# The Unapologetic Mathematician

## Oriented Manifolds with Boundary

Let’s take a manifold with boundary $M$ and give it an orientation. In particular, for each $p\in M$ we can classify any ordered basis as either positive or negative with respect to our orientation. It turns out that this gives rise to an orientation on the boundary $\partial M$.

Now, if $p\in\partial M$ is a boundary point, we’ve seen that we can define the tangent space $\mathcal{T}_pM$, which contains — as an $n-1$-dimensional subspace — $\mathcal{T}_p(\partial M)$. This subspace cuts the tangent space into two pieces, which we can distinguish as follows: if $(U,x)$ is a coordinate patch around $p$ with $x(p)=0$, then the image of $\partial M$ near $p$ is a chunk of the hyperplane $x^n=0$. The inside of $M$ corresponds to the area where $x^n>0$, while the outside corresponds to $x^n>0$.

And so the map $x_{*p}$ sends a vector $v\in\mathcal{T}_pM$ to a vector in $\mathcal{T}_0\mathbb{R}^n$, which either points up into the positive half-space, along the hyperplane, or down into the negative half-space. Accordingly, we say that $v$ is “inward-pointing” if $x_{*p}(v)$ lands in the first category, and “outward-pointing” if if lands in the last. We can tell the difference by measuring the $n$th component — the value $\left[x_{*p}(v)\right](u^n)=v(u^n\circ x)=v(x^n)$. If this value is positive the vector is inward-pointing, while if it’s positive the vector is outward-pointing.

This definition may seem to depend on our choice of coordinate patch, but the division of $\mathcal{T}_pM$ into halves is entirely geometric. The only role the coordinate map plays is in giving a handy test to see which half a vector lies in.

Now we are in a position to give an orientation to the boundary $\partial M$, which we do by specifying which bases of $\partial M$ are “positively oriented” and which are “negatively oriented”. Specifically, if $v_1,\dots,v_{n-1}$ is a basis of $\mathcal{T}_p(\partial M)\subseteq\mathcal{T}_pM$\$ then we say it’s positively oriented if for any outward-pointing $v\in\mathcal{T}_pM$ the basis $v,v_1,\dots,v_{n-1}$ is positively oriented as a basis of $\mathcal{T}_pM$, and similarly for negatively oriented bases.

We must check that this choice does define an orientation on $\partial M$. Specifically, if $(V,y)$ is another coordinate patch with $y(p)=0$, then we can set up the same definitions and come up with an orientation on each point of $V\cap\partial M$. If $U$ and $V$ are compatibly oriented, then $U\cap\partial M$ and $V\cap\partial M$ must be compatible as well.

So we assume that the Jacobian of $y\circ x^{-1}$ is everywhere positive on $U\cap V$. That is $\displaystyle\det\left(\frac{\partial(y^i\circ x^{-1})}{\partial u^j}\right)>0$

We can break down $x$ and $y$ to strip off their last components. That is, we write $x(q)=(\tilde{x}(q),x^n(q))$, and similarly for $y$. The important thing here is that when we restrict to the boundary $U\cap\partial M$ the $\tilde{x}$ work as a coordinate map, as do the $\tilde{y}$. So if we set $u^n=0$ and vary any of the other $u^j$, the result of $y^n(x^{-1}(u))$ remains at zero. And thus we can expand the determinant above: \displaystyle\begin{aligned}0&<\det\left(\frac{\partial(y^i\circ x^{-1})}{\partial u^j}\bigg\vert_{u^n=0}\right)\\&=\det\begin{pmatrix}\displaystyle\frac{\partial(y^1\circ x^{-1})}{\partial u^1}\bigg\vert_{u^n=0}&\cdots&\displaystyle\frac{\partial(y^1\circ x^{-1})}{\partial u^{n-1}}\bigg\vert_{u^n=0}&\displaystyle\frac{\partial(y^1\circ x^{-1})}{\partial u^n}\bigg\vert_{u^n=0}\\\vdots&\ddots&\vdots&\vdots\\\displaystyle\frac{\partial(y^{n-1}\circ x^{-1})}{\partial u^1}\bigg\vert_{u^n=0}&\cdots&\displaystyle\frac{\partial(y^{n-1}\circ x^{-1})}{\partial u^{n-1}}\bigg\vert_{u^n=0}&\displaystyle\frac{\partial(y^{n-1}\circ x^{-1})}{\partial u^n}\bigg\vert_{u^n=0}\\{0}&\cdots&0&\displaystyle\frac{\partial(y^n\circ x^{-1})}{\partial u^n}\bigg\vert_{u^n=0}\end{pmatrix}\end{aligned}

The determinant is therefore the determinant of the upper-left $n\times n$ submatrix — which is the Jacobian determinant of the transition function $\tilde{y}\circ\tilde{x}^{-1}$ on the intersection $(U\cap\partial M)\cap(V\cap\partial M$ — times the value in the lower right.

If the orientations induced by those on $U$ and $V$ are to be compatible, then this Jacobian determinant on the boundary must be everywhere positive. Since the overall determinant is everywhere positive, this is equivalent to the lower-right component being everywhere positive on the boundary. That is: $\displaystyle\frac{\partial(y^n\circ x^{-1})}{\partial u^n}\bigg\vert_{u^n=0}$

But this asks how the $n$th component of $y$ changes as the $n$th component of $x$ increases; as we move away from the boundary. But, at least where we start on the boundary, $y^n$ can do nothing but increase! And thus this partial derivative must be positive, which proves our assertion.

September 16, 2011 - Posted by | Differential Topology, Topology

## 1 Comment »

1. […] say that the surface is the boundary of some -dimensional submanifold of , and that it’s outward-oriented. That is, we can write . Then our hypersurface integral looks […]

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