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Is the following true:

For a representation of a finite group $G$ on $\mathbb{C}^n$, the quotient $\mathbb{C}^n/G$ is a topological manifold if and only if $G$ is generated by pseudo-reflections.

( A pseudo-reflection is an element of $G$ whose fixed set is of codimension $\leq 1$)

Nico Bellic
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    For (1), did you mean to write that $G$ is not generated by pseudo-reflections? If $G$ is generated by pseudo-reflections, then the Chevalley-Shephard-Todd theorem says that the quotient space is a manifold. – Jason Starr Jul 08 '15 at 17:49
  • Jason Starr: yes, you are right. I made the corrections – Nico Bellic Jul 09 '15 at 18:41
  • I believe the answer is "no". First, use Chevalley-Shephard-Todd to reduce to the case that $G$ contains no pseudo-reflections. Thus, for the quotient $q:\mathbb{C}^n\to Q$, the branch locus $B$ in $Q$ has complex codimension $\geq 2$, i.e., real codimension $\geq 4$. But now if you take a small sphere about the origin (assuming locally Euclidean), and remove this real analytic subset of real codimension $\geq 4$, that should still leave a simply connected topological space, contradicting the existence of the cover $q:(\mathbb{C}^n\setminus q^{-1}(B)) \to (Q\setminus B)$. – Jason Starr Jul 09 '15 at 19:11
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    Jason Starr: can you please shed any light on why should it so? – Nico Bellic Jul 09 '15 at 19:36
  • It "should" follow from van Kampen's theorem, and the existence of a stratification of $B$ by manifolds. If I remove a submanifold of real codimension $\geq 3$ from a manifold, then the fundamental group of the punctured tubular neighborhood of the submanifold projects isomorphically to the fundamental group of the submanifold. So, by van Kampen's theorem, the fundamental group of the open complement of the submanifold equals the fundamental group of the original manifold. Now iterate this with each stratum of $B$. – Jason Starr Jul 09 '15 at 19:44
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    To me it seems to be a very important statement. Isn't it surprising if there is no reference for it? – Nico Bellic Jul 09 '15 at 21:37
  • "Isn't it surprising if there is no reference for it?" In the 2-dimensional case, the local fundamental group of the origin in $Q$ is nontrivial, so $Q$ is not locally Euclidean near the origin. I suppose that in the case of higher dimensions, there is the possibility that the homeomorphism of $Q$ with $\mathbb{R}^{2n}$ might restrict to a wild embedding on the strata of $B$. Presumably if the homeomorphism is piecewise linear, this is easy to rule out. – Jason Starr Jul 09 '15 at 22:08
  • Let us continue this discussion in chat. – Nico Bellic Jul 10 '15 at 18:35
  • I don't know why there is a bounty ... it seems Jason has already answered the question! – Geordie Williamson Jul 14 '15 at 20:37
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    Geordie: How many seems, believes and shoulds were in your last math paper? – Nico Bellic Jul 15 '15 at 02:46
  • @Nico: Sorry, I wrote that comment after a glass of wine. – Geordie Williamson Jul 15 '15 at 07:44
  • @Nico: so now I think I understand better: the quotient $\mathbb{C}^n \to \mathbb{C}^n/\Gamma$ is a quotient both in the category of topological spaces, and algebraic varieties. In particular, if $\Gamma$ is generated by pseudoreflections then it will be a topological manifold after all (by classical Chevalley-Shephard-Todd). – Geordie Williamson Jul 15 '15 at 07:45
  • So now we arrive at the converse, and Jason's argument. I agree there is something left to show... – Geordie Williamson Jul 15 '15 at 08:32
  • The topological Chevalley-Shephard-Todd theorem is false in dimensions $n\geq 3$. I just ran across counterexamples in an old MO answer by Greg Kuperberg. – Jason Starr Jul 22 '15 at 19:23

2 Answers2

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Since the bounty is now ended, I will post my comment above as an answer. When the dimension $n$ equals $2$, the quotient of $\mathbb{C}^n$ by a finite group $G$ of holomorphic automorphisms is a topological manifold if and only if $G$ is generated by pseudoreflections. Up to using the positive direction (and much harder direction) of Chevalley-Shephard-Todd, it suffices to consider the case when $G$ contains no pseudoreflections. Thus, the branch locus $B$ of the quotient morphism, $$q : \mathbb{C}^n \to Q,$$ has real codimension $4$. When $n$ equals $2$, this implies that $B$ is isolated. The local fundamental group of $(\mathbb{R}^m,0)$ is trivial for $m\geq 3$, in particular for $m=2n=4$. This contradicts the existence of the covering map, $$ q^*: \mathbb{C}^n\setminus q^{-1}(B)\to Q\setminus B.$$

For all $n$, the natural PL structure on $\mathbb{C}^n/G$ is PL-equivalent to a PL-manifold if and only if $G$ is generated by pseudoreflections. The point here is that there is a triangulation of $Q$, as a PL-space, such that $B$ is a polyhedral subcomplex, cf. the main theorem, p. 170, of Hironaka's classic article.

