# Symplectic map and Poisson bracket notation

I'm having trouble trying to work with a certain notation.

Def: A diffeomorphism $$\Phi$$ of $$\mathbb{R}^{2n}$$ is symplectic if, for all $$f,g\in C^{\infty}(\mathbb{R}^{2n})$$, $$\{f\circ\Phi,g\circ\Phi\}=\{f,g\}\circ\Phi.$$

Now I found an example in a book where they consider a the map $$\Phi$$:$$\mathbb{R}^{2n}\rightarrow\mathbb{R}^n$$, $$(q^i,p_i)\mapsto x^i$$, defined by: $$x^i=\mathrm{e}^{p_i-\frac{1}{2}\sum_{j=1}^na_{ij}q^j}$$ then they write $$\{x^i\circ\Phi,x^j\circ\Phi\}_{\mathbb{R}^{2n}}=\{\mathrm{e}^{p_i-\frac{1}{2}\sum_{k=1}^na_{ik}q^k},\mathrm{e}^{p_i-\frac{1}{2}\sum_{l=1}^na_{il}q^l}\}_{\mathbb{R}^{2n}},$$ then they go on with the calculation. But then with the composition notation this means that $$x^i(\Phi(q,p))$$ but shouldn't this be $$x^i(x^i)$$ since $$\Phi(q,p)=x$$?

Then there is another example where $$f=\cos q$$, and $$g=\sin p$$, calculating the Hamilton flow generated $$f$$ and $$g$$ is $$\Phi^{f}_t=(q_0,t\cos (q_0)+p_0)$$ and $$\Phi^{g}_t=(q_0-t\sin(p_0),p_0)$$ according to the above definition $$f\circ\Phi=\cos(q_0)$$ and $$g\circ\Phi=\sin(p_0)$$ would one then calculate the Poisson bracket using the variables $$q_0,p_0$$?

Some clarification on this matter would be greatly appreciated! And some simple line by line calculations would also be helpful! thanks guys!

• I think when they write $x^i \circ \Phi$, the $x^i$ means the function $\Bbb{R}^n \to \Bbb{R}$ which just takes the $i^\mathrm{th}$ coordinate. Then this composition makes sense as a map $\Bbb{R}^{2n} \to \Bbb{R}$. – Nick Jun 9 at 14:12