# isomorphism of extensions by abelian varieties

Let $A$ and $G$ be abelian varieties over $\mathbb{C}$. An element $P$ of $\text{Ext}(A, G)$ is an exact sequence

$0 \to G \to P \to A \to 0$,

here one can give $P$ the structure of an abelian variety. And $P$ can be viewed as principal $G$-bundle over $A$. In general, is there a way to determine if two given elements $P$ and $P'$ in $\text{Ext}(A, G)$ are isomorphic?

In particular, assume that $A$ is an elliptic curve. Given two extensions $0 \to G \to P \to A \to 0$ and $0 \to G \to P' \to A \to 0$ with morphisms $g : G \to G$, $f : P \to P'$, and $h : A \to A$ so that the resulting diagram commutes, $g$ is an isomorphism, and $h$ is an isogeny with kernel $(\mathbb{Z}/n\mathbb{Z})^2$. Can we show that $P$ and $P'$ are isomorphic or is there an example otherwise?

-
In your example in the last paragraph, wouldn't you expect only an isogeny? Take for instance $P$ to be the trivial extension. If $h$ is an isogeny, so should be $f$. Right? –  Sándor Kovács Mar 9 '11 at 5:56
...I mean that if $h$ is not an isomorphism, then neither is $f$. –  Sándor Kovács Mar 9 '11 at 5:56
In the example, $f$ is an isogeny with kernel $(\mathbb{Z}/n\mathbb{Z})^2$. I just wonder if there is a way to construct an isomorphism between $P$ and $P'$ (this isomorphism does not necessarily fit into the given diagram with $g$ and $h$). –  Tuan Mar 9 '11 at 16:00

The answer to your second question is no. Take $E_1,E_2$ non-isogenous elliptic curves without complex endomorphisms. Take $P_1,P_2$ 5-torsion points in $E_1,E_2$ resp. Now let $P=E_1\oplus E_2/(P_1-P_2)$ and $P'=E_1\oplus E_2/(P_1-2P_2)$. Now, there is clearly a map of the type you described with $n=2$ descending from the endomorphism of $E_1\oplus E_2$ which is multiplicatino by 1 on $E_1$ and $2$ on $E_2$.
However, $P$ and $P'$ are not isomorphic: as there are only $2$ injections from each of $E_1,E_2$ to each of $P,P'$, they can each only be written as 5-quotients of $E_1\oplus E_2$ in essentially one way.
In general, in your setup the class of $P'$ is $n$ times the class of $P$ in $\textrm{Ext}(A,G)$. This is because $P$ is the fiber product of $P$ with $A$ over $A$ under the multiplication by $n$ map. Thus, any class of $\textrm{Ext}(A,G)$ which is not $(n-1)$-torsion OR $(n+1)$ - torsion, and there are lots of them, will yield a distinct $P$ from $P'$. You have to include $(n+1)$ since the $-1$ automorphism flips the Ext class.
Of course, if $A,G$ are isogenous and have automorphisms you have to be a little more careful, as you might have distinct elements of $\textrm{Ext}(A,G)$ which are abstractly isomorphic.