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?


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.


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