# Relation between Lie bracket and Poisson bracket

For any vector field $$X$$ on a smooth manifold $$Q$$, define $$f_X : T^* Q \to \mathbb{R}, \omega \mapsto \omega(X_x)$$ for $$\omega \in T_x^* Q$$.

We also have that $$\{ \cdot ,\cdot\}$$ is an arbitrary Poisson bracket and $$[\cdot,\cdot]$$ is the Lie bracket, i.e. in coordinates we have

$$[X,Y] = \sum_{i,j=1}^n \left( X^i \frac{\partial Y^j}{\partial q^i} - Y^i \frac{\partial X^j}{\partial q^i} \right) \frac{\partial}{\partial q^j}.$$

Now, we want to show that for any vector fields $$X$$ and $$Y$$, $$\{ f_X, f_Y \} = - f_{[X,Y]},$$ if and only if $$\{ \cdot ,\cdot\}$$ is the canonical Poisson bracket. I was able to show the forward direction without too much trouble. For the converse, I wanted to show that $$\{ q^i , p_j\} = \delta^i_j$$, $$\{ q^i , q^j\} = \{ p_i , p_j\} =0$$. To do this, I am trying to write $$p$$ and $$q$$ in the form $$f_X$$ and $$f_Y$$ for some vector fields $$X$$ and $$Y$$. This is where I am currently stuck. Any hints or answers are appreciated!

The functions on $$T^{*}Q$$ that are of the form $$f_{X}$$ for some $$X\in\mathfrak{X}(Q)$$ are exactly the fiberwise linear functions $$C^{\infty}_{lin}(T^{*}Q)$$ on $$T^{*}Q$$. So the $$p_{i}$$ are of this form, but the $$q^{j}$$ are not (as they are fiberwise constant).
a) We have that $$p_{i}=f_{\partial_{q^{i}}}$$, since $$p_{i}(dq^{j})=\delta_{i,j}=f_{\partial_{q^{i}}}(dq^{j}).$$ Hence, $$\label{1}\tag{1} \{p_{i},p_{j}\}=\{f_{\partial_{q^{i}}},f_{\partial_{q^{j}}}\}=-f_{[\partial_{q^{i}},\partial_{q^{j}}]}=-f_{0}=0.$$
b) As said, $$q^{j}$$ is not fiberwise linear, but $$q^{j}p_{i}$$ is. Indeed, we have $$q^{j}p_{i}=f_{q^{j}\partial_{q^{i}}}$$. So $$\tag{2}\label{2} \{q^{j}p_{i},p_{k}\}=\{f_{q^{j}\partial_{q^{i}}},f_{\partial_{q^{k}}}\}=-f_{[q^{j}\partial_{q^{i}},\partial_{q^{k}}]}=f_{\frac{\partial q^{j}}{\partial q^{k}}\partial_{q^{i}}}=f_{\delta_{j,k}\partial_{q^{i}}}=\delta_{j,k}f_{\partial_{q^{i}}}=\delta_{j,k}p_{i}.$$ On the other hand, using the Leibniz rule of the Poisson bracket $$\{\cdot,\cdot\}$$, we have $$\tag{3}\label{3} \{q^{j}p_{i},p_{k}\}=q^{j}\{p_{i},p_{j}\}+p_{i}\{q^{j},p_{k}\}=p_{i}\{q^{j},p_{k}\},$$ using \eqref{1} in the last equality. Comparing \eqref{2} and \eqref{3} then gives $$\tag{4}\label{4} \{q^{j},p_{k}\}=\delta_{j,k}.$$
c) At last, for the fiberwise linear vector fields $$q^{k}p_{i}=f_{q^{k}\partial_{q^{i}}}$$ and $$q^{l}p_{j}=f_{q^{l}\partial_{q^{j}}}$$, we get \begin{align} \{q^{k}p_{i},q^{l}p_{j}\}&=\{f_{q^{k}\partial_{q^{i}}},f_{q^{l}\partial_{q^{j}}}\}=-f_{[q^{k}\partial_{q^{i}},q^{l}\partial_{q^{j}}]}=-f_{q^{k}\frac{\partial q^{l}}{\partial q^{i}}\partial_{q^{j}}-q^{l}\frac{\partial q^{k}}{\partial q^{j}}\partial_{q^{i}}} =-\delta_{l,i} q^{k} f_{\partial_{q^{j}}}+\delta_{k,j}q^{l}f_{\partial_{{q}^{i}}}\\ &=-\delta_{l,i} q^{k} p_{j}+\delta_{k,j}q^{l}p_{i}.\tag{5}\label{5} \end{align} But also, by the Leibniz rule: \begin{align} \{q^{k}p_{i},q^{l}p_{j}\}&=\{q^{k},q^{l}\}p_{i}p_{j}+\{q^{k},p_{j}\}p_{i}q^{l}+\{p_{i},q^{l}\}q^{k}p_{j}+\{p_{i},p_{j}\}q^{k}q^{l}\\ &=\{q^{k},q^{l}\}p_{i}p_{j}+\delta_{k,j}p_{i}q^{l}-\delta_{i,l}q^{k}p_{j},\tag{6}\label{6} \end{align} using \eqref{1} and \eqref{4} in the last equality. Comparing \eqref{5} and \eqref{6} then gives $$\tag{7}\label{7} \{q^{k},q^{l}\}=0.$$ The equalities \eqref{1},\eqref{4} and \eqref{7} now show that $$\{\cdot,\cdot\}$$ is indeed the canonical Poisson bracket.