Examples of Macdonald polynomials type A₁

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

Last update: 23 December 2012

Type $A_1$

The Weyl group $W_0=⟨s_1\mid s_1^2=1⟩$ has order two and acts on the lattices $$\begin{array}{cc}{𝔥}_{\mathbb{Z}}=\mathbb{Z}{\omega }^{\vee }\text{and}{𝔥}_{\mathbb{Z}}^{*}=\mathbb{Z}\omega \text{by}{s}_1{\omega }^{\vee }=-{\omega }^{\vee }\text{and}{s}_1\omega =-\omega ,& \text{(4.1)}\end{array}$$ and $$\begin{array}{cc}{\phi }^{\vee }={\alpha }^{\vee }=2{\omega }^{\vee },\phi =\alpha =2\omega ,\text{and}⟨{\omega }^{\vee },\alpha ⟩=1\text{.}& \text{(4.2)}\end{array}$$ $$g∨s0∨s1∨s0∨ g∨s0∨s1∨ g∨s0∨ g∨ g∨s1∨ g∨s1∨s0∨ g∨s1∨s0∨s1∨ X-3ω X-ωs1 X-ω Xωs1 Xω X3ωs1 X3ω s1∨s0∨s1∨ s1∨s0∨ s1∨ 1 s0∨ s0∨s1∨ s0∨s1∨s0∨ X-αs1 X-α s1 1 Xαs1 Xα X2αs1 𝔥α∨+2d 𝔥α∨+d 𝔥α∨ 𝔥-α∨+d 𝔥-α∨+2d 𝔥-α∨+3d$$ 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 $$\begin{array}{cc}\begin{array}{c}{T}_0=g{T}_1{g}^{-1},g^2=1,q=X^{\delta },\\ g{X}^{\omega }=q^{\frac{1}{2}}{X}^{-\omega }g,T_1{X}^{\omega }{T}_1=X^{-\omega },\text{and}{T}_0{X}^{-\omega }{T}_0=q^{-1}{X}^{\omega }.\end{array}& \text{(4.3)}\end{array}$$ 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 $$\begin{array}{c}{Y}^d=q^{-1},{\left(g^{\vee }\right)}^2=1,T_0^{\vee }=g^{\vee }{T}_1^{\vee }{\left(g^{\vee }\right)}^{-1},\\ g^{\vee }{Y}^{{\omega }^{\vee }}=q^{-\frac{1}{2}}{Y}^{-{\omega }^{\vee }}{g}^{\vee },T_1^{-1}{Y}^{{\omega }^{\vee }}{T}_1^{-1},\text{and}{\left(T_0^{\vee }\right)}^{-1}{Y}^{-{\omega }^{\vee }}{\left(T_0^{\vee }\right)}^{-1}=q{Y}^{{\omega }^{\vee }}\text{.}\end{array}$$

		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)$\Rightarrow$(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 $$\begin{array}{c}{g}^{\vee }{Y}^{{\omega }^{\vee }}=g^{\vee }g{T}_1=q^{-\frac{1}{2}}{T}_1^{-1}g{g}^{\vee }=q^{-\frac{1}{2}}{Y}^{-{\omega }^{\vee }}{g}^{\vee },\\ T_1^{-1}{Y}^{{\omega }^{\vee }}{T}_1^{-1}=T_1^{-1}g{T}_1{T}_1^{-1}=T_1^{-1}g=Y^{-{\omega }^{\vee }},\end{array}$$ and $${\left(T_0^{\vee }\right)}^{-1}{Y}^{-{\omega }^{\vee }}{\left(T_0^{\vee }\right)}^{-1}=g^{\vee }{T}_1^{-1}{g}^{\vee }{Y}^{-{\omega }^{\vee }}{g}^{\vee }{T}_1^{-1}{g}^{\vee }=q^{\frac{1}{2}}{g}^{\vee }{T}_1^{-1}{Y}^{{\omega }^{\vee }}{T}_1^{-1}{g}^{\vee }=q^{\frac{1}{2}}{g}^{\vee }{Y}^{-{\omega }^{\vee }}{g}^{\vee }=q{Y}^{{\omega }^{\vee }},$$ respectively. (5.7)$\Rightarrow$(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 $$\begin{array}{c}{g}^2=Y^{{\omega }^{\vee }}{T}_1^{-1}{Y}^{{\omega }^{\vee }}{T}_1^{-1}=Y^{{\omega }^{\vee }}{Y}^{-{\omega }^{\vee }}=1,\\ g{T}_1{g}^{-1}=Y^{{\omega }^{\vee }}{T}_1^{-1}{T}_1{T}_1{Y}^{-{\omega }^{\vee }}=Y^{{\omega }^{\vee }}{T}_1{Y}^{-{\omega }^{\vee }}=Y^{{\omega }^{\vee }}{T}_1{Y}^{-{\omega }^{\vee }}{T}_1{T}_1^{-1}=Y^{2{\omega }^{\vee }},\end{array}$$ and $$T_1{g}^{\vee }g=T_1{g}^{\vee }{Y}^{{\omega }^{\vee }}{T}_1^{-1}=q^{-\frac{1}{2}}{T}_1{Y}^{-{\omega }^{\vee }}{g}^{\vee }{T}_1^{-1}=q^{-\frac{1}{2}}g{g}^{\vee }{T}_1^{-1},$$ respectively. $\square$

