## Examples of Macdonald polynomials type A1

Last update: 23 December 2012

## Type ${A}_{1}$

The Weyl group ${W}_{0}=⟨{s}_{1}\phantom{\rule{0.2em}{0ex}}\mid \phantom{\rule{0.2em}{0ex}}{s}_{1}^{2}=1⟩$ has order two and acts on the lattices $𝔥ℤ=ℤω∨and 𝔥ℤ*=ℤω bys1ω∨=- ω∨ands1 ω=-ω, (4.1)$ and $φ∨= α∨ =2ω∨, φ= α=2ω, and ⟨ω∨,α⟩=1. (4.2)$ $g∨s0∨s1∨s0∨ g∨s0∨s1∨ g∨s0∨ g∨ g∨s1∨ g∨s1∨s0∨ g∨s1∨s0∨s1∨ X-3ω X-ωs1 X-ω Xωs1 Xω X3ωs1 X3ω s1∨s0∨s1∨ s1∨s0∨ s1∨ 1 s0∨ s0∨s1∨ s0∨s1∨s0∨ X-αs1 X-α s1 1 Xαs1 Xα X2αs1 𝔥α∨+2d 𝔥α∨+d 𝔥α∨ 𝔥-α∨+d 𝔥-α∨+2d 𝔥-α∨+3d$ The double affine braid group $\stackrel{\sim }{ℬ}$ is generated by ${T}_{0}$, ${T}_{1}$, $g$, ${X}^{\omega }$, and ${q}^{\frac{1}{2}}$, with relations $T0= gT1g-1, g2=1, q=Xδ, gXω= q12 X-ωg, T1 Xω T1 = X-ω, and T0 X-ω T0 = q-1 Xω. (4.3)$ In the double affine braid group At this point, the following Proposition, which is the Type ${A}_{1}$ case of Theorem 2.1, is easily proved by direct computation.

(Duality). Let ${Y}^{d}={q}^{-1}$. The double affine braid group $\stackrel{\sim }{ℬ}$ is generated by ${T}_{0}^{\vee },{T}_{1}^{\vee },{g}^{\vee },{Y}^{{\omega }^{\vee }}$ and ${q}^{\frac{1}{2}}$ with relations $Yd =q-1 , (g∨)2 =1, T0∨ =g∨ T1∨ (g∨) -1, g∨Yω∨ = q-12 Y-ω∨ g∨ , T1-1 Yω∨ T1-1 ,and (T0∨) -1 Y-ω∨ (T0∨) -1 =qYω∨.$

 Proof. We prove that the presentation in (5.7) is equivalent to the presentation in (5.3). The proof that the presentation in (5.6) is equivalent to the presentation in (5.3) is similar. (5.3)$⇒$(5.7): Use (5.4) to define ${Y}^{\omega }$ in terms of $g$ and ${T}_{1}\text{.}$ The first and second relations in (5.7) are the third and fourth relations (5.3). The proof of the third, fourth and fifth relations in (5.7) are $g∨Yω∨= g∨gT1= q-12 T1-1gg∨= q-12 Y-ω∨g∨, T1-1Yω∨ T1-1=T1-1 gT1T1-1= T1-1g= Y-ω∨,$ and $(T0∨)-1 Y-ω∨ (T0∨)-1 =g∨T1-1 g∨Y-ω∨ g∨T1-1g∨ =q12g∨ T1-1 Yω∨T1-1 g∨=q12 g∨Y-ω∨ g∨=qYω∨,$ respectively. (5.7)$⇒$(5.3): Define $g={Y}^{{\omega }^{\vee }}{T}_{1}^{-1}$ and ${T}_{0}={Y}^{2{\omega }^{\vee }}{T}_{1}^{-1}\text{.}$ The third and fourth relations of (5.3) are exactly the first and second relations of (5.7). The proof of the first, second and fifth relations in (5.3) are $g2=Yω∨ T1-1Yω∨ T1-1= Yω∨ Y-ω∨=1, gT1g-1= Yω∨T1-1 T1T1 Y-ω∨= Yω∨T1 Y-ω∨= Yω∨T1 Y-ω∨T1 T1-1= Y2ω∨,$ and $T1g∨g=T1g∨ Yω∨T1-1 =q-12T1 Y-ω∨g∨ T1-1= q-12gg∨ T1-1,$ respectively. $\square$

The double affine Hecke algebra $\stackrel{\sim }{H}$ is $ℂ\stackrel{\sim }{ℬ}$ with the additional relations $Ti2= ( t1/2- t-1/2 ) Ti+1 ,for i=0,1, and t0= t1= t= qc. (4.5)$

