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Consider the Shimura datum $(GSp_{2g},\mathcal{H}_g)$. Let $G$ be a reductive $\mathbb{Q}$-subgroup of $Sp_{2g}$. I want to know under what condition there exists a point $x\in\mathcal{H}_g$ such that $h_x\colon\mathbb{S}\rightarrow GSp_{2g,\mathbb{R}}$ factors through $Z(GSp_{2g})G$.

There are some easy necessary conditions: $G$ shouldn't be of mixed type $D_n$, $G(\mathbb{R})$ should be non compact and $Z(G)(\mathbb{R})$ should be compact. My first question is whether these conditions are sufficient.

My second question is: can a $\mathbb{Q}$-reductive subgroup of $Sp_{2g}$ be of mixed type $D_n$?

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First question: No, these conditions are not sufficient. For details see: Deligne, Pierre, Variétés de Shimura: interprétation modulaire, et techniques de construction de modèles canoniques. (French) Automorphic forms, representations and L-functions (Proc. Sympos. Pure Math., Oregon State Univ., Corvallis, Ore., 1977), Part 2, pp. 247–289, Proc. Sympos. Pure Math., XXXIII, Amer. Math. Soc., Providence, R.I., 1979.

Second question: Yes, any linear algebraic $\mathbb Q$-group can be embedded into $GL_n$ for some $n$, and $GL_n$ can be embedded into $Sp_{2n}$.

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  • $\begingroup$ Do you mind zooming in a little in the reference? I forgot to say that $G(\mathbb{R})$ is non compact (I'll add this to the original post). $\endgroup$
    – user70064
    Commented Apr 2, 2015 at 21:15
  • $\begingroup$ E.g. $G$ cannot be of type $G_2$, or of type $B_n$ $(n\ge 3)$ with any representation different from the spinor representation, etc.. $\endgroup$
    – Mikhail Borovoi
    Commented Apr 2, 2015 at 21:32
  • $\begingroup$ I see. So there are plenty of counterexamples because any algebraic $\mathbb{Q}$-group can be embedded into $Sp_{2n}$ for some $n$. Thanks. $\endgroup$
    – user70064
    Commented Apr 2, 2015 at 21:48
  • $\begingroup$ Another reference (in addition to Deligne): Milne, J. S. Shimura varieties and moduli. Handbook of moduli. Vol. II, 467--548, Adv. Lect. Math. (ALM), 25, Int. Press, Somerville, MA, 2013. $\endgroup$
    – anon
    Commented Apr 2, 2015 at 23:17
  • $\begingroup$ Exactly. Noncompact $G_2$ can be embedded into $Sp_{2n}$, but its symmetric space does not admit an invariant complex strusture. Furthermore, the symmetric space of $Spin(11,2)$ admits an invariant complex structure, but if you consider any homomorphism into $Sp_{2n}$ other than the spinor representation, then the corresponding embedding of symmetric spaces will not be holomorphic. $\endgroup$
    – Mikhail Borovoi
    Commented Apr 3, 2015 at 6:30

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