Chevalley–Shephard–Todd theorem - MathOverflow

Chevalley–Shephard–Todd theorem

2010-07-19T06:42:04Z

The wikipedia article claims that the theorem "was first proved by G. C. Shephard and J. A. Todd (1954) who gave a case-by-case proof. Claude Chevalley (1955) soon afterwards gave a uniform proof". I read the paper by Chevalley and it seems that he only proves the implication: "If the group is generated by pseudo-reflections, then the ring of invariants is polynomial". I wonder whether there is a uniform proof of the inverse implication? Where is it written?

Answer by Torsten Ekedahl for Chevalley–Shephard–Todd theorem

2010-07-19T07:33:34Z

There are indeed many presentations (if I remember correctly Bourbaki has it) but the proof is very elegant and short so that I find it hard to refrain from giving it. Let $H$ be the normal subgroup of the finite $G\subset \mathrm{GL}_n$ generated by the pseudo-reflections. By the other direction $X:=\mathbb{A}^n/H$ is again affine space and in particular is smooth. We have an action of $G/H$ on $X$ and a moment's thought reveals that it acts freely in codimension $1$ (as a point fixed by a non-identity element would lie below a reflection hyperplane of $\mathbb{A}^n$ and the fixing element below a pseudo-reflection). Hence $X \to X/(G/H)=\mathbb{A}^n/G$ is étale in codimension $1$. If $\mathbb{A}^n/G$ were smooth, purity of the branch locus would imply that the map were étale. However, that forces $G/H$ to act freely on $X$ but the image of the origin is fixed by all of $G/H$ and therefore $G=H$.

Answer by David Wehlau for Chevalley–Shephard–Todd theorem

2011-07-24T02:37:28Z

Chevalley was interested in the action of (real) Weyl groups and so a reflection to him had determinant -1 and so was a real reflection, i.e. order 2. My understanding is that Serre had seen the paper by Shepard and Todd and so he knew that pseudo-reflections were relevant. He pointed out that Chevalley's proof was valid for pseudo-reflections.