Let $\mathfrak{g}_0$ be a noncompact simple Lie algebra, and fix a Cartan involution $\theta$ of $\mathfrak{g}_0$, which gives a Cartan decomposition $\mathfrak{g}_0=\mathfrak{k}_0+\mathfrak{p}_0$. Since the automorphism group $G=\mathrm{Aut}\mathfrak{g}_0$ has finite center ($\textbf{Right?}$), there is an automorphism $\Theta$ of $G$ whose differential is $\theta$, and the subgroup $K$ of the fixed points of $\Theta$ on $G$ is a maximal compact subgroup of $G$. Moreover, $G=Ke^{\mathfrak{p}_0}$ as a polar decomposition of $G$.

Let $\mathfrak{g}=\mathfrak{g}_0\oplus\sqrt{-1}\mathfrak{g}_0$ be the complexification of $\mathfrak{g}_0$, and denote by $G_\mathbb{C}$ the universal complexification of $G$. Set $\mathfrak{u}_0=\mathfrak{k}_0+\sqrt{-1}\mathfrak{p}_0$ which is a compact Lie algebra. Let $U$ be the subgroup of $G_\mathbb{C}$ corresponding to Lie algebra $\mathfrak{u}_0$.


1) Is $U$ a compact dual of $G$?

2) Does the polar decomposition $U=Ke^{\sqrt{-1}\mathfrak{p}_0}$ hold? In particular, does $K\subseteq G\cap U$ hold?

  • 1
    $\begingroup$ What do you mean by the universal complexification of $G$? $\endgroup$ Mar 29 '17 at 16:58
  • $\begingroup$ @MikhailBorovoi Let $G$ be a real Lie group. A universal complexification of $G$ is a complex Lie group $G_\mathbb{C}$ and a continuous group homomorphism $i:G\rightarrow G_\mathbb{C}$ with the property that, if $f$ is an arbitrary continuous group homomorphism from $G$ to a complex Lie group $H$, then there exists a unique continuous group homomorphsim $\tilde{f}:G_\mathbb{C}\rightarrow H$ such that $f=\tilde{f}\circ i$. $\endgroup$
    – Hebe
    Mar 30 '17 at 9:23
  • $\begingroup$ I think @MikhailBorovoi's point may have been that what you are calling the universal complexification is usually just called the complexification, with no modifier (although your terminology is eminently logical). $\endgroup$
    – LSpice
    Apr 4 '18 at 2:04

(0) The group ${\rm Aut}(\mathfrak{g}_0)$ does not have to be connected (even over $\mathbb C$), take $\mathfrak{g}_0=\mathfrak{su}(2,2)$ as a counter-example. So let $G={\rm Inn}(\mathfrak g_0)$, that is, the identity component (over $\mathbb C$) of ${\rm Aut}(\mathfrak{g}_0)$. Then right, $G$ has trivial center.

(1) Since $\mathfrak u_0$ is a compact semisimple Lie algebra, yes, $U$ is a compact semisimple Lie group. Since $\mathfrak u_0$ is a real form of $\mathfrak g$, yes, $U$ is a real form of $G_{\mathbb C}$. Thus yes, $U$ is a compact real form of $G_{\mathbb C}$.

(2a) No, this polar decomposition does not hold. Indeed, otherwise the compact group $U$ would be homeomorphic to the direct product of $K$ and $\mathfrak p_0$, hence it would be noncompact, which is a contradiction.

(2b) However, it is true that $K=G\cap U$. Indeed, we have $K\subset G$. Let us show that $K\subset U$. Let $\rho\colon G_{\mathbb C}\to G_{\mathbb C}$ denote the complex conjugation corresponding to $G$, that is, such that $G=(G_{\mathbb C})^\rho$. Then $\theta$ commutes with $\rho$. We have $$ K=G^\theta=\{g\in G_{\mathbb C}\ |\ \rho(g)=g,\ \theta(g)=g.\} $$ On the other hand $$ U=(G_{\mathbb C})^{\theta\circ\rho}=\{g\in G_{\mathbb C}\ |\ \theta(\rho(g))=g.\} $$ We see that $K\subset U$, as required. Thus $K\subset G\cap U$. Since $G\cap U$ is a compact subgroup of $G$ containing the maximal compact subgroup $K$ of $G$, we conclude that $K=G\cap U$.

  • $\begingroup$ Thank you for your answer! I have two questions. (1) I understand that $\mathrm{Aut}\mathfrak{g}_0$ is not necessary connected, but since $\mathrm{Inn}\mathfrak{g}_0$ has trivial center and the quotient group $\mathrm{Aut}\mathfrak{g}_0/\mathrm{Inn}\mathfrak{g}_0$ is finite, it is follows that $\mathrm{Aut}\mathfrak{g}_0$ has finite center. Right? $\endgroup$
    – Hebe
    Mar 30 '17 at 9:13
  • $\begingroup$ (2) I do not quite understand the complex conjugation $\rho$. Of course, on the complex Lie algebra $\mathfrak{g}$, one may define the complex conjugation $\rho$ corresponding to the real form $\mathfrak{g}_0$. However, we do not assume that $G_\mathbb{C}$ is simply connected, in which way is $\rho$ lifted from $\mathfrak{g}$ to $G_\mathbb{C}$? Thank you. $\endgroup$
    – Hebe
    Mar 30 '17 at 9:15
  • $\begingroup$ @Hebe: We have $G_{\mathbb C}={\rm Inn}(\mathfrak g)\subset{\rm Aut}(\mathfrak g)\subset {\rm GL}(\mathfrak g)$. The complex conjugation on $\mathfrak g$ induces a complex conjugation on $\rm GL$, $\rm Aut$ and $\rm Inn$. $\endgroup$ Mar 30 '17 at 9:42
  • 1
    $\begingroup$ Title of @Holonomia's reference: Eschenburg - Lecture notes on symmetric spaces. $\endgroup$
    – LSpice
    Dec 5 '20 at 19:06
  • 1
    $\begingroup$ @LSpice: OK you added the title in case the pdf is no longer available in the link I posted. $\endgroup$
    – Holonomia
    Dec 5 '20 at 20:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.