Jacobi identity for Poisson bracket in local coordinates

Suppose a bivector field $$\pi^{ij}$$ such that $$\pi^{ij}=-\pi^{ji}$$, $$\pi^{ij}\partial_{i}f\partial_{k}g=\{f, g\}$$ defines a Poisson bracket $$\{,\}$$ on a smooth manifold (Einstein's summation is implied). The task is to find the condition on $$\pi^{ij}$$ following from the Jacobi identity for Poisson brackets: $$\{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\}=0$$ Note that this is the inverse problem to that which I was able to find asked before, namely from proving Jacobi identity for Poisson brackets. I know that the answer is supposed to be $$\sum_{\langle(ikl)\rangle}\pi^{ij}\frac{\partial \pi^{kl}}{\partial x^{j}}=0$$ where $$\langle (ijk) \rangle$$ denotes all cyclic permutations of indices $$i,j,k$$, but I can't find the way to it. My attempt to solution was the following:

$$\newcommand{\p}{\partial}$$ $$\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}$$ $$\newcommand{\bra}{\langle}$$ $$\newcommand{\ket}{\rangle}$$

The Jacobi identity translates to $$\pi^{ij}\p_{i}f\p_{j}(\pi^{kl}\p_{k}g\p_{l}h)+\text{cyclic permutations of }f,g,h=0$$ $$\p_{j}(\pi^{kl}\p_{k}g\p_{l}h)=\p_{j}\pi^{kl}\p_{k}g\p_{l}h+\pi^{kl}\p_{j}(\p_{k}g\p_{l}h)$$ Denote $$\p_{i}f:=f_{i}$$. Then for arbitrary $$f_{i}, g_{i},h_{i}$$ $$\pi^{ij}\frac{\p\pi^{kl}}{\p x^j}\sum_{\bra(fgh)\ket}f_{i} g_{k}h_{l}+\pi^{ij}\pi^{kl}\sum_{\bra (fgh)\ket}f_{i}\p_{j}(g_{k}h_{l})=0$$ We can rewrite summation over $$\bra(fgh)\ket$$ as summation over $$\bra (ikl)\ket$$. I guess, the sum in right term should somehow be symmetric w.r.t. transpositions $$k\leftrightarrow l$$ or $$i\leftrightarrow j$$ and always disappear, being contracted with skew-symmetric $$\pi^{kl}$$, leaving only the left term, the equivalence to zero of which, given the fact $$f_{i},g_{i},h_{i}$$ are arbitrary functions, implies that the differential expression in it is zero. But I can't see why the sum in the right term is symmetric. I tried substituting $$f=g=h$$, which, of course, kills the right term, but then the equality does not imply that the differential expression in the left term is zero, it would suffice for it to be completely skew-symmetric.

Hints:

1. Let $$\pi=\pi^{ij}\partial_i\wedge\partial_j$$ be an antisymmetric bivector field [i.e. a $$(2,0)$$ contravariant tensor field], and $$\{\cdot,\cdot\}:C^{\infty}(M)\times C^{\infty}(M)\to C^{\infty}(M)$$ the corresponding antisymmetric bracket, which does not necessarily satisfy the Jacobi identity.

2. Recall that in a local coordinate system $$(x^1,\ldots,x^n)$$ we have $$\{x^i,x^j\}=\pi^{ij}$$.

3. Define the Jacobiator $$J(f,g,h) ~:=~ \sum_{{\rm cycl.} f,g,h} \{f,\{g,h\}\}.$$

4. Show that $$J$$ is a totally antisymmetric trivector field [i.e. a $$(3,0)$$ contravariant tensor field].

5. Show that the tensor $$J=0$$ vanishes identically iff in every local coordinate system $$(x^1,\ldots,x^n)$$ we have $$\forall i,j,k\in\{1,\ldots,n\}: J(x^i,x^j,x^k)=0$$.

• I've solved the problem in a different way (just writing down all the expressions in the right term explicitly, they really do cancel out), but thank you for this alternative answer. Mar 6 at 16:02