Let $G$ be a connected and compact Lie Group and $\mathfrak{g}$ be its Lie Algebra. We define an action of $\mathfrak{g}$ on the space of smooth functions $C^{\infty} (G)$ on G as:
$$X f(g) = \frac{d}{dt}f(ge^{tX} ) |_{t=0}$$
It is claimed that the mapping
$$ \phi: \mathfrak{g} \to \mathfrak{gl}(C^{\infty}(G))$$
defined as above is a Lie Algebra homomorphism. My attempt is as follows:
Essentially we need to show that $$ \psi([X,Y]) f(g) = ((\psi(X) \circ \psi(Y)) - (\psi(Y) \circ \psi(X)))f(g) $$
The right hand side equals $$(\psi(X) \circ \psi(Y))f(g) - (\psi(Y) \circ \psi(X))f(g)$$ $$= \psi(X)(\frac{d}{dt}f(ge^{tY} ) |_{t=0}) - \psi(Y)(\frac{d}{dt}f(ge^{tX} ) |_{t=0})$$
Whereas the left-hand side is $$\frac{d}{dt} f(ge^{t[X,Y]})|_{t=0}$$
I'm not sure how the left-hand and right-hand sides of the equations are meant to equal each other especially since it seems like the right-hand side would involve taking two derivatives.
Any help would be useful.
