# The Unapologetic Mathematician

## Homotopic Maps Induce Identical Maps On Homology

The first and most important implication of the Poincaré lemma is actually the most straightforward.

We know that a map $f:M\to N$ induces a chain map $f^*:\Omega^k(N)\to\Omega^k(M)$, which induces a map $f^*:H^k(N)\to H^k(M)$ on the de Rham cohomology. This is what we mean when we say that de Rham cohomology is functorial.

Now if $H:f\to g$ is a homotopy, then the Poincaré lemma gives us a chain homotopy from $f^*$ to $g^*$ as chain maps, which tells us that the maps they induce on homology are identical. That is, passing to homology “decategorifies” the 2-categorical structure we saw before and makes two maps “the same” if they’re homotopic.

As a great example of this, let’s say that $M$ is a contractible manifold. That is, the identity map $i:M\to M$ and the constant map $p:M\to\{p\}$ for some $p\in M$ are homotopic. These two maps thus induce identical maps on homology. Clearly, by functoriality, $H^k(i)$ is the identity map on $H^k(M)$. Slightly less clearly, $H^k(\{p\})$ is the trivial map sending everything in $H^k(M)$ to $0\in H^k(M)$. But this means that the identity map on $H^k(M)$ is the same thing as the zero map, and thus $H^k(M)$ must be trivial for all $k$.

The upshot is that contractible manifolds have trivial homology. And — as an immediate corollary — we see that any compact, oriented manifold without boundary cannot be contractible, since we know that they have some nontrivial homology!

December 6, 2011

## Compactly Supported De Rham Cohomology

Since we’ve seen that all contractible spaces have trivial de Rham cohomology, we can’t use that tool to tell them apart. Instead, we introduce de Rham cohomology with compact support. This is just like the regular version, except we only use differential forms with compact support. The space of compactly supported $k$-forms on $M$ is $\Omega_c^k(M)$; closed and exact forms are denoted by $Z_c^k(M)$ and $B_c^k(M)$, respectively. And the cohomology groups themselves are $H_c^k(M)$.

To see that these are useful, we’ll start slowly and compute $H_c^n(\mathbb{R}^n)$. Obviously, if $\omega$ is an $n$-form on $\mathbb{R}^n$ its exterior derivative must vanish, so $Z_c^n(\mathbb{R}^n)=\Omega_c^n(\mathbb{R}^n)$. If $\omega\in B_c^n(\mathbb{R}^n)$, then we write $\omega=d\eta$ for some compactly-supported $n-1$-form $\eta$. The support of both $\omega$ and $\eta$ is contained in some large $n$-dimensional parallelepiped $R$, so we can use Stokes’ theorem to write

$\displaystyle\int\limits_{\mathbb{R}^n}\omega=\int\limits_Rd\eta=\int\limits_{\partial R}\eta=0$

I say that the converse is also true: if $\omega$ integrates to zero over all of $\mathbb{R}^n$ — the integral is defined because $\omega$ is compactly supported — then $\omega=d\eta$ for some compactly-supported $\eta$. We’ll actually prove an equivalent statement; if $U$ is a connected open subset of $\mathbb{R}^n$ containing the support of $\omega$ we pick some parallelepiped $Q_0\subseteq U$ and an $n$-form $\omega_0$ supported in $Q_0$ with integral $1$. If $\omega$ is any compactly supported $n$-form with support in $U$ and integral $c$, then $\omega-c\omega_0=d\eta$ for some compactly-supported $\eta$. It should be clear that our assertion is a special case of this one.

To prove this, let $Q_i\subseteq U$ be a sequence of parallelepipeds covering the support of $\omega$. Another partition of unity argument tells us that it suffices to prove this statement within each of the $Q_i$, so we can assume that $\omega$ is supported within some parallelepiped $Q$. I say that we can connect $Q$ to $Q_0$ by a sequence of $N$ parallelepipeds contained in $U$, each of which overlaps the next. This follows because the set of points in $U$ we can reach with such a sequence of parallelepipeds is open, as is the set of points we can’t; since $U$ is connected, only one of these can be nonempty, and since we can surely reach any point in $Q_0$, the set of points we can’t reach must be empty.

