Closed mpham26uchicago closed 1 year ago
We know that $\mathfrak{g} = \mathfrak{m} \oplus \mathfrak{k}$, and this gives $G = KM$. Does this also mean G = MK?
Yes, by adjoint orbit theorem, we have the KAK decomposition $G = K_1 A K_2$. Let $M = K_1 A K_1^\dagger$, then $K = K_1 K_2$. Thus $G = (K_1 A K_1^\dagger) (K_1 K_2) = MK$
We know that $\mathfrak{g} = \mathfrak{m} \oplus \mathfrak{k}$, and this gives $G = KM$. Does this also mean G = MK?