Lie Bialgebras

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

Last updates: 21 March 2011

Lie bialgebras

1.1 A Lie bialgebra is a Lie algebra with bracket

[,]

and cobracket

φ : 𝔤 \to 𝔤 \otimes 𝔤

such that

$𝔤^{*}$ with bracket $φ^{*} : 𝔤^{*} \otimes 𝔤^{*} \to 𝔤^{*}$ is a Lie algebra, and
$φ$ satisfies the $1$ -cocycle condition, $φ ([x, y]) = [x \otimes 1 + 1 \otimes x, φ (y)] - [y \otimes 1 + 1 \otimes y, φ (x)], for all x, y \in 𝔤 .$

In 1),

φ^{*} : {(𝔤 \otimes 𝔤)}^{*} \to 𝔤^{*}

is the induced mapping on the dual spaces. Given an element

x^{*} \in 𝔤^{*}

let us let

⟨ x^{*}, a ⟩ = x^{*} (a)

denote the evaluation of

x^{*}

on the element

a \in 𝔤

. Then

φ^{*} : {(𝔤 \otimes 𝔤)}^{*} \to 𝔤^{*}

is given explicitly by

⟨ φ^{*} (x^{*} \otimes y^{*}), a \otimes b ⟩ = ⟨ x^{*}, a ⟩ ⟨ y^{*}, b ⟩,

for all

x^{*}, y^{*} \in 𝔤^{*}

. Note that

φ^{*} : 𝔤^{*} \otimes 𝔤^{*} \to 𝔤^{*}

is a well defined map since

𝔤^{*} \otimes 𝔤^{*} \subseteq {(𝔤 \otimes 𝔤)}^{*}

(even if

𝔤

is infinite dimensional).

1.2 An element $r \in 𝔤 \otimes 𝔤 = V = C^{0} (𝔤, 𝔤 \otimes 𝔤)$ determines a $1$ -coboundary $d r \in B^{1} (𝔤, 𝔤 \otimes 𝔤)$ given by $d r (x) = [1 \otimes x + x \otimes 1, r]$ , for all $x \in 𝔤$ . Since $d^{2} = 0$ we know that every $1$ -coboundary is a $1$ -cocycle. Thus $r \in 𝔤 \otimes 𝔤$ determines a 1-cocycle and posibly a bialgebra structure on $𝔤$ . Thus it is a natural question:

Given an element $r \in 𝔤 \otimes 𝔤$ , what conditions should we place on $r$ to guarantee that the map $δ : 𝔤 \to 𝔤 \otimes 𝔤$ determined by $δ (x) = [1 \otimes x + x \otimes 1, r]$ , determines a Lie bialgebra structure on $𝔤$ ?

This question is answered by the following proposition:

Let $𝔤$ be a Lie algebra, let $r \in 𝔤 \otimes 𝔤$ and define a map $d r : 𝔤 \to 𝔤 \otimes 𝔤$ by $d r (x) = [1 \otimes x + x \otimes 1, r]$ . Let $ρ = \frac{1}{2} (r^{12} - r^{21})$ and let $P = \frac{1}{2} (r^{12} + r^{21})$ so that $r = P + ρ$ .

The map ${d r}^{*} : 𝔤^{*} \otimes 𝔤^{*} \to 𝔤^{*}$ satisfies the skew-symmetric condition if and only if $({ad}^{\otimes 2}) (r^{12} + r^{21}) = 0$ for all $x \in 𝔤$ .
Assume that $P$ is ${ad}^{\otimes 2}$ invariant. Then ${d r}^{*}$ satisfies the Jacobi identity if and only if $({ad}^{\otimes 3}) ([r^{12}, r^{13}] + [r^{12}, r^{23}] + [r^{13}, r^{23}]) = 0$ for all $x \in 𝔤$ .

If $r = \sum_{i} a_{i} \otimes b_{i}$ then the notation above is $\begin{matrix} r^{12} + r^{21} = \sum_{i} (a_{i} \otimes b_{i} + b_{i} \otimes a_{i}), \\ [r^{12}, r^{13}] + [r^{12}, r^{23}] + [r^{13}, r^{23}] = \sum_{i, j} ([a_{i}, a_{j}] \otimes b_{i} \otimes b_{j} + a_{i} \otimes [b_{i}, a_{j}] \otimes b_{j} + a_{i} \otimes a_{j} \otimes [b_{i}, b_{j}]) . \end{matrix}$ Condition a) states that $r^{12} + r^{21}$ is $𝔤$ -invariant and condition b) states that $[r^{12}, r^{13}] + [r^{12}, r^{23}] + [r^{13}, r^{23}]$ is $𝔤$ -invariant.

1.4 A quasitraingular Lie bialgebra is a pair $(𝔤, r)$ such that $𝔤$ is a Lie bialgebra, $r \in 𝔤 \otimes 𝔤$ , the cobraket in $𝔤$ is equal to $d r$ and $r$ satisfies $r^{12} + r^{21} = 0,$ i.e, $r \in ⋀^{2} 𝔤$ .

