Bracket satisfies Jacobi-identity?

Question

For all $F,G \in C^{\infty}(\mathbb{R}^2)$, define the following bracket:

$$\lbrace F, G \rbrace= y\Big(\frac{\partial F}{\partial x}\frac{\partial G}{\partial y}-\frac{\partial F}{\partial y}\frac{\partial G}{\partial x}\Big)$$

I have to verify that this defines a Poisson-bracket. I've already checked that it is anti-symmetric, linear in both variables and that it satisfies the Leibniz-rule. So I still have to check that the Jacobi-identity is satisfied. I started calculating all the brackets but I got stuck since the terms don't cancel out. Is there a direct way to see that the Jacobi-identity holds?

Answer 1

You definitely made a mistake. There are several ways to see this. One is to only consider the terms in the triple bracket linear in y since the combinatorics of the ones bilinear in y are identical to the flat metric one. The ones linear in y, $$ y\Bigl(H_x(F_xG_y-F_yG_x)+ F_x(G_xH_y-G_yH_x)+G_x(H_xF_y-H_yF_x) \Bigr) =0, $$ have the proper pairwise cancellations.

But it is even more trivial than that. You may absorb the loose y into the $\partial y$ denominators, converting them to $\partial \ln y$, or, equivalently, changing variables from y to $z\equiv \ln y$. The PB is then the conventional form in x and z, whose Jacobi compliance you have already checked.