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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.

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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 $\forall i,j,k\in\{1,\ldots,n\}: J(x^i,x^j,x^k)=0$.

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  • $\begingroup$ 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. $\endgroup$ Mar 6 at 16:02

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