Topology. Computational exercise.

Let $\displaystyle U = \mathbb{R}^2 \setminus \{(0,0)\}$; consider the $\displaystyle 1-$form in $\displaystyle U$ defined by

$\displaystyle \omega = \dfrac{x dx + ydy}{x^2+y^2}$

(a) Show that $\displaystyle d\omega =0$

(b) Show that there is a $\displaystyle 0$-form $\displaystyle \theta$ on $\displaystyle U$, such that $\displaystyle d\theta = \omega$

Please help!!!

Re: Topology. Computational exercise.

(a) $\displaystyle d\omega=\frac{2xy dy\wedge dx + 2xy dx \wedge dy}{x^2+y^2}=0$

(b) since $\displaystyle d\omega=0$, for any close loop $\displaystyle \gamma$ not enclosing 0, we have $\displaystyle \oint_{\gamma} w ds = \int_D dw=0$

so we can define $\displaystyle \theta(p) = \int_{p_0}^{p} w$, where $\displaystyle p_0 \ne 0 $ is fixed and the line integral goes on any curve that not crossing 0.

Choose $\displaystyle p_0=(1,0)$, and the curve from (1, 0) to p(x0,y0) be from (1,0) to (x0,0), then from (x0, 0) to (x0,y0)

we get $\displaystyle \theta(x_0,y_0)=\int_{(1,0)}^{(x_0, y_0)} w = \int_1^{x_0}\frac{xdx}{x^2+y^2} + \int_0^{y_0} \frac{ydy}{x^2+y^2}$

=$\displaystyle \frac{1}{2}log(x_0^2+y_0^2)$