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Is there a big vector bundle $E$ on a complex abelian variety $A$ with $H^0(E\otimes P)=0$ for general $P\in Pic^0(A)$?

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Why do you want to know this? In other words, please provide some background information (e.g. what you know, in what context did this come up), so that people can decide how best to answer this question. –  David Roberts Nov 14 '11 at 6:24
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2 Answers

How about the following example in Debarre's paper "ON COVERINGS OF SIMPLE ABELIAN VARIETIES" available on his web page. On page 5, it says: "...if L is an ample line bundle on an abelian variety A of dimension g, a general map (L^{−d})^{⊕g} → (L^{−1})^{⊕2g} is injective for d ≫ 0 and its cokernel is an ample vector bundle E ([L2], Theorem 6.3.65). If g ≥ 2, we have H^0(A,E ⊗ P_ξ) = 0 for all ξ ∈ Pic0(A)..."

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Welcome on MathOverflow! –  Arend Bayer Nov 19 '11 at 3:37
    
Thank you very much! The example is helpful. –  Hao Sun Nov 20 '11 at 4:05
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I agree with David that a bit more background/explanation would help. For instance, I'm not sure what you mean by a "big vector bundle". Let me suppose that $E$ is a big line bundle, i.e. a line bundle with positive self-intersection. Then the answer is no.

To see this, note that $E$ is nef i.e. $E\cdot C\ge 0$ for any curve $C$, since $C$ can be moved into general position without affecting the intersection number because $A$ is homogenous. It follows that $E\otimes P$ is nef and big for any $P\in Pic^0(A)$ because $E\otimes P$ and $E$ are numerically equivalent. Therefore $H^i(E\otimes P)=0$ for $i>0$ by Kawamata-Viehweg vanishing. If this held for $i=0$, then the euler characteristic $\chi(E\otimes P)=\chi(E)=0$. On an abelian variety, Riemann-Roch gives $\chi(E) = E^n/n!$ where $n=\dim A$ (see Mumford's Abelian varieties). Therefore $E^n=0$ contradicting bigness.

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Very nice argument! –  Henri Nov 14 '11 at 22:45
    
Thanks. $\mbox{}$ –  Donu Arapura Nov 15 '11 at 1:46
    
Big means the anti-tautological bundle $O_{P(E)}(1)$ is big. I know that the answer is negative for a line bundle and a splitting vector bundle. I just want to know whether there is a big vector bundle but not ample on $A$. –  Hao Sun Nov 15 '11 at 3:16
    
OK, that's more subtle. –  Donu Arapura Nov 15 '11 at 12:56
    
Let $E_1$ and $E_2$ be elliptic curves. Let $L_i$ be ample line bundles on $E_i$. $L_1\oplus L_2$ is a big and nonample vector bundle. –  Hao Sun Nov 22 '11 at 3:38
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