User dmitri - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T07:22:46Z http://mathoverflow.net/feeds/user/943 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/124899/solid-angles-of-a-tetrahedron/124902#124902 Answer by Dmitri for Solid angles of a tetrahedron Dmitri 2013-03-18T23:42:36Z 2013-03-18T23:47:59Z <p>I think this statement does not hold. The following tetrahedron should give a counterexample:</p> <p>$A=(-0.5, 0, 0)$, $B=(1,0,0)$, $C=(0,\varepsilon^2, \varepsilon)$, $D=(0,\varepsilon^2, -\varepsilon)$, $1>>\varepsilon>0$.</p> <p>Here $BCD$ has largest area, but the spherical angles at $C$ and $D$ should be significantly larger than the one at $A$. (I have done the calculation only approximatively, but I believe it is true).</p> http://mathoverflow.net/questions/7733/the-work-of-thurston The work of Thurston Dmitri 2009-12-04T00:34:43Z 2013-03-16T04:00:46Z <p>I seem to remember written or said somewhere that at some point Thurston decided to stop writing down his theorems in order not to repel mathematicians from his field (maybe this is not correct?). I am really curious if now 25-30 years later there is some nice source, book, or notes, where it is possible to learn some basic ideas about the proof of the fact that Haken manifolds admit a hyperbolic structure? Maybe some of his ideas got a more accessible explanation? Of course his beautiful notes <a href="http://www.msri.org/publications/books/gt3m/" rel="nofollow">http://www.msri.org/publications/books/gt3m/</a> exist, but they don't go so far. </p> http://mathoverflow.net/questions/123606/bolza-curve-admits-no-anticonformal-fixedpointfree-involution/124165#124165 Answer by Dmitri for Bolza curve admits no anticonformal fixedpointfree involution Dmitri 2013-03-10T17:31:42Z 2013-03-10T22:30:58Z <p>Here is a reasoning which uses (as Peter's one) the fact that on every hyperelliptic curve the hyper-elliptic involution $\sigma$ commutes with every conformal self-map (since hyper-elliptic involution is unique). </p> <p>Suppose $\sigma_1$ is an orientation reversing fixed point free involution of $B$. Then since $\sigma\sigma_1=\sigma_1\sigma$ the action of $\sigma_1$ on $B$ descends to an orientation reversing involution (call it $\sigma_2$) on $\mathbb CP^1=B/\sigma_1$. Clearly $\sigma_2$ permutes $6$ vertices of the octahedron and it does not fix any of them. So $\sigma_2$ is the central symmetry of (octahedral) $\mathbb CP^1$. Now, take on $\mathbb CP^1$ any big circle $S^1$ (invariant under $\sigma_2$) that does not pass through the vertices. Let $S'^1$ be its preimage on $B$. $S'^1$ is a circle the double covers $S^1$ (because $S^1$ splits $6$ vertices on $\mathbb CP^1$ into two groups of $3$ verities). Since $\sigma_2$ makes the half-turn of $S^1$ it follows that $\sigma_1$ makes a quoter turn of $S'^1$. So it is not an involution of $B$. Contradiction.</p> http://mathoverflow.net/questions/123183/what-can-one-say-about-differentiable-topological-structure-of-cy3s/123204#123204 Answer by Dmitri for What can one say about (differentiable) topological structure of CY3s? Dmitri 2013-02-28T09:31:18Z 2013-02-28T19:22:20Z <p>There is at least one pair of examples of diffeomorphic but not deformation equivalent three-dimensional Calabi-Yau manifold (Ruan-Gross). The example is explained on pages 47-48 of this paper: <a href="http://arxiv.org/pdf/math/9806111.pdf" rel="nofollow">http://arxiv.org/pdf/math/9806111.pdf</a></p> <p>Otherwise it is of course natural to try to distinguish Calabi Yau three folds by their diffo type. Note that in dimension six two smooth compact manifolds that are homeomorphic are necessarily diffeomorphic, so the classifications up to homeo and diffeo are the same. Classification of simply connected 6-manifolds with torsion free homology according to diffeo is given by a theorem of Wall (the essential bit here is the cubic intersection form on $H^2(M^6,\mathbb Z)$). I am not aware of (current) classification work in this direction for Calabi-Yau 3-fold. But I think someone who would like to do this should use computer (the majority of examples of CY 3-folds are an outcome of a certain computer program). And it seems to me that it should be possible in principle to improve the existing algorithm so that it computes not only betti numbers, but also multiplication on $H^*$ and so the type as well.