[proofplan] We verify the identity $[V_f, V_g] = -V_{\{f,g\}}$ by direct computation in coordinates. The Hamiltonian vector field generated by $f$ has components $(V_f)_i = \sum_k J_{ik} \, \partial_k f$. We expand the $i$-th component of the Lie bracket $[V_f, V_g]_i = \sum_j \bigl((V_f)_j \, \partial_j (V_g)_i - (V_g)_j \, \partial_j (V_f)_i\bigr)$, substitute the definitions, and use the constancy of $J$ and the symmetry of mixed partial derivatives to identify the result as $-(V_{\{f,g\}})_i = -\sum_k J_{ik} \, \partial_k \{f,g\}$. [/proofplan]

[step:Write out the $i$-th component of the Lie bracket using the definition of $V_f$ and $V_g$]Recall that the Hamiltonian vector field generated by a smooth function $f: M \to \mathbb{R}$ is $V_f = J \, \partial f / \partial x$, so the $i$-th component is \begin{align*} (V_f)_i = \sum_{k=1}^{2n} J_{ik} \, \partial_k f. \end{align*} The Lie bracket of two vector fields has $i$-th component \begin{align*} [V_f, V_g]_i = \sum_{j=1}^{2n} \bigl( (V_f)_j \, \partial_j (V_g)_i - (V_g)_j \, \partial_j (V_f)_i \bigr). \end{align*} Substituting $(V_g)_i = \sum_k J_{ik} \, \partial_k g$ and $(V_f)_i = \sum_k J_{ik} \, \partial_k f$, and using the fact that $J$ is a constant matrix so $\partial_j J_{ik} = 0$, the Lie bracket becomes \begin{align*} [V_f, V_g]_i = \sum_{j=1}^{2n} \sum_{k=1}^{2n} J_{ik} \bigl( (V_f)_j \, \partial_j \partial_k g - (V_g)_j \, \partial_j \partial_k f \bigr). \end{align*}[/step]

[guided]The Lie bracket $[V_f, V_g]$ measures the failure of the flows of $V_f$ and $V_g$ to commute. For general vector fields, expanding the Lie bracket involves derivatives of both the coefficient functions and the base coordinates. Here, the key simplification is that $J$ is a constant matrix, so differentiating $(V_f)_i = \sum_k J_{ik} \, \partial_k f$ with respect to $x_j$ produces only $\sum_k J_{ik} \, \partial_j \partial_k f$ --- the matrix $J$ passes through the derivative. This is why the computation reduces to expressions involving only second derivatives of $f$ and $g$. Explicitly, we compute \begin{align*} \partial_j (V_g)_i = \partial_j \Bigl( \sum_{k=1}^{2n} J_{ik} \, \partial_k g \Bigr) = \sum_{k=1}^{2n} J_{ik} \, \partial_j \partial_k g, \end{align*} and similarly $\partial_j (V_f)_i = \sum_k J_{ik} \, \partial_j \partial_k f$. Substituting into the Lie bracket formula gives \begin{align*} [V_f, V_g]_i = \sum_{j=1}^{2n} \sum_{k=1}^{2n} J_{ik} \bigl( (V_f)_j \, \partial_j \partial_k g - (V_g)_j \, \partial_j \partial_k f \bigr). \end{align*}[/guided]

[step:Expand $(V_f)_j$ and $(V_g)_j$ and rewrite using the Poisson bracket] Substituting $(V_f)_j = \sum_l J_{jl} \, \partial_l f$ into the first sum: \begin{align*} \sum_{j,k} J_{ik} (V_f)_j \, \partial_j \partial_k g = \sum_{j,k,l} J_{ik} \, J_{jl} \, \partial_l f \, \partial_j \partial_k g. \end{align*} Similarly, substituting $(V_g)_j = \sum_l J_{jl} \, \partial_l g$ into the second sum: \begin{align*} \sum_{j,k} J_{ik} (V_g)_j \, \partial_j \partial_k f = \sum_{j,k,l} J_{ik} \, J_{jl} \, \partial_l g \, \partial_j \partial_k f. \end{align*} Therefore \begin{align*} [V_f, V_g]_i = \sum_{j,k,l} J_{ik} \, J_{jl} \bigl( \partial_l f \, \partial_j \partial_k g - \partial_l g \, \partial_j \partial_k f \bigr). \end{align*} [/step]

[step:Compute $-(V_{\{f,g\}})_i$ and show it equals the Lie bracket] The target is $-(V_{\{f,g\}})_i = -\sum_k J_{ik} \, \partial_k \{f,g\}$. We compute $\partial_k \{f,g\}$ using the Poisson bracket definition $\{f,g\} = \sum_{j,l} J_{jl} \, \partial_j f \, \partial_l g$: \begin{align*} \partial_k \{f,g\} = \sum_{j,l} J_{jl} \bigl( \partial_k \partial_j f \cdot \partial_l g + \partial_j f \cdot \partial_k \partial_l g \bigr). \end{align*} Therefore \begin{align*} -(V_{\{f,g\}})_i &= -\sum_{k} J_{ik} \sum_{j,l} J_{jl} \bigl( \partial_k \partial_j f \cdot \partial_l g + \partial_j f \cdot \partial_k \partial_l g \bigr) \\ &= -\sum_{j,k,l} J_{ik} \, J_{jl} \, \partial_l g \, \partial_k \partial_j f - \sum_{j,k,l} J_{ik} \, J_{jl} \, \partial_j f \, \partial_k \partial_l g. \end{align*} By symmetry of mixed partial derivatives ($\partial_k \partial_j f = \partial_j \partial_k f$ and $\partial_k \partial_l g = \partial_l \partial_k g$), and relabelling the dummy index $l \leftrightarrow j$ in the second sum (noting that $J_{jl}$ becomes $J_{lj} = -J_{jl}$ by the antisymmetry $J^\top = -J$), the second sum becomes \begin{align*} -\sum_{j,k,l} J_{ik} \, J_{jl} \, \partial_j f \, \partial_k \partial_l g = -\sum_{j,k,l} J_{ik} \, J_{lj} \, \partial_l f \, \partial_k \partial_j g = \sum_{j,k,l} J_{ik} \, J_{jl} \, \partial_l f \, \partial_k \partial_j g. \end{align*} Combining both terms: \begin{align*} -(V_{\{f,g\}})_i &= -\sum_{j,k,l} J_{ik} \, J_{jl} \, \partial_l g \, \partial_j \partial_k f + \sum_{j,k,l} J_{ik} \, J_{jl} \, \partial_l f \, \partial_j \partial_k g \\ &= \sum_{j,k,l} J_{ik} \, J_{jl} \bigl( \partial_l f \, \partial_j \partial_k g - \partial_l g \, \partial_j \partial_k f \bigr), \end{align*} where in the last line we used $\partial_k \partial_j g = \partial_j \partial_k g$. This is identical to the expression obtained for $[V_f, V_g]_i$, completing the proof that $[V_f, V_g] = -V_{\{f,g\}}$. [/step]