## The double affine Hecke algebra

Last update: 17 September 2012

## The double affine Hecke algebra

With ${T}_{0}^{\prime }$ as in (2.4), the double affine Hecke algebra ${\stackrel{\sim }{H}}_{n}$ is the quotient of $ℂ\stackrel{\sim }{ℬ}$ by

$( T0′-u012 ) ( T0′+u0-12 ) =0, ( T0∨-un12 ) ( T0∨+un-12 ) =0, ( T0-t012 ) ( T0+t0-12 ) =0, ( Tn-tn12 ) ( Tn-tn-12 ) =0,and (3.17) ( Ti-t12 ) ( Ti+t-12 ) =0,for i=1,…,n-1.$

Let ${X}^{{\epsilon }_{1}},\dots ,{X}^{{\epsilon }_{n}}$ and ${Y}^{{\epsilon }_{1}^{\vee }},\dots ,{Y}^{{\epsilon }_{n}^{\vee }}$ be the elements of $\stackrel{\sim }{ℬ}$ defined in (2.3). The following theorem establishes the "Bernstein presentation" of the double affine Hecke algebra. Often this presentation is used as the definition of the double affine Hecke algebra. The affine Weyl group

$Xwμ=wXμ w-1,for μ∈𝔥ℤ*and w∈W. (3.18)$

Similarly, thhe affine Weyl group

$W∨= { Xμv∣μ∈ 𝔥ℤ*,v∈W0 } acts onY∼= { qk/2Yλ∨ ∣k∈ℤ,λ∨∈ 𝔥ℤ } ,by conjugation.$

Write

$Ywλ∨=w Yλ∨w-1, forw∈W∨and λ∨∈𝔥ℤ. (3.19)$

In particular,

$Ys0∨ε1∨ = s0∨Yε1∨ s0∨=q-1 Y-ε1∨and Ys0∨εj∨ s0∨Yεj∨ s0∨=Yεj∨ ,forj=2,…,n . (3.20)$

1. Let ${t}_{i}^{\frac{1}{2}}={u}_{i}^{\frac{1}{2}}={t}^{\frac{1}{2}}$ for $i=1,\dots ,n-1\text{.}$ The algebra ${\stackrel{\sim }{H}}_{n}$ is the quotient of $ℂ{\stackrel{\sim }{ℬ}}_{n}$ by the relations

$Ti-Ti-1= ti12- ti-12,for i=0,1,…,n.$

and

$TiXμ=Xsiμ Ti+ ( ( ti12- ti-12 ) + ( ui12- ui-12 ) Xαi ) ( Xμ- Xsiμ ) 1-X2αi ,$

for $i=0,\dots ,n,$ where ${X}^{{s}_{i}\mu }$ is defined in (3.18),

$Xα0=q12 X-ε1, αn=εn,and αi=εi- εi+1,for i=1,…,n-1.$
2. The algebra ${\stackrel{\sim }{H}}_{n}$ is the quotient of $ℂ{\stackrel{\sim }{ℬ}}_{n}$ by the relations

$T0∨- (T0∨)-1= un12- un-12, Tn-Tn-1= tn12- tn-12, Ti-Ti-1= t12-t-12, fori=1,…,n-1,$

and

$TiYλ∨ = Ysiλ∨Ti+ ( t12- t-12 ) Yλ∨- Ysiλ∨ 1-Y-αi∨ ,fori=1,…, n-1, TnYλ∨ = Ysnλ∨Tn+ ( ( tn12- tn-12 ) + ( t012- t0-12 ) Y-εn∨ ) 1-Y-2εn∨ ( Yλ∨- Ysnλ∨ ) ,and T0∨Yλ∨ = Ys0∨λ∨ T0∨+ ( ( un12- un-12 ) + ( u012- u0-12 ) Y-α0∨ ) 1-Y-2α0∨ ( Yλ∨- Ys0∨λ∨ ) .$

