# Thread: Boundary of the Upper Hemisphere

1. ## Boundary of the Upper Hemisphere

Hello.

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).

Notation:
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.

Thanks.

2. 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.