1
$\begingroup$

I asked this question on MSE https://math.stackexchange.com/questions/4475382/extension-of-an-involution-on-g-to-an-involution-on-g-mathbbc but didn't receive any answer so far. My question is the following:

Let $G$ be a compact connected Lie group and $ \sigma :G \rightarrow G $ be an involution on $G$. Let $G^\sigma :=\lbrace g \in G, \sigma(g)=g \rbrace$.

Denote by $G_\mathbb{C}$ the complexification of $G$. Does there exist an involution $\tilde{\sigma}$ on $G_\mathbb{C}$ which coïncides with $\sigma$ on $G$ and such that ${(G^{\sigma})}_\mathbb{C} = {(G_\mathbb{C})}^\tilde{\sigma}$ ?

My though was to define $\tilde{\sigma}$ to be $\sigma$ on G and to be the identity on the complement of $G$ on $G_\mathbb{C}$, but don't see whether this imply the ${(G^{\sigma})}_\mathbb{C} = {(G_\mathbb{C})}^\tilde{\sigma}$ or not, any help please!

$\endgroup$
16
  • 3
    $\begingroup$ You should add a cross-reference to your question on MSE, and also in MSE to MO. $\endgroup$ Commented Jun 23, 2022 at 0:12
  • 4
    $\begingroup$ First you need a definition of the complexification of a compact real Lie group $G$. See Serre, Galois Cohomology, Section III.4.5. $\endgroup$ Commented Jun 23, 2022 at 0:19
  • 2
    $\begingroup$ Your group $G^\sigma$ might be nonconnected. For example, take $G=T$, a one-dimensional compact real torus, and take $\sigma(t)=t^{-1}$ for $t\in T$. $\endgroup$ Commented Jun 23, 2022 at 0:57
  • 2
    $\begingroup$ The correspondence $G\leadsto G_{\Bbb C}$ is a functor, and an involution $\sigma$ of $G$ induces an involution of $G_{\Bbb C}$ (I denote it again by $\sigma$). $\endgroup$ Commented Jun 23, 2022 at 1:08
  • 1
    $\begingroup$ If $G^\sigma$ is connected, then, since $${\rm Lie}\big((G_{\Bbb C})^\sigma\big)=\big({\rm Lie}(G_{\Bbb C})\big)^\sigma =\big(({\rm Lie}\, G)^\sigma\big)_{\Bbb C}=\big({\rm Lie}(G^\sigma)\big)_{\Bbb C}, $$ the complexification of $G^\sigma$ is $(G_{\Bbb C})^\sigma$. $\endgroup$ Commented Jun 23, 2022 at 1:31

0

You must log in to answer this question.

Browse other questions tagged .