Non-semisimple symmetric subgroups of simply connected simple algebraic groups

Let $G$ be a simply connected simple algebraic group over the field of complex numbers $\mathbb C$. Let $H$ be a symmetric subgroup of $G$. This means that there exists an automorphism of order 2 $\sigma\colon G\to G$ such that $H$ is the group of fixed points of $\sigma$ in $G$. It is known that $H$ is connected.

Let us assume that $H$ is not semisimple. It is known that then the center of $H$ is one-dimensional. Consider the derived subgroup $H^{\rm der}:=[H,H]$.

Question. Is it true that the semisimple group $H^{\rm der}$ is always simply connected?

It seems that this is true when $G$ is a classical group (say, when $G={\rm Sp}_{2n}$, $H=U_n$). What about $E_6$ and $E_7$? If the answer is always "yes", I would prefer a classification-free proof.

• Though it may not be directly relevant to your research, it's worth considering the more general Chevalley definition of "simply connected" which applies to semisimple groups in any characteristic. The algebraic approach to your question might be easier in some ways (as well as more general). See for example $\S4$ in the old Borel-Tits paper numdam.org/item/PMIHES_1972__41__253_0 – Jim Humphreys Mar 28 '17 at 13:54
• @JimHumphreys, of course, one problem with the more general definition is that fixed points of involutions can be much worse behaved over fields of characteristic 2. Probably at least one has to restrict to odd characteristic. – LSpice Mar 29 '17 at 1:04
• @L Spice: Yes, I should have specified here that all the literature I'm aware of avoids characteristic 2 when dealing with such involutions and fixed point groups. Still, the Chevalley notion of "simply connected" does allow for more use of algebraic methods even while including the topological notion in Lie groups. What I don't see is whether that helps here. – Jim Humphreys Mar 29 '17 at 13:17
• @JimHumphreys: Of course I need the algebraic notion of "simply connected". I need the fact that $[H,H]$ is simply connected in order to describe $H^1(k,H)$ when $k$ is a number field. – Mikhail Borovoi Mar 29 '17 at 13:24

2 Answers

In your case, $H$ is a Levi subgroup and the derived subgroup of a Levi subgroup in a simply connected group is always simply connected (since all fundamental weights are characters).

• I think that it is not true that $H$ must be a Levi subgroup. Consider, for example, the case where $G = \mathrm{Sp}_4$ and $\sigma$ is conjugation by $s = \beta^\vee(-1)$, where $\beta^\vee$ is the short simple coroot, so that $H = \mathrm{SL}_2 \times \mathrm{SL}_2$. – LSpice Mar 28 '17 at 1:51
• Oops, that's semisimple. Hmm. Is your claim about being a Levi subgroup obvious? – LSpice Mar 28 '17 at 1:53

The answer is YES. Such a subgroup $H$ is called a Hermitian symmetric subgroup (because then for suitable real forms of $G$ and $H$, the quotient $G/H$ is a Hermitian symmetric space). It follows from the theory of Victor Kac that any Hermitian symmetric subgroup of a simple group is a "diagrammatic" subgroup (Levi subgroup) corresponding to the Dynkin subdiagram obtained by removing one vertex corresponding to a simple root appearing with coefficient one in the expression of the highest root as a linear combination of simple roots. See e.g. Table 7, type II in the book by Onishchik and Vinberg, or case (B) in the text after Lemma 5.17 on page 513 in Ch. X of the book by Helgason "Differential Geometry, Lie Groups, and Symmetric Spaces". It is well known that the derived group of any Levi subgroup of a simply connected semisimple group is simply connected, see e.g. Proposition 12.14 in the book by G. Malle and D. Testerman "Linear Algebraic Groups and Finite Groups of Lie Type".

• For what it's worth, although I don't know an earlier reference, the fact about Levi subgroups certainly predates Malle–Testerman (not that you claimed otherwise). For example, Kottwitz uses it on p. 294 of "Sign changes in harmonic analysis on reductive groups" in 1983, regarding it there as obvious. – LSpice Mar 29 '17 at 1:03
• @LSpice: Thank you, I have edited my answer and have written that this fact is well known. – Mikhail Borovoi Mar 29 '17 at 5:26