# Dmitri

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## Registered User

 Name Dmitri Member for 3 years Seen 14 hours ago Website Location London Age 36
 May14 awarded ● Enlightened May13 awarded ● Nice Answer Apr7 awarded ● Popular Question Mar30 comment How many polynomial Morse functions on the sphere?If you consider the case $n=2$, I believe the precise answer to your question should be known for all $d$. V.I. Arnol'd was interested in such type of questions, originally in the case when you replace $S^1$ by $\mathbb R^1$ and homogeneous polynomials by inhomogeneous. Probably one should chase the references to Arnold's article : mathnet.ru/php/… You might also want to have a look on the article of Barannikov "On the space of real polynomials without multiple critical values" Mar29 awarded ● Nice Answer Mar23 comment finite dimensional real division algebrasThere is a readable proof in the book of Shafarevich "basic algebraic geometry" of the fact that these algebras have dimension $2^n$. The proof indeed uses Bezout's theorem. Mar23 comment cohomology of a normal crossing divisorThis is not true, consider for example a divisor $D$ on a surface that is a wheel of $\mathbb P^1$'s, i.e, each $\mathbb P^1$ intersects two neighbouring $\mathbb P^1$'s. Then $\pi_1(D)=\mathbb Z$, so $H^1(D)=\mathbb Z$. Mar21 awarded ● Nice Question Mar19 awarded ● Enlightened Mar19 accepted Solid angles of a tetrahedron Mar19 awarded ● Nice Answer Mar18 revised Solid angles of a tetrahedronadded 29 characters in body Mar18 answered Solid angles of a tetrahedron Mar16 awarded ● Popular Question Mar10 revised Bolza curve admits no anticonformal fixedpointfree involutionadded 117 characters in body Mar10 answered Bolza curve admits no anticonformal fixedpointfree involution Mar1 comment Darboux SurfaceNoam, sure I want lines to vary too, so this becomes YQ's question in dimension one less. Mar1 comment Darboux SurfaceThis a very nice question! I wonder if a similar statement holds in $\mathbb P^2$ - if you take five lines, $10$ intersection points of them and consider quartics that contain these $10$ points, is it true that double conic is not in the Zariski closure of the space of such quartics? Feb28 comment What can one say about (differentiable) topological structure of CY3s?Dear Kim, by Bogomolov-Beuaville theorem every CY manifold has a finite cover that is a product of Tori, hyperkahler manifolds and manifolds $M^n$ such that $H^k(M^n,O)=0$ for $k\ne 0,n$. So for some people "proper" CY manifolds are only those that satisfy the last condition: $H^k(M^n,O)=0$ for $k\ne 0,n$. Such manifolds also have the property that the holonomy group of CY metrics on them coincide with $SU(n)$ (and not smaller than this). Such manifolds do have finite fundamental groups. Maybe for Oguis-Sakurai a Kahler manifold is $CY$ iff it has a holomorphic volume form... Feb28 revised What can one say about (differentiable) topological structure of CY3s?the answer is corrected and expanded Feb28 answered What can one say about (differentiable) topological structure of CY3s? Feb23 accepted Betti numbers of Proper nonprojective varieties Feb23 comment Betti numbers of Proper nonprojective varietiesDonu, thanks for the reference, I'll have a look (and will try to see if indeed I have an alternative proof :) ). LMN, you are welcome :) Feb23 revised Betti numbers of Proper nonprojective varietiesadded 32 characters in body Feb23 answered Betti numbers of Proper nonprojective varieties Feb14 comment Properties of quotient varietyConsider the following example: $(x,y)\to (x^2,y)$. Then the preimage of the curve $x=y^2$ under this map is $x=\pm y$. It is singular at $(0,0)$. It seems to me that you need to make the question a bit more specific... Feb6 revised Birational Automorphisms and infinite divisibilityadded 4 characters in body Feb6 revised Birational Automorphisms and infinite divisibilityadded 102 characters in body Feb6 comment Birational Automorphisms and infinite divisibilityYves, thanks I was a bit sloppy :) . But for \mathbb Q this is ture :) Feb5 revised Birational Automorphisms and infinite divisibilityadded 165 characters in body; added 10 characters in body Feb5 comment Birational Automorphisms and infinite divisibilityDaniel, the claim is that any homomorphism from $\mathbb Z[1/2]$ to $GL(n,\mathbb Z)$ sends $\mathbb Z[1/2]$ to $1$, since $1$ in $GL(n,\mathbb Z)$ is the only infinitely divisible element. Feb5 revised Birational Automorphisms and infinite divisibilitydeleted 2 characters in body Feb5 revised Birational Automorphisms and infinite divisibilityadded 202 characters in body; added 36 characters in body Feb5 answered Birational Automorphisms and infinite divisibility Feb4 accepted singular divisors in a complete linear system Feb4 comment Are rational varieties simply connected?Thank you Vesselin. The property of been rationally connected is a birational invariant, I guess? Feb4 comment Are rational varieties simply connected?Sandor, thank you :), I completely agree with you, I added missing words. In fact I was meaning "rational complex projective varieties". I don't know what is the definition of rationally connected projective varieties in the case they are singular. For example, if you consider a cone over a genus $g>0$ curve, every to points can be connected by a two $\mathbb P^1$'s (through the center of the cone), but I don't think this variety should be called rationally connected... Feb4 revised Are rational varieties simply connected?deleted 24 characters in body Feb4 revised Are rational varieties simply connected?added 42 characters in body Feb4 comment Are rational varieties simply connected?Dear Laurent I decided to check the reference and it looks to me that the proof of the fact is not really there. It is proven in two ways that projective spaces over algebraically closed fields are simply connected SGA 1. XI. Prop. 1.1, ( arxiv.org/pdf/math/0206203v2.pdf) and then comes corollary 1.2 without an actual proof. It is just said there that the proof is the same as for projective space :). Could you indicate how to make this an actual proof? I am asking this because I want to see how to make a proof over C without Hironaka's resolution of singularities. Feb3 answered another diameter-perimeter-area inequality Feb3 comment Why are the holomorphic line bundle sections finite dimensional?I really like this reasoning with Montel theorem :) Feb3 answered Why are the holomorphic line bundle sections finite dimensional? Feb2 accepted a diameter-perimeter-area inequality for convex figures Feb2 revised singular divisors in a complete linear systemadded 132 characters in body Feb2 comment singular divisors in a complete linear systemJames, thank you! Of course this is what I meant :) Feb2 revised singular divisors in a complete linear systemadded 3 characters in body Feb2 answered singular divisors in a complete linear system Feb2 comment a diameter-perimeter-area inequality for convex figuresConnor, that is of course correct. In fact, I was thinking of exactly this example but for some reason (I guess to make the answer as short :) ) as possible put the vertices of the rombus in $\pm 1, \pm varepsilon$) instead of what I had in mind. Feb2 comment a diameter-perimeter-area inequality for convex figuresI also called rectangle what should be called a rombus :)