The quantum double

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

Last updates: 16 July 2011

The quantum double

([D1] §13) Let $A$ be a finited dimensional Hopf algebra and let $A^{* opp}$ denote the Hopf algebra $A^{*}$ except with the opposite comultiplication. Then there exists a unique quasitriangular Hopf algebra $D (A, R)$ such that

$D (A)$ contains $A$ and $A^{* opp}$ as Hopf subalgebras.
$R$ is the image of the canonical element of $A \otimes A^{* opp}$ under $A \otimes A^{* opp} \to D (A) \otimes D (A),$ i.e. if $e_{i}$ is a basis of $A$ and ei is the dual basis in $A^{* opp}$ then $R = \sum e_{i} \otimes e^{i} \in D (A) \otimes D (A) .$
The linear map $\begin{matrix} A \otimes A^{* opp} & \to & D (A) \\ a \otimes b & \to & a b \end{matrix}$ is bijective.

2.2 Remark. If $A$ is infinite dimensional then one may be abl to apply the theorem if there is a suitable way of completing the tensor product $D (A) \otimes D (A)$ so that the element $R = \sum e_{i} \otimes e^{i}$ is a well defined element of the completion $D (A) \hat{\otimes} D (A)$ .

		Proof of Theorem 1.1.

□

2.3 Let the algebra $A$ be the Hopf algebra with basis $\{e_{r}\}$ and multiplication, comultiplication, and skew antipode given by $\begin{array}{rcl} e_{r} e_{s} & = & \sum_{t} m_{r s}^{t} e_{t}, \\ Δ (e_{t}) & = & \sum_{r, s} μ_{t}^{r s} e_{r} \otimes e_{s}, \\ σ (e_{t}) & = & \sum_{r} σ_{t}^{r} e_{r} . \end{array}$ The unit and counit will ge given by $1 = \sum_{t} E^{t} e_{t}$ and $ϵ (e_{r}) = e_{r}$ . Recall that the skew antipode is the inverse $S^{- 1}$ of the antipode of $A$ and is the antipode for the Hopf algebra $A^{opp}$ which is the same as the algebra $A$ except with the opposite comultiplication.

2.4 The algebra $A^{* opp}$ has basis $\{e^{r}\}$ which is dual to the basis $\{e_{r}\}$ of $A$ and has multiplication and comultiplication given by $\begin{array}{rcl} e^{r} e^{s} & = & \sum_{t} μ_{t}^{r s} e_{t}, \\ Δ (e^{t}) & = & \sum_{r s} e^{s} \otimes e^{r} . \end{array}$ Then the algebra $A \otimes A^{* opp}$ has basis $\{e^{r} e_{s}\}$ and has multiplication given by

(e^{r} e_{s} \otimes e^{p} e_{q}) (e^{k} e_{l} \otimes e^{m} e_{n}) = (e^{r} e_{s} e^{k} e_{l} \otimes e^{p} e_{q} e^{m} e_{m})

LABEL

and comultiplication given by

\begin{array}{rcl} Δ (e^{r} e_{s}) & = & Δ (e^{r}) Δ (e_{s}) \\ = & (\sum_{u, v} m_{u v}^{r} e^{v} \otimes e^{u}) (\sum_{p, q} μ_{s}^{p q} e_{p} \otimes e_{q}) \\ = & \sum_{u, v, p, q} m_{u v}^{r} μ_{s}^{p q} e^{v} e_{p} \otimes e^{u} e_{q} . \end{array}

Alternatively, we could have chosen to use the basis

\{e_{r} e^{s}\}

instead of the basis

e^{r} e_{s}

. It is clear from (LABEL) that we need to describe a product

e_{s} e^{k}

in terms of the basis

e^{p} e_{q}

. The relation is

e_{r} e^{s} = \sum_{α, β, γ, δ, p} μ_{r}^{γ β α} σ_{α}^{p} m_{p δ γ}^{s} e^{δ} e_{β} .

