Calculating $d(xdx + z^2dy + xydz)$

Question

Inspired by this answer, I am trying to learn about differential forms. I am going through these notes were (on page 3) $d(xdx + z^2dy + xydz)$. I believe that the formula that I should use is $$ df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz. $$ So for example I get $$ d(xy) = ydx + xdy. $$ The notes seem to say that $d(xy) = dx + dy$ so I am a bit confused. What am I doing wrong?

Answer 1

It is a typo. It is indeed true that $d(xy)=ydx+xdy$. In short, to calculate with differential forms you take total derivatives paired with the wedge product. The wedge product has $dx_i \wedge dx_j= -dx_j\wedge dx_i $ hence $dx_i \wedge dx_i=0$ for all $i$. Then, you extend that rule naturally. In the same way, we could say complex numbers are just real numbers paired with some new element $i$ such that $i^2=-1$ then multiplication and addition proceeds as usual.

Answer 2

Hint. Let $f$ and $g$ be functions, $\phi$ and $\psi$ $1$-forms. Then $$ \begin{align*} (1)&\;\mathrm{d}(fg)=\mathrm{d}f\;g+f\;\mathrm{d}g\\ (2)&\;\mathrm{d}(f\phi)=\mathrm{d}f\wedge\phi\\ (3)&\;\mathrm{d}(\phi\wedge\psi)=\mathrm{d}\phi\wedge\psi-\phi\wedge\mathrm{d\psi}. \end{align*} $$

Notice that $\wedge$ denotes the wedge product.