User alex gavrilov - MathOverflow

Answer by Alex Gavrilov for Sufficient conditions for inverse-positivity

2013-01-11T13:18:44Z

Sorry for promoting my own results, but I think the condition in my old paper "A sufficient condition for the monotonicity of a positive definite matrix" (Computational Mathematics and Mathematical Physics vol. 41, No 9., pp. 1237-1238) may be of help. (Unfortunately, I lost the file years ago). I don't know whether it actually works, for you didn't say much about your $Q$, but it looks promising.

Answer by Alex Gavrilov for Poincare line bundle

2013-01-02T13:24:16Z

[Not exactly an answer. Still, it may be helpful, I hope.]

I think there are two obscure points in this proof: apart from the homology question, why the map is holomorphic? (Is it really obvious?). Of course, all this can be settled. However, I prefer a more explicit construction of the Poincare bundle, which can be found, e.g., in Hindry&Silverman, "Diophantine Geometry", Exercise A.5.6. (I like it, but even if you won't, it is definitely worth reading).

Answer by Alex Gavrilov for Is there any techniques for solving a differential equation including iterated function? f ' (x) = f( f( x ) )

2012-10-30T12:37:45Z

For what I know, the standard method is the Taylor series expansion at a fixed point, i.e. at a point $x=a$ such that $f(a)=a$.

How to prove a $\Pi_2$ statement properly?

2012-10-30T11:41:23Z

Consider the following situation. In a parallel world (let's hope not in this one), in 2020 a clever guy proved $P\neq NP$ in a theory $T_1=\{ZFC+some\, reasonable\, new\, axioms\}$ . Then, in 2021 a clever gal proved $P=NP$ in $T_2=\{ZFC+other\, reasonable\, axioms\}$ . We assume that the both theories are consistent and the both proofs are correct. Still, the situation is theoretically possible, because $P\neq NP$ is a $\Pi_2$ statement.

Question: How to clear the mess?

Actually, this question is not exactly about $P\neq NP$. Rather, it is about $\Pi_2$ statements in general. If such a statement is false then it contradicts some true $\Pi_1$ statement; all we have to do is to find and prove the latter. (Which may be not so easy, but this is a different problem). The question is, if the statement is true, how do we know this, even if we have a correct proof in a consistent theory? In my opinion, this question is not purely academic: after all, a lot of
mathematical problems belong to this class, and we cannot possibly know what new axioms (or whole theories) will be proposed in the future.

What software one needs to solve a big linear system on a small computer?

2012-10-19T13:07:34Z

A time ago I was intrigued by the following remark: "... one ends up with a (non-sparse) system of equations in about 10000 real variables. One important practical point is that solving such systems is now well within the capabilities of a standard desctop PC;...". The above is a quote from Notices of AMS 55 no. 9 (October 2008), page 1093 (left). Occasionally, I have to solve a system or two which are not the size of an undergraduate excercise in linear algebra. In my experience, Maple become pretty useless for the purpose as soon as the number of variables approaches 1000.

QUESTION: What kind of software I need to solve such a big system on a "standard desctop PC"?

P.S. It's a pity that in the quoted article ("Uncovering a New L-function" by A.R.Booker) there is no single word about this.

The Hodge numbers of a covering

2012-10-16T13:07:44Z

Let $X$ be a Kahler manifold and $Z\subset X$ be a smooth hypersurface. How to compute the Hodge diamond of the double covering $Y\to X$ ramified over $Z$? (And what I have to know? Would the map $H^*(X)\to H^*(Z)$ be enough?)

P.S. I tried the Gysin sequence, but it looks like there are many loose ends.

Usual vs. cohomological Brauer groups of Calabi-Yau threefolds

2012-10-05T13:06:41Z

In the preprint arXiv:math/0505432v1 by Batyrev and Kreuser I have found (on pages 2 and 10) the claim that "by a recent result of Kresch and Vistoli [arXiv:math/0301249]" the (usual) Brauer group of a Calabi-Yau threefold is isomorphic to its cohomological Brauer group. However, in that preprint of Kresch and Vistoli I have not found a word about Calabi-Yau or anything. (Admittedly, it is about Brauer groups). Could anyone help me to clear the mess?

P.S. For what I know, if there is always an isomorphism is a big open problem which is only settled in a few special cases. So, I suppose, if this is known indeed for Calabi-Yau threefolds (for 10 years by now), then every expert must be aware of it.

The Gysin map for a singular hypersurface

2012-08-09T10:05:26Z

Let $X$ be a projective complex manifold and $Y\subset X$ be an irreducible hypersurface. If $Y$ is smooth, there is a well known Gysin sequence. However, even if $Y$ is not smooth, a kind of Gysin map can still be devised.