This relation is derived as follows.

\begin{array}{rcl} ⟨ e^{v} e_{b}, e_{j} e^{l} ⟩ & = & ⟨ e^{v} e_{b}, m \circ σ (e^{l} \otimes e_{j}) ⟩ \\ = & ⟨ σ \circ Δ (e^{v} e_{b}), e^{l} \otimes e_{j} ⟩ \\ = & ⟨ Δ' (e^{v} e_{b}), e^{l} \otimes e_{j} ⟩ \\ = & ⟨ R Δ (e^{v} e_{b}) R^{- 1}, e^{l} \otimes e_{j} ⟩ \\ = & ⟨ R Δ (e^{v} e_{b}) (id \otimes S^{- 1}) (R), e^{l} \otimes e_{j} ⟩ by (1.3e) \\ = & ⟨ R \otimes Δ (e^{v} e_{b}) \otimes (id \otimes S^{- 1}) (R), {(Δ^{\otimes})}^{2} (e^{l} \otimes e_{j}) ⟩ . \end{array}

Let us expand the left hand factor of this inner product.

\begin{array}{rcl} R \otimes Δ (e^{v} e_{b}) \otimes (id \otimes S^{- 1}) (R) & = & \sum_{k, p} e_{k} \otimes e^{k} \otimes Δ (e^{v} e_{b}) \otimes (id \otimes S^{- 1}) (R) (e_{p} \otimes e^{p}) \\ = & \sum_{\begin{matrix} k, p, q \\ r, s, t, u \end{matrix}} e_{k} \otimes e^{k} \otimes m_{r s}^{v} μ_{b}^{u t} e^{r} e_{u} \otimes e^{s} e_{t} \otimes e_{p} \otimes σ_{q}^{p} e^{q} . \end{array}

LABEL2

The right hand factor of the inner product expands in the form

\begin{array}{rcl} {(Δ^{\otimes})}^{2} (e^{l} \otimes e_{j}) & = & (Δ^{\otimes} \otimes {id}^{\otimes}) \circ Δ^{\otimes} (e^{l} \otimes e_{j}) \\ = & (Δ^{\otimes} \otimes {id}^{\otimes}) (\sum_{x, y, w, z} m_{x y}^{l} μ_{j}^{w z} e^{y} \otimes e_{w} \otimes e^{x} \otimes e^{z}) \\ = & \sum_{\begin{matrix} x, y, w, z \\ m, n, c, d \end{matrix}} m_{m n}^{y} μ_{w}^{c d} m_{x y}^{l} μ_{j}^{w z} e^{n} \otimes e_{c} \otimes e^{m} \otimes e_{d} \otimes e^{x} \otimes e_{z} \\ = & \sum_{\begin{matrix} m, n, x \\ c, d, z \end{matrix}} m_{x m n}^{l} μ_{j}^{c d z} e^{n} \otimes e_{c} \otimes e^{m} \otimes e_{d} \otimes e^{x} \otimes e_{z} . \end{array}

Now let us evaluate the inner product. Thi inner product picks out only the terms when

m = n, k = c, v = m, b = d, p = x, q = z,

and this term appears with coefficient

m_{x m m}^{l} μ_{j}^{c d z} σ_{q}^{p} = m_{p v k}^{l} μ_{j}^{k b q} σ_{q}^{p} .

It follows that

⟨ e^{v} e_{b}, e_{j} e^{l} ⟩ = \sum_{p, q, k} m_{p v k}^{l} μ_{j}^{k b q} σ_{q}^{p} .

The multiplication rule follows.

