Poisson-Lie Group, Lie bialgebra and Manin triple

General Idea

A Poisson-Lie group is a Lie group that is a Poisson manifold, with the multiplication map Poisson, in other words a Lie group with a multiplicative Poisson structure. The idea is particularly natural:

  1. For a Poisson-Lie group any Poisson submanifold $H\subset G$ that is also a Lie subgroup is also a Poisson-Lie group.
  2. The quotient $G/H$ is has a unique Poisson manifold structure s.t. $G\to G/H$ is Poisson.
  3. The dual $\mathfrak{g}^\ast$ of the Lie algebra $\mathfrak{g}$ of a Poisson-Lie group has a canonical Lie algebra structure (tangent Lie algebra). This gives rise to $\mathfrak{g}$ a natural Lie bialgebra structure.

The infinitesimal structure of a Poisson-Lie group is controlled by its tangent Lie bialgebra. There is a correspondence similar to Lie group-Lie algebra correspondence between Poisson-Lie groups and their tangent Lie bialgebras.

Closely related to the notion of Lie bialgebra is that of Manin triple. Every Manin triple can be made into a Lie bialgebra, and from any Lie bialgebra a Manin triple can be constructed.

Poisson-Lie group

Recall that given Poisson manifolds $(M,\pi_M)$ and $(N,\pi_N)$. The structure of product Poisson manifold on $M\times N$ is the one given by $$ \pi = \pi_M \oplus \pi_N \in \Gamma(\wedge^2 T(M\times N)). $$ In the bracket notation this is $$ \{f,g\}(x,y) = \{f(\cdot,y),f(\cdot,y)\}_M (x) + \{f(x,\cdot),f(x,\cdot)\}_N (y). $$

Definition(Poisson-Lie group) . A Poisson-Lie group is a Lie group $G$ with a Poisson bivector $\pi$, s.t. the product $$ m:G\times G\to G $$ is a Poisson map, where the Poisson structure on $G\times G$ is the product Poisson structure.
Remark. This does not mean that regarded as an action $m_g:G\to G: h\mapsto gh$ is a Poisson map (preserves the Poisson structure on $G$).

The Poisson structure is called multipliative. There is an equivalent definitions for the Poisson-Lie group.

Proposition. Let $G$ be a Lie group with a Poisson bivector $\pi$, the following are equivalent
  1. The product $m:G\times G\to G$ is a Poisson map.
  2. $\pi(g_1 g_2) = L_{g_1,\ast}\pi(g_2)+R_{g_2,\ast}\pi(g_1)$ where $L_{g,\ast}$ is the derivative of the left translation, and similar for $R_{g,\ast}$.
  3. For every $X$ right invariant vector field on $G$ and $Y$ left invariant, the Lie derivative $L_X L_Y \pi = 0$, and $\pi(1)=0$ where $1\in G$ is the neutral element.

The third definition comes from the fact that $$ \pi(g_1g_2) = L_{g_1,\ast}\pi(g_2) + R_{g_2,\ast}\pi(g_1) + L_{g_1,\ast}R_{g_2,\ast}\pi(e). $$ When $\pi(e)\neq 0$, the bivector is no longer multiplicative, and is called an affine Poisson structure.

Remark. For a Poisson-Lie group $G$ the inversion map $g\mapsto g^{-1}$ is not a Poisson map. It is anti-Poisson.


A non-trivial example. Let $G$ be a Lie group, and $\mathfrak{g}$ any Lie algebra. Its dual $\mathfrak{g}^\ast$ is naturally a Poisson manifold with the Lie-Poisson structure. Then $\mathfrak{g}^\ast$ regarded as an abelian Lie group under addition is a Poisson-Lie group.

If $\mathfrak{h}$ is another Lie algebra, the homomorphism of Poisson-Lie groups $\mathfrak{h}^\ast\to\mathfrak{g}^\ast$ are exactly the duals of the Lie algebra homomorphisms $\mathfrak{g}\to\mathfrak{h}$.

