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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 $dx \wedge dx = 0$).

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  • $\begingroup$ 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. $\endgroup$
    – Dlmn
    Jan 22, 2020 at 9:31
  • $\begingroup$ what examples of exterior derivatives calculation do you know?? do you know the definition of exterior derivative? $\endgroup$ Jan 22, 2020 at 9:33
  • $\begingroup$ 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. $\endgroup$
    – Dlmn
    Jan 22, 2020 at 9:49
  • $\begingroup$ for a function $\omega_I$ you mentioned above, what is $d\omega_I$?? $\endgroup$ Jan 22, 2020 at 10:06

2 Answers 2

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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.

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  • $\begingroup$ How do you get this using the definition of $df_p$ ? $\endgroup$
    – Dlmn
    Jan 22, 2020 at 9:52
  • $\begingroup$ I don't know what you have in mind as "definition of $df_p$". Take a look en.wikipedia.org/wiki/Exterior_derivative $\endgroup$
    – Gabrielek
    Jan 22, 2020 at 10:02
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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 $f:\mathbb{R}^n\rightarrow \mathbb{R}$.

Is it clear till here?

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  • $\begingroup$ 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. $\endgroup$
    – Dlmn
    Jan 22, 2020 at 12:41
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    $\begingroup$ 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? $\endgroup$ Jan 22, 2020 at 13:02

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