algebra of functions? Is its closure, with respect to topology of uniform convergence on compact subsets, equal to space of all holomorphic functions from $M$ to $\mathbb{C}$?
Hm, does $A$ contain any nonconstant functions if $M$ is the Riemann surface of $w^2 = \sin z$ (a.k.a. the double cover of $\bf C$ branched on $\pi\bf Z$)?
Counterexample to the first question ("is $A$ an algebra of functions?"):
Let $M$ be a vertical strip such as {$x + iy : 0 < x < 1$},
and define $f_1,f_2$ as the restriction to $M$ of the
entire functions
$$
f_1(z) = \exp((1+i)z), \quad f_2(z) = \exp((1-i)z).
$$
Then $f_1,f_2 \in A$ but $f_1 f_2 \notin A$.
Indeed if $f_1(z) = f_1(z')$ or $f_2(z) = f_2(z')$ then
the real part of $z-z'$ is a multiple of $\pi$, so
all the level sets of $f_1$ and $f_2$ are of bounded size
(indeed size at most $1$ for our width-$1$ strip).
But $f_1 f_2$ is the function $z \mapsto e^{2z}$,
whose level sets are translates of $\pi i \bf Z$
and thus can contain infinitely many points of $M$.
It follows that $A$ is not closed under addition either:
if it were, then $A$ would contain $f_1+f_2$ and $f_1-f_2$,
and since $A$ is closed under squaring this would imply that
$A \ni (f_1 + f_2)^2 - (f_1 - f_2)^2 = 4 f_1 f_2$.
Thank you very much for your very interesting answer
|
2025-03-21T14:48:29.678530
| 2020-01-27T19:44:37 |
351273
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625684",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351273"
}
|
Stack Exchange
|
How to properly verify that $E\times E'$ has no non-trivial effective divisors with Kodaira dimension zero
Let $E$ and $E'$ be elliptic curves over $\mathbb{C}$. I am pretty confident that the only effective divisor $D\subset E\times E'$ with Kodaira dimension zero is the trivial divisor.
How to prove this in a simple manner?
If $D$ has Iitaka dimension zero, then $\dim H^0 ( n E) =1$ for all $n$, because if it were any larger than $\dim H^0(k ne) \geq k+1$.
If we have an automorphism $\sigma$ with $[ n \sigma (D)] = [n D]$ in the divisor class group, we would have two linearly independent sections unless in fact $\sigma(D)= D$.
For $\sigma$ translation by an $n$-torsion point, for any $D$, $\sigma(D)- D$ is $n$-torsion, and so $[n \sigma(D) ] =[nD]$. (To see this, one can use the exponential exact sequence $H^1( A, \mathcal O_X) \to H^1(A, \mathcal O_X^\times) \to H^2(A, \mathbb Z)$. Translation acts trivially on the first and last terms, meaning that $\sigma (\sigma(D)-D) =\sigma(D)-D$ and thus $\sigma^n=1$ implies $\sigma(D)- D$ is $n$-torsion.)
So to have Kodaira dimension $0$, $D$ would have to be invariant under translation by $n$-torsion for all $n$. Thus, it must be either empty or Zariski dense, and therefore it is empty.
|
2025-03-21T14:48:29.678646
| 2020-01-27T20:14:26 |
351278
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Bertram Arnold",
"SUDEEP PODDER",
"https://mathoverflow.net/users/151571",
"https://mathoverflow.net/users/35687",
"https://mathoverflow.net/users/43326",
"user43326"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625685",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351278"
}
|
Stack Exchange
|
Chern classes of complex vector bundle
I'm reading characteristic classes form the book Differential forms in Algebraic Topology by Bott and Tu. The Chern classes are defined as follows:
$E\xrightarrow{\rho} M$ is a vector bundle and $E_p$ be the fiber over $p$. Then $P(E)$, the projectivization of $E$ is a vector bundle with fiber $P(E_p)\colon=\{\text{1-dim subspaces of $E_p$}\}$ over $\ell_p\in P(E)$. It's then discussed that the first Chern class $x$ of the dual of the universal subbundle over $P(E)$ restricted to a fiber is the first Chern class of the fiber. Therefore the $1,x,\ldots,x^{n-1}$ are global classes on $P(E)$, whose restriction to each fiber $P(E)$ freely generates the cohomology of the fiber and hence by Leray-Hirsch $H^*(P(E))$ is a free module over $H^*(M)$ with basis $\{1,x,\ldots,x^{n-1}\}$ and then the Chern classes are the coefficients in the expression for $x^n$.
Also by Leray-Hirsch, $H^*(P(E))\cong H^*(M)\otimes H^*(Fiber)$. The fiber $P(E_p)$ has cohomology $\frac{\mathbb{R}[x]}{x^n}$, as $E_p$ is a vector space of complex dimension $n$. My question is, why $x^n\neq 0$.
Note: for the trivial bundle $M\times V$, $P(E)=M\times P(V)$ and by Künneth formula $H^*(P(E))\cong H^*(M)\otimes H^*(P(V))$ and then $x^n=0$ as $H^*(P(V))=\frac{\mathbb{R}[x]}{x^n}$. What is the difference in the general case and the trivial bundle?
The short answer is that Leray-Hirsch doesn't say (much) about the product structure on cohomology. It might help to work out a sufficiently generic example, eg take $E_n\to Gr_n(\mathbb C^{n+1})$ the tautological bundle over the Grassmanian; then $P(E_n)$ is the set of flags $L\subset E\subset \mathbb C^{n+1}$ , and $x$ is dual to the (real) codimension $2$ submanifold where $L\subset V$ for some codimension $1$ complex subspace $V$. Then $x^n$ is dual to the submanifold where $L$ is some fixed line, which is the inclusion of $Gr_{n-1}(\mathbb C^n)$ and thus not zero in homology.
(I initially considered the general Grassmannian $Gr_k(\mathbb C^n)$ for general $k\neq n$; for $k +1 = n$, this is of course just $\mathbb{CP}^n$, with the tautological bundle the orthogonal complement of the tautological line bundle, so that its total Chern class is $(1+x)^{-1}$ as predicted by this calculation.)
Basically what you are saying that for trivial bundle, because of the Kunneth formula, the chern class is 0. In other words the Chern class is the first obstruction to the trivialization of the bundle.
I realized that Leray-Hirsch is about the additive structure, thanks @BertramArnold .
Bertram already mentioned this in the comments but I thought I'd write an answer for completness's sake.
The Leray-Hirsch theorem says that $H^{*}P(E)\cong H^{*}M\otimes H^{*}(Fiber)\ \ $ $\textit{as $H^{*}M$ modules}$.
So if $x$ is the first chern class of the tautological line bundle over $P(E)$, there's no reason to expect $x^{n}=0$, even if its pullback to the fiber satisfies that equation. Because if the fiber "twists around," there's no reason to expect that the behavior of a class on one fiber is indicative of its behavior globally.
On the other hand, when the bundle is trivial, the Kunneth formula (in this setting) tells you that $H^{*}P(E)\cong H^{*}M\otimes H^{*}(Fiber)\ \ $ $\textit{as rings}$.
Here's another fun example to illustrate the difference, where you can even draw pictures! You'll have to use cohomology over $\mathbb{Z}/2$ and you'll be calculating Stiefel-Whitney classes instead of chern classes, but the procedure is exactly the same: Leray-Hirsch, read off coefficients from a basis. Form $E\rightarrow S^1$ by taking the Mobius strip as a vector bundle over $S^1$ plus a copy of the trivial line bundle. Then $P(E)$ will be a familiar non-orientable surface (Klein bottle) whose cohomology ring you can compute from a delta-decomposition. Comparing it to the cohomology ring of the torus (which would be $P(E)$ for the trivial rank 2 bundle) where every positive degree class squares to zero, you can see exactly how the "twist" in the Klein bottle allows one of those classes to no longer square to zero!
|
2025-03-21T14:48:29.678961
| 2020-01-27T20:34:38 |
351280
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jason Starr",
"Jürgen Böhm",
"Mohan",
"R. van Dobben de Bruyn",
"Richard Thomas",
"Will Sawin",
"https://mathoverflow.net/users/13265",
"https://mathoverflow.net/users/18060",
"https://mathoverflow.net/users/21940",
"https://mathoverflow.net/users/7653",
"https://mathoverflow.net/users/82179",
"https://mathoverflow.net/users/9502"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625686",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351280"
}
|
Stack Exchange
|
Local versus global embedding dimension
Fix a complex projective scheme $X$ and a closed point $x\in X$.
Let $d_x$ denote the dimension of the Zariski tangent space at $x$.
This is the local embedding dimension of $X$ at $x$ -- the minimal dimension of a smooth scheme containing an open neighbourhood of $x$.
In a paper I blithely asserted that $d(X):=\max_{x\in X}d_x$ is the global embedding dimension -- the minimal dimension of a smooth scheme containing $X$.
I had assumed we could embed $X\subset\mathbb P^N$ and then take an intersection of $N-d(X)$ generic sufficiently positive hypersurfaces containing $X$.
But since a troublesome referee has quite unreasonably asked me for a proof (I'm joking) I checked more carefully and saw this construction does not work everywhere at once (even when $N-d(X)=1$ and $X$ is smooth!).
Can anyone suggest another construction, maybe by suitable projections, or a reference, or...?
For a reduced $k$-scheme $X$ of pure dimension $n$ and local embedding dimension $n+1$, one obstruction to finding a global embedding into a smooth $(n+1)$-fold with invertible conormal sheaf $\mathcal{I}$ is the connecting map $$H^0(X,\mathcal{Ext}^1_{\mathcal{O}X}(\Omega{X/k},\mathcal{I})) \to H^2(X,\mathcal{Hom}_{\mathcal{O}X}(\Omega{X/k},\mathcal{I})).$$ If there exists an embedding, the kernel has a section that everywhere locally generates the domain $\mathcal{O}_X$-module (which is locally principal).
Wonderful, thanks Jason. At first I thought we could kill this obstruction by choosing $\mathcal I$ to be a sufficiently positive line bundle, so the $H^2$ vanishes. But then the global sections of $\mathcal Ext^1(\Omega_X,\mathcal I)$ would have zeros and so not generate (assuming the singular locus has dimension $\ge1$). So I think you've probably give a method to debunk my claim. I.e. using this one should be able to find an example of a variety with at worst hypersurface singularities which is nonetheless not a hypersurface (in something smooth)?
I am trying to work this out explicitly for a threefold obtained from a $\mathbb{P}^1$-bundle over a surface where a bisection is glued to itself. There is a class in $H^2$ of the bisection that I need to compute . . .
@Jason Starr: Your formula above (from the local-global-ext-ss) and the argument looks like it is part of a general theory of embeddings of algebraic varieties into each other. What would be a good article or book to look for, where this is explained in more detail? (I tried with Google already, but found nothing especially pertaining).
@JürgenBöhm: Jason is describing the Kodaira-Spencer class that classifies first order thickenings of $X$ by a sheaf $\mathcal I$ (that becomes the square-zero ideal of $X$ inside the thickening). So he's describing the (obstruction to the existence of the) first order part of the ambient space I was after. "Kodaira-Spencer class" is probably the thing to google.
I had a different strategy in mind, but was unable to carry it out completely. I asked a follow-up question about this.
For an affine variety a precise bound was given by V. Srinivas in his paper in math.ann in 1991.
It seems there is a counterexample. This is based on Jason Starr's suggestion in the comments.
If we have a surface $S$ with two smooth disjoint curves $C_1$ and $C_2$, which are isomorphic, and let $X$ be obtained by gluing $C_1$ and $C_2$ along that isomorphism $i: C_1\to C_2$, then $X$ is projective if there is an ample line bundle on $X$ whose restrictions to $C_1$ and $C_2$ are equal (under $i$).
$X$ has singularities locally isomorphic to a nodal curve cross a smooth curve, thus has local embedding dimension $3$. Can $X$ be embedded as a hypersurface in a smooth $3$-fold? If so, then by (part of) Jason Starr's obstruction, the sheaf
$$\mathcal{Ext}^1_{\mathcal O_X} (\Omega_{X/k}, \mathcal I)= \mathcal{Ext}^1_{\mathcal O_X} (\Omega_{X/k}, \mathcal O_X) \otimes \mathcal I$$
must be globally generated, where $\mathcal I$ is the conormal line bundle. This sheaf is clearly supported on the glued curve $C$, and we can calculate that it is isomorphic to $\mathcal I $ tensored with the normal bundle of $C_1$ and the normal bundle of $C_2$ there. (It suffices to work, carefully, locally in $k[x,y]/xy$, where $\Omega$ is generated by $dx$ and $dy$ with relation $xdy+ ydx=0$ and the generator of the $\mathcal{Ext}^1$ is precisely the linear map that sends $xdy+ydx$ to $1$, which the automorphism group acts on the same way it acts on the tensor rpoduct of the normal bundles.)
So for this sheaf to have a nonvanishing section, the conormal bundle $\mathcal I$ of $X$, restricted to $C$ must be isomorphic to the tensor product of the conormal bundle of $C_1$ to the conormal bundle of $C_2$.
So here's what we're going to do. We will take $E_1$ and $E_2$ two distinct, but isomorphic, elliptic curves in $\mathbb P^1$. In fact, we will take them to be two isomorphic curves appearing in the Dwork family, so their intersection points will be $3$-torsion. We will blow up all $9$ intersection points, plus two points $P_1, Q_1$ on $E_1$ and two points $P_2, Q_2$ on $E_2$. We choose $P_1, Q_1, P_2, Q_2$ very general, subject to the condition that $i(P_1) + 2i(Q_1) = P_2 + 2 Q_2$ in the group law on $E_2$.
To make our ample class, we'll just take a sufficiently high multiple of the hyperplane class, minus the sum of the exceptional divisors at all $9$ intersection points, minus the exceptional divisors over $P_1$ and $P_2$, minus twice the exceptional divisors over $Q_1$ and $Q_2$. Because of our assumption on the group law, this restricts to the same line bundle on $E_1$ and $E_2$, as each exceptional divisor corresponds to that point in the Picard group.
However, the Picard class of the tensor product of the two conormal bundles on $E_2$ will be some multiple of the hyperplane class, plus twice the sum of all the $3$-torsion points, plus $i(P_1) + i(Q_1) + P_2 + Q_2$. If this class comes from a global line bundle, then it must come from a sum of hyperplane classes and exceptional divisors, which means (projecting to Pic) it must come from a sum of $3$-torsion points, $P_2$ and $Q_2$. The exceptional divisors over $P_1$ and $Q_1$ don't contribute because they don't intersect $E_2$. Thus, it can only happen if we have some relation that $i(P_1) + i(Q_2) = a P_2 + b Q_2$ for $a,b\in \mathbb Z$, up to $3$-torsion. But there are countably many such relations, and none of them is forced by our condition on $P_1,P_2, Q_1,Q_2$, so none of them will hold for our very general choice.
Magnificent, thanks Will and Jason. Can I summarise the overall method as follows? Make $X$ by gluing a smooth surface $S$ along two disjoint curves $C_1\cong C_2$. If $X$ lived in a smooth 3-fold $Y$, we know by elementary geometry that the normal bundle $N_{X/Y}$ must restrict on $C_1\cong C_2$ to the line bundle $L:=N_{C_1/S}\otimes N_{C_2/S}$. But you found an example where $L$ is not the restriction of any line bundle on $X$ (though of course it would be on its normalisation $\overline{X}=S$).
@RichardThomas Yes, with the caveat that the line bundle that is that restriction on the normalization is not the tensor product of $N_{C_1/S}$ and $N_{C_2/S}$, but rather the line bundle that is $N_{C_1/S}$ on $C_1$ and $N_{C_2/S}$ on $C_2$.
I wonder if the following example, adapted from section 18 of Kollár's Links of complex analytic singularities, would also work: Let $E_i := V(x_i^3 + y_i^3 + z_i^3) \subset \mathbb{P}_i^2$ for $i = 1, 2$, let $\tau: E_1 \to E_2$ be an isomorphism corresponding to a translation of the elliptic curve $V(x^3 + y^3 + z^3)$ and use it to glue the 2 copies of $\mathbb{P}^2$, to get $X(\tau) := \mathbb{P}_1^2 \cup_{\tau} \mathbb{P}_2^2$. Let $E \subset X$ denote the common image of $E_1, E_2$.
Then using $\mathcal{O}_{\mathbb{P}_1^2}(1)|_{E}$ as a basepoint, we can make the identification $\mathrm{Pic}^3(E) \simeq \mathrm{Pic}^0(E) \simeq E$. Under this identification $\mathcal{O}_{\mathbb{P}_2^2}(1)|_{E} = \tau^*(\mathcal{O}_{\mathbb{P}_1^2}(1)|_{E} = \tau \in \mathrm{Pic}^3(E)$, and more generally $\mathcal{O}_{\mathbb{P}_2^2}(d)|_{E} = \tau^d \in \mathrm{Pic}^{3d}(E)$ for $d \in \mathbb{Z}$. So $X(\tau)$ is projective if and only if $\tau$ is torsion, in which case $\mathcal{O}_{\mathbb{P}_1^2}(d), \mathcal{O}_{\mathbb{P}_2^2}(d)$ glue to form a line bundle on $X(\tau)$ if and only if $\tau^d = 1$.
On the other hand, $N_{E \subset \mathbb{P}_i^2}^\vee = \mathcal{O}_{\mathbb{P}_i^2}(-3)|_E$ for $i = 1, 2$ so that $N_{E \subset \mathbb{P}_1^2}^\vee \otimes N_{E \subset \mathbb{P}_2^2}^\vee = \mathcal{O}_{\mathbb{P}_1^2}(-3)|_E \otimes \mathcal{O}_{\mathbb{P}_2^2}(-3)|_E$, corresponding to $\tau^{-3}\in \mathrm{Pic}^{-18}(E)\simeq E$. Hence $X(\tau)$ is an snc divisor if and only if $\tau^3=1$.
Same idea but simpler, I like it, thanks.
|
2025-03-21T14:48:29.679504
| 2020-01-27T21:20:56 |
351284
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625687",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351284"
}
|
Stack Exchange
|
Inequalities involving Gamma function
I am having some difficulty in proving the following inequality:
\begin{equation*}
\frac{1-e^{-\gamma b}}{b^\eta}-\frac{1-e^{-\gamma s}}{s^\eta}\geq \gamma(1-\eta)\int^b_sy^{-\eta}e^{-\gamma y}dy
\end{equation*}
where $\gamma,\ \eta\in(0,1)$, $0\leq s\leq b$ and $b$ satisties:
\begin{equation*}
\frac{(1-e^{-\gamma b})\eta}{e^{-\gamma b}}=(1-\eta)\gamma b
\end{equation*}
Any suggestions on how to approach this would be appreciated. Any suggested literature on inequalities involving the gamma function would also be really appreciated. Thanks!
Let $D(b)$ denote the difference between the left-hand side and the right-hand side of your displayed inequality. Then $D(s)=0$, and the inequality $D'(b)\le0$ can be easily shown to be equivalent to $e^{\gamma b}\ge1+\gamma b$, which latter is true. So, $D(b)\le0$ if $b\ge s\ge0$; that is, the opposite to your proposed inequality holds.
|
2025-03-21T14:48:29.679606
| 2020-01-27T21:26:23 |
351286
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alex Kruckman",
"Frode Alfson Bjørdal",
"Noah Schweber",
"Todd Trimble",
"Wojowu",
"Zuhair Al-Johar",
"https://mathoverflow.net/users/2126",
"https://mathoverflow.net/users/2926",
"https://mathoverflow.net/users/30186",
"https://mathoverflow.net/users/37385",
"https://mathoverflow.net/users/8133",
"https://mathoverflow.net/users/95347"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625688",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351286"
}
|
Stack Exchange
|
Weak power set - what strength may it have?
In The Consistency of Classical Set Theory Relative to a Set Theory with Intuitionistic Logic in THE JOURNAL OF SYMBOLIC LOGIC Volume 38, Number 2, June 1973 page 316 Harvey Friedman's axiom 8* $Weak \ Power \ Set$ is:
$(\forall a)(\exists x)(\forall y)(\exists z\in x)(z=y\cap a)$
What do we know about weak power set? I am curious about what if anything weak power set can do when added to $KP\omega$ or $ZF-Power \ Set$.
what does $E$ mean? Possibly you mean $\exists$
I believe in presence of extensionality and some weak separation ($KP\omega$ should have enough) this axiom is equivalent to full power set.
@ZuhairAl-Johar Yes, I edited
@Wojowu It would be interesting if that could be whown
@ I think if you add this to ZF - Power, then you recover full ZF.
@ZuhairAl-Johar OK
Of course I'm speaking about classical ZF. But with intuitionistic logic, I don't know.
@ZuhairAl-Johar Friedman showed that a weaker intuitionistic theory interprets , but that is a different situation. I don't at this point believe that can be recovered in the sense that the power set axiom is a theorem of −+; but I may change my mind.
I’m voting to close this question on request from the OP.
The asserted set $x$ is just a set that contains all overlap sets between the set $a$ and any set, among its elements. Weak Power as written above is simply:
$(\forall a)(\exists x)(\forall y)(a \cap y\in x) $
Now in classical ZF all subsets of $a$ are overlaps with $a$, i.e. $z \subseteq a \to z \cap a=z$, so all of them would be included in the weak power of $a$ (just substitute each subset $z$ of $a$ instead of $y$ in the above formula), then by separation one can easily recover full $P(a)$ by separating on the weak power of $a$ using the property of being a subset of $a$.
@FrodeAlfsonBjørdal Friedman in the cited article seems to work in set theory without extensionality. Without it it makes no sense to write $a\cap y$, as there may not be a unique set satisfying its defining property. If you wish to also work in a theory without extensionality (which I'm not sure is the case), you should say so explicitly in your question.
@Wojowu Extensionality is part of $KP\omega$ and $ZF-Power+Weak Power$, so I already made clear that I presuppose extensionality in the cases I expressed interest in.
@FrodeAlfsonBjørdal In that case, what is unconvincing about this answer?
@NoahSchweber Perhaps it is not unconvincing. It is getting late here, so let me see what I think tomorrow.
@FrodeAlfsonBjørdal To clarify, what logic are you working in - e.g. does "$KP\omega$" mean $KP\omega$ in classical or intuitionsitc logic?
@NoahSchweber I presuppose classical logic
@Zuhair Al-Johar There are some problems with your rendering of the Weak Power Set Axiom.
@FrodeAlfsonBjørdal What problems? The sentence $(\forall a)(\exists x)(\forall y)(\exists z\in x)(z = y\cap a)$ is tautologically equivalent to the sentence $(\forall a)(\exists x)(\forall y) (y\cap a\in x)$.
@Alex Kruckman Sorry, you are right.
@Zuhair Al-Johar Thanks, I am convinced.
|
2025-03-21T14:48:29.680207
| 2020-01-27T21:36:31 |
351287
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Arthut",
"https://mathoverflow.net/users/151471"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625689",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351287"
}
|
Stack Exchange
|
Density of integers related to the size of its order of appearance in the Fibonacci sequence
Let $z(n)=\min\{k>0 : n\mid F_k\}$. This function is known as the Fibonacci entry point (for example). A result of Sallé gives the sharpest upper bound for $z(n)$, namely, $z(n)\leq 2n$, for all $n$ (the equality holds for all $n=6\cdot 5^k$). Also, it is well-known that $\liminf_{n\to \infty}z(n)/n=0$.
By using the Mathematica software, I am almost convinced that $z(n)$ is not too small, in density. For example, I do believe that the set
$$
A:=\{n\geq 1: z(n)>n/4\}
$$
has positive upper density, i.e.,
$$
\lim_{x\to \infty}\sup\displaystyle\frac{|A\cap [1,x]|}{x}>0.
$$
However, I was not able to prove that. Someone can give me some suggestion?
Thanks in advance.
@MaxAlekseyev Yes!
Thank you Gerry Myerson for bringing this problem to WCNT 2021: Problems in Number Theory
The answer to this question is No as we can prove that the density of $A$ is zero. Simon Rubinstein-Salzedo outlined a solution with one step to be confirmed. Then I filled in the proof of the one step.
Theorem $A$ has density 0. (Simon Rubinstein-Salzedo)
The argument relies on the following known facts:
Fact The Fibonacci numbers form a strong divisibility sequence, i.e. $\gcd(F_m,F_n)=F_{\gcd(m,n)}$. It follows that $z(\text{lcm}(m,n))=\text{lcm}(z(m),z(n))$.
Fact If $p$ is a prime, then $z(p)\mid p-\left(\frac{p}{5}\right)$.
Thus if $p_1.\ldots,p_k$ are distinct primes different from 2 and 5, then $\frac{z(p_1\cdots p_k)}{p_1\cdots p_k}\le\frac{1}{2^{k-1}}\prod_{i=1}^k \left(1+\frac{1}{p_i}\right)$, and this tends to 0 as $k\to\infty$. Let's write
$$
m(k)=\frac{1}{2^{k-1}} \prod_{i=1}^k \left(1+\frac{1}{p_i}\right),
$$
where the product is over the first $k$ odd primes other than 5.
For an integer $n$, let
$$
r(n)=\prod_{\substack{p\text{ prime} \\ p\neq 2,5 \\ p\mid n \\ p^2\nmid n}} p
$$
and $s(n)=\frac{n}{r(n)}$. I believe (to be confirmed) that for any $k$,
$$
\lim_{x\to\infty} \frac{\#\{n:1\le n\le x, \omega(r(n))\ge k\}}{x}=1.
$$
For any $n$ with $\omega(r(n))\ge k$, we have
$$
\frac{z(n)}{n}\le \frac{z(r(n))}{r(n)}\cdot\frac{z(s(n))}{s(n)}\le m(k)\cdot 2,
$$
which tends to 0 as $k\to\infty$. Thus the upper density of $A$ is 0.
However, this says nothing about what happens if we restrict to Fibonacci entry points of primes.
Confirming the Density 1 result (Sungjin Kim)
The constant $C>0$ may appear several times, not necessarily the same everytime.
Let $P_k$ be the set of positive integers with $< k$ distinct prime factors. Let $A_k(t)=\sum_{n\le t, n\in P_k} 1$ be the counting function of $P_k$. By Hardy-Ramanujan, we have an estimate
$$
A_k(t) \le C \Psi_k(t):=C\frac{t(\log\log (t+C)+C)^{k-2}}{\log (t+C)}.
$$
The numbers satisfying $\omega(r(n))<k$ can be decomposed as
$$
n=my
$$
with $m=2^{\nu_2(n)}5^{\nu_5(n)}r(n)$ so that $\omega(m)<k+2$ and $y$ is power-full, that is, $p|y \Rightarrow p^2|y$.
Let $\mathcal{F}$ be the set of power-full numbers. The estimate of the number of power-full numbers is obtained by Bateman and Grossward in 1958 (as a stronger form than below),
$$
\sum_{y\le x, y\in\mathcal{F}}1 \le C \sqrt x.
$$
Combining these to estimate the numbers $n\leq x$ with $\omega(r(n))<k$,
$$
\leq C\sum_{m\leq x, m\in P_{k+2}} \sum_{y\leq \frac xm, y\in\mathcal{F}} 1 \leq C\sum_{m\leq x, m\in P_{k+2}} \sqrt{\frac xm}.
$$
Applying the partial summations to the last sum,
$$
\sum_{m\leq x, m\in P_{k+2}} \sqrt{\frac xm} \leq C\Psi_{k+2}(x) +C\sqrt x
\int_{1}^x \frac{\Psi_{k+2}(t)}{t\sqrt t} dt\leq C\Psi_{k+2}(x).
$$
Hence,
$$
\sum_{n\leq x, \omega(r(n))<k}1\leq C\Psi_{k+2}(x).
$$
|
2025-03-21T14:48:29.680459
| 2020-01-27T22:18:46 |
351290
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dan Christensen",
"Ivan Di Liberti",
"Jiří Rosický",
"Martin Brandenburg",
"Yang",
"https://mathoverflow.net/users/104432",
"https://mathoverflow.net/users/2841",
"https://mathoverflow.net/users/529138",
"https://mathoverflow.net/users/73388",
"https://mathoverflow.net/users/952"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625690",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351290"
}
|
Stack Exchange
|
functors $\text{Vect} \to \text{Vect}$ that preserve filtered and sifted colimits
I'm considering various functors from the category $\text{Vect}$ of real vector spaces to itself, and would like to know that they preserve filtered colimits and possibly even sifted colimits. The functors I'm interested in send $V$ to the power $V^k$, the tensor power $V^{\otimes k}$, the exterior power $\Lambda^k(V)$, and the free vector space $F(V)$ on the underlying set of $V$. I have proved by hand that some of these preserve filtered colimits or sifted colimits, but I'm looking for references and/or conceptual arguments.
Of course, I'd also like to know if some of these don't preserve filtered and/or sifted colimits.
(I doubt it matters that I'm working over the reals.)
I suggest to try and apply the equivalent characterizations of finitary functors provided by Adamek et al. in "On finitary functors": http://www.tac.mta.ca/tac/volumes/34/35/34-35abs.html.
Thm. 3.12 and 3.13 in the abovementioned paper apply to your situation, as explained in 3.19. Also, the criterion of boundedness is extremely easy to check in your case.
@IvanDiLiberti Thanks, those (recent!) theorems do give a straightforward way to check that the functors on my list preserve filtered colimits. I'd still be interested in a proof that requires less technology, and in whether those functors also preserve sifted colimits.
In the case of sifted colimits, recall that by Thm. 3.1 in "What are Sifted Colimits?" by the usual suspects show that an endofunctor of a locally presentable category preserves sifted colimits iff it preserves directed colimits and reflexive coequalizers.
Since you are looking for references, maybe the reference-tag would be suitable. I also know that these facts are true, but always wonder about references ...
Following 3.3 in https://arxiv.org/pdf/1207.2732.pdf, any functor ${\bf Vect}\to{\bf Vect}$ preserving filtered colimits preserves sifted colimits.
@MartinBrandenburg I added the reference-request tag. Thanks for pointing it out.
@JiříRosický That proposition exactly handles the remaining part of my question. Thanks! For others reading along, the cited paper is "Strongly complete logics for coalgebras" by A. Kurz and J. Rosický.
@MartinBrandenburg, is this true for modules over a commutative ring? By the way, I see that the functors mentioned here not include the symmetric power, which didn't preserve even finite colimit as shown in this answer.
Maybe the following tools can help.
$G: \mathcal{C}_1 \times \cdots\times \mathcal{C}_k\to \mathcal{D}$ preserves sifted colimits separately in each variable if and only if it preserves sifted colimits. Indeed, given a diagram $p: K \to \mathcal{C}_1\times\cdots \times \mathcal{C}_k$ with $K$ sifted, observe that $p= (q_i)\circ \delta$ where $(q_i): \prod_{i=1}^kK \to \prod_{i=1}^k\mathcal{C}_i$ is the product of $q_i=\mathrm{proj}_i\circ p$ and $\delta$ is the diagonal. Since $K$ is sifted, $\delta$ is final, and the colimit may be computed over $K^{\times k}$ instead of over $K$. But then we can apply the hypothesis $k$ times to see that this colimit is preserves by $G$.
The diagonal map $\mathcal{C} \to \mathcal{C}^{\times j}$ preserves all colimits.
The forgetful functor $\mathbf{Vect} \to \mathbf{Set}$ preserves all sifted colimits (since it preserves filtered colimits and reflexive coequalizers by inspection).
The free vector space functor preserves all colimits, being a left adjoint.
Combining (4) and (3) gives the "F(V)" example you wanted, so let's move on to the others.
Notice that the tensor power functor $V \mapsto V^{\otimes n}$ is a composite of the diagonal and the tensor product functor, which preserves all colimits separately in each variable, so it preserves sifted colimits. Moreover, there is an action of $\Sigma_n$ on $V^{\otimes n}$ so this functor refines to $\mathbf{Vect} \to \mathbf{Fun}(\mathrm{B}\Sigma_n, \mathbf{Vect})$. This also preserves sifted colimits since colimits in functor categories are computed pointwise.
It's also true that the product $V \mapsto V^{\times n}$, as a functor to vector spaces with a $\Sigma_n$ action, preserves sifted colimits, since the cartesian product of sets preserves all colimits of sets in each variable, and sifted colimits of vector spaces may as well be computed on the underlying set, so the same reasoning applies (even though the cartesian product functor does not preserve all colimits of vector spaces separately in each variable).
So now, given a functor $\mathbf{Fun}(\mathrm{B}\Sigma_n, \mathbf{Vect})\to \mathbf{Vect}$ that preserves sifted colimits, we can apply it to the tensor power functor to get new functors. For example, we can take $\Sigma_n$-coinvariants (i.e. symmetric powers). We can tensor with the sign representation and then take $\Sigma_n$-coinvariants (which gives the exterior power). Any Schur functor works too, since we're just tensoring over $\mathbb{C}[\Sigma_n]$ (or $\mathbb{R}[\Sigma_n]$ over the reals) with the corresponding representation, and that preserves colimits.
|
2025-03-21T14:48:29.680815
| 2020-01-27T22:45:56 |
351295
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben Wieland",
"Benjamin Steinberg",
"HJRW",
"IJL",
"Santana Afton",
"W4cc0",
"YCor",
"https://mathoverflow.net/users/124004",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/1463",
"https://mathoverflow.net/users/15934",
"https://mathoverflow.net/users/4639",
"https://mathoverflow.net/users/66046",
"https://mathoverflow.net/users/95243"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625691",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351295"
}
|
Stack Exchange
|
Infinitely generated non-free group with all proper subgroups free
Is there any example of group $G$ satisfying the following properties?
$G$ is non-abelian, infinitely generated (i.e. it is not finitely generated) and not a free group.
$H< G$ implies that $H$ is a free group.
Clearly such an example should be torsion-free and countable.
(Added, from comments) For completeness, here's why it has to be countable. First, it is enough to construct a subgroup of index 2. Indeed, a theorem of Swan asserts that a torsion-free group with a free finite-index subgroup is free. Next, if a group $G$ has no subgroup of index 2 (i.e. every element is product of squares), it is easy to construct a nontrivial countable subgroup $H$ with the same property). In the current setting, $G$ is uncountable so $H$ is a proper subgroup, and hence has to be free, contradiction.
Olshanskii constructed a non-abelian group all of whose proper subgroups are infinite cyclic and so this group meets your property in a somewhat degenerate way since they are free on one-generator.
@BenjaminSteinberg To complement Mark's comment, groups with Olshanskii's property are generated by every non-commuting pair.
@PaulPlummer I don't really understand your first comment. A free group itself has plenty of generating families under which it is not free, so using non-freeness of a given generating family will not help much.
About the countability issue, I think I'm not aware of a non-free group in which every countable subgroup is free. (By analogy, $\mathbf{Z}^\mathbf{N}$ is not free abelian but all its countable subgroups are. Also, the Hawaiian earring group is locally free but has non-free countable subgroups.)
Sorry. I missed the infinitely Generated.
$\mathbb Z[\frac12]$
@MarkSapir Oh, yeah, but probably better to call the subgroup $3\cdot\mathbb Z[\frac12]$
@MarkSapir $\mathbf{Z}[3/2]=\mathbf{Z}[1/2]$, so it's not a proper subgroup. Actually, the only unital subrings of $\mathbf{Z}[1/2]$ are itself and $\mathbf{Z}$. Also, every subgroup of $\mathbf{Z}[1/2]$ is ${0}$, infinite cyclic, or has finite index and isomorphic to $\mathbf{Z}[1/2]$ (namely equals $k\mathbf{Z}[1/2]$ for some unique positive odd $k$.)
@BenWieland a torsion-free abelian group in which every proper subgroup is cyclic, is itself cyclic, by an easy argument.
I should have been more clear, I meant free non-abelian.
It feels as though no such group exists. Since every finitely-generated subgroup is both proper and free by assumption, how could the group have any nontrivial relations?
@SantanaAfton -- see Mark Sapir's comment above.
@HJRW Are you referring to the HNN construction? In that example, if you take the relation $a^t=ab$ and the group generated by all letters involved, you don’t get a proper subgroup. This never happens in OP’s situation.
@W4cc0 no, you don't: there are always free abelian subgroups. So the question was correctly stated, and wouldn't if "non-abelian" were added. Also as already mentioned, every group with your property has a non-abelian free subgroup.
@SantanaAfton It's overly simplistic, if not senseless, to think of a free group as a group with no relations.
Higman constructed non-free groups of cardinality $\aleph_1$ in which every countable subgroup is free. But every proper subgroup being free sounds hard.
@YCor, I meant $G$ non-abelian (even if it makes sense in this case, it is easy to see that there are no abelian such groups), of course it would have no sense in the other case; but unfortunately I always write the opposite of what I would like to, sorry.
@W4cc0 on the other hand do you maintain your claim that such a group has to be countable? or do you want to require the group to be countable? (I guess the countable case and uncountable case are quite distinct questions.)
@SantanaAfton -- Mark explained how to construct a non-free group in which every finitely generated subgroup is free. This contradicts your intuition that since "every finitely-generated subgroup is both proper and free" ... the group must itself be free.
