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The Murphy cellular basis of H(S_3)

Example

Work over any commutative ring $R$ with $q \in R^{\times}$ and let $H = H_{R, q} (S_{3})$ with generators $T_{1}, T_{2}$ , the braid relation $T_{1} T_{2} T_{1} = T_{2} T_{1} T_{2}$ and quadratics $T_{i}^{2} = (q - 1) T_{i} + q$ . The Murphy cellular basis (The Murphy tableau elements form a cellular basis) is indexed by the partitions $(3), (2, 1), (1^{3})$ ordered by dominance. Explicitly, with $*$ the antiautomorphism $T_{w}^{*} = T_{w^{- 1}}$ (Murphy elements and Murphy tableau elements): (a) for $λ = (3)$ , $S t d (3) = {s}$ and $m_{s s} = m_{(3)} = 1 + T_{1} + T_{2} + T_{1} T_{2} + T_{2} T_{1} + T_{1} T_{2} T_{1}$ ; (b) for $λ = (2, 1)$ , $S t d (2, 1) = {t, u}$ with $t = \begin{matrix} 1 & 2 \\ 3 \end{matrix}$ , $u = \begin{matrix} 1 & 3 \\ 2 \end{matrix}$ , and $m_{(2, 1)} = 1 + T_{1}$ (the row symmetrizer of $S_{(2, 1)} = ⟨ (1 2) ⟩$ ), so that $m_{t t} = 1 + T_{1}$ , $m_{t u} = (1 + T_{1}) T_{2}$ , $m_{u t} = T_{2} (1 + T_{1})$ and $m_{u u} = T_{2} (1 + T_{1}) T_{2}$ ; (c) for $λ = (1^{3})$ , $S t d (1^{3}) = {v}$ and $m_{v v} = m_{(1^{3})} = 1$ . In total $1 + 4 + 1 = 6 = 3! = ∣ S_{3} ∣$ elements; the middle cell carries $∣ S t d (2, 1) ∣^{2} = 4$ basis elements, its two off-diagonal elements satisfy $m_{t u}^{*} = m_{u t}$ , and the largest cell is spanned by $m_{s s}$ , illustrating the cellular anti-involution. At $q = 1$ this is the Murphy basis of $R [S_{3}]$ , obtained by base change from the integral basis over $Z$ (Polytabloid specialization and the exact signed multiplicity dictionary).

Facts & Assumptions

Given: A commutative ring $R$ , $q \in R^{\times}$ , $H = H_{R, q} (S_{3})$ with standard basis ${T_{w}}$ and Murphy elements (Murphy elements and Murphy tableau elements); the cellular axioms (Cellular bases and cell modules); the Murphy cellular basis theorem (The Murphy tableau elements form a cellular basis).

[F1]

(Murphy elements and Murphy tableau elements) $m_{μ} = \sum_{w \in S_{μ}} T_{w}$ , $m_{s t} = T_{d (s)}^{*} m_{μ} T_{d (t)}$ , $m_{s t}^{*} = m_{t s}$ ; for the shape $(2, 1)$ the distinguished representatives are $d (t) = 1$ and $d (u) = s_{2}$ , and for $(3)$ and $(1^{3})$ the Young subgroups are all of $S_{3}$ and trivial respectively.

[F2]

(The Murphy tableau elements form a cellular basis) The six elements $m_{s t}$ form an $R$ -basis of $H$ indexed by the three partitions ordered by dominance.

[F3]

At $q = 1$ , $T_{w} \mapsto w$ identifies $H_{R, 1} (S_{3})$ with $R [S_{3}]$ over every commutative $R$ (Hecke Jucys-Murphy elements and q-contents, Definition, final paragraph). The integral Murphy basis specializes by [F2].

Verification

1.1givenF1

Cell $(2, 1)$ : $S_{(2, 1)} = ⟨ s_{1} ⟩$ , so $m_{(2, 1)} = 1 + T_{1}$ ; the two standard tableaux of shape $(2, 1)$ are $t = \begin{matrix} 1 & 2 \\ 3 \end{matrix}$ (row reading word $1 2 3$ , so $d (t) = 1$ ) and $u = \begin{matrix} 1 & 3 \\ 2 \end{matrix}$ (row reading word $1 3 2$ , so $d (u) = s_{2}$ ). Hence $m_{t t} = 1 \cdot (1 + T_{1}) \cdot 1 = 1 + T_{1}$ , $m_{t u} = (1 + T_{1}) T_{2}$ , $m_{u t} = T_{2}^{*} (1 + T_{1}) = T_{2} (1 + T_{1})$ and $m_{u u} = T_{2}^{*} (1 + T_{1}) T_{2} = T_{2} (1 + T_{1}) T_{2}$ , using $T_{2}^{*} = T_{2}$ and $m_{(2, 1)}^{*} = m_{(2, 1)}$ .

1.2givenF1

Cells $(3)$ and $(1^{3})$ : $S_{(3)} = S_{3}$ has six elements with reduced expressions $1, s_{1}, s_{2}, s_{1} s_{2}, s_{2} s_{1}, s_{1} s_{2} s_{1}$ , so $m_{(3)} = 1 + T_{1} + T_{2} + T_{1} T_{2} + T_{2} T_{1} + T_{1} T_{2} T_{1} = T_{d (s)}^{*} m_{(3)} T_{d (s)}$ for the unique standard tableau; $S_{(1^{3})}$ is trivial, so its group algebra sum is the identity, $m_{(1^{3})} = 1$ , and $m_{v v} = 1$ .

2.1step 1.1step 1.2F1F2

Count and anti-involution: the three cells contribute $1^{2} + 2^{2} + 1^{2} = 6 = 3! = ∣ S_{3} ∣$ elements, in agreement with $H$ being free of rank $6$ ; and $m_{t u}^{*} = m_{u t}$ is the instance of the general identity $m_{s t}^{*} = m_{t s}$ of [F1], the middle cell being the one with two indices.

3.1step 1.1step 1.2F3∎

Specialization: setting $q = 1$ turns $T_{w}$ into $w$ , so the displayed elements become the Murphy basis of $R [S_{3}]$ ; this is the $n = 3$ case of [F3].

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Sources

Andrew Mathas, Hecke algebras and Schur algebras of the symmetric group, complete prepublication manuscript (1998; expanded published edition: AMS University Lecture Series 15, 1999); complete 79-page manuscript, SHA-256 8cfeb611a95a3ff25cb95ad22bfd3a68475818301c34b04d131a8480a1b997e0 (standard reference, not scraped)