So now for each $i$ we can pick $\nu_i$ supported in the intersection of the $i$th and $i+1$st parallelepipeds and with integral $1$. The difference $\nu_i-\nu_{i-1}$ is supported in the $i$th parallelepiped and has integral $0$; since the parallelepiped is contractible, we can conclude that $\nu_{i-1}$ and $\nu_i$ differ by an exact form. Similarly, $\omega_0-\nu_1$ has integral $0$, as does $\omega-c\nu_N$, so these also give us exact forms. And thus putting them all together we find that

$\displaystyle(\omega-c\nu_N)+c(\nu_N-\nu_{N-1})+\dots+c(\nu_2-\nu_1)+c(\nu_1-\omega_0)$

is a finite linear combination of a bunch of exact $n$-forms, and so it’s exact as well.

The upshot is that the map sending an $n$-form $\omega$ to its integral over $\mathbb{R}^n$ is a linear surjection whose kernel is exactly $B_c^n(\mathbb{R}^n)$. This means that $H_c^n(\mathbb{R}^n)=Z_c^n(\mathbb{R}^n)/B_c^n(\mathbb{R}^n)\cong\mathbb{R}$.

December 6, 2011

## The Poincaré Lemma (proof)

We can now prove the Poincaré lemma by proving its core assertion: there is a chain homotopy between the two chain maps $\iota_0^*$ and $\iota_1^*$ induced by the inclusions of $M$ into either end of the homotopy cylinder $M\times[0,1]$. That is, we must define a map $I:\Omega^k(M\times[0,1])\to\Omega^{k-1}(M)$ satisfying the equation

$\displaystyle\iota_1^*\omega-\iota_0^*\omega=d(I\omega)+I(d\omega)$

Before defining the map $I$, we want to show that any $k$-form $\omega$ on the homotopy cylinder can be uniquely written as $\omega'+dt\wedge\eta$, where $\omega'$ is a $k$-form and $\eta$ is a $k-1$-form, both of which are “constant in time”, in a certain sense. Specifically, we can pull back the canonical vector field $\frac{d}{dt}$ on $[0,1]$ along the projection $M\times[0,1]\to[0,1]$ to get a “time” vector field $T$ on the cylinder. Then we use the interior product to assert that $\iota_T\omega'=0$ and $\iota_T\eta=0$.

But this should be clear, if we just define $\eta=\iota_T\omega$ then we definitely have $\iota_T\eta=\iota_T\iota_T\omega=0$, since interior products anticommute. Then we can define $\omega'=\omega-dt\wedge\eta$, and calculate $\iota_T\omega'=\iota_T\omega-\iota_T(dt\wedge\eta)=\eta-\eta=0$, since the pairing of $dt$ with $T$ is $1$. The uniqueness should be clear.

So now let’s define

$\displaystyle[[I\omega](p)](v_1,\dots,v_{k-1})=\int\limits_0^1[\eta(p,t)](\iota_{t*}v_1,\dots,\iota_{t*}v_{k-1})\,dt$

where $\iota_t$ is the inclusion of $M$ into the homotopy cylinder sending $p$ to $(p,t)$.

Now to check that this is a chain homotopy, which is purely local around each point $p\in M$. This means that we can pick some coordinate patch $(U,x)$ on $M$, which lifts to a coordinate patch $(U\times[0,1],(\bar{x},t))$ on $M\times[0,1]$, where $\bar{x}=x\circ\pi_M$. Since everything in sight is linear we will consider two cases: $\omega=fd\bar{x}^I$, where $I$ is some multi-index of length $k$; and $\omega=fdt\wedge d\bar{x}^I$, where $I$ is some multi-index of length $k-1$.

