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2025-03-21T14:48:29.651772
| 2020-01-16T16:06:22 |
350563
|
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|
Stack Exchange
|
Different definitions of formality for groups
Let $X$ be a space with fundamental group $G$. Recall that the de Rham fundamental group of $X$ is the inverse limit of the Malcev completions of the nilpotent truncations of $G$. This has a Lie algebra, which I will denote by $\mathfrak{g}$. The Lie algebra $\mathfrak{g}$ has a natural filtration, and thus has an associated graded Lie algebra $\text{gr}\ \mathfrak{g}$. In the literature, I have seen two different definitions of what it means for $G$ to be formal:
The Lie algbera $\mathfrak{g}$ is isomorphic to its associated graded $\text{gr}\ \mathfrak{g}$.
The Lie algebra $\mathfrak{g}$ has a presentation with only quadratic relations.
There is also a notion of $X$ being formal:
The minimal model of $X$ is quasi-isomorphic to the cohomology algebra (with the trivial differential).
I assume that 3 implies 1 and 2 (though presumably it is stronger).
Question: What is the precise relationship between the above notions? Is there a good place to read proofs of whatever implications exist?
Clearly 2 implies 1, and the converse doesn't hold (for instance the 3-dimensional Heisenberg Lie algebra satisfies 1 and not 2).
In addition to Denis excellent answer you might want to have a look at the very nice survey by Suciu-Wang in https://arxiv.org/abs/1504.08294
There is a notion of $q$-equivalence between cdgas: it is a chain of morphisms each being isomorphism on cohomology in degrees $\leq q$ and monomorphism in degree $q+1$. Now, we call something $q$-formal if it is $q$-equivalent to its cohomology. Obviously, a space $X$ is $1$-formal if and only if $K(\pi_1(X), 1)$ is $1$-formal because killing homotopy groups map $X \to K(\pi_1(X), 1)$ is evidently an 1-equivalence.
Define holonomy algebra $hol(X)$ of a cdga as quotient of free Lie algebra on $(H^1)^*$ by ideal spanned by image of comultiplication $\mu^*: (H^2)^* \to (H^1)^* \wedge (H^1)^*$. It's obviously a graded quadratic Lie algebra.
Notice that 1-equivalences give isomorphisms on holonomy Lie algebras. So, for spaces, it depends only on fundamental group, because if you have $G \cong \pi_1(X) \cong \pi_1(Y)$, then both map into $K(G, 1)$ with maps being $1$-equivalences.
Malcev completion of a group $\mathfrak m (G)$ is defined as Lie algebra of primitive elements in $\lim_{\leftarrow} \Bbb QG/IG^n$ where $IG$ is augmentation ideal. There is a homomorphism from Magnus algebra of a group $L(G) := \bigoplus \gamma_kG/\gamma_{k+1}G \otimes \Bbb Q$ to Malcev algebra which becomes isomorphism after taking associated graded quotients, as proven by Quillen in paper On the associated graded ring of a group ring, 1968.
We have a theorem due to Sullivan.
$\square$ Space $X$ is $1$-formal if and only if degree completion of holonomy Lie algebra $hol(X)$ is filtered isomorphic to $\mathfrak m (\pi_1(X))$. $\square$
Good reference is two books by Felix, Halperin, Thomas, called (unsurprisingly) Rational homotopy theory I/II. 1-formality is very clearly explained in the beginning of second tome.
Now, to your question. If a space is formal, it is indeed $q$-formal for all $q$, in particular, 1, so your (3) implies both (2) and (1).
YCor mentioned in comments that (1) is strictly weaker as witnessed by integral Heisenberg group. More generally, every two-step nilpotent Lie algebra is isomorphic to its associated lower central quotient, but only virtually abelian nilpotent groups are 1-formal. (If some commutator $[a, b]$ is nontrivial and there are no relations in weight 3, then $[a, [a, b]]$ will be nontrivial).
Let's prove that (2) implies (3).
Suppose $\mathfrak m(G)$ is a completion of some quadratic algebra $E = FreeLie(V)/(R), R \subset \wedge^2 V$. Now, going back to Quillen, we have a homomorphism $L(G) \to \mathfrak m(G)$ which becomes isomorphism on associated graded (in particular, it's injective). As Malcev algebra was graded from the beginning, $L(G)$ inherits this grading, and the only thing it can be is $E$. From Sullivan's Infinitesimal computations in topology we know that kernel of Lie bracket induced by group commutator on $G_{ab} \wedge G_{ab}$ rationally isomorphic to image of comultiplication $\mu^*$, so $E \cong hol(G)$ q.e.d.
|
2025-03-21T14:48:29.652059
| 2020-01-16T16:41:10 |
350565
|
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|
Stack Exchange
|
Relate the solid angle and surface measure of a surface
Let $M$ be a 2-dimensional embedded $C^1$-submanifold of $\mathbb R^3$ with a global chart$^1$ $(U,\phi)$. If $u\in U$ and $x=\phi^{-1}(u)$, let $\nu_M(x)$ denote the unique unit normal vector of $M$ with $$\det\left({\rm D}\phi(u),\nu_M(x)\right)>0\tag1.$$ Moreover, let $\sigma_M$ denote the surface measure$^2$ on the Borel $\sigma$-algebra $\mathcal B(M)$ and $$\pi:\mathbb R^3\setminus\{0\}\to S^2\;,\;\;\;x\mapsto\frac x{|x|}$$ denote the projection onto the unit 2-sphere $S^2$.
If $0\not\in B\in\mathcal B(M)$, are we able to express $\sigma_{S^2}(\pi(B))$ as an integral with respect to $\sigma_M$?
For clarity of exposition, let $$\omega_{x\to y}:=\pi(y-x)\;\;\;\text{for }x,y\in\mathbb R^3\text{ with }x\ne y.$$ There are plenty of (mathematically vaguely) references$^3$ claiming that $$\sigma_{S^2}({\rm d}\omega_{x\to y})=\sigma_M({\rm d}y)\frac{\left|\langle\nu_M(y),\omega_{x\to y})\rangle\right|}{|x-y|^2}\tag2,$$ which is reasonable from geometric inspection. However, I struggle to state and prove this in a measure-theoretic rigorous way.
Idea 1: Noting that $\pi=\nabla\rho$, where $\rho(x):=|x|$ for $x\in\mathbb R^3$, $(2)$ may be related to the divergence theorem. To be precise, if $K\subseteq M\setminus\{0\}$ is compact and has a $C^1$-boundary$^2$, $$\langle\nu_{\partial K},\pi\rangle=\langle\nu_{\partial K},\nabla\rho\rangle=:\frac{\partial\rho}{\partial\nu_{\partial K}}.\tag3$$ However, I'm unsure how the "outer" normal field$^5$ $\nu_{\partial K}$ and $\nu_M$ are related.
Idea 2: My guess is that $$\left(\sigma_{S^2}\circ\pi\right)(B)=\int_B\sigma_M({\rm d}y)\frac{\left|\langle\nu_M(y),\pi(y)\rangle\right|}{|y|^2}\tag4$$ (this would include to show that $\pi(B)\in\mathcal B(S^2)$). Since $\mathcal B(M)$ is generated by the open balls with center in $M$, it should be sufficient to prove $(4)$ for $B=B_\varepsilon(x)$ for some fixed $x\in M$ and $\varepsilon>0$. Now, $B$ is compact and has a $C^1$-boundary. Moreover, $$\nu_{\partial B}(y)=\frac{y-x}\varepsilon\;\;\;\text{for all }y\in\partial B\tag5$$ and $$\int\sigma_{\partial B}({\rm d}y)\frac{\left|\langle\nu_{\partial B}(y),\pi(y-x)\rangle\right|}{|y-x|^2}=\frac1{\varepsilon^2}\sigma_{\partial B}(\partial B)=\sigma_{S^2}(S^2)=\left(\sigma_{S^2}\circ\pi\right)(\partial B).\tag6$$ However, I don't know if $(6)$ is helpful for showing $(4)$. (But intuitively, the solid angle subtended by $\partial B$ should be the same as the solid angle subtended by $B$.)
EDIT: Isn't there a "visibility" function $v:M\times M\to\{0,1\}$ missing on the right-hand side of $(2)$ indicating whether $y$ is "visible" as seen from $x$ ($v(x,y)=1$) or is occluded by another point $z\in M$ with $\omega_{x\to y}=\omega_{x\to z}$ and $|x-z|<|x-y|$?
$^1$ i.e. $U\subseteq\mathbb R^2$ is open and $\psi:U\to M$ is an immersion and a topological embedding of $U$ onto $M$.
$^2$ Let $\lambda$ denote the Lebesgue measure on $\mathcal B(\mathbb R)$, $J_\phi$ denote the Jacobian of $\phi$ and $g_\phi:=\det J_\phi^TJ_\phi$. Then $$\sigma_M=\sqrt{g_\phi}\left.\lambda^{\otimes 2}\right|_U\circ\phi^{-1}.$$
$^3$ for example, [1, p. 14], [2, p. 53 (PDF numbering)] or [3, p. 6].
$^4$ i.e. for all $x\in\partial K$ there is an open neighborhood $V$ of $x$ and a continuously differentiable $\psi:V\to\mathbb R$ with $K\cap V=\{\psi\le0\}$ and $\nabla\psi\ne0$.
$^5$ i.e. if $\psi$ is as in footnote 4, then $$\nu_{\partial K}(x):=\frac{\nabla\psi(x)}{\left|\nabla\psi(x)\right\|}.$$
What would be one of the more commonly encountered lecture books which make this claim?
What is $\mathcal{B}(M)$? The expression $\sigma_{S^2}(d\omega_{x\to y})$ appears to apply a measure on a surface to the exterior derivative of a function on $\mathbb{R}^3\times \mathbb{R}^3 - \text{diagonal}$. Can you clarify that notation a little please?
I am guessing $\mathcal{B}(M)$ is the Borel sigma algebra on $M$.
I share @BenMcKay's confusion about your notation. So first, forget about $x$, which seems to just describe the center of the sphere. Let $\omega$ be the projection map $M \to \mathbb{S}^2$ where $0\not\in M$ is an embedded 2-manifold of $\mathbb{R}^3$ given by $y \mapsto y / |y|$. Are you just asking about how to relate the surface measure on $M$ and the surface measure on $\mathbb{S}^2$ via the function $\omega$?
@BenMcKay $\mathcal B(M)$ is the Borel $\sigma$-algebra on $M$. You can find the claim, for example, in this PHD thesis (on page 15): https://backend.orbit.dtu.dk/ws/portalfiles/portal/51112612/phd240_awk_vers2.pdf. But I can give you way more references, if I search for them.
@WillieWong I've tried to simplify the question.
@BenMcKay I've edited the question and think it is clearer now what I'm asking. It would be great if you could take a look.
The density you are asking about is precisely the Jacobian of the central projection of $M$ to the unit sphere centered at the origin (as was already mentioned by Willie Wong). For an explicit expression (for instance, in the form given on p.14 in the reference you quote) it is enough to look at the picture on p.15 (which is apparently what you mean by "geometric inspection"). If you wish, one can do it in two steps by first replacing $M$ with the corresponding tangent plane (which does not change the Jacobian). I really fail to see what is the "measure-theoretic rigor" one needs in addition to this observation (which does not really seem to be at the research level). However, if you want a mathematical reference, this formula is, for instance, derived in great detail (in a somewhat different notation though) in Palamodov's book Reconstructive integral geometry, (Section 3.1, Example 5).
Talking about notation, the way you reproduce the (already somewhat confused) formula from p.14 (not p.15 as you write) of your reference is completely erroneous (which was also mentioned by Willie Wong). In that reference a point $y\in M$ is projected to the unite sphere centered at $x$, whereas in your notation it is $x\in M$ that is projected to the unite sphere centered at 0 (and the role of $y$ is completely unclear).
The right expression for the density is $|\langle x, \nu_M(x) \rangle|/|x|^3$. Geometrically, the numerator is the distance from the origin to the tangent plane to $M$ at the point $x$.
Thank you for your answer. You're right, there was a typo in my equation (2). However, my problem with (2) is that the left-hand side is not really well-defined, since the differential ${\rm d}\omega_{x\to y}$ does depend on $y$. So, since it is still not clear to me from your answer, is (4) the correct formula (assuming $x=0$) or is it $$\sigma_{S^2}(\Omega)=\int\sigma_M({\rm d}y)1_\Omega(\pi(y))\frac{\left|\langle\nu_M(y),\pi(y)\rangle\right|}{|y|^2}\tag7$$ for all $\Omega\in\mathcal B(S^2)$? In either case, it seems like a "visibility function" $v:M\to{0,1}$ is missing indicating
whether $y$ is occluded ($v(y)=0$) by another point in $M$ as seen from the origin $0$ or not ($v(y)=1$). That's what I mean with "measure-theoretic rigor". Besides that, shouldn't we be able to derive the formula without reference to the geometric picture? And most importantly: What kind of regularity assumption do we need to impose on $M$ (does it need to be the boundary of a compact set?) and how precisely is $Ξ½_M$ (is it given by $\frac{β_1\phiΓβ_2\phi}{|β_1ΟΓβ_2\phi|}$ (in the submanifold case) or, in the boundary case, by footnote 5?) defined?
The geometric part of your question boils dows to finding the Jacobian of the projection map. All the rest is not specific for your concrete setup and is covered by standard textbook expositions on change of variables in surface integrals; the same textbooks usually discuss normal vectors to surfaces as well.
You did not answer my question. Is $(4)$ or $(7)$ the correct formula? And don't we need the visibility function?
The change of variables formulas are always written locally, under the assumption that the considered transformation is locally a bijection. If this is not the case, and there are several preimages, then the image measure is the sum of several measures obtaned by applying the corresponding local bijections. See any multivariable calculus textbook for the details. If $\pi$ is not an injection, then the LHS of (4) is not defined, whereas (7) is only true if you restrict to a subset of $M$ where $\pi$ is a bijection (for instance, by using your "visibility function").
(a) So, I guess the assumption we need is that the restriction $\left.\pi\right|M$ is injective. Alright, that makes sense to me. (b) But I don't get what you wrote about (4). $\sigma{S^2}$ is a finite measure on $\mathcal B(S^2)$. And $\pi$ is a Borel measurability preserving map, i.e. $\pi(A)\in\mathcal B(S^2)$ for all $A\in\mathcal B(\mathbb R\setminus{0})$. So, in total, the left-hand side of (4)$ is always well-defined.
(c) I get the point that only the local properties of the area under consideration are relevant. But it's still not clear to me what kind of regularity we need to assume. Must $y$ in (2) belong to a surface (i.e. 2-dimensional submanifold), to the boundary of a compact set (which is a surface) or anything else? We certainly need orientability.
(b) What I meant is that $\pi$ maps $M$ to $S^2$, and therefore it induces a "push forward" map from measures on $M$ to measures on $S^2$. Unless $\pi$ is bijective there is no way to "pull back" measures from $S^2$ to $M$.
(c) Everything is local, so that your surface can be presented as the graph or as the level set of a certain locally defined smooth function. There is no need to consider it as the boundary of something. Since all considerations are local and we are talking about measures rather than forms, orientability is not needed either (notice that the expressions which appear in all formulas contain the absolute value of the scalar product and do not depend on the orientation of the normal vector).
(b) Okay, what you mean is that the pushforward measure $\sigma_M\circ\left.\pi\right|_M^{-1}$ needs to be well-defined, right?
No - once again, measures are naturally mapped in the same direction as points. Therefore, if you have a map $\pi:M\to S^2$, then measures can be naturally moved from $M$ to $S^2$, but not the other way round.
(b) But $\sigma_M\circ\left.\pi\right|_M^{-1}$ is a measure on $\mathcal B(S^2)$. What am I missing? (c) So, okay, I would argue that we need orientability to ensure that $\nu_M$ is well-defined, but I guess you again assume that we just need to replace $M$ by a local representation, right? The things that confused me are the assumptions made in http://graphics.stanford.edu/papers/veach_thesis/thesis.pdf on page 76.
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2025-03-21T14:48:29.652833
| 2020-01-16T16:54:14 |
350566
|
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|
Stack Exchange
|
Existence of a unitary fusion category with this relation ruled out on finite groups
In this answer, Geoff ruled out the existence of a finite group $G$ such that the fusion category $\mathrm{Rep}(G)$ has simple objects $5_1$ and $7_1$ of FPdim $5$ and $7$ resp., with (for some object $X$): $$ 5_1 \otimes 5_1 = 1 \oplus 5_1 \oplus 7_1 \oplus X$$
and in this comment he thinks that it is also probably true if $(5,7)$ is replaced by any twin prime pair $(p,q)$ with $p>3$. This leads to ask about an extension to (unitary) fusion category:
Question: Is there a (unitary) fusion category having two simple objects $p_1$ and $q_1$ of FPdim $p$ and $q$ resp., with $(p,q)$ a twin prime pair, $p>3$ and (for some object $X$) satisfying the following? $$ p_1 \otimes p_1 = 1 \oplus p_1 \oplus q_1 \oplus X$$
Specifically interested in $(p,q) = (5,7)$ and $X = 5_2 \oplus 7_2$, so if one such example is known (or cannot exist), a proof or a reference would be welcome in answer.
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2025-03-21T14:48:29.652921
| 2020-01-16T16:59:01 |
350567
|
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|
Stack Exchange
|
Is the $\mathbb A^1$ homotopy type of a punctured curve independent of the choice of puncture?
Let $C$ be a smooth connected algebraic curve over $\mathbb C$. Assume that $C$ admits no automorphisms. Let $p_1, p_2 \in C$ be two distinct points.
Question: Are the $\mathbb A^1$ homotopy types of $C - p_1$ and $C- p_2$ the same?
I think that the answer is "no", but I do not know how to construct an invariant that distinguishes the two. Of course, the ordinary homotopy types of $C-p_1$ and $C-p_2$ are the same. I also think that the stable $\mathbb A^1$ homotopy types are the same because of the purity cofiber sequence $$C-p_{i} \to C \to (\mathbb A^1, \mathbb A^1 - 0).$$
Is there a way of seeing that these two are not $\mathbb A^1$ homotopic, or of constructing an equivalence between them?
If I understand it correctly, this follows from the results of Severitt's master's thesis. (I am not an expert on this, so it is possible I misread one of the statements.)
Indeed, Lemma 9.1.1 shows that smooth projective curves of genus $> 0$ are $\mathbf A^1$-rigid. But the proof actually shows the following:
Lemma. Let $X$ be a smooth variety such that every map $\mathbf A^1 \to X$ is constant. Then $X$ is $\mathbf A^1$-rigid. $\square$
Now if $X = C - p$ for some smooth projective curve $C$ with $\operatorname{Aut}(C) = 1$, then in particular $g(C) > 0$ so every map $\mathbf A^1 \to C$ (or $\mathbf A^1 \to C \setminus p$) is constant. Thus, the lemma implies $X$ is $\mathbf A^1$-rigid. Then Theorem 6.3.7 implies the result: if $C - p$ and $C - q$ are (weakly) $\mathbf A^1$-homotopy equivalent, then the theorem gives $C - p \cong C - q$, which implies that there is a nontrivial automorphism of $C$ taking $p$ to $q$. $\square$
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2025-03-21T14:48:29.653062
| 2020-01-16T17:27:27 |
350570
|
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"RobPratt",
"Robert Israel",
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|
Stack Exchange
|
Necessary optimality condition for quadratic programming: a solution of a constrained QAP is a solution of a LP
I have a concern about a result given by Murty in [1] and also written by Floudas and Visweswaran in [2]
They consider a QP:
\begin{array}{ll}{\min _{x} Q(x)} & {=c^{T} x+\frac{1}{2} x^{T} D x} \\ {\text {s.t.} \quad A x} & {\geq b} \\ {x} & {\geq 0}\end{array}
with no particular assumption on $D$ (appart from symmetric). Murty argues that if $x_{*}$ is a solution of the QP it is also solution of the following LP:
\begin{array}{cl}{\min _{x}} & {\left(c+x_{*}^{T} D\right) x} \\ {\text {s.t.}} & {A x \geq b} \\ {x} & {\geq 0}\end{array}
The proof of Murty is the following:
Let $\hat{x}$ be any feasible solution of the LP and $x_{\lambda}=\lambda \hat{x}+(1-\lambda) x_{*}$ for $\lambda \in ]0,1[$. It is also feasible since the set of feasible solutions is a convex polyhedron.
Then by optimality of $x_{*}, Q\left(x_{\lambda}\right)-Q(x_{*}) \geq 0$ which implies:
$$\lambda\left(c+x_{*}^{T} D\right)(\hat{x}-x_{*})+(1 / 2)\lambda^{2}(\hat{x}-x_{*})^{T} D(\hat{x}-x_{*}) \geq 0 $$
And so dividing by $\lambda$ leads to:
$$\left(c+x_{*}^{T} D\right)(\hat{x}-x_{*}) \geq(-\lambda / 2)(\hat{x}-x_{*})^{T} D(\hat{x}-x_{*})$$
At this part Murty argues that it "obviously implies":
$$\left(c+x_{*}^{T} D\right)(\hat{x}-x_{*}) \geq 0 $$
and hence $\hat{x}$ must be an optimum feasible solution of the LP.
It seems to be false at first sight except when the QP is concave (and D neg definite) so that the rightmost term of the inequality is negative. Am I missing something ?
Ref:
[1] Linear Complementarity, Linear and non Linear Programming, 1988
[2] Quadratic Optimization by Floudas and Visweswaran, 1995
Murty is correct. Take the limit of the inequality for $(c + x_^\top D)(\hat{x} - x_)$
as $\lambda \to 0+$.
$Ax \le b$ or $Ax \ge b$?
But if it is correct then would it imply that any solution of a QP lies in an extremal point of the polyhedron ? This is true for concave QP but in general it seems strong right ?
In [2] it is $\leq$ for the definition of the QP and then $\geq$ for the theorem and in the original [1] it is $\geq$ everywhere. I edit my post
No it does not imply an optimal solution is an extreme point. For example, the optimal solution of the QP could be an interior point of the feasible region, in which case the objective of the LP is $0$, and every point of the feasible region is optimal for the LP.
thank you very much for your answers I got it!
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2025-03-21T14:48:29.653245
| 2020-01-16T17:43:17 |
350571
|
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|
Stack Exchange
|
Why do we study complex orientable cohomology theories
It seems that much of the literature in stable homotopy theory seems to study complex orientable cohomology theories. What is the reason of restricting to this class of multiplicative cohomology theories? Is it simply that they are more computable? Is there a good a priori reason that this is an important class of cohomology theories to study?
It's perhaps not the most important reason but "complex oriented" beyond the existence of chern classes means you can "integrate" cohomology classes along submersions of almost complex manifolds. I believe this was a crucial part in Hirzebruch's original proof of HRR theorem.
There is a sort of a priori reason why one would consider the cohomology theory $MU$, without first knowing of its connection to manifold geometry, to formal groups, β¦ .
Since complex-oriented cohomology theories are those cohomology theories with a ring map from $MU$, perhaps having a sufficiently strong interest in $MU$ would suffice to then motivate them.
$\newcommand{\co}{\colon\thinspace}\newcommand{\F}{\mathbb F}\newcommand{\Z}{\mathbb Z}\newcommand{\A}{\mathcal A}\newcommand{\from}{\leftarrow}$
There are a variety of constructions in homotopy theory that allow one to delete a class, where "class" can have different meanings and "delete" different levels of finesse.
For instance, if you have a map $f\co S^n \to X$ which on homology sends the fundamental class $\iota_n \in H_n S^n$ to some class $f_* \iota_n \in H_n X$, then the mapping cone $C(f)$ has the property that $f_* \iota_n$ pushes forward to $0$ in $H_n C(f)$.
In fact, if $f_* \iota_n$ is not torsion, then $H_* C(f)$ will be exactly $(H_* X) / f_* \iota_n$, as one can see by writing out the homology long exact sequence.
By the same token, if $f_* \iota_n$ is $m$βtorsion, the same long exact sequence will produce a fresh class in $H_{n+1} C(f)$, due to the death of $m \iota_n$ under $f_*$.
This is a general truism in homotopy theory: if a class has been deleted "twice", then you get a fresh class one degree higher.
For a slight twist on this same idea, consider the modβ$p$ cohomology of a space, which is related to its homology by the universal coefficient sequence $$ 0 \to \operatorname{Ext}(H_{n-1}(X; \Z), \F_p) \to H^n(X; \F_p) \to \operatorname{Hom}(H_n(X; \Z), \F_p) \to 0. $$
A torsion-free class in $H_n(X; \mathbb Z)$ contributes a class only to $H^n(X; \F_p)$, but a $p$βpowerβtorsion class contributes classes both to $H^n(X; \F_p)$ and $H^{n+1}(X; \F_p)$.
There is actually a process by which one can marry these pairs of classes and reconstruct integral cohomology: the Bockstein spectral sequence has signature $$H^*(X; \F_p) \otimes \F_p[b] \Rightarrow H^*(X; \Z_p),$$ it converges for connected $X$ of finite type, and its differentials perform exactly these marriages.
Its construction relies on the facts that $\Z_p$ is $p$βcomplete and that it participates in the short exact sequence $$0 \to \Z_p \xrightarrow{p \cdot -} \Z_p \to \F_p \to 0,$$ and for us it suffices to leave its explanation at that.
The first differential in the Bockstein spectral sequence takes the form of a stable, additive homomorphism $$\beta\co H^n(X; \F_p) \to H^{n+1}(X; \F_p).$$
In fact, this map is natural in $X$, and such natural transformations in general are called stable cohomology operations.
The Steenrod algebra is the collection of all stable modβ$p$ cohomology operations, given by $$\A^* := [H\F_p, H\F_p]_* = H^*(H\F_p; \F_p).$$
This associative algebra is calculable from first principles [1]: $$\A^* = \F_p\langle \beta, P^n \mid j \ge 1 \rangle \, / \, (\text{various relations}),$$ where the angle brackets indicate that these generators do not commute; where $P^n$ is the "$p$th power operation", which sends a cohomology class of degree $2n$ to its $p$th power [2]; and where $\beta$ is the same Bockstein operation as above.
Taking this calculation for granted, one can then make a further calculation of a cousin of these operations: $$H^*(H\Z_p; \F_p) = [H\Z_p, H\F_p]_*.$$
Starting with same the quotient sequence $$H\Z_p \xrightarrow{p} H\Z_p \to H\F_p,$$ and applying $[-, H\F_p]_*$, the multiplication-by-$p$ map induces zero on cohomology, and hence the going-around map participates in a short-exact sequence of $\A$βmodules $$0 \from [H\Z_p, H\F_p]_* \from \A^* \from [H\Z_p, H\F_p]_{*+1} \from 0.$$
The first map is onto, hence our $\A^*$βmodule of interest is cyclic; the second map presents it as a submodule of $\A^*$ generated by a class in degree $1$; and, since $\beta$ is the only operation in degree $1$, we learn $$[H\Z_p, H\F_p]_* = \A^* / (\beta \cdot \A^*).$$
In this sense, $\beta$ witnesses the double-quotient of $H\Z_p$ by $p$, in that the quotient appears on the left- and on the right-hand sides of $[H\F_p, H\F_p] = [H\Z_p / p, H\Z_p / p]$.
By removing the quotient from the left and instead studying $[H\Z_p, H\Z_p/p] = [H\Z_p, H\F_p]$, we avoid killing $p$ twice, and $\beta$ disappears.
However, longer monomials in which $\beta$ appears in the middle still survive, and they contribute other odd-degree cohomology operations.
One might wonder whether every such odd-degree cohomology operation belongs to some spectrum $X$ with a non-nilpotent endomorphism $v\co \Sigma^n X \to X$ whose quotient recovers $H\F_p$ and which admits an associated Bockstein spectral sequence.
It turns out that a version of this is true: $[MU, H\F_p]_*$ is (almost [3]) the quotient of $\A^*$ with $\beta$ fully deleted.
In this sense, $MU$ is the "maximally unquotiented" version of $H\F_p$ within even-concentrated ring spectra.
Even more than this, there are multiple theorems along these lines, coming at the same problem from different angles.
The study of the modβ$p$ cohomology of $MU$ (and its various features, including its relation to that of $H\F_p$) is originally due to Milnor. The homotopy of $MU$ is given by $$\pi_* MU = \Z[x_1, x_2, \ldots],$$ and the generators $x_n$ (again, almost [3]) iteratively play the role of $v$ in the above fantasy resolution of $H\F_p$.
Starting with the $p$βlocal sphere, one can iteratively remove its odd-degree homotopy while retaining even-concentrated homology.
Priddy showed that this ultimately leads to a spectrum called $BP$, which is an indivisible $p$βlocal summand of $MU$.
Starting with the $p$βlocal sphere, one can iteratively remove its odd-degree homotopy by $A_\infty$βalgebra maps.
This leads to a sequence of spectra $X(n)_{(p)}$ with $X(\infty)_{(p)} = MU_{(p)}$.
The original study of this sequence of spectra is due to Ravenel, then taken up by Hopkins, Devinatz, and Smith, and this perspective in terms of iterated quotients is due to Beardsley.
Each of these objects begins with some desirable properties: the sphere spectrum has pleasant homology (but very knotty homotopy), and the EilenbergβMac Lane spectrum has pleasant homotopy (but knotty co/homology).
By trying to correct the unpleasant part, one keeps ending up at (a chunk of) $MU$, which has even-concentrated homotopy, even-concentrated homology, even-concentrated co/operations, β¦ .
All I mean to point out by this is that $MU$ is an extremely natural object to bump into, especially if one has a preference for even-concentration, whether due to a preference for commutativity over graded-commutativity or due to an aversion toward unnecessarily killing classes twice.
ββββββββββ
[1] - Here I'm restricting to odd primes, but you can say all these same words with slightly different formulas at $p = 2$.
[2] - $P^n$ does other, more mysterious things in degrees other than $2n$.
[3] - The precise statement is that the $p$βlocalization $MU_{(p)}$ splits as a sum of shifts of copies of a ring spectrum $BP$.
This new spectrum has homotopy given by $$\pi_* BP = \Z_{(p)}[v_1, v_2, \ldots],$$ $[BP, H\F_p]$ is the submodule of $\A^*$ where $\beta$ is totally deleted, and these $v_j$s are the desired self-maps $v$.
(In terms of $\pi_* MU$, $v_j$ is equivalent to $x_{p^j}$ modulo decomposables.)
Although $BP$ has all these nice properties, it depends on the prime $p$, and $MU$ is to be thought of as the best integral object capturing all of them at once.
Incidentally: I feel there are totally separate and very enlightening answers as to why people study oriented cohomology theories. Unfortunately, that has no outward indications as to why people then fixate on the complex case.
A surface-level motivation for complex oriented cohomology theories is that they are precisely those that admit a theory of generalized chern classes.
More surprisingly, the universal complex oriented theory $MU$ tends to see quite a lot of information about the stable homotopy groups. I am a novice, but there is a ton of history here-- you should really read some of the literature. Novikov used $MU$ in his version of the Adams spectral sequence. Quillen discovered the connection between $MU$ and formal group laws. Landweber exploited this to construct other interesting cohomology theories. Ravenel made a series of conjectures, which were mostly proved by Devinatz, Hopkins and Smith. Broadly these imply that $MU$ sees all non-nilpotent phenomena in stable homotopy theory. Thus, knowledge of $MU$ allows one to compute the Balmer spectrum of the stable homotopy category (analogous to the prime numbers for abelian groups). Because of the connection between formal group laws and $MU$, this spectrum can be interpreted in terms of formal group laws. To each point of the spectrum, there are cohomology theories, originally constructed by Morava $K(n,p)$ and $E(n,p)$. Various properties of these cohomology theories can be interpreted in terms of the algebraic geometry of the space (stack) of formal group laws. This is just a superficial glimpse of the picture of chromatic homotopy theory that we have today.
|
2025-03-21T14:48:29.653920
| 2020-01-16T18:05:27 |
350574
|
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|
Stack Exchange
|
Why is the divergence theorem used in the Eells-Sampson paper slightly different from that in a textbook?
I am reading Harmonic Mappings of Riemannian Manifolds by Eells and Sampson. In chapter 2, the author(s) used the divergence theorem, which does not look like the usual divergence theorem for manifolds, and I can't figure out why it is correct.
Simply speaking, in this paper, a family of quantities $\xi^j\ (j=1,\cdots,m)$ is defined using local coordinates $(x^1,\cdots,x^m)$ on a Riemannian manifold $M$. And one can check $\xi^j$ vary contravariantly under coordinate changes, and therefore define a tangent vector field on $M$ (Einstein's summation convention used):
$$X=\xi^j\frac{\partial}{\partial x^j}$$
Then the paper says and I quote:
...whence by the Green's divergence theorem
$$\int_M\xi^j_jdM=0.$$
where $dM$ is the volume element on $M$.
There are two things I am not sure of.
I am not sure what $\xi^j_j$ here means. My best guess would be $\xi^j_j=\frac{\partial\xi^j}{\partial x^j}$.
Suppose $\xi^j_j$ does mean as above. What I don't get is why $\int_M\xi^j_jdM=0$ follows from the divergence theorem. According to Proposition 2.46 of Lee's book Introduction to Riemannian Manifolds, the coordinate representation of the divergence of $X=\xi^j\frac{\partial}{\partial x^j}$ should be ($\det g$ is the determinant of the matrix for the metric tensor under local coordinates)
$${\rm div} X=\frac{1}{\sqrt{\det g}}\frac{\partial}{\partial x^j}(\xi^j\sqrt{\det g})$$
The divergence theorem then takes the form ($M$ is assumed to be compact without boundary so that the RHS vanish):
$$\int_M(\xi^j_j+\frac{\xi^j}{\sqrt{\det g}}\frac{\partial\sqrt{\det g}}{\partial x^j})dM=0$$
There's clearly an extra term.
Any help is appreciated.
I haven't read the paper bu I would bet $\xi_j^j$ means $\nabla_j\xi^j$ which is exactly $\mathrm{div}, X$.
@ThomasRichard Thank you! I think you are right. I redid the calculation assuming the subscript means covariant derivative and everything works now.
|
2025-03-21T14:48:29.654114
| 2020-01-16T18:57:03 |
350578
|
{
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|
Stack Exchange
|
K-theory of free $G$-sets and the classifying space, and generalization
$\newcommand\Sets{\mathrm{Sets}}\DeclareMathOperator\Nerve{Nerve}$Let $G$ be a finite group, $\mathcal{G}^0$ be the category of finite free
$G$-sets and isomorphisms between them. Then $\mathcal{G}^0$ is a symmetric monoidal category with respect to the disjoint union, so we can talk about its $K$-theory, which is homotopy equivalent to $\Sigma ^{\infty}BG_+$.
My first question is, where in the literature can I find this, preferably explicitly stated in this way?
Now, consider the category $\mathcal{G}$ who has unique object and morphisms are elements of $G$. Then we can identify $G$-sets with functors from $\mathcal{G}$ to $\Sets$. Thus the statement above can be reformulated as follows.
$$\Sigma ^{\infty}\lvert\Nerve(\mathcal{G})\rvert_+\simeq K(F^0(\mathcal{G},\Sets)) $$
where $F^0(A,B)$ denote the category whose objects are "free" functors from
$\mathcal{G}$ to $\Sets$. And by free, we mean the objects that are in the essential image of the left adjoint of the "forgetful" functor to $\Sets$.
Now my second question is: is there any known sufficient condition on the category $\mathcal{C}$ and its object $C$ so that we have $$\Sigma ^{\infty}\lvert\Nerve(\mathcal{C})\rvert_+\simeq K(F^0(\mathcal{C},\Sets)), $$ where the notation is just as in above except we use the evaluation at $C$ instead of the forgetful functor?
I thought $K$-theory of a category is the one of its classifying space. But, it seems you are identifying the suspension spectrum of the classifying space as the $K$-theory of your category?!
c.f. my comment on the answer by John Klein.
The general point is just that if $\mathcal{U}$ is equivalent to the free symmetric monoidal category $F\mathcal{C}$ generated by $\mathcal{C}$ then $K(\mathcal{U})\simeq\Sigma^\infty_+\mathcal{C}$. Here $F\mathcal{C}$ can be constructed as the category of pairs $(X,C)$, where $X$ is a finite set and $C\in\prod_{x\in X}\text{obj}(\mathcal{C})$. A morphism from $(X,C)$ to $(Y,D)$ consists of a bijection $\sigma\colon X\to Y$ together with a family of $\mathcal{C}$-morphisms $C_x\to D_{\sigma(x)}$ for all $x\in X$. It's not hard to identify the groupoid $\mathcal{F}G$ of finite $G$-sets with the free symmetric monoidal category generated by subgroupoid $\text{Orb}(G)$ of transitive $G$-sets. We can choose subgroups $H_1,\dotsc,H_m$ containing one representative of each conjugacy class, and let $W_k=N_G(H_k)/H_k$ denote the Weyl group, considered as a one-object groupoid. Then $\text{Orb}(G)$ is equivalent to the coproduct of the groupoids $W_k$, so we get
$$ K(\mathcal{F}G) \simeq \Sigma^\infty_+ B\text{Orb}(G)
\simeq \bigvee_k \Sigma^\infty_+ BW_k $$
This is the tom Dieck splitting. If we restrict attention to the subcategory $\mathcal{F}_1G$ of finite free $G$-sets, we find that this is freely generated by the free orbit $G/1$, whose $G$-equivariant automorphism group is the Weyl group $W1=G$, so we get $K(\mathcal{F}_1G)=\Sigma^\infty_+BG$. There are quite a few interesting results in this vein, but unfortunately I do not think that there is a good source in the literature.
Maybe I am asking something dumb, but how does "The general point" work?
I believe you meant to write $Q(BG_+)$ in the first paragraph of your post, where $Q = \Omega^\infty\Sigma^\infty$. The this result is really a folk theorem and is sometimes called the "BarrattβPriddyβQuillenβSegal" theorem. This is not a folk theorem but rather it is a theorem with a folk authorship in that none of these people wrote down the theorem in this context as far as I know.
One way to deduce it is to use the the Group Completion Theorem (see e.g., McDuff, D.; Segal, G. Homology fibrations and the βgroup-completionβ theorem. Invent. Math. 31 (1975/76), no. 3, 279β284).
The classifying space the category of finite free $G$-sets and their isomorphisms defines a topological monoid $M$. The group completion theorem tells us in this case that $\Omega B M$ coincides with $Q(BG_+)$.
Actually I really meant $BG_+$ because some author use the word $k$-theory to mean the associated spectrum (infinite delooping) and not the infinite loop space.
@user43326 the K-theory of finite free $G$-sets is not $BG$, it is $Q(BG_+)$. Here is a simple reason why it can't be $BG$: the latter is not in general an infiinite loop space (but the $K$-theory is an infinite loop space). By the way, $G$ does not need to be finite in the Barratt-Priddy-Quillen-Segal theorem.
@user43326: every connective spectrum determines an infinite loop space and vice-versa.
that is exactly what is I am saying, for some authors it is the spectrum and not the infinite loop space...
The expression "$BG$" refers to a space, not a spectrum. You are using incorrect notation. If you want to associate a spectrum to $BG$, you should be writing $\Sigma^\infty (BG_+)$--the suspension spectrum of $BG$. Again, I am saying there is an incorrect statement in the first paragraph of your post.
Oh, thank you, I had missed that. In most of my writing, I identify a based space with its suspension spectrum... Corrected now.
@JohnKlein different people use different notation. I would quite often leave the $\Sigma^\infty$ implicit in this kind of context.
@user43326 even with the notation you say you like to use (and which I don't), it would have been more correct to write $BG_+$ rather than $BG$.
$\DeclareMathOperator\End{End}\newcommand\Set{\mathrm{Set}}\DeclareMathOperator\Nerve{Nerve}\DeclareMathOperator\Fun{Fun}$Here's perhaps a way to understand the general case that Neil mentioned, and a way to apply it to your question at the end.
Suppose you have a category $C$. Then there's a free symmetric monoidal category on $C$, $FC$, which Neil described in his answer.
Now if you're looking at $F^0(C,\Set)$ with respect to some $c\in C$, the category that you get only depends on $B{\End(c)}\subset C$, the full subcategory on $c$ (which is a one-object category, so you can see it as a monoid somehow), and in fact its core-groupoid (which is the only thing that $K$-theory depends on) is exactly $F(B{\End(c)^\times})$, i.e. finite free $\End(c)^\times$-sets, where $\End(c)^\times$ is the subgroup of invertible elements of $\End(c)$.
