User nikita kalinin - MathOverflow

Diameter-area ratio for affine tranformations.

2013-05-13T18:02:22Z

Consider an convex plane figure $F$. How to prove that there is an affine transformation $a$ such that $\sqrt{3}$ diameter$(a(F))^2\leq 4$ area$(a(F))$?

I found only one reference, to "Über einige Affininvarianten konvexer Bereiche", but unfortunately it is in German.

Added: formula (12) there looks like desirable. After I found a solution myself, I can understand German. The proof there in the pages 734 (corresponding to considering $D'$ below) and 735 (considering $D''$). The author estimated $f/d_u^2$, $f$ is an area (Flacheninhalt) and $d_u$ is a diameter(Durchmesser).

So, emergency over, thank you))

Answer by Nikita Kalinin for Diameter-area ratio for affine tranformations.

2013-05-14T15:10:05Z

It seems that I found a proof. Consider a figure $A$. Consider a figure $F$ with minimal ratio $S/d^2$ among all affine transforms of $A$.

Lemma. There are two diameters of $F$ with angle at least $\pi/3$ between them. Note that Lemma implies the estimation because $\sin(\pi/3) =\sqrt 3/2$ and area of $F$ is at least $d^2\sin(\pi/3)/2$.

Proof.

Consider a diameter $D$. Try to squeeze $F$ in the direction of $D$ and stretch out in the perpendicular direction. It is not possible, therefore there is an other diameter $D'$ with angle at least $\pi/4$ and less than $\pi/3$ with $D$. Well, among all pairs of diameters chose the pair $D,D'$ with the biggest shapr angle between them.

Now try to perform an affine shift parallel to $D$ in the direction decreasing $D'$.

It is not possible, therefore there is a diameter $D''$ which is a kind in a "symmetric" position with $D'$. So, now either the angle between $D$ and $D'$ is big enough, or the angle between $D'$ and $D''$.

Asymptotics vs Puiseux series

2013-05-01T07:36:34Z

Define asymptotic as a class of sequences {$ x_i$},$_{i\in\mathbb N}$ modulo equivalence {$x_i$}={$y_i$} if $\lim_{i\to\infty} (x_i/y_i)=c\in\mathbb R,c\ne 0$.

More, we define $X= \{x_i\} \lt Y= \{ y_i \}$ if $\lim_{i\to\infty} (x_i/y_i)=0$.

We can interpret $x_i$ as $f(1/i)$ and say that these asymptotics are precisely aymptotics of functions at 0.

There is a classic object - Puiseux series ${\sum\limits_{\alpha\in I}a_\alpha t^{\alpha}}$ (where $I$ is well-ordered set).

Asymptotics and Puiseux series look quite similar, isn't it?

My question is a bit vague.

Suppose one has a finite set of Puiseux series $X_1\leq X_2\leq\dots X_n$ , linear dependencies between them(so, one can build matroids etc). One proves something about this set - more or less using only valuation map $val(\sum\limits_{\alpha\in I}) = -\min\limits_{\alpha\in I}\alpha$, which is just "order" for an asymptotics.

How to argue that all these arguments which work for Puiseux series work for asymptotics as well?

In fact one can rewrite all proofs but the problem is that it is not clear what is "order" of an asymptotic (and order of a Puiseux series is just a real number).

Area of a lattice polygon in terms of its width

2013-03-27T09:35:30Z

Let $M$ be a lattice polygon on a plane (i.e. its vertices are integer points $(i,j)\in\mathbb Z^2$).

Let us define lattice width in a direction $v=(m,n)\in\mathbb Z^2$ as $w_v(M)=\max\limits_{x,y\in M} v\cdot(x-y)$.

Suppose the $minimal$ lattice width of $M$ equals $d$. It is clear that the area of $M$ should be proportional to $d^2$.

One can prove an inequalities $area\geq d^2/4$ as is written in comments. But it seems far from the best one.

So, the question is what is the best $\alpha$ such that $area(M)\geq \alpha d^2$ for each $M$ with minimal lattice width equals $d$. From my comment below one can extract that $\alpha\leq 3/8$

Answer by Nikita Kalinin for Area of a lattice polygon in terms of its width

2013-03-29T17:26:30Z

This solution is not true, sorry. ~~I can prove that $area(M)\geq 3d^2/8$. First of all find $v\in \mathbb Z^2$ such that $w_v(M)$ is minimal. Than by an affine transform $a$ put $v$ into vector $(1,1)$.~~

Now draw lines which give us widths of $a(M)$ in the directions $(1,0),(0,1)$. Now $a(M)$ is inside a horizontal stripe of width at least $d$, inside a vertical stripe of width at least $d$ and inside a diagonal (in the direction $(1,-1)$) stripe of width exactly $d/{\sqrt{2}}$.

