Global Cartan decomposition for a connected finite center semisimple Lie group

$K$ is the preimage of $e$ under the continuous map $G\to G$, $g\mapsto \mathrm{\Theta }(g)g^{-1}$, hence closed; it is a subgroup because $\mathrm{\Theta }$ is an involutive automorphism. By [L3] $K$ is an embedded Lie subgroup with Lie algebra $\{X:\exp (tX)\in K\text{ for all }t\}$.

The Lie algebra of $K$ is ${\mathfrak{k}}_0$: for $X\in {\mathfrak{g}}_0$ one has $\exp (tX)\in K$ for all $t$ if and only if $\exp (t{\theta }_{\ast }X)=\mathrm{\Theta }(\exp tX)=\exp (tX)$ for all $t$, by [L2], which holds if and only if ${\theta }_{\ast }X=X$, by differentiating at $t=0$; this is exactly the condition $X\in {\mathfrak{k}}_0$. Also $Z\subseteq K$, because $\mathrm{\Theta }$ fixes $Z$ pointwise by definition of a global Cartan involution.

For $A\in \mathrm{Aut}({\mathfrak{g}}_0)$ let $A^{\ast }$ denote its adjoint for $B_{{\theta }_{\ast }}$. Then $(A^{\ast }{)}^{-1}={\theta }_{\ast }A{\theta }_{\ast }$, and consequently $A^{\ast }={\theta }_{\ast }{A}^{-1}{\theta }_{\ast }\in \mathrm{Aut}({\mathfrak{g}}_0)$. Indeed, for $Z,W$ one computes, using that $B$ is invariant under both automorphisms ${\theta }_{\ast }$ and $A$, so that $B({\theta }_{\ast }U,{\theta }_{\ast }V)=B(U,V)$ and $B(AU,AV)=B(U,V)$, and using $B(AU,V)=B(U,A^{-1}V)$: $B_{{\theta }_{\ast }}({\theta }_{\ast }A{\theta }_{\ast }Z,W)=-B({\theta }_{\ast }A{\theta }_{\ast }Z,{\theta }_{\ast }W)=-B(A{\theta }_{\ast }Z,W)=-B({\theta }_{\ast }Z,A^{-1}W)=B_{{\theta }_{\ast }}(Z,A^{-1}W),$ so $({\theta }_{\ast }A{\theta }_{\ast }{)}^{\ast }=A^{-1}$, equivalently $A^{\ast }={\theta }_{\ast }{A}^{-1}{\theta }_{\ast }$. In particular $\mathrm{Ad}(G)$ is closed under taking adjoints.

For $W\in {\mathfrak{g}}_0$ one has $(\mathrm{ad}W{)}^{\ast }=-\mathrm{ad}({\theta }_{\ast }W)$, where the adjoint is taken for $B_{{\theta }_{\ast }}$: for $Y,Z$ one has $B_{{\theta }_{\ast }}((\mathrm{ad}{\theta }_{\ast }W)Y,Z)=-B([{\theta }_{\ast }W,Y],{\theta }_{\ast }Z)=B(Y,[{\theta }_{\ast }W,{\theta }_{\ast }Z])=B(Y,{\theta }_{\ast }[W,Z])=-B_{{\theta }_{\ast }}(Y,(\mathrm{ad}W)Z)$. Hence $\mathrm{ad}W$ is self-adjoint for $B_{{\theta }_{\ast }}$ if and only if $W\in {\mathfrak{p}}_0$.

Define $T:=\{v\in G:\mathrm{\Theta }(v)v^{-1}\in Z\}$. This is a subgroup containing $K$: it is the preimage of the finite set $Z$ under the continuous map $v\mapsto \mathrm{\Theta }(v)v^{-1}$, hence closed, and for $u,v\in T$ one has $\mathrm{\Theta }(uv)(uv{)}^{-1}=\mathrm{\Theta }(u)\mathrm{\Theta }(v)v^{-1}{u}^{-1}=\mathrm{\Theta }(u)u^{-1}\mathrm{\Theta }(v)v^{-1}\in Z$ because $\mathrm{\Theta }(v)v^{-1}$ is central and $\mathrm{\Theta }(u)u^{-1}\in Z$; inverses are handled by reversing the order of the same computation. Its Lie algebra is $\{X:\exp (tX)\in T\text{ for all }t\}=\{X:\mathrm{ad}({\theta }_{\ast }X)=\mathrm{ad}X\}={\mathfrak{k}}_0$, since $\exp (tX)\in T$ is equivalent to $\mathrm{Ad}(\mathrm{\Theta }(\exp tX))=\mathrm{Ad}(\exp tX)$, that is, to $e^{t\mathrm{ad}{\theta }_{\ast }X}=e^{t\mathrm{ad}X}$ for all $t$, and differentiating at $t=0$ gives $\mathrm{ad}{\theta }_{\ast }X=\mathrm{ad}X$, which by injectivity of $\mathrm{ad}$ is ${\theta }_{\ast }X=X$.