The reason for this is that left Kan extension along $\{c\}\to C$ factors as left Kan extension along $\{c\}\to B{\End(c)}$, and then left Kan extension along $B{\End(c)}\to C$, but the latter is a full subcategory inclusion, therefore left Kan extension along it is one as well, so if you're looking at the category of people that are left Kan extended from $\{c\}$ (from a finite set I would assume, to avoid Eilenberg swindle type phenomena), you might as well look at the same subcategory, but in $\Fun(B{\End(c)},\Set)$, so you might as well assume $C = B{\End(c)}$. Then you can notice that a good description of this category is just finite copies of $\End(c)$ with its $\End(c)$-action, and the core-groupoid of that will just be the same thing but for $\End(c)^\times$.
Therefore you get exactly the same situation as for a group : $K(\Fun^0(C,\Set)) = \Sigma^\infty_+(B(\End(c)^\times))$ where my $B$ is your $\lvert\Nerve({-})\rvert$.
Now the reason for that thing is that when $C$ is a groupoid (e.g. $B(\End(c)^\times)$), (the nerve of) $FC$ happens to also be the free symmetric monoidal $\infty$-groupoid on $C$, i.e. the free $E_\infty$-space on $\lvert\Nerve(C)\rvert$; hence if you apply group completion to it, you get the free grouplike $E_\infty$-space on it, in other words (up to delooping) the free connective spectrum on it. But that is precisely $\Sigma^\infty_+\lvert\Nerve(C)\rvert$. As Neil pointed out, this gets you the tom Dieck splitting, and in fact with the right setup for $G$-spectra it can give you an "equivariant BarrattβPriddyβQuillen" theorem.
Certainly this idea is present in Segal's Categories and cohomology theories β where he gives a proof "along those lines" of the BarrattβPriddyβQuillen theorem. A modern reference where this idea is explicitly used that way to prove the tom Dieck splitting is Barwick's Spectral Mackey functors and equivariant algebraic $K$-theory (I), specifically theorem A.9. His argument can be generalized to describe the free $E_\infty$-space on a $1$-groupoid.
|
2025-03-21T14:48:29.654857
| 2020-01-16T19:29:12 |
350579
|
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|
Stack Exchange
|
Reconstructing probability distribution with high probability
Sample $m$ times from unknown probability distribution $p=(p_1,p_2,\cdots,p_n)$, we can construct a probability distribution $q=(q_1.q_2,\cdots,q_n)$.
How large $m$ should be to achieve that the probability of $||p-q||_2<\epsilon$ is at least $1-\delta$? We are particularly interested in the case of $\delta=1/n^2$.
$\newcommand\ep{\epsilon}$
$\newcommand\ch{\operatorname{ch}}$
This answer is based on Theorem 3 and formula (2), which yield
$$P(|S_m|\ge x)
\le2\exp\Big(-tx+\sum_{i=1}^m E(e^{t|X_i|}-t|X_i|-1)\Big), \tag{0}
$$
where $x$ and $t$ are any nonnegative real numbers, $S_m:=X_1+\dots+X_m$, and $X_1,\dots,X_m$ are any independent zero-mean random vectors in any separable Hilbert space $H$ with norm $|\cdot|$. Inequality (0) is also a special case of Theorem 3.1 for martingales in $(2,D)$-smooth separable Banach spaces; note that a Hilbert space is $(2,1)$-smooth.
In our case, let $H=\mathbb R^n$ with $|\cdot|:=\|\cdot\|_2$ and let
$$X_i:=(1_{Y_i=1}-p_1,\dots,1_{Y_i=n}-p_n),$$
where $Y_1,\dots,Y_m$ are iid random variables (r.v.'s) such that $P(Y_i=j)=p_j$ for $j\in[n]:=\{1,\dots,n\}$ and $p_1+\dots+p_n=1$. Then we can write
$$S_m=m(q-p),$$
where $p$ and $q$ are as in the OP.
So, for all $\ep\in(0,1/2]$ and all real $t\ge0$,
$$P(|q-p|\ge\ep)=P(|S_m|\ge m\ep) \\
\le2\exp\Big(-tm\ep+\sum_{i=1}^m E(e^{t|X_i|}-t|X_i|-1)\Big). \tag{1}
$$
Next, almost surely (a.s.)
$$|X_i|^2=\sum_{j=1}^n1_{Y_i=j}-2\sum_{j=1}^np_j1_{Y_i=j}+\sum_{j=1}^np_j^2
\le1-2\times0+\sum_{j=1}^np_j^2\le2 $$
and
$$E|X_i|^2=1-\sum_{j=1}^np_j^2\le1.$$
Also, $r(u):=\frac{e^u-u-1}{u^2}$ is increasing in real $u$ (with $r(0):=1/2$), whence
$$r(u)\le r(\sqrt 2/2)=:c=0.64\ldots$$
for $u\le\sqrt 2/2$. Letting now $t_*:=\frac\ep{2c}<\ep\le1/2$, we have $t_*|X_i|\le\sqrt2/2$ a.s. So, a.s. for all $i\in[m]$
$$e^{t_*|X_i|}-t_*|X_i|-1=r(t_*|X_i|)t_*^2|X_i|^2\le c t_*^2|X_i|^2.
$$
Hence, by (1),
$$P(|q-p|\ge\ep)\le2\exp\Big(-t_*m\ep+\sum_{i=1}^m c t_*^2E|X_i|^2\Big) \\
\le2\exp(-t_*m\ep+mct_*^2)
=2\exp\Big(-\frac{m\ep^2}{4c}\Big).
$$
So, for any $\delta>0$,
$$P(|q-p|<\ep)\ge1-\delta
$$
if
$$m\ge\frac{4c}{\ep^2}\,\ln\frac2\delta $$
and $\ep\in(0,1/2]$. (Replacing $1/2$ in the condition $\ep\in(0,1/2]$ by a small enough positive real number, we can replace $c$ by a constant factor however close to $1/2$.)
Based on this cstheory post's argument from ClΓ©ment Canonne[1], for $\delta = O(1)$ it suffices to draw $m = O(1/\epsilon^2)$ samples. This can be extended to show that
$$ m = O\left(\frac{\log(1/\delta)}{\epsilon^2}\right) $$
samples suffices.
Slightly more details, which are worked out thanks to ClΓ©ment / folklore in Theorem D.2 of my paper[2]: We can use the cstheory argument to show if $m \geq \frac{4}{\epsilon^2}$, then $\mathbb{E}\|p-q\|_2^2 \leq \frac{\epsilon^2}{4}$. Jensenβs gives $\mathbb{E}\|p-q\|_2 \leq \frac{\epsilon}{2}$. We can use McDiarmid's concentration inequality to show that if $m \geq \frac{4\ln(1/\delta)}{\epsilon^2}$, then $\|p-q\|_2$ is within $\frac{\epsilon}{2}$ of its expectation w.p. $1-\delta$. This combines to show $\|p-q\|_2$ is at most $\epsilon$ w.p. $1-\delta$.
Itβs surprising at first that this doesnβt depend on the support size n (unlike L1 learning), which was the motivation for [2].
[1] https://cstheory.stackexchange.com/a/18498/8243
[2] https://arxiv.org/abs/1412.2314
Thanks for answering!
Indeed, the result may well be folklore. If you're looking for earlier published results, then there's Iosif's paper linked below in his answer, and also
https://www.sciencedirect.com/science/article/abs/pii/S0167715213000242
.
I am assuming you are using the empirical distribution as in the comment by @MattF.
This is a very broad question, and the answer changes quite a bit according to what properties the true distribution $p$ has. Note that if $\min_{i} p_i=\epsilon,$ which is very small, then you need a very large number of samples before you actually obtain the value $i,$ by sampling. If there are a few large probabilities, they will dominate the samples.
A good place to start is Valiant and Valiant's paper Instance Optimal Learning of Discrete Distributions available here. Also have a look at the references there. Most people focus on the distance $\mid\mid p-q\mid\mid_1$ which is directly related to the maximal query probability of distinguishing $p$ from $q.$ Related work includes estimating entropy and other properties of distributions.
Historically, the Good-Turing method is where all this stuff originates.
Edit: The paper by Indyk et al here seems to provide an answer which is to a question very close to your question. The techniques there should help. Basically they state that, for $\ell_2$ distance $\leq \varepsilon$ between $p$ and $q$ one needs $\widetilde{O}((n/\varepsilon)^2 \log n)$ samples and similar running time. The definition of successful approximation there is the standard one in complexity theory, of probability of error being $\leq 1/3.$
|
2025-03-21T14:48:29.655203
| 2020-01-16T19:55:12 |
350581
|
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|
Stack Exchange
|
Confused about A. Kosinski's description about surgery in his book "differential manifolds"
Please excuse me, if MO is not the proper place for this question. I aksed the same question on M.SE
https://math.stackexchange.com/questions/3511134/confused-about-a-kosinskis-description-of-surgery-in-his-book-differential-ma
but i am not sure, whether MO is not the better place to adress this. I am really struggling to understand Kosinski's description of surgery on a manifold.
On p.112 in Kosinski's "Differential Manifolds" he introduces Surgery on a $(\lambda-1)$-Sphere in a manifold $M^m$. He says
Surgery on a $(\lambda-1)$-Sphere in a manifold $M^m$ is a special
case of pasting. We paste $M$ and $S^m$ along $S$ and $S^{\lambda-1}$.
The resulting manifold will be denoted $\chi(M,S)$. (...) it can be
described as follows:
Let $T' = \{x \in S^m \mid x_\lambda^2 > 0\};$ we view $T'$ as a
tubular neighborhood of $S^{\lambda-1}$ in $S^m$. Let $h: T' \to M$ be
a diffeomorphism, $h(S^{\lambda-1}) = S$. Then $$\chi(M,S) =
(M\setminus S) \cup_{h\alpha} (S^m \setminus S^{\lambda-1})$$
remark: $\alpha$ is the composition of the diffeomorphism $D^m \setminus S^{\lambda-1} \to \mathring{D}^\lambda \times D^{m-\lambda}$ and the involution on $(\mathring{D}^\lambda \setminus \boldsymbol{0})\times D^{m-\lambda}$
He then continues:
Note that the operation of attaching a $\lambda$-handle along $S$
becomes, when restricted to the boundaries, precisely surgery on $S$.
This can be conveniently stated as follows. Consider $h$ as an
embedding of $T'$ in $M \times \{1\} \subset M \times I$ and attach a
$\lambda$-handle to $M\times I$ along $S$. Let $W = (M\times I)\cup
H^\lambda;$ $W$ is called the trace of the surgery.
My question: whenever i read about surgery on a $m$-Manifold $M$, it's always described as cutting out $S^n\times D^{m-n}$ and gluing in $D^{n+1}\times S^{m-n-1}$ (see Ranicki's Surgery Theory) or any other source about surgery theory.
I simply can't get behind the way Kosinski describes this procedure. Where exactly do we remove $S^n\times D^{m-n}$ and glue in $D^{n+1}\times S^{m-n-1}$ ?
The way I understand Kosinski's approach is that we remove the embedded $(\lambda-1)$-sphere from $M$ and $S^m$ simultaneously and past them along the tubular neighborhoods of the embedded sphere $S^{\lambda-1}$... whilst i do recognize $S^m$ being $$S^m =\partial D^{m+1} = \partial (D^\lambda\times D^{m-\lambda+1}) = S^{\lambda-1}\times D^{m-\lambda+1} \cup D^\lambda \times S^{m-\lambda}$$
i'm still failing to see the link between Kosinski's definition of surgery and the common definition (as of Ranicki) i've stated.
On p.142 Kosinski even mentions himself:
"Surgery is informarlly described as "taking out $S^k\times D^{n+1}$
and gluing in $D^{k+1}\times S^n$."
But i fail to see how this relates to his definition i've stated.
Can anyone help me understanding how they're related or what i might not seeing here?
thank you very much
The purpose of Kosinski's approach is that the classical handle attachment language puts you in a category of manifolds with corners, and he wants to avoid that. I would suggest reading Kosinski's description of handle attachment first. Once you are comfortable with that, the surgery description follows.
thank you @RyanBudney. I will do so.
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2025-03-21T14:48:29.655452
| 2020-01-16T21:15:58 |
350586
|
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|
Stack Exchange
|
Imprecise Definition of a $\sigma$-algebra
I'm reading some works on the hierarchical model in statistical mechanics and I came across an strange definition, which I need to clarify. Consider a finite set $\Lambda \subset \mathbb{Z}^{d}$. The set of all functions $\varphi : \Lambda \to \mathbb{R}$ is isomorphic to $\mathbb{R}^{|\Lambda|}$, so that these functions are represented as vectors $\varphi = (\varphi_{x})_{x\in \Lambda} \in \mathbb{R}^{|\Lambda|}$, and are called fields. Given $\xi_{1},...,\xi_{N} \in \mathbb{R}^{|\Lambda|}$, we set $\varphi_{j} := \sum_{k> j}\xi_{k}$ for each $j=0,...,N-1$ and $\varphi_{N} = 0$.
Now the author states the following: "Given $X\subset \Lambda$, let $\mathcal{N}_{j}(X)$ be the algebra of functions measurable with respect to the $\sigma$-algebra generated by $\{\varphi_{j}(x), \hspace{0.1cm} x \in X\}$. In more down to earth terms, an element of $\mathcal{N}_{j}(X)$ is a function only of fields at points $x \in X$."
I don't understand what is this $\sigma$-algebra. It seems confusing to me. What does it mean?
Edit: The text can be found in
math.ubc.ca/~db5d/Seminars/PCMILectures/lectures.pdf
As a rule for this kind of questions it is good to indicate exactly the reference you are reading. I guess this is the book by Bauerschmidt, Brydges and Slade, correct?
Actually is the Brydges lecture notes on RG. I will add it to the post.
$\newcommand\vpi{\varphi}$
$\newcommand\La{\Lambda}$
$\newcommand\R{\mathbb R}$
You misunderstood the notes: For each $j\in\{0,\dots,N\}$ and each $x\in\La$, $\vpi_j(x)$ is (not a real number but) a real-valued random variable (r.v.).
Indeed, (i) formula (2.17) on page 25 in the notes implies that $\vpi_j=\sum_{k=j+1}^N\zeta_j$ (with $\vpi_N=0$), (ii) line 2 on page 22 there tells us that $\zeta_j\sim N(C_j)$, and (iii) by formula (1.20) on page 10, $\zeta_j\sim N(C_j)$ means that $\zeta_j=(\zeta_j(x)\colon x\in\La)$ is a random zero-mean Gaussian vector (with values in $\R^\La$ and) with the covariance matrix $C_j=(C_j(x,y))_{(x,y)\in\La^2}$.
So, for each $j\in\{0,\dots,N\}$ and each $X\subseteq\La$, the $\sigma$-algebra in question is the $\sigma$-algebra generated by the set $\{\varphi_j(x)\colon x\in X\}$ of real-valued r.v.'s, that is, the smallest $\sigma$-algebra with respect to which all the r.v.'s $\varphi_j(x)$ with $x\in X$ are measurable.
Thanks for your answer! Now I can spot my misunderstanding. But then, for each $j $, $\varphi_{j}(x) $ is a function, right? What is that function? I don't seem to understand how this emerge from $\varphi_{j}=\sum_{k>j} \xi_{k}$.
@Willy.K : For each $j$ and each $x$, $\zeta_j(x)$ is a real-valued random variable defined on a probability space $(\Omega,\mathscr F,\mathbb P)$. That is, $\zeta_j(x)$ is an $\mathscr F$-measurable function from $\Omega$ to $\mathbb R$: $\Omega\ni\omega\mapsto\zeta_j(x)(\omega)\in\mathbb R$. Therefore, for each $j$ and each $x$, $\varphi_j(x)$ is an $\mathscr F$-measurable function from $\Omega$ to $\mathbb R$. (Just as in these notes, oftentimes the probability space $(\Omega,\mathscr F,\mathbb P)$ remains in the background and is not mentioned explicitly.)
It's getting clearer to me. There is just one thing that is bothering me: he seems to be using variables $\varphi$ interchangeably as a vector in $\mathbb{R}^{|\Lambda|}$ and as a random vector. For example, in page 10 he writes that, for a positive-definite matrix $A$ in $\mathbb{R}^{|\Lambda|}$, $\mu := \mbox{const} e^{-\frac{1}{2}\langle \varphi, A^{-1}\varphi\rangle}$ is a Gaussian measure. Thus $\varphi \in \mathbb{R}^{|\Lambda|}$ as I understand. Then, he writes $\varphi \sim N(C)$, where $C=A^{-1}$ and here $\varphi$ is a random vector $\varphi: \Omega \to \mathbb{R}^{\Lambda}$.
Is this just abuse of notation or both things are essentially the same?
@Willy.K : I also noticed that. It is indeed abuse of notation, which may confuse readers.
Oh, good to know! I thought I was missing something!
Disclaimer: maybe this would be more appropriate as a comment since I do not know the cited text, but I am posting this as an answer because I do cannot yet comment:
My understanding would be that it is meaning the $\sigma$-algebra generated by the restriction $\phi_j|_{X}$, that is the $\sigma$-algebra generated by all the sets $C \subseteq \Lambda$ such that $C=\{x\,: \, \phi_j|_X(x) \in C\}$.
By a standard result, a real function $g$ is measurable with respect to a $\sigma$-algebra generated by another real function $f$ if and only if there exist a measurable function $h:\mathbb{R} \to \mathbb{R}$ such that $g=h \circ f$.
So in this case, if the definition is the one I conjectured, then $f \in \mathcal{N}_j$ if and only if there is a measurable $h$ such that $f=h \circ \phi_j|_X$, that is, $f$ would depend only on the values of $\phi_j$ on the set $X$, as the text says. Cannot help with the physics though :)
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2025-03-21T14:48:29.656056
| 2020-01-16T23:02:47 |
350590
|
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|
Stack Exchange
|
Lifting $G$-invariants from characteristic $p\gg 0$ to characteristic 0 for a reductive algebraic group $G$
Let $S\subset \mathbb{C}$ be a finitely generated ring, let $R$ be a finitely generated commutative ring over $S$. Let $G$ be a linear algebraic group over $S$, such that $G_{\mathbb{C}}$ is reductive. Suppose that Spec$(R)$ is equipped with a $G$-action over $S$.
In this setting, I hope that the following statement holds.
For any large enough prime $p$, given a base change
$S\to \bf{k}$ to an algebraically closed field of characteristic $p,$ then $G_{\bf{k}}$-invariants of $R_{\bf{k}}$ are generated by
the image of $G$-invariants of $R.$
It feels that the above statement is either explicitly well-known, or at least should follow from a well-known result. Any suggestions or references will be greatly appreciated.
The hypothesis that $G_{\mathbb{C}}$ is reductive looks too weak. One would prefer to have $G$ reductive over $S$ in the sense of SGA3. That is, one wants $G$ to be smooth over $S$ with geometric fibers that are connected reductive.
Yes, thanks! I am certainly fine with making the assumptions you suggested.
Perhaps this should be viewed as a question about representation theory of $G$; here much but not everything is known, especially for $p$ large enough.
Related to https://mathoverflow.net/questions/176393/git-over-integers/176562#176562
Indeed it is! Thank you very much!
We offer two facts and a Theorem.
Let $S$ be a commutative noetherian ring containing $\mathbb Z$ and let $G=G_S$ be
reductive over $S$ in the sense of SGA3. That is, $G$ is smooth over $S$
with geometric fibers that are connected reductive.
Let $R$ be a finitely generated commutative $S$-algebra.
Suppose that $\mathrm{Spec}(R)$ is equipped with a right
$G$-action over $S$.
Fact 1 Let us be given a base change $S\to \bf k$ to a field of positive characteristic $p$.
For every $x\in (R\otimes_S {\bf k})^G$ there is an $r\geq1$ so that $x^{p^r}$ lies in the $\bf k$-span of the image
of $R^G$.
Fact 2 For almost all primes $p$ the map $R^G\to (R/pR)^G$ is surjective.
Remark. We do not need to distinguish between $(R/pR)^G$ and $(R/pR)^{G_{S/pS}}$. Base change of the group
is unnecessary when computing invariants or cohomology. (See 1.13. Restriction in our survey
Reductivity properties over an affine base
.)
Theorem Assume further that $R$ is flat over $S$ and that $S$ is of finite global homological dimension.
Then there is an integer $n\geq1$ so that if $S[1/n]\to \bf k$ is a base change to a field, then the map
$R^G\otimes_S{\bf k}\to (R\otimes_S{\bf k})^{G_{\bf k}}$ is surjective.
To prove Fact 1, let $D$ be the image of $S$ in $\bf k$. As $D\to \bf k$ is flat, we have $(R\otimes_S {\bf k})^G=(R\otimes_S D)^G\otimes_D {\bf k}$,
so it suffices to show that for every $x\in (R\otimes_S D)^G$ there is an $r\geq1$ so that $x^{p^r}$ lies in the the image
of $R^G$. Now $G$ is power reductive, so we may apply Proposition 41
of our paper
Power Reductivity over an Arbitrary Base
.
See also my survey
Reductivity properties over an affine base
.
To prove Fact 2, recall from Theorem 10.5 of
Good Grosshans filtration in a family
that $H^1(G,R)$ is a finitely generated module over $R^G$.
This module is also $\mathbb Z$-torsion.
To see this, first take an fppf base change to reduce to the case that $G$ is split over $S$ (see SGA3).
Then $G_{\mathbb Q}$ makes sense and
$H^1(G,R)\otimes_{\mathbb Z}{\mathbb Q}=H^1(G_{\mathbb Q},R\otimes_{\mathbb Z}{\mathbb Q})=0$.
Choose $n\geq1$ so that $n$ annihilates the generators of $H^1(G,R)$. Choose $m\geq1$ so that $m$
annihilates the generators of the $\mathbb Z$-torsion ideal of $R$.
Now if $p$ does not divide $mn$, then $\partial$ vanishes in the exact sequence
$0\to R^G\stackrel{\times p}\to R^G\to (R/pR)^G\stackrel\partial\to H^1(G,R)$.
Now we turn to the proof of the Theorem. If $G$ is split over $S$ then we may apply
Remark 31 and Theorem 33 of
Power Reductivity over an Arbitrary Base
to obtain $n$ so that $H^i(G,R[1/n])$ vanishes for $i\geq1$.
If $G$ is not yet split the same result is true. Indeed we may by SGA3 do an fppf base change $S\to T$ so that $G_T$ is split over $T$.
Then we may arrange that $H^i(G,R[1/n])\otimes_ST=H^i(G_T,R[1/n]\otimes_ST)$ vanishes for $i\geq1$. This implies that $H^i(G,R[1/n])$ vanishes for $i\geq1$.
Having chosen $n$ this way we now claim that for every $S$-module $N$ with trivial $G$ action, the
$H^i(G,R[1/n]\otimes_SN)$ also vanish for $i\geq1$. This is clear if $N$ is free and then it follows by induction on the projective dimension of the $S$-module $N$.
(If $0\to N'\to F \to N\to0$ is exact, with $F$ free, consider the long exact sequence for $G$-cohomology associated with the exact sequence
$0\to R[1/n]\otimes_SN'\to R[1/n]\otimes_SF \to R[1/n]\otimes_SN\to0$.)
Now let us be given a base change $S[1/n]\to \bf k$ to a field. Let $N$ be the kernel of $S[1/n]\to \bf k$ and let $D$ be its image.
Note that $0\to R\otimes_SN\to R\otimes_SS[1/n] \to R\otimes_SD\to0$ is exact and that $R\otimes_SN=R[1/n]\otimes_SN$.
As $H^1(G,R[1/n]\otimes_SN)=0$ we have a surjection $(R\otimes_SS[1/n])^G\to (R\otimes_SD)^G$.
As $D\to \bf k$ is flat, $(R\otimes_S{\bf k})^{G_{\bf k}}=(R\otimes_SD)^G\otimes_D{\bf k}$.
We see that $(R\otimes_SS[1/n])^G\otimes_S{\bf k}$ maps onto $(R\otimes_S{\bf k})^{G_{\bf k}}$. But $(R\otimes_SS[1/n])^G\otimes_S{\bf k}$ equals
$R^G\otimes_SS[1/n]\otimes_S{\bf k}=R^G\otimes_S{\bf k}$. The result follows.
Thanks, this is great!
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2025-03-21T14:48:29.656413
| 2020-01-16T23:16:47 |
350592
|
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|
Stack Exchange
|
Do topos-valued sheaves form a topos?
Let $\bf C$ be a category, $\mathcal S$ an (elementary) topos.
If $\mathcal S$ is a presheaf category over $\bf D$, then it's easy to see $[\mathbf C^{\rm op},\, \mathcal{S}] \cong [(\mathbf C \times \mathbf D)^{\rm op},\, \mathcal{Sets}]$ is still a topos. In more general situations I struggle to see an easy reason for it to be true as well. Thus:
When is the category $[\mathbf C^{\rm op},\, \mathcal S]$ of contravariant functors from $\bf C$ to $\mathcal S$ a topos?
If $\mathcal{S}$ is a Grothendieck topos, or more generally has large enough disjoint universal coproduct you are always fine. But If $\mathcal{S}$ is an elementary topos in general it is not going to work, but I do not know if there are nice conditions under which it works.
Can you elaborate/give references for these claims?
If $\mathbf C$ were an internal category in $\mathcal S$, then the category of internal $\mathcal S$-valued presheaves on $\mathbf C$ would be a topos. (This is surely in Johnstone's "Topos Theory".) I think the point of @SimonHenry's comment is that good coproducts let you regard any genuine small category as an internal category in $\mathcal S$.
That was indeed where I was going. But the question seem much more interesting that this observation: For example if the category C is a groupoid I think that S-valued presheaf will be an elementary topos without needing any assumption on S, so the general question definitely do not reduce to the case I was refering too.
@SimonHenry presumably there's something happening with colimits indexed by each connected component of $\mathbf{C}$?
@DavidRoberts : Maybe, but that shouldn't be enough: $C$ conected isn't enough I think.
@SimonHenry I mean that even if the category is connected and even skeletal, then colimits indexed by it require (if one is being naive) a coproduct indexed by the objects. Though I'm slightly worried about getting the full coqualiser diagram, so maybe needing the set of arrows of each connected component to be small enough to get coproducts (or maybe colimits) of that size in the topos ...?
The category of $G$-sets for a large group $G$ is a cocomplete elementary topos but is not a Grothendieck topos. [1]
@ZhenLin good to see you again! Yes, that's definitely relevant, and that example generalises a lot to other special large limits of toposes.
Right. I think it settles the groupoid case, anyway. For the category case, we probably want the category to be "locallly essentially small" in the sense that each slice is equivalent to a "small" category (relative to $\mathcal{S}$). Just setting up a context where we can compare "large" external categories and internal categories in $\mathcal{S}$ seems to be tedious but I think it can be done if $\mathcal{S}$ has pullback-stable unions of "large" disjoint families of subobjects β we use extensivity to replace morphisms $A \to \coprod_X 1$ with decompositions of $A$ indexed by $X$, etc.
(cont.) An example of a "locally essentially small" category in this sense would be the poclass of ordinal numbers $\mathbf{ON}$, and $[\mathbf{ON}^\mathrm{op}, \mathbf{Set}]$ is another example of a non-Grothendieck topos.
Rather than a full-fledged answer, this is a sketch of a plan:
As you yourself pointed out, if $\mathcal S=[\bf D^{\rm op},\, \mathcal \cal{Sets}] $, the result is trivial.
Now, suppose you consider an arbitrary elementary topos $\mathcal S$. The steps would be:
Use the representation theorem of Joyal-Tierney to embed your target topos as the equivariant sheaves over a localic topos, see here.
$E:\mathcal S \to Set^{\mathcal{L}^{op}}$
Now each map $[\bf C^{\rm op},\, \mathcal S]$ composes with the embedding $E$ and thus lands into a localic topos.
The last step would be to see how $[\bf C^{\rm op},\, E(\mathcal S)]$ sits inside $[\bf C^{\rm op},\, Set^{\mathcal{L}^{op}}]$.
Conjecture: under some some conditions (to be determined, but see this post When is a reflective subcategory of a topos a topos? ) there is a reflection which is sufficiently exact to ensure that it is indeed a subtopos of $[(\bf C \times \bf\mathcal{L})^{\rm op},\, \cal{Sets}]$
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2025-03-21T14:48:29.656831
| 2020-01-16T23:31:36 |
350593
|
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Stack Exchange
|
Conformal mappings for domains whose complement is totally disconnected
Consider a conformal homeomorphism $f \colon \Omega \to \Omega'$ between domains $\Omega, \Omega'$ in the plane. Let $E = \mathbb{R}^2 \setminus \Omega$ and $E' = \mathbb{R}^2 \setminus \Omega'$. My understanding is that you can have a situation where $E$ is totally disconnected and yet $E'$ is not. If we think of the map $f$ as extending to a map between the connected components of $E$ and the connected components of $E'$, then $f$ would take a point $x \in E$ to a non-trivial continuum in $E'$.
Is there a standard, more-or-less explicit example of such a set $E$ and map $f$?
The only example I know of a conformal map that "stretches" a point boundary component to a nondegenerate continuum was constructed by Gehring and Martio in Quasiextremal distance domains and extension of quasiconformal mappings, J. Analyse Math. 45 (1985), 181β206. See Theorem 4.1.
The authors use earlier results of Ahlfors and Beurling to construct a Cantor set $E$ of positive area whose complement is mapped conformally to a domain whose boundary components are single points and the unit circle. The construction is based on the following classical result relating the existence of such a map with extremal length considerations:
Theorem
Let $\Omega \subset \mathbb{C}$ be a domain and let $z_0 \in \partial \Omega$ be a point boundary component. The following are equivalent :
There is a conformal map on $\Omega$ that sends $z_0$ to non-degenerate boundary component of $f(\Omega)$;
There exists $r>0$ such that $\operatorname{mod}(\Gamma) < \infty$, where $\Gamma$ is the family of all closed curves in $\Omega \cap B(z_0,r)$ having nonzero winding number around $z_0$.
The idea is to start with a single point, say $0$, and surround it by sufficiently many point boundary components in order to block the curves winding around $0$ so that the modulus is finite. This implies that $0$ is mapped by a nondegenerate continuum under some conformal map $f$.
Lastly, let me mention that your question, more specifically the existence of such a set $E$ which is small (at least zero area), is related to a very difficult conjecture of He and Schramm on the rigidity of circle domains in Koebe's uniformization conjecture. See Section 9 of my paper with Ntalampekos
D. Ntalampekos and M. Younsi, Rigidity theorems for circle domains, Invent. Math. 2019
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2025-03-21T14:48:29.657014
| 2020-01-17T00:24:24 |
350595
|
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|
Stack Exchange
|
Optimization algorithm sought
Suppose I have $N$ pairs of positive numbers $(a_1, b_1), (a_2, b_2), \dotsc, (a_N, b_N).$ and I want to find a subset of $M$ of them maximizing
$$
\frac{\sum_{j=1}^M a_{i_j}}{\sum_{j=1}^M b_{i_j}}.
$$
Can this be done in polynomial time?
This is a mediant sum. Gerhard "Pick The M Biggest Ratios" Paseman, 2020.01.16.
@GerhardPaseman: "Pick The M Biggest Ratios" doesn't always work. In the case N=3, M=2, and pairs (80,100), (20,100), and (1,20), it is best to pick the first and third pairs, even though the first and second pairs have the M biggest ratios. See my answer for a link to a good discussion.
The paper "Dropping Lowest Grades" by Daniel M. Kane and Jonathan M. Kane addresses this question in the context of dropping $r$ quiz grades from a collection of weighted grades.
The solution described there is fundamental the same as that in David Eppstein's answer. However, there is also a discussion of a practical implementation that may be even faster in practice.
Great reference, thanks!
I had forgotten, but your answer reminded me: there's a linear time algorithm for the original question (motivated by the same context of dropping lowest grades) in D. Eppstein and D. S. Hirschberg, "Choosing subsets with maximum weighted average", J. Algorithms 24: 177β193, 1997, https://doi.org/10.1006/jagm.1996.0849, https://www.ics.uci.edu/~eppstein/pubs/EppHir-TR-95-12.pdf
It is equivalent to look for the largest positive value $x$ such that, for some $M$-subset, $\sum (a_i-x b_i)\ge 0$.
Plot the $n$ lines $y = a_i - x b_i$ in the plane. The $M$-subset that maximizes $\sum (a_i-x b_i)$ for a given $x$ is the one defined by the $M$ lines from this arrangement that have the highest crossings with a vertical line through $x$. (If you trace out, for each $y$, the $M$th-from-top line in this arrangement, the trace is a polygonal curve describing the optimal subset for each $x$.) You can use a binary search among the crossings of the arrangement to find the largest $x$ whose optimal subset is good enough.
This immediately gives you $O(n^2)$ time but by using techniques from https://en.wikipedia.org/wiki/Theil%E2%80%93Sen_estimator to find medians of sets of crossings more quickly you can reduce it to $O(n\log^2 n)$.
I don't know about polynomial time, but here's an integer linear programming formulation. For $i\in\{1,\dots,N\}$, let binary decision variable $x_i$ indicate whether pair $(a_i,b_i)$ is selected. The problem is to maximize $$\frac{\sum_{i=1}^N a_i x_i}{\sum_{i=1}^N b_i x_i}$$ subject to
$$\sum_{i=1}^N x_i = M.$$
Now linearize the objective via the Charnes-Cooper transformation, as discussed here.
|
2025-03-21T14:48:29.657248
| 2020-01-17T02:15:40 |
350599
|
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|
Stack Exchange
|
A type of principal ideal theorem of class field theory for ramified primes
Let $K$ be a number field and $\mathcal{O}_K$ be its ring of integers. Also let $p$ be a prime number, $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_K$ and $\zeta_{m}$ be a primitive mth root of unity. I have two questions:
When we work over $β\mathbb{Q}$, the (finite) abelian extensions of $β\mathbb{Q}$ which are unramified outside prime $p$ are subfields of $L=β\mathbb{Q}(\zeta_{p^{r}})$ for some integer $r$. In fact, the ray class field of $β\mathbb{Q}$ associated to the modulus $p^r$ is $L=β\mathbb{Q}(\zeta_{p^{r}})$. Then, by a generalization of classical principal ideal theorem for Hilbert class field (see for example this page of MO), any unramified prime of $\mathbb{Q}$ (all primes other than $p$) becomes principal in $L=β\mathbb{Q}(\zeta_{p^{r}})$. But for $p$, which is ramified, we have $p\mathcal{O}_L = (1-\zeta_{p^{r}})^{[L:β\mathbb{Q}]}$, i.e. $p\mathcal{O}_L$ is principal and totally ramified.
I think in any number field $K$ instead of $\mathbb{Q}$, for any prime power modulus $\mathfrak{p}^r$, the ray class field modulo $\mathfrak{p}^r$, $L=K(\mathfrak{p}^r)$, has this property: the unique ramified prime of $K$ becomes principal and totally ramified in $L$, i.e. $\mathfrak{p}\mathcal{O}_L=(a)^{[L:K]}$ for some $a \in L$.
But I can't prove this, or find a theorem that say this.
When we work over $β\mathbb{Q}$, the ray class field for any modulus $m=p_1^{r_1}p_2^{r_2}...p_n^{r_n}$ is $L=\mathbb{Q}(\zeta_m)$. In this case ramified primes $p_i$ maybe don't become principal in $L$, but the product of all prime ideals of $L$ above $p_i$ is principal. In fact this product is equal to $(1-\zeta_{p_i^{r_i}})\mathcal{O}_L$. Remember that we saw unramified primes become principal in $L$.
I think in any number field $K$ instead of $\mathbb{Q}$, for any modulus $\mathfrak{m}$, the ray class field modulo $\mathfrak{m}$, $L=K(\mathfrak{m})$, has this property: for any (finite) $\mathfrak{p}|\mathfrak{m}$, the product of all prime ideals of $L$ above $\mathfrak{p}$ is principal.
But I can't prove this, or find a theorem that say this.
See https://mathoverflow.net/questions/63465/where-does-the-principal-ideal-theorem-from-cft-go and https://www.jstage.jst.go.jp/article/jjm1924/7/0/7_0_315/_pdf
@Franz Lemmermeyer: Thank you. I have read the page of MO that you mentioned and in my question a have sited to it. For Iyanaga's paper, I can't read German but it seems that Iyanaga prove the theorem for unramified primes, as you have said in mentioned page of MO. But, my question is about ramified primes.
My mistake. But if all primes except finitely many are principal, then all ideals are principal by the standard density theorems.
@Franz Lemmermeyer: Yes. But my question is about principalization in extensions. By Iyanaga's generalization of PIT, all but finitely many of primes of ground field become principal in any of its ray clas fields, because all but finitely many of primes of ground field are unramified. For ramified primes, if we take the ground field to be the rationals, when in a cyclotomic field at leat two primes ramified, then the prime ideals over them are not in general principal (but their product is, see my second question). Also for my first question, totally ramifiedness is not solved for me. Thanks
@Franz Lemmermeyer: Also if we adjoin the square root of -5 to rationals, then the only nonprincipal prime in its ring of integers is the one above 2, I think.
The answer to the first question is negative since the ray class field contains the Hilbert class field, which is unramified (and nontrivial for fiields with class number $> 1$). I strongly believe that the second claim is also not true in general; I'd look for counterexamples with quadratic base fields.
@Franz Lemmermeyer: Excuse me, I can't understand your argument for firts question. Please explan it. Also I mention that we have only one ramified prime in first question. (Sorry for my poor english)
It is not true that the prime ideal ${\mathfrak p}$ is totally ramified in the ray class field modulo ${\mathfrak p}^r$.
Yes. I think I should correct my question.
It is not true that the prime ${\mathfrak p}$ is totally ramified in the ray class field of a number field defined modulo ${\mathfrak p}$. For example, the ray class field modulo $3$ over the rationals is trivial, and the ray class field of $K = {\mathbb Q}(\sqrt{-6})$ modulo the prime ideal $(2, \sqrt{-6})$ above $2$ is equal to the Hilbert class field $K^1 = {\mathbb Q}(\sqrt{2}, \sqrt{-3})$ of $K$: it has conductor $1$, and $(2, \sqrt{-6})$ is not ramified.
For the second question, choose a number field $K$ and a prime ideal ${\mathfrak p}$ such that the prime ideal ${\mathfrak P}$ above ${\mathfrak p}$ in the Hilbert class field of $K$ is not principal. You are claiming that ${\mathfrak P}$ becomes principal in the ray class field modulo ${\mathfrak P}$ of $K$; to me it seems unlikely to be true for every choice of $K$ and ${\mathfrak P}$.
I think ray class field modulo 3 of rationals is cyclotomic field by 3rd root of unity. For $K$ and $K^1$ you'r right. So I think I should correct my question. I think I should say "conductor" instead of "modulus". In fact in my first question, I want unique ramified prime, and in second question I want arbitrary ramified primes (not unramified). Because in unramified case the Iyanaga's theorem (and other things related to my work) solve any thing that I want.
So, should I edit my question? (excuse me for my unknowing and my poor English).
Replacing modulus by conductor does not save the first question. Even in the class field with conductor ${\mathfrak p}^r$, the prime ideal {\mathfrak p}$ need not be totally ramified since the ray class field modulo any ideal contains the Hilbert class field.
Sorry for my inaccuracy. Yes and thank you very much. But if it be principal, it is still very good for me.
In fact I think this sentence is not true: "since the ray class field modulo any ideal contains the Hilbert class field". For example, in your example $K=\mathbb{Q}(\sqrt{-6})$, any extension (with prime power conductor, or not) of $K$ that its modulus is prime to ideals above $2$ and $3$ is not contains the Hilbert class field of $K$.
The Hilbert class field is the ray class field with conductor $(1)$. It is contained in any ray class field by the inclusion theorem for ray class fields.
Thank you (I don't remember a refrence for it. Please say one if possible. Thanks). But, as I said before, in my first question the principality is the most important thing for me and I know nothing about it.
Also, because, as you said, the only ramified prime $\mathfrak{p}$ of $K$ need not be totally ramified in any ray class field of K with conductor $\mathfrak{p}^a$ for some positive integer $a$, there is a new question: how many primes are there above $\mathfrak{p}$ in the ray class field? I hope there is only one.
If the prime ideal is principal, it will split in the unramified subextension of the ray class field.
Yes and thank you. But, there is still one question, the principality of our unique ramified prime in its ray class field. This is (if be true) the generalization of PIT for ramified primes (as I said it title). Also until now I can't find a soure for the inclusion theorem for ray class fields and if possible please introduce one.
|
2025-03-21T14:48:29.657698
| 2020-01-17T03:42:54 |
350605
|
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|
Stack Exchange
|
Looking for an efficient way of maximising 'pair scores' for subset of 30 selected from 50 to 10 000 objects
Context: I have a tiling program that uses a directed breadth first search algorithm. It is 'directed' by what I call 'pair scoring'. There are $N$ polyforms (pieces) used in the tiling. I have an $N\times N$ array with a score in every location. While tiling I seek to maximise the sums of pair scores of unused pieces. At the start I sum all the scores, find the mean score and subtract the mean from all scores so they sum to zero. As I tile I can find the new total just by adjusting the current total for the node by removing scores belonging to the last placed piece.
