Canonical transformation and Hamilton's equations I was trying to prove, that for a transformation to be Canonical, one must have a relationship:
$$
\left\{ Q_a,P_i \right\} = \delta_{ai}
$$
Where $Q_a = Q_a(p_i,q_i)$ and $P_a = P_a(p_i,q_i)$.
Now to do the proof I started with $\dot{Q_a}$:


*

*Chain rule and Hamilton's equation for initial coordinates $q_i,p_i$
$$
\dot{Q_a} = 
\frac{\partial Q_a}{\partial q_j} \dot{q_j} + 
\frac{\partial Q_a}{\partial p_j} \dot{p_j}
=
\frac{\partial Q_a}{\partial q_j} \frac{\partial H_a}{\partial p_j} -
\frac{\partial Q_a}{\partial p_j} \frac{\partial H_a}{\partial q_j} 
$$

*Then I apply chain rule for the Hamiltonian derivatives:
$$
\dot{Q_a} = 
\frac{\partial Q_a}{\partial q_j} 
\left( 
\frac{\partial H}{\partial Q_i} \frac{\partial Q_i}{\partial p_j} +
\frac{\partial H}{\partial P_i} \frac{\partial P_i}{\partial p_j} 
\right) -
\frac{\partial Q_a}{\partial p_j}
\left( 
\frac{\partial H}{\partial Q_i} \frac{\partial Q_i}{\partial q_j} +
\frac{\partial H}{\partial P_i} \frac{\partial P_i}{\partial q_j} 
\right)
$$

*Now reordering the terms yields us:
$$
\dot{Q_a} = 
\frac{\partial H}{\partial Q_i} \left\{ Q_a,Q_i \right\} +
\frac{\partial H}{\partial P_i} \left\{ Q_a,P_i \right\}
$$
Now here the problem, for the transformation from a coordinate system $(q_i,p_i)$ to a coordinate system $(Q_a(q_i,p_i), P_a(q_i,p_i))$ to be canonical we require:
$$\left\{ Q_a,P_i \right\} = \delta_{ai}$$
But why we have as well the following requirement, or is it just too obvious or true because of some property of any coordinate transformation?
$$\left\{ Q_a,Q_i \right\} = 0$$

The problem I am having is as follows. I agree, that the following two are true (if I used the covariant notation correctly):
$$
\left\{ q^i,q_j \right\}_{q,p} = 
\frac{\partial q^i}{\partial q^k}
\frac{\partial q_j}{\partial p_k}
-
\frac{\partial q^i}{\partial p^k}
\frac{\partial q_j}{\partial q_k}
= 0
$$
$$
\left\{ Q^i,Q_j \right\}_{Q,P} = 0
$$
But why its the case that the following is also true?
$$
\left\{ Q^i,Q_j \right\}_{q,p} = 0
$$
 A: A canonical transformation $(q^i,p_j) \to (Q^i,P_j)$ preserves the form of Hamilton's equations.
Similarly, a symplectic transformation$^1$ $(q^i,p_j) \to (Q^i,P_j)$ preserves the Poisson structure, aka. as a symplectomorphism. In other words, all the fundamental Poisson brackets (PB)
$$ \{  q^i,p_j \} ~=~ \delta^i_j, \qquad   \{q^i,q^j \}~=~0, \qquad \{  p_i,p_j \} ~=~ 0,\qquad  i,j \in\{1, \ldots, n\},$$
have the same form in the new coordinates
$$ \{  Q^i,P_j \} ~=~ \delta^i_j, \qquad   \{Q^i,Q^j \}~=~0, \qquad \{  P_i,P_j \} ~=~ 0,\qquad  i,j \in\{1, \ldots, n\}. $$
In particular, to answer OP's question(v2), the relations $\{Q^i,Q^j \}=0$ and $\{P_i,P_j\} = 0$ are only trivial if $n=1$, because of skewsymmetry of PB.
As is well-known, canonical and symplectic transformations are the same.
For a proof [at least in the case of restricted transformations, i.e. transformations without explicit time dependence], see e.g. Ref.1, which uses so-called symplectic notation. An important point is that the Jacobian matrix of a symplectic transformation must be a symplectic matrix.
References:


*

*H. Goldstein, Classical Mechanics, Section 9.4 in eds. 3 or Section 9.3 in eds. 2.


--
$^1$ In this answer we will for simplicity only discuss non-degenerate Poisson brackets in finite dimensions using globally defined coordinates.
