Poisson manifolds

The map $f\mapsto X_f$ takes values in $\Der(\Coo(M))=\gX^1(M)$. Thus we can write
$$\begin{align*} \Cas(M) &=\{f\in\Coo(M) : \{f,g\}=0\quad\forall g\in\Coo(M) \} = ... ...),\\ \Ham(M) &=\text{Hamiltonian vector fields}=\im(f\mapsto X_f). \end{align*}$$
To put it another way we have the short exact sequence

$$0\to \Cas(M)\to \Coo(M) \to \Ham(M)\to 0.$$

The Cartesian product of Poisson manifolds is a Poisson manifold

$$\Coo(M_1\times M_2)\supset \Coo(M_1)\ox \Coo(M_2).$$

There is a Poisson structure on $\Coo(M_1)\ox \Coo(M_2)$ and it extends to $\Coo(M_1\times M_2)$ by

$$\{f(x_1, x_2), g(x_1,x_2)\}=\{f_{x_2}, g_{x_2}\}_2(x_1)+ \{f_{x_1}, g_{x_1}\}_1(x_2),\quad\text{where}$$

$$f_{x_1}\:x_2\mapsto f(x_1, x_2),\quad f_{x_1}\in\Coo(M_2),$$

$$f_{x_2}\:x_1\mapsto f(x_1, x_2),\quad f_{x_2}\in\Coo(M_1).$$

Consider special case of the previous example, $(\bR^{2n}, \om=\sum dp_i\land dq_i)$. Every symplectic manifold is locally symplectomorphic to this one (but that does not mean that this is unique symplectic structure on $\bR^{2n}$ !).

Let $f:=f(p_i, q_i)$, $\om(X_f, Y)=-Yf$. Then for $Y=\del_{q_i}$ and $Y=\del_{p_i}$ we have respectively
$$\begin{align*} -i _{\del_{q_i}}\om &= \om(-, \del_{q_i})=-dp_i \\ -i _{\del_{p_... ...p_i})=-dq_i \\ X_f &= \sum_{i=1}^n a_i\del_{q_i} + b_i\del_{p_i}. \end{align*}$$
Now $\om(X_f, \del_{q_i})=-b_i$, $\om(X_f, \del_{p_i})=-a_i$ and
$$\begin{align*} X_f &= \sum_{i=1}^n -\del_{p_i} f \del_{q_i} + \del_{q_i} f \del_... ...sum_{i=1}^n -\del_{p_i} f \del_{q_i} g + \del_{q_i} f \del_{p_i} g. \end{align*}$$








Fix on $M$ a coordinate chart $(U; x_1,\hdots, x_n)$. Then the bivector $\Pi$ is locally given by

$$\Pi_U=\sum_{i<j}\Pi_{ij}\del_{x_i}\land\del_{x_j}$$

where the coefficients $\Pi_{ij}$ are functions on $U$ explicitely given by $\Pi_{ij}=\{x_i, x_j\}$. Therefore $\Pi$ is determined once you know brackets of local coordinate functions

$$\{f,g\}=\sum_{i,j=1}^n\{x_i, x_j\}\del_{x_i}f\del_{x_j}g.$$

Let $\Pi:= \sum_{i<j}\Pi_{ij}\del_{x_i}\land \del_{x_j}$ be a bivector field, where $\Pi_{ij}=\{x_i, x_j\}$. In many examples a Poisson structure on $\bR^{2n}$ will be given simply by lifting brackets of coordinants.

Let $V$ be a real $n$-dimensional vector space. Consider coordinates $x_1,\hdots, x_n$. Then $\sum_{i<j}\Pi_{ij}\del_{x_i}\land \del_{x_j}$ is Poisson tensor if and only if () holds.

Another way to obtain the same result is to take a Lie algebra $\gerg$, $V=\gerg^*$ linear functionals on $\gerg$ If one knows a Poisson bracket on a basis of $\gerg^*$, then knows it on $V$. Let $X_1,\hdots, X_n$ be basis of $\gerg$, $[X_i, X_j]=\sum_{k=1}^n c_{ij}^k X_k$. Let $\xi_1, \hdots, \xi_n$ be the dual basis of $\gerg^*$. Say $\al\in\gerg^*$, $f,g\in \Coo(\gerg^*)$. Then $d_{\al}f\in (\gerg^*)^*=\gerg$ and

$$\{f,g\}(\al)=\scalar{\al}{[d_{\al}f, d_{\al}g]}.$$

For example if $f\simeq X_i$, $g\simeq X_j$, $X_i(\xi_j)=\dl_{ij}$
$$\begin{align*} \{X_i, X_j\}(\xi_k) &= \scalar{\xi_k}{[X_i, X_j]} \\ &= \scalar{... ...^n c_{ij}^h X_h} \\ &= c_{ij}^k, \\ \{X_i, X_j\} &= c_{ij}^k X_k. \end{align*}$$
Thus $\Pi=\sum_{k=1}^n c_{ij}^k X_i\land X_j$ is a linear Poisson tensor on $\gerg^*$. The dual of a Lie algebra has always a canonically defined Poisson tensor.