I generate the $N\times N$ 'pair score' array by tiling a small shape with around 6 to 10 pieces, for every tiling found I look at the border cells of every piece '$a$' in the tiling, and wherever it touches piece '$b$' I increment $[a,b]$ and $[b,a]$. And wherever it touches the edge of the shape I increment $[a,a]$.
This works fairly well, but I suspect that I could optimise the 'end game' further by encouraging a tendency to end with one of the 'local maxima', being the group(s) of pieces that 'score the best as a set'.
In order to know whether my algorithm is finding such maxima, I need to first find them myself so I know when the node scores are missing these maxima.
So my question is, how do I efficiently find the highest scoring sets of 1 through $M$ out of $N$ pieces, where $M$ ranges up to about 30 and $N$ up to 1000 or so?
There is some redundancy available in the array, I currently just repeat $[a,b]$ in $[b,a]$. I could use $[a,b]$ for the pair score when $a < b$ and $[b,a]$ for something else. I already use $[a,a]$ for keeping track of how many times piece a touches an edge square in a tiling of the smaller shape. I can adjust the importance of this edge score by using a simple 'edge score factor'.
It would be nice to have an algorithm that was fast enough to track these local maxima as they change due to pieces being used, without adversely impacting tiling speed too much. Then I would know pretty quickly that I had 'gone wrong' i.e. even though I had a nice high node score, I had used a set of pieces which would reduce my best possible end game, and could use that to direct the BFS in addition to the simple node score.
Or is it the case that the node score already gives me that?
Is the score symmetric, i.e. is the score of (a,b) the same as for (b,a)?
The question reduces to fixed cardinality maximum weight matching .
@ManfredWeis yes it's symmetric currently, but I could conceivably use the wasted space in the array for something else, in which case I would have to add a test for a > b when accessing the array. Thanks for the pointer to what to search for.
Bumped $16$ times by the Community Bot.
The stated problem is equivalent to determine a maximimum weight general matching of fixed cardinality; that problem can be solved efficiently as indicated in the answer to complexity of finding optimal matchings of given fixed size.
If however only a small number $M$ of pairs with minimum weight-sum has to be determined from a vast number $N$ of individuals, the $O(N^3)$ time complexity can render the exact algorithm not usable and heuristics are needed.
There are many heuristics that can be envisaged; the simplest one being the greedy strategy of choosing as tne next pair the one with maximal weight and then remove its elements from the set of candidates.
A more elaborate heuristic based on vertex weights could be:
determine vertex weights, that reflect the vertice's contribution to the edge weights
select the $2M$ top ranked vertices $T\subset V$
calculate the general maximum weight perfect matching $\mathfrak{M}_T$ of the subgraph induced by those $2M$ top ranked vertices
define $U$ to be set of vertices that have the smaller label of the pairs that are adjacent to an edge in $\mathfrak{M}_T$
determine the maximum weight maximal bipartite matching $\mathfrak{M}_{(U,\,V\setminus U)}$
explanations:
to have the vertex weights reflect their contribution to edge weights, they are choosen so that summing over the weights of edges, that are adjacent to the same vertex, equals the weights of their adjacent vertices; those conditions result in a system of equations that is explicitly solvable.
calculating $\mathfrak{M}_T$ only yields a lower bound on the optimal solution; an illustrative situation is a set of points chosen from a closed disk, with sufficiently many points on the boundary and we want find the $M$ disjoint edges with maximal length sum; it is clear the edges will connect points on the boundary, but choosing $2M$ points from the boundary is only a necessary criterion for finding $M$ approximately antipodal pairs.
The construction of the bipartite matching $\mathfrak{M}_{(U,\,V\setminus U)}$ from the edges of $\mathfrak{M}_T$ guarantees that the edge set of the bipartite matching contains the edges of $\mathfrak{M}_T$ and thus $\mathfrak{M}_{(U,\,V\setminus U)}$ is at least as good.
this answer is admittedly a bit sketchy, but a more specific description can be provided.
|
2025-03-21T14:48:29.658130
| 2020-01-17T05:58:03 |
350608
|
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|
Stack Exchange
|
Can base-change be non-surjective on Brauer groups?
Is there a finite-degree separable field extension $\mathbb{K} \subset \mathbb{L}$ such that the induced map on Brauer groups $\operatorname{Br}(\mathbb{K}) \to \operatorname{Br}(\mathbb{L})$ is not a surjection?
I assume the answer is yes. What is an example?
Can it ever happen for finite fields? For number fields?
For finite fields, the Brauer group is zero ( It comes from Wedderburn's theorem), so the answer is NO.
For number fields, the answer is YES. Following RP's question in the comments, I will prove the stronger statement that the map $Br(K)\to Br(L)^{Gal(L/K)}$ is not necessarily surjectve when $L/K$ is a Galois extension of number fields.
Take $K=\mathbb{Q}$, $L=\mathbb{Q}(i)$, and let $Q=(1+i,3)_L$.
We have $\overline{Q}\simeq (1-i, 3)_L$, and thus $Q\otimes_L\overline{Q}\simeq (2,3)_L\simeq (-2,3)_L$, since $-1$ is a square in $L$. Now $3=1^2-(-2)1^2$ is a norm in $L(\sqrt{-2})$ so $(-2,3)_L$ is split. Therefore, $Q\otimes_L\overline{Q}\sim 0$, and since $Q\otimes_LQ\sim 0$ (it is a quaternion algebra), we get $Q\sim \overline{Q}$ , thus $Q\simeq \overline{Q}$ for degree reason.
Now it is a well-known fact that $Q$ is defined over $K$ if and only if $Cor_{L/K}(Q)\sim 0$ (it is a result specific to quadratic extensions and algebras of exponent 2: in fact, if $L/K=K(\sqrt{d})$, we have an exact sequence
$H^1(K,\mu_2)\to H^2(K,\mu_2)\to H^2(L,\mu_2)\to H^2(K,\mu_2)$, where the maps are respectively cup-product by $(d),$ restriction, and corestriction.)
Now, we have $Cor_{L/K}(Q)\sim(N_{L/K}(1+i),3)_K=(2,3)_K$. Since $2$ is not a square mod $3$, the residue of $(2,3)_K$ at $3$ is non zero, hence $(2,3)_K$ is not split.
Consequently, the Brauer class of $Q$ lies in $Br(L)^{Gal(L/K)}$, but does not come from $Br(K)$.
So a more subtle question could be whether there are examples such that $\operatorname{Br}(K) \to \operatorname{Br}(L)^\Gamma$ is non-surjective, where of course $\Gamma$ is the Galois group.
I think there are counterexamples even for quadratic extensions. For exemple, if $L/K$ is quadratic with non trivial automorphism $$ and $Q$ is a quaternion $L$-algebra, being in $Br(L)^\Gamma$ means $Q\otimes Q^$ splits, that is $Res_{L/K}((Cor_{L/K})(Q))=0\in Br(L)$, or again that $ Cor_{L/K}(Q)$ is split by $L$, while being in the image of the restriction map means $Cor_{L/K}(Q)=0\in Br(K)$ (this is because $Q$ is a quaternion algebra, It does not work in general), that is $Cor_{L/K}(Q)$ splits over $K$. So the two things are different.
I am probably being dense, but why is the Galois invariance of the Brauer class the same as Q tensor Q* splitting? It's sort of counterintuitive to me, since by Hochschild-Serre I'd expect the cokernel of Br(K) -> Br(L)^\Gamma to live in some H^3, but your criterion would make it seem as if it was in some H^2 instead. But that's only using some general nonsense, you are probably using the specifics of the situation in some way I do not understand.
It is because quaternion algebras have exponent at most $2$, meaning $Q\otimes Q=0$ in the Brauer group.
I have edited my answer and gave some details, in order to answer your question.
Great answer, thank you!
i think you could leave your original answer and add the answer to the refined question.
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2025-03-21T14:48:29.658370
| 2020-01-17T06:06:25 |
350609
|
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|
Stack Exchange
|
Reference for computation of $K_8(\mathbb{Z})$
In an Oberwolfach report from 2016 [1, page 2] it is said that $K_8(\mathbb{Z})$ has recently been computed. Does anyone know a reference for the computation?
[1] https://orbilu.uni.lu/bitstream/10993/29499/1/preliminary_OWR_2016_52.pdf
There are two preprints available:
https://arxiv.org/abs/1910.11598
https://www.utsc.utoronto.ca/people/kupers/wp-content/uploads/sites/50/2021/01/k8zshorter.pdf
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2025-03-21T14:48:29.658426
| 2020-01-17T07:50:52 |
350614
|
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|
Stack Exchange
|
What's the role of commutation relations in stochastic mechanics?
In a stochastic context, we can understand a term like
$$ \int_0^T \frac{d q(t)}{dt} dq $$
either as the (Ito) limit
$$ \lim_{N\to\infty} \sum_{i}^N dq(t_i) \frac{d q(t_i)}{dt} $$
or the (Anti-Ito) limit
$$ \lim_{N\to\infty} \sum_{i}^N \frac{d q(t_i)}{dt} dq(t_i+\epsilon_i) , $$
where we divided the interval $[0,T]$ in equal slices with width $\epsilon$. Moreover, I've written the terms in a time-ordered manner and we have
$$\frac{d q(t_i)}{dt} \equiv \lim_{\epsilon \to 0 } \frac{q(t_i+\epsilon_i) - q(t_i)}{\epsilon} \, . $$
For Brownian motion we also have
$$ \langle \Delta q(t_i) \rangle \equiv \langle q(t_i+ \epsilon) - q(t_i) \rangle = \sqrt{\epsilon} \, . $$
This implies
\begin{align}
1 &= \lim_{\epsilon \to 0 } \frac{\epsilon}{\epsilon} \\
&= \lim_{\epsilon \to 0 } \langle \frac{(\Delta q(t_i) )^2}{\epsilon} \rangle \\
&= \lim_{\epsilon \to 0 } \langle \frac{( q(t_i+ \epsilon) - q(t_i) )^2}{\epsilon} \rangle \\
&=\lim_{\epsilon \to 0 } \langle \frac{ \Big( q(t_i+ \epsilon) - q(t_i) \Big)q(t_i+ \epsilon) - q(t_i) \Big( q(t_i+ \epsilon) - q(t_i) \Big) }{\epsilon} \rangle \\
&= \langle \frac{d q(t_i)}{dt}q(t_i) - q(t_i) \frac{d q(t_i)}{dt} \rangle
\end{align}
We can read off here
$$ \langle [\frac{d q(t_i)}{dt},q(t_i)] \rangle \equiv \langle \frac{d q(t_i)}{dt}q(t_i) - q(t_i) \frac{d q(t_i)}{dt} \rangle = 1 $$
This is quite similar to the canonical commutation relations in quantum mechanics. Thus I was wondering if commutation relations like this play a role in other (stochastic) contexts too? If yes, is there something analogous to the uncertainty relation for more general stochastic processes?
Yes, indeed, there is a close relationship between Ito's formula for stochastic integrals and the Heisenberg uncertainty principle. This connection is outlined in https://www.sciencedirect.com/science/article/pii/S0304414910000256 (Itoβs stochastic calculus and Heisenberg commutation relations, by P. Biane). For a generalization of such kind of uncertainty relation, see https://www.sciencedirect.com/science/article/pii/S0375960118303633 (Generalization of uncertainty relation for quantum and stochastic systems, by T. Koide and T. Kodama).
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2025-03-21T14:48:29.658582
| 2020-01-17T08:25:21 |
350615
|
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|
Stack Exchange
|
Holomorphic Spectrum
My question: is it true that we can define a spectrum of $\mathbb{C}$-algebra $A$ in such a way that it becomes a complex manifold with the algebra of holomorphic functions $A$?
Maybe it will work if we restrict to algebras that already are an algebra of holomorphic functions for some manifold?
I'm aware of the construction of Gelfand spectrum but it gives us only continuous functions on the spectrum and not holomorphic functions...
It is a cross-post from math.exchange https://math.stackexchange.com/questions/3510203/holomorphic-spectrum
If you crosspost on mathstackexchange and mathoverflow, please say so in your post and provide a link (on both sites).
@MaoWao thank you so much for explaining this rule!
In some cases the spectrum of a commutative Banach algebra $\mathcal A$ may contain "analytic disks" on which the Gelfand transforms of members of $\mathcal A$ correspond to analytic functions. You might look at section 1.5 of Andrew Browder, "Introduction to Function Algebras".
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2025-03-21T14:48:29.658825
| 2020-01-17T10:02:55 |
350619
|
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|
Stack Exchange
|
Is this a criterion for uniform distribution modulo one?
For all $k \in \mathbb N$ let $a_k$ be strictly positive bounded weights, i.e. there are constants $C_1$ and $C_2$ such that $0<C_1\le a_k \le C_2$. Now a real valued sequence $(x_k)_{k \in \mathbb N}$ satisfies $\lim_{n \to \infty}\frac{1}{n} \sum_{k=1}^n (a_k)^m e^{2\pi i x_km}=0$ for all non-zero integers $m$. Does this imply that the sequence $x_k$ is uniformly distributed modulo one?
It is similar to the Weyl criterion and would surely not be true if we consider just weights $a_k$ where the condition would only imply weighted uniform distribution modulo one. But with the weights $(a_k)^m$ depending on the parameter $m$ this could be true.
Any thoughts or potential counterexamples would be highly appreciated.
If the "for all" applies to the "strictly positive bounded weights $a_k$," then in particular you can take $a_k=1$ for all $k$ and just recover Weyl. So maybe you mean "there exist strictly positive bounded weights $a_k$"?
Yes, the "for all" applies only to the integers m, so I added "there exist" to be clear, thx.
The quantifier on $a_k$ should come before they are used. Can you restate more carefully?
Thx, I restated and tried to be clearer on the weights $a_k$.
I provide a sequence $(x_k)_k$ that is not uniformly distributed but satisfies $\frac{1}{N}\sum_{k \le N} a_k^m e^{2\pi i x_k m} \to 0$ as $N \to \infty$ for each $m \ge 1$, where $a_k = 1$ if $k$ is odd and $a_k = 3$ if $k$ is even. Let $x_k = 0$ if $k$ is odd. Then, for any $m \ge 1$, $\sum_{\substack{k \le N \\ 2 \nmid k}} e^{2\pi i x_k m} = \frac{1}{2}N+o(N)$, so it suffices to have, for any $m \ge 1$, $\sum_{\substack{k \le N \\ 2 \mid k}} e^{2\pi i x_k m} = \frac{-1}{2\cdot 3^m}N+o(N)$.
Rescaling, our general problem is to find a sequence $(x_k)_k$ such that $$\sum_{k \le N} e^{2\pi i x_k m} = \frac{-1}{3^m}N+o(N)$$ for each $m \ge 1$. I feel like there's a reference that guarantees this result, but let me say what I came up with (which has probably already been done somewhere). Let $f : \mathbb{T} \to \mathbb{C}$ be such that $\hat{f}(0) = 1$, $\hat{f}(m) = -\frac{1}{3^m}$ for $m \ge 1$, and $\hat{f}(m) = \overline{\hat{f}(-m)}$ for $m \le -1$. Since the fourier coefficients decay rapidly, $f$ is continuous. Also, $f$ is real, due to how the negative fourier coefficients are defined. In fact, $f(x) = 1-2\sum_{m \ge 1} \frac{1}{3^m}\sin(2\pi m x)$, which in particular shows $f$ is non-negative. Therefore, for each $N \ge 1$, we may take rational approximations $\frac{j_0}{M_N},\dots,\frac{j_{N-1}}{M_N}$ to $f(\frac{0}{N}),f(\frac{1}{N}),\dots,f(\frac{N-1}{N})$ with $j_i \in \mathbb{Z}^{\ge 0}$ for each $i$. Let $s_N$ denote a block of length $T_N := j_0+\dots+j_{N-1}$ that has $\frac{t}{N}$ appearing $j_t$ times. Define our sequence $(x_k)_k$ by $2^{2^1+T_1}$ consecutive $s_1$'s, followed by $2^{2^2+T_2}$ consecutive $s_2$'s, followed by $2^{2^3+T_3}$ consecutive $s_3$'s, etc.. Then, for any $m \ge 1$, if $N$ is large enough, $$\frac{1}{N}\sum_{k \le N} e^{2\pi i x_k m} \approx \frac{1}{N_0}\sum_{t=0}^{N_0-1} f(\frac{t}{N_0})e^{2\pi i \frac{t}{N_0}m} \approx \int_0^1 f(x)e^{2\pi i xm}dx = \frac{-1}{3^m},$$ for some large $N_0$, the first approximation holding due to $f$ having mean $1$ (I'll leave details/quantifiers to you).
Thank you very much for this construction! Very interesting example.
ok @mathworker21 will definitely do, I think I got the general idea. But just a question on the exact construction of the sequence $x_k$: How can one deal with negative values of $f$ which I think occur? Then $j_t$ can get negative which should be the number that $t/N$ appears. Also how can we ensure that the block length is $M_N$ which is the denominator of the approximations?
@Schmelli for your second question, the definition of the length of the block is $M_N$. The definition of $M_N$ is just a common denominator of all of the rationals. your first question is a good one: now thinking about my answer, I don't think I use that $\hat{f}(0) = 1$; I think we just need $\hat{f}(m) = \frac{-1}{2^m}$ for $m \ge 1$, which should imply that $f$ is bounded I think (they imply $f$ is $C^\infty$ in a quantitative sense, and we also know $f$ is $L^2$), so we can just translate $f$ upwards to make it non-negative (which is fine, since I don't think we care about $\hat{f}(0)$).
Hi @mathworker21, thanks for the very quick reply yesterday. I think the upward translation of $f$ could work even though I think that would change the ratios between the $j_t$. Maybe its because I still struggle a bit with the role of the common denominator $M_N$. I understand its needed to make the $j_t$ comparable and to assign the "right portion" of $t/N$ in our approximation of $f$. But aren't the numbers $j_t$ entirely determining the length of the blocks $s_N$ in the above construction, so can we get to a block of length $M_N$ if for example the sum of $j_t$ are much higher?
@Schmelli sorry about that! For some reason, I had in my mind that $j_0+\dots+j_{N-1} = M_N$; I don't know why. See the updated answer. I think it addresses all your concerns.
Thanks @mathworker21, that was very useful!
@Schmelli I think boundedness of $f$ is obvious. By fourier inversion, $f(\theta) = \sum_m \frac{1}{2^m}\cos(2\pi m \theta)$ (or something like that).
Hi @mathworker21, just realized that the shift of $f$ could be a problem. If we skip of your initial condition $\hat{f}(0)=1$ we lose $1/N_0 \sum_{t=0}^{N_0-1} f(\frac{t}{N_0})= 1$ which I think is essential in the first approximation step of the last equation chain. Otherwise we could shift the bounded function very high up and that would imply a almost uniform distribution of the second sequence which I think wouldn't work.
@Schmelli thanks again, nice catch! It seems the limit would instead be $\frac{\int_0^1 f(x)e^{2\pi i mx}dx}{\int_0^1 f(x)dx}$, so we do indeed want $\int_0^1 f(x)dx = 1$. However, I think my last comment might save the day. The real-valued function $f$ with $\hat{f}(0) = 1$ and $\hat{f}(m) = \frac{-1}{2^m}$ for $m \ge 1$ is $f(\theta) = 1-\sum_{m \ge 1}\frac{1}{2^m}\cos(2\pi m\theta)$ (please check this), which is non-negative (since $\sum_{m \ge 1} \frac{1}{2^m} = 1$). If you agree with all of this, then that's good; however, I'd feel a bit bad about how lucky we got...
Hi @mathworker21, thanks, unfortunately considering the Fourier coefficients for negative $m$ which we need in order to obtain a real-valued function I think we get a function that gets negative. Using $\exp(2\pi ix)+\exp(-2\pi ix)=2 \cos(2 \pi x)$ I arrived at $1-\cos(2\pi x)-\frac{1}{2}\cos(4\pi x)-\frac{1}{4}\cos(6\pi x)...$ where $f(0)=-1$.
@Schmelli ah, I did consider the negative fourier coefficients, but forgot the factor of $2$. Ok, but it looks like we're close enough to getting lucky. In particular, if we redo the whole answer with $a_k = 1$ if $k$ is odd and $a_k = 3$ if $k$ is even, then I think the corresponding $f$ will be non-negative. It should be $1-\frac{2}{3}\cos(2\pi x)-\frac{2}{9}\cos(4\pi x)-\dots$, which is non-negative. Do you concur with all of this? If so, I'll retype the answer to fix everything.
Hi @mathworker21, yes that looks good. Just one more question: In the initial problem I also considered negative $m$ (all non-zero integers). Do you think a similar example could be constructed that $\lim_{n \to \infty} \sum_{k=1}^{n} (a_k)^m e^{2 \pi i x_k m}=0$ holds also for negative $m$? In the above example I think we wouldn't have this due to the definition of the Fourier coefficients for negative values.Thanks a lot again!
@Schmelli You would have to convince me of something first. In Weyl's criterion, negative $m$ is the same as positive $m$, due to conjugation. In the problem you're asking about now, negative $m$ is different from positive $m$, which makes the problem very unnatural to me and therefore uninteresting. Note that the construction of $f$ in my answer also explains why we want negative $m$ to be the same as positive $m$.
Hi @mathworker21, will try to convince you, but first thanks again for the great construction and the amount of time you looked into this, it made a lot much clearer for me. The problem I want to solve stems initially from analytic number theory and is concerned with the distribution of imaginary parts of nontrivial zeros of zeta functions. Essentially Landau proved that in the case of the Riemann zeta function for these zeros $\rho_k=\beta_k+ i\gamma_k$ the following holds:
$\lim_{n \to \infty} \sum_{k=1}^{n} x^{\rho_k}=0$
for a free parameter $x>0$ but $x \neq 1$.
Part2 @mathworker21: If we now set $x=e^{2 \pi m}$ and separate real and imaginary part we arrive at basically the problem I posted, since it is known that the real parts $\beta_k$ are bounded. But that holds also for negative $m$ where $ e^{2 \pi m}$ stays positive.
Now it is shown that if the real parts are concentrated at one point (for example as assumed in the Riemann Hypothesis) then we can immediately apply Weyls criterion. I want to show that a similar assumption is necessary or that the Landau condition is enough to get uniform distribution
Part3 @mathworker21
which would be nice as a Landau type formula applies to other zeta functions as well. I think, also given what I got from you that there is a counterexample but would be great to have an explicit one.
I understand the point that negative $m$ need not to be looked at if a sequence is uniform distributed due to conjugation as you indicated. But looking the other way round couldnβt it still be possible that a condition like stated above which is valid for all $m \neq 0$ implies uniform distribution?
@Schmelli thanks! indeed interesting. However, I still see no reason why having the condition for negative $m$ will now yield a criterion for uniform distribution. But finding a counterexample seems much harder now. I'll think about it more.
Thanks @mathworker21, I think one reason that this condition could suffice is that it yields additional information for negative $m$, as we get $1/n \sum_{k \leq n} (a_k)^m \sin(2 \pi m x_k)$ and $1/n \sum_{k \leq n} 1/{(a_k)^m} \sin(2 \pi m x_k)$ tending to $0$ (analouge for $\cos$) which is in case of Weyl basically the same if we take constant weights $a_k=1$ for all $k$.
@Schmelli I thought I should let you know that I recently realized that the method used in the answer to construct a sequence with prescribed "fourier coefficients" is sharp. Namely, for $(c_k){k \in \mathbb{Z}}$ in $l^2(\mathbb{Z})$ (say), there exists a sequence $(x_n)n$ with $\frac{1}{N}\sum{n \le N} e(kx_n) \to c_k$ for each $k \in \mathbb{Z}$ if and only if the function $\sum{k \in \mathbb{Z}} c_k e^{2\pi i kx}$ is non-negative for all $x \in \mathbb{T}$. The answer shows the "if" direction. To see the "only if" direction, let $\mu$ be a weak-$*$ limit of a subsequence of the measures
$\frac{1}{N}\sum_{n \le N} \delta_{x_n}$. Then $\hat{\mu}(k) = c_k$ for all $k \in \mathbb{Z}$, but also $\hat{f}(k) = c_k$ for all $k \in \mathbb{Z}$, where $f(x) := \sum_{k \in \mathbb{Z}} c_ke^{2\pi i kx}$, so we must have $\mu = fdx$. If $f$ were negative at some point, then since it is continuous, it is negative in some open set $U$, which yields a contradiction: $0 > \int_U f \ge \liminf_k \frac{1}{N_k}\sum_{n \le N_k} \delta_{x_n}(U) \ge 0$.
Thanks @mathworker21 for this hint. Do you know if there are necessary and sufficient conditions on the Fourier coefficients $c_k$ such that $f$ is non-negative? I am not an expert on Fourier analysis, but from what I read this is not easy to determine.
@Schmelli I don't think there's a nice criterion.
|
2025-03-21T14:48:29.659582
| 2020-01-17T13:04:00 |
350626
|
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"url": "https://mathoverflow.net/questions/350626"
}
|
Stack Exchange
|
connection on principal bundles over algebraic/geometric stacks
Is there a notion of connection on a principal bundle over an algebraic or geometric stack?
By a geometric stack, I mean a stack over category of manifolds that is representable by a Lie groupoid; that is of the form $B\mathcal{G}$ for some Lie groupoid $\mathcal{G}$.
As far as I know, the paper (https://arxiv.org/abs/math/0401420) discuss the notion of principal bundles over Lie groupoid and connection on principal bundle over Lie groupoid but does not mention if this notion can be turned into a notion of connection on a principal bundle over a geometric stack.
Are there other versions of connection on a principal bundle over stack?
see section 4.4 here ://arxiv.org/pdf/0806.1357.pdf
On first reading, it looks slightly different from what the setup I mentioned above. But, I hope it would be of some use. I will read carefully and respond.
I fail to see how your notion of principal fibered category is related to the notion of principal bundle over lie groupoid mentioned above... Can you please explain something along those lines
I saw it late... In page 18, after corollary 3.15, it does say that "Therefore, this allows us to speak about connections and flat connections of a principal bundle over a differentiable stack". But, I still want to look for other references that discuss noton of connection on principal bundle over differentiablr stack.. For that matter, they do not say explicitly what does it mean by a principal bundle over differentiable stack $B\mathcal{G}$. From cor 2.12 I am assuming that, by principal bundle over the stack $B\mathcal{G}$, they mean principal bundle over the Lie groupoid $\mathcal{G}$.
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2025-03-21T14:48:29.659719
| 2020-01-17T13:21:32 |
350627
|
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|
Stack Exchange
|
Relationship between "infinitely unequal" and "eventually different"
Suppose that $\kappa$ has the property that for every family $A\subseteq\omega^{\omega}$, if $|A|<\kappa$, then there exists some $g\in\omega^{\omega}$ such that for any $f\in A, \exists^{\infty}n\;f(n)\neq g(n)$. Does it then follow that for any family $A\subseteq\omega^{\omega}$ of size $<\kappa$, there exists a $g\in\omega^{\omega}$ such that for any $f\in A, \forall^{\infty}n\;f(n)\neq g(n)$?
It does not follow. In fact, it is not true in the Cohen model. Let $X$ be a set of $\aleph_2$ mutually generic Cohen reals over $V$. (Recall: ``mutually generic'' means that if $x \in X$ and $Y \subseteq X$, then $x \in V[Y]$ if and only if $x \in Y$, and otherwise $x$ is Cohen-generic over $V[Y]$.)
In $V[X]$, if $A \subseteq \omega^\omega$ with $|A| < \aleph_2$, then (because $A$ is an object that is hereditarily of size $\leq\!\aleph_1$) there is some $X_0 \subseteq X$ with $|X_0| = \aleph_1$ such that $A \subseteq V[X_0]$. If $g \in X \setminus X_0$, then $g$ is Cohen-generic over $V[X_0]$, and hence over any $f \in A$. In particular, for any $f \in A$, $\exists^\infty n f(n) \neq g(n)$.
On the other hand, let $A \subseteq X$ with $|A| = \aleph_1$. If $g \in \omega^\omega$, then (because $g$ is a hereditarily countable object) there is some countable $A_0 \subseteq A$ such that $g \in V[A_0]$. If $f \in A \setminus A_0$, then it is Cohen-generic over $V[A_0]$, and hence $\exists^\infty n f(n) = g(n)$.
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2025-03-21T14:48:29.659845
| 2020-01-17T14:16:10 |
350632
|
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|
Stack Exchange
|
integral of the laplacian to some power
I want to know the space of functions where the following quantity is uniformly bounded from above
$$\int_{K} (\Delta u)^j d\lambda< C,$$
where K is a compact and j is an integer number greater than 1.
You mean some Sobolev space or something different? It is a bit confusing what you write.
I know this inequality is true for u in $ u\in W^{2,p} for some p, but I want my constant C to be independently of u. Then i am looking in which space of functions this is true.
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2025-03-21T14:48:29.659905
| 2020-01-17T14:41:58 |
350633
|
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|
Stack Exchange
|
Is finite verbal subgroup equivalent to finite index of marginal subgroup?
There is a well known fact:
If $G$ is a finitely generated group. Then $|Gβ| < \infty$ iff $[G:Z(G)]<\infty$.
Suppose $\mathfrak{U}$ is a group variety. Letβs denote the corresponding verbal subgroup as a $V_{\mathfrak{U}}(G)$ and the corresponding marginal subgroup as $M_{\mathfrak{U}}(G)$. Note, that for the variety of all abelian groups $\mathfrak{A}$ (defined for the word $[x, y]$) we have $V_{\mathfrak{A}}(G) = Gβ$ and $M_{\mathfrak{A}}(G) = Z(G)$.
My question is:
Can the aforementioned statement be generalized to the following one:
Suppose $G$ is a finitely generated group and $\mathfrak{U}$ is a finitely based variety. Then $|V_{\mathfrak{U}}(G)| < \infty$ iff $[G:M_{\mathfrak{U}}(G)]<\infty$.
?
This question on MSE with several useful comments
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2025-03-21T14:48:29.659984
| 2020-01-17T15:22:15 |
350635
|
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"Amir Mafi",
"Hailong Dao",
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|
Stack Exchange
|
Adding first generator to Cohen-Macaulay monomial ideal
Let $I$ be a Cohen-Macaulay monomial ideal of $R=K[x_1,...,x_n]$, where $K$ is a field. Can we say the ideal $(x_1)+I$ is Cohen Macaulay?
Not even when $I$ is square-free. Let $I$ be the ideal $(x_2x_4,x_2x_5,x_3x_4,x_3x_5, x_1x_2,x_1x_5)$. It is Cohen-Macaulay because it is the Stanley-Reisner ideal of a path of $5$ vertices (with $x_1$ in the middle). Killing $x_1$ will disconnect the graph, and indeed the ideal $(x_2x_4,x_2x_5,x_3x_4,x_3x_5)$ is not Cohen-Macaulay. You can check both statements by Macaulay 2.
In fact, you can get rid of $x_5$ to make a smaller example.
Thank you so much for your nice example
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2025-03-21T14:48:29.660057
| 2020-01-17T16:17:48 |
350638
|
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"0xbadf00d",
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|
Stack Exchange
|
If $M$ is a manifold, $xβM$ and $d(x,Ο)=\inf\{t>0:x+tΟβM\}$, does the pushforward of the solid angle measure under $S^2βΟβ¦x+d(x,Ο)Ο$ admit a density?
Let $S^2$ denote the unit 2-sphere, $M$ be a 2-dimensional oriented embedded $C^1$-submanifold of $\mathbb R^3$ with $$d_M(x,\omega):=\inf\left\{t>0:x+t\omega\in M\right\}<\infty\;\;\;\text{for all }x\in M\text{ and }\omega\in S^2\tag1,$$ $\nu_M$ denote the outer normal field on $M$, $\sigma_M$ denote the surface measure on $\mathcal B(M)$, $$S^2(x):=\left\{y\in\mathbb R^3:|x-y|=1\right\}=x+S^2\;\;\;\text{for }x\in\mathbb R^3$$ and $$\omega_{x\to y}:=\frac{y-x}{|y-x|}\in S^2\;\;\;\text{for }x,y\in\mathbb R^3\text{ with }x\ne y.$$
Let $x\in M$. Can we show that the pushforward measure of $\sigma_{S^2}$ under $$\varphi:S^2\to M\;,\;\;\;\omega\mapsto x+d_M(x,\omega)\omega$$ has a density with respect to $\sigma_M$?
The desired the claim should be true. I guess we need to use $$\sigma_{S^2}(B)=\int\sigma_M({\rm d}y)1_B\left(\omega_{x\to y}\right)\frac{\left|\langle\nu_M(y),\omega_{x\to y}\rangle\right|}{|x-y|^2}\;\;\;\text{for all }B\in\mathcal B(S^2)\tag2$$ (I've asked for that separetely: Can we prove this relation between the solid angle measure and the surface measure of a smooth manifold?). Clearly, if $x\in\mathbb R^3$, $\varepsilon>0$ and $f:\varepsilon S^2(x)\to\mathbb R$ with $f\ge0$ or $\int|f|\:{\rm d}\sigma_{\varepsilon S^2(x)}<\infty$, then $$\int f\:{\rm d}\sigma_{\varepsilon S^2(x)}=\varepsilon^2\int\sigma_{S^2}({\rm d}\omega)f(\varepsilon(x+\omega))\tag3.$$ However, if $B\in\mathcal B(M)$, we cannot immediately apply this to $$\left(\sigma_{S^2}\circ\varphi^{-1}\right)(B)=\int\sigma_{S^2}({\rm d}\omega)1_B\left(d_M(x,\omega)\left(\frac x{d_M(x,\omega)}+\omega\right)\right)\tag4.$$
What is $\mathcal B(M)$? Is $M$ a surface, i.e. a 2-dimensional manifold, or can $M$ be 1-dimensional or 0-dimensional, or 3-dimensional?
@BenMcKay $\mathcal B(M)$ is the Borel $\sigma$-algebra on $M$. And $M$ is a surface.
|
2025-03-21T14:48:29.660191
| 2020-01-17T16:52:27 |
350640
|
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|
Stack Exchange
|
Less regular version of the Gaussian free field
One can define (continuous) Gaussian free field as follows: one can consider some orthonormal basis $(\psi_k)_{k=1}^{\infty}$ in the Sobolev space $H^1(\Omega)$ (here $\Omega \subset \mathbb{R}^d$) and then consider $\sum_{k=1}^{\infty} \xi_k \psi_k$ where $\xi_k$ are independent, identically distributed (normalized) normal random variables. When $d=1$ this sum makes sense as a continuous function: however when $d>1$ this random sum only makes sense as a distribution. I wonder what will be the difference if we consider $L^2(\Omega)$ instead of $H^1(\Omega)$ (for example $L^2(\mathbb{R}^d)$ with Hermite polynomials as a basis). Is this object well defined/was it studied? I will welcome any references
In general, a sum of the form $\sum \xi_k \psi_k$, where $\psi_k$ is an orthonormal set in a Hilbert space $H$, will diverge almost surely in the topology of $H$. It is always possible to find some larger space $W$ with a weaker topology (possibly again Hilbert or Banach), with a dense embedding of $H$ into $W$, such that the sum converges i.p. in $W$. The resulting "object" depends entirely on the choice of $W$, though.
To have some receptacle which is a bit more canonical, one can also take $W=\mathcal{S}'(\mathbb{R}^d)$ for $\Omega=\mathbb{R}^d$.
Also, the case of $L^2$ with the $\psi$'s given by Hermite functions leads to white noise rather than the GFF.
@AbdelmalekAbdesselam thank you for your comment-could you please give me some reference for this statement regarding white noise?
I didn't use a reference. It's easy to figure out by hand.
|
2025-03-21T14:48:29.660322
| 2020-01-17T18:04:10 |
350643
|
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"Jochen Glueck",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/350643"
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|
Stack Exchange
|
Growth bounds for the exponential of an operator
Let $X$ be a complex Banach space and $A:X \to X$ a compact operator. It spectrum is the set $\sigma(A)=\lbrace \lambda \in \mathbb{C}, \ A-\lambda I \text{ is not invertible}\rbrace$. Let $L=\sup\lbrace re(\lambda), \lambda \in \sigma(A)\rbrace$.
Let $e^A$ denote its exponential operator, defined by the series $$\sum_{n=0}^{\infty} \frac{K^n}{n!}.$$
It is true, in the finite dimensional case, that if $L< 0$, then $$\Vert e^{tA}\Vert \xrightarrow{t \to \infty}{} 0.$$
Is it also true in the general case?
Sketch of a proof of the finite-dimensional case:
Let's prove it for $(\mathbb{R}^n, \Vert \ \Vert_{\infty})$ (we'll omit the subscript $\infty$). First let's recall that $$\Vert M \Vert =\max_{1\leq i \leq n}\lbrace \sum_{j=1}^{n}\vert a_{ij}\vert \rbrace.$$ In particular $\Vert \operatorname{diag}(d_1,d_2,...,d_n)\Vert\leq max\lbrace \vert d_i \vert \rbrace$.
Let's asume $A$ can be writen in Jordan form $A=C(N+D)C^{-1}$ with $N$ nilpotent and $D$ diagonal. Then, $$e^{tA}=Ce^{tN}e^{tD}C=CG(t)H(t)C.$$The matrix $G(t)$ has polynomials in $t$ in its entries, so $\Vert G(t) \Vert=g(t)$, with $g(t)$ a real valued function of polynomial order in $\infty$. The matrix $H(t)=\operatorname{diag}(e^{t\lambda_1},...,e^{t\lambda_n})$, so $\Vert H(t) \Vert\leq \max\lbrace\vert e^{t\lambda_i} \vert\rbrace=\max\lbrace e^{t \ re(\lambda_i)} \rbrace=e^{tL}$. So, if $L<0$,
$$\Vert e^{tA} \Vert \leq \Vert C \Vert p(t) e^{tL} \Vert C^{-1} \Vert \xrightarrow{t \to \infty} 0.$$
There are several problems in the infinite-dimensional case to adapt this proof:
there is no Jordan form, the eigenvalues are not necessarily associated with eigenvectors. There is an extra issue: informally, even if we had some sort of representation, the $g(t)$ factor would not be a finite degree polynomial and could outweigh the $e^{tL}$ term.
Welcome to MathOverflow! This is even true if $A$ is not compact but merely bounded. It is, for instance, a consequence of the spectral mapping theorem for the holomorphic functional calculus.
Hi, to continue the comment of Jochen, use Cauchy integral formula instead of power series to represent the exponential function.
|
2025-03-21T14:48:29.660494
| 2020-01-17T19:16:27 |
350644
|
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"Stephen McKean",
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"url": "https://mathoverflow.net/questions/350644"
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|
Stack Exchange
|
Is the Hilbert series of an ideal related to the Hilbert series of its homogenization?
Suppose we have a field $k$ of characteristic 0, let $I$ be an ideal of $R=k[x_1,...,x_n]$, and let $H$ be the homogenization of $I$ in $S=R[z]$. Is there any relationship between the Hilbert series of $R/I$ and the Hilbert series of $S/H$?
I know that the Hilbert series for $R/I$ can be computed via GrΓΆbner bases, but I'm hoping to gain information about a GrΓΆbner basis for $I$ from the Hilbert series of $R/I$. In particular, if $R/I$ has Krull dimension 0, then I would like to compute $\dim_k(R/I)$ without relying on GrΓΆbner bases. The hope is that if the Hilbert series for $R/I$ and $S/H$ are related, then I can use some of the techniques for homogeneous ideals to work in the non-homogeneous case.
Sorry for asking a stupid question, but if $I$ is not homogeneous then what is the Hilbert series of $R/I$?
Good question! The first answer is that I had just been using Macaulay2 to compute this, so I didnβt know off the top of my head. The second answer is that $R/I$ is filtered by degree, and the Hilbert series is the Hilbert series of the associated graded $\mathrm{gr}(R/I)$.
When I was double checking this definition, I saw a post that mentioned that the Hilbert series of $R/I$ is the same as the Hilbert series of $R/\mathrm{in}(I)$, but it feels like I would again need a GrΓΆbner basis to exploit this.
|
2025-03-21T14:48:29.660616
| 2020-01-17T19:21:22 |
350645
|
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"Noah Riggenbach",
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"myzhang24"
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|
Stack Exchange
|
When are complex polynomial maps surjective?
Consider a complex polynomial map $f: \mathbb{C}^p \to \mathbb{C}^q$ for some $p \geq q \geq 1$ (not necessarily equal).
