The Unapologetic Mathematician

Mathematics for the interested outsider

An Example (part 3)

Now we can take our differential form and our singular cube and put them together. That is, we can integrate the 1-form \omega over the circle c_a.

First we write down the definition:


to find this pullback of \omega we must work out how to push forward vectors from [0,1]. That is, we must work out the derivative of c_a.

This actually isn’t that hard; there’s only the one basis vector \frac{d}{dt} to consider, and we find

\displaystyle {c_a}_{*t}\left(\frac{d}{dt}\right)=-2\pi a\sin(2\pi t)\frac{\partial}{\partial x}+2\pi a\cos(2\pi t)\frac{\partial}{\partial y}

We also have to calculate the composition

\displaystyle\begin{aligned}\omega(c_a(t))&=-\frac{a\sin(2\pi t)}{(a\cos(2\pi t))^2+(a\sin(2\pi t))^2}dx+\frac{a\cos(2\pi t)}{(a\cos(2\pi t))^2+(a\sin(2\pi t))^2}dy\\&=\frac{1}{a}\left(-\sin(2\pi t)dx+\cos(2\pi t)dy\right)\end{aligned}

This lets us calculate

\displaystyle\begin{aligned}\int\limits_{c_a}\omega&=\int\limits_{[0,1]}{c_a}^*\omega\\&=\int\limits_{[0,1]}\left[\omega(c_a(t))\right]\left({c_a}_{*t}\left(\frac{d}{dt}\right)\right)\,dt\\&=\int\limits_{[0,1]}\left[\frac{1}{a}\left(-\sin(2\pi t)dx+\cos(2\pi t)dy\right)\right]\left(2\pi a\left(-\sin(2\pi t)\frac{\partial}{\partial x}+\cos(2\pi t)\frac{\partial}{\partial y}\right)\right)\,dt\\&=2\pi\int\limits_0^1\sin(2\pi t)^2+\cos(2\pi t)^2\,dt=2\pi\int\limits_0^1\,dt=2\pi\end{aligned}

So, what conclusions can we draw from this? Well, Stokes’ theorem now tells us that the 1-form \omega cannot be the differential of any 0-form — any function — on \mathbb{R}^2\setminus{0}. Why? Well, if we had \omega=df, then we would find

\displaystyle\int\limits_{c_a}\omega=\int\limits_{c_a}df=\int\limits_{\partial c_a}f=\int\limits_{0}f=0

which we now know not to be the case. Similarly, c_a cannot be the boundary of any 2-chain, for if c_a=\partial c then

\displaystyle\int\limits_{c_a}\omega=\int\limits_{\partial c}\omega=\int\limits_{\partial c}d\omega=\int\limits_{\partial c}0=0

It turns out that there’s a deep connection between the two halves of this example. Further, in a sense every failure of a closed k-form to be the differential of a k-1-form and every failure of a closed k-chain to be the boundary of a k+1-chain comes in a pair like this one.


August 24, 2011 - Posted by | Differential Topology, Topology

1 Comment »

  1. You were right. This example was really illustrative. Thank-you. You have no idea how much your input has helped me, so keep posting. Tchuss!

    Comment by Alice | April 27, 2015 | Reply

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