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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!

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    $\begingroup$ 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 $\Bbb{R}^{2n} \to \Bbb{R}$. $\endgroup$ – Nick Jun 9 at 14:12

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