# Lie algebra stability implies Lie group stability for a certain representation

Suppose $G$ is a connected complex algebraic group with lie algebra $\mathfrak{g}$ and $V$ a (potentially infinite-dimensional) vector space with actions of $G$ and $\mathfrak{g}$ which are 'compatible', i.e. for all $g\in G$, $X\in\mathfrak{g}$, $v\in V$, we have $$g\cdot X\cdot v=Ad(g)(X)\cdot g\cdot v$$ Note that I am not assuming that $V$ is a 'nice' representation of $G$, I'm only assuming it is a group action by its complex points, $G(\mathbb{C})$. Although see below for where the action originates.

My question is: suppose that $\mathfrak{g}$ stabilizes some finite-dimensional subspace $W\subseteq V$. Then must $G$ also stabilize $W$?

Actual motivation and setting of question: I have an action of $G$ on an algebraic variety $X$, and there is a finite-dimensional subspace $W$ of the field of rational functions on $X$, that the action of $\mathfrak{g}$ preserves, and I want to know if that implies $G$ must preserve it as well.

• Your definition of compatibility is a bit weird. It says that the action of the Lie algebra on V is G-equivariant, and that is quite not good enough to get what you want. – Mariano Suárez-Álvarez Mar 29 '18 at 5:06
• In your situation, you really have a lot more than you are using. Perhaps you should state precisely the situation you are actually interested in, instead of trying to axiomatize it? – Stephen Mar 29 '18 at 17:33
• Thanks for your responses. I was away from technology for several days and so I'm only seeing this now. I think you are both right, and Dunay's answer below is correct as well. Clearly more hypotheses are needed... – freeRmodule Apr 2 '18 at 16:06

I think the answer is no. Just consider an any Lie group $G$ action on a vector space and have its lie algebra $\mathfrak{g}$ act trivially on it. Then the actions will be compatible. But, any subspace will be stabilized by $\mathfrak{g}$, although some may not be stabilized by $G$.