@YCor such a group must be countable because if it is not, then it would be residually finite, but then a very well known theorem of Swan would imply that it is free
@W4cc0 thanks; I wouldn't have said "clearly"! I didn't know Swan's theorem, what does it say exactly? Here's an argument to show residual finiteness: it's enough to check that there is a proper subgroup of index 2. Indeed, the latter being free by assumption, it has to be res. finite and hence so is the whole group. Now, if by contradiction there is no subgroup of index 2, then every element is a finite product of squares. From this, it is easy to construct a nontrivial countable subgroup in which every element is a finite product of squares, and hence this is a non-free countable subgroup.
@YCor You're probably right I should not have have said "clearly" but in my mind it was a very direct application of some known facts (to me at least) in infinite group theory. Swan Theorem says that if your group is torsion-free and it is a finite extension of a free group, then it must be free itself.
Since a partly equivalent question was just asked, I have added the short argument to discard the uncountable case.
|
2025-03-21T14:48:29.681161
| 2020-01-27T23:08:43 |
351297
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Math604",
"https://mathoverflow.net/users/66623"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625692",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351297"
}
|
Stack Exchange
|
solutions of a pde smooth with respect to a parameter
I have a generic type question, but I will pose it with a specific example. Suppose $1<p<\frac{N+2}{N-2}$ and $B_1$ is the unit ball in $ R^N$ for $N \ge 3$. Consider the pde
$$-\Delta u(x) = t u + u^p$$ in $B_1$ with $u=0$ on $ \partial B_1$ where $ 0<t< \lambda_1$ (the first eigenvalue of $ -\Delta$ in $H_0^1(\Omega)$). Lets assume $ 1< \lambda_1$ and lets index the solutions $u=u_t$. By some variational techniques we prove the existence for all $t$ is the desired range ((so lets assume they are some sort of ground state solutions)). One would expect some smoothness in $t$.
So this is exactly my question.
``Can we typically assume the map $ t \mapsto u_t$ has some smoothness in $t$ for $t$ near $1$?
My end goal is to try and show the solution corresponding to $t=1$ is nondegenerate (in the sense that the linearized kernel is trivial) and hence of course I don't want to try and invoke the implicit function theorem (since if I could ...then i would already have proven my end goal).
With some fairly standard methods I seem to be able to proof the following: if the solution is smooth in $t$ (say existence of one derivative) then in fact the kernel is empty at $ t=1$.
If I look for ground state solutions and let $ \mu_t$ denote the minimal energy iti appears I can show the energy is differentiable (well at least the quotients are bounded). So this is halfway to showing the solution is differentiable in $t$ I guess I should delete the question for now.
|
2025-03-21T14:48:29.681339
| 2020-01-27T23:21:36 |
351298
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexandre Eremenko",
"Ben McKay",
"Chris Sangwin",
"Gerry Myerson",
"Hauke Reddmann",
"JCK",
"Piyush Grover",
"Todd Trimble",
"Tom Copeland",
"https://mathoverflow.net/users/11504",
"https://mathoverflow.net/users/12178",
"https://mathoverflow.net/users/130706",
"https://mathoverflow.net/users/13268",
"https://mathoverflow.net/users/25510",
"https://mathoverflow.net/users/2926",
"https://mathoverflow.net/users/30684",
"https://mathoverflow.net/users/3684",
"https://mathoverflow.net/users/87779"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625693",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351298"
}
|
Stack Exchange
|
Quirky, non-rigorous, yet inspiring, literature in mathematics
In contrast with such lucid, pedagogical, inspiring books such as Visualizing Complex Analysis by Needham and Introduction to Applied Mathematics by Strang, I've had the pleasure of coming across the non-rigorous, thought-provoking/stimulating, somewhat quirky works of Heaviside on operational calculus and divergent series; of Ramanujan on series, in particular, his use of his master theorem/formula (as explicated by Hardy); and the relatively unknown posthumous notes of Bernard Friedman on distributions and symbolic/operational calculus Lectures on Applications-Oriented Mathematics.
These works just blindside you. You think, "What the hey?" and slowly they grow on you and you start to understand them after further research using other texts, translating the terminology/concepts, and working out details. You're left with a deeper understanding and appreciation of the originality and applicability of the work--much like Hardy professed the day after he received Ramanujan's letter, no doubt.
(Friedman's Applied Mathematics, in contrast, is of a very different nature and an immediately enlightening intro to its topics.)
Any similar experiences with other mathematical works?
(This question is not research-level per se and may be more appropriate for MSE, but certainly the works cited have inspired and continue to inspire much advanced research into the related topics, and the question falls in a similar category to MO-Q1 and the MO-Qs that pop up in the Related section of that question).
Rota may have had a similar experience. He speaks of "the eerie witchcraft" of the umbral calculus as originally presented by Blissard and sucessors, before being tamed by Rota and his colleagues.
You consider Strang's and Needham's books "rigorous" ??
@AlexandreEremenko, not in the sense that they begin with axiomatics and give full proofs but in the sense that the operations they build upon are standard/mainstream in analysis, geometry, and linear algebra.
E.g., I know mathematicians who still balk at using umbral notation or fractional calculus, taking the derivative of the Heaviside step function, or are unfamiliar with the utility of divergent series.
Strang's books that I know (and use in my teaching) lack not only rigorous proofs but even clear, unambiguous definitions.
Okay, let's put it this way. I doubt that Strang's or Needham's books inspired the construction of cottage industries to explain them in contrast to work on the Laplace transform, distributions, convolutional algebras, etc. for Heaviside operational calculus, whole theories in analysis to explain Fourier's claims, Ramanujan's, Feynman's (ongoing), etc.
Cartier's "Mathemagics"?
Anthing in QFT would qualify
@PiyushGrover, physics is mathematics' muse.
Although Dyson in a review (https://physicstoday.scitation.org/doi/abs/10.1063/1.3062600?journalCode=pto) of "Introduction to Fourier Analysis and Generalized Functions" by Lighthill claims it is fully rigorous, I found the style refreshingly distinct from the dry, axiomatic, approach of Schwartz with many surprising (to me on first reading) connections made. For me, in its uniqueness among books on distribution theory, it lies somewhere between the two categories of books I've tried to characterize.
@MattF. nice paper. Why not write it up as an answer?
Most of Blass's paper however is simply brimming with meaningful and rigorous combinatorial (and later, logical) considerations. I think of it as being more "fun" than actually "quirky".
Related https://mathoverflow.net/questions/115032/non-rigorous-reasoning-in-rigorous-mathematics?rq=1
Newton's Philosophiæ Naturalis Principia Mathematica must have provoked similar reactions historically concerning the derivative (fluxions) and infinitesimals with various approches to understanding, summarized by Thurston in "On proof and progress in mathematics" on page 3 (https://arxiv.org/abs/math/9404236).
Would "Goedel, Escher, Bach" count? It's been several decades since I've read it but I remember it as being 'quirky', 'non-rigorous', 'inspiring' (to me anyway), and at least math-adjacent.
@JCK, maybe so. From Wiki: Douglas Hofstadter's books, especially Metamagical Themas and Gödel, Escher, Bach, play with many self-referential concepts and were highly influential in bringing them into mainstream intellectual culture during the 1980s.
Personally, I learnt from "Group Theory" (i.e. Birdtracks) by Predrag Cvitanovic more about Lie algebra than from all textbooks in my math lib (and I think the potential of graphical methods hasn't even scratched the surface) - but then, I'm a living quirk either.
@HaukeReddmann, thanks I wasn't aware of that book. I'll give it a look. You reminded of a book by Cvi and his colleagues that I've always wanted to get back to--the Chaos Book, or Chaos: Classical and Quantum . And, it has a fitting quote from Alice: (for those with ambiguity tolerance or delight in ambiguity)
It seems very pretty,” she said when she had finished it, “but it’s rather hard to understand!” (You see she didn’t like to confess, even to herself, that she couldn’t make it out at all.) “Somehow it seems to fill my head with ideas — only I don’t exactly know what they are!”
Lewis Carroll, Through the Looking Glass
Andrew Granville and Jennifer Granville's "Prime Suspects" is a comic book which despite being phrased as a murder mystery does a surprisingly lucid job connecting ideas involving primes with ideas involving permutations. This paper Andrew Granville outlining the basic idea is pretty non-rigorous but fascinating.
Nice. (The paper is "THE ANATOMY OF INTEGERS AND PERMUTATIONS" by Andrew Granville. // Links often are broken, so ... .)
To elaborate on Piyush Grover's comment, large swaths of theoretical physics can be considered "non-rigorous yet inspiring." The adjective "quirky" might not be so apposite, though. But perhaps somewhat in the spirit of the question is Richard Feynman's semi-popular book QED: The Strange Theory of Light and Matter. There are still some open questions about the mathematical consistency of QED.
Quarky, not so quirky.
From Wikipedia (in agreement with recall of readings many years ago): Historically, as a book-keeping device of covariant perturbation theory, the graphs were called Feynman–Dyson diagrams or Dyson graphs, because the path integral was unfamiliar when they were introduced, and Freeman Dyson's derivation from old-fashioned perturbation theory was easier to follow for physicists trained in earlier methods. Feynman had to lobby hard for the diagrams, which confused the establishment physicists trained in equations and graphs.
(Cont.) From "Julian Schwinger: Nuclear Physics, the Radiation Laboratory, Renormalized QED, Source Theory, and Beyond" by Kimball Milton: The formal solution of Schwinger’s differential equations was Feynman’s functional integral; yet while the latter was ill-defined, the former could be given a precise mean-
ing, and for example, required the introduction of fermionic variables, which
initially gave Feynman some difficulty.
(Cont.) It may be fair to say that while the path integral formulation to quantum field theory receives all the press, the most precise exegesis of field theory is provided by the functional differential equations of Schwinger resulting from his action principle.
Quirky, as initially received historically , seems à propos. (Familiarity, in this case, doesn't breed contempt.)
Feyman's popularizing book QED is a lucid jewel.
This book is very elementary, but has a lot of excellent material.
@BOOK{Polya54,
author = {Polya, G.},
title = {Mathematics and Plausible Reasoning. Vol.1: Induction and Analogy in Mathematics. Vol 2. Patterns of Plausible Inference},
publisher = {Princeton University Press},
year = {1954},
}
OK, it's a fine book, but is it quirky? is it non-rigorous? does it blindside you?
Lots of it is non rigours, indeed the whole section on analogy. Polya was very keen on guessing and pattern spotting. Some bits did blindside me when I first read it. I commend the book, particularly for less experienced mathematics students.
I enjoyed reading the book 'Quantization, Classical and Quantum Field Theory and Theta Functions' by Andrej Tyurin very much. It is certainly not rigorous but it was very inspiring for me.
Available at https://arxiv.org/abs/math/0210466
|
2025-03-21T14:48:29.681963
| 2020-01-28T04:05:12 |
351309
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Nemo",
"Paramanand Singh",
"https://mathoverflow.net/users/15540",
"https://mathoverflow.net/users/82588"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625694",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351309"
}
|
Stack Exchange
|
Reference request: proof of Ramanujan's Cos/Cosh Identity
The Ramanujan Cos/Cosh Identity, as stated here, is
$$\left[1+2\sum_{n=1}^{\infty}\frac{\cos n\theta}{\cosh n\pi}\right]^{-2}+
\left[1+2\sum_{n=1}^{\infty}\frac{\cosh n\theta}{\cosh n\pi}\right]^{-2}=
\frac{2\Gamma^4\left(\frac34\right)}{\pi}.$$
I am looking for a proof, preferably from a reputable source. I hoped I would find something in Ramanujan's Notebooks, but have so far found no mention of it.
You should check out Ramanujan Notebooks vol 3 by Bruce C. Berndt. Look in chapter dealing with elliptic functions. See chapter 18, page 162, Entry 11.
https://math.stackexchange.com/questions/517409/extensions-of-ramanujans-cos-cosh-identity
I expand my comment into an answer.
The key here is the Fourier series for the elliptic function $\operatorname {dn} (u, k) $ given as $$\operatorname {dn} (u, k) =\frac{\pi} {2K}\left(1+4\sum_{n=1}^{\infty} \frac{q^n} {1+q^{2n}}\cos\left(\frac{n\pi u} {K} \right) \right) $$ where $K$ is the complete elliptic integral of first kind corresponding to modulus $k$ and $q=\exp(-\pi K'/K) $ is the nome corresponding to modulus $k$.
Let $\theta=\pi u/K$ then we have $$\operatorname {dn} \left(\frac{K\theta} {\pi}, k\right) =\frac{\pi} {2K}\left(1+2\sum_{n=1}^{\infty}\frac{\cos n\theta} {\cosh (n\pi K'/K)} \right) $$ Putting $K'=K$ so that $k=1/\sqrt{2}$ and $K=\dfrac{\Gamma^2(1/4)}{4\sqrt{\pi}}$ we get $$1+2\sum_{n=1}^{\infty}\frac{\cos n\theta} {\cosh n\pi} =\frac{2K}{\pi}\operatorname {dn} \left(\frac{K\theta} {\pi}, \frac{1}{\sqrt{2}}\right)$$ Let the above expression be denote by $A$ and the expression obtained from it by replacing $\theta$ with $i\theta$ be denoted by $B$. Then we have to show that $$\frac{1}{A^2}+\frac{1}{B^2}=\frac{2\Gamma^{4}(3/4)}{\pi}$$ Note that we have $$\Gamma(1/4)\Gamma (3/4)=\sqrt{2}\pi$$ and therefore $$\frac{2K} {\pi} =\frac{\sqrt{\pi}}{\Gamma ^2(3/4)}$$ and we have then $$A^{-2}+B^{-2}=\frac{\Gamma ^{4}(3/4)}{\pi}\left(\operatorname {dn} ^{-2}\left(\frac{K\theta}{\pi}\right)+\operatorname {dn} ^{-2}\left(\frac{iK\theta}{\pi}\right)\right)$$ The expression in parentheses is easily seen to be $2$ if we note that $$\operatorname {dn} (iu, k) =\frac{\operatorname {dn} (u, k')} {\operatorname {cn} (u, k')} $$ and here $k=k'=1/\sqrt{2}$.
|
2025-03-21T14:48:29.682119
| 2020-01-28T06:27:02 |
351314
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Puzzler",
"dohmatob",
"https://mathoverflow.net/users/151588",
"https://mathoverflow.net/users/36687",
"https://mathoverflow.net/users/78539",
"passerby51"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625695",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351314"
}
|
Stack Exchange
|
Hanson-Wright inequality with random matrix
I'm interested in bounding the tail probabilities of a quadratic form
$x^t A x$ where $x\in \mathbb{R}^n$ is a sub-Gaussian vector with independent entries. $A\in \mathbb{R}^{n\times n}$ is a matrix. So I'm exactly in the setup of the Hanson-Wright inequality. In fact, I wish I could use it because if it would apply, it would give me exactly the bounds I'm looking for.
My problem is that in my case, the matrix $A$ is random, too. Worse even, I don't have independence of $A$ and $x$. However, there are two special properties in my case which can be used:
$A$ and $x$ are uncorrelated, i.e. $\mathbb{E}[Ax]=\mathbb{E}[A]\mathbb{E}[\mathbb{x}]=\mathbb{E}[A]0=0$.
$A$ is an orthogonal projection of rank $r < n$.
So my question is: Does somebody know a generalization of the Hanson-Wright inequality which would apply in this case?
[I am asking this question is because I study the finite-sample performance of OLS and other linear estimators. In the case of OLS, one can think of $A$ as orthogonal projection on the column space of the regressors and of $x$ as the error term. If the regressors were fixed, then Hanson-Wright would do the job immediately, but I need to allow for the regressors to be random, too.]
Uncorrelatedness alone is most likely insufficient for saying much.
@Puzzler Nice! Upvoted. Similar question here https://mathoverflow.net/q/385586/78539
Let $x \sim N(0,I_n)$. For any independent rank-1 projection $A$, conditioned on $A$, we have
$$x^T A x \sim \chi^2_1.$$ So unconditionally, $x^T A x = O(1)$ with high probability.
Now, let $A = \frac{x x^T}{\|x\|_2^2}$. Then, $A$ is a rank-1 projection and we have $\mathbb E [A x] = \mathbb E[x] = 0 = \mathbb E[A] \mathbb E[x]$. So, $A$ and $x$ are uncorrelated (in the sense stated in the question). But
$$
x^T A x = \frac{x^T x x^T x}{\|x\|_2^2} =\|x\|_2^2 \sim \chi_n^2
$$
so $x^T Ax \approx n$ with high probability. (This rank-1 projection behaves like the full rank projection when applied to $x$.)
Assuming independence of $A$ and $x$, one can condition on $A$ and apply the Hanson--Wright inequality. Since the bound does not depend on $A$ (it only depends on $\|A\|_F = r$ and $\|A\| = 1$), the same bound would hold unconditionally. It would be as if $A$ was deterministic.
Wow, what a fantastic example! Would you expect a positive answer with independence of A and x? Or should I ask a new question for that and accept your answer straight away? I'm new here.
@Puzzler, yes, the answer would be positive with independence. I will add a comment to the answer, no need for a new question.
Thank you, @passerby51 ! I'll try to see whether I can justify independence.
@Puzzler, You are welcome.
@passerby51 Nice! Upvoted. Similar question here https://mathoverflow.net/q/385586/78539
@dohmatob, thanks. The question is deleted though.
@passerby51 Question moved here https://math.stackexchange.com/q/4050099/168758
|
2025-03-21T14:48:29.682443
| 2020-01-28T08:05:33 |
351317
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625696",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351317"
}
|
Stack Exchange
|
Strong data-processing inequality ? Upper bound on a certain modified total-variation metric
Let $\mathcal X=(\mathcal X,d)$ be a Polish space equipped with the Borel sigma-algebra. Let $p\ge 1$ and $P_1,P_2$ be probability distributions on $\mathcal X$ such that $\max_{k=1,2}\int d(x,x_0)^pdP_k(x) < +\infty$ for some (and therefore for every!) $x_0 \in \mathcal X$. A fancy way to say this is that, $P_1$ and $P_2$ are in the $p$-Wasserstein space over $\mathcal X$, usually denoted $\mathcal P_p(\mathcal X)$.
Now, for every $\varepsilon \ge 0$ and coupling $(X,X')$ of $P_1$ and $P_2$, one can use Markov's inequality to get the bound
$$
\mathbb P(d(X,X') > \varepsilon) \le \mathbb P(d(X,X')^p > \varepsilon^p) \le \frac{\mathbb E[d(X,X')^p]}{\varepsilon^p}.
$$
Optimizing over the coupling $(X,X')$, we get
$$\inf_{X,X'}\mathbb P(d(X,X') > \varepsilon) \le \left(\frac{W_p(P_1,P_2)}{\varepsilon}\right)^p.
$$
Now, define $TV_\varepsilon(P_1,P_2) :=\inf_{X,X'}\mathbb P(d(X,X') > \varepsilon)$ and note that $TV_\varepsilon(P_1,P_2) \le TV_0(P_1,P_2) = \inf_{X,X'}\mathbb P(X \ne X')$ corresponds to the usual total-variation metric. Thus $TV_\varepsilon$ extends the usual total-variation metric. BTW, $TV_\varepsilon$ is a Wasserstein distance induced by the cost $c_\varepsilon(x,x') = \begin{cases}1,&\mbox{ if }d(x,x') > \varepsilon,\\0,&\mbox{ else.}\end{cases}$
We have proved that
[Weak bound] For every $\varepsilon > 0$ and $P_1,P_2 \in \mathcal P_p(\mathcal X)$, we have
$$
TV_\varepsilon(P_1,P_2) \le \left(\frac{W_p(P_1,P_2)}{\varepsilon}\right)^p, \tag{1}
$$
where $W_p(P_1,P_2)$ is the $p$-Wasserstein distance between $P_1$ and $P_2$.
Of course, the above inequality is only meaningful for $\varepsilon \ge W_p(P_1,P_2)$. In particular, it breaks down completely for $\varepsilon > 0$. This issue comes at no surprise (at least to me!), as we have only imposed any important constraints on the distributions $P_1$, $P_2$.
Question (stronger bounds).
How can the bound (1) be improved under further restrictions (for example, sub-Gaussianness, etc.) on the distributions $P_1$ and $P_1$ ?
For example, the holy grail for me would be to exhibit a function $F:[0, 1] \rightarrow [0, 1]$, such that $TV_\varepsilon(P_1,P_2) \le F(TV(P_1,P_2))$, uniformly on $P_1$ and $P_2$. Of course, I'm open to other kinds of bounds.
Notes
For simplicity, I've ignored issues related to the possible non-measurability of the set $\{(x,x') \in \mathcal X^2 \mid d(x,x') > \varepsilon\}$.
|
2025-03-21T14:48:29.682606
| 2020-01-28T08:37:43 |
351318
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Daniele Tampieri",
"fibonatic",
"https://mathoverflow.net/users/103296",
"https://mathoverflow.net/users/113756",
"https://mathoverflow.net/users/143783",
"lrnv"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625697",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351318"
}
|
Stack Exchange
|
Can this quadratic program be solved analyticaly?
I have a convex quadratic program wich is structured as follows :
\begin{align*}
\operatorname{argmin}_{p} &\hspace{0.5em} p' A p - 2 p' b \\
\mathrm{s.t.} &\hspace{0.5em} Ep=f \\
& \hspace{0.5em} p_i \ge 0 \;\forall i
\end{align*}
Some hypothesis :
The matrix $A$ is a $n \times n$ diagonal matrix, with diagonal elements strictly positives and summing to one (it is then pos def and invertible)
The vector $b$ is non-negative
The matrix $E$ is a $m\times n$ matrix such that $m<n$, making the number of constraints smaller than the number of variables. Elements in $E$ are inside $[0,1]$. The dimension of the affine subspace spanned by $Ep=f$ is $n-m$.
The solution of the unconstraint QP, $p^* = A^{-1}b$, is positive (inside the inequality constraints) but not inside the constraint $Ep = f$.
I do know a point, let's call it $p_{eq}$, such that $Ep_{eq} = f$ and such that $p_{ep,i} \ge 0\; \forall i$.
$\operatorname{diag}(p_{eq}) = A^{-1}$
If you look closely at the previous points, you'll see that we have a convex problem, whithin a non-empty convex domain, and hence existence and unicity of a minimum.
Some of these informations might not be relevant, and some might, so i gave you all i have ;)
I know i can reexpress the problem by using the block matrix $[[A,E'],[E,0]]$, but this did not help me.
What do you think? Will these hypothesis be enough to find a closed-form expression for the solution? If no closed-form solution, can you think of a smarter way of solving it than using a standard QP solver?
Edit : A similar problem was asked a few years back, but with less hypothesis (more general), and the answer was No, see there. My question is, are the hypothesis i have changing this or not. Especialy, i think that the fact that the (unconstraint) optimal solution fullfills the inequality constraint and that we have a point fullfilling all constraints (but not the minimum) might be relevant.
Hi, there's a small typo at the end of the sentence at the third point: it should be "Elements", not "Ellements".
Thanks for the typo. You know that you can correct it yourself ? Usualy it's good practice to correct it directly.
Yes, I know, and I also know that +2 points of reputation are awarded for this, if the edit is accepted. However, when the changes are few and very small, I prefer to tell the Author of the post: to me it is a nonsense to get experience points for correcting insignificant typos on a research level mathematics web site.
Did not now for the points. Makes sense now. Thanks anyway
Does the solution of the KKT system (as shown in the linked question) also satisfy the inequalities?
@fibonatic No !
|
2025-03-21T14:48:29.682814
| 2020-01-28T11:16:26 |
351323
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625698",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351323"
}
|
Stack Exchange
|
Frobenius dimensions of Nakayama algebras
The Frobenius dimension $F(A)$ of an Artin algebra $A$ is given by the dimension of $Hom(D(A),A)$ (see the answer in On nearly Frobenius algebras ).
Question 1: Is it true that $F(A) \geq gldim(A)$ for any Nakayama algebra $A$ of finite global dimension?
Question 2: What is the maximal Frobenius dimension of a quasi-hereditary Nakayama algebra with $n$ simple modules? For $n \geq 2$, the sequence starts with 5,10,17,26,37,50,65. It seems like the answer is $n^2+1$ but I do not see an easy reason why such a simple result might be true.
For general Nakayama algebras with finite global dimension the maximal Frobenius dimension seems to be more complicated, at least it does not appear in the oeis.
|
2025-03-21T14:48:29.682890
| 2020-01-28T11:26:30 |
351324
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Michael Stoll",
"https://mathoverflow.net/users/21146"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625699",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351324"
}
|
Stack Exchange
|
Division by two in the jacobian
Let $X$ be an hyperelliptic complex curve of genus $3$, $J$ its jacobian,
and $P,Q$ two Weierstrass points on $X$. Can one write explicitly
a class $x$ in $J$ such that $2x = P-Q$?
In Section 5 of my paper http://www.mathe2.uni-bayreuth.de/stoll/schrift.html#AG54 there is an algorithm that can compute "halves" of points. We can assume that $Q$ is at infinity (and also that $P$ has zero $x$-coordinate) and apply Cor. 5.4. The input includes a choice of square root of the class of $x - x(P)$ in the algebra ${\mathbb C}[x]/\langle f(x) \rangle$, where the curve is given by $y^2 = f(x)$. This works for any genus.
Expanding on my comment above, here is what the algorithm in my paper gives for a genus 3 Jacobian. We write the curve as
$$ y^2 = f(x) = x(x-a_1^2) \cdots (x - a_6^2) $$
and take $P = (0,0)$ and $Q = \infty$, the point at infinity.
We cannot directly apply Cor. 5.4 (it requires the point to
have non-zero $y$-coordinate), but we can still run the
algorithm. The result is that the point on the Jacobian
with Mumford representation
$$ (x^3 - \sigma_2 x^2 + \sigma_4 x - \sigma_6,
(\sigma_1 \sigma_2 - \sigma_3) x^2
- (\sigma_1 \sigma_4 - \sigma_5) x + \sigma_1 \sigma_6) $$
is one of the ``halves'' of $P - Q$. Here $\sigma_j$ denotes
the $j$th elementary symmetric polynomial in the $a_i$.
The other solutions are obtained by changing the signs
of some of the $a_i$.
(Let $(a(x), b(x))$ be the Mumford representation above.
The the point is represented by the divisor
$D - (\deg a) \cdot\infty$, where $D$ is the sum of points
$(\xi, b(\xi))$, where $\xi$ runs through the roots of $a$,
counted with multiplicities.)
|
2025-03-21T14:48:29.683054
| 2020-01-28T11:29:39 |
351325
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gro-Tsen",
"Joel David Hamkins",
"Martin Sleziak",
"Nate Eldredge",
"Ramiro de la Vega",
"Robert Furber",
"Will Brian",
"https://mathoverflow.net/users/17064",
"https://mathoverflow.net/users/17836",
"https://mathoverflow.net/users/1946",
"https://mathoverflow.net/users/4832",
"https://mathoverflow.net/users/61785",
"https://mathoverflow.net/users/70618",
"https://mathoverflow.net/users/8250"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625700",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351325"
}
|
Stack Exchange
|
Is the lexicographic ordering on the unit square perfectly normal?
It's known that order topology is completely normal, so the lexicographic ordering on the unit square is also completely normal. It's also known that the lexicographic ordering on the unit square is not metrizable. I am interested in whether it is perfectly normal. (A space is perfectly normal if for any two disjoint nonempty closed subsets, there is a continuous function $f$ to $[0,1]$ that "precisely separates" the two sets, meaning that the two closed sets are $f^{-1}(0)$ and $f^{-1}(1)$.) And how do we prove it?
Wikipedia on this space: https://en.wikipedia.org/wiki/Lexicographic_order_topology_on_the_unit_square. (Very interesting question!)
You should clarify whether you mean the closed unit square or the open unit square (or either one indifferently, or some other variant).
He means the closed unit square, for otherwise it won't be compact.
But one can be completely normal without being compact, can't one? Anyway, even if there's a reason, this sort of things should always be clarified.
Steen and Seebach, Counterexamples in Topology, 2e, say that it isn't, but the proof is left as an exercise (problem 71, page 209).
I'm not sure about your definition of "perfectly normal". Doesn't Urysohn's lemma say that, in any $T_4$ space, any disjoint closed sets are separated by a continuous function? The definition I know of "perfectly normal" is that every closed set is a $G_\delta$. This is equivalent to "every nonempty closed set can be separated from its complement by a continuous function onto $[0,1]$". Maybe this is the definition/characterization you're thinking of?
He's using the same definition as in Wikipedia: https://en.wikipedia.org/wiki/Normal_space.
Thanks @JoelDavidHamkins. The wikipedia entry uses the phrase "precisely separated" which I think makes a difference here -- I'll edit the question to make it more clear.
It is without proof also in the pi-base.
@Gro-Tsen, the open unit square would be metrizable (it is just a disjoint union of open intervals).
As Nate Eldredge points out in the comments, the book Counterexamples in Topology states that this space is not perfectly normal, but does not provide a proof. Here is the idea for a proof.
A perfectly normal space is a normal space in which every closed set is a $G_\delta$. So to prove this space is not perfectly normal, we'd like to find a closed set that is not a $G_\delta$. I claim that the subspace $C = [0,1] \times \{0,1\}$ works. ($C$ is the two horizontal lines on the top and bottom.) The reason is that if $U \subseteq [0,1] \times [0,1]$ is an open set containing $C$, then
$$A_U = \{x \in [0,1] \ : \ \{x\} \times [0,1] \subseteq U\}$$
(i.e., the set of all vertical lines contained in $U$) is a co-countable subset of $[0,1]$. (This is because for every $p \in (0,1]$, $U$ must contain a neighborhood of $(p,1)$, and therefore $A_U$ must contain an interval of the form $(p-\varepsilon,p)$. Likewise, $U$ must contain a neighborhood of $(p,0)$, and therefore $A_U$ must contain an interval of the form $(p,p+\varepsilon)$. So every point of $[0,1] \setminus A_U$ is isolated.) Therefore any countable intersection of open neighborhoods of $C$ contains a horizontal line.
EDIT: Thanks to Nate Eldredge for helping me to simplify my original argument.
I think you missed one: "horizontal line" at the end should be "vertical line"?
Just to note a slight variation (since I had been thinking about a similar proof before I saw yours): the complement of $A_U$ is actually at most countable, since every point of it is isolated in the usual topology of $[0,1]$. So any $G_\delta$ containing $C$ must in fact contain co-countably many vertical lines, and you don't need Baire.
@NateEldredge: Good observation! It's not exactly using a nuke to kill a fly, but it is using more than just the flyswatter. I've edited to simplify things.
In the the Moore plane/Niemytskii plane every closed set is $G_\delta$, but it is not normal. Call me old-fashioned, but I think "perfectly normal" should imply "normal". It is true, via a proof using Urysohn's lemma countably many times and carefully summing up the resulting functions, that a perfectly normal space is the same thing as a normal space in which every closed set is $G_\delta$. Of course, this does not affect the argument for the lexicographic ordering not being normal, because only the direction "perfectly normal $\Rightarrow$ all closed sets are $G_\delta$" is used.
@RobertFurber: You're perfectly right, of course. A perfectly normal space is a normal space in which every closed set is $G_\delta$. (I was thinking of this whole problem in the context of normal spaces, but apparently that's not clear. I'll edit to make it clear.)
It is a known fact that any perfectly normal (countably) compact space $X$ is ccc. The proof is what you would try: start with an uncountable cellular family $\mathcal{U}$, choose a point $p_U \in U$ for each $U \in \mathcal{U}$ and consider the closed set $C=X \setminus \bigcup \mathcal{U}$. Since the space is perfectly normal, $C$ is a $G_\delta$ (say $C= \bigcap_nV_n$) and since $\mathcal{U}$ is uncountable you can find $n$ such that infinitely many of the $p_U$'s are outside of $V_n$. Now this infinite set of $p_U$´s has no limit points, contradicting the compactness of $X$.
Since $I^2_{lex}$ is compact and not ccc (both standard facts), it cannot be perfectly normal.
For a compact Hausdorff $X$ it is equivalent that $X$ is hereditarily Lindelöf or that $X$ is perfectly normal. Sketch: $X$ is hereditarily Lindelöf implies that every open set is an $F_\sigma$ (as $X$ is regular) so dually every closed set is a $G_\delta$. OTOH if $X$ is perfectly normal, every open set is an $F_\sigma$ so $\sigma$-compact and Lindelöf, making all open sets Lindelöf and $X$ hereditarily Lindelöf again.
$[0,1]^2$ in the lexicographic order topology has a discrete subset $[0,1]\times \{\frac12\}$ so is definitely not hereditarily Lindelöf, so not perfectly normal either.
|
2025-03-21T14:48:29.683450
| 2020-01-28T11:35:29 |
351326
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"Yellow Pig",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/12395"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625701",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351326"
}
|
Stack Exchange
|
Applications of quantum representations of the mapping class group to quantum computers
Quantum representations of the mapping class group of a surface are certain representations constructed from the data of a TQFT and described, for example, in and 1 and 2.
The following sources 3 and 4 imply that there are applications of quantum representations of the mapping class group to quantum computing.
What are some of these applications? Which concrete mathematical questions (e.g. from low dimensional topology or quantum algebra) have applications/are relevant to quantum computing?
The context is topological quantum computation, where quantum information is stored nonlocally in a physical system, so that it is protected from decoherence by local sources of noise. The nonlocal degree of freedom is a socalled non-Abelian anyon, a particle-like excitation which is described by a (2+1)-dimensional topological quantum field theory. Arxiv:1705.06206 provides a survey of the mathematics of topological quantum computing, with a list of conjectures and open problems.
I should add, as a physicist, that I am uncertain whether any of these mathematical questions are relevant in the quest to actually build and operate a quantum computer. The key challenge there is to identify a physical system that has these exotic particles. The fractional quantum Hall effect was a primary candidate for several decades, but this system has now been largely abandoned in favor of superconducting systems, where the energy gaps can be much larger (allowing for operation in a realistic temperature range). Microsoft is heavily invested in the design of a topological quantum computer using anyons in the Ising universality class (Majorana fermions). These are "trivial" from a mathematical perspective, since they only implement a Clifford algebra and do not provide access to the full unitary group. There are no realistic options for Fibonacci anyons (which would cover all unitary operations).
Thanks a lot! Are you aware of any mathematical questions which are relevant to a more realistic version of a quantum computer (either the one Microsoft is building or other possibilities)?
For a computer scientist there are many issues on how to manipulate topologically encoded quantum information in a way that requires the least amount of resources, see for example Majorana-based fermionic quantum computation. My impression that the mathematical groundwork is complete, it's the step from concept to reality that is the main obstacle at present.
But no mathematical questions about the mapping class group/low-dimensional topology?
none that are relevant for quantum computers, AFAIK.
Thanks, sounds a bit disappointing for people like me who like topology...
|
2025-03-21T14:48:29.683660
| 2020-01-28T11:56:45 |
351327
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Pedro Kaufmann",
"Yemon Choi",
"https://mathoverflow.net/users/27566",
"https://mathoverflow.net/users/763"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625702",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351327"
}
|
Stack Exchange
|
On a specific Fourier series on solenoids
In [1], Hewitt and Ritter define a sequence of operators $S_n$ on $\mathcal{L}_1(\sum_{a})$, where $\sum_a$ is a compact abelian group called $a$-adic solenoid. This operators can be realized as convolutions with certain trigonometric polynomials:
$$
S_nf(x,t)= \sum_{\alpha\in C_j}\hat f(\chi_\alpha)\chi_\alpha(t,x)= f\ast P_n(x,t),
$$
where $P_n = \sum_{\alpha\in C_j}\chi_\alpha$. My question is:
Is there a $C>0$ such that $\|P_n\|_1\leq C$, for all $n$?
I'd be also interested in knowing if moreover $\{P_n\}_n$ form an approximate identity.
My motivation is to try to obtain Young-type convolution inequalities for these groups. More specifically, if one fixes an invariant and compatible metric on $\sum_a$, I would like to obtain an inequality of the type
$$
\|f\ast P_n\|_{Lip} \leq C \|f\|_{Lip},
$$
for all $n$ and all Lipschitz function $f$. Here, $\|g\|_{Lip}$ denotes the smallest Lipschitz constant for $g$.
[1] Hewitt, Ritter, Fourier series on certain solenoids, Math. Ann., 1981.
This is not a definite answer, but by analogy with usual Fourier series indexed by ${\bf Z}$, these $P_n$ look like Dirichlet kernels rather than Fejer kernels, and hence my first guess is that they will not form a bounded sequence in the $L^1$-norm. In general, summability kernels for Fourier analysis on LCA groups will have bad properties if they are just unweighted sums of characters.
Thank you for your comment @YemonChoi. I understand, it seems unlikely that in this case we'd have the desired bound. For the torus we use the very specific properties of the group to produce, out of the Dirichlet Kernels, the approximate identity formed by the Fejér kernels. It'd be interesting to know if in this case it's also possible to combine the $P_n$ in order to obtain better kernels in this sense.
I am reasonably confident (but not certain without checking) that you can get analogues of the Fejer kernel on any LCA group G, by taking the "Dirichlet kernels" and convolving them with themselves. The basic idea is that if U is a symmetric compact neighbourhood of the origin in the dual group $\widehat{G}$, then you let $D_U$ be the characteristic function/indicator function of U, normalised to have $L^2$-norm equal to $1$, and consider $D_U*D_U$. For solenoids, maybe there is something under the name "Bochner-Fejer kernel" which IIRC occurs in some old work on almost-periodic functions?
