Why differential operator preserves $K$-finiteness of automorphic form?

Question

Let $G$ be a reductive group over a number field $\mathbb{Q}$ and $K$ be a maximal compact subgroup of $G$. Let $\Gamma$ be an arithmetic subgroup of $G(\mathbb{Q})$.

Let $\mathfrak{g}$ be the complexfied Lie alogebra of $G$ and $Z(\mathfrak{g})$ the center of the universal enveloping algebra $U(\mathfrak{g})$ of $\mathfrak{g}$.

Then one can define the automorphic form on $\Gamma \backslash G(\mathbb{R})$ as function on $\Gamma \backslash G(\mathbb{R})$ satisfying 'smooth', 'right $K$-finite', 'moderate growth', '$Z(\mathfrak{g})$-finite' conditions.

Let $X$ be an element in $\mathfrak{g}$ and $\phi$ is an automorphic form $\Gamma \backslash G(\mathbb{R})$.

Then some book says that $X\phi$ satisfies also $K$-finite conditions. But there is no proof on this and I can't check it well.

Would you let me know why $X\phi$ is also $K$-finite?

It seems that it would use some compatibility property of $\mathfrak{g}$-actions and $K$-actions, (i.e., $kX\phi=((Ad(k)X))k\phi$ for $k\in K$, $X\in \mathfrak{g}$.)

Any comments are appreciated!

Answer 1

Consider the map $\mathfrak{g}\otimes C^\infty(\Gamma\backslash G)\to C^\infty(\Gamma\backslash G)$, $X\otimes f\mapsto Xf$. The group $K$ acts on both sides and the map is equivariant. If $f$ lies in a finite-dimensional $K$-module $M$, then $Xf$ lies in the finite-dimensional $K$-module that is the image of $\mathfrak{g}\otimes M$.