</p> <p>Concerning the topology of CY 3-folds, on can say at least that the fundamental group of CY manifolds is finite (or, depending on definition of what you call CY manifold, virtually Abelian). At the same time general Kahler manifolds can have very sophisticated fundamental groups. </p> http://mathoverflow.net/questions/122689/betti-numbers-of-proper-nonprojective-varieties/122718#122718 Answer by Dmitri for Betti numbers of Proper nonprojective varieties Dmitri 2013-02-23T13:20:04Z 2013-02-23T14:06:21Z <p>I think, that many basic restrictions, that you have for complex projective varieties still hold for proper smooth complex varieties. Let me show that $b_2>0$, and $b_{2n-2}>0$.</p> <p>Suppose that $X^{2n}$ is a proper smooth complex variety. Then there is a birational morphism $\phi: Y^{2n}\to X^{2n}$ from a <em>projective variety</em> $Y$ to $X$. Let $E_1,...,E_n$ be all the exceptional divisors of $\phi$. Since $Y$ is projective there is a divisor $D\subset Y$ that has a point $x$ in $Y^{2n}\setminus (E_1,...E_n)$ and there is a curve $C$ in $Y$ passing through $x$ and not contained in $D$. Consider finally the divisor $\phi(D)$ and the curve $\phi(C)$ in $X^{2n}$. Note that they intersect positively (since they have common point $\phi(x)$ and $\phi(C)$ does not belong to $\phi(D)$). So they represent non-zero cycles in $H_2(X^{2n})$ and $H_{2n-2}(X^{2n})$.</p> <p>Also, it is very easy to see that $b_1$ should be even. Indeed the fundamental group of a smooth complex variety is a birational invariant, and $b_1$ depends only on $\pi_1$.</p> <p>I have an idea how using <em>weak factorization of birational maps</em> one can prove that all other $2k+1$ Betty numbers are even. But this requires some work, so I'll write this if I manage to work out details (I believe this should be a very well known fact).</p> http://mathoverflow.net/questions/120818/birational-automorphisms-and-infinite-divisibility/120896#120896 Answer by Dmitri for Birational Automorphisms and infinite divisibility Dmitri 2013-02-05T18:52:24Z 2013-02-06T12:59:06Z <p>Edit. I have not realised that this question is on <em>Birational</em> automorphisms, and not on automorphisms, so what I wrote below does not really answer the question. But I will leave this for a background to the question.</p> <p>This answer concerns <em>automorphisms</em> of complex projective manifolds (and more generally Kahler ones). </p> <p><strong>Statement.</strong> If $X$ is Kahler then $\mathbb Z[1/2]$ action on $X$ extends to an action of $\mathbb R$ on $X$ unless it factors through the action of a finite group (thanks to Yves). This follows immediately from Lieberman's-Fujiki theorem (thanks to YangMills for a refernece to Fujiki, see his comment), which I will state now (D. I. Lieberman. Compactness of the Chow scheme: applications to automorphisms and deformations of Kahler manifolds, 1978).</p> <p>Denote by $Aut_0(X)$ the connected component of identity map in the group of automorphisms of $X$.</p> <p><strong>Lieberman-Fujiki Theorem.</strong> Consider the action of $Aut(X)$ on $H^*(X,\mathbb Z)$, </p> <p>$\phi: Aut(X)\to GL(H^*(X,\mathbb Z))$. </p> <p>Then the group $Aut_0(X)$ has finite index in $ker(\phi)$.</p> <p><strong>Proof of the statement.</strong> Clearly if $\mathbb Z[1/2]$ belongs to $Aut(X)$, then a finite index subgroup of it belong to $ker(\phi)$ (indeed $GL(H^*(X,\mathbb Z))$ does not have infinitely $2$-divisible elements apart form $Id$). So by Lieberman-Fujiki theorem it belongs to $Aut_0(X)$. But $Aut_0(X)$ is a Lie group. This finishes the proof.</p> <p>I don't know if the same reasoning can work in the real case (one might first look closer into the proof of Liebermann's result).</p> http://mathoverflow.net/questions/120442/are-rational-varieties-simply-connected/120467#120467 Answer by Dmitri for Are rational varieties simply connected? Dmitri 2013-01-31T21:26:50Z 2013-02-04T11:58:12Z <p>I decided to rewrite the answer so it is more clear.</p> <p>Smooth complex rational projective varieties are indeed simply-connected as is explained in the first answer to the question. My point is that from this fact it follows immediately that normal complex rational projective varieties are always simply-connected. Here is a proof.</p> <p><strong>Proof.</strong> Let $X$ be an normal complex projective variety and $X'\to X$ be a resolution of singularities. Then we have a homomorphism $\pi_1(X')\to \pi_1(X)$. I claim that is is surjective. In order to see this one just needs to know that every loop on $X$ can be lifted to $X'$ and this is true since all the fibres of $X'\to X$ are connected since $X$ is normal.