Using (4.5), the relations in Proposition 4.1 give $g∨Yω∨= q-1/2 Y-ω∨ g∨ , T1Yω∨= Y-ω∨T1+ ( t1/2- t-1/2 ) Yω∨- Y-ω∨ 1-Y-α∨ ,and T0∨Yω∨= q-1Y-ω∨ T0∨+ ( t1/2- t-1/2 ) ( Yω∨-q-1 Y-ω∨ 1-qYα∨ ) .$ With ${Y}^{{\alpha }_{0}^{\vee }}=q{Y}^{{\alpha }^{\vee }}$ and ${Y}^{{\alpha }_{1}^{\vee }}={Y}^{{\alpha }^{}},$ then $τg∨=g∨,and τi∨=Ti∨- ( t1/2- t-1/2 ) ( 1 1-Y-αi∨ ) ,fori=0,1.$

To illustrate Theorem 2.2, note that ${X}^{-2\omega }={s}_{1}^{\vee }{s}_{0}^{\vee }$ is a reduced word and

$τ1∨τ0∨ = ( T1∨+ t-1/2(1-t) 1-Y-α1∨ ) τ0∨ = T1∨T0∨+ T1∨ t-1/2(1-t) 1-Y-α0∨ +(T0∨)-1 t-1/2(1-t) 1-Y-s0α1∨ + ( t-1/2(1-t) 1-Y-s0α1∨ ) ( t-1/2(1-t) Y-α0∨ 1-Y-α0∨ ) = X-2ω + T1∨ t-1/2(1-t) 1-Y-α0∨ +X2ω T1∨ t-1/2(1-t) 1-Y-s0α1∨ + ( t-1/2(1-t) 1-Y-s0α1∨ ) ( t-1/2(1-t) Y-α0∨ 1-Y-α0∨ )$ The corresponding paths in $ℬ\left(1,\stackrel{\to }{-2\omega }\right)=B\left(\stackrel{\to }{-2\omega }\right)$ are $X-2ω T1∨ t-1/2(1-t) 1-Y-α0∨ X2ω T1∨ t-1/2(1-t) 1-Y-s0α1∨ t-1/2(1-t) 1-Y-s0α1∨ t-1/2(1-t) Y-α0∨ 1-Y-α0∨$

The polynomial representations is defined by $g1=1,T0 1=t121, andT11= t121.$ In this case $ρc=12cα andW0= { X-ℓw∣ ℓ∈ℤ≥0 } ∪ { Xℓωs1∨ ∣ℓ∈ℤ>0 } , (4.6)$ is the set of minimal length coset representatives of ${W}^{\vee }/{W}_{0}$.

Applying the expansion of ${\tau }_{1}^{\vee }{\tau }_{0}^{\vee }$ to $1$ and using

$Y-α0∨1=q Yα∨1=qqc 1=qt1,and Y-s0α1∨1= Yα∨+2d1= q2Yα∨1= q2t,$

gives

$E-2ω = τ1∨τ0∨1 = X-2ω+t1/2 t-1/2(1-t) 1-qt +X2ωt1/2 t-1/2(1-t) 1-q2t + ( t-1/2(1-t) 1-q2t ) ( t-1/2(1-t)qt 1-qt ) = X-2ω+ 1-t1-qt +X2ω 1-t1-q2t+ (1-t1-q2t) ((1-t)q1-qt) .$ Since ${1}_{0}={T}_{1}^{\vee }+{t}^{-1/2}$ the symmetric Macdonald polynomial ${P}_{2\omega }={1}_{0}{E}_{2\omega }={1}_{0}{\tau }_{0}^{\vee }1$ is

$P2ω = 10E2ω= (T1∨+t-1/2) τ0∨1 = ( T1∨T0∨+ T1∨ t-1/2(1-t) 1-Y-α0∨ +t-1/2 (T0∨)-1+ t-1/2 t-1/2(1-t) Y-α0∨ 1-Y-α0∨ ) 1 = ( X-2ω+ t1/2 t-1/2(1-t) 1-qt +t-1/2 X2ωT1∨+ t-1/2 t-1/2(1-t) qt 1-qt ) 1 = ( X-2ω+ t1/2 1-t 1-qt +X2ω+ t-1/2 (1-t) q 1-qt ) 1 = ( X2ω+ X-2ω+ t1/2+ (1+q) 1-t 1-qt ) 1.$ The corresponding paths in $𝒫\left(\stackrel{\to }{2\omega }\right)$ are $X-2ω 1-t1-tq X2ω q (1-t)1-tq$

## Notes and References

This page is section 4.1 from the paper of A. Ram and M. Yip entitled A combinatorial formula for Macdonald polynomials.