# The Unapologetic Mathematician

## 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.