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G = KM = MK? #10

Closed mpham26uchicago closed 1 year ago

mpham26uchicago commented 1 year ago

We know that $\mathfrak{g} = \mathfrak{m} \oplus \mathfrak{k}$, and this gives $G = KM$. Does this also mean G = MK?

mpham26uchicago commented 1 year ago

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 $G = (K_1 A K_1^\dagger) (K_1 K_2) = MK$