3
$\begingroup$

The definition of a "connected Shimura datum" (as in Milne's notes) is a pair $(G, X)$, where $G$ is a reductive algebraic group and $X$ is a $G(\mathbb{R})$-conjugacy class of morphisms $$ x: \mathbb{S}^1 \to G_\mathbb{R}, $$ where $\mathbb{S}^1$ is the norm one subtorus of $\text{Res}_{\mathbb{C}/\mathbb{R}}$ satisfying a short list of axioms. Given such a morphism $x$, one gets a family of morphisms $$ x_n: \text{Res}_{\mathbb{C}/\mathbb{R}} \mu_n \to G_\mathbb{R} $$ compatible with the natural inclusions $\mu_n \hookrightarrow \mu_{mn}$, and if I'm not mistaken, by Zariski density, $x$ is determined uniquely by the $x_n$, and moreover, two maps $x$, $x'$ are $G(\mathbb{R})$ conjugate if and only if their associated families $x_n$, $x_n'$ are. It also makes sense to demand that Deligne's axioms hold for the $x_n$, and one sees that if they hold for $x$, they hold for $x_n$.

From the perspective of special points and canonical models, it is not clear to me where one uses the morphism $x$; on the level of real points, it seems that only the action of roots of unity are used, and so much of the theory should be recoverable from only the $x_n$; however, there is a lot of it that I haven't understood yet.

My questions are:

  1. if we define a generalized Shimura datum to be a family of $x_n$ compatible with the natural inclusions (equivalently, a conjugacy class of maps from the direct limit of the $\mu_n$), do we get generalized Shimura varieties?

  2. If so, are there generalized Shimura varieties which do not come from Shimura varieties?

$\endgroup$

2 Answers 2

3
$\begingroup$

Any reasonable interpetation of the first axiom (on the weights of the action on the adjoint representation) will force the homomorphisms $\mu_n \to G$ to canonically factor through $\mathbb G_m$, and so a generalized Shimura variety is just a Shimura variety.

The reason is that any representation of $\mu_\infty$ is a sum of one-dimensional representations. One-dimensional representations are classified by $\hat{\mathbb Z}$, and the ones that factor through $\mathbb Z$ are classified by $\mathbb Z$. The ones appearing in the adjoint representation lie in $\{-1,0,+1\}$, hence in $\mathbb Z$. Since a tensor power of any representation is contained in a tensor power of the adjoint representation, a power of any character appearing in the representation will be classified by $\mathbb Z$, so every character will be classified by an element of $\mathbb Q$, but $\mathbb Q \cap \hat{\mathbb Z}= \mathbb Z$.

$\endgroup$
1
  • $\begingroup$ Thank you Will. I should maybe hint the context of my naive question. I was playing with these roots, and (re)proved the existence of Special poins in Shimura varieties. $\endgroup$
    – Tirelda
    Commented Mar 28, 2018 at 21:19
2
$\begingroup$

The axioms imply that $X$ has a natural structure of a hermitian symmetric domain, and hence the quotients are algebraic varieties in a natural way. Moreover, the homomorphisms $x$ have a natural interpretation in terms of the complex structure on the tangent space at the point of the hermitian symmetric domain. If you only take a projective system of homomorphisms, I don't see how you will get an algebraic variety in any natural way.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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