Timeline for Basis of coinvariant algebra on which reflection group acts as regular representation
Current License: CC BY-SA 3.0
15 events
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| May 1, 2018 at 22:40 | comment | added | Vladimir Dotsenko | @SamHopkins no, the recipe I have in mind certainly does not produce Grothendieck polynomials since the Grothendieck polynomial for the longest element you mention is a monomial, while the realisation of coinvariants I am talking about is the space of harmonic polynomials (polynomials annihilated by all $S_n$-invariant differential operators with constant coefficients), and a monomial cannot be harmonic. | |
| May 1, 2018 at 21:27 | comment | added | Chris McDaniel | @SamHopkins Regarding your edit: note that your basis is the same basis I give in my conjecture below. In fact I claim my formula will hold whenever W is abelian. | |
| May 1, 2018 at 17:30 | comment | added | Sam Hopkins | @LSpice: This is the usual notion of coinvariant algebra for finite reflection groups, as far as I know. In the case of e.g. $W=S_n$ this algebra is isomorphic to the cohomology of the full flag manifold $Fl_n$. | |
| May 1, 2018 at 17:19 | comment | added | LSpice | Is this the usual usage of 'coinvariant' in this context? I am used to it meaning "largest quotient on which $W$ acts trivially", rather than "quotient by a module of $W$-invariants". | |
| May 1, 2018 at 16:45 | comment | added | Sam Hopkins | Right, I should’ve also mentioned Chevalley’s original proof uses Galois theory (which also feels “nonconstructive”) | |
| May 1, 2018 at 16:42 | comment | added | Vladimir Dotsenko | Another remark: while indeed many/most proofs compute characters, for $S_n$ this can alternatively be proved using Galois theory. Namely, take $L=\mathbb{C}(x_1,\ldots,x_n)$. The field of $S_n$-invariants of this would be rational symmetric functions which are $K=\mathbb{C}(e_1,\ldots,e_n)$. We see that $[L:K]=n!$ and in fact is the regular representation by the normal basis theorem. Passing from that result to the statement about the rings before taking fields of fractions is an easy exercise :-) | |
| May 1, 2018 at 16:35 | comment | added | Vladimir Dotsenko | I think that maybe deforming the harmonic polynomials / space of all iterated partial derivatives of the Vandermonde in the case of $S_n$ is promising. At least it is easy to see how to try and make things inhomogeneous, by doing things like $\partial_i+a$ instead of $\partial_i$. | |
| May 1, 2018 at 16:18 | history | edited | Sam Hopkins | CC BY-SA 3.0 |
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| May 1, 2018 at 15:03 | comment | added | darij grinberg | True... I guess this is why those bases are so hard to find. | |
| May 1, 2018 at 13:56 | comment | added | Sam Hopkins | @darijgrinberg: the descent basis still has the problem that it is homogenous, correct? | |
| Apr 30, 2018 at 20:52 | answer | added | Chris McDaniel | timeline score: 1 | |
| Apr 30, 2018 at 19:07 | history | edited | Sam Hopkins | CC BY-SA 3.0 |
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| Apr 30, 2018 at 18:51 | answer | added | Richard Stanley | timeline score: 6 | |
| Apr 30, 2018 at 18:10 | comment | added | darij grinberg | Even for $W = S_n$, this question is extremely interesting! The descent basis seems to be a step in the right direction, but still not what we're looking for. | |
| Apr 30, 2018 at 17:40 | history | asked | Sam Hopkins | CC BY-SA 3.0 |