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Question: Let $S$ be a 0-dimensional Shimura variety. Does $S$ necessarily admit a morphism (in the category of Shimura varieties) to $\mathcal{A}_g$ for some $g\geq 1$? Here $\mathcal{A}_g$ is the moduli space of principally polarized abelian varieties.

Motivation: Basically, I am trying to understand something for all Shimura varieties. This reduces to some statement about 0-dimensional Shimura varieties, and I wanna see if I can apply techniques from the theory of Abelian Varieties to do so.

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  • $\begingroup$ Is there any analogue of "hermitian condition" on the morphism of groups (to $Sp_n$)? If so, then there's an obstruction for the hermitian-type groups $E_6,E_7$, which admit no hermitian-type imbeddings to $Sp_n(\R)$, as Satake observed in the 1960s. Could you clarify? $\endgroup$
    – paul garrett
    Commented Dec 15, 2014 at 18:50
  • $\begingroup$ I'm afraid I don't understand your question, paul. zero dimensional Shimura varieties just come from Tori, so how would groups of type E_6,E_7 come up? I also don't know the "hermitian condition" you are referring to. Could you clarify that, please? To clarify my own question, I'm only curious about Tori and their appearance in $\mathcal{A}_g$. $\endgroup$
    – jacob
    Commented Dec 16, 2014 at 16:33
  • $\begingroup$ Ah! Ok, zero-dimensional in that sense, and no "hermitian" condition. Then I guess I don't know what condition(s) must be satisfied for a legitimate morphism "of Shimura varieties". The "hermitian condition" is about morphisms $G\to H$ over $\mathbb Q$ so that on non-compact factors over $\mathbb R$ the induced map on (hermitian) symmetric spaces $G_v/K_v$ is holomorphic. In that case, there is a "physical" induced map of complex analytic varieties... But this is evidently not what you're asking about! :) Thanks for clarifying. $\endgroup$
    – paul garrett
    Commented Dec 16, 2014 at 16:37
  • $\begingroup$ Hmm. Well, working with the definition in terms of a conjugacy class of maps h:S-->G, I just want a map T-->G which maps the conjugacy classes to each other. This will still induce a "physical" map of complex analytic varieties, but now the first space is just a finite collection of points so holomorphicity is guaranteed. Does this make sense or am I confused somehow? $\endgroup$
    – jacob
    Commented Dec 16, 2014 at 16:42
  • $\begingroup$ It certainly at-least-potentially makes sense, just as in case $G_v$ (for $v$ archimedean) might happen to be compact, the holomorphy condition is vacuous. I am not practiced with induction scenarios in which this is the correct "bottom" stage, unfortunately. A presumably bad answer to your original question would be that, sure, because $Sp_n$ contains a copy of $GL_n$ (depending on indexing...), any torus can be imbedded there (if there're really no further conditions...) I'd wager this is not what you wanted, but I don't know. $\endgroup$
    – paul garrett
    Commented Dec 16, 2014 at 16:48

1 Answer 1

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You can always map $(T,h)$ to the trivial Shimura datum and then this into the Siegel one. I assume this isn't what you want however. May I therefore modify the question to ask whether a zero dimensional variety can be embedded in a Siegel variety. I.e. You want to know if all zero dimensional Shimura varieties are of Hodge type.

I think this is certainly false and can give two quick reasons.

1) every Hodge type datum has weight defined over $\mathbb{Q}$ and has compact centre modulo the image of this weight. In general neither condition need hold for a zero dimensional Shimura datum.

2) consider the 'cyclotomic' Shimura datum $T=\mathbb{G}_m$ and $h(z) = z\bar{z}$. Any representation of T will induce a hodge structure of even weight but if it admits a Siegel embedding then this must yield a representation giving the weight 1 Hodge structure defining an abelian variety.

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