What is wrong with this modification of the definition of Shimura datum?

Question

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?

Answer 1

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 $\mathbb Q \cap \hat{\mathbb Z}= \mathbb Z$.

Answer 2

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.