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The rational skew group algebra for type A

Definition

Keep $R_{0}$ , $A$ , and the action of $S_{n}$ from The type-A weight lattice, Laurent ring, and divided differences. Let $K : = Frac (A)$ , the fraction field supplied by Multivariate polynomial and Laurent rings over commutative rings, domains and fraction fields. Each $w \in S_{n}$ acts on $K$ by the unique extension

w (f / g) : = w (f) / w (g) (f, g \in A, g \neq 0) .

The rational skew group algebra $K ⋊ S_{n}$ is the $K$ -vector space of finite sums $\sum_{w \in S_{n}} a_{w} [w]$ , with multiplication determined by

(a [w]) (b [v]) = a w (b) [w v] (a, b \in K, w, v \in S_{n}) .

Its unit is $1 [1]$ and its $K$ -basis is $([w])_{w \in S_{n}}$ . It acts on $K$ by

(a [w]) \cdot f : = a w (f) (a, f \in K, w \in S_{n}) .

Facts & Assumptions

Given: The domain $A$ , its fraction field $K$ , and the permutation action of $S_{n}$ on $A$ from the preceding definition.

[F1]

The type-A weight lattice, Laurent ring, and divided differences: $A$ is a commutative integral domain and each $w \in S_{n}$ permutes its Laurent variables.

[F2]

Multivariate polynomial and Laurent rings over commutative rings, domains and fraction fields: the fraction field of an integral domain is a field containing that domain injectively.

Proof

Given: The data of the Definition.

1.1F1F2algebra

For $f / g \in K$ with $g \neq 0$ , the denominator $w (g)$ is nonzero because the automorphism $w$ of $A$ is injective. If $f / g = f^{'} / g^{'}$ , then $f g^{'} = f^{'} g$ , so applying $w$ gives $w (f) w (g^{'}) = w (f^{'}) w (g)$ ; hence $w (f) / w (g) = w (f^{'}) / w (g^{'})$ . Thus the displayed formula is well-defined on $K$ and defines a field automorphism, with inverse the action of $w^{- 1}$ .

2.1step 1.1algebra

The finite-support sums form a $K$ -module with basis symbols $[w]$ . Multiplication is closed because $w v \in S_{n}$ and $a w (b) \in K$ . For $a, b, c \in K$ and $w, v, u \in S_{n}$ , $((a [w]) (b [v])) (c [u]) = a w (b) (w v) (c) [w v u]$ , while $(a [w]) ((b [v]) (c [u])) = a w (b v (c)) [w v u] = a w (b) (w v) (c) [w v u]$ . These agree because $w$ acts as a field automorphism and the action is a group action; distributivity gives associativity for finite sums.

3.1step 2.1algebra

The symbol $1 [1]$ is a two-sided identity since the identity permutation fixes every coefficient and $1 w (b) = b$ . By construction the symbols $[w]$ are a $K$ -basis, and multiplication also gives $[w] b = w (b) [w]$ . Since every permutation fixes $R_{0}$ , the displayed operations define a unital associative $R_{0}$ -algebra.

4.1F1step 2.1algebra∎

The stated action is an algebra representation: $(a [w]) \cdot ((b [v]) \cdot f) = a w (b v (f)) = a w (b) (w v) (f) = ((a [w]) (b [v])) \cdot f$ , using the multiplication from step 2.1. Additivity extends this identity to finite sums, so multiplication in $K ⋊ S_{n}$ agrees with composition of its operators on $K$ .

Depends on

Used by

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Sources

Alexander A. Kirillov, Jr., Lectures on the affine Hecke algebras and Macdonald conjectures (Harvard lectures, 1994; arXiv:math/9501219v1) (standard reference, not scraped)
I. G. Macdonald, Affine Hecke algebras and orthogonal polynomials, Séminaire Bourbaki 797, Astérisque 237 (1996), pp. 189–207 (standard reference, not scraped)