Bialgebra structures on Lie algebras with triangular decomposition

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

Last updates: 24 March 2011

Lie algebras with triangular decomposition

1.1 Let $I$ be a finite set. Let $Q = ℤ [I]$ , $Q^{+} = ℕ [I]$ , and $Q^{-} = - Q^{+}$ .
Let $𝔤$ be a Lie algebra that satisfies the following:

$𝔤$ is $Q$ -graded, i.e., $𝔤$ is a direct sum of spaces $𝔤_{α}, α \in Q$ , and $[𝔤_{α}, 𝔤_{β}] \subseteq 𝔤_{α + β}$ for all $α, β \in Q$ .
The Lie algebra $𝔤$ has a triangular decomposition with respect to $Q$ , i.e., $𝔤 = 𝔫^{-} \oplus 𝔥 \oplus 𝔫^{+}$ where $𝔫^{+} = ⨁_{α \in Q^{+} ∖ {0}} 𝔤_{α}, 𝔫^{-} = ⨁_{α \in Q^{-} ∖ {0}} 𝔤_{- α}, 𝔥 = 𝔤_{0} .$
For each $α \in Q$ the vector space $𝔤_{α}$ is finite dimensional.
For each $i \in I$ , $dim (𝔤_{i}) = 1$ .
The subspaces $𝔤_{i}$ , $i \in I$ generate $𝔤$ as a lie algebra.
There is a nondegenerate invariant symmetric bilinear form ⟨ , ⟩ :𝔤 × 𝔤→k on 𝔤 such that
1. The restriciton $⟨, ⟩ : 𝔤_{α} \times 𝔤_{- α} \to k$ is nondegenerate for each $α \in Q$ such that $𝔤_{α} \neq 0$ . In particular $⟨, ⟩$ is a nondegenerate form on $𝔥$ .
2. If $α, β \in Q$ , $α \neq β$ and $x \in 𝔤_{α}$ and $y \in 𝔤_{β}$ then $⟨ x, y ⟩ = 0$ .
There is a linear map θ:𝔤→𝔤 called the Chevalley involution, such that
1. $θ$ is a Lie algebra automorphism.
2. $θ (h) = - h$ for all $h \in 𝔥$ .
3. $θ (𝔤_{α}) = θ (𝔤_{- α})$ for all $α \in Q$ .
4. $⟨ θ (x), θ (y) ⟩ = ⟨ x, y ⟩$ for all $x, y \in 𝔤$ .

1.2 For each $i \in i$ let us fix an element $X_{i} \in g_{i}$ . Let $Y_{i} \in 𝔤_{- i}$ be dual to $X_{i}$ with respect to the form $⟨, ⟩$ and let $H_{i} = ⟨ X_{i}, Y_{i} ⟩ \in 𝔥$ .

1.3 We shall let $𝔤^{+}$ and $𝔟^{-}$ be the Lie subalgebras of $𝔤$ given by $\begin{matrix} 𝔟^{+} = 𝔥 \oplus 𝔫^{+}, \\ 𝔟^{-} = 𝔥 \oplus 𝔫^{-} . \end{matrix}$

1.4 Given an element $x \in 𝔤$ we write $x = x^{-} + x_{𝔥} + x^{+},$ where $x^{-} \in 𝔫^{-}, x_{𝔥} \in 𝔥, x^{+} \in 𝔫^{+}$ .

1.5 The finite dimensional simple Lie algebras over $ℂ$ and the Kac-Moody Lie algebras $𝔤^{'} (A)$ in [Kc] are Lie algebras that satisfy the conditions in (1.1). See Theorems 1.2 and 2.2 of [Kc].

