This question may be too naive, in which case I apologise in advance. Anyway, it is a well-known fact (see e.g. Milne's notes) that any abelian variety A has only finitely many direct factors up to automorphisms of A. (Here a direct factor of A is an abelian subvariety B for which there exists another abelian subvariety C of A such that $A \cong B \times C$.)

My question is: how much is known about the corresponding question for arbitrary abelian subvarieties, rather than direct factors? That is, is it known whether every abelian variety A has finitely many abelian subvarieties, up to automorphisms of A? If not, what's the best known result in this direction?

I've asked a couple of people about this, and their opinion seems to be that it's "more or less" known. But I would like something a little more concrete, if possible. Any relevant references would be appreciated!

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    $\begingroup$ Over the complex numbers this is just a purely elementary linear algebra question in disguise. Did you try thinking about it this way? $\endgroup$ Dec 3, 2010 at 16:21
  • $\begingroup$ According to Poincaré's Complete reducibility theorem every abelian variety is isogenous to the product of simple abelian varieties with the reasonably expected uniqueness condition on the factors. You should be able to find this theorem in any standard book on abelian varieties, say, Mumford's or Birkenhake-Lange. (I moved this here since it is more of a comment than an answer and it certainly was not meant to be an answer.) $\endgroup$ Dec 3, 2010 at 19:56
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    $\begingroup$ Here is a copy of BCnrd's comment that would probably disappear with the deletion of the answer that he made the comment to. So this is a comment to the above comment: $$\quad $$ Since the question is sensitive to the distinction between End(A) and End0(A), even reducing the problem to the isotypic cases seems an unwise strategy. But the Poincare reducibility theorem does show that the abelian subvarieties are the images of endomorphisms, so the problem reduces to a question about orders in finite-dimensional semisimple Q-algebras (and more specifically, Albert algebras). – BCnrd 1 hour ago $\endgroup$ Dec 3, 2010 at 19:58
  • $\begingroup$ Thanks for all the helpful comments. @BCnrd: I got that far, but the sticking point was that it seems necessary to know that the endomorphisms involved have bounded eigenvalues. I don't see how to get that yet, so I'd better think some more. $\endgroup$
    – user5117
    Dec 6, 2010 at 8:44

1 Answer 1


For the benefit of others who might look at this question, let me mention that I found the following reference proving exactly what I wanted. (More precisely, I was told about it by David Ploog.)

Lenstra, H; Oort, F; Zarhin, Yu. Abelian subvarieties. J. Algebra 180 (1996), no. 2, 513–516.

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    $\begingroup$ math.leidenuniv.nl/~hwl/PUBLICATIONS/1996b/art.pdf $\endgroup$ Dec 13, 2010 at 9:56
  • $\begingroup$ Now, Kani shows in (mast.queensu.ca/~kani/papers/hum-msm.pdf) a result that was already well known, that if an abelian surface contains at least 3 elliptic subgroups (i.e. abelian subvarieties), then it contains infinitely many. $\endgroup$
    – rfauffar
    Mar 11, 2013 at 18:51
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    $\begingroup$ An example of this would be the self product of an elliptic curve $E$; just take $E_n:=\{(z,nz):z\in E\}$ in $E^2$; this gives an abelian subvariety for every $n$. $\endgroup$
    – rfauffar
    Mar 11, 2013 at 22:31

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