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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:

  1. 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?

  1. 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.
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The usual way of presenting elements of the Neron-Severi group of an abelian variety is as Hermitian forms on the tangent space at the identity (subject to the condition that the imaginary part of the pairing of any two elements of the fundamental group is an integer). A line bundle is ample if and only if it corresponds to a positive definite Hermitian form. (see Mumford, Abelian Varieties, Theorem of Appell-Humbert on p. 20 and Theorem of Lefschetz on p. 29-30.)

Since you want to fix a basis for this space and work explicitly with $2\times 2$ matrices, we may as well choose a basis for which the Hermitian form associated to the fixed principal polarization is the standard Hermitian form on $\mathbb C^2$. (We can always do this since the Hermitian form is positive definite).

Having done this, the matrix of the endomorphism is the usual Hermitian matrix associated to the Hermitian form. (Furthermore the Rosatti involution will just be the restriction of the conjugate transpose to the subalgebra of endomorphisms.)

Thus your first question has a positive answer, because of the classical fact that a $2\times 2$ Hermitian matrix is positive definite if and only if its trace and determinant are positive.

I believe your second question has a negative answer. The right criterion to be principal is that the determinant is equal to $1$. The norm-squared of the determinant gives the degree of the corresponding endomorphism, the polarization is principal if the degree is $1$, and Hermitian positive definite matrices have real positive determinants so the only possibility with norm-squared $1$ is $1$.

Thus your question has a positive answer if and only if all Rosati-invariant positive-definite elements of the endomorphism algebra can be expressed as sums of Rosati-invariant positive-definite elements with norm $1$. This will not happen if, e.g. the endomorphism algebra is the set of integral elements in a typical real quadratic field, so that the Rosati involution is trivial, an element is positive definite if and only if it and its conjugates are both positive, and the determinant is the norm so the norm one elements are units. It is not true in general that the ring of integers of a real quadratic field is generated by units.

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    $\begingroup$ For a simple counter-example to 2., take $X=E\times F$, where $E,F$ are elliptic curves. Let $e\in E$, $f\in F$; the divisor $\{e\} \times F+2E\times \{f\} $ is not a sum of 2 ample divisors, but does not define a principal polarization. $\endgroup$
    – abx
    Apr 24 at 14:54
  • $\begingroup$ @abx Nice example! When I first looked at the question, the abelian variety was required to be simple, so that wasn't relevant, but it has been changed and this is no longer the case. $\endgroup$
    – Will Sawin
    Apr 24 at 15:05
  • $\begingroup$ Thanks so much! I'm mainly interested in the simple case, but I realized most of what I was saying didn't make use of that assumption so I dropped it. "The endomorphism algebra is a typical element of a real quadratic field" doesn't parse so a typo must have snuck in. Btw, the det = 1 criterion for principalness suits my needs much better than the lame one I gave (even if it were correct). $\endgroup$
    – John Baez
    Apr 24 at 16:33
  • $\begingroup$ @JohnBaez You're welcome! I fixed that sentence. $\endgroup$
    – Will Sawin
    Apr 24 at 17:14

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