Lie bialgebra structure on $𝔟^{+}$ and $D (𝔟^{+})$

Let $𝔤$ be a Lie algebra with invariang form ${⟨, ⟩}_{𝔤}$ satisfying the conditions of Section 1. Let ${⟨, ⟩}_{𝔥}$ denote the restriction of the form ${⟨, ⟩}_{𝔤}$ to $𝔥$ . Given an element $x \in 𝔤$ we write we write $x = x^{-} + x_{𝔥} + x^{+},$ where $x^{-} \in 𝔫^{-}, x_{𝔥} \in 𝔥, x^{+} \in 𝔫^{+}$ . The triple $(𝔤 \times 𝔥, 𝔭_{1}, 𝔭_{2}), where 𝔭_{1} = \{(x, x_{𝔥}) ∣ x \in 𝔟^{+}\} and 𝔭_{2} = \{(y, - y_{𝔥}) ∣ y \in 𝔟^{-}\}$ with the scalar product and bracket on $𝔤 \times 𝔥$ given by $⟨ (x, x_{𝔥}), (y, y_{𝔥}) ⟩ = ⟨ x, y ⟩_{𝔤} - ⟨ x_{𝔥}, y_{𝔥} ⟩_{𝔥}, and [(x, x_{𝔥}), (y, y_{𝔥})] = ([x, y], [h_{𝔥}, y_{𝔥}]) = ([x, y], 0)$ respectively, is a Manin triple.

		Proof:
	We must show that $𝔭_{1}$ and $𝔭_{2}$ are isotropic Lie subalgebras of $𝔤 \otimes 𝔥$ . $𝔤 \otimes 𝔥 = 𝔭_{1} \oplus 𝔭_{2}$ . The form $⟨, ⟩$ on $𝔤 \times 𝔥$ is invariant. The form $⟨, ⟩$ on $𝔤 \times 𝔥$ is nondegenerate. The proofs of these facts follow in 2.2-2.5. $□$

2.2 If $x, y \in 𝔟^{+}$ then $[x, y] \in 𝔫^{+}$ and $[(x, x_{𝔥}), (y, y_{𝔥})] = ([x, y], 0) \in 𝔭_{1}$ So $𝔭_{1}$ is a Lie subalgebra of $𝔤 \otimes 𝔥$ . Using the fact that $x, y \in 𝔟^{+}$ , so that $⟨ x, y ⟩ = ⟨ x_{𝔥}, y_{𝔥} ⟩$ , We have that $⟨ (x, x_{𝔥}), (y, y_{𝔥}) ⟩ = 0 .$ Thus $𝔭_{1}$ is isotropic.
Similarly one shows taht $𝔭_{2}$ is an isotropic Lie algebra of $𝔤 \times 𝔥$ .

2.3 If $(x, y) \in 𝔤 \times 𝔥$ then $(a, h) = (a^{+} + \frac{a_{𝔥} + h}{2}, \frac{a_{𝔥} + h}{2}) + (a^{-} + \frac{a_{𝔥} - h}{2}, \frac{- a_{𝔥} + h}{2}) \in 𝔭_{1} + 𝔭_{2} .$ So $𝔭_{1} + 𝔭_{2} = 𝔤 \times 𝔥$ .
If $(a, h) \in 𝔭_{1} \cap 𝔭_{2}$ then $a = a_{𝔥}$ and $h = a_{𝔥}$ and $h = - a_{𝔥}$ . It follows that $a = a_{𝔥} = h = 0$ . Thus $𝔭_{1} \cap 𝔭_{2} = 0$ and we have that $𝔤 \times 𝔥 = 𝔭_{1} \oplus 𝔭_{2}$ .

2.4 Recall that for $(x, h), (y, h^{'}) \in 𝔤 \times 𝔥$ , $⟨ (x, h), (y, h^{'}) ⟩ = {⟨ x, y ⟩}_{𝔤} - ⟨ h, h^{'} ⟩_{𝔥}$ . It follows that $\begin{array}{rcl} ⟨ [(x, h), (y, h^{'})], (z, h^{''}) ⟩ & = & ⟨ [x, y], z ⟩_{𝔤} - ⟨ [h, h^{'}], h^{''} ⟩ \\ = & ⟨ y, [z, x] ⟩_{𝔤} - ⟨ h^{'}, [h^{''}, h] ⟩_{𝔥} \\ = & ⟨ (y, h^{'}), ([z, x], [h^{''}, h]) ⟩ \\ = & ⟨ (y, h^{'}), [(z, h^{''}), (x, h)] ⟩ . \end{array}$ It follows that the form $⟨, ⟩$ is an invariant form on $𝔤 \times 𝔥$ .

