The Unapologetic Mathematician

Mathematics for the interested outsider

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 nth 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 nth 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:


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 Posted by | Differential Topology, Topology | 7 Comments

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 nth 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 nth component of y changes as the nth 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