Lie bialgebra

Poisson-Lie group and Lie bialgebra

Definition(Lie bialgebra) . Let $\mathfrak{g}$ be a Lie algebra. A Lie bialgebra structure on $\mathfrak{g}$ is an antisymmetric linear map $\delta_{\mathfrak{g}}: \mathfrak{g}\to\mathfrak{g}\otimes\mathfrak{g}$, called the cocommutator, s.t.
  1. $\delta^\ast_{\mathfrak{g}}:\mathfrak{g}^\ast\otimes\mathfrak{g}^\ast\to\mathfrak{g}^\ast$ is a Lie bracket on $\mathfrak{g}^\ast,$
  2. $\delta_{\mathfrak{g}}$ is a 1-cocylce of $\mathfrak{g}$ with values in $\mathfrak{g}\otimes\mathfrak{g}$.

A homomorphism of Lie bialgebras $\varphi:\mathfrak{g}\to\mathfrak{h}$ is a homomorphism of Lie algebras s.t. $$ (\varphi\otimes\varphi)\circ \delta_{\mathfrak{g}} = \delta_{\mathfrak{h}}\circ\varphi $$

This is motivated by the infinitesimal structure of Poisson-Lie groups. There is a theorem similar to Lie group–Lie algebra correspondence theorems:

  1. If $G$ is a Poisson-Lie group, the Lie algebra $\mathfrak{g}$ has a natural Lie bialgebra structure (called tangent Lie bialgebra).
  2. If $G$ is connected and simply connected, every Lie bialgebra structure on $\mathfrak{g}$ is the tangent Lie algebra of a unique Poisson structure on $G$ that makes $G$ into a Poisson-Lie group.
  3. The derivative $\varphi:\mathfrak{g}\to\mathfrak{h}$ of a homomorphism of Poisson-Lie groups $\Phi:G\to H$ at $e\in G$ is a homomorphism of Lie bialgebras.
  4. If $\varphi:\mathfrak{g}\to\mathfrak{h}$ is a homomorphism of Lie bialgebras and $H$ is a Poisson-Lie group with tangent Lie algebra $\mathfrak{h}$, then there is a unique homomorphism of Poisson-Lie groups $\Phi:G\to H$ with the derivative at $d_e \Phi = \varphi$.

The tangent Lie bialgebra structure

Given a Poisson-Lie group $G$ with Lie algebra $\mathfrak{g}$, if $\xi_1,\xi_2\in\mathfrak{g}^\ast$, choose $f_1,f_2\in C^\infty(G)$ s.t. $(df_i)(e) = \xi_i$ and put $$ [\xi_1,\xi_2]_{\mathfrak{g}^\ast} = (d\{f_1,f_2\})(e). $$ The bracket $[\cdot,\cdot]_{\mathfrak{g}^\ast}$ defined is antisymmetric and satisfies the Jacobi identity, i.e. it is a Lie bracket. It is well defined, i.e. the R.H.S. only depends on $\xi_1,\xi_2$. This can be seen by the following:

Let $\pi^R: G\to \mathfrak{g}\otimes \mathfrak{g}$ be the right translate of the bivector field $\pi:G\to TG\otimes TG$ to the identity, in detail this is, since $R_{g}: G\to G: h \mapsto hg$ and the differential is $R_{g,\ast}: T_h G \to T_{hg} G$, so $$ \pi^R(g) = R_{g^{-1},\ast}\pi(g). $$ Now $$ \langle \pi(g), df_1(g) \otimes df_2(g) \rangle = \langle (R_{g,\ast} \pi^R(g), df_1(g)\otimes df_2(g)\rangle = \langle \pi^R(g), (R_{g,\ast})^\ast(df_1(g)\otimes df_2(g))\rangle. $$ Here the notation is bad since $(\cdot)^\ast$ is taking dual while $(\cdot)_\ast$ is the pushforward/differential. Differentiate at $g=e$ in the direction $X\in\mathfrak{g}$, by $\pi^R(e) = 0$ this yields $$ \langle X,d\{f_1,f_2\}(e)\rangle = \langle X, d\langle \pi^R(e),df_1(e)\otimes df_2(e)\rangle\rangle. $$ Now put $\delta: \mathfrak{g}\to \mathfrak{g}\otimes \mathfrak{g}$ to be the differential of $\pi^G$ at $e$, switching notations for the same operations, $$ \langle X,d\{f_1,f_2\}(e)\rangle = \langle X, d\langle \pi^R(e),df_1(e)\otimes df_2(e)\rangle\rangle = \langle \delta(X),\xi_1\otimes \xi_2\rangle. $$ In fact, $$ \delta^\ast(\xi_1\otimes\xi_2) = [\xi_1,\xi_2]_{\mathfrak{g}^\ast}. $$


