will be smooth, and we can define $(Xf)(g_\infty,g_f)$ to be the derivative of this function at $t=0$. Is this the correct way to get the $\mathfrak{g}$-action and hence the $(\mathfrak{g},K)$-module structure on the subspace of $K$-finite vectors? If so, is there a less ad hoc way to get the $\mathfrak{g}$-action? The fact this isn't an actual vector-valued derivative, like what you'd have with a smooth vector in e.g. a Banach space representation of $G_\infty$, is somehow unsatisfying to me, but perhaps I shouldn't expect anything like that since the point of view here does not involve a locally convex topology on the space $C^\infty(G(\mathbf{A}))$...of which I'm aware.