Here I use usual widths.

Now look at the picture - we see a 6-gon (or 3,4-gon) $B$ as the intersection of the strips, we know that on each side of $B$ there is a vertex of $a(M)$. By playing with vertices one can find all the extremal cases (so, vertices of $a(M)$ must coincide with vertices of $B$).

~~Finally, in all the extremal cases the area of $a(M)$ is no less than $3d^2/8$~~

Answer by Nikita Kalinin for Family of hypersurfaces in (C^*)^2 corresponding to tropical family

2013-03-27T17:32:13Z

If you ask is it true that a tropical nodal curve can be presented such a way as a limit of nodal curves then the answer is no.

What is true:

Curves in a family which tropicalises to tropical curve of genus g has genus at least g.

So, tropical nonsingular curve may be presented as a limit of nonsingular curves and a tropical limit of singular curves is singular too.

Sum of two tangent bundles of $S^{2n}$

2013-03-26T21:31:24Z

I was wondering if the sum $TS^{2n}\oplus TS^{2n}$ is a trivial bundle? The same is true for spheres of odd dimension (one can find a nowhere zero section of the second bundle, add it to the first, the first becomes trivial and the rest of second bundle plus trivial bundle of rk 2 is trivial too).

It seems that one should take $2n$ sections $v=(v_1,\dots,v_{2n})$ of $TS^{2n}$ (for example projections of coordinate vector fields from $\mathbb R^{2n+1}$ to $S^{2n}$), $u=(u_1,\dots,u_{2n})$ for the second part and than perturb a little u=u+av, v=v+bu.

Nevertheless I can not prove that it works. From the other point of view I see no reasons for this bundle to be non-trivial.

Answer by Nikita Kalinin for Video lectures of mathematics courses available online for free

2013-03-08T12:25:25Z

There are many good quality math lectures (mostly in Russian but sometimes in English) http://www.lektorium.tv/ they are groupped by courses (for example http://www.lektorium.tv/course/?id=22876)

Thom polynomial for contact algebraic structures

2011-10-18T16:10:40Z

Let's consider a algebraic contact structure $P$ on $\mathbb CP^3$ and a algebraic curve $C$ degree $d$ and genus $g$. Let's assume that contact structure has degree $p$ (see http://mathoverflow.net/questions/58000/polynomial-contact-structures-on-rp3 about algebraic contact structure).

It seems that there is some constant $f(d,g,p)$ (maybe, even polynomial!) such that if $C$ is tangent to $P$ at $f$ points then $C$ is tangent to $P$ everywhere.

For example, it is easy to prove that $f(d,0,0)$ equals $2d-1$.

Somebody can expect that this question is about some homological conditions : generic curve is tangent to $P$ at $a$ points, so, if it is tangent to $P$ at $a+1$ points then it is tangent everywhere. It seems that it is true because pull-back of contact form to tangent bundle of $C$ is a holomorphic form, so it has some degree..

So, my questions are:

1) How can we prove that $f(d,g,p)$ exists? At least for some values $d,g,p$? It seems that the case $p=0$ is the most easy.

2) Is it true that $f$ is a polynomial? (Thom polynomial of something...) I'm sure that it is known or similar constructions are already examined.

Answer by Nikita Kalinin for Thom polynomial for contact algebraic structures

2012-11-14T18:06:32Z

there is an article of Quo-Shin Chi "The dimension of the moduli space of superminimal surfaces of a fixed degree and conformal structure in the 4-sphere" where the dimension of contact curves moduli space is computed.

It lays between $2d-4g+4$ and $2d-g+4$ for a fixed complex structure on a curve

Polynomial contact structures on $RP^3$

2011-03-09T21:43:57Z

Let us consider polynomial contact structures on $\mathbb RP^3$, i.e. contact structures on $\mathbb R^3$ defined by a form $w=Pdx+Qdy+Rdz,\ P,Q,R\in \mathbb R[x,y,z]\ $ in an affine part and then extended to $\mathbb RP^3$, and $ w \wedge dw \ne 0$ everywhere.

One can find all such forms $w$ that $deg P, deg Q, deg R \leq 1$ by direct calculation:

$w=(qy-rz+a)dx+ (pz-qx+b)dy + (rx-py+c)dz,\ a,b,c,p,q,r\in \mathbb R;$ $ap+br+cq \ne 0$.

But I can't do anything for greater degrees. Do you know any criteria for coefficients of $P,Q,R$?

Does anybody know any contact polynomial form with $deg P, deg Q, deg R \geq 2$?

