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We define the so-called rigid body Poisson bracket as $\{F,G\}(\Pi) = -\Pi \cdot(\nabla{F} \times \nabla{G}) $. I want to prove Jacobi's identity, which is : $\{F,\{G,H\}\} + \{G,\{H,F\}\}+\{H,\{F,G\}\} =0.$

My (naive) approach is to go ahead and use the definition of the bracket mentioned above and write: $\{F,\{G,H\}\}= -\Pi\cdot(\nabla{F} \times \nabla{\{G,H\}})$. I can then write that $\nabla{\{G,H\}}= \nabla({-\Pi}\cdot(\nabla{G}\times\nabla{H})).$ I then use the vector identity $\nabla({A \cdot B}) = \nabla{A}\cdot B + \nabla{B}\cdot A$.

This gives me: $\nabla{\{G,H\}}= \nabla({-\Pi}\cdot(\nabla{G}\times\nabla{H})) = -\nabla{\Pi}\cdot(\nabla{G}\times\nabla{H})+ \nabla{(\nabla{G}\times\nabla{H)}}\cdot(-\Pi).$

I am almost tempted to say that the second term vanishes, so $\nabla{(\nabla{G}\times\nabla{H)}}\cdot(-\Pi) =0$. But I am not so sure we can do that. I lack a good argument for this part. I also know that $\nabla{\Pi} = I$, since the coordinates are $\Pi = (\Pi_1,\Pi_2,\Pi_3)$.

Can anyone help point me in the right direction? Also, I am trying to avoid using coordinates (I don't want to turn to the dark side of the force...)

Update: I am open to using coordinates (yes, I am a Sith lord now). But since I am not so familiar with this approach, can someone please enlighten me?

Update 2: The operation above $\nabla{(\nabla{G}\times\nabla{H)}}$ does not make sense unless we think of $\nabla$ as a Jacobian. This makes me even more confused...

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In coordinates $\Pi = (\Pi_1,\Pi_2,\Pi_3)$ we have $$\{\Pi_1,\Pi_2\} = -\Pi\cdot (\nabla \Pi_1 \times \nabla\Pi_2) = -\Pi \cdot ((1,0,0)\times (0,1,0))=-\Pi\cdot (0,0,1) = -\Pi_3.$$ Similarly $\{\Pi_2,\Pi_3\} = -\Pi_1$ and $\{\Pi_1,\Pi_3\} = \Pi_2$.

Since $\{-,-\}$ is a bi-vector field (e.g. by using Taylor's theorem), the coordinate expression $$\{F,G\} = \sum_{1\leq i < j\leq 3} \{\Pi_i,\Pi_j\}\bigg(\frac{\partial F}{\partial \Pi_i}\frac{\partial G}{\partial \Pi_j} - \frac{\partial F}{\partial \Pi_j}\frac{\partial G}{\partial \Pi_i}\bigg)$$ implies that the Jacobi identity (the left-hand side of which is a tri-vector field) reduces to $$\{\Pi_1,\{\Pi_2,\Pi_3\}\} + \{\Pi_2,\{\Pi_3,\Pi_1\}\} + \{\Pi_3,\{\Pi_1,\Pi_2\}\} = 0.\label{jac}\tag{1}$$

But each of these three terms vanishes individually, because the inner bracket is $\pm$ the first argument of the outer bracket (so apply bi-linearity and skew-symmetry).


Since the Poisson structure is linear (the structure coefficients $\{\Pi_i,\Pi_j\}$ are linear functions), the Jacobi identity is equivalent to the Jacobi identity for the Lie algebra, which is $\mathfrak{so}(3)$ with signs reversed here.


This is a Jacobian determinant Poisson structure with Casimir $C = -\frac{1}{2}(\Pi_1^2 + \Pi_2^2 + \Pi_3^2)$: $$\{F,G\} = \bigg|\frac{\partial(F,G,C)}{\partial(\Pi_1,\Pi_2,\Pi_3)}\bigg|, \qquad \text{or}\quad \{\Pi_i,\Pi_j\} = \epsilon_{ijk}\partial_k C.$$


Introducing $P = -\Pi$, the Jacobi identity $\eqref{jac}$ is equivalent to $$P\cdot \operatorname{curl} P = 0,$$ which is satisfied because a direct calculation gives $\operatorname{curl} P = 0$.

Alternatively since $P = \operatorname{grad} C$ it follows from the classical identity $\operatorname{curl} \operatorname{grad} C = 0$.

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  • $\begingroup$ Thank you for your input and time. I am learning more about using tensor notation. As a mathematician, I always like to find coordinate-free explanations/arguments, but this is very nice! $\endgroup$ – LordVader007 Apr 14 at 0:44

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