diverietti
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Registered User
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I am a researcher in complex analytic and algebraic geometry at the Institut de Mathématiques de Jussieu.
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May 17 |
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There are many varieties with ample canonical bundle An irreducible compact quotient $X$ of a polydisc has ample canonical bundle but if its dimension is greater than one then $H^1(X,T_X)=0$, so that $X$ is rigid! See [Y. Matsushima and G. Shimura, "On the cohomology groups attached to certain vector valued differential forms on the product of the upper half planes". Ann. of Math. (2) 78 1963 417–449]. |
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May 9 |
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pencils on varieties of general type I don't understand what you exactly mean with your question. Do you want a fiber space? In this case I don't understand why the blow up is a counterexample. Do you want just a pencil? For instance, does an elliptic surface embedded in a projective manifold of general type answer your question? |
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May 3 |
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Hyperbolic Riemann Surface A priori, in the non-compact setting, it is not enough to exclude the existence of entire curves in order to establish Kobayashi hyperbolicity (this is a necessary but not a sufficient condition). But for Riemann surfaces, of course, it is necessary and sufficient to be uniformized by the disc (so that in this case it suffices to check non-existence of entire curves). |
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Apr 30 |
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Euler Sequence on Homogeneous Spaces added 221 characters in body |
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Apr 29 |
accepted | Euler Sequence on Homogeneous Spaces |
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Apr 29 |
revised |
Euler Sequence on Homogeneous Spaces deleted 15 characters in body |
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Apr 29 |
answered | Euler Sequence on Homogeneous Spaces |
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Apr 16 |
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Algebraic machinery for algebraic geometry I vote to male this question community wiki! |
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Apr 9 |
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constructing a projection onto a variety Not at all, I wasn't ironic at all in my comment! Re-cheers |
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Apr 8 |
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constructing a projection onto a variety Dear Artie, thanks for your kind comment. Indeed, I think I understood Martin's comment. I just wanted to point out that such an adjective is so relative (and sometimes could even be offensive) that should be used a little bit more carefully. Cheers |
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Apr 8 |
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constructing a projection onto a variety boring? lot of people thinks that math itself is boring. what kind of comment is this? |
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Mar 30 |
awarded | ● Popular Question |
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Mar 18 |
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Holomorphic objects associated with a compact complex manifold? en.wikipedia.org/wiki/Intermediate_Jacobian, but this is for the Kähler case. Aren't you mostly interested in the more general case of compact complex manifolds? |
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Feb 20 |
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generic cubic surface I voted to close since it didn't seem to me a research-level question. |
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Feb 12 |
awarded | ● Civic Duty |
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Feb 5 |
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Symmetric power of tangent space The question is not clear at all... |
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Feb 1 |
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When does $Aut(X)=Bir(X)$ hold? Hi Koppa. Of course if X is projective then the rational function field equals the meromorphic function field. From this it follows easily that any meromorphic mapping $f\colon X\to Y$ between complex projective varieties is indeed rational. When I wrote "more generally", I meant that I was looking at general abstract compact complex manifolds, where the notion of rational mapping is not even defined... |
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Jan 31 |
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When does $Aut(X)=Bir(X)$ hold? Yes, of course, I misunderstood what you wrote! Cheers ! |
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Jan 31 |
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When does $Aut(X)=Bir(X)$ hold? $Y$ containing no rational curves ? or $X$? |
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Jan 31 |
answered | When does $Aut(X)=Bir(X)$ hold? |
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Jan 29 |
revised |
Normal sheaf of non-reduced space curves fixed typos.; deleted 1 characters in body |
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Jan 22 |
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Complete intersections in complex and algebraic geometry Hi Francesco, why Chow's theorem? Sorry, I don't understand. |
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Jan 22 |
awarded | ● Nice Question |
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Jan 18 |
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Sections of a normal subsheaf. You are welcome Ginevra! |
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Jan 18 |
accepted | Sections of a normal subsheaf. |
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Jan 18 |
answered | Sections of a normal subsheaf. |
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Jan 18 |
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Sections of a normal subsheaf. I don't understand your question. The quotient of a vector bundle by a vector subbundle is always a vector bundle. Maybe you mean that you have a sheaf morphism $\mathcal O_X(F)\hookrightarrow\mathcal O_X(E)$? |
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Jan 17 |
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Irreducible divisors containing an arbitrary closed set For me the nonsingular case it's fine now. $H^0(X,\mathcal O_X(mA)\otimes\mathcal I_V)$ is globally generated for large $m$. Moreover, tensoring the short exact sequence $0\to\mathcal I_V\to\mathcal O_X\to\mathcal O_V\to 0$ by $\mathcal O_X(mA)$ gives $0\to H^0(\mathcal O_X(mA)\otimes\mathcal I_V)\to H^0(\mathcal O_X(mA))\to H^0(\mathcal O_V(mA))\to 0$ for $m\gg 1$, by Serre's vanishing. Thus, $h^0(\mathcal O_X(mA)\otimes\mathcal I_V)$, which is the dimension of the relevant linear system, growths like $m^{\dim X}−m^{\dim X−2}$, so that eventually $h^0(\mathcal O_X(mA)\otimes\mathcal I_V)>2$. |
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Jan 17 |
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Irreducible divisors containing an arbitrary closed set Thanks quim! I'll wait for that! |
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Jan 17 |
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Irreducible divisors containing an arbitrary closed set Could you explain me a little bit better, please? I am sorry but I don't understand completely! Suppose $X$ is smooth. Coll $\mathfrak d_{m,V}$ the linear system of divisors in $|mA|$ which contain $V$. Bertini tells us that if $\mathfrak d_{m,V}$ has no base points then its generic element is non-singular (in fact irreducible and non singular provided the dimension of the image of the corresponding map to the projective space is at least two). So how do you conclude? And what about the reduction to the non singular case? |
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Jan 17 |
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Irreducible divisors containing an arbitrary closed set I don't really understand. Which kind of Bertini are you applying? |
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Jan 11 |
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Higher dimensional version of the Hurwitz formula? Great answer Sándor. |
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Dec 31 |
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Embedding a surface in a projective space What do you mean "resulting surface smooth"? If the starting surface is smooth and you embed it, of course it is smooth again... Maybe you have a softer definition of embedding? Anyhow, a first obstruction is given by topology: by the Lefschetz hyperplane section theorem, any (smooth or not) hypersurface in $\mathbb P^3$ must be simply connected. |
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Dec 19 |
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Pseudoeffective but anti-nef divisor @ACL: I think all this is just a matter of reflexes... As you and also Sándor said, torsion non-trivial classes give of course examples of pseudoeffective "anti-nef" divisors. My personal point of view (more "curvature", as you guess) makes me instinctively consider "zero" a line bundle endowed with a flat metric, as long as "positivity" concepts like pseudoeffective or nef are concerned. Cheers! |
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Dec 18 |
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Pseudoeffective but anti-nef divisor Hehe! You're welcome, I was afraid you weren't able to answer all alone... :) |
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Dec 18 |
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Generalized Euler sequence on a projective scheme I completely agree with Olivier. |
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Dec 18 |
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Pseudoeffective but anti-nef divisor Antoine, I think that Sándor is talking about numerical equivalence classes of (rational, or real Cartier) divisors: there, there is no torsion! The concept itself of pseudoeffective (Cartier) divisor is usually defined for $\mathbb Q$-numerical equivalence classes, since you have to take limits... |
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Dec 17 |
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Upper bound on the dimension of linear series on a smooth hypersurface fixed the problem in the short exact sequence |
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Dec 17 |
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Preliminaries to BPV’s compact complex surfaces I expanded the reference. |
