Suppose $G$ is a connected complex algebraic group acting on a variety $X$. Write $\mathfrak{g}$ for the Lie algebra of $G$. Then both $G$ and $\mathfrak{g}$ act on $\mathbb{C}(X)$, the ring of rational functions on $X$.
We clearly have an inclusion $\mathbb{C}(X)^{G}\subseteq\mathbb{C}(X)^{\mathfrak{g}}$, since $\mathfrak{g}$ is acting by infinitesimal translations. My question is, do we also have the opposite inclusion, i.e. do we get $\mathbb{C}(X)^{G}=\mathbb{C}(X)^{\mathfrak{g}}$?