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I just picked up the paper "The classification of quotient singularities which are complete intersections" by Haruhisa Nakajima and Kei-Ichi Watanabe, which is in the book

  • Greco, Silvio, and Rosario Strano, eds. Complete intersections: lectures given at the 1st 1983 session of the Centro Internationale Matematico Estivo (CIME) held at Acireale (Catania), Italy, June 13-21, 1983. Vol. 1092. Springer, 2006.

Nakajima and Watanabe's paper is pp. 102-120. On p. 103, the authors make an assertion (without comment) that I find surprising or at least very non-obvious. (Here, $V$ is a complex finite-dimensional vector space, $G$ is a finite subgroup of $\mathrm{GL}(V)$, and $S$ is the symmetric algebra on $V$, i.e., a ring of polynomials whose degree 1 component is $V$.)

Theorem C: If $G$ does not contain pseudo-reflections, $S^G$ is a Gorenstein ring if and only if $G\subset \mathrm{SL}(V)$.

The condition "$G$ does not contain pseudo-reflections" is not so serious. If $G$ does contain pseudo-reflections, let $H$ be the subgroup of $G$ generated by all the pseudo-reflections of $G$. Then $S^H$ is a polynomial ring by Theorem A, and the action of $G/H$ is linear and does not contain pseudo-reflections. So we can apply Theorem C to this action.

Emphasis added. Theorem A here is the Chevalley-Shepard-Todd theorem. The assertion seems to be that there is a set $f_1,\dots,f_n$ of polynomial generators for $S^H$ (i.e., a set of algebraically independent generators) such that the action of the group $G/H$ on the ring $S^H$ is induced from a linear action of $G/H$ on the vector space $W=\langle f_1,\dots,f_n\rangle_{\mathbb{C}}$, so that $S^H$ can be seen as the symmetric algebra on $W$.

This is not at all clear to me. The degrees of the generators $f_1,\dots,f_n$ are determined by $H$, and it's possible that they're all distinct (e.g. if $H$ is the reflection group $A_n$). Meanwhile, the action of $G/H$ on $S^H$ is certainly graded. In this case, for the action of $G/H$ on $f_1,\dots,f_n$ to be linear in the above sense, it is necessarily diagonal. This seems like a lot to know for sure.

Question: Why do Nakajima and Watanabe feel sure that the action of $G/H$ on $S^H$ is linear?

Remark: Four years ago I asked a related question which is still unanswered as of today, but Nakajima and Watanabe's situation is more specific than the setting of this previous question. In particular, for them, the graded subring is always the invariant ring of a reflection group. (Also, the ground field is $\mathbb{C}$ as opposed to a general field of characteristic 0.)

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I think one should consider the ideal $I = S^H_+$ of invariants with zero constant coefficient. To give algebra-generators for $S^H$ is the same as to give generators of $I/I^2$ as a vector space. So here dim$_K(I/I^2) = n$. $G/H$ acts linearly on $I$, $I^2$ and on $I/I^2$. Since we are working on the nonmodular case, $I^2$ has a complement $W$ in $I$ (as $G/H$-spaces). So a basis of $W$ yields suitable generators $f_i$. Now there is an isomorphism $S(W) \to S^H$ (with $S(W)$ the symmetric algebra). The isomorphism is $G/H$-equivariant, but NOT graded.

I hope this makes sense. I agree that Nakajima and Watanabe's comment raises a lot of questions, and I haven't answered them all. For instance, is there a problem with applying Theorem C to $S(W)$ since the isomorphism is not graded?

But Amiram Braun has answered them and carried this argument to the modular case by proving that if you assume that $S^H$ is a polynomial ring, then $S^G$ is quasi-Gorenstein iff $G/H$ acts on $I//I^2$ by transformations of determinant 1. See Corollary 3.9.12 in the book of Derksen and Kemper.

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  • $\begingroup$ This is great! ... But okay, $I$ is certainly a homogeneous ideal, therefore so is $I^2$, and since the action of $G/H$ is graded, doesn't it follow that $W$ can be taken to be graded (i.e. possessed of a homogeneous basis)? [I'm thinking we can take a graded basis for $I$ over $\mathbb{C}$, take the standard dot product with respect to that basis, and average it over $G/H$, to get a $G/H$-equivariant dot product, and use that to take the complement.] And if this is true, can't we grade $S(W)$ according to the degrees of a basis for $W$, in which case the isomorphism will be graded? $\endgroup$ – benblumsmith Mar 11 at 22:08
  • $\begingroup$ ... if that works, it completely answers the question in the affirmative. $\endgroup$ – benblumsmith Mar 11 at 22:13
  • $\begingroup$ Incidentally, I can't find the reference to the result of Amiram Braun in Derksen & Kemper? I have the 2002 edition. 3.9.12 seems to be a remark on degree bounds, and Amiram Braun doesn't appear in the index or bibliography? $\endgroup$ – benblumsmith Mar 11 at 22:15
  • $\begingroup$ Yes, $W$ is graded. But when talking about $S(W)$ and doing its invariant theory, the elements of $W$ are always taken to be of degree 1. That's why I said the isomorphism is not graded. Braun's result is mentioned in the new 2015 edition of our book. The result has never been published (as far as I know). $\endgroup$ – Gregor Kemper Mar 12 at 9:39
  • $\begingroup$ I see. So the answer to the question "does $S^H$ have a graded linear subspace $W$ that is $G/H$-invariant and such that a basis for $W$ generates $S^H$ as a polynomial ring?" is "yes", but an arbitrarily chosen result from invariant theory won't necessarily apply to the action of $G/H$ on $S^H$, because $S^H$ is not standard-graded with $W$ as its degree-1 component. $\endgroup$ – benblumsmith Mar 12 at 12:26

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