# The Unapologetic Mathematician

## Homotopies as 2-Morphisms

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 $M$ and $N$, we now don’t just have a set of continuous maps $\hom(M,N)$, 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 $f_1,f_2:M\to N$, maps $g_1,g_2:N\to P$, and homotopies $F:f_1\to f_2$ and $G:g_1\to g_2$. From this we can build a homotopy $G\circ F:g_1\circ f_1\to g_2\circ f_2$. The procedure is obvious: for any $t\in[0,1]$ and $m\in M$, we just define

$\displaystyle[G\circ F](m,t)=G(F(m,t),t)$

That is, the time-$t$ frame of the composed homotopy is the composition of the time-$t$ 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 $f_3:M\to N$ and $g_3:N\to P$, change $F$ and $G$ to $F_1$ and $G_1$, and add homotopies $F_2:f_2\to f_3$ and $G_2:g_2\to g_3$. Then we have to check that

$(G_2*G_1)\circ(F_2*F_1)=(G_2\circ F_2)*(G_1\circ F_1)$

That is, if we stack $G_2$ onto $G_1$ and $F_2$ onto $F_1$, and then compose them as defined above, we get the same result as if we compose $G_2$ with $F_2$ and $G_1$ with $F_1$, 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

\displaystyle\begin{aligned}{}[(G_2*G_1)\circ(F_2*F_1)](m,t)&=[G_2*G_1]([F_2*F_1](m,t),t)\\&=\left\{\begin{array}{lr}G_1([F_2*F_1](m,t),2t)&0

Meanwhile, on the right we have

\displaystyle\begin{aligned}{}[(G_2\circ F_2)*(G_1\circ F_1)](m,t)&=\left\{\begin{array}{lr}[G_1\circ F_1](m,2t)&0

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.

November 30, 2011

## Homotopies as 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 $f$ to map $g$, then run the one that takes you from map $g$ to map $h$.

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 $0$ to time $1$, and then the second from time $1$ to time $2$. But this gives us a function $H:M\times[0,2]\to N$ instead of $H:M\times[0,1]\to N$. The usual way to handle this is by rescaling — run the first movie twice as fast from $t=0$ to $t=\frac{1}{2}$, then the second movie twice as fast from $t=\frac{1}{2}$ to $t=1$.

The problem with this is that it makes associativity weird. Let’s say we have homotopies $F:f\to g$, $G:g\to h$, and $H:h\to k$ we want to compose. If we write

$\displaystyle [G*F](p,t)=\left\{\begin{array}{lr}F(p,2t)&0

Then we have

$\displaystyle [H*(G*F)](p,t)=\left\{\begin{array}{lr}F(p,4t)&0

and

$\displaystyle [(H*G)*F](p,t)=\left\{\begin{array}{lr}F(p,2t)&0

these two are indeed homotopies from $f$ to $k$, 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 $r:[0,1]\to[0,1]$ so that

$\displaystyle[H*(G*F)](p,t)=[(H*G)*F](p,r(t))$

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 $f$? Clearly it’s just $I_f(p,t)=f(p)$, independent of $t$. 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.

November 29, 2011

## Homotopy

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 $f_0,f_1:M\to N$ 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 $H:M\times[0,1]\to N$ is a continuous homotopy from $f_0$ to $f_1$ if $H(p,0)=f_0(p)$ and $H(p,1)=f_1(p)$ for all $p\in M$. For any time $t\in[0,1]$, the map $p\mapsto H(p,t)$ is a continuous map from $M$ to $N$, which is sort of like a “frame” in the movie that takes us from $f_0$ to $f_1$. 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 $M$ at that time is included into the larger space $N$. 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 $B^n\subseteq\mathbb{R}^n$ is contractible. Explicitly, we define a homotopy $H:B^n\times[0,1]\to B^n$latex H(p,t)=(1-t)p\$, which is certainly smooth; we can check that $H(p,0)=p$ and $H(p,1)=0$, so at one end we have the identity map of $B^n$ into itself, while at the other we have the constant map sending all of $B^n$ to the single point at the origin.

We should be careful to point out that homotopy only requires that the function $H$ be continuous, and not invertible in any sense. In particular, there’s no guarantee that the frame $p\mapsto H(p,t)$ for some fixed $t$ is a homeomorphism from $M$ onto its image. If it turns out that each frame is a homeomorphism of $M$ onto its image, then we say that $H$ is an “isotopy”.

November 29, 2011

## Compact Oriented Manifolds without Boundary have Nontrivial Homology

If we take $M$ to be a manifold equipped with an orientation given by an orientation form $\omega$. Then $\omega$ is nowhere zero, and $\omega(v_1,\dots,v_n)>0$ for any positively oriented basis $\{v_i\}$ of $\mathcal{T}_pM$ at any point $p\in M$.

