# Integral of a differential k-form on a closed surface

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.

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 $0$.

• Ah of course! The sign change and intermediate value. Thanks. – rubikscube09 Mar 3 '18 at 21:58
• Actually, a question for you: what enables us to write the form in the above way? Say we were integrating over a 2 - dimensional closed surface in $\mathbb{R}^3$. Then because $d \omega$ is a 2-form in $\mathbb{R}^3$, wouldn't it necessarily be written as $d \omega = f_1 dy \wedge dz + f_2 d_x \wedge d_z + f_3 d_x \wedge d_y$? I understand the notion of a top form, but I guess it's the ambient space that is tripping me up. – rubikscube09 Mar 3 '18 at 22:29
• You need to be looking in the coordinates of the manifold that you are on. Call that 2-D surface $S$. Then $x$, $y$, $z$ will not be the local coordinates of $S$. So if the local coordinates of $S$ are given by $\phi^1, \phi^2$ you can see that we can write $dx \wedge dy$ as $g_1 d \phi^1 \wedge d \phi^2$ $dx \wedge dz$ as$g_2 d\phi ^1 \wedge d \phi^2$ and $dy \wedge dz$ as $g_3 d\phi^1 \wedge d\phi^2$ on $S$ – Memeboy Inc. Mar 3 '18 at 22:46
• Sorry, I had to edit that a bunch of times I don't know if comments can be previewed haha – Memeboy Inc. Mar 3 '18 at 22:51
• No worries. Thanks for the help. – rubikscube09 Mar 3 '18 at 22:53