The Lie-Poisson property for $\pi$ translates, for $\pi^R$, into $$ \begin{gather*} R_{g_1 g_2,\ast}\pi^R(g_1 g_2) = L_{g_1,\ast} R_{g_2,\ast}\pi^R(g_2) + R_{g_2,\ast}R_{g_1,\ast}\pi^R(g_1)\\ \pi^R(g_1 g_2) = R_{g_2^{-1},\ast} R_{g_1^{-1},\ast} L_{g_1,\ast} R_{g_2,\ast}\pi^R(g_2) + R_{(g_1 g_2)^{-1},\ast}R_{g_1 g_2,\ast}\pi^R(g_1) \\ \pi^R(g_1 g_2) = R_{g_2^{-1},\ast} \operatorname{Ad}_{g_1} R_{g_2,\ast}\pi^R(g_2) + \pi^R(g_1) \\ \pi^R(g_1 g_2) = \operatorname{Ad}_{g_1} \pi^R(g_2) + \pi^R(g_1). \end{gather*} $$ This means $\pi^R$ is a 1-cocylce of $G$ with values in $\mathfrak{g}\otimes \mathfrak{g}$ on which $G$ acts by the adjoint representation in each factor. The derivative of $\pi^R$ at $e$, namely $\delta$, is then a 1-cocycle of $\mathfrak{g}$ with values in $\mathfrak{g}\otimes\mathfrak{g}$: $$ \delta[X,Y] = (\operatorname{ad}_X\otimes 1 + 1\otimes \operatorname{ad}_X)\delta(Y) - (\operatorname{ad}_Y\otimes 1 + 1\otimes \operatorname{ad}_Y)\delta(X). $$

Manin triple

Definition(Manin triple) . A Manin triple is a triple of Lie algebras $(\mathfrak{p},\mathfrak{p}_+,\mathfrak{p}_-)$ together with a non-degenerate symmetric bilinear form $(\cdot,\cdot)$ on $\mathfrak{p}$, invariant under the adjoint action (Lie bracket) of $\mathfrak{p}$, s.t.
  1. The Lie algebras $\mathfrak{p}_+,\mathfrak{p}_-$ are Lie subalgebras of $\mathfrak{p}$,
  2. As vector spaces $\mathfrak{p}=\mathfrak{p}_+\oplus\mathfrak{p}_-$,
  3. The restriction of $(\cdot,\cdot)$ to $\mathfrak{p}_+$ and $\mathfrak{p}_-$ vanishes.

When the dimensions of the vector spaces are finite, Manin triples are in 1-to-1 correspondence with Lie bialgebras.

Lemma. Suppose that $\mathfrak{g}$ and its dual $\mathfrak{g}^\ast$ are both finite-dimensional Lie algebras. Define an inner product $(\cdot,\cdot)$ on the vector space direct sum $\mathfrak{g}\oplus\mathfrak{g}^\ast$ by $$ (X,\xi)_{\mathfrak{p}} = \langle X,\xi\rangle,\quad (X,X)_{\mathfrak{p}} = (\xi,\xi)_{\mathfrak{p}} =0 $$ for $X\in\mathfrak{g},\xi\in\mathfrak{g}^\ast$. The followings are equivalent:
  1. The pair $(\mathfrak{g},\mathfrak{g}^\ast)$ is a Lie bialgebra.
  2. There is a unique Lie algebra structure on $\mathfrak{g}\oplus\mathfrak{g}^\ast$ s.t. $(\mathfrak{g}\oplus\mathfrak{g}^\ast,\mathfrak{g},\mathfrak{g}^\ast)$ together with $(\cdot,\cdot)$ is a Manin triple.