Consider a desingularization $f:\widetilde{Y}\to Y$ and the inclusion $i:Y\hookrightarrow X$. We have two maps

$H^i(Y,\mathbb{Q})\to H^i(\widetilde{Y},\mathbb{Q})\to H^{i+2}(X,\mathbb{Q});$

the first one is $f^*$ and the second one is the Poincare dual to $(i\circ f)^*$ (essentially the Gysin map). I am convinced that the composition does not depend on desingularization, though I do not know a rigorous proof of this.

QUESTION: Is the sequence

$H^i(Y,\mathbb{Q})\to H^{i+2}(X,\mathbb{Q})\to H^{i+2}(X\setminus Y,\mathbb{Q})$

exact (as it is in the smooth case)?

All I know about it is a result of Deligne [Theorie de Hodge III. Publ. Math. IHES 44 (1974) pp. 5–77.; Corollary 8.2.8] that

$H^i(\widetilde{Y},\mathbb{Q})\to H^{i+2}(X,\mathbb{Q})\to H^{i+2}(X\setminus Y,\mathbb{Q})$

is exact, but this is much weaker than what I need.

Are all the theorems true?

2012-08-01T17:49:30Z

The title sounds a bit philosophical, but it is
about mathematics. Let me explain.

Consider a first order theory $T$, which is an extension of Peano Arithmetic. Call this theory "good" if it is consistent and satisfies the following

Property: For any $\phi\in\Sigma^0_{n+1}$ such that $T\vdash \phi$ there exists $\psi\in\Pi^0_{n}$ such that $T\vdash\psi$ and $PA\vdash\psi\to\phi$ .

Question 1. Is $ZFC$ "good"?

Question 2. The same for $ZFC+something$ (from the lot of proposed new axioms).

Motivation.

If $ZFC$ is not "good" then there (EDIT) may be theorems which can be proved in $ZFC$ despite they are false (in the standard model of PA). I believe that $ZFC$ is "good". However, I would like to know if there is a formal proof. (Admittedly, I don't have a slightest idea what a proof may be like). By the way, "goodness" implies consistency, hence a proper proof requires some new axioms (a large cardinal, perhaps).

(EDIT). As Andreas Blass pointed out correctly, even if a theory is not "good" in the above sense, it does not yet follow that some of the theorems are wrong (an obvious fact which I have missed somehow). Still, the question if ZFC is "good" may be of some interest, in my opinion.

Question 3. Is "goodness" equivalent to consistency? (I doubt this).

EDIT: (Clarification). In this question, the theory $T$ is supposed to be "good" and at least as strong as $ZFC$. (Thus, the answer to Question 1 must be yes). The question is, whether $T\vdash Con(T)\to Good(T)$ , where $Good(T)$ is a formalization of "goodness"; note that $Good(T)\in \Pi^0_{2}$ .

P.S. Is there a standard term for "good"?

The differential of the exponential map: reductive homogeneous space

2012-07-20T10:55:58Z

The differential of the exponential map on a symmetric space can be expanded (abusing some notation) as

$d{\rm Exp}_X=\sum_{n=0}^{\infty}\frac{({\rm ad}X)^{2n}}{(2n+1)!}.$

This is an old (1958) result of Helgason.

Question (EDITED): Is there any generalization to reductive homogeneous spaces?

Reference Request: de Rham vs. Dolbeault

2012-07-24T06:49:49Z

Hi everyone. I need the following statement:

For a Kahler manifold $X$, the natural map $H^n(X,\mathbb{C})\to H^n(X,\mathcal{O})$ (from the sheaf extension) coincides with the Hodge projection $\Pr_{0,n}$, up to the de Rham isomorphism and the Dolbeault isomorphism.

Does anybody know a good reference?

P.S. Surely there must be a reference. I am much less interested in proofs: I think I know one.

Does the absolute Galois group act transitively on the trees with 3 terminal vertices?

2012-06-04T17:08:23Z

Hi everyone. My question is about the absolute Galois group action on the set of the Grothendieck dessins. The dessins I am interested in are trees with only one vertex of valency more then 2. (I don't know if there is a generally accepted term in graph theory. Starlike trees?). What exactly is known about them? Is the action transitive, at least for trees with 3 terminal vertices (with the same valency lists)?

