example for calculation of the exterior derivative

I don't understand how to calculate exterior derivatives. For the form $$ \theta = \frac{x\, dy - y\, dx}{x^2 + y^2} $$ I arrive at the following solution: $$ \begin{align*} d\theta &= d\left(\frac{x}{x^2 + y^2}\right) \wedge dy + d\left(\frac{-y}{x^2 + y^2}\right) \wedge dx \\ &= \frac{-x^2 + y^2}{(x^2 + y^2)^2} dx \wedge dy - \frac{x^2 - y^2}{(x^2 + y^2)^2}dy \wedge dx \\ &= 0. \end{align*} $$ I don't understand the second step. Or also in these examples Exterior Derivatives I don't see how we get the solutions.

I know that $df_p$ is a function from $T_pM$ to $\mathbb{R}$. But here we don't have a specific point $p$ and I don't think I can use this definition to calculate the exterior derivative of $x/(x^2 + y^2)$. Is there another definition for $df$?

I was also wondering if there are general rules that could help with this kind of calculations (such as $dx \wedge dx = 0$).

Solution 1:

If you apply the definition you have: $$ \begin{align*} d\left(\frac{x}{x^2 + y^2}\right) \wedge dy &= \partial_x\left(\frac{x}{x^2 + y^2}\right) dx \wedge dy + \partial_y\left(\frac{x}{x^2 + y^2}\right) dy \wedge dy \\ &= \partial_x\left(\frac{x}{x^2 + y^2}\right) dx \wedge dy \\ &= \frac{-x^2 + y^2}{(x^2 + y^2)^2} dx \wedge dy \end{align*} $$ I'll let you conclude the exercise.

Solution 2:

I suggest you to read definition of exterior derivative.

Consider the manifold $\mathbb{R}^n$.

Let $f:\mathbb{R}^n\rightarrow \mathbb{R}$ be a differential $0$-form on $\mathbb{R}^n$. The exterior derivative of $f$ is a differential $1$-form on $\mathbb{R}^n$. Recall that, a differential $1$ form is an expression of the form $$\sum_{i=1}^n g_i dx_i$$ where $g_i:\mathbb{R}^n\rightarrow \mathbb{R}$ is a smooth map. So, for $f$, the exterior derivative $df$ is defined as $$\sum_{i=1}^n g_i dx_i$$ where $g_i$ is the $i^{th}$ partial derivative of the function $f:\mathbb{R}^n\rightarrow \mathbb{R}$.