Let $\omega$ be a $2$-form on $\mathbb{R}^3\setminus\{0\}$ defined by $ \omega = \frac{x\,dy\wedge dz+y\,dz\wedge dx +z\,dx\wedge > dy}{(x^2+y^2+z^2)^{\frac{3}{2}}} $ Show that $\omega$ is closed but not exact.
What I have tried so far: In order to show that $\omega$ is closed, I need to show that $d\omega=0$. I'm having some problems getting all of the calculus right and somewhere along the way I'm messing up. I started by rewriting $\omega$ as $ \omega = (x\,dy\wedge dz+y\,dz\wedge dx +z\,dx\wedge dy)(x^2+y^2+z^2)^{-\frac{3}{2}} $ Now I should be able to use the product rule to evaluate (I think). Then $ d\omega = (dx\wedge dy\wedge dz+dy\wedge dz\wedge dx +dz\wedge dx\wedge dy)(x^2+y^2+z^2)^{-\frac{3}{2}} + (\ast) $ where $ (\ast) = (x\,dy\wedge dz+y\,dz\wedge dx +z\,dx\wedge dy)\left(-\frac{3}{2}(2x\,dx+2y\,dy+2z\,dz)\right)(x^2+y^2+z^2)^{-\frac{5}{2}} $ Even after trying to simplify everything, I can't get it to cancel. This makes me think that perhaps I can't apply the product rule like this.
What should I do to calculate $d\omega$?
If $\omega$ is a globally defined smooth form and if $d\omega=0$, then $\omega$ is exact because there is some other form $\alpha$ with $d\alpha=\omega$ and $d^2\alpha=d\omega=0$. Because $\omega$ is not defined at $(0,0,0)$, it makes sense that it isn't exact.
Is there a way to use the above reasoning to show that there can't be an $\alpha$ such that $d\alpha=\omega?$