EDIT: I see some clarification is necessary. In fact, I consider the alternating dessins, so there are two valency lists for each of them. (It is convenient to assume that the "center" is always, say, black). As Will Savin pointed out, a dessin with rotational symmetry cannot turn into one with less symmetry. I confess I missed this obvious fact, but certainly this is not what I had in mind. The question was actually about nontrivial obstacles to transitivity. (I know there are some for general trees).

Then, "what is known" part of the question. The best answer would be a long list of references.

Answer by Alex Gavrilov for Stability of Dirichlet data for Helmholtz equation

2012-06-05T16:28:21Z

The solutions of the Helmholtz equation are somewhat difficult to estimate. In particular, the constants depend heavily on the geometry of the domain. To see what I mean let us imagine for a moment the set $D$ to be with an interior void. The boundary value problem inside the void would be ill-posed for those $k$ which are square roots of the (positive) eigenvalues of the Laplacian.

Now, if there is a hole connecting the void with the exterior, the problem become well-posed (with appropriate boundary conditions, including the infinity). However, if the hole is small enough, the constant $C$ can be made as big as you wish. (The things are better if $D$ is convex. Unfortunately, I do not remember any appropriate reference now).

References for some analogs of the Picard group.

2011-05-11T08:28:16Z

Let $X$ be a compact complex manifold. By definition, $Pic(X)={\rm H^1}(X,\mathcal{O}^\times)$. We know a lot about this group. What is known about the groups ${\rm H^n}(X,\mathcal{O}^\times)$ for $n\ge 2$?

A bit more specialized question. It is well known that for a nonsingular projective complex variety $X$ the natural map $${\rm H^1}(X,\mathcal{O}^\times)\to{\rm H^1}(X,\mathcal{M}^\times)$$ is trivial. What is known about the kernel of the same map for $n=2$ or $n=3$? (Here $\mathcal{M}^\times$ is the sheaf of nonzero meromorphic functions, and the topology is the strong one).

Answer by Alex Gavrilov for Is complex analytic extension of real-analytic diffeomorphism a diffeomorphism ?

2011-07-05T04:52:06Z

Indeed, a real-analytic map can always be extended to some complex neighbourhood. The problem is, the neighbourhood may be very small. Consider, for example, the map $$f:I\to I,\, I=[-1,1],$$ defined by $$f:x\mapsto x+\frac{a^3x(x-1)^2}{x^2+a}.$$ For small $a>0$ this is a diffeomorphism of $I$, but it cannot be extended very much due to the poles near $x=0$. The similar map (though a bit more contrived) can be designed for a disc. So, the answer to the question is no.

Of course, all this is well known but what is a proper reference I cannot say.

P.S. This is an answer to the question as I understand it. There are some points I do not understand. $\Omega$, as it is defined, is a sphere. And, I hope, the restriction is not defined by $F(z,0)=(f(z),0)$: if it is, you can always take $F(z,w)=(f(z),w)$.

Answer by Alex Gavrilov for Does the Bergman kernel always arise as the Jacobian of a biholomorphism?

2011-05-22T08:17:31Z

Consider the case $n=1$. In this case the question actually is:

"For a fixed $w\in D$, is the antiderivative of $K(z,w)$ a well defined univalent function on $D$?"

For a simply connected domain the answer is yes, because there exists a biholomorphic map to a disk. But for a general domain the answer is no. To see this, let $D$ be an annulus. It is not difficult to see that the antiderivative of $K(z,w)$ is not even well defined, because
$\int K(z,w)dz\neq 0$ for any path which is not homotopic to zero. (This follows directly from the Laurent series of the kernel).

I am convinced that for $n\gt 1$ your conjecture is not true even for a domain homeomorphic to a ball, but I do not know how to construct a counterexample.

Are undecidable consequences of Con recursively enumerable?

2011-05-02T10:05:26Z

Let $X\subset\Pi_1^0$ be the set of statements which are provable in PA$+$Con(PA) but independent of PA. Is $X$ recursively enumerable?

Answer by Alex Gavrilov for Complex structure on $L^2(\mathbb R)$ generalizing the Hilbert transform.

2011-04-30T06:26:45Z

EDIT: This solution does not satisfy the third condition, which rules out the Hilbert transform itself. So, this is an answer to different question. I do not delete it in hope it may be useful for someone.

Let $\phi(x)$ be a smooth monotone function such that $x-\phi(x)$ has compact support. This is a diffeomorphism of the real line and the pullback $\phi^*$ is a linear operator acting on $L^2(\mathbb R)$. The singular integral operator $$(\phi^*)^{-1}{\mathcal H}\phi^*$$ has all the properties you need.

Answer by Alex Gavrilov for The disjunction property in Peano Arithmetic?