2.5 Let $x, h) \in 𝔤 \times 𝔥$ such that $x, h) \neq(0,0)$ . Since $⟨, ⟩_{𝔥}$ is nondegenerate, if $h \neq 0$ , there exists some $h^{'} \in 𝔥$ such that $⟨ h, h^{'} ⟩_{𝔥} \neq 0$ . Thus if $h \neq 0$ , $⟨ (x, h), (0, h^{'}) ⟩ = - ⟨ h, h^{'} ⟩_{𝔥} \neq 0 .$ If $h = 0$ then $x \neq 0$ , and by the nondegeneracy of $⟨, ⟩_{𝔤}$ there exists some $y \in 𝔤$ such that $⟨ x, y ⟩_{𝔤} \neq 0$ . Thus, if $h = 0$ , $⟨ (x, 0), (y, 0) ⟩ = ⟨ x, y ⟩_{𝔤} \neq 0 .$ Thus $⟨, ⟩$ is a nondegenerate form on $𝔤 \times 𝔥$ .

2.6 For each $i \in I$ let us fix an element $X \in 𝔤_{i}$ . Let $Y_{i} \in 𝔤_{- i}$ be dual to $X_{i} \in 𝔤_{i}$ with respect to the form $⟨, ⟩$ and let $H_{i} = [X_{i}, Y_{i}] \in 𝔥$ . The elements $X_{i}, Y_{i}, H_{i}, i \in I$ generate $𝔤$ . The manin triple in (2.1) determines a Lie algebra structure on $𝔟^{+}$ with cocommututator $φ : 𝔤_{A} \to ⋀^{2} 𝔤_{A}$ given by $\begin{array}{l} φ (H_{i}) = 0, \\ φ (X_{i}) = \frac{1}{2} X_{i} \land H_{i} . \end{array}$

