The Unapologetic Mathematician

Mathematics for the interested outsider

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 »

  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 An Example (part 3) « The Unapologetic Mathematician | August 24, 2011 | Reply

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: