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?

  • 3
    $\begingroup$ Under that definition, the answer is no, as the example of an abelian surface with quaternion algebra multiplication already shows. $\endgroup$
    – Will Sawin
    Apr 18, 2018 at 20:09
  • 2
    $\begingroup$ 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 $\endgroup$
    – jmc
    Apr 18, 2018 at 20:33

1 Answer 1


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.


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