		Proof:
	Thus the cobracket on $𝔭_{1}$ is determined by the equation $⟨ φ ((z, z_{𝔥})), (y, - y_{𝔥}) \otimes (f, - f_{𝔥}) ⟩ = ⟨ (z, z_{𝔥}), [(y, - y_{𝔥}), (f, - f_{𝔥})] ⟩$ where $y, f \in 𝔟^{-}$ and $z \in 𝔟^{+}$ . We have $\begin{array}{rcl} ⟨ φ ((z, z_{𝔥})), (y, - y_{𝔥}) \otimes (f, - h_{𝔥}) ⟩ & = & ⟨ (z, z_{𝔥}), [(y, - y_{𝔥}), (f, - f_{𝔥})] ⟩ \\ = & ⟨ (z, z_{𝔥}), ([y, f], [- y_{𝔥}, - f_{𝔥}] ⟩ \\ = & ⟨ z, [y, f] ⟩ - ⟨, z_{𝔥} [- y_{𝔥}, - f_{𝔥}] ⟩ . \end{array}$ Let $H_{i} = [X_{i}, Y_{i}] \in 𝔥$ . It follows that $⟨ φ ((H_{i}, H_{i})), (y, x) \otimes (f, e) ⟩ = ⟨ H_{i}, [y, f] ⟩ - ⟨ X_{i}, [x, e] ⟩ = 0 - 0 = 0$ , since $[y, f] \in 𝔫^{-}$ and $[x, e] \in 𝔫^{+}$ . Thus $φ ((H_{i}, H_{i})) .$ $\begin{array}{rcl} ⟨ φ ((X_{i}, 0)), (y, - y_{𝔥}) \otimes (f, - f_{𝔥}) ⟩ & = & ⟨ X_{i}, [y, f] ⟩ - ⟨ 0, [- y_{𝔥}, - f_{𝔥}] ⟩ \\ = & ⟨ X_{i}, [y, f] ⟩ - ⟨ 0, [- y_{𝔥}, - f_{𝔥}] ⟩ \\ = & ⟨ X_{i}, [y, f] ⟩ \\ = & ⟨ [X_{i}, y], f ⟩ \\ = & ⟨ [X_{i}, ⟨ y, X_{i} ⟩ Y_{i} + y_{𝔥}], f ⟩ \\ = & ⟨ y, X_{i} ⟩ ⟨ [X_{i}, Y_{i}], f ⟩ + ⟨ [X_{i}, y_{𝔥}], f ⟩ \\ = & ⟨ y, X_{i} ⟩ ⟨ H_{i}, f ⟩ - ⟨ [X_{i}, f], y_{𝔥} ⟩ \\ = & ⟨ X_{i}, y ⟩ ⟨ H_{i}, f ⟩ - ⟨ [X_{i}, ⟨ f, X_{i} ⟩ Y_{i} + f_{𝔥}], y_{𝔥} ⟩ \\ = & ⟨ X_{i}, y ⟩ ⟨ H_{i}, f ⟩ - ⟨ f, X_{i} ⟩ ⟨ [X_{}, Y_{i}], y_{𝔥} ⟩ + ⟨ [X_{i}, f_{𝔥}], y_{𝔥} ⟩ \\ = & ⟨ X_{i}, y ⟩ ⟨ H_{i}, f ⟩ - ⟨ f, X_{i} ⟩ ⟨ H_{i}, y ⟩ + 0 . \end{array}$ On the other hand, $\begin{array}{rcl} ⟨ (X_{i}, 0) \land (H_{i}, H_{i}), (y, - y_{𝔥}) ⟩ & = & ⟨ (X_{i}, 0) \otimes (H_{i}, H_{i}) - (H_{i}, H_{i}) \otimes (X_{i}, 0), (y, - y_{𝔥}) \otimes (f, - f_{𝔥}) ⟩ \\ = & ⟨ X_{i}, y ⟩ ⟨ H_{i}, f_{𝔥} + f_{𝔥} ⟩ - ⟨ H_{i}, y + y_{𝔥} ⟩ ⟨ X_{i}, f ⟩ \\ = & ⟨ X_{i}, y ⟩ ⟨ H_{i}, f_{𝔥} + f_{𝔥} ⟩ - ⟨ H_{i}, y_{𝔥} + y_{𝔥} ⟩ ⟨ X_{i}, f ⟩ \\ = & 2 (⟨ X_{i}, y ⟩ ⟨ H_{i}, f_{𝔥} ⟩ - ⟨ H_{i}, y_{𝔥} ⟩ ⟨ X_{i}, f ⟩) . \end{array}$ It follows that $φ ((X_{i}, 0)) = \frac{1}{2} ((X_{i}, 0) \land (H_{i}, H_{i})) .$ $□$

The double $D (𝔟^{+}) ≅ 𝔤 \times 𝔥$ .

		Proof:
	This is clear from the form of the Manin triple in (2.1). $□$

Quasitriangular Lie bialgebra structure on 𝔤

Let $𝔤$ be a Lie algebra with invariant form $⟨, ⟩$ satisfying the conditions of Section 1.