where ${Y}^{{s}_{i}{\lambda }^{\vee }}$ is as defined in (3.19),

$Yα0∨=q-12 Y-ε1∨, αn∨=εn∨, andαi∨=εi∨ -εi+1∨,for i=1,…,n-1.$

 Proof. Following [M03, p. 81]: Since ${T}_{0}^{\prime }={q}^{-\frac{1}{2}}{X}^{{\epsilon }_{1}}{T}_{0}^{-1},$ $T0Xα0- X-α0T0 = T0q12X-ε1 -q-12Xε1T0 = T0q12X-ε1 -q-12Xε1 ( T0-1+ t012- t0-12 ) = (T0′)-1- T0′- ( t012- t0-12 ) q-12Xε1 = - ( u012- u0-12 ) - ( t012- t0-12 ) q-12Xε1 = ( ( t012- t0-12 ) + ( u012- u0-12 ) q12X-ε1 ) q12X-ε1 -q-12Xε1 1-qX-2ε1 = ( ( t012- t0-12 ) + ( u012- u0-12 ) Xα0 ) Xα0- X-α0 1-X2α0 ,$ and, since ${X}^{-{\epsilon }_{n}}{T}_{n}^{-1}={T}_{n}\dots {T}_{1}{T}_{0}^{\vee }{T}_{1}^{-1}\dots {T}_{n}^{-1}$ is conjugate to ${T}_{0}^{\vee },$ $TnXαn- X-αnTn = TnXεn- X-εnTn = TnXεn- X-εn ( Tn-1+tn12 -tn-12 ) = - ( un12- un-12 ) - ( tn12- tn-12 ) X-εn = ( ( tn12- tn-12 ) + ( un12- un-12 ) Xεn ) Xεn- X-εn 1-X2εn .$ Since ${Y}^{{\alpha }_{0}^{\vee }}={q}^{-\frac{1}{2}}{Y}^{-{\epsilon }_{1}^{\vee }}$ and ${T}_{0}^{\prime }={\left({T}_{0}^{\vee }\right)}^{-1}{q}^{-\frac{1}{2}}{Y}^{-{\epsilon }_{1}^{\vee }},$ $T0∨Yα0∨- Y-α0∨T0∨ = ( (T0∨)-1 + ( un12- un-12 ) ) q-12 Y-ε1∨- q12 Yε1∨ T0∨ = (T0∨)-1 q-12 Y-ε1∨- q12 Yε1∨ T0∨+ ( un12- un-12 ) Yα0∨ = ( u012- u0-12 ) + ( un12- un-12 ) Yα0∨ = ( ( u012- u0-12 ) Y-α0∨ + ( un12- un-12 ) ) Yα0∨ - Y-α0∨ 1- Y-2α0∨ ,$ and, since ${T}_{n}^{-1}{Y}^{{\epsilon }_{n}^{\vee }}={T}_{n}^{-1}{T}_{n-1}^{-1}\dots {T}_{1}^{-1}{T}_{0}{T}_{1}\dots {T}_{n}$ conjugate to ${T}_{0},$ $TnYαn∨- Y-αn∨ Tn = TnYεn∨- Y-εn∨Tn = ( Tn-1+ ( tn12- tn-12 ) ) Yεn∨- Y-εn∨ Tn = ( Tn-1 Yεn∨ ) - ( Tn-1 Yεn∨ ) -1 + ( tn12- tn-12 ) Yεn∨ = ( ( t012- t0-12 ) Y-εn∨ + ( tn12- tn-12 ) ) Yεn∨- Y-εn∨ 1-Y-2εn .$ $\square$

(Duality) [M03, (4.7.6)] The involution $\iota :\phantom{\rule{0.2em}{0ex}}{\stackrel{\sim }{ℬ}}_{n}\to {ℬ}_{n}$ in Proposition 2.4 descends to an involution $\iota :\phantom{\rule{0.2em}{0ex}}\stackrel{\sim }{H}\to \stackrel{\sim }{H}$ given by