</p> <p>So if $X$ is rational then $X'$ is smooth rational and $\pi_1(X)=\pi_1(X')=0$ </p> http://mathoverflow.net/questions/120676/another-diameter-perimeter-area-inequality/120692#120692 Answer by Dmitri for another diameter-perimeter-area inequality Dmitri 2013-02-03T18:03:59Z 2013-02-03T18:03:59Z <p>Let us call a <em>circular segment</em> a thing like this: (| : <a href="http://en.wikipedia.org/wiki/Circular_segment" rel="nofollow">http://en.wikipedia.org/wiki/Circular_segment</a></p> <p>I think that the best possible constant will be achieved on doubled circular segments, i.e. things like this : ()</p> <p>Indeed suppose you fix the area and diameter of you domain, and try to minimise its perimeter. Then the diameter cuts your domain into two halves. If you fix the area of one half (and minimise its preimiter), then this is a classical fact that half will be a circular segment (|. </p> <p>So one juts needs to take all the formulas from wiki (on the area and perimeter of circular segments) and find the minimum.</p> http://mathoverflow.net/questions/120033/why-are-the-holomorphic-line-bundle-sections-finite-dimensional/120678#120678 Answer by Dmitri for Why are the holomorphic line bundle sections finite dimensional? Dmitri 2013-02-03T15:50:03Z 2013-02-03T15:50:03Z <p>Here is an algebraic version of Margaret's answer that gives an explicit bound on the dimension of the space of section. This reasoning can be also adapted to the case of complex (non-algebraic) manifolds, but without an explicit bound.</p> <p><strong>Claim.</strong> Suppose $X^k\subset \mathbb CP^n$ is a $k$-dimensional projective variety. Let $D$ be the zero divisor of a section of a line bundle $L$ on $X^k$. Then the dimension of the space of sections of $L$ is at most the binomial coefficient: $$\binom{k+ deg(D)\cdot deg(X)}{deg(D)\cdot deg(X)}$$</p> <p>The proof of this statement uses just Bezout theorem. Indeed from Bezout it follows that at a fixed (say smooth) point $x\in X^k$ a section of $L$ can vanish at most to order $deg(D)\cdot deg(X)$ (to see this cut $X^k$ through $x$ by a generic plane $\mathbb CP^{n-k+1}$ and intersect the obtained curve with $D$). Now, the binomial coefficient is the dimension of the space of all $deg(D)\cdot deg(X)$-jets at $x$.</p> http://mathoverflow.net/questions/120565/singular-divisors-in-a-complete-linear-system/120584#120584 Answer by Dmitri for singular divisors in a complete linear system Dmitri 2013-02-02T11:16:41Z 2013-02-02T12:06:12Z <p>The answer is yes. Here is an example generalising the one of Serge and giving arbitrary codimension. </p> <p>Consider line bundle $L=O(1)\boxtimes O(1)$ on $\mathbb CP^1\times \mathbb CP^n$. A section of such a bundle has shape $$\sum_{i=0,j=0}^{i=1,j=n}a_{ij}x_iy_j=0,$$ i.e. $|L|=\mathbb C^{2n+2}$. A section is singular iff the matrix $a_{ij}$ has rank $1$. This happen in codimension $n$. If the rank of $a_{ij}$ is two the section is smooth and is a $\mathbb CP^{n-1}$ bundle over $\mathbb P^1$.</p> http://mathoverflow.net/questions/120529/a-diameter-perimeter-area-inequality-for-convex-figures/120563#120563 Answer by Dmitri for a diameter-perimeter-area inequality for convex figures Dmitri 2013-02-01T23:59:30Z 2013-02-02T00:44:42Z <p>This inequality is not true. Consider the rectangle on $\mathbb R^2$ with vertices $(\pm 1, 0)$, $(0, \pm \varepsilon)$. Then on the left you have $2\varepsilon(4-8/\pi)$ on the right you have approximatively $4\varepsilon^2$.</p> http://mathoverflow.net/questions/120139/homotopy-groups-of-k3/120141#120141 Answer by Dmitri for Homotopy groups of K3 Dmitri 2013-01-28T19:06:59Z 2013-01-28T19:12:10Z <p>Hurewicz theorem says that for a simply connected space $X$, $\pi_2(X)\cong H_2(X,\mathbb Z)$. So $\pi_2(K3)\cong H_2(K3,\mathbb Z)\cong \mathbb Z^{22}$. Here is a link:</p> <p><a href="http://en.wikipedia.org/wiki/Hurewicz_theorem" rel="nofollow">http://en.wikipedia.org/wiki/Hurewicz_theorem</a></p> http://mathoverflow.net/questions/118251/is-the-cup-product-of-holomorphic-n-forms-with-a-fixed-class-injective/118256#118256 Answer by Dmitri for Is the cup product of holomorphic $n$-forms with a fixed class injective? Dmitri 2013-01-07T09:23:24Z 2013-01-07T09:23:24Z <p>The answer to this question is negative in dimensions $\ge 3$. For example, take a quintic in $\mathbb CP^4$ and consider its blow up $X$ in $10^{100}$ points (just to be safe). Then the space $H^1(X, T_X)$ will be huge, since it parametrises deformations of the blown up variety and you can move points as you wish. So there will be non-zero $u\in H^1(X, T_X)$ so that you map is trivial. </p> <p>Note that when you blow up the quintic $H^{2,1}$ does not change. </p> <p>Also, this trick with blow ups will not work for Kahler surfaces as is explained for example, in appendix 1 in <a href="http://arxiv.org/pdf/1301.0478.pdf" rel="nofollow">http://arxiv.org/pdf/1301.0478.pdf</a></p> http://mathoverflow.net/questions/118139/smooth-projective-toric-varieties-which-are-quotients-of-product-of-spheres-and-t/118217#118217 Answer by Dmitri for Smooth projective toric varieties which are quotients of product of spheres and torii by a free torus action? Dmitri 2013-01-06T20:19:35Z 2013-01-06T20:19:35Z <p>Let us consider the case of toric varieties of real dimension $4$ and prove they can not be represented as such a quotient unless they have second Betti number $1$ or $2$.</p> <p>Proof.</p> <p>Let us introduce some notations. Let $B$ be the toric manifold of real dimension $4$, $n=b_2(B)$. Denote by $E$ the product $\Pi_i S^{n_i}\times T^k$. Let $k_2=b_2(E)$, $k_3=b_3(E)$. Finally, denote by $m$ the dimension of the torus that is acting on $E$.</p> <p>Suppose that $T^m$ is acting freely on $E$ and $B=E/T^m$. Then we have the following obvious relation on dimensions:</p> <p>$$4=dim(E)-m\ge k+2k_2+3k_3-m$$</p> <p>We will explain now that we must have $n\le 2$ (recall $n=b_2(B)$). For this purpose we will consider the long exact sequence of homotopy groups for the fibration $E\to B$.</p> <p>$$0\to \pi_3(E)\to \pi_3(B)\to 0 \to \pi_2(E)\to \pi_2(B)\to \mathbb Z^m\to \mathbb Z^k\to 0$$</p> <p>Since $\pi_2(B)=\mathbb Z^n$, from the second half of the sequence we get </p> <p>$$-k_2+n-m+k=0$$</p> <p>substituting this in the inequality on dimensions we get </p> <p>$$4\ge 3k_2+3k_3-n$$</p> <p>To prove finally that $n\le 2$ we use the classical statement that $\pi_3(B)$ contains sub-group $\mathbb Z^{(n^2+n)/2-1}$. It follows that</p> <p>$$rk (\pi_3(E))=k_3+k_2\ge (n^2+n)/2-1$$</p> <p>so </p> <p>$$4\ge 3((n^2+n)/2-1)-n$$</p> http://mathoverflow.net/questions/117420/hamiltonian-actions-and-contractible-loops/117482#117482 Answer by Dmitri for Hamiltonian actions and contractible loops Dmitri 2012-12-29T06:56:00Z 2012-12-30T11:36:39Z <p>There are counter-examples, hope they answer your question completely, just take any non-simply connected $G$ and consider its action on $T^*G$. The simplest case is:</p> <p>Let $M$ be the cylinder $S^1\times \mathbb R$ with the symplectic form $ds \wedge dt$. Then the Hamiltonial $H=t$ defines an $S^1$-action on the cylinder.</p> <p><em>One more counterexamle</em>. Consider just the action of $SO(3)$ on its cotangent space. Clearly this action is Hamiltonian. Let us take the subgroup $S^1\subset SO(3)$ that represents the non-zero element of $\pi_1(SO(3))$. Obviously all the orbits of the action of this $S^1$ on $T^*(SO(3))$ will not be contractible. </p> <p>So we see that in the case the Lie group is not simply-connected it always admits a "bad" action. </p> http://mathoverflow.net/questions/94110/linear-equivalence-and-hilbert-function/117260#117260 Answer by Dmitri for Linear equivalence and Hilbert function Dmitri 2012-12-26T15:27:16Z 2012-12-26T15:27:16Z <p>Let me show that the answer to this question is positive for $d>3$. Indeed, for a general surface $X$ of degree $d>3$ its Picard group is $\mathbb Z$ and is generated by $O(1)$. It follows that both curves $C_1$ and $C_2$ are complete intersections, and so they have the same Hibert function (see for example Section 13 pages 172-173 in book of Harris "first course in algebraic geometry"). Hence the statement is proved. </p> http://mathoverflow.net/questions/117249/convex-polyhedral-decomposition-of-spheres/117250#117250 Answer by Dmitri for Convex polyhedral decomposition of spheres Dmitri 2012-12-26T12:54:17Z 2012-12-26T13:43:19Z <p>For any even $k>4$ there is a decomposition of $S^2$ into $k$ congruent triangles with angles $\pi/2,\pi/2, 4\pi/k$. </p> <p>For $k=n+2$ in order to get a decomposition of $S^n$ into $k$ congruent simplexes you should just inscribe in $S^n$ the regular simplex and project its hyper-faces to the sphere from its centre.</p> <p>In general, a sphere of arbitrary dimension $n$ can be decomposed in arbitrary large number of congruent simplexes, for example into $k2^{n-1}$ simplexes with any $k>2$. This can be done by induction. The $n=1$ case is obvious, to go from $S^n$ to $S^{n+1}$ just put $S^n$ into $S^{n+1}$ as the equatorial sphere and consider the suspension of the simplicial decomposition of $S^n$ (a hypersimplex of such decomposition is the convex hull of a union of a hypersimplex in $S^n$ with one of the poles of $S^{n+1}$).