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

  • 1
    $\begingroup$ 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 $\omega(v_{1},...,v_{p})=0$. $\endgroup$ – 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.


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.