<s>[INST] Error with Differential Forms I'm learning a little bit about differential forms in my Complex Analysis course, and I'm not sure if I'm making an error in what follows. I wonder if anyone could offer any insight into this. Thank you.
So, the question is to show that $\alpha = \frac{xdy - ydx}{x^2 + y^2}$ is closed. To show this, I need to show that $d\alpha = 0$, which I can show as follows (according to my understanding).
Let $f = \frac{1}{x^2 + y^2}, g = xdy - y dx$. Then $d \alpha = f \wedge dg + df \wedge g$.
So, calculate $df = \frac{-1}{(x^2+y^2)^2}2x dx + \frac{-1}{(x^2 + y^2)^2}2y dy = \frac{-2}{(x^2+y^2)^2} (x dx + y dy)$ and $dg = 1 dx \wedge dy - 1 dy \wedge dx = 2 dx \wedge dy$.
Then, substituting these in the right places, we get
\begin{align*}
d \alpha &= \left(\frac{1}{x^2 + y^2}\right) \wedge (2 dx \wedge dy) + \left(\frac{-2}{(x^2+y^2)^2} (x dx + y dy)\right) \wedge (xdy - y dx)\\ 
&= \frac{2}{x^2+y^2} dx \wedge dy - \frac{2}{(x^2+y^2)^2}(x^2 dx \wedge dy - y^2 dy \wedge dx)\\ 
&= \frac{2}{x^2 + y^2} dx \wedge dy - \frac{2}{x^2 + y^2} dx \wedge dy\\ 
&= 0.
\end{align*}
This seems to work.
However, when I interchange $f$ and $g$, I seem to run into a problem, and I'm not sure why. Here's what happens. I keep $d \alpha = f \wedge dg + df \wedge g$. However, now I let $f = xdy - ydx$ and $g = \frac{1}{x^2 + y^2}$.
In this case, $df$ and $dg$ are no different (except of course for interchanging them), so we have $df = 2 dx \wedge dy$ and $dg = \frac{-2}{(x^2+y^2)^2} (x dx + y dy)$.
Now, substituting in, we get
\begin{align*}
d \alpha &= (xdy - ydx) \wedge \left(\frac{-2}{(x^2+y^2)^2} (x dx + y dy)\right) + (2 dx \wedge dy) \wedge \left(\frac{1}{x^2 + y^2}\right)\\ 
&= \frac{-2}{(x^2+y^2)^2} (xdy - ydx) \wedge (xdx + ydy) + \frac{2}{x^2 + y^2} dx \wedge dy\\ 
&= \frac{-2}{(x^2+y^2)^2}(x^2 dy \wedge dx - y^2 dx \wedge dy)+ \frac{2}{x^2 + y^2} dx \wedge dy\\ 
&= \frac{-2}{(x^2+y^2)^2}(-x^2 dx \wedge dy - y^2 dx \wedge dy)+ \frac{2}{x^2 + y^2} dx \wedge dy\\ 
&= \frac{-2}{(x^2+y^2)^2}((-x^2 -y^2)dx \wedge dy)+ \frac{2}{x^2 + y^2} dx \wedge dy\\ 
&= \frac{-2}{(x^2+y^2)^2}(-(x^2 +y^2)dx \wedge dy)+ \frac{2}{x^2 + y^2} dx \wedge dy\\ 
&= \frac{2}{x^2+y^2} dx \wedge dy + \frac{2}{x^2+y^2} dx \wedge dy\\ 
&= \frac{4}{x^2 + y^2}\\ 
&\neq 0.
\end{align*}
I've done this several times by hand and keep having this issue, so I assume the issue is not a simple calculation error on my part but a real conceptual misunderstanding about how to do these calculations.
It seems like just interchanging which function is $f$ and which is $g$ might change the sign I wind up with, but I don't see how it could change the result from 0 (which is what it's supposed to be) to something nonzero.
Any insights would be appreciated! Please remember that I have very little experience with differential forms.
 [/INST] If $\phi$ is a $p$-form and $\psi$ is a $q$-form, then $d(\phi\wedge\psi) = d\phi\wedge\psi + (-1)^p\phi\wedge d\psi$. In the second case, $f$ is a one-form and $g$ is a $0$-form, so $d\alpha = d(f\wedge g) = df\wedge g + (-1)^1f\wedge dg = df\wedge g \color{red}{-} f\wedge dg$.
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