A triangular Lie bialgebra is a pair $(𝔤, r)$ such that $𝔤$ is a Lie bialgebra, $r \in 𝔤 \otimes 𝔤$ , the cobracket in $𝔤$ is equal to $d r$ and $r$ satisfies $\begin{matrix} r^{12} + r^{21} = 0, and \\ [r^{12}, r^{13}] + [r^{12}, r^{23}] + [r^{13}, r^{23}] = 0 . \end{matrix}$ The equation $[r^{12}, r^{13}] + [r^{12}, r^{23}] + [r^{13}, r^{23}] = 0$ is the classical Yang-Baxter equation (CYBE).

1.5 If $𝔤$ is a semisimple Lie algebra over a field of characteristic $0$ , then it follows from Whitehead's lemma ([J] III §7 Lemma 3 p.77 and Thm. 13 p.95) that $H^{1} (𝔤, r)$ =0 for all finite dimensional $𝔤$ -modules $V$ . Thus all $1$ -cocycles are coboundaries; in this case, if $φ : 𝔤 \to 𝔤 \otimes 𝔤$ is any linear map which satisfies the $1$ -cocycle condition then there is an element $r \in 𝔤 \otimes 𝔤$ such that $φ (x) = [1 \otimes x + x \otimes 1, r]$ for all $x \in 𝔤$ .

Manin triples and the double

2.1 Let $⟨, ⟩ : 𝔭 \otimes 𝔭 \to 𝔭$ be a symmetric bilinear form on a vector space $𝔭$ , i.e. $⟨ x, y ⟩ = ⟨ y, x ⟩$ for all $x, y \in 𝔭$ .

The form $⟨, ⟩$ is nondegenerate if the map given by $\begin{matrix} * : & 𝔭 & \to & 𝔭^{*} \\ x & \mapsto & ⟨ x, \cdot ⟩ \end{matrix}$ is an isomorphism. Alternatively, the form $⟨, ⟩$ is invariant if for every $x \in 𝔭$ , $x \neq 0$ there is a $y \in 𝔭$ such that $⟨ x, y ⟩ \neq 0$ . A third way to say it is that $⟨, ⟩$ is invariant if the null space of the form $N = \{x \in 𝔭 ∣ ⟨ x, y ⟩ = 0 for all y \in 𝔭\}$ is $0$ . A fourht way to say it is that the matrix of the form (with respect to any fixed basis of $𝔭$ ) has nonzero determininant.

A subspace $𝔭^{'}$ of $𝔭$ is isotropic if $⟨ x, x ⟩ = 0$ for all $x \in 𝔭^{'}$ .

Given vector spaces $𝔭_{1}$ and $𝔭_{2}$ with symmetric bilinear forms ${⟨, ⟩}_{1}$ and ${⟨, ⟩}_{2}$ respectively then the vector space $𝔭_{1} \otimes 𝔭_{2}$ has a bilinear form $⟨, ⟩$ given by $⟨ x \otimes y, z \otimes w ⟩ = {⟨ x, z ⟩}_{1} {⟨ y, w ⟩}_{2}$ for all $x, z \in 𝔭_{1}$ and $y, w \in 𝔭_{2}$ . In particular, if $𝔭$ is a vector space with a bilinear form $⟨, ⟩$ then $𝔭 \otimes 𝔭$ has a bilinear form given by $⟨ x \otimes y, z \otimes w ⟩ = ⟨ x, z ⟩ ⟨ y, w ⟩$ for all $x, y, z, w \in 𝔭$ .

Suppose that the vector space $𝔭$ is a Lie algebra. The form $⟨, ⟩$ is invariant if $⟨ (ad z) x, y ⟩ = - ⟨ x, (ad z) y ⟩,$ for all $x, y, z \in 𝔭 .$

2.2 A Manin triple is a triple $(𝔭, 𝔭_{1}, 𝔭_{2})$ such that

$𝔭$ is a Lie algebra with a nondegenerat invariant symmetric bilinear form $⟨, ⟩$ and
$𝔭_{1}$ and $𝔭_{2}$ are isotropic Lie subalgebra of $𝔭$ .
$𝔭 = 𝔭_{1} \oplus 𝔭_{2}$ as vector spaces.

Let $(𝔭, 𝔭_{1}, 𝔭_{2})$ be a Manin triple. Then $𝔭_{1}$ is a Lie algebra with cobracket $δ : 𝔭_{1} \to 𝔭_{1} \oplus 𝔭_{2}$ determined by the equation $⟨ δ (x), y_{1} \otimes y_{2} ⟩ = ⟨ x, [y_{1}, y_{2}] ⟩,$ for all $x \in 𝔭_{1}$ and all $y_{1}, y_{2} \in 𝔭_{2}$ . In other words, $δ$ is the adjoint of the bracket on $𝔭_{2}$ .

