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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!

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

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