A p-form taking discrete values on p-chains must be 0.

I want to show that if $w$ is a $p$-form such that its induced cochain on $p$-chains:

$w(\gamma)= \int_{\gamma} w \in S$

takes values in a discrete set $S \subset \mathbb{R}$ then $w$ must be zero.

• My idea is to say that we know some chains (i.e. the zero chain) will integrate to zero and that there is a way to continuously vary the chain so that the value $w(\gamma)$ should vary continuously as well; hitting values outside of $S$.

Any suggestions or ideas are greatly appreciated!

Thanks, CM

-
It suffices to do this in $\mathbb{R}^{n}$, in which case it is just the fundamental theorem of calculus: choose vectors $v_{1},...,v_{p}$, and now integrate $\omega$ over the parallelpiped boxed in by $\epsilon v_{1},...,\epsilon v_{p}$: this will give you $\omega(v_{1},...,v_{p})\epsilon + o(\epsilon)$ by the FTC. So by taking $\epsilon$ small, we conclude that $\omega(v_{1},...,v_{p})=0$. – David Cohen Dec 13 '11 at 17:56

You can determine the value of the $p$-form at a point as the limit if integrals over very small $p$-simplices and rescaling. If the integral takes values in a discrete subgroup of $\mathbb{R}$, then you get zero.