MR0374131 (51 #10331) Reviewed
Hironaka, Heisuke
Triangulations of algebraic sets.
Algebraic geometry (Proc. Sympos. Pure Math., Vol. 29, Humboldt State Univ., Arcata, Calif., 1974), pp. 165–185. Amer. Math. Soc., Providence, R.I., 1975.
14B99 (32B20 57C15)

If $Q$ were PL-equivalent to a PL-manifold, then we prove, inductively on the strata of $B_k$ of $B$ that have dimension $\leq k$, that excising $B_k$ with $k\leq 2n-3$ does not affect the fundamental group of $Q$. Since $\text{dim}_{\mathbb{R}}(B)\leq 2n-4$, finally we conclude that the pushforward homomorphism, $$\pi_1(Q\setminus B) \to \pi_1(Q),$$ is an isomorphism. As above, this contradicts the existence of the covering $q^*$.

For $n>3$, there do exist wild embeddings of $(n-3)$-dimensional PL spaces in $n$-manifolds whose open complement has nontrivial local fundamental groups. If I have properly understood the following article, every such fundamental group $\pi$ is perfect and has trivial $H_2(\pi)$.

MR0994411 (90f:57025) Reviewed
Ferry, Steven C.(1-KY); Pedersen, Erik Kjaer(1-KY); Vogel, Pierre(1-KY)
On complements of codimension-3 embeddings in $S^n$.
Topology Appl. 31 (1989), no. 2, 197–202.
57N35 (57N45 57Q35)

This rules out many cases. If there is a good reason for considering other wild cases, then I recommend "localizing" on the components of $B$ of maximal dimension. There is a relation between the strata of $B$ and the corresponding inertia subgroups of $G$, which seems to distinguish this case from the general case of Ferry-Pedersen-Vogel.

Edit. Apparently the topological Chevalley-Shephard-Todd theorem is FALSE in dimensions $\geq 3$, and the issue is precisely that an algebraic variety that is not a PL-manifold may nonetheless be a topological manifold. I just learned all of this from Greg Kuperberg's answer (and the articles linked in his answer) for this MO question: Algebraic varieties which are topological manifolds.

In particular, for the icosahedral group $\Gamma \subset \textbf{SL}_2(\mathbb{C})$, extending the action to $\mathbb{C}^2\times \mathbb{C}$ by taking the trivial action on the $\mathbb{C}$-factor, the quotient $\mathbb{C}^3/\Gamma$ is a topological manifold.

Community
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Jason Starr
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    Yeah, binary icosahedral group! The unique perfect fixed-point-free group, also known as $SL_2(\mathbb{F}_5)$-- best finite group ever – Nico Bellic Jul 23 '15 at 20:19
  • Jason: Do you by any chance have a reference (or a proof?..) for the statement that $\mathbb{C}^3/\Gamma$ is a topological manifold? I definitely believe this. I think that $\mathbb{C}^2/\Gamma$ is a homology sphere, from which the desired result would follow by Milnor's Double suspension problem. But I don't have a proof/reference that $\mathbb{C}^2/\Gamma$ is a homology sphere. – Nico Bellic Jul 25 '15 at 19:34
  • In Kuperberg's answer (linked above), he includes a reference. – Jason Starr Jul 25 '15 at 20:04
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To elaborate on Jason's great answer.

The answer is no and the counter-example is the binary icosahedral group together with the sum of an irreducible representation of degree $2$ and a unit representation, over $\mathbb{C}$.

Let $\Gamma$ be the binary icosahedral group--the unique perfect and fixed-point-free finite group, and consider one of it irreducible representations of degree $2$ over complex numbers (it is fixed-point free), for example the one resulting from embedding of $\Gamma$ in $SL_2(\mathbb{C})$. Then the quotient $\mathbb{C}^2/\Gamma$ is called the Poincare Homology Sphere. Now by the affirmative answer to Milnor's double suspension problem, the double suspension of a homology sphere is a true sphere. Now the image of the origin in $(\mathbb{C}^2\times \mathbb{C})/\Gamma$, where the second factor comes with a trivial $\Gamma$-action, looks locally like the vertex of the cone over double suspension of the Poincare Homology Sphere, hence the quotient space is locally Euclidean at the image of the origin. But then, given the scaling action, it is locally Euclidean everywhere.

EDIT That said, I claim the following sufficient condition:

Suppose no sub-quotient of a finite group $\Gamma$ is isomorphic to the binary icosahedral group. Then $\Gamma$ satisfies the topological Chevalley-Shephard-Todd Theorem:

Given a linear representation of $\Gamma$, finite degree over $\mathbb{C}$, the quotient is a topological manifold iff $\Gamma$ is generated by pseudo-reflections.

Nico Bellic
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