# Is there a topological Chevalley-Shephard-Todd Theorem?

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 $\leq 1$)

• 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 8 '15 at 17:49
• Jason Starr: yes, you are right. I made the corrections – Nico Bellic Jul 9 '15 at 18:41
• Jason Starr: can you please shed any light on why should it so? – Nico Bellic Jul 9 '15 at 19:36
• To me it seems to be a very important statement. Isn't it surprising if there is no reference for it? – Nico Bellic Jul 9 '15 at 21:37
• Geordie: How many seems, believes and shoulds were in your last math paper? – Nico Bellic Jul 15 '15 at 2:46

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.

• 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

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.