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.
If we have a manifold with boundary , then at all the interior points it looks just like a regular manifold, and so the tangent space is just the same as ever. But what happens when we consider a point ?
Well, if is a chart around with , then we see that the part of the boundary within — — is the surface . The point has a perfectly good tangent space as a point in : . We will consider this to be the tangent space of at zero, even though half of its vectors “point outside” the space itself.
We can use this to define the tangent space . Indeed, the function goes from to and takes the point to ; it only makes sense to define as .
This is all well and good algebraically, but geometrically it seems that we’re letting tangent vectors spill “off the edge” of . But remember our geometric characterization of tangent vectors as equivalence classes of curves — of “directions” that curves can go through . Indeed, a curve could well run up to the edge of at the point in any direction that — if continued — would leave the manifold through its boundary. The geometric definition makes it clear that this is indeed the proper notion of the tangent space at a boundary point.
Now, let be the function we get by restricting to the boundary . The function sends the boundary to the boundary — at least locally — and there is an inclusion . On the other hand, there is an inclusion , which then sends to — again, at least locally. That is, we have the equation
Taking the derivative, we see that
But must be the inclusion of the subspace into the tangent space . That is, the tangent vectors to the boundary manifold are exactly those tangent vectors on the boundary that sends to tangent vectors in whose th component is zero.
Ever since we started talking about manifolds, we’ve said that they locally “look like” the Euclidean space . We now need to be a little more flexible and let them “look like” the half-space .
Away from the subspace , is a regular -dimensional manifold — we can always find a small enough ball that stays away from the edge — but on the boundary subspace it’s a different story. Just like we wrote the boundary of a singular cubic chain, we write for this boundary. Any point that gets sent to by a coordinate map must be sent to by every coordinate map. Indeed, if is another coordinate map on the same patch around , then the transition function must be a homeomorphism from onto , and so it must send boundary points to boundary points. Thus we can define the boundary to be the collection of all these points.
Locally, is an -dimensional manifold. Indeed, if is a coordinate patch around a point then , and thus the preimage is an -dimensional coordinate patch around . Since every point is contained in such a patch, is indeed an -dimensional manifold.
As for smooth structures on and , we define them exactly as usual; real-valued functions on a patch of containing some boundary points are considered smooth if and only if the composition is smooth as a map from (a portion of) the half-space to . And such a function is smooth at a boundary point of the half-space if and only if it’s smooth in some neighborhood of the point, which extends — slightly — across the boundary.
Let’s say we have a diffeomorphism from one -dimensional manifold to another. Since is both smooth and has a smooth inverse, we must find that the Jacobian is always invertible; the inverse of at is at . And so — assuming is connected — the sign of the determinant must be constant. That is, is either orientation preserving or orientation-reversing.
Remembering that diffeomorphism is meant to be our idea of what it means for two smooth manifolds to be “equivalent”, this means that is either equivalent to or to . And I say that this equivalence comes out in integrals.
So further, let’s say we have a compactly-supported -form on . We can use to pull back from to . Then I say that
where the positive sign holds if is orientation-preserving and the negative if is orientation-reversing.
In fact, we just have to show the orientation-preserving side, since if is orientation-reversing from to then it’s orientation-preserving from to , and we already know that integrals over are the negatives of those over . Further, we can assume that the support of fits within some singular cube , for if it doesn’t we can chop it up into pieces that do fit into cubes , and similarly chop up into pieces that fit within corresponding singular cubes .
But now it’s easy! If is supported within the image of an orientation-preserving singular cube , then must be supported within , which is also orientation-preserving since both and are, by assumption. Then we find
In this sense we say that integrals are preserved by (orientation-preserving) diffeomorphisms.
If we have an oriented manifold , then we know that the underlying manifold has another orientation available; if is a top form that gives its orientation, then gives it the opposite orientation. We will write for the same underlying manifold equipped with this opposite orientation.
