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.

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September 13, 2011 - Posted by | Differential Topology, Topology

7 Comments »

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  4. Thank you for the explanation. I have one question: say I have a patch U in M. what is the boundary of this patch, dU? why must it be equal to dM intersection with U ?
    if M has a topology, then open sets, or patches, can have boundary that doesn’t have to be in M’s boundary. Take for example S2 – 2D sphere. this manifold is boundaryless, however an open set is a “dome”, and it has a boundary (the circle surrounding it).

    Comment by hilabar | January 7, 2014 | Reply

  5. The problem is that the open set inside M will not contain any of its boundary. The open upper hemisphere that you suggest does not actually contain any of the bounding circle. You can get as close as you like, but as soon as you go to the boundary you leave the patch itself.

    Comment by John Armstrong | January 7, 2014 | Reply

  6. prehaps I’m confusing with boundary from topology. the upper hemisphere (“dome”), as an open set of the topological space S2, has a boundary – the circle around it. But you say, that as an open set of the manifold S2, it doesn’t have a boundary?
    And I will ask in addition – if I take a closed sub manifold of S2 – the upper hemisphere with the unit circle – is it a manifold with boundary inside a boundaryless manifold, S2?

    Comment by hilabar | January 7, 2014 | Reply

  7. So it’s a question of whether the patch not just “has” a boundary, but contains that boundary. The closed upper hemisphere is a manifold-with-boundary inside the sphere, as a manifold-with-(empty)-boundary.

    But I didn’t make that statement about submanifolds; I made it about coordinate patches. The closed upper hemisphere is not a valid coordinate patch for the sphere.

    Comment by John Armstrong | January 7, 2014 | Reply


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