Let $G$ be a finite subgroup of $SL(n,\mathbb{C})$. Let $G$ act on $\mathbb{C}^{n}$ with the action induced by $SL(n,\mathbb{C})$ and the induced action of $G$ on the unit sphere $S^{2n-1}$ is free. My question is the following: is there a $N>0$ s.t.there exist a $G$-invariant holomorphic map $i\_{N}$  

$$i\_{N}:\mathbb{C}^{n}\rightarrow \mathbb{C}^{N}$$ 
and 
$$i\_{N}:(\mathbb{C}^{n}\setminus\left\{0\right\})/G\rightarrow \mathbb{C}^{N}$$
is a smooth embedding? If it is not the case in general, are there conditions that guarantee the existence of such a map?


Does the situation change if the ambient group is $U(n)$ instead of $SL(n,\mathbb{C})$?



Embeddings of of quotient singularities
Yes. By Hilbert's theorem, the ring of $G$-invariant functions on $\mathbb C^n$ is finitely generated. Say the number of generators is $N$, then this gives a $G$-invariant holomorphic map to $\mathbb C^N$.


If one defines the quotient $\mathbb C^n/ G$ as the spectrum of this ring, then the map is clearly an embedding, and since, by computing the tangent spaces, the space being embedded is smooth, it is a smooth embedding. But if we take a complex-geometry perspective on what the quotient is, there is someting to check. 


I claim this map is a smooth embedding. We will check this by checking that the map separates points and tangent vectors. If $P$ and $Q$ are two nonzero points of $\mathbb C^n$ that are not in the same $G$-orbit, then let $f$ be a function that vanishes at $P$ but not $Q$. The product of all the $G$-conjugates of $f$ is invariants, and separates $P$ and $Q$.


Similarly, if two tangent vectors at a nonzero point $P$ are distinct, we can find a function $f$ that vanishes at $P$ but not any $G$-conjugate point, and with a derivative that vanishes along one tangent vector but not the other. Here we use the fact that $G$ acts freely on $\mathbb C^n - 0$. Then multiplying all the $G$-conjugates of $f$ produces a function that vanishes on one vector but not the other. (Alternately, if the tangent vectors are in the same direction, we just need to find a map with nonvanishing derivative in that direction - the same construction works.


