# How the Poisson bracket transform when we change coordinates?

I'm studying the book Geometric Mechanics by Darryl D. Holm and there's one exercise in the book I'm not quite getting what has to be done. The same discussion the author makes in the book is made on page 7 of these solutions to some exercises provided by himself. This exercise, though, is not dealt with there.

The point is that he defines one kind of Possion bracket in the following way:

For any smooth functions $$F,H\in C^\infty(\mathbb{R}^3)$$ of coordinates $$\mathbf{X}\in \mathbb{R}^3$$ with volume element $$d^3X$$, the Nambu bracket $$\{F,H\}$$ defined by

$$-dC\wedge dF\wedge dH = \{F,H\}d^3X$$

is a Poisson bracket for any choice of distinguished smooth function $$C: \mathbb{R}^3\to \mathbb{R}$$.

So if I understood the definition, we pick one function $$C$$ to be a distinguished function, that is, $$C$$ will comute with any other function and then, the bracket $$\{F,H\}$$ in the coordinate system $$(X_1,X_2,X_3)$$ is obtained by expressing $$-dC\wedge dF\wedge dH$$ in terms of $$d^3X$$. Since $$-dC\wedge dF\wedge dH$$ is a multiple of $$d^3X$$ we set $$\{F,H\}$$ to be that scalar.

After that he applies this to geometric optics. Considering cartesian coordinates $$(Y_1,Y_2,Y_3)$$ the author does the following: he considers the distinguished function $$S^2 = Y_1^2-Y_2^2-Y_3^2$$ (which here is know as the skewness) and he introduces the coordinate system

$$Y_1^2 = S\cosh u, \quad Y_2= S\sinh u \cosh \psi, \quad Y_3 = S\sinh u \sinh \psi.$$

With this he presents the equality

$$\{F,H\}dY_1 \wedge dY_2\wedge dY_3 = -\{F,H\}_{\mathrm{Hyperb}}S^2 dS\wedge d\psi \wedge d \cosh u$$

Verify that the above equation transforms the $$\mathbb{R}^3$$ bracket from Cartesian to hyperboloidal coordinates, by using the definitions of $$Y_1,Y_2,Y_3$$ in terms of $$S,\psi,u$$.

I think I didn't get what has to be done. Indeed, I transformed the volume element $$dY_1\wedge dY_2\wedge dY_3$$ so that we have

$$dY_1\wedge dY_2\wedge dY_3 = -S^2\sinh u dS\wedge d\psi \wedge du,$$

thus we have

$$\{F,H\} dY_1\wedge dY_2\wedge dY_3 =-\{F,H\} S^2 dS\wedge d\psi \wedge d\cosh u,$$

notice, however, that by definition of the bracket, the $$\{F,H\}$$ term is just the bracket in cartesian coordinates. On the right hand side then it doesn't appear any $$\{F,H\}_{\mathrm{Hyperb}}$$ but rather, the same $$\{F,H\}$$ which was on the other side.

On the other hand, if we simply apply the definition using hyperboloidal coordinates we get

$$-dS^2\wedge dF\wedge dH = \left(-2S\dfrac{\partial F}{\partial \psi}\dfrac{\partial H}{\partial u}+2S\dfrac{\partial F}{\partial u}\dfrac{\partial H}{\partial \psi}\right)dS\wedge d\psi\wedge du,$$

so that in the end the bracket in these coordinates would be

$$\{F,H\}_{\mathrm{Hyperb}}=\left(-2S\dfrac{\partial F}{\partial \psi}\dfrac{\partial H}{\partial u}+2S\dfrac{\partial F}{\partial u}\dfrac{\partial H}{\partial \psi}\right).$$

In that case what is happening here? I think I didn't get what the author wants. What really has to be done here?

My point here is that simply transforming the differentials, and comparing with the expression found in the book I end up getting $$\{F,H\} = \{F,H\}_{\mathrm{Hyperb}}$$ which on my opinion is not right. So what has to be done more here?

How just transforming the differentials and using the definition of hyperboloidal coordinates in cartesian coordinates shows the transformation between the brackets?

Every Poisson bracket $\{\cdot\,,\cdot\}$ is associated with a bivector field $\Lambda$ (see, for instance, this book chapter 4.3):
$$\{f,\,h\}\,=\,\Lambda(\mathrm{d}f,\,\mathrm{d}h)\,.$$ What you need to do is to compute $\Lambda$ on the holonomic basis $\frac{\partial}{\partial Y_{j}}$ and then apply the rule of transformation for the bivector field $\Lambda$ when you pass from Cartesian coordinates $(Y_{1},Y_{2},Y_{3})$ to hyperboloidal ones.