2011-04-28T06:21:59Z

This is not actually an answer but rather a comment to Joel's answer. I am not very good in models, so here is an idea how to do without them. There is a theorem of Kreisel: if a $\Pi_1^0$ statement is provable in $T+\neg Con(T)$, then it is provable in $T$. In $PA+\neg Con(PA)$ we may prove that there exists the smallest code of a proof of $\phi$ and the smallest code of a proof of $\psi$. Denote them by $n_{\phi},n_{\psi}$. Then $\phi$ asserts that $n_{\psi}<n_{\phi}$ while $\psi$ asserts that $n_{\phi}<n_{\psi}$ . Then $\phi\vee\psi$ means $n_{\phi}\neq n_{\psi}$ which is provalbe if $\phi$ and $\psi$ are syntactically different. By Kreisel, $\phi\vee\psi$ is provalbe in $PA$. (Note that the numbers $n_{\phi},n_{\psi}$ do not really exist).

The disjunction property in Peano Arithmetic?

2011-04-27T12:49:38Z

Let $\phi,\psi\in\Pi_1^0$ be independent of PA. Is the disjunction $\phi\vee\psi$ independent of PA?

Is there a generalization of Brouwer's fixed point theorem?

2011-03-31T08:51:00Z

In essence, this is the same problem as in “The generalization of Brouwer's fixed point theorem?”. But now I am determined to be careful. The main question is the following:

Is there any generalization of Brouwer's fixed point theorem in terms of general topology? (That is, without triangulations, or vector spaces, or anything else).

The question is a bit vague, but, I hope, it admits a precise answer. Now, I am trying my best to propose a candidate.

Let $X$ be a contractible locally contractible Hausdorff second countable compact topological space. Let $f\colon X\to X$ be a continuous map. Has then $f$ a fixed point?

The generalization of Brouwer's fixed point theorem?

2011-03-28T04:05:56Z

Let $X$ be a contractible compact [edit: locally connected] topological space (Hausdorff and second countable). Let $f\colon X\to X$ be a continuous map. Then (I suppose) $f$ has a fixed point. Personally, I cannot think of a better generalization of Brouwer's fixed point theorem, but is it true?

Comment by Alex Gavrilov

2013-01-22T11:11:33Z

Thanks. Good luck.

Comment by Alex Gavrilov

2013-01-11T15:16:53Z

You may try this: maik.rssi.ru/cgi-perl/…

Comment by Alex Gavrilov

2013-01-11T15:14:01Z

This journal is translated, so there must be an English copy somewhere.

Comment by Alex Gavrilov

2012-11-06T10:42:01Z

Zeta-function is not entire: it has a pole.

Comment by Alex Gavrilov

2012-10-20T14:28:31Z

Yes, you are perfectly right. There is no canonical map; I missed it.

Comment by Alex Gavrilov

2012-10-19T15:00:24Z

Thank you. I don't have MATLAB at hand; probably I will try Octave for the first.

Comment by Alex Gavrilov

2012-10-19T12:57:44Z

Let me explain what I mean. There exists a unique map $H_0(Z)\to \mathbb{Z}\slash 2$ which is nontrivial on all the connected components of $Z$. There exists a (unique) map $H_1(X\setminus Z)\to H_0(Z)$. Thus, we have the map $\pi_1(X\setminus Z)\to \mathbb{Z}\slash 2$. It determines an (unramified) covering of $X\setminus Z$ which can be compactified to get a covering of $X$. (Certainly, there may be many other coverings as well). I may be mistaken about the uniqueness of this construction but your example did not convince me.

Comment by Alex Gavrilov

2012-10-18T13:41:25Z

Thank you, especially for the reference.

Comment by Alex Gavrilov

2012-10-18T13:39:58Z

Thank you much. Apparently, it is what I need.

Comment by Alex Gavrilov

2012-10-18T13:36:44Z

You misunderstood the question a bit. I consider a specific covering (the covering) which comes from $H_1(X\setminus Z)\to H_0(Z)$.

Comment by Alex Gavrilov

2012-10-16T14:50:42Z

@yuwei: Oops! I fixed it.

Comment by Alex Gavrilov

2012-10-06T14:22:51Z

Thank you for the reference

Comment by Alex Gavrilov

2012-08-10T10:13:16Z

Perhaps, but I hope that the statement may be true as it is. I do not need the Whole Gysin sequence after all.

Comment by Alex Gavrilov

2012-08-10T02:51:33Z

Thank you. Though I am not sure if it may be helpful.

Comment by Alex Gavrilov

2012-08-02T16:00:33Z

I got it. It's Theorem 4, is it not? Thank you much.