Thank you again @YemonChoi for your comment, I will look into this. I'll post if I find something.
|
2025-03-21T14:48:29.683862
| 2020-01-28T12:31:30 |
351328
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dylan Wilson",
"Eric Peterson",
"Robert Bruner",
"Tyler Lawson",
"https://mathoverflow.net/users/1094",
"https://mathoverflow.net/users/360",
"https://mathoverflow.net/users/6872",
"https://mathoverflow.net/users/6936"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625703",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351328"
}
|
Stack Exchange
|
$KO_*$ groups of $\mathbb{R}P^\infty$, "Snaiths" theorem for $KO$
I posted this question some days ago at math.stackexchange, but didn't receive an answer.
I have two questions:
I wonder whether anyone has taken the time to compute $KO_*(\mathbb{R}P^\infty)$?
The standard tools to compute these Groups in the complex case rest on the requirement for the cohomology theories $E$ to be complex orientable. Naturally, I looked up whether something like real orientable cohomology theories exist in the literature but found out that $KO$ is not real-oriented. Anyway, there is a way to "circumvent" Snaiths theorem for the spectrum $K$ if one is only interestd in the algebra of cooperations, in the sense that one can show that
$$K_*(\mathbb{C}P^\infty) \xrightarrow{i_*} K_*K$$ is an injection of rings, where $i$ is induced from the inclusion $\mathbb{C}P^\infty \simeq BU(1) \hookrightarrow BU$. In fact, one only needs to invert the Bott element $\beta$ to turn it into an isomorphism, so it is a localization. This can be concluded from
Robert M. Switzer. Algebraic topology—homotopy and homology. Classics in Mathematics. Springer-Verlag, Berlin, 2002. Reprint of the
1975 original [Springer, New York; MR0385836 (52 #6695)].
17.33, which states
$K_*K$ is generated over $\mathbb{Z}[u,u^{-1},v^{-1}]$ by the polynomials $\{p_1,p_2,\ldots\}$.
By a process reminiscent of
J. F. Adams. Stable homotopy and generalised homology. University of
Chicago Press, Chicago, Ill.-London, 1974. Chicago Lectures in Mathematics.
p. 44 we can describe the relations of the generators of $\beta_i$ of $K_*(\mathbb{C}P^\infty) = K_* \{\beta_0 , \beta_1 , \ldots \}$ such that
$$\beta_1\beta_n = n \beta_n +(n+1)\beta_{n+1}$$
and by setting
$$\binom{x}{i} = \frac{x(x-1)\cdots (x-(i-1))}{i!} \in \mathbb{Q}[x]$$
with $x:=\beta_1$ one can see that
$K_*(\mathbb{C}P^\infty)\otimes \mathbb{Q}$ is the polynomial algebra $K_* \otimes \mathbb{Q}[x]$ over $K_*\otimes \mathbb{Q} = \mathbb{Q} [t,t^{-1}]$ and $K_*(\mathbb{C}P^\infty)$ can be identified with the subalgebra of $K_* \otimes \mathbb{Q}[x]$ generated by $\binom{x}{i}$ for $i=0,1,2, \ldots$,
where we set $\binom{x}{0}=1$.
While snaiths theorem works on the spectrum level and the aforementioned result follows, I wonder whether a similar result holds in the real case, i.e.
$$ KO_*(\mathbb{R}P^\infty)[\alpha^{-1}] \cong KO_*KO$$ for some element $\alpha \in KO_*(\mathbb{R}P^\infty)$?
Regarding Snaith's theorem: the direct analogue of Snaith's construction, using the generator you might expect from $\Bbb{RP}^\infty$ by analogy, will result in a ring spectrum $R$ that has an orientation for real vector bundles. This means that it can't quite be $KO$. For example, it would mean that it would have a map of ring spectra $MO \to R$. Because of known structure about $MO$, this would force $R$ to be a sum of shifts of the Eilenberg-Mac Lane spectrum $H\Bbb{F}_2$.
The $p$–completion of $KO_* KO$ is known and easy to state: it is the algebra of continuous functions from the $p$–adic group $(\mathbb Z_p^\times / C_2)$ to $(KO^\wedge_p)_*$. Given the size of domain and codomain, I believe this to be too hopelessly large to be caught by inverting an element in (the $p$–completion of) Bob's answer.
There are many ways to do this. One elementary approach is to use the Adams spectral sequence
$Ext_A(H^*ko \wedge RP^\infty, F_2) \cong Ext_{A(1)}(H^*RP^\infty,F_2) \Rightarrow ko_*(RP^\infty) $
and invert the Bott map. An $A(1)$ resolution giving the answer (since $E_2 = E_\infty$) can be seen at the bottom of the page
http://www.rrb.wayne.edu/art/index.html
We see the groups, starting with $KO_0$, are $0, Z/2, Z/2, Z/2^\infty, 0, 0, 0, Z/2^\infty$ and repeat by Bott periodicity, with the evident action of the coefficients $KO_*$. (This action follows from the homological algebra of the $Ext$ calculation.)
A cute trick for computing $ko_*(\mathbb{R}P^{\infty})$ is explained in section 4 of Mahowald-Milgram's "Operations which detect..." (https://academic.oup.com/qjmath/article/27/4/415/1531880)
Right: this uses the fact that the cofiber of the reduced transfer $RP^\infty \to S$ has cohomology which is 'made up of' $A(1)//A(0)$'s, so that smashing it with $ko$ gives a wedge of $HZ$'s in degrees 0,4,8, .... As a $ko$-module, however, the Bott map and hence also the 4 dim class in $ko_$ act nontrivially, so the $Z$ summands in $ko_$ map by an Adams filtration preserving map to this cofiber. The $\eta$'s in the kernel, and the finite quotients in degrees 4n shifted down one, then assemble to give the answer.
There is one extension to determine, from filtration 1 to filtration 2, easily shown to be non-zero, and then Bott periodicity says all the others are non-zero as well.
|
2025-03-21T14:48:29.684212
| 2020-01-28T13:11:16 |
351333
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Aravind A",
"Emil Jeřábek",
"JMP",
"Michael Hardy",
"Nate Eldredge",
"https://mathoverflow.net/users/12705",
"https://mathoverflow.net/users/129086",
"https://mathoverflow.net/users/147924",
"https://mathoverflow.net/users/4832",
"https://mathoverflow.net/users/6316",
"https://mathoverflow.net/users/70355",
"none"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625704",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351333"
}
|
Stack Exchange
|
Is there any way to solve this equation without knowing the inverse modulo?
Suppose I have an arbitrary 256 bit number $m$ another secret number $k$ of the same bit length, and then I multiply them both modulo a 256 bit prime number $p$ to get $c$ as follows:
$$
c = (m\cdot k) \mod p
$$
Is there any way to get $m$ back without knowing $k$?
Is this problem as hard as the discrete log problem?
How can this task be made more computationally difficult?
Without knowing $k$, $m$ can be literally anything. So discrete log is not going to help you, the problem is simply impossible.
@Emil Jeřábek supports Monica Can this be used as a good crypto system ?
It’s just as good and just as bad as any other one-time pad. (The most common construction uses XOR, which is slightly easier to implement, but mathematically speaking any abelian group will work just the same, such as the one you use.)
@Emil Jeřábek supports Monica Thankyou very much ! Is this easy for a Quantum computer to break ?
Neither quantum computers nor any other kind can break a one-time pad.
@Nate Eldredge Can this be used as a good crypto system ?
This is the wrong site to ask about that, and I am the wrong person. But it would appear to have all the same advantages and disadvantages as any other one-time pad. Whether that's "good" depends on your requirements. It certainly doesn't seem to be any significant improvement on what's well known.
@Nate Eldredge Thankyou very much ! But can you please tell me if a same key k is used for different m . How can an attacker exploit the key repetition to break this one time pad ? Is there any "mathematical exploit" to this construction ?
Sorry, I'm not really interested in discussing this any further.
@Nate Eldredge Thankyou, your help is very much appreciated!
Since p is prime, each element except 1 has a multiplicative inverse, so k can be anything you want, and $m=c\cdot k^{-1}$ and you can compute $k^{-1}$ with the extended Euclidean algorithm. Usually you'd use exponentiation by repeated squaring whose inverse is discrete log, or point multiplication on an elliptic curve by repeated doubling, etc. I downvoted because this is not a research math question.
$c$ represents a congruence class, and there are $p$ of them. However both $m$ and $k$ belong to the same complete residue system, and so for any given $c$ there are $2^{256}$ pairs of $m,k$ that satisfy the equation. So, the answer is no.
Is it wise to use this a good crypto system ?
That's a good question for https://crypto.stackexchange.com/.
Thankyou very much for your answer ! How can I move this question to get more answers from there regarding crypto?
This question isn't very crypto, so belongs here. Ask different questions over at crypto and see what responses you get.
Thankyou, that's the same reason why I felt it should be posted it here. I am very much happy with your explanation. By the way can you please tell me how you calculated that there are 2 raised to 256 pairs for any given c ?
By using complete residues as developed by Gauss in Disquisitiones Arithmeticae, which states the number of (non-zero) residues modulo a prime $p$ is $p-1$.
Do you think this construction is weak like a Affine cipher as pointed out by Nemo in the second answer?
@jmp $$ \pi U_n=\bigcup_t ;\pi U_n(t) $$ First time someone deleted a posting within seconds after I corrected a typesetting issue in it. But for future reference you may want to know that among the advantages of \bigcup over \large\cup are the positions of the subscripts and superscripts, as in $\displaystyle \bigcup_{n=1}^\infty, $ etc. (Besides the other advantages.) $\qquad$
@MichaelHardy; I found too many errors in it, so I'm going to look at it over the week, maybe even come up with a fancy name! Thanks for the MathJax tip.
This is a special case of an affine cipher. See the Wikipedia article on affine ciphers (https://en.wikipedia.org/wiki/Affine_cipher) for a discussion on weaknesses and cryptanalysis. It is not considered a good system of encryption.
I read that page, but I don't think that it will be easy to guess the key here, because of its large bit size. Further more I am thinking that if we use a different k, frequency analysis won't be able to crack it. I don't think my question is along the lines of a mono alphabet substitution cipher, where each letter is mapped to an element in the field {0, ..., p-1}. Please can you explain a little more, of why you thought my method, is weak like the Affine cipher...
|
2025-03-21T14:48:29.684678
| 2020-01-28T13:13:50 |
351334
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Willie Wong",
"Zurab Silagadze",
"https://mathoverflow.net/users/32389",
"https://mathoverflow.net/users/3948"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625705",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351334"
}
|
Stack Exchange
|
Gravity, connection, and curvature
Starting with Synge and Fock, many modern authors identify gravity with curvature. On the other hand, Einstein always emphasized that gravity should be equated with a connection, but not with curvature. For example, in a September 1950 letter to Max von Laue, Einstein explicitly stated that, from an empirical point of view, the presence of a gravitational field means that the Christoffel symbols are different from zero, and not the Riemann tensor (a fragment of the letter can be found in the article by J.J. Stachel, "How Einstein discovered general relativity: a historical tale with some contemporary morals").
Einstein’s position may seem a little strange, because, as Peter Havas noted in 1967, Christoffel’s symbols can be made non-zero simply by choosing curvilinear coordinates without any mention of an accelerated frame of reference. For me, the best definition, reconciling both positions, was given by Jürgen Ehlers in the "Survey of General Relativity Theory" https://link.springer.com/chapter/10.1007/978-94-010-2639-0_1 "gravity is a non-integrable, symmetric connection, whose geodesics are the free fall trajectories".
An interesting argument showing that the curvature provides too narrow perspective on gravity was given by Kevin Brown (a mysterious author of excellent site MathPages https://www.mathpages.com/ ) in https://www.mathpages.com/home/kmath622/kmath622.htm
he writes:
It’s possible to arrange masses in such a way as to produce a situation such as shown in this figure. There is no doubt that this situation requires the existence of some curvature somewhere, but the interior region can be intrinsically flat, even though free objects in the interior accelerate relative to the flat exterior spacetime. Would we really say there is no gravitational field in the “flat” interior region? This shows that it is too simplistic to identify gravity with curvature. It is analogous to claiming an absolute distinction between electric and magnetic fields. The approach advocated by Einstein – regarding the metric tensor as both the gravitational and the inertial field, regardless of whether there is curvature at any given point – is much more intelligible.
I wonder if the space-time example shown in the figure has ever been explicitly constructed.
If your question is the final sentence: there are a few problems. (a) what does it mean to have "homogeneous acceleration relative to exterior Minkowski space-time? How is this any different from just choosing a weird coordinate system on the flat space-time where the coordinate system "travels" in the middle? (b) Unless you actually require an equation to be satisfied, in the smooth category you can write down pretty much any pseudo-Riemannian manifold you want. Do you actually mean you want a solution to Einstein's equation with an appropriate matter model that looks like your picture? (ctd)
... If so, for any reasonable matter model your picture is ruled out by the positive mass theorem. (Flat on the exterior means automatically asymptotically flat, but reasonable matter model gives non-negative matter energy density, and so flat at infinity means globally flat.)
In Newtonian gravity if one has a spherical cavity inside a ball, the gravitational field is homogenous inside the cavity (Newtonian space-time is flat inside cavity). However outside the cavity gravitational field is not homogenous - Newtonian space-time is curved outside the cavity except at infinity where it becomes asymptotically flat. I'm asking whether something similar exists in general relativity.
Yes. But note that in the Newtonian picture you can just assume there exists a rigid shell of matter; in the Relativistic picture you have to think about an actual matter model and prove that the matter will not collapse under self gravitation. For certain hyperelastic matter you can have shells. https://arxiv.org/abs/gr-qc/0603103 The exterior metric is that of Schwarzschild.
If you are willing for the shell to be dynamics, then there are many solutions with fluid/dust matter in a shell with a flat interior. // final remark: the Newtonian spherical cavity has 0 gravitational field inside the cavity. It has homogeneous (constant) gravitational potential. But in any case I fail to see how these cases connect with the passage quoted from "MathPages" in yellow in your original post.
If the Newtonian spherical cavity is off-centered (its center doesn't coincide with the center of the ball), the gravitational field inside the cavity will be not zero (but homogeneous/constant).
|
2025-03-21T14:48:29.685011
| 2020-01-28T13:19:27 |
351335
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Goupil",
"fedja",
"https://mathoverflow.net/users/1131",
"https://mathoverflow.net/users/151601"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625706",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351335"
}
|
Stack Exchange
|
Joint optimization of the order 1 moment of a function and its Fourier transform
For the purpose of a quantum optics experiment, I come to the following problem :
Let $X,P \in \mathbf{R}^2$
\begin{equation}
J(\psi) = | \int_{-X}^{X} x |\psi^2(x) | dx + \int_{-P}^{P} p |Tf(\psi)^2(p) | dp |
\end{equation}
Find:
\begin{equation}
\left\{
\begin{aligned}
\sup_{ \psi } J(\psi) \\
\text{with constraint} \int_{\mathbf{R}} |\psi^2(x) | dx = 1
\end{aligned}
\right.
\end{equation}
where Tf is the Fourier transform
$\psi$ functions are supposed to be smooth enough for any purpose. That's why I don't explicitly state in what space they live.
I tried in this way :
1) If i upper bound each term independantly, I can obtain X+P with a $\psi^2(x) = \delta(X) $ and $Tf(\psi^2(p)) = \delta(p)$ . From now on, the aim is to make the best of the relation between $\psi$ and its Tf
2) In quantum physics, the problem can be rewritten in the following form :
\begin{equation}
J(\psi) = | \int_{-X}^{X} x |\psi^2(x) | dx + \int_{-P}^{P} -i \psi^*(x) \frac{\partial \psi}{\partial x} dx |
\end{equation}
I tried to differentiate J with respect to $\psi$ and $\psi^*$ in order to apply Lagrange multipliers on the functional space of $\psi$ functions.
With X = P and the constraint changed
\begin{equation}
\left\{
\begin{aligned}
\sup_{ \psi } J(\psi) \\
\text{with constraint} \int_{-X}^X |\psi^2(x) | dx = 1
\end{aligned}
\right.
\end{equation}
I found:
\begin{equation}
\exists \lambda \text{ such that } \\
i \frac{\partial \psi}{\partial x} + (x - \lambda) \psi = 0
\end{equation}
which I solved with
\begin{equation}
\psi = C \exp \left( i(x^2 - \lambda x) \right)
\end{equation}
Could you please help me with the general problem ? Would you be aware of any litterature treating it ? It seems to me this is linked to the Pauli Problem, yet slightly different.
Thanks in advance.
Should we think of $X$ and $P$ as small or large (or you need the whole range)?
X and P should be between 1 and 4, for typical values
|
2025-03-21T14:48:29.685170
| 2020-01-28T14:45:14 |
351341
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/16934",
"kaleidoscop"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625707",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351341"
}
|
Stack Exchange
|
distributional divergence of the gravitational / Coulomb force close to the boundary
First of all, I am not sure of the terminology here, I am interested in the function $$F(x)=x|x|^{-d},x\in \mathbb{R}^d\setminus \{0\}$$ in dimension $d\geq 2$. I read somewhere that this is called the gravitational force (and I agree in dimension 3, but I am not sure for other dimensions), and elsewhere that this comes from the Coulomb potential.
Anyway, the divergence of this force is $-c_d\delta_0$ in the distributional sense (for some constant $c_d$), i.e. for any sufficiently smooth domain $D\subset \mathbb{R}^d$ not having $0$ in the boudnary and with normal vector $n_D$ on the boundary,
$$
\int_{\partial D}n_D(x)\cdot F(x)S(dx)=-c_d1_{0\in D}.
$$
First question: does anyone knows a good keyword to look upon this function and this kind of facts in the literature? Or a textbook?
Second question: consider now the force of mass spread uniformly over some smooth set $A\subset \mathbb{R}^d$:
$$
F_A(x)=\int F(y-x)1_A(y)dy.
$$
I'm pretty sure this is well defined, at least outside $\partial A$.
Has the divergence of $F_A$ also a good interpretation in the distributional sense? Ultimately I would like to replace $1_A(y)dy$ by some finite measure $\mu(dy)$...
$\text{div},F_A(x)=1_A(x)$, so well-defined for $x$ not on the boundary of $A$.
That is surprising because if $A$ is a ball, then $F_A$ is the force of the corresponding "planet", and I think I remember from physics class that $F_A=0$ inside $A$, which would give $div F(x)=0$ for $x$ in the interior of $A$. I was more expecting some measure lying on $\partial A$...
no, the force of gravity inside a planet is not zero, see, for example, Journey through the center of the Earth
I see, thanks, I was mistaken between the mass on a sphere and the mass on a ball. When you say $div F_A=1_A$, this is true in any dimension? You think it extrapolates to any measure $\mu$?
yest, it holds in any dimension, and for any measure $d\mu(y)=\rho(y)dy$, see answer box.
In $d$ dimensions, with a charge/mass density $\rho(\mathbf{x})$, we have
\begin{align}
&F(\mathbf{x})=\frac{1}{2-d}\,\text{grad}\,\frac{1}{|\mathbf{x}|^{d-2}},\;\;\text{div}\,F(\mathbf{x})=-S_{d}\,\delta(\mathbf{x}),\\
&F_\rho(\mathbf{x})=\int F(\mathbf{y}-\mathbf{x}) \rho(\mathbf{y})\,d\mathbf{y},\\
&\Rightarrow\text{div}\,F_\rho(\mathbf{x})=S_d\int\delta(\mathbf{y}-\mathbf{x}) \rho(\mathbf{y})\,d\mathbf{y}=S_d\,\rho(\mathbf{x}).
\end{align}
Here $S_d$ is the surface area of the $d$-dimensional unit sphere.
In the electrostatic context, $F_\rho$ is the electric field from an electrical charge density $\rho$, and the statement that the divergence of the electric field equals the charge density (times a constant) is the first Maxwell equation (the differential form of Gauss's law).
To more specifically answer the question in the OP: the divergence of the force field is discontinuous but finite at the boundary $\partial A$ of a uniform charge density in $A$.
Ok thanks, but how do you justify switching $\int$ and $div$?
Well, there is an issue here beause $F$ is not absolutely integrable
I am probably just missing your point, but you are basically asking for Gauss' law of electrostatics, that the divergence of the electric field equals the charge density; I thought that relation was true even for singular charge densities, at least I have never seen it conditioned on smooth densities.
Basically I agree with all you said, but I am looking maybe for a textbook doing things in a mathematically rigourous fashion in any dimension. I guess gravitation and electrostatics are more something that happen in dimension 3.
This is an addendum to Carlo's answer, about how to make it rigorous, as per the OP's comments.
Although one can probably do it by hand or brute force, the commutation of ${\rm div}$ and $\int$ is best shown using Fubini's Theorem for distributions as explained in my answer to Can distribution theory be developed Riemann-free?
Careful also about what ${\rm div}$ means. Inside the integral it is made of derivatives in the sense of distributions, while outside these are classical derivatives. In the notations used for my answer to the other MO question, $F_{\rho}$ is an element of $\mathscr{O}_{\rm M}$, the space of temperate smooth functions which embeds into the space of temperate distributions $\mathscr{S}'$. Moreover, one has a commutative diagram, with this embedding turning classical derivatives in $\mathscr{O}_{\rm M}$ into derivatives in the sense of distributions in $\mathscr{S}'$.
Summary:
In the first line of Carlo's answer, everything in sight (and in particular $F$ and ${\rm div}$) is to be understood in the sense of distributions. However, in the second line $F_{\rho}$ is to be understood as a classical function while $\int$ is in the sense of distributions (I prefer the notation $\langle\ ,\ \rangle_{\mathbf{y}}$).
Finally in the third line ${\rm div}$ is first understood as a classical derivative in $\mathscr{O}_{\rm M}$, then in the sense of distributions in $\mathscr{S}'$, before commutation with the formal integral sign.
Remark: In the above I took $\rho$ in $\mathscr{S}$.
|
2025-03-21T14:48:29.685783
| 2020-01-28T14:47:47 |
351343
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Daniele Tampieri",
"https://mathoverflow.net/users/113756"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625708",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351343"
}
|
Stack Exchange
|
Basic operation geometrical meaning
What is the geometrical meaning of doing
$x^TAx \;$?
$Ax \; $ is trivially "applying A to x", but then, what the multiplication for $x^T$ stands for?
Mattia, this question would be better received in our sister site Mathematics.SE than here. Mathoverflow is devoted to mathematical research questions.
In $x^TAx$, you should not think of $A$ as a linear operator; you should think of $A$ as defining a bilinear form or quadratic. If you take $A=I$, the identity matrix, then $x^Tx$ is the dot product of the vector $x$ with itself and $x^Ty$ is the dot product of the vectors $x$ and $y$; taking a different $A$ is replacing the dot product by a new bilinear form in which the product of the $i$th and $j$th standard basis vectors is $a_{ij}$.
|
2025-03-21T14:48:29.685889
| 2020-01-28T15:05:30 |
351346
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"D_S",
"Friedrich Knop",
"https://mathoverflow.net/users/38145",
"https://mathoverflow.net/users/89948"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625709",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351346"
}
|
Stack Exchange
|
Rosenlicht's theorem and fundamental domain for unipotent group acting on $\mathbb A_k^n$
I have a question about unipotent group actions. I was referred to Rosenlicht's papers, but I had trouble getting much out of these because I don't understand the old algebraic geometry language very well. From what I can tell Rosenlicht's results do not imply what I'm looking for. I expect the answer or a reference should be known by people more knowledgeable than myself.
Let $G$ be a connected, unipotent algebraic group over a field $k$ of characteristic zero, acting as a group of $k$-vector space automorphisms on $X = \mathbb A_k^n$.
Rosenlicht's theorem says that there exists a $k$-open set $X'$ of $X$ which is $G$-stable and for which the geometric quotient $G \backslash X'$ exists. It also says that a "cross section $k$-morphism" $G \backslash X' \rightarrow X'$ exists.
My question: let $d$ be the largest dimension of a closed orbit in $X$.
Does there exist a $k$-open $G$-stable set $X'$ of $X$, and a $k$-closed set $W$ of $X'$, such that $W$ is a fundamental domain for the action of $G$ on $X'$, i.e. each the orbit of each $x \in X'$ meets $W$ at exactly one point?
If so, can $X'$ and $W$ be chosen so that the map taking $x \in X'$ to its unique orbit representative in $W$ is the geometric quotient of Rosenlicht's theorem?
Does $W$ always arise from the intersection of $X'$ and a "generic" $n-d$-dimensional subspace of $\mathbb A^n$?
My question is motivated by a general collection of examples I've been looking at for the past couple of years where I have always found a positive answer.
Example:
Let
$$G = \{ \begin{pmatrix} g \\ & h \end{pmatrix} : \textrm{ $g, h \in \operatorname{GL}_n$ are upper triangular unipotent} \}$$
$$X = \begin{pmatrix} 0 & x \\ 0 & 0 \end{pmatrix} : x \in \operatorname{Mat}_n\} \cong \mathbb A_k^{n^2}$$
Then $G$ acts on $X$ by conjugation. The largest dimension of a closed orbit in $X$ is $\operatorname{Dim} G$. A fundamental domain for the action of $G$ on an open set in $X$ is the set of nonzero antidiagonal matrices
$$W = \begin{pmatrix} & & & \ast \\& & \ddots \\ & \ast \\ \ast \end{pmatrix}$$
and the conjugation map $G \times W \rightarrow X$ is an isomorphism of varieties onto an open set in $X$.
More general class of examples where I expect nice results
Let $H$ be a quasi-split group over $k$ with maximal parabolic subgroup $P = MN$ and Borel subgroup $B = TU$ with Levi factors $T \subset M$ and unipotent radicals $N \subset U$. Then $G = U \cap M$ acts as Lie algebra automorphisms of $\mathfrak n = \operatorname{Lie}(N)$ by conjugation. In many examples I have determined that a fundamental domain for the action of $G$ on an open subset $\mathfrak n$ arises from looking at subspaces spanned by certain root vectors.
The affirmative answer to the first two question is indeed well-known. The existence of the section $W$ boils down to the vanishing of $H^1(X,\mathbf G_a)$ on an affine variety $X$.
A generic $n-d$-dimensional subspace will intersect a $G$-orbit in $D$ points where $D$ is the degree of that orbit. So, the answer to the third question is clearly negative if the generic orbits is not itself an affine subspace.
In general, Rosenlicht's theorem is very weak since it doesn't provide any control over $X'$ and $W$. For general actions of unipotent groups one cannot hope for anything better, though. Judging from your examples I presume, though, that you are interested in a much more special (and important) situation namely where $G$ is the unipotent radical of a parabolic $P$ of a reductive group $H$ and the action of $G$ is the restriction of an $H$-action. That' s what the "Local Structure Theorem" is for. It was first proved by Brion-Luna-Vust (Espaces homogènes sphériques, Invent. Math. 84 (1986) 617–632) and goes in its simplest form as follows:
Let $H$ be connected reductive acting on an affine variety $X$. Let $f\in\mathcal O(X)$ be a highest weight vector. Let $P$ (a parabolic) be the stabilizer of $\mathbb Cf$. Let $P_u$ be its unipotent radical and $M\subseteq P$ a Levi complement. Let $P_u^-\subseteq H$ be the opposite subgroup with respect to $M$. Now let
$$X':=\{x\in X\mid f(x)\ne0\},$$ a $P$-stable open subset of $X$, and $$W:=\{x\in X'\mid (\xi f)(x)=0\text{ for all }\xi\in{\rm Lie P_u^ -}\},$$
a closed affine $M$-stable sub variety of $X'$. Then the canonical morphism $$P_u\times W=P\times^MW\to X':[p,x]\mapsto px$$ is an isomorphism of $P$-varieties. In particular, the action of $P_u$ on $X'$ is free and $W$ is a slice.
The big advantage of the LST over Rosenlicht's theorem is that $X'$ and $W$ are completely explicit. For example if $X$ is a vector space and $f$ is linear then $W$ is the open part of a linear subspace, as you have observed in your examples.
The LST applies a priori only to a specific unipotent radical but one can either move to a "generic" semiinvariant $f$ or iterate the construction for the action of $M$ on $W$. At the end one can manage that all unipotent elements of $M$ act trivially on $W$. But then the $U$-orbits in $X'$ are the $P_u$-orbits where $U\subseteq H$ is maximal unipotent.
This way one can show that indeed $W$ can be chosen open in a linear subspace if $X$ is a vector space. For details see my paper "Some remarks on multiplicity free spaces" (Representation theories and algebraic geometry (Montreal, PQ, 1997), 301–317, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 514, Kluwer Acad. Publ., Dordrecht, 1998.)
Thanks very much for your answer, this is exactly the kind of thing I'm looking for. What is meant by saying that a global section $f \in \mathcal O(X)$ is a highest weight vector?
That $f$ is $B$-semiinvariant.
I am trying to recover the fundamental domain of my example in $\operatorname{GL}_{2n}$ using what you have said. $X$ is acted on by the reductive group $H = \operatorname{GL}_n \times \operatorname{GL}_n$, and I am trying to determine a fundamental domain in $X$ for the action of the unipotent radical of the standard Borel of $H$.
The only Borel semi-invariant function I think of on $X$ is the determinant map. The corresponding parabolic subgroup of $H$ is $H$ itself, so this unfortunately tells me nothing. How would one obtain proper parabolic subgroups for this example?
The determinant is of no use since it is even an $H$-semiinvariant. In that case, the parabolic $P$ is all of $H$. Take as $f$ the coefficient $x_{n1}$ in the lower left corner of $x$ which is $B$- but not $H$-semiinvariant. This yields a reduction $n\to n-1$. Repeat $n-1$ times and you obtain $W$ from your example.
The first two questions have an affirmative answer. See my paper https://arxiv.org/abs/1712.03838 for a constructive proof. The results are even true for connected solvable groups, also in positive characteristic. In the paper there are further references for special cases that should cover your setting. Your third questions is not covered, though, since the setting in the papers is an action on an affine or quasi-affine variety. I'd have to think about the third question, but I have doubts. Even for an additive group action, it would mean that a local slice can be found "generically" in degree 1.
|
2025-03-21T14:48:29.686380
| 2020-01-28T15:15:57 |
351347
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Mateusz Kwaśnicki",
"Rajesh D",
"https://mathoverflow.net/users/108637",
"https://mathoverflow.net/users/14414"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625710",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351347"
}
|
Stack Exchange
|
Relation between two matrices associated with a positive definite function
Let $f:\mathbb{R}^N \to \mathbb{R}$ be a positive definite function. Let $$g(h) = \int_{\mathbb{R}^N}f(x)f(h-x)\mathrm{d}x$$ Due to Bochner's and Parseval's theorems, $g$ is also a positive definite function.
Define the set $S$ = $\{x_1,x_2,...x_n\}$ containing $n$ distinct points in $\mathbb{R}^N$
Define matrices $F$ and $G$ as $F = [f_{i,j}]_{n\times n}$ and $G = [g_{i,j}]_{n\times n}$, where $f_{i,j} = f(x_i-x_j)$ and $g_{i,j} = g(x_i-x_j)$.
Question : What is the relation between the matrices $F$ and $G$? If there isn't any simple relation, then how can I determine matrix $G$ from matrix $F$?
Is this question already solved? If someone has any quick reference works or a book, appreciate it.
I doubt there is any link between $F$ and $G$. Just think about $n = 1$: the quantities $f(0)$ and $g(0) = f * f(0)$ are not related in any meaningful way.
|
2025-03-21T14:48:29.686474
| 2020-01-28T16:51:03 |
351355
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625711",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351355"
}
|
Stack Exchange
|
References for numerical approach of Hilbert uniqueness method (HUM)
Finding of the control that achieves the exact controllability of the wave equation (Neumann boundary conditions) using the HUM method (see: J.L. Lions, Controlabilité exacte perturbations et stabilisation des systèmes distribués, Tome 1: controlabilité exacte, Masson, Paris, 1988) depends to the resolution of the equation $\Lambda (\phi_0 , \phi_1 ) = (\psi_t (x,0). \psi(x,0))$.
How can one solve this equation numerically?
How can one find the initial conditions of the adjoint system?
|
2025-03-21T14:48:29.686548
| 2020-01-28T17:05:13 |
351357
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"A. Batsis",
"Alexandre Eremenko",
"Jochen Glueck",
"Nate Eldredge",
"https://mathoverflow.net/users/102946",
"https://mathoverflow.net/users/130361",
"https://mathoverflow.net/users/25510",
"https://mathoverflow.net/users/4832"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625712",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351357"
}
|
Stack Exchange
|
Matrix iteration for non-negative matrices. Does it converge to some eigenvector?
Let $A$ be a non-negative (entrywise) matrix such that $A(1,1)>0$. Set $u=(1,0,0,...,0)^T$. Is it always true that there exists a non-negative eigenvector $v$ of $A$ such that $\lim_{n\rightarrow\infty}\frac{A^nu}{||A^nu||_1}=\frac{v}{||v||_1}$?
$A = \left(\begin{smallmatrix}0 & 1 \ 1 & 0 \end{smallmatrix} \right)$, $u = (1,0)$
@NateEldredge thank you for your comment I added some more details.
Search Perron's theorem on Google.
@AlexandreEremenko (and at those who voted to close): I think the problem is more subtle than just a standard applications of Perron-Frobenius theory. If $A$ were irreducible, it would indeed be quite simple, but as $A$ is not assumed to be irreducible, various problems occur: for instance, $A$ can have more than one eigenvalue on the spectral circle, and these eigenvalues might not be semi-simple. Still, the assumption that $A(1,1) > 0$ and that $u(1) = 1$ gives some additional structure to the problem - so the answer might still be "yes", but I don't think it's in the standard references.
The statement is not true. Let $a>1$ and define
\begin{equation}
A:=\begin{bmatrix}1&0&0\\1&0&a\\0&a&0 \end{bmatrix}.
\end{equation}
Suppose there is $v\in\mathbb{R}^n\backslash\{0\}$ such that $\lim_{n\rightarrow\infty}\frac{A^nu}{||A^nu||_1}=\frac{v}{||v||_1}$. By a trivial induction argument we can prove that
\begin{equation}
A^nu=\left(1,\frac{a^{n+1}-1}{a^2-1},\frac{a^{n}-a}{a^2-1}\right)
\end{equation}
when $n$ is odd bigger than $2$ and
\begin{equation}
A^nu=\left(1,\frac{a^{n}-1}{a^2-1},a\frac{a^{n}-1}{a^2-1}\right)
\end{equation}
when is $n$ even bigger than 3. Observe that $\frac{v(1)}{||v||_1}=\lim_{n\rightarrow\infty}\frac{A^nu(1)}{||A^nu||_1}=0$ so $v(1)$=0. That means that $u(2)\neq 0 $ or $u(3)\neq 0$.
From that we get
\begin{equation}
\lim_{n\rightarrow\infty}\frac{A^nu(3)}{A^nu(2)}=\begin{cases}\frac{v(3)}{v(2)}\quad v(2)\neq 0\\\infty \quad v(2)=0\end{cases}.
\end{equation}
This gives a contradiction since
\begin{equation}
\lim_{n\rightarrow\infty}\frac{A^{2n+1}u(3)}{A^{2n+1}u(2)}=a^{-1}
\end{equation}
and
\begin{equation}
\lim_{n\rightarrow\infty}\frac{A^{2n}u(3)}{A^{2n}u(2)}=a.
\end{equation}
|
2025-03-21T14:48:29.686820
| 2020-01-28T17:58:33 |
351359
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/29697",
"usul"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625713",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351359"
}
|
Stack Exchange
|
Nash equilibria for "presidential election" game
Suppose, in a country there are $m$ different social issues, positions on which are being indexed with numbers $[-1; 1]$, with radicals on the opposing ends and moderates in the center. In this country there are also $n$ candidates running for president. Before they run, they have to choose their position on all those social issues by placing themselves into some point of this $[-1;1]^m$ «political compass». Suppose they have chosen their positions $a_1, … , a_m$ (as vectors in $\mathbb{R}^n$). Then define $Pref(x)$ as the set of the preferable candidates for voters with position $x$ (the candidates, the Euclidean distance from whose position to $x$ is minimal). Suppose, the voter positions are uniformly distributed on $[-1;1]^m$ and that each voter in the position $x$ chooses one of the candidates from $Pref(x)$ with equal probability. (So the total percentage of votes received by $k$-th candidate is $\int_{[-1;1]^m} \frac{I_{Pref(x)}(k)}{2^m|Pref(x)|}dx$). After the votes are counted, the president is chosen at random with equal probability from those candidates, that received the maximal numbest of votes. The candidates do not have their own opinion on those $m$ issues and only want to win the election (thus their final payoff is the exact probability of them winning). Is there some sort of classification of Nash equilibria in this class of games (for different $m$-s and $n$-s)?
For $n = 1$, every position is a Nash equilibrium as the only candidate will win the election no matter what they do.
For $n = 2$ both candidates should take the extremely moderate position that is $0$. That is due to the «geometric» fact, that if the positions of two candidates are distinct, then the one who is closer to the center wins. So if $a_1 \neq 0$, then the second candidate can adopt the position $\frac{a_1}{2}$ to win the election for sure, and vice versa.
However, I do not know, how to deal with the general problem.
This question on MSE. There is also a very useful partial answer by @antkam, with analysis of the particular case when $m = 1$ and $n$ is even.