Let $(𝔤 δ)$ be a Lie bialgebra. Then the triple $(𝔤 \oplus 𝔤^{*}, 𝔤, 𝔤^{*})$ is a Manin triple where

The bilinear form $⟨, ⟩$ on $𝔤 \oplus 𝔤^{*}$ is given by $⟨ x_{1} + y_{1}^{*}, x_{2} + y_{2}^{*} ⟩ = ⟨ x_{1}, y_{2}^{*} ⟩ + ⟨ x_{2}, y_{1}^{*} ⟩ = y_{2}^{*} (x 1) + y_{1}^{*} (x 2),$ for all $x_{1}, x_{2} \in 𝔤$ and all $y_{1}^{*}, y_{2}^{*} \in 𝔤^{*}$ , and
The bracket on $𝔤 \oplus 𝔤^{*}$ is determined by the formulas $\begin{array}{rcl} ⟨ [y_{1}^{*}, y_{2}^{*}], p ⟩ & = & \{\begin{cases} ⟨ y_{1}^{*} \otimes y_{2}^{*}, δ (p) ⟩, & if p \in 𝔤, \\ 0, & if p \in 𝔤^{*}, \end{cases} \\ ⟨ [x, y^{*}], p ⟩ & = & \{\begin{cases} ⟨ y^{*}, [p, x] ⟩, & if p \in 𝔤, \\ ⟨ x, [y^{*}, p] ⟩, & if p \in 𝔤^{*}, \end{cases} \end{array}$ where $x \in 𝔤$ and $y, y_{1}, y_{2}^{*} \in 𝔤^{*}$ .

This is essentially the only way to define things so that the bracket on $𝔤^{*}$ is the dual of the cobracket on $𝔤$ and so that the bilinear form is invariant. The Lie algebra $𝔤 \oplus 𝔤 = D (g)$ constructed from the the Lie bialgebra $𝔤$ is called the double corresponding to the Lie algebra $𝔤$ .

The previous two propositions show that there is a one-to-one correspondence between Lie bialgebras and Manin triples.

Let $𝔤$ be a Lie bialgebra and $D (𝔤) = 𝔤 \oplus 𝔤^{*}$ be the double of $𝔤$ . Let $\{a_{i}\}$ be a basis of $𝔤$ and let $\{a^{i}\}$ be the dual basis in $𝔤^{*}$ . Define $r = \sum_{i} a_{i} \otimes a^{i} \in 𝔤 \otimes 𝔤^{*} \subseteq D (𝔤) \otimes D (𝔤) .$ Then

The element $r$ does not depend on the choice of basis $a_{i}$ of $𝔤$ .
$r$ satisfies the CYBE.
$(D (𝔤), r)$ is a quasitriangular Lie bialgebra.

Proofs

The map ${d r}^{*} : 𝔤^{*} \otimes 𝔤^{*} \to 𝔤^{*}$ satisfies the skew-symmetric condition if and only if $({ad}^{\otimes 2}) (r^{12} + r^{21}) = 0$ for all $x \in 𝔤$ .
Assume that $P$ is ${ad}^{\otimes 2}$ invariant. Then ${d r}^{*}$ satisfies the Jacobi identity if and only if $({ad}^{\otimes 3}) ([r^{12}, r^{13}] + [r^{12}, r^{23}] + [r^{13}, r^{23}]) = 0$ for all $x \in 𝔤$ .

Proof:

The following computation shows that $({ad}^{\otimes 2} z) (r^{12} + r^{21}) = 0$ if and only if $δ^{*} (x^{*} \otimes y^{*}) + δ^{*} (y^{*} \otimes x^{*}) = 0$ for all $x^{*}, y^{*} \in 𝔤^{*}$ . $\begin{array}{rcl} 0 & = & ⟨ δ^{*} (x^{*} \otimes y^{*}) + δ^{*} (y^{*} \otimes x^{*}), z ⟩ \\ = & ⟨ x^{*} \otimes y^{*} + y^{*} \otimes x^{*}, δ (z) ⟩ \\ = & ⟨ x^{*} \otimes y^{*} + y^{*} \otimes x^{*}, ({ad}^{\otimes 2} z) \sum_{i} a_{i} \otimes b_{i} ⟩ \\ = & ⟨ x^{*} \otimes y^{*}, ({ad}^{\otimes 2} z) \sum_{i} (a_{i} \otimes b_{i} + b_{i} \otimes a_{i}) ⟩ \\ = & ⟨ x^{*} \otimes y^{*}, ({ad}^{\otimes 2} z) (r^{12} + r^{21}) ⟩ . \end{array}$
The proof will be done in several steps.
Step 1. $d r = d ρ$ .
Step 2. $C (r) = C (ρ) + C (P)$ .
Step 3. If $P$ is ${ad}^{\otimes 2}$ invariant then $C (P)$ is ${ad}^{\otimes 3}$ invariant.

r \in ⋀^{2} 𝔤

{d r}^{*}

C (r)

{ad}^{\otimes 3}

Punchline. Let us show that these 4 steps are sufficient to prove the result. By step 3,

C (P)

{ad}^{\otimes 3}

invariant. Thus by step 2,

C (r)

{ad}^{\otimes 3}

invariant if and only if

C (ρ)

is. Now

ρ \in ⋀^{2}

and so by step 4

{d ρ}^{*}

satisfies the Jacobi identity if and only if

C (ρ)

{ad}^{\otimes 3}

invariant. Since

d ρ = d r

, it follows that

{d r}^{*}

satifies the Jacobi identity if and only if

C (r)

{ad}^{\otimes 3}

invariant.

