## Oriented Manifolds with Boundary

Let’s take a manifold with boundary and give it an orientation. In particular, for each 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 .

Now, if is a boundary point, we’ve seen that we can define the tangent space , which contains — as an -dimensional subspace — . This subspace cuts the tangent space into two pieces, which we can distinguish as follows: if is a coordinate patch around with , then the image of near is a chunk of the hyperplane . The inside of corresponds to the area where , while the outside corresponds to .

And so the map sends a vector to a vector in , which either points up into the positive half-space, along the hyperplane, or down into the negative half-space. Accordingly, we say that is “inward-pointing” if lands in the first category, and “outward-pointing” if if lands in the last. We can tell the difference by measuring the th component — the value . 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 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 , which we do by specifying which bases of are “positively oriented” and which are “negatively oriented”. Specifically, if is a basis of $ then we say it’s positively oriented if for any outward-pointing the basis is positively oriented as a basis of , and similarly for negatively oriented bases.

We must check that this choice does define an orientation on . Specifically, if is another coordinate patch with , then we can set up the same definitions and come up with an orientation on each point of . If and are compatibly oriented, then and must be compatible as well.

So we assume that the Jacobian of is everywhere positive on . That is

We can break down and to strip off their last components. That is, we write , and similarly for . The important thing here is that when we restrict to the boundary the work as a coordinate map, as do the . So if we set and vary any of the other , the result of remains at zero. And thus we can expand the determinant above:

The determinant is therefore the determinant of the upper-left submatrix — which is the Jacobian determinant of the transition function on the intersection — times the value in the lower right.

If the orientations induced by those on and 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:

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

[…] 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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