The Unapologetic Mathematician

Mathematics for the interested outsider

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 »

  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 […]

    Pingback by The Divergence Theorem « The Unapologetic Mathematician | November 22, 2011 | Reply

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