Step 1. Since $P$ is ${ad}^{\otimes 2}$ invariant we have that $\begin{array}{rcl} d ρ (x) & = & ({ad}^{\otimes 2} x) (ρ) \\ = & ({ad}^{\otimes 2} x) (r - \frac{1}{2} P) \\ = & ({ad}^{\otimes 2} x) (r) - \frac{1}{2} ({ad}^{\otimes 2} x) (P) \\ = & ({ad}^{\otimes 2} x) (r) \\ = & d r (x) . \end{array}$

Step 2. Using the fact that $P$ is ${ad}^{\otimes 2}$ invariant and the fact that $ρ = \sum ρ^{(1)} \otimes ρ^{(2)} = - \sum ρ^{(2)} \otimes ρ^{(1)}$ , we get $\begin{array}{rcl} C (r) & = & [r^{12}, r^{13}] + [r^{12}, r^{23}] + [r^{13}, r^{23}] \\ = & [ρ^{12} + P^{12}, ρ^{13} + P^{13}] + [ρ^{12} + P^{12}, ρ^{23} + P^{23}] + [ρ^{13} + P^{13}, ρ^{23} + P^{23}] \\ = & C (ρ) + C (P) + [ρ^{12}, P^{13}] + [ρ^{12}, P^{23}] + [ρ^{13}, P^{23}] [P^{12}, ρ^{13}] + [P^{13}, ρ^{23}] \\ = & C (ρ) + C (P) + \sum [ρ^{(1)}, P^{(1)}] \otimes ρ^{(2)} \otimes P^{(2)} + ρ^{(1)} \otimes [ρ^{(2)}, P^{(1)}] \otimes P^{(2)} + ρ^{(1)} \otimes P^{(1)} \otimes [ρ^{(2)}, P^{(2)}] \\ + [P^{(1)}, ρ^{(1)}] \otimes P^{(2)} \otimes ρ^{(2)} + P^{(1)} \otimes [P^{(2)}, ρ^{(1)}] \otimes ρ^{(2)} + P^{(1)} \otimes ρ^{(1)} \otimes [P^{(2)}, ρ^{(2)}] \\ = & C (ρ) + C (P) + \sum [ρ^{(1)}, P^{(1)}] \otimes ρ^{(2)} \otimes P^{(2)} \\ = & C (ρ) + C (P) + \sum [ρ^{(1)}, P^{(1)}] \otimes ρ^{(2)} \otimes P^{(2)} \\ + ρ^{(1)} \otimes [ρ^{(2)}, P^{(1)}] \otimes P^{(2)} + ρ^{(1)} \otimes P^{(1)} \otimes [ρ^{(2)}, P^{(2)}] \\ + [P^{(1)}, ρ^{(1)}] \otimes P^{(2)} \otimes ρ^{(2)} + P^{(1)} \otimes [P^{(2)}, ρ^{(1)}] \otimes ρ^{(2)} \\ + P^{(1)} \otimes ρ^{(1)} \otimes [P^{(2)}, ρ^{(2)}] \\ = & C (ρ) + C (P) + \sum [ρ^{1}, P^{1}] \otimes ρ^{2} \otimes P^{2} + 0 + 0 - P^{1} \otimes ρ^{2} \otimes [P^{2}, ρ^{1}] \\ = & C (ρ) + C (P) + \sum [ρ^{(1)}, P^{(1)}] ρ^{(2)} \otimes P^{(2)} + P^{(1)} \otimes ρ^{(2)} \otimes [ρ^{(1)}, P^{(2)}] \\ = & C (ρ) + C (P) . \end{array}$

Step 3. Suppose that $P = \sum P_{(1)} \otimes P_{(2)} = \sum_{j} P^{(1)} \otimes P^{(2)}$ . Then $\begin{array}{rcl} ({ad}^{\otimes 3} x) [P^{12}, P^{13}] & = & ({ad}^{\otimes 3} x) \sum [P_{(1)}, P^{(1)}] \otimes P_{(2)} \otimes P^{(2)} \\ = & \sum [x, [P_{(1)}, P^{(1)}]] \otimes P_{(2)} \otimes P^{(2)} + [P_{(1)}, P^{(1)}] \otimes [x, P_{(2)}] \otimes P^{(2)} + [P_{(1)}, P^{(1)}] \otimes P_{(2)} \otimes [x, P^{(2)}] \\ = & \sum - [P^{(1)}, [x, P_{(1)}]] \otimes P_{(2)} \otimes P^{(2)} + [P_{(1)}, P^{(1)}] \otimes [x, P_{(2)}] \otimes P^{(2)} - [P_{(1)}, [P^{(1)}, x]] \otimes P_{(2)} \otimes P^{(2)} \\ + [P_{(1)}, P^{(1)}] \otimes P_{(2)} \otimes [x, P^{(2)}] \\ = & \sum [[x, P_{(1)}], P^{(1)}] \otimes P_{(2)} \otimes P^{(2)} + [P_{(1)}, P^{(1)}] \otimes [x, P_{(2)}] \otimes P^{(2)} + [P_{(1)}, [x, P^{(1)}]] \otimes P_{(2)} \otimes P^{(2)} \\ + [P_{(1)}, P^{(1)}] \otimes P_{(2)} \otimes [x, P^{(2)}] \\ = & 0 . \end{array}$ By arguing similarly for the other terms one shows that $C (P) = [P^{12}, P^{13}] + [P^{12}, P^{23}] + [P^{13}, P^{23}]$ is an invariant element of $𝔤^{\otimes 3}$ .

