# Endomorphisms of abelian varieties with real multiplication

Let us work over $\mathbb{C}$ to make life easier.

I've came across to the following definition. Let $F$ be a totally real number field of degree $g$, with ring of integers $\mathcal{O}_F$. An abelian variety $A$ of dimension $g$ has real multiplication by $\mathcal{O}_F$ if there is a ring embedding $\mathcal{O}_F \to {\rm End}(A)$.

Question: Is it always the case that $F$ maps into the center of ${\rm End}^0(A)$? What if $A$ is simple?

In general, can the structure of ${\rm End}^0(A)$ be completely understood?

• Under that definition, the answer is no, as the example of an abelian surface with quaternion algebra multiplication already shows. Apr 18, 2018 at 20:09
• The structure of $\mathrm{End}^0(A)$, for $A$ simple, is described by the Albert classification of division algebras with involution. See Mumford's book on abelian varieties, or §12 of math.ru.nl/~bmoonen/research.html#bookabvar
– jmc
Apr 18, 2018 at 20:33

The definition of real multiplication given needs a few more assumptions. First, the Abelian variety should be polarized; second, the action of $${\cal O}_F$$ should be by self-adjoint transformations, e.g. with respect to the induced inner product on the space of 1-forms Omega(A).
A nice example is to take $$A = E \times E$$ where $$E$$ is an elliptic curve; then $$End(A)$$ contains $$M_2(Z)$$, and the symmetric matrix $$S=((2,1),(1,1))$$ gives an action of $$Z[S]$$ by real multiplication on $$A$$ such that $$S$$ is not in the center. Moreover $$Z[S]$$ is isomorphic to the maximal order in $$Q(\sqrt{5})$$, so this gives a negative answer to the Q.