$ι(q12)= q-12, ι(t12)= t-12, ι(tn12)= tn-12, ι(u012)= u0-12, ι(un12)= t0-12, ι(T0∨)= T0-1, ι(T0′)= (T0′)-1 ,ι(Ti)= Ti-1,for i=1,…,n.$

 Proof. A straightforward check shows that the relations in (3.17) are preserved. $\square$

In an attempt to relation the notations in [M03, §4.7], [S99, §3] and [C03, Def. 2.1] let

$τ0′=u012, τn′=un12 ,τ0=t012, τn=tn12, andτi=τi′ =t12fori=1 ,…,n-1.$

The summary of (1.5.1), (4.4.1), (4.4.2), (4.4.3), and (5.1.4) in [M03] is that

$τa= (tat2a)12 =q12κa= q 12 (k(a)+k(2a)) ,and τa′=ta12= q12κa′= q 12 (k(a)+k(2a)) .$

In our situation

$τn = tn12 = q12κn = tεn12 t2εn12 = q 12k(εn)+ 12k(2εn) = q 12k1+ 12k2 , τn′ = un12 = q12κn′ = tεn12 = q 12k(εn)- 12k(2εn) = q 12k1- 12k2 , τ0 = t012 = q12κ0 = t -ε1+ 12δ 12 t -2ε1+δ 12 = q 12k (ε1+12δ) +k (-2ε1+δ) = q 12k3+ 12k4 , τ0′ = u012 = q12κ0′ = t -ε1+ 12δ 12 = q 12k (ε1+12δ) -k (-2ε1+δ) = q 12k3- 12k4 , τi = t12 = q12κ = t εi-εi+1 12 = q12k5 ,$

for $i=1,\dots ,n-1,$ and the formulas in [M03, (1.5.1)] correspond to interchanging ${\kappa }_{0}$ and ${\kappa }_{n}^{\prime }\text{.}$

The other classical affine root systems are obtained from this one by setting some of the ${t}_{a}$ equal to 1. EXPAND ON THIS STATEMENT!

## Intertwiners

REPLACE THE ${Y}^{{\alpha }_{0}^{\vee }}$s IN THIS SUBSECTION.

From Theorem 3.3(b) it follows that

$τi∨Yλ∨= Ysiλ∨ τi∨,for i=0,1,2,…,n.$

if

$τi∨ = Ti+ t-12 (1-t) 1-Y-αi∨ =Ti-1+ t-12 (1-t) 1-Y-αi∨ 1-Y-αi∨ ,fori=1,…, n-1,and τn∨ = Tn+ tn-12 (1-tn)+ t0-12 (1-t0) Y-εn∨ 1-Y-2εn∨ = Tn-1+ ( tn-12 (1-tn)+ t0-12 (1-t0) Yεn∨ ) Y-2εn∨ 1-Y-2εn∨ , τ0∨ = T0∨+ un-12 (1-un)+ u0-12 (1-u0) Y-α0∨ 1-Y-2α0∨ = (T0∨)-1 ( un-12 (1-un)+ u0-12 (1-u0) Yα0∨ ) Y-2α0∨ 1-Y-2α0∨ ,$

The operators ${\tau }_{0}^{\vee },\dots ,{\tau }_{n}^{\vee }$ satisfy the relations

and

$(τ0∨)2 = un-1 ( u012 un12 -Yα0∨ ) ( u0-12 un12 +Yα0∨ ) ( u012 un12 -Y-α0∨ ) ( u0-12 un12 +Y-α0∨ ) (1-Yα0∨) (1+Yα0∨) (1-Y-α0∨) (1+Y-α0∨) , (τn∨)2 = tn-1 ( t012 tn12 -Yεn∨ ) ( t0-12 tn12 +Yεn∨ ) ( t012 tn12 -Y-εn∨ ) ( t0-12 tn12 +Y-εn∨ ) (1-Yεn∨) (1+Yεn∨) (1-Y-εn∨) (1+Y-εn∨) , (τi∨)2 = ( t12- t-12 Y-αi∨ ) ( t12- t-12 Yαi∨ ) (1-Y-αi∨) (1-Yαi∨) , fori=1,…,n-1.$

