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Math Help - orientation of manifolds

  1. #1
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    orientation of manifolds

    Hello,

    I have a question about the topic in the headline.
    We have defined a oriented manifold as a manifold with a atlas, s.t. the determinant of the differential of chart changes is >0, i.e. det d(x\circ y^{-1})>0 forall x,y charts.

    I think it is true that if we have a non orientable manifold, then the der above has to be 0 for some charts.
    To put it another way, if the manifold is not or., then it can't appear that der(...)<0.

    My questions are:
    1)Is my conjecture correct?
    2)Do you know a argument, why there has to be a atlas, s.t. det(..)>0 if we have a atlas s.t. det(..)<0?

    I try to put a "-" to each chart. putthen the "-" cancel out, since we have the composition of thwo charts.

    Regards
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  2. #2
    Senior Member Tinyboss's Avatar
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    The requirement is that if your atlas is \{(U_\alpha,h_\alpha)\}_{\alpha\in A}, then for every \alpha,\beta\in A and every p\in U_\alpha\cap U_\beta, you have \mathrm{det}(D(h_\beta\circ h_\alpha^{-1})|_{h_\alpha(p)})>0. It sounds like you might be forgetting that the matrix (and hence its determinant) depends on which point in the chart intersection you look at it from.

    Now, since determinant is continuous and we never have determinant zero (these are diffeomorphisms from \mathbb{R}^m\to\mathbb{R}^m), the determinant is either all positive or all negative on each connected component of U_\alpha\cap U_\beta. But of course there might be several components--for example, the usual construction of the Mobius bundle over S^1.

    To answer #2 (if I understand it correctly), consider S^n with the atlas given by stereographic projection. There are two charts, their intersection is S^n-\{N,S\} which is connected for n>1, and the determinant can be +1 or -1 depending on how you define your projection.
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  3. #3
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    Thank you
    Last edited by Sogan; February 3rd 2011 at 02:56 PM.
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  4. #4
    Senior Member Tinyboss's Avatar
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    Go the other way--assuming you have an oriented atlas for AxB, show you can "project" it to an oriented atlas of A. And yes, that's what the charts on the product manifold look like.
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  5. #5
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    Hello tinyboss,

    i think it is not so obvious to "project" the atlas. What what you mean? I try to solve a similar problem. I mean any coordinate map of AxB is in general of the form:
    (f x g)(x,y):=(f(x,y) x g(x,y)). That is we habe functions which depends on both elements! How can i project this?

    Regards
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  6. #6
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    Can nobody help me?
    Once again:
    Let M,N be manifolds of dimension m,n respectively.
    We know that M is not orientable. And want to show that MxN is also not orientable.

    If we assume that MxN is not orientable, we have to conclude that M has to be orientable,too.

    But why?


    Regards
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