Added: What is the form (I mean form coefficients in $\mathbb R^3\subset \mathbb RP^3$) defines polynomial contact structure constructed by plurisubharmonic function $f=x^4+y^4+z^4+t^4$?

Answer: $f=x^4+y^4+z^4+t^4$ is not strictly plurisubharmonic (see on the plane $x=y=0$ on subspace generated by $dx,dy$). So it does not produce a contact structure.

Answer by Nikita Kalinin for Polynomial contact structures on $RP^3$

2012-11-14T18:03:35Z

There is a very good article "Complex contact threefolds and their contact curves" of Yun-Gang Ye where on can find a classification of complex contact structures on threefolds

Answer by Nikita Kalinin for Upper bound on the number of generators of a local complete intersection curve in $\mathbb{P}^3$

2012-11-14T17:54:27Z

"every locally complete intersection curve in P3 can be defined by four equations."

look at http://www.math.binghamton.edu/somnath/Notes/curves.pdf and http://www.math.tifr.res.in/~publ/ln/tifr62.pdf

How to describe deformation space of a reducible curve ?

2012-10-29T13:04:59Z

One can describe deformations of a smooth curve by the sections of its normal bundle.

In which terms can one describe deformation space of a reducible curve, for example, union of three intersecting lines in $\mathbb C P^3$ into a cubic ?

One can compute local deformations at point of line's intersection, so the complete construction, I think, is about some sheaf stuff...

Riemann-Roch and dim of deformation space.

2012-06-18T16:46:30Z

Let's consider curve $C\subset \mathbb P^n$ of degree $d$ and genus $g$. We want to calculate dimension of deformation space of $C$, i.e. $h^0(C,L)$ where $L$ is the normal bundle.

We can decompose $L$ as $L_1\subset L_2\dots \subset L$, such as $dim L_{i+1}/L_i =1 \ $ and apply Riemann-Roch to each $L_{i+1}/L_i$.

I heard that this gives us $(n-1)(d-g+1)+2g-2 + 2d\ $ (as expected to be dimension of deformation space of $C$ or the number of points need to fix to count curves degree $d$ and genus $g$ passing through them).

But I can't calculate it ! Could you help me with this or give me a reference ?

Quadratic reciprocity and Weil reciprocity theorem

2012-06-07T11:14:24Z

I was told that Weil reciprocity theorem (one has two meromorphic function $f,g$ on a complex curve $C$, so $\prod\limits_{x\in C} g(x)^{ord_xf}=\prod\limits_{x\in C}f(x)^{ord_xg} \ $ where $ord_xf$ is the smallest degree in Taylor expansion of $f$ at $x$, product is taken only by points in divisors of $f,g$, we assume that these divisors are not intersected with each other) was introduced by Weil after thinking about quadratic reciprocity. Could you explain me the connection between them?

Answer by Nikita Kalinin for Hypersurfaces in Toric Varieties, Help understand a proof from Mikhalkin's paper

2012-04-05T20:23:14Z

Let's consider an example. Triangle $(0,0),(0,d),(d,0)\ $ gives us $\mathbb CP^2$ and each integer point ${(k,l) |k,l\geq 0, k+l\leq d}\ $ corresponds to monomial $x^ky^l$ (or $x^ky^lz^{d-k-l}$ in projective coordinates). There is the same situation for any toric variety - ring of function is generated by monomials corresponding to integer points inside the polytop $P$.

So, it is evident where is the zero set of each monomial -if a point $(k,l)$ belongs to interior of $P$ then zero set is just the union of boundary divisors. If no, for example the point is $(k,0) \to x^kz^{d-k}$, zero set is all boundary divisors without that where our point lies.

Therefore, if you truncate your polynomial, then its zero set doesn't change - throwed off monomials are equal zero on you boundry divisor.

Families of three dimensional algebraic curves

2012-04-03T16:14:54Z

Let's consider spatial algebraic curve $C\subset \mathbb P^3$. How could I describe a family of such curves, for example the set of all curves genus $g$ passing through $k$ points?

I'd like to some "computationable" description(and then use macaulay2 ), so in case of rational curves we can parametrize them and say that my set is some linear subspace of coefficient space.

For example, how can I describe a pencil of elliptic cubics degree 4 (3,5,6 ?) passing through 4 points?

