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Let $Y$ be an abelian surface. Is it true that for every general point $P \in Y$, there exists an elliptic curve passing through $P$?

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So as explained by Sandor, if $Y$ is a simple abelian surface, the anwser is no. If $Y$ contains an elliptic curve $E$, then by any point $y\in Y$, it passes a curve $y+E$ of genus $1$. If we want a real elliptic curve (abelian subvariety) passing through a general $P$, then it is easy to see that $Y$ is isogeneous to the square of an elliptic curve. I think that even when $Y=E^2$, the answer is no for a general $P$. –  Qing Liu Nov 3 '11 at 20:20
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@Qing Liu: your last statement is true at least over the complex numbers. An abelian variety has at most countably many subvarieties, hence by Baire's theorem the union of all the abelian subvarieties cannot be all the space. –  rita Nov 3 '11 at 20:49
    
@Rita: the same arguments hold over uncountable fields. –  Qing Liu Nov 3 '11 at 21:07
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@rita: "An abelian variety has at most countably many subvarieties" - I'm missing something: what about the translates of a fixed subvariety? –  Qfwfq Nov 4 '11 at 12:01
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Qfwfq: insert the word "abelian" before the word "subvarieties" in the quote. –  Artie Prendergast-Smith Nov 4 '11 at 12:41
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3 Answers 3

up vote 13 down vote accepted

In general, if an abelian variety $A$ contains an abelian subvariety $B\subseteq A$, then $A$ contains another abelian subvariety $B'\subseteq A$ such that $A$ is isogenous to $B\times B'$. This is Poincaré's reducibility theorem. (See also Poincaré's complete reducibility theorem, same book, next page).

An abelian variety is called simple if it does not contain any nontrivial abelian subvarieties. Simon's argument shows that there exist simple complex tori of dimension 2. One could also count moduli and conclude that not every abelian surface (or abelian variety of arbitrary dimension $>1$) can be isogenous to a product.

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@anon: If you don't say why $\mathrm{End}(A)=\mathbb Z$, then this statement is as good as saying that "a general abelian variety is simple". –  Sándor Kovács Nov 4 '11 at 4:01
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@anon: It is not true that on an open subset of moduli space $End(A) = \mathbb{Z}$; for example, in characteristic zero, abelian varieties isogenous to a fixed abelian variety form a dense subset. –  ulrich Nov 4 '11 at 12:38
    
@anon: What do you mean by "the correct condition"? The question was about elliptic curves on surfaces. We're not "working", we're "discussing". –  Sándor Kovács Nov 4 '11 at 17:27
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No. In general, there are no elliptic curves on an Abelian surface.

Thinking in terms of lattices, if there is an elliptic curve on $A$, then there is a rank 2 sublattice of $\Lambda$ (Where $A = \mathbb{C}^2/\Lambda$) which is invariant under multiplication by $\sqrt{-1}$. But it is easy to see that there are many rank four lattices in $\mathbb{C}^2$ for which this is not true.

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Thank you... it was as I suspected! –  fds Nov 3 '11 at 17:06
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Of course, most lattices do not give abelian varieties. –  Angelo Nov 3 '11 at 17:41
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Note that this is true even if $A$ is algebraic (which, unlike the case for elliptic curves, is not the case for the quotient of ${\bf C}^2$ by a generic lattice). –  Noam D. Elkies Nov 3 '11 at 17:42
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You may have a look to:

Ernst Kani, Elliptic curves on Abelian surfaces

(the credit for this reference goes to Dan Petersen who already suggested it in a comment to this question)

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