# Poisson bracket of function of coordinates in terms of canonical brackets

Suppose that we work with a general Poisson structure where we are given the Poisson brackets of the individual coordinates. In practise, how would we be able to use these to determine the Poisson bracket of some more complicated function of the coordinates? To illustrate the question let us work with coordinates $$(x, y, z ,t)$$. Suppose we are given the relations $$\{ x, y \} = a, \{z, t \} = b$$ for some constants $$a$$ and $$b$$ and all brackets other than permutations of the above are taken to be zero. Define the functions $$F = x^2 + y^2 + z^2 + t^2$$ and $$G = e^{(x-y)^2} + e^{(z-t)^2}$$ - how would we go about calculating $$\{F, G \}$$?

Since we are only given the abstract definition of the Poisson bracket, we cannot use the standard approach that applies to functions on phase space and must only proceed from the axioms (bilinearity, skew symmetry, Jacobi identity and the Leibniz property). In the case that the given functions are polynomials in the coordinates I imagine we could repeatedly apply the Leibniz rule to eventually express $$\{ F, G \}$$ in terms of the given brackets but I can't see how to proceed in the more general case other than attempting to express everything in terms of power series - this definitely doesn't seem like the most elegant approach to take...

• If you do the power series, you'll see the expression simplify so it is not as inelegant as it might seem. May 18, 2020 at 18:43
• @AHusain You were right, it seemed a bit clunky at first but after following everything through using the suggestion in the answer below, it all worked out surprisingly nicely! May 19, 2020 at 0:24

The axioms of the Poisson bracket imply that for $$f\in C^\infty(M)$$, the mapping $$\lbrace f,\cdot\rbrace:C^\infty(M)\to C^\infty(M)$$ is a vector field (i.e. a derivation on the ring of $$C^\infty(M)$$ functions). Therefore like all vector fields, it satisfies $$\lbrace f,\cdot \rbrace = \sum_j\lbrace f,x^j\rbrace \frac{\partial}{\partial x^j}$$ (this is a standard result about vector fields, that essentially follows from Taylor's theorem). So $$\lbrace f,g\rbrace = \sum_i\lbrace f,x^j\rbrace \frac{\partial g}{\partial x^j}.$$ However since $$\lbrace\cdot,\cdot\rbrace$$ is antisymmetric, the same applies to $$f$$, so $$\lbrace f,g\rbrace = \sum_{ij}\lbrace x^i,x^j\rbrace \frac{\partial f}{\partial x^i}\frac{\partial g}{\partial x^j}.$$
Proof: Define the vector field $$\mathbf{X}_g$$ such that $$\{ f, g \} = \mathbf{X}_g f$$. Then the $$i$$th component of $$\mathbf{X}_g$$ is given by $$(\mathbf{X}_g)_i = \mathbf{X}_g x_i = \{ x_i, g \} = - \{ g, x_i \} = - \mathbf{X}_{x_i} g$$.
Proceeding in a similar fashion, we can show that the $$j$$th component of $$\mathbf{X}_{x_i}$$ is given by $$( \mathbf{X}_{x_i} )_j = \mathbf{X}_{x_i} x_j = \{ x_j, x_i \} = - \{ x_i, x_j \}$$ and hence $$\mathbf{X}_{x_i} = - \sum_{j} \{x_i, x_j \} \dfrac{\partial}{\partial x_j}.$$
Substituting this into the back into the previous result for $$\mathbf{X}_g$$ tells us that $$\mathbf{X}_g = \sum_{i,j} \{x_i, x_j \} \dfrac{\partial g}{\partial x_j} \dfrac{\partial}{\partial x_i}$$ and finally we conclude that $$\{f, g \} = \mathbf{X}_g f = \sum_{i,j} \dfrac{\partial f}{\partial x_i} \{ x_i, x_j \} \dfrac{\partial g}{\partial x_j}.$$