## Examples of Macdonald polynomials

Last update: 25 September 2012

## Type ${A}_{2}$

The Weyl group ${W}_{0}=⟨{s}_{1},{s}_{2}\phantom{\rule{0.2em}{0ex}}\mid \phantom{\rule{0.2em}{0ex}}{s}_{1}^{2}={s}_{2}^{2}=1,\phantom{\rule{0.2em}{0ex}}{s}_{1}{s}_{2}{s}_{1}={s}_{2}{s}_{1}{s}_{2}⟩$ acts on the lattices

$𝔥ℤ=ℤω1∨+ ℤω2∨and 𝔥ℤ*=ℤω1+ ℤω2, (4.7)$

where ${s}_{1}$ and ${s}_{2}$ are the reflections in the hyperplanes determined by

$α1∨=2ω1∨ -ω2∨, α2∨=-ω1∨ +2ω2∨ ,α1= 2ω1-ω2, andα2=-ω1 +2ω2, (4.8)$

with $⟨{\omega }_{i}^{\vee },{\alpha }_{j}⟩={\delta }_{ij},$ and $⟨{\omega }_{i}^{\vee },{\alpha }_{j}^{\vee }⟩={\delta }_{ij}\text{.}$ In this case,

$φ∨=α1∨+ α2∨,and φ=α1+α2. (4.9)$

The double affine braid group $\stackrel{\sim }{ℬ}$ is generated by ${T}_{0},{T}_{1},{T}_{2},g,{X}^{{\omega }_{1}},{X}^{{\omega }_{2}},$ and ${q}^{1/3},$ with relations

$TiTjTi= TjTiTj, fori≠j, XμXλ= Xμ+λ= XλXμ, forμ,λ∈ 𝔥ℤ*, T1Xω2= Xω2T1, T2Xω1= Xω1T2, T1Xω1T1 =X-ω1+ω2, T2Xω2 T2= Xω1-ω2, g3=1,g Xω1= q1/3 X-ω1+ω2g, gXω2= q2/3 X-ω1g, gT0g-1=T1, gT1g-1=T2, gT2g-1=T0. (4.10)$

The formula (2.27) gives

$g=Yω1∨ T1-1 T2-1, g2=Yω2∨ T2-1T1-1, T0=Yφ∨ T1-1T2-1 T1-1, (4.11)$ $\begin{array}{cc}{g}^{\vee }={X}^{{\omega }_{1}}{T}_{1}^{\vee }{T}_{2}^{\vee },\phantom{\rule{2em}{0ex}}{\left({g}^{\vee }\right)}^{2}={X}^{{\omega }_{2}}{T}_{2}^{\vee }{T}_{1}^{\vee },\phantom{\rule{2em}{0ex}}{\left({T}_{0}^{\vee }\right)}^{-1}={X}^{\phi }{T}_{1}^{\vee }{T}_{2}^{\vee }{T}_{1}^{\vee }\text{.}& \text{(4.12)}\end{array}$

At this point, the following Proposition, which is the type ${A}_{2}$ case of Theorem 2.1, is easily proved by direct computation.

(Duality). Let ${Y}^{d}={q}^{-1}\text{.}$ The double affine braid group $\stackrel{\sim }{ℬ}$ is generated by ${T}_{0}^{\vee },\phantom{\rule{0.2em}{0ex}}{T}_{1}^{\vee },\phantom{\rule{0.2em}{0ex}}{T}_{2}^{\vee },\phantom{\rule{0.2em}{0ex}}{g}^{\vee },\phantom{\rule{0.2em}{0ex}}{Y}^{{\omega }_{1}^{\vee }},\phantom{\rule{0.2em}{0ex}}{Y}^{{\omega }_{2}^{\vee }},\phantom{\rule{0.2em}{0ex}}$ and ${q}^{1/3},$ with relations

$(T1∨)-1 Yω1∨ (T1∨)-1= Y -ω1∨+ ω2∨ , (T2∨)-1 Yω2∨ (T2∨)-1= Y ω1∨- ω2∨ , (T1∨)-1 Yω2∨= Yω2∨ (T1∨)-1, (T2∨)-1 Yω1∨= Yω1∨ (T2∨)-1, (g∨)3=1, g∨Yω1∨= q-1/3 Y-ω1∨+ω2∨ g∨,g∨ Yω2∨=q-2/3 Y-ω1∨g∨, g∨T0∨ (g∨)-1= T1∨,g∨ T1∨ (g∨)-1= T2∨and g∨T2∨ (g∨)-1= T0∨.$

