example for calculation of the exterior derivative

I don't understand how to calculate exterior derivatives. For example for the form $$\theta = \frac{xdy - ydx}{x^2 + y^2}$$ I have the following solution:

$$d\theta = d(\frac{x}{x^2 + y^2}) \wedge dy + d(\frac{-y}{x^2 + y^2}) \wedge dx = \frac{-x^2 + y^2}{(x^2 + y^2)^2} dx \wedge dy - \frac{x^2 - y^2}{(x^2 + y^2)^2}dy \wedge dx = 0$$

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 for example $$d(\frac{x}{x^2 + y^2})$$. 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 $$dx \wedge dx = 0$$).

• I just realised that $d(\frac{x}{x^2 + y^2}) = \frac{d}{dx}(\frac{x}{x^2 + y^2}) dx$. But I don't understand why we can do this. – Debs Jan 22 at 9:31
• what examples of exterior derivatives calculation do you know?? do you know the definition of exterior derivative? – Praphulla Koushik Jan 22 at 9:33
• The definition I have is $d: \Omega^k(M)\to \Omega^{k+1}(M)$ with $d(\sum \omega_Idx^I) := \sum d\omega_I\wedge dx^I$ but this only helps with the first step in the example. – Debs Jan 22 at 9:49
• for a function $\omega_I$ you mentioned above, what is $d\omega_I$?? – Praphulla Koushik Jan 22 at 10:06

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 $$f:\mathbb{R}^n\rightarrow \mathbb{R}$$.
• Yes, for 1-forms I see how it's done. I think part of the problem was the sum over $d\omega_I$, as you said in the comments. – Debs Jan 22 at 12:41
• But $\omega_I$ is also of the form $\mathbb{R}^n\rightarrow \mathbb{R}$.. Isn't it? What ever I have mentioned for $f$, that is $df$, should be carried to do for $\omega_I$, that is d\omega_I$... Do you see any problem? – Praphulla Koushik Jan 22 at 13:02 If you apply the definition you have for example: $$\begin{split} d( \frac{x}{x^2 + y^2}) \wedge dy &= \quad \partial_x(\frac{x}{x^2 + y^2}) dx \wedge dy + \partial_y(\frac{x}{x^2 + y^2}) dy \wedge dy \\ & = \quad \partial_x(\frac{x}{x^2 + y^2}) dx \wedge dy \\ & = \quad \frac{-x^2 + y^2}{(x^2 + y^2)^2} dx \wedge dy \end{split}$$ I let you conclude the exercise • How do you get this using the definition of$df_p$? – Debs Jan 22 at 9:52 • I don't know what you have in mind as "definition of$df_p\$". Take a look en.wikipedia.org/wiki/Exterior_derivative – Gabrielek Jan 22 at 10:02