</p> http://mathoverflow.net/questions/117098/compact-complex-n-folds-with-betti-numbers-b-1b-2b-n0-for-n-3/117178#117178 Answer by Dmitri for Compact Complex n-folds with Betti numbers $b_1=b_2=b_n=0$ for $n >3$ Dmitri 2012-12-25T01:09:56Z 2012-12-25T02:32:37Z <p>The product of two spheres of odd dimensions admits a complex structure (Calabi-Eckmann) </p> <p><a href="http://en.wikipedia.org/wiki/Calabi%E2%80%93Eckmann_manifold" rel="nofollow">http://en.wikipedia.org/wiki/Calabi%E2%80%93Eckmann_manifold</a></p> http://mathoverflow.net/questions/117173/genus-zero-gromov-witten-invariants/117176#117176 Answer by Dmitri for genus-zero Gromov-Witten invariants Dmitri 2012-12-25T00:42:21Z 2012-12-25T01:35:36Z <p>I will just consider the simplest example, maybe someone will give the answer in complete generality. So let $M=\mathbb CP^n$. Then $4n-2c_1(M)(A)=2n-2$. This means that we are in good shape. Basically we can take for $X$ and $Y$ any complex submanifolds of $M$ satisfying your condition. Indeed in this case for a generic point $p$ in $\mathbb CP^n$ there will be $deg X\cdot deg Y$ lines that contain $p$ and intersect both $X$ and $Y$. I assumed $X$ (or $Y$) is not zero dimensional, in which case there is only one line and also that $X$ and $Y$ are in general position, but this does not matter for GW, of course.</p> http://mathoverflow.net/questions/117169/reasons-for-the-arnold-conjecture/117177#117177 Answer by Dmitri for Reasons for the Arnold conjecture Dmitri 2012-12-25T01:00:32Z 2012-12-25T01:00:32Z <p>In a certain sense, symplectic geometry (or safer to say symplectic topology) as we know it now was not existing before Arnold formulated these conjectures. So many would say that Arnold conjectures gave birth to symplectic geometry. At the time Arnold made this conjecture only one non-trivial statement in this directions was known - Poincaré's last geometric theorem <a href="http://en.wikipedia.org/wiki/Poincar%C3%A9%E2%80%93Birkhoff_theorem" rel="nofollow">http://en.wikipedia.org/wiki/Poincar%C3%A9%E2%80%93Birkhoff_theorem</a></p> http://mathoverflow.net/questions/117170/a-partial-differential-equation-on-mathbbcp1/117175#117175 Answer by Dmitri for A partial differential equation on $\mathbb{CP}^1$ Dmitri 2012-12-25T00:20:33Z 2012-12-25T00:20:33Z <p>It sounds like the only non-trivial solution is $ae^{i(z+\bar z)/2r}=ae^{-y/r}$. Taking difference of two equations we see that all solution depend only on $y$ and if you restrict to $y$ axis the equation is easy to solve.</p> http://mathoverflow.net/questions/103358/moishezon-manifolds-with-vanishing-first-chern-class/116494#116494 Answer by Dmitri for Moishezon manifolds with vanishing first Chern class Dmitri 2012-12-15T23:45:06Z 2012-12-16T10:43:55Z <p><strong>Added.</strong> I just realised that the statement concerning Moishenzon manifolds holds in dimension up to $4$. In dimension three this is a corollary of minimal model programme and in dimension $4$ this follows from Theorem 0.4 here:</p> <p><a href="http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/coneduality.pdf" rel="nofollow">http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/coneduality.pdf</a></p> <p>In dimensions higher than four the statement would follow from the main conjecture of minimal model programm it states that a projective manifold with pseudoeffective canonical class has non-negative Kodaira dimension. This conjecture holds if dimension less than $4$ and for dimension four the above article can be used instead of it for our purpose.</p> <p>Indeed, every Moishenson manifold admits a blow up that is projective. According to the condition that you state the canonical class of the blow up is <em>pseudoeffective</em>. So according to the conjecture a power of the canonical bundle on the blow up has a non-zero section. Such a section should vanish on the exceptional divisors of the blow up. So, I guess you should be able to push it down to the original manifold (again to a section of the power of the canonical bundle). This section would not vanish since it could vanish only on a hypersurface and this would mean the $c_1\ne 0$ (since there are plenty of curves on Moishenson manifolds).</p> <p><strong>More details.