Step 4. Assume that $r \in ⋀^{2} 𝔤$ so that $r^{12} + r^{21} = 0$ or equivalently that $r = \sum a_{i} \otimes b_{i} = - \sum b_{i} \otimes a_{i}$ . Let us first calculate $\begin{array}{rcl} ⟨ δ^{*} (x^{*} \otimes δ^{*} (x^{*} \otimes z^{*})), w ⟩ & = & ⟨ x^{*} \otimes δ^{*} (y^{*} \otimes z^{*}), δ (w) ⟩ \\ = & ⟨ x^{*} δ^{*} (y^{*} \otimes z^{*}), \sum [w, a_{i}] \otimes b_{i} + a_{i} \otimes [w, b_{i}] ⟩ \\ = & ⟨ x^{*} \otimes y^{*} \otimes z^{*}, \sum_{i} [w, a_{i}] \otimes δ (b_{i}) + a_{i} \otimes δ ([w, b_{i}]) ⟩ \\ = & ⟨ x^{*} \otimes y^{*} \otimes z^{*}, \sum_{i} [w, a_{i}] \otimes (\sum_{j} [b_{i}, a_{j}] \otimes b_{j} + a_{j} \otimes [b_{},]) \\ + a_{i} \otimes (\sum_{j} [[w, b_{i}], a_{j}] \otimes b_{j} + a_{j} \otimes [[w, b_{i}], b_{j}]) ⟩ \\ = & ⟨ x^{*} \otimes y^{*} \otimes z^{*}, \sum_{i, j} [w, a_{i}] \otimes [b_{i}, a_{j}] \otimes b_{j} + [w, a_{i}] \otimes a_{j} \otimes [b_{i}, b_{j}] \\ + a_{i} \otimes [[w, b_{i}], a_{j}] \otimes b_{j} + a_{i} \otimes a_{j} \otimes [[w, b_{i}], b_{j}] ⟩ & . \end{array}$ Now consider the sum $⟨ δ^{*} (x^{*} \otimes δ^{*} (y^{*} \otimes z^{*})), w ⟩ + ⟨ δ^{*} (z^{*} \otimes δ^{*} (x^{*} \otimes y^{*})), w ⟩ + ⟨ δ^{*} (y^{*} \otimes δ^{*} (z^{*} \otimes x^{*})), w ⟩ .$ For simplicity, write only the terms in this sum which have a $w$ in the first tensor slot, $\begin{matrix} [w, a_{i}] \otimes [b_{i}, a_{j}] \otimes b_{j}, \\ [w, a_{i}] \otimes a_{j} \otimes [b_{i}, b_{j}], \\ [[w, a_{i}], a_{j}] \otimes b_{j} \otimes a_{i}, \\ [[w, b_{i}], b_{j}] \otimes a_{i} \otimes a_{j} . \end{matrix}$

The first term is equal to

(ad w \otimes 1 \otimes 1) [r^{12}, r^{23}]

and the second term is equal to

(ad w \otimes 1 \otimes 1) [r^{13}, r^{23}]

. For the third term, applying the skew symmetry relation and then switching the indices

i

and

j

[[w, b_{i}], b_{j}] \otimes a_{i} \otimes a_{j} = - [[w, a_{i}], a_{j}] \otimes b_{j} \otimes b_{j} = - [[w, a_{j}], a_{i}] \otimes b_{i} \otimes b_{j} = [[a_{j}, w], a_{i}] \otimes b_{i} \otimes b_{j} .

For the fourth term, apllying the skew-symmetry relation twice gives

[[w, b_{i}], b_{j}] \otimes a_{i} \otimes a_{j} = - [[w, a_{i}], b_{j}] \otimes b_{i} \otimes a_{j} = [[w, a_{i}], a_{j}] \otimes b_{i} \otimes b_{j} .

Now it is clear that the sum of the third and fourth terms is, by the Jacobi identity

- [[a_{i}, a_{j}], w] \otimes b_{i} \otimes b_{j} = [w, [a_{i}, a_{j}]] \otimes b_{i} \otimes b_{j} = (ad w \otimes 1 \otimes 1) [r^{12}, r^{13}]

. The result now follows by symmetry.
This completes the proof of Proposition (1.1).

□

Let $𝔭, 𝔭_{1}, 𝔭_{2}$ be a Manin triple. Then $𝔭_{1}$ is a Lie bialgebra with cobracket $δ : 𝔭_{1} \to 𝔭_{1} \otimes 𝔭_{1}$ determined by the equation $⟨ δ (x), y_{1} \otimes y_{2} ⟩ = ⟨ x, [y_{1}, y_{2}] ⟩$ for all $x \in 𝔭_{1}$ and all $y_{1}, y_{2} \in 𝔭_{2}$ . In other words, $δ$ is the adjoint of the bracket on $𝔭_{2}$ .

Proof:

It is clear from the fact that delta is the adjoint of the bracket on $𝔭_{2}$ that $δ$ is a cobracket on $𝔭_{1}$ . It remains to check the cocycle condition.

