Let $E \to X$ be a Lie algebroid over the manifold $X$. Let $x_1,...x_n$ be local coordinates on $X$ and $e_1,...e_m$ be the basis of local sections of $E$. In terms of these coordinate functions Lie bracket and the anchor map $\rho$ are described like this: $$ [e_i,e_j]_E = \sum\limits_k c_{ijk}e_k $$ $$ \rho(e_i) = \sum\limits_j b_{ij}\frac{\partial}{\partial x_j} $$

Let $\xi_1,..\xi_m$ be the basis dual to $e_1,...e_m$. Now we define Poisson structure on $E^*$ by setting: $$ \{x_i,x_j\} = 0 $$ $$ \{\xi_i,\xi_j\} = \sum\limits_k c_{ijk}\xi_k $$ $$ \{\xi_i, x_j\} = -b_{ij} $$

This Poisson structure is independent of the choice of local coordinates and basis of local sections on $E$, so it's very possible that there exists a coordinate free expression of the Poisson structure like the one we have for dual Lie algebras. How does this expression look like?


1 Answer 1


Depending on you sign convention, this goes as follows. First you denote the bundle projection by $pr\colon E^* \longrightarrow X$. For a section $s \in \Gamma^\infty(E)$ you have a linear function $J(s) \in C^\infty(E^*)$ defined by pointwise evaluation, i.e. $J(s)(\alpha_p) = \alpha(s(p))$ where $\alpha_p \in E^*_p$ and $p \in X$. Then the linear Lie bracket on $E^*$ is uniquely determined by \begin{equation} \{pr^*f, pr^*g\} = 0, \quad \{pr^*f, J(s)\} = pr^*(\varrho(s)f), \quad \{J(s), J(t)\} = - J([s, t]), \end{equation} where $f, g \in C^\infty(X)$ are functions on the base and $s, t \in \Gamma^\infty(E)$ are sections of the Lie algebroid.

In fact, one can show that this Poisson structure is linear in the sense that for its Poisson tensor $\pi$ one has $L_\xi \pi = - \pi$ where $\xi$ is the Euler vector field on $E^*$. Conversely, every linaer Poisson structure on $E^*$ defines a Lie algebroid this way. It is a nice exercise the express anchor and Lie algebroid bracket in terms of the Poisson bracket for a linear Poisson tensor.


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