# Definition of boundary, Stokes' theorem

1. Nov 2, 2014

### tom.stoer

this is a rather stupid question regarding preliminaries for the definition of boundaries

the question is whether every closed n-1 dim. closed submanifold $C$ of an arbitrary n-dim. manifold defines a volume $V$; i.e. whether $\partial V = C$ can be turned around such that V is defined as the "interior" of $C$ (instead of defining the boundary $\partial V$ in terms of V)

my observation was that this seems to fail for the torus $T^2$ and a closed loop $\gamma$ on $T^2$ if $\gamma$ has a non-trivial winding number, i.e. if the winding number $w(\gamma) \neq (0,0)$; a loop with $(1,0)$ does not define an "interior" on the torus (the interior is the whole torus) and therefore seems not to be the boundary of a volume (the torus itself has no boundary);

so my question is which conditions for a closed manifold must hold such that it can be used as a boundary of a volume in the sense of integration of differential firms and stokes theorem

2. Nov 5, 2014

### lavinia

Generally a hypersurface is not a boundary as with those circles on the torus. In the Euclidean plane it is true. This is the Jordan curve theorem.I believe it is also true for higher dimensional Euclidean space.

3. Nov 5, 2014

### tom.stoer

from Wikipedia (Jordan theorem, Jordan Schoenflies theorem) I do not get the impression that much is known for general topologies

thanks anyway

4. Nov 6, 2014

### tom.stoer

unfortunately these two theorems do not talk about manifolds different from R2; are there generalizations for other 2-dim. manifolds? are there results regarding necessary conditions for the surface and the embedding manifold?

Last edited: Nov 6, 2014
5. Nov 6, 2014

### Terandol

To prove the Jordan curve theorem in n-dimensions, one way is to compute the homology of $S^n-S$ where S is any subspace homeomorphic to $S^{n-1}$ and observe that $H_0(S^n-S) =\mathbb{Z^2}$ so $S^n-S$ has two components, an interior and an exterior. This can obviously be extended to $\mathbb{R}^n$ since removing a single point won't change the number of components. Similarly given a (connected) n-manifold M and some n-1 dimensional manifold N the condition $H_0(M-N)=\mathbb{Z}^2$ would guarantee you have an interior and an exterior though I'm not sure that this helps you any since I don't know any general method to compute $H_0(M-N)$. I've never heard of generalizations of Jordan curve theorem to more general manifold so there probably is no simple way to compute it.

I may be misinterpreting you here(I'm not sure if you still have some ambient manifold you are considering here or not) but if you mean given an arbitrary closed n-manifold M, when does there exist some n+1 dimensional manifold with boundary N such that $\partial N=M$ then this is exactly the same as asking when a closed manifold is null-cobordant. Oriented cobordism classes are determined by Stiefel-Whitney and Pontrjagin numbers so a closed manifold is a boundary (with matching orientation) if and only if these numbers are all the same as for the empty manifold (ie. equal to zero.) If you don't want to keep track of orientations then I believe all that matters are the Stiefel-Whitney numbers.

6. Nov 6, 2014

### tom.stoer

Terandol, many thanks for you reply; I have to read something regarding null-cobordisms, but I guess this is exactly what I am looking for.

7. Nov 6, 2014

### mathwonk

I agree the relation of interest is bordism, but it seems to me you asked when a closed submanifold bounds a piece of the given larger manifold M, not whether it bounds some other abstract manifold. Of course that would be a necessary condition. I don't know the precise answer, but it seems as if a necessary condition would be that the submanifold represent the zero homology class in the larger manifold, in n-1 homology group. For this I suppose it would be sufficient that the integral of every closed n-1 form on the given larger manifold over it be zero, or perhaps that it have intersection number zero with every closed curve in M.

By the way, the Jordan curve theorem seems a little weaker than what you want, which is apparently called Alexander's theorem. I.e. just knowing a sphere separates Euclidean space into 2 components does not yet guarantee that the bounded component is an embedded disc, and that the sphere is the boundary of that disc

In manifold theory, a manifold is called irreducible if every embedded sphere bounds a ball. There are various results generalizing the Jordan curve and Schoenflies theorems, e.g. every embedded torus in the 3 sphere bounds a solid torus. You may find these notes of Hatcher of interest:

http://www.math.cornell.edu/~hatcher/3M/3Mfds.pdf

Last edited: Nov 6, 2014