This is a location game, a class of games that was well studied in game theory. Look at Exercise 4.49 in Maschler-Solan-Zamir, which contains one of the results in the link you provided. You can find other results by searching the web.
Also called Voronoi games (as any strategy profile induces a Voronoi diagram).
|
2025-03-21T14:48:29.687031
| 2020-01-28T19:10:51 |
351366
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625714",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351366"
}
|
Stack Exchange
|
Morphisms of hammocks in the simplicial localization
Let $\mathcal{C}$ be a category together with a wide subcategory $\mathcal{W} \subset \mathcal{C}$.
In Calculating Simplicial Localizations by Dwyer and Kan, a morphism of hammocks is defined to be a natural transformation between the zigzags that fixes the endpoints and with the extra property that vertical maps are in $\mathcal{W}$.
I have seen claims in the literature and in this post, stating that we can relax the second condition in the definition of morphism of hammocks and consider any morphism of $\mathcal{C}$ and still obtain equivalent simplicially enriched categories modelling the localization.
I have not found a reference for this fact and I would like to know if there is one available.
The link in the OP leads to a thread with an answer by Charles Rezk, who wrote
For each "shape" of zig-zag, there is a "hammock category" for it... whose objects are functors $f\colon Z\to C$ ($Z$ is an abstract zig-zag of a particular shape) such that the backwards arrows of $Z$ are sent into $W$. The morphisms are natural transformations $f\to f'$ which are identities at the ends (and which in the original formulation of Dwyer and Kan are such that the vertical arrows of the transformation must also be in $W$, though this condition turns out not to really be necessary, so it is nowadays often dropped)...Go look at the original Dwyer-Kan paper, or at the book by Dwyer-Hirschhorn-Kan-Smith.
Consider a hammock without the assumption that the vertical morphisms are weak equivalences. It might help to look at the picture here or on page 8 of this pdf. In the first link, since the backwards arrows $X \gets K_1$ and $X\gets L_1$ are weak equivalences, then so is the vertical arrow $K_1\to L_1$, by the two-out-of-three property. However, for a hammock with many intermediate layers, the two-out-of-three property does not imply all vertical arrows are weak equivalences.
Now consider the situation from 3.2 of the book Homotopy Limit Functors on Model Categories and Homotopical Categories by Dwyer-Hirschhorn-Kan-Smith. This is about hammocks coming from model categories. In that context, of a model structure, the authors prove that elements of $Ho M(X,Y)$ are in one-to-one correspondence with equivalence classes of zig-zags of the form $X \gets \bullet \to \bullet \to Y$ (let's call any hammock with more non-composable arrows than this "wide"). Note that, for a hammock of this form, working from the outside to the inside, you could deduce that the vertical maps in a map between zig-zags are weak equivalences, knowing that the backward maps are. And the authors point this out in their 3.2. Later, in 7.7, the authors point out that this argument didn't really need anything to do with the cofibrations and fibrations, so should remain true in any homotopical category that admits a 3-arrow calculus. That's proven in their 11.2. I strongly suspect this is what Rezk had in mind, given the quote above.
Now, note that when you drop the assumption about the 3-arrow calculus, then you cannot automatically reduce from an arbitrary "wide" zig-zag to one of the form displayed above (with two backwards arrows and one forwards arrow). In Dwyer-Hirschhorn-Kan-Smith, 34.2, they take extra care to work with arbitrary zig-zags and for these they do need to assume the vertical maps are weak equivalences, in a map between zig-zags. In the original Dwyer-Kan paper, they do require the vertical maps to be weak equivalences when they deal with "wide" zig-zags.
Now, the natural question is "does every relative category admit a 3-arrow calculus?" And the answer is no. You can read more about this in the following papers: 1, 2, 3, 4. These papers illustrate the yoga of the 3-arrow calculus, and give several examples of models of $(\infty,1)$-theory that do admit the 3-arrow calculus, including partial model categories. If someone hands you a relative category, you can replace it by a partial model category of the same homotopy type and use the 3-arrow calculus there, which is perhaps also something Rezk had in mind when he gave his answer.
|
2025-03-21T14:48:29.687348
| 2020-01-28T19:31:49 |
351367
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Noah Schweber",
"Zuhair Al-Johar",
"https://mathoverflow.net/users/8133",
"https://mathoverflow.net/users/95347"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625715",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351367"
}
|
Stack Exchange
|
Does MK prove internally that there are more proper classes than sets?
Is the following provable in MK?
$\not \exists S: \\ \text{ } \\1. \ \ \forall s \in S \exists a,b (s=\langle a,b \rangle) \\ \text{ } \\2. \ \ \forall x (set(x) \to \exists! X (\neg set(X) \land \{x\} \times X \subset S \land \forall k(\langle x,k \rangle \in S \rightarrow k \in X))) \\ \text{ } \\3. \ \ \forall X (\neg set(X) \to \exists! x ( \{x\} \times X \subset S \land \forall k(\langle x,k \rangle \in S \rightarrow k \in X))) $
In English: there exists a class $S$ of ordered pairs, such that for every set $x$ there exist a unique proper class $X$ such that $\{x\} \times X$ is a maximal subclass of $S$ that is a cartesian product of $\{x\}$ by a class. And the opposite direction also holds, i.e. for every proper class $X$ there exists a unique set $x$ such that $\{x\} \times X$ is a maximal sublass of $S$ that is a cartesian product of $\{x\}$ by a class.
Now it is known that there can exist a class $S^*$ that fulfills only the first two conditions, simply take the union of all cartesian product classes of singletons by their complements.
The idea here is to express that there are more proper classes than sets, internally within MK. Since existence of $S^*$ can be taken to mean that there are at least as many proper classes as sets are there. But failure of condition 3, tell's us that we don't have as many sets as proper classes are there, so from both situations we infer that there are more proper classes than sets. However is non existence of $S$ provable in MK? What's the proof?
Maybe I'm missing something, but isn't this just the diagonal argument?
Suppose $S$ were as in your question. Let $E=\{\langle x,y\rangle: set(x)\wedge y\in x\}$, and consider the set $$Z=\{\langle \langle x,0\rangle, y\rangle: \langle x,y\rangle\in S\}\cup\{\langle\langle x,1\rangle, y\rangle: \langle x,y\rangle\in E\}.$$ $Z$ has the property that for each class (proper or non) $C$ there is some $a$ with $$C=\{y: \langle a,y\rangle\in Z\}.$$
We now apply diagonalization to this new $Z$: the class $$D=\{x: \langle x,x\rangle\not\in Z\}$$ cannot appear as any of the "rows" of $Z$.
Yes but $D$ need not be a proper class?
@ZuhairAl-Johar The emphasis on proper classes isn't essential - we can always "merge" two enumerations, and since we have an enumeration of all sets on hand the issue remains. (I've edited to address this.)
what does the "rows" of $Z$ mean?
@ZuhairAl-Johar The classes indexed by $Z$ - the classes of the form ${y: \langle a,y\rangle\in Z}$ for some fixed $a$.
Yes! That works! Although it is NOT "just" the diagonal argument. It is a version of it, but not the usual one! So as I expected MK can indeed prove having more proper classes than sets! Thanks!
@ZuhairAl-Johar "it is NOT "just" the diagonal argument. It is a version of it, but not the usual one!" I disagree with this - the tweak is really cosmetic, and the underlying idea is exactly the same.
|
2025-03-21T14:48:29.687563
| 2020-01-28T19:45:03 |
351368
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625716",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351368"
}
|
Stack Exchange
|
Physical interpretation of the Manifold Hypothesis
Motivation:
Most dimensionality reduction algorithms assume that the input data are sampled from a manifold $\mathcal{M}$ whose intrinsic dimension $d$ is much smaller than the ambient dimension $D$. In machine learning and applied mathematics circles this is typically known as the manifold hypothesis.
Empirically, this is observed to be true for many kinds of data including text data and natural images. In fact, Carlsson et al. found that the high-dimensional space of natural images has a two-dimensional embedding that is homeomorphic to the Klein bottle [1].
Question:
Might there be a sensible physical interpretation of this phenomenon? My current intuition, from the perspective of dynamical systems, is that if we can collect large amounts of data for a particular process then this process must be stable. Might there be a reason why stable physical processes would tend to have low-dimensional phase spaces?
Might there be a better physical perspective for interpreting this phenomenon?
References:
Carlsson, G., Ishkhanov, T., de Silva, V. et al. On the Local Behavior of Spaces of Natural Images. Int J Comput Vis 76, 1–12. 2008.
Charles Fefferman, Sanjoy Mitter, and Hariharan Narayanan. TESTING THE MANIFOLD HYPOTHESIS. JOURNAL OF THE AMERICAN MATHEMATICAL SOCIETY
Volume 29, Number 4, October 2016, Pages 983–1049. 2016.
Bastian Rieck, Markus Banagl, Filip Sadlo, Heike Leitte. Persistent Intersection Homology for the Analysis of Discrete Data. Arxiv. 2019.
D. Chigirev and W. Bialek, Optimal manifold
representation of data : an information theoretic approach, in Advances in Neural Information Processing Systems 16 161–168, MIT Press, Cambridge MA. 2004.
T. Roweis and L. K. Saul. Nonlinear dimensionality reduction by locally linear embedding. Science, 290(5500):2323–2326, 2000.
Henry W. Lin, Max Tegmark, and David Rolnick. Why does deep and cheap learning work so well? Arxiv. 2017.
Q: Might there be a reason why stable physical processes would tend to have low-dimensional phase spaces.
Yes. One reason is physical processes have dissipation. E.g., turbulence is "known" to be chaotic dynamics on a low dimensional manifold (i.e., strange attractor) in the infinite dimensional phase space (of $L^2$ velocity fields). Even its dimension can be estimated. See for example:
Doering, Charles R., and John D. Gibbon. "Note on the Constantin-Foias-Temam attractor dimension estimate for two-dimensional turbulence." Physica D: Nonlinear Phenomena 48.2-3 (1991): 471-480.
.
An example I was told one time was to imagine data recording the location of a pen tip in space, being held by a human hand (or a robot arm). The pen tip will trace out smooth curves as a function of passing through the angles describing the wrist, elbow, fingertips, etc.
A quick Google search turned up this PDF, which makes the analogy more explicit in terms of a "configuration space".
|
2025-03-21T14:48:29.687785
| 2020-01-28T20:06:37 |
351369
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Hailong Dao",
"R. van Dobben de Bruyn",
"Sasha",
"Yosemite Stan",
"https://mathoverflow.net/users/17630",
"https://mathoverflow.net/users/2083",
"https://mathoverflow.net/users/4428",
"https://mathoverflow.net/users/82179"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625717",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351369"
}
|
Stack Exchange
|
Global obstructions for being a quotient of a rank $d$ vector bundle
In this recent question (which now has an answer), Richard Thomas asked whether any projective $k$-scheme $X$ of (local) embedding dimension $d(X)$ can be embedded in a smooth $k$-scheme of dimension $d(X)$. If $i \colon X \hookrightarrow Y$ is such an embedding, then in particular we get a surjection $i^*\Omega_Y \twoheadrightarrow \Omega_X$. My (so far unsuccessful) strategy was to obstruct such a surjection from existing.
For a coherent sheaf $\mathscr F$, write $d_x(\mathscr F) = \dim_{\kappa(x)} \mathscr F_x \otimes_{\mathcal O_{X,x}} \kappa(x)$ and
$$d(\mathscr F) = \max \left\{d_x(\mathscr F)\ |\ x \in X\right\}.$$
Question. If $X$ is a quasi-projective $k$-scheme, and $\mathscr F$ a coherent sheaf, does there exist a surjection $\mathscr E \twoheadrightarrow \mathscr F$ from a locally free sheaf of rank $d(\mathscr F)$?
Already if $X = \mathbf A^n$ this seems false to me; for example there should exist finite modules $M$ with $d(M) = 2$ that cannot be generated by $2$ elements (here I am using the Quillen–Suslin theorem that a finite projective module on $\mathbf A^n$ is free). But I don't know so many ways to prove that something is not generated by $2$ elements, except for a local obstruction $d_x(\mathscr F) > 2$.
I think it should be possible to give a negative answer to Thomas's question along these lines, by exhibiting a finite flat cover $\pi \colon X \to \mathbf A^n$ such that $\pi_*\Omega_X$ does not admit a surjection from a vector bundle of rank $\deg(\pi) \cdot d(\Omega_X)$. A great answer would incorporate something like this, but I would already be very happy with some global obstruction to surjecting from a vector bundle of a given rank.
You may want to look at the literature around Foster-Swan theorem.
Let me explain a simple example.
Let $C \subset \mathbb{P}^3$ be a twisted cubic curve. It is a locally complete intersection of codimension 2, hence its ideal $I_C$ is locally generated by two sections. Let me show that there are no surjections $E \twoheadrightarrow I_C$ from a locally free sheaf $E$ of rank 2.
Indeed, assume such a surjection exists. Its kernel is a reflexive sheaf of rank 1, hence is a line bundle, so we have an exact sequence
$$
0 \to L \to E \to I_C \to 0.
$$
Restricting to $C$ we obtain an exact sequence
$$
0 \to \det N^* \to L\vert_C \to E\vert_C \to N^* \to 0,
$$
where $N^*$ is the conormal bundle. But $N^*$ is locally free of rank 2, hence the surjection $E\vert_C \to N^*$ is an isomorphism, hence the middle arrow is zero, hence
$$
\det N^* \cong L\vert_C.
$$
But the adjunction fromula shows that $\det N^* \cong \mathcal{O}_C(-10)$, and this line bundle does not restrict from $\mathbb{P}^3$ (because 10 is not divisible by 3). This contradiction proves that no surjection from $E$ as above exists.
Of course, the same argument works for many other lci of codimension 2.
Could you say a word about where $\det N^*$ comes from? (I can do a $\mathscr Tor$ computation locally, but the Koszul resolution doesn't globalise by the very statement you're trying to prove.)
In fact, it does (sometimes this is called the fundamental local isomorphism): $Tor_i(\mathcal{O}_Z,\mathcal{O}_Z) \cong \wedge^iN^*$ for any lci scheme $Z$.
I don't mean that the resolution generalizes, but its consequence (the isomorphism for $Tor_i$) does.
Ah, that clarifies it (but I think your indexing might be off).
What is the problem with my indexing?
Apologies; I misread the terms inside the $\mathscr Tor$ and thought that this statement contradicted the isomorphism $\mathscr Tor_1(\mathcal O_Z, \mathcal I_Z) \cong \wedge^2 N^*$ that you used.
Maybe it's worth mentioning that in many "practical" situations, by the Serre construction, the existence of an isomorphism $\text{det} N^*\cong L|_C$ is exactly what is needed to give a global resolution of $I_C$.
@YosemiteStan: doesn't the Serre construction mostly refer to codimension $2$ situations? Or is there a more general thing?
@R.vanDobbendeBruyn Yes! I was referring to the context of the answer not the context of the original question :)
|
2025-03-21T14:48:29.688074
| 2020-01-18T04:43:53 |
350671
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Brendan McKay",
"VS.",
"https://mathoverflow.net/users/136553",
"https://mathoverflow.net/users/9025"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625718",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350671"
}
|
Stack Exchange
|
Sum of squares of middle binomial sums or 'Truncated mean' of binomial coefficients under binomial distribution
$\mu=1+\epsilon$ where $\epsilon>0$ holds.
1.Is there a good bound for $$T=\frac{\sum_{i=-\sqrt{\mu n\ln n}}^{\sqrt{\mu n\ln n}}\binom{n}{\frac n2 +i}^2}{2^n}?$$
This quantity can be interpreted as $$\sum_{i=-\sqrt{\mu n\ln n}}^{\sqrt{\mu n\ln n}}\binom{n}{\frac n2 +i}\mathbb P(\frac n2+i)$$
where $\mathbb P(\frac n2+i)$ is under bionmial distribution and thus has probability $\frac{\binom{n}{\frac n2 +i}}{2^n}$ which is 'trucated expected value of $\binom{n}{\frac n2 +i}$'.
Computing few values suggests $\log_2T< n-\log_2\sqrt{\mu n\ln n}$ at $\mu\rightarrow1^+$.
For example at $n=1000$ to $50000$ gives such margin.
How large can $\mu$ be for this $\log_2T< n-\log_2\sqrt{\mu n\ln n}$ bound to hold up?
Naively I can get $$<\frac{2^n}{\sqrt{n\pi/2}}(1-o(1))$$ by using $$2^nT<\binom{n}{\frac n2}\sum_{i=-\sqrt{\mu n\ln n}}^{\sqrt{\mu n\ln n}}\binom{n}{\frac n2 +i}.$$
I also know we can prove $$\binom{n}{\frac n2 +i}\asymp\frac{2^{nH(\frac12+\frac in)}}{\sqrt{n\pi/2}}$$ approximation.
I don't know if $\mu$ is something special or just a parameter, but it is odd that increasing $\mu$ makes your sums larger and your conjectured answers smaller.
$\mu=1+\epsilon$ at $\epsilon>0$. That is correct. Perhaps then the relation is flawed. On the $\log_2$ scale I see $\ll n-\log_2 f(n)$ where $f(n)$ seems to be $\Omega(\sqrt{\mu n\ln n})$ when $\mu=1.001$.
The normal approximation of the binomials will give you accurate values.
can you explain the n-log_2f(n) and what f(n) should i expect?
Your main conjecture is not quite correct.
Indeed, for each natural $j$, let $B_j$ be a random variable (r.v.) with the binomial distribution with parameters $j$ and $1/2$, and let $C_j$ be an independent copy of $B_j$. Let also $u:=\sqrt{\mu n\ln n}$. Then for all even natural $n$
$$T/2^n=U_n:=\sum_{k\colon\,|k-n/2|<u}P(B_n=k,C_n=n-k)
=\sum_{k=0}^nP(B_n=k,C_n=n-k)-R_n=P(B_{2n}=n)-R_n,
$$
where
$$R_n:=\sum_{k\colon\, n/2\ge|k-n/2|\ge u}P(B_n=k,C_n=n-k)\le P(B_n\ge n/2+u)^2
\le e^{-4u^2/n}=1/n^{4\mu^2},
$$
where, in turn, the latter inequality is an instance of an exponential Hoeffding inequality. On the other hand, by Stirling's formula,
$$P(B_{2n}=n)\sim1/\sqrt{\pi n}$$
(the asymptotics everywhere here are as $n\to\infty$). So,
$$T_n\sim2^n/\sqrt{\pi n}$$
and hence
$$\log_2T=n-\log_2\sqrt{(1+o(1))\pi n}.$$
So, the inequality $\log_2T<n-\log_2\sqrt{\mu n\ln n}$ does not hold for any real $\mu\ge1$ if $n$ is large enough.
|
2025-03-21T14:48:29.688253
| 2020-01-18T05:44:46 |
350674
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Nik Weaver",
"Robert Furber",
"https://mathoverflow.net/users/151245",
"https://mathoverflow.net/users/23141",
"https://mathoverflow.net/users/61785",
"user151245"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625719",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350674"
}
|
Stack Exchange
|
Proof of uniqueness of predual of von Neumann algebra
I am currently reading Jesse Peterson's lecture notes on von Neumann algebras. I'm confused by lemma 4.4.2. In particular, it seems to me that Hahn Banach theorem here can only conclude the map $\phi$ is in the X** not in X. Without this it seems the subsequent proof of uniqueness of predual of von Neumann algebra cannot go through. Can somebody help? Thank you very much.
I haven't read Peterson's notes, but I can tell you that Sakai's original paper proving uniqueness of the predual is pretty readable.
The proof is using the version of the Hahn-Banach theorem which works for locally convex spaces. We have a $C^*$-algebra $A$ and a Banach space $X$ with $X^*=A$. We then equip $A$ with the weak$^*$-topology, that is, $\sigma(A,X)$. As the proof notes, $(A)_1 \cap A_+$ is $\sigma(A,X)$ compact, so if $a<0$ we can find a $\sigma(A,X)$-continuous linear functional which separates. But $\sigma(A,X)$-continuous linear functionals are simply elements of $X$.
I agree that this is a little unclear in the proof. What's a good book for the Hahn-Banach theorem at this level of generality? You could look at Rudin, Functional Analysis. Or I really like Conway's Functional Analysis. I think Theorem 3.9 of Chapter IV in the 2nd edition is enough. I'm afraid I don't know a good online resource (can anyone suggest one?)
I don't know an online resource either, but these statements can be found in Schaefer's Topological Vector Spaces as II.9.2 and IV.1.2, and also somewhere in the first volume of Dunford and Schwartz. The statement that $\sigma(A,X)$-continuous linear functionals all come from elements of $X$ is itself not trivial unless you know to use the definition of continuity of linear maps in terms of neighbourhoods of $0$.
Thank you very much for the help. I understand the proof now.
|
2025-03-21T14:48:29.688406
| 2020-01-18T06:06:51 |
350675
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"YCor",
"https://mathoverflow.net/users/14094"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625720",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350675"
}
|
Stack Exchange
|
Can matrix factorizations be nonsquare?
I originally asked this question on math stackexchange, but got no attention. As such, I am asking here as well.
$\textbf{Background}$
On page 49 of a 1980 paper by Eisenbud, he defines matrix factorizations. Let $R$ be a (unital) commutative ring. Given $x\in R$, Eisenbud defines a matrix factorization for $x$ as a pair of $R$-linear maps $(\varphi:F\to G,\psi:G\to F)$ where $G$ and $F$ are free $R$-modules such that $\varphi\circ\psi=x\cdot id_G$ and $\psi\circ\varphi=x\cdot id_F$. Note that some authors require in their definition that $\text{rank}(F)=\text{rank}(G)$, but I am trying to work with Eisenbud's definition. In the case that $F$ and $G$ have finite rank, the definition can be stated equivalently with matrices instead of $R$-linear maps.
It's not hard to see that $F$ and $G$ must have the same rank if $x$ is a non-zero divisor of $R$ since $x$ being a non-zero divisor implies $\varphi$ and $\psi$ must be injective. Eisenbud shows in Corollary 5.4 of his paper that $\text{rank}(F)=\text{rank}(G)$ whenever $R$ is Noetherian and $(x)/(x^2)$ is free over $R/(x)$ (although I think he is missing an assumption here since this appears to be false when $x=0$).
$\textbf{Question}$
Eisenbud does not include any examples of matrix factorizations with free modules of different ranks in his paper, nor have I encountered them elsewhere. As such, I am looking for an example of a matrix factorization for a nonzero element of a ring using free modules of different ranks. By the earlier statements, the matrix factorization would need to be for a zero divisor, $x\in R$, such that either $R$ is not Noetherian, or $(x)/(x^2)$ is not free over $R/(x)$. If possible, I would prefer for $R$ to be Noetherian and for the matrix factorization to be between modules of finite rank, so that the maps can be expressed as matrices.
Written in terms of matrices, I am looking for a Noetherian ring, $R$, a nonzero element, $x\in R$, an $n\times m$ ($n\neq m$) matrix, $A$, and an $m\times n$ matrix, $B$, such that $$AB=x\cdot I_{n\times n}$$ $$BA=x\cdot I_{m\times m}$$
If you have such a 6-tuple $(R,m,n,A,B,x)$, then you also have one $(R',m,n,A',B',y)$ for some finite local ring $R'$ and $y^2=0$, by an easy argument.
Note that the equality of traces yields $|m-n|x=0$, and in particular this is impossible for $|m-n|=1$.
Still it is possible for $(m,n)=(1,3)$. Let $R$ be the ring $\mathbf{F}_2[a_1,a_2,a_3,b_1,b_2,b_3]/J$, where the ideal $J$ is generated by all $a_ib_j$ for $i\neq j$ and $a_1b_1-a_2b_2$, $a_2b_2-a_3b_3$. Write $x=a_1b_1=a_2b_2=a_3b_3$ in $R$. Note that $x\neq 0$. Indeed, grading $R$ with all generators of degree 1, $J$ is graded and $J_2$ is linearly generated by the given family, which is linearly independent, and we readily see $a_1b_1\notin J_2$.
Let $A$ be the $3\times 1$ matrix $(a_1,a_2,a_3)^\dagger$ and $B$ the $1\times 3$ matrix $(b_1,b_2,b_3)$. Then $AB=xI_3$, and $BA=3xI_1=xI_1$.
|
2025-03-21T14:48:29.688615
| 2020-01-18T10:02:33 |
350678
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Martin Brandenburg",
"YCor",
"Yemon Choi",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/142808",
"https://mathoverflow.net/users/2841",
"https://mathoverflow.net/users/44191",
"https://mathoverflow.net/users/763",
"phdstud",
"user44191"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625721",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350678"
}
|
Stack Exchange
|
Closed submonoid of $(\mathbb{C}^*)^n$
The answer of this question might be known but I was not able to find any answer. Let $n\geq 1$ and $S$ be a closed submonoid of $(\mathbb{C}^*)^n$, that is, a closed and stable by product subset of $(\mathbb{C}^*)^n$ which also contains the unit $(1,\ldots,1)$. Let also $R_1,\ldots,R_d\geq1$, and denote by $A:=\{(r_ie^{i\theta_i})_{i=1}^{d}:\text{ for every }i=1,\ldots,d\ , 1\leq r_i\leq R_i, \theta_1,\ldots,\theta_d\in[0;2\pi]\}$.
We assume that (the product of the two sets has to be understood component by component):$$A\times S=(\mathbb{C}^*)^n.$$
Whatever $R_1,\ldots,R_d$ and $n$, is it true that $S$ has to be a subgroup?
How about $n = 1, S = 2^a 3^{-b}, (a, b) \in \mathbb{N}$?
@user44191 This is not a closed subset. Indeed its log is the subsemigroup generated by $\log(2)$ and $-\log(3)$. But for any $u<0<v$ with $u/v$ irrational, the additive subsemigroup generated by $u$ and $v$ is not discrete (it's dense in $\mathbf{R}$).
@YCor Ah, I thought "closed" here was algebraic, not topological. That does make more sense.
@MarkSapir I agree that the OP should clarify some of his or her terminology, but I think a reasonable guess would be that he or she seeks a closed subset of $({\mathbb C}^)^n$ with its usual topology, which is also a submonoid for the natural pointwise product on $({\mathbb C}^)^n$
With the new modification, I'm pretty sure the problem reduces to the question of whether every closed submonoid of a torus is a subgroup.
@user44191 if that is the case, then the answer to the problem stated in your comment is "yes" by invoking general macinery: such a submonoid would be a compact semigroup which is cancellative (ax=bx implies a=b and similarly on the left) and compact cancellative semigroups are known to be groups. In the case of tori this should be provable in a more direct way, I guess
@YemonChoi: That's exactly what I meant, thank you!
@Mark Sapir: I changed this mysterious set. Is it clearer now?
@MarkSapir Presumably, it means that the subset is closed under multiplication, i.e. it forms a monoid.
@user44191:How do you transfer this to a torus as you mentionned earlier?
My idea in the above comments didn't quite work as stated, but the generality of the result Yemon Choi mentioned bridges the gap.
Define $\log^n |\cdot|: \mathbb{C}^{*n} \rightarrow \mathbb{R}^n, \log^n |(z_1, z_2, \dots, z_n)| = (\log |z_1|, \log |z_2|, \dots, \log |z_n|)$. This is a map of topological groups, where the "multiplication" on $\mathbb{R}^n$ is addition. Then the $A_R$ condition can be rewritten as: $$\log^n |S| + [0, \log R]^n = \mathbb{R}^n \left(1\right)$$
Let $f \in S$; we want to prove that $f^{-1} \in S$. Condition $\left(1\right)$ can be used to show that there is some set $\{g_i\}$ such that $\{log^n |g_i|\}$ forms a basis of $\mathbb{R}^n$ and such that the coordinates of $\log^n |f|$ are negative with respect to that basis. Then I claim that $f^{-1} \in \overline{\{\prod_i g_i^{a_i} f^b | a_i, b \in \mathbb{Z}_{\geq 0}\}}$. Equivalently, $e \in \overline{\{\prod_i g_i^{a_i} f^b | a_i \in \mathbb{Z}_{\geq 0}, b \in \mathbb{Z}_{\geq 1}\}}$.
Assume otherwise. Consider the map $p: \mathbb{C}^{*n} \rightarrow \mathbb{C}^{*n}/\{\prod g_i^{a_i} | a_i \in \mathbb{Z}\} \simeq \mathbb{T}^{2n}$. This is a map of topological groups, so $p(\{f^b | b \in \mathbb{Z}_{\geq 1}\})$ is a subsemigroup. It is not necessarily closed; however, its closure is - so by Yemon Choi's comment, it must be a group. Specifically, it must include the identity.
Let $e \in U \subseteq \mathbb{C}^{*n}$ be an open neighborhood of the identity; we want to show that $U$ contains an element of $\{\prod_i g_i^{a_i} f^b | a_i \in \mathbb{Z}_{\geq 0}, b \in \mathbb{Z}_{\geq 1}\}$. We can assume WLOG that U is both symmetric and "small enough" (more on this later). Then $p(U)$ is an open neighborhood of the identity in $\mathbb{T}^{2n}$, so it must contain some element of $p(\{f^b | b \in \mathbb{Z}_{\geq 1}\})$. Equivalently, $U$ must contain some element of the form $\prod_i g_i^{a_i} f^b$ such that $a_i \in \mathbb{Z}, b \in \mathbb{Z}_{\geq 1}$. But because the coordinates of $\log^n |f|$ are all negative with respect to $\log^n |g_i|$, by choosing $U$ small enough, we can guarantee that all of the $a_i$ are positive - so we have proven that $U$ contains an element of $\{\prod_i g_i^{a_i} f^b | a_i \in \mathbb{Z}_{\geq 0}, b \in \mathbb{Z}_{\geq 1}\}$. We are therefore done: $e \in \overline{\{\prod_i g_i^{a_i} f^b | a_i \in \mathbb{Z}_{\geq 0}, b \in \mathbb{Z}_{\geq 1}\}}$, so $f^{-1} \in \overline{\{\prod_i g_i^{a_i} f^b | a_i, b \in \mathbb{Z}_{\geq 0}\}} \subseteq S$.
It seems that this set of products is used 6 times, maybe use a symbol for it.
@MartinBrandenburg Most of the sets are different, because of differing restrictions on the exponents; I thought that defining new sets with different indices would be more complicated (and take longer), so I chose not to. I think only on set is used 3 times.
|
2025-03-21T14:48:29.689047
| 2020-01-18T10:59:49 |
350682
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"JCM",
"https://mathoverflow.net/users/33849"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625722",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350682"
}
|
Stack Exchange
|
Limit for series of Bessel functions evaluated at zeros
The following series arises in an electrostatics problem for a conducting cylinder:
$$
V=\sum_{n=1}^\infty\frac{J_0(k_n\rho)e^{-k_nz}}{k_nJ_1(k_n)^2}
$$
where $J_i$ is the Bessel function of $i^{th}$ order, and $k_n$ is the location of the $n^{th}$ zero of $J_0$. $V$ can be proven to converge for $z>0$, and from numerical tests, converges conditionally also at $z=0$ except for poles at $\rho=0,2,4,6...$
Is there any kind of analytic or asymptotic expression as a function of $\rho$ in the limit as $z\rightarrow0$, for $\rho>2$ in particular?
In relation to your question, it seems that there would potentially be something helpful in Watson's Bessel functions book. I can't say for sure though.
Employing the asymptotics of large zeroes of Bessel functions and the large-argument asymptotics of the Bessel functions, it can be shown that the $n$th term of the series behaves like
$$
\frac{1}{\sqrt{2n\rho}}\cos\left(\rho\left(n-\tfrac{1}{4} \right)\pi-\tfrac{\pi}{4}\right)\mathrm{e}^{ -\left(n-\tfrac{1}{4}\right)\pi z} + \mathcal{O}_{z,\rho}\!\left(\frac{1}{n^{3/2}}\right).
$$
Thus, in terms of the polylogarithm,
$$ V = \frac{\mathrm{e}^{\frac{\pi}{4}z}}{\sqrt {2\rho }}\operatorname{Re} \left[ \mathrm{e}^{ - \frac{\pi}{4}(\rho + 1)\mathrm{i}} \operatorname{Li}_{1/2} \left( \mathrm{e}^{\pi \rho \mathrm{i} - \pi z} \right) \right] + \mathcal{O}_{z,\rho } (1).
$$
Yo may proceed from here.
|
2025-03-21T14:48:29.689169
| 2020-01-18T11:58:33 |
350683
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"John Murray",
"Mihawk",
"https://mathoverflow.net/users/20764",
"https://mathoverflow.net/users/68983"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625723",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350683"
}
|
Stack Exchange
|
Tableaux switching
I'm reading the article Tableau Switching: Algorithms and Applications by Benkart, Sottile, and Stroomer. Do you know if there are any articles or books that talk more about the properties of tableau switching further than Theorem 2.2, 2.3 and 3.1? Or give another point of view for tableau switching?
https://hal.inria.fr › documentPDF
A variation on the tableau switching and a Pak-Vallejo's conjecture - HAL-Inria by Olga Azenhas (and other papers by same author)
Thank you so much. I will look at this!
|
2025-03-21T14:48:29.689243
| 2020-01-18T13:06:11 |
350685
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"CNS709",
"Gustavo Granja",
"Peter May",
"https://mathoverflow.net/users/137622",
"https://mathoverflow.net/users/14447",
"https://mathoverflow.net/users/33141"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625724",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350685"
}
|
Stack Exchange
|
Postnikov tower for $S^2$
I am learning Postnikov towers from this lecture. Here is the first part of the proof that I am studying
Why is true the marked statement?
For example, let be $X = S^2$.
To build $Y_1$ (i.e, with killed $\pi_2$, $, \pi_3$...) I add to $X$ a $3$-cell glued on a generator of $\pi_2(S^2) = \mathbb{Z}\cdot[id]$. So $Y_1^{(1)} = D^3$. We have no more homotopy groups to kill, then $Y_1 = Y_1^{(1)} = D^3$.
To build $Y_2$ I add to $X$ a $4$-cell to a generator of $\pi_3(S^2) = \mathbb{Z}\cdot[hopf]$ and then maybe $5$-cells, $6$-cells...
Then what is the "canonical inclusion" $Y_2 \to Y_1$? I had not adjoined more cells for $Y_1$ than for $Y_2$ as claimed!
I suppose that in the construction of the spaces $Y_n^{(k)}$ you don't just attach cells for a set of generators of the homotopy group in question but rather attach a cell for each continuous map from $S^{k}$ to $Y_n^{(k-1)}$. This is the canonical thing to do (it is functorial as it involves no choices). With this construction it is indeed true that $Y_{n+1}$ is contained in $Y_n$ for the reason given.
@Gustavo, right you are, but we (or at least I) don't have access to Lemma 2 in the source: the question has two answers, depending on that. Of course, rectification of this first step to a Postnikov tower is where the real math lies either way.
@GustavoGranja https://www.math.ru.nl/~mgroth/teaching/htpy13/Section11.pdf
@PeterMay https://www.math.ru.nl/~mgroth/teaching/htpy13/Section11.pdf
I have no idea what source you are quoting, but you are quite right that it is wrong, unless we are both screwing up. One builds $\phi_n$ rigorously by inducting on the stages of the construction of $Y_{n+1}$ from $X$, using null homotopies of attaching maps. The resulting tower is then corrected to a Postnikov tower. A glib outline construction is given in Section 22.4 of Concise
http://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf
A less concise but tediously careful proof along the lines you are considering is given in Section 3.5 of More Concise
http://www.math.uchicago.edu/~may/TEAK/KateBookFinal.pdf
where the construction is generalized to nilpotent spaces.
|
2025-03-21T14:48:29.689437
| 2020-01-18T13:22:51 |
350686
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Maxime Ramzi",
"Valery Isaev",
"https://mathoverflow.net/users/102343",
"https://mathoverflow.net/users/136909",
"https://mathoverflow.net/users/62782",
"lab"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625725",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350686"
}
|
Stack Exchange
|
pullback and fiber sequence
Let $A\rightarrow D\leftarrow C$ a diagram of connected pointed toplogical space where $A\rightarrow D$ is a fibration. Denote $P=A\times_{D}C$. We obtain a homotopy fiber sequence $$ \Omega D\rightarrow P\rightarrow A\times C $$
If we suppose that $D=\Omega X$ for some pointed topological space $X$. Do we obtain a homotopy fiber sequence $$ P\rightarrow A\times C\rightarrow D ?$$
where the map $A\times C\rightarrow D$ is obtained as a composition $A\times C\rightarrow D\times D\rightarrow D$ (the second map is a concatenation of loops)
The hofiber of $A\times C\to D$ has a model consisting of points $(a,c)$ together with a homotopy $f(a)g(c)^{-1} \sim *$, so points $(a,c)$ together with a homotopy $f(a)\sim g(c)$, so it amounts to a point in a model of the homotopy pullback $A\times^h_D C$, but $P$ is precisely that homotopy pullback, because $A\to D$ is a fibration.
@Max if i understand correctly your comment then the answer is YES ?
This is true if one of the loops is inverted before the concatenation in the definition of the map $D \times D \to D$ and the chosen point of $D$ is the trivial loop.
@ValeryIsaev Now I see thanks!
Yes. Here are some details.
The space $P$ sits in homotopy pullback diagram
$\require{AMScd}$
$$
\begin{CD}
P @>>> D \\
@VVV@VVV \\
A\times C @>>> D\times D
\end{CD}
$$
where the the right vertical map is the diagonal. In fact, one can see this by replacing the latter map with the free path fibration $D^I \to D\times D$. After this replacement the diagram becomes a pullback and a pullback with one of the terminal maps a fibration is always a homotopy pullback.
