A form $ \omega $ is said to be exact if $ \omega=df $ for some smooth function $ f:M\to \mathbb{R} $. Let $ \omega \in \Omega^1\left( \mathbb{R}^2\backslash \{ (0,0) \} \right) $ be defined by $$ \omega(x,y)=\frac{-ydx+xdy}{x^2+y^2}, $$ and let $ \gamma:[0,2\pi]\to \mathbb{R}^2\backslash \{(0,0)\} $ be the curve $ \gamma(t)=(\cos t,\sin t) $. Calculate $$ \int_\gamma \omega$$ and deduce that $ \omega $ is not exact.
Let's calculate $ (\gamma^{-1})^*\omega $. Notice first that $ dx=d(\cos t)=-\sin t dt$ and $ dy=d(\sin t)=\cos t dt$. \begin{align*} (\gamma^{-1})^*\omega&=\frac{-(\gamma^{-1})^*y(\gamma^{-1})^*dx+(\gamma^{-1})^*x(\gamma^{-1})^*dy}{(\gamma^{-1})^*(x^2+y^2)}\\ &=\frac{-\sin t d(\cos t)+\cos t d(\sin t)}{1}\\ &=-\sin t (-\sin t)dt+\cos t \cos t dt\\ &=(\sin^2 t +\cos ^2 t )dt\\ &=1\cdot dt\\ &=dt \end{align*} Hence, since $ (\gamma^{-1})^*\omega=dt $, $$ \omega=\gamma^*dt=d\gamma (t). $$ Therefore, \begin{align*} \int_{\gamma} \omega &=\int_{\gamma} d \gamma (t)\\ &=\Big[ \gamma(t) \Big]_0^{2\pi}\\ &=\gamma(2\pi)-\gamma(0)\\ &=(\cos 2\pi,\sin 2\pi )-(\cos 0,\sin 0)\\ &=(1,0)-(1,0)\\ &=(0,0) \end{align*} Since $ \omega=d\gamma(t) $, $ \omega $ is exact, so something must be wrong. Can you help me to correct my mistakes?
If $\alpha:[a,b]\to\Gamma$ is a (piecewise smooth) parameterization of the curve $\Gamma$, then $\int_\Gamma\omega=\int_a^b\alpha^*\omega$. Here $\Gamma$ is the unit circle $S^1$ with counterclockwise orientation. We’ve been given the parameterization $\gamma:[0,2\pi]\to S^1$ to use and you computed $\gamma^*\omega$ correctly, but called it $(\gamma^{-1})^*\omega$ instead. Setting aside the existence of $\gamma^{-1}$, the latter expression doesn’t even make sense since the codomain of $\gamma^{-1}$ is $[0,2\pi]$, not $\mathbb R^2\setminus\{(0,0)\}$. Perhaps you had in mind computing the inverse image of $S^1$ under $\gamma$, which would be the new domain of integration, but that’s something else entirely. After that, you went off even farther into the weeds. Instead, what you should have had is $\int_{S^1}\omega=\int_0^{2\pi}\gamma^*\omega=\int_0^{2\pi}dt=2\pi$.