# Proving Jacobi identity for Poisson bracket using antisymmetric matrix

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?

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.