# The Unapologetic Mathematician

## 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