The Unapologetic Mathematician

Mathematics for the interested outsider

Stokes’ Theorem (proof part 2)

Now that we’ve proven Stokes’ theorem in the case of the standard cube, we can now tackle the general case.

Of course, we only have to handle the case of a general singular cube, since we defined the boundary operator \partial to be additive; if c is a general chain — a formal sum of singular cubes — then \partial c is the same formal sum of the boundaries of these cubes. Since integration is also defined to be additive on the chain over which we integrate, everything works out:

\displaystyle\begin{aligned}\int\limits_cd\omega&=\int\limits_{a_1c_1+\dots+a_lc_l}d\omega\\&=a_1\int\limits_{c_1}d\omega+\dots+a_l\int\limits_{c_l}d\omega\\&=a_1\int\limits_{\partial c_1}\omega+\dots+a_l\int\limits_{\partial c_l}\omega\\&=\int\limits_{a_1\partial c_1+\dots+a_l\partial c_l}\omega\\&=\int\limits_{\partial(a_1c_1+\dots+a_lc_l)}\omega\\&=\int\limits_{\partial c}\omega\end{aligned}

where we have used the special case of singular cubes to pass from the second line to the third.

So let’s tackle this special case:

\displaystyle\begin{aligned}\int\limits_{\partial c}\omega&=\sum\limits_{i=1}^k\sum\limits_{\alpha=0,1}(-1)^{i+\alpha}\int\limits_{c\circ I^{k-1}_{i,\alpha}}\omega\\&=\sum\limits_{i=1}^k\sum\limits_{\alpha=0,1}(-1)^{i+\alpha}\int\limits_{[0,1]^{k-1}}(c\circ I^{k-1}_{i,\alpha})^*\omega\\&=\sum\limits_{i=1}^k\sum\limits_{\alpha=0,1}(-1)^{i+\alpha}\int\limits_{[0,1]^{k-1}}{I^{k-1}_{i,\alpha}}^*c^*\omega\\&=\sum\limits_{i=1}^k\sum\limits_{\alpha=0,1}(-1)^{i+\alpha}\int\limits_{I^{k-1}_{i,\alpha}}c^*\omega\\&=\int\limits_{\partial([0,1]^k)}c^*\omega\\&=\int\limits_{[0,1]^k}dc^*\omega\\&=\int\limits_{[0,1]^k}c^*d\omega\\&=\int\limits_{[0,1]^k}c^*d\omega\\&=\int\limits_cd\omega\end{aligned}

Basically, it all works out for the same reason parameterization invariance and the change of variables formula do. Passing from the boundary of the singular cube back to the boundary of the standard one transforms the integral one way; passing from the standard cube itself back to the singular cube undoes this transformation. And so Stokes’ theorem is proved.

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August 20, 2011 - Posted by | Differential Topology, Topology

1 Comment »

  1. Your bog is very interesting. I would be up to date.

    Comment by Clairgironet | September 13, 2011 | Reply


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