Dmitri
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Registered User
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May 14 |
awarded | ● Enlightened |
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May 13 |
awarded | ● Nice Answer |
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Apr 7 |
awarded | ● Popular Question |
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Mar 30 |
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" |
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Mar 29 |
awarded | ● Nice Answer |
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Mar 23 |
comment |
finite dimensional real division algebras There 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. |
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Mar 23 |
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cohomology of a normal crossing divisor This 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 $H^1(D)=\mathbb Z$. |
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Mar 21 |
awarded | ● Nice Question |
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Mar 19 |
awarded | ● Enlightened |
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Mar 19 |
accepted | Solid angles of a tetrahedron |
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Mar 19 |
awarded | ● Nice Answer |
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Mar 18 |
revised |
Solid angles of a tetrahedron added 29 characters in body |
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Mar 18 |
answered | Solid angles of a tetrahedron |
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Mar 16 |
awarded | ● Popular Question |
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Mar 10 |
revised |
Bolza curve admits no anticonformal fixedpointfree involution added 117 characters in body |
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Mar 10 |
answered | Bolza curve admits no anticonformal fixedpointfree involution |
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Mar 1 |
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Darboux Surface Noam, sure I want lines to vary too, so this becomes YQ's question in dimension one less. |
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Mar 1 |
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Darboux Surface This 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? |
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Feb 28 |
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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... |
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Feb 28 |
revised |
What can one say about (differentiable) topological structure of CY3s? the answer is corrected and expanded |
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Feb 28 |
answered | What can one say about (differentiable) topological structure of CY3s? |
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Feb 23 |
accepted | Betti numbers of Proper nonprojective varieties |
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Feb 23 |
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Betti numbers of Proper nonprojective varieties Donu, 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 :) |
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Feb 23 |
revised |
Betti numbers of Proper nonprojective varieties added 32 characters in body |
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Feb 23 |
answered | Betti numbers of Proper nonprojective varieties |
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Feb 14 |
comment |
Properties of quotient variety Consider 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... |
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Feb 6 |
revised |
Birational Automorphisms and infinite divisibility added 4 characters in body |
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Feb 6 |
revised |
Birational Automorphisms and infinite divisibility added 102 characters in body |
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Feb 6 |
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Birational Automorphisms and infinite divisibility Yves, thanks I was a bit sloppy :) . But for \mathbb Q this is ture :) |
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Feb 5 |
revised |
Birational Automorphisms and infinite divisibility added 165 characters in body; added 10 characters in body |
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Feb 5 |
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Birational Automorphisms and infinite divisibility Daniel, 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. |
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Feb 5 |
revised |
Birational Automorphisms and infinite divisibility deleted 2 characters in body |
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Feb 5 |
revised |
Birational Automorphisms and infinite divisibility added 202 characters in body; added 36 characters in body |
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Feb 5 |
answered | Birational Automorphisms and infinite divisibility |
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Feb 4 |
accepted | singular divisors in a complete linear system |
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Feb 4 |
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Are rational varieties simply connected? Thank you Vesselin. The property of been rationally connected is a birational invariant, I guess? |
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Feb 4 |
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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... |
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Feb 4 |
revised |
Are rational varieties simply connected? deleted 24 characters in body |
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Feb 4 |
revised |
Are rational varieties simply connected? added 42 characters in body |
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Feb 4 |
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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. |
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Feb 3 |
answered | another diameter-perimeter-area inequality |
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Feb 3 |
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Why are the holomorphic line bundle sections finite dimensional? I really like this reasoning with Montel theorem :) |
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Feb 3 |
answered | Why are the holomorphic line bundle sections finite dimensional? |
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Feb 2 |
accepted | a diameter-perimeter-area inequality for convex figures |
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Feb 2 |
revised |
singular divisors in a complete linear system added 132 characters in body |
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Feb 2 |
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singular divisors in a complete linear system James, thank you! Of course this is what I meant :) |
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Feb 2 |
revised |
singular divisors in a complete linear system added 3 characters in body |
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Feb 2 |
answered | singular divisors in a complete linear system |
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Feb 2 |
comment |
a diameter-perimeter-area inequality for convex figures Connor, 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. |
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Feb 2 |
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a diameter-perimeter-area inequality for convex figures I also called rectangle what should be called a rombus :) |
