# Jacobi Identity for Poisson Bracket

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...

## 1 Answer

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 $$\operatorname{curl} \operatorname{grad} C = 0$$.

• 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! – LordVader007 Apr 14 at 0:44