If $z \in 𝔭$ then $z_{1}$ and $z_{2}$ shall denote the elements of $𝔭_{1}$ and $𝔭_{2}$ respectively such that $z = z_{1} + z_{2}$ . Then we have that, for any $x_{1} \in 𝔭_{1}, x_{2} \in 𝔭_{2}$ , $\begin{array}{rcl} ⟨ ({ad}^{\otimes 2} x_{1}) δ y_{1}, x_{1} \otimes y_{2} ⟩ & = & - ⟨ δ (y_{1}), (ad x_{1} \otimes 1 + 1 \otimes ad x_{1}) (x_{1} \otimes y_{2}) ⟩ \\ = & - ⟨ δ (y_{1}), {[x_{1}, x_{2}]}_{2} \otimes y_{2} + x_{2} \otimes {[x_{1}, y_{2}]}_{2} ⟩ \\ = & - ⟨ y_{1}, [{[x_{1}, x_{2}]}_{2}, y_{2}] + [x_{2}, {[x_{1}, y_{2}]}_{2}] ⟩ \\ = & ⟨ {[y_{1}, y_{2}]}_{1}, {[x_{1}, x_{2}]}_{2} ⟩ + ⟨ {[x_{2}, y_{1}]}_{1}, {[x_{1}, y_{2}]}_{2} ⟩ \end{array}$ Similarly, $\begin{array}{rcl} ⟨ ({ad}^{\otimes 2} y_{1}) δ (x_{1}), x_{2} \otimes y_{2} ⟩ & = & - ⟨ δ (x_{1}), (ad y_{1} \otimes 1 + 1 \otimes ad y_{1}) (x_{2} \otimes y_{2}) ⟩ \\ = & - ⟨ δ (x_{1}), {[y_{1}, x_{2}]}_{2} \otimes y_{2} + x_{2} \otimes {[y_{1}, y_{2}]}_{2} ⟩ \\ = & - ⟨ x_{1}, [{[y_{1}, x_{2}]}_{2}, y_{2}] + [x_{2}, {[y_{1}, y_{2}]}_{2}] ⟩ \\ = & ⟨ {[x_{1}, y_{2}]}_{1}, {[y_{1}, y_{2}]}_{2} ⟩ + ⟨ {[x_{2}, x_{1}]}_{1}, {[y_{1}, y_{2}]}_{2} ⟩ \\ = & - ⟨ {[x_{1}, x_{2}]}_{1}, {[y_{1}, y_{2}]}_{2} ⟩ + ⟨ {[x_{2}, y_{2}]}_{1}, {[x_{2}, y_{1}]}_{2} ⟩ . \end{array}$

Thus

\begin{array}{rcl} ⟨ ({ad}^{\otimes 2} x_{1}) δ (y_{1}) - ({ad}^{\otimes 2} y_{1}) δ (x_{1}), x_{2} \otimes y_{2} ⟩ & = & ⟨ {[y_{1}, y_{2}]}_{1}, {[x_{1}, x_{2}]}_{2} ⟩ + ⟨ {[x_{2}, y_{1}]}_{1}, {[x_{1}, y_{2}]}_{2} ⟩ \\ + ⟨ {[x_{1}, x_{2}]}_{1}, {[y_{1}, y_{2}]}_{2} ⟩ + ⟨ {[x_{2}, y_{2}]}_{1}, {[x_{2}, y_{1}]}_{2} ⟩ \\ = & ⟨ [x_{1}, y_{2}], [x_{2}, y_{1}] ⟩ + ⟨ [x_{1}, x_{2}], [y_{1}, y_{2}] ⟩ . \end{array}

On the other hand we have that for any

x_{1}, y_{1} \in 𝔭_{1}

and

x_{2}, y_{2} \in 𝔭_{2}

\begin{array}{rcl} ⟨ δ ([x_{1}, y_{1}]), x_{2} \otimes y_{2} ⟩ & = & ⟨ [x_{1}, y_{1}], [x_{2}, y_{2}] ⟩ \\ = & - ⟨ [[x_{1}, y_{1}], y_{2}], x_{2} ⟩ \\ = & ⟨ [[y_{1}, x_{1}], y_{1}], x_{2} ⟩ + ⟨ [[y_{1}, y_{2}], x_{1}], x_{2} ⟩ \\ = & - ⟨ [y_{2}, x_{1}], [x_{2}, y_{1}] ⟩ + ⟨ [y_{1}, y_{2}], [x_{1}, x_{2}] ⟩ \\ = & ⟨ [x_{1}, y_{2}], [x_{2}, y_{1}] ⟩ + ⟨ [x_{1}, x_{2}], [y_{1}, y_{2}] ⟩ . \end{array}

It follows from this computation oand the previous one that

δ ([x_{1}, y_{1}]) = ({ad}^{\otimes 2} x_{1}) δ (y_{1}) - ({ad}^{\otimes 2} y_{1}) δ (x_{1}),

for all

x_{1}, y_{1} \in 𝔭_{1}

. It follows that

𝔭_{1}

is a Lie bialgebra.