For $g\in G$ put $A:=\mathrm{Ad}(g{)}^{\ast }\mathrm{Ad}(g)$. By step 1.3, $A\in \mathrm{Aut}({\mathfrak{g}}_0)$, and $A$ is self-adjoint and positive definite for $B_{{\theta }_{\ast }}$: $B_{{\theta }_{\ast }}(AZ,Z)=B_{{\theta }_{\ast }}(\mathrm{Ad}(g)Z,\mathrm{Ad}(g)Z)\ge 0$, with equality only for $Z=0$. Define the endomorphism $D:=\log A$ by requiring $D$ to act as $\log \lambda$ on the eigenspace of $A$ with eigenvalue $\lambda >0$; by [L5] these eigenspaces span ${\mathfrak{g}}_0$, and $e^D=A$. If $X,Y$ are eigenvectors of $A$ with eigenvalues $\lambda ,\mu$, then $A[X,Y]=[AX,AY]=\lambda \mu [X,Y]$, so $[X,Y]$ lies in the eigenspace for $\lambda \mu$, and $D[X,Y]=(\log \lambda +\log \mu )[X,Y]=[DX,Y]+[X,DY]$; by bilinearity $D$ is a derivation of ${\mathfrak{g}}_0$. By [L4] there is a unique $X\in {\mathfrak{g}}_0$ with $D=\mathrm{ad}X$, and $X\in {\mathfrak{p}}_0$ because $D$ is self-adjoint and step 1.4 applies. Then $A=e^{\mathrm{ad}X}=\mathrm{Ad}(\exp X)$ by [L3], and $\mathrm{ad}X=D$ is determined by $A$, hence so is $X$ by injectivity of $\mathrm{ad}$.

Let $g\in G$ and let $X\in {\mathfrak{p}}_0$ be as in step 2.2, and put $u:=g\exp (-X\mathrm{/}2)\in G$. Then $\mathrm{Ad}(u)$ is orthogonal for $B_{{\theta }_{\ast }}$: $\mathrm{Ad}(u{)}^{\ast }\mathrm{Ad}(u)=e^{-\mathrm{ad}X\mathrm{/}2}A{e}^{-\mathrm{ad}X\mathrm{/}2}=e^{-\mathrm{ad}X\mathrm{/}2}{e}^{\mathrm{ad}X}{e}^{-\mathrm{ad}X\mathrm{/}2}=1$, since all factors are functions of the self-adjoint endomorphism $\mathrm{ad}X$ and therefore commute. Consequently ${\theta }_{\ast }\mathrm{Ad}(u){\theta }_{\ast }=(\mathrm{Ad}(u{)}^{\ast }{)}^{-1}=\mathrm{Ad}(u)$ by step 1.3.