What is a sufficient condition for $f$ to be surjective?
I am aware of some necessary conditions. For example, one must of course assume that the components $f_1, \dotsc, f_q : \mathbb{C}^p \to \mathbb{C}$ are algebraically independent. (If this is not the case, then $f(\mathbb{C}^p)$ is contained in an affine algebraic variety). When they are indeed independent, one knows that $f(\mathbb{C}^p)$ is dense in $\mathbb{C}^q$ (see When are complex polynomial maps almost surjective?). However, I wonder if conditions are known ensuring that the image is equal to $\mathbb{C}^q$. The algebraic independence is not sufficient by itself, as shown by the counter-example $f(z_1, z_2) = (z_1, z_1 z_2)$.
The Ax-Grothendieck theorem gives that when $p=q$, then injectivity implies surjectivity.
Thank you. Is something known when $p > q$, e.g. something about common roots of the $f_i$?
not that I know of, but I am far from an expert. Sorry.
This arises as a map of polynomial rings $F:k[x1..xq]\to k[x1..xp]$, the image scheme is defined by the kernel of $F$. For $f$ to be surjective you want $F$ to be injective. If you have concrete polynomials $f_i$, then you can compute the kernel of $F$ using computer algebra systems such as Macaulay2.
|
2025-03-21T14:48:29.660845
| 2020-01-17T19:48:14 |
350647
|
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|
Stack Exchange
|
Riemannian metric over moduli space of Riemann spheres with n punctures
In the paper `Tessellations of moduli spaces and the mosaic operad' by Devadoss (https://arxiv.org/pdf/math/9807010.pdf), on page 5-6, the author identifies hyperbolic planar tree space (or the compactification of moduli space of $\mathbb{RP}^1$ with n punctures $\overline{\mathcal{M}_{0}^n(\mathbb{R})}$) with the space of all ideal $n$-polygons on the Poincare disk with $k$ marked diagonals for some $k$.
He then claims that "this perspective naturally gives a Riemann metric to $\overline{\mathcal{M}_{0}^n(\mathbb{R})}$" and it really escapes me what this Riemannian metric actually is.
Can some one explain and/or give reference to the description of this metric? Thanks in advance!
Where do the quotation marks end in your second sentence?
oops, edited to indicate where the quotation ends. thanks for catching that!
|
2025-03-21T14:48:29.660940
| 2020-01-17T19:56:30 |
350648
|
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|
Stack Exchange
|
Given a vertex $u$ (of bounded degree $k$) and another vertex $v$ in a planar graph, what is the smallest number of "curves"?
Given a vertex $u$ (of bounded degree $k$) and another vertex $v$ in a planar graph $G$, what is the smallest number of "curves" in the plane drawn from $u$ to $v$ such that no $u$--$v$ path in $G$ intersects each curve at a point other than $u$ or $v$?
(any finite bound is good as well)
Also, can anyone recommend a reference for graph drawings?
Sorry if this is a stupid question.
Unfortunately, there is no finite bound, even if both vertices have bounded degree. To see this, consider a large grid graph $G$ with $u$ a degree-$4$ vertex in the 'left half' of the grid and $v$ a degree-$4$ vertex in the 'right half' of the grid. Let $\mathcal{I}$ be a family of $u$--$v$ curves such that there is no $u$--$v$ path in $G$ which intersects each $I \in \mathcal I$ at a point other than $u$ or $v$.
Let $C$ be the cycle in $G$ consisting of the 'middle' vertical path $P$ of $G$ and the part of the boundary of $G$ to the left of $P$. Since $C$ topologically separates $u$ from $v$, every curve $I \in \mathcal I$ must intersect an edge of $C$ (for this proof, an edge contains its endpoints). Choose such an edge $e(I)$ for each $I \in \mathcal I$. Note that $E(P) \subseteq \{e(I) \mid I \in \mathcal I\}$, because for each $e \in E(P)$, there is a $u$--$v$ path in $G$ using all edges of $E(C) \setminus \{e\}$. Thus, $|\mathcal I| \geq |E(P)|$, which can be arbitrarily large if we increase the size of the grid.
|
2025-03-21T14:48:29.661063
| 2020-01-17T20:22:45 |
350649
|
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|
Stack Exchange
|
Reference for the positive probability of convergence to a stable point of a stochastic approximation algorithm
Consider a stochastic approximation process with
$$x_{t+1} = x_t + \frac{1}{t} (g(x_t)+u_t)$$
where $(u_s)_s$ is a sequence of i.i.d. shocks.
Assume $g$ is Lipschitz, $u_t$ has finite variance, and that
$(x_s)_s$ is bounded with probability one.
Furthermore, assume that
$$
C = \{ x \in \mathbb{R} \colon g(x)=0\}
$$
is finite.
In this case it follows from the results in stochastic approximation theory that $x$ converges a.s. to a point in $C$ (see for example Kushner and Yin, ``Stochastic Approximation and Recursive Algorithms and Applications'' Theorem 2.1 page 127).
Question: Do you know a reference that provides conditions such that $x$ converges to a given point in $C$ with strictly positive probability.
There seems to be an error in your definition of stable points.
Thanks a lot, corrected.
I still don't think it is correct.
Thanks a lot Piyush, could you explain what the problem is, I am sorry if I miss something obvious.
I don't understand it. The term g(x)(x-y) cannot be positive for all y close to x, since suppose $y_1>x$, then it requires g(x)<0, but then for $y_2<x$ close to x, it gives $g(x)(x-y_2)<0$.
Thanks sorry this was a typo I confused y and x. It should be g(y) not g(x) corrected.
Can you point out where in the Kushner/Yin book is this definition of stability given ?
You are completely right this is not in Theorem 2.1, in the book section 5.8 briefly discusses ruling out unstable points, but this needs an additional continuity assumption on the distribution of $u$. I adjusted the question accordingly.
|
2025-03-21T14:48:29.661188
| 2020-01-17T20:53:59 |
350651
|
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|
Stack Exchange
|
When is the iso-comma category $F \wr G $ a reflective subcategory of the comma category $F \downarrow G$?
Let $F \wr G$ be the iso-comma category. This is basically the same construction as the comma category $F \downarrow G$ except that we require the morphism $F(x) \to G(y)$ to be an isomorphism instead. Observe that $F \wr G$ is a full subcategory of $F \downarrow G$, my question is: under what conditions does the inclusion admit a left (or possibly right) adjoint? i.e. when is $F \wr G$ a (co)reflective subcategory of $F \downarrow G$?
I'm looking for any references which consider this question as well.
Edit: 4 years later and rereading this question, I've noticed that perhaps it is a bit dry. I will try to add some context to why I was thinking about it in the first place.
If we think of an $\Omega$-prespectrum as a sequence of pointed spaces together with pointed maps between them, then you can see that the category of $\Omega$-prespectra is a comma category where the objects of the source of $F$ and $G$ are sequences of pointed spaces and the $G$ functor shifts the indexing of a pointed space by 1. Then the triple in the comma category consists of a sequence of morphisms.
In the same way, such an iso-comma category would require a pointed isomorphism between the spaces. This would then be the category of $\Omega$-spectra. Obviously, this is a bit nonsense since we should be asking for weak-homotopy equivalences rather than isomorphisms, but pretend, even if it is not true, that such a category is possible. (My intuition being that it will hold in some higher setup).
There is an $\Omega$-spectrification functor from prespectra to spectra, therefore the inclusion functor has a left adjoint, using the universal property of spectrification.
Hence, in this poorly constructed example, I have reason to believe that something about the category of spectra allows for these inclusion maps to be adjoint.
Spectra are obviously a very special category, where all limits and colimits exist, and furthermore they are equivalent over the same shapes.
My question can therefore be rephrased as follows: Is there a general reason for this phenomenon?
It seems to me that such an adjoint acts on objects of $(F/G)$ yielding an isomorphism $Fx\cong Gy$ between the same objects; this is because a natural requesto on it is that it commutes with the forgetful functors to the domains of $F,G$. Now, I wonder what is a natural choice for such an isomorphism between the same objects...
If we put the additional restriction that @Fosco suggested and also the restriction that it acts in the same way on morphisms, then the adjoint will exist only in trivial cases when $F \wr G$ coincides with $F \downarrow G$.
If the domains of $F$ and $G$ are locally presentable, their codomain is accessible, and $F$ and $G$ preserve colimits, then both $F \wr G$ and $F \downarrow G$ are locally presentable and $F \wr G$ is closed under colimits in $F \downarrow G$. It follows that it is a coreflective subcategory.
Cool, so it is too restrictive to ask that the adjoint is an adjunction in ${\sf Cat}/(A\times B)$. I guess the existence follows from AFT or Freyd-Kelly (which is still AFT :D ); is there any chance to get a grip on how the adjoint is defined in this case? It looks like a controlled way of localizing (in the sense of homotopy categories) $(F\downarrow G)$.
|
2025-03-21T14:48:29.661439
| 2020-01-17T21:05:29 |
350653
|
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|
Stack Exchange
|
Intuition and analogue of Wraith axiom from synthetic differential geometry
In synthetic differential geometry, an object $M$ verifies the Wraith axiom if for all functions $\tau:D\times D\to M$ which are constant on the axes $\tau(d,0)=\tau(0,d)=\tau(0,0)$ for all $d\in D$, there's a unique factorization through the multiplication map, i.e there's a unique function $t:D\to M$ such that $t(d_1\cdot d_2)=\tau(d_1,d_2)$.
What is the geometric/physical intuition behind this axiom? What is the analogue in the category of smooth manifolds?
Update. Following the answers I think I should add some motivation. Lavendhomme's book defines the commutator of vector fields as $\tau(d_1,d_2)=Y_{-d_2}X_{-d_1}Y_{d_2}X_{d_1}$. Because this is constant on axes, it factors through the multiplication map to give a vector field $t$ charaterized by $t(d_1d_2)=\tau(d_1,d_2)$. I understand $t$ is desirable since it is a vector field, but I don't know how to geometrically interpret its characterizing property. For instance, why not consider the vector field $\tau(d_1,d_1)$ given by precomposing the diagonal? This motivated my question.
Axiom W is about the behaviour of the second tangent bundle - it ensures that the vertical bundle of the tangent bundle, $V(M) \subseteq T\circ T(M)$, where $V(M) = T(p)^{-1}(0)$, decomposes as the pullback of the projection $p_M: T(M) \to M$ along itself. The map $[\bullet, M]:[D,M] \to [D \times D, M]$ would be written fiberwise as $\ell(v) = \frac{d}{dt}_{t = 0} (vt)$. If you look at Robin Cockett and Geoff Cruttwell's first paper on tangent categories, you can see they spend a fair bit of time talking about the universality of the vertical lift and its relationship with the Lie bracket.
You may also want to look at the definition of the core of a double vector bundle - axiom W can also be read as saying the core of the second tangent bundle of M is $TM$.
Edit: If you look at the answer here, you can see how the lift from the fibred product $TM \times_M TM \to T^2(M)$ is written. Using infinitesimals, you would write $\gamma,\beta:D\to M, \gamma(0)=\beta(0)$ is sent to the map $D\times D \to M$ which is given by $d_1,d_2 \mapsto \gamma(d_1) + \beta(d_1d_2)$. The condition that this be the kernel of $T(p)$ is provable from property W (and a good exercise).
The square root business you mentioned looks to be how you would show property $W$ holds in the category of smooth manifolds (when rewritten as the universality of the vertical lift from the tangent category axioms).
Dear Ben, thanks for this answer. I don't quite see how the Wraith axiom gives the canonical isomorphism between the vertical bundle and its kernel pair. Indeed the classical analogue is straightforward. It seems much more blatantly relevant for the Lie bracket due to the square root business, which is much more confusing. The paper by Cockett and Cruttwell is fairly impenetrable for me, so I would enormously appreciate some more details on the geometric explanation!
Hi Arrow, I've expanded my answer a bit. Hopefully that helps.
Dear Ben, I asked a follow up question here.
|
2025-03-21T14:48:29.661654
| 2020-01-17T21:26:35 |
350655
|
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"Jochen Wengenroth",
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|
Stack Exchange
|
Span of a nonlinear function
Fix vectors $x,y\in\mathbb{R}^d$ and a smooth function $\phi:\mathbb{R}\rightarrow \mathbb{R}$. Define $\phi^d: \mathbb{R}^d \rightarrow \mathbb{R}^d$ as applying $\phi$ entrywise (i.e. $\phi^d(x_1, x_2, \dots) = (\phi(x_1),\phi(x_2), \dots)$). Under what conditions do we have the equality
$$\mathbb{R}^d=\text{span}(\{\phi^d(\alpha x+y)~\big |~\alpha\in \mathbb{R}\})$$
Clearly if $\phi$ is the identity operator $\phi(x)=x$ or if any two entries of $x$ are equal, this equality does not hold. Main claim is equality holds for generic (i.e. almost all) $x,\phi$ choices.
Suppose the span is not the full $\mathbb{R}^d$, then there exists a normal vector to the span which we can call $\nu$. This means that $\sum \nu_i \phi(\alpha x_i) = 0$ for all $\alpha$. Take $k$ derivatives with respect to $\alpha$ you still get 0. Evaluate at $\alpha = 0$ you get that $\phi^{(k)}(0) \sum \nu_i (x_i)^k = 0$. So provided not too many derivatives of $\phi$ vanish at zero (generically true), such $\nu$ cannot exist for generic $x_i$.
For the final claim, see https://en.wikipedia.org/wiki/Vandermonde_matrix
There could be some typo around, because the question does not make any sense, as it is.
@Pietro Majer I think that $\phi (x)=(\phi (x_1),\ldots, \phi (x_d)) $.
So maybe it is a map $\phi:\mathbb{R}^d\to\mathbb{R}^d$?
@PietroMajer It's an induced map that the original question also labeled $\phi$ by abuse of notation; I've edited to remove the abuse of notation.
Thanks. This answer is very clean thanks to Vandermonde connection. Would a similar approach work for
$\text{span}(\phi^d(\alpha x+y)~\big |~\alpha \in \mathbb{R})$ where $y$ is a fixed vector?
|
2025-03-21T14:48:29.661800
| 2020-01-17T21:31:02 |
350656
|
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|
Stack Exchange
|
Recognizing reflection subgroups of Coxeter groups
Given a Coxeter system $(W,S)$ with reflections $T$, and any subset $A \subseteq T$, it is known that the reflection subgroup $W_A$ generated by $A$ has a canonical choice $S_A$ of generators so that $(W_A,S_A)$ is a Coxeter system (a result of Deodhar and of Dyer).
Question. Given two Coxeter graphs $G, G'$, is there some way to tell whether the Coxeter group $W$ associated to $G$ has a reflection subgroup $W'$ with Coxeter graph $G'$? I suspect this is hard in general, but are there nontrivial necessary or sufficient conditions known? In particular, is there some general principle by which I could recognize that the Coxeter group of type $B_n$ has a reflection subgroup of type $D_n$ just from the graphs?
Clearly $G'$ being an induced subgraph of $G$ is sufficient, but not necessary. I'm hoping there are more interesting things that can be said in general.
Another question asks for a stronger thing (an algorithm finding all "Coxeter subgroups"; as far as I can tell, that asker does not even require them to be reflection subgroups).
The B/D example is explained by Borel-de Siebenthal theory, I believe.
(Borel-de Siebenthal theory says that sub-root systems of maximal rank of a given root system correspond to Dynkin diagrams obtained from the affine Dynkin diagram corresponding to the original root system by deleting a single node.)
As an extension of the process described in Sam's comment, Dyer gives a description of the reflection subgroups in affine type. The Coxeter graphs of reflection subgroups are those which can be constructed from the original affine Coxeter diagram by successively removing vertices and converting finite type diagrams to the corresponding affine diagrams as desired. See the end of this article.
Though in general (i.e., outside affine type) we will have infinite rank reflection subgroups (see, for instance, David Speyer's answer here). So a general algorithm of the form "convert one type of diagram to another and delete points" would need to address this.
@ChristianGaetz: For right-angled Coxeter groups, the article arxiv:1910.04230 might interest you. It can be deduced from it an algorithm which determines whether or not a right-angled Coxeter group $C(\Gamma_1)$ embeds as a reflection subgroup into another right-angled Coxeter group $C(\Gamma_2)$.
|
2025-03-21T14:48:29.662219
| 2020-01-17T21:54:24 |
350657
|
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|
Stack Exchange
|
On a interpolation inequality for the SchrΓΆdinger unitary group (NLS)
I'm trying to understand scattering for the classical nonlinear SchrΓΆdinger equation and for that i'm studying a scattering criterion on Tao's paper. At Lema 3.1 he states that $$\left\|e^{it\Delta}f\right\|_{L^4_tL^{\infty}_x}\lesssim\left\|\nabla f\right\|_{L^2},$$
where $\left\|g\right\|_{L^4_tL^{\infty}_x}:=\left\|\left\|g(x,t)\right\|_{L^{\infty}_x}\right\|_{L^4_t}$.
Problem is for the proof he uses Lorentz and Besov spaces, which i'm not familiar with.
Is there a way to prove this inequality by using just basics about Sobolev spaces and interpolation inequalities? I've studied the Riesz-Thorin and the Marcinciewicz theorems, Hardy-Littlewood-Sobolev, the Strichartz estimates and some Sobolev embeddings. Thanks in advance.
|
2025-03-21T14:48:29.662315
| 2020-01-17T21:59:21 |
350658
|
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|
Stack Exchange
|
The intersection of all normal ultrafilters on a measurable cardinal
Suppose $\kappa$ is a measurable cardinal. Let $W$ be the intersection of all normal ultrafilters on $\kappa$.
I am interested in a precise characterization of the filter $W$.
One sure way to conclude that a set $A$ is in all normal measures, is to find a regressive function $f$ on $\kappa\setminus A$ such that $\forall\beta<\kappa:\space\space |f^{-1}(\beta)| <\kappa$.
So we can define $W'$ to be the set of all such subsets $A$, i.e. subsets of $\kappa$ such that there exist such a regressive function $f$.
The main question is as follow:
Question 1: Is it consistent that $W=W'?$ Is it consistent that $W \neq W'?$
It is also interesting to know the consistency strengths.
The following is an equivalent form of a privet case where there is only one normal measure.
Question 2: Suppose that there is only one normal measure on $\kappa$. Consider the statement: For every $A\subset\kappa$ there is a regressive function $f:\kappa\to\kappa$ s.t for all ordinals $\beta,\gamma<\kappa$, $|f^{-1}(\beta)\cap A|<\kappa$ or $|f^{-1}(\gamma)\cap (\kappa\setminus A)|<\kappa$. Is this statement and it's negation consistent?
I would be very happy if you can give me some advice or references for those questions. Thank you very much for every help.
Note: By regressive function I mean a function which is regressive on all of its domain but a number $<\kappa$ of ordinals.
Concerning question 1: The set $W'$ is usually called the club filter, and the complements of sets in $W'$ are called nonstationary sets.
I do not understand question 2.If $A$ is unbounded then you can take $f$ constant on $A$ .
I have edited it so it's ok now
Vaguely related: https://mathoverflow.net/questions/336511/stationary-sets-and-kappa-complete-normal-ultrafilters
The answer to Question 1 is no. Your filter $W$ contains the set of inaccessible cardinals below $\kappa$, but the club filter $W'$ does not.
why not? we can take $f$ of a non inaccessible ordinal to be its cardinality if it's not a cardinal, it's processor if it is not a limit cardinal, and it's cofinality if it's singular.
why does that preclude the existence of some $\beta$ such that $|f^{-1}(\beta)|=\kappa$? In fact there exists one since the non-inaccessible ordinals is stationary.
@Otto You're right. For example, below any measurable (or even just inaccessible) $\kappa$, there are $\kappa$ singular cardinals $\lambda$ of cofinality $\omega$. So all $\kappa$ of these would have $f(\lambda)=\omega$ for the proposed $f$.
|
2025-03-21T14:48:29.662508
| 2020-01-17T22:38:47 |
350660
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/350660"
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|
Stack Exchange
|
What is the minimal $m$ for which the independence graph is $n$-universal?
Suppose, an $m$ sided die is rolled. Let's define the independence graph $I_m$ as a graph with the set of all possible events as vertices, and edges between two events iff they are independent.
Suppose $\Gamma(V, E)$ is a finite simple graph. Letβs call a finite simple graph $\Gammaβ(Vβ, Eβ)$ an induced subgraph of $\Gamma$ iff $Vβ \subset V$ and $Eβ = (Vβ \times Vβ) \cap E$.
Letβs call a finite simple graph $\Gamma$ $n$-universal, iff any finite simple graph on $n$ vertices is isomorphic to some induced subgraph of $\Gamma$.
My question is:
What is the minimal $m$ for which $I_m$ is $n$-universal?
I already know, that $I_n^2$ is $n$-universal, as for any graph with $n$-vertices one can chose $n$ $n$-element subsets (indexed with vertices of that graph) of an $n^2$-element set, such that their their intersection is one-element if there is an edge between the corresponding vertices, and $0$ otherwise.
I also know, that $m$ should be larger that $n + o(1)$ as $I_m$ has $2^m$ vertices and the least possible $n$-universal graph has $(1 + o(1))2^n$ vertices
This question on MSE
I think your $O(1)$ should be a $o(1)$.
|
2025-03-21T14:48:29.662627
| 2020-01-17T23:15:37 |
350662
|
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|
Stack Exchange
|
Reference request for educational material In source format, for blind accessibility purposes
Introduction
I am a blind undergraduate studen in mathematics. I use screen reading software, which uses synthesized speech to read aloud the contents of the screen, to read and write math.
Due to the limitations in presentation-focused formats like PDF and MathJax, screen readers can't properly handle most mathematical content contained in them.
This means that for the most part, I can only properly access math documents in their source format, like LaTeX; this severely limits the resources have access to for studying.
Thankfully, there are some excellent online math resources that include content sources, like Wikipedia and NLab, but sometimes these aren't enough on their own to learn a topic well.
Sometimes I've had luck reaching out to kind lecturers and publishers about particular textbooks, which they were able to provide in source format;
but in general it's very hard to find other good learning resources available in source form.
One other good resource is Arxiv, which often allows you to download the source of an article, but it's very hard to search through the mountains of articles specifically for educational resources.
The Request
What would be really nice is a comprehensive listing by topic of educational math resources that people have made available online in source form;
and this is what I would like this question to become.
So: if you have (or know of) lecture notes, textbooks, or the like, that are available online in source form, via Arxiv, or somewhere else, then post it as an answer, and I will keep the question updated with a categorized list.
(Note: There is no special processing needed to make LaTeX screen-reader-friendly.)
Thank you in advance for making the lives of blind mathematics students easier.
P.S. I hope that this is an appropriate place to make this post; sorry in advance if it isn't.
Do you have a reasonably robust workflow to turn random LaTeX code from arXiv into screen reader friendly material? It'd be great if I could make available accessible versions of my lecture notes without messing with the TeX (too much). (Most of my notes in English are on arXiv anyway, but the German ones might be of interest to somebody as well.)
Also, you can try this search for lecture notes on arXiv: https://arxiv.org/search/advanced?advanced=1&terms-0-operator=AND&terms-0-term=lecture+notes&terms-0-field=comments&classification-mathematics=y&classification-physics_archives=all&classification-include_cross_list=exclude&date-filter_by=all_dates&date-year=&date-from_date=&date-to_date=&date-date_type=submitted_date&abstracts=show&size=50&order=-announced_date_first
There's nothing you need to do, really. If you use macros defined in a separate file then it helps to make that available as well (not a problem on Arxiv), but it's otherwise already screen-reader friendly. I've read tons of unprocessed LaTeX lecture notes from my professors and never had a screen reader or comprehension issue with them.
More likely this finds a better audience at matheducators.stackexchange, but let's see what turns up here in the next little while. Gerhard "Can Always Migrate It Later" Paseman, 2020.01.17.
Perhaps of interest: a very recent (one day ago) announcement about producing math texts in Braille: https://aimath.org/aimnews/braille_full/.
I'm not sure this is technically on topic, but I think that it's a great question and I hope it survives. (It's at least as much on topic as the digital-pen question, for example.)
Concerning arXiv, it has far too many entries every day to permit easy access; but the subject labels provide a good short list if your interests are specific enough.
I think this question is great: not only could it lead to a useful resource, but it made me aware of a zero-effort way to make my lecture notes more accessible which I hadn't thought of before. I'll share this idea with my colleagues.
Out of curiosity, what is the specific problem with MathJax? Are there other math-enabled HTML formats that are usable? What about Unicode math characters? I personally find the mathematics community's dependence on compiled PDFs frustrating and I had assumed that HTML-like formats would be more disability-accessible.
To be clear, MathJax is a lot better than PDF.
The screen reader I use (Gnome Orca), for example, will properly read through most expressions on Wikipedia as a single unit, which is good enough for a lot of less equation-involved content.
The problems arrise when you try to read through expressions at a more granular level, at which point the support gets spotty between the different screen readers.
Orca stops presenting structure information, for example;
(1/)
For example, if I'm trying to read through a long solution of an equation in an aligned environment, with nested fractions, roots, and super/subscripts; I can either listen to the whole thing at once with the structure properly presented (this is not great as you can imagine),
or I can read one cell at a time but not get any structure information (announcements of subscripting, fractions, etc).
(2/)
One great thing about MathJax is that the source is necessarily there for you to read it; but then you either have to copy and clean up the page into a text editor and move back and forth
(when you copypaste MathJax pages into a text editor, it pulls out flattened and newline-stripped LaTeX source right next to the flattened unicode in a jumbled mess);
or go into the page source and read the math alongside the HTML (often messy).
In short: MathJax is much better, and it's getting better all the time, but at this point it's still often a frustrating reading experience.
(3/)
Unicode math characters individually are pretty well-supported by screen readers now.
One great way to represent math in HTML pages that some blogs use (Gowers's and Tao's, in particular) is to embed it as pictures with the LaTeX source as the alt-tag. When you copy and paste the whole page into a text editor, or just disable images, you get a pretty seemless reading experience. I've found this to be the most pleasant HTML math presentation method, actually.
(4/4)
Also, thanks all for the support and the helpful Arxiv searching tips.
One observation about Arxiv that some have already pointed out is that even most of the lecture notes on there are on very specific topics, like the kind that would be presented at shorter seminar courses.
If you do have lecture notes about full semester-long or year-long courses, which are much rarer, that you are willing to share the source of (or already have), then that would be especially helpful for undergraduate students.
As a personal example: I'm currently on the lookout for a good introductory learning resource on representation theory, and I haven't found any so far on Arxiv.
@Zach Teitler That looks like a really awesome project!
Though, on a related note, it might be a useful observation to make for the curious that Braille isn't a very convenient learning medium for university students.
It takes up a lot of space, and takes longer to read, which are both significant problems for students having to lug books around campus and keep up with a fast-paced multiple-class workload.
(1/)
For example, I was able to get Dummit and Foot's Abstract algebra Brailled through my university, but it took up more than 30 volumes that I had to somehow transport to and fit in a small off-campus shared apartment (It was not easy or pleasant).
My university also only fulfills Brailling requests for required course textbooks, so I'm out of luck in that regard for recommended readings, alternate references, or topics I want to self-study.
(2/2)
I feel sorry that there's still no answer for this important question after 1.5 years. I lack reputation for adding a bounty, but pherhaps another user could do so?
|
2025-03-21T14:48:29.663320
| 2020-01-22T17:48:02 |
350956
|
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|
Stack Exchange
|
Computations of Bredon homology of $S(1+\sigma)$ with Universal Coefficient S.S
What I am trying to do is to compute $\mathbb{Z}$-graded Bredon homology of $S(1+\sigma)$ over $Q\times\Sigma_2$, where
$Q$ is a cyclic group of order 2
$\sigma$ is its real sign representation
$\Sigma_2$ is acting on $S(1+\sigma)$ by antipodal action.
In particular, different notation for a cyclic group of order 2 is used to emphasize different actions. To make notation more legible, let's put $S:=S(1+\sigma)$. Its Bredon homology with $\underline{\mathbb{F}}_2$ coefficients is given by $H^{Q\times\Sigma_2}_0(S;\underline{\mathbb{F}}_2)=\mathbb{F}_2\oplus\mathbb{F}_2$ and $H^{Q\times\Sigma_2}_1(S;\underline{\mathbb{F}}_2)=\mathbb{F}_2$.
I may compute this homology also by the Universal Coefficient Spectral Sequence, which is given by
$$
E^2_{p,q}=Tor^{\mathcal{O}_{Q\times\Sigma_2}}_p\left(H_q(S^\bullet),\underline{\mathbb{F}}_2\right)\Rightarrow H^{Q\times\Sigma_2}_{p+q}(S;\underline{\mathbb{F}}_2)
$$
Singular homology here is taken with integral coefficients. (Using this method is a little bit of overkill, but this is supposed to be a toy example for further calculations).
I have inputs of the $E^2$ page computed - they are given by:
$$
Tor_i(H_0(S^\bullet),\underline{\mathbb{F}}_2)=
\left\{
\begin{array}{cc}
\mathbb{F}_2\oplus\mathbb{F}_2 & i=0 \\
\mathbb{F}_2 & i=1 \\
\mathbb{F}_2^{i-1} & i>1
\end{array}
\right.
$$
and
$$
Tor_i(H_1(S^\bullet),\underline{\mathbb{F}}_2)=
\mathbb{F}_2^{i+1}
$$
for $i\geq 0$. Everything else vanish.
So I have my second page of the spectral sequence, but I am struggling with identifying the differentials. From the calculations "by hand" I see that they are supposed to be isomorphisms. But I have got no clue how to show that.
Could anybody give me a hand on that?
The best I can suggest is that, if you give $S$ an equivariant cell structure, it only has nonzero cells in dimensions $0$ and $1$, so there is an exact sequence $$0 \to H_1(S) \to C_1(S) \to C_0(S) \to H_0(S) \to 0.$$ You can deduce (using long exact sequences or spectral sequences) that there is a map $Tor_{i+2}(H_0(S),M) \to Tor_i(H_1(S),M)$, induced by cap product with this Yoneda extension. The middle two terms are projective Mackey functors, so this cap is an isomorphism for large enough $i$. But perhaps this is just shuttling around information you already know...
Well, it's exactly opposite :) How you can deduce existence of these maps?
Which maps? The ones in the exact sequence above, or the ones on Tor? (Or both?)
The ones on Tor, I can see the ones in the exact sequence.
Ok, I have Weibel's Homological Algebra in front of me, I suppose I should be able to deduce these maps - but how to see that they are the ones fitting in the spectral sequence?
If you un-splice into short exact sequences $0 \to H_1(S) \to C_1(S) \to B_0(S) \to 0$ and $0 \to B_0(S) \to C_0(S) \to H_0(S) \to 0$ then the map on $Tor$ is just applying the associated connecting homomorphisms twice.
So far as the spectral sequence goes, you need a specific construction of the universal coefficient spectral sequence to give a precise argument of how the differentials happen. One construction is exactly that you filter $C_\bullet(S)$ by "connective covers" and then put together the long exact sequences on $Tor$ precisely so that the differentials are these iterative connecting homomorphisms, which makes the above description easier -- but a different construction of the spectral sequence might make it harder. Do you have a particular construction in mind?
Yes, I was thinking about construction similar to one given in Weibel's book - i.e. we choose a projective resolution $P_\bullet$ of $\underline{\mathbb{F}}2$, we form a double complex by tensoring with $C\bullet(S^\bullet)$ and filter it by rows and by columns. However, I was struggling wth getting anything out of this construction. It seems that you have better way of constructing it - could you give me any references?
In this case, since it comes from a double complex the d_2 differential is explicit: given an element, lift it to a representing cycle, apply the horizontal differential, observe that it lifts along the vertical differential, and apply the horizontal differential to that.
So start with an element in $Tor_{i+2}(H_0(S),M)$, represented by an element in $H_0(S) \square P_{i+2}$. This element is represented by lifting to some element in $C_0(S) \square P_{i+2}$. The horizontal boundary in $C_0(S) \square P_{i+1}$ has image zero in $H_0(S) \square P_{i+1}$, so (by the 4-term exact sequence, box-product with $P_{i+1}$) it lifts to an element in $C_1(S) \square P_{i+1}$. Applying the horizontal boundary map gives us an element in $C_1(S) \square P_i$ whose vertical differential is zero, so it actually comes from $H_1(S) \square P_i$.
However, this has precisely the same effect as the double application of the connecting homomorphism that I mentioned earlier, except that we have omitted some unnecessary steps (pushing down to $B_0(S) \square P_{i+1}$ and then immediately lifting back to $C_1(S) \square P_{i+1}$).
(I believe that this expression for the $d_2$-differential in a double complex is explicit in Weibel, but I do not know where.)
Thank you so much, this looks like very nice solution to my problem!
|
2025-03-21T14:48:29.663646
| 2020-01-22T18:02:58 |
350958
|
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|
Stack Exchange
|
Best two round game organization
Consider some game (say, table tennis) with two players, in each play outcome is random and independent from previous outcomes. Assume also that probability of winning is fixed for first player and equals to $p$.
If $n$ is odd integer, probability that first player wins more than half of games is
$$
f_n(p) = \sum_{k\ =\frac{n+1}2}^{n} \binom n k p^k(1-p)^{n-k}.
$$
So, which assume there are another odd integer $m$. If there are $m$ rounds with $n$ plays each, probability that first player wins is exactly $f_m(f_n(p))$.
The question is: how to compare $f_n(f_m(p))$ and $f_m(f_n(p))$? Which scheme has better "polorazing" properties? I'm particulary interested in case when $p$ is close to $\frac 1 2$, say $p = \frac 1 2 + \varepsilon$. How to compare values in this points?
I'm also interested in comparison with function $f_{nm}(p)$.
I have a feeling this is may be connected to analysis of boolean functions and Fourier transform on Boolean cube, but is didn't found appropriate tag.
This is an answer only to the very easy part β- namely, we show that, if $m,n>1$ are odd numbers and $p>1/2$, then $f_{mn}(p)>f_m(f_n(p))$.
In both cases, all elementary events are indexed by subsets $I\subseteq 2^{[mn]}$, their probabilities being equal to
$$
P_I=p^{|I|}(1-p)^{mn-|I|}.
$$
Consider the multiset $S$ of all such elementary probabilities $P_I$. Both $f_{mn}(p)$ and $f_m(f_n(p))$ are sums of exactly half of elements in $S$. But $f_{mn}(p)$ is the sum of the largest half in $S$, while $f_m(f_n(p))$ is the sum of some half (it includes at least one element from the smallest half!). Hence the result.
|
2025-03-21T14:48:29.663791
| 2020-01-22T18:47:13 |
350960
|
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|
Stack Exchange
|
Express an integer as ac + bd
Which positive integers $n$ can be expressed in the form $n = ac + bd$, where $a,b,c,d$ are positive integers and satisfy $\gcd(a,b) = \gcd(c,d) = 1$? I know that all numbers that are not divisible by 4 or by a prime $p$ of the form $p = 4m + 3$ can be represented this way due to the "primitive" sum of two squares theorem (see https://math.stackexchange.com/questions/1282550/asymptotic-for-primitive-sums-of-two-squares).
Edit: The interesting case is $a \neq b, c \neq d$.
You probably meant some additional condition, since $a=n-1$ and $b=c=d=1$ satisfies $n=ac+bd$ and your gcd condition! Maybe you want $a,b,c,d$ to be at least $2$?
@JoeSilverman yes you are correct. I have edited the question.
@JoeSilverman actually the weaker condition $a \neq b, c \neq d$ is the case I want.
If cd is allowed to be prime, then all sufficiently large integers should be so expressible. Gerhard "Primes Make So Many Numbers" Paseman, 2020.01.22.
More elementarily, letting a=d=1 turns the problem into representing numbers as b+c with b and c larger than 1, giving all numbers greater than 3 as possibilities.
If only d is allowed to be 1, then let bd be one of 2,4, or 6. For odd n bigger than 9, n-bd is odd and composite,for one of these choices and often not a multiple of three. The problem is if it is a power of three, in which case make a=1, c the power of three, and d=2 and b=3. For even n pick bd a prime not dividing n, then n-bd will sometimes be composite and coprime to bd (otherwise there are too many primes). There should be no n over thirty that do not succumb to such a decomposition if one of the numbers is allowed to be 1 and the two gcd conditions and inequalities apply.
When all numbers are required to be greater than 1 and mutually coprime, there should also be all but finitely many numbers so decomposed. I do not have a construction to cover all cases, but picking two coprime composites near n/2 would be a good start.
Update 2020.01.23:
So I wrote a program to generate positive integers which are not a prime nor a power of a prime.
I then ran a routine to pick a small number diff, and if d and diff were coprime and d and d+diff were both numbers generated above, then record the sum d+d+diff if it had not already been recorded.
The smallest such sum is 29, and I expected few small numbers below 100 to be represented as sums. I also expected to hit most if not all sums by keeping diff below 10 or 20.
That wasn't the case. I needed to have diff go up to 300 before I could find that numbers like 570 and 870 were sums of two coprime and properly composite numbers. I am finding 168,180,204,210,240,270,330, and 420 not of this form. (Since I am excluding prime powers and summands sharing a prime, the list of non representable numbers is larger than suggested in the original post.)
Currently I am finding numbers above 420 representable, and if I constrain diff to less than 200, all but 570 and 870 remain so within the limits of the computation. (If a number near 2*LIM is not realized, I consider that an artifact from not generating enough summands.)
This suggest the conjecture that not only are there finitely many such nonrepresentable positive integers, but that diff does not have to exceed a finite bound to realize such a sum. It may be possible to modify the answer by GH from MO to prove both statements.
End Update 2020.01.23.
Gerhard "Trying To Balance All Sides" Paseman, 2020.01.22.
this answer is very helpful. Can you explain a bit how to show that at most finitely many numbers cannot be represented this way (when all numbers are required to be greater than 1 and mutually coprime)?
For odd n, if n-2 and n-4 are both prime, then n-6 is a nontrivial multiple of three. Set a=3^k, c=(n-6)/a, b=2, and d=3. Choose k so that the gcd conditions hold, and c is the only number that might be 1. For even n, choose odd numbers ac which are composite and coprime to n. If n is larger than 30, this should be possible. Then n-ac = bd is coprime to ac and sometimes is composite. If it isn't set d=1. Gerhard "It Really Is That Simple" Paseman, 2020.01.22.
The program listed in the updates has some curious artifacts. For example (taking sums in increasing order of diff), no sums with diff = 15,21, 27, or 33 are needed, and only 2 with diff 25 are recorded. Numbers needing diff greater than 150 are 570,660,870,4830, and 57330, with LIM=400000. Gerhard "Surprising As Covering With NearPrimes" Paseman, 2020.01.23.
Every $n\neq 3$ can be written as $n=ac+bd$ with $a\neq b$ and $c\neq d$. Simply, take $a=2$, $b=1$, $c=1$, $d=n-2$.
It is more interesting to ask how many representations $n=ac+bd$ there are with $a\neq b$ and $c\neq d$. Let $d(m)$ and $\sigma(m)$ denote the number of divisors and sum of divisors of a number $m$. Then the number of representations of the form $n=ac+bd$ equals
$$\sum_{k=1}^{n-1}d(k)d(n-k)=\left(\frac{6}{\pi^2}+o(1)\right)\sigma(n)(\log n)^2$$
by a result of Ingham (J. London Math. Soc. 2 (1927), 202-208). The number of representations with $a=b$ or $c=d$ is less than $2\sigma(n)$, hence in fact the above asymptotic formula also counts the number of representations with $a\neq b$ and $c\neq d$. In particular, the number of representations tends to infinity as $n\to\infty$.
many thanks for these references though. I certainly find this very interesting!
Are there a lot of terms remaining (do the asymptotics remain) if a) we let k run from n/2 -1000 to n/2 + 1000, and b) also use only terms with k coprime to n? If you get in addition that each k have at least two distinct prime factors, that would also help. Gerhard "Wondering About More Restricted Sums" Paseman, 2020.01.24.
@GerhardPaseman: If you restrict $k$ to be coprime to $n$, then surely the right hand side needs to be modified. This is because there are more ways to write $n$ as a sum of two positive integers than as a sum of two positive integers coprime to $n$.
Yes, but maybe the modification is to drop the o(1) when leaving out k not coprime to n? Gerhard "That's How I'm Seeing It" Paseman, 2020.01.24.
@GerhardPaseman: No, I meant that the asymptotic behaviour will be different. In particular, I expect that $\sigma(n)$ would need to be adjusted as well (which is more dramatic than fiddling with the constant).
|
2025-03-21T14:48:29.664256
| 2020-01-22T18:55:58 |
350961
|
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|
Stack Exchange
|
A $\mathsf{ZF}$ example of two Baire spaces whose product is not Baire?
