Let $G$ be a Lie group, $K$ be a maximal compact subgroup, and $\frak{g}, \frak{k}$ be their respective Lie algebras.
A $({\frak{g}},K)$-module is defined as a complex vector space $V$ that $\frak{g}$ and $K$ both acts on, such that
- For any $Y \in \frak{k}$, $v \in V$ $$Yv = \left(\exp(tY)v\right)'(0)$$
- $V$ is $K$-finite, and
- For any $k \in K$, $X \in \frak{g}$, we have $$(Ad(k)X)v = k(X(k^{-1}v))$$
I am confused about the third condition. According to Knapp-Trapa's "Representation theory of semi-simple Lie groups" or other sources, the third condition is really addressing the compatibility of the two actions when $K$ is disconnected.
How to prove that 3 follows from 1 and 2 when $K$ is connected?
It's clear that we can reduce to the case where $k = \exp(Y)$ for some $Y \in \frak{k}$. I am not sure how I can proceed afterwards - I tried to look at $(Ad(\exp(tY))X)v$ versus $\exp(tY) \cdot X \cdot \exp(-tY)v$ and hope that some uniqueness results of ODE may work, but then I realize that I cannot guarantee the map $$t \to (Ad(\exp(tY))X) v$$ is even smooth to start with.
Any help, hint, or reference would be very appreciated.