Does the self-product of a $g$-dimensional abelian variety contain an abelian variety of dimension smaller than $g$ at some point - MathOverflow

Does the self-product of a $g$-dimensional abelian variety contain an abelian variety of dimension smaller than $g$ at some point

2012-08-01T15:10:53Z

Let me be more precise than the title. (This will be my last attempt to do something with abelian varieties. Sorry for all the basic questions. The answers have been great!)

Let $A$ be a simple abelian variety over a field $k$. Let $g\geq 2$ be the dimension of $A$.

Does there exist an integer $n\geq 1$ such that $A^n = A\times_k A\ldots\times_k A$ contains an abelian variety of dimension less than $g$?

It suffices to prove that $A^n$ contains a curve of genus strictly smaller than $ g$ for some $n\geq 1$.

I'm afraid that this is not true. In fact, if $B\subset A^n$, then $B$ is isogenous to $A^m$ probably. Therefore, $\dim B =mg$. I'm just asking to be sure.

Answer by Damian Rössler for Does the self-product of a $g$-dimensional abelian variety contain an abelian variety of dimension smaller than $g$ at some point

2012-08-01T15:40:38Z

No (I suppose that $k$ is algebraically closed). This is because Poincaré's complete reducibility theorem contains a unicity statement for the intervening factors (up to isogeny). See Mumford, Abelian varieties, p. 173-174.

Answer by Donu Arapura for Does the self-product of a $g$-dimensional abelian variety contain an abelian variety of dimension smaller than $g$ at some point

2012-08-01T18:54:47Z

In fact, it is no for completely elementary reasons. If $A$ is simple and $B\subset A^n$ is an abelian variety with $\dim B < g$, then $Hom(B,A^n)=Hom(B,A)^n$ is necessarily zero. So $B=0$.