# what does it mean for a differential form to be well defined on a manifold?

What does it mean for a differential form to be well defined on some manifold. In particular, why the $2$-form $\omega=d\psi\wedge d\theta$ is well defined on $S^{2}$?

Thank you in advance.

## 1 Answer

Sometimes a differential form is defined using coordinate functions that are defined only locally. For example, the polar angle $\theta$ cannot be defined as a smooth function on all of $S^1$. It can be defined (with a choice of an additive constant) on every proper arc of $S^1$, it's just that those pieces can't be glued together into something that is smooth on $S^1$. Yet, $d\theta$ is a well-defined form. Why? because different local versions of $\theta$ agree up to an additive constant, and such constants do not affect $d\theta$.

This shows the process of checking that a differential form is well-defined: define it on coordinate patches covering the space, then check that the definitions agree where the patches overlaps.

Similar in your example, where I assume $\theta$ is longitude and $\psi$ is latitude. Exception: $d\theta$ is not defined at the poles, and neither is your form. The volume form on $S^2$ is $\sin\psi \, d\psi\wedge d\theta$ (taking $0\le \psi\le \pi$), which avoids the issue with $d\theta$ blowing up at the poles by virtue of the factor $\sin \psi$.