## The Quantum Double of ${𝔘}_{h}{𝔟}^{+}$

Last updates: 19 June 2011

## The Quantum Double of ${𝔘}_{h}{𝔟}^{+}$

1.1 Recall that ${𝔘}_{h}{𝔟}^{+}$ is the algebra over $k\left[\left[h\right]\right]$ with generators $X$ and $Y$ and product and coproduct satisfying $[ H , X ]= HX-XH=2X Δ(X)= H⊗1+1⊗H,and Δ(X)=X⊗ e h 4 H + e h 4 H⊗X$ Let ${\left({𝔘}_{h}{𝔟}^{+}\right)}^{\ast }$ be the dual of ${𝔘}_{h}{𝔟}^{+}$ with basis ${\left({H}^{r}{X}^{a}\right)}^{\ast }$ which is dual to the basis ${H}^{r}{X}^{a}$ of ${𝔘}_{h}{𝔟}^{+}$ We shall use the notation $⟨ (Hr)∗ , HsXb ⟩= δrsδab.$ For notational convenience we shall set $Y={X}^{\ast }$. We shall determine the algebra structure on ${\left({𝔘}_{h}{𝔟}^{+}\right)}^{\ast }$ which is dual to the coalgebra structure on ${𝔘}_{h}{𝔟}^{+}$. This is given by defining $⟨ a∗b∗ , c ⟩= ⟨ a∗⊗b∗ , Δ(c) ⟩,$ where ${\ast }^{},{b}^{\ast }\in {\left({𝔘}_{h}{𝔟}^{+}\right)}^{\ast }$ and $c\in {𝔘}_{h}{𝔟}^{+}.$ The coalgebra structure on ${\left({𝔘}_{h}{𝔟}^{+}\right)}^{\ast }$ is dual to the algebra structure except with the opposite multiplication. More specifically, $⟨ Δ(a∗) , b⊗c ⟩ = ⟨ a∗ , cb ⟩$ for all ${a}^{\ast },{\ast }^{}\in {\left({𝔘}_{h}{𝔟}^{+}\right)}^{\ast }$ and $c\in {𝔘}_{h}{𝔟}^{+}$.

1.2 The identity $⟨ (H∗)r , Hs ⟩ = ⟨ H∗⊗⋯⊗H∗ , Δ(r) (Hs) ⟩ = ⟨ H∗⊗⋯⊗H∗ , ( ∑ i=1 r 1⊗⋯⊗1⊗H∗⊗1⋯ ⊗1 )s ⟩ = δrs ⟨ H∗⊗⋯⊗H∗ , r! ( H⊗H⊗⋯⊗H ) ⟩ = δrsr!,$ shows that $r! (Hr)∗= (H∗)r.$

Recall that there is a grading on ${𝔘}_{h}{𝔟}^{+}$ given by setting $deg\left(X\right)=1$ and $deg\left(H\right)=0.$ Since the comultiplication $\Delta$ on ${𝔘}_{h}{𝔟}^{+}$ preserves this grading there will be a similar grading on ${\left({𝔘}_{h}{𝔟}^{+}\right)}^{\ast }.$ Thus, one will have that $⟨ H∗Y , HrXa ⟩=0$ if $a\ne 0$, sinc eht left hand side is degree $1$ and the right hand side is degree $a$. Consider the equation $⟨ H∗Y , HrX ⟩ = ⟨ H∗⊗Y , Δ(Hr)Δ(X) ⟩ = ⟨ H∗⊗Y , ( ∑ i+j=r ( r i Hi⊗Hj ( X⊗ e h 4 H + e - h 4 H ⊗X ) ) ⟩.$ The right hand side is $0$ unless $r=0$ or $1$. In the case $r=0$ we have $⟨ H∗Y , HrX ⟩ = ⟨ H∗⊗Y , X⊗e H 4 + e - H 4 ⊗X ⟩ = ⟨ H∗⊗y , - h 4 H⊗X ⟩= - h 4 .$ In the case $r=1$ we have $⟨ H∗Y , HrX ⟩ = ⟨ H∗⊗Y , ( H⊗1+1⊗H ) ( X⊗e H 4 + e - H 4 ⊗X ) ⟩ = ⟨ H∗⊗Y , H⊗X ⟩=1.$ It follows that ${H}^{\ast }Y={HX}^{\ast }-\frac{h}{4}Y$. A similar calculation shows that $Y{H}^{\ast }={\left(HX\right)}^{\ast }+\frac{h}{4}Y$. Thus we have that $[ H∗ , Y ]= H∗Y-YH∗= ( - h 4 - h 4 )Y =- h 2 Y .$

