The exercise ask to prove that for every 1-form closed on the $2$ sphere there was a function defined on the sphere such that its differential is the form. Then it asks if this function is unique.

Well, I use the fact that the $1$ de Rham cohomology on the $2$ sphere is trivial to claim the existence of such function.

Then, I use Stoke's theorem with the fact the the sphere is a compact regular domain to claim it uniqueness.

Is there anything wrong with this argumentation?


The first part is fine. By definition, $H^1(S^2) = 0$ means that any element in $\ker(d : \Omega^1(S^2) \to \Omega^2(S^2))$, i.e. any closed $1$-form, is actually in $\operatorname{im}(d : \Omega^0(S^2) \to \Omega^1(S^2))$, i.e. it is exact, i.e. it is the differential of some function.

However, I don't really see what Stokes's theorem has to do with anything for the second part. Your argument is not convincing. In fact, it cannot be, because the claim is false.

Suppose that you have a closed $1$-form $\alpha$. You know it's exact, so $\alpha = df$ for some $f$. Now if you have another function $g$ such that $\alpha = dg$, then you get $df = dg \implies d(f-g) = 0$. The question is does $f=g$, or equivalently does $f-g = 0$. So now you are essentially asking: is the zero function the only closed $0$-form (function) on $S^2$?

By definition $H^0(S^2) = \ker(d : \Omega^0(S^2) \to \Omega^1(S^2))$, and you don't quotient by $\operatorname{im}(d)$, because there is no such thing as $\Omega^{-1}$. So you are asking if $H^0(S^2) = 0$. The answer is no, because $H^0(S^2) = \mathbb{R}$, which basically follows from the fact that $S^2$ is (path-)connected. In fact it's easy to see that $H^0(S^2)$ is spanned by the constant functions on $S^2$.

More concretely, given some $f$ such that $df = \alpha$, you can always add a constant $c \in \mathbb{R}$ to $f$, and you still get $d(f+c) = df + dc = df = \alpha$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.