There is a Lie bialgebras on $𝔤$ determined by the Manin triple $(𝔭_{1}, 𝔭_{1}, 𝔭_{2})$ given by $\begin{array}{rcl} 𝔭 & = & 𝔤 \times 𝔤, \\ 𝔭_{1} & = & \{(x, y) \in 𝔤 \times 𝔤_{A} ∣ x = y\} ≅ 𝔤 \\ 𝔭_{2} & = & \{(x, y) \in 𝔟_{-} \times 𝔟_{+} ∣ x_{𝔥} + y_{𝔥} = 0\}, \end{array}$ and Lie bracket and invariant scalar product $\begin{array}{rcl} [(x_{1}, y_{1}), (x_{2}, y_{2})] & = & ([x_{1}, x_{2}], [y_{1}, y_{2}]), \\ ⟨ (x_{1}, y_{1}), (x_{2}, y_{2}) ⟩ & = & ⟨ x_{1}, x_{2} ⟩ - ⟨ y_{1}, y_{2} ⟩, \end{array}$ for all $x_{1}, x_{2}, y_{1}, y_{2} \in 𝔤_{A}$ respectively.
The Lie bialgebra structure on $𝔤$ determined by the Manin triple in 1) is given by the cocommutator $φ : 𝔤_{A} \to ⋀^{2} 𝔤_{A}$ determined by $\begin{matrix} φ (H_{i}) & = & 0, \\ φ (X_{i}) & = & \frac{1}{2} X_{i} \land H_{i}, \\ φ (Y_{i}) & = & \frac{1}{2} Y_{i} \land H_{i}, \end{matrix}$ for all $i \in I$ .
For each $α \in Q^{+} ∖ \{0\}$ , let $\begin{array}{l} \{X_{α (j)} ∣ 1 \leq j \leq dim (𝔤_{α})\} \\ \{X_{- α (j)} ∣ 1 \leq j \leq dim (𝔤_{α})\} \end{array}$ be a basis of $𝔤_{α}$ and a dual basis in $𝔤_{- α}$ . Let $\{{\tilde{H}}_{i}\}$ be an orthonormal basis of $𝔥$ . The Lie algebra $𝔤$ is a quasitriangular Lie bialgebra with $r$ -matrix given by $r = \sum_{α \in Q^{+} ∖ \{0\}} \sum_{j = 1}^{dim (𝔤_{α})} X_{α (j)} \otimes X_{- α (j)} + \frac{1}{2} \sum {\tilde{H}}_{i} \otimes {\tilde{H}}_{i} .$

		Proof:
	We must show that $𝔭_{1}$ is an isotropic Lie subalgebra of $𝔤$ . $𝔭_{2}$ is an isotropic Lie subalgebra of $𝔤$ . $𝔭 = 𝔤 \times 𝔤 = 𝔭_{1} \oplus 𝔭_{2}$ . The bilinear form $⟨, ⟩$ is invariant. The bilinear form $⟨, ⟩$ is nondegenerate. These are proved in (3.2)-(3.6). This is proved in (3.7). This follows in (3.8). $□$

3.3 $𝔭_{1}$ is an isotropic Lie subalgebra of $𝔤 \times 𝔤$ . THe fact that $𝔭_{1}$ is an isotropic Lie subalgebra of $𝔤 \times 𝔤$ follows from the following computations. $\begin{matrix} ⟨ (x, x), (y, y) ⟩ ⟨ x, y ⟩ - ⟨ x, y ⟩ = 0 \\ [(x, x), (y, y)] = ([x, y], [x, y]) \in 𝔭_{1} . \end{matrix}$

3.3 Let $(y, x), (f, e) \in 𝔭_{2}$ so that $\begin{matrix} y = y^{-} + y_{𝔥}, \\ f = f^{-} + f_{𝔥}, \end{matrix} and \begin{matrix} x = x^{+} + x_{𝔥}, \\ e = e^{+} + e_{𝔥}, \end{matrix}$ and $x_{𝔥} = - y_{𝔥}$ and $f_{𝔥} = e_{𝔥}$ . Then $⟨ (y, x), (f, e) ⟩ = ⟨ y, f ⟩ - ⟨ x, e ⟩ = ⟨ y_{𝔥}, f_{𝔥} ⟩ - ⟨ x_{𝔥}, e_{𝔥} ⟩ = ⟨ y_{𝔥}, f_{𝔥} ⟩ - ⟨ - y_{𝔥}, - f_{𝔥} ⟩ = 0 .$ Thus $𝔭_{2}$ is isotropic. Note that since $[y, f] \in 𝔫^{-}$ , and $[x, e] \in 𝔫^{+}$ , it follows that $[y, f]_{𝔥} = [x, e]_{𝔥} = 0$ , and thus $[(y, x), (f, e)] = ([y, f], [x, e]) \in 𝔭_{2} .$ Thus $𝔭_{2}$ is a Lie subalgebra.

