1
$\begingroup$

Let $G$ be a connected real simple Lie group, and $\tau$ be an involutive automorphism of $G$. Then $\tau$ defines a symmetric subgroup $G^\tau$ of $G$. In general, $G^\tau$ is not necessarily connected. (RIGHT?)

Consider the symmetric space $M=G/G^\tau$. If $M$ is simply connected, then by the short exact sequence $0=\Pi_1(M)\rightarrow\Pi_0(G^\tau)\rightarrow\Pi_0(G)\rightarrow\Pi_0(M)=0$, $G^\tau$ is connected because $G$ is supposed to be connected. (RIGHT?) Else suppose that $M$ is not simply connected. Denote by $\widetilde{M}$ the universal covering of $M$, which is again a symmetric space (RIGHT?), and then $\widetilde{M}=\widetilde{G}/\widetilde{G}^\widetilde{\tau}$ for the universal cover $\widetilde{G}$ of $G$.

Let $\mathfrak{g}$ be the Lie algebra of $G$, and denote by $\mathrm{Int}(\mathfrak{g})$ the group of inner automorphims of $\mathfrak{g}$. Then $\mathrm{Int}(\mathfrak{g})$ is simply connected (RIGHT?) and hence is the universal cover of $G$; namely, $\mathrm{Int}(\mathfrak{g})\cong\widetilde{G}$.

I shall be grateful if any expert helps me to confirm that all the four points marked with (RIGHT?) are correct or not.

QUESTION

Let $G=\mathrm{Int}(\mathfrak{g})$ for some real simple Lie algebra $\mathfrak{g}$, which is simply connected. Let $\tau\in G$ be an element of order 2, i.e., an involutive inner automorphism of $\mathfrak{g}$. Write $G^\tau=\{g\in G\mid \tau^{-1}g\tau=g\}$. Then, is $G^\tau$ always connected?

Improve this question
$\endgroup$
8
  • $\begingroup$ What do you mean by ${\rm Int}\frak g$? A real algebraic group, a real Lie group, and if yes, then which real Lie group? $\endgroup$
    – Mikhail Borovoi
    Jul 27, 2017 at 16:30
  • 5
    $\begingroup$ If $G={\rm SL}(2,\Bbb R)$, then (the identity component of) ${\rm Int}\frak g$ is $G/\{\pm1\}$, which is clearly not simply connected. $\endgroup$
    – Mikhail Borovoi
    Jul 27, 2017 at 16:38
  • 2
    $\begingroup$ If $G$ is a connected Lie group that is topologically simply connected (e.g., the underlying real Lie group of $H(\mathbf{C})$ for a connected semisimple algebraic group $H$ over $\mathbf{C}$ whose root datum is simply connected, such as ${\rm{Sp}}_{2g}$ or ${\rm{SL}}_n$) and $\tau$ is an order-2 automorphism of $G$ then the closed subgroup $G^{\tau}$ of $\tau$-fixed points is connected. This was proved by E. Cartan using Riemannian geometry via reducing to the compact case whose main content is Theorem 8.2 in Ch. VII of Helgason's Differential Geometry, Lie Groups, and Symmetric Spaces. $\endgroup$
    – nfdc23
    Jul 27, 2017 at 22:53
  • $\begingroup$ @MikhailBorovoi Yes, you are right. Thank you. I was confused with the definition before. The definition for $\mathrm{Int}\mathfrak{g}$ is the analytic subgroup of $\mathrm{Aut}\mathfrak{g}$ with the Lie algebra $\mathrm{ad}\mathfrak{g}$, and hence it is the identity component of $\mathrm{Ad}G$. $\endgroup$
    – Hebe
    Jul 28, 2017 at 4:37
  • 2
    $\begingroup$ It holds for simply connected Lie groups (both compact and non-compact). It fails in many cases when $G$ isn't simply connected (but simple in the sense of connected semisimple Lie groups). For instance, consider $G=H(\mathbf{C})$ (as a connected real Lie group with semisimple Lie algebra) for a connected semisimple $\mathbf{R}$-group $H$ with absolutely simple Lie algebra and $\tau$ complex conjugation. In such cases, $G^{\tau}=H(\mathbf{R})$ is often disconnected when $H$ isn't simply connected in the sense of semisimple algebraic groups (e.g., $H={\rm{PGL}}_{2m}, {\rm{SO}}_{n,n}$, ...) $\endgroup$
    – nfdc23
    Jul 28, 2017 at 5:29

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.