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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 $P$?

flag	asked Nov 3 2011 at 16:00 fds 217●4

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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 2011 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 2011 at 20:49

@Rita: the same arguments hold over uncountable fields. – Qing Liu Nov 3 2011 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 2011 at 12:01

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Qfwfq: insert the word "abelian" before the word "subvarieties" in the quote. – Artie Prendergast-Smith Nov 4 2011 at 12:41

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Sándor Kovács · Accepted Answer · 2011-11-03 17:04:32Z UTC

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.

Simon Rose · Answer 2 · 2011-11-03 16:22:13Z UTC

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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.

edited Nov 3 2011 at 16:22

answered Nov 3 2011 at 16:15

Simon Rose
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Thank you... it was as I suspected! – fds Nov 3 2011 at 17:06

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Of course, most lattices do not give abelian varieties. – Angelo Nov 3 2011 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 2011 at 17:42

Qfwfq · Answer 3 · 2011-11-04 11:56:36Z UTC

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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)

answered Nov 4 2011 at 11:56

Qfwfq
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Elliptic curves on abelian surface

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