</strong> The canonical class of the blow up is pseudoeffective beacuse it is positive on every coverening family of curves. In particular the blow up is not unirulled. So it should have non-negative Kodaira dimension according to conjecture 1.6 here (page 6): <a href="https://www.dpmms.cam.ac.uk/~cb496/birgeom-paris-public.pdf" rel="nofollow">https://www.dpmms.cam.ac.uk/~cb496/birgeom-paris-public.pdf</a></p> http://mathoverflow.net/questions/116476/where-does-the-notion-of-pseudoholomorphic-curve-come-from/116478#116478 Answer by Dmitri for Where does the notion of pseudoholomorphic curve come from? Dmitri 2012-12-15T19:23:06Z 2012-12-15T19:23:06Z <p>I think it would not be wrong to say that pseudo-holomorphic curves became really popular thanks to the work of Gromov <em>Pseudo-holomorphic curves in complex manifolds</em> http://www.ihes.fr/~gromov/PDF/9[45].pdf</p> <p>In this work Gromov shows that certain facts about holomorphic curves in complex manifolds survive when the holomorphic structure is replaced by merely an almost complex structure provided the almost complex structure is tamed by a symplectic one.<br> For example there is a statement that every mathematician knows - for every two points in a plane there is a unique line that passes through them. This statement survives and becomes a difficult theorem in symplectic geometry that helps in particular to classify symplectic structures on $\mathbb CP^2$.</p> <p>Pseudo-holomorphic curves were appearing here and there also without (an apparent) relation to Gromov's work. For example Elles-Salamon noticed that minimal surfaces in hyperbolic 4-space are in correspondence with almost complex curves in the twistor space of the hyperbolic space: J. Eells and S. Salamon. <em>Twistorial construction of harmonic maps of surfaces into four-manifolds.</em></p> http://mathoverflow.net/questions/32938/surfaces-in-mathbbp3-with-isolated-singularities/116300#116300 Answer by Dmitri for Surfaces in $\mathbb{P}^3$ with isolated singularities Dmitri 2012-12-13T17:18:12Z 2012-12-14T11:20:51Z <p>This answer is completely rewritten. This is not an actual answer but a thought related to the question. I decided to leave it hear since it is short.</p> <p>Note first that if there is a <em>regular</em> map from a surface $X$ to $\mathbb P^3$ whose image has only isolated singularities, then $X$ has curves with negative self-intersection. In particular, if $X$ has no such curves then its image in $\mathbb P^3$ is smooth.</p> <p>Now, suppose we have a surface $X$ with isolated singularities in $\mathbb CP^3$, say of general type and consider the question:</p> <p><strong>Question.</strong> Let $X'$ be the minimal resolution of singularities on $X$. Can we say something about $X$ if $X'$ contains rational $-1$ curves?</p> http://mathoverflow.net/questions/114849/isometric-embedding-of-a-real-analytic-riemannian-manifold-in-a-compact-kahler-ma/114859#114859 Answer by Dmitri for Isometric embedding of a real-analytic Riemannian manifold in a compact Kähler manifold Dmitri 2012-11-29T08:36:40Z 2012-11-29T08:36:40Z <p>I think the answer to your question should be positive and below is a sketch of what should work (I think).</p> <p>Any real analytic manifold can be realised as the real part of a complex projective manifold. I.e. one should embedd the real analytic manifold smoothly in $\mathbb R^n$ and then approximate by a real algebraic submanifold, whose complexification will be the Kahler manifold you are thinking about. Now, the emdedding can be done almost isometrically. I think, using this one should be able to show that the standard (say Fubini-Study) metric on the complexification can be adjusted a bit to make the embedding in the complexification an isometry.</p> http://mathoverflow.net/questions/114806/are-there-countably-many-diffeomorphism-classes-of-finite-radius-balls-of-complet/114826#114826 Answer by Dmitri for Are there countably many diffeomorphism classes of finite radius balls of complete Riemannian manifolds? Dmitri 2012-11-28T23:24:16Z 2012-11-28T23:34:28Z <p>The answer to this question should be negative for $M=S^2$. I will sketch a tentative proof of this claim that would use a couple of simple lemmas.</p> <p><strong>Lemma 1.</strong> The set of closed totally disconnected subsets $X_{dis}$ in the interval $[0,1]$ modulo diffeos of $[0,1]$ is uncountable.</p> <p><strong>Lemma 2.</strong> The set of surfaces $S^2\setminus X_{dis}$ where $X_{dis}$ as above is uncountable (we consider here $X_{dis}$ lying in a segment $I\cong [0,1]$ in $S^2$, $I\subset S^2$.)</p> <p><strong>Lemma 3.