# The Unapologetic Mathematician

## Stokes’ Theorem on Manifolds

Now we come back to Stokes’ theorem, but in the context of manifolds with boundary.

If $M$ is such a manifold of dimension $n$, and if $\omega$ is a compactly-supported $n$-form, then as usual we can use a partition of unity to break up the form into pieces, each of which is supported within the image of an orientation-preserving singular $n$-cube. For each singular cube $c$, either the image $c([0,1]^n)$ is contained totally within the interior of $M$, or it runs up against the boundary. In the latter case, without loss of generality, we can assume that $c([0,1]^n)\cap M$ is exactly the face $c_{n,0}([0,1]^{n-1})$ of $c$ where the $n$th coordinate is zero.

In the first case, our work is easy:

$\displaystyle\int\limits_Md\omega=\int\limits_cd\omega=\int\limits_{\partial d}\omega=\int\limits_{\partial M}\omega$

since $\omega$ is zero everywhere along the image of $\partial c$, and along $\partial M$.

In the other case, the vector fields $\frac{\partial}{\partial u^i}$ — in order — give positively-oriented basss of the tangent spaces of the standard $n$-cube. As $c$ is orientation, preserving, the ordered collection $\left(c_*\frac{\partial}{\partial u^1},\dots,c_*\frac{\partial}{\partial u^n}\right)$ gives positively-oriented bases of the tangent spaces of the image of $c$. The basis $\left(c_*\left(-\frac{\partial}{\partial u^n}\right),c_*\frac{\partial}{\partial u^1},\dots,c_*\frac{\partial}{\partial u^{n-1}}\right)$ is positively-oriented if and only if $n$ is even, since we have to pull the $n$th vector past $n-1$ others, picking up a negative sign for each one. But for a point $(a,0)$ with $a\in[0,1]^{-1}$, we see that

$\displaystyle c_{*(a,0)}\left(\frac{\partial}{\partial u^i}\right)=(c_{n,0})_{*a}\left(\frac{\partial}{\partial u^i}\right)$

for all $1\leq i\leq n-1$. That is, these image vectors are all within the tangent space of the boundary, and in this order. And since $c_*\left(-\frac{\partial}{\partial u^n}\right)$ is outward-pointing, this means that $c_{n,0}:[0,1]^{n-1}\to\partial M$ is orientation-preserving if and only if $n$ is even.

Now we can calculate

\displaystyle\begin{aligned}\int\limits_Md\omega&=\int\limits_cd\omega\\&=\int\limits_{\partial c}\omega\\&=\int\limits_{(-1)^nc_{n,0}}\omega\\&=(-1)^n\int\limits_{c_{n,0}}\omega\\&=(-1)^n(-1)^n\int\limits_{\partial M}\omega\\&=\int\limits_{\partial M}\omega\end{aligned}

where we use the fact that integrals over orientation-reversing singular cubes pick up negative signs, along with the sign that comes attached to the $(n,0)$ face of a singular $n$-cube to cancel each other off.

So in general we find

\displaystyle\begin{aligned}\int\limits_{\partial M}\omega&=\sum\limits_{\phi\in\Phi}\int\limits_{\partial M}\phi\omega\\&=\sum\limits_{\phi\in\Phi}\int\limits_Md(\phi\omega)\\&=\sum\limits_{\phi\in\Phi}\int\limits_M\left(d\phi\wedge\omega+\phi d\omega\right)\\&=\sum\limits_{\phi\in\Phi}\int\limits_Md\phi\wedge\omega+\int\limits_Md\omega\end{aligned}

The last sum is finite, since on of the support of $\omega$ all but finitely many of the $\phi$ are constantly zero, meaning that their differentials are zero as well. Since the sum is (locally) finite, we have no problem pulling it all the way inside:

$\displaystyle\sum\limits_{\phi\in\Phi}\int\limits_Md\phi\wedge\omega=\int\limits_Md\left(\sum\limits_{\phi\in\Phi}\phi\right)\wedge\omega=\int\limits_Md\left(1\right)\wedge\omega=0$

so the sum cancels off, leaving just the integral, as we’d expect. That is, under these circumstances,

$\displaystyle\int\limits_Md\omega=\int\limits_{\partial M}\omega$

which is Stokes’ theorem on manifolds.

September 16, 2011

## 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