The upper bound of number of the automorphism of principal polarization of abelian variety over algebraically closed field

Question

I would like to find a upper bound of principal polarization of abelian variety in the following stiution:

Suppose $A$ is an abelian variety over a $char=0$ algebraically closed field. And for any two principal polarization $\lambda_1$, $\lambda_2$:$A \to A^t$, we we say they are eqivalent if and only if there exsits an isomorphism $\sigma\in Aut(A)$ such that $\lambda_1=\sigma^t \circ\lambda_2\circ\sigma$.

Now, let $T$ be the collection of the principal polarization of $A$, is there any upper bound (which is just relative to the dimension $g$) of the cardnarity of $T$/~?

Answer 1

No, not without some extra assumptions. Take for example $A = E \times E'$ where $E$ and $E'$ are generic elliptic curves connected by an isogeny $E \to E'$ with kernel $\mathbb{Z}/m\mathbb{Z}$, for some squarefree integer $m$. For each factorization $dk = m$, you can find elliptic curves $E_k, E_d$ on $A$ which are quotients of $E$ of degree $k, d$ respectively and such that $A \simeq E_k \times E_d$. This gives you something like $2^{a-1}$ non-isomorphic (product) principal polarizations, where $a$ is the number of primes dividing $m$.

On the other hand, for a generic abelian variety, the endomorphism ring of $A$ is $\mathbb{Z}$ and the number of principal polarizations is 1.

In general, you can translate this question into a purely algebraic question about the number of conjugacy classes of elements in the endomorphism ring of A. Here, the conjugacy will be defined in a slightly weird way in terms of the Rosati involution. The answer to this algebraic question should ultimately be connected to class numbers of various commutative rings which embed in the endomorphism ring. The case I describe above roughly corresponds to $\mathbb{Z}[x]/(x^2 - m^2)$ embedding in $\mathrm{Mat}_2(\mathbb{Z})$.