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.


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