# The Unapologetic Mathematician

## Stokes’ Theorem (proof part 1)

We turn now to the proof of Stokes’ theorem.

$\displaystyle\int\limits_cd\omega=\int\limits_{\partial c}\omega$

We start by considering the case where $c$ is the standard cube $[0,1]^k\subseteq\mathbb{R}^k$. Whipping out the definition of the boundary operator $\partial$, the integral on the right proceeds as follows:

$\displaystyle\int\limits_{\partial([0,1]^k)}\omega=\sum\limits_{j,\alpha}(-1)^{j+\alpha}\int\limits_{[0,1]^{k-1}}\left[{I^k_{j,\alpha}}^*\omega\right]\left(\frac{\partial}{\partial u^1},\dots,\frac{\partial}{\partial u^{k-1}}\right)$

Now any $k-1$-form $\omega$ can be written out as

$\displaystyle\omega=\sum\limits_{i=1}^kf^idu^1\wedge\dots\wedge\widehat{du^i}\wedge\dots\wedge du^k$

where each term omits exactly one of the basic $1$-forms. Since everything in sight — the differential operator and both integrals — is $\mathbb{R}$-linear, we can just use one of these terms. And so we can calculate the pullbacks:

\displaystyle\begin{aligned}\left[{I^k_{j,\alpha}}^*f^idu^1\wedge\dots\wedge\widehat{du^i}\wedge\dots\wedge du^k\right]\left(\frac{\partial}{\partial u^1},\dots,\frac{\partial}{\partial u^{k-1}}\right)&=\\(f^i\circ I^k_{j,\alpha})du^1\wedge\dots\wedge\widehat{du^i}\wedge\dots\wedge du^k\left({I^k_{j,\alpha}}_*\frac{\partial}{\partial u^1},\dots,{I^k_{j,\alpha}}_*\frac{\partial}{\partial u^{k-1}}\right)&=\\(f^i\circ I^k_{j,\alpha})\det\left(\frac{\partial(v^p\circ I^k_{j,\alpha})}{\partial u^q}\right)&\end{aligned}

It takes a bit of juggling with the definition of $I^k_{j,\alpha}$, but we can see that this determinant is $1$ if $j=i$ and $0$ otherwise. Roughly this is because $I^k_{j,\alpha}$ takes the $k-1$ basis vector fields of $\mathcal{T}[0,1]^{k-1}$ and turns them into all of the basis vector fields of $\mathcal{T}[0,1]^k$ except the $j$-th one. If $i\neq j$ then some basis $1$-form has to line up against some basis vector field with a different index and everything goes to zero, while if $i=j$ then they can all pair off in exactly one way.

The upshot is that only the two faces of the cube in the $i$ direction contribute anything at all to the boundary integral, and we find

\displaystyle\begin{aligned}\int\limits_{\partial([0,1]^k)}\omega=&(-1)^{i+1}\int\limits_{[0,1]^{k-1}}f(u^1,\dots,u^{i-1},1,u^i,\dots,u^{k-1})\,d(u^1,\dots,u^{k-1})\\&+(-1)^i\int\limits_{[0,1]^{k-1}}f(u^1,\dots,u^{i-1},0,u^i,\dots,u^{k-1})\,d(u^1,\dots,u^{k-1})\end{aligned}

On the other side, we can calculate the differential of $\omega$:

\displaystyle\begin{aligned}d\left(f^idu^1\wedge\dots\wedge\widehat{du^i}\wedge\dots\wedge du^k\right)&=df^i\wedge du^1\wedge\dots\wedge\widehat{du^i}\wedge\dots\wedge du^k\\&=\left(\sum\limits_{j=1}^k\frac{\partial f^i}{\partial u^j}du^j\right)\wedge du^1\wedge\dots\wedge\widehat{du^i}\wedge\dots\wedge du^k\\&=\left(\frac{\partial f^i}{\partial u^i}du^i\right)\wedge du^1\wedge\dots\wedge\widehat{du^i}\wedge\dots\wedge du^k\\&=(-1)^{i-1}\frac{\partial f^i}{\partial u^i}du^1\wedge\dots\wedge du^k\end{aligned}

The tricky bit here is that when $j\neq i$ there’s nowhere to put this brand new $du^j$, since it must collide with one of the other basic $1$-forms in the wedge. But when $j=i$ then it can slip right into the “hole” where we’ve left out $du^i$, at a cost of a factor of $(-1)^{i-1}$ to pull the $du^i$ across the first $i-1$ terms in the wedge.

With this result in hand, we calculate the interior integral:

$\displaystyle\int_{[0,1]^k}d\omega=(-1)^{i-1}\int_{[0,1]^k}\frac{\partial f^i}{\partial u^i}\,d(u^1,\dots,u^k)$

We can turn this into an iterated integral, which Fubini’s theorem tells us we can evaluate in any order we want:

\displaystyle\begin{aligned}\int_{[0,1]^k}d\omega=&(-1)^{i-1}\int_{[0,1]^{k-1}}\int\limits_0^1\frac{\partial f^i}{\partial u^i}\,du^i\,d(u^1,\dots,\widehat{u^i},\dots,u^k)\\=&(-1)^{i-1}\int_{[0,1]^{k-1}}f^i(u^1,\dots,u^{i-1},1,u^{i+1},\dots,u^k)\,d(u^1,\dots,\widehat{u^i},\dots,u^k)\\&-(-1)^{i-1}\int_{[0,1]^{k-1}}f^i(u^1,\dots,u^{i-1},0,u^{i+1},\dots,u^k)\,d(u^1,\dots,\widehat{u^i},\dots,u^k)\end{aligned}

which it should be clear is the same as our answer for the boundary integral above. Thus Stokes’ theorem holds for the standard cube.

August 18, 2011 - Posted by | Differential Topology, Topology