The double affine Hecke algebra $\stackrel{\sim }{H}$ is $ℂ\stackrel{\sim }{ℬ}$ with the additional relations $$\begin{array}{cc}{T}_i^2=(t^{1/2}-t^{-1/2})T_i+1,\text{for}i=0,1,\text{and}{t}_0=t_1=t=q^c.& \text{(4.5)}\end{array}$$

Using (4.5), the relations in Proposition 4.1 give $$\begin{array}{c}{g}^{\vee }{Y}^{{\omega }^{\vee }}=q^{-1/2}{Y}^{-{\omega }^{\vee }}{g}^{\vee },T_1{Y}^{{\omega }^{\vee }}=Y^{-{\omega }^{\vee }}{T}_1+(t^{1/2}-t^{-1/2})\frac{Y^{{\omega }^{\vee }}-Y^{-{\omega }^{\vee }}}{1-Y^{-{\alpha }^{\vee }}},\text{and}\\ T_0^{\vee }{Y}^{{\omega }^{\vee }}=q^{-1}{Y}^{-{\omega }^{\vee }}{T}_0^{\vee }+(t^{1/2}-t^{-1/2})\left(\frac{Y^{{\omega }^{\vee }}-q^{-1}{Y}^{-{\omega }^{\vee }}}{1-q{Y}^{{\alpha }^{\vee }}}\right)\text{.}\end{array}$$ With $Y^{{\alpha }_0^{\vee }}=q{Y}^{{\alpha }^{\vee }}$ and $Y^{{\alpha }_1^{\vee }}=Y^{{\alpha }^{}},$ then $${\tau }_g^{\vee }=g^{\vee },\text{and}{\tau }_i^{\vee }=T_i^{\vee }-(t^{1/2}-t^{-1/2})\left(\frac{1}{1-Y^{-{\alpha }_i^{\vee }}}\right),\text{for}i=0,1\text{.}$$

To illustrate Theorem 2.2, note that $X^{-2\omega }=s_1^{\vee }{s}_0^{\vee }$ is a reduced word and

$$\begin{array}{ccc}{\tau }_1^{\vee }{\tau }_0^{\vee }& =& (T_1^{\vee }+\frac{t^{-1/2}(1-t)}{1-Y^{-{\alpha }_1^{\vee }}}){\tau }_0^{\vee }\\ & =& T_1^{\vee }{T}_0^{\vee }+T_1^{\vee }\frac{t^{-1/2}(1-t)}{1-Y^{-{\alpha }_0^{\vee }}}+{\left(T_0^{\vee }\right)}^{-1}\frac{t^{-1/2}(1-t)}{1-Y^{-s_0{\alpha }_1^{\vee }}}+\left(\frac{t^{-1/2}(1-t)}{1-Y^{-s_0{\alpha }_1^{\vee }}}\right)\left(\frac{t^{-1/2}(1-t)Y^{-{\alpha }_0^{\vee }}}{1-Y^{-{\alpha }_0^{\vee }}}\right)\\ & =& X^{-2\omega }+T_1^{\vee }\frac{t^{-1/2}(1-t)}{1-Y^{-{\alpha }_0^{\vee }}}+X^{2\omega }{T}_1^{\vee }\frac{t^{-1/2}(1-t)}{1-Y^{-s_0{\alpha }_1^{\vee }}}+\left(\frac{t^{-1/2}(1-t)}{1-Y^{-s_0{\alpha }_1^{\vee }}}\right)\left(\frac{t^{-1/2}(1-t)Y^{-{\alpha }_0^{\vee }}}{1-Y^{-{\alpha }_0^{\vee }}}\right)\end{array}$$ The corresponding paths in $ℬ(1,\overrightarrow{-2\omega })=B\left(\overrightarrow{-2\omega }\right)$ are $$\begin{array}{cccc}& & & \\ X^{-2\omega }& T_1^{\vee }\frac{t^{-1/2}(1-t)}{1-Y^{-{\alpha }_0^{\vee }}}& X^{2\omega }{T}_1^{\vee }\frac{t^{-1/2}(1-t)}{1-Y^{-s_0{\alpha }_1^{\vee }}}& \frac{t^{-1/2}(1-t)}{1-Y^{-s_0{\alpha }_1^{\vee }}}\frac{t^{-1/2}(1-t)Y^{-{\alpha }_0^{\vee }}}{1-Y^{-{\alpha }_0^{\vee }}}\end{array}$$

