The Unapologetic Mathematician

Mathematics for the interested outsider

Manifolds with Boundary

Ever since we started talking about manifolds, we’ve said that they locally “look like” the Euclidean space \mathbb{R}^n. We now need to be a little more flexible and let them “look like” the half-space H^n=\{x\in\mathbb{R}^n\vert x^n\geq0\}.

Away from the subspace x^n=0, H^n is a regular n-dimensional manifold — we can always find a small enough ball that stays away from the edge — but on the boundary subspace it’s a different story. Just like we wrote the boundary of a singular cubic chain, we write \partial H^n=\{x\in\mathbb{R}^n\vert x^n=0\} for this boundary. Any point p\in M that gets sent to \partial H^n by a coordinate map x must be sent to \partial H^n by every coordinate map. Indeed, if y is another coordinate map on the same patch U around p, then the transition function y\circ x^{-1}:H^n\to H^n must be a homeomorphism from x(U) onto y(U), and so it must send boundary points to boundary points. Thus we can define the boundary \partial M to be the collection of all these points.

Locally, \partial M is an n-1-dimensional manifold. Indeed, if (U,x) is a coordinate patch around a point p\in\partial M then x(p)\in\partial H^n\cap x(U), and thus the preimage x^{-1}(\partial H^n) is an n-1-dimensional coordinate patch around p. Since every point is contained in such a patch, \partial M is indeed an n-1-dimensional manifold.

As for smooth structures on M and \partial M, we define them exactly as usual; real-valued functions on a patch (U,x) of M containing some boundary points are considered smooth if and only if the composition f\circ x^{-1} is smooth as a map from (a portion of) the half-space to \mathbb{R}. And such a function is smooth at a boundary point of the half-space if and only if it’s smooth in some neighborhood of the point, which extends — slightly — across the boundary.


September 13, 2011 Posted by | Differential Topology, Topology | 7 Comments