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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 $SL(n,\mathbb{C})$?

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The ambient group is $\text{U}(n)$ without loss of generality ($G$ preserves some inner product on $\mathbb{C}^n$ by the standard averaging argument). –  Qiaochu Yuan Jun 29 '13 at 10:02
    
What do you mean exactly by "the origin is the only singular point" ? Does $G$ act freely on $S^{2n-1}$ ? –  BS. Jun 29 '13 at 11:55
    
@BS: Yes, i assume that $G$ acts freely on $S^{2n-1}$. If in the question is not clear i assume the action of $G$ is linear (the one of the matrix group). –  Italo Jun 29 '13 at 11:59

1 Answer 1

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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.

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