Motivated by this question I'm looking for a pair of Baire topological spaces whose product is not Baire and whose construction does not need the axiom of choice.
The example of such spaces I'm familiar with, which is due to Cohen, is to consider $\Bbb P_{S_1}\times\Bbb P_{S_2}$, where $S_1,S_2\subseteq\omega_1$ are disjoint stationary subsets of $\omega_1$ and $\Bbb P_{S_i}$ is the usual $\sigma$-distributive forcing that adds to $\omega_1$ a club contained in $S_i$, but this construction needs some choice.
I'm not too familiar with other examples of Baire spaces which are not productively Baire, but those that I could find in the literature all seemed to need some form of choice. Is there a construction in $\mathsf{ZF}$ of two Baire spaces $X,Y$ such that $X\times Y$ is not Baire? Or at the opposite, is it consistent with $\mathsf{ZF}$ that every finite product of Baire spaces is Baire?
I think that your example can be stretched even more. If we assume DC, then for all regular $\kappa$,
the club filter restricted to $S^{\kappa}{\omega}$ is an ultrafilter (if $S_1, S_2 \subseteq S^{\kappa}{\omega}$ are both stationary and disjoint, then the forcings $Col(\omega_1, \kappa) * \mathbb{P}{S_1}$ and $Col(\omega_1, \kappa) * \mathbb{P}{S_2}$ will have non-$\sigma$-distributive product). This looks like a ZF variant of the HOD-conjecture...
@Yair: If you assume DC, might as well just $\operatorname{Add}(\omega_1,2)$ and then take $S_i$ to be the $i$th Cohen subset (or some other configuration of easily definable set from each Cohen subset).
@Asaf: I don't understand your example. I think that the product that you describe is going to be $\sigma$-distributive (under DC), because those two Cohen sets are going to have a large intersection. In the example that I gave, I'm using the fact that $S_1$, $S_2$ are from the ground model.
@YairHayut: Okay, then consider the almost disjoint forcing for $\omega_1$ (so a condition is a countable initial segment, and a countable family of NS subsets), then iterate it twice. Since we have DC, and these are $\sigma$-closed, each step also preserves DC. And yes yes, I know what you're going to say, why are these stationary? Well, they are meeting every club in the ground model, and I think there is enough structure there to ensure that new clubs are not too many, and are also met. This can probably be significantly tweaked and improved, too.
Alessandro, the real question is whether or not it is consistent that there are no $\sigma$-distributive forcings to begin with. For example, what is the situation in Gitik's model where every set is a countable union of smaller sets? Unclear. This somewhat relates to Foreman's maximality principle as well (without choice), that every forcing adds a real or collapses cardinals. But without choice.
That's the first time I hear of Gitik's model, what is a good reference to read its construction? @Asaf
Well, Ioanna Dimitriou wrote a nice review in her PhD thesis, but the second one is easier to read.
@Asaf What is the "second" referring to? Thanks for the reference, I downloaded a pdf of Dimitriou's thesis, the construction looks interesting but way longer than I can read in detail before the exams, I added it to my infinite reading list...
Moti wrote two papers. The original one with a class. Of strongly compact cardinals, and one a couple years later with a huge cardinal.
|
2025-03-21T14:48:29.664514
| 2020-01-22T20:22:41 |
350969
|
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|
Stack Exchange
|
About the Moore composition of paths
1) QUESTION (EDIT: 04/28/2020 to remove a possible counterexample)
I work with weak Hausdorff $k$-spaces (so all spaces are $T_1$). The internal hom is denoted by $\mathbf{TOP}(-,-)$. Let $\mathcal{G}$ be the topological group of nondecreasing homeomorphisms from $[0,1]$ to itself, the composition law being the composition of maps. I denote by $\mu_\ell:[0,\ell]\to [0,1]$ the homothetie $x\mapsto x/\ell$ with $\ell>0$.
I consider a triple $(X,P_{a,b},P_{b,c})$ where $X$ is a topological space, $a,b,c$ are three distinct points of $X$, $P_{a,b}$ is an arbitrary subspace of $\mathbf{TOP}([0,1],X)$ such that for all $\gamma\in P_{a,b}$, $\gamma(0)=a$ and $\gamma(1)=b$, $P_{b,c}$ is an arbitrary subspace of $\mathbf{TOP}([0,1],X)$ such that for all $\gamma\in P_{b,c}$, $\gamma(0)=b$ and $\gamma(1)=c$. I suppose $P_{a,b}$ and $ P_{b,c}$ closed under the reparametrization by $\mathcal{G}$.
I consider the space of paths $P_{a,b}*P_{b,c} \subset \mathbf{TOP}([0,1],X)$ defined as follows: every path $\gamma$ of $P_{a,b}*P_{b,c}$ is of the form :
We choose $(\ell_1,\ell_2)$ such that $0<\ell_1,\ell_2<1$, $\ell_1+\ell_2=1$
$\gamma = (\gamma_1.\mu_{\ell_1}) * (\gamma_2.\mu_{\ell_2})$ with $\gamma_1\in P_{a,b}$ and $\gamma_2\in P_{b,c}$ where $*$ is the Moore composition.
I consider another triple $(X,Q_{a,b},Q_{b,c})$ as above with $P_{a,b} \subset Q_{a,b}$ and $P_{b,c} \subset Q_{b,c}$ and such that the two inclusions are weak homotopy equivalences.
Is it true that the continuous map $g:P_{a,b}*P_{b,c} \subset Q_{a,b}*Q_{b,c}$ is a weak homotopy equivalence ?
The difficulty of this question is to understand $P_{a,b}*P_{b,c}$. There is a continuous map $$\mathrm{Int}(\Delta^1) \times P_{a,b}\times P_{b,c} \to P_{a,b}*P_{b,c} \ \ (1)$$ where $\mathrm{Int}(\Delta^1)$ is the interior of the $1$-simplex which takes $((\ell_1,\ell_2),\gamma_1,\gamma_2)$ to $(\gamma_1.\mu_{\ell_1}) * (\gamma_2.\mu_{\ell_2})$ which is surjective. The map (1) is actually a quotient map. It is not necessarily one-to-one because the paths of $P_{a,b}*P_{b,c}$ may stop at $b$. If none of the paths of $P_{a,b}*P_{b,c}$ stops at $b$ (i.e. the inverse image is one point of $]0,1[$, then the map (1) above is a homeomorphism. And if none of the paths of $Q_{a,b}*Q_{b,c}$ stops at $b$ as well, then the map $g$ is a weak homotopy equivalence.
2) MOTIVATION (EDIT 02/13/2020)
I add the motivation in the hope that someone could have a suggestion by reading it. The proof of the left properness of the q-model category of multipointed $d$-spaces given in Left properness of multipointed d-spaces is incomplete. I forgot to treat the case of the generating cofibration $R:\{0,1\}\to \{0\}$ identifying two states, i.e. that the pushout of a weak equivalence along $R:\{0,1\}\to \{0\}$ is a weak equivalence. The problem above is a particular case. The similar fact for the category of flows is trivial. The generating cofibration $R:\{0,1\}\to \{0\}$ is not necessary to build the cellular objects so it would not be a real problem if the category of multipointed $d$-spaces was only "almost" left proper.
3) ABOUT THE WEAK HAUSDORFF CONDITION (EDIT 02/13/2020)
If the spaces are not $T_1$, then I run into several complicated point-set topology obstacles because there are three non-homeomorphic topologies on two points: the discrete one, the indiscrete one and the Sierpinski topology. To avoid pointless complicated mathematical arguments, it is sufficient to work with the correct separability condition for $\Delta$-generated spaces as explained in the model category structure and its left determinedness (Section 3). The question above is written down in the framework of weak Hausdorff $k$-spaces because I do not think that the local presentability condition has any role in this problem.
I don't understand how $\mathcal{G}$ is a group? Shouldn't the inverse of a nondecreasing function be nonincreasing?
@JeffStrom $\mathcal{G}$ is a group for the composition of maps. It contains only homeomorphisms.
Of course. Too silly.
|
2025-03-21T14:48:29.664781
| 2020-01-22T21:37:48 |
350973
|
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|
Stack Exchange
|
Relative property (T) and normal closure
I am in a situation where a discrete, finitely generated group $H$ satisfies property (T), and was wondering if I was able to conclude anything about the pair $(G,H^G)$, where $G$ is a finitely generated discrete group containing $H$, and $H^G$ is the normal closure of $H$ in $G$. In particular, does this pair necessarily also satisfy (T)? If not, are there conditions which would guarantee that it did?
Thanks for your time!
Clearly no: for instance when $H$ is finite $H^G$ can be large. For instance in a free product $G=H\ast K$ with $H,K\neq 1$ and $K$ having Property T the answer is always no. It's even false when $H$ is a commensurated subgroup, using slightly more refined counterexamples, such as $\mathbf{Z}[1/2]^2\rtimes (\langle 2\rangle \mathrm{SL}_2(\mathbf{Z}))$.
Also in my paper Relative Kazhdan Property, Proposition 1.13: I showed that there is a group $\Gamma$ (lattice in some semisimple group) with some infinite subgroup $\Lambda$ such that $(\Gamma,\Lambda)$ has relative Property T, but such that for every subgroup $\Gamma'$ of $\Gamma$ and normal infinite subgroup $\Lambda'$ of $\Gamma$, the pair $(\Gamma',\Lambda')$ does not have relative Property T.
Meant normal subgroup of $\Gamma'$, sorry.
|
2025-03-21T14:48:29.664895
| 2020-01-22T22:19:19 |
350976
|
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|
Stack Exchange
|
Pseudo-holomorphic disk which is constant along boundary
Let $(M,J,\omega)$ be a symplectic manifold with a compatible almost complex structure, $D$ be the closed unit disk in $\mathbb{C}$, and $u:(D,i)\to (M,J)$ be a $(J,i)$-holomorphic map.
Question: Assume $u|_{\partial D}$ is constant, does this imply $u$ is a constant map?
Extend $u$ to get a $C^1$ pseudoholpmorphic map defined on $\mathbb{C}$ by setting $u$ constant outside the unit disc. It's $C^1$ because you know the derivative of $u$ along the unit circle vanishes (by assumption), so the Cauchy-Riemann equations satisfied by $u$ on the disc tell you that $du$ vanishes along the unit circle; clearly $du$ continues to vanish outside the disc, hence it's $C^1$. It's pseudoholomorphic because this is a pointwise condition on derivatives which clearly holds piecewise for this map. Now by unique continuation, $u$ is constant.
Thank you Jonny! I now understand. Actually I am thinking about a seemingly harder question: if $u:D\to M$ is nonconstant pseudoholomorphic, then does $u|_{\partial D}$ have only (finitely many) isolated critical points? May I ask if you have an idea about this?
My guess is that the answer is yes, but I don't know for sure.
I realized that boundary critical points are isolated if we add Lagrangian/ totally real boundary condition. This is due to a similar similarity principle as interior points. Since I only care about holomorphic disks bounded by a Lagrangian submanifold, itβs all done.
|
2025-03-21T14:48:29.665015
| 2020-01-23T00:43:30 |
350977
|
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"Leo Herr",
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|
Stack Exchange
|
When do generizations ("generalizations") lift uniquely?
If $f : X \to Y$ is proper, then specializations lift along $f$, and uniquely.
(This means, if $R$ is a discrete valuation ring with fraction field $K$ and I choose a factorization $\text{Spec}K \to \text{Spec} R \to Y$ through $X$, I get a unique factorization of $\text{Spec} R \to Y$ through $X$ inducing it. )
If $f : Z \to Y$ is flat, then generizations lift (Stacks Project 040F).
(Let $k$ be the residue field of the DVR $R$ from before. Subtlety: I have to allow $k \subseteq k'$ an extension and choose a factorization $\text{Spec}k' \to \text{Spec} k \in \text{Spec} R \to Y$ through $X$. Then I get a DVR $R'$ with residue field $k'$ and a factorization of $\text{Spec}R' \to \text{Spec} R \to Y$ through $X$. This is ''going down.'')
Is there a condition on a scheme/map so that generizations lift uniquely? You have to exclude the trivial generization, which seems ad hoc. I'm not sure how to make this precise. Here is an example to highlight the difference:
Consider $S = \mathbb{A}^1$ and $T = (xy = 0) \subseteq \mathbb{A}^2$ over a base field $\text{Spec} L$. Both are flat over the base automatically. Yet for $S$, there's only one nontrivial generization of any closed point, the generic point. For $T$, the generic points of each component both generize the point $x = y = 0$.
Generizations happen within an open neighborhood of a point, so it seems like I'm asking for the topological space to be locally irreducible.
(Due diligence statement: I tried a handful of different searches to find something like this question. Hensel's Lemma is related. The phrase "generalizations lift" yields too many google responses, so my apologies if I've missed something.)
edit: Strong conditions on dimensions of points to be generized seem necessary. If we require such lifting for codimension-one points, then ``local irreducibility'' seems right. On $\mathbb{P}^2$, there should be infinitely many curves specializing through each point given by lines $\mathbb{P}^1$ through the origin. This may indicate that the expectation is unreasonable or uninteresting, but I'd still like to hear if the notion has any utility with some dimensional restrictions.
Your condition is equivalent to stating that, for any map $\operatorname{Spec} R \to X \times_Y X$, if the special point factors through the diagonal $\Delta: X \to X \times_Y X$, then the whole map factors through the diagonal.
A sufficient condition is clearly that the diagonal is an open immersion, because if the special point is in an open set than the whole local ring will be. This is one of the conditions for a morphism to be unramified, so being unramified is also a sufficient condition.
I'm not sure, but I think it's probably not a necessary condition.
If X \to Y is quasiseparated, the diagonal is compact and is open if and only if stable under generization (stacks 0542). If X \to Y is further locally of finite type, unramified and open diagonal are equivalent 02GE. I'm fine with these minor finiteness hypotheses, and I think yours is the answer.
|
2025-03-21T14:48:29.665361
| 2020-01-23T01:03:10 |
350979
|
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|
Stack Exchange
|
Permuting $n$ points in a $2$-manifold
Given $n$ points on a connected $2$-manifold $M$, I'd like to consider the homotopy classes of paths that "permute" these points.
Edit (Clarifying what I mean by this):
Given a set of $n$ distinct points $T=\{x_{1},\ldots,x_{n}\}\subset M$, to each point we assign a continuous simple curve $\gamma_{i}:[0,1]\to M$ such that $\gamma_{i}(0)=x_{i}, \gamma_{i}(1) \in T$ and $\gamma_{i}(s)\neq\gamma_{j}(s)$ for all $s\in[0,1]$ (where $i\neq j$). I'd like to consider the homotopy classes of all such possible curves.
It seems obvious that these homotopy classes should constitute the elements of a group. Is that right? If so, what's the name of this group? I'm inclined to simply call this the motion group $\text{Mot}_{n}(M)$ of the $n$ points on $M$. Does this coincide with the mapping class group of $n$ points in $M$? Also, do I need any more restrictions on $M$? If so, why?
E.g. considering $3$-space for a moment, it is obvious that $\text{Mot}_{n}(\mathbb{R}^{3})\cong S_{n}$ (where $S_{n}$ is the permutation group).
It is also obvious that $\text{Mot}_{n}(\mathbb{D}_{2})\cong \text{Mot}_{n}(\mathbb{R}^{2})\cong B_{n}$, where $\mathbb{D}_{2}$ is the $2$-disk with boundary and $B_{n}$ is the braid group.
Consider a presentation of $\text{Mot}_{n}(M)$ with relations $R$.
(i) Is it true that $\text{Mot}_{n}(M)\cong B_{n}(M)$, where $B_{n}(M)$ is the surface braid group for $M$?
(ii) Under what conditions will it be true that the generator relations $G$ of $B_{n}$ will be a subset of $R$?
For instance, I'm sure that $\text{Mot}_{n}(S^{2})\cong B_{n}(S^{2})$ : in which case, we do have $G\subset R$ (in fact, $B_{n}(S^{2})$ is a quotient of $B_{n}$).
Relevant Resources:
A survey of surface braid groups and the lower algebraic K-theory of their group rings
I think my notion of $\text{Mot}_{n}(M)$ coincides with the Definition in Section 2.2 of the above paper. If so, then the answer to 2(i) is yes (according to the paper). Building on this, I believe Theorems 12 and 13 of Bellingeri (for $m=0$) may provide a partial answer to 2(ii).
Questions about the definition. Let $T\subset M$ be a finite subset of cardinality $n$.This gives a basepoint for the configuration space of $n$ unordered points in $M$. Call this configuration space $C_n(M)$. Then it seems to me what you are describing is $\pi_0$ of the loop space $\Omega C_n(M)$. Is that right? If so, then your group is nothing more than $\pi_1(C_n(M))$. Right?
Without further care in formulating definitions, I do not know what it might mean for a path to permute points. I presume by a "path" you mean a continuous function $f : [0,1] \to M$, in which case "permuting points" is not something a path ordinarily does.
I've clarified what I meant in the post now. @JohnKlein I don't have a clear understanding of why $\pi_{1}(C_{n}(M))$ doesn't restrict to the trivial permutation?
*I would identify $Ο_1(C_n(M))$ with the homotopy equivalence classes of curves that fix $\gamma_i(0)=\gamma_i(1)$ in my description.
What you wrote in your post is still not clear: you have a curve in $M$ and you also have one in $M\times [0,1]$. Please commit and make it precise.
Isn't your description just an equivalence class of path $\gamma: [0,1] \to C_n(M)$ such that $\gamma(0) = T = \gamma(1)$, where $C_n(M)$ is the space of subsets of cardinality $n = |T|$?
Ah, I see. That makes sense if $T$ is an unordered set (which is implicit in your description). So these $n$ curves in $M$ just look like one loop in $C_{n}(M)$. If $T$ was ordered, then you'd only get a loop for the trivial permutation? I guess there are some subtleties about $M$ being path-connected meaning $C_{n}(M)$ being path-connected too (for $T$ unordered), allowing us to forget about the basepoint.
@MarkSapir That's unfortunate - the description I gave was intended to match that of the Definition given in Section 2.2 of the linked paper. I believe $\text{Mot}{n}(M)$ is then $\pi{1}(C_{n}(M)/S_{n})$ (which the above discussion helped me understand). According to the paper, this is indeed $B_{n}(M)$. I guess that just leaves question 2(ii).
The comments seem to have answered questions 1 and 2i, to show that the group $\operatorname{Mot}_n(M)$ is indeed the surface braid group $B_n(M)$.
To answer 2ii, consider a disk $D \subset M$ such that $T \subset D$. Then the inclusion map $D \hookrightarrow M$ induces a homomorphism $B_n \to B_n(M)$, so the relations in $B_n$ always hold in $B_n(M)$.
Simple enough !
Bump... quick question: for your (induced homomorphism) argument to be true, doesn't this also mean there has to be a continuous map from the (unordered) configuration space of $D$ to that of $D\subset M$? That doesn't seem immediately obvious to me. Sorry for the posthumous enquiry!
It's not immediately obvious, but It does induce a continuous map between unordered configuration spaces. One way to see this is to first check it on the ordered configuration spaces, and then make sure the induced inclusion respects the quotient maps.
However, it's clearer if you use the viewpoint you presented in your original question, of looking at elements of the surface braid group as a collection of paths on $D$ or $M$. Then a collection of paths in $D$ is clearly a collection of paths in $M$ once you include $D$ into $M$.
Got it, thanks for your help!
|
2025-03-21T14:48:29.665741
| 2020-01-23T01:17:10 |
350980
|
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|
Stack Exchange
|
Scaling in Mehta's integral
The following expression is known as Mehta's integral and deeply connected to random matrix theory:
$$\frac{1}{(2\pi)^{n/2}}\int_{-\infty}^{\infty} \cdots \int_{-\infty}^{\infty} \prod_{i=1}^n e^{-t_i^2/2}
\prod_{1 \le i < j \le n} |t_i - t_j |^{2 \gamma} dt_1 \cdots dt_n =\prod_{j=1}^n\frac{\Gamma(1+j\gamma)}{\Gamma(1+\gamma)}.$$
An interesting question is what happens if one assumes $\gamma$ to be a function of $n.$
For example by choosing $\gamma=1/n$ one finds that as $n$ tends to infinity, the value of the integral tends to zero whereas for $\gamma=1/n^2$ the value of the integral approaches a positive constant value as $n$ tends to infinity.
These properties one can deduce from the asymptotics of the product of gamma functions. I would like to ask:
It is not too surprising that for some suitable scaling $\gamma=1/n^{\alpha}$ one approaches a constant value, as $\vert t_i-t_j \vert^{1/n} \xrightarrow 1$ for fixed $t_i,t_j$ and
$$\frac{1}{(2\pi)^{n/2}}\int_{-\infty}^{\infty} \cdots \int_{-\infty}^{\infty} \prod_{i=1}^n e^{-t_i^2/2}
dt_1 \cdots dt_n =1.$$
Can one also conclude these two properties from the integral directly without evaluating it?
Yes, this follows by the de la VallΓ©e-Poussin necessary and sufficient condition for the uniform integrability. Indeed, suppose that
\begin{equation}
\gamma n^2\to a
\end{equation}
(as $n\to\infty$) for some real $a\ge0$.
Your integral is
$$E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma},$$
where the $X_i$'s are independent standard normal random variables.
Introducing $N:=n(n-1)/2$, $X:=(X_1,\dots,X_n)$, and $\|X\|:=\sqrt{\sum_1^n X_i^2}$, and then using the arithmetic-geometric-mean inequality, we have
$$\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}
\le\Big(\frac1N\,\sum_{1\le i<j\le n}|X_i-X_j|^2\Big)^{N\gamma} \\
=O\Big(\frac{\|X\|^2}n\Big)^{N\gamma}=O\Big(1+\frac{\|X\|^2}n\Big)^C
$$
for $C:=a/2+1$ and all large enough $n$.
Note also that $\|X\|^2$ has the gamma distribution with parameters $n/2$ and $2$ and hence $E\|X\|^{2C}=O(n^C)$. So,
$$E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}=O(1)$$
and, similarly,
$$E\Big[\Big(\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}\Big)^2\Big]=O(1).$$
Also, obviously, $t^2/t\to\infty$ as $t\to\infty$. So, we have the uniform integrability.
So, we only need to establish the convergence of
\begin{equation*}
{2\gamma}\sum_{1\le i<j\le n}\ln|X_i-X_j|=2\gamma NU_n
\end{equation*}
in probability, where
\begin{equation*}
U_n:=\frac1N\,\sum_{1\le i<j\le n}h(X_i,X_j)
\end{equation*}
is a so-called U-statistic with kernel $h(X_i,X_j):=\ln|X_i-X_j|$, and still $N=\binom n2=n(n-1)/2$. It is easy to see (cf. e.g. page 20) that $Var\,U_n=O(1/n)=o(1)$, whereas
$$EU_n=m:=E\ln|X_1-X_2|.$$
So, $U_n\to m$ in probability, whence
\begin{equation*}
2\gamma NU_n\to am
\end{equation*}
in probability and thus, by the uniform integrability,
$$E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}\to e^{am}=\exp\{a\,E\ln|X_1-X_2|\}$$
as $n\to\infty$.
interesting, but how did you estimate this product? Also, the dimension of the integral changes, which is perhaps a bit different from the usual DCT.
@SolidStatePhysicist : I have added the requested details.
@SolidStatePhysicist : You were right. I had indeed overlooked that the dimension of the integral, $n$, goes to infinity. This is now fixed. Instead of the integration over $\mathbb R^n$, with the variable $n$, we now use the expectation, which is the integration over a fixed background probability space. Instead of dominated convergence, we now use uniform integrability, which complicates the reasoning just a bit.
@IosifPinelis I think there is an issue with this answer. The OP claims and asking for a proof that this property holds for $\gamma=1/n^2$. In your answer however, you assume that $\gamma=o(1/n^2).$ This seems to be irrelevant for the uniform integrability, although the property is mentioned there, but it is clearly used to show convergence in probability. I'd be curious to hear what you think?
@XinWang good point, I guess this point would require some clarification. I think I looked through the proof too fast when I read it and was satisfied to see that it was essentially uniform integrab. + convergence in measure
@IosifPinelis Do you see a way to extend your argument to $\gamma=\mathcal O(1/n^2)$ rather than $o(1/n^2)$?
@XinWang : I have added the modification you requested.
@SolidStatePhysicist : I have added the modification you requested.
thank you very much.
I have now combined the cases $\gamma=o(1/n^2)$ and $\gamma\sim a/n^2$ for real $a>0$ into one case: $\gamma n^2\to a\ge0$.
|
2025-03-21T14:48:29.666045
| 2020-01-23T01:19:29 |
350981
|
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"Benighted",
"Will Sawin",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350981"
}
|
Stack Exchange
|
Prove category of constructible sheaves is abelian
Let $X$ be a nice enough topological space, perhaps a complex algebraic variety with its analytic topology. I'm hoping someone could help me prove that the category $\text{Constr}(X)$ of constructible sheaves on $X$ is abelian.
I think one could use that subsheaves and quotients of local systems are local systems. However, conceptually what is confusing me is that given two constructible sheaves $\mathcal{F}, \mathcal{G}$, they might be constructible with respect to completely different Whitney stratifications of X, right?
So given a morphism $\mathcal{F} \to \mathcal{G}$ of constructible sheaves, can you always "refine" the two Whitney stratifications to get one such that the morphism restricts to a morphism of local systems?
Or should I be thinking along another lines?
There is always a stratification that refines any two given stratifications. Isnβt this almost obvious from the definition?
@WillSawin I suppose, but is it clear the refinement will be Whitney? Or is it not true that a constructible sheaf must be locally constant over a Whitney stratification? I've been confused whether you need Whitney in the definition of constructible, because I've seen it both ways.
I think every stratification of a reasonable space has a refinement that is a Whitney stratification. This would make the two definitions nicely equivalent, and answer this question as well.
For instance Theorem 2.2 of Verdier's paper seems to be the desired statement https://eudml.org/doc/142424
@WillSawin Great that makes sense. Thanks a lot. If it matters, I would accept it if you wrote this into an answer.
Verdier has proved (for multiple classes of spaces, including in particular complex varieties) that for any finite set of analytic subsets, there exists a Whitney stratification for which all these analytic subsets are unions of strata.
In particular, given two constructible sheaves, and a stratification on which they are lisse, there exists a Whitney stratification that refines both of them.
This suffices to check that kernels and cokernels of maps of constructible sheaves are constructible, as well as direct sums of constructible sheaves.
|
2025-03-21T14:48:29.666208
| 2020-01-23T01:22:20 |
350982
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350982"
}
|
Stack Exchange
|
Derived manifold and real virtual dimension
In https://arxiv.org/pdf/1504.00690.pdf, it seems like the "derived manifold structure" given on a certain complex analytic space seems to have the real virtual dimension the same as the complex virtual dimension of the original (derived) scheme over $\mathbb{C}$. I have little knowledge in anything that's behind this, so what does this result really mean? Particularly, what does it mean for a derived manifold over a complex analytic space to have an odd real virtual dimension?
|
2025-03-21T14:48:29.666270
| 2020-01-23T01:33:46 |
350983
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350983"
}
|
Stack Exchange
|
Is there a lower bound in the opposite direction of the data processing inequality?
Let $f_1$ and $f_2$ be the distributions of random variables $X_1$ and $X_2$ respectively. Let $g_1$ and $g_2$ be the distributions of $T(X_1)$ and $T(X_2)$ respectively, where $T(\cdot)$ is some function. For any f-divergence, $D_{f}(\cdot\|\cdot)$, the data processing inequality gives us $$ D_{f}(g_{1}\|g_{2}) \leq D_{f}(f_{1}\|f_{2}).$$ Are there any known results in the opposite direction in the form $D_{f}(g_{1}\|g_{2}) \geq h(D_{f}(f_{1}\|f_{2})) $ for some function $h(\cdot)$?
|
2025-03-21T14:48:29.666339
| 2020-01-23T05:00:50 |
350988
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andy Jiang",
"Denis Nardin",
"Matthew Pressland",
"Sam Gunningham",
"https://mathoverflow.net/users/136287",
"https://mathoverflow.net/users/21483",
"https://mathoverflow.net/users/43054",
"https://mathoverflow.net/users/7762"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350988"
}
|
Stack Exchange
|
Is the category of spectra on $\mathbb{P}^1$ a module category?
I cannot really state my question in an incredibly precise way as I'm very new to this area, but I hope what I'm asking will be clear. Let $\mathcal{C}$ be the infinity category of sheaves of quasi-coherent spectra on $\mathbb{P}^1$ as a scheme. Then, I think this should be a stable infinity category. Higher Algebra <IP_ADDRESS> says that if this category is presentable and has a compact generator, it is the module category for some ring ($\mathbb{E}_1$ ring). My question is, are the conditions satisfied in this case? I suspect it shouldn't be because I don't think it should be a module category. However, I'm not sure why the structure sheaf is not a compact generator and I feel that the category of quasi-coherent sheaves ought to be presentable. What is the issue here? Is there actually a ring which represents projective space?
If I understand the question correctly,$M = \mathcal O \oplus \mathcal O(1)$ should be a compact generator for $\mathcal C$, so $\mathcal C$ is equivalent to modules for $RHom(M,M)$. Note that $RHom(\mathcal O, \mathcal O(-1)) = 0$, so the structure sheaf is not a generator for $\mathcal C$.
Does taking homotopy categories recover the equivalence of derived categories of coherent sheaves on $\mathbb{P}^1$ and modules over the Kronecker algebra? Asking both as a sanity check for me, not being an infinity categories person, and because if yes it would be a reason to have suspected the higher categorical statement in advance.
@SamGunningham Ah I see, you're definitely right about the structure sheaf. but you're saying that there is some ring whose module category is the modules on P1! Can you describe this ring ? (I mean E1 ring everywhere)
@davik Its homotopy is rather easy to describe (this is basically the computation of the sheaf cohomology of line bundles on $\mathbb{P}$), is there anything else you want to know about it in particular?
@davik The standard answer, due to Beilinson, and as mentioned by Mathhew Pressland above, is that the derived endomorphism ring of $\mathcal O \oplus \mathcal O(1)$ on $\mathbb P^1$ is equivalent to the path algebra of the Kronecker quiver (so in particular is concentrated in degree $0$). I guess you are asking whether this remains true of we work over the sphere spectrum rather than over a field (which is how I have usually seen this stated). I would guess that the answer is yes, but I am by no means certain.
$\newcommand{\PP}{\mathbf{P}} \newcommand{\QCoh}{\mathrm{QCoh}} \newcommand{\cf}{\mathcal{F}} \newcommand{\cg}{\mathcal{G}} \newcommand{\Map}{\mathrm{Map}} \newcommand{\co}{\mathcal{O}}$ The ordinary abelian category of quasicoherent sheaves on $\PP^1$ is not the category of modules over any ring. We shall prove the following more general result (which in the discrete case is precisely the result of Beilinson's mentioned in the comments): let $R$ be a connective $\mathbf{E}_\infty$-ring, and let $f:\PP^n_R\to \mathrm{Spec}(R)$ denote the flat projective $n$-space over $R$ (see Section 5.4.2 of SAG --- in particular, if $R$ is discrete or a $\mathbf{Q}$-algebra, then this is the usual projective space). Then $\QCoh(\PP^n_R)$ is equivalent to the ($\infty$-)category of right modules over an $\mathbf{E}_1$-ring. For non-higher-categorical people: if $R$ is discrete, then taking homotopy categories recovers Beilinson's result.
(In the course of the proof below, I'll appeal to some results from SAG, which are easy to prove in the classical setting; the argument below is essentially the usual one used to deduce Beilinson's result.)
Since $\QCoh(\PP^n_R)$ is stable and presentable, the Schwede-Shipley theorem you mentioned implies that it suffices to prove that there is a compact generator for $\QCoh(\PP^n_R)$. As in the classical case, the sheaf $\cf = \co \oplus \co(1) \oplus \cdots \oplus \co(n)$ is such a compact generator. Indeed, $\cf$ is obviously compact. Suppose $\cg\in \QCoh(\PP^n_R)$ is such that $\Map(\cf, \cg) = 0$ (in the classical setting, this means derived Hom). Since $\Map(\cf, \cg) = \bigoplus_{k=0}^n f_\ast(\cg\otimes \co(-k))$, the hypothesis on $\cg$ implies that $f_\ast(\cg\otimes \co(k))$ vanishes for all $0\leq k\leq n$. Lemma <IP_ADDRESS> of SAG implies by induction that $f_\ast(\cg\otimes \co(k))$ vanishes for all $k$. In the classical setting, this lemma from SAG amounts to the statement that there is an exact sequence
$$0\to \co\to \co(1)^{\oplus n+1}\to \co(2)^{\binom{n+1}{2}} \to \cdots \to
\co(n+1)\to 0.$$
In particular, we conclude that $\Map(\co(k), \cg) = 0$ for all $k$.
By Lemma <IP_ADDRESS> of SAG, there is a map $\bigoplus_\alpha \co(d_\alpha)\to \cg$ which induces a surjection on $\pi_0$. In the classical setting, this amounts to the statement that $\cg$ can be resolved by line bundles. The above discussion implies that this morphism is null, and therefore that $\pi_0 \cg = 0$. By replacing $\cg$ with a shift, we conclude that $\pi_d \cg = 0$ for all integers $d$, and hence that $\cg = 0$, as desired.
This argument shows that $\QCoh(\PP^n_R) \simeq \mathrm{Mod}(A)$, where $A$ is the $\mathbf{E}_1$-ring $\mathrm{End}\left(\bigoplus_{k=0}^n \co(k)\right)$, just as in the classical setting. When $n=1$, this decomposition amounts to the semiorthogonal decomposition of $\QCoh(\PP^1_R)$: namely, the full subcategory of $\QCoh(\PP^1_R)$ determined by the functor $\mathrm{Mod}(R)\to \QCoh(\PP^1_R)$ sending $\cf$ to $f^\ast(\cf)\otimes \co(1)$ determines a semiorthogonal decomposition
$$\mathrm{Mod}(R) \underset{f_\ast(-\otimes \co(-1))}{\stackrel{f^\ast}{\rightleftarrows}} \QCoh(\PP^1_R) \underset{f^\ast(-)\otimes \co(1)}{\stackrel{f_\ast}{\rightleftarrows}}
\mathrm{Mod}(R),$$
which is a categorification of the projective bundle formula. In the discrete case, the $\mathbf{E}_1$-ring $A$ is the path algebra of the Beilinson quiver (for general $n$), which when $n=1$ is the Kronecker quiver $\bullet \rightrightarrows \bullet$. The semiorthogonal decomposition above is just restrictions to each of the vertices. (One comment: the equivalence proved above is not monoidal: the tensor product of two representations of the Kronecker quiver is not just the usual tensor product of quasicoherent sheaves on $\PP^1$.)
|
2025-03-21T14:48:29.666736
| 2020-01-23T07:54:52 |
350990
|
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"ABIM",
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|
Stack Exchange
|
Probabilistic Approximation of non-linear Dynamical System by Diffusion Process
Setting
Suppose I have a discrete dynamical system given by:
$$
X^{n+1} = f(X^{n})
\qquad X^0 =x
,
$$
where $f$ is some diffeomorphism from $\mathbb{R}^{d}$ to itself, and some $x \in \mathbb{R}^d$.
Question
Given a $d$-dimensional Brownian motion $W_t$ and integer $n$.
When does there exist, functions $\tilde{f}:\mathbb{R}^{d\times n}\rightarrow \mathbb{R}^d$, $\rho:[0,\infty)\rightarrow [0,\infty)$ and a predictable-process $u_t$ such that for small $\sigma > 0$:
$$
Z_t^{\sigma}= \int_0^t\tilde{f}(Z_{s}^{\sigma},u_s)ds + \sigma W_t \qquad Z_0^{\sigma} = x
$$
satisfies the "small deviation-type principle:"
(Approximation of $X^n$ by $Z_t$) for every $\epsilon>0$ there is some $n_{\epsilon}>0$ satisfying:
$$
\limsup_{\sigma \to 0}
\mathbb{P}\left\{
\|Z_t^{\sigma} - X^n\|<\epsilon:\, (\forall t \in [n,n+1))
\right\}>\rho(\epsilon) \qquad (\forall n\geq n_{\epsilon})
$$
(Rate function $\rho$ decays to $0$) $$\lim\limits_{t\to\infty} \rho(t) = 0.$$
Ideally:
Can $\tilde{f}$ be written down explicitly (I don't need $u_t$ to be explicit)?
I made some edits since there was no explicit need for the control $U^n$ in the discrete system.
You're right, I didn't need the control. Thanks!
|
2025-03-21T14:48:29.666848
| 2020-01-23T08:40:17 |
350991
|
{
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"authors": [
"Ramiro de la Vega",
"Taras Banakh",
"Todd Trimble",
"https://mathoverflow.net/users/17836",
"https://mathoverflow.net/users/2926",
"https://mathoverflow.net/users/61536"
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|
Stack Exchange
|
The diamond principle for functors
Let $F:\mathbf{Comp}\to\mathbf{Set}$ be a continuous functor from the category of compact Hausdorff spaces to the category of sets such that $|Fn|\le\mathfrak c$ for any finite ordinal $n$. The continuity of $F$ means that $F$ preserves limits of inverse spectra.
Typical examples of such a functor $F$ are the functors of countable power, of the hyperspace, or of spaces of probability measures.
I am interested in validity of the following functorial version of the Jensen Diamond principle:
$\diamondsuit_F$: there exists a transfinite sequence $\langle \mu_\alpha:\alpha\in\omega_1\rangle$ such that
$\bullet$ $\mu_\alpha\in F(2^{\alpha})$ for every $\alpha\in\omega_1$;
$\bullet$ for every $\mu\in F(2^{\omega_1})$ the set $\{\alpha\in\omega_1:\mu_\alpha=F\pi_\alpha(\mu)\}$ is stationary in $\omega_1$.
Here $\pi_\alpha:2^{\omega_1}\to 2^\alpha$, $\pi_\alpha:f\mapsto f{\restriction}\alpha$, is the projection onto the $\alpha$th face of $2^{\omega_1}$.
Observation. The classical Jensen Diamond Principle is just $\diamondsuit_{Id}$ for the identity functor $Id$. It is equivalent to the Principle $\diamondsuit_{Id^\omega}$ for the functor $Id^\omega$ of countable power. Using the Parovichenko Theorem, it can be shown that the Jensen Diamond Principle is equivalent to $\diamondsuit_{\exp}$ for the functor $\exp$ of hyperspace.
I am interested in $\diamondsuit_P$ for the functor $P$ of probability measures.
Problem. Does $\diamondsuit_P$ follow from the Jensen diamond principle? Or it is a stronger statement (still holding in the Constructible Universe)?
What does "normal" mean in this context?
@ToddTrimble, I would guess Taras meant it in ShchepinΒ΄s sense (i.e. a functor that preserves: inverse limits, weight, injectivity of maps, surjectivity of maps, intersections, inverse images, a point and the empty set). Shchepin introduced this notion precisely to encompass functors like $\exp$, $P$ and symmetric powers.
Indeed, I had in mind normal functors in the sense of Shchepin. However the definition of $\diamondsuit_F$ can be given for any functor from the category $\mathbf{Comp}$ to the category $\mathbf{Set}$. So, I have changed my problem accordingly.
The answer to this question is affirmative and follows from a more general
Theorem. Let $X_{\omega_1}$ be the limit of a continuous well-ordered spectrum $\langle X_\alpha,p_\alpha^\beta:\alpha\le \beta<\omega_1\rangle$ in the category of sets such that each set $X_\alpha$, $\alpha<\omega_1$ has cardinality $\le\mathfrak c$. The Jensen Diamond Principle $\diamond$ implies that there exists a transfinite sequence $(\mu_\alpha)_{\alpha\in\omega_1}\in\prod_{\alpha\in\omega_1}X_\alpha$ such that for any $\mu\in X_{\omega_1}$ the set $\{\alpha\in\omega_1:\mu_\alpha=p^{\omega_1}_\alpha(\mu)\}$ is stationary in $\omega_1$.
Proof. Let $C=2^\omega$ be the Cantor cube. For ordinals $\alpha\le\beta\le\omega_1$ let $\pi_\alpha^\beta:C^{\beta}\to C^\alpha$, $\pi_\alpha:x\mapsto x{\restriction}\alpha$, the projection onto the $\alpha$th face of the cube $C^\beta$.
It is well-known that $\diamondsuit$ implies the existence of a transfinite sequence $(x_\alpha)_{\alpha\in\omega_1}\in C^{\omega_1}$ such that
for any $x\in C^{\omega_1}$ the set $\{\alpha\in\omega_1:x_\alpha=\pi^{\omega_1}_\alpha(x)\}$ is stationary in $\omega_1$.
