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Thread: Boundary of the Upper Hemisphere

  1. #1
    Junior Member
    May 2007

    Boundary of the Upper Hemisphere


    I am trying to do a problem from Munkres "Analysis on Manifolds", pg. 209 problem 4.

    The problem asks to show that E^(n-1) = S^(n-1)(a) n H^n is an n-1 manifold. It that asks you to find the boundary of E^(n-1).

    S^(n-1)(a) = the n-1 sphere of radius a = { x in R^n :|x| = a }
    H^n = the upper half-space in R^n = { x in R^n : x_k > 0 }

    Now I figure that I am suposed to use the following theorem:

    Let O be open in R^n; let f: O -> R be of class C^r. Let M be the set op points x for which f(x) = 0; let N be the set of points for which f(x) >= 0. Supposed M in non-empty and Df(x) has rank 1 at each point of M. Then N is an n-manifold in R^n and the boundary of N = M.

    However my problem is finding such a function. Am I approaching this problem the wrong way? I have been racking my brain on this for a while. Any help would be appreciated.

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  2. #2
    Super Member Rebesques's Avatar
    Jul 2005
    My house.
    Munkres... Ah, the memories... (actually never went deeper than page 4)

    Take $\displaystyle r=|x|, f(x)=a-r$. Then $\displaystyle M=E^{n-1}, N=$the closed unit ball of H^n.

    We check $\displaystyle <df(x^1,...,x^n),(h_1,...,h_n)>=\sum_k \frac{x^k}{r}h_k$, h in $\displaystyle T_{f(x)}E^{n-1}$ (tangent space). Since this generates an n-1 dimensional subspace - alternatively, because its kernel is only the zero vector - df(x) has rank 1 at all points of M.

    So N is an n-dimensional manifold, and its boundary is E^{n-1} - of dimension n-1.
    Last edited by Rebesques; Jun 6th 2007 at 04:27 PM.
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