Relation of the Hamiltonian vector fields and the Lie bracket I have to proof that
$$X_{\{f,g\}}=[X_f,X_g]$$
Where $X_f=\{f, .\}$ is the Hamiltonian vector field acting on $C^\infty(\mathbb{R}^{2n})$ and $[ \ ,\ ]$ is the commutator (or the Lie bracket).
This exercise can be done just doing the computations by hand, but that is not the way I would like to do it, instead of that I saw that $X_{q_j}=\dfrac{\partial}{\partial p_j}$ nd $X_{p_j}=-\dfrac{\partial}{\partial q_j}$. With this the Hamiltonin vector field $X_f$ take the form
$$X_f=\sum_{j=1}^n\dfrac{\partial f}{\partial q_j}X_{q_j}+\dfrac{\partial f}{\partial p_j}X_{p_j}$$
And this can be simpliyed as
$$X_f=\sum_{j=1}^m \dfrac{\partial f}{\partial e_j}X_{e_j}$$
Where $e_j=q_j$ and $e_{n+j}=p_j$ for $j=1,\ldots,n$. At this point I just wrote the vector field $X_f$ in a simpler form, but with this we can compute better the commutator:
$$[X_f,X_g]=\left[\sum_{j=1}^m \dfrac{\partial f}{\partial e_j}X_{e_j},\sum_{k=1}^m \dfrac{\partial g}{\partial e_k}X_{e_k}\right]=\sum_{j=1}^m\left[\sum_{k=1}^m \dfrac{\partial f}{\partial e_k}\dfrac{\partial^2g}{\partial e_k\partial e_j}-\dfrac{\partial g}{\partial e_k}\dfrac{\partial^2f}{\partial e_k\partial e_j}\right]X_{e_j}$$
(I used the representation of the conmutator as a vector field, the elments between bracket are the coefficient of this vector field) and we also have
$$X_{\{f,g\}}=\sum_{j=1}^m\dfrac{\partial\{f,g\}}{\partial e_j}X_{e_j}$$
So to prove our equlity we just need to verify that the coefficients under bracket are equal to $\dfrac{\partial\{f,g\}}{\partial e_j}$.
Writing the Poisson bracket as
$$\{f,g\}=\sum_{j=1}^m\dfrac{\partial f}{\partial e_j}X_{e_j}(g)$$
And use the product rule of the derivative, the problem is when I do that I have
$$\sum_{j=1}^m\dfrac{\partial^2f}{\partial e_k\partial e_j}X_{e_j}(g)+\dfrac{\partial f}{\partial e_j}\dfrac{\partial}{\partial e_k}X_{e_j}(g)$$
That is very similar to the expresion for the conmutator except that in the expresion for the commutator it appears a minus sing and I don't know what I have to do for the two expression being equeals.
 A: I can't spot your mistake but hopefully this helps.
Define
$$
\dfrac{\partial}{\partial b_k} :=
\begin{cases} 
      \dfrac{\partial}{\partial p_k} &, k\leq n \\
      -\dfrac{\partial}{\partial q_{k-n}} &, k >n .
\end{cases}
$$
Now $$X_f=(X_f^1,\dots,X_f^{2n})=(\dfrac{\partial f}{\partial p_1},\dots,\dfrac{\partial f}{\partial p_n},-\dfrac{\partial f}{\partial q_1},\dots,-\dfrac{\partial f}{\partial q_n})=(\dfrac{\partial f}{\partial b_1},\dots,\dfrac{\partial f}{\partial b_{2n}}) \tag{*}\label{*}$$.
Let $\xi=(q_1,\dots,q_n,p_1,\dots,p_n)$. The $k$th element of the commutator $[X_f,X_g]$ is by definition
$$
[X_f,X_g]_k = \sum_{i=1}^{2n}(X_f^i \dfrac{\partial X_g^k}{\partial \xi_i}-X_f^i \dfrac{\partial X_g^k}{\partial \xi_i}).
$$
Splitting the sum $\sum_{i=1}^{2n} = \sum_{i=1}^{n} + \sum_{i=n+1}^{2n}$and using $\eqref{*}$ gives
$$
[X_f,X_g]_k = \sum_{i=1}^{n}\bigg((\dfrac{\partial f}{\partial p_i})\dfrac{\partial^2 g }{\partial q_i\partial b_k}+(-\dfrac{\partial f}{\partial q_i})\dfrac{\partial^2 g }{\partial p_i\partial b_k}-(\dfrac{\partial g}{\partial p_i})\dfrac{\partial^2 f }{\partial q_i\partial b_k}-(-\dfrac{\partial g}{\partial q_i})\dfrac{\partial^2 f }{\partial p_i\partial b_k}\bigg)\\
= \sum_{i=1}^{n}\bigg(\dfrac{\partial f}{\partial p_i}\dfrac{\partial^2 g }{\partial b_k\partial q_i}+\dfrac{\partial g}{\partial q_i}\dfrac{\partial^2 f }{\partial b_k\partial p_i}-\dfrac{\partial f}{\partial q_i}\dfrac{\partial^2 g }{\partial b_k\partial p_i}-\dfrac{\partial g}{\partial p_i}\dfrac{\partial^2 f }{\partial b_k\partial q_i}\bigg),
$$
where in the last step we changed order of the four terms and order of partial derivates of $g$. Product rule yields
$$
[X_f,X_g]_k = \dfrac{\partial}{\partial b_k} \sum_{i=1}^{n} \bigg(\dfrac{\partial f}{\partial p_i}\dfrac{\partial g}{\partial q_i}-\dfrac{\partial f}{\partial q_i}\dfrac{\partial g}{\partial p_i}\bigg) \\
=  \dfrac{\partial}{\partial b_k} \{f,g\}
= X_{\{f,g\}}^k.
$$
The element $k$ was arbitrary, and hence
$$[X_f,X_g] = X_{\{f,g\}}.$$
