# A question about canonical transformation

I have posted this question in math.stackexchange before with no answer till now. It may be more suitable to post here.

There is a problem in Arnold's Mathematical Methods of Classical Mechanics which says that:

Show that the map $A: \mathbb{R}^{2n} \rightarrow \mathbb{R}^{2n}$ sending $(p, q) \rightarrow (P(p,q), Q(p,q))$ is canonical(p206) if and only if the Poisson brakets of any two functions in the variables $(p,q)$ and $(P,Q)$ coincide: $$(F,H)_{p,q} = \frac{\partial H}{\partial p} \frac{\partial F}{\partial q} - \frac{\partial H}{\partial q} \frac{\partial F}{\partial p} = \frac{\partial H}{\partial P} \frac{\partial F}{\partial Q} - \frac{\partial H}{\partial Q} \frac{\partial F}{\partial P} = (F,H)_{P,Q}.$$

I cannot solve this problem and think about it as following: From $(F,H)_{p,q} = (F,H)_{P,Q}$ I can induce that $$\sum_i \det\left( \frac{\partial(P_j, P_k)}{\partial(p_i, q_i)} \right) = \sum_i \det\left( \frac{\partial(Q_j, Q_k)}{\partial(p_i, q_i)} \right) = 0, \sum_i \det\left( \frac{\partial(P_j, Q_k)}{\partial(p_i, q_i)} \right) = \delta_{j,k},$$ and $$\sum_i \det\left( \frac{\partial(p_j, p_k)}{\partial(P_i, Q_i)} \right) = \sum_i \det\left( \frac{\partial(q_j, q_k)}{\partial(P_i, Q_i)} \right) = 0, \sum_i \det\left( \frac{\partial(p_j, q_k)}{\partial(P_i, Q_i)} \right) = \delta_{j,k}.$$ But in the other hand, to induce $dP\wedge dQ = dp \wedge dq$ I need that $$\sum_i \det\left( \frac{\partial(p_i, q_i)}{\partial(P_j, P_k)} \right) = \sum_i \det\left( \frac{\partial(p_i, q_i)} {\partial(Q_j, Q_k)} \right) = 0, \sum_i \det\left( \frac{\partial(p_i, p_i)}{\partial(P_j, Q_k)} \right) = \delta_{j,k},$$ or $$\sum_i \det\left( \frac{\partial(P_i, Q_i)}{\partial(p_j, p_k)} \right) = \sum_i \det\left( \frac{\partial(P_i, Q_i)}{\partial(q_j, q_k)} \right) = 0, \sum_i \det\left( \frac{\partial(P_i, Q_i)}{\partial(p_j, q_k)} \right) = \delta_{j,k}.$$ Is there something wrong in the above reasoning? Can you show me how to solve this problem?

1. OP's sought-for proof becomes shorter if we first introduce some symplectic notation. Consider therefore a symplectic manifold $(M;\omega)$. In a local chart $U\subseteq \mathbb{R}^{2n}$, the symplectic two-form reads $$\tag{1} \omega~=~\frac{1}{2} \omega_{IJ}~\mathrm{d}x^I \wedge \mathrm{d}x^J ,$$ and the corresponding Poisson bi-vector $$\tag{2} \pi~=~\frac{1}{2} \pi^{IJ}~\partial_I \wedge \partial_J,$$ where $$\tag{3} \omega_{IJ} ~\pi^{JK}~=~\delta_I^K.$$ The Poisson bracket reads $$\tag{4} \{f,g\}_{PB} ~=~ (\partial_I f) \pi^{IJ} (\partial_J g).$$ See also e.g. this and this Phys.SE posts.

2. In a local Darboux chart $U\subseteq \mathbb{R}^{2n}$, the symplectic two-form reads $$\tag{4} \omega~=~\frac{1}{2}(J^{-1})_{IK}~\mathrm{d}x^I \wedge \mathrm{d}x^K~=~-\frac{1}{2}J^{IK}~\mathrm{d}x^I \wedge \mathrm{d}x^K,$$ and the corresponding Poisson bracket reads $$\tag{5} \{x^I,x^K\}_{PB}~=~J^{IK},$$ where $$\tag{6} J ~=~\begin{bmatrix} 0_n & I_n \cr -I_n & 0_n \end{bmatrix}, \qquad J^2~=~-I_{2n}.$$

3. Now let us return to OP's question with some hints. The definition of a canonical transformation$^1$ $\phi:M\to M$ given in Ref. 1 is equivalent to the notion of a symplectomorphism $$\tag{7} \phi^{\ast}\omega ~=~\omega.$$ Let us choose a local Darboux neighboorhood. Let the Jacobi matrix for the symplectomorphism be $$\tag{8} M^I{}_J ~:=~\frac{\partial \phi^I(x)}{\partial x^J}.$$

4. Show that the condition (7) implies that the matrix $M$ is a symplectic matrix $$\tag{9} M^t J^{-1} M~=~ J^{-1},$$ or equivalently, $$\tag{10} M J M^t~=~ J.$$

5. Show that the symplectic condition (10) is equivalent to the preservation of the Poisson bracket $$\tag{11} \{ \phi^I(x),\phi^J(x)\}_{PB}~=~\{x^I,x^J\}_{PB}.$$

References:

1. V.I. Arnold, Mathematical methods of Classical Mechanics, 2nd eds., 1989; p. 206.

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$^1$ Note that different authors give different definitions of a canonical transformation, cf. e.g. this Phys.SE post.