3.4 Let $(y, x) \in 𝔭_{1} \cap 𝔭_{2}$ . It follows that $y \in 𝔟^{-}, x \in 𝔟^{+}, y = x$ and $x_{𝔥} = y_{𝔥}$ and $x_{𝔥} + y_{𝔥} = 0$ . Thus $x^{+} = 0$ and $y^{-} = 0$ giving also that $x_{𝔥} = y_{𝔥}$ . Thus if $char (k) \neq 2$ we have that $x_{𝔥} = y_{𝔥} = x = y = 0$ . So $𝔭_{1} \cap 𝔭_{2} = 0$ . Given two elements $x, y \in 𝔤$ we may write $(x, y) = (x^{+} - y^{+} + \frac{x_{𝔥} - y_{𝔥}}{2}, - x^{-} + y^{-} - \frac{x_{𝔥} - y_{𝔥}}{2}) + (x^{-} + y^{+} + \frac{x_{𝔥} + y_{𝔥}}{2}, x^{-} + y^{+} - \frac{x_{𝔥} + y_{𝔥}}{2})$ to see that $(x, y) \in 𝔭_{1} + 𝔭_{2}$ . It follows that $𝔤 \times 𝔤 = 𝔭_{1} + 𝔭_{2}$ .

3.5 Recall that $⟨ (x_{1}, x_{2}), (x_{1}, y_{2}) ⟩ = ⟨ x_{1}, y_{1} ⟩ - ⟨ x_{2}, y_{2} ⟩$ . It follows that $\begin{array}{rcl} ⟨ [(x_{1}, y_{1}), (x_{2}, y_{2})], (x_{3}, y_{3}) ⟩ & = & ⟨ [x_{1}, x_{2}], x_{3} ⟩ - ⟨ [y_{1}, y_{2}], y_{3} ⟩ \\ = & ⟨ x_{2}, [x_{3}, x_{1}] ⟩ - ⟨ y_{2}, [y_{3}, y_{1}] ⟩ \\ = & ⟨ (x_{2}, y_{2}), ([x_{3}, x_{1}], [y_{3}, y_{1}]) ⟩ \\ = & ⟨ (x_{2}, y_{2}), [(x_{3}, y_{3}), (x_{1}, y_{1})] ⟩ . \end{array}$ Thus, the form $⟨, ⟩$ is invariant.

3.6 Let $(x, y) \in 𝔤 \times 𝔤$ such that $(x, y) \neq (0, 0)$ . If $x \neq 0$ then, since the form on $𝔤$ is nondegenerate there exists some $z \in 𝔤$ such that $(x, z) \neq 0$ . Thus $⟨ (x, y), (z, 0) ⟩ = ⟨ x, 0 ⟩ - ⟨ y, 0 ⟩ = ⟨ x, z ⟩ \neq 0 .$ If $x = 0$ then $y \neq 0$ and there exists $w \in 𝔤$ such that $⟨ y, w ⟩ \neq 0$ . It follows that $⟨ (x, y), (0, w) ⟩ = ⟨ x, 0 ⟩ - ⟨ y, w ⟩ = - ⟨ y, w ⟩ \neq 0 .$ Thus, the form on $𝔤 \times 𝔤$ is nondegenerate.