Now it turns out that the integrals over the same manifold with the two different orientations are closely related. Indeed, if is any -form on the oriented -manifold , then we find
Without loss of generality, we may assume that is supported within the image of a singular cube . If not, we break it apart with a partition of unity as usual.
Now, if is orientation-preserving, then we can come up with another singular cube that reverses the orientation. Indeed, let . It’s easy to see that sends to and preserves all the other . Thus it sends to its negative, which shows that it’s an orientation-reversing mapping from the standard -cube to itself. Thus we conclude that the composite is an orientation-reversing singular cube with the same image as .
But then is an orientation-preserving singular cube containing the support of , and so we can use it to calculate integrals over . Working in from each side of our proposed equality we find
We know that we can write
for some function . And as we saw above, sends to its negative. Thus we conclude that
meaning that when we calculate the integral over we’re using the negative of the form on that we use when calculating the integral over .
This makes it even more sensible to identify an orientation-preserving singular cube with its image. When we write out a chain, a positive multiplier has the sense of counting a point in the domain more than once, while a negative multiplier has the sense of counting a point with the opposite orientation. In this sense, integration is “additive” in the domain of integration, as well as linear in the integrand.
The catch is that this only works when is orientable. When this condition fails we still know how to integrate over chains, but we lose the sense of orientation.
Well, paradoxically, we start by getting smaller. Specifically, I say that we can always find an orientable open cover of such that each set in the cover is contained within the image of a singular cube.
We start with any orientable atlas, which gives us a coordinate patch around any point we choose. Without loss of generality we can pick the coordinates such that . There must be some open ball around whose closure is completely contained within ; this closure is itself the image of a singular cube, and the ball obviously contained in its closure. Hitting everything with we get an open set — the inverse image of the ball — contained in the image of a singular cube, all of which contains . Since we can find such a set around any point we can throw them together to get an open cover of .
So, what does this buy us? If is any compactly-supported form on an -dimensional manifold , we can cover its support with some open subsets of , each of which is contained in the image of a singular -cube. In fact, since the support is compact, we only need a finite number of the open sets to do the job, and throw in however many others we need to cover the rest of .
We can then find a partition of unity subordinate to this cover of . We can decompose into a (finite) sum:
which is great because now we can define
But now we must be careful! What if this definition depends on our choice of a suitable partition of unity? Well, say that is another such partition. Then we can write
so we get the same answer no matter which partition we use.
We’ve defined how to integrate forms over chains made up of singular cubes, but we still haven’t really defined integration on manifolds. We’ve sort of waved our hands at the idea that integrating over a cube is the same as integrating over its image, but this needs firming up. In particular, we will restrict to oriented manifolds.
To this end, we start by supposing that an -form is supported in the image of an orientation-preserving singular -cube . Then we will define
Indeed, here the image of is some embedded submanifold of that even agrees with its orientation. And since is zero outside of this submanifold it makes sense to say that the integral over the submanifold — over the singular cube — is the same as the integral over the whole manifold.
What if we have two orientation-preserving singular cubes and that both contain the support of ? It only makes sense that they should give the same integral. And, indeed, we find that
where we use to reparameterize our integral. Of course, this function may not be defined on all of , but it’s defined on , where is supported, and that’s enough.
Of course now that we have more structure, we have more structured maps. But this time it’s not going to be quite so general; we will only extend our notion of an embedding, and particularly of an embedding in codimension zero.
That is, let be an embedding of manifolds where each of and has dimension . Since their dimensions are the same, the codimension of this embedding — the difference between the dimension of and that of — is . If and are both oriented, then we say that preserves the orientation if the pullback of any -form on which gives the chosen orientation gives us an -form on which gives its chosen orientation. We easily see that this concept wouldn’t even make sense if and didn’t have the same dimension.
More specifically, let and be oriented by -forms and , respectively. If for some smooth, everywhere-positive , we say that is orientation-preserving. The specific choices of and don’t matter; if gives the same orientation on then we must have for some smooth, everywhere-positive , and ; if gives the same orientation on then we must have for some smooth, everywhere-positive , and .
