# The product of $Z(\mathfrak{g})$-finite functions is also $Z(\mathfrak{g})$-finite?

Let $$G$$ be a classical group defined over $$\mathbb{Q}$$.

Let $$\mathfrak{g}$$ be the Lie algebra of $$G(\mathbb{R})$$ and $$U(\mathfrak{g}_{\mathbb{C}})$$ its universal enveloping algebra of $$\mathfrak{g}_{\mathbb{C}}$$.

Let $$Z(\mathfrak{g})$$ be the center of $$U(\mathfrak{g}_{\mathbb{C}})$$. We regard the elements of $$U(\mathfrak{g}_{\mathbb{C}})$$ as differential operators on $$C^{\infty}(G)$$, the space of smooth functions on $$G(\mathbb{R})$$, acting by right infinitesimal translation.

Let $$f,g \in C^{\infty}(G)$$ be $$Z(\mathfrak{g})$$-finite. (I.e. $$\langle z\cdot f | z\in Z(\mathfrak{g})\rangle$$, $$\langle z\cdot g | z\in Z(\mathfrak{g})\rangle$$ are finite dimensional vector spaces.)

Then I am wondering whether $$f \cdot g$$ is also $$Z(\mathfrak{g})$$-finite.

Any comments are appreciated!

• "classical group over $Q$" means "classical form" or "form of classical group over $C$"? (in particular does it allow triality forms of $SO_8$)?
– YCor
Apr 17, 2021 at 9:45
• Shouldn't it follow from $z(f\cdot g) = (zf)\cdot g + f\cdot (zg)$? Apr 18, 2021 at 22:12
• @Sunhajit, No! $Z$ may not be of first order differential operator! Apr 20, 2021 at 6:19