4
$\begingroup$

I know that if $A$ is a simple abelian variety over a number field $k$ with all endomorphisms defined over $K$ then $\mbox{End}(A_K)\otimes \mathbb{Q}$ is a division algebra with a positive involution. Albert's classification tells us that these types of division algebras come in four types

Type I: Totally real number field

Type II/III: Central simple algebra $L$ over a totally real field such that simple components of $L\otimes \mathbb{R}$ are isomorphic to $M_2(\mathbb{R})$ or $\mathbb{H}$ (depending on Type II or III respectively)

Type IV: Central simple algebra over a CM-field.

My question is, for a fixed dimension, $g$, can we find simple abelian varieties of dimension $g$ such that it's endomorphism ring falls into each of these types?

I ask because in the paper Sato-Tate distributions and Galois endomorphism modules in genus 2 (by Fite, Kedlaya, Rotger and Sutherland), they say that for simple abelian varieties of dimension 2, then $\mbox{End}(A_K)\otimes \mathbb{R}$ can be one of

$$\mathbb{R}, \mathbb{R}\times\mathbb{R}, \mathbb{C}\times\mathbb{C}, M_2(\mathbb{R})$$

If it were true that for a fixed dimension all types appear, shouldn't it be possible to get an abelian variety with endomoprhism ring $\mathbb{H}$? If it is the case that not all types can appear for a fixed dimension, is it known when they can appear?

$\endgroup$
5
$\begingroup$

Albert's classification works/is good enough for (algebraically closed) fields of characteristic zero. For the complete list of all possibilities for a given $g$ (including the case of prime characteristic) see a survey of Frans Oort:

``Endomorphism algebras of abelian varieties". Algebraic geometry and commutative algebra, Vol. II, 469–502, Kinokuniya, Tokyo, 1988. (MR0977774 (90j:11049) ).

In particular, if $g=2$ then you cannot get type III.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.