□

Let $(𝔤, δ)$ be a Lie bialgebra. Then the triple $(𝔤 \oplus 𝔤^{*}, 𝔤, 𝔤^{*})$ is a Manin triple where

The bilinear form $⟨, ⟩$ on $𝔤 \oplus 𝔤^{*}$ is given by $⟨ x_{1} + y_{1}^{*}, x_{2} + y_{2}^{*} ⟩ = ⟨ x_{1}, y_{2}^{*} ⟩ + ⟨ x_{2}, y_{1}^{*} ⟩ = y_{2}^{*} (x_{1}) + y_{1}^{*} (x_{2}),$ for all $x_{1}, x_{2} \in 𝔤$ and all $y_{1}^{*}, y_{2}^{*} \in 𝔤^{*}$
The bracket on $𝔤 \oplus 𝔤^{*}$ is determined by the formulas $\begin{array}{rcl} ⟨ [y_{1}^{*}, y_{2}^{*}], p ⟩ & = & \{\begin{cases} ⟨ y_{1}^{*} \otimes y_{2}^{*}, δ (p) ⟩, & if p \in 𝔤, \\ 0, & if p \in 𝔤^{*}, \end{cases} \\ ⟨ [x, y^{*}], p ⟩ & = & \{\begin{cases} ⟨ y^{*}, [p, x] ⟩, & if p \in 𝔤, \\ ⟨ x, [y^{*}, p] ⟩, & if p \in 𝔤^{*}, \end{cases} \end{array}$ where $x \in 𝔤$ and $y, y_{1}, y_{2}^{*} \in 𝔤^{*}$ .

Proof:

The facts, $𝔤^{*}$ is Lie subalgebra of $𝔤 \oplus 𝔤^{*}$ , the bilinear form is invariant, the bracket of $𝔤 \oplus 𝔤^{*}$ satisfies the skew-symmetry condition, all follow directly form the definitions. Only the Jacobi identity really needs a proof.

If all three elements in the Jacobi sum are in $𝔤$ or all three are in $𝔤^{*}$ then the Jacobi identity follows immediately from the Jacobi identity for these Lie subalgebras. Suppose two indices are in $𝔤^{*}$ and the third is in $𝔤$ . THen we have that, for any $x \in 𝔤$ and $y_{1}^{*}, y_{2}^{*}, z^{*} \in 𝔤^{*}$ , $\begin{array}{rcl} ⟨ [y_{2}^{*}, [x, y_{1}^{*}]] + [y_{1}^{*}, [y_{2}^{*}, x]] + [x, [y_{1}^{*}, y_{2}^{*}]], z^{*} ⟩ & = & ⟨ x, [y_{1}^{*}, [z^{*}, y_{2}^{*}]] + [[z^{*}, y_{1}^{*}], y_{2}^{*}] + [[y_{1}^{*}, y_{2}^{*}], z^{*}] ⟩ \\ = & ⟨ x, [[y_{2}^{*}, z^{*}], y_{1}^{*}] + [[z^{*}, y_{1}^{*}], y_{2}^{*}] + [[y_{1}^{*}, y_{2}^{*}], z^{*}] ⟩ \\ = & 0, \end{array}$ by the Jacobi identity in $𝔤^{*}$ . In the other case we have, for any $x, z \in 𝔤$ and $y_{1}, y_{2} \in 𝔤^{*}$ , $\begin{array}{rcl} ⟨ [y_{2}^{*}, [x, y_{1}^{*}]] + [y_{1}^{*}, [y_{2}^{*}, x]] + [x, [y_{1}^{*}, y_{2}^{*}]], z^{*} ⟩ & = & ⟨ [x, y_{1}^{*}], [z, y_{2}^{*}] ⟩ + ⟨ [y_{2}^{*}, x], [z, y_{2}^{*}] ⟩ + ⟨ [y_{1}^{*}, y_{2}^{*}], [z, x] ⟩ \\ = & - ⟨ ({ad}^{\otimes 2} x) δ (x) - ({ad}^{\otimes 2} z) δ (x), y_{1}^{*} \otimes y_{2}^{*} ⟩ + ⟨ y_{1}^{*} \otimes y_{2}^{*}, δ ([z, x]) ⟩ \\ = & 0 . \end{array}$ In this computation we are using the $1$ -cocycle condition on $𝔤$ and the fact that $⟨ [x, y_{1}^{*}], [z, y_{2}^{*}] ⟩ + ⟨ [y_{2}^{*}, x], [z, y_{1}^{*}] ⟩ = ⟨ ({ad}^{\otimes 2} x) δ (x) - ({ad}^{\otimes 2} z) δ (x), y_{1}^{*} \otimes y_{2}^{*} ⟩$ which was proved in the proof of Proposition (2.3).

The case of the Jacobi identity where two of the indices are in $𝔤$ and one is in $𝔤^{*}$ is proved similarly. Thus the bracket on $𝔤 \oplus 𝔤^{*}$ satisfies the Jacobi identity and $𝔤 \oplus 𝔤^{*}$ is a Lie algebra with an invariant bilinear form.

□

Let $𝔤$ be a Lie bialgebra and let $D (𝔤) = 𝔤 \oplus 𝔤^{*}$ be the double of $𝔤$ . Let $\{a_{i}\}$ be a basis of $𝔤$ and $\{a^{i}\}$ be the dual basis in $𝔤^{*}$ . Define $r = \sum_{i} a_{i} \otimes a^{i} \in 𝔤 \otimes 𝔤^{*} \subseteq D (𝔤) \otimes D (𝔤) .$ Then

The element $r$ does not depend on the choice of basis $\{a_{i}\}$ of $𝔤$ .
$r$ satisfies the CYBE.
$(D (𝔤), r)$ is a quasitriangular Lie bialgebra.

