Suppose that $X$ is a complex algebraic variety, i.e. it is integral, separated, and of finite-type over $\mathbb{C}$. Let $\mathfrak{g}$ be a finite-dimensional complex Lie algebra.


Suppose that $\mathfrak{g}$ acts on $X$, i.e. we have a homomorphism $\phi:\mathfrak{g}\to \mathrm{Vec}(X)$, where $\mathrm{Vec}(X)$ is the Lie algebra of vector fields on $X$. Suppose further that for every closed point $x\in X$, the canonical morphism
$$ \mathfrak{g}\to T\_xX$$
is surjective. This should be the way to say the action is 'homogeneous'. My question is: does this imply $X$ is smooth? Or at least that the tangent sheaf is locally free? Feel free to assume $X$ is, say, normal if that is necessary. 



Lie algebra action on complex algebraic variety
It seems to me that the answer should be positive. 


Suppose by contradiction, that there is a singular point $x\in X$. Let us take an affine neighbourhood $U$ of $x$. Let $f\_1,\ldots, f\_n$ be the regular functions that generate the ring of functions on $U$. This functions give us an embedding $\varphi: U\to \mathbb C^n$ and the lie algebra action on $U$ lifts to an action on $\mathbb C^n$. Now, locally (say in analytic topology) this action should preserve $\varphi(U)$. But this is a contradiction. Indeed, take at $x$ a vector $v$ that lies in the tangent space to $\varphi(U)$ at $x$, but not in the tangent cone. Then by our assumptions there should exist a vector field whose local flow close to $\varphi(x)$ preserves $\varphi(U)$ and which is equal to $v$ at $x$. But such a field clearly can not preserve $\varphi(U)$ since it pushes $x$ out of $\varphi(U)$, contradiction.