Next we take $c:[0,1]^n\to M$ to be an orientation-preserving embedding — a singular cube of top dimension. Then the pullback $c^*\omega=fdu^1\wedge\dots\wedge du^n$ for some strictly-positive function $f$. We conclude that

$\displaystyle\int\limits_c\omega=\int\limits_{[0,1]^n}fdu^1\wedge\dots\wedge du^n=\int\limits_{[0,1]^n}f(u^1,\dots,u^n)\,d(u^1,\dots,u^n)>0$

The integral of $\omega$ over all of $M$ must surely be even greater than the integral over the image of $c$, since we can cover $M$ by orientation-preserving singular $n$-cubes, and none of them can ever contribute a negative to the integral.

If we further suppose that $M$ is compact, we can cover $M$ 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 $M$ exists and, further, must be strictly positive. In particular it’s not zero.

But now suppose that $M$ also has an empty boundary. Since $\omega$ is a top form, we know that $d\omega=0$ — it’s closed in the de Rham cohomology. But we know that it cannot also be exact, for if $\omega=d\eta$ for some $n-1$-form $\eta$ then Stokes’ theorem would tell us that

$\displaystyle\int\limits_M\omega=\int\limits_Md\eta=\int\limits_{\partial M}\eta=0$

since $\partial M$ is empty.

And so if $M$ is a compact, oriented $n$-manifold without boundary, then there must be some $n$-forms which do not arise from taking the exterior derivatives of $n-1$-forms. If $M$ is pseudo-Riemannian, so we have a Hodge star to work with, this tells us that we always have some functions on $M$ which are not the divergence of any vector field on $M$.

November 24, 2011

## The Classical Stokes Theorem

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 $n=3$, where the differential can take us straight from a line integral over a $1$-dimensional region to a surface integral over an $n-1$-dimensional region.

So, let’s say that $S$ is some two-dimensional oriented surface inside a three-dimensional manifold $M$, and let $c=\partial S$ be its boundary. On the other side, let $\alpha$ be a $1$-form corresponding to a vector field $F$. We can easily define the line integral

$\displaystyle\int\limits_c\alpha$

and Stokes’ theorem tells us that this is equal to

$\displaystyle\int\limits_{\partial S}\alpha=\int\limits_Sd\alpha$

Now if we define $\beta=*d\alpha$ as another $1$-form then we know it corresponds to the curl $\nabla\times F$. But on the other hand we know that in dimension $3$ we have $*^2=1$, and so we find $*\beta=**d\alpha=d\alpha$ as well. Thus we have

$\displaystyle\int\limits_c\alpha=\int\limits_S*\beta$

which means that the line integral of $F$ around the (oriented) boundary $c$ of $S$ is the same as the surface integral of the curl $\nabla\times F$ through $S$ itself. And this is exactly the old Stokes theorem from multivariable calculus.

November 23, 2011

## The Divergence Theorem

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 $S$ be a hypersurface with dimension $n-1$ in an $n$-manifold $M$, and let $F$ be a vector field on some region containing $S$, so we can define the hypersurface integral

$\displaystyle\int\limits_SF\cdot dS$

And if $F$ corresponds to a $1$-form $\alpha$, we can write this as

$\displaystyle\int\limits_S\langle\alpha,*\omega_S\rangle\omega_S$

where $\omega_S$ is the oriented volume form of $S$ and $*\omega_S$ is a $1$-form that “points perpendicular to” $S$ in $M$. We take the given inner product and integrate it as a function against the volume form of $S$ itself.

A little juggling lets us rewrite:

$\displaystyle\int\limits_S*\alpha$

where we take our $1$-form and flip it around to the “perpendicular” $n-1$ form $*\alpha$. Integrating this over $S$ involves projecting against $\omega_S$, which is basically what the above formula connotes.

Now, let’s say that the surface $S$ is the boundary of some $n$-dimensional submanifold $E$ of $M$, and that it’s outward-oriented. That is, we can write $S=\partial E$. Then our hypersurface integral looks like

$\displaystyle\int\limits_{\partial E}*\alpha$

Next we’ll jump over to the other end and take the divergence $\nabla\cdot F$ and integrate it over the region $E$. In terms of the $1$-form $\alpha$, this looks like

$\displaystyle\int\limits_E(*d*\alpha)\omega=\int\limits_E(d*\alpha)$

But Stokes’ theorem tells us that

$\displaystyle\int\limits_{\partial E}*\alpha=\int\limits_E(d*\alpha)$

which tells us in our vector field notation that

$\displaystyle\int\limits_SF\cdot dS=\int\limits_E\nabla\cdot F\,dV$

This is the divergence — or Gauss’, or Gauss–Ostrogradsky — theorem, and it’s yet another special case of Stokes’ theorem.

November 22, 2011