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Suppose we have a principally polarized abelian variety X over the complex number field. Given two ample, effective divisors D_1, D_2 such that the global sections of both line bundles are 1 dimensional vector spaces. If D_1 and D_2 have the same first Chern class, then is it true that D_1 and D_2 differ by a translation, i.e., there is x in X such that D_1 + x = D_2 as divisors? Why or why not?

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Is that a homework problem? Why not read Mumford's "Abelian varieties" or, really, any book on abelian varieties? –  VA. Apr 3 '10 at 4:25
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@Theo: I'm always a bit ambivalent about comments such as yours. I can see your point on one level, but on another level, if you don't what a ppav is then probably you're not going to be answering this question. I could take any question about functional analysis at MO and 'complain' that some of the terms are unknown to me, but even if they were explained to me I wouldn't be able to answer the question. –  Kevin Buzzard Apr 3 '10 at 8:02
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Writing things out have other benefits: helping people who don't know that much but interested to learn more easily, making it easier to find the relevant questions when you search. Plus more carefully written questions are more likely to get answers and also sort of morally force the responders to put in more efforts as well, etc. –  Hailong Dao Apr 3 '10 at 14:06
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I'm going to agree with both Kevin and Hailong. For the asker to write all that out won't really make it more likely that say, Theo (who doesn't know ppav's) will answer, but it makes the asker think things through a bit, and it also establishes the common notation, as to which definitions are the ones that the asker thinks we should use. –  Charles Siegel Apr 3 '10 at 15:21
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Questions on MO are not just written for the people who might answer, they're also written for people searching on the internet. Defining an "effective divisor" isn't necessary in my opinion, though I would encourage it; its definition is easily found on Wikipedia. On the other hand, it's rude to use an acronym if you can't find out what it means by Googling. There's no math results (other than this question) for PPAV in the first 3 pages of Google hits. I've gone ahead and edited the question to expand the acronym. –  Ben Webster Apr 4 '10 at 0:16
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1 Answer

This is true. For $A_{/\mathbb{C}}$ an abelian variety, $L$ an ample line bundle on $A$, then any line bundle $M \in \operatorname{Pic}^0(A)$ -- over $\mathbb{C}$, this is equivalent to having first Chern class zero -- is of the form $T_x^{*} L \otimes L^{-1}$ for some $x \in A$. (e.g. Theorem 1 on p. 77 of Mumford's Abelian Varieties).

Applying this theorem with $L = L(D_2)$, $M = L(D_1) - L(D_2)$, we get that

$L_1 - L_2 = T_x^*(L_2) - L_2$, so

$L_1 = T_x^*(L_2)$.

So $D_1$ and $x+D_2$ (meaning translation of $D_2$ by $x$!) must be linearly equivalent, but by your assumption $h^0(L(D_1)) = h^0(L(D_2)) = 1$, they are each the unique effective divisors in their linear equivalence classes, so we must have $D_1 = x + D_2$.

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