From the equality we get that $Δ(H∗)= H∗⊗1+1⊗H∗.$ In order to compute ${\Delta }^{op}\left(Y\right)$ we have Noting that $XH=-\left[H,X\right]+HX=-2X+HX=\left(H-2\right)X$ we have that It follows from the previous computations that $Δop(Y) = 1⊗Y+ ∑ b≥0 (-2)bY⊗ (Hb)∗ = 1⊗Y+ ∑ b≥0 (-2)bY⊗ (H∗)b b! = 1⊗Y+Y⊗e-2H∗.$

1.4 In summary, the bialgebra ${\left({𝔘}_{h}{𝔟}^{+}\right)}^{\ast }$ is given by generators ${H}^{\ast }$ and $Y$ with operations given by $[ H∗ , Y ]= - h 2 Y, Δ(H∗)= H∗⊗1+1⊗H∗, Δ(Y)= Y⊗1+ e-2H⊗Y.$

1.5 Recall that the multiplication in $D\left({𝔘}_{h}{𝔟}^{+}\right)={𝔘}_{h}{𝔟}^{+}\otimes {\left({𝔘}_{h}{𝔟}^{+}\right)}^{\ast }$ is determined by $∑ α,β,δ,γ,p μrγβα σαp mpδγs eδeβ$ where ${e}_{r}$ is a basis of ${𝔘}_{h}{𝔟}^{+}$, ${e}^{s}$ is a basis of ${\left({𝔘}_{h}{𝔟}^{+}\right)}^{\ast }$, and $m$, $\mu$, $\sigma$ denote the matrices of the multiplication, comultliplication and skew antipode of ${𝔘}_{h}{𝔟}^{+}$ respectively. Write $Δ(2)(er) = ∑ α,β,γ μrγβα eγ⊗eβ⊗eα, Δ(2)(es)= ∑ p,δ,γ mpδγs eγ⊗eδ⊗ep, ( id⊗id⊗S-1 ) Δ(2)(er) = ∑ α,β,γ γrγβα σαp eγ⊗eβ⊗eα.$ Then define a map $M:{\left({𝔘}_{h}{𝔟}^{+}\right)}^{\otimes 3} × {\left({\left({𝔘}_{h}{𝔟}^{+}\right)}^{\ast }\right)}^{\otimes 3}\to D\left({𝔘}_{h}{𝔟}^{+}\right)$ which

1. takes the inner product of the first components,
2. takes the inner product of the third components,
3. multiplies the middle components.
An alternate way of writing (1.5) is to say that the multiplication in $D\left({𝔘}_{h}{𝔟}^{+}\right)$ is determined by the equation $eres= M ( Δ(2)(es), id⊗id⊗S-1 Δ(2)(er) ).$

1.6 Let us collect together some useful relations. The skew antipode ${S}^{-1}$ is determined by $0=ϵ(X) = m(id⊗S-1) Δop(X) = m(id⊗S-1) (X⊗e - h 4 H + e h 4 H ⊗X) = XS-1 (e - h 4 H ) +e h 4 H S-1(X) = Xe h 4 H + e h 4 H S-1(X).$ Thus ${S}^{-1}\left(X\right)=-{e}^{-\frac{h}{4}H}X{e}^{\frac{h}{4}H}$. It follows from the equality $XH=\left(H-2\right)X$ that

 $S-1(X)=- e - h 4 H X e h 4 H = - e - h 4 H e h 4 (H-2) X = - e - h 2 X.$ (1.6a)
A similar calculation shows that
 $S-1(H)=-H.$ (1.6b)
From the definition of the coproduct we have
 $Δ(2)(H) =H⊗1⊗1+ 1⊗H⊗1+1⊗1⊗H, Δ(2)(X)= X⊗ e h 4 H ⊗ e h 4 H + e - h 4 H ⊗ X ⊗ e h 4 H + e - h 4 H ⊗ e - h 4 H ⊗ X, Δ(2)(H∗)= H∗ ⊗1⊗1 + 1⊗H∗⊗1 + 1⊗1⊗H∗, Δ(2)(Y)= Y⊗1⊗1+ e-2H∗⊗Y⊗1 + e-2H∗⊗ e-2H∗⊗Y.$ (1.6c)
It follows that
 $( id⊗id⊗S-1 ) Δ(2)(H)= H⊗1⊗1+1⊗H⊗1- 1⊗1⊗H, ( id⊗idS-1 ) Δ(2)(X)= X⊗ e h 4 H ⊗ e - h 4 H + e - h 4 H ⊗ X ⊗ e - h 4 H - e - h 4 H ⊗ e - h 4 H ⊗ e - h 4 H X.$ (1.6d)

