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Group algebra and finite-field specializations of the generic Hecke algebra

Statement

Let $H_{v} (n)$ be the generic type-A Hecke algebra over $A = Z [v^{\pm 1}]$ and let $q$ be a prime power. (1) The specialization $v \mapsto 1$ is an isomorphism of $C$ -algebras $C \otimes_{A, v \mapsto 1} H_{v} (n) ≅ C [S_{n}],$ carrying $T_{w}$ to $w$ ; (2) the specialization $v \mapsto q$ is an isomorphism of $C$ -algebras $C \otimes_{A, v \mapsto q} H_{v} (n) ≅ H = e_{B} C [G] e_{B}, G = {GL}_{n} (F_{q}),$ carrying the generic generator $T_{i}$ to the standard basis element $T_{s_{i}}$ ; (3) both specializations are semisimple $C$ -algebras, and the isomorphisms are compatible with the standard bases ( ${T_{w}}$ in each case). No choice principle is used.

Facts & Assumptions

Given: The generic type-A Hecke algebra $H_{v} (n)$ over $A = Z [v^{\pm 1}]$ with generators $T_{i}$ , the symmetric group $S_{n}$ with simple transpositions $s_{i}$ , the finite Hecke algebra $H = e_{B} C [G] e_{B}$ of $G = {GL}_{n} (F_{q})$ , and the specializations $v \mapsto 1$ and $v \mapsto q$ .

[F1]

$H_{v} (n)$ is the quotient of the free unital associative $A$ -algebra on $T_{1}, \dots, T_{n - 1}$ by the relations $T_{i}^{2} = (v - 1) T_{i} + v$ , the braid relations and the distant commutations; for every unit $v_{0} \in R^{\times}$ the specialization $R \otimes_{A} H_{v} (n)$ is presented over $R$ by the same relations with $v$ replaced by $v_{0}$ . It is free over $A$ with basis the products $T_{w}$ along reduced words, so its rank is $n!$ (The generic type-A Hecke algebra, The standard basis of the generic type-A Hecke algebra).

[F2]

The specialization of $H_{v} (n)$ at $v \mapsto q$ is isomorphic to $H$ ; the isomorphism carries the generator $T_{i}$ to the standard basis element $T_{s_{i}} = q e_{B} {\dot{s}}_{i} e_{B}$ and the generic basis element $T_{w}$ to the standard basis element $T_{w}$ of $H$ (The type-A Iwahori-Hecke presentation of the finite Hecke algebra).

[F3]

$S_{n} = ⟨ s_{1}, \dots, s_{n - 1} ∣ s_{i}^{2} = 1, s_{i} s_{i + 1} s_{i} = s_{i + 1} s_{i} s_{i + 1}, s_{i} s_{j} = s_{j} s_{i} (∣ i - j ∣ > 1) ⟩$ ; for $n = 0, 1$ the trivial group has the empty presentation (The symmetric group has the Coxeter presentation).

[F4]

The group algebra of a finite group has dimension equal to the group order, so $\dim_{C} C [S_{n}] = n!$ (If $G$ is finite then $\dim_{k} k [G] = ∣ G ∣$ ).

[F5]

$C [S_{n}]$ is semisimple, because $char C = 0$ does not divide $∣ S_{n} ∣ = n!$ (Maschke's theorem for finite groups over fields whose characteristic does not divide $∣ G ∣$ ).

[F6]

$H$ is a semisimple finite-dimensional $C$ -algebra of dimension $n!$ (The finite spherical Hecke algebra is semisimple with nondegenerate trace form).

[F7]

Tensoring over a commutative ring preserves cokernels and surjections, so base change of a quotient presentation of a free algebra is the quotient of the base-changed free algebra by the images of the relators (Tensoring is right exact).

Proof

technique · direct

1.1F1F3F4F7algebra

At $v \mapsto 1$ the relators of [F1] become $T_{i}^{2} = 1$ together with the braid and commutation relators, so by [F7] the specialization $C \otimes_{A, v \mapsto 1} H_{v} (n)$ is the $C$ -algebra with generators $τ_{i}$ and these relations, and it has $C$ -basis the images of $T_{w}$ by [F1], hence dimension $n!$ . By the Coxeter presentation [F3] the assignment $τ_{i} \mapsto s_{i}$ extends to a unital algebra homomorphism onto $C [S_{n}]$ , which is surjective because the $s_{i}$ generate $S_{n}$ ; both algebras have dimension $n!$ by [F4], so it is an isomorphism. A reduced word $w = s_{i_{1}} \dots s_{i_{ℓ}}$ gives $τ_{i_{1}} \dots τ_{i_{ℓ}} \mapsto s_{i_{1}} \dots s_{i_{ℓ}} = w$ , so the basis element $T_{w}$ is carried to $w$ : clause (1).

1.2F2

By [F2] the specialization at $v \mapsto q$ is isomorphic to $H$ with $T_{i} \mapsto T_{s_{i}}$ and $T_{w} \mapsto T_{w}$ : clause (2).

2.1F2F5F6step 1.1step 1.2

The specialization at $v \mapsto 1$ is $C [S_{n}]$ , which is semisimple by [F5], and the specialization at $v \mapsto q$ is $H$ , which is semisimple of dimension $n!$ by [F6]; in both cases the isomorphisms of steps 1.1 and 1.2 match the standard bases $T_{w}$ , so the specializations are semisimple and basis-compatible: clause (3).

3.1step 1.1step 1.2step 2.1∎

Step 1.1 proves clause (1) with the basis compatibility, step 1.2 proves clause (2), and step 2.1 proves the semisimplicity and basis compatibility of clause (3). Both specializations are base changes of a free finite-rank algebra along explicit ring homomorphisms, and all dimensions and index sets are finite, so no choice principle is used.

Depends on

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Sources

Ivan Losev, Lecture 8: Representations of GL_n(F_q) - Section 2.2 ($H_{\mathbb C,z}(n)$ and $H_{\mathbb C,1}(n)\cong\mathbb C S_n$), PDF pp. 4-5 (standard reference, not scraped)
Jay Taylor, Finite Reductive Groups - Proposition 5.16 (specializations at $u_s=1$ and $u_s=q_s$), printed p. 45 (standard reference, not scraped)
Olivier Dudas and Jean Michel, Lectures on Finite Reductive Groups and Their Representations - Sections 11.1-11.2 (the specialization $q\mapsto1$ is the group algebra), printed pp. 46-47 (standard reference, not scraped)