9
$\begingroup$

Short version: if $G$ is a Coxeter group and $H \subset G$ is a parabolic subgroup, both acting on a space $V$, is it true that the invariant-coinvariant algebra $(S(V)_G)^H$ has a natural bilinear form induced by taking coefficients in the top component?

Now the long, detailed version. Let $G$ be a Coxeter group acting on a real vector space $V$ with a fixed generating set of reflections $S$. We can form the symmetric algebra $S(V)$, the invariant subalgebra $S(V)^G \subset S(V)$, and the ideal $I_G \subset S(V)$ generated by elements of $S(V)^G$ of positive degree. The coinvariant algebra is defined as $S(V)_G = S(V)/I_G$. By the Cehvalley-Shephard-Todd theorem it is isomorphic to the regular representation of $G$ as a $G$-module. If $H \subset G$ is a parabolic subgroup then the $H$-invariants $(S(V)_G)^H$ form a subalgebra of $S(V)_G$, which is finite dimensional and whose top nonzero component is $1$-dimensional over $\mathbb R$, say spanned by $s$. Then $(S(V)_G)^H$ has a nice bilinear pairing given by taking $(f,g)$ to be the coefficient of $s$ in the product $fg$ for all $f, g \in (S(V)_G)^H$.

If $G$ is a Weyl group, the algebra $(S(V)_G)^H$ is isomorphic to the cohomology ring of a generalized flag variety by a famous theorem of Borel. Plus the bilinear form described above is the pullback of the Poincaré pairing in said cohomology, and in particular it is nondegenerate. Questions:

  1. Can the non-degeneracy of the bilinear form be proved by purely combinatorial methods, without appealing to Borel's theorem?

  2. Can it be extended to all pairs $(G,H)$ where $G$ is just a finite Coxeter group?

$\endgroup$
10
$\begingroup$

Yes. By Chevalley-Shepard-Todd, $S(V)^G$ and $S(V)^H$ are polynomial rings. Let $S(V)^G=\mathbb{R}[g_1, \ldots, g_n]$ and $S(V)^H = \mathbb{R}[h_1,\ldots, h_n]$ where the $g_i$ and $h_i$ are homogenous. Then $$S(V)_G^H = \mathbb{R}[h_1,\ldots,h_n]/\langle g_1,\ldots, g_n \rangle.$$ Here the denominator is the ideal of $\mathbb{R}[h_1,\ldots,h_n]$ generated by the $g_i$, and $g_i \in \mathbb{R}[h_1,\ldots,h_n]$ because $S(V)^G \subseteq S(V)^H$.

So $S(V)^H_G$ is a complete intersection, and therefore Gorenstein. Saying $R$ is a finite dimensional Gorenstein $k$ algebra (for $k$ a field) exactly means that there is a linear functional $\int: R \to k$ such that $\langle f,g \rangle = \int fg$ is a perfect pairing. If $R$ is graded, then this functional is "take the top degree piece". See Eisenbud, Commutative Algebra with a view toward Algebraic Geometry Chapter 21.2 for a good overview of finite dimensional Gorenstein algebras.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.