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.
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 and let . Now, is the boundary of the component of , such that , 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?
Thanks in advance.
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.
It's tricky. Help please!
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 of N around p such that is a diffeomorphism, and . However since M is open in N, there is an open set V of N around p such that . The image is an open set around 0 in , so must contain a point with , which contractdicts the fact that