The map $\psi \text{\hspace{0.17em}⁣}:T\times {\mathfrak{p}}_0\to G$, $\psi (v,X)=v\exp X$, is smooth. Its differential is invertible at every point: since $\psi (v_{0}v,X)=v_0\psi (v,X)$, left translation in the first variable reduces the computation to the points $(e,X_0)$. Put $A=\mathrm{ad}{X}_0$. For a direction $S$, apply $\mathrm{Ad}$ to the curve $\exp (-X_0)\exp (X_0+tS)$. By [L3] its image is $e^{-A}{e}^{A+t\mathrm{ad}S}$, an ordinary finite-dimensional matrix exponential. Differentiating its convergent matrix series gives ${\frac{d}{dt}\mid }_0{e}^{-A}{e}^{A+t\mathrm{ad}S}={\int }_0^1{e}^{-sA}(\mathrm{ad}S)e^{sA}ds={\mathrm{ad}}_{{\int }_0^1{e}^{-sA}Sds}\mathrm{.}$ Because $d{\mathrm{Ad}}_e=\mathrm{ad}$ and $\mathrm{ad}$ is injective by [L3] and [L4], the left-trivialized derivative of the group exponential is therefore ${\mathrm{dexp}}_{X_0}^L(S)={\int }_0^1{e}^{-sA}Sds=\frac{1-e^{-A}}{A}S,$ where the quotient denotes the entire series ${\sum }_{n\ge 0}(-A{)}^n\mathrm{/}(n+1)!$. Consequently directions $(U,S)\in {\mathfrak{k}}_0\times {\mathfrak{p}}_0$ give $d{\psi }_{(e,X_0)}(U,S)=e^{-A}U+\frac{1-e^{-A}}{A}S\mathrm{.}$ This derivation uses only the matrix exponential of $\mathrm{ad}X$, whose identity with ${\mathrm{Ad}}_{\exp X}$ is [L3], and does not assume a series formula for the arbitrary Lie-group exponential.

For every $g\in G$ one has $\mathrm{Ad}(\mathrm{\Theta }(u))={\theta }_{\ast }\mathrm{Ad}(u){\theta }_{\ast }$, because $\mathrm{\Theta }\circ C_g=C_{\mathrm{\Theta }(g)}\circ \mathrm{\Theta }$ for the conjugations, and differentiating at $e$ gives this identity. Hence step 3.1 yields $\mathrm{Ad}(\mathrm{\Theta }(u))=\mathrm{Ad}(u)$, that is, $\mathrm{\Theta }(u)u^{-1}\in \ker \mathrm{Ad}=Z$, so $u\in T$ by step 2.1. Therefore $g=u\exp (X\mathrm{/}2)$ with $u\in T$ and $X\mathrm{/}2\in {\mathfrak{p}}_0$: every element of $G$ lies in $T\cdot \exp ({\mathfrak{p}}_0)$.

The factorization is unique: suppose $u\exp X=u^{\mathrm{\prime }}\exp X^{\mathrm{\prime }}$ with $u,u^{\mathrm{\prime }}\in T$ and $X,X^{\mathrm{\prime }}\in {\mathfrak{p}}_0$. Applying $\mathrm{Ad}$ gives $\mathrm{Ad}(u)e^{\mathrm{ad}X}=\mathrm{Ad}(u^{\mathrm{\prime }})e^{\mathrm{ad}{X}^{\mathrm{\prime }}}$; multiplying on the left by $\mathrm{Ad}(u{)}^{-1}$ and on the right by $e^{-\mathrm{ad}{X}^{\mathrm{\prime }}}$ gives $Q:=\mathrm{Ad}(u{)}^{-1}\mathrm{Ad}(u^{\mathrm{\prime }})=e^{\mathrm{ad}X}{e}^{-\mathrm{ad}{X}^{\mathrm{\prime }}}\mathrm{.}$ Here $Q$ is orthogonal, because $\mathrm{Ad}(u)$ and $\mathrm{Ad}(u^{\mathrm{\prime }})$ are orthogonal by step 3.1; computing $Q^{\ast }Q=1$ with $Q^{\ast }=e^{-\mathrm{ad}{X}^{\mathrm{\prime }}}{e}^{\mathrm{ad}X}$ gives $e^{-\mathrm{ad}{X}^{\mathrm{\prime }}}{e}^{2\mathrm{ad}X}{e}^{-\mathrm{ad}{X}^{\mathrm{\prime }}}=1$, hence $e^{2\mathrm{ad}X}=e^{2\mathrm{ad}{X}^{\mathrm{\prime }}}$. Both exponents are self-adjoint, and the logarithm of a positive definite self-adjoint operator is unique because the eigenvalue $2\lambda$ is recovered as the logarithm of $e^{2\lambda }$; hence $\mathrm{ad}X=\mathrm{ad}{X}^{\mathrm{\prime }}$ and, by injectivity of $\mathrm{ad}$, $X=X^{\mathrm{\prime }}$. Then $Q=e^{\mathrm{ad}X}{e}^{-\mathrm{ad}X}=1$, so $\mathrm{Ad}(u)=\mathrm{Ad}(u^{\mathrm{\prime }})$ and $u^{\mathrm{\prime }}{u}^{-1}\in Z$; writing $u^{\mathrm{\prime }}=zu$ with central $z\in Z$ and substituting into $u\exp X=u^{\mathrm{\prime }}\exp X=uz\exp X$ gives $z=e$ and $u=u^{\mathrm{\prime }}$.

