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Does anybody know how to show the Jacobi identity for the Nambu bracket in $\mathbb{R}^3$? The Nambu bracket with respect to $c \in \mathcal{F}(\mathbb{R}^3)$ is defined as $$\{F,G\}_c = \langle\nabla c , \nabla F \times\nabla G\rangle $$ where $F,G \in \mathcal{F}(\mathbb{R}^3).$

I don't know if this will help but, if one shows that the homomorphism $$ F \rightarrow \{F, \bullet\}_c = \langle\nabla \bullet , \nabla c \times \nabla F\rangle,$$ between $\mathcal{F}(\mathbb{R}^3)$ and the divergenceless vector fields $\nabla c \times \nabla F,$ preserve the lie algebra structure then it is done...

Thank you very much for the help!

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I think that your observation "the map $F\to\nabla c\times\nabla F$ translates the Nambu bracket into the Lie bracket" is indeed a necessary and sufficient condition for the Nambu to satisfy the Jacobi identity. – Giuseppe Apr 21 '12 at 17:14
@Rppacheco, it may be useful to give the bibliographic reference: Geometric Mechanics, by Darryl Holm, part II. – matgaio Apr 21 '12 at 18:28
Oh yes! Sorry about that. I think that this is also an exercise in Marsden, Ratiu's Introduction to Mechanics and Symmetry. – Rppacheco Apr 21 '12 at 19:06

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