When $D = \Omega X$, there the diagonal is induced by the map $m:D\times D \to D$ given by $(\gamma_1,\gamma_2) \mapsto \gamma_1 \cdot \bar\gamma_2$, where the bar means loop inversion. This means that there is a commutative homotopy pullback
$$
\begin{CD}
D @>\text{diag} >> D\times D \\
@VVV @VV m V \\
* @>>> D
\end{CD}
$$
where $*$ is some contractible space. (Perhaps the easiest way to see this is to note that the diagonal $X\to X \times X$ has homotopy fiber $D$.)
The base change of a map which is induced in also induced: this means that there is a commutative homotopy pullback
$$
\begin{CD}
P @>>> A \times C \\
@VVV @VVV \\
* @>>> D
\end{CD}
$$
where the bottom map is the same as in the previous diagram.
|
2025-03-21T14:48:29.689622
| 2020-01-18T14:40:45 |
350690
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexandre Eremenko",
"Ben McKay",
"Jimmy_the_analyst",
"https://mathoverflow.net/users/13268",
"https://mathoverflow.net/users/151260",
"https://mathoverflow.net/users/25510"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625726",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350690"
}
|
Stack Exchange
|
Real analytic function: zero set of the gradient is a subset of the zero set of the function
I had this question when reading Bierstone and Milman's famous paper "Semianalytic and subanalytic sets". In their proof of the Łojasiewicz gradient inequality (Proposition 6.8 in the paper), they used the following: for any function $g$ real analytic in a neighborhood of the origin such that $g(\mathbf{0})= \nabla g (\mathbf{0}) = 0$, there exists a small ball $K$ centered at $\mathbf{0}$ in which the zero set of the gradient $\nabla g$ is contained in the zero set of $g$. Does anyone know how to prove this statement or can provide some reference? Thanks!
But what if $\nabla g$ is not 0 or does not have isolated zero at the origin.
Id the zero set is not isolated at 0, you can integrate along it and show that the function is zero on this set, as stated.
You need something like the Weierstrass Preparation Theorem to see that the zero locus of the gradient is smooth enough to integrate along, i.e. that any two points of that locus are connected by a rectifiable path, if you work inside a small enough ball. To me, that is not immediately obvious. Then you apply Alexandre Eremenko's comment: if the gradient is normal to a set connected by rectifiable paths, then you can integrate and see that the function is constant on that set.
|
2025-03-21T14:48:29.689764
| 2020-01-18T15:27:09 |
350691
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625727",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350691"
}
|
Stack Exchange
|
Categorical description of log as approximate rig homomorphism?
Summary
The base-$\beta$ logarithm gives an isomorphism of topological spaces
$$
\log_\beta\colon\mathbb{R}_{\geq0}\xrightarrow{\cong}[-\infty,\infty).
$$
This continuous map preserves some algebraic structure, e.g. sending multiplication to addition. As is well-known in tropical geometry, it also approximately sends the binary operation addition to the binary operation $\max$,
$$\log_\beta(a+b)\approx\max\left(\log_\beta(a),\log_\beta(b)\right),$$
and this approximation gets better as the base $\beta\in(1,\infty)$ increases.
Question: Is there a category-theoretic way to tell this story?
Background
A commutative rig ("ring without negatives") is a tuple $(S, 0, +, 1, *)$, where $(S,0,+)$ and $(S,1,*)$ are both commutative monoids, and where multiplication distributes over addition, $a*(b+c)=(a*b)+(a*c)$. Both
$$
\big(\mathbb{R}_{\geq0},0,+,1,*\big)\quad\text{and}\quad\big([-\infty,\infty), -\infty, \max, 0, +\big)
$$
are rigs; in fact, they are topological rigs. That is, each of the sets $\mathbb{R}_{\geq0}$ and $[-\infty,\infty)$ has a well-known topology (e.g. in $[-\infty,\infty)$, the open sets containing $-\infty$ are generated under union by those of the form $[-\infty,x)$ for $x\in \mathbb{R}$), and in each case the two binary operations, namely $+,*$ and $\max,+$, are continuous with respect to that topology.
As mentioned above, the $\log_\beta$ function is an isomorphism of topological spaces, but it is not a homomorphism of rigs. It preserves almost all of the structure ($0\mapsto-\infty, 1\mapsto 0, *\mapsto +$), the one exception being that the addition operation is not preserved.
However, we do have simple bounds on the difference between $\log_\beta(a+b)$ and $\max(\log_\beta a,\log_\beta b)$ for all $a,b\in\mathbb{R}_{\geq0}$. One checks easily (using the two exhaustive cases $a\leq b$ and $b\leq a$) that
$$
0\leq \log_\beta(a+b)-\max\big(\log_\beta(a),~\log_\beta(b)\big)\leq\log_\beta 2.
$$
Thus as $\beta$ gets bigger—one might think of $\beta$ physically as "coldness", so "as the system gets colder"—the approximation gets better: $\log_\beta$ becomes closer and closer to an isomorphism of topological rigs.
Motivation
Tropical geometry is widely studied field, and I imagine that nothing in this section is particularly novel in that context. But let me just say what motivates me.
Having an almost-isomorphism (asking precisely what that should mean is the goal of this post) between these two topological rigs is interesting because, for one thing, $\max,+$ matrix multiplication is much easier for humans and faster to compute than $+,*$ matrix multiplication. Moreover, lots of ideas port over.
For example, one can discuss what I might call "log-stochastic matrices" where the max of each column is 0. These form a category that is similar in certain ways to the usual category Stoch of finite sets and stochastic maps, but of course having key differences. Working a bit in this category, I could imagine it being relevant in behavioral economics, where there may be multiple "best choices" in a given situation (multiple 0's in a given column), and calculations need to be easy (again max, + being easier than +,*).
Categorical issues
My goal is to be able to say, category-theoretically, both that $\log_\beta$ is an approximate rig map for each $\beta\in(1,\infty)$, and also that these approximations get better as $\beta$ increases. To tell this story categorically, one needs to define what "better approximation" means.
One attempt could be to give something like a distance between $\log(a+b)$ and $\max(\log a, \log b)$, and for this we would need something like a metric. The usual category of metric spaces has short maps (distance non-decreasing functions) as its morphisms. This doesn't work too well for us, because $(\mathbb{R},0,+,1,*)$ is not an internal rig in that category; in particular, the multiplication operation $*\colon\mathbb{R}_{\geq0}\times\mathbb{R}_{\geq0}\to\mathbb{R}_{\geq 0}$ is not a short map. In fact, it is not even Lipschitz. But once one moves to the category of topological spaces or sets—where our two structures do become internal rigs—how will one measure the distances between their distance addition operations?
The goal is not to necessarily stick with the viewpoint of the previous paragraph, but to come up with a categorical viewpoint in which the successive approximations story is most at home.
Summary
While it could certainly be prettied up, it seems that the basic idea is actually fairly straightforward: use each $\log_\beta$ to transport the rig structure from $\mathbb{R}_{\geq0}$ to the space $[-\infty,\infty)$ and use the metric structure there.
Transporting the rig structure
For any $\beta$, consider the operation $P_\beta\colon[-\infty,\infty)\times[-\infty,\infty)\to[-\infty,\infty)$ given by
$$P_\beta(a,b):=\log_\beta\left(\beta^a+\beta^b\right).$$
This is obtained by transporting $+$ along the $\log_\beta$ bijection. As such, it is easy to see that $P_\beta$ is unital with respect to $-\infty$ and associative, and that $+$ distributes over $P_\beta$. We can also see this directly, e.g.:
\begin{align*}
a+P_\beta(b,c)&=
\log_\beta(\beta^a)+\log_\beta(\beta^b+\beta^c)\\&=
\log_\beta(\beta^{a+b}+\beta^{a+c})\\&=
P_\beta(a+b,a+c).
\end{align*}
The limit of these operations $P_\beta$ as $\beta$ increases is
$$
\lim_{\beta\to\infty}P_\beta(a,b)=
\lim_{\beta\to\infty}\log_\beta\left(\beta^a+\beta^b\right)=
\max(a,b).
$$
As an aside, a similar phenomenon occurs for the family of $\ell_p$ norms, and there seems to be a relationship. Namely, the variables in the exponential terms have switched roles ($\beta\leftrightarrow p$ and $\ell_p\leftrightarrow P_\beta$). Writing $\ell$ for $\ell_p$ and $P$ for $P_\beta$, one sees the resemblance immediately:
$$
\ell(a,b)^p=a^p+b^p
\qquad\text{vs.}\qquad
\beta^{P(a,b)} = \beta^a + \beta^b.
$$
We'll save this curiosity for another time.
A family of metric rigs
Going back to the main story, for any $\beta\in(1,\infty)$ we have a rig
$$
R_\beta:=\big([-\infty,\infty),-\infty,~P_\beta~,0,+\big).
$$
Each of these is of course isomorphic to the original rig $(\mathbb{R}_{\geq0},0,+,1,*)$, but the point is to show that these are rig objects in the category of (Lawvere) metric spaces. That is, we need to show that both operations are short with respect to the usual metric on $[-\infty,\infty)$.
Checking shortness for the "multiplication" operation $+$, we choose $a,b_1,b_2$ and see immediately that we have $d(a+b_1,a+b_2)=d(b_1,b_2)$ and hence the necessary inequality holds
$$d(a+b_1,a+b_2)\leq d(b_1,b_2).$$
Checking shortness for the "addition"operation $P_\beta$ is slightly more involved. It uses the fact that for nonnegative reals $0\leq x$ and $0<y_2\leq y_1$ we have $\frac{x+y_1}{x+y_2}\leq\frac{y_1}{y_2}$. For any $a,b_1,b_2\in[-\infty,\infty)$, if either $b_1=-\infty$ or $b_2=-\infty$ it is easy to show that $d(P_\beta(a,b_1),P_\beta(a,b_2))\leq d(b_1,b_2)$. So we assume without loss of generality that $-\infty<b_2\leq b_1$, i.e. $0<\beta^{b_2}\leq\beta^{b_1}$, and compute
\begin{align*}
d\big(P_\beta(a,b_1),P_\beta(a,b_2)\big)&=
\log_\beta(\beta^a+\beta^{b_1})-\log_\beta(\beta^a+\beta^{b_2})\\&=
\log_\beta\left(\frac{\beta^a+\beta^{b_1}}{\beta^a+\beta^{b_2}}\right)\\&\leq
\log_\beta\left(\frac{\beta^{b_1}}{\beta^{b_2}}\right)\\&=
b_1-b_2=d(b_1,b_2).
\end{align*}
We have thus shown that $R_\beta$ is an internal rig in the category of Lawvere metric spaces. Indeed, in the case of both operations, we implicitly used the fact that the category of (Lawvere) metric spaces is monoidal closed, so a function $P\colon X_1\times X_2\to Y$ is short iff for each $a\in X_1$ the function $P_a\colon X_2\to Y$ given by $P_a(b):=P(a,b)$ is short.
Thus we have a whole family of operations $P_\beta$, one for each $\beta\in(1,\infty]$. Each of these fulfills the role of addition in the metric rig $R_\beta=([-\infty,\infty),-\infty,P_\beta,0,+)$.
where we put $P_\infty:=\max$. And thus we also have a whole family of rig structures, all on the same space and all with the same units and the same multiplication.
There may be more to say, e.g. perhaps the $P_\beta$ satisfy certain laws for varying $\beta$, but I won't pursue anything further for now.
|
2025-03-21T14:48:29.690288
| 2020-01-18T15:28:23 |
350692
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Anton Sorokovskiy",
"efs",
"https://mathoverflow.net/users/109085",
"https://mathoverflow.net/users/151262"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625728",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350692"
}
|
Stack Exchange
|
Reference request: Ito formula for function $G(t, x)$ when $G$ depend on $\omega$
There is proved Lemma in book : Let the function $G(t,x)$ is defined when $t\in [0,T], x\in(-\infty,\infty)$, $G$ has continuous derivative with respect to $t$ and twice continuously diferentiable with respect to $x$. Then $$dG(t,w(t))=[G'_t(t,w(t))+1/2G''_{xx}(t,w(t))]+G'_x(t,w(t))dw(t).$$
Here $w(t)$ is brownian motion.
After that without proof there is Claim, that Lemma is true if $G(t,x)$ depend on $\omega$ , $G(t,w(t))$ is measurable with respect to $\sigma$-algebra $F_t$ and for each $\omega$ conditions of the lemma are satisfied.
My question: How to prove this Claim? Could you provide reference to concise proof of it?
Could you provide the book your question is about?
This book is Stochastic Differential Equations, Authors: Gihman, Iosif I., Skorohod, Anatolij V. I have only russian edition of this book.
|
2025-03-21T14:48:29.690390
| 2020-01-18T16:52:38 |
350697
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Anixx",
"Christian Remling",
"Michael Engelhardt",
"Wojowu",
"https://mathoverflow.net/users/10059",
"https://mathoverflow.net/users/134299",
"https://mathoverflow.net/users/30186",
"https://mathoverflow.net/users/48839",
"https://mathoverflow.net/users/74539",
"lcv"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625729",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350697"
}
|
Stack Exchange
|
Is there a sense in which one could expand $\frac{\sin (x/\epsilon )}{x} $ in powers of $\epsilon $?
A standard representation of the $\delta $-distribution is
$$
\pi \delta (x) = \lim_{\epsilon \searrow 0} \frac{\sin (x/\epsilon )}{x}
$$
Is there a sense in which this could be seen as the leading term in an expansion of $\sin (x/\epsilon ) /x$ in positive powers of $\epsilon $, presumably with distributions as coefficients? If so, is it possible to give the expansion explicitly?
Do you see the sign of the expression depends on the sign of $\epsilon$, so the direction of the limit is important? .
For what it worth, at $x\to0$ the function becomes $\frac1\epsilon$.
Of course, I had in mind the usual $\epsilon \searrow 0$. I've edited the question.
Your distribution is the Dirichlet kernel, so the discrepancy $f(0)-(D_{1/\epsilon}*f)(0) = \int_{|t|>1/\epsilon} \widehat{f}(t), dt$ goes to zero faster than any power of $\epsilon$, thanks to the rapid decay of $\widehat{f}$.
@ChristianRemling - ah yes, I see what you're saying, this provides some clarity. The behavior is manifestly determined by the behavior of the Fourier transform of the test function at infinity, and strictly speaking, that should decay faster than any power. Now, I might get physicist notions like applying this on less well-behaved spaces than bona-fide test functions - so the Fourier transform might only decay as a power, and then I'd get nonvanishing coefficients in the expansion in powers of $\epsilon $. I wonder how this would look in $x$-space, that's another way of putting the question.
I've asked the same question on MSE but it went essentially unanswered. Here is the link https://math.stackexchange.com/questions/463724/sinc-to-delta-function-error-term
If you are interested in Laurent series, the expansion is
$$\frac{\sin (x/\epsilon)}x=\frac{1}{\epsilon }-\frac{x^2}{6 \epsilon ^3}+\frac{x^4}{120 \epsilon ^5}-\frac{x^6}{5040 \epsilon ^7}+\frac{x^8}{362880 \epsilon ^9}+O\left(\frac{1}{\epsilon^{11} }\right)=\sum_{n=0}^\infty \frac {(-1)^{2n}x^{2n}}{(2n+1)!\epsilon^{2n+1}}$$
Ah, I see I've again not been specific enough in the formulation of the question. No, by power series I meant positive powers, not the Laurent series, which is of course not very useful. I've once more edited the question.
The expansion around point $a$ in positive powers exists, but what is the general form of the term and whether it is convergent requires further research.
$$\frac{\sin (x/\epsilon)}x=\frac{\sin \left(\frac{x}{a}\right)}{x}-\frac{(\epsilon -a) \cos \left(\frac{x}{a}\right)}{a^2}+\frac{(\epsilon -a)^2 \left(2 a \cos \left(\frac{x}{a}\right)-x \sin \left(\frac{x}{a}\right)\right)}{2 a^4}$$
$$+\frac{(\epsilon -a)^3 \left(-6 a^2 \cos \left(\frac{x}{a}\right)+x^2 \cos \left(\frac{x}{a}\right)+6 a x \sin \left(\frac{x}{a}\right)\right)}{6 a^6}$$
$$+\frac{(\epsilon -a)^4 \left(24 a^3 \cos \left(\frac{x}{a}\right)-36 a^2 x \sin \left(\frac{x}{a}\right)+x^3 \sin \left(\frac{x}{a}\right)-12 a x^2 \cos \left(\frac{x}{a}\right)\right)}{24 a^8}$$
$$+\frac{(\epsilon -a)^5 \left(-120 a^4 \cos \left(\frac{x}{a}\right)+240 a^3 x \sin \left(\frac{x}{a}\right)+120 a^2 x^2 \cos \left(\frac{x}{a}\right)-x^4 \cos \left(\frac{x}{a}\right)-20 a x^3 \sin \left(\frac{x}{a}\right)\right)}{120 a^{10}}$$
$$+O\left((\epsilon -a)^6\right)$$
This of course works for $a\neq 0$. Can we make sense of this as $a\searrow 0$? Do the coefficients perhaps converge in a suitable (distribution?) sense to give a series convergent to $\sin(x/\epsilon)/x$?
|
2025-03-21T14:48:29.690641
| 2020-01-18T17:33:16 |
350700
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"AHusain",
"Andy Sanders",
"Praphulla Koushik",
"https://mathoverflow.net/users/118688",
"https://mathoverflow.net/users/49247",
"https://mathoverflow.net/users/69850"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625730",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350700"
}
|
Stack Exchange
|
Chern -Weil map for topological principal G bundles
Let $G$ be a Lie group.
In the book Curvature and Characteristic classes, the author (Johan L. Dupont) mentiones in beginning of chapter 5 the following :
The notion of a topological principal $G$-bundle $\pi:E\rightarrow X$ on a topological space $X$ is defined exactly as in Definition $3.1$ (the definition of usual principal bundle over manifold), only the words "differentiable" and "diffeomorphism" are replaced by "continuous" and "homomorphism". The purpose of this and the following section is to show that the Chern-Weil morphism defines characteristic classes of topological G-bundles.
But does not mention (or I could not find) anything about characteristic classes of topological principal bundles. Are there other places that discuss the notion of characteristic classe of topological principal bundles using Chern-Weil theory?
At the risk of saying something stupid, Chern-Weil theory involves the notions of connections and curvature, so requires at least a $C^{2}$-regularity on your bundle and manifold for their naive definition. If you want to relax this regularity, you're back in the world of classifying spaces and their cohomology to define characteristic classes in this language. I'm sure you could work much harder to make headway in either direction, but already you don't define what you mean by topological, so doing so would be a good start.
I have mentioned what does a topological principal bundle mean to the author.
fair enough, I apologize for asking this: but my comment still stands.
I think we're basically in agreement, Chern-Weil theory for bundles that are not $C^{2}$ over a $C^{2}$-manifold are at most a niche area, and at worst, not sensibly defined. This doesn't mean that such an investigation isn't important, it just means that it goes from being "standard material" to a "research area."
I feel the same... He does discuss Chern Weil map for Lie groups, $I(G)\rightarrow H^*(BG,\mathbb{R})$.. He use the notion of simplicial manifold associated to the topological space $BG$ to do that.. Thus, he is able to write Chern Weil map for topological principal bundles $EG\rightarrow BG$, but still it does not explain how one would do for an arbitrary topological principal bundle $E\rightarrow X$
There may well be a way to circumvent the issues I mention above. Some very smart people often find a way to redefine various objects so that their statements make sense. As I said before, this usually places you out of the world of standard constructions, and into the world of current ideas/research. You'd be better served by asking why they do this, then how it works. Of course, it's also possible I'm an idiot, and am just missing something trivial...
Usually comment sections are not used for discussion, so feel free to email me at the email listed in my profile if you want to discuss this more.
@AndySanders You can also move comments to chat here.
I think "Chern–Weil theory" here is being used in a slightly nonstandard way. The main result of the chapter, Theorem 5.5, is really just a topological statement that elements of $H^*(BG,\Lambda)$ correspond exactly to characteristic classes (Definition 5.1), which are defined to be functorial associations of ($\Lambda$-valued) singular cohomology classes to isomorphism classes of topological bundles. This is combined with the calculation that the cohomology of $BG$ is calculated to be a certain invariant polynomial ring (assuming, eg, $G$ is compact), i.e. the Chern–Weil homomorphism, since the group $G$ is still a Lie group.
In traditional Chern–Weil theory (for smooth bundles on manifolds), one also has the isomorphism of de Rham with singular cohomology, and, most importantly, that the invariant polynomials can be evaluated on curvature forms to give de Rham classes.
But in the topological case at hand, one still has invariant polynomials giving rise to characteristic classes, which is what I think Dupont means here.
Thanks for the answer. Yes, It looks more like a non standard term.. I will wait for some more time to see if any one has anything else to say and then accept this answer..
|
2025-03-21T14:48:29.691037
| 2020-01-18T18:00:38 |
350701
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625731",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350701"
}
|
Stack Exchange
|
Choosing the best submatrix
Let $\mathbf{A}_{m\times n}$ be a matrix with non-negative elements. Assume that a submatrix $\mathbf{B}$ from $\mathbf{A}$ is defined as
\begin{align}
B_{i,j} =
\begin{cases}
A_{i,j}, & i\in\mathcal{I},\\
0, & \text{otherwise},
\end{cases}
\end{align}
where $\mathcal{I}$ is a subset of $\{1,2,\cdots,m\}$.
For a constant size of $\mathcal{I}$, how does one should select $\mathcal{I}$ such that the following quantity becomes maximum
$$\|\mathbf{B}^\mathrm{T}\mathbf{B}\|_1+\|\mathbf{B}\mathbf{B}^\mathrm{T}\|_1,$$
where $\|.\|_1$ is defined as the sum of the all elements of the matrix, and superscript $(.)^{\mathrm{T}}$ denotes the transpose operatrion.
You can solve the problem via integer programming as follows. For $i\in\{1,\dots,m\}$, let binary decision variable $x_i$ indicate whether row $i$ is selected. Then $B_{i,j}=A_{i,j}x_i$, so the problem is to maximize
$$\sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^m A_{k,i} A_{k,j} x_k + \sum_{i=1}^m \sum_{j=1}^m \sum_{k=1}^n A_{i,k} A_{j,k} x_i x_j$$
subject to $\sum_{i=1}^m x_i = s$, where $s$ is the desired cardinality of $\mathcal{I}$.
Now use a mixed integer quadratic programming (MIQP) solver.
Alternatively, you can linearize the quadratic objective by introducing a new decision variable $y_{i,j} \ge 0$, with $1 \le i < j \le m$, to represent the product $x_i x_j$. The resulting mixed integer linear programming (MILP) problem is to maximize
$$\sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^m A_{k,i} A_{k,j} x_k + \sum_{i=1}^m \sum_{j=1}^m \sum_{k=1}^n A_{i,k} A_{j,k} y_{i,j}$$ subject to linear constraints:
\begin{align}
\sum_{i=1}^m x_i &= s\\
y_{i,j} &\le x_i \\
y_{i,j} &\le x_j \\
y_{i,j} &\ge x_i + x_j - 1
\end{align}
Optionally, you can use the cardinality constraint to strengthen the formulation by including an additional constraint
$$\sum_{i:i<j} y_{i,j} + \sum_{i:i>j} y_{j,i} = (s-1) x_j$$
for $j\in\{1,\dots,m\}$.
|
2025-03-21T14:48:29.691172
| 2020-01-18T18:11:16 |
350703
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"archipelago",
"https://mathoverflow.net/users/27004",
"https://mathoverflow.net/users/32022",
"https://mathoverflow.net/users/40804",
"mme",
"wonderich"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625732",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350703"
}
|
Stack Exchange
|
Self diffeomorphism of $S^2\times S^2$
The main question is motivated from the answer of this question https://math.stackexchange.com/questions/2481200/finite-groups-gs-which-acts-freely-on-s2-times-s2
Is it true that every self diffeomorphism of $S^2\times S^2$ either fiber preserving or exchange the two spheres (upto isotopy)?
It is certainly true that the homology classes of $S^2 \times \ pt$ and $\ pt \times S^2$ must be preserved or switched since you can easily check they are the only homology classes with self-intersection 0. And also the complement of an $\epsilon$-nebighbourhood of them is $B^4$. So from here can we conclude anything?
Can anyone share an idea of how to prove or disprove it?
I am sure your question is open. Very little is known about mapping class groups in 4 dimensions. Some things to look at are work of Ruberman and Konno. These use gauge theory techniques for manifolds with $b_2 >>0$ that are not likely applicable to $S^2 \times S^2$. More hands-on (and for manifolds with relatively few handles) is recent work of Gabai, Watanabe, and Budney-Gabai, which says something about mapping class groups of $D^2 \times S^2, S^4, S^1 \times S^3$ in that order. I've missed some things here, so chase references to see other work.
Isnt the mapping class group of $^2×^2$ known already?
No, the mapping class group of no closed 4-manifold is known.
|
2025-03-21T14:48:29.691639
| 2020-01-18T18:59:34 |
350707
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dror Speiser",
"Jared Weinstein",
"https://mathoverflow.net/users/2024",
"https://mathoverflow.net/users/271"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625733",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350707"
}
|
Stack Exchange
|
An isogeny between Jacobians of hyperelliptic curves
Let $\mathbf{F}_q$ be a finite field of odd characteristic. Let $X_t$ be the hyperelliptic curve over $\mathbf{F}_{q^2}(t)$ with affine equation
$$y^2 = \left((x^{(q+1)/2}-(x-1)^{(q+1)/2})^2 - t\right)
\left((x^{(q+1)/2}+(x-1)^{(q+1)/2})^2 - t\right).$$
This family of hyperelliptic curves is smooth of genus $q-1$ away from $t=0,1,\infty$. For theoretical reasons I expect the following claim to be true:
The Jacobians of $X_t$ and $X_{1-t}$ are isogenous. Equivalently, there is a nontrivial correspondence between $X_t$ and $X_{1-t}$.
It is easy to verify (with MAGMA for instance) that the zeta functions of $X_t$ and $X_{1-t}$ agree for small values of $q$ and specific values of $t$, to the point that I'm quite convinced of the claim. To prove it, one could try to construct the correspondence explicitly. There is some literature about pairs of hyperelliptic curves with isogenous Jacobians (for instance this paper of Mestre), but it seems to involve finding a congruence between bivariate polynomials, which comes out of nowhere.
Short of constructing the actual correspondence, is there an algorithm for deciding whether two hyperelliptic curves admit an isogeny of given degree between their Jacobians?
Did you have luck with this?
Hi Dror, sorry, I didn't check this for awhile. The curves $X_t$ and $X_{1-t}$ have a meaning: they are a Drinfeld modular curve and a Drinfeld-Shimura curve, respectively. I thought it was cute that they had such a simple formula. The Jacquet-Langlands correspondence combined with the Tate conjecture for divisors on abelian varieties over function fields produces the desired isogeny. I wanted to know if there was an explicit correspondence, but maybe this is too much to hope for.
In the most easy case $q=3$, the curve $X_t$ is bielliptic (the bielliptic involution given by $x\mapsto 1-x$), and the Jacobian of $X_t$ is then $(2,2)$-isogenous to the product
$E_{1,t}\times E_{2,t}$, where $E_{1,t}$ and $E_{2,t}$ are the elliptic curves
$$E_{1,t}: y^2=x(x^2 - (t+1)x + (t- 1)^2)$$
and
$$E_{2,t}: y^2=x(x^2 + x - (t^3 - 1))$$
Now, $E_{1,t}$ is 2-isogenous to
$$E_{1,t}': y^2= x(x+1)(x-(t-1))$$
and one has that
$$E_{1,t}'\cong E_{1,1-t}$$
by sending $x\mapsto 1-x$ and then twisting by $-1$, which is an isomorphic curve over $\mathbb{F}_9$ (this is the only place where we use we are over $\mathbb{F}_9$ and not over $\mathbb{F}_3$).
while $E_{2,t}$ is 2-isogenous to
$$E_{2,t}': y^2 = x(x^2 + x + t^3)$$
which in turn is clearly equal to $E_{2,1-t}$.
Thank you Xarles! I am attempting to see whether your construction generalizes. The Jacobian of $X_t$ is isogenous to a product of Jacobians of hyperelliptic curves $C_{1,t}$ and $C_{2,t}$. Evidence suggests that your pattern persists: the Jacobians of $C_{1,t}$ and $C_{1,1-t}$ are isogenous over $\mathbf{F}{q^2}(t)$, and the Jacobians of $C{2,t}$ and $C_{2,1-t}$ are isogenous already over $\mathbf{F}_q(t)$.
The Tate conjecture is known for function fields (Zarhin). To check whether the Tate modules are isomorphic you need to check that the image of Frobenius match for a finite set of places that can be bounded a priori (the argument is in Faltings's Mordell paper) although the bound is probably not very small. Checking that the image of Frobenius match at a place is essentially your computation of specializing $t$ and compute zeta functions. This gives, in principle, an algorithm to check that the Jacobians are isogenous for a given $q$. It's probably not practical except for very small $q$. It doesn't give any information of the degree of the isogeny, though.
|
2025-03-21T14:48:29.691892
| 2020-01-18T19:59:45 |
350709
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jean Marie Becker",
"https://mathoverflow.net/users/88984"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625734",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350709"
}
|
Stack Exchange
|
Relationship between eigenvectors of projected and original matrix
Let $A = \mathrm{Diag}(\lambda_1, \dots, \lambda_n)$ where $\lambda_1 \le \lambda_2 \dots \le \lambda_n$. Let $P = I - ww^T$ be a projection operator on an arbitrary $n$-dimensional hyperplane. Let $B = PAP$ have eigenvalues $\mu_1 \le \mu_2 \le \dots \mu_{n - 1}$ and (unit norm) eigenvectors $v_1, \dots, v_{n - 1}$. From Cauchy's interlacing theorem it follows that $\lambda_1 \le \mu_1 \le \lambda_2 \le \dots \mu_{n-1} \le \lambda_n$.
Is there anything that can be said about the relationship between inner products $\langle e_i, v_j \rangle$ depending on the eigenvalues of $A$ and $B$?
Thinking in 3D, the projection of relationship $AV=\lambda V$ gives rise to $BV'=\mu V'$ with $\mu=\lambda$... which is a particular case of the interlacing property.
|
2025-03-21T14:48:29.691993
| 2020-01-18T20:16:05 |
350711
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Daniele Tampieri",
"J. Doe",
"Nik Weaver",
"Pietro Majer",
"Yemon Choi",
"https://mathoverflow.net/users/113756",
"https://mathoverflow.net/users/142929",
"https://mathoverflow.net/users/151270",
"https://mathoverflow.net/users/23141",
"https://mathoverflow.net/users/6101",
"https://mathoverflow.net/users/763",
"user142929"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625735",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350711"
}
|
Stack Exchange
|
What are some problems for research in functional analysis that can possibly be solved by someone with basic knowledge of the subject?
I wanted to know are there any problems in Functional Analysis (FA) that can possibly be successfully tackled by someone like me who does not have any expertise in this area but is only familiar with a few basic topics that you would find in most undergraduate level courses?
I wanted to mention that I did look around the web before posting here. There doesn't seem to be much left that an undergrad can do in this area (or almost any other area), but I have seen sometimes papers by other researchers who in the end of their papers mention how their work can be used to do something (usually these are concrete applications or suggestions to work on specific examples), but the author didn't find the time or hadn't the resources to carry out the work and it's left to the interested reader. I wanted someone to help me find these kind of problems that would be easy to work on if I give it some time.
My goal is to write a research paper and get it published in a suitable journal. I'm out of school at the moment and would like to get admission into a good PhD program. It is very hard for someone like me to get the attention of a professor to take me as a doctoral student without having proven first that I am motivated to and can do the work in FA.
Thank you for your time and help.
In general, it's hard to find good research problems for undergraduates. You just don't know enough at that stage. It might be possible in more "elementary" subjects like combinatorics, but I think you are unlikely to find something in functional analysis that does what you want.
"these kind of problems that would be easy to work on if I give it some time" - this is not going to be a way "to get admission into a good PhD program" and won't be a way to succeed on such a program if one gets in.
To be more constructive: it's unusual to have publications as an undergrad. Math departments are looking for people who have the potential to do research. This means having a good undergraduate record and (in the US) good GRE scores. If you don't have this your best bet is probably to apply to a master's program somewhere, do well, and use that as a springboard into a PhD program.
Being an engineer, my mathematical training was perhaps not as complete as it was necessary: therefore I started learning the subjects I needed by reading their history. The History a subject explains its genesis, the problems that triggered its development, the problems it solved and furthermore opened, the basic persons involved in its development.
So my advice is: survey the subject by reading papers on the history of functional analysis, understand by yourself what are the topic(s) you like to work to. Then, basing on you taste and interest, write a survey, not aimed to publication, but meat as a relation motivating your will to do research, and submit to someone who works on that topics and ask for their advice.
I'm voting to close this question because it seems to me to be based on a misreading of how one can or should get into a PhD program, and any problems listed here are unlikely to yield the desired results for the OP
Despite my comments above: good luck getting into a PhD program, perhaps by the route(s) suggested by @NikWeaver
I add this comment after your question was closed, with the purpose to add these words. My belief (I'm not a professor) is that without the help of a professor your project is impossible (I am no longer young and I only know one of such case of a person, a secondary school teacher, who thanks to his effort and research articles later get a research career in number theory). The project to publish research articles, without a background as a researcher in mathematics, seems to me impossible. A different thing is to try fight as a non-professional mathematician, but it is as an infinite battle.
Since your aim is to show your abilities and get the attention of a professor that would take you in a PhD program, the best and most natural thing would be, getting the problems from the professors themselves. Read their papers, find their open problems that you may like, and then write to them (with discretion) for clarifications or for suggestions of "easy cases" to start with. After all, this is how great Sofja Kovalevski got the attention of Weierstrass: solving his test problems in a week. Good luck!
Thank you for your answer. I have found that when I email professors whom I would love to work with, most of the time they don't want to take on new students, already have too many students to supervise, or don't have funding to take on new students (these are the most common responses to my emails and I'm beginning to suspect that it's usually just the professor's polite way of saying I lack the subject knowledge required to carry out research). I might spend an awful amount of energy reading their papers and familiarizing myself with the work they're doing only to be rejected by them.
@ChristianRemling You are right, of course this is not the usual route; the OP explicitly says they would prefer not to follow the usual route (otherwise I guess they would have followed it, without need of asking)
|
2025-03-21T14:48:29.692391
| 2020-01-18T21:36:26 |
350712
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Anixx",
"Dave L Renfro",
"Pietro Majer",
"Wojowu",
"Yaakov Baruch",
"https://mathoverflow.net/users/10059",
"https://mathoverflow.net/users/15780",
"https://mathoverflow.net/users/2480",
"https://mathoverflow.net/users/30186",
"https://mathoverflow.net/users/6101"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625736",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350712"
}
|
Stack Exchange
|
A set whose Hausdorff dimension gradually changes?
Can there be a set whose Hausdorff dimension gradually changes?
For instance, a set of real numbers contained in an interval, whose Hausdorff dimension is 0 at the beginning and 1 closer to the end, and changes without jumps?
Take a sequence of Cesaro fractals
Related questions: How can dimension depend on the point? AND What results exist for functions with regionally fluctuant fractal dimension?
I assume you want a set $A\subseteq [0,1]$ such that $\dim (A\cap [0,x])=x$ for all $x$. We can define $A_1$ by taking the union of a (Borel) subset of dimension $0$ of $[0,1/2]$ with a subset of dimension $1/2$ of $[1/2,1]$.To obtain $A_2$, we then make our sets larger on $[1/4,1/2]$, $[3/4,1]$ and again, we make the dimension on each current interval equal to its left endpoint. Then $A_1\subseteq A_2\subseteq \ldots$, and $A=\bigcup A_n$ works: by monotone convergence, $h^d(I\cap A)=\lim h^d(I\cap A_n)$, so $A\cap [0,x]$ has the right dimension at each $x=k2^{-n}$ and thus everywhere.
Can we Lebesgue-integrate over such set?
@Anixx Each $A_n$ has Hausdorff dimension below $1$, so their (1-dimensional) Lebesgue measure is zero. So $A$ has Lebesgue measure zero. You can certainly integrate functions over null sets, but the results are not too interesting...
In fact for all $0<x\le1$ $A\cap[0,x]$ is an $x$-dimensional set of $x$-dimensional null measure, for the same reason.
Cool. I think a similar idea (reversing subintervals of $[k/2^n,(k+1)/2^n]$ as needed, for $n\rightarrow \infty, 0\le k<n$) can also be used to produce a set $A$ with density $x$ at $x$, i.e. $\mu(A\cap[0,x])=x^2/2$.
@YaakovBaruch Are you sure? Doesn't that contradict Lebesgue Density Theorem?
@Wojowu you are right. So it must be that the procedure I had in mind doesn't converge (a.e. pointwise) to a well defined set.
