# different Shimura data with common underlying group?

A pure Shimura datum is of the form $(G,X)$ with $G$ a connected reductive $\mathbb{Q}$-group, and $X$ a homogeneous space under $G(\mathbb{R})$, subject to Deligne's conditions in terms of Hodge types, Cartan involutions, etc. cf.Deligne, "Varietes de Shimura: interpretation modulaire, et techniques de constructions de modeles canoniques"

One may ask to a given connected reductive $\mathbb{Q}$-group, how many pure Shimura data could one obtain of the form $(G,X)$. As remarked by M.Borovoi, if $G$ is a compact $\mathbb{Q}$-torus, then any homomorphism $h:\mathbb{C}^\times/\mathbb{R}^\times\rightarrow G_{\mathbb{R}}$ defines a pure Shimura datum $(G,\{h\})$. Thus for a finiteness answer, one should restrict to the case where $G$ is semi-simple.

For simplicity one even assume that $G$ is of adjoint type. Then as is pointed out in Deligne's article, the existence of $X$ is characterized by the special nodes in the Dynkin diagram of $G_\mathbb{C}$ (plus certain condition so that the node gives rise to an $\mathbb{R}$-homomorphism $\mathbb{S}\rightarrow G_\mathbb{R}$. For the adjoint $G$ chosen, there are at most finitely many special nodes (possibly empty in certain cases). And thus the finiteness is clear.

My first question is: for a given adjoint $\mathbb{Q}$-group $G$ and some pure shimura datum $(G,X)$, how many Shimura data can one obtain to be of the form $(G,X')$ (with the same $G$)$up to isomorphism? When is it unique? (added: M.Borovoi has answered this in detail, see below). Secondly, what about Shimura subdatum in a fixed$(G,X)$can one find to share the common underlying$\mathbb{Q}$-group? By a Shimura subdatum is meant a pair$(G_1,X_1)$where$G_1$is a$\mathbb{Q}$-subgroup,$X_1$a$G_1(\mathbb{R})$-orbit in$X$such that$(G_1,X_1)$is a Shimura datum itself. it is also easy to check that to obtain such a aubdatum, it suffices to (1) find a point$x$in$X$such that the corresponding homomorphism$h:\mathbb{S}\rightarrow G_\mathbb{R}$has image in$G_{1, \mathbb{R}}$. And my second question is: if a connected reductive$\mathbb{Q}$-subgroup$G'$of$G$is given, how many subdatum of$(G,X)$can one find to be of the form$(G',X')$? If there are such subdata, are they unique up to isomorphism (or isomorphism induced by inner automorphism of$G$)? thanks a lot! - for the second question, the case when$G'=T$being a$\mathbb{Q}$-torus is different. Say one has a point$x$in$X$such that$x(\mathbb{S})\subset T_\mathbb{R}$for some$\mathbb{Q}$-torus$T$in$G$, which is also minimal for such inclusions, then$(T,x)$is necessarily unique up to isomorphism: if$(T,x')$is a second subdatum , then$x$is conjugate to$x'$by some$g\in G(\mathbb{R})$, and$g$should stabilize$T$under the assumption of existence. – genshin Feb 25 '11 at 13:36 add comment ## 1 Answer The question is essentially about${\mathbf{R}}$-groups, so we shall assume that$G$is defined over$\mathbf{R}$. It is not true that for a given connected reductive${\mathbf{R}}$-group$G$, there are at most finitely many Shimura data of the form$(G,X)$up to isomorphism. Indeed, assume that$G$is a compact (i.e. anisotropic)${\mathbf{R}}$-torus. Then for any homomorphism$h\colon \mathbf{C}^* / \mathbf{R}^* \to G$, the pair$(G,h)$is a Shimura datum. Now assume that$G$is adjoint and simple. I assume that the question is about classification of Shimura data$(G,X)$up to isomorphism which is the identity on$G$. The possible Shimura data$(G,X)$are described in Section 1.2 of the paper: P. Deligne, Variétés de Shimura: interprétation modulaire, et techniques de construction de modèles canoniques. Automorphic forms, representations and$L$-functions, Part 2, pp. 247–289, Proc. Sympos. Pure Math., XXXIII, Amer. Math. Soc., Providence, R.I., 1979. To a$G({\mathbf{R}})$-conjugacy class$X$of homomorphisms$h \colon \mathbf{C}^* / \mathbf{R}^* \to G$satisfying conditions 1.2.1(i),(ii),(iii) of the paper, Deligne associates a special vertex$s=s(X)$of the Dynkin diagram$D$of$G_{\mathbf{C}}$. In Proposition 1.2.7 he says that the set of all such homomorphisms has two connected components, interchanged by$h\mapsto h^{-1}$. In Corollary 1.2.8 he says: (i) If$s(X)$is not fixed by the opposition involution, then$G({\mathbf{R}})$and$X$are connected (so I conclude that there exist two such conjugacy classes). (ii) If$s(X)$is fixed by the opposition involution, then$G({\mathbf{R}})$and$X$have two connected components (so I conclude that there is only one conjugacy class). Table 1.3.9 shows that there is one conjugacy class for the following adjoint${\mathbf{R}}$-groups of Hermitian type:$PSU(p,p)$,$B_l$,$C_l$,$D_l^{\mathbf{R}}$,$D_{2l}^{\mathbf{H}}$,$E_7$. Note that for$D_{2l}^{\mathbf{H}}$the opposition involution is trivial, see e.g. A.L. Onishchik and E.B. Vinberg, Lie Groups and Algebraic Groups, Springer-Verlag 1990, Table 3 on page 298. EDIT: I answer the second part of the question. Let$G=PGL_{2,\mathbf{R}}$, and let$T\subset G$be a compact maximal torus. There exists exactly one conjugacy class$X$of homomorphisms satisfying conditions 1.2.1(i),(ii),(iii) of Deligne's paper. Let$i \colon T \hookrightarrow G$be the inclusion homomorphism. There exists a homomorphism$h \colon \mathbf{C}^* / \mathbf{R}^* \to T$such that$i \circ h \in X$. Then$i \circ h^{-1}$also satisfies conditions 1.2.1(i),(ii),(iii) of Deligne's paper, hence$i \circ h^{-1}\in X$. We see that$(T,h)$and$(T,h^{-1})$are different Shimura subdata of$(G,X)$with the same subgroup$T\$.

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Thanks a great deal for the explanation in detail as well as correcting the formulation. As for the second part, does it seems that the question is independent of the embedding of Shimura subdata? I didn't expect the problem to be so intrinsic. And on the other hand, such embedding of reductive subgroups might not necessarily be recognized as Levi subgroups or through operations on Dynkin diagrams, or should one go to the tables of Dynkin to see if more can be said about these subgroups? –  genshin Feb 22 '11 at 21:32
@algchen I tried my best to write my answer clearly. Could you make a small effort to write your questions clearly? I don't understand a word in your comment (except "thanks a great deal")..... –  Mikhail Borovoi Feb 22 '11 at 22:04