Chevalley–Shephard–Todd theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T08:52:05Z http://mathoverflow.net/feeds/question/32450 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32450/chevalleyshephardtodd-theorem Chevalley–Shephard–Todd theorem Roman Fedorov 2010-07-19T06:42:04Z 2011-07-24T02:37:28Z <p>The <a href="http://en.wikipedia.org/wiki/Chevalley%E2%80%93Shephard%E2%80%93Todd_theorem#Statement_of_the_theorem/%22wikipedia%20article%22" rel="nofollow">wikipedia article</a> 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? </p> http://mathoverflow.net/questions/32450/chevalleyshephardtodd-theorem/32456#32456 Answer by Torsten Ekedahl for Chevalley–Shephard–Todd theorem Torsten Ekedahl 2010-07-19T07:33:34Z 2010-07-21T22:04:55Z <p>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 $G=H$.</p> http://mathoverflow.net/questions/32450/chevalleyshephardtodd-theorem/71097#71097 Answer by David Wehlau for Chevalley–Shephard–Todd theorem David Wehlau 2011-07-24T02:37:28Z 2011-07-24T02:37:28Z <p>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.</p>