</strong> For any $X_{dis}\subset [0,1]$ there is a smooth non-negative function $f$ on $[0,1]$ whose set of zeros is $X$.</p> <p>All these lemmas are very straightforward (Lemma 2 requires a bit of thinking but it should be correct). I want to use them to construct for any $X_{dis}\subset S^2$ a metric on $S^2$ such that for a certain point $O\in S^2$ the ball $B(O,1)$ is $S^2\setminus X_{dis}$. In order to do this start with a metric on $S^2$ such that $B(O,1)$ is $S^2\setminus [0,1]$, where $[0,1]$ is an interval smoothly embedded in $S^2$. Note that the equidistants emanating from $O$ come to $[0,1]$ from two sides. So just "speed up" them from one side so that they hit all the points from $[0,1]\setminus X_{dis}$ before the "other side" of equidistant hits them; at the same time do so that all points of $X_{dis}$ are hit at the same time from both sides. (to do this explicitly I would use something like Lemma 3).</p> <p>I am afraid that this sketch is not so well phrased, but I hope that the idea is more-less clear. </p> http://mathoverflow.net/questions/114745/monic-polynomial-with-integer-coefficients-with-roots-on-unit-circle-not-roots-r/114748#114748 Answer by Dmitri for Monic polynomial with integer coefficients with roots on unit circle, not roots root of unity? Dmitri 2012-11-28T10:01:45Z 2012-11-28T11:20:23Z <p>There exist irreducible monic polynomials such that all their roots apart from two lie on the unique circle (and are not roots of unity). Such polynomials can be chosen among Salem polynomials and they exit in arbitrary high degree. By definition a Salem polynomial $S(x)\in \mathbb Z[x]$ is a monic irreducible reciprocal polynomial with exactly two roots off the unit circle, both real and positive. Of course non of the roots of these polynomials are roots of unity, since these polynomials are irreducible.</p> <p>See for example theorem 1.6 in the article</p> <p>Automorphisms of even unimodular lattices and unramified Salem numbers of Gross and Mcmullen: </p> <p><a href="http://www.math.harvard.edu/~ctm/home/text/papers/unim/unim.pdf" rel="nofollow">http://www.math.harvard.edu/~ctm/home/text/papers/unim/unim.pdf</a></p> <p><strong>Theorem.</strong> For any odd integer $n\ge 3$ there exist infinitely many unramified Salem polynomials of degree $2n$.</p> http://mathoverflow.net/questions/114414/is-there-any-holomorphic-version-of-the-tubular-neighborhood-theorem/114417#114417 Answer by Dmitri for Is there any holomorphic version of the tubular neighborhood theorem? Dmitri 2012-11-25T12:42:12Z 2012-11-25T12:42:12Z <p>To start with something positive, it is indeed true that whenever you have a $\mathbb CP^1$ with negative self-intersection in a complex surface it has a standard holomorphic neighbourhood. </p> <p>On the other hand the comparison in 2) between symplectic and complex geometry is not correct unless you consider $2$-spheres with negative self-intersection in symplectic $4$-manifolds. </p> <p>Namely in symplectic geometry whenever we have a symplectic submanifold $N^{2k}$ in a symplectic manifold $M^{2n}$ a symplectic neighbourhood of $N^{2k}$ is always standard. I.e. such a neighbourhood depends only on type of almost complex structure on the normal bundle to $N^{2k}$. On the other hand in complex geometry the existence of a standard neighbourhood is extremely rare especially if the normal bundle of $N^{2k}$ in $M^{2n}$ is not negative.</p> <p><strong>Example.</strong> Consider a smooth quadric $Q$ in $\mathbb CP^n$ and let us show that its neighbourhood is not byholomorphic to a neighbourhood of the zero section of the $O(4)$ bundle over $Q$. Indeed if this were the case, one would be able to realise the normal bundle to $Q$ in $\mathbb CP^n$ as holomorphic sub-bunlde $N$ of $T\mathbb CP^n$ restricted on $Q$. But the later is impossible. Indeed, in this where possible through each point of $Q$ one would be able to draw a line that would be tangent to this normal sub-bundle $N$. Such a family of lines would generate an involution of $Q$ without fixed point, which is absurd. </p> http://mathoverflow.net/questions/111650/rational-curved-lying-in-the-boundary-of-deligne-mumford-compactification-bar-m Rational curved lying in the boundary of Deligne-Mumford compactification $\bar M_g$ Dmitri 2012-11-06T14:56:58Z 2012-11-06T15:16:57Z <p>Let $\bar M_g$ be the Deligne-Mumford compactifiction of the moduli space of complex genus $g$ curves $M_g$. Is this correct that through every point of the boundary $\bar M_g\setminus M_g$ passes a rational curve that lies in the boundary $\bar M_g\setminus M_g$?