Inversion of Lie subspace in Cartan decomposition

Question

Consider a matrix Lie group $G\subseteq GL$ with Lie algebra $\mathfrak{g}$. Using the Cartan decomposition, we can write $\mathfrak{g}$ as $(\mathfrak{k},\mathfrak{p})$. Let now $k\in\mathfrak{k}$ and $p\in\mathfrak{p}$. Let: $$ M:=\exp(k+p)\in G. $$

Using the polar decomposition, we can write $M$ as: $$ M=\exp(k')\exp(p'), $$ where $k'\in\mathfrak{k}$ and $p'\in\mathfrak{p}$ are in general different from $k,p$, as $k,p$ in general do not commute (right?).

The question now is: even if $k,p$ do not commute, is it still true that, for the same quantities as above (i.e. if $\exp(k+p)=\exp(k')\exp(p')$): $$ \exp(k-p)=\exp(k')\exp(-p')\,? $$

If not, what are the hypotheses for this to be true (e.g. commutativity)?

Thanks.

Answer 1

Okay, I have figured it out!

Let $\Theta$ be the Cartan involution at the group level, and $\theta$ its differential (i.e. Cartan involution of Lie algebra). Then: $$ \exp(k-p) = \exp(\theta(k+p)) = \Theta(\exp(k+p)). $$

Now if $\exp(k+p)=\exp(k')\exp(p')$, since $\Theta$ is a morphism: $$ \exp(k-p) = \Theta(\exp(k'))\Theta(\exp(p'))=\exp(\theta(k'))\exp(\theta(p')). $$ We have that $k'\in\mathfrak{k}$ is a fixed point of $\theta$, while $p'$ flips a sign, so: $$ \exp(k-p) =\exp(k')\exp(\theta(p')) =\exp(k')\exp(-p'). $$

Please correct me if I am wrong!