# Trick Question?

1. Feb 6, 2008

### Kreizhn

1. The problem statement, all variables and given/known data

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

3. 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?

2. Feb 6, 2008

### 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.

3. Feb 6, 2008

### 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.

4. Feb 6, 2008

### 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).

5. Feb 6, 2008

### 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.

6. Feb 6, 2008

### 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.

7. Feb 6, 2008

### 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.

8. Feb 6, 2008

### 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.

9. Feb 6, 2008

### 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.

10. Feb 6, 2008

### 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.