Is there a non-orientable compact submanifold in $\mathbb{R}^{m+1}$?

[Kreizhn]

Homework Statement

Is there a non-orientable compact m-dimensional boundry-less submanifold of \mathbb{R}^{m+1}?

The Attempt at a Solution

It should be noted that in the context of the situation, we've assumed that the manifolds we're dealing with are Hausdorff.

But I'm wondering if this isn't a trick question since all compact subspaces of a Hausdorff space are necessarily closed, and as such have a boundary.

And if it isn't a trick question, does anybody have any clues as how I can go about showing that there is/isn't one. I suspect that there isn't since m-dimensional submanifolds are often hypersurfaces that can be expressed as the preimage of a regular point of a homogeneous polynomial, and are therefore automatically orientable.

I've also thought about defining an orientation preserving map between the submanifold, say M, and \mathbb{R}^{m+1} via \{v_1, ... , v_m\} \rightarrow \{v_1, ..., v_m, N(p) \} where N(p) is a normal vector field at a point p mapping M to the tangent bundle. This would show that non-orientability in M would imply non-orientability of \mathbb{R}^{m+1}, a contradiction. The only problem here is that we would require N(p) to be everywhere non-vanishing.

Any ideas?

 

[Kreizhn]

After looking around a bit, I've seen that you can have boundaryless compact manifolds, and so that relieves the issue of it being a trick question - though I'm still somewhat uncertain as to why the whole compact -> closed -> boundaryless doesn't work. But I'm still not sure as how to proceed with the actual question.

 

[Ben Niehoff]

I'm unfamiliar with "Hausdorff", but a sphere is compact, closed, and boundaryless. It is orientable, however.

An example of a nonorientable compact, unbounded 2-manifold is the Klein bottle; but it is a submanifold of R4, not R3.

I suppose that might be a clue, but I'm not familiar enough with topology to tell for certain.

 

[Ben Niehoff]

After Wiki-ing "Hausdorff", it appears that a manifold being Hausdorff ought to imply that it has no self-intersections. I think that completes the clue (Klein bottle has self-intersections in R3, but not R4).

 

[Kreizhn]

It's been awhile since I've done any topology, so I must be screwing something up in my association between closed and having a boundary.

 

[Kreizhn]

I'm not entirely sure as to what clue you're suggesting, since haven't you just shown that the Klein bottle under our demand that manifolds (and consequently submanifolds) be Hausdorff isn't a submanifold of \mathbb{R}^3? It can obviously be embedded in \mathbb{R}^4 via the Whitney embedding theorem, but that doesn't help since the codimension would be 2, and we want a hypersurface of codim = 1.

 

[Ben Niehoff]

There is a piece missing from the question: namely, is this for all m, or for any m?

My clue would suggest that "no", there is no m for which one can satisfy the question.

 

[Kreizhn]

I'm pretty sure the question pertains to any m, but your clue isn't really a clue in that case. I too suspect that the answer is no, but providing an example that doesn't work isn't really a clue.

It's like saying

"Is there a countable non-standard model of arithmetic?"

And the clue being "Well, here's a non-standard model, but it's uncountable, so it's a clue that the answer is no."

Edit: I apologize for the example, I couldn't think of anything more relevant at the time.

 

[Ben Niehoff]

I meant, the Klein bottle might provide some clue as to how to prove the statement in higher dimensions by induction. I can't say for sure, though.

 

[Kreizhn]

Well, I think we may have deliberately been given \mathbb{R}^{m+1} since it is very easy to consider an ordered basis orientation over the tangent space. Thus, if I could define a non-vanishing normal vector, I could show that all such hypersurfaces are orientable - this is clearly not possible by the Hairy Ball Theorem. Thus, I need to find some other way to find a natural orientation on any submanifold.