Taking into acount that the spectrum $\langle X_\alpha,p_\alpha^\beta:\alpha\le \beta<\omega_1\rangle$ is continuous and consists of sets of cardinality $\le\mathfrak c=|C|$, it is possible to construct inductively a transfinite sequence of injective maps $(f_\alpha:X_\alpha\to C^\alpha)_{\alpha\le\omega_1}$ such that for any $\alpha<\beta\le\omega_1$ we have the equality $f_\alpha\circ p_\alpha^\beta=\pi^\beta_\alpha\circ f_\beta$.
Let $\Omega=\{\alpha\in\omega_1:x_\alpha\in f_\alpha(X_\alpha)\}$.
Let $(\mu_\alpha)_{\alpha\in\omega_1}\in\prod_{\alpha\in\omega_1}X_\alpha$ be any transfinite sequence such that $f_\alpha(\mu_\alpha)=x_\alpha$ for any $\alpha\in\Omega$. We claim that this sequence has the required property.
Given any $\mu\in X_{\omega_1}$ consider the element $x=f_{\omega_1}(\mu)\in C^{\omega_1}$. The choice of the transfinite sequence $(x_\alpha)_{\alpha\in\omega_1}$ ensures that the set $S=\{\alpha\in\omega_1:x_\alpha=x{\restriction}\alpha\}$ is stationary in $\omega_1$. For every $\alpha\in S$ we have $$x_\alpha=x{\restriction}\alpha=\pi_\alpha^{\omega_1}(x)=\pi_\alpha^{\omega_1}\circ f_{\omega_1}(\mu)=f_\alpha\circ p_\alpha^{\omega_1}(\mu)\in f_\alpha(X_\alpha)$$ and hence $\alpha\in\Omega$ and $f_\alpha(\mu_\alpha)=x_\alpha=f_\alpha\circ p_\alpha^{\omega_1}(\mu)$. Now the injectivity of the map $f_\alpha$ ensures that $\mu_\alpha=p_\alpha^{\omega_1}(\mu)$.
Therefore, the set $\{\alpha\in\omega_1: \mu_\alpha=p_\alpha^{\omega_1}(\mu)\}\supset S$ is stationary in $\omega_1$.
|
2025-03-21T14:48:29.667281
| 2020-01-23T09:32:15 |
350992
|
{
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"Igor Belegradek",
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"student"
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|
Stack Exchange
|
Multisignature and homeomorphism type
In classical surgery theory, there is a map
$$L_{n+1}(\pi_1M)\to S(M^n)$$
Element in $L_{n+1}(\pi_1M)$ is realized as surgery obstruction of a surgery problem to $M\times I$ with one boundary piece the identity map and the other a homotopy equivalence (Wall's realization). The map is defined by sending the element in $L$ to the boundary piece which is a homotopy equivalence (a structure on $M$).It is NOT clear to me if the domain manifold of the structure is homeomorphic to $M$.
In Homology spheres and fundamental group
(when $n+1$ is even) Danny Ruberman commented that "The effect of the action is to change the multisignature, and hence it changes the homeomorphism type of M"
What's the correct reference if i want to understand some details about the effect of Wall's realization on multisignature?
I would look at Wall's book (search for "multisignature" in the index) and also Wall's paper "On the classification of hermitian forms VI. Group rings".
There is perhaps some confusion over the terminology. Wall (chapter 13A) uses the term multisignature to denote a collection of invariants of certain Hermitian forms over group rings, giving rise to a function from $L_{2k}(\pi) \to \mathbb{Z}^n$. In that chapter, he interprets the multisignature in terms of equivariant signatures. Such signatures occur in many places in geometric topology, for instance as the Tristram-Levine signatures and Casson-Gordon invariants of knots.
One main use of the multisignature comes in the study of odd-dimensional manifolds. Given such a manifold $Y^{2k-1}$ and a finite regular covering with covering group $G$, classified by a homomorphism $\phi:\pi_(Y) \to G$, then one shows that some multiple (disjoint union) $N\cdot(Y,\phi)$ extends to a $2k$-manifold $(X,\Phi)$. Then (suitably normalized) the equivariant signature of $(X,\Phi)$ is an invariant $\rho(Y,\phi)$.
This can be packaged in various ways, see for instance chapter 14E in Wall, where for a manifold with finite fundamental group $G$, one obtains an invariant $\rho$. Technically one has to choose an identification of $\pi_1(Y)$ with G, and $\rho$ depends on this identification. A priori, the fact that one gets an invariant depends on the Atiyah-Singer index theorem, and so is really an invariant up to diffeomorphism. But a bordism argument (Wall 14B) shows that it is actually a homeomorphism invariant.
Putting these notions together, the way that the invariant $\rho$ is defined tells you that if one acts on (the identity map from $Y$ to itself) by an element $A \in L_{2k}(\pi_1(Y))$ then the invariant $\rho$ of the manifold $Y'$ at the other end of the resulting normal cobordism satisfies $\rho(Y') - \rho(Y) = $ multisignature of $A$. If this multisignature is non-trivial, then there is no homeomorphism of $Y$ and $Y'$ that respects the identification of their fundamental groups given by the normal cobordism. If you want to argue that they are simply not homeomorphic you need to look a little more closely.
Is there an explicit example in the literature s.t. the action of $L(\pi_1 M)$ on $S(M)$ is nontrivial, but the homeomorphism type of the domain manifold is not changed?
|
2025-03-21T14:48:29.667569
| 2020-01-23T10:00:35 |
350993
|
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"Abdelmalek Abdesselam",
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|
Stack Exchange
|
Smoothness of family of distributions
Let $X$ be a compact manifold. Denote by $\mathscr{D}^\prime(X \times X)$ the space of tempered distributions on the cartesian product $X \times X$. Given two test functions $\varphi, \psi \in \mathscr{D}(X)$, an element $T \in \mathscr{D}^\prime(X \times X)$ can be evaluated at the function $\varphi \otimes \psi$ on $X \times X$ defined by $(\varphi \otimes \psi)(x, y) := \varphi(x) \psi(y)$.
Suppose now that $T_\lambda \in \mathscr{D}^\prime(X \times X)$, $\lambda \in\mathbb{R}$, is a family of distributions such that
$$\lambda \longmapsto T_\lambda[\varphi \otimes \psi]$$
is a smooth function from $\mathbb{R}$ to $\mathbb{R}$ for any two $\varphi, \psi \in \mathscr{D}(X)$.
Q: Does it follow that also the function
$$ \lambda \longmapsto T_\lambda[\Phi]$$
is smooth for every $\Phi \in \mathscr{D}(X \times X)$?
My rapid guess is the answer is yes. I think the compact hypothesis is a distraction and should be removed. I would try first $X=\mathbb{R}$ and if it works, it should adapt to reasonable smooth manifolds (second countable). For $X=\mathbb{R}$, I would also first consider the analogue for $\mathscr{S},\mathscr{S}'$ instead of $\mathscr{D},\mathscr{D}'$. The condition on the functions of $\lambda$ being smooth, i.e., being in $\mathscr{E}$, should be replaced in the analogue by being in $\mathscr{O}_{\rm M}$, then the multiplier space characterization of the latter and the Kernel Theorem...
..might be enough.
If the answer is yes, then the family $T_\lambda$ for $\lambda$ in a compact set should be pointwise bounded on $\mathscr D(X\times X)$ and hence equicontinuous, which roughly means that the continuity estimates for $T_\lambda$ only depend on a fixed number of derivatives with uniform constants. On the other hand, if you have this kind of equicontinuity a priori for $T_\lambda$ and all of its derivatives w.r.t. $\lambda$, an approximation argument might yield the desired result.
Another guess: for good behavior one needs the kernel $T_{\lambda}(x,y)$ to be a distribution in $(\lambda,x,y)$. I don't yet see how this follows from the hypotheses. Perhaps using the uniform boundedness principle?
In other words: the hypotheses define a bilinear map $\mathscr{D}(X)^2\rightarrow \mathscr{E}(\mathbb{R})$, but do they imply this bilinear form is continuous?
Your question has been answered and this is rather OT but it might be of interest that your result has only marginally to do with distributions and more with criteria for smoothness of operator valued functions and with properties of nuclear spaces which have been known for over 60 years. Here is an example: Let $E$ and $F$ be nuclear $F$-spaces, $T$ a mapping, say from the line, into the tensor product of the dual spaces.Then $T$ is smooth ($C^\infty$) if and only if the same is true of the scalar functions $s\mapsto T(s)(x\otimes y)$ for each $x$ and $y$ in $E$ and $F$.
Here is a proof using convenient analysis:
By the kernel theorem $\mathscr{D}^\prime(X \times X) = L(\mathscr{D}(X),\mathscr{D}^\prime(X))$ and by
The Convenient setting of Global Analysis, 5.18 (which is just the uniform boundedness principle) and 2.14 we have
\begin{align*}
&\lambda\mapsto T_\lambda(\Phi) \in \mathbb R \text{ is }C^\infty
\quad\forall \Phi \in \mathscr{D}(X \times X)
\\
\iff &\lambda\mapsto T_\lambda \in \mathscr{D}^\prime(X \times X)
\text{ is }C^\infty \quad &\text{by 2.14}
\\
\iff &\lambda\mapsto T_\lambda \in L(\mathscr{D}(X),\mathscr{D}^\prime(X))
\text{ is }C^\infty
\\
\iff & \lambda\mapsto T_\lambda(\varphi) \in \mathscr{D}^\prime(X)
\text{ is }C^\infty\quad \forall \varphi\in \mathscr{D}(X) &\text{by 5.18}
\\
\iff & \lambda\mapsto T_\lambda(\varphi)(\psi) \in \mathbb R
\text{ is }C^\infty \quad\forall \varphi,\psi\in \mathscr{D}(X) \quad &\text{by 2.14}
\end{align*}
Up to Frechet spaces convenient smoothness equals each other reasonable notion, but beyond it differs. A short description of convenient analysis can be found in
Wikipedia.
Interesting. I thought you would contribute an answer because, if I understood correctly, convenient analysis builds on the notion of smooth curves in LCTVS's. Do you know if $T_{\lambda}(x,y)$ belongs to $\mathscr{D}'(\mathbb{R}\times X\times X)$?
@ Abdelmalek Abdesselam $T_\lambda(x,y)$ belongs to $C^\infty(\mathbb R, \mathscr{D}^\prime(X \times X) = C^\infty(\mathbb R)\bar\otimes \mathscr{D}^\prime(X \times X)\subset \mathscr{D}^\prime(\mathbb R\times X \times X)$.
|
2025-03-21T14:48:29.667876
| 2020-01-23T11:33:26 |
350996
|
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|
Stack Exchange
|
How to relate toric data and group character?
If I collect the Cartan generators of su(3) or sl(3,C), then the diagonal entries can be arranged as
$$
\begin{pmatrix}
1 &1& -2 \\\
0& 1 &-1
\end{pmatrix}
$$
after a row operation this becomes
$$
\begin{pmatrix}
1 & 0 & -1\\\
0 & 1 & -1
\end{pmatrix}
$$
which is the toric data of $CP^2$. I can then write the semigroup action/coordinate ring through the characters.
In fact, this generalizes to show the connection between $SU(n+1)$ and $CP^n$.
Now, how to relate this character to the character of repersentation of the group? this should be related to the highest weight rep.
|
2025-03-21T14:48:29.667964
| 2020-01-23T12:18:11 |
350998
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/350998"
}
|
Stack Exchange
|
On variants of the abc conjecture in terms of Lehmer means
In this post we denote the Lehmer mean of a tuple $\text{x}$ of positive real numbers as $$L_p(\text{x})={\sum_{k=1}^nx_k^p\over\sum_{k=1}^nx_k^{p-1}},$$
see the reference Wikipedia Lehmer mean.
The idea of this post arises when I've consider the inequality $L_p(a,b,a+b)<a+b$ whenever $p$ is enough large, and the limit $\lim_{p\to\infty}L_p(a,b,a+b)=a+b$, where $L_p(\text{x})$ denotes the Lehmer mean for the tuple of positive integers $\text{x}=(a,b,a+b)$ in the context of variants of the abc conjecture. We use in this post the formulation ABC conjecture II from the Wikipedia abc conjecture, thus as usual we denote $$\operatorname{rad}(n)=\prod_{\substack{p\text{ prime}\\p\mid n}}p$$
for integers $n>1$ with the definition $\operatorname{rad}(1)=1$. From this scenario we've the following obvious claim.
Claim. On assumption of the abc conjecture, for all $\varepsilon>0$ and pairwise coprime integers $1\leq a$, $1\leq b$, we assume $a<b$, and such that $c=a+b$, there exists a positive constant $\mu(\varepsilon)$ such that
$$L_p(a,b,a+b)<c<\mu(\varepsilon)\operatorname{rad}(ab(a+b))^{1+\varepsilon}\tag{1}$$
where we consider here $p\geq 2$ integer. Additionally we know $\lim_{p\to\infty} L_p(a,b,a+b)=c$.
Question. I would like to know if for each large enough integer $p\geq 2$ and, thus after this choice of a $p$ sufficiently large, for all $\varepsilon>0$ there exists a positive constant $\mu(\varepsilon,p)$ (thus that depends on our fixed integer $p$) such that $$\frac{a^p+b^p+(a+b)^p}{a^{p-1}+b^{p-1}+(a+b)^{p-1}}<\mu(\varepsilon,p)\operatorname{rad}(ab(a+b))^{1+\varepsilon}\tag{2}$$
for all those positive integers $1\leq a<b$ such that $\gcd(a,b)=1$. Many thanks.
I'm asking about what work can be done about it unconditionally, computations or reasonings with the intention to accept a suitable answer as soon is possible. I don't know how approach this interesting variant of the abc conjecture, I don't know if there is a relationship between the $\mu(\varepsilon,p)$ in $(2)$ and the $\mu(\varepsilon)$, in $(1)$, of the abc conjecture as $p\to\infty$.
I add the reference [2] where is added also the OesterlΓ© and Masserβs abc-conjecture, and subsequent references.
References:
[1] P. S. Bullen, Handbook of means and their inequalities, Springer (1987).
[2] Andrew Granville and Thomas J. Tucker, Itβs As Easy As abc, Notices of the AMS, Volume 49, Number 10 (2002).
(1/2) In the spirit to explore inequalities involving $\operatorname{rad}(n)$ explained in the slide at minute 21' from the video of YouTube What is the abc conjecture? by Keith Conrad, lecture notes from the official channel UConn Mathematics (date March, 23th 2014) I tried to study the case $p=2$, the contraharmonic mean, with the script written in Pari/GP that you can evaluate from the web Sage Cell Server (choosing as language GP) for(a=1, 100, for(b=1,100,if(a<b&&gcd(a,b)==1&&(a^2+b^2+a*b)/(a+b)>factorback(factorint(a*b*(a+b))[, 1])^(1),print(a," ",b))))
(2/2) As you see I've encoded explicitly the exponent $1$ in the expression $\operatorname{rad}(ab(a+b))^1$. The code for the radical of an integer $\operatorname{rad}(n)$ is due to Andrew Lelechenko (May, 09th 2014) as you can see from the section PROG of the article dedicated to the sequence A007947 from the On-Line Encyclopedia of Integer Sequences.
Many thanks for your help @GerryMyerson
All, I've edited two posts on Mathematics Stack Exchange with titles and identificators, respectively, Weaker than abc conjecture invoking the inequality between the arithmetic and logarithmic means (MSE 3580506, Mar 14' 20) and A weak form of the abc conjecture involving the definition of HΓΆlder mean (MSE 3648776, asked yesterday Apr 29' 20). Please, feel free if you or some of yours colleagues (professors studying the abc conjecture) want provide feedback about if this kind of inequalities of this MO or MSE posts are interesting in comments or as companion of your answer(s).
|
2025-03-21T14:48:29.668613
| 2020-01-23T12:31:31 |
351000
|
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|
Stack Exchange
|
Solve congruence equation where unknown variable is in both sides of congruent operator
I am trying to solve the following equation:
$(a*n + c) \mod (b-n) \equiv 0$
and $n$ must be the lowest value in $[0, b-1]$
for example $a=17$, $c=-59$ and $b=128$, the solution is $n=55$
$n=b-1$ will be always a solution, because $m \mod 1 \equiv 0$
Equivalently, you want to solve in integers
$$ y (b-n) = an+c$$
This is equivalent to
$$ (y+a)(b-n) = ab+c $$
You want $b-n$ to be as large as possible subject to $b-n \le b$. Thus you want to
factor $ab+c$ and take its largest divisor $\le b$. In your example, $ab+c = 2117 = 29 \cdot 73$ whose largest divisor $\le 128$ is $73$, so $n = 128 - 73 = 55$.
|
2025-03-21T14:48:29.668699
| 2020-01-23T13:11:42 |
351001
|
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"authors": [
"Asaf Karagila",
"THC",
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|
Stack Exchange
|
Self-embeddings of uncountable total orders, 2
Let $S = (\Omega,\leq)$ be an uncountable dense total order, such that for all positive integers $m$ and all finite ordered sequences $a_1 < a_2 < \ldots < a_m$ and $b_1 < b_2 < \ldots < b_m$, we have an order automorphism of $S$ mapping the former sequence to the latter.
Does there exist a proper subset $\Omega' \subset \Omega$ such that the induced total order $S'$ is order isomorphic to $S$ ?
If so, can $\Omega'$ be chosen to be an interval ?
(I guess the answer will follow from some general theorem on total orders, and I would love to know what that is.)
Thanks !
When you say interval, do you mean a bounded interval, or also a ray?
@AsafKaragila : for me, rays are also intervals.
For the first question, the answer is YES, even if we just assume that $S$ is not rigid. Indeed, fix $f\in\mathrm{Aut}(S)$ and $a\in S$ such that $a<f(a)$. Then $\bigl(\mathrm{id}\restriction(-\infty,a]\bigr)\cup\bigl(f\restriction(a,+\infty)\bigr)$ is an isomorphism of $S$ to its proper suborder $(-\infty,a]\cup(f(a),+\infty)$.
For the second question, the answer is NO in general. One counterexample is the lexicographic product $S=(\omega_1^*+\omega_1)\times\mathbb Q$.
To see that $S$ has the required property (which, by the way, is what model theorists call being strongly $\omega$-homogeneous), notice that any given $a_1,\dots,a_m,b_1,\dots,b_m$ are included in a countable open subinterval of $S$. Such a subinterval must be isomorphic to $\mathbb Q$, which is strongly $\omega$-homogeneous, and an automorphism of the subinterval extends to an automorphism of $S$ by extending it with the identity.
On the other hand, if $S'\subseteq S$ is isomorphic to $S$, it has upwards and downwards cofinality $\omega_1$, which can only happen if it is an (upwards and downwards) cofinal subset of $S$. Thus, it cannot be a proper subinterval.
|
2025-03-21T14:48:29.668855
| 2020-01-23T14:09:17 |
351003
|
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"authors": [
"Martin Brandenburg",
"Mike Shulman",
"Peter LeFanu Lumsdaine",
"Saal Hardali",
"Simon Henry",
"Tom Leinster",
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|
Stack Exchange
|
An adjunction between monads on $\mathcal{C}$ and presentable categories under $\mathcal{C}$
Fix a regular cardinal $\kappa$ and let $\mathcal{C}$ be a $\kappa$-presentable $\infty$-category (comments about the 1-categorical case are welcome as well!).
I'm looking for a reference for the following statement (which in particular I hope is true).
There exists a ("canonical") adjunction of $\infty$-categories
$$ Alg_{(-)}(\mathcal{C}): Monad_{\kappa}(\mathcal{C}) \rightleftharpoons Pr^{L,\kappa}_{\mathcal{C} /}:T_{(-)} $$
Where on the left we have $\kappa$-accessible monads on $\mathcal{C}$ and on the right $\kappa$-presentable categories equipped with a left adjoint (whose right adjoint is $\kappa$-accessible of course) from $\mathcal{C}$.
The right adjoint of the adjunction should send a left adjoint functor to it's corresponding monad on $\mathcal{C}$ and the left adjoint of the adjunction should send a monad to it's category of algebras equipped with the free algebra functor from $\mathcal{C}$.
Similarly (in fact more pressingly) I would like to know if there's also a dual statement for an adjunction of the form:
$$S_{(-)} : Pr^{L, \kappa}_{/ \mathcal{C}} \rightleftharpoons CoMonad_{\kappa}(\mathcal{C}) :CoAlg_{(-)}(\mathcal{C})$$
Aside: The reason I phrased the second adjunction in a more suspicious tone is I do not even know a reference for the statement that the category of coalgebras over a $\kappa$-accesible comonad on a $\kappa$-presentable category is $\kappa$-presentable (I am not even sure whether or not this should be true).
EDIT: I just realized that the statement for coalgebras with $\kappa = \omega$ can't be true. Indeed there are many counterexamples in categories of comodules. In light of that I could change the question to be about $\kappa > \omega$ or just remove the cardinal restriction all together. Since no answers were given yet I think it's better to just leave the question as is.
I recently worked on the exact same question and I couldn't find a reference. For the first question I needed it in a paper, and it does not seem to be treated anywhere, so I'm currently writting a proof. The second question I recently asked a reference request for the 1-categorical case of your aside on MO as https://mathoverflow.net/questions/350351/presentability-rank-of-categories-of-coalgebras and it didn't get an answer. I have written a proof myself that the category of co-algebra is $\kappa$-presentable when $\kappa$ is uncountable (I don't think it is true for $\kappa=\omega$), but...
It will be difficult to upgrade this proof to the $\infty$-categorical case (it is fairly technical...)
Maybe I'm misunderstanding, but do you really mean "under C" rather than "over C"? As I understand it, one of the functors in your first adjunction takes a monad T on C and outputs its category of algebras C^T together with the forgetful functor C^T --> C (which of course has a left adjoint). That's a category over, not under, C.
Basically you are asking for a universal property of the presentable category of T-algebras, right? For the 1-categorical setting and $\kappa=\aleph_0$ it should follow from Prop. 6.2.15 in my thesis, right? (For which I also couldn't find a reference either.) https://arxiv.org/pdf/1410.1716.pdf
In the 1-categorical case, a result close to what you want is Theorem 6.5 of http://www.tac.mta.ca/tac/volumes/28/13/28-13abs.html, which is due to Dubuc (see Proposition 6.2 there).
@TomLeinster I think what I wrote is ok. The functor sends a monad to the corresponding free functor into the category of algebras which is a left adjoint under $\mathcal{C}$. Perhaps the confusion stems from the fact that my adjunction is covariant whereas in the result you cite it is contravariant.
Oh I see. In that case, maybe my reference to Dubuc's theorem is less relevant. Not sure. What confused me was the part of your question where you wrote "categories with left adjoints". Since a category can't have a left adjoint (!), I assumed that was an error, and that you'd meant to say that the functor from your category to C had a left adjoint. But now I guess what you meant is that the objects of your right-hand category are categories, and the maps in it are the functors that are left adjoints (i.e. have right adjoints).
Like Tom, Iβd find it somehow more natural to take the maps of presentable clothes categories as following the direction of their right adjoints (not left), and so would see the RHS of your adjunction as categories over C, not under. This direction is more standard in the 1-categorical case, at least, and fits with conventions like the direction of geometric morphisms of (infinity-)toposes, and so on. But of course, this is just a cosmetic issue.
@TomLeinster Thanks, I tried to fix the phrasing a bit to make it clearer.
A keyword to look for that describes this situation is "semantics-structure adjunction", e.g. https://ncatlab.org/nlab/show/monadic+adjunction#semanticsstructure_adjunction. However, I think the most subtle part of the question is the preservation of the cardinal of accessibility, which (perhaps surprisingly) frequently turns out not to be the case in locally-presentable-category theory.
@MikeShulman "... frequently turns out not to be the case in locally-presentable-category theory." What are some good examples demonstrating this?
See https://mathoverflow.net/a/353475/22131 for references to the fact that categories of coalgebras for a $\kappa$-accessible comonad on a $\kappa$-presentable category (with $\kappa$-uncountable) is $\kappa$-presentable. The key reference is https://www.sciencedirect.com/science/article/pii/S0304397503003785 but some details need to be added.
To get an $\infty$-categorical version, the hard point is to generalize theorem 4.2 of the second link, whose proof is an explicit construction and is a bit technical, the rest should follows from the work of Lurie.
|
2025-03-21T14:48:29.669254
| 2020-01-23T14:13:51 |
351004
|
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|
Stack Exchange
|
A non-recursive, explicit formula for the Fabius function
The Fabius function $F\colon\mathbb R\to[-1,1]$ may be defined as the unique solution of the functional integral equation
$F(x)=\int_0^{2x}F(t)\,dt$ for all real $x$ such that $F(1)=1$.
The recent MO post provided a link to the MathSE question, asking to confirm a conjectured "non-recursive, self-contained formula for the Fabius function". The MO post has been overall negatively received and may get closed. I think the mentioned MathSE question may be of interest.
On this page, whereas the mentioned conjectured formula will not be confirmed so far, a simpler non-recursive, explicit formula for the Fabius function will be offered, which is expressed in terms similar to, but simpler than, those in the conjectured formula.
So, a question yet remains: Can one use the simpler formula below to confirm the conjecture on MathSE? Or maybe one could do that otherwise?
Is there a question, or is this just a way to provide a place to put your answer?
@LSpice : Indeed, I may have to think of a better way to present the MathSE question.
@LSpice : Now I have an explicit question.
As noted in the linked Wikipedia article on the Fabius function, on the interval $I:=[0,1]$ the Fabius function coincides with the cumulative distribution function (cdf) of
$$\sum_{j=1}^\infty 2^{-j}U_j,$$
where the $U_j$'s are independent random variables uniformly distributed on $I$. So, for each $x\in I$
$$F(x)=\lim_{n\to\infty} F_n(x),\tag{1}$$
where $F_n$ is the cdf of $\sum_{j=1}^n 2^{-j}U_j$.
Next (see e.g. formula (2.2)), for any real $x$
$\newcommand\vp{\varepsilon}$
\begin{equation}
F_n(x)=\text{vol}_n(I^n\cap H_{n;c^{(n)},x})
=\frac1{n!\prod_1^n c_i}\,\sum_{\vp\in\{0,1\}^n}(-1)^{|\vp|}\,
\big(x-c^{(n)}\cdot\vp\big)_+^n,
\end{equation}
where $\text{vol}_n$ is the Lebesgue measure on $\mathbb R^n$, $H_{n;b,x}:=\{v\in\mathbb R^n\colon b\cdot v\le x\}$, $c^{(n)}:=(c_1,\dots,c_n)$, $c_j:=2^{-j}$, $|\vp|:=\vp_1+\dots+\vp_n$, $\cdot$ denotes the dot product, and $t_+^n:=\max(0,t)^n$. So, for $x\in I$
\begin{equation}
F_n(x)=\frac{2^{n(n+1)/2}}{n!}\,\sum_{y\in D_{n,x}}(-1)^{s(y)}\,
\big(x-y\big)^n, \tag{2}
\end{equation}
where $D_{n,x}$ is the set of all dyadic numbers in $[0,x]$ of the form $m2^{-n}$ for integers $m$, and $s(y)$ is the sum of the binary digits of $y$.
Formulas (1) and (2) provide the answer to the question.
Some of the differences between (1)--(2) and the formula conjectured in the linked MathSE post are as follows:
In the MathSE post, the main conjectured formula for $F(x)$ is stated only for dyadic numbers $x$, and then extended to all values of $x$ by the continuity of $F$.
The expression in that post for $F(x)$ for dyadic $x$ contains a double summation and a number of $q$-Pochhammer symbols (I have had no experience with those symbols).
I'm having trouble seeing how this formula terminates at a finite sum for dyadic x, or perhaps it doesn't. This seemed to be one of the goals of the original MathSE post.
I don't know what you mean by "this formula terminates". Anyhow, the sum in (2) is over a finite set.
I mean that this approach takes in a value x in I and produces a sequence of numbers which converge to F(x). However, if x is a dyadic rational then F(x) is known to be rational, and it is not clear if the sequence of values that this method produces will return that value exactly after a finite number of steps. I think this would be a desirable quality of a "solution" to the problem of finding a fast approximation to F(x).
@GregZitelli : This sequence is strictly increasing for each $x$ in $(0,1)$.
|
2025-03-21T14:48:29.669533
| 2020-01-23T15:33:22 |
351006
|
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|
Stack Exchange
|
Gluing filtered object from associated graded pieces
So, I believe the following result is correct but do not know the exact reference (and not sure to what extent what I'm saying is true). If anyone could give a reference for this it would be great.
1) Consider three objects in some abelian category, $F_1, F_2, F_3$. Then objects with filtrations $F^1 \subset F^2 \subset F^3$ such that $F^1 = F_1, F^2/F^1 = F_2, F^3/F^2 = F_3$ are classified by triples $(\alpha, \beta, \gamma)$ of extension elements, $\alpha \in Ext^1(F_2, F_1), \beta \in Ext^1(F_3, F_2), \gamma \in Ext^1(F_3, F_1)$, such that the Yoneda product $\alpha \beta \in Ext^2(F_3, F_1)$ vanishes.
2) Similar result should hold for any amount of objects at least for the case of linear category in characteristic zero - the ways of gluing objects $F_1, ..., F_n$ into a filtered object $F^1 \subset F^2 \subset ... \subset F^n$ such that $F^k / F^{k-1} = F_k$ should be classified by Maurer-Cartan elements in the algebra $\bigoplus_{j>i}RHom^{\bullet}(F_j, F_i)$ (considered as either dgla or L-infinity algebra).
I also know that this type of questions frequently appear in the theory of mixed hodge structures (but unable to find any direct reference, too).
Edit: added forgotten $\gamma$
It is not precisely correct. You need to specify how it is zero and this is an extra piece of information. For example you might have a case in which both the first pair and the second are glued trivially but the third object is glued no trivially to the first. In this case all the information lies in the trivialization
Excuse me, I've definitely forgot that I also need an element in Ext^1(F_3, F_1), but I think in case of three objects this is all data I need: you are talking about some kind of Massey product and for three objects there won't be any.
One comment is that you can think of the equivalence you are defining as Koszul duality for representations of the $A_n$ quiver. The Koszul dual (linear) category to the $A_n$ quiver is the category $Ch_n$ whose representations are chain complexes. The structure you get on the associated gradeds is exactly a homotopy coherent chain complex $Ch_n \to D(A)$ which takes the $k$th object to $F_k [k]$.
@LevSoukhanov This "more data you need" is precisely there, from my perspective, to indicate the different options for trivializing $\alpha \beta$: every two such trivializations differ by a class in $Ext^1$. Note that it has still some ambiguity in this class of the form $\alpha \xi$ for $\xi:F_3\to F_2$, as in Kuhn's answer.
Lets choose some resolutions of our objects. Then I think we both agree that the data specified is an element $(\alpha, \beta, \gamma) \in \bigoplus_{j>i} RHom^{\bullet}(F_j, F_i)$, satisfying Maurer-Cartan condition $\alpha \beta = d \gamma$. The solutions of such equations are in $1-1$ correspondense to solutions of this equation in any quasiisomorphic dg-algebra.
However, in this case this algebra is formal (quasiisomorphic to its cohomology), and the equation reduces to $\alpha \beta = 0$.
This works only for char=0 however.
However, the answer to this question suggests I'm doing something wrong but I can not quite recognize what: https://mathoverflow.net/questions/301641/spelling-out-explicitly-the-data-of-a-two-step-filtration-in-terms-of-pieces-and?rq=1
The argument why this algebra is formal is just that it should be quaiisomorphic to its cohomology with $A_{\infty}$-structure (transferred by Kadeishvili's theorem) and this $A_{\infty}$-structure doesn't have any higher operations (aka Massey products) because of the grading reasons.
@LevSoukhanov We definitely agree on that, except that I would prefer to say "the equation reduces to $\alpha \beta= 0$ and $d\gamma = 0$. This is essentially the same as choosing a chain killing $\alpha \beta$ in advance (in the non-formal case) and consider $\gamma$ as the difference from our's and the fixed one. And again, it is not unique but rather defined inside the ideal of $\alpha$ and $\beta$ because we can change $\alpha$ and $\beta$ by a coboundary, even in the formal case, and this affects $\gamma$, even in the formal case!
In formal case we have no d?...
(I believe my argument proves that 2-step extensions with fixed isomorphism to the associated graded module ARE classified by triples $\alpha, \beta, \gamma \in$ their respective $Ext$ groups with $\alpha \beta = 0$ and no additional conditions. do we agree on that statement?)
No, excuse me, $\gamma$ is even defined up to addition of elements of the form $\alpha \psi$ and $\phi \beta$, where $\psi \in Hom(F_3, F_2), \phi \in Hom(F_2, F_1)$. But is still does not leave Ext-groups.
We agree on everything.
@LevSoukhanov Leva, hi. One comment: all Maurer-Cartan solutions aren't invariant under quasiisomorphism, only those close to the identity are (meaning a functor on f.d. nilpotent algebras). Consider the case of an acyclic dg-Lie algebra which is one-dimensional in degrees 1 and 2 and zero everywhere else, with the differential an identity map and a multiplication of two numbers for the bracket. It has two Maurer-Cartan solutions.
@GrishaPapayanov hi! I didn't think about it - but it seems correct for me now, the thing which is invariant is the germ of solutions mod gauge group equivalence.
Here is how I would think about this: $\alpha \in Ext^1(F_2,F_1)$ corresponds to a short exact sequence $$0 \rightarrow F_1 \xrightarrow{i} F(1,2) \xrightarrow{p} F_2 \rightarrow 0.$$
Similarly $\beta \in Ext^1(F_3,F_2)$ corresponds to a short exact sequence $$0 \rightarrow F_2 \rightarrow F(2,3) \rightarrow F_3 \rightarrow 0.$$
The first short exact sequence induces a long exact sequence including
$$ Hom(F_3,F_2) \xrightarrow{\alpha \circ} Ext^1(F_3,F_1) \xrightarrow{i_*} Ext^1(F_3,F(1,2)) \xrightarrow{p_*} Ext^1(F_3,F_2) \xrightarrow{\alpha \circ} Ext^2(F_3,F_1),$$
and from this one sees that an $F(1,2,3) \in Ext^1(F_3,F(1,2))$ exists such that $F(1,2,3)/F_1 = F(2,3)$ if and only if $\alpha \circ \beta = 0 \in Ext^2(F_3,F_1)$. Furthermore, choices correspond to the image of $i_*$. Perhaps this is the classification you desire. I don't know of a reference, but the argument is just using basic triangulated category/homological algebra techniques.
Understanding filtered objects with 4 or more composition factors leads one quickly to Massey products. Ambiguities tend to get out of hand unless one has something special going on in the case in hand.
Yes, I have something similar in mind (and for general result it should follow from formal deformation theory). I actually search for a reference though so I won't accept the answer for now (while I agree what you are saying is correct and answers the first question).
|
2025-03-21T14:48:29.670101
| 2020-01-23T15:58:03 |
351008
|
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"authors": [
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"url": "https://mathoverflow.net/questions/351008"
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|
Stack Exchange
|
When do completely positive maps have a closed image?
Let $\mathcal{A}, \mathcal{B}$ be C*-algebras. A map $\phi \colon \mathcal{A} \rightarrow \mathcal{B}$ is completely positive (cp) if it's linear, * preserving and all of its' coordinatewise extensions to matrices $\phi^{(n)} \colon M_n(\mathcal{A}) \rightarrow M_n(\mathcal{B})$ map positive matrices to positive matrices. For instance, by Stinesprings' Representation, completely positive maps are just of the form $V \pi V^*$, where $\pi$ is a *-homomorphism.
My question is under which conditions do these cp maps have a closed image (see here for a proof that homomorphisms do have a closed image).
Of course this is true if the map is isometric, but I can't seem to prove that, for instance, injective cp maps of norm 1 are isometric. The usual proof for homomorphisms uses that the spectrum remains unchanged under a homomorphism, but does this also hold for cp maps? It does seem false to me.
Thank you for any help you may give. I must be overlooking something.
You should be able to easily find counterexamples for commutative A and B, when cp is the same as usual positivity by Naimark's theorem (see Effros-Ruan or Paulsen's book). I think this is a useful exercise for you to figure out.
Based on the uiquity of counterexamples (i.e. cp maps with non-closed range) I don't think "under which conditions to cp maps have closed range" is sufficiently focused to make a good MO question: they have closed range when, erm, they have closed range.
In the Stinespring case, you want $\mathcal B = B(H)$ where $H$ is a Hilbert space, $\pi$ is a *-homomorphism into $B(K)$ for some Hilbert space $K$, and $V: K \to H$. I think in this case you should have closed image if $V$ has closed range.
@YemonChoi that, erm, makes a lot of sense. What I meant was whether injective cp maps of norm 1 are isometric, but that is also false. Thank you for pointing it out.
@RobertIsrael if you ever find the time and energy to provide a sketch of the proof that'd be great, because that'd mean that unital cp maps, for instance, do have closed range. Thank you as well
@RobertIsrael If the cp map sends 1 to 1 then as Diego points out, the V constructed in Stinespring's representation is an isometry. But I think that by choosing suitable integral operators on C[0,1] one can get positive operators that send 1 to 1 yet do not have closed range.
Hmm, I take that back. I think it is not enough for $V$ to have closed range.
This is not a complete answer to the original question, since the original question has a rather open-ended phrasing; but I think it addresses Diego's main points, and shows that even quite well-behaved cp maps won't have the desired properties.
First of all: if we restrict attention to cases where either $A$ or $B$ is commutative, then every positive linear map $A\to B$ is automaticaly completely positive: see Lemma 5.1.4 and Theorem 5.1.5 in the book of Effros--Ruan, for instance. To keep things simple I will take both $A$ and $B$ to be commutative ${\rm C^*}$-algebras; I'm sure one can create more fancy examples by modifying the constructions below.
Q1: "Are injective norm-one cp maps isometric?"
Counterexample: take $A=B = \ell_\infty^2$ (viewed as $C(\{0,1\})$) and take the linear map $T:A\to B$ defined by $T(x,y) = (x, y/3)$.
Q2: "Do ucp maps have closed range?"
This counterexample requires a bit more work than I originally thought, so it seems to deserve writing down as an answer rather than merely a comment. We work with $A=B=C[0,1]$ and consider positive, unit-preserving operators $T:C[0,1]\to C[0,1]$ of the form
$$ (Tf)(x) = \int_0^1 K(x,t) f(t)\,dt $$
for a suitably chosen positive kernel function $K$. The condition $T1 = 1$ translates into requiring $\int_0^1 K(x,t)\,dt=1$ for all $x$, so that each "row" of $K$ should have total mass 1, and then on the analysts' general principle that "to make things go wrong, get things to pile up more and more" it is natural to try an Ansatz like
$$ K(x,t) = \begin{cases} x^{-1} & \hbox{ if $0\leq t\leq x$} \\ 0 & \hbox{if $x< t \leq 1$} \end{cases} $$
With this choice, our operator $T$ becomes
$$ (Tf)(x) = \frac{1}{x} \int_0^x f(t)\,dt $$
(sometimes known as the Hardy operator, if I recall correctly) and then we can see directly that $T$ is positive and unit-preserving.
On the other hand, because $T$ is averaging the behaviour of a given $f$ over the initial parts of the interval $[0,1]$, one intuitively feels that $T$ has a good chance of shrinking the norms of functions which are small near $0$ and big near $1$, which will mean that $T$ can take elements of norm $1$ to elements with small norm. Therefore we consider $Tf_n$, where $f_n(t)=t^n$, and calculate that
$$ (Tf_n)(x) = \frac{1}{x} \int_0^x t^n\,dt = \frac{1}{n+1} x^n $$
so that each $f_n$ is in fact an eigenvector of $T$ with eigenvalue $(n+1)^{-1}$. In particular, $T$ is not bounded below.
To conclude that $T$ does not have closed range, it only remains to note that $T$ is injective. This is immediately plausible as we are basically integrating and then dividing by a weight, and taking the indefinite integral of a non-zero function should give a non-zero function. More precisely: suppose $f\in C[0,1]$ and $Tf=0$. Then $\int_0^x f(t)\,dt=0$. So by the fundamental theorem of calculus $f=0$.d
Thank so much for this answer! It really does seem that ccp maps cannot be expected to have closed range. I'll proceed to close the question. Again, thank you.