To give a concrete example of Theorem 3.4 let us compute the symmetric Macdonald polynomial ${P}_{\rho }$ where $\rho ={\alpha }_{1}+{\alpha }_{2}\text{.}$ Since

$10= Xs1s2s1+ t-1/2Xs1s2 +t-1/2Xs2s1 +t-2/2Xs1+ t-2/2Xs2+ t-3/2,$

and ${X}^{\rho }m={s}_{0}^{\vee }$ is the minimal length element of the coset ${X}^{\rho }{W}_{0},$

$Pρ = 10Eρ=10 τ0∨1 = ( Xs1s2s1+ t-1/2 Xs1s2+ t-1/2 Xs2s1 ) ( T0∨+ t-1/2(1-t) 1-Y-α0∨ ) 1 + ( t-2/2Xs1+ t-2/2Xs2+ t-3/2 ) ( (T0∨)-1+ t-1/2(1-t) Y-α0∨ 1-Y-α0∨ ) 1 = ( Xs1s2s1s0 +t-1/2 Xs1s2s0+ t-1/2 Xs2s1s0+ t-2/2Xs1s0 +t-2/2Xs2s0 +t-3/2Xs0 ) 1 + ( Xs1s2s1+ t-1/2Xs1s2 +t-1/2Xs2s1 ) t-1/2 (1-t) 1-Y-α0∨ 1 + ( t-2/2Xs1+ t-2/2Xs2+ t-3/2 ) t-1/2 (1-t) Y-α0∨ 1-Y-α0∨ 1.$

Since ${Y}^{-{\alpha }_{0}^{\vee }}1={Y}^{{\phi }^{\vee }-d}1=q{Y}^{{\alpha }_{1}^{\vee }+{\alpha }_{2}^{\vee }}1={t}^{2}q1,$

$Pρ = ( Xw0ρ+ t-1/2 Xs1s2ρ T2∨+ t-1/2 Xs2s1ρ T1∨ +t-2/2 Xs1ρ T2∨T1∨+ t-2/2 Xs2ρ T1∨T2∨+ t-3/2Xρ T1∨T2∨ T1∨ ) 1 + ( T1∨T2∨ T1∨+t-1/2 T1∨T2∨+ t-1/2T2∨ T1∨ ) t-1/2 (1-t) 1-t2q 1 + ( t-2/2T1∨+ t-2/2T2∨+ t-3/2 ) t-1/2 (1-t) t2q 1-t2q 1 = ( Xw0ρ+ Xs1s2ρ+ Xs2s1ρ+ Xs1ρ+ Xs2ρ+ Xρ ) 1 + ( t3/2+ t1/2+t1/2 ) t-1/2 (1-t) 1-t2q 1+ ( t-1/2+ t-1/2+ t-3/2 ) t-1/2 (1-t)t2q 1-t2q 1 = ( Xw0ρ+ Xs1s2ρ+ Xs2s1ρ+ Xs1ρ+ Xs2ρ+ Xρ+ ( t+2+2tq+q ) 1-t 1-t2q ) 1.$

The set $𝒫\left(\stackrel{\to }{\rho }\right)$ contains 12 alcove walks,

The Hall-Littlewood polynomial and the Weyl character are

$Pρ(0,t)=mρ+ (2+t)(1-t) andsρ=Pρ (0,0)=mρ+2,$

where ${m}_{\rho }={X}^{{w}_{0}\rho }={X}^{{s}_{1}{s}_{2}\rho }+{X}^{{s}_{2}{s}_{1}\rho }+{X}^{{s}_{1}\rho }+{X}^{{s}_{2}\rho }+{X}^{\rho }\text{.}$

