Integral of a differential k-form on a closed surface

Question

Let us say we have a $k$ dimensional closed, compact, and orientable surface $S$ with a differential form $\omega$ defined on it. Stokes theorem tells us that: $$ \int_S \mathrm{d} \omega = \int_{\partial S} \omega $$ Now, because this is a closed surface, $\partial S$ is empty, hence: $$ \int_S \mathrm{d} \omega = \int_{\partial S} \omega =0 $$ Now, I would like to use this fact to show that there exists a point $p$ on S such that: $$ \mathrm{d}\omega_p (v_1,...,v_k) =0 $$ for any tangent vectors $v_1,...,v_k$ at $p$. I understand that the whole integral above is $0$, but I feel like this could occur due to positive and negative components cancelling out.

Answer 1

Let $\nu$ denote the volume form on $M$, and because $d\omega$ is a top form there must be a function $f:M \to \mathbb{R}$ so that we can write $$d\omega = f \cdot \nu$$ and in particular the integral can be written as

\begin{align} \int_M d\omega = \int_M f \cdot \nu= 0 \end{align}

Now the cases to consider are either $f$ is changing sign or $f$ is not changing sign. If $f$ is changing sign then by continuity there must be some point $p$ so that $f(p) = 0$ and thus $d\omega$ is $0$ at $p$.

If $f$ is not changing sign then the fact that the integral is $0$ will imply that $f$ is equal to $0$ and therefore we again obtain a point $p$ where $f$ and consequently $d\omega$ is $0$.