Timeline for Why do Nakajima and Watanabe claim the induced action of a finite linear group on the invariant subring of the reflection subgroup is linearizable?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 12, 2020 at 12:33 | vote | accept | benblumsmith | ||
| Mar 12, 2020 at 12:33 | comment | added | benblumsmith | It seems like the "if" direction of Theorem C goes through regardless, because Gorensteinness of $S^G$ doesn't depend on the grading, thus, if $G/H\subset SL(W)$, the isomorphism of rings $S(W)^{G/H}\cong S^G$ implies the latter is Gorenstein. On the other hand, the "only if" direction is less clear without Braun's result. Do you agree? | |
| Mar 12, 2020 at 12:26 | comment | added | benblumsmith | 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. | |
| Mar 12, 2020 at 9:39 | comment | added | Gregor Kemper | 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). | |
| Mar 11, 2020 at 22:15 | comment | added | benblumsmith | 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? | |
| Mar 11, 2020 at 22:13 | comment | added | benblumsmith | ... if that works, it completely answers the question in the affirmative. | |
| Mar 11, 2020 at 22:08 | comment | added | benblumsmith | 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? | |
| Mar 11, 2020 at 7:56 | history | answered | Gregor Kemper | CC BY-SA 4.0 |