# Polemic problem in differential geometry

• Nov 30th 2012, 03:21 PM
ModusPonens
Polemic problem in differential geometry
Hello

I was asked in a test to do an integration of an exact form on a manifold. The first thing you think is to apply the Stokes theorem. The problem here is to determine the boundary of the manifold. Let $S=\{(x,y,z,w):x^2+y^2+z^2+w^2=2\}$ and let $T^2=\{(x,y,z,w):x^2+y^2=z^2+w^2=1\}$. Now, is the boundary of the component of $S-T^2$, such that $x^2+y^2<1$, empty?

My teacher says it's not empty, but a fellow PhD of mine says it's empty. I don't know who's right. Can anyone find a way to decide this without any doubt remaining?

• Dec 1st 2012, 08:59 AM
xxp9
Re: Polemic problem in differential geometry
T^2 is closed. So S-T^2 is open. What is the definition of the boundary of an open manifold? It must be empty.
• Dec 1st 2012, 11:32 AM
ModusPonens
Re: Polemic problem in differential geometry
S is compact in R^4 and thus closed, so you can consider S, as a space itself with the subspace topology. In it, S is open and T^2 is closed, thus S-T^2 is open. However, it's open in the subspace topology, not necessarily in R^4. For example, I^2=[0,1]X[0,1] is open as the whole space. Take [1/2,1]X[0,1] out of [0,1]X[0,1]. The result is an open set in I^2, but not in R^2. A similar example with disks can be made to fit in the context of smooth manifolds.

• Dec 1st 2012, 02:57 PM
xxp9
Re: Polemic problem in differential geometry
You don't need to( or you should not always) consider a manifold as an embedded sub-manifold of R^n. So my argument from the first reply works well.
For your example, what is the boundary of the result?
• Dec 1st 2012, 03:31 PM
ModusPonens
Re: Polemic problem in differential geometry
In my example, the manifold isn't smooth. But consider this example: [0,1) is a smooth manifold. [0,1/2) is open in [0,1). Its boundary is {0}.
• Dec 2nd 2012, 04:36 AM
xxp9
Re: Polemic problem in differential geometry
This is because that [0,1) itself has a boundary. So an open sub-manifold of [0,1) can have a boundary.
However S itself has no boundary.
• Dec 2nd 2012, 07:08 AM
ModusPonens
Re: Polemic problem in differential geometry
I'm sorry, but you haven't proved what was asked so far.
• Dec 2nd 2012, 07:15 AM
xxp9
Re: Polemic problem in differential geometry
Let me prove the statement: an open sub-manifold M of a (boundaryless) manifold N has no boundary.
Suppose M has a boundary and p is a point on the boundary, there is a chart $(U, \phi)$ of N around p such that $\phi: U\cap M \rightarrow R^n_+=\{(x^1, ..., x^n) \in R^n | x^n \ge 0 \}$ is a diffeomorphism, and $\phi(p)=0$. However since M is open in N, there is an open set V of N around p such that $V \subset M\cap U$. The image $\phi(V)$ is an open set around 0 in $R^n$, so must contain a point with $x^n < 0$, which contractdicts the fact that $\phi(V) \subset R^n_+$
• Dec 2nd 2012, 07:53 AM
ModusPonens
Re: Polemic problem in differential geometry
Thank you very much for your effort and help. :)