2.6 We shall need the following calculation in our proof that $D (A)$ is quasitriangular. We shall need the identities in §4 of the notes on co-Poisson Hopf algebras. $\begin{array}{rcl} \sum_{γ, α, p, s} μ_{a}^{γ β α s} σ_{α}^{p} m_{s p δ γ}^{v} & = & \sum_{γ, α, n, k, s, p} μ_{a}^{γ β n} μ_{n}^{α s} σ_{α}^{s} m_{s p}^{k} m_{k δ γ}^{v} by 4.1 and 4.4 \\ = & \sum_{γ, α, n, k} μ_{a}^{γ β n} ε_{n} E^{k} m_{k δ γ}^{v} by 4.11 \\ = & \sum_{γ, α, n, k} μ_{a}^{γ β n} ε_{n} E^{k} δ_{δ n} m_{m γ}^{v} by 4.2 \\ = & \sum_{γ, n} μ_{a}^{γ β n} ε_{n} E^{k} m_{δ γ}^{v} \\ = & \sum_{γ, n, k} μ_{a}^{γ k} μ_{k}^{β n} ε_{n} m_{δ γ}^{v} by 4.4 \\ = & \sum_{γ, k} μ_{a}^{γ k} δ_{k β} m_{δ γ}^{v} by 4.5 \\ = & \sum_{γ} μ_{a}^{γ β} m_{δ γ}^{v} . \end{array}$

2.7 Now we prove that $A \otimes A^{* opp}$ satisfies the condition (1.2a) for a quasitriangular Hopf algebra. $\begin{array}{rcl} ((σ \circ Δ) (e^{v} e_{b})) R & = & \sum_{δ, γ, r, s, m} m_{δ γ}^{v} μ_{b}^{r s} (e^{δ} e_{s} \otimes e^{γ} e_{r}) (e_{m} \otimes e^{m}) \\ = & \sum_{δ, γ, r, s, m} m_{δ γ}^{v} μ_{b}^{r s} (e^{δ} e_{s} e_{m} \otimes e^{γ} e_{r} e^{m}) \\ = & \sum_{δ, γ, r, s, m, λ} m_{δ γ}^{v} μ_{b}^{r s} m_{s m}^{λ} (e^{δ} e_{λ} \otimes e^{γ} e_{r} e^{m}) \\ = & \sum_{\begin{matrix} δ, γ, r, s, m, λ \\ u, t, α, p, β \end{matrix}} m_{δ γ}^{v} μ_{b}^{r s} μ_{r}^{u t α} σ_{α}^{p} m_{p β u}^{m} m_{s m}^{λ} (e^{δ} e_{λ} \otimes e^{γ} e^{β} e_{t}) \\ = & \sum_{\begin{matrix} δ, γ, r, s, m, λ \\ u, t, α, p, β, a \end{matrix}} m_{δ γ}^{v} μ_{a}^{γ β} μ_{r}^{u t α} σ_{α}^{p} m_{p β u}^{m} m_{s m}^{λ} (e^{δ} e_{λ} \otimes e^{a} e_{t}) \\ = & \sum_{\begin{matrix} δ, γ, s, m, λ \\ u, t, α, p, β, a \end{matrix}} m_{δ γ}^{v} μ_{a}^{γ β} μ_{b}^{u t α s} σ_{α}^{p} m_{s p β u}^{λ} (e^{δ} e_{λ} \otimes e^{a} e_{t}) \\ = & \sum_{\begin{matrix} δ, γ, s, λ \\ u, t, α, p, β, a \end{matrix}} m_{δ γ}^{v} μ_{a}^{γ β} μ_{b}^{u t α s} σ_{α}^{p} m_{s p β u}^{λ} (e^{δ} e_{λ} \otimes e^{a} e_{t}) \\ = & \sum_{δ, γ, u, t, β, a, λ} m_{δ γ}^{v} μ_{a}^{γ β} μ_{b}^{u t} m_{β u}^{λ} (e^{δ} e_{λ} \otimes e^{a} e_{t}) . \end{array}$ A similar calculation on the right hand side gives $\begin{array}{rcl} R Δ (e^{v} e_{b}) & = & \sum_{r, s, t, u, m} m_{s r}^{v} μ_{b}^{u t} (e_{m} \otimes e^{m}) (e^{r} e_{u} \otimes e^{s} e_{t}) \\ = & \sum_{r, s, t, u, m} m_{s r}^{v} μ_{b}^{u t} (e_{m} e^{r} e_{u} \otimes e^{m} e^{s} e_{t}) \\ = & \sum_{r, s, t, u, m, a} m_{s r}^{v} μ_{b}^{u t} μ_{a}^{m s} (e_{m} e^{r} e_{u} \otimes e^{a} e_{t}) \\ = & \sum_{\begin{matrix} r, s, t, u, m, a \\ α, β, γ, δ, p \end{matrix}} m_{s r}^{v} μ_{b}^{u t} μ_{a}^{m s} μ_{m}^{γ β α} σ_{α}^{p} m_{p δ γ}^{r} (e^{d} e_{β} e_{u} \otimes e^{a} e_{t}) \\ = & \sum_{\begin{matrix} r, s, t, u, m, a \\ α, β, γ, δ, p, λ \end{matrix}} m_{s r}^{v} μ_{b}^{u t} μ_{a}^{m s} μ_{m}^{γ β α} σ_{α}^{p} m_{p δ γ}^{r} m_{β u}^{λ} (e^{δ} e_{λ} \otimes e^{a} e_{t}) \\ = & \sum_{\begin{matrix} s, t, u, m, a \\ α, β, γ, δ, p, λ \end{matrix}} μ_{b}^{u t} μ_{a}^{γ β α s} σ_{α}^{p} m_{s p δ γ}^{v} m_{β u}^{λ} (e^{δ} e_{λ} \otimes e^{a} e_{t}) \\ = & \sum_{δ, γ, u, t, β, a λ} m_{δ γ}^{v} μ_{a}^{γ β} μ_{b}^{u t} m_{β u}^{λ} (e^{δ} e_{λ} \otimes e^{a} e_{t}) . \end{array}$