		Proof:
	Suppose that $\{b_{j}\}$ is another basis of $𝔤$ and that $\{b^{j}\}$ is the dual basis in $𝔤^{}$ . Then $\begin{array}{rcl} r & = & \sum_{i} a_{i} \otimes a^{i} \\ = & \sum_{i, j} ⟨ a_{i}, b^{j} ⟩ b_{j} \otimes a^{i} \\ = & \sum_{i, j} b_{j} \otimes a^{i} ⟨ a_{i}, b^{j} ⟩ \\ = & \sum_{j} b_{j} \otimes b^{j} . \end{array}$ Thus the element $r$ is independent of the choice of basis. Let us check that $r$ satisfies the CYBE. $\begin{array}{rcl} [r^{12}, r^{13}] + [r^{12}, r^{23}] + [r^{13}, r^{23}] & = & \sum_{i, j} [a_{i}, a_{j}] \otimes b_{i} \otimes b_{j} + a_{i} \otimes [b_{i}, a_{j}] \otimes b_{j} + a_{i} \otimes a_{j} \otimes [b_{i}, b_{j}] \\ = & \sum_{i, j} [a_{i}, a_{j}] \otimes a^{i} \otimes a^{j} + a_{i} \otimes [a^{i}, a_{j}] \otimes a^{j} + a_{i} \otimes a_{j} \otimes [a^{i}, a^{j}] \\ = & \sum_{i, j, s} ⟨ [a^{i}, a^{j}], a^{s} ⟩ a_{s} \otimes a^{i} \otimes a^{j} + ⟨ [a^{i}, a_{j}], a_{s} ⟩ a_{i} \otimes a^{s} \otimes a^{j} \\ + ⟨ [a^{i}, a_{j}], a^{s} ⟩ a_{i} \otimes a_{s} \otimes a^{j} + ⟨ [a^{i}, a^{j}], a_{s} ⟩ a_{i} \otimes a^{j} \otimes a^{s} \\ = & \sum_{i, j, s} ⟨ [a^{i}, a^{j}], a^{s} ⟩ \otimes a_{s} \otimes a^{i} \otimes a^{j} - ⟨ a^{i}, [a_{s}, a_{j}] ⟩ a_{i} \otimes a^{s} \otimes a^{j} \\ - ⟨ [a^{i}, a^{s}], a_{j} ⟩ a_{a} \otimes a_{s} \otimes a^{j} + ⟨ [a^{i}, a^{j}], a_{s} ⟩ a_{i} \otimes a_{j} \otimes a^{s} \\ = & 0 . \end{array}$ In view of Proposition () we only need to show that $({ad}^{\otimes 2} x) (r^{12} + r^{21}) = 0$ for all $x \in 𝔤 \oplus 𝔤^{}$ . Let $x \in 𝔤$ . Then $\begin{array}{rcl} [x \otimes 1 + 1 \otimes x, r^{12} + r^{21}] & = & \sum_{i} [x, a_{i}] \otimes a^{i} + [x, a^{i}] \otimes a_{i} + a^{i} \otimes [x, a_{i}] \\ = & \sum_{i, k} ⟨ [x, a_{i}], a^{k} ⟩ a_{k} \otimes a_{i} + ⟨ [x, a^{i}], a^{k} ⟩ a_{k} \otimes a_{i} + ⟨ [x, a^{i}], a_{k} ⟩ a^{k} \otimes a_{i} \\ + ⟨ [x, a^{i}], a_{k} ⟩ a_{i} \otimes a^{k} + ⟨ [x, a^{i}], a^{k} ⟩ a_{i} \otimes a_{k} + ⟨ [x, a_{i}], a^{k} ⟩ a^{k} \otimes a_{i} . \end{array}$ After reindexing we have $\begin{array}{rcl} [x \otimes 1 + 1 \otimes x, r^{12} + r^{21}] & = & \sum_{i, k} ⟨ [x, a_{i}], a^{k} ⟩ a_{k} \otimes a^{i} + ⟨ [x, a^{i}], a^{k} ⟩ a_{k} \otimes a_{i} + ⟨ [x, a^{i}], a_{k} ⟩ a^{k} \otimes a_{i} \\ + ⟨ [x, a^{k}], a_{i} ⟩ a_{k} \otimes a^{i} + ⟨ [x, a^{k}], a^{i} ⟩ a_{k} \otimes a_{i} + ⟨ [x, a_{k}], a^{i} ⟩ a^{k} \otimes a_{i} . \end{array}$ Now note that, by invariance of the inner product we have that $\begin{matrix} ⟨ [x, a_{i}], a^{k} ⟩ = - ⟨ [x, a^{k}], a_{i} ⟩, \\ ⟨ [x, a^{i}], a^{k} ⟩ = - ⟨ [x, a^{k}], a^{i} ⟩, \\ ⟨ [x, a^{i}], a_{k} ⟩ = - ⟨ [x, a_{k}], a^{i} ⟩ . \end{matrix}$ It follows that $[x \otimes 1 + 1 \otimes x, r^{12} + r^{21}] = 0$ . The case when $x \in 𝔤^{*}$ is proved similarly. The result follows. $□$

References

The motivating reference is

[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

There is a detailed exposition of Lie bialgebras in the following article

[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

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

page history