 $[ H , H∗ ]=0.$ (1.7a)
Similarly, $HH∗ = M ( Δ(2)(H∗), (id⊗id⊗S-1) Δ(2)(X) ) = M ( H∗⊗1⊗1+ 1⊗H∗⊗1+ 1⊗1⊗H∗, X⊗ e h 4 H ⊗ e - h 4 H + e - h 4 H ⊗ X ⊗e - h 4 H - e - h 4 H ⊗ e - h 4 H ⊗ e - h 2 X ) = 0 - h 4 X -0+0+H∗X-0+0- h 4 X -0 = H∗X- h 2 X,$ which shows that
 $[ H∗ , X ]= h 2 X.$ (1.7b)
To compute the product $HY$ we have $HY = M ( Δ(2)(Y), (id⊗id⊗S-1) Δ(2)(H) ) = M ( Y⊗1⊗1+ e-2H∗⊗Y⊗1+ e-2H∗⊗ e-2H∗⊗Y, H⊗1⊗1+ 1⊗H⊗1- 1⊗1⊗H ) = 0+0+0-2Y+YH-0+0 +0-0 = YH-2Y,$ which shows that
 $[ H , Y ]=-2Y.$ (1.7c)
The last case is the product $XY$. $XY = M ( Δ(2)(Y), (id⊗id⊗S-1) Δ(2)(X) ) = M ( Y⊗1⊗1+ e-2H∗⊗Y⊗1+ e-2H∗⊗ e-2H∗⊗Y, X⊗ e h 4 H ⊗ e - h 4 H + e - h 4 H ⊗X⊗ e - h 4 H - e - h 4 H ⊗ e - h 4 H ⊗ e - h 2 X ) = e h 4 H +0-0+0+ ⟨ e-2H∗ , e - h 4 H ⟩ YX-0+0-0- ⟨ e-2H∗ , e - h 4 H ⟩e-2H∗ e - h 4 H e h 2 .$ Evaluating the inner product gives $⟨ e-2H , e - h 4 H ⟩ = ⟨ ∑ r≥0 (-2)r (H∗)r r! , ∑ s≥0 (-h)s 4s Hs s! ⟩ = ⟨ ∑ r≥0 (-2)r (Hr)∗ r! , ∑ s≥0 (-h)s 4s Hs s! ⟩ = ∑ r≥0 (-2)r (-h)r 4rr! = e h 2 .$ Thus we get that
 $xY = e h 4 H + e h 2 YX - e h 2 e-2H∗ e - h 4 H e - h 2 = e h 4 H + e h 2 YX- e-H∗ e - h 4 H .$ (1.7d)

1.8 The coproduct evaluated at $Y$ is given by $Δ(Y)= e-2H∗ ⊗Y+Y⊗1.$ We shall renormalize $Y$ to make this coproduct more symmetric. Let $Y ˆ = YeH∗.$ Then $Δ ( Y ˆ ) = ( e-2H∗⊗Y+Y⊗1 ) ( eH∗⊗eH∗ ) = e-H∗⊗ Y ˆ + Y ˆ ⊗ e-H∗.$ Furthermore, with this normalization we have that $X Y ˆ = ( e h 4 H - e-2H∗ e - h 4 H + e h 2 YX ) eH∗ = e h 4 H+H∗ - e - h 4 H+H∗ + e h 2 Y e ( H∗- h 2 ) X = Y ˆ X + e h 4 H+H∗ - e - ( h 4 H+H∗ ) ,$ since $X{H}^{\ast }=-\left[{H}^{\ast },X\right]+{H}^{\ast }X=-\left(h/2\right)X+{H}^{\ast }X=\left(H-\left(h/2\right)\right)X.$