The Manin triple associated to a Lie algebra $(\mathfrak{g},\mathfrak{g}^\ast)$ is $(\mathfrak{g}\oplus\mathfrak{g}^\ast,\mathfrak{g},\mathfrak{g}^\ast)$ with $(\cdot,\cdot)$ defined by the pairing $\langle\cdot,\cdot\rangle$. The Lie bracket on $\mathfrak{g}\oplus\mathfrak{g}^\ast$ is determined by the invariance of the bilinear form under adjoint $$ ([\eta,X],\xi) := -(X,[\eta,\xi]), \quad (X,[Y,\xi]) := -([Y,X],\xi) $$ where the minus sign comes from switching from adjoint to coadjoint action.

From a Manin triple with $\mathfrak{p}_+ =\mathfrak{g}$ and the symmetric bilinear form $(\cdot,\cdot)$, there is an isomorphism of vector spaces $\mathfrak{g}^\ast \cong \mathfrak{p}_-$ determined by the bilinear form. This makes $\mathfrak{g}^\ast$ into a Lie algebra. By the above lemma $(\mathfrak{g},\mathfrak{g}^\ast)$ is a Lie bialgebra.

Hence the notion of Manin triple is, for finite-dimensional Lie algebras, another way of defining Lie bialgebras, and it is a convenient way for constructing examples of Lie bialgebras.

Dual Poisson-Lie group, the classical double

Let all the Lie algebras be finite-dimensional in this subsection.

Dual Poisson-Lie group

Manin triple also makes explicit the self-duality of Lie bialgebras: if $\mathfrak{g}$ determines a Lie bialgebra, sodoes $\mathfrak{g}^\ast$. Lie bialgebras integrates to Poisson-Lie groups. If $\mathfrak{g}$ integrates to $G$, then call the group $G^\ast$ that integrates $\mathfrak{g}^\ast$ the dual Poisson-Lie group of $G$.

The classical double

For a Lie bialgebra $(\mathfrak{g},\mathfrak{g}^\ast)$, the Lie algebra $\mathfrak{g}\oplus\mathfrak{g}^\ast$ whose Lie algebra structure is given by the Manin triple is also canonically a Lie bialgebra, s.t. the inclusions into the summands $$ \mathfrak{g} \hookrightarrow \mathfrak{g}\oplus\mathfrak{g}^\ast \hookleftarrow (\mathfrak{g}^\ast)^{\operatorname{op}} $$ are homomorphisms of Lie bialgebras. Here $(\cdot)^{\operatorname{op}}$ means taking the opposite Lie algebra, i.e. the Lie algebra with the bracket $[a,b]^{\operatorname{op}} := [b,a]$.

The Lie bialgebra $\mathfrak{g}\oplus\mathfrak{g}^\ast$ is called the classical double $\mathcal{D}(\mathfrak{g})$ of $\mathfrak{g}$. The cocommutator is given by $$ \delta_{\mathcal{D}}(x) =(\operatorname{ad}_x\otimes \operatorname{id} + \operatorname{id}\otimes\operatorname{ad}_x) (r) $$ where $x\in\mathcal{D}(g)$ and $r\in \mathfrak{g}\otimes\mathfrak{g}^\ast\subset \mathcal{D}(\mathfrak{g})\otimes\mathcal{D}(\mathfrak{g})$ corresponds to the identity map $\mathfrak{g}\to\mathfrak{g}$. This integrates to the double $\mathcal{D}(G)$ of a Poisson-Lie group $G$.