Proof of Liouville's Theorem

Question

The Wikipedia article on Canonical Transformations has a section on Liouville's Theorem. It makes the following argument:

$$J=\frac{\partial(\mathbf{Q},\mathbf{P})}{\partial(\mathbf{q},\mathbf{p})}$$ The Jacobian has the ``division'' property: $$J=\frac{\partial(\mathbf{Q},\mathbf{P})}{\partial(\mathbf{q},\mathbf{p})} = \frac{\partial(\mathbf{Q},\mathbf{P})}{\partial(\mathbf{q},\mathbf{P})}\Big/ \frac{\partial(\mathbf{q},\mathbf{p})}{\partial(\mathbf{q},\mathbf{P})}$$ Eliminating the repeated variable gives $$J= \frac{\partial(\mathbf{Q})}{\partial(\mathbf{q})}\Big/ \frac{\partial(\mathbf{p})}{\partial(\mathbf{P})}$$ Then, the conditions derived in the section above give $J=1.$

Why is it legal to "eliminate the repeated variables?" Basically, we have shown that $$J= \frac{\partial(\mathbf{Q})}{\partial(\mathbf{q})}\cdot \frac{\partial(\mathbf{P})}{\partial(\mathbf{p})}$$ Wouldn't applying this type of reasoning allow us to conclude $$J=\frac{\partial(Q_1)}{\partial(q_1)} \frac{\partial(Q_2)}{\partial(q_2)} \frac{\partial(Q_3)}{\partial(q_3)} ... \frac{\partial(P_1)}{\partial(p_1)} \frac{\partial(P_2)}{\partial(p_2)} \frac{\partial(P_3)}{\partial(p_3)}?$$

That is, that the determinant is just equal to the diagonal of the entries (something not true in general).

Answer 1

As pointed out in the linked question by Qmechanic, the expression should read $$J= \left(\frac{\partial(\mathbf{Q})}{\partial(\mathbf{q})}\right)_P\cdot \left(\frac{\partial(\mathbf{P})}{\partial(\mathbf{p})}\right)_q$$ where the subscript indicates which variable was held constant. Applying this type of reasoning will lead to the equation $$J= \frac{\partial Q_1}{\partial q_1} \frac{\partial Q_2}{\partial q_2} ... \frac{\partial P_1}{\partial p_1} \frac{\partial P_2}{\partial p_2}$$ but this is not equal to the product of the diagonal entries, because care needs to be taken as to which variables are held constant for each partial derivative.