For an example of a different sort, you can consider the record set $R$ of a fractional Brownian motion with varying Hurst parameter $H(t)$. In this short paper it is shown that when $H(t)$ is constant, it equals the Hausdorff dimension of $R$. The properties of the fBM with varying Hurst parameter $H(t)$ are given here for measurable functions $H(t)$ taking values in $(\tfrac12,1)$. One expects that the Hausdorff dimension of $R\cap (t-\epsilon,t+\epsilon)$ converges to $H(t)$ as $\epsilon\to 0$.
|
2025-03-21T14:48:29.692569
| 2020-01-18T22:07:28 |
350713
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625737",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350713"
}
|
Stack Exchange
|
Proof of Prop 1.1 in Wiles' "Modular elliptic curves and Fermat's last theorem"
On p. 459 of "Modular elliptic curves and Fermat's last theorem", proof of Prop. 1.1, where it says "Since $H^2(G,\mu_{p^r}) \rightarrow H^2(G,\mu_{p^s})$ is injective for $r \leq s$...", is there any chance that that's a typo and he really means to say $H^1$ both times rather than $H^2$?
He uses it to conclude that the tensor product of one $H^1$ with a certain module is isomorphic to another $H^1$, so if the hypothesis really is with an $H^2$ both times then that presumably must involve exact sequences in some way. I'd appreciate any help in understanding the argument.
Well, although there is a typo (Wiles forgot to close his parenthesis, and wrote $H^2(G,\mu_{p^r}\to H^2(G,\mu_{p^s})$ in his proof), his claim is correct. Let, as ibid. $F$ be the finite extension of $\mathbb{Q}_p$ fixed by $G$, so that your arrow can be written $H^2(F,\mu_{p^r})\to H^2(F,\mu_{p^s})$. Consider the Kummer sequence (for any $n\geq 1$)
$$
1\longrightarrow \mu_{p^n}\longrightarrow \overline{F}^\times\overset{(\cdot)^{ p^n}}{\longrightarrow}\overline{F}^\times\longrightarrow 1.
$$
Taking Galois cohomology and using Hilbert '90, which tells you $H^1(F,\overline{F}^\times)=1$, you find
$$
1\longrightarrow H^2(F,\mu_{p^n})\longrightarrow H^2(F,\overline{F}^\times)\overset{\cdot p^n}{\longrightarrow }H^2(F,\overline{F}^\times).
$$
Now, local class field theory tells you that the Brauer group $H^2(F,\overline{F}^\times)$ is isomorphic to $\mathbb{Q}/\mathbb{Z}$, so the above sequence identifies $H^2(F,\mu_{p^n})$ with the group $\mathbb{Z}/p^n$, seen as the kernel of multiplication by $p^n$ on $\mathbb{Q}/\mathbb{Z}$. If now you apply this for $r\leq s$, you see that the arrow you were firstly interested in is the injection $\mathbb{Z}/p^r\hookrightarrow \mathbb{Z}/p^s$ or, if you prefer, the injection
$$
\Big(\ker(\cdot p^r)\hookrightarrow \ker(\cdot p^s)\Big)\subseteq \mathbb{Q}/\mathbb{Z}.
$$
Added Jan 24th Concerning the second exact sequence (that involving $H^1$), there, there is a typo! What Wiles wanted to write was that the natural map
$$\tag{1}
H^1(G,\mu_{p^n})\otimes M\longrightarrow H^1(G,M(1))
$$
is an isomorphism, but in his paper he made (again!) a mistake with parenthesis and ibid he replaces the first term by $H^1(G,\mu_{p^n}\otimes M)$. You can see (1) in the quoted paper by Diamond and, moreover, it is (1) which shows up later in Wiles' proof: indeed, he deduces from (1) something about another sequence where $H^1(G,\mu_{p^n})\otimes M$ is replaced, by Kummer theory, by $((F^\times/(F^\times)^{p^n})\otimes M$, which he could not do had he only some result about $H^1(G,\mu_{p^n}\otimes M)$.
To prove (1), observe that Wiles has managed to make the action of $G$ on $M$ trivial, and $n$ is such that $p^nM=0$. So, $M$ is a finite abelian $p$-group of exponent bounded by $n$, and isomorphic (as $G$-module!) to a finite number of copies of $\mathbb{Z}/p^a\mathbb{Z}$ for $a\leq n$. It thus suffices to prove (1) assuming $M=\mathbb{Z}/p^a$ for $a\leq n$. Consider, for any $s\geq 0$, the exact sequence
$$
0\longrightarrow \mu_{p^s}{\longrightarrow}\mu_{p^{s+a}}\longrightarrow \mu_{p^a}\longrightarrow 0.
$$
By the first result on $H^2$ it gives rise to a surjective map
$$
H^1(G, \mu_{p^s}){\longrightarrow}H^1(G,\mu_{p^{s+a}})\longrightarrow H^1(G,\mu_{p^a})\longrightarrow 0
$$
and, by taking projective limit over $s$,
$$
H^1(G, \mathbb{Z}_p(1))\overset{p^a}{\longrightarrow}H^1(G,\mathbb{Z}_p(1)\longrightarrow H^1(G,\mu_{p^a})\longrightarrow 0
$$
which means $H^1(G, \mathbb{Z}_p(1))/p^aH^1(G, \mathbb{Z}_p(1))\cong H^1(G,\mu_{p^a})$. But this can be rewritten as
$$
H^1(G, \mathbb{Z}_p(1))\otimes\mathbb{Z}/p^a\cong H^1(G, \mathbb{Z}/p^a(1)).
$$
This is (1) for the module $\mathbb{Z}/p^a$ and finished the proof.
|
2025-03-21T14:48:29.692819
| 2020-01-18T23:26:52 |
350716
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Iosif Pinelis",
"VS.",
"https://mathoverflow.net/users/136553",
"https://mathoverflow.net/users/36721"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625738",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350716"
}
|
Stack Exchange
|
A conjecture on 'truncated joint moments' of binomial coefficients under binomial distribution
This is similar in spirit to Sum of squares of middle binomial sums or 'Truncated mean' of binomial coefficients under binomial distribution but gives some total estimates. Though the other one was amenable to computations this by nature looks formidable. I do not have much intuition on how fast $f(m,n)$ can grow?
$\mu=1+\epsilon$ where $\epsilon>0$ holds.
$n<m$ holds.
Is there a good bound for $$\log_2\Bigg({\sum_{i_1,\dots,i_{m/n}=-\sqrt{\mu n\ln n}}^{\sqrt{\mu n\ln n}}\binom{n}{\frac n2 +i_1}\dots\binom{n}{\frac n2 +i_{m/n}}}{\mathbb P(\frac n2+i_1)\dots\mathbb P(\frac n2+i_{m/n})}\Bigg)?$$
where $\mathbb P(\frac n2+i)$ is under bionmial distribution and thus is $\frac{\binom{n}{\frac n2 +i}}{2^n}$ and thus this expression is 'truncated joint moment $\binom{n}{\frac n2 +i_j}$ from $j\in\{1,\dots,{m/n}\}$'.
Clearly this is $m-f(m,n)$ (seen from similar nature of terms and bounds from Sum of squares of middle binomial sums or 'Truncated mean' of binomial coefficients under binomial distribution) for some function $f$. How fast can $f(m,n)$ grow?
Conjecture: $\exists c>1: f(m,n)>c\frac mn$.
Question: Can $c>c'\ln n$ hold in above at some $c'>1$?
The OP has changed the original question. This change invalidates my previous answer. The answer below is to the changed question.
For any natural $m$ and $n$ such that $m/n$ is also natural, the expression now to be bounded is
\begin{equation}
L:=\frac mn\,\log_2 T,
\end{equation}
where $T$ is just as in the question here. In the corresponding answer there, it was shown that
$$\log_2T=n-\log_2\sqrt{(1+o(1))\pi n};$$
(the asymptotics everywhere here are as $n\to\infty$).
Hence,
$$L=m-\frac mn\,\log_2\sqrt{(1+o(1))\pi n}.$$
I was reading $\ln\sqrt n$ as $\sqrt{\ln n}$.
For any natural $m$ and $n$ such that $m/n$ is also natural, the expression to be bounded is
\begin{equation}
L:=\frac mn\,\log_2(U_n/2^n),
\end{equation}
where
\begin{equation}
U_n:=\sum_{k\colon\,|k-n/2|\le u}\binom nk^3
\end{equation}
and
\begin{equation}
u:=\sqrt{\mu n\ln n}.
\end{equation}
Letting
\begin{equation}
h_k:=\frac{k-n/2}{n/2}\quad\text{and}\quad t_k:=h_k\sqrt{2n}=\frac{k-n/2}{\sqrt{n/8}},
\end{equation}
by Stirling's formula we have
\begin{align}
\binom nk&\sim\frac{2^n}{\sqrt{\pi n/2}}\exp\{-n[(1+h_k)\ln(1+h_k)+(1-h_k)\ln(1-h_k)]\} \\
&\sim\frac{2^n}{\sqrt{\pi n/2}}\exp\{-n[h_k^2+O(h_k^3)]\} \\
& \sim\frac{2^n}{\sqrt{\pi n/2}}\exp\{-t_k^2/2\};
\end{align}
everywhere here, the asymptotics are for $n\to\infty$ and $k$ such that $|k-n/2|\le u$.
Hence,
\begin{align*}
U_n&\sim\frac{2^{3n}}{\pi^{3/2}n}\;\sum_{k\colon\,|t_k|\le u/\sqrt{n/8}}e^{-3t_k^2/2}(t_{k+1}-t_k) \\
&\sim\frac{2^{3n}}{\pi^{3/2}n}\;\int_{|t|\le u/\sqrt{n/8}}e^{-3t^2/2}\,dt \\
&\sim\frac{2^{3n}}{\pi n}\sqrt{\frac23}.
\end{align*}
Here the transition from the sum to the integral is possible because $t_{k+1}^2-t_k^2=(t_{k+1}-t_k)(t_{k+1}+t_k)\le\frac1{\sqrt {n/8}}\,(2u/\sqrt{n/8}+\frac1{\sqrt{n/8}})=o(1)$.
So,
\begin{align}
L&=\frac mn\,\Big(2n+\log_2\Big(\frac1{(\pi+o(1)) n}\sqrt{\frac23}\,\Big)\Big) \\
&=2m-\frac mn\,\log_2\frac{(\pi+o(1)) n}{\sqrt{2/3}}.
\end{align}
Thus, your conjecture will hold if (and only if) you replace $m-f(m,n)$ by the correct expression $2m-f(m,n)$, with the main term $2m$ rather than $m$.
I think the quantity inside the logarithm is at most $2^m$.
Sorry I made a mistake. If you look at the other problem I either had a square form or a moment form. I mixed the two and I removed the square now and so for this do we have $m-\Omega(\frac mn\ln n)$?
I see only $m\sqrt{\ln n}/n$ when combined with other problem. Perhaps $\ln n$ is not correct.
@VS. : You should not change the question so as to invalidate an answer, especially if the answer required a non-negligible effort. If you have any additional questions, you should ask them in separate posts. Anyway, I have now given an answer to the changed question as well. If any specific step in either answer is unclear, please let me know.
I like your answer. Only thing I made a typo mistake from copying in problem. Essentially what happens if there are no squares?
|
2025-03-21T14:48:29.693054
| 2020-01-19T02:52:01 |
350719
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625739",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350719"
}
|
Stack Exchange
|
Extended double 2-cocycle conditions: Mathematical structure behind?
Note: For experts, to save your time, you can just read the highlighted texts and Eqs directly.
The ordinary group 2-cocycle condition:
Let us remind the usual so-called homogeneous group 2-cocycle $\mu(a,b,c)$ of cohomology group $H^2(G,U(1))$ where $U(1)=\mathbb{R}/\mathbb{Z}$ is given by
$$
\frac{\mu(b,c,d)\mu(a,b,d)}{\mu(a,c,d)\mu(a,b,c)}=1.
$$
where all $a,b,c,d \in G$.
We can focus on the case $G$ is a finite group (or even finite Abelian group if you want to simplify further.)
See References on group cohomogy:
Dijkgraaf-Witten (1990)
Wang-Gu-Wen (2014) Appendices on cocycle conditions
The homogeneous group 2-cocycle $\mu(a,b,c)$ can be coverted to a homogeneous group 2-cocycle via
$$
\omega(A,B):=\omega(a d^{-1},b d^{-1})= \mu(a d^{-1},b d^{-1},1).
$$
so if we define $a d^{-1}=A$ and $b d^{-1}=B$, $c d^{-1}=C$, then
$$
\frac{\mu(bd^{-1},cd^{-1},1)\mu(ad^{-1} , b d^{-1},1)}{\mu(a d^{-1},c d^{-1},1)\mu(a c^{-1},b c^{-1},1)}=\frac{\mu(B,C,1)\mu(A , B,1)}{\mu(A,C,1)\mu(AC^{-1}, BC^{-1},1)}=
\frac{\omega(B,C)\omega(A,B)}{\omega(A,C)\omega(A C^{-1},BC^{-1})}=1.
$$
or equivalently the 2-cocycle condition is:
$$
\frac{\mu(A,B,1)}{\mu(AC^{-1}, BC^{-1},1)}
=\frac{\mu(A,C,1)}{\mu(B,C,1)}
\Leftrightarrow\frac{\omega(A,B)}{\omega(AC^{-1}, BC^{-1})}
=\frac{\omega(A,C)}{\omega(B,C)}.
$$
Extended double 2-cocycle condition: Mathematical structure behind?
Let us define a new object call $F$ which is related to the ordinary homogeneous group 2-cocycles $\mu_1$ and $\mu_2$ (also inhomogeneous group 2-cocycles $\omega_1$ and $\omega_2$ ) with two tensor product inputs:
$$
F(A,B,\alpha ,\beta) :=\mu_1(A \otimes \alpha,B \otimes \beta,1)
=\omega_1(A \otimes \alpha,B \otimes \beta)
$$
also
$$
F(A,B,\alpha ,\beta) :=\mu_2(A \otimes B,\alpha \otimes \beta,1)=\omega_2(A \otimes B,\alpha \otimes \beta).
$$
The 2-cocycle condition for a homogeneous group 2-cocycle $\mu_1$ (also an inhomogeneous group 2-cocycle $\omega_1$ )
becomes:
$$
\frac{\omega_1(A \otimes \alpha,B \otimes \beta)}{\omega_1(AC^{-1} \otimes \alpha \gamma^{-1},
BC^{-1} \otimes \beta \gamma^{-1})}
=\frac{\omega_1(A \otimes \alpha,C \otimes \gamma)}{\omega_1(B \otimes \beta,C \otimes \gamma)}
$$
$$\Rightarrow\boxed{
\frac{F(A,B,\alpha ,\beta)}{
F(AC^{-1},BC^{-1},\alpha\gamma^{-1} ,\beta \gamma^{-1})
}
=
\frac{F(A,C,\alpha ,\gamma)}{
F(B,C,\beta, \gamma)
}}
\tag{1}
$$
The 2-cocycle condition for a homogeneous group 2-cocycle $\mu_2$ (also an inhomogeneous group 2-cocycle $\omega_2$ )
becomes:
$$
\frac{\omega_2(A \otimes B, \alpha \otimes \beta)}{\omega_2(AC^{-1} \otimes B \gamma^{-1} ,
\alpha C^{-1} \otimes \beta \gamma^{-1})}
=\frac{\omega_2(A \otimes B,C \otimes \gamma)}{\omega_2(\alpha \otimes \beta,C \otimes \gamma)}
$$
$$\Rightarrow\boxed{
\frac{F(A,B,\alpha ,\beta)}{
F(AC^{-1},B\gamma^{-1},\alpha C^{-1} ,\beta \gamma^{-1})
}
=
\frac{F(A , B,C ,\gamma)}{
F(\alpha , \beta,C ,\gamma)
}} \tag{2}
$$
Here all $A,B,C,\alpha,\beta,\gamma \in G$.
My puzzle for you: Is there any known mathematical structures behind these two extended double 2-cocycle conditions in Eq.(1) and Eq.(2)? If so, what does the corresponding cocycle class solution form? Are there certain modified notions of cohomology group counting distinct classes of these cocycles $F(A,B,\alpha ,\beta)$?
(Thanks in advance for your answer.)
|
2025-03-21T14:48:29.693348
| 2020-01-19T08:43:23 |
350725
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"GGT",
"Martin Rubey",
"https://mathoverflow.net/users/3032",
"https://mathoverflow.net/users/45170"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625740",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350725"
}
|
Stack Exchange
|
Combinatorial bijection on monotone sequences
Let $(n),\mu$ be the partition of $n$ define $H_g^{m}((n);\mu)$ count's the number of tuples $(\tau_1,\ldots,\tau_r)$ of transposition in symmetric group $S_n$ with the following conditions
$$ (1,2,\ldots,n)\tau_1\ldots \tau_r =\mu$$
$\tau_i$ are transposition written as $(a_i , b_i)$ and $a_i <b_i$ such that
$$ b_1\leq b_2\leq\ldots \leq b_r. $$
$$r=2g-1+\text{len}(\mu) $$
A result in the paper that is Theorem 2 states
https://arxiv.org/pdf/1005.0151.pdf
state that when $\mu= 1^n$ then
$$H_g^m((12..n), 1^n )= Cat_{n-1}T(n+g-1,n-1).....(**) $$
Notice in this particular case $r=2g-1+n$.
I have the following expression the numbers which are called central factorial numbers
$$T(a,b)= 2 \sum_{j={0}}^{b}(-1)^{b-j} \frac{j^{2a}}{(b-j)!(b+j)!}$$
where $Cat_n$ dentoe the catalan number.
Even the paper gives a combinatorial description of $T(a,b)$.
$T(a,b)$ has the following combinatorial description. It is the number of ways to put $\{1,2,\ldots a, 1^{'} , \ldots , a^{'}\}$ in $b$ non empty disjoint box such that if $i$ be the least integer in the box then $i^{'}$ belong the box as well.
$Cat_{n-1}$ has the combinatorial description of writing $(1,2\ldots,n)$ cycle as monotonic $(n-1)$ tuples of transposition.
My question is if there exists a combinatorial bijection? That is given a the seq in LHS of (**) I would construct an element belong to $Cat_{n-1}$ and $T(n-1+g,n-1)$ ?
I am also placing an example in the case $n=3, g=1$
The LHS set is given as follows
$$[[(2,3),(2,3),(2,3),(1,3)],[(2,3),(1,3),(2,3),(2,3)],[(2,3),(1,3),(1,3),(1,3)],[(1,3),(2,3),(1,3),(2,
3)],[(1,3),(1,3),(2,3),(1,3)],[(1,2),(2,3),(2,3),(2,3)],[(1,2),(2,3),(1,3),(1,3)],[(1,2),(1,3),(1,3),
(2,3)],[(1,2),(1,2),(2,3),(1,3)],[(1,2),(1,2),(1,2),(2,3)]]
$$
The above gives a set of 10 tuple of length 4 each. it monotonic
Now $$Cat_2 =2$$ and $T(3,2)=5$ The elements of $Cat_2$ can be given as follows 2 cyles
$$[[(2,3),(1,3)],[(1,2),(2,3)]]$$
I have no idea how to construct the set $T(3,2)$ to make an combinatorial bijection?
This looks like an interesting question, but it is extremely hard to read currently. Maybe you could polish it a little bit. For a start: you introduce $\nu$ in the very first line, but it seems it is unused. Also, "Let define $H^m_g((n); \mu)$ count the number of tuples $(\tau_1,\dots \tau_r)$ in symmetric group $S_n$." doesn't make sense on its own. I guess the description in the paragraph that follows explains what it means, but you should say so. Also, $\mu$ doesn't appear.
It would also help if you could cite the equation in the paper, because apparently your notation is different.
(It's Theorem~2 in the paper.)
I made some edits. I extremely sorry about the bad write up initially. I tried to fix it. Yes it's Theorem 2
|
2025-03-21T14:48:29.693537
| 2020-01-19T10:31:37 |
350728
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625741",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350728"
}
|
Stack Exchange
|
Are locally integrable functions almost completely determined by their approximate modulus of continuity?
This is a follow up to this question, which was answered in the affirmative: Are continuous functions almost completely determined by their modulus of continuity?
Note: We do not identify functions that agree a.e.
Given $f: R \to R$ in $L^{1}_{loc}$, define the approximate left modulus of continuity, $L_f (x, e): \mathbb R \times \mathbb R^+ \to [0, \infty]$ by
$$L_{f} (x, e)=\sup \{ d \geq 0\: \big|\, A(f, r) \leq e, \forall r \text{ s.t. } 0 \leq r \leq d \}.$$
where $A(f, r) := \frac{1}{r} \int\limits_{[x-r, x]} |f(t) - f(x)| {d}t.$
Similarly define the approximate right modulus of continuity by
$$R_{f} (x, e)=\sup \{ d \geq 0\: \big| I(f, r) \leq e, \forall r \text{ s.t } 0 \leq r \leq d \}.$$
where $I(f, r) := \frac{1}{r} \int\limits_{[x, x + r]} |f(t) - f(x)| {d}t.$
Note that if $f = g$ a.e., then $L_f = L_g$ a.e. and $R_f = R_g$ a.e., by which we mean for almost all $x$, for all $e$, $L_f (x, e) = R_f (x, e).$
Do $L_f$ and $R_f$ almost determine f uniquely? In the following sense:
Suppose $f$ and $g$ in $L^{1}_{loc}$ are such that $L_f = L_g$ a.e. and $R_f = R_g$ a.e. Does it follow that $f = g + c$ a.e. or $f = -g + c$ a.e. for some a.e. constant function $c$?
|
2025-03-21T14:48:29.693639
| 2020-01-19T10:53:28 |
350729
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Lev Soukhanov",
"Lior Bary-Soroker",
"https://mathoverflow.net/users/2042",
"https://mathoverflow.net/users/22377",
"https://mathoverflow.net/users/33286",
"https://mathoverflow.net/users/38068",
"spin",
"verret"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625742",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350729"
}
|
Stack Exchange
|
Is there a 2 transitive finite group with transitive normal subgroup having a cyclic quotient other than $A_n$ and $S_n$?
Let $G \leq S_n$ be $2$-transitive other than $A_n$ and $S_n$. Is it possible that there exists $N\lhd G$ with $N\neq G$, $N$ transitive and $G/N$ cyclic?
I am interested mostly in the answer when $n$ is large and also when the group $G$ is $3$-transitive.
$G = S_n$ and $N = A_n$?
Other than that, I forgot to mention it. I will fix the question
I suppose you want $G \neq N$ as well. For all $n \geq 2$ there is a classification of $n$-transitive groups, so I guess a starting point would be looking at these lists for examples.
Thanks, I fixed that too. Sorry for being sloppy.
$PGL_n(F_q)$ and $PSL_n(F_q)$ should do the trick I think, the determinant on $PGL_n(F_q)$ is a well-defined element of the group $(F_q^) / (F_q^)^n$, which can be non-trivial (and always non-trivial for n=2, q odd).
For $n=2$ you will also get $3$-transitivity of $G$, not $N$ though.
Now that you've added $N\neq G$, you can remove $A_n$ as an exception..
@GeoffRobinson: Good point, but I think you mean non-trivial normal subgroup.
For $G = \operatorname{Aut}(M_{22})$ and $N = M_{22}$, with the action of $M_{22}$ on $22$ points you have $N \triangleleft G < S_{22}$. Here both $N$ and $G$ are $3$-transitive, and $G/N \cong C_2$.
The most obvious family of examples is $AGL(1,q)$ for $q$ a prime power.
As spin said in the comments, finite $2$-transitive groups are classified. They are all almost simple or of affine type (like the example I gave). The almost simple ones are quite explicitly listed, so you would just have to go through the list. You should get plenty more examples. The classification of affine ones is a little less explicit (see Have finite doubly transitive groups been classified?)
|
2025-03-21T14:48:29.693783
| 2020-01-19T11:23:22 |
350730
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Zhi-Wei Sun",
"https://mathoverflow.net/users/124654"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625743",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350730"
}
|
Stack Exchange
|
Primitive roots modulo primes related to Fibonacci numbers or Lucas numbers
The Fibonacci numbers $F_0,F_1,F_2,\ldots$ and the Lucas numbers $L_0,L_1,L_2,\ldots$ are given by
$$F_0=0,\ F_1=1,\ \text{and}\ F_{n+1}=F_n+F_{n-1}\ (n=1,2,3,\ldots)$$
and
$$L_0=2,\ L_1=1,\ \text{and}\ L_{n+1}=L_n+L_{n-1}\ (n=1,2,3,\ldots).$$
Here I report my following conjecture on primitive roots modulo primes related to Fibonacci or Lucas numbers.
Conjecture.
(i) For any prime $p$, there are two Fibonacci numbers $F_k$ and $F_m$ with $F_kF_m<p$ such that $F_kF_m$ is a primitive root modulo $p$.
(ii) For any prime $p$, there are two Lucas numbers $L_k$ and $L_m$ with $L_kL_m<p$ such that $L_kL_m$ is a primitive root modulo $p$.
Clearly, part (i) of the conjecture implies that for each odd prime $p$ there exists a Fibonacci number $F_k<p$ which is a quadratic nonresidue modulo $p$. Similarly, part (ii) of the conjecture implies that for each odd prime $p$ there exists a Lucas number $L_k<p$ which is a quadratic nonresidue modulo $p$.
It seems that the first part of the above conjecture can be strengthened as follows: For any prime $p$, there is a primitive root $g < p$ modulo $p$ such that one of
$$\frac g{F_2} = g,\ \ \frac g{F_3}=\frac g2,\ \ \frac g{F_4}=\frac g3$$
is a Fibonacci number. I have verified this strong version for all primes $p < 10^9$. For example,
$F_4F_5=3\times5=15$ is a primitive root modulo the prime $439$. I have also verified the second part of the conjecture for all primes $p<10^9$.
QUESTION. How to prove the conjecture? Can one find a concrete counterexample to the strong version of the first part?
See also http://oeis.org/A331506.
I have entended the verification of the strong version of the first part of the conjecture for all primes $p<5\times10^9$. In my opinion, the strong version might be true.
|
2025-03-21T14:48:29.693897
| 2020-01-19T12:22:11 |
350731
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Chris Wuthrich",
"S. Carnahan",
"https://mathoverflow.net/users/121",
"https://mathoverflow.net/users/5015"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625744",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350731"
}
|
Stack Exchange
|
Flat representations: trying to understand Wiles' proof
Trying to understand the proof of Fermat's last theorem, can anyone explain to me how exactly a finite flat group scheme over $\mathbb{Z}_{p}$ gives rise to a Galois representation over a finite field?
Look at the generic fiber. There is a finite extension $K/\mathbb{Q}_p$ such that base change yields a constant group over $K$ with a descent datum. If this constant group scheme is isomorphic to a vector space over a finite field $F$, then the descent datum may yield an $F$-linear Galois action.
From this and your previous question, I make the guess that you are trying to read through Wiles' paper lacking some basic knowledge. I would suggest that you read parallely Saito's books on Fermat's Last Theorem, which contains a very large part of the proof and much of the background needed to understand it.
|
2025-03-21T14:48:29.693987
| 2020-01-19T13:19:48 |
350735
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gerhard Paseman",
"Nandakumar R",
"https://mathoverflow.net/users/142600",
"https://mathoverflow.net/users/3402"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625745",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350735"
}
|
Stack Exchange
|
On dissecting a triangle into another triangle
It is easy to see that an equilateral triangle can be cut into 2 identical 30-60-90 degrees right triangles which can then be patched together to form a 30-30-120 degrees triangle. So, via 2 intermediate pieces, we can dissect an equilateral triangle into the 30-30-120 triangle of the same area.
Question: Find two equal area, non-congruent triangles T1 and T2 such that T1 can be cut into 3 (and not less than 3) pieces which can be reassembled into T2. Not sure if such a T1-T2 pair exists.
One can readily generalize and ask for equal area {T1-T2} pairs which can be dissected into each other via n (and not less than n) intermediate pieces (n cannot be arbitrarily large, from the Wallace-Bolyai-Gerwein theorem).
Further question: What can one say about equal area pairs of triangles which are the worst for mutual dissection - ie for which the number of intermediate pieces is the highest possible?
Guess: a triangle pair with equal area and equal perimeter is bad for mutual dissection.
Further Note - Going to Pairs of Quadrilaterals: the problem is trivial for rectangles - indeed, for any rectangle R and any n, there is a different rectangle with same area as R and n times as long which can be got by cutting R into n equal, length-parallel strips and attaching them end to end. For general quadrilaterals, things seem less trivial. For any cyclic quad Q (each pair of opposite vertices of Q add to 180 degrees), it appears that we can cut Q into 2 pieces and patch these to get another quad Q' different from Q. But for non-cyclic Q's, one is not sure of finding such a Q' even for n =2. For general n, things appear harder.
If the problem is solvable for rectangles for arbitrary n, then it should be solvable for right triangles for arbitrary n. Gerhard "Just Use Half A Rectangle" Paseman, 2020.01.20.
Speaking of rectangles, here is a quick solution to the question. From a rectangle corner, draw lines to the midpoints or the sides not adjacent to a corner. This gives a three piece dissection which can form two non congruent triangles. I suspect there is a proof that this cannot be replaced by a two piece dissection. Gerhard "Ready Rectangles To The Rescue!" Paseman, 2020.01.20.
Just checked your construction. Yes, it does yield two non-congruent right triangles which can be dissected into each other via 3 intermediate pieces. And no 2-piece dissection seems to work! But I don't readily see how it generalizes to n greater than 3.
The three piece construction doesn't; the 1 by mn to m by n rectangle extends to triangles, although it is not clear that exactly n pieces are needed. Gerhard "Perhaps Something Close To N" Paseman, 2020.01.20.
|
2025-03-21T14:48:29.694167
| 2020-01-19T16:43:31 |
350746
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ali Enayat",
"Dmytro Taranovsky",
"https://mathoverflow.net/users/113213",
"https://mathoverflow.net/users/9269"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625746",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350746"
}
|
Stack Exchange
|
Elementary self-embeddings conservative over ZFC
Question: Is the following theory conservative over ZFC? And if not, what is its strength?
Language: $∈$, $j$ (unary function symbol)
Axioms:
1. ZFC (without separation and replacement for formulas using $j$).
2. (schema) $j$ is a nontrivial elementary embedding $(V,∈)→(V,∈)$.
3. (schema) $S∩φ ≡ \{s∈S:φ(s)\}$ exists whenever $φ$ is a formula with parameters and one free variable and set $S$ is definable (allowing $j$).
Note that (3) implies existence of $S'∩φ$ for all $S'$ with $|S'|≤|S|$ for some $S$ as above. Thus, (3) simply asserts the full Separation Schema for sets that are not too large, including all definable sets.
I am also interested in the extension (its strength is above $n$-Mahlo):
3a. Allow any $S$ definable (allowing $j$) from some set $T$ with $T=j(T$).
Background
An important phenomenon in mathematical logic is that if we extend a theory with new symbols (and sometimes new types) and add reasonable new axioms, the new theory will sometimes be conservative over the original one; and such correspondences often give key structural insight into both theories. A special case, relevant here, is that the new theory is seemingly stronger and captures a part of the structure of a stronger theory, but is missing a key ingredient.
Under various reasonable formalizations (that imply replacement for $j$-formulas), existence of a nontrivial elementary embedding $V→M$ ($M$ transitive) is equivalent to existence of a measurable cardinal, and $V=M$ is inconsistent with ZFC.
However, (1) + (2) is conservative over ZFC: Add Skolem functions for $V$ and add $ω$ constant symbols for indiscernible ordinals, use compactness to find a model, take the Skolem hull of the indiscernibles, and add a nontrivial order preserving injection between the indiscernibles, which will then extend to the desired elementary embedding.
Such a model (above) may be ill-founded, but we can try to make it well-behaved with respect to small sets. Every countable ZFC model $M$ has a nontrivial elementary end extension $N$; by elementarity, $N$ is also a top extension, that is for all $α∈\mathrm{Ord}^M$, $V_α^M$ and $V_α^N$ have the same elements. We can even require $N$ to have a nontrivial automorphism fixing all elements of $M$. And perhaps some way of incorporating $j$ there will give a positive answer to the question.
Under (1)-(3a), $\{s:s=j(s)\}$ (under '$∈$') is a rank-initial proper elementary substructure of $(V,∈)$, but it does not exist as a set unless its cardinality is $n$-huge for every $n$. Also, under (1)-(3a), if we let $I = \{β∈\mathrm{Ord}:j(β)=β\}$ and $C=\{x∩V_I:x∈V\}$, then $(V_I,C,∈)$ satisfies NBG + "$\mathrm{Ord}$ is weakly compact". NBG + "$\mathrm{Ord}$ is weakly compact" is equiconsistent with ZFC + $\{n\text{-Mahlo}\}_{n∈ω}$. A related theory is discussed in Automorphisms, Mahlo Cardinals, and NFU by Ali Enayat.
As an aside, there is a rich hierarchy theories based on how 'close' $j$ is to $V$. For example (see this question), if (3) is replaced with the critical point axiom (i.e. the least ordinal moved by $j$ exists), the resulting theory is conservative over ZFC + {there is $n$-ineffable cardinal}$_{n∈ℕ}$.
Your self-embedding axiom seems close to Corazza's Wholeness axiom, see https://en.wikipedia.org/wiki/Wholeness_axiom
@AliEnayat Unlike the Wholeness Axiom (which it superficially resembles), the theory only appears to require separation below sets moved by $j$ since without the critical point it appears that none of the sets moved by $j$ need to be definable.
|
2025-03-21T14:48:29.694399
| 2020-01-19T19:09:02 |
350752
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625747",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350752"
}
|
Stack Exchange
|
(Co)tensoring of enriched slice categories
In an answer to this question: Enriched slice categories, a description of the enrichment of the slice category in an enriched category is given. I'm interested in going a bit further. If we assume that the original enriched category, say $\mathcal{C}$, is $\mathcal{V}$-enriched but also $\mathcal{V}$-tensored and cotensored, and that $\mathcal{V}$ is semi-Cartesian (i.e. the monoidal unit is terminal) how can we describe the tensoring and cotensoring of a slice $\mathcal{C}_{/X}$ for $X\in\mathcal{C}$? It seems to me that the tensoring should probably be some kind of weighted colimit, which I believe would mean we just compute it on the "underlying object." My immediate guess about what this would look like is the following: given an object $f:A\to X\in\mathcal{C}_{/X}$, the tensoring with $K\in\mathcal{V}$ is $$f\otimes K:A\otimes K\overset{A\otimes\varepsilon_K}\longrightarrow A\otimes 1_{\mathcal{V}}\cong A\overset{f}\to X$$
But first of all, I think this is probably wrong, and second of all, I'm wondering if there is an explicit description for the cotensoring as well, such that we have the usual isomorphisms
$$\mathit{Map}_{\mathcal{C}_{/X}}(f\otimes K, g)\cong \mathit{Hom}_\mathcal{V}(K,Map_{\mathcal{C}_{/X}}(f,g))\cong \mathit{Map}_{\mathcal{C}_{/X}}(f, g^K)$$
I happen to be only interested in the case that $\mathcal{V}=sSet$, but perhaps there's a more general answer. I'm not even sure we need the condition that $\mathcal{V}$ is semi-Cartesian honestly, since it seems like the statement about tensors being colimits, and the forgetful functor preserving colimits, wouldn't require it.
Your guess for the copower (née tensor) is correct. You can check it by giving it the right universal property with respect to the enriched homs as described in the question you linked. Similarly, for the power (née cotensor) you can check that
$g^K$ for $g:B\to X$ can be defined by the pullback of the induced map $B^K \to X^K$ along $X \cong X^{1_V} \to X^K$.
However, I don't know offhand how to generalize away from the semicartesian case. Possibly the right point of view in that case is that the slice $C/X$ is not enriched over $V$ itself but over $V/I$; I think that it can be given powers and copowers for that enrichment.
|
2025-03-21T14:48:29.694555
| 2020-01-19T19:53:55 |
350753
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Emil Jeřábek",
"Sam Hopkins",
"Tobias Fritz",
"https://mathoverflow.net/users/12705",
"https://mathoverflow.net/users/25028",
"https://mathoverflow.net/users/27013"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625748",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350753"
}
|
Stack Exchange
|
Name for partial orders which are total on connected components
In my context, I encounter a lot of partial orders with the distinguished property that the order is total on connected components. Equivalently, they satisfy the condition
$$x \le y,z \enspace \lor \enspace y,z \le x \qquad \Longrightarrow \qquad y \le z \enspace \lor \enspace z \le y.$$
Is there an established term for partial orders with this property?
Context: I'm looking for a single adjective that will allow me to say "[adjective] ordered group" by analogy with "totally ordered group". My current choice is to say "fully ordered group" in the draft, but I'd like to switch to the established term if there is one.
I think you need to add the dual condition too it to be equivalent.
@SamHopkins: of course, thanks, I'll make the edit.
Anyways, "disjoint union of total orders" is concise enough that I doubt there is another name for these.
@SamHopkins: I'm looking for a single adjective that would capture this property, so that I can say "(whatever) order" in a similarly convenient way as I can say "total order". More concretely, I want to be able to say "(whatever) ordered group", just as I can say "totally ordered group". Saying "partially ordered group which is a union of total orders" sounds quite clumsy. I'm currently going with "full order" in the draft, but I'd like to change it to the established term if there is one.