Answer by Nikita Kalinin for Polynomial contact structures on $RP^3$

2011-11-12T00:02:02Z

Here is a code in Macaulay for checking that some plurisubharmonic function determines a contact structure

R = QQ[x,y,z,t]

degf=4

f=x^degf+y^degf+z^degf+t^degf + 3*x^2*z^2 + x^2*t^2

--f2 = x^degf+y^degf+z^degf+t^degf

nR = diff(y,f)*x-diff(x,f)*y+diff(t,f)*z-diff(z,f)*t

cdx = degf*f*diff(y,f)-nR*diff(x,f)

cdy = -degf*f*diff(x,f)-nR*diff(y,f)

cdz = degf*f*diff(t,f)-nR*diff(z,f)

cdt = -degf*f*diff(z,f)-nR*diff(t,f)

cxyz=cdx*(diff(y,cdz)-diff(z,cdy))+cdy*(diff(z,cdx)-diff(x,cdz))+cdz*(diff(x,cdy)-diff(y,cdx))

re = sub(cxyz,t=>1)

factor re

So, $f2=x^2+y^2+z^2+t^2\ $ DOES not produce contact structure. Is convex but not strictly(!) plurisubharmonic (on the plane $x=y=0$). Here (http://www.math.ethz.ch/~evansj/lecture9.pdf) there is a good explanation why induced structure is contact ($d\eta$ tames complex structure on $\mathbb R^4$ )

Can any numerical polynomial be a Hilbert polynomial of something?

2012-03-26T13:32:46Z

is it true that any numerical polynomial , i.e. $f(t)\in \mathbb Q[t], f(n)\in\mathbb Z\ \forall n\in\mathbb Z\ $ can be presented as Hilbert polynomial of some variety?

Topological type $x_0^2+x_1^2+x_2^2+ x_3^2+x_4^2 = 0$ in $\mathbb P^4$

2012-03-09T17:16:12Z

Let's consider projective variety $V$ given by th equation $x_0^2+x_1^2+x_2^2+ x_3^2+x_4^2 = 0 \ $ in $\mathbb CP^4$.

I was wondering what is the Picard group of $V$ ? Or cohomology ring of $V$ ?

Nagata's conjecture, Seshadri constant

2011-12-28T19:05:19Z

What is it known now about Nagata's conjecture and Seshadri constant (http://en.wikipedia.org/wiki/Nagata%27s_conjecture_on_curves and http://en.wikipedia.org/wiki/Seshadri_constant) for toric surfaces? It seems that it should be some lower bounds in terms of fans or polytops. Is it true?

Does there exist some simple examples where exact inequalities are proved?

Minimal genus, adjunction inequality

2011-12-12T20:25:30Z

Let's consider closed simply-connected 4-manifold $M$ and some $a\in H^2(M)$. It is very natural question to estimate minimal $g$ that $a$ can be presented as embedded surface of genus $g$.

As I know there is the adjunction inequality for estimation of minimal genus via Seiberg-Witten theory.

Question 1: Does there exist other methods to estimate minimal genus ?

I heard that there are homeomorphic but not diffeomorphic 4-manifolds $M,N$ such that for some $a\in H^2$, $a$ has different minimal genus in $M,N$.

Question 2: Could you give me such examples? As I understand it should be some manipulations with Seiberg-Witten invariants...

Answer by Nikita Kalinin for What is a continuous path?

2011-11-20T08:30:04Z

There is something like Cech cohomology. It is Alexander-Kolmogorov cohomology (see Spenyer "Algebraic topology", Alexander cohomology). Simplixes in this theory are collections of near points. As I understand it works well for locally contracted spaces (and coincides with Cech cohomology). But it seems that in your situation it works well too.

Answer by Nikita Kalinin for Three Variable Hilbert Polynomials

2011-11-16T21:08:11Z

Your question isn't correct. Such $W_1$ are parametrized by $\mathbb P (V\setminus V_1)$ not by $\mathbb P (V/V_1)$ as you wrote. And $\mathbb P (V\setminus V_1)$ is not a closed variety.

In any case, $W_1\cap W_2\cap W_3\ne 0$ means that there is vector $v\in W_1\cap W_2\cap W_3$. So, you can parametrize almost all such triples by vectors $v$. So, we found one component of $P$ (it is $\mathbb P(V) $). The other ones are when $W_1\cap W_2\subset V_3$ and the same triples with permuting indices (and if $V_1\cap V_2\cap V_3 = 0$ then these triples are parametrized by $\mathbb P(V_3)$). Etc. It is easy to compute Hilbert polynomial for linear subspaces. But if $V_1\cap V_2\cap V_3\ne 0$ then situation is more complicated, and< I think it is strange to expect a good answer.

Answer by Nikita Kalinin for On the smooth structure of the spaces of $k$-jets

2011-11-16T20:30:11Z

As I understand your question, the answer is: for any open set $U'\subset M, 'V\subset N$ such that $U,V$ are diffeomorphic to $U',V'$ you can identify $J^k(U,V)$ with subset of $J^k(M,N)$. So, it is enough that all such identivication are diffeomorphisms.

