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Thread: Orientation of submanifold

  1. #1
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    Orientation of submanifold

    Hello,

    I try to solve this problem:

    Let M, N be orientable Manifolds (of dimension m,n) and f:M->N a smooth submersion. =>$\displaystyle G:=f^{-1}(p)$ is orientable.

    Ok i have a proof, but it can't be correct, since the same "proof" would also work for any embedded submanifold.:

    We know that $\displaystyle G$ is an embedded submanifold. And therefore we have an canonical atlas for G. If $\displaystyle (U_i,f_i)$ is an atlas for M, then$\displaystyle (U_i \cap S, \pi \circ f_i) $is an atlas for G, whereas $\displaystyle \pi $is the projection on the first (m-n) coordinates.

    Now we have to show that det $\displaystyle d(\pi \circ f_i) \circ(\pi \circ f_j)^{-1})>0$

    And this follows, since the f_i are orientable.(?) This can't be. (because not every submanifold is orient.)
    But why is it >0?? We know also that the projection is in some nbh. equal to the map:
    $\displaystyle g_j\circ f \circ (f_i)^{-1}$ whereas $\displaystyle g_j$ is a coordinate map of N and $\displaystyle f_i$ a coordinate maß of M.
    Therefore we can put this forumula to the one above and get:
    det $\displaystyle d( (g_j\circ f \circ (f_i)^{-1})\circ f_i) \circ((g_j\circ f \circ (f_i)^{-1}) \circ f_j)^{-1})>0$

    Do you see, why this is greater than 0?

    Regards
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  2. #2
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    Use the local normal form of a submersion, f can be written as
    $\displaystyle f( x^1, ..., x^m) = (x^1, ..., x^n)$. And $\displaystyle (x^{n+1}, ..., x^m)$ are the coordinates of the sub-manifold G.

    Suppose $\displaystyle (u^1, ..., u^m)$ is another local chart of M and $\displaystyle (u^1, ..., u^n)$ the chart of N with non-empty intersection with the above charts respectively. Since $\displaystyle (u^1, ..., u^n)$ and $\displaystyle (x^1, ..., x^n)$ both are coordinates of N, $\displaystyle (u^1, ..., u^n)$ is only the function of $\displaystyle (x^1, ..., x^n)$, i.e. it does not depend on $\displaystyle (x^{n+1}, ..., x^m)$. Also $\displaystyle (u^{n+1}, ..., u^m)$ is only the function of $\displaystyle (x^{n+1}, ..., x^m)$.

    Since M and N are both orientable we can choose the sign of the coordinates so that the Jocobian $\displaystyle \frac{du^1 \wedge ... \wedge du^m}{dx^1 \wedge ... \wedge dx^m} > 0$ and $\displaystyle \frac{du^1 \wedge ... \wedge du^n}{dx^1 \wedge ... \wedge dx^n} > 0$, which implies that $\displaystyle \frac{du^{n+1} \wedge ... \wedge du^m}{dx^{n+1} \wedge ... \wedge dx^m} > 0$
    Last edited by xxp9; Feb 9th 2011 at 03:34 PM.
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