When is the ring of functions of a quotient variety a smooth algebra? Let $V$ be a finite-dimensional complex vector space over an algebraically closed field $k$ of characteristic $0$. Let $k[V]$ be the ring of functions on V. Since $V$ is a smooth (non-singular) variety, we know that $\mathbb C[V]$ is a smooth $k$-algebra.
Assume that $G$ is a finite subgroup of $GL(V)$. In general, the quotient variety $V/G$ is not smooth. However, what can we say about the ring of functions $k[V]^G$ of $V/G$? Is it smooth or non-smooth?
If it is non-smooth under what conditions it is smooth? Thank you.
 A: The definition of smooth you state in the comments can be phrased geometrically as saying the diagonal $\Delta : X \hookrightarrow X \times_k X$ is a regular embedding. Note the flatness is automatic since $k$ is a field. Using the fact that $\Delta^*(I/I^2) \cong \Omega_{X/k}$ where $I$ is the ideal of $\Delta(X)$ and some standard results on formal smoothness, its not hard to see that this definition is equivalent to saying that $X/k$ is formally smooth.
If $X$ is a variety (or more generally for any morphism of finite presentation) formal smoothness is the same as smoothness so the question is really to characterize when is $V/G = \mathrm{Spec} k[V]^G$ smooth when $V$ is a finite dimensional vector space over an algebraically closed field of characteristic $0$ and $G$ a finite group acting linearly. This is answered by the following well known theorem. First we need some background.
Definition: Let $G \subset GL(V)$. An non-identity element $g \in G$ is called a pseudoreflection if it fixes a codimension $1$ subspace of $V$.
Theorem: (Chevalley-Shephard-Todd) With $G$ and $V$ as above, the following are equivalent:

*

*$G$ is generated by pseudoreflections.

*$V/G$ is smooth.

*$k[V]^G$ is isomorphic to a polynomial algebra.

*$k[V]$ is flat as a $k[V]^G$ algebra.