3.7 The cobracket on $𝔭_{1}$ is determined by $⟨ φ ((z, z)), (y, x) \otimes (f, e) ⟩ = ⟨ (z, z), [(y, x), (f, e)] ⟩$ where $y, f \in 𝔟^{-}$ and $x, e \in 𝔟^{+}$ . We have $⟨ φ ((z, z)), (y, x) \otimes (f, e) ⟩ = ⟨ (z, z), [(y, x), (f, e)] ⟩ = ⟨ (z, z), ([y, f], [x, e]) ⟩ = ⟨ z, [y, f] ⟩ - ⟨ z, [x, e] ⟩ .$ Let $H_{i} \in 𝔥$ . Then, since $[y, f] \in 𝔫^{-}$ and $[x, e] \in 𝔫^{+}$ , it follows that $⟨ φ ((H_{i}, H_{i})), (y, x) \otimes (f, e) ⟩ = ⟨ H_{i}, [y, f] ⟩ - ⟨ H_{i}, [x, e] ⟩ = 0 - 0 = 0$ . Thus, $φ ((H_{i}, H_{i})) = 0 .$ The computations to determine $φ (X_{i})$ are as follows. $\begin{array}{rcl} ⟨ φ ((X_{i}, X_{i})), (y, x) \otimes (f, e) ⟩ & = & ⟨ X_{i}, [y, f] ⟩ - ⟨ X_{i}, [x, e] ⟩ \\ = & ⟨ X_{i}, [y, f] ⟩ - 0 \\ = & ⟨ [X_{i}, y], f ⟩ \\ = & ⟨ [X_{i}, ⟨ y, X_{i} ⟩ Y_{i}] + y_{𝔥} ⟩ \\ = & ⟨ y, X_{i} ⟩ ⟨ [X_{i}, Y_{i}], f ⟩ + ⟨ [X_{i}, y_{𝔥}], f ⟩ \\ = & ⟨ y, X_{i} ⟩ ⟨ H_{i}, f ⟩ - ⟨ [X_{i}, f], y_{𝔥} ⟩ \\ = & ⟨ X_{i}, y ⟩ ⟨ H_{i}, f ⟩ - ⟨ [X_{i}, ⟨ f, X_{i} ⟩ Y_{i} + f_{𝔥}], y_{𝔥} ⟩ \\ = & ⟨ X_{i}, y ⟩ ⟨ H_{i}, f ⟩ - ⟨ f, X_{i} ⟩ ⟨ [X_{i}, Y_{i}], y_{𝔥} ⟩ + ⟨ [X_{i}, f_{𝔥}], y_{𝔥} ⟩ \\ = & ⟨ X_{i}, y ⟩ ⟨ H_{i}, f ⟩ - ⟨ f, X_{i} ⟩ ⟨ H_{i}, y ⟩ + 0 . \end{array}$ On the other hand, since $y_{𝔥} = - x_{𝔥}$ and $f_{𝔥} = - e_{𝔥}$ it follows that $⟨ (X_{i}, X_{i}) \land (H_{i}, H_{i}), (y, x) \otimes (f, e) ⟩ & = & ⟨ (X_{i}, X_{i}) \otimes (H_{i}, H_{i}) - (H_{i}, H_{i}) \otimes (X_{i}, X_{i}) \otimes, (y, x) \otimes (f, e) ⟩ = & ⟨ X_{i}, y - x ⟩ ⟨ H_{i}, f - e ⟩ - ⟨ H_{i}, y - x ⟩ ⟨ X_{i}, f - e ⟩ = & ⟨ X_{i}, y - x ⟩ ⟨ H_{i}, f_{𝔥} + f_{𝔥} ⟩ - ⟨ H_{i}, y_{𝔥} + y_{𝔥} ⟩ ⟨ X_{i}, f - e ⟩ = & 2 (⟨ X_{i}, y - x ⟩ ⟨ H_{i}, f ⟩ - ⟨ H_{i}, y ⟩ ⟨ X_{i}, f - e ⟩) .$ Since $X_{i} \in 𝔫^{+}$ and $x \in 𝔟^{+}$ it follows that $⟨ X_{i}, x ⟩ = 0$ . Similarly, $⟨ X_{i}, e ⟩ = 0$ . Thus $⟨ (X_{i}, X_{i}) \land (H_{i}, H_{i}), (y, x) \otimes (f, e) ⟩ = 2 (⟨ X_{i}, y ⟩ ⟨ H_{i}, f ⟩ - ⟨ H_{i}, y ⟩ ⟨ X_{i}, f ⟩) .$ It follows that $φ ((X_{i}, X_{i})) = \frac{1}{2} ((X_{i}, X_{i}) \land (H_{i}, H_{i})) .$ The formula $φ ((Y_{i}, Y_{i})) = \frac{1}{2} ((Y_{i}, Y_{i}) \land (H_{i}, H_{i})) .$ is proved similarly.

