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The center of the universal affine Hecke algebra of type A

Statement

Let $H_{n}^{a f f}$ be the universal algebra over $R_{0} = Z [Q, Q^{- 1}]$ from The multiplicative Bernstein presentation of the affine Hecke algebra of type A, and let $A = R_{0} [X_{1}^{\pm 1}, \dots, X_{n}^{\pm 1}]$ . Write $Z (H_{n}^{a f f}) : = {z \in H_{n}^{a f f} : z h = h z for every h \in H_{n}^{a f f}}$ and $A^{S_{n}} : = {f \in A : w (f) = f for every w \in S_{n}}$ . Then, over this universal coefficient ring,

Z (H_{n}^{a f f}) = A^{S_{n}} .

This theorem is stated only over $R_{0}$ ; it makes no assertion that the center of an arbitrary scalar specialization is the scalar extension of this center.

Facts & Assumptions

Given: The universal affine algebra, its Bernstein generators, its PBW basis, and the rational skew group representation.

[F1]

Bernstein PBW basis for the affine Hecke algebra of type A: every element of $H_{n}^{a f f}$ has a unique finite expansion $\sum_{w \in S_{n}} f_{w} T_{w}$ with $f_{w} \in A$ , and the coefficient map $A \to H_{n}^{a f f}$ is injective.

[F2]

Demazure–Lusztig operators satisfy the affine type-A relations: the map $ρ : H_{n}^{a f f} \to K ⋊ S_{n}$ is an algebra homomorphism, with each simple operator of the form $c_{i} [1] + d_{i} [s_{i}]$ and $d_{i} \neq 0$ .

[F3]

The rational skew group algebra for type A: $K ⋊ S_{n}$ has basis $[u]$ over $K$ and multiplication $(a [u]) (b [v]) = a u (b) [u v]$ ; in particular $[u] X_{j} = u (X_{j}) [u]$ .

[F4]

The multiplicative Bernstein presentation of the affine Hecke algebra of type A: $T_{i} f - s_{i} (f) T_{i} = (Q - 1) D_{i} (f)$ for $f \in A$ , and $D_{i} (f) = (f - s_{i} (f)) / (1 - X_{i} / X_{i + 1})$ .

[F5]

The type-A weight lattice, Laurent ring, and divided differences: $A$ is an integral domain with Laurent monomial basis $(X^{λ})_{λ \in Z^{n}}$ and embeds injectively in its fraction field $K$ .

Proof

Given: The data of the Statement.

1.1F1F3F5givenalgebra

Let $z \in Z (H_{n}^{a f f})$ and write its unique PBW expansion $z = \sum_{w} f_{w} T_{w}$ using [F1]. Its image $p = ρ (z)$ commutes with each $X_{j} [1]$ . Write $p = \sum_{u \in S_{n}} g_{u} [u]$ in the skew-group basis of [F3]. The coefficient of $[u]$ in $p X_{j} - X_{j} p$ is $g_{u} (u (X_{j}) - X_{j})$ by [F3]. Since this is zero for every $j$ , any $u$ with $g_{u} \neq 0$ fixes each Laurent variable. The variables are distinct basis monomials in $A$ and remain distinct in its fraction field $K$ by [F5], so a permutation fixing them all is the identity. Therefore $p = g_{1} [1] \in K [1]$ .

2.1F1F2F5step 1.1algebra

Suppose some $f_{w} \neq 0$ for a nonidentity permutation. Choose the maximum $L > 0$ among lengths of such $w$ . For a reduced expression $w = s_{i_{1}} \dots s_{i_{L}}$ , expansion of the factors $c_{i_{k}} [1] + d_{i_{k}} [s_{i_{k}}]$ from [F2] shows that every term in $ρ (T_{w})$ has length at most $L$ , and its coefficient at $[w]$ is the nonzero product $b_{w} = \prod_{k = 1}^{L} (s_{i_{1}} \dots s_{i_{k - 1}}) (d_{i_{k}})$ : omitting any reflection factor leaves fewer than $L$ simple transpositions, so cannot produce $w$ . A different $v$ of length $L$ contributes to $[w]$ only if $v = w$ , and one of smaller length cannot contribute. But step 1.1 shows $p$ has no nonidentity permutation coefficient. Thus $f_{w} b_{w} = 0$ in $K$ , and since $b_{w} \neq 0$ and $A ↪ K$ by [F5], every such $f_{w} = 0$ , contradicting the choice of $L$ . It follows that $z = f_{1} \in A$ .

3.1F1F4step 2.1algebra

Since $z$ is central and $z \in A$ by step 2.1, [F4] gives $0 = T_{i} z - z T_{i} = (s_{i} (z) - z) T_{i} + (Q - 1) D_{i} (z)$ . The PBW basis of [F1] is an $A$ -basis, so the coefficients of $T_{s_{i}} = T_{i}$ and $T_{i d} = 1$ vanish separately. Thus $s_{i} (z) = z$ for each $i$ , and the adjacent transpositions generate $S_{n}$ , so $z \in A^{S_{n}}$ .

4.1F1F4step 3.1∎

If $f \in A^{S_{n}}$ , then $s_{i} (f) = f$ and $D_{i} (f) = 0$ for every $i$ by [F4], so the Bernstein relation gives $T_{i} f = f T_{i}$ . Since $A$ is commutative and the algebra is generated by $A$ and the $T_{i}$ , $f$ commutes with every generator and is central. This reverse inclusion with step 3.1 proves $Z (H_{n}^{a f f}) = A^{S_{n}}$ over $R_{0}$ . No specialization or Choice principle is used.

Depends on

Used by

Symmetric Laurent polynomials in the center Example

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

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)
Eric Opdam and Maarten Solleveld, Homological algebra for affine Hecke algebras (2007) (standard reference, not scraped)