Showing that $\int_{c} \omega =0$ when $\partial c =0$

Question

Let $\omega$ be a $k$-form on $\mathbb{R}^n$ and suppose that $\omega=d\alpha$ for some $(k-1)$-form $\alpha$. Show that, for any singular $k$-cube $c$ on $\mathbb{R}^n$ with $\partial c=0$,

$$\displaystyle \int_{c}\omega=0$$

I get that $\displaystyle \int_{c}\omega = \int_{c}d\alpha= \int_{\partial c}\alpha=0?$

I just cannot explain why the last equality should equal $0$, ie $\int_{\partial c}\alpha=0$.

Note that : $\displaystyle \partial c = \sum_{j=1}^{k} \sum_{\alpha=0,1}(-1)^{j+\alpha}c_{(j,\alpha)} $ and

$\displaystyle \int_{\partial c} \omega = \sum_{j=1}^k \sum_{\alpha=0,1} (-1)^{j+\alpha}\int_{c(j,\alpha)}\omega$

Please comment if I have left unexplained terminology you might need.

Answer 1

This is really just a matter of algebra. Fixing a $k$-form $\omega$, $$\phi:c\mapsto\int\limits_c\omega$$ is a linear map from the free abelian group on singular $k$-cubes, whose elements we call $k$-chains, to $\mathbb{R}$, where we are considering $\mathbb{R}$ as a $\mathbb{Z}$-module. In particular, as a homomorphism of modules, $$\phi\left(\sum_ia_ic_i\right)=\sum_ia_i\phi(c_i)=\sum_ia_i\int\limits_{c_i}\omega.$$

Since the $(k-1)$-chain $\partial c$ is $0$, we thus obtain

$$\int\limits_{\partial c}\alpha= 0 \cdot \left( \ \int\limits_{\text{any chain}}\alpha \right)=0$$

(If you are unfamiliar with modules, just think of them as vector spaces over rings, such as $\mathbb{Z}$.)

Answer 2

Note that they are talking about a "singular $k$-cube", i.e., the image $c:=f(\hat c)\subset {\mathbb R}^n$ of a cube $\hat c:=[0,1]^k\subset{\mathbb R}^k$. It is assumed that $\partial c=0$, which means that $f$ closes up $\hat c$ on itself somehow. Consider the following example: The $c$ produced by $$f:\quad [0,1]\to{\mathbb R}^2,\qquad t\mapsto(\cos t, \sin t)$$ has $\partial c=0$.

The stated claim can then be proven by considering pullbacks.