Last time, while talking about homotopies as morphisms I said that I didn’t want to get too deeply into the reparameterization thing because it could get too complicated. But since when would I, of all people, shy away from 2-categories? In case it wasn’t obvious then, it’s because we’re actually going to extend in the other direction.
Given any two topological spaces and , we now don’t just have a set of continuous maps , we have a whole category consisting of those maps and homotopies between them. And I say that composition isn’t just a function that takes two (composable) maps and gives another one, it’s actually a functor.
So let’s say that we have maps , maps , and homotopies and . From this we can build a homotopy . The procedure is obvious: for any and , we just define
That is, the time- frame of the composed homotopy is the composition of the time- frames of the original homotopies. It should be straightforward to verify that this composition is (strictly) associative, and that the identity map — along with its identity homotopy — acts as an (also strict) identity.
What we need to show is that this composition is actually functorial. That is, we add maps and , change and to and , and add homotopies and . Then we have to check that
That is, if we stack onto and onto , and then compose them as defined above, we get the same result as if we compose with and with , and then stack the one onto the other.
This is pretty straightforward from a bird’s-eye view, but let’s check it in detail. On the left we have
Meanwhile, on the right we have
And so we do indeed have a 2-category with topological spaces as objects, continuous maps as 1-morphisms, and continuous homotopies as 2-morphisms. Of course, if we’re in a differential topological context we get a 2-category with differentiable manifolds as objects, smooth maps as 1-morphisms, and smooth homotopies as 2-morphisms.
We can think of homotopies between maps as morphisms in a category that has the maps as objects. In terms of the movie analogy, the composition is obvious: run the movie that takes you from map to map , then run the one that takes you from map to map .
In practice, the way we make these intuitive concepts explicit tend to get in the way. For one thing, the naïve approach would be to run the first movie from time to time , and then the second from time to time . But this gives us a function instead of . The usual way to handle this is by rescaling — run the first movie twice as fast from to , then the second movie twice as fast from to .
The problem with this is that it makes associativity weird. Let’s say we have homotopies , , and we want to compose. If we write
Then we have
these two are indeed homotopies from to , but they’re not the same homotopy! Associativity doesn’t seem to work for this composition.
The easy answer is to wave our hands and say they’re the same “up to reparameterization”. That is, there is some (invertible) function so that
It’s not hard to figure it out as an exercise.
The fact that we’re talking about two different things being “really the same” is a clue that there’s some higher categorical structure here that we’re “decategorifying” and forgetting about. In particular, we could flesh out the idea of reparameterizations as morphisms between homotopies, but that will quickly become more complicated than I want to get into.
Still, it’s worth pointing out that the reparameterization in the above exercise behaves like an associator, like we talked about in the context of monoidal categories. And, just like in that case, we will find left and right identity reparameterizations.
What’s the obvious homotopy to use as the identity on a map ? Clearly it’s just , independent of . I’ll leave the identity reparameterizations as another exercise. The upshot is that we have identity homotopies — “up to reparameterization” — for each map, which completes the definition of our category.
The common layman’s definition of topology generally involves rubber sheets or clay, with the idea that things are “the same” if they can be stretched, squeezed, or bent from one shape into the other. But the notions of topological equivalence we’ve been using up until now don’t really match up to this picture. Homeomorphism — or diffeomorphism, for differentiable manifolds — is about having continuous maps in either direction, but there’s nothing at all to correspond to the whole stretching and squeezing idea.
Instead, we have homotopy. But instead of saying that spaces are homotopic, we say that two maps are homotopic if the one can be “stretched and squeezed” into the other. And since this stretching and squeezing is a process to take place over time, we will view it sort of like a movie.
We say that a continuous function is a continuous homotopy from to if and for all . For any time , the map is a continuous map from to , which is sort of like a “frame” in the movie that takes us from to . As time passes over the interval, we highlight one frame at a time to watch the one function transform into the other.
To flip this around, imagine starting with a process of stretching and squeezing to turn one shape into another. In this case, when we say “shape” we really mean a subspace or submanifold of some outside space we occupy, like the three-dimensional space that contains our idiomatic doughnuts and coffee mugs. The maps in this case are the inclusions of the subspaces into the larger space.
Anyway, next imagine carrying out this process, but with a camera recording it at each step. Then cut out all the frames from the movie and stack them up. We see in each layer of this flipbook how the shape at that time is included into the larger space . That is, we have a homotopy.
Now, for an example: we say that a space is “contractible” if its inclusion into itself is homotopic to a map of the whole space to a single point within the space. As a particular example, the unit ball is contractible. Explicitly, we define a homotopy latex H(p,t)=(1-t)p$, which is certainly smooth; we can check that and , so at one end we have the identity map of into itself, while at the other we have the constant map sending all of to the single point at the origin.
We should be careful to point out that homotopy only requires that the function be continuous, and not invertible in any sense. In particular, there’s no guarantee that the frame for some fixed is a homeomorphism from onto its image. If it turns out that each frame is a homeomorphism of onto its image, then we say that is an “isotopy”.
If we take to be a manifold equipped with an orientation given by an orientation form . Then is nowhere zero, and for any positively oriented basis of at any point .
The integral of over all of must surely be even greater than the integral over the image of , since we can cover by orientation-preserving singular -cubes, and none of them can ever contribute a negative to the integral.
If we further suppose that is compact, we can cover by finitely many such singular cubes, and the integral on each is well-defined. Using a partition of unity as usual this shows us that the integral over all of exists and, further, must be strictly positive. In particular it’s not zero.
But now suppose that also has an empty boundary. Since is a top form, we know that — it’s closed in the de Rham cohomology. But we know that it cannot also be exact, for if for some -form then Stokes’ theorem would tell us that
since is empty.
And so if is a compact, oriented -manifold without boundary, then there must be some -forms which do not arise from taking the exterior derivatives of -forms. If is pseudo-Riemannian, so we have a Hodge star to work with, this tells us that we always have some functions on which are not the divergence of any vector field on .
At last we come to the version of Stokes’ theorem that people learn with that name in calculus courses. Ironically, unlike the fundamental theorem and divergence theorem special cases, Stokes’ theorem only works in dimension , where the differential can take us straight from a line integral over a -dimensional region to a surface integral over an -dimensional region.
So, let’s say that is some two-dimensional oriented surface inside a three-dimensional manifold , and let be its boundary. On the other side, let be a -form corresponding to a vector field . We can easily define the line integral
and Stokes’ theorem tells us that this is equal to
which means that the line integral of around the (oriented) boundary of is the same as the surface integral of the curl through itself. And this is exactly the old Stokes theorem from multivariable calculus.
I’ve been really busy with other projects — and work — of late, but I think we can get started again. We left off right after defining hypersurface integrals, which puts us in position to prove the divergence theorem.
So, let be a hypersurface with dimension in an -manifold , and let be a vector field on some region containing , so we can define the hypersurface integral
And if corresponds to a -form , we can write this as
where is the oriented volume form of and is a -form that “points perpendicular to” in . We take the given inner product and integrate it as a function against the volume form of itself.
A little juggling lets us rewrite:
where we take our -form and flip it around to the “perpendicular” form . Integrating this over involves projecting against , which is basically what the above formula connotes.
Now, let’s 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 like
Next we’ll jump over to the other end and take the divergence and integrate it over the region . In terms of the -form , this looks like
But Stokes’ theorem tells us that
which tells us in our vector field notation that
This is the divergence — or Gauss’, or Gauss–Ostrogradsky — theorem, and it’s yet another special case of Stokes’ theorem.