Generic Mumford Tate group and algebraic points

Question

I will stick with a concrete example for this question, but it should probably be cast in a more general framework.

Let $Sym_g(\mathbf{C})$ be the space of symmetric matrices of order $g$ with coefficients in $\mathbf{C}$, and consider the Siegel upper half-space $\mathcal{H}_g\subset Sym_g(\mathbf{C})$.

First question. I've read that the Mumford-Tate group of a 'generic' principally polarized complex abelian variety is $GSp_{2g}$. What does 'generic' means here? For instance, how is this genericity formulated in terms of $\mathcal{H}_g$?

Let's say that a point in $\mathcal{H}_g$ is generic if the corresponding principally polarized abelian variety has the above property.

Second question. If $g=1$, the 'non-generic' points in $\mathcal{H}_1\subset \mathbf{C}$ are quadratic imaginary. In particular, there are many generic $z\in \mathcal{H}_1\cap \overline{\mathbf{Q}}$. Does this also holds for general $g$, i.e., the existence of many generic $Z \in \mathcal{H}_g\cap Sym_g(\overline{\mathbf{Q}})$?

Answer 1

For Question 1, there is a countable union of proper analytic subsets in $\mathcal{H}_g$ (so meagre in the sense of Baire category) such that anything in the complement has Mumford-Tate $GSp_{2g}$. Unfortunately, it's hard to make a more precise statement.

Your formulation of Question 2 is a bit off, you probably want to ask whether there are many "generic" points in $A_g(\overline{\mathbb{Q}})$, where $A_g(\mathbb{C})= \mathcal{H}_g/Sp_{2g}(\mathbb{Z})$. It think this is true, although I don't have a reference right now. Let me add that the relationship with the Mumford-Tate group and the endomorphism algebra of an abelian variety is more complicated in higher dimensions. (Added (1) I understand now that the OP's question 2 was as intended as written. Will Sawin's answer seems to be the way deal with this. (2) There is more than one convention about what the Mumford-Tate group means, depending on it either $SP_{2g}$ or $GSp_{2g}$ would give the generic group.)

Answer 2

It's possible to be more precise. The non-generic points form a countable family of subsets defined by algebraic equations of bounded degree.

To see this, we use Larsen's alternative - if the Mumford-Tate group is not $Sp_{2g}$, then the action on the adjoint representation of $Sp_{2g}$ must be reducible (as the Lie algebra of the Mumford-Tate group is a nontrivial proper subspace - nontriviality because it contains at least a torus.)

To check for irreducible components of the adjoint representation, we first construct its Hodge structure. This is given by taking the natural map $\mathbb Q^{2g} \to \mathbb C^g$, and then applying $\operatorname{Sym}^2$, obtaining $\mathbb Q^{g(2g+1)} \to \mathbb C^{g(g+1)/2}$. This defines a Hodge structure whose $H^{2,0}$ is that $\mathbb C^{g(g+1)/2}$, whose $H^{0,2}$ is its complex conjugate, and whose $H^{1,1}$ is the remainder. Because this Hodge structure is necessarily semisimple, it is reducible if and only if there is an idempotent projector on the rational vector space which preserves the Hodge structure, or equivalently an idempotent projector that preserves the kernel of this map tensored up to $\mathbb C$ (and its complex conjugate, but that's the same).

For each possible rational idempotent projector, this defines a set of algebraic equations in the Siegel coordinates. By construction, these algebraic equations are of degree bounded in terms of $g$.

Hence indeed we can find sufficiently generic algebraic numbers that don't satisfy them - indeed it is possible to find algebraic numbers such that all monomials of degree at most $n$ in the numbers are linearly independent over $\mathbb Q$, for any $n$.