3.8 We know that $D (𝔟^{+}) ≅ 𝔤 \times 𝔥$ from Corollary (2.7). The set $\{(X_{α (j)}, 0) ∣ α \in Q^{+} ∖ \{0\}, 1 \leq j \leq dim (𝔤_{α})\} \cup {\frac{1}{\sqrt[]{2}} ({\tilde{H}}_{i}, {\tilde{H}}_{i})}$ is a basis of $𝔭_{1}$ with dual basis (with respect to the form on $𝔤 \times 𝔥$ ) $\{(X_{- α (j)}, 0) ∣ α \in Q^{+} ∖ \{0\}, 1 \leq j \leq dim (𝔤_{α})\} \cup {\frac{1}{\sqrt[]{2}} ({\tilde{H}}_{i}, - {\tilde{H}}_{i})}$ in $𝔭_{2}$ . The $r$ -matrix $\hat{r} = \sum_{α \in Q^{+} ∖ \{0\}} \sum_{j = 1}^{dim (𝔤_{α})} (X_{α (j)}, 0) \otimes (X_{- α (j)}, 0) + \frac{1}{2} \sum {\tilde{H}}_{i} \otimes {\tilde{H}}_{i}$ gives a quasitriangular Lie bialgebra structure on $𝔤 \times 𝔥$ . The subspace $0 \times 𝔥$ of $𝔤 \times 𝔥$ is an ideal of $𝔤 \times 𝔥$ and $𝔤 ≅ (𝔤 \times 𝔥) / (0 \times 𝔥)$ . Thus the projection $\begin{matrix} r \in (𝔤 \times 𝔥) / (0 \times 𝔥) \otimes (𝔤 \times 𝔥) / (0 \times 𝔥) \\ r = = \sum_{α \in Q^{+} ∖ \{0\}} \sum_{j = 1}^{dim (𝔤_{α})} (X_{α (j)}, 0) \otimes (X_{- α (j)}, 0) + \frac{1}{2} \sum (H_{i}, 0) \otimes (H_{i}, 0) \end{matrix}$ of the $r$ -matrix $\hat{r}$ gives a quasitriangular Lie bialgebra structure on $𝔤$ .
It remains to check that htis bialgebra structure is the same as that given by the Manin triple in 1). By (2.6) we know that the cocommutator determined by $r$ satisfies $\begin{matrix} φ (H_{i}) = 0 \\ φ (X_{i}) = \frac{1}{2} X_{i} \land H_{i}, \end{matrix}$ on the subalgebra $𝔟^{+}$ . We can calculate $φ (Y_{i})$ by using the Cartan involution $θ$ . $\begin{array}{rcl} ({ad}^{\otimes 2} (Y_{i})) (θ \otimes θ) (r) & = & (θ \otimes θ) ({ad}^{\otimes 2} (X_{i})) (r) \\ = & (θ \times θ) (\frac{1}{2} X_{i} \land H_{i}) \\ = & - \frac{1}{2} Y_{i} \land H_{i} . \end{array}$ Since $\begin{array}{rcl} (θ \otimes θ) (r) & = & \sum_{α \in Q^{+}} \sum_{j = 1}^{dim (𝔤_{α})} θ (X_{α (j)}) \otimes θ (X_{- α (j)}) + \frac{1}{2} \sum θ (H_{i}) \otimes θ (H_{i}) \\ = & \sum_{α \in Q^{+} ∖ \{0\}} \sum_{j = 1}^{dim (𝔤_{α})} X_{α (j)} \otimes X_{- α (j)} + \frac{1}{2} \sum H_{i} \otimes H_{i} \\ = & r^{21}, \end{array}$ and $r^{21} + r^{12}$ is invariant, it follows that $({ad}^{\otimes 2} (Y_{i})) (r) = - ({ad}^{\otimes 2} (Y_{i})) (r^{21}) = \frac{1}{2} Y_{i} \land H_{i} .$ Thus we have shown that the bialgbera structure on $𝔤$ determined by the $r$ -matrix and the bialgebra structure on $𝔤$ determined by the Manin triple given in 1) are identical.

References

The definitions and proofs of the fact that Kac-Moody Lie algebras satisfy the properties given in Section 1 are contained in the first few pages of the following standard book.

[Kc] V. Kac, Infinite dimensional Lie algebras, Third Ed., Cambridge University Press 1990. MR1104219

The examples of bialgebra structures geven here appear in example 3.2 of the following paper by Drinfel'd.

[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

[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.

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