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Let $dx^I$ be local coordinates and let $$\omega := \frac{x^1dx^2 - x^2dx^1}{(x^1)^2+(x^2)^2}.$$ Compute $d\omega$ for $(x^1,x^2) \ne 0$.

Remark: $d\omega$ means the exterior derivative, which we defined in the lecture as

$$d(f \ dx^I) = df \wedge dx^I$$

, where $df$ is the differential of $f$.

I can see that transfering $(x^1,x^2)$ into polar coordinates $y^I:=(r, \alpha)$ it holds $d\alpha = \omega$, so we are left to compute $d\omega = dd\alpha$. However, I do not see how to evaluate $d\alpha$.

If I am not mistaken $\alpha$ should be a $1$-form, so

$$d(\alpha \ dy^I) = d\alpha \wedge dy^I = 1 \wedge dr \wedge d\alpha,$$

but I am not sure what to do now. Could you please give me hint?

3nondatur
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  • You may also want to study my answer here which is discussing the question to what extend this form is / is not *exact*. – Kurt G. Jan 09 '23 at 19:50

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I shall write $x=x^1$ and $y=x^2\,.$ Your one-form is $$\tag{1} \omega=f\,dx+g\,dy\,,\quad f(x,y):=\frac{-y}{x^2+y^2}\,,\quad g(x,y):=\frac{x}{x^2+y^2}\,. $$ and, in Cartesian coordinates, \begin{align} d\omega&=\partial_yf\,dy\wedge dx+\partial_xg\,dx\wedge dy\\[2mm] &=\frac{(y^2-x^2)\,dy\wedge dx+(y^2-x^2)\,dx\wedge dy}{(x^2+y^2)^2}\\ &=0\,.\tag{2} \end{align} In polar coordinates, \begin{align}\tag{3} dx&=\cos\alpha\,dr-r\sin\alpha\,d\alpha\,,\quad &dy&=\sin\alpha\,dr+r\cos\alpha\,d\alpha\,,\\[2mm] dr&=\cos\alpha\,dx+\sin\alpha\,dy\,,\quad &r\,d\alpha&=-\sin\alpha\,dx+\cos\alpha\,dy\,.\tag{4} \end{align} so that indeed, from the expression for $r\,d\alpha$ in (4),

\begin{align} \omega=d\alpha\,. \end{align} This leads to \begin{align} d\omega=0 \end{align} as it should.

Another way to see that $\omega=d\alpha$ is to differentiate $$ \alpha=\arccos\Big(\frac{x}{\sqrt{x^2+y^2}}\Big) $$ w.r.t. $x$ and $y\,.$

Kurt G.
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    @3nondatur I don't understand what that $\omega,dx^I$ is when $\omega=f,dx+g,dy$. When you apply the differentiation rule of a $k$-form $\varphi$ to the one-form $\omega$ you get what I wrote. – Kurt G. Jan 10 '23 at 13:24
  • I am sorry, but I still do not get it. When I use the definition from my lecture I only get $d(\omega \ dx^I) = d\omega \wedge dx^I = d(f \ dx + g \ dy) \wedge dx^I = d(f \ dx)\wedge dx^I + d(g \ dy) \wedge dx^I $, but I do not see how to compute the differentials $d(f \ dx)$ and $d(g \ dy)$ now. – 3nondatur Jan 10 '23 at 15:10
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    By your own definition in OP, $\omega$ is of the form $f,dx^1+g,dx^2,.$ What do you want with $dx^I,?$ You can omit it. The Wikipedia link I provided shows that $d\omega=d(f,dx^1+g,dx^2)=(\partial_2 f),dx^2\wedge dx^1+(\partial_1 g),dx^1\wedge dx^2,.$ – Kurt G. Jan 10 '23 at 18:58
  • Thanks a lot for your help, I understand it now (I got confused with the notation). – 3nondatur Jan 10 '23 at 21:01