</p> http://mathoverflow.net/questions/111550/hilbert-scheme-of-2-points-on-an-elliptic-curve/111551#111551 Answer by Dmitri for Hilbert scheme of 2 points on an elliptic curve Dmitri 2012-11-05T12:01:27Z 2012-11-05T12:01:27Z <p>In this case the corresponding locally free sheaf or rank $2$ on $C$ is the unique indecomposable one.</p> http://mathoverflow.net/questions/125944/how-many-polynomial-morse-functions-on-the-sphere Comment by Dmitri Dmitri 2013-03-30T10:29:59Z 2013-03-30T10:29:59Z 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 : <a href="http://www.mathnet.ru/php/archive.phtml?wshow=paper&amp;jrnid=rm&amp;paperid=4470&amp;option_lang=rus" rel="nofollow">mathnet.ru/php/&hellip;</a> You might also want to have a look on the article of Barannikov &quot;On the space of real polynomials without multiple critical values&quot; http://mathoverflow.net/questions/125897/ideal-generated-by-elements Comment by Dmitri Dmitri 2013-03-29T10:18:24Z 2013-03-29T10:18:24Z What about the following: note that a polynomial in $I(X\cup Y)$ can not have linear terms. Then count the dimension of the space of degree $2$ polynomials in $I(X\cup Y)$? http://mathoverflow.net/questions/105478/finite-dimensional-real-division-algebras Comment by Dmitri Dmitri 2013-03-23T21:08:25Z 2013-03-23T21:08:25Z There is a readable proof in the book of Shafarevich &quot;basic algebraic geometry&quot; of the fact that these algebras have dimension $2^n$. The proof indeed uses Bezout's theorem. http://mathoverflow.net/questions/125373/cohomology-of-a-normal-crossing-divisor Comment by Dmitri Dmitri 2013-03-23T13:56:08Z 2013-03-23T13:56:08Z 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 $H^1(D)=\mathbb Z$. http://mathoverflow.net/questions/123120/darboux-surface Comment by Dmitri Dmitri 2013-03-01T23:33:49Z 2013-03-01T23:33:49Z Noam, sure I want lines to vary too, so this becomes YQ's question in dimension one less. http://mathoverflow.net/questions/123120/darboux-surface Comment by Dmitri Dmitri 2013-03-01T21:57:13Z 2013-03-01T21:57:13Z 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? http://mathoverflow.net/questions/123183/what-can-one-say-about-differentiable-topological-structure-of-cy3s/123204#123204 Comment by Dmitri Dmitri 2013-02-28T21:36:34Z 2013-02-28T21:36:34Z 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 &quot;proper&quot; 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... http://mathoverflow.net/questions/122689/betti-numbers-of-proper-nonprojective-varieties/122718#122718 Comment by Dmitri Dmitri 2013-02-23T18:34:03Z 2013-02-23T18:34:03Z 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 :) http://mathoverflow.net/questions/121799/properties-of-quotient-variety Comment by Dmitri Dmitri 2013-02-14T15:34:43Z 2013-02-14T15:34:43Z 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... http://mathoverflow.net/questions/120818/birational-automorphisms-and-infinite-divisibility/120896#120896 Comment by Dmitri Dmitri 2013-02-06T09:07:11Z 2013-02-06T09:07:11Z Yves, thanks I was a bit sloppy :) . But for \mathbb Q this is ture :) http://mathoverflow.net/questions/120818/birational-automorphisms-and-infinite-divisibility/120896#120896 Comment by Dmitri Dmitri 2013-02-05T20:26:53Z 2013-02-05T20:26:53Z 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. http://mathoverflow.net/questions/120442/are-rational-varieties-simply-connected/120467#120467 Comment by Dmitri Dmitri 2013-02-04T12:35:27Z 2013-02-04T12:35:27Z Thank you Vesselin. The property of been rationally connected is a birational invariant, I guess? http://mathoverflow.net/questions/120442/are-rational-varieties-simply-connected/120467#120467 Comment by Dmitri Dmitri 2013-02-04T12:02:17Z 2013-02-04T12:02:17Z Sandor, thank you :), I completely agree with you, I added missing words. In fact I was meaning &quot;rational complex projective varieties&quot;. 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 $g&gt;0$ 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... http://mathoverflow.net/questions/120442/are-rational-varieties-simply-connected/120659#120659 Comment by Dmitri Dmitri 2013-02-04T11:12:21Z 2013-02-04T11:12:21Z 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. http://mathoverflow.net/questions/120033/why-are-the-holomorphic-line-bundle-sections-finite-dimensional/120035#120035 Comment by Dmitri Dmitri 2013-02-03T15:54:35Z 2013-02-03T15:54:35Z I really like this reasoning with Montel theorem :)