Thomas Bauer shows in http://arxiv.org/pdf/alg-geom/9712019v1.pdf that for a complex abelian variety a nef line bundle is numerically equivalent to an effective divisor (this is shown in Lemma 1.1). It seems to me (by comments in other papers) that this is known over a general algebraically closed field, but I have yet to find a proof. Since Bauer's proof deals with the eigenvalues of the Hermitian form associated to the (polarized) abelian variety, I'm not quite sure how a proof would proceed over an arbitrary algebraically closed field. Any suggestions or references?


Here is a sketch of a purely algebraic proof based on the theory developed in Chapter 3 of Mumford's "Abelian Varieties".

Let $L$ be a nef line bundle on the abelian variety $A$ of dimension $g$. If $K(L)$ is finite, then the index of $L$ is a well defined integer between $0$ and $g$ and we need to show that $g=0$. By the theorem on p. 155, the index is the number of positive roots of the polynomial $P(n) = \chi(L \otimes M^n)$, where $M$ is any ample line bundle. Using Riemann-Roch and the nefness of $L$ one sees that all the coefficients of the polynomial are non-negative so it has no positive roots.

We now reduce the general case to the one above. We may assume, by replacing $L$ by $L \otimes (-1)^*L$, that $L$ is symmetric. If $K(L)$ is not finite, i.e. $T_x^*L \cong L$ for $x$ in a positive dimensional subvariety $B$ of $A$ we claim that $L^{2}$ descends to a line bundle $L'$ on $A/B$. This follows as we can get descent data for $L^2$ by choosing symmetric isomorphisms $T_x^*L \cong L$ for all $x \in B$. (Note that one cannot always descend $L$ as shown by the example of a line bundle of order $2$ in $Pic(A)$.)

Since $L$ is nef it follows that $L'$ is also nef, so it is numerically equivalent to an effective line bundle by induction.

(In the sketch above, the class of $L$ modulo numerical equivalence is replaced with a multiple. One can perhaps avoid this by an additional argument.)

  • $\begingroup$ Where can I read about descent, in this sense? Why does the data you gave assure me that $L^2$ descends to a line bundle on $A/B$? $\endgroup$ – rfauffar Jul 11 '13 at 21:16
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    $\begingroup$ This is a special case of flat descent; see, for example, Theorem 4 in Chapter 6 of the book Neron Models by Bosch, Lutkebohmert and Raynaud. (The situation here is simpler though since the morphism $A \to A/B$ is a smooth homomorphism so you should be able to rewrite the condition for the existence of descent datum). After fixing an isomorphism of $L$ with $(-1)^*L$, there is a symmetric isomorphism $T_x^*L \cong L$ which is well defined up to sign. The sign ambiguity disappears if you square these isomorphisms. $\endgroup$ – naf Jul 12 '13 at 13:11
  • $\begingroup$ Ok great, I'm going to take a look at that book. Thanks a lot! $\endgroup$ – rfauffar Jul 12 '13 at 15:11
  • $\begingroup$ After reading a bit, I see that one of the conditions necessary for $L$ to descend is that the commutator map $e^{L^2}:K(L^2)\times K(L^2)\to k^\times$ be trivial on $K(L)\times K(L)$. I don't see why this should happen by only asking for $L$ to be symmetric... $\endgroup$ – rfauffar Jul 18 '13 at 20:53
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    $\begingroup$ I don't see why you need this condition. It suffices to descend to $A/B$ where $B$ is the identity component of $K(L)$; the pullback of the theta group to $B$ is then commutative and what you need to show is that the corresponding extension splits. Assuming $L$ symmetric allows one to define a symmetric isomorphism $T_x^*L \cong L$. By this I mean that this isomorphism is preserved by $(-1)^*$ (after using the fixed identification of $L$ with $(-1)^*L$ and translating by $x$). This is well defined after squaring and should give the required splitting. $\endgroup$ – naf Jul 21 '13 at 5:59

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