Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Fix a CM-field $K$ of degree $2g$, and a natural number $n$ which is a multiple of $g$. 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. Let $\Phi$ be an $n$-dimensional complex representation of $K$ in the form

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

Consider the moduli space of pairs $(A, \theta)$ where $A$ is an $n$-dimensional abelian variety and $\theta : K \hookrightarrow End(A) \otimes_{\mathbb{Z}} \mathbb{Q}$ is an injection such that the $n$-dimensional complex representation of $K$ defined by the tangent space to the identity of $A$ is isomorphic to $\Phi$.

In terms of $(r_1, s_1, \ldots, r_g, s_g)$, what is the minimal field over which this moduli space is defined? Moreover, what is the minimal extension of this field over which all of its irreducible components are defined?

EDIT: As pointed out by Kevin Buzzard below, what I really want is the minimal field not just where the moduli space is defined, but where the corresponding moduli functor is also defined. Moreover (as pointed out by Keerthi below), we should probably also fix a polarization of $A$, in order to guarantee that this moduli space exists.

share|improve this question
2  
Aare you really sure you want to ask about the minimal field over which the space is defined, rather than the minimal field over which the functor is defined? Say I have a natural moduli problem which only makes sense for schemes over $F$, $F$ some number field (e.g. it mentions something like "scheme plus action of $F$ plus blah such that the two induced $F$-actions on (something) agree" (one action coming from the fact that the scheme is defined over $F$). Say this problem happens to be represented by projective 1-space over $F$. Then the functor is defined over $F$ but... –  Kevin Buzzard Feb 12 '11 at 9:38
2  
...but the space is defined over the rationals, and indeed there will probably be several non-isomorphic spaces over the rationals whose pullback to $F$ give projective 1-space, and there is no way of determining which, if any, is the "right" model for the space---until you have extended the functor. –  Kevin Buzzard Feb 12 '11 at 9:40
    
By the way, you might want to impose the additional structure of a polarization to ensure representability. In this case, I would think the minimal field of definition of the moduli problem is the minimal field to which the isomorphism class of the representation of $K$ (as a semi-simple $\mathbb{Q}$-algebra) given over $\mathbb{C}$ by $(r_i,s_i)$ descends. One way to find it is to take the number field generated by the traces of the elements of $K$ with respect to this representation. –  Keerthi Madapusi Pera Feb 12 '11 at 17:18
    
Call this field $F$: then one can make sense of a moduli problem of triples $(A,\lambda,\theta)$ over an $F$-algebra $R$, where $A$ is an ab. sch. over $R$, $\lambda$ a pol. and $\theta$ a $K$-action on the tangent space making it isomorphic as a $K\otimes_QR$-algebra to the one above. –  Keerthi Madapusi Pera Feb 12 '11 at 17:20
    
Sorry, $\theta$ a $K$-action on the isogeny class of $A$, making the tangent space of $A$... –  Keerthi Madapusi Pera Feb 12 '11 at 17:21

1 Answer 1

This problem is solved completely (and in significantly more generality) in a paper of Deligne (Travaux de Shimura, in Seminaire Bourbaki 1971).

share|improve this answer
    
Actually, before that, it was already in the papers of Shimura. –  mephisto Feb 28 '11 at 23:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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