# Manifold with Boundary

1. Mar 6, 2009

### eok20

Because of boundary points, I can sort of see intuitively why Euclidean half-space, i.e. {(x_1, ... , x_n) : x_n >= 0} is not a manifold, but is there a simple rigorous argument for why Euclidean half-space is not homeomorphic to an open set of R^n. I do not know too much topology and the topological properties I am familiar with (e.g. compactness, connectedness, fundamental group) are the same for both spaces.

Thanks.

2. Mar 6, 2009

### yyat

The usual proof uses http://en.wikipedia.org/wiki/Invariance_of_domain" [Broken] as follows:

Denote the half-space in R^n by H. Assume that f:R^n->H is a homeomorphism. Invariance of domain implies that f is an open map, in particular H=f(R^n) is open in R^n. This is false, hence such a homeomorphism can not exist.

In the case where f is a diffeomorphism, invariance of domain is an easy consequence of the implicit function theorem. This gives an elementary proof that R^n and H can not be diffeomorphic.

I do not know of an elementary proof that R^n and H are not homeomorphic.

In the case n=2 you can use the fundamental group to prove the result, since removing a boundary point from H does not change its fundamental group, but removing any point from R^2 does. This argument generalizes to higher dimensions with the higher homotopy groups (or homology theory).

Last edited by a moderator: May 4, 2017
3. Mar 7, 2009

### WWGD

Doesn't this imply that the boundary of the boundary of a manifold M with boundary is
the boundary itself? :

If H^n ={(x_1,..,x_n) in R^n : x^n >=0 } , as eok20 said, then:

DelM = { x in M : x has a neighborhood homeomorphic to a relatively open
subset of H^n } .

Then Del(DelM) = { y in DelM: y has a neighborhood homeo. to
a relatively open subset of H^n . }

(or should the dimension go down by 1 in Del(DelM)? )

Then this is satisfied, by definition, by every x in DelM , so Del(DelM)=DelM .

Seems to go against the usual Del^2 =0 , and even seems to contradict

the statement for, e.g., simplicial complexes, etc. unless the two uses of

the term boundary are different. Could anyone comment on that?.

4. Mar 7, 2009

### yyat

The boundary of a manifold with boundary is a manifold without boundary, so the boundary of the boundary of a manifold with boundary is the empty set. :tongue2:

Does this not apply to all points of M, by the definition of manifold with boundary?

Some different uses of the term "boundary" that I know of:

- The closure minus the interior of a subset of a topological space.

- The points of a topological manifold with boundary which do not have a neighbourhood homeomorphic to an open ball.

- An element in the image of the boundary operator of a chain complex.

5. Mar 7, 2009

### WWGD

O.K , yyat . I am not disagreeing with the truth of the claim, but it does not seem

to follow from the defs. : ( I cannot find the quote button again, for some reason)

Some different uses of the term "boundary" that I know of:

- The closure minus the interior of a subset of a topological space.

- The points of a topological manifold with boundary which do not have a neighbourhood homeomorphic to an open ball.

- An element in the image of the boundary operator of a chain complex.

Let's use definition number 2 , but instead , changing "manifold with boundary" for

manifold, and allowing a manifold to have an empty boundary if there are no such

points . I think this is still a correct definition of a boundary point for a general manifold.

( and I think this is actually how we determine that a manifold has no boundary: every

point has a 'hood homeomorphic to an open ball . Otherwise: how do we determine whether

a manifold has boundary or not ?)

Would you agree?

Then the boundary of the boundary is the set of points in the boundary that do not

have neighborhoods homeomorphic to R^n ( or an open ball ). This contains every

point in the boundary.

I think there is some substance here, and it is not just some sort of "legalese".

6. Mar 7, 2009

### yyat

Yes. The term "manifold with boundary" is often used (somewhat awkwardly) for a manifold which may or may not have a boundary.

Think of the unit disk $$D^2$$, which is a manifold with boundary $$S^1$$. All points in $$S^1$$ have a neighbourhood in $$S^1$$ that is homeomorphic to an open interval, so $$S^1$$ has no boundary.

The non-trivial part is exactly the invariance of domain I mentioned earlier in this thread. It ensures that a boundary point is not the same as an interior point.

7. Mar 9, 2009

### WWGD

Just a quick followup: An idea for a chart for the bdry. of a manifold with boundary.

It seems, from the few examples I know, that for points in the boundary of M

(with boundary charts (U,Phi) ) , that Phi(U) is open in R^(n-1) . Is this true

in general ? (so we could at least have C^0 charts ). OWise: how do we give

the bdry. points a chart?.

Also, my confusion was with bdry points of an n-manifold being an (n-1)-manifold,

which seemed to violate invariance of dimension, i.e., that the dimension of a mfld.

is constant in each component. I think the explanation for this is that the boundary

is a manifold _ under a different topology_ than the overlying ( if the word exists--

I vote it should :) ) manifold. Please ignore this if it is too confused; I have some

ideas here I am trying to work out.

Another result (locally constant dimension ) brought to you courtesy of

Invariance of Domain.

8. Mar 9, 2009

### yyat

Yes, the natural topology on R^(n-1) is the same as the induced topology from the halfspace H^n (basically because the metric (euclidean distance) in R^(n-1) is just the restriction of the metric on H^n). Hence, by the definition of the induced topology the intersection of an open subset of H^n with R^(n-1) is an open subset in R^(n-1).