In fact, we have a convenient way of coming up with test forms. Let be a coordinate patch on around whose native orientation agrees with that of , and let be a similar coordinate patch on around . Now we have neighborhoods of and between which is a diffeomorphism, and we have top forms and in and , respectively. Pulling back the latter form we find
That is, the pullback of the (local) orientation form on differs from the (local) orientation form on by a factor of the Jacobian determinant of the function with respect to these coordinate maps. This repeats what we saw in the case of transition functions between coordinates. And so if whenever we pick local coordinates on and we find an everywhere-positive Jacobian determinant of , then preserves orientation.
If we orient a manifold by picking an everywhere-nonzero top form , then it induces an orientation on each coordinate patch . Since each one also comes with its own orientation form, we can ask whether they’re compatible or not.
And it’s easy to answer; just calculate
and if the answer is positive then the two are compatible, while if it’s negative then they’re incompatible. But no matter; just swap two of the coordinates and we have a new coordinate map on whose own orientation is compatible with .
This shows that we can find an atlas on all of whose patches have compatible orientations. Given any atlas at all for , either use a coordinate patch as is or swap two of its coordinates depending on whether its native orientation agrees with or not. In fact, if we’re already using a differentiable structure — containing all possible coordinate patches which are (smoothly) compatible with each other — then we just have to throw out all the patches which are (orientably) incompatible with .
The converse, as it happens, is also true: if we can find an atlas for such that for any two patches and the Jacobian of the transition function is everywhere positive on the intersection , then we can find an everywhere-nonzero top form to orient the whole manifold.
Basically, what we want is to patch together enough of the patches’ native orientations to cover the whole manifold. And as usual for this sort of thing, we pick a partition of unity subordinate to our atlas. That is, we have a countable, locally finite collection of functions so that is supported within the patch . Then we define the -form on by
and by outside of . Adding up all the gives us our top form; the sum makes sense because it’s locally finite, and at each point we don’t have to worry about things canceling off because each orientation form is a positive multiple of each other one wherever they’re both nonzero.
Any coordinate patch in a manifold is orientable. Indeed, the image is orientable — we can just use to orient — and given a choice of top form on we can pull it back along to give an orientation of itself.
But what happens when we bring two patches and together? They may each have orientations given by top forms and . We must ask whether they are “compatible” on their overlap. And compatibility means each one picks out the same end of at each point. But this just means that — when restricted to the intersection — for some everywhere-positive smooth function .
Another way to look at the same thing is to let be the pullback , and . Then we must ask what this function is. It must exist even if the orientations are incompatible, since is never zero, but what is it?
A little thought gives us our answer: is the Jacobian determinant of the coordinate transformation from one patch to the other. Indeed, we use the Jacobian to change bases on the cotangent bundle, and transforming between these top forms amounts to taking the determinant of the transformation between the -forms and .
So what does this mean? It tells us that if the Jacobian of the coordinate transformation relating two coordinate patches is everywhere positive, then the coordinates have compatible orientations. On the other hand, if the coordinate transformation’s Jacobian is everywhere negative, then the coordinates also have compatible orientations. Why? because even though the sample orientations differ, we can just use and , which do give the same orientation everywhere.
The problem comes up when the Jacobian is sometimes positive and sometimes negative. Now, it can never be zero, but if the intersection has more than one component it may be positive on one and negative on the other. Then if you pick orientations which coincide on one part of the overlap, they must necessarily disagree on the other part, and no coherent orientation can be chosen for the whole manifold.
I won’t go into this example in full detail yet, but this is essentially what happens with the famous Möbius strip: glue two strips of paper together at one end and we can coherently orient their union. But if we give a half-twist to the other ends before gluing them, we cannot coherently orient the result. The Jacobian is positive on the one overlap and negative on the other.