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

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 $$k\in K$$, $$X\in \mathfrak{g}$$.)

• I know close to nothing about the subject but are you sure you want to denote your number field by $\mathbb{Q}$? Dec 13, 2021 at 14:17
• It's a bit odd, as $\mathbb{Q}$ usually denotes the field of rational numbers, and not an arbitrary finite extension thereof. Dec 13, 2021 at 14:36
• @SylvainJULIEN, You re right! For simplicity, I took the rational number field $\mathbb{Q}$ as a number field here. But my question can be samely asked to arbitrary number field. Dec 13, 2021 at 15:47
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 $$\mathfrak{g}\otimes M$$.
• Oh, I see! Since $\mathfrak{g}$ is finite dimension, the image should be also fin. dimension. Thank u very much! Dec 13, 2021 at 19:48