Representations of dihedral Hecke algebras

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 10 June 2010

Representations of dihedral Hecke algebras

The dihedral Hecke algebra $H = H_{m, m, 2} (p, q)$ can be given by generators $T_{1}, T_{2}$ and relations $T_{1}^{2} = (p - 1) T_{1} + p,$ $T_{2}^{2} = (q - 1) T_{2} + q,$ and $T_{1} T_{2} T_{1} \dots (m factora) = T_{2} T_{1} T_{2} \dots (m factora) .$ The braid relations given that ${(T_{1} T_{2})}^{m} \in Z (H) .$

If $ρ : H \to M_{2} (ℂ)$ is an irreducible representation of $H$ then $ρ {(T_{1} T_{2})}^{m} = c . id and c^{2} = \det {(T_{1} T_{2})}^{m} = {(- p)}^{m} {(- q)}^{m} = {(p q)}^{m} .$ So $c = \pm {(p q)}^{\frac{m}{2}}$ and, when $p = q = 1, ρ {(s_{1} s_{2})}^{m} = 1 .$ So $c = {(p q)}^{m} .$ So $ρ (T_{1} T_{2} p^{- \frac{1}{2}} q^{- \frac{1}{2}}) has root of unity eigenvalues, and$ $\det (ρ (T_{1} T_{2} p^{- \frac{1}{2}} q^{- \frac{1}{2}})) = (- p) (- q) p^{- 1} q^{- 1} = 1,$ since, because $ρ$ is irreducible, $ρ (T_{1})$ must have two distinct eigenvalues which must be $p$ and $- 1$ because of the equation $(T_{1} - p) (T_{1} + 1) = 0,$ and similarly $ρ (T_{1})$ has eigenvalues $q$ and $- 1 .$ So, with appropriate choice of basis, $ρ (T_{1} T_{2}) = p^{\frac{1}{2}} q^{\frac{1}{2}} (\begin{matrix} ξ^{k} & 0 \\ 0 & ξ^{- k} \end{matrix}), 0 < k < ⌊ \frac{m}{2} ⌋ .$ If $ρ (T_{2}) = (\begin{matrix} a & b \\ c & d \end{matrix}) with a + d = q - 1 and a d - b c = - q,$ then we may assume that $ρ (T_{2}) = (\begin{matrix} a & q + a d \\ 1 & d \end{matrix}),$ in which case $ρ (T_{1}) = ρ (T_{1} T_{2} T_{2}^{- 1}) = p^{\frac{1}{2}} q^{\frac{1}{2}} (\begin{matrix} ξ^{k} & 0 \\ 0 & ξ^{- k} \end{matrix}) (\begin{matrix} - d & q + a d \\ 1 & - a \end{matrix}) q^{- 1} = p^{\frac{1}{2}} q^{\frac{1}{2}} (\begin{matrix} - d ξ^{k} & (q + a d) ξ^{k} \\ ξ^{- k} & - a ξ^{- k} \end{matrix}) .$ Then $a + d = q - 1$ and $p^{\frac{1}{2}} q^{- \frac{1}{2}} (- d ξ^{k} - a ξ^{- k}) = p - 1 .$ Solving for $a$ and $d$ gives $a = \frac{(q - 1) ξ^{k} + (p - 1) p^{- \frac{1}{2}} q^{\frac{1}{2}}}{ξ^{k} - ξ^{- k}} and d = \frac{(q - 1) ξ^{- k} + (p - 1) p^{- \frac{1}{2}} q^{\frac{1}{2}}}{ξ^{- k} - ξ^{k}} .$ If $p = q$ then $a = \frac{(q - 1) (ξ^{k} + 1)}{ξ^{k} - x i^{- k}} and d = \frac{(q - 1) (ξ^{- k} + 1)}{ξ^{- k} - ξ^{k}} .$

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

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