2.8 It remains to prove the identities $(id \otimes Δ) (R) = R^{13} R^{12}$ and $(Δ \otimes id) (R) = R^{13} R^{23}$ . $\begin{array}{rcl} (id \otimes Δ) (R) & = & \sum_{k} e_{k} \otimes Δ (e^{k}) \\ = & \sum_{k, r, s} m_{r s}^{k} e_{k} \otimes e^{s} \otimes e^{r} \\ = & \sum_{r, s} e_{r} e_{s} \otimes e^{s} \otimes e^{r} \\ = & \sum_{r, s} (e_{r} \otimes 1 \otimes e^{r}) (e_{s} \otimes e^{s} \otimes 1) \\ = & R^{13} R^{12} . \end{array}$ Similarly we have that $\begin{array}{rcl} (Δ \otimes id) (R) & = & \sum_{k} Δ (e_{k}) \otimes e^{k} \\ = & \sum_{k, r, s} μ_{k}^{r s} e_{r} \otimes e_{s} e^{k} \\ = & \sum_{r, s} e_{r} \otimes e_{s} \otimes e^{r} e^{s} \\ = & \sum_{r, s} (e_{r} \otimes 1 \otimes e^{r}) (1 \otimes e_{s} \otimes e^{s}) \\ = & R^{13} R^{23} \end{array}$ This completes the proof of Theorem 1.1.

References

The quantum double seems to have appeared first in the following paper.

[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

Some further proofs and hints appear in the following.

[D1] V.G. Drinfelʹd, Almost cocommutative Hopf algebras, (Russian) Algebra i Analiz 1 (1989), 30–46; translation in Leningrad Math. J. 1 (1990), 321–342. MR1025154

[Re] N. Yu. Reshetikhin, Quantized universal enveloping algebras, the Yang-Baxter equation and invariants of links I, LOMI preprint no. E–4–87, (1987).

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