The differential of step 3.2 is injective, hence an isomorphism. Suppose $e^{-A}U+f(A)S=0$ with $A:=\mathrm{ad}{X}_0$, $f(A)=(1-e^{-A})A^{-1}$, $U\in {\mathfrak{k}}_0$, $S\in {\mathfrak{p}}_0$. Applying ${\theta }_{\ast }$, which satisfies ${\theta }_{\ast }A{\theta }_{\ast }=-A$, ${\theta }_{\ast }U=U$ and ${\theta }_{\ast }S=-S$, and using ${\theta }_{\ast }{e}^{-A}{\theta }_{\ast }=e^A$ and ${\theta }_{\ast }f(A){\theta }_{\ast }=f(-A)$ gives $e^{A}U-f(-A)S=0$. Multiplying the original equation by $e^A$ gives $U=-g(A)S$ with $g(A)=(e^A-1)A^{-1}=f(-A)$, and substituting into the transformed equation gives $-(e^A+1)g(A)S=0$. Since $A$ is self-adjoint and $e^{\lambda }+1\mathrm{\ne }0$ for real $\lambda$, the factor $e^A+1$ is invertible, so $g(A)S=0$. The eigenvalues of $g(A)$ are $(e^{\lambda }-1)\mathrm{/}\lambda \mathrm{\ne }0$ for the eigenvalues $\lambda \mathrm{\ne }0$ of $A$ and $1$ on the kernel of $A$, so $g(A)$ is invertible and $\ker g(A)=0$; hence $S=0$ and then $U=-g(A)S=0$.

By the inverse function theorem, the bijection $\psi$ of step 3.2 is a diffeomorphism: it is a local diffeomorphism because its differential is everywhere invertible by step 4.3, and a bijective local diffeomorphism has smooth inverse.

The diffeomorphism $\psi$ exhibits $G$ as diffeomorphic to $T\times {\mathfrak{p}}_0$, and ${\mathfrak{p}}_0$ is a real vector space; since $G$ is connected, $T$ is connected. Since $K\subseteq T$ by step 2.1, $\mathrm{Lie}(K)={\mathfrak{k}}_0=\mathrm{Lie}(T)$ by steps 1.2 and 2.1, and $T$ is connected, [L3] gives $T=K$.

$T$ is compact. Let $F:=\mathrm{Ad}(T)=\{A\in \mathrm{Ad}(G):{\theta }_{\ast }A{\theta }_{\ast }=A\}$ by step 2.1. Every element of $\mathrm{Ad}(G)$ is an automorphism of ${\mathfrak{g}}_0$, the group $\mathrm{Aut}({\mathfrak{g}}_0)$ is closed in $\mathrm{GL}({\mathfrak{g}}_0)$ with Lie algebra $\mathrm{ad}({\mathfrak{g}}_0)$, and $\mathrm{Ad}(G)$ is a connected Lie subgroup with the same Lie algebra; hence $\mathrm{Ad}(G)$ is an open subgroup of the connected group $\mathrm{Aut}({\mathfrak{g}}_0{)}^0$ and therefore equals it, so $\mathrm{Ad}(G)$ is closed in $\mathrm{GL}({\mathfrak{g}}_0)$. The condition ${\theta }_{\ast }A{\theta }_{\ast }=A$ is closed, so $F$ is closed in the compact orthogonal group of $B_{{\theta }_{\ast }}$ and hence compact. The adjoint map $T\to F$ is a covering with finite fibre $T\cap Z=Z$, so $T$ is a finite union of compact sets and is compact. Hence $K=T$ is compact, closing statement 1, and the map of statement 2 is the diffeomorphism $\psi$ of step 5.1 with $T=K$.