Fix a CM-field $K$ of degree $2g$. Let $A$ be a polarized abelian variety of dimension $n$ over $\mathbb{C}$, with an isomorphism $\theta : K \to End_{\mathbb{C}}(A) \otimes_{\mathbb{Z}} \mathbb{Q}$. (So $n$ is a multiple of $g$.)

Then, the tangent space to the identity of $A$ defines an $n$-dimensional complex representation $\Phi$ of $K$ (which Shimura calls the type of $(A, \theta)$). Write

$\tau_1, \tau_2, \ldots, \tau_g, \rho \tau_1, \rho \tau_2, \ldots, \rho \tau_g$

for the different embeddings of $K$ into $\mathbb{C}$, where $\rho$ denotes complex complex conjugation. Then (as shown by Shimura), the representation $\Phi$ decomposes as a direct sum

$\Phi = \bigoplus_{\nu = 1}^g (r_\nu \cdot \tau_\nu \oplus s_\nu \cdot \rho \tau_\nu)$ where $r_1 + s_1 = r_2 + s_2 = \cdots = r_g + s_g = \frac{n}{g}$.

What are the possible values of $(r_1, s_1, r_2, s_2, \ldots, r_g, s_g)$ as $A$ ranges over all complex abelian varieties, while $K$ remains fixed? Can they be arbitrary nonnegative integers satisfying the above constraint, or does fixing $K$ impose further conditions?


3 Answers 3


There are certain additional conditions on ``multiplicities" that are spelled out in the original Shimura's paper. Let me discuss a couple of interesting cases.

The case of CM type, i.e. when $n=g$. Let $\Phi$ be the corresponding CM type, which is a $n$-element set of embeddings of $K$ into the field $C$ of complex numbers. For each pair of complex-conjugate embeddings $K \to C$ exactly one of them lies in $\Phi$. If $K$ does not contain a proper CM subfield then everything is fine, i.e., the endomorphism algebra of $A$ is $K$. However, if $K$ contains a proper CM subfield $L$ then a ``bad" choice of $\Phi$ would imply that the endomorphism algebra of $A$ is a matrix algebra over $L$ rather than $K$. In order to guarantee that the endomorphism algebra is $K$, one should require that if $\tau_1, \tau_2 \in \Phi$ and their restrictions to the maximal totally real subfield $L_0$ of $L$ coincide then their restrictions to $L$ also coincide. (Of course, one should require it for all CM subfields.) For example, if $n=g=2$ and a quartic CM field $K$ contains an imaginary quadratic subfield then $K$ must contain another imaginary quadratic subfield as well! In this case one may check that the endomorphism algebra of $A$ is always bigger than $K$ (and $A$ is isogenous to a square of a CM elliptic curve).

The case of an abelian surface $A$ and an imaginary quadratic field $K$, i.e., $g=1, n=2$. It is known that the endomorphism algebra of an abelian surface is never an imaginary quadratic field. (Actually, as was pointed out by Frans Oort, this assertion remains true in characteristic p.) More precisely, suppose we are given an embedding of $K$ in the endomorphism algebra of $A$. Then either $A$ is isogenous to a square of an elliptic curve with multiplication by $K$ or the endomorphism algebra of $A$ is an indefinite quaternion algebra over the rationals.


A CM type for K is a choice of one out of every pair of complex conjugate embeddings of K. For each CM type on K, there is an abelian variety of that type. This is a complex abelian variety B of dimension g with an action of K (rather, its integers) such that K acts on the tangent space of B through the g chosen complex embeddings. Now you can take A to be a product of B's corresponding to whichever CM types you choose. That certainly gives you a lot of possibilities for the multiplicities.

Added: In fact, that gives all multiplicities subject to the obvious condition that the multiplicity of an embedding and its conjugate add to n/g. However, the questioner points out that he wants an abelian variety A such that $K=End(A)\otimes Q$. Take an A as constructed in the first paragraph. Then $V=H_1(A,Q)$ is a K-vector space with a Hodge structure and a compatible Riemann form, which we can use to define a Shimura variety (= moduli variety). The problem now is to prove that (as expected), the family of abelian varieties on this variety contains one satisfying the condition. For this check that each abelian variety with endomorphism algebra strictly larger than K lies in a (one of countably many) Shimura subvarieties of strictly lower dimension.

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    $\begingroup$ It's more accurate to say that there is a (unique) abelian variety up to isogeny of that type. This also clarifies your comment about there being 'an action of $K$' on $B$. $\endgroup$ Feb 1, 2011 at 5:43
  • $\begingroup$ Such an abelian variety doesn't satisfy the hypothesis above since $\theta : K \to End_{\mathbb{C}}(A) \otimes_{\mathbb{Z}} \mathbb{Q}$ will be an inclusion, but will not necessarily be an isomorphism. (The condition of it being an isomorphism implies that A must be simple.) $\endgroup$ Feb 1, 2011 at 14:27
  • $\begingroup$ Dear mephisto--Could you elaborate a bit on the punchline? What would be the minimally painful way to formalize it? $\endgroup$ Feb 1, 2011 at 16:12
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    $\begingroup$ There are some cases in which the family fails to contain a variety with the desired endomorphism algebra "as expected" - see Yuri Zarhin's answer. These cases are determined in Shimura's paper. This is also done in Birkenhake-Lange IIRC. $\endgroup$
    – Martin Orr
    Feb 2, 2011 at 0:23

I posted the following answer yesterday after only a quick skim of the question. When I read it with more care, it seemed to me to be the answer to a different question entirely. After having looked at some of the other answers, I find myself less sure of whether my answer is relevant or not! Anyway, here it is: make of it what you will.

I believe that a complete account of the Albert-Shimura classification of endomorphism algebras of complex abelian varieties may be found in Chapter 9 of Birkenhake and Lange (Complex Abelian Varieties).

As someone who has worked in the area of abelian varieties with endomorphism structure, I have always found the complete classification to be rather daunting. One particularly subtle point is that after you set up a moduli space of abelian varieties having endomorphisms at least by a certain order in a simple algebra, you are still left with the task of showing that a sufficiently general element of the family has precisely that order as the ring of endomorphisms and not something larger. But the discussion in Section 9.9 (I am looking at the second edition) successfully negotiates this subtlety and does, I think, in fact give a complete classification.

Good luck!


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