I want to prove that the Poisson bracket from Hamiltonian mechanics satisfies the Jacobi identity and I want to do so using the matrix
$$(J^{ij})=\begin{pmatrix}0 & -I_2 \\ I_2 & 0\end{pmatrix},$$
where $I_2$ is the 2$\times$2 identity matrix. It is obvious to see that $J^{ij}=-J^{ji}$. The matrix $J$ can be used to write Hamilton's equations for position and conjugate momentum compactly as
$$\dot{x}=J\cdot\nabla H;\quad x:=(p_1,\cdots,p_n,q_1,\cdots,q_n).$$
For any two observables $A,B$, the Poisson bracket may be written as
$$\{A,B\}=J^{ij}\partial_iA\partial_jB.$$
Now I want to use this to prove the Jacobi identity
$$\{A,\{B,C\}\}=-\{B,\{C,A\}\}-\{C,\{A,B\}\}.$$
Using the formula above for writing the Poisson bracket in terms of $J$, we get: $$\{A,\{B,C\}\}=J^{ij}J^{kl}\partial_iA\partial_j(\partial_kB\partial_lC)=J^{ij}J^{kl}(\partial_iA\partial^2_{jk}B\partial_lC+\partial_iA\partial_kB\partial^2_{jl}C).$$
Now we use:
$$\partial_iA\partial^2_{jk}B=\partial_k(\partial_iA\partial_{j}B)-\partial^2_{ik}A\partial_jB,\\
\partial_iA\partial^2_{jl}C=\partial_l(\partial_iA\partial_{j}C)-\partial^2_{il}A\partial_jC.$$
Substituting this into our result above leaves us with:
$$J^{ij}J^{kl}\left(\partial_k(\partial_iA\partial_{j}B)\partial_lC
+\partial_l(\partial_iA\partial_{j}C)\partial_kB\right)
-J^{ij}J^{kl}\left(\partial^2_{ik}A\partial_jB\partial_lC
+\partial^2_{il}A\partial_kB\partial_jC\right).$$
The first term is easily transformed into:
$$J^{kl}\partial_k(J^{ij}\partial_iA\partial_{j}B)\partial_lC
+J^{kl}\partial_kB\partial_l(J^{ij}\partial_iA\partial_{j}C)=\{\{A,B\},C\}+\{B,\{A,C\}\}\\
=-\{B,\{C,A\}\}-\{C,\{A,B\}\}.$$
But from here on, I have a bit of a problem to show, that the other term vanishes:
$$J^{ij}J^{kl}\left(\partial^2_{ik}A\partial_jB\partial_lC
+\partial^2_{il}A\partial_kB\partial_jC\right)=0.$$
First, I pull $\partial^2_{ik}A$ out of the bracket by switching the indices $l,k$ in the second term. Using $J^{kl}=-J^{lk}$, we obtain:
$$J^{ij}\partial^2_{ik}A\left(J^{kl}\partial_jB\partial_lC
-J^{lk}\partial_kB\partial_lC\right).$$
Notice that $\partial_kB$ becomes $\partial_lB$. Now we switch $j,k$, so we can pull $\partial_jB\partial_lC$ out of the bracket as well. We are left with:
$$\partial^2_{ik}A\partial_jB\partial_lC\left(J^{ij}J^{kl}
-J^{ik}J^{lj}\right).$$
How do I now show that this vanishes? Any hints or ideas?
Best Answer
When you pull $\partial^2_{ik}A$ out of the brackets you actually get $$J^{ij}\partial^2_{ik}A\big(J^{kl}\partial_j B\partial_l C + J^{lk} \partial_l B \partial_j C\big).$$ Then you use $J^{lk} = -J^{kl}$ and switch $l \rightleftarrows j$ in the second term to get $$\partial^2_{ik}A \partial_j B\partial_l C\big(J^{ij}J^{kl}-J^{il}J^{kj}\big).$$ Finally use that $\partial_i$ and $\partial_k$ commute and do $i \rightleftarrows k$ to one of the terms to conclude the result.