Another argument for $T$ not having closed image is the following: The image of $T$ is dense by Weierstrass's approximation theorem, but $T$ is not surjective because for all $f \in C([0,1])$, $T(f)$ is differentiable on $(0,1)$.
|
2025-03-21T14:48:29.670487
| 2020-01-23T18:42:58 |
351016
|
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"authors": [
"AndrΓ©s E. Caicedo",
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|
Stack Exchange
|
Models of ZF intermediate between a model of ZFC and a generic extension
Let $M$ be a countable model of $ZFC$ and $M[G]$ be a (set) generic extension of $M$. Suppose $N$ is a countable model of $ZF$ with
$$M\subseteq N \subseteq M[G]$$
and that $N=M(x)$ for some $x\in N$; i.e., it is the smallest inner model of $M[G]$ which contains $x$ and $M$.
Is $N$ a symmetric extension of $M$?
The answer is no! For a rather dramatic example, see for instance MR3878470 Karagila, Asaf The Bristol model: an abyss called a Cohen real. J. Math. Log. 18 (2018), no. 2, 1850008, 37 pp. (Take a look at section 7.2 for a brief comment on this matter.)
Thanks. I was aware of this result and had it in mind, but I'm not sure it does answer the question. This is why I added the condition that M[G] is a generic extension of M. Grigorieff's Theorem B shows that in such a case N=M[x] for some x \in N. If I understand the remarks on the Bristol model correctly, then it cannot be of the form M[x] (assuming that L=M). I probably should have mentioned the Theorem B think in the OP.
Toby, a Cohen real is the most basic of generic extensions...
Hmmm .. I'll edit the question and see if you still think it's solved. Thanks for looking at it.
@Toby I edited $M[x]$ to $M(x)$. In modern notation, the former is by design a model of choice, so it would be confusing to use it in this setting.
Yes, if $N=M(x)$ (taking the modern notation over Grigorieff's $M[x]$), then it is a symmetric extension. This is a very recent result of Toshimichi Usuba (see this and that).
However, by the Bristol model construction, this symmetric extension need not be obtained by the forcing that was used to get $M[G]$.
The Bristol model is pretty intimidating. Can you elaborate a little on your second comment? Happy to move to email if that is preferable.
The point is that $L(V_\alpha^M)$ where $M$ is the Bristol model are all symmetric models.
But regardless, feel free to drop me an email if you have any followups or any other choiceless questions...
@AsafKaragila I have a question. In Grigorieff's Intermediate submodels and generic extensions in set theory he proves that every symmetric submodel of $M[G]$ is of the form $(\mathsf{HOD} (M(x)))^{M[G]}$ for some $x \in M[G]$ (Theorem C), and then he observes that every inner model of this kind is equal to $M(y)$ for some $y \in M[G]$ with $y\subset M$ (Corollary 2, Section 9). Thus every $M(y)$ for some $y \in M[G]$ with $y\subset M$ is a symmetric extension of $M$. The novelty of Usuba's result is that $M(y)$ is a symmetric extension of $M$ even when $y$ is not a subset of $M$?
@Lorenzo Send me an email. This question has a long answer.
|
2025-03-21T14:48:29.670708
| 2020-01-23T19:06:52 |
351017
|
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|
Stack Exchange
|
Affine equivalence of Coxeter permutahedra?
Suppose that $W=\langle s_1,\ldots, s_d\mid (s_is_j)^{m_{ij}}=e\rangle$ is a finite reflection group and consider its standard $d$-dimensional geometric realization (i.e., the Tits representation) $\rho: W\rightarrow V$.
For any $a\in V$, one can consider the convex hull $P(W;a)$ of the orbit $W\cdot a$, often called a (Coxeter) permutahedron in the literature. It is a standard fact that if $a$ and $b$ are generic points in $V$ (i.e., neither is fixed by any reflection), then the resulting permutahedra are necessarily combinatorially equivalent (essentially, this is because, without loss of generality, $a$ and $b$ can be taken to lie in the same fundamental region/Coxeter chamber).
My question is this: in the literature I have read, authors seem to say that $P(W;a)$ and $P(W;b)$ are only combinatorially equivalent. Why aren't they actually affinely equivalent (and in fact $W$-equivariantly so)? It would seem that sending $a$ to $b$ and each $s_i\cdot a$ to $s_i\cdot b$ would fit the bill for this, since $\{s_i\cdot x\mid 1\leq i\leq d\}$ forms a basis for $V$ for all generic $x$. What am I missing?
Note that the term reflection group already refers to this "standard realization". You probably meant Coxeter group.
Consider the image below.
These polygons are permutahedra of the reflection group $I_2(3)$ (symmetry group of the triangle), but they are not affinely equivalent.
Let $T$ be the affine transformation that maps $a\mapsto b$ and $s_i\cdot a\mapsto s_i\cdot b$.
Usually there are further $s\in W$ other than the generators $s_1,...,s_d$.
The vertex $s\cdot a$ is mapped to $s\cdot b$ if $s$ and $T$ commute, but this is most often not the case.
OK I see it now. Thanks!
|
2025-03-21T14:48:29.670863
| 2020-01-23T19:40:29 |
351021
|
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|
Stack Exchange
|
Isomorphism between finite algebras over ${\Bbb Z}_p$
Let $\pi \colon R \twoheadrightarrow {\Bbb T}$ be a surjective ring homomorphism between finite algebras over ${\Bbb Z}_p$. Further, we suppose the following three conditions$\colon$
$R$ is a complete intersection, i.e. $R = {\Bbb Z}_p[[X_1,\ldots,X_d]]/(f_1,\ldots,f_d)$.
$\pi$ induces an isomorphism $\pi^{*} \colon {\mathrm{Hom}}_{{\Bbb Z}_p}({\Bbb T}, \overline{{\Bbb Z}_p}) = {\mathrm{Hom}}_{{\Bbb Z}_p}(R, {\overline{\Bbb Z_p}})$, where we denote by $\overline{{\Bbb Z}_p}$ the integral closure of ${\Bbb Z}_p$ in the algebraic closure $\overline{{\Bbb Q}_p}$.
${\Bbb T}$ is reduced.
We shall denote by $R^{\mathrm{red}}$ the reducification of $R$. Then, I would like to ask
Q. Do the above three conditions imply the isomorphism $\pi \colon R^{\mathrm{red}} \cong {\Bbb T}$?
In the case where $\overline{{\Bbb Z}_p}$ is replaced with ${\Bbb Z}_p$,
the question was answered in the negative by Professor Spivakovsky.
The answer is ``no''. Let $R=\frac{\mathbb Z_p[X]}{(X^2-p)(X^3-p)}$ and $\mathbb T=\frac{\mathbb Z_p[X]}{(X^2-p)}$. The ring $R$ is a hypersurface, hence a complete interserction; in addition, it is reduced, so $R=R^{red}$. The ring $\mathbb T$ is also reduced. The map $\pi^*$ is the zero map between two modules, each of which is equal to $(0)$, so it is an isomorphism. Yet, $R=R^{red}$ is not isomorphic to $\mathbb T$.
Why is T a quotient of R?
Because the ideal $(X^2-p)(X^3-p)$ is contained in the ideal $(X^2-p)$. There is a natural surjective homomorphism from $R$ to $\mathbb T$ that sends every element of $(X^2-p)$ to 0.
The answer to the modified question (that is, with $\overline{\mathbb Z_p}$ instead of $\mathbb Z_p$) is "yes".
The prime $p$ is not a zero divisor in $R$. Replacing $R$ by $R^{red}$ does not change the problem (we no longer claim that $R$ is a complete intersection, but $p$ is still not a zero divisor). We will prove that $\pi$ is an isomorphism. Assume the contrary, aiming for contradiction. Let $I=Ker(\pi)$. Then there exists a minimal prime $P$ of $R$ such that $I\not\subset P$, so $\pi$ induces a map $\frac RP\rightarrow\frac R{I+P}$ with $\frac R{I+P}$ zero-dimensional.
The ring $\frac RP$ contains $\mathbb Z_p$ and is integral over it, hence admits a non-zero homomorphism to $\overline{\mathbb Z_p}$. This homomorphism does not factor through $\mathbb T$, contradicting the hypothesis 2.
Dear Mark, I would like you teach me how you deduce the map from minimal primes of T to those of R? Is it obvious? As far as I know, the intersection of all minimal primes coincides with nilpotent elements. However, if R is reduced from the beginning, it must be $0$. I don't see where you used the condition 1. (or 3.) Could you please explain the body of the proof more closely?
Proof that $p$ is not a zero divisor in $R$. The ring $R$ is a one-dimensional Cohen-Macaulay ring. Hence all of its associated primes are minimal and have coheight 1. The ring $\frac R{(p)}$ is finite over $\mathbb F_p$, hence zero-dimensional. Thus $p$ is not contained in any minimal prime of $R$, hence in no associated prime of $R$. This proves that $p$ is not a zero divisor in $R$.
I use hypothesis 1 to prove that $p$ is not a zero divisor in $R$. The intersection of ALL minimal primes of $R$ is $(0)$; the intersection of SOME of them need not be $(0)$. If $\pi$ is not an iso, its kernel is such an intersection. I use hypothesis 1 once again to conclude that $\frac RP$ admits a map to $\mathbb Z_p$ (since $p\notin P$). I use hypothesis 3 in the sentence "Replacing ... does not change the problem"
|
2025-03-21T14:48:29.671129
| 2020-01-23T23:00:00 |
351029
|
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|
Stack Exchange
|
"Sameness" of dg and A-infinity categories
Let $k$ be a field.
A folklore theorem states that dg-categories (over $k$), $A_{\infty}$-categories (over $k$) and stable ($k$-linear) $(\infty, 1)$-categories are "the same" (see for example
Stable infinity categories vs dg-categories).
I would like to understand in what sense (and why) dg- and $A_{\infty}$-categories are "the same".
There is a canonical inclusion functor from the category of dg categories to the category of $A_{\infty}$ categories. However, this functor is not full, so it does not define an equivalence of categories.
Any $A_\infty$-algebra is quasiequivalent to a dg-algebra through the cobar-bar construction.
Ah, so perhaps the equivalence of categories I'm looking for sends an $A_{\infty}$ algebra $A$ to the dg algebra $\Omega B A$. I need to think a bit more, but I think this answers the question at the level of algebras.
By Corollary 9.2.1 in the paper https://arxiv.org/abs/1410.5675
the model category of small A_β-categories
is Quillen equivalent to the model category of small categories
(with a fixed set of objects for simplicity, but see
also Proposition 9.2.3 for the general case),
where both types of categories are enriched over a fixed monoidal tractable model category,
such as chain complexes over a commutative ring,
simplicial sets,
simplicial modules over a commutative ring,
symmetric spectra, etc.
The proof there also gives explicit formulas for the rectification operation.
Could you please clarify how Corollary 9.2.1. relates to dg-categories? (I apologize if this is obvious -- I have very little background in this area).
@user142700: If C=V=Ch, then algebras over the operad Cat^O_W are A_β-dg-categories respectively dg-categories if O=A_β respectively As (the associative operad) with W as a set of objects.
But A_infinity categories don't have a model structure!
@FrancescoGenovese they do.
@FrancescoGenovese: The easiest way to construct a model structure on A_β-categories is to fix a set of objects first. A_β-categories with a fixed set of objects are algebras over a certain nonsymmetric colored operad, and algebras over nonsymmetric colored operads in chain complexes or simplicial sets always admit a transferred model structure (see Fernando Muro's paper Homotopy theory of nonsymmetric operads, and its errata).
@DmitriPavlov no, and the reason is that A_infty categories are not complete/cocomplete. Check Β§1.5 here https://arxiv.org/pdf/1811.07830.pdf If you say that there is some "variant" of this, then maybe... but definitely not according to the usual definition
@FrancescoGenovese: The paper you referenced is talking about a very different category, the category of A_β-categories and A_β-functors (not ordinary enriched functors) between them. As far as I am aware, we don't know how to define A_β-functors for arbitrary A_β-operads in arbitrary monoidal model categories, which is the context of my answer.
I think that the best answer so far comes from this paper: https://arxiv.org/abs/1811.07830 , where it is proved that homotopy categories of dg-categories and various flavours of A-infinity categories are equivalent.
|
2025-03-21T14:48:29.671363
| 2020-01-23T23:57:12 |
351031
|
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|
Stack Exchange
|
Lipschitz functions that saturate the Lipschitz inequality on the average (part 1)
Consider a 1-Lipschitz function $f: \mathbb R^n \to \mathbb R$ satisfying the inequality
\begin{align*}
|f(x) - f(y)| \le \|x-y\|_2, \;\forall x,y \in \mathbb R^n.
\end{align*}
For $n \ge 2$, can we find a 1-Lipschitz function that saturates the above inequality on the average?
To make the notion of "on the average" precise, let $x$ and $y$ be independent standard Gaussian vectors, i.e., $x,y \sim N(0,I_n)$. One can show that
$$\mathbb E \big|\|x\|_2 - \|y\|_2\big| \asymp 1.$$
while
$$\mathbb E\|x-y\|_2 \asymp \sqrt{n}.$$
Is there a 1-Lipschitz function $f : \mathbb R^n \to \mathbb R$ such that
$$\mathbb E|f(x) - f(y)| \asymp \sqrt{n}?$$
Here $\asymp$ means inequalities go in both directions up to constants.
A related question is determining the order of
$$
\sup_{f \in \text{Lip}(1)}\mathbb E|f(x) - f(y)|
$$
where $\text{Lip}(1)$ is the set of $1$-Lipschitz functions from $\mathbb R^n$ to $\mathbb R$.
What do you mean by $|x|$ as opposed to $|x|_2$?
@AnthonyQuas, yes, the $\ell_2$ norm.
All the norms are the $\ell_2$ norm. I have edited the question.
There is no $1$-Lipschitz function $f\colon \mathbb R^n \to \mathbb R$ such that
$$\mathbb E|f(x) - f(y)| \asymp \sqrt{n}.$$
Indeed, for any such function, by the Gaussian concentration for Lipschitz functions (see e.g. Theorem 2.4, page 31),
$$P(|f(x)-Ef(x)|\ge t)\le2e^{-t^2/2}$$
for all $t\ge0$. So,
$$E|f(x)-Ef(x)|=\int_0^\infty P(|f(x)-Ef(x)|\ge t)\,dt\le\sqrt{2\pi},$$
and hence also $E|f(y)-Ef(x)|=E|f(y)-Ef(x)|\le\sqrt{2\pi}$, so that
$$E|f(x)-f(y)|\le E|f(x)-Ef(x)|+E|f(y)-Ef(x)|\le2\sqrt{2\pi}=o(\sqrt n).$$
It also follows that
$$\sup_{f\in\text{Lip}(1)}E|f(x) - f(y)|\asymp1.$$
Ah, you are right! How I did not notice that.... Now how about general sub-Gaussian distributions. I have updated the question.
@passerby51 : If you have additional questions, please ask them in separate posts, especially when your original question has been fully answered.
OK, fair enough.
|
2025-03-21T14:48:29.671519
| 2020-01-24T01:18:01 |
351033
|
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|
Stack Exchange
|
Continuity of Lipschitz constant over multi-variable function
Let $f\in C^0(I\times \mathbb{R})$,$I=[\xi,\xi+a]$,$a>0$, and suppose that $\max_{x\in I}|f(x,\eta)|=M<\infty$ for some given $\eta\in\mathbb{R}$. Also, suppose $f$ is differentiable with respect to $y$ with $|f_y(x,y)|\leq K|f(x,y)|$ uniformly on $x\in I$. Then for any given $x\in I$, $f(x,\cdot)\in C^1(\mathbb{R})$ is locally Lipshitz, i.e. $$\exists \delta_x>0 \exists L_x\geq 0,(y_1,y_2\in[\eta-\epsilon_x,\eta+\epsilon_x]\rightarrow\\\forall(y_1,y_2)\in\mathbb{R}^2(|f(x,y_1)-f(x,y_2)|\leq L_x|y_1-y_2|)).$$
The question is ,
Is such $L_x$ continuous with respect to $x$?
and I suppose such $L_x$ is taken as a infimum in kind that are possible.
The function $f$ given by $f(x,y)=e^y$ satisfies all the conditions (say with $\xi=0,a=1,\eta=0,M=1,K=1$), but is not Lipschitz in $y$.
@losif Yes, you are right, I changed my argument into locally Lipschitz.
The answer is no.
E.g., let $\xi:=-1$ and $a:=2$, so that $I=[-1,1]$. Let $f(x):=1+y(1-|y|/x)^2\,1_{x>0,\,|y|<x}$ for $x\in I$ and real $y$. Let
$\eta:=0$, so that $M=0$. Let $K=1$. Then all your conditions on $f$ hold.
Yet, $L_x=1_{x>0}$ is not continuous in $x\in I$.
Here is the graph $\{(x,y,f(x,y))\colon -0.2\le x\le1,\,-1.2\le y\le1.2\}$:
Perfect, thank you!
|
2025-03-21T14:48:29.671623
| 2020-01-24T03:20:44 |
351037
|
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|
Stack Exchange
|
The collection of mean value abscissas in the Mean value theorem
The integral mean value theorem for continuous f on [0,b] and finite positive continuous measure $\mu$ we have
$$\frac{1}{\mu[a,b]}\int_{a}^{b}f(x)d\mu(x)=f(c)(*)$$
for at least one $c\in [a,b]$. We want to study the properties of such $r\in C_{f,a,b}:=\{c: c \text{ that satisfy (*)} \}=f^{-1}\big(\frac{1}{\mu[a,b]}\int_{a}^{b}f(x)d\mu(x)\big)$.
In our particular problem, we want to compare to it a different $c'$ from
$$\frac{1}{\mu[0,a]}\int_{0}^{a}f(x)d\mu(x)=f(c')$$, by estimating their difference from below:
$$c-c'>\delta.$$
(Our f is positive continuous but nondifferentiable and non-monotone and so we cannot simply invert).
Q1: How to express the condition $c-c'>\delta$ in terms of properties of f and $\mu$? What are some distinct requirements on $f,\mu$ for this lower bound to be true? Conversely, suppose that the MVT set is contained within a small neighbourhood $C_{f,a,b}\subset [x_{0}-\epsilon,x_{0}+\epsilon]$. What does that imply for the behaviour of f (at least when $\mu=$Lebesgue)?
Q2: But I am curious if there is a framework for studying the properties of $C_{f,a,b}$? Is there any literature out there on putting bounds on the inf and sup of this set for various types of functions?
Q3: A geometric interpretation for such c is as the values that give matchings area in (*). I am curious if there have been any more theoretical results using that idea. Maybe studying some minimization problem:
$$arginf\{|\frac{1}{\mu[a,b]}\int_{a}^{b}f(x)d\mu(x)-f(r)|^{2}\}$$
using some optimization algorithm.
(from MVT's wiki)
If $|c-c'|\leq \delta$ for all $c\in C_{f,a,b},c'\in C_{f,0,a}$ then by the geometric interpretation and the Second-MVT, we intuitively see that the function must be like a trapezoid reaching a plateau inside the $\delta-$neighbourhood of $x=a$. Inside that plateau f must have a few high peaks that have enough area underneath them to give a mean value equality. This is assuming we work with $\mu$ Lebesgue; for more general measures, we can get different shapes.
Some work on exploring this same reverse question: given c what can we say about f: "On Functions Whose Mean Value Abscissas Are Midpoints, with Connections to Harmonic Functions" and "When Are There Continuous Choices for the Mean Value Abscissa?".
Partial Converse to the mean value theorem
|
2025-03-21T14:48:29.671920
| 2020-01-24T03:21:56 |
351038
|
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|
Stack Exchange
|
Bijections of Littlewood-Richardson coefficients
Let $c^{\lambda}_{\mu\nu}$ be the Littlewood-Richardson coefficients, where $\lambda,\mu,\nu$ are partitions. We know that $c^{\lambda}_{\mu\nu}= c^{\lambda}_{\nu\mu}$. Up to now, what are the bijections we known to prove this equality?
I found two papers that might be helpful: https://doi.org/10.1137/17M1162834, https://mathscinet.ams.org/mathscinet-getitem?mr=1773572.
Do you know how to download the first one ? I can not do this.
Sci-Hub helps with the first one. Also, there is a preprint. Moreover, see this (not the same paper).
thank you so much Darij!
|
2025-03-21T14:48:29.671995
| 2020-01-24T04:17:24 |
351042
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351042"
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|
Stack Exchange
|
What's the meaning of the nontrivial zeros of Selberg zeta function?
In the case of arithmetic variety over finite field, the zero points of the Hasse-Weil zeta function reflect the pure weights (i.e. dimension). On the other hand, in the case of the Selberg zeta function, the nontrivial zero points are deduced from the Laplace operator and it seems to me that they are almost trivial. Is there any
good interpretation of these zero points?
What do you mean by "they are almost trivial"?
|
2025-03-21T14:48:29.672072
| 2020-01-24T05:39:20 |
351045
|
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|
Stack Exchange
|
Jacobian change of variables formula for $p$-adic valued integration?
Let $k$ be a $p$-adic field. It's possible to make sense of the Haar measure $\mu_{\operatorname{Haar}}$ on $k^n$ as a $k$-valued measure and define integrals
$$\int\limits_{k^n} f(x_1, ... , x_n) d\mu_{\operatorname{Haar}}(x)$$
for suitable analytic $k$-valued functions $f$ on $k^n$. If $\varphi: U \rightarrow V$ is a $k$-analytic isomorphism between open sets $U$ and $V$ in $k^n$, then for suitable complex valued functions $f: k^n \rightarrow \mathbb C$, we have a change of variables formula
$$\int\limits_V f(x)dx_1 \cdots dx_n = \int\limits_U f \circ \varphi(x) |D_{\varphi}(x)|dx_1 \cdots dx_n$$
where $D_{\varphi}$ is the determinant of the Jacobian of $\varphi$, and $dx_1 \cdots dx_n$ is the (real valued) Haar measure on $k^n$. Is there an analogue of this formula for $k$-valued measures and $k$-analytic maps $f: k^n \rightarrow k$? Something like
$$\int\limits_V f(x) d\mu_{\operatorname{Haar}}(x) = \int\limits_U f \circ \varphi(x) D_{\varphi}(x) d\mu_{\operatorname{Haar}}(x)$$
(without the absolute value)?
What kind of "p-adic Haar measure" are you using here? Obviously whether or not such a thing exists depends on what properties that you want it to have, but for most reasonable definitions, such a thing doesn't exist.
I don't really know. The motivation behind this is I'm repeatedly encountering two $p$-adic Jacobian determinants whose absolute values are equal (which I determine via the usual complex change of variables formula). I want to say that they are actually equal, so I was hoping I could see this by some $p$-adic valued integration.
So you "don't really know" how to justify the assertion you make in the first sentence of your own question?
No I don't $\space$
@David Loeffler: Maybe a more interesting question would be whether there is some sort of p-adic measure / distribution / something for which the stated change of variables formula holds, at least for some sufficiently nice class of functions?
|
2025-03-21T14:48:29.672228
| 2020-01-24T07:10:00 |
351047
|
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"Gerry Myerson",
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|
Stack Exchange
|
Triangulating the plane using edges of unique rational lengths
Basic question: Can the Euclidean plane be divided into a vertex-to-vertex arrangement of non-overlapping triangles such that every edge has a unique rational length that lies between 1 and some specific rational R greater than 1?
Note: By vertex-to-vertex layout, we mean any vertex of the layout is necessarily an end point of all the edges meeting at that vertex. IOW, exactly two triangles meet at each edge in the layout. Also note that the length of each edge has to be different from other edge lengths.
(Note added on August 27th, 2022: I think "edge-to-edge" is the standard terminology for what I have referred to as "vertex-to-vertex").
If this is possible, one can apply further constraints such as "all triangles should have equal area (OR equal perimeter)". Alternatively, one can relax the vertex-to-vertex requirement. One can also replace the requirement that every edge has a unique length with (say) the triangles being pairwise non-congruent.
Note: Requiring the lengths of all edges to be integers rather than rationals would lead to the lengths of the edges being unbounded even if a triangulation with all edges having unique lengths is possible (not sure if this is possible).
Can you explain what is "vertex-to-vertex" arrangement? You can triangulate the plane into equal triangles with sides 3,4,5.
Why not equilateral triangles with edge length=1 ?
thanks. added explanation to vertex-to-vertex and 'uniqueness' of edge lengths.
If a scalene triangle ABC with rational lengths contains a similar copy DEA, also of rational lengths, and the lengths BD, DC, CE are also rational, then (generically) one should be able to solve the problem by gluing together various rotated and dilated versions of this configuration (the final triangulation resembles a discretised logarithmic spiral). This may be few enough constraints that the relevant Diophantine problem actually admits solutions.
@Terry, will this meet the condition of all side lengths coming from a rational subset of a bounded interval (1+ epsilon, 1+delta)? I'm not seeing how to partition the spiral to do that with small epsilon and delta. Gerhard "I Could Be Missing Something" Paseman, 2020.01.24.
Ah, right, I missed that requirement. No, these sorts of self-similar constructions will definitely not keep the lengths bounded.
Is there a link somewhere to a clear definition of edge-to-edge or vertex-to-vertex triangulation?
Strikes me as a fishing expedition. The triangles have to be edge-to-edge, or they don't, and the edge-lengths must all be different, or maybe it's enough for the triangles to be non-congruent, and maybe we should ask for all the areas to be the same, or maybe all the perimeters.... What do you actually want to know?
Here is a reduction to a more compact problem. (Actually, it isn't. However, it is closely related to an open problem, and maybe the second step involving "every other" vertex can be salvaged. Oops.)
Given a circle of radius R with R rational and an inscribed regular hexagon, there is K less than R and a concentric circle of radius K, such that this circle contains a countable infinite collection of points x, such that:
a) the distance between x and any vertex of the hexagon is a rational number, and
b) for any two different points x and y in this collection, the twelve distances between one of the two points and one of the hexagon vertices are distinct.
If we have this, take a regular hexagonal tessellation (with sides of length R), and find a point x in side each hexagon and tweak those sides, and then tweak "every other" hexagon vertex to get all distinct edge lengths. (This does not work as stated. We need c) which is a condition that preserves rationality of certain tweaked hexagons.)
Unfortunately, the compact problem resembles the open problem of finding a rational point inside a square, so the quadrilateral version of this problem reduces to a known open problem.
Gerhard "But At Least It's Bounded" Paseman, 2020.01.24.
|
2025-03-21T14:48:29.672539
| 2020-01-24T07:41:40 |
351049
|
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|
Stack Exchange
|
Isometries between two convex bodies
Let $A,B$ be two convex compact subsets with non-empty interior in a Euclidean space $\mathbb{R}^n$. Let $f\colon A\to B$ be a bijective isometry between them.
Does there exist an isometry $F\colon \mathbb{R}^n\to \mathbb{R}^n$ such that its restriction to $A$ is equal to $f$?
Remark. Necessarily $F$ is a composition of translation and orthogonal transformation.
A reference would be helpful.
convexity is not needed: non-empty interior is enough, moreover, having $n+1$ linearly independent points is enough. distances to these points determine the point in a space uniquely, so knowing images of these points you determine the images of every point of space.
@LevSoukhanov: Thanks, you are right. This is a final answer in fact.
Even without $n+1$ linearly independent points, there exists an isometry such that its restriction is $f$; but in that case it is not unique.
|
2025-03-21T14:48:29.672636
| 2020-01-24T07:50:27 |
351050
|
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|
Stack Exchange
|
Simple examples of colimits of affine schemes (evaluated in the presheaf category) which are not affine schemes
Notation and Setting: let $\operatorname{Aff}$ denote the category of affine schemes whose objects are covariant representable functors $\operatorname{X}:\operatorname{Ring}\rightarrow\operatorname{Set}$ and $\operatorname{Spec}:\operatorname{Ring^{op}}\rightarrow\operatorname{Func(Ring, Set)} $ be the contravariant Yoneda embedding of $\operatorname{Ring^{op}}$ in its category of presheaves so that $\operatorname{Aff}\simeq\operatorname{Ring^{op}}$.
In addition, let $\mathcal{O}:\operatorname{Func(Ring, Set)}\rightarrow\operatorname{Ring^{op}}$ be the functor that sends a functor $\operatorname{X}:\operatorname{Ring}\rightarrow\operatorname{Set}$ to the ring of maps $\operatorname{X}\rightarrow \mathbb{A}^1$ (where $\mathbb{A}^1$ is the forgetful functor) so that $\operatorname{Spec}$ and $\mathcal{O}$ are inverse of one another.
Let $\widehat{\operatorname{Aff}}$ be the indization of $\operatorname{Aff}$, i.e. the category whose objects are functors $\operatorname{X}:\operatorname{Ring}\rightarrow\operatorname{Set}$ that are small filtered colimits of affine schemes.
My question:
I am looking for (simple) examples of functors which are objects of $\widehat{\operatorname{Aff}}$ but are not affine schemes.
I am particularly interested in examples of the following form:
Let $\operatorname{X}$ be an affine scheme, $I\subseteq\mathcal{O}_{X}$ an ideal and consider the following diagram in $\operatorname{Func(Ring, Set)}$ over $\mathbb{Z_{\geq0}}$
$0=\operatorname{Spec(\mathcal{O}_{X}/I^{0})}\hookrightarrow\ldots\hookrightarrow\operatorname{Spec(\mathcal{O}_{X}/I^{n-1})}\hookrightarrow\operatorname{Spec(\mathcal{O}_{X}/I^{n})}\hookrightarrow\ldots$
Since $\operatorname{Func(Ring, Set)}$ admits small colimits, $\mathrm{colimit}_{n\in\mathbb{Z}_{\geq0}}\operatorname{Spec(\mathcal{O}_{X}/I^{n})}$ exists. Thus I am looking for examples of affine schemes $\operatorname{X}$ and ideals $I\subseteq\mathcal{O}_{X}$ for which $(\mathrm{colimit}_{n\in\mathbb{Z}_{\geq0}}\operatorname{Spec(\mathcal{O}_{X}/I^{n})})\neq \mathrm{colimit}_{n\in\mathbb{Z}_{\geq0}}(\operatorname{Spec(\mathcal{O}_{X}/I^{n})})$
The only example that I could find so far was that of $\operatorname{Spec(\mathbb{Z}[x])}$ and the ideal $(x)$ which give
the functor $\operatorname{Nil}\simeq\mathrm{colimit}_{n\in\mathbb{Z}_{\geq0}}\operatorname{Spec(\mathbb{Z}[x]/(x)^{n})}\neq(\mathrm{colimit}_{n\in\mathbb{Z}_{\geq0}}\operatorname{Spec(\mathbb{Z}[x]/(x)^{n})})\simeq\operatorname{Spec}(\mathbb{Z}[\![ x ]\!])$.
To see that you can either show that $\operatorname{Nil}$ is not representable1 or check, for example, that $\operatorname{Spec}(\mathbb{Z}[\![ x ]\!])(\mathbb{Z})\neq\operatorname{Nil}(\mathbb{Z})$.
It looks like you are asking about formal schemes. Your example Nil is the formal spectrum $\operatorname{Spf}\mathbb{Z}[[x]]$. They are discussed at the end of EGA1.
I don't think that the question is really about formal schemes (since here no "global space" is given). A better keyword is "ind-schemes".
A basic standard example is the colimit of $\mathbb{A}^0 \to \mathbb{A}^1 \to \mathbb{A}^2 \to \cdots$ with transition maps $x \mapsto (x,0)$. The $R$-valued points are finite sequences in $R$. This functor is not representable.
More generally, let $A$ be a commutative ring with a sequence of ideals $I_0 \supseteq I_1 \supseteq I_2 \supseteq \cdots$. (In the mentioned example, $A = \mathbf{Z}[x_0,x_1,\dotsc]$ and $I_n = \langle x_n,x_{n+1},\dotsc\rangle$.) Then the colimit $X$ of $\mathrm{Spec}(A/I_0) \to \mathrm{Spec}(A/I_1) \to \cdots$ is the subfunctor of $\mathrm{Spec}(A)$ whose $R$-valued points are those $A \to R$ whose kernel contains some $I_n$. We have $\mathcal{O}(X) = \lim_n A/I_n =: \widehat{A}$.
Then $X$ is representable aka affine iff the canonical morphism $X \to \mathrm{Spec}(\widehat{A})$ is an isomorphism. It is injective anyway, and it is surjective on $R$-valued points iff every homomorphism $\widehat{A} \to R$ factors through some projection $\widehat{A} \to A/I_n$. So $X$ is representable iff the identity $\widehat{A} \to \widehat{A}$ factors through some projection $\widehat{A} \to A/I_n$. But this clearly implies $I_n = I_{n+1} = \dotsc$ and the sequence is stationary.
Conversely this means that for every non-stationary sequence $I_0 \supseteq I_1 \supseteq I_2 \supseteq \cdots$ the functor $X$ is not representable.
|
2025-03-21T14:48:29.672875
| 2020-01-24T09:21:39 |
351053
|
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|
Stack Exchange
|
Distance between value function of deterministic and stochastic control problems
Suppose that one wants to control a diffusion process
$$
dX_t^u = \mu(X_t^u,u)dt + \sigma dW_t; \qquad X_0^u=x
$$
in order to optimize a stochastic control problem with value function
$$
V_T(u)=\mathbb{E}[f(X^u_T) + \int_0^T g(X_t^u)dt].
$$
Where $f,g$ are bounded smooth functions and $u$ is an admissible control.
Now suppose that one wants to optimize the deterministic control problem:
$$
\partial_t x_t^u = \mu(x_t^u,u)dt ; \qquad X_0^u=x
$$
with value function
$$
v_T(u)=f(X^u_T) + \int_0^T g(X_t^u)dt.
$$
For small $\epsilon$, when can one guarantee that the optimal policies of both problems are "close" (in expectation or some other appropriate sense)?
See this: https://www.icts.res.in/sites/default/files/ldt2017-2017-08-17-Vivek-S-Borkar.pdf
On page 18, it says "Under suitable technical conditions" where could I find those?
Maybe start here: http://www.princeton.edu/~moll/viscosity_slides.pdf
|
2025-03-21T14:48:29.672965
| 2020-01-24T10:17:13 |
351054
|
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"Dominic van der Zypen",
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"url": "https://mathoverflow.net/questions/351054"
}
|
Stack Exchange
|
Meetability of $\pm 1$-functions on $\omega$
If ${\cal S}$ is a collection of functions $f:\omega\to\omega$ we say that ${\cal S}$ is meetable if there is a "global function" $g:\omega\to \omega$ such that for every $f\in {\cal S}$ there is $n\in\omega$ such that $g(n) = f(n)$.
Let ${\cal S}^{\pm 1}$ denote collection of functions $f:\omega\to\omega$ such that $|f(n) - f(n+1)| = 1$ for all $n\in\omega$ and for $k\in\omega \setminus \{0\}$, let ${\cal S}^{\pm 1}_{\leq k} = \{ f\in {\cal S}^{\pm 1}: f(0) \leq k\}$.
Is ${\cal S}^{\pm 1}$ meetable? If not, is ${\cal S}^{\pm 1}_{\leq k}$ meetable for all $k\in\omega\setminus\{0\}$?
(User Wojowu pointed out in a comment below that ${\cal S}^{\pm 1}_{\leq 1}$ is meetable.)
${\cal S}^{\pm 1}_{01}$ is met by the constant $1$ function.
Right - thanks!
Am including Wojowu's remark in the post and editing it slightly
$\newcommand\om{\omega}$
The set $S:={\cal S}^{\pm 1}$ is meetable by any function $g$ of the following form:
Let $l_0,l_1,\dots$ be odd numbers in $\om=\{0,1,\dots\}$ such that $l_j/n_j\to\infty$ (as $j\to\infty$), where $n_0=-1$ and $n_{j+1}=n_j+l_j$ for $j\in\om$. (E.g., take $l_j=1+j!$ for $j\ge2$.) For each $j\in\om$ and all $n\in\om$ such that $n_j<n\le n_{j+1}$, let $g(n)=n_{j+1}-n$.
Indeed, take any $f\in S$ and suppose that $f$ is not met by $g$. Since $l_j/n_j\to\infty$, we have $f(n_j+1)\le f(0)+n_j+1\le l_j-1=g(n_j+1)$ for all large enough $j$. For all such $j$, there exists
$$m_j:=\max\{n\in\{n_j+1,\dots,n_{j+1}\}\colon f(n)\le g(n)\}.$$
Since $f$ is not met by $g$ and $|f(n+1)-f(n)|=1$, we see that
$f(m_j)=g(m_j)-1=n_{j+1}-m_j-1$, whence $n_{j+1}-1=f(m_j)+m_j=f(0)\text{ mod }2$ for all large enough $j$. This is a contradiction, because the $l_j$'s are all odd and hence the $n_j$'s alternate $\text{ mod }2$.
Here are the list plots $\{(n,g(n))\colon n\in\{1,\dots,n_5\}\}$ (blue) for $l_j\equiv1+j!$ and $\{(n,f(n))\colon n\in\{1,\dots,n_5\}\}$ (red) for a "random" $f\in S$, showing how $g$ eventually "catches/meets" $f$ -- catches only by the last in the picture (negatively sloped) dotted blue line segment ($f$ was able to get through the previous blue barrier):
Amazing construction - thanks Iosif!
|
2025-03-21T14:48:29.673111
| 2020-01-24T10:36:47 |
351055
|
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|
Stack Exchange
|
Adjunctions between $\mathcal{A}_{\infty}$-categories
Is there a theory of homotopy coherent adjunctions between $\mathcal{A}_{\infty}$-categories? By this I mean at least a definition of what an adjunction is and a construction of the corresponding unit/counit.
I couldn't find any reference in the literature. Thank you very much in advance.
Riehl and Verity study the free homotopy coherent adjunction as a simplicial category, see below; it should be possible to make the correct definition via this and by taking an appropriate nerve of the $A_\infty$-category or an appropriate realization of the simplicial category. https://arxiv.org/abs/1310.8279
|
2025-03-21T14:48:29.673182
| 2020-01-24T13:50:55 |
351062
|
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"giulio bullsaver",
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|
Stack Exchange
|
Trasforming a system of rational equations into an equivalent system of polynomial equations
Suppose that a system of rational equations $r_1=0, r_2=0, \dots, r_m=0$ defines a zero dimensional variety $V$.
Is there an algorithm to produce polynomials $p_i$, starting from the rational functions $r_i$, such that the polynomial system $p_1 = 0, p_2 = 0, \dots, p_n = 0$ defines the SAME zero dimensional variety $V$?
All of the functions involved can be considered in $\mathbb{C}^k$ for some $k \ge 1$.
EXAMPLE:
Let $r_1 = x^2 - 1, r_2 = \frac{x y}{x-1}$. The naive thing I would do would be to take the numerators of each $r_i$, that way I get $p_1 = x^2-1$ and $p_2 = x y$. This system of polynomial equations defines $y=0, x = \pm 1$, while the system of rational defines $y=0, x = -1$.
but this is not an algorithm to go from the $r_i$ to the $p_i$, because there is no algorithm to solve for the equations in general
Could you elaborate more? It seems like you are saying that there is an algorithmic way to find the roots of a rational system, in particular you could the algorithmically find the roots of an arbitrary high degree polynomial which I thought was impossible. (https://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem)
The AbelβRuffini theorem concerns solvability in radicals. It has nothing to do whatsoever with algorithmic root finding.
I see. Well, then I could perhaps change the question in "what are the algorithms to go from r to p"
What you do is add equations that force the denominators to be non-zero. For example, if you add $r_3$ and the equation $r_3(x-1) = 1$, this rules out the $x = 1$ solution.
I think there exists no such an algorithm in the multivariate case. At least, the Kronecker software aimed at solving system of polynomial equations and non-equalities treats the latter separately.
arsmath, make it an answer and I will accept it. It's clearly the thing to do
|
2025-03-21T14:48:29.673342
| 2020-01-24T14:38:11 |
351066
|
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"Ben McKay",
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|
Stack Exchange
|
Bezout's theorem for finite fields
I would like to know if the following statements holds:
Suppose $K$ is a finite field. Is it true then that for any two polynomials $f,g\in K[x,y]$ that have no common factors, the number of solutions of $f(x,y)=g(x,y)=0$ is at most $\deg(f)\times \deg(g)$?
I know a proof of this statement when $K$ is infinite, but I don't see how to adapt it to the case when $K$ is finite.
By the way, is it true that if $K$ is finite, $f,g\in K[x,y]$ and $f$ and $g$ have a common factor over an algebraic closure of $K$, then $f$ and $g$ also have a common factor over $K$?
Peoples who voted to close the question, could you please answer the question at least? The comment left by Daniel Lourghan doesn't answer this question. Indeed, why, when you pass to $\bar k$ it can not happen that suddenly $f$ and $g$ have a common factor? (By the way, this is what I have already asked above...)
The version given as a equality for plane projective curves is almost always stated and proved over an arbitrary algebraically closed field (e.g. here https://en.wikipedia.org/wiki/B%C3%A9zout%27s_theorem). The version you want (upper bound) then trivially holds over an arbitrary field $k$, since the number of solutions over $k$ is less than or equal to the number of solutions over $\bar{k}$.
