# The Unapologetic Mathematician

## 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:

$\int\limits_{c_a}\omega=\int\limits_{[0,1]}{c_a}^*\omega$

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

## 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 $t=1$:

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

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