Problem on deriving canonical transformation condition

Question

I'm trying to compute how a canonical transformation should be, given that preserve the symplectic form and trying to recover the condition on the Poisson Bracket. I then start with $$\omega=\stackrel{\scriptscriptstyle i=1...n}{\sum}dP_{i}\wedge dQ^{i}=\\$$ $$=\stackrel{\scriptscriptstyle i,j=1...n} {\sum}\left(\frac{\partial P_{i}}{\partial q^{j}}dq^{j}+\frac{\partial P_{i}}{\partial p_{j}}dp_{j}\right)\wedge\left(\frac{\partial Q^{i}}{\partial q^{j}}dq^{j}+\frac{\partial Q^{i}}{\partial p_{j}}dp_{j}\right)$$ And therefore I end with this $$=\stackrel{\scriptscriptstyle i,j=1...n} {\sum}\left\{ P_{i},Q^{i}\right\} dp_{j}\wedge dq^{j},$$ While I think it should be this $$=\stackrel{\scriptscriptstyle j=1...n} {\sum}\left\{ P_{j},Q^{j}\right\} dp_{j}\wedge dq^{j},$$ What did I do wrong?

Answer 1

Hint: The symplectic 2-form should stay on canonical form

$$ \omega ~=~ \sum_{i=1}^n \mathrm{d}p_i \wedge \mathrm{d}q^i~=~\sum_{i=1}^n \mathrm{d}P_i \wedge \mathrm{d}Q^i $$

under a canonical transformation (CT): $(q,p)\to (Q,P)$.

Answer 2

You seem to be summing over the same index $j$ in the transformation formulae for both $dP_i$ and $dQ^i$. That seems wrong to me. I would use a different index for each, so you produce a triple sum over $i,j,k$, and proceed on from there.