It does not look very true as stated. Say $f=x$ and $g=y$.... Don't you want $f=0=g$ instead??
Daniel Loughran, if you say this is trivial, could you explain then why when you pass to $\bar k$ the polynomials still don't have a common factor? You agree that if polynomials have a common factor over $\bar k$, the number of solutions will be infinite, so your argument will break?
The existence of a common factor is computed using the resultant, which can be computed over any division ring which contains the coefficients. See my lecture notes for more: https://github.com/Ben-McKay/concrete-algebra/blob/master/algebra.pdf
Sorry Ben, are you speaking about common factors of polynomials of two variables? In your notes as far as I can see you speak about resultants of polynomials in one variable? However here we are dealing with two variables...
If your polynomials have a common factor $h$, they are also divisible by ${}^{\sigma }h$ for each $\sigma $ in $\operatorname{Gal}(\bar{K}/K) $. It follows easily that they have a common factor in $K[x,y]$.
(@abx is, I think, also using implicitly the existence of a unique monic gcd, so that Galois is not just permuting around common divisors but actually fixing the greatest.)
Dear abx, thanks for the comment. This answers the question indeed.
|
2025-03-21T14:48:29.673535
| 2020-01-24T14:42:20 |
351067
|
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"Vahid Shams",
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|
Stack Exchange
|
An upper bound for second type of Reidemeister move
Suppose there are tow diagram $D_1$ and $D_2$ of knot $K$ with $c_1$ and $c_2$ crossing. Are there any bound of second type of Reidemeister move in term of $c_1$ and $c_2$? In other words, Are there Reidemeister moves like $\Omega_1,...,\Omega_n$ which transfer $D_1$ to $D_2$ and the number of second type of Reidemeister move in $\Omega_1,...,\Omega_n$ is bounded by $f(c_1,c_2)$? What is the best $f$?
I know Marc Lackenby found an upper bound for number of Reidemeister move but is there a better upper bound for the number of second type of Reidemeister move?
@MarkSapir It was wrong, I edited it.
|
2025-03-21T14:48:29.673598
| 2020-01-24T14:52:47 |
351068
|
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|
Stack Exchange
|
Whitney stratification for proper morphisms
Let $f: X \to \Delta$ be a flat, projective morphism, smooth over the punctured disc $\Delta^*:=\Delta \backslash \{0\}$ and central fiber $f^{-1}(0)$ is a reduced, simple normal crossings divisor. Does there exist a Whitney stratification of $X$ with one strata consisting of $f^{-1}(\Delta^*)$? For example, in the case $X$ is of relative dimension $1$ (i.e., $X$ is a family of curves) and the central fiber is the union of two smooth, irreducible curves $Y_1, Y_2$, is $$\{f^{-1}(\Delta^*)\} \coprod \{(Y_1 \cup Y_2)\backslash (Y_1 \cap Y_2)\} \coprod \{Y_1 \cap Y_2\}$$
a Whitney stratification of $X$? Any idea/reference will be most welcome.
EDIT: Assume $X$ is a smooth, complex manifold.
|
2025-03-21T14:48:29.673779
| 2020-01-24T15:04:41 |
351070
|
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|
Stack Exchange
|
Is there a restrain on the total number of proper classes strictly smaller than the universe in variants of MK?
In an old discussion thread at sci.math, Herman Rubin said:
"
There exist models where all proper classes have the same cardinality;
i.e., the universe is equinumerous with the class of ordinal numbers.
There exist models where there are proper classes which are neither
larger nor smaller than the class of all ordinal numbers.
I am almost certain that there exist models with proper classes
strictly larger than the class of all ordinal numbers, and all
comparable.
"
On the other hand Asaf Karagila in a comment to this posting to MO, have said:
"
in ZF we can at least prove that every subset of $V_\kappa$ is an
element of it, or maps onto $\kappa$.
" ($\kappa$ is meant to be inaccessible).
I take that to mean in absence of choice [since its stated in ZF alone]. Now in some sense $V_\kappa$ can be viewed as being the standard model of MK.
Clearly the second and third of Rubin's statements indicate the existence of a model of a variant of MK\NBG in which there do exist a proper class that is strictly less in size than $V$ namely the class $\small \sf [ON]$ of all ordinal numbers (that are sets of course).
I'm interested in the last of Rubin's statements. He said "all being comparable". I don't know if that entails choice [not global of course].
In the same referred thread Aatu Koskensilta answeres to the third of Rubin's statements by:
"
Using the consistency of ZFC + "there is an inaccessible" (and
standard well known independence results) it's easy to show there are
such models. As noted, by some tweaking we can remove the inessential
inaccessible, bringing us back to ZFC and MK in terms of consistency
strength.
"
Questions:
What are the models of those variants of MK in which there are proper classes of sizes strictly smaller than the universe $V$ of all sets, that are closest to the standard model of MK. Provided that the total number of proper classes in them is higher than that of sets.
Is there a restrain on the total number of such proper classes (i.e. strictly smaller than $V$) in such models.
Can those models satisfy choice for sets.
|
2025-03-21T14:48:29.673953
| 2020-01-24T15:22:13 |
351071
|
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|
Stack Exchange
|
Law of large numbers for random Dirac measures
Suppose $\{X_1,...X_n\}:\Omega \to \mathbb{R}^p$ be i.i.d. random vectors with common probability law/measure $p$, i.e. $Prob(X_i^{-1}(E))=p(E) \forall E \subset \mathbb{R}^p $ Borel measurable.
Consider the random Dirac measures $\delta_{X_i}$, and their average, which is a random probability measure on $\mathbb{R}^p$, defined by $\frac{1}{n}\sum_{i=1}^{n} \delta_{X_i}$. I'd like to know if $\frac{1}{n}\sum_{i=1}^{n} \delta_{X_i}$ weakly converges to the deterministic measure $p$,
i.e. for every continuous, bounded function $f: \mathbb{R}^p \to \mathbb{R}$, must we have:
$$\frac{1}{n}\sum_{i=1}^{n} {f(X_i)} \to \int_{\mathbb{R}^p } f(x)dp(x)$$
as a convergence in probability of a sequence of random variables $\frac{1}{n}\sum_{i=1}^{n} {f(X_i)}$?
ON a related note, I'd also like to know if the following is true or not:
If a sequence of random measures converges to a deterministic measure in probability, is it equivalent to have the same convergence almost surely? This question is motivated by the fact a when a sequence of random variables converge to a sonstant in probability, the convergence is a.s.
P.S. I understand that this question might be elemenrtary to many of you, so some references would be greatly apreciated!
If I'm not mistaken, this is exactly the Glivenko-Cantelli theorem.
@NateEldredge Honestly I'm not familiar with the many theorems in probability, but I just looked up Glivenko-Cantelli theorem, and you're right, it's what I was looking for. I can add "reference request" as a topic and modify the question, of you're free to vote to close it.
BTW: what's the answer to the second question which concerns the equivalence of convergence in probability and a.s. convergence of a sequence of random measures?
I guess Glivenko-Cantelli shows something a little stronger: that the cdfs actually converge uniformly (for weak convergence, you only need pointwise convergence at points of continuity). This only makes any difference if the distribution $p$ has a discrete part (i.e. atoms). It also gives a.s. convergence.
By the way, you've probably found this by now, but the usual term for $\frac{1}{n} \sum^n \delta_{X_i}$ is "empirical distribution" or "empirical measure".
@NateEldredge sorry for the late reply in comments. Yes I agree that the the Glivenko-Cantelli theorem proves something stronger that weak convergence of measures, and yes I've found that it's called the "empirical distribution".
But looking at all these "law of large number type" theorems of random measures, I wonder if there's a lecture note/blog post that'd nicely combine and compare them. So the kind of note/post I'm looking for is something that'll first state the WLLN/SLLN for r.v., and then would do the same for random measures, i.e. measure-valued random variables (contd.)
@NateEldredge (contd.) and will state a general theorem of metric (or some other topological) space (say $S$) valued random variable, and will state in full generality the laws of large numbers and the central limit theorems, and then will show that the SLLN/WLLN and the question I asked here is really a special case for these generalized law of large numbers for $S$-valued random variables. (contd.)
Related but somewhat different: I also keep seeing several literature that compares the convergence of the random matrix $\frac{1}{n} \sum_{i=1}^{n} k(x,x_i)f(x_i)$ to the random integral $\int k(x,y)f(y)dP(y)$, where $x_i$'s are iid with commmon law $P$. I've seen a paper of GinΓ© and Koltchinkii (https://projecteuclid.org/euclid.bj/1082665383) that talks about the convergence of corresponding eigenvalues, and also some papers by stat/machine learning people which seem to lack rigor. I'll ask a follow up question on the reference request for this, where I can get an integrated vision.
For your question to make sense, you have to actually have an infinite sequence $X_1,X_2,\dots$ of iid random vectors. Then, by the strong law of large numbers (SLLN) ,
$$\frac1n\,\sum_{i=1}^n f(X_i)\to Ef(X_1)=\int_{\mathbb R^p} f\, dp$$
almost surely (a.s.), which, as desired, implies the weak law of large numbers (WLLN) , that is, the convergence in probability (which is always implied by the a.s. convergence).
Concerning your second question, it is of course not true that "when a sequence of random variables converge[s] to a [constant] in probability, the convergence is a.s." -- That would be the same as to say that the convergence in probability always implies the a.s. convergence, because $Y_n\to Y$ a.s. (respectively, in probability) if and only if $Y_n-Y$ converges to the constant $0$ a.s. (respectively, in probability).
|
2025-03-21T14:48:29.674503
| 2020-01-24T18:48:44 |
351079
|
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|
Stack Exchange
|
complex of psuedo-tensorial differential forms
Let $P(M,G)$ be a principal bundle. Let $V$ be a finite dimensional vector space (over $\mathbb{R}$). Let $\rho:G\rightarrow GL(V)$ be a representation.
A $V$-valued differential $r$-form $\varphi$ on $P$ is said to be a
psuedo-tensorial $r$-form of type $(\rho,V)$ if
$(R_a)^*\varphi=\rho(a^{-1}).\varphi$ for all $a\in G$; that is,
$$(R_a^*\varphi)(p)(v_1,\cdots,v_r)=\rho(a^{-1})\big(\varphi(p)(v_1,\cdots,v_r)\big)$$
for all $a\in G$, $p\in P$ and $v_i\in T_pP$ with $1\leq i\leq r$.
Let $\varphi$ be a psuedo-tensorial differential $r$-form on $P$. Then, the usual exterior derivative $d\varphi$ is a psuedo-tensorial differential $r+1$-form on $P$ (for reference, see Kobayashi and Nomizu, Volume $1$, page $76$). Thus, one can consider the complex $$\cdots\Omega_{(\rho,V)}^{r-1}(P)\xrightarrow{d}\Omega_{(\rho,V)}^r(P)\xrightarrow{d}\Omega_{(\rho,V)}^{r+1}(P)\rightarrow \cdots$$ of psuedo-tensorial forms of type $(\rho,V)$ on $P$. We can also consider the corresponding cohomology groups.
Question : Does choosing $(\rho, V)$ tactically and considering the complex of psuedo-tensorial forms and the corresponding cohomology groups give an extra insight about the principal bundle $P(M,G)$?
Side question : Only place I have seen psuedo-tensorial forms is when defining the curvature form $\Omega$. Given a connection $\Gamma\subseteq TP$ on the principal bundle $P(M,G)$, the corresponding connection form $\omega:P\rightarrow \Lambda^1_{\mathfrak{g}}T^*P$ is a psuedo-tensorial $1$-form on $P$ (of type $(ad,\mathfrak{g})$) and the corresponding curvature form $\Omega:P\rightarrow \Lambda^2_{\mathfrak{g}}T^*P$ is a tensorial $2$-form on $P$. Are there other places where tensorial differential forms arise?
|
2025-03-21T14:48:29.674640
| 2020-01-24T19:07:32 |
351080
|
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|
Stack Exchange
|
Direct proof that Chern-Weil theory yields integral classes
Suppose $E$ is a complex vector bundle of rank $n$ on a compact oriented manifold (both assumed smooth). Let $h$ be a Hermitian metric on $E$, and let $A$ be a Hermitian connection on $E$ and $F_A$ its curvature form, an element of $\Omega^2(M,\text{End }E)$. Chern-Weil theory produces a closed even differential form $c(A) = \det(1 + \frac i{2\pi}F_A) = c_0(A)+c_1(A) + \cdots + c_n(A)$.
These classes have the property that for all compact oriented submanifolds $\Sigma\subset M$ of dimension $2k$, the expression $\int_\Sigma c(A)$ is an integer. The usual way to prove this seems to be to first show that the de Rham cohomology class of $c(A)$ is independent of the connection $A$ and functorial under pullbacks. We then check that it has the desired integrality property for the tautological bundles on Grassmannians by direct computation (which says this class lies in singular cohomology with $\mathbb Z$ coefficients) and then use the fact that every bundle on $M$ is a pullback of the tautological bundle from a sufficiently large Grassmannian. Alternatively, we show that the above definition satisfies the axiomatic characterization of Chern classes and then we identify it with the image in $H^\bullet(-,\mathbb C)$ of Chern classes constructed in $H^\bullet(-,\mathbb Z)$ by using a computation of the cohomology ring of infinite Grassmannians.
Is there a direct proof that $\int_\Sigma$ c(A) is an integer for all compact oriented submanifolds $\Sigma\subset M$?
By a direct proof, I mean something which only uses only local reasoning about curvature and connections and doesn't have to pass through much algebraic topology. For instance for the case $n=1$ (i.e., a line bundle) and taking $\Sigma = M$ to be a compact oriented surface, we note by elementary local computations (using Stokes' theorem) that for any disc $D$ inside $\Sigma$ with oriented boundary circle $\gamma$, the number $e^{\pm\int_D F_A}\in U(1)$ measures the holonomy of the parallel transport along the loop $\gamma$. From this, it easily follows that we can extend the above statement to the case when $D$ is a more general region with smooth oriented boundary $\gamma$. As $\Sigma$ has no boundary, applying this to $D = \Sigma$ gives $\int_\Sigma F_A\in \frac{2\pi}i\mathbb Z$ as desired. (This proof is similar to the moral proof of Stokes' theorem given by checking it at the infinitesimal level -- essentially the definition of the exterior derivative -- and then triangulating the domain of integration and noting that boundary terms with opposite orientations cancel.) The $c_1$ case seems to be simple because of $U(1)$ being abelian.
Is the splitting principle too indirect for you? It should reduce the general case to the abelian case.
This seems to be related to finding a "direct proof" of the following assertion: given $n$ line bundles $L_i$ ($1\le i\le n$) with connections $A_i$, how do we show that $\int_\Sigma\prod_i c(A_i)\in\mathbb Z$ for each compact oriented $\Sigma\subset M$? I'd like to avoid using singular (co)homology with $\mathbb Z$ coefficients in the proof.
This seems surprisingly non-trivial to me, essentially requiring the Kunneth formula mod torsion for integral homology...
@MohanSwaminathan could you please describe the elementary local computation that justifies: $e^{\pm\int_D F_A}\in U(1)$ measures the holonomy of the parallel transport along the loop $\gamma$?
@MohithRaju here is a reference. Theorem 1 in the beginning of section 7.3 of Singer and Thorpe's "Lecture Notes on Elementary Topology and Geometry". To make sense of S&T's notation, you might have to also look at sections 7.1 and 7.2. Also, S&T seem to prove the theorem only in the case of the Levi-Civita connection on an oriented 2-manifold but the proof is general and works for any principal $U(1)$ bundle with a $U(1)$ connection (with any smooth 2-simplex in the manifold).
Yes, the ChernβWeil homomorphism lifts to differential cohomology,
which guarantees that periods are integral.
See the original paper by Cheeger and Simons, or the paper by Hopkins and Singer.
The (modernized) construction of such a refinement relies on the computation of the de Rham complex of the stack B_β(G) of principal G-bundles with connection and their isomorphisms, which is isomorphic (and not just quasi-isomorphic) to the algebra of G-invariant polynomials on the Lie algebra of G.
This is proved by Freed and Hopkins,
and their argument is a local argument that uses computations
with differential forms.
There is a bit of linear algebra and invariant theory at the very end.
This argument is essentially an abstract higher-dimensional version of the n=1 argument above with the Stokes theorem.
The lift to differential cohomology is already explained in the original paper by Cheeger and Simons. But they use a result of Narasimhan-Ramanan, that every vector bundle with a metric connection can be pulled back (as a vector bundle with metric connection) from a sufficiently large Grassmannian. So their proof is not much better than those the OP mentions. Is the paper by Hopkins and Singer in any way more of a "local" nature?
@SebastianGoette: There is a modernized version of this argument, due to Freed and Hopkins (ChernβWeil forms and abstract homotopy theory), which computes instead the de Rham complex of the stack B_β(G) of principal G-bundles with connections and their isomorphisms, and show that it is isomorphic (and not just quasi-isomorphic) to the G-invariant polynomials on the Lie algebra g. Their proof consists mostly of local constructions with differential forms, like the OP wanted. There is a small bit of linear algebra and invariant theory at the very end.
I'm not at all familiar with differential cohomology and I think I will need some time to take a look at this paper to see if it is what I want. Thanks for the reference!
I am pretty sure that Cheeger and Simons use integrality of CW classes in the construction of their lift. In fact, the lift itself is pretty trivial (modulo Narasimhan-Ramanan and modulo the main general long exact sequence for differential characters that they prove, which a priori has nothing to do with Chern-Weil) so it would be strange if it produced integrality by itself.
|
2025-03-21T14:48:29.675037
| 2020-01-24T19:23:21 |
351081
|
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|
Stack Exchange
|
Pointwise vs. local homotopy equivalences of continuous and smooth complexes of real vector bundles
Let $(E^\bullet,d_E)$ and $(F^\bullet,d_F)$ be two complexes of real vector bundles on a topological manifold $X$, and let $f^\bullet\colon E^\bullet\to F^\bullet$ be a morphism of complexes, i.e. a collection of degreewise morphism of vector bundles $f^n\colon E^n\to F^n$ compatible with the differentials. Then at every point $x$ of $X$ we have a morphism $f^\bullet_x\colon E_x^\bullet\to F_x^\bullet$ of chain complexes of finite dimensional real vector spaces, and we say that $f^\bullet$ is a pointwise homotopy equivalence at $x$ if there exist a morphism of chain complexes of vector spaces $g_x^\bullet\colon F_x^\bullet \to E_x^\bullet$ and morphisms of graded vector spaces $H_x^\bullet\colon E_x^\bullet\to E_x^{\bullet-1}$ and $K_x^\bullet\colon F_x^\bullet\to F_x^{\bullet-1}$ such that $g_x^\bullet\circ f_x^\bullet= Id-[d,H_x^\bullet]$ and $f_x^\bullet\circ g_x^\bullet= Id-[d,K_x^\bullet]$.
Similarly, we say that $f^\bullet$ is a homotopy equivalence on an open subset $U$ of $X$ if in the above one can choose $g,H,K$ to be morphisms of chain complexes of vector bundles and of graded vector bundles over $U$, respectively. In other words, if we restrict our attention to a trivializing open set $U$ for our bundles, then $f^\bullet$ is a homotopy equivalence on $U$ if we can choose the pointwise defined functions $g_x^\bullet,H_x^\bullet,K_x^\bullet$ to vary continuously over $U$. Clearly, if $f^\bullet$ is a homotopy equivalence over $U$, then it is a pointwise quasi-isomorphism at every point $x$ of $U$, but one would expect the converse not to be true. That is, I would expect that the pointwise existence of a solution $g_x^\bullet,H_x^\bullet,K_x^\bullet$ to the homotopy equivalence equations alone cannot imply that one can find also a continuous solution $\tilde{g}_x^\bullet,\tilde{H}_x^\bullet,\tilde{K}_x^\bullet$ (possibly different from the given pointwise one).
However, I'm finding it much harder than expected to produce a counterexample. Namely, the equations $g_x^\bullet,H_x^\bullet,K_x^\bullet$ are just linear equations once $f^\bullet$ and the differentials are given, so that the problem is a particular instance of the problem of determining the existence of a continuous solution for a parameter-dependent linear system $A(x)\cdot v_x=b_x$ (with $A(x)$ and $b_x$ continuously depending on $x$) once one knows that for every $x$ there exist at least one solution. And it is very easy to produce examples of $A(x)$ and $b_x$ where no continuos solution can exist. Yet,I have been so far failing in my attempts of producing an explicit example with no continuous solution of the specific form given by the homotopy equivalence equations.
What has been frustrating my very low-rank few-terms complexes attempts so far has been the existence of partition of units. So I guess that despite what I was expecting, pointwise homotopy equivalence could actually imply local homotopy equivalence in the continuous setting (and then arguably also in the smooth setting but not the real analytic setting).
Apart from my low-rank few-terms I have been searching the literature, but also here I have not been successful so far.
|
2025-03-21T14:48:29.675240
| 2020-01-24T19:54:48 |
351082
|
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|
Stack Exchange
|
Roots of a family of 4-parameter polynomials
Let $k, \ell, p$ and $q$ be positive integers, with $q>p>1$ and $\gcd(p,q)=1$. Let $f(x)$ the polynomial given by
$$
f(x)=x^q-kx^{q-p}-\ell.
$$
This polynomial is related to a family of two-parameters binary sequences.
Now, I would like to prove that $f$ has a dominant root and that all the roots are simples. I was able to solve for some particular cases (e.g., $p=2$ and $k=q=1$).
Someone has some suggestion for proving this, in the general case? Thanks in advance.
if $q$ and $p$ are both divisible by 3, then with each root $\theta$ there are roots $e^{2\pi i/3}\theta$ and $e^{-2\pi i/3}\theta$, which of them is "dominant"?
Thanks for your comment. In fact, I forgot to say that, clearly, $p$ and $q$ are coprime.
That assumption seems to render "$f$ is not an even function" redundant, no?
@user44191 yes, I modified now.
@FedorPetrov Yes, you are right. I forgot to include that p and q are coprime.
"dominant" = "of the largest absolute value" or "of the smallest absolute value" (or anything else)?
Sorry I keep adding and deleting an answer, but I keep noticing small different criteria that I think I missed. I do think that your claim of simple roots is incorrect; integer solutions to $\ell^p = \frac{(q - p)^{q - p} k^q p^p}{q^q}$ for a given even $p$, odd $q$ will imply a double root.
For an example: $x^5 - 15x^3 - 162$ has a double root at $-3$.
|
2025-03-21T14:48:29.675375
| 2020-01-27T09:56:23 |
351232
|
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|
Stack Exchange
|
Kauffman bracket polynomial algorithm
I am doing a project on the invariants of knots and links specifically working on Kauffman bracket polynomial but drawing the states for knots with high crossings is tedious.
I am searching for a working code that can generate the states and the type of smoothings used.
https://tex.stackexchange.com/
Hi Martha, welcome to MO. I think your question might not misinterpreted as question about putting diagrams in a paper. Are you looking for something along these lines: http://katlas.org/wiki/The_Kauffman_Polynomial?
Bar-Natan's KnotTheory` Mathematica package can compute the Kauffman polynomial. If you need the actual state diagrams, I recommend looking here
Sage has also some code about knots, but maybe not what you need.
ββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββ
β SageMath version 9.1.beta1, Release Date: 2020-01-21 β
β Using Python 3.7.3. Type "help()" for help. β
ββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββ
ββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββ
β Warning: this is a prerelease version, and it may be unstable. β
ββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββ
sage: K = Knots()
sage: J = K.from_table(8,10)
sage: J.
J.alexander_polynomial J.gauss_code J.orientation
J.arcs J.genus J.oriented_gauss_code
J.arf_invariant J.homfly_polynomial J.parent
J.base_extend J.is_alternating J.pd_code
J.base_ring J.is_colorable J.plot
J.braid J.is_idempotent J.powers
J.cartesian_product J.is_knot J.regions
J.category J.is_one J.rename
J.colorings J.is_zero J.reset_name
J.connected_sum J.jones_polynomial J.save
J.determinant J.khovanov_homology J.seifert_circles
J.dowker_notation J.mirror_image J.seifert_matrix
J.dt_code J.n J.signature
J.dump J.number_of_components J.subs
J.dumps J.numerical_approx J.substitute
J.fundamental_group J.omega_signature J.writhe
The documentation for knots/links may be useful here.
|
2025-03-21T14:48:29.675587
| 2020-01-27T10:10:44 |
351233
|
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|
Stack Exchange
|
The multipartition of the dual of a $C_m \wr S_n$ module
I know that the irreducible modules of $C_m \wr S_n$ over $\mathbb{C}$ are parametrised by m-multipartitions. The parts of the multipartition are indexed by the elements of $C_m$.
My question now is: If $V$ is an irreducible module over $\mathbb{C}$ with multipartition $\underline{\lambda} = (\lambda_0, \lambda_1, ..., \lambda_{m-1})$, what is the multipartition of $V^*$?
I feel like it should be a permutation of $\underline{\lambda}$ corresponding to the map $i \mapsto m-i$, but I cannot find that statement anywhere. It should be fairly easy to prove but I have not managed to do so yet.
If the proof turns out to be too complicated, any source would suffice.
Thanks in advance!
This is probably closely related to the Wreath product Schur functions, https://www.math.upenn.edu/~peal/polynomials/schurMisc.htm#schurWreath
I assume that by a multipartition you mean here that $\lambda_i$ is a partition of $k_i$ and $\sum_i k_i = n$. In this case the correspondence you described is indeed the duality correspondence. I believe the easiest way to see this is via Clifford Theory: Consider the short exact sequence $$1\to C_m^n\to G\to S_n\to 1$$ where $G=C_m\wr S_n$.
Let $\zeta$ be an $m$-th root of unity and let $g$ be a generator of $C_m$. Every irreducible representation of $C_m^n$ is of the form $$\rho_{t_1,\ldots,t_n}\big((g^{a_1},\ldots,g^{a_n})\big)=\zeta^{\sum_i t_ia_i}$$
where $t_i\in \{0,\ldots, m-1\}$.
Every irreducible representation is $S_n$-conjugate to a unique representation in which $t_1\leq t_2\ldots \leq t_{n-1}$. Fix such a tuple. For $i\in \{0,\ldots, m-1\}$ write $k_i:= |\{j| t_j=i\}|$. The stabilizer of $\rho_{t_1,\ldots, t_n}$ in $S_n$ will then be isomorphic to $S_{k_0}\times\cdots\times S_{k_{m-1}}$. If $(\lambda_i,\ldots, \lambda_m)$ is a multipartition in the sense that $\lambda_i$ is a partition of $k_i$, then the corresponding tensor product of Specht modules $\mathbb{S}_{\lambda_0}\otimes\cdots\otimes \mathbb{S}_{\lambda_{m-1}}$ is an irreducible representation of $S_{k_0}\times\cdots\times S_{k_{m-1}}$. By letting $C_m^n$ act via the character $\rho_{t_1\ldots,t_n}$ on this vector space you get a representation $V$ of $$H:=C_m^n\ltimes (S_{k_1}\times\cdots\times S_{k_m}).$$ By taking the induced representation $\text{Ind}^G_H V$ you get a representation of $G$. Clifford Theory asserts that this is an irreducible representation of $G$, and that you get all the irreducible representations of $G$ this way.
Now for the duality: it holds that $$(\text{Ind}^G_H V)^* \cong \text{Ind}^G_H (V^*).$$
Since all the irreducible representations of $S_{k_0}\times\cdots\times S_{k_{m-1}}$ are self dual, you are left with the same representation of this group. However, $C_m^n$ acts now via the dual character of $\rho_{t_1,\ldots, t_n}$, which is $\rho_{m-t_1,\ldots, m-t_n}$. When you calculate the numbers $k'_i$ which correspond to this representation you will get now that $k'_i=k_{m-i}$.
So the multipartition which corresponds to the dual representation is indeed $(\lambda_0,\lambda_{m-1},\ldots, \lambda_1)$
|
2025-03-21T14:48:29.675908
| 2020-01-27T10:16:44 |
351234
|
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|
Stack Exchange
|
Ultrafilter on the ordinal $\omega^\omega$
For any ultrafilter $\mathcal{U}$ on $\omega$ and any finite $k$ we can construct tensor power $\mathcal{U}^{\otimes k}$ which is ultrafilter on $\omega^k$. Does there exist some natural extension of this construction for the ordinal $\omega^\omega$?
Edit: my suggestion:
$$
\mathcal{U}^{\otimes\omega}=\{B\subset\omega^\omega~|~\{k<\omega~|~B\cap\omega^k\in\mathcal{U}^{\otimes k}\}\in\mathcal{U}\}
$$
But is it good idea?
Note that it is consistent with ZF that the reals are well ordered (and CH holds) but there is no uniform ultrafilter on the continuum, so it is likely that there is no natural construction.
@Yair: I think that a "natural extension" would be on the ordinal $\omega^\omega$, which may exist as some sort of ultralimit of $\mathcal U^{\otimes k}$ with $\cal U$ being the ultrafilter used for the limit.
Also, to clarify Yair's comment, the fact that the continuum is well-ordered means that there are free ultrafilters on $\omega$, so there is essence to this question. I'd go on to add that the above fact means also that there is no definable way of having such extension (so in particular no real natural extension).
By curiosity, is it consistent with ZF that there exists a non-principal ultrafilter on $\omega$ while every ultrafilter on $\omega^\omega$ has countable support?
@YCor: Yes, of course, the model Yair mentioned satisfies just that. Since there are free ultrafilters on $\omega$ (as the continuum is well-orderable), these extend to free ultrafilters on any set with a countable subset, e.g. the continuum. But of course, if there are no uniform ultrafilters, so every free ultrafilter contains a small subset. Now if you add CH (which holds in the model mentioned by Yair), then easily enough such small set must be countable.
The edit suggests that the $\omega^\omega$ in the question is, in fact, ordinal exponentiation, not cardinal exponentiation.
@EmilJeΕΓ‘beksupportsMonica: yes, of course, ordinal exponentiation
You write $
\mathcal{U}^{\otimes\omega}={B\subset\omega^\omega~|~{k<\omega~|~B\cap\omega^k\subset\mathcal{U}^{\otimes k}}\subset\mathcal{U}}
$ --> shouldn't we replace $\subset$ by $\in$ in the last two instances, so $
\mathcal{U}^{\otimes\omega}={B\subset\omega^\omega~|~{k<\omega~|~B\cap\omega^k\in\mathcal{U}^{\otimes k}}\in\mathcal{U}}
$ ?
One little side question. For $n\in\omega$ let $\text{eval}_n:\omega^\omega\to \omega$ be given by $f\mapsto f(n)$. Is ${\mathcal U}^{\oplus\omega}_1:=\big{B\subseteq \omega^\omega: {n\in\omega: \text{im}(\text{eval}_n|_B )\in {\cal U}}\in{\cal U}\big}$ an ultrafilter, and if yes, equal to what the OP is suggesting?
@DominicvanderZypen: Edited. Thanks.
@DominicvanderZypen: your suggested family hasn't finite intersections property imho. Let us consider $B={f~|~\forall i:~f(i)\neq 1}\subset\omega^\omega$ and $B'=\omega^\omega\setminus B$. Both $B$ and $B'$ are in $\mathcal{U}_1^{\otimes \omega}$
@ar.grig Right - thanks for your example!
The relevant general construction is the sum of a family $\{\mathcal V_i:i\in I\}$ of an indexed family of ultrafilters, with respect to an ultrafilter $\mathcal U$ on the index set $I$. If $\mathcal V_i$ is an ultrafilter on $X_i$, then the sum is the ultrafilter $\mathcal W$ on the disjoint union $\bigsqcup_{i\in I}X_i$ defined by
$$
\mathcal W=\{A:\{i\in I:A\cap X_i\in\mathcal V_i\}\in\mathcal U\}.
$$
In your situation, taking $\mathcal V_i$ to be $\mathcal U^{\otimes i}$, you get a sum ultrafilter on $\bigsqcup_{i\in\omega}\omega^i$, which can be identified with the ordinal $\omega^\omega$ to produce the ultrafilter $\mathcal U^{\otimes\omega}$ in the question.
Thank you for your answer. What about construction for $(\omega^\lambda,~\mathcal{U}^{\otimes\lambda})$, where $\lambda<\omega_1$ is an arbitrary limit ordinal ?
Such a $\mathcal U^{\otimes\lambda}$ would depend on how you rearrange the relevant index set $\lambda$ in an $\omega$-sequence, in order to use the ultrafilter $\mathcal U$ as an ultrafilter on $\lambda$.
By the way, if we replace disjoint union with union in your definition of the sum, we also get ultrafilter. Particularly, in case $X_i=X$ we get ultrafilter on $X$.
@ar.grig Yes. But in the special case where all $X_i=X$ I would all this ultrafilter the limit (not the sum) of the $\mathcal V_i$ along $\mathcal U$. (It is, in fact, the limit in the usual topology of $\beta X$.)
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|
2025-03-21T14:48:29.676210
| 2020-01-27T10:44:05 |
351235
|
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"authors": [
"Chris Wuthrich",
"GH from MO",
"Geoff Robinson",
"Luis Ferroni",
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}
|
Stack Exchange
|
find all of rational solutions of the quartic equation?
Consider the equation $a^4+6v^2a^2-8a+v^4=0$ over the rationals.
Note that the following are solutions: $(a,v)=(1,1),(0,0),(2,0)$. Are there any other rational solutions?
Not sure if it helps, but this can be rewritten as $(a^{2} + 3v^{2})^{2} = 8(v^{4}+a).$
Thank you very much for your comments.My general question is how we can determine the coefficients of a quartic equation(or higher degree eqtation) ax^4+bx^+cx^3+dx^2+e=0 such that it has rational solutions. A trivial method is first determine the roots (x_i), then the coefficients!!!... I mean as example how I determine the coefficients a,b in the quartic equation 3x^4+2^x^3+ax^2+x+ b=0 such that it has the rational roots?
@user151553: There is no general algorithm to decide if a given quartic equation has a rational solution. This follows from Matiyasevich's theorem, because every system of polynomial equations can be encoded into a single quartic equation (using sums of squares of quadratic equations, including simple variable changes like $x=yz$).
For $y=v^2+3a^2$, the curve $y^2= 8a^4+8a$ is an elliptic curve; so it should not be hard to find its rank. If $0$, the answer is easy, otherwise one needs harder method for genus 3 curves.
What is the motivation for the question?
@ChrisWuthrich: That curve has rank 1. See my comment below Luis Ferroni's answer.
thank you very much for comments. if the above elliptic curve has positive rank, now how i can search the solutions that for them we have v^2=y-3a^2(with v rational number)?
by letting $2/a=X$, $y/a^2=Y$, we get $Y^2=X^3+8$ with the rank=1. now note that $v^2=y-3a^2=4(Y-3)/X^2$, then $ Y-3(=t^2)$ must be square number.Thus by letting $Y=t^2+3$ in the equaion $Y^2=X^3+8$, the solution reduces to finding the rational solution in the equation $X^3=t^4+6t^2+1$? note that the solution $(t,X)=(0,1),(1,2),(-1,2)$ are trivial.... Now the question is Dose the equation any other rational solution????
This is not a complete solution but rather a line of action. I think this method may work completely, but it is too tedious to make step by step.
Note that you can think of your equation as a quadratic with variable $v^2$. Said that, for $v^2$ to be rational, the discriminant of the quadratic has to be the square of a rational number, this is:
$$36a^4 - 4(a^4-8a)=\left(\frac{m}{n}\right)^2$$
$$2a(a^3+1) = \left(\frac{m}{4n}\right)^2$$
Writing $a=\frac{p}{q}$ with $\gcd(p,q)=1$, and replacing above:
$$2p(p^3+q^3) = \left(\frac{q^2m}{4n}\right)^2$$
Which in turn, since the LHS is an integer number, implies that $4n$ divides $q^2m$. Notice that:
$$d:=\gcd(p,p^3+q^3) = \gcd(p,q^3) = 1$$
In particular $d:=\gcd(2p,p^3+q^3)\in \{1,2\}$.
If $d=1$, then one has that $2p$ and $p^3+q^3$ have both to be
squares of integers. Furthermore, notice that $p^3+q^3=(p+q)(p^2-pq+q^2)$,
and also
$$d':=\gcd(p+q,p^2-pq+q^2) \in \{1,3\}$$
In any case, one has that $\frac{p+q}{d'}$ and $\frac{p^2-pq+q^2}{d'}$ have to be squares of integers. A bit of manipulation converts the equation $p^2-pq+q^2=u^2d'$, into the Pell equation with variables $x,y\in\mathbb{Q}$:
$$ x^2 + 3y^2 = d'$$
where $x=\frac{2p-q}{2u}$ and $y=\frac{q}{2u}$. If $d'=1$ it can be solved by standard methods. If $d'=3$ it is very similar.
Once you've got the solutions of these Pell equations, you can parametrize all possible values of $p$ and $q$. Plugging them into our conditions that $\frac{p+q}{d'}$ and $2p$ had to be squares of integers, we will probably get a complete parametrization of the feasible couples $(p,q)$. This gives you all possible values of $a$ for this case.
From them you get all possible rational values of $v^2$.
Its similar to Case 1, but with $d=2$. The steps should be similar (probably this one will require more care).
SAGE tells that your second display defines an elliptic curve, which is isomorphic to the one defined by $y^2=x^3+8$. SAGE also tells that this elliptic curve has infinitely many rational points. Relevant SAGE commands: "a, b = polygen(QQ, 'a, b')", "E = Jacobian(b^2-2a^4-2a)", "print E", "print E.mwrank()".
Thank you for the above comments. But I do not understsnd it completely..Does it yield a solution to the problem?
Well I only guided you through a path that can be useful (at least to try to find other non-trivial solutions). It's up to you to fill the details and see if it works.
|
2025-03-21T14:48:29.676502
| 2020-01-27T11:49:50 |
351238
|
{
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}
|
Stack Exchange
|
What is the largest possible probability that a random matrix over $\mathbb{F}_2$ is non-singular?
Suppose $A(p, n)=(a_{ij}(p))_{i, j \leq n}$ is an $n\times n$ random matrix over $\mathbb{F_2}$, with all its entries being i.i.d. and such that $P(a_{ij}(p) = 1) = p$, where $p$ is some real number from $[0; 1]$. What is the largest possible probability, that $A(p, n)$ is non-singular and with what $p$ is it reached?
Note, that $A(p, n)$ is non-singular iff $\det(A(p, n)) = 1$.
Solution for $n=1$:
$\det(A(p, 1)) = 1$ with probability $p$. The maximum of $\det(A(p, 1))$ is $1$ and it is reached with $p = 1$.
Solution for $n = 2$:
$\det(A(p, 2)) = 1$ with probability $2p^2(1 - p^2)$. The maximum of $P(\det(A(p, 2))=1)$ is $\frac{1}{2}$ and it is reached with $p = \frac{1}{\sqrt{2}}$.
However, I would like to know some sort of general formula (or at least asymptotics).
After I failed to solve this problem using determinants, I tried to prove this using the fact that a square matrix is non-singular iff its rows are linearly dependent. As there exists only one non-zero element in $\mathbb{F_2}$, we can write linear dependence of the vector system $\{v_i\}_{i \leq n}$ in $\mathbb{F}_2^n$ as $\forall S \subset \{1, ... , n\}$ such that $S \neq \emptyset$ we have $\sum_{i \in S} v_i \neq \overline{0}$. I know the probability that a given set of vectors with i.i.d. random entries Bernoulli distributed with parameter $p$ $\{v_i\}_{i \leq k}$ over $\mathbb{F}_2^n$ satisfy $\sum_{i = 1}^k v_i \neq \overline{0}$ is $(1 - \frac{p((1 - 2p)^k - 1)}{1 - 2p})$. However, I do not know how to proceed further in this direction.
This question on MSE
For $n\rightarrow\infty$ the probability ${\cal P}_\infty$ that $A(p,n)$ is nonsingular becomes independent of $p\in(0,1)$, given by
$${\cal P}_\infty=\prod_{i=1}^\infty(1-2^{-i})=0.2887880951$$
See theorem 3.2 in Properties of random matrices and applications (2007).
For finite $n$, there is a result that could be instructive, which is the expectation value $E(p,n)$ of the number of linear dependencies among the rows of $A(p,n)$. This is given by theorem 4.5,
$$E(p,n)=2^{-n}\sum _{j=1}^n \binom{n}{j} \left(1+(1-2 p)^j\right)^n.$$
A plot of $E(p,n)$ as a function of $p$ becomes flatter and flatter with increasing $n$ (see below for $n$ up to 20), consistent with the understanding that for large $n$ the probability that the matrix is singular no longer depends on $p$.
|
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