Suppose $X$ is a complex abelian variety of dimension 2. Then I believe the ring of endomorphisms $\mathrm{End}(X)$, tensored with $\mathbb{C}$, is isomorphic to a subalgebra $M_2(\mathbb{C})$ of $2 \times 2$ complex matrices. This let us choose an inclusion $\mathrm{End}(X) \subset M_2(\mathbb{C})$, which lets us talk about the determinant and trace of any endomorphism $f: X \to X$. (Maybe we can also this more intrinsically, but I want to think in terms of $2 \times 2$ matrices.)

If we pick a principal polarization of $X$, I believe $\mathrm{End}(X)$ gets an involution called the Rosati involution, and I think we can choose the inclusion $\mathrm{End}(X) \subset M_2(\mathbb{C})$ so that this involution extends to the usual 'complex conjugate transpose' or 'adjoint' involution $A \mapsto A^\ast$ on $M_2(\mathbb{C})$.

I believe that using the principal polarization, we can then identify self-adjoint elements of $\mathrm{End}(X)$ with elements of the Néron-Severi group of $X$, i.e. the group of connected components of the Picard scheme of $X$, in such a way that the group structure corresponds to addition in $\mathrm{End}(X)$. If so, we can think of self-adjoint elements of $\mathrm{End}(X)$ as equivalence classes of holomorphic line bundles on $X$. Given a holomorphic line bundle $L$ on $X$, I'll write $[L]$ for the self-adjoint element of $\mathrm{End}(X)$ corresponding to its equivalence class.

If I'm wrong about any of these things please let me know! I have some questions, but I'll try to whittle them down to two:

- Does $[L]$ lie in the cone

$$ K = \lbrace A \in \mathrm{M}_2(\mathbb{C}) \vert A = A^*, \det(A) > 0, \mathrm{tr}(A) > 0 \rbrace $$

if and only if $L$ is ample?

- Any ample line bundle determines a polarization; assuming the answer to question 1) is "yes", does an ample line bundle $L$ determine a
*principal*polarization if and only if $[L]$ is not the sum $[L'] + [L'']$ for some ample line bundles $L', L''$? I'm trying to say that $[L]$ is 'minimal' in some way, but I'm not sure I've got it right.