ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passaudited 2026-10-08

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Rank-one Hecke multiplication in both normalizations

Example

Let $S = {s}$ with $m (s, s) = 1$ , so that $W = {1, s} ≅ Z / 2$ with $ℓ (1) = 0$ and $ℓ (s) = 1$ (Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups). The odd-edge graph has one vertex and one component, so $R = Z [v^{\pm 1}]$ with $v : = v_{s}$ is the coefficient ring of Universal parameters, the generic Coxeter Hecke algebra and generator conjugacy, and $H$ is the quotient of the free associative $R$ -algebra $R ⟨ T_{s} ⟩$ by the single relation $(T_{s} - v) (T_{s} + v^{- 1}) = 0$ .

Normalized table. ${1, T_{s}}$ is an $R$ -basis of $H$ (The standard basis of the generic Hecke algebra and base change), and $1 \cdot 1 = 1, 1 \cdot T_{s} = T_{s} = T_{s} \cdot 1, T_{s}^{2} = (v - v^{- 1}) T_{s} + 1,$ equivalently $(T_{s} - v) (T_{s} + v^{- 1}) = 0$ . Moreover $T_{s}^{- 1} = T_{s} - (v - v^{- 1})$ , so $T_{s}^{- 1} T_{s} = T_{s} T_{s}^{- 1} = 1$ (The reversal anti-involution, generator invertibility, the bar operator and the multiplicative normalization).
Multiplicative table. Put $Q : = v^{2}$ and $S_{s} : = v T_{s}$ . Then ${1, S_{s}}$ is again an $R$ -basis, $S_{s}$ is a unit, and $1 \cdot 1 = 1, 1 \cdot S_{s} = S_{s} = S_{s} \cdot 1, S_{s}^{2} = (Q - 1) S_{s} + Q,$ equivalently $(S_{s} - Q) (S_{s} + 1) = 0$ ; indeed $S_{s}^{- 1} = Q^{- 1} (S_{s} + 1 - Q)$ .
Conversion. The two tables are interconverted by $S_{s} = v T_{s}$ , $T_{s} = v^{- 1} S_{s}$ : substituting in $T_{s}^{2} = (v - v^{- 1}) T_{s} + 1$ gives $v^{- 2} S_{s}^{2} = (1 - v^{- 2}) S_{s} + 1$ , i.e. $S_{s}^{2} = (Q - 1) S_{s} + Q$ ; conversely $T_{s}^{2} = v^{- 2} S_{s}^{2} = v^{- 2} ((Q - 1) S_{s} + Q) = (v - v^{- 1}) T_{s} + 1$ . In particular $R$ is a domain, both bases persist after every base change by The standard basis of the generic Hecke algebra and base change, part 4, and no choice or finiteness hypothesis beyond $∣ S ∣ = 1$ is used.

Facts & Assumptions

Given: The one-element generator set $S = {s}$ , the group $W = {1, s}$ , the Laurent ring $R = Z [v^{\pm 1}]$ and the algebra $H = R ⟨ T_{s} ⟩ / ((T_{s} - v) (T_{s} + v^{- 1}))$ .

[F1]

$R = Λ_{Z, 1}$ is a commutative ring in which $v$ is a unit, $H$ is presented by the generator $T_{s}$ and the single quadratic relation $(T_{s} - v) (T_{s} + v^{- 1}) = 0$ , and each $v_{s}$ is a unit; there are no braid relations because $∣ S ∣ = 1$ . (Universal parameters, the generic Coxeter Hecke algebra and generator conjugacy)

[F2]

For $S = {s}$ the presentation of $W$ has the single generator $s$ and the relation $s^{2} = 1$ . Its universal property extends any assignment of $s$ to an involution in a group, and $ℓ$ is the least word length. (Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups)

[F3]

${T_{w} : w \in W}$ is an $R$ -basis of $H$ , and after base change along any ring homomorphism the specialized family is an $R^{'}$ -basis; in particular $T_{1} = 1$ and ${1, T_{s}}$ is a basis. (The standard basis of the generic Hecke algebra and base change)

[F4]

$T_{s}$ is a unit with $T_{s}^{- 1} = T_{s} - (v - v^{- 1})$ ; with $Q_{s} = v_{s}^{2}$ and $S_{s} = v_{s} T_{s}$ , the element $S_{s}$ is a unit, $T_{s} = v_{s}^{- 1} S_{s}$ , and $(S_{s} - Q_{s}) (S_{s} + 1) = 0$ . (The reversal anti-involution, generator invertibility, the bar operator and the multiplicative normalization)

[F5]

The integers are a commutative ring (The integers form a commutative ring) with no zero divisors (The integers have no zero divisors; multiplicative cancellation), and $0 \neq 1$ because the natural-number embedding is injective (The naturals embed in the integers). Thus $Z$ is an integral domain, and the Laurent construction over a domain makes $Λ_{Z, 1} = Z [v^{\pm 1}]$ an integral domain (Multivariate polynomial and Laurent rings over commutative rings, domains and fraction fields, part 2).