1.9 Summarizing, we have that $D\left({𝔘}_{h}{𝔟}^{+}\right)$ is generated by $X,\stackrel{ˆ}{Y},H,{H}^{\ast }$ with multiplication given by $[ H , H∗ ]=0 [ H , X ]=2X, [ H , Y ˆ ]=-2 Y ˆ , [ H∗ , X ]= h 2 X, [ H∗ , Y ˆ ]= - h 2 Y ˆ , X Y ˆ - Y ˆ X= e(h/2)J1 - e-(h/2)J2,$ where $J1= (1/h) ( 2H∗+ h 2 H ), and J1= (1/h) ( 2H∗- h 2 H ).$ Then we have that $[ J1 , X ] = (1/h) [ 2H∗+ h 2 H , X ] = (1/h) (2(h/2)X+ (h/2)2X ) =2X, [ J1 , Y ] = (1/h) [ 2H∗+ h 2 H , Y ] = (1/h) ( -2((h/2))Y -(h/2)2Y ) =-2Y, [ J2 , X ] = (1/h) [ 2H∗- h 2 , X ] = (1/h) ( 2(h/2)X- (h/2)2X )=0, [ J2 , Y ] = (1/h) [ 2H∗- h 2 H , Y ] = (1/h) ( -2(h/2)Y- (h/2)(-2)Y )=0.$ Furthermore, $Δ ( Y ˆ ) = e-H∗ ⊗ Y ˆ + Y ˆ ⊗ e-H∗ = e h 4 (J1+J2) ⊗ Y ˆ + Y ˆ ⊗ e h 4 (J1+J2) .$ Set $Y ‾ = Y ˆ e h 2 - e - h 2 .$ Then $[ J1 , X ]=2X, [ J2 , Y ‾ ]=-2 Y ‾ , [ X , Y ‾ ]= e h 2 J - e - h 2 J e h 2 - e - h 2 , [ J1 , X ]=0, [ J2 , Y ‾ ]=0, [ J1 , J2 ]=0.$ It follows that ${J}_{2}$ generates an ideal in $D\left({𝔘}_{h}{𝔟}^{+}\right)$.

1.10 Let us define $𝔘h𝔰𝔩(2)= 𝔘h𝔟+ ⟨J2⟩$ where $⟨{J}_{2}⟩$ is the ideal of ${𝔘}_{h}{𝔟}^{+}$ generated by ${J}_{2}$. Let $K= e h 4 J1 , and q= e h 2 .$ Then ${𝔘}_{h}𝔰𝔩\left(2\right)$ is the Hopf algebra over $\left[\left[h\right]\right]$ generate by ${J}_{1},X,\stackrel{}{Y}$ with the relations $[ J1 , X ]= 2X, [ J1 , Y ‾ ] =-2 Y ‾ , [ X , Y ‾ ] = K2-K-2 q-q-1 .$ It follows from the fact that $X{J}_{1}=\left({J}_{1}-2\right)X$ and $\stackrel{}{Y}{J}_{1}=\left({J}_{1}+2\stackrel{}{Y}\right)\stackrel{}{Y}$ that $KXK-1=qX, and KY‾ K-1= q-1KY‾.$ The coproduct is given by $Δ(J1)= J1⊗1+1⊗J1, Δ(K)=K⊗K, Δ(X)= X⊗K+K-1⊗X, Δ (Y‾) = Y‾⊗K+ K-1⊗Y‾.$ The counit and the antipode are given explicitly by $ϵ(J1)=0, ϵ(K)=1, ϵ(Y‾)=0, S(J1)= -J1, S(K)=K-1, S(X)=-qX, S(Y‾)= -q-1Y‾.$

## References

The fact that ${𝔘}_{h}𝔰𝔩\left(2\right)$ is a quotient of the double of ${𝔘}_{h}{𝔟}^{+}$ is stated in §13 of [
D] along with the formula for the universal $R$-matrix of ${𝔘}_{h}𝔰𝔩\left(2\right)$

[Bou] N. Bourbaki, Lie groups and Lie algebras, Chapters I-III, Springer-Verlag, 1989.

[D] V.G. Drinfeld, Quantum Groups, Vol. 1 of Proccedings of the International Congress of Mathematicians (Berkeley, Calif., 1986). Amer. Math. Soc., Providence, RI, 1987, pp. 198–820. MR0934283

The theorem that a Lie bialgebra structure determinse a co-Poisson Hopf algebra structure on its enveloping algebra is due to Drinfel'd and appears as Theorem 1 in the following article.

[D1] V.G. Drinfel'd, Hopf algebras and the quantum Yang-Baxter equation, Soviet Math. Dokl. 32 (1985), 254–258. MR0802128

[DHL] H.-D. Doebner, Hennig, J. D. and W. Lücke, Mathematical guide to quantum groups, Quantum groups (Clausthal, 1989), Lecture Notes in Phys., 370, Springer, Berlin, 1990, pp. 29–63. MR1201823

There is a very readable and informative chapter on Lie algebra cohomolgy in the forthcoming book

[HGW] R. Howe, R. Goodman, and N. Wallach, Representations and Invariants of the Classical Groups, manuscript, 1993.

The following book is a standard introductory book on Lie algebras; a book which remains a standard reference. This book has been released for a very reasonable price by Dover publishers. There is a short description of Lie algebra cohomology in Chapt. III §11, pp.93-96.

[J] N. Jacobson, Lie algebras, Interscience Publishers, New York, 1962.

A very readable and complete text on Lie algebra cohomology is

[Kn] A. Knapp, Lie groups, Lie algebras and cohomology, Mathematical Notes 34, Princeton University Press, 1998. MR0938524