I'm reading Milne's Introduction to Shimura varieties (http://www.jmilne.org/math/xnotes/svi.pdf) and there is something I don't get.

Let $G$ be a connected semisimple algebraic group $G$ over $\mathbb{Q}$ and let $D$ be an hermitian symmetric domain such that $G(\mathbb{R})^+$ (the $^+$ denote the identity component) acts on $D$ via a surjective homomorphism $Hol(D)^+$ with compact kernel.

Let $G^{ad}_\mathbb{R} \simeq H_1 \times \ldots \times H_r$ be the decomposition into $\mathbb{R}$-simple factors. Denote by $H_{c}$ the product of $H_i$'s such that $H_i(\mathbb{R})$ is compact. Denote by $H_{nc}$ the product of the others (which we assume is not empty !).

Do we have an isomorphism $H_{nc}(\mathbb{R})^+ \simeq Hol(D)^+$ ?

  • $\begingroup$ Please define the notation. $\endgroup$
    – Marc Palm
    May 20, 2014 at 21:18
  • $\begingroup$ You can probably find what you're looking for in Chapter VIII of Helgason's book "Differential geometry, Lie groups,.." (but I haven't checked). $\endgroup$
    – naf
    May 21, 2014 at 9:17
  • $\begingroup$ As you have assumed, $H_{nc}({\mathbb R})^+$ maps onto $Hol (D)^+$ with finite kernel; however, they need not be isomorphic. For example, $H=SU(1,1)$ maps onto the group of identity component of holomorphic automorphisms of the unit disc, with kernel $\pm 1$. $\endgroup$ Aug 26, 2014 at 17:29
  • $\begingroup$ @Venkataramana: $H_{nc}$ is defined to be the product of noncompact factors of the adjoint group so has no centre. $\endgroup$
    – Martin Orr
    Aug 27, 2014 at 10:59
  • $\begingroup$ @Orr. Sorry; I did not notice that the groups were adjoint; then of course it is true that $H_{nc}({\mathbb R}^+$ is $Hol (D)^+$ for $D$ of non-compact type. $\endgroup$ Aug 27, 2014 at 13:44

1 Answer 1


You will need to add a condition that $D$ has noncompact type.

Since we have a surjective homomorphism $G(\mathbb{R})^+ \to Hol(D)^+$ with compact kernel, it suffices to show that $Hol(D)^+$ has trivial centre (so this homomorphism factors through $H_{nc}(\mathbb{R})^+$) and that $Hol(D)^+$ has no compact factors. The second part is essentially the definition of $D$ having noncompact type.

I could not find the fact that $Hol(D)^+$ has trivial centre explicitly stated in Helgason (but I might not have looked hard enough). It is part of Theorem 8.7.9 in Wolf, with the following proof:

Let $K \subset Hol(D)^+$ be the stabiliser of a point $p \in D$ and let $u_p \colon U_1 \to Hol(D)^+$ be the homomorphism from Theorem 1.9 of Milne. The first paragraph of the proof of Milne's Theorem 1.21 shows that $Lie(K)$ is the subspace of $Lie(Hol(D)^+)$ on which $u_p(-1)$ acts trivially.

Let $L \subset Hol(D)^+$ be the centraliser of $u_p$. The uniqueness of $u_p$ implies that $K \subset L$. Since $u_p(-1)$ acts trivially on $Lie(L)$, the previous paragraph implies that $L^+ = K^+$. Since $L$ is the centraliser of a torus, it is connected, and so $L = K$.

The centre of $Hol(D)^+$ is contained in $L$, and so in $K$. But the intersection of $K$ with the centre of $Hol(D)^+$ is trivial by Helgason, chapter IV, Theorem 3.3(ii). So the centre of $Hol(D)^+$ is trivial.


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