Verification

technique · direct

1.1F1F2F3F4

By [F2] every word in the single generator reduces using $s^{2} = 1$ to $1$ or $s$ . The assignment $s \mapsto - 1$ extends by the universal property to a homomorphism $W \to {\pm 1}$ , so $s \neq 1$ ; thus $W = {1, s}$ and the reduced expressions are the empty word and the one-letter word $s$ , with lengths $0$ and $1$ ; by [F3] the family ${T_{1}, T_{s}}$ is an $R$ -basis of $H$ and $T_{1} = 1$ is the empty product. The unit axioms give $1 \cdot T_{s} = T_{s} = T_{s} \cdot 1$ , and expanding $(T_{s} - v) (T_{s} + v^{- 1}) = 0$ from [F1] gives $T_{s}^{2} = (v - v^{- 1}) T_{s} + 1$ ; the inverse formula $T_{s}^{- 1} = T_{s} - (v - v^{- 1})$ and $T_{s}^{- 1} T_{s} = T_{s} T_{s}^{- 1} = 1$ are [F4]. This is the normalized table of part 1.

1.2F1F3F4algebra

Put $Q : = v^{2}$ and $S_{s} : = v T_{s}$ . Since $v$ is a unit of $R$ ([F1]) and multiplication by it is an invertible $R$ -linear map, ${1, S_{s}}$ is again an $R$ -basis of $H$ ([F3]); $S_{s}$ is a unit with $S_{s}^{- 1} = v^{- 1} T_{s}^{- 1}$ as a product of units ([F4]). Multiplying $T_{s}^{2} = (v - v^{- 1}) T_{s} + 1$ by $v^{2}$ gives $S_{s}^{2} = v (v - v^{- 1}) S_{s} + v^{2} = (Q - 1) S_{s} + Q$ , because $v (v - v^{- 1}) = v^{2} - 1 = Q - 1$ ; equivalently $(S_{s} - Q) (S_{s} + 1) = S_{s}^{2} + (1 - Q) S_{s} - Q = 0$ . Finally $S_{s} (S_{s} + 1 - Q) = S_{s}^{2} + S_{s} - Q S_{s} = ((Q - 1) S_{s} + Q) + S_{s} - Q S_{s} = Q$ , so $S_{s}^{- 1} = Q^{- 1} (S_{s} + 1 - Q)$ . This is the multiplicative table of part 2.

2.1F3F4F5step 1.1step 1.2∎

The two tables are interconverted by $S_{s} = v T_{s}$ and $T_{s} = v^{- 1} S_{s}$ ([F4]). Substituting $T_{s} = v^{- 1} S_{s}$ into $T_{s}^{2} = (v - v^{- 1}) T_{s} + 1$ gives $v^{- 2} S_{s}^{2} = (v - v^{- 1}) v^{- 1} S_{s} + 1$ , and multiplying by the unit $v^{2}$ gives $S_{s}^{2} = v (v - v^{- 1}) S_{s} + v^{2} = (Q - 1) S_{s} + Q$ ; conversely, substituting $S_{s} = v T_{s}$ into $S_{s}^{2} = (Q - 1) S_{s} + Q$ and multiplying by $v^{- 2}$ returns $T_{s}^{2} = (v - v^{- 1}) T_{s} + 1$ . The ring $R = Z [v^{\pm 1}]$ is a domain by [F5], both bases persist under every base change by [F3], and no choice is used: all identities are explicit polynomial identities in $v$ involving no selection. This completes the conversion of part 3 and with 1.1 and 1.2 all three parts of the example.

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Sources

George Lusztig, Lectures on Hecke Algebras with Unequal Parameters (MIT Fall 1999 lecture notes, arXiv:math/0108172v1) (standard reference, not scraped)
Meinolf Geck, Modular Representations of Hecke Algebras (EPFL course notes, arXiv:math/0511548v2) (standard reference, not scraped)
Anders Bjorner and Francesco Brenti, Combinatorics of Coxeter Groups, Springer GTM 231 (2005), complete book PDF (standard reference, not scraped)