The polynomial representations is defined by $$g1=1,T_{0}1=t^{\frac{1}{2}}1,\text{and}{T}_{1}1=t^{\frac{1}{2}}1\text{.}$$ In this case $$\begin{array}{cc}{\rho }_c=\frac{1}{2}c\alpha \text{and}{W}^0=\{X^{-\ell w}\mid \ell \in {\mathbb{Z}}_{\ge 0}\}\cup \{X^{\ell \omega }{s}_1^{\vee }\mid \ell \in {\mathbb{Z}}_{>0}\},& \text{(4.6)}\end{array}$$ 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

$$Y^{-{\alpha }_0^{\vee }}1=q{Y}^{{\alpha }^{\vee }}1=q{q}^{c}1=qt1,\text{and}{Y}^{-s_0{\alpha }_1^{\vee }}1=Y^{{\alpha }^{\vee }+2d}1=q^2{Y}^{{\alpha }^{\vee }}1=q^{2}t,$$

gives

$$\begin{array}{ccc}{E}_{-2\omega }& =& {\tau }_1^{\vee }{\tau }_0^{\vee }1\\ & =& X^{-2\omega }+t^{1/2}\frac{t^{-1/2}(1-t)}{1-qt}+X^{2\omega }{t}^{1/2}\frac{t^{-1/2}(1-t)}{1-q^{2}t}+\left(\frac{t^{-1/2}(1-t)}{1-q^{2}t}\right)\left(\frac{t^{-1/2}(1-t)qt}{1-qt}\right)\\ & =& X^{-2\omega }+\frac{1-t}{1-qt}+X^{2\omega }\frac{1-t}{1-q^{2}t}+\left(\frac{1-t}{1-q^{2}t}\right)\left(\frac{(1-t)q}{1-qt}\right)\text{.}\end{array}$$ 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

$$\begin{array}{ccc}{P}_{2\omega }& =& 1_0{E}_{2\omega }=(T_1^{\vee }+t^{-1/2}){\tau }_0^{\vee }1\\ & =& (T_1^{\vee }{T}_0^{\vee }+T_1^{\vee }\frac{t^{-1/2}(1-t)}{1-Y^{-{\alpha }_0^{\vee }}}+t^{-1/2}{\left(T_0^{\vee }\right)}^{-1}+t^{-1/2}\frac{t^{-1/2}(1-t)Y^{-{\alpha }_0^{\vee }}}{1-Y^{-{\alpha }_0^{\vee }}})1\\ & =& (X^{-2\omega }+t^{1/2}\frac{t^{-1/2}(1-t)}{1-qt}+t^{-1/2}{X}^{2\omega }{T}_1^{\vee }+t^{-1/2}\frac{t^{-1/2}(1-t)qt}{1-qt})1\\ & =& (X^{-2\omega }+t^{1/2}\frac{1-t}{1-qt}+X^{2\omega }+t^{-1/2}\frac{(1-t)q}{1-qt})1=(X^{2\omega }+X^{-2\omega }+t^{1/2}+(1+q)\frac{1-t}{1-qt})1\text{.}\end{array}$$ The corresponding paths in $𝒫\left(\overrightarrow{2\omega }\right)$ are $$\begin{array}{cccc}& & & \\ X^{-2\omega }& \frac{1-t}{1-tq}& X^{2\omega }& q\frac{(1-t)}{1-tq}\end{array}$$

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.

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