I believe Sam Hopkins is right, but if you want a more concise adjective, how about “component-wise totally ordered group”?
|
2025-03-21T14:48:29.694680
| 2020-01-19T19:57:04 |
350754
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexander Kalmynin",
"Blue",
"Bogdan Ion",
"https://mathoverflow.net/users/101078",
"https://mathoverflow.net/users/148184",
"https://mathoverflow.net/users/151281"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625749",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350754"
}
|
Stack Exchange
|
Singularities of power series
The power series $\sum_{n=1}^\infty \ln(n)z^n$ has radius of convergence $1$ and $z=1$ is a singular point. Is $z=1$ an isolated singularity? If yes, what kind of isolated singularity?
I am only able to deduce that $z=1$ cannot be a pole.
Such type of questions appear naturally when one tries to relate the singularities of the power series and those of the Dirichlet series associated to the same sequence.
Let $f(z)$ be your function. Then $g(z)=f(z)(1-z)$ is equal to
$$
(1-z)\sum_n \ln(n)z^n=\sum_{n\geq 2} (\ln(n)-\ln(n-1))z^n
$$
Now, $\ln(n)-\ln(n-1)=\frac{1}{n}+O\left(\frac{1}{n^2}\right)$, which gives us
$$
g(z)=\sum_{n≥1} \left(\frac{1}{n}+g_n\right)z^n,
$$
where $g_n=\ln(n)-\ln(n-1)-\frac{1}{n}=O\left(\frac{1}{n^2}\right)$ for $n>1$ and $g_1=-1$, so that
$$
f(z)=\frac{\ln(1-z)}{z-1}+\frac{h(z)}{z-1}.
$$
Here $h(z)$ is holomorphic in the unit disc and continuous on its boundary.
Hope this answers your question on the type of singularity in $z=1$.
Edit:
Let me also present a slightly less elementary way to study properties of this series, based on my favorite method of lots of contour integration. We will use the derivative of Riemann zeta-function, so this is more in the spirit of the question.
Let $x\in \mathbb C$ be a number with positive real part. Using the formula
$$
e^{-nx}=\frac{1}{2\pi i}\int_{3/2-i\infty}^{3/2+i\infty} \Gamma(s)(nx)^{-s}ds,
$$
we obtain
$$
f(e^{-x})=\frac{1}{2\pi i}\int_{3/2-i\infty}^{3/2+i\infty}\Gamma(s)\zeta'(s)x^{-s}ds.
$$
From this we easily get
$$
f(e^{-x})=\mathrm{Res}_{s=1}\,\Gamma(s)\zeta'(s)x^{-s}+\sum_{n\geq 0}\mathrm{Res}_{s=-n}\,\Gamma(s)\zeta'(s)x^{-s}.
$$
The first summand is actually a bit different from all the other, because we get double pole. From expansions
$$
\Gamma(s)=1-\gamma(s-1)+O((s-1)^2), \zeta'(s)=\frac{1}{(s-1)^2}+O(1)
$$
and
$$
x^{-s}=\frac{1}{x}-\frac{(s-1)\ln x}{x}+O((s-1)^2)
$$
(here $\gamma$ is the Euler-Mascheroni constant) we get
$$
\mathrm{Res}_{s=1}\,\Gamma(s)\zeta'(s)x^{-s}=-\frac{\ln x+\gamma}{x},
$$
which corresponds to the first part of my answer and also gives $h(1)=-\gamma$. The rest is way easier to compute and we obtain
$$
f(e^{-x})=-\frac{\ln x+\gamma}{x}+\sum_{n\geq 0}\frac{(-1)^n\zeta'(-n)x^n}{n!}.
$$
Now, from this answer about derivative of zeta we see that this series has a nonzero radius of convergence (namely, $2\pi$) and we can even see singularities at $x=2\pi i n$ for $n\in \mathbb Z$, which is of course what one should expect because of singularity of $f$ at $z=1$.
Hello! So just to make sure, the singularity is a pole, right?
@Blue, you mean at $2\pi i n$? No, this is still a logarithmic singularity (would be super strange if $f(e^{-x})$ had a logarithmic singularity at $x=0$ but only a pole at $x=2\pi i$), because the factor $\psi(2n)$ in $\zeta'(1-2n)$ grows logarithmically in $n$ and for $x=2\pi i$ we get essentially the same series with logs
I think that the second part of the answer is more satisfactory. The first part, while it reveals a branch point at $z=1$, does not clarify whether $h(z)$ is holomorphic in a neighborhood of $z=1$. The second explanation, if the statement about the radius of convergence checks out, does not have this flaw.
|
2025-03-21T14:48:29.694890
| 2020-01-19T21:56:47 |
350757
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625750",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350757"
}
|
Stack Exchange
|
Analogous theorem for Hilbert modular forms
I have studied modular forms and saw a correspondence like a newform correspond to a automorphic representation of $\mathrm{GL}_n(\mathbb{A_Q})$. Does any similar result holds for Hilbert modular forms?
Also we know $S_k(\Gamma_1(M))= \bigoplus\nolimits_{\epsilon:(\mathbb{Z}/n\mathbb{Z})^*\to\mathbb{C}^*} S_k(\Gamma_0(M),\epsilon)$ and $\dim_{\mathbb{C}}S_k(\Gamma_1(M))$ ~ $M^2$ where $\Gamma_1(M)$ and $\Gamma_0(M)$ are well-known congruence subgroups of $\mathrm{SL}_2(\mathbb{Z})$. $S_k(\Gamma_1(M))$ is the subspace of holomorphic cusp forms and $S_k(\Gamma_0(M),\mathbb{\epsilon})$ is the subspace of holomorphic cusp forms having character $\epsilon$. I want to know again if there are similar results for Hilbert modular forms or Hilbert Cusp forms.
If yes, please suggest some good references for this.
|
2025-03-21T14:48:29.695080
| 2020-01-19T21:58:10 |
350758
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625751",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350758"
}
|
Stack Exchange
|
Characteristic function and moments
Let $X\in L^1(\Omega)$ and $\phi_X$ the corresponding characteristic function.
We know that: $\phi_X$ is $n$ times differentiable (at $u=0$) iff $\mathbb{E}[X^n]<\infty$. (This depends a bit on if $n$ is even or odd but that's not important for my question). In fact, the derivatives of $\phi_X$ give a way of computing moments, i.e. $\mathbb{E}[X^n]=i^{-n}\phi_X^{(n)}(0)$.
My questions are about avoiding to compute these derivatives and working in the complex plane instead, i.e. so-called ''generalized'' characteristic functions defined on strips of the complex plane (''strips of regulatory'').
Questions
Suppose $X=\ln(Y)$ has a nice $C^\infty$ characteristic function. Can I compute the moments of $Y=e^X$ by simply evaluating $\phi_X$ on the imaginary axis, i.e. $$\mathbb{E}[Y^n]=\mathbb{E}[e^{i(-in)\ln(Y)}]=\phi_{\ln(Y)}(-in).$$
If one knows that $\mathbb{E}[X^n]<\infty$ up to some $N$ (potentially infinity) due to the differentiability of $\phi_X$, does one then know that the lower half of the complex plane (up to $N$) is a subset of the stip of regularity?
Example
An example would be the Variance Gamma process (a subordinated Brownian motion).
Here,
\begin{align*}
\phi_{X_t}(u)&=\left(\frac{1}{1-\theta\mu iu+\frac{1}{2}\mu\sigma^2u^2}\right)^{\frac{t}{\mu}} \\
&= \exp\left(-\frac{t}{\mu}\ln\left(1-\theta\mu iu+\frac{1}{2}\mu\sigma^2u^2\right)\right).
\end{align*}
Let $Y_t=e^{X_t}$ be an exponentiated VG process. Do we then really have
\begin{align*}
\mathbb{E}[Y^n] &=\phi_X(-in) \\
&=\exp\left(-\frac{t}{\mu}\ln\left(1-\theta\mu n-\frac{1}{2}\mu\sigma^2n^2\right)\right)<\infty
\end{align*}
for all $n\in\mathbb{N}$?
The answers to your questions are no and no.
Question 1: Let $X$ have the standard double exponential distribution, so that the pdf $p_X$ of $X$ is given by $p_X(x)=\frac12\,e^{-|x|}$ for real $x$. Then the characteristic function $f_X$ of $X$ is in $C^\infty$, since $f_X(t)=\frac1{1+t^2}$ for real $t$. However, all the moments $EY^n$ for $Y=e^X$ and $n\ge1$ equal $\infty$.
Question 2: Let $X$ have the pdf $p_X$ given by $p_X(x)=ce^{-|x|^{1/2}}$ for some real $c>0$ and all real $x$. Then $E|X|^n<\infty$ for all real $n>0$. However, the characteristic function $f_X$ of $X$, given by the formula $f_X(t)=\int_{-\infty}^\infty e^{itx}p_X(x)\,dx$ for real $t$, cannot be extended to any horizontal strip in the lower half of the complex plane.
|
2025-03-21T14:48:29.695237
| 2020-01-19T23:45:30 |
350765
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625752",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350765"
}
|
Stack Exchange
|
How close are inverses of sub-matrices of positive definite matrices?
Suppose that $A,B \in \mathbb{R}^{p\times p}$ are two positive definite matrices, and let $S\subset \{1,2,\ldots,p\}$ be a subset of indices. I suspect that the following is true, and was wondering if someone can point me to a reference or a proof:
$$\|(A_S)^{-1} - (B_S)^{-1}\|_{\rm max} \lesssim \left\| A^{-1} - B^{-1} \right\|_{\rm max},$$
where $\lesssim$ is hiding constants that do not depend on $p$, $A_S$ is the submatrix of $A$ indexed by the elements in $S$, and $\left\| \right\|_{\rm max}$ is the element-wise absolute maximum of a matrix.
|
2025-03-21T14:48:29.695301
| 2020-01-20T01:04:40 |
350767
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625753",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350767"
}
|
Stack Exchange
|
Questions about a return map
Consider the following map in the interval $u\in[-1,1]$ ($U\in[-1,1]$ also)
$$U=u-\frac{64}{3}u^{3}+64u^{5}-64u^{7}+\frac{64}{3}u^{9}$$
It has 3 fixed points at $u=0,\pm 1$. If we compute the derivative of the map
$$\frac{dU}{du}=1-64u^{2}+5\times 64u^{4}-7\times 64 u^{6}+3\times 64u^{8}$$
we see that it is $1$ at the fixed points. So we cant determine whether the fixed points are stable/unstable linearly.
Numerically we can easily see that $0$ is a stable point and $\pm 1$ are unstable globally.
Is there a way to show/prove this analytically?
If I compute the Lyapunov exponent for several initial values of $u$ in the $-1,1]$ interval and plot it, together with the iteration map I get the following graph
The exponent is positive in certain ranges where the map seems to bounce around an infinite number of times. In other ranges the Lyapounov exponent is negative and orbits in this range seem to end in the fixed point 0. The fact the exponent is positive would seem to indicate the map is chaotic in those ranges but looking at the maps it seems it just bounces around in a square.
For example for $u_{0}=0.39$ the orbit goes to $u=0$
but for $u_{0}=0.4$ it seems to move around in a square
Is the map chaotic in any range? Or all the orbits for positive Lyapunov exponent are just periodic (2,4,6, etc cycles)? Is there a way to show that all orbits with negative Lyapunov exponent approach the fixed point $u=0$?
We have $f(u) < u$ for $0 < u < 1$ and $f(u) > u$ for $u > 1$, so the fixed point $1$ is unstable. Similarly $f(u) < u$ for $u < -1$ and $f(u) > u$ for $-1 < u < 0$ implies $-1$ is unstable. Since $|f(u)| < |u|$ for $u$ close to $0$, $0$ is stable.
|
2025-03-21T14:48:29.695432
| 2020-01-20T01:38:53 |
350770
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"MyNinthAccount",
"https://mathoverflow.net/users/140496"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625754",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350770"
}
|
Stack Exchange
|
Software for finding conjugates in the braid group
The conjugacy problem for the braid group was solved by Garside, and gives an algorithm for determining whether two braids are conjugate. Since this algorithm is rather tedious, I was wondering if anybody knows of any software that can tell me if any two given braids (given their braid word) are conjugate.
Magma claims to have this. https://magma.maths.usyd.edu.au/magma/handbook/text/866#10067 "Note that testing elements for conjugacy is a hard problem and may require significant amounts of memory and CPU time." https://magma.maths.usyd.edu.au/magma/handbook/text/869#10152
The program flipper can solve the conjugacy problem for pseudo-Anosov elements of mapping class groups. In principle this allows one to solve the conjugacy problem in the braid group $B_n$ for many pairs of words. Suppose one has two braids whose monodromy is pseudo-Anosov in the $n$-punctured plane mapping class group. Flipper can solve the conjugacy problem in the mapping class group of the $n+1$-punctured sphere for this pair of braids, and then one must determine if the conjugacy preserves the point at infinity (this actually requires one to compute the centralizer of the mapping class group element, which is possible using the veering triangulation data flipper produces). Since the braid group is a central extension of a subgroup of the mapping class group of the $n$-punctured plane fixing infinity, two braids whose mapping class representatives are conjugate will be conjugate in the braid group if they have the same writhe.
Braidlab, a matlab package: https://arxiv.org/abs/1410.0849
(see p.19 of the PDF manual linked)
|
2025-03-21T14:48:29.695551
| 2020-01-20T04:06:57 |
350775
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625755",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350775"
}
|
Stack Exchange
|
2-quotient of integer partition
This question is mostly about understanding the notation used in the following article:
Alex Eskin, Andrei Okounkov, Pillowcases and quasimodular forms, in: Victor Ginzburg (ed.), Algebraic Geometry and Number Theory in Honor of Vladimir Drinfeld's 50th Birthday, Birkhäuser 2006.
page 8 about the 2-quotients of a partition. It claims that every partition ${\bf{\lambda}}:=(\lambda_1, \lambda_2,\ldots,)$ has two partitions
$\bf{\alpha, \beta}$ such that
$$\Big\{\lambda_i -1 +1/2\Big\}=\Big\{\alpha_i -i+1/2 +\bar{p}_0(\lambda) \Big\}\sqcup \Big\{\beta_i -i+1/2+\bar{p}_0(\lambda) \Big\} $$
I mostly interested in balanced partition so I can forget about $\bar{p}_0(\lambda) $. I don't understand what is the definition of $\alpha_i, \beta_i$. Is it related to Frobenius coordinates?
I suspect $\alpha_i$ and $\beta_i$ refer to heights in the Russian way to describe Young diagrams (rotate the English notation 135 degrees), this makes it into a piecewise linear function, and this interpretation has a few nice applications.
Information regarding quotients, (and its relation to character values in particular), can be found here.
There is the article
Olivier Brunat and Rishi Nath. Cores and quotients of partitions through the Frobenius symbol. ArXiv e-prints, 2019.
which apparently does what the title suggest, in case the Frobenius description of partitions is relevant.
|
2025-03-21T14:48:29.695662
| 2020-01-20T05:02:24 |
350778
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Noah Schweber",
"https://mathoverflow.net/users/8133"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625756",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350778"
}
|
Stack Exchange
|
Improving a Lindstrom-y fact about $\mathcal{L}_{\omega_1,\omega}$?
See e.g. the last section of Ebbinghaus/Flum/Thomas for the relevant background on abstract model theory. Below, all languages are finite for simplicity. "$HC$" is the set of hereditarily countable sets.
There's a silly version of Lindstrom's theorem for the infinitary logic $\mathcal{L}_{\omega_1,\omega}$ gotten by replacing compactness with Barwise compactness:
$(*)\quad$ Suppose that $\mathcal{L}=(Sent_\mathcal{L}, \models_\mathcal{L})$ is a regular logic such that:
$Sent_\mathcal{L}\subseteq HC$;
$Sent_{\mathcal{L}_{\omega_1,\omega}}\subseteq Sent_\mathcal{L}$ and for each $\varphi\in Sent_{\mathcal{L}_{\omega_1,\omega}}$ we have $Mod_{\mathcal{L}_{\omega_1,\omega}}(\varphi)=Mod_\mathcal{L}(\varphi)$;
$\mathcal{L}$ has the downward Lowenheim-Skolem property for single sentences: if $\theta\in Sent_\mathcal{L}$ has a model, then it has a countable model; and
$\mathcal{L}$ is Barwise compact: for every countable admissible $\mathbb{A}$ and every $\mathbb{A}$-c.e. theory $\mathfrak{T}\subseteq\mathcal{L}\cap\mathbb{A}$, if every $\mathbb{A}$-finite subset of $\mathfrak{T}$ has a model then $\mathfrak{T}$ has a model.
Then $\equiv_\mathcal{L}$ coarsens $\equiv^{EF}_{\omega_1}$: if Duplicator has a winning strategy in the EF-game of length $\omega_1$ between $\mathcal{A}$ and $\mathcal{B}$, then $\mathcal{A}\equiv_\mathcal{L}\mathcal{B}$.
(Note that the above would be trivial if we restricted attention to countable structures, by Scott's isomorphism theorem.)
Now, this is suboptimal in a couple different ways. The most obvious is that we don't get that $\equiv_\mathcal{L}$ and $\equiv_{\omega_1,\omega}$ coincide (since EF-games don't characterize infinitary equivalence nicely). However, the annoyance I want to focus on is the "implementation-specific" aspect: while some focus on implementation (the first bulletpoint) is necessary to make sense of Barwise compactness as a property of an abstract logic, the second bulletpoint in my opinion emphasizes the specific details of $\mathcal{L}$ too much.
Question: Is $(*)$ still true if we replace the second bulletpoint in $(*)$ with "$\mathcal{L}\ge\mathcal{L}_{\omega_1,\omega}$" - that is, if we require merely that for each $\varphi\in Sent_{\mathcal{L}_{\omega_1,\omega}}$ there be some $\psi\in Sent_\mathcal{L}$ with $Mod_{\mathcal{L}_{\omega_1,\omega}}(\varphi)=Mod_\mathcal{L}(\psi)$?
In the next two sections, I sketch the proof of $(*)$ and point out the key obstacle.
Proving $(*)$
Suppose $\mathcal{A}\equiv_{\omega_1}^{EF}\mathcal{B}$ but $\mathcal{A}\models\varphi$ and $\mathcal{B}\models\neg\varphi$ for some $\varphi\in \mathcal{L}$. Let $\varphi^A,\varphi^B$ be the relativizations of $\varphi$ to two new unary predicates, and let $r$ be a real coding these sentences, $\mathcal{A}$, and $\mathcal{B}$. Inside $\mathbb{A}=L_{\omega_1^{CK}(r)}[r]$ - which contains $\varphi^A$, $\varphi^B$, $\mathcal{A}$, and $\mathcal{B}$ since we can build a set in $HC$ from a real coding it using $\Sigma_1$-recursion - we consider the $\mathbb{A}$-c.e. theory $\mathfrak{T}\subseteq\mathcal{L}\cap\mathbb{A}$ describing the following type of structure:
The "design" sentence $(D)$, which says that we have sorts $A,B$ which describe structures with $A\models\varphi$ and $B\models\neg\varphi$ and sorts $S$ and $I$ where $I$ is a linear order and $S$ is a back-and-forth system of finite partial isomorphisms between $A$ and $B$ appropriately indexed by $I$.
For each $\alpha<\omega_1^{CK}(r)$, the "extension" sentence $(E)_\alpha$: "$\alpha$ embeds as an initial segment of $I$."
The "object" sentence $(O)$: "$I$ itself is an $r$-computable relation on $\omega$."
The point is that $\mathcal{A}$ and $\mathcal{B}$ generate in the obvious way a model $\mathcal{M}$ of $\{(D)\}\cup\{(E)_\alpha:\alpha<\omega_1^{CK}(r)\}$. This doesn't satisfy $(O)$ of course, but fixing $\alpha<\omega_1^{CK}$ we can "cut off" the $I$-part of $\mathcal{M}$ to get a structure $\mathcal{M}_\alpha$ which only captures $\mathcal{A}\equiv^{EF}_\alpha\mathcal{B}$ but which is an element of $\mathbb{A}$ - that is, $\mathcal{M}_\alpha\models\{(D),(O)\}\cup\{(E)_\beta:\beta<\alpha\}$.
Barwise compactness then gives us a model of the whole theory $\mathfrak{T}$. From this in turn we get a structure with an $A$-piece satisfying $\varphi$, a $B$-piece satisfying $\neg\varphi$, and an $\omega^*$-indexed back-and-forth system of finite partial isomorphisms between them. That whole situation is described by a single $\mathcal{L}$-sentence $\theta$, and so we get a countable model of $\theta$; but then the $A$- and $B$-pieces of that model are isomorphic, contradicting their $\mathcal{L}$-inequivalence.
The obstacle
So what breaks down if we try to run the argument above to prove the hoped-for stronger result?
Well, ignoring $(D)$ and the "one-and-done" requirements that a real coding $\varphi^A,\varphi^B,\mathcal{A},\mathcal{B}$ be in $\mathbb{A}$, the first thing we needed was for $\mathbb{A}$ to contain the sentences $(E)_\alpha$ for each $\alpha<\mathbb{A}\cap Ord$. When the $\mathcal{L}_{\omega_1,\omega}$-sentences is contained in $\mathcal{L}$ a la the undesired second bulletpoint above, this is trivial; but if the embedding $\mathcal{L}_{\omega_1,\omega}\le \mathcal{L}$ is sufficiently weird we might need to build this $\mathbb{A}$ as a big union of admissible sets to "catch our tail." That's doable ...
... But it ruins the second requirement, which is that we need to be able to get something like $(O)$. Basically, we need an $\mathcal{L}$-sentence (or at least an $\mathbb{A}$-c.e. $\mathcal{L}$-theory) which only describes things in $\mathbb{A}$ but does describe well-orderings of arbitrarily large ordertype $<\mathbb{A}\cap Ord$. This is a sense in which $\mathbb{A}$ would have to be "similar to" $L_{\omega_1^{CK}}$" ... which we'd have no reason to expect here.
That is, the only admissible sets big enough to have their fragments of $\mathcal{L}$ talk about all ordinals smaller than their height might be too big to do anything like the $(O)$-trick above. But $(O)$ was crucial: it was the only way we could conclude that after applying Barwise compactness we would get something whose back-and-forth-indexer was ill-founded.
Actually, the second bulletpoint I'm annoyed at here can be phrased in a much more palatable way by shifting from arbitrary to effective reductions of logics - e.g. if the translation from $\mathcal{L}{\omega_1,\omega}$ into $\mathcal{L}$ is $E$-recursive in some parameter $p\in HC$ then dLS + BC gives $\equiv\mathcal{L}\supseteq\equiv^{EF}_{\omega_1}$ because we can still trap things in good admissibles. So arguably it's the "boldface" comparison between logics which isn't appropriate in this setting. But, I'm still interested in whether the stronger implication hoped for above is true.
|
2025-03-21T14:48:29.696077
| 2020-01-20T08:43:08 |
350783
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Rony Bitan",
"https://mathoverflow.net/users/17828"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625757",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350783"
}
|
Stack Exchange
|
When is a group scheme a twisted form of a Bruhat-Tits model of its generic fiber?
Let $R$ be a compact DVR with field of fractions $K$.
Let $\underline{G}$ be an affine, flat and smooth group scheme defined over $\text{Spec}(R)$, whose generic fiber $G:=\underline{G} \otimes_{\text{Spec}(R)}K$ is (connected) reductive.
My question is:
Under what assumption is $\underline{G}$ isomorphic over some finite etale extension of $R$ to a Bruhat-Tits $R$-model $\underline{G}_{\text{BT}}$ of $G$, stabilizing some simplex in the Bruhat-Tits building associated to $G$ ?
For example, suppose $\underline{G}(R)$ is a maximal compact subgroup of $G(K)$. May I assume that the Bruhat-Tits model $\underline{G}_{\text{BT}}$ stabilizing a special vertex in the building of $G$ such that $\underline{G}_{\text{BT}}(R) = \underline{G}(R)$ is such a twist of $\underline{G}$ ?
Thank you very much !
Rony
This question maybe watered down by restricting to the small site of functors:
Namely: we know by the Bruhat-Tits fixed point Lemma that any compact subgroup of G(K) necessarily fixes a simplex in the building, so if $\underline{G}(R) = \underline{G}{\text{BT}}(R)$ (for some $\underline{G}{\text{BT}}$), and this isomorphism holds true for any finite etale extension of $R$, then $H^1_{\text{et}}(R,\underline{G})$ is bijective to $H^1_{\text{et}}(R,\underline{G}_{\text{BT}})$, right ?
|
2025-03-21T14:48:29.696178
| 2020-01-20T08:54:22 |
350784
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Fedor Petrov",
"Zhi-Wei Sun",
"https://mathoverflow.net/users/124654",
"https://mathoverflow.net/users/4312"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625758",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350784"
}
|
Stack Exchange
|
Is it true that $|\{k^{k+1}+(k+1)^k\pmod p:\ k=0,\ldots,p-1\}|=(1-e^{-1})p+O(\sqrt{p})\ ?$
For each prime $p$, let us define
$$w_p:=|\{k^{k+1}+(k+1)^k\pmod p:\ k=0,\ldots,p-1\}|,$$
where $a\pmod p$ denotes the residue class $a+p\mathbb Z$.
Based on my computation, I conjecture that
$$w_p=(1-e^{-1})p+O(\sqrt{p}).\tag{1}$$
For the $2\times 10^6$-th prime $p=32452843$, we have
$$w_p=20519206,\ \ w_p-(1-e^{-1})p\approx 5096.75\ \ \text{and}\ \ \sqrt{p}\approx 5696.74.$$
Question. Does $(1)$ hold? If it is true, how to prove it?
Your comments are welcome!
In contrast, it seems that $|{k^k\pmod p:\ k=1,\ldots,p-1}|/p$ does not have a limit.
let $x_1,\ldots,x_p$ be random independent residues modulo $p$. Then each residue $s$ appears in this sequence with probability $1-(1-1/p)^p\approx 1-e^{-1}$, so the average size of the set ${x_1,\ldots,x_p}$ is about $(1-1/e)p$. So your computations confirm that the value of the map $k^{k+1}+(k+1)^k$ do behave as random, and the values of $k^k$ do not. The first is not surprising, but should be difficult to prove, the second comes from restrictions like "each second term is a quadratic residue", and these restrictions come from divisors of $p-1$, thus there is no limit along all primes.
|
2025-03-21T14:48:29.696308
| 2020-01-20T09:15:53 |
350785
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexandre Eremenko",
"Andreas Blass",
"Doriano Brogioli",
"HenrikRüping",
"LSpice",
"Michael Bächtold",
"Pietro Majer",
"SBK",
"Yemon Choi",
"https://mathoverflow.net/users/122587",
"https://mathoverflow.net/users/138060",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/25510",
"https://mathoverflow.net/users/33741",
"https://mathoverflow.net/users/3969",
"https://mathoverflow.net/users/6101",
"https://mathoverflow.net/users/6794",
"https://mathoverflow.net/users/745",
"https://mathoverflow.net/users/763",
"leo monsaingeon"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625759",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350785"
}
|
Stack Exchange
|
A function in $\mathbb{R}^n$ is equal to its linearization in each point
I have a function $P: \mathbb{R}^n \to \mathbb{R}^n$. This function satisfies:
$$ P(\vec{x}) = J_P(\vec{x}) \cdot \vec{x}$$
where $\vec{x}\in \mathbb{R}^n$, $J_P$ is the Jacobian of $P$ and "$\cdot$" is
the matrix-vector product. I would rougly describe it as in the title.
I guess that it is a well known property. Do you know what is its name? Can you
suggest some literature about it, or any known property?
Also, I would like to to know about the symplectic equivalent:
$$ P(\vec{x}) = -\Omega \cdot J_P(\vec{x}) \cdot \Omega \cdot \vec{x}$$
where $\Omega$ defines the symplectic form.
Aren't all such functions just linear ?
@FedericoPoloni but the derivative does not exist at zero. I thought the upper condition means differentiable everywhere and that equality holds. If it does not mean that, what are the implied differentiability assumptions for higher $n$.
I can't work out right now but this is closely related to the equation for homogeneous degree one functions: see "Euler's homogeneous function theorem"
in fact maybe it is exactly that
We should take each component of $P$ separately, right? So I see that, if all the $P_i(\vec{x})$ functions are homogeneous, then my condition holds. This shows that my condition does not imply that $P$ is linear, but maybe must have a discontinuity! However, this is the reverse of what I was asking: can you prove that my condition implies that all the components are homogeneous?
The first PDE is a decoupled system of first order scalar equations, so one may analyse its solutions by the method of characteristics.
@T_M : can you please write your comment on homogeneous functions as an answer, so that I can mark it as accepted? Although I already got a "rigorously correct" answer, this one looks more helpful for me.
The characteristic lines for this PDE are just the rays from the origin, and solving the corresponding ODE is exactly the proof of Euler Th.m on homogeneous functions.
As for the symplectic case. Let $J$ be the symplectic matrix $J:=\left[ \matrix{ 0 & I_n \\ -I_n & 0 }\right]$. The characteristic lines for the (first order, linear, partial differential, vector) equation
$$P(x)=-J \,{\rm d}P(x)J x, \qquad x\in\mathbb{R}^{2n}\setminus\{0\},$$
are the solutions of the ODE $\dot \xi=J\xi$, that is circles $\xi(t)=e^{tJ}x_0$. Along these characteristic lines, the equation is
$\partial_t P(\xi(t))= {\rm d}P(\xi(t))J \xi(t)=JP(\xi(t))$, meaning that $P\circ \xi$ satisfies the same ODE, so that we have that $P(e^{tJ}x)=e^{tJ}P(x)$, for all $t\in\mathbb{R}$ and $x\in\mathbb{R}^{2n}\setminus\{0\}$, is a necessary and sufficient condition, for a differentiable map $P:\mathbb{R}^{2n}\setminus\{0\}\to\mathbb{R}^{2n}$, to satisfy your equation.
We may use complex notation and identify $\mathbb{R}^{2n}$ with $\mathbb{C}^n$, $J$ with the multiplication by $-i$, and the operators $e^{tJ}$ with multiplication by the complex scalars of modulus $1$ (Warning: in doing so, we still consider real differentiability, not complex differentiability, i.e. the Fréchet differential ${\rm d}P(x)$ is an $\mathbb{R}$-linear map, not necessarily $\mathbb{C}$-linear, i.e. not assumed to commute with $J$). The condition then reads: $P$ is an equivariant map w.r.to the
group action of $\mathbb{S}^1$:
$$P(\theta x)=\theta P(x)\quad\text{for all } x\in\mathbb{C}^n\setminus\{0\} \text{ and } \theta\in\mathbb{C} \text{ with } |\theta|=1.$$
for instance, in dimension $2$ (i.e. $n=1$) these differentiable maps $P:\mathbb{R}^2\setminus\{0\}\to\mathbb{R}^2$ are exactly those that in polar coordinates write as $P(r e^{it})=e^{it}\phi(r)$, where $\phi:\mathbb{R}_+\to\mathbb{R}$ is a differentiable map.
Rmk If one takes $J$ to be the identity map, and $P$ differentiable for any $x\ne0$, the same computation gives $P(tx)=tP(x)$ for all $t>0$ and $x\ne0$, that is, $P$ is (positively) homogeneous of degree $1$, (equivariant w.r.to the action of homoteties) If $P$ is also assumed to be defined and differentiable at $0$, this implies $P$ linear, as observed in comments.
I modified a bit and removed the origin from the domain, which seems a more natural setting .
I think this is the equation for a homogeneous degree one function; see "Euler's homogeneous function theorem".
If such a function is differentiable at the origin, then it has to be linear.
Thank you and all the others who contributed, in particular @Alexandre Eremenko. Do you have any suggestion about the "symplectic version", maybe inspired by the analogy with the homogeneous functions?
Such functions are linear. In dimension $1$, your equation means
$$P(x)=P'(x)x$$
Solving this differential equation we obtain $P(x)=cx$. Now in arbitrary dimension, your condition can be written as
$$x_j\sum_j \partial P/\partial x_j=P$$
for each coordinate $P$, which means that $P$ is a homogeneous function of degree $1$,
by Euler's theorem,
http://www.its.caltech.edu/~kcborder/Notes/EulerHomogeneity.pdf
This is false. In 2D, take $P_i(x,y)=x^2/(x+y)$. It satisfies my condition, but it is not linear. Moreover, along any line $y=\alpha x$, it is linear, but not over the plane $x-y$.
@DorianoBrogioli Your example is not differentiable at the origin (which was implicitly assumed in the question). If $P:\mathbb{R}^n\to\mathbb{R}^n$ (or $P:X\to Y$ between normed spaces, more in general) is homogeneous of degree $1$ and differentiable at the origin, the remainder $P(x)-dP(0)x$ is also $1$-homogeneous, and since it is $o(x)$ for $x\to0$, it is identically zero, that is $P=dP(0)$ is linear.
@Doriano Brogioli: I did not understand your example: you have a subscript $i$ in the left hand side, but there is no $i$ in the right hand side. Where is the map from $R^2$ to $R^2$?
As highlighted in the comments to the question, the problem can be posed for each component. For writing a 2D version we can take $P_1(x,y)=P_2(x,y)=x^2/(x+y)$, two equal components. My example is not defined in 0, so I must admit that you are right. Rigorously... but the more general meaning of my question was better answered by the connection to homogeneous functions.
@DorianoBrogioli Your original question talked of a function from $R^n$ to $R^n$ and the usual understanding among analysts is that "from $X$ to $Y$" means that the domain of the function is $X$. Maybe you had a slightly different question in mind, which could be asked separately?
No, it is this. $X=Y=\mathbb{R}^n$. In my example above, the domain is in $\mathbb{R}^2$, i.e. couples of numbers $x,y$. The image is (possibly a subset of) $\mathbb{R}^2$, i.e. the couple of numbers $P_1, P_2$. Could you explain better where is the misunderstanding?
@DorianoBrogioli I don't see that your example has domain $\mathbb R^2$. It's undefined along the line $x=-y$.
@DorianoBrogioli: the misunderstanding that Yemon Choi had in mind (I guess?) was that, when one writes "$f$ is a map from $X$ to $Y$", one usually means that $f$ is defined on the WHOLE space $X$.
Yes, your reasoning is correct! Rigorously, my question asked about functions defined everywhere in $\mathbb{R}^n$. So, rigorously, the functions $P$ are linear. However, I feel that the most complete answer was the one given in the comments to the question, which captures the more general meaning: any $P_i$ is a homogeneous function. This implies: either $P_i$ is linear, or it is not defined everywhere in $\mathbb{R}^n$. Thank you all for your contribution.
Homogeneous function of degree $1$ can be defined everywhere on $R^n$, for example $x^3/(x^2+y^2)$, but they are usually not differentiable at the origin.
@AlexandreEremenko, that function is not defined at the origin.
@LSpice: Why not? It is $0$ at the origin.
|
2025-03-21T14:48:29.697053
| 2020-01-20T11:12:29 |
350789
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Henri Cohen",
"MyNinthAccount",
"Ptino",
"Sylvain JULIEN",
"https://mathoverflow.net/users/13625",
"https://mathoverflow.net/users/140496",
"https://mathoverflow.net/users/151127",
"https://mathoverflow.net/users/81776"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625760",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350789"
}
|
Stack Exchange
|
Proving that central zero of an L-function is multiple
Central point of critical strip, ie. $s=1/2$, is conjectured to be the only argument at which an L-function can have multiple zero. This is interesting ia. for proving effective lower bounds for class numbers of quadratic number fields (probably only degree 2 L-functions work for this purpose) in light of the work of Goldfeld. As far as I'm aware, the only proof method that an L-function of a modular form vanishes at $s=1/2$ uses the fact that it coincides with L-function of an elliptic curve with positive rank. The question is whether it is possible to study multiplicity of central zeros of L-functions of modular forms directly, that is without recourse to arithmetical objects whose L-functions coincide with the former. This question seems to be natural in view of difficulty of Birch and Swinnerton-Dyer conjecture and the fact the former class of L-functions is perhaps wider than the latter, ie. perhaps there exist modular forms without corresponding arithmetical objects.
"As far as I'm aware, the only proof method that an $L$-function of a modular form vanishes at (the central point) uses the fact that it coincides with $L$-function of an elliptic curve with positive rank." $ $ One can use modular symbols to show central vanishing (including in the even parity case), and this is significantly easier IMO than using something like Kolyvagin or Coates-Wiles.
Can you give a reference to any paper which shows/employs this? Thank you!
I guess you mean a primitive L-function? Cause $\zeta^{2}$ certainly has multiple zeros outside the central argument. Otherwise, it may be the case that the more symmetries of a given L-function preserve the argument, the higher the multiplicity of a zero at this argument can be. Under RH only the critical line can contain non trivial zeros and this line is preserved under both the identity and $s\mapsto 1-\bar{s}$. Similarly the real axis is preserved under both the identity and $s\mapsto\bar{s}$.
Naturally, a primitive L-function to ever hope for a nontrivial multiple zero.
I am not sure that I understand some comments: there are numerous examples of $L$ functions of modular forms of weight $4$, say, which have a double zero at $s=2$, and as mentioned by MyNinthAccount this can be proved using modular symbols.
If it was at least a triple zero and if it was possible to prove this via modular symbols, then a simpler route to effective lower bounds for class number problem would follow, right? So this works only for double zeros so far, right? If yes, then what are the main obstacles to extending?
