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Homework Help: Trick Question?

  1. Feb 6, 2008 #1
    1. The problem statement, all variables and given/known data

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

    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 [itex]\mathbb{R}^{m+1} [/itex] via [itex]\{v_1, ... , v_m\} \rightarrow \{v_1, ..., v_m, N(p) \}[/itex] 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 [itex]\mathbb{R}^{m+1}[/itex], a contradiction. The only problem here is that we would require N(p) to be everywhere non-vanishing.

    Any ideas?
     
  2. jcsd
  3. Feb 6, 2008 #2
    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.
     
  4. Feb 6, 2008 #3

    Ben Niehoff

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    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.
     
  5. Feb 6, 2008 #4

    Ben Niehoff

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    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).
     
  6. Feb 6, 2008 #5
    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.
     
  7. Feb 6, 2008 #6
    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 [itex]\mathbb{R}^3[/itex]? It can obviously be embedded in [itex]\mathbb{R}^4[/itex] via the Whitney embedding theorem, but that doesn't help since the codimension would be 2, and we want a hypersurface of codim = 1.
     
  8. Feb 6, 2008 #7

    Ben Niehoff

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    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.
     
  9. Feb 6, 2008 #8
    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.
     
  10. Feb 6, 2008 #9

    Ben Niehoff

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    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.
     
  11. Feb 6, 2008 #10
    Well, I think we may have deliberately been given [itex]\mathbb{R}^{m+1}[/itex] 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.
     
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