Answer by Nikita Kalinin for Can one approximate "close" smooth functions?

2011-11-16T20:00:13Z

You have some Riemannian metric on $M$, so, for each point there is it small neighbourhood where any two points can be connected by unique geodesic. So, if you have $n$ points in this neighbourhood you can canonically map simplex to it.

Answer by Nikita Kalinin for degenerating immersion

2011-11-15T19:53:01Z

The answer is no. Two 2-dim smooth immersed in $\mathbb R^3$ objects generically intersect by line, so if intersection is a point then it can be eliminated. But it is clear that near $z^2$ there are no embeddings.

Therefore what do you want it is a immersions with self-intersections as a small circles and these circles collapse to points when $k\to\infty$. But if a selfintersection is a small circle, it can be eliminated too. Large circles in selfintersection can't disappear in limit.

added. Sorry, this answer is about absolutely different problem.

First cohomology of the space of long knots in R^4

2011-11-14T17:34:58Z

Let's consider the space of long knots in $\mathbb R^n, n>3$. I know that there are many results (Vassiliev, Turchin, Sinha, Kontsevich) about different expressions of cohomology of this space. I think the last result is about convergency in $E^1$ term (http://palmer.wellesley.edu/~ivolic/pdf/Papers/VassilievCollapseFinal-G%26T.pdf).

But my question is about what these cohomologies are precise. So, is it true that $H^1(long\ knots\ in\ \mathbb R^4)=0$ ?

Does there exist some table with $H^i(space\ of\ long\ knots\ in\ \mathbb R^j)$ at least for small $i,j$ ?

Answer by Nikita Kalinin for What is the simplest way to show that a section of a vector bundle is transverse to the zero set

2011-11-04T07:54:10Z

It sounds like a multijet transversality theorem (see for example M. Golubitsky, V. Guillemin, Stable mappings and their singularities) in context of algebraic geometry. So, the answer is true --- a proof of this theorem uses only polynoms for perturbations which achieves general position.

Comment by Nikita Kalinin

2013-05-16T18:16:37Z

yes sure. Triangle with sides equal $d$. Its diameter is $d$, its area is $\sqrt{3}d^2/4$. But any figure spanned on two intervals of length $d$ and angle $\pi/3$ between them works as well.

Comment by Nikita Kalinin

2013-05-02T10:27:57Z

That is true, but in a question I have a finite set of pairwise comparable asymptotics, so, I evoke for properties of "good" asymptotics

Comment by Nikita Kalinin

2013-04-22T12:05:12Z

yes, but before I thought that this map can be extended on [0,1]

Comment by Nikita Kalinin

2013-04-19T09:20:08Z

I think if they are not transversal at a point $a(t)$ then for any $t_0$ they are not transversal at the point $a(t_0)$

Comment by Nikita Kalinin

2013-04-02T00:32:34Z

@Ilya: it is not a simple problem, my solution is not true.

Comment by Nikita Kalinin

2013-03-31T07:11:32Z

@robot: I was looking for an estimation $area(M)\geq cd^2$

Comment by Nikita Kalinin

2013-03-30T12:44:17Z

@Ilya: I had started this bounty before I realized that it is a simple problem, and I was very nervous =)) Now I can not cancel it. Concerning the first comment: an affine transformation preserves the lattice width, it is enough for me.

Comment by Nikita Kalinin

2013-03-29T12:08:58Z

Thay gives us $area(M)\geq d^2/4$, it is worse than $g\geq d^2/{2\sqrt{3}}$

Comment by Nikita Kalinin

2013-03-27T21:55:11Z

at least you should always distinguish real bundles (and you have diffeomorphism here) and complex ones (here you have Chern classes)

Comment by Nikita Kalinin

2012-08-26T15:05:36Z

$\pi_1($ any suspension) is trivial, isn't it ?

Comment by Nikita Kalinin

2012-06-08T10:07:42Z

@ François thanks, I know)

Comment by Nikita Kalinin

2012-06-07T12:52:55Z

Thanks for correction!

Comment by Nikita Kalinin

2012-04-26T16:27:20Z

How did you define flatness (curvature) in a non-smooth manifold?

Comment by Nikita Kalinin

2012-04-05T20:26:39Z

for this situation, you are right, but what about other situations ? Or there is no general construction and in each situation approach have to be different? In this case, could you give me some good reference about construction of spatial algebraic curves?

Comment by Nikita Kalinin

2012-04-03T21:18:02Z

sorry, I mean "spatial"