# 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.

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Posted by John Armstrong | Differential Topology, Topology

## 1 Comment	»

1. [...] we can take our differential form and our singular cube and put them together. That is, we can integrate the -form over the circle [...]

Pingback by | August 24, 2011 | Reply

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## About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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