# The Unapologetic Mathematician

## The Tangent Space at the Boundary

If we have a manifold with boundary $M$, then at all the interior points $M\setminus\partial M$ it looks just like a regular manifold, and so the tangent space is just the same as ever. But what happens when we consider a point $p\in\partial M$?

Well, if $(U,x)$ is a chart around $p$ with $x(p)=0\in H^n$, then we see that the part of the boundary within $U$ $U\cap\partial M$ — is the surface $\{q\in U\vert x^n(q)=0\}$. The point $0\in H^n\subseteq\mathbb{R}^n$ has a perfectly good tangent space as a point in $\mathbb{R}^n$: $\mathcal{T}_0\mathbb{R}^n$. We will consider this to be the tangent space of $H^n$ at zero, even though half of its vectors “point outside” the space itself.

We can use this to define the tangent space $\mathcal{T}_pM$. Indeed, the function $x^{-1}$ goes from $H^n$ to $M$ and takes the point $0$ to $p$; it only makes sense to define $\mathcal{T}_pM$ as $(x^{-1})_{*0}\left(\mathcal{T}_0\mathbb{R}^n\right)$.

This is all well and good algebraically, but geometrically it seems that we’re letting tangent vectors spill “off the edge” of $M$. But remember our geometric characterization of tangent vectors as equivalence classes of curves — of “directions” that curves can go through $p$. Indeed, a curve could well run up to the edge of $M$ at the point $p$ in any direction that — if continued — would leave the manifold through its boundary. The geometric definition makes it clear that this is indeed the proper notion of the tangent space at a boundary point.

Now, let $y$ be the function we get by restricting $x$ to the boundary $\partial M$. The function $y^{-1}$ sends the boundary $\partial H^n\cong\mathbb{R}^{n-1}\times\{0\}$ to the boundary $\partial M$ — at least locally — and there is an inclusion $i:\partial M\to M$. On the other hand, there is an inclusion $j:\mathbb{R}^{n-1}\times\{0\}\to\mathbb{R}^n$, which $x^{-1}$ then sends to $U$ — again, at least locally. That is, we have the equation $\displaystyle i\circ y^{-1}=x^{-1}\circ j$

Taking the derivative, we see that $\displaystyle i_*\left((y^{-1})_*\left(\mathcal{T}_0\mathbb{R}^{n-1}\right)\right)=(x^{-1})_*\left(j_*\left(\mathcal{T}_0\mathbb{R}^{n-1}\right)\right)$

But $i_*$ must be the inclusion of the subspace $\mathcal{T}_p(\partial M)$ into the tangent space $\mathcal{T}_pM$. That is, the tangent vectors to the boundary manifold are exactly those tangent vectors on the boundary that $x_*$ sends to tangent vectors in $\mathcal{T}_0\mathbb{R}^n$ whose $n$th component is zero.

September 15, 2011