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

An Example (part 2)

We follow yesterday’s example of an interesting differential form with a (much simpler) example of some 1-chains. Specifically, we’ll talk about circles!

More specifically, we consider the circle of radius a around the origin in the “punctured” plane. I used this term yesterday, but I should define it now: a “punctured” space is a topological space with a point removed. There are also “twice-punctured” or “n-times punctured” spaces, and as long as the space is a nice connected manifold it doesn’t really matter much which point is removed. But since we’re talking about the plane \mathbb{R}^2 it comes with an identified point — the origin — and so it makes sense to “puncture” the plane there.

Now the circle of radius a will be a singular 1-cube. That is, it’s a curve in the plane that never touches the origin. Specifically, we’ll parameterize it by:

\displaystyle c_a(t)=(a\cos(2\pi t),a\sin(2\pi t))

so as t ranges from 0 to 1 we traverse the whole circle. There are two 0-dimensional “faces”, which we get by setting t=0 and t=1:


When we calculate the boundary of c, these get different signs:

\displaystyle\begin{aligned}\partial c_a&=(-1)^{1+0}c_a(0)+(-1)^{1+1}c_a(1)\\&=-(a,0)+(a,0)=0\end{aligned}

We must be very careful here; these are not vectors and the addition is not vector addition. These are merely points in the plane — 0-cubes — and the addition is purely formal. Still, the same point shows up once with a positive and once with a negative sign, so it cancels out to give zero. Thus the boundary of c_a is empty.

On the other hand, we will see that this circle cannot be the boundary of any 2-chain. The obvious thing it might be the boundary of is the disk of radius a, but this cannot work because there is a hole at the origin, and the disk cannot cross that hole. However this does not constitute a proof; maybe there is some weird chain that manages to have the circle as its boundary without crossing the origin. But the proof will have to wait.

August 24, 2011 Posted by | Differential Topology, Topology | 1 Comment