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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?

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The map $A$ is canonical if and only if $A^\star\omega^2=\omega^2$, where $\omega^2=dp_i\wedge dq^i$ that gives the sympletic structure to $R^{2n}$.

$$A^\star\omega^2=A^\star(dp_i\wedge dq^i)=(A^\star dp_i) \wedge (A^\star dq^i)=d(p_i\circ A)\wedge d(q^i\circ A)=dP_i\wedge dQ^i\quad (1)$$

Therefore, $A$ is canonical if and only if $$dP_i\wedge dQ^i=dp_i\wedge dq^i \qquad (2)$$

Eq. (2) is equivalent to $$\Lambda^T\Omega\Lambda=\Omega \qquad (3)$$ where $\Omega=\begin{pmatrix} 0\, & E\, \\ -E\, & 0\,\end{pmatrix}$ is the coefficient matrix of $\omega^2$, $\Lambda=\left(\frac{\partial x^i}{\partial x'^{i'}}\right)=\begin{pmatrix} p_P & p_Q \\ q_P & q_Q\end{pmatrix}$ is the coorinate transform matrix, and the submatrix $p_P$, for example, is short for the matrix $\Big(\frac{\partial p_i}{\partial P_j}\Big)$.

Eq. (3) is equivalent to $$\ I=\Omega^{-1}=\big(\Lambda^T \Omega \Lambda\big)^{-1}=\Lambda^{-1}I(\Lambda^{-1})^T\qquad (4)$$ where $I$ is the coefficient matrix of Poisson tensor (Poisson bivector).

Substituting $I=\begin{pmatrix} 0\, & -E\, \\ E\, & 0\,\end{pmatrix}, \ \Lambda^{-1}=\begin{pmatrix}P_p & P_q \\ Q_p &Q_q\end{pmatrix}$ into Eq. (4) yields $$\begin{pmatrix} 0\, & -E\, \\ E\, & 0\,\end{pmatrix}=\begin{pmatrix} P_q P_p^T-P_p P_q^T & P_q Q_p^T-P_p Q_q^T \\ Q_q P_p^T-Q_p P_q^T & Q_q Q_p^T-Q_p Q_q^T \end{pmatrix}\qquad (5)$$ which is actually the Poisson brackets of $P,\ Q$: $$\begin{align}P_q P_p^T-P_p P_q^T =&(P,P)_{p,q}=0\\[5pt]Q_q Q_p^T-Q_p Q_q^T =&(Q,Q)_{p,q}=0\qquad (6)\\[5pt]Q_q P_p^T-Q_p P_q^T =& (Q,P)_{p,q}=E\end{align}$$ Therefor, $A$ is canonical if and only if $(Q,P)_{p,q}=E,\ (P,P)_{p,q}=(Q,Q)_{p,q}=0$.

For arbitrary functions $F,\ H$, $$\begin{split}(F,H)_{p,q}=&H_pF_q^T-H_qF_p^T\\ =&(H_PP_p+H_QQ_p)(F_PP_q+F_QQ_q)^T-(H_PP_q+H_QQ_q)(F_PP_p+F_QQ_p)^T\\ =&H_P(P,P)_{p,q}F_P^T+H_P(Q,P)_{p,q}F_Q^T +H_Q(P,Q)_{p,q}F_P^T+H_Q(Q,Q)_{p,q}F_Q^T\end{split}$$ Thus $(F,H)_{p,q}=H_PF_Q^T-H_QF_P^T=(F,H)_{P,Q}$ if and only if Eqs. (6) hold. Accordingly, $A$ is canonical if and only if $(F,H)_{p,q}=(F,H)_{P,Q}$


P.S. The equations you shows in the question are essentially Eq. (3), i.e. $$\begin{pmatrix} 0\, & E\, \\ -E\, & 0\,\end{pmatrix}=\begin{pmatrix} p_P^T q_P-q_P^Tp_P & p_P^T q_Q-q_P^Tp_Q \\ p_Q^T q_P-q_Q^Tp_P & p_Q^T q_Q-q_Q^Tp_Q \end{pmatrix}\qquad(7)$$ It is easy to verify that $$\begin{pmatrix} p_P^T q_P-q_P^Tp_P & p_P^T q_Q-q_P^Tp_Q \\ p_Q^T q_P-q_Q^Tp_P & p_Q^T q_Q-q_Q^Tp_Q \end{pmatrix}\begin{pmatrix} P_q P_p^T-P_p P_q^T & P_q Q_p^T-P_p Q_q^T \\ Q_q P_p^T-Q_p P_q^T & Q_q Q_p^T-Q_p Q_q^T \end{pmatrix}=\begin{pmatrix} E\, & 0\, \\ 0\, & E\,\end{pmatrix}=E_{2n}$$

Therefor, to calculate the inverse of both sides of Eq. (7), the result should be Eq. (5), then the statement will be proved. However, it seems very complicated to calculate the inverse of the right side of Eq. (7) directly. Eq. (4) thus can be treated as the trick to calculate this inverse.

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It looks like a triviality: by definition symplectic form is preserved under canonical transformation and, thus, poisson bivector is preserved.

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  • $\begingroup$ Thanks! We need to prove if and only if, and the post is aimed at the oppisite direction too. $\endgroup$ – user25607 Apr 9 '14 at 13:27
  • $\begingroup$ In other direction it also trivial: if poisson bivector is non-degenerate then it comes from some symplectic structure, in coordinates it is given as inverse matrix of that bivector. In very basic terms this exercise means that diffeomorphism preserves matrix (of symplectic form) if and only if it preserves inverse matrix (of poisson bivector). $\endgroup$ – Alex Apr 9 '14 at 23:22
  • $\begingroup$ @Alex Can you write it down more explicitly as an answer? $\endgroup$ – Eden Harder Apr 10 '14 at 12:02

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