The expression ${X}^{{s}_{1}{s}_{2}\rho }{s}_{2}={s}_{1}^{\vee }{s}_{2}^{\vee }{s}_{0}^{\vee }$ is a reduced word for the minimal length element in the coset ${X}^{{s}_{1}{s}_{2}\rho }{W}_{0}$ and Theorem 2.2 is illustrated by

$τ1∨τ2∨τ0∨ = ( T1∨+ t-1/2 (1-t) 1-Y-α1∨ ) τ1∨τ0∨= ( T1∨τ2∨+ τ2∨ t-1/2 (1-t) 1-Y-s2α1∨ ) τ0∨ = ( T1∨T2∨+ T1∨ t-1/2 (1-t) 1-Y-α2∨ +T2∨ t-1/2 (1-t) 1-Y-s2α1∨ + t-1/2 (1-t) 1-Y-α2∨ t-1/2 (1-t) 1-Y-s2α1∨ ) τ0∨ = T1∨T2∨τ0∨ +T1∨τ0∨ t-1/2 (1-t) 1-Y-s0α2∨ +T2∨τ0∨ t-1/2 (1-t) 1-Y-s0s2α1∨ +τ0∨ t-1/2 (1-t) 1-Y-s0α2∨ t-1/2 (1-t) 1-Y-s0s2α1∨ = T1∨T2∨T0∨ +T1∨T2∨ t-1/2 (1-t) 1-Y-α0∨ +T1∨ (T0∨)-1 t-1/2 (1-t) 1-Y-s0α2∨ + T1∨ t-1/2 (1-t) Y-α0∨ 1-Y-α0∨ t-1/2 (1-t) 1-Y-s0α2∨ +T2∨ (T0∨)-1 t-1/2 (1-t) 1-Y-s0s2α1∨ + T2∨ t-1/2 (1-t) Y-α0∨ 1-Y-α0∨ t-1/2 (1-t) 1-Y-s0s2α1∨ +(T0∨)-1 t-1/2 (1-t) 1-Y-s0α2∨ t-1/2 (1-t) 1-Y-s0s2α1∨ + t-1/2 (1-t) Y-α0∨ 1-Y-α0∨ t-1/2 (1-t) 1-Y-s0α2∨ t-1/2 (1-t) 1-Y-s0s2α1∨ ,$

where the eight terms in this expansion correspond to the eight alcove walks in $ℬ\left(1,{s}_{1}^{\vee }{s}_{2}^{\vee }{s}_{0}^{\vee }\right)=ℬ\left(\stackrel{\to }{{s}_{1}{s}_{2}\rho }\right)$ pictured below. Applying the expansion of ${\tau }_{1}^{\vee }{\tau }_{2}^{\vee }{\tau }_{0}^{\vee }$ to $1$ and using

$Y-α0∨1= Yφ∨-d1= t2q1, Y-s0α2∨ 1=Yα1∨-d 1=tq1, and Y-s0s2α1∨ 1=Yφ∨-2d1 =t2q21, (4.13)$

computes

$Es1s2ρ = ( Xs1s2ρ t1/2+t t-1/2 (1-t) 1-t2q +Xs1ρt t-1/2 (1-t) 1-tq +t1/2 t-1/2 (1-t) t2q 1-t2q t-1/2 (1-t) 1-tq + Xs2ρt t-1/2 (1-t) 1-t2q2 +t1/2 t-1/2 (1-t) t2q 1-t2q t-1/2 (1-t) 1-t2q2 + Xρt3/2 t-1/2 (1-t) 1-tq t-1/2 (1-t) 1-t2q2 + t-1/2 (1-t) t2q 1-t2q t-1/2 (1-t) 1-tq t-1/2 (1-t) 1-t2q2 ) 1. = t1/2 ( Xs1s2ρ+ (1-t) 1-t2q +Xs1ρ (1-t) 1-tq +t (1-t)q 1-t2q (1-t) 1-tq +Xs2ρ (1-t) 1-t2q2 +t (1-t)q 1-t2q (1-t) 1-t2q2 +Xρ (1-t) 1-tq (1-t) 1-t2q2 + (1-t)q 1-t2q (1-t) 1-tq (1-t) 1-t2q2 ) 1.$

## Notes and References

This page is taken from a paper entitled A combinatorial formula for Macdonald Polynomials by Arun Ram and Martha Yip.