 Proof. $(τn∨)2 = τn∨ ( Tn- ( tn12- tn-12 ) + ( t012- t0-12 ) Y-εn∨ 1-Y-2εn∨ ) = τn∨Tn- ( tn12- tn-12 ) + ( t012- t0-12 ) Yεn∨ 1-Y2εn∨ τn∨ = Tn2- ( tn12- tn-12 ) + ( t012- t0-12 ) Y-εn∨ 1-Y-2εn∨ Tn- ( tn12- tn-12 ) + ( t012- t0-12 ) Yεn∨ 1-Y2εn∨ Tn + ( tn12- tn-12 ) + ( t012- t0-12 ) Yεn∨ 1-Y2εn∨ + ( tn12- tn-12 ) + ( t012- t0-12 ) Y-εn∨ 1-Y-2εn∨ = ( tn12- tn-12 ) Tn+ (1-Y2εn∨) (1-Y-2εn∨) (1-Y2εn∨) (1-Y-2εn∨) + ( ( tn12- tn-12 ) + ( t012- t0-12 ) Yεn∨ ) ( 1-Y2εn∨ ) ( ( tn12- tn-12 ) + ( t012- t0-12 ) Y-εn∨ ) ( 1-Y-2εn∨ ) + ( - ( tn12- tn-12 ) + ( t012- t0-12 ) Y-εn∨ 1-Y-2εn∨ + ( tn12- tn-12 ) Y-2εn∨ + ( t012- t0-12 ) Y-εn∨ 1-Y-2εn∨ ) Tn = ( 1-Y2εn∨ -Y-2εn∨+ 1+ ( tn12- tn-12 ) 2 + ( t012- t0-12 ) 2 + ( t012- t0-12 ) ( tn12- tn-12 ) Y-εn∨ + ( tn12- tn-12 ) ( t012- t0-12 ) Yεn∨ ) (1-Y2εn∨) (1-Y-2εn∨) = ( tn+tn-1 +t0+t0-1 -1-1-Y2εn∨ -Y-2εn∨ + ( t012 tn12- t012 tn-12- t0-12 tn12+ t0-12 tn-12 ) Y-εn∨ + ( t012 tn12- t012 tn-12- t0-12 tn12+ t0-12 tn-12 ) Yεn∨ ) (1-Yεn∨) (1+Yεn∨) (1-Y-εn∨) (1+Y-εn∨) = tn-1 ( t012 tn12- Yεn∨ ) ( t0-12 tn12+ Yεn∨ ) ( t012 tn12- Y-εn∨ ) ( t0-12 tn12+ Y-εn∨ ) (1-Yεn∨) (1+Yεn∨) (1-Y-εn∨) (1+Y-εn∨)$ The formula for ${\left({\tau }_{i}^{\vee }\right)}^{2}$ is established by exactly the same computation, except with ${t}_{0}^{\frac{1}{2}}$ and ${t}_{n}^{\frac{1}{2}}$ both replaced by ${t}^{\frac{1}{2}}\text{.}$ $\square$

Question: Is the affine Hecke algebra a quotient of DAHA? This seems to be true on the braid group. NO?!? The obvious morphism would take ${T}_{0}^{\vee }↦1$ which would violate $\left({T}_{0}^{\vee }-{u}_{n}^{\frac{1}{2}}\right)\left({T}_{0}^{\vee }+{u}_{n}^{\frac{1}{2}}\right)-0\text{.}$ Not clear that reps of affine Hecke would lift to DAHA. The 1 dimensional representations lift or do not lift?

## Notes and References

This page is taken from a paper entitled Double affine braid groups and Hecke algebras of classical type by Arun Ram, March 5, 2009.