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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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    $\begingroup$ 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. $\endgroup$ Nov 14 '11 at 6:24
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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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  • $\begingroup$ Welcome on MathOverflow! $\endgroup$ Nov 19 '11 at 3:37
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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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  • $\begingroup$ Very nice argument! $\endgroup$
    – Henri
    Nov 14 '11 at 22:45
  • $\begingroup$ 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 $A$. $\endgroup$ Nov 15 '11 at 3:16
  • $\begingroup$ OK, that's more subtle. $\endgroup$ Nov 15 '11 at 12:56
  • $\begingroup$ 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. $\endgroup$ Nov 22 '11 at 3:38

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