Modular symbols allow you to compute the central $L$-value (writing it as a rational divided by an appropriate period, and you can bound the height of the rational, thus giving a criterion for it being zero), but not its derivative. See Section 2.8 of Cremona's book.
|
2025-03-21T14:48:29.697390
| 2020-01-20T13:43:53 |
350793
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Albert",
"https://mathoverflow.net/users/19189",
"https://mathoverflow.net/users/40804",
"mme"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625761",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350793"
}
|
Stack Exchange
|
Caratheodory's theorem in any compact Riemann surface
The classical Caratheodory theorem states that if $G$ is a simply connected domain in the plane, whose boundary is a Jordan curve, then the Riemann uniformization extends continuously to a homeomorphism between the boundary of $G$ and the unit circle.
Question: does this hold if the domain is embedded in a compact Riemann surface of higher genus? More precisely, is it true that if $G$ is a simply connected domain in a compact Riemann surface $X$, and if the boundary of $G$ is a simple closed curve, then the uniformizing map extends as a homeomorphism between the boundary of $G$ and the unit circle?
The uniformization theorem applied to $X$ gives that its universal cover is one of $\Bbb P^1, \Bbb C, $ or the unit disc. The compact simply connected domain $G$ lifts to the universal cover, which exhibits it as a subset of $\Bbb C$. So this is just the classical Caratheodary theorem.
@MikeMiller I thought of that, but a priori lifting to the universal cover will change the boundary. Is there a simple argument why the closure of $G$ would lift to the closure of a Jordan domain?
I said "compact domain" for a reason --- you should include the boundary of $G$ when talking about lifting it, so that $G$ is homeomorphic to a closed unit disc. Then the whole closed unit disc lifts to the universal cover, giving a closed unit disc embedded upstairs (the boundary lifting the boundary, the interior lifting the interior).
@MikeMiller I see, but if I may ask one more stupid question: how do you know that the closure of $G$ is homeomorphic to the closed unit disk? this seems sort of close to proving what we want (keeping in mind that in general the closure of a simply connected domain needs not be simply connected, of course)
This seems trickier than I thought at first, but still not too bad. A proof uses that any embedded curve $C \subset S$ in a surface has, for any point $p \in C$, a neighborhood $U$ and a homeomorphism $f: U \to \Bbb R^2$ with $f(C \cap U) = \Bbb R$ and $f(p) = 0$. Then $f(G \cap U)$ is an open set in $\Bbb R^2$ with boundary $\Bbb R$. Some fiddling shows that it must either be one open half-plane or the union of both; a little more shows that whether it's one half-plane or two is independent of the point $p \in C$. But if it was both, then we would have $G = S \setminus C$, which is not s.c.
So $\overline G$ is a manifold with interior an open disc, and so it is homeomorphic to the closed unit disc.
|
2025-03-21T14:48:29.697578
| 2020-01-20T14:12:53 |
350795
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625762",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350795"
}
|
Stack Exchange
|
Confidence Intervals for strongly mixing stochastic processes
I have $s(t)$, a stationary stochastic process that we know is strongly mixing - and I also know that samples from $s(t)$ are definitely correlated over time. I want to estimate the mean of $s(t)$, $\mathbb{E}[s(t)]$. Since strong mixing implies ergodicity, I can simply calculate sample means to find $\mathbb{E}[s(t)]$.
Does anyone know how I can go about estimating empirical variance so I can get confidence intervals for the estimated mean?
If $s(t)$ was Markovian, for example, I could use the Strong Markov Property to find empirical variance, but I have no clue how to do this for my general, non-Markovian $s(t)$.
See e.g. Confidence interval estimation of the mean of stationary stochastic processes: A comparison of batch means and weighted batch means approach and Using standardized time series to estimate confidence intervals for the difference between two stationary stochastic processes
and references therein.
|
2025-03-21T14:48:29.697676
| 2020-01-20T14:33:54 |
350798
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Emil Jeřábek",
"Marcin Wrochna",
"https://mathoverflow.net/users/12481",
"https://mathoverflow.net/users/12705",
"https://mathoverflow.net/users/151388",
"joro"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625763",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350798"
}
|
Stack Exchange
|
Is Hamiltonian cycle fixed parameter tractable with parameter clique cover?
Let $G$ be connected simple graph.
Clique cover of graph $G$ is partition of the vertices of $G$
into $k$ disjoint cliques $D'_i$.
Given $G$ and $k$-clique cover, can we solve Hamiltonian cycle
in time $O(\mathrm{polynomial}(n) n^{O(k)}$)?
Let $k$ be fixed, is it true that for all graphs with given $k$-clique
cover Hamiltonian cycle is polynomial in $n$?
The basic idea is that all paths in cliques are easy.
One possible approach is to merge the cliques to single vertex
and then enumerate walks allowing visiting vertex more than once.
For $k=2$ the answer is easily true.
For $k=3$ the merged graph is either $K_3$ or $P_3$
and we can try to extend walk to Hamiltonian cycle in $G$.
For $k=4$ the merged graph might be claw, so we need
more complicated algorithm.
Small cliques need not be on the merged walk, which complicates
the algorithm.
Cross-posted to https://cstheory.stackexchange.com/q/46218 , and to make things worse, with no cross-links from one post to the other. You’ve been here for many years, so you are well aware that this is unacceptable and rude behaviour.
It's FPT. The idea is that one never needs to move between different cliques too many times.
Lemma If there is a hamiltonian cycle, then there is a hamiltonian cycle which, for all $i\neq j\in\{1,\dots,k\}$, uses at most one edge from $D_i$ to $D_j$.
Proof
Consider a hamiltonian cycle which minimizes the number of edges between different cliques. Suppose it uses some edges from $D_i$ to $D_j$ twice, for some $i,j$. That is, the sequence of vertices is of the form $P a_1 b_1 Q a_2 b_2 R$, where $a_1,a_2 \in D_i$, $b_1,b_2 \in D_j$ and $P,Q,R$ are whatever, possibly empty. Then (assuming the graph is undirected) you can "flip the square" to get the hamiltonian cycle $P a_1 a_2 Q^{-1} b_1 b_2 R$. But this uses two non-clique edges less (by adding in-clique edges $a_1 a_2$ and $b_1 b_2$ instead). $\square$
So in particular there is a hamiltonian cycle that enters and leaves each clique $D_i$ at most $2k$ times (it might need to go from $D_i$ to $D_j$ and then back). This already gives an easy $n^{O(k^2)}$ algorithm: guess the sequence of movements between cliques (that is, try all sequences of clique names of length at most $k^2$) and for each step in the sequence guess which vertex you leave from and which you enter into. Then, for each such guess, it's trivial to check if that sequence can be extended with walks inside cliques to cover all vertices.
To get an FPT algorithm you can e.g. kernelize, i.e. remove things that are redundant, in a sense. Look at the bipartite graph between $D_i$ and $D_j$. If there is any vertex $v \in D_i$ of degree $>2k$, then select any $2k$ incident edges $F$ and remove the rest; the supposed hamiltonian cycle can be easily changed to use edges in $F$ in place of the removed ones (because it enters and leaves the clique $D_j$ at most $2k$ times), so that doesn't change the answer.
After this, in the bipartite graph between $D_i$ and $D_j$, every vertex has bounded degree. Now take any maximal matching (e.g. obtain it greedily). If it has more than $4k$ edges, then select any subset $F$ of $4k$ edges and remove the rest between $D_i$ and $D_j$. As before you can easily argue this won't change the answer.
After such a removal, the bipartite graph has a maximal matching of at most $4k$ edges, so their endpoints make a vertex cover of at most $8k$ vertices. Since they have bounded degree, the number of edges between $D_i$ and $D_j$ is bounded by $16k^2$. Repeat for all $i\neq j$; then the total number of non-clique edges is $\leq 16k^4$. Any vertex that is not incident to any non-clique edge can be removed without changing the answer, so you're left with a graph on $\leq 32 k^4$ vertices, which you can solve brutally in time factorial in that, so $2^{O(k^4 \log k)}$. The above kernelization itself is easily implemented in time $\mathrm{poly}(k) \cdot m$ (linear in the number of edges).
I guess with some technical effort $2^{O(k^2 \log k)} \cdot m$ or even $2^{O(k^2 \log k)} + \mathrm{poly}(k) \cdot m$ should be possible, but it's hard to say whether $2^{O(k)} \mathrm{poly}(n)$ or $n^{O(k)}$ is possible.
Thanks. Don't you need at least TWO edges and vertices between cliques, e.g. for 2K_3 + two edges?
At most one from $D_i$ to $D_j$, at most one other from $D_j$ to $D_i$. "It might need to go from $D_i$ to $D_j$ and then back." Or did I miss another factor of 2 somewhere?
I misread the definition of Lemma in the beginning. There is similar known result on cstheory: https://cstheory.stackexchange.com/questions/46218/is-hamiltonian-cycle-fixed-parameter-tractable-with-parameter-clique-cover
|
2025-03-21T14:48:29.698005
| 2020-01-20T15:12:13 |
350800
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"George Shakan",
"Oliver Roche-Newton",
"https://mathoverflow.net/users/23951",
"https://mathoverflow.net/users/50426"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625764",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350800"
}
|
Stack Exchange
|
Packing almost-subgroups into a group
We consider a group finite $G$. We say a set $A\subset G$ injects a set $B$ if $|A+B| = |A||B|$, and let $I(B) = \max \{|A| :A\text{ injects } B\}$.
For a subgroup $H$, it is well-known that $I(H) = |G|/|H|$. Now, if we consider $H' = H \setminus \{e\}$, we still have that $I(H') = I(H)$, if $|H| > 2$, as then we can only fit 1 copy of $H'$ into each coset of $H$.
However, given a second subgroup, $F$, $I(F+H')$ is not always equal to $I(F+H)$.
Letting $G$ be the symmetric group on $n$ symbols, we have cases where $I(F+H') = |G|/|F+H'|$. For example, letting $\sigma = (1,2,3\dots n), \tau = (1,2)$, this occurs when $F$ is generated by $\sigma^{-1}\tau$, and $H$ is generated by $\sigma$. (this can be derived from this paper of Aaron Williams)
In my research I'm hoping to establish upper bounds of $I(F+H')$ for particular $G, F$ and $H$. Is there some literature about such problems? Are techniques of Ruzsa Calculus well-suited, or is some other topic in additive combinatorics more related?
You can see this previous post for a related question https://mathoverflow.net/questions/244567/small-set-such-that-1-ldots-n-cdot-a-mathbbz-p-mathbbz
Inequality (1.1) in this paper may be of some use to you...https://arxiv.org/pdf/1611.00529.pdf
|
2025-03-21T14:48:29.698122
| 2020-01-20T15:47:39 |
350801
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Deane Yang",
"Jakob Möller",
"https://mathoverflow.net/users/146998",
"https://mathoverflow.net/users/613"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625765",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350801"
}
|
Stack Exchange
|
Taylor expansion of exponential of a Lie derivative
In this paper on page 8 the author claims that the Taylor expansion for the expression $e^{tD_V}$ where $D_V$ is the Lie derivative with respect to a vector field $V$ (defined by $(D_VG)(x) = \frac{d}{dt}|_{t=0}V(\phi^t_V(x))$ and $\phi^t_V(x)$ is the flow of the differential equation $\dot{\psi} = V(\psi)$ looks like this
$$e^{tD_V} = I +tD_V +t^2\int_0^1(1-\theta)e^{\theta t D_V}D_V^2 d\theta $$
I can't wrap my head around the remainder term. Shouldn't it be of third oder in $D_V$ because the integrand involves the third derivative of the exponential? I was thinking that a change of variables is performed but it didn't work. A related question might be this which is concerned with the same paper but ask a different question, still.
Maybe I'm missing something, but it seems to me you're overthinking this. Since
$$
\int_{0}^{1} d\theta (1-\theta ) \theta^{n} = \frac{1}{n+1} -\frac{1}{n+2} = \frac{n!}{(n+2)!}
$$
the third term in your expression is
$$
\sum_{n=0}^{\infty } \frac{t^{n+2} D_V^{n+2} }{n!} \frac{n!}{(n+2)!}
= \sum_{n=2}^{\infty } \frac{t^n D_V^n }{n!}
$$
exactly as it should be.
It's also, I believe, the integral remainder formula for the first order Taylor polynomial of $f(x) = e^x$, evaluated at $x = tD_V$. See https://en.wikipedia.org/wiki/Taylor%27s_theorem#Explicit_formulas_for_the_remainder
So I figured that the integral led my onto a false lead because I thought this would be the integral formula for the remainder and I unsuccessfully tried to fit in the "false" order of $D_V$.
|
2025-03-21T14:48:29.698252
| 2020-01-20T15:52:37 |
350802
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Bjørn Kjos-Hanssen",
"https://mathoverflow.net/users/4600"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625766",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350802"
}
|
Stack Exchange
|
Small set in partition-large class
A collection $\mathcal{A}\subseteq \mathcal{P}(X)$
is $k$-large in $X$
if for every $k$-partition
of $X$ namely
$X_1,\cdots,X_k$, there exists an $i\leq k$ such that $X_i\in \mathcal{A}$;
$\mathcal{A}$ is upward closed if for every $X\subseteq Y$,
$X\in \mathcal{A}$ implies $Y\in \mathcal{A}$.
Fix a $0<\delta<1/2$, $k\in\mathbb{N}$. Let $N$ be very large, $\mathcal{A}\subseteq\mathcal{P}(\{1,\cdots,N\})$ is $k$-large in $\{1,\cdots,N\}$ and upward closed. Let $Z$ be a uniformly random $N/k$-element subset of $\{1,\cdots,N\}$.
Question: Is it true that there exists a $\delta N$-element subset $Z'$ of $\{1,\cdots,N\}$ such that with high probability (as $N\rightarrow\infty$): $Z\cup Z'\in\mathcal{A}$.
Katona, Gyula (1975) The Hamming-sphere has minimum boundary. STUDIA SCIENTIARUM MATHEMATICARUM HUNGARICA, 10. pp. 131-140. ISSN 0081-6906
Theorem 3 of [KG1975] gives the following:
for every $\mathcal{A}'\subseteq 2^N$ with $|\mathcal{A}'|/2^N=c>0$,
let $\rho\in 2^N$ be uniformly random, then $d_H(\rho,\mathcal{A}) = O_p(\sqrt{N})$. Where $d_H$ denote the Hamming distance. This implies a weaker "answer" of above question:
with high probability, there exists a $Z'$ (depending on $Z$)
such that $Z\cup Z'\in \mathcal{A}\wedge |Z'|<\delta N$.
I wonder if the following (which implies a positive answer to above question) holds:
For every $\mathcal{A}'\subseteq 2^N$ with $|\mathcal{A}'|/2^N\geq c>0$, every $\delta>0$,
there exists a set $Z'\subseteq N$ with $|Z'|<\delta N$ such that let
$\rho\in 2^N$ be uniformly random, then with hight probability:
there exists a $\sigma\in \mathcal{A}'$ such that $\rho$ agree with $\sigma$ on
$N\setminus Z'$.
I don't know the answer in general, but it's true if $\mathcal A$ consists of all sets of measure at least $1/k$, for some measure $\mu$ on $[N]$...
|
2025-03-21T14:48:29.698388
| 2020-01-20T16:07:37 |
350805
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Chen",
"Yellow Pig",
"https://mathoverflow.net/users/12395",
"https://mathoverflow.net/users/38832"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625767",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350805"
}
|
Stack Exchange
|
Minimal extension of local systems
Let $M$ be a complex manifold of dimension $2$, $D \subset M$ be a connected, simple, normal crossings divisor and $L$ be a $\mathbb{C}$-local system defined over $M\backslash D$. Denote by $j: M\backslash D \to M$ and $i:D \to M$, the natural inclusions. Recall, there exists a minimal extension $j_{!*}L$ of $L$ to $M$ (this is the image of $j_!L \to j_*L$). Denote by $i^{-1}(j_{!*}L)$ the restriction of $j_{!*}L$ to $D$. Is there a necessary and sufficient condition under which the restriction $i^{-1}(j_{!*}L)$ is a local system on $D$?
Sorry, in my limited experience people usually want to prove that a constructible sheaf is a local system on a $\it smooth$ subspace. What is the motivation for having a local system on a the whole $\it singular$ subspace $D$ rather than on the smooth open part of $D$?
@YellowPig Do you have an answer for the question, if we replace $D$ by the smooth part of $D$?
I thought that on the smooth part of $D$ it's a local system because, looking at a transversal slice to $D$, you are taking the minimal extension of the monodromy around the divisor
@YellowPig can you suggest some reference/literature.
|
2025-03-21T14:48:29.698505
| 2020-01-20T16:22:01 |
350807
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexandre Eremenko",
"Iosif Pinelis",
"Jon",
"https://mathoverflow.net/users/151319",
"https://mathoverflow.net/users/25510",
"https://mathoverflow.net/users/36721",
"https://mathoverflow.net/users/44191",
"user44191"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625768",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350807"
}
|
Stack Exchange
|
Characterising the number of turning points of a 'generalised' polynomial
Is it possible to find the number of turning points of a power function whose largest exponent is some real number known to lie between $(n,n+1)$ for some $n\in\mathbb{Z}$?
To give an example
Consider the function $f:(0,1)\rightarrow\mathbb{R}$ with:
$$
f(z)=A(1-z)^{\gamma+1}+Bz(1-z)^{\gamma}+Cz^{\gamma+1}
$$
where A,B,C are real constants and $\gamma\in(0,1)$. This implies that the highest power of $z$ lies in (1,2), and my initial intuition was that this should imply that the function has at most 1 turning point. Any suggetsions on how to go about proving this?
I think this should clearly be false. Let $P(x)$ be a polynomial with $k$ turning points, all strictly positive. Then $P(x^\alpha)$ is a generalized polynomial with $k$ turning points, no matter how small $\alpha$ is. .
What is the definition of "turning point"?
If by a turning point you mean a point where the function switches from increasing to decreasing or vice versa, then the function $f$ given by $f(z)=A(1-z)^{\gamma+1}+Bz(1-z)^{\gamma}+Cz^{\gamma+1}$ with $A=27,B=4,C=18,\gamma=2/5$ has (not one but) two turning points, near $0.74$ and near $0.98$.
Here is the graph of $f$:
And here are relevant values of $f$:
$$f(0)=27>f(8/10)\approx17.7<f(95/100)\approx18.3>f(1)=18.$$
Unless I'm missing something, the function with the given values only has one turning point close to 0.78. The function is then increasing and approaches 18 from below as z approaches 1.
@Jon : Oops! I had copied the value of $B$ from a wrong place. This is now corrected, and everything else holds.
|
2025-03-21T14:48:29.698636
| 2020-01-20T16:27:32 |
350808
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Anthony Quas",
"Gilles Mordant",
"aurora_borealis",
"https://mathoverflow.net/users/11054",
"https://mathoverflow.net/users/125260",
"https://mathoverflow.net/users/78326"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625769",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350808"
}
|
Stack Exchange
|
Distance between two sample quantiles
Let $X_1,\dots X_n$ be i.i.d. samples from an unknown distribution. We know the distribution has uniformly bounded probability density function $f(x)$. Let $1>\tau_1>\tau_2>0$ be two quantile levels and $\hat{X}^{\tau_1}$ and $\hat{X}^{\tau_2}$ be the corresponding sample quantile points. Note that the sample quantile is not unique. Now I want to give a sharp bound for the distance
\begin{equation}
|\hat{X}^{\tau_1}-\hat{X}^{\tau_2}|.
\end{equation}
This problem also has a special case like the finding distance between two neighboring quantiles (i.e. $\tau_1-\tau_2=1/n$).
For simplicity we may assume $X_i$'s all sampled from standard normal distribution. So my question is, are there any elegant proofs for this problem? I would be grateful if somebody provide any reference about concentration results for median-type estimators.
I think you are looking for the results on the distribution of order statistics (https://en.wikipedia.org/wiki/Order_statistic).
@AnthonyQuas Thanks, but I think I need sharper inequalities, like exponential-type inequality or even Bahadur representation, the results like variance is not enough
Would an asymptotic bound serve your needs ?
|
2025-03-21T14:48:29.698739
| 2020-01-20T16:30:35 |
350809
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625770",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350809"
}
|
Stack Exchange
|
Rank 1 valuations that are not discrete on finite transcendental extensions of the rationals
Suppose $K=\mathbb{Q}(X_1,\dots,X_n)$ is a purely transcendental extension of the rationals on finitely many indeterminates. Can anyone give an example of a rank $1$ valuation on $K$ that fails to be discrete?
If not, is there a theorem that shows that such a rank $1$ valuation must be discrete?
Example: $val(X_1^{e_1}X_2^{e_2})=e_1 + e_2 \sqrt{2}$.
|
2025-03-21T14:48:29.698800
| 2020-01-20T16:58:57 |
350810
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Asaf Karagila",
"https://mathoverflow.net/users/7206"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625771",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350810"
}
|
Stack Exchange
|
Forcing square introduces diamond
Let $\mathbb S_\kappa$ be the standard forcing for $\square_\kappa$ by initial segments. This is $(\kappa+1)$-strategically closed.
Observation: Let $T \subseteq \kappa^+$ be stationary. If $T$ concentrates on $\mathrm{cof}({<}\kappa)$,then $\mathbb S_\kappa$ forces $\diamondsuit(T)$. If $T$ concentrates on $\mathrm{cof}(\kappa)$ and $T$ is approachable, then $\mathbb S_\kappa$ forces $\diamondsuit(T)$.
The proof is like the argument that adding a Cohen subset of a regular cardinal $\kappa$ forces $\diamondsuit_\kappa$. The coding goes by fixing bijections $f_\alpha : \alpha \to \kappa$ for $\alpha <\kappa^+$ and taking the diamond sequence to be $a_\alpha = \{ \beta<\alpha: \alpha+f_\alpha(\beta)+1 \in C_{\alpha+\kappa} \}$. It works because of the freedom we have in choosing $C_{\alpha+\kappa}$. In the $\mathrm{cof}(\kappa)$-concentration case, we use an elementary submodel argument plus approachability so that we can work with a chain of conditions of length only $\kappa+1$.
Question 1: This seems like it might have been observed before. Does it appear in a paper?
Question 2: Can we eliminate the approachability assumption?
I mean, he is sitting two offices over. Just ask him.
Regarding Question 2, the assumption of approachability is necessary (or, more precisely, the assumption that $T$ has a stationary subset that is approachable), at least if we have $2^\kappa = \kappa^+$ in the ground model. The reason is that, if $T$ does not have a stationary subset that is approachable, then it becomes non-stationary after forcing with $\mathbb{S}_\kappa$, so $\diamondsuit(T)$ necessarily fails.
To see this, note that if $2^\kappa = \kappa^+$, then we can fix in $V$ an enumeration $\vec{a} = \langle a_\alpha \mid \alpha < \kappa^+ \rangle$ of all bounded subsets of $\kappa^+$. Then a set $S \subseteq \kappa^+$ is in $I[\kappa^+]$ if and only if there is a club $C \subseteq \kappa^+$ such that $\gamma$ is approachable with respect to $\vec{a}$ for every $\gamma \in S \cap C$. This is because, if $\vec{b} = \langle b_\alpha \mid \alpha < \kappa^+ \rangle$ is any other sequence of bounded subsets of $\kappa^+$, then there is a club of $\delta < \kappa^+$ such that all entries in $\langle b_\alpha \mid \alpha < \delta \rangle$ appear in $\langle a_\alpha \mid \alpha < \delta \rangle$. But now the assumption that $T$ does not have a stationary subset in $I[\kappa^+]$ means that there is in fact a club $C \subseteq \kappa^+$ such that, for every $\gamma \in T \cap C$, $\gamma$ is not approachable with respect to $\vec{a}$. Forcing with $\mathbb{S}_\kappa$ does not add any new bounded subsets to $\kappa^+$, so if $G$ is generic, then $\vec{a}$ remains an enumeration of all bounded subsets of $\kappa^+$ in $V[G]$ and every ordinal $\gamma < \kappa^+$ is approachable with respect to $\vec{a}$ in $V[G]$ if and only if it is approachable with respect to $\vec{a}$ in $V$. However, $\square_\kappa$ holds in $V[G]$, so also $\mathrm{AP}_\kappa$ holds. Then, in $V[G]$, $T \cap C \in I[\kappa^+]$, but every ordinal in this set is not approachable with respect to $\vec{a}$ so, there must be a club $D \subseteq \kappa^+$ disjoint from $T \cap C$. Then $C \cap D$ is disjoint from $T$, so $T$ is non-stationary.
Regarding Question 1, I made a similar but slightly different observation in my paper "Aronszajn trees, square principles, and stationary reflection" (see Lemma 3.11). I was forcing $\square(\lambda)$ rather than $\square_\kappa$, which slightly simplifies things. I just state the lemma for $\diamondsuit(\lambda)$, but the same argument works for $\diamondsuit(T)$ for any stationary $T \subseteq \lambda$. Approachability is not needed there, since the forcing notion for $\square(\lambda)$ is fully $\lambda$-strategically closed. I'm sure this had been recognized before, but I haven't seen it elsewhere in print.
|
2025-03-21T14:48:29.699031
| 2020-01-20T17:23:49 |
350814
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625772",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350814"
}
|
Stack Exchange
|
special classes of ideals (eg. toric) that admit faster Buchberger algorithm?
I have heard that toric ideals allow one to speed up the Buchberger algorithm considerably (see Grobner bases of toric ideals, Remark 2,3). My question is two-fold:
What are the precise complexity-theoretic bounds known for the Buchberger algorithm for toric ideals?
Are there other non-trivial classes of ideals on whom a grobner basis can be computed efficiently? Can you please provide a reference to them?
For context, I wish to use a grobner basis as a way to encode dataflow analysis problems in compiler construction, in a manner that allows mutiple analyses to "share" information easily. So, knowing special types of ideals and fast algorithms to compute their Grobner basis woud help design specific dataflow analyses.
|
2025-03-21T14:48:29.699231
| 2020-01-20T18:53:53 |
350818
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Lennart Meier",
"Nicolas Hemelsoet",
"Will Sawin",
"Yellow Pig",
"https://mathoverflow.net/users/104742",
"https://mathoverflow.net/users/12395",
"https://mathoverflow.net/users/18060",
"https://mathoverflow.net/users/2039"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625773",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350818"
}
|
Stack Exchange
|
Areas of algebraic geometry useful for geometric representation theory
What topics/areas of algebraic geometry (aside from perverse sheaves/D-modules, etale cohomology, and possibly derived algebraic geometry) is it useful to learn/master if one is interested in doing research in geometric representation theory? What are the best sources to learn these topics? How does one know that one understands the material on a deep enough level to do interesting research?
My impression is that reading almost all of Hartshorne and solving a large portion of problems there (+ reading some specialized literature in geometric representation theory) is insufficient to solve interesting research problems. How different is preparation for research in geometric representation theory from preparation for research in other areas of algebraic geometry? Does one need a deep understanding of differential geometry and analysis and how to gain it, or are there specific areas/topics in geometry and analysis which are more useful to learn/understand?
What about symplectic geometry ? I think symplectic resolutions are really useful in geometric representation theory.
@Nicolas Hemelsoet But I think symplectic geometry is a really big/deep area with a lot of very analytic aspects, so what exactly should one learn from it? I've read Chriss and Ginzburg/Ginzburg's lectures on various subjects, they have symplectic geometry to some extent. What other things in symplectic geometry should one learn?
I was thinking to what is described here : https://conferences.cirm-math.fr/1956.html
@Nicolas Hemelsoet Thanks a lot, I heard of this conference at Luminy, but it is all about very current research, not some reading background, right? Also, the link does not open when I press it
I guess, you already had a look at the book of Chriss and Ginzburg?
@Lennart Meier Yes, I read some of it, thanks
Are the basics of the sheaf-function dictionary (in particular, over finite fields) also part of the assumed material?
@Will Sawin Thanks, yes, I read a book on etale cohomology.
|
2025-03-21T14:48:29.699387
| 2020-01-20T20:23:06 |
350820
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Noah Snyder",
"Sebastien Palcoux",
"https://mathoverflow.net/users/22",
"https://mathoverflow.net/users/34538"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625774",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350820"
}
|
Stack Exchange
|
Is there a non-irreducible maximal subfactor other than two-sided TLJ?
A subfactor $N \subseteq M$ is called:
irreducible if $N' \cap M = \mathbb{C}$,
maximal if for any intermediate subfactor $N \subseteq P \subseteq M$ then $P=\{N,M \}$.
The two-sided Temperley-Lieb-Jones (TLJ) subfactors $A_{\infty}^{(1)}$ are both maximal and non-irreducible.
Question: Is there a non-irreducible maximal subfactor other than $A_{\infty}^{(1)}$?
We assume that the non-irreducible subfactors $A_{n}^{(1)}$, with $n< \infty$, are not maximal, but if some of them are maximal, then we also exclude them in the question.
Can't you just use diagonal subfactors for n>2?
@NoahSnyder You are talking about the subfactors studied in Popa’s Classification of amenable subfactors of type II, Subsection 5.1.5, right? The principal graph is a Cayley graph (with multiplicities) starting from the neutral element e, so it is non-irreducible, ok. Now why should we have maximal examples with $n>2$?
|
2025-03-21T14:48:29.699476
| 2020-01-20T20:49:10 |
350821
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625775",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350821"
}
|
Stack Exchange
|
How to determine a highest weight corresponding to a parabolic subgroup?
Let $G$ be a simply connected, semisimple algebraic group over $\mathbb C$ with maximal torus $T$ and Borel subgroup $B$ containing $T$. If $(V,\pi)$ is an irreducible representation of $G$, then $(V,d\pi)$ is an irreducible representation of the Lie algebra $\mathfrak g$ which has a unique highest weight $\lambda \in \mathfrak t^{\ast}$. I have read that if we identify the one-dimensional weight space $V_{\lambda} \subset V$ with a point in projective space $\mathbb P(V)$, then under the action $$ G \xrightarrow{\pi} \operatorname{GL}(V) \rightarrow \operatorname{Aut}(\mathbb P(V))$$
the stabilizer of $V_{\lambda}$ is a parabolic subgroup of $G$, and every parabolic subgroup of $G$ arises this way.
How does one go in the opposite direction? If $P$ is a (let's say maximal) parabolic standard subgroup of $G$, how does one find a dominant integral weight whose corresponding irreducible representation determines $P$ in the above sense? Can the highest weight occur as the highest root of $T$ in the unipotent radical of $P$?
This is done in e.g. Baston-Eastwood (1989, pp. 40, 55): with $P$ characterized as usual by a subset $\mathcal S_{\mathfrak p}$ of the simple roots (for a maximal parabolic, $\mathcal S_{\mathfrak p}$ is all but one simple root), take $\lambda=$ sum of the fundamental weights $\varpi_i$ corresponding to simple roots $\alpha_i$ not in $\mathcal S_{\mathfrak p}$. Then $G/P$ is the coadjoint orbit of $\lambda$ under the compact real form, and $V$ is its geometric quantization.
$\lambda$ can be the highest root (giving $V=$ adjoint representation) but then $P$ is usually not maximal.
|
2025-03-21T14:48:29.699609
| 2020-01-20T20:59:28 |
350823
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Hussain Rashed",
"M. Dus",
"https://mathoverflow.net/users/111917",
"https://mathoverflow.net/users/147032"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625776",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350823"
}
|
Stack Exchange
|
Topology on the boundary compactification $X^{-}=\partial X\cup X$ of a Gromov-hyperbolic space
Consider a proper geodesic $\delta$-hyperbolic space $X$ (in the sense of Gromov). Let ∂ be its Gromov boundary. In the book "Geometric Group Theory" by Cornelia Druţu and Michael Kapovich https://www.math.ucdavis.edu/~kapovich/EPR/ggt.pdf
page 391, for $k>2\delta$ and $a\in X$, they define the shadow topology $\Im _{a,k}$ by declaring the basis:
$$\begin{Bmatrix}
B_{r}(x): x\in X ,r>0
\end{Bmatrix}\cup \begin{Bmatrix}
U_{\rho (t),k}(\xi ):t\geq 0, \xi\in\partial X
\end{Bmatrix}$$ where $\rho$ is a geodesic ray asymptotic to $\xi$ and initiating from $a$, and $$U_{\rho (t),k}(\xi )=\begin{Bmatrix}
z\in X^{-}:[a,z]\cap B_{k}(\rho(t))\neq \phi
\end{Bmatrix},$$ where $[a,z]$ is any geodesic connecting $a$ and $z$.
I need to clarify the following:
1) I know what $[a,z]$ means when $z\in X$, but what does that mean when $z\in \partial X$?
2) Let $\gamma \in \partial X$, when do we say that $\gamma \in U_{\rho (t),k}(\xi )$?
To answer your first question, I'm a bit confused because you seem to be okay with the existence of$\rho$ which is a geodesic asymptotic to $\xi$. Actually, $[a,z]$ is nothing else than a geodesic asymptotic to $z$, if $z$ is in the boundary.
Maybe you need a bit of clarification though. I guess it depends on your definition of the Gromov boundary. I assume that you use the definion with equivalence classes of sequences converging in the sense of Gromov. Then, for any proper geodesic Gromov hyperbolic space $X$, for any point $o\in X$ and any point $\gamma\in \partial X$, there exists a geodesic ray $\alpha:\mathbb{R}_{\geq 0}\to X$ such that $\alpha(0)=o$ and $\alpha(t)$ converges to $\gamma$ in the sense of Gromov. The proof is basically a use of Arzelà-Ascoli lemma. You can actually define the Gromov boundary as the equivalence classes of geodesic rays, declaring that two geodesic are equivalent if they stay within a bounded distance of each other. This is proved in Section 11.11 of the book you refer too. You can also find a detailed proof of all this (and in particular of the fact that these two definitions are equivalent) in Section 7 of de la Harpe and Ghys's book Sur les Groupes Hyperboliques d’après Mikhael Gromov. You can also look at Boundaries of hyperbolic groups by Ilya Kapovic and Nadia Benakli. In particular, they prove there that without the properness assumption, this statement is still true replacing geodesic by (1,10$\delta$)-quasi-geodesic.
For your second question, maybe you have some specific thoughts, but I can only suggest a reformulation. Letting $\gamma\in \partial X$, $\gamma\in U_{\rho(t),k}(\xi)$ if any geodesic from $a$ to $\gamma$ passes within $k$ of the point at distance $t$ from $a$ on the fixed geodesic $\rho$ from $a$ to $\xi$. In other words, the geodesics from $a$ to $\gamma$ and from $a$ to $\xi$ start diverging more than $k$ only after time $t$.
Notice that two geodesics asymptotic to $\gamma$ stay within a uniform distance of each other, so that up to changing $k$, you could also ask in the definition that some geodesic asymptotic to $\gamma$ passes in $B_k(\rho(t))$. This answer the question in your comment. Note also that because of this property, the choice of the geodesic $\rho$ is not important.
Maybe the following will help and will explain the terminology shadow. If you think of geodesics as light rays, then put an opaque ball around $\rho(t)$ and a light source at $a$. Then, the shadow will exactly be the part which is not enlightened.
Also, to understand why this is a system of neighborhood, note the following. Take a tree (which is 0-hyperbolic) and take $k=0$. Then, $U_{\rho(t),0}(\xi)$ is exactly all the points in the boundary that will be "after $\rho(t)$", in the sense that a geodesic from $a$ to $\gamma$ has to pass through $\rho(t)$. If you make $t$ go to infinity, then you only have $\xi$ left.
Thank you, @M. Dus! I appreciate your detailed answer. Just to be sure I understand your answer to my second question: $\gamma \in U_{\rho (t),k}(\xi )$ if and only if there exists a geodesic ray initiating from $a$ and asymptotic to $\gamma$ intersecting the ball of radius $k$ around $\rho (t))$.
@HussainRashed According to the definition you have, it is every geodesic ray asymptotic to $\gamma$ that needs to intersect the ball $B_k(\rho(t))$. However, you can replace every by some, using $\delta$-hyperbolicity. I added details in the answer.
To have the freedom of choosing our $k> 2\delta$, it would make sense if we declare the basis $$\begin{Bmatrix}
B_{r}(x): x\in X ,r>0
\end{Bmatrix}\cup \begin{Bmatrix}
U_{\rho (t)}(\xi ):t\geq 0, \xi\in\partial X
\end{Bmatrix}$$ where $\rho$ is a geodesic ray asymptotic to $\xi$ and initiating from $a$, and $\$$U_{\rho (t)}(\xi )$={$z\in X^{-}$:$[a,z]\cap B_{k}(\rho(t))\neq \phi$, for some $k>2\delta$} where $[a,z]$ is any geodesic connecting $a$ and $z$. Otherwise, I cannot see why for some geodesic ray implies that for all geodesic rays.
@HussainRashed I'm not saying that you have the freedom of choosing $k$. Let me clarify. I denote by $U_{\rho(t),k}(\xi)$ the neighborhood defined as above (with any geodesic $[a,z]$ in the definition) and I denote by $U'{\rho(t),k}(\xi)$ the neighborhood defined in the same way, but replacing "any geodesic $[a,z]$" by "some geodesic $[a,z]$". Then I'm saying that for every $k$, there exists $k'$ such that $U'{\rho(t),k}(\xi)\subset U_{\rho(t),k'}(\xi)$.
Great! and $k^{'}$ can be taken to be $2k$.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.