In the first case we have $I\omega=0$, while $d\omega=df\wedge d\bar{x}^I$, which we can write as a bunch of terms not involving $t$ at all plus $\frac{\partial f}{\partial t}dt\wedge d\bar{x}^I$. Therefore we calculate:

\displaystyle\begin{aligned}{}[[I(d\omega)](p)]\left(\frac{\partial}{\partial x^{j_1}},\dots,\frac{\partial}{\partial x^{j_k}}\right)&=\int\limits_0^1\frac{\partial f}{\partial t}\bigg\vert_{(p,t)}[d\bar{x}^I(p,t)]\left(\iota_{t*}\frac{\partial}{\partial x^{j_1}},\dots,\iota_{t*}\frac{\partial}{\partial x^{j_k}}\right)\,dt\\&=\int\limits_0^1\frac{\partial f}{\partial t}\bigg\vert_{(p,t)}[dx^I(p)]\left(\frac{\partial}{\partial x^{j_1}},\dots,\frac{\partial}{\partial x^{j_k}}\right)\,dt\\&=\left(\int\limits_0^1\frac{\partial f}{\partial t}\bigg\vert_{(p,t)}\,dt\right)[dx^I(p)]\left(\frac{\partial}{\partial x^{j_1}},\dots,\frac{\partial}{\partial x^{j_k}}\right)\\&=(f(p,1)-f(p,0))[dx^I(p)]\left(\frac{\partial}{\partial x^{j_1}},\dots,\frac{\partial}{\partial x^{j_k}}\right)\\&=[[\iota_1^*\omega](p)-[\iota_0^*\omega](p)]\left(\frac{\partial}{\partial x^{j_1}},\dots,\frac{\partial}{\partial x^{j_k}}\right)\end{aligned}

and we conclude that $\iota_1^*\omega-\iota_0^*\omega=I(d\omega)+d(I\omega)$, as asserted.

Now, as to the other side. This time, since $\iota_{t_0}^*(dt)=0$ for any $t_0\in[0,1]$, we know that both terms on the left hand side of the chain homotopy equation is zero. Meanwhile, we calculate

\displaystyle\begin{aligned}{}[I(d\omega)](p)&=\left[I\left(-\sum\limits_{\alpha=1}^n\frac{\partial f}{\partial\bar{x}^\alpha}dt\wedge d\bar{x}^\alpha\wedge d\bar{x}^I\right)\right](p)\\&=-\sum\limits_\alpha\left(\int\limits_0^1\frac{\partial f}{\partial\bar{x}^\alpha}\bigg\vert_{(p,t)}\,dt\right)dx^\alpha\wedge dx^I\end{aligned}

and

\displaystyle\begin{aligned}{}[d(I\omega)](p)&=d\left(\left(\int\limits_0^1f(p,t)\,dt\right)dx^I\right)\\&=\sum\limits_\alpha\frac{\partial}{\partial x^\alpha}\left(\int\limits_0^1f(p,t)\,t\right)dx^\alpha\wedge dx^I\end{aligned}

so $I(d\omega)+d(I\omega)=0$ as well, just as asserted.

December 3, 2011

## The Poincaré Lemma (setup)

Now we’ve seen that differentiable manifolds, smooth maps, and homotopies form a 2-category, but it’s not the only 2-category around. The algebra of differential forms — together with the exterior derivative — gives us a chain complex. Since pullbacks of differential forms commute with the exterior derivative, they define a chain map between two chain complexes.

And now I say that a homotopy $H:f_0\to f_1$ between two maps $f_0,f_1:M\to N$ induces a chain homotopy between the two chain maps $f_0^*$ and $f_1^*$. And, indeed, if the homotopy is given by a smooth map $H:M\times[0,1]\to N$ then we can write $f_i=H\circ\iota_i$, where $\iota_0(p)=(p,0)$ and $\iota_1(p)=(p,1)$ are the two boundary inclusions of $M$ into the “homotopy cylinder” $M\times[0,1]$, and we will work with these inclusions first.

Since $\iota_i:M\to M\times[0,1]$, we have chain maps $\iota_i^*:\Omega^k(M\times[0,1])\to\Omega^k(M)$, and we’re going to construct a chain homotopy $I:\Omega^k(