Skip to main content
18 events
when toggle format what by license comment
Jun 23, 2022 at 15:17 comment added Mikhail Borovoi I have answered your question only in the case when $G^\sigma$ is connected. Therefore, let the comments remain comments.
Jun 23, 2022 at 4:39 comment added Mira @MikhailBorovoi, your comments have answered my question, so thanks a lot! If you don't mind, you could collect them in the answer section in MSE, I'll accept your answer and offer the bounty .
Jun 23, 2022 at 3:08 comment added Mikhail Borovoi The connected compact groups (like $G$) have been classified, and their involutions have been classified too (for semisimple compact groups by Elie Cartan, Victor Kac, and others). Instead of proving this, you can compute $(G_{\Bbb C})^\sigma$ and $G^\sigma$, and check the above assertion about the points in connected components.
Jun 23, 2022 at 3:05 comment added Mira No problem! Your comments were of a great help, thank you very much :)
Jun 23, 2022 at 3:03 comment added Mikhail Borovoi I think that you need to prove that any connected component of $(G_{\Bbb C})^\sigma$ contains a point of $G^\sigma$. I cannot do that now.
Jun 23, 2022 at 2:58 comment added Mira I think that I understand a lot of what you have explained (even though I have to work it on more details ). Maybe last question: What we can do in case $G^\sigma$ is not connected ?
Jun 23, 2022 at 2:56 comment added Mira @MikhailBorovoi, I actually Find this reference and it is for Free, thanks a lot for pointing this out !
Jun 23, 2022 at 2:50 comment added Mikhail Borovoi Anyway, you need the definition of the complexification of Borel and Serre in terms of algebraic groups. An alternative reference is the paper of Borel and Serre "Théorèmes de finitude...", Section 6.7, but it is not for free either.
Jun 23, 2022 at 2:45 comment added Mikhail Borovoi The equality $\big({\rm Lie}(G_{\Bbb C})\big)^\sigma=\big(({\rm Lie}\, G)^\sigma\big)_{\Bbb C}$ follows from the fact the the involutive automorphism $(d\sigma)_{\Bbb C}$ of ${\rm Lie}(G_{\Bbb C})$ is the complexification of the involutive automorphism $d\sigma$ of ${\rm Lie}\, G$.
Jun 23, 2022 at 1:31 comment added Mikhail Borovoi 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 $(G_{\Bbb C})^\sigma$.
Jun 23, 2022 at 1:25 comment added Mira @MikhailBorovoi, could you please explain why ${(Lie (G_\mathbb{C}))}^\sigma = {({(LieG)}^\sigma)}_\mathbb{C}$ ? I didn't get it.
Jun 23, 2022 at 1:08 comment added Mikhail Borovoi 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 $\sigma$).
Jun 23, 2022 at 0:57 comment added Mikhail Borovoi 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 $t\in T$.
Jun 23, 2022 at 0:48 comment added Mira @MikhailBorovoi, thanks for your comments! Unfortunately I can't access to this reference since it's not free online. The definition of the complexification of a compact connected real Lie group $G$ is that it is a Lie group $G_\mathbb{C}$ which contains $G$ and such that its Lie algebra is the complexification of the Lie algebra of $G$.
Jun 23, 2022 at 0:27 history edited Mira CC BY-SA 4.0
added 111 characters in body
Jun 23, 2022 at 0:19 comment added Mikhail Borovoi First you need a definition of the complexification of a compact real Lie group $G$. See Serre, Galois Cohomology, Section III.4.5.
Jun 23, 2022 at 0:12 comment added Mikhail Borovoi You should add a cross-reference to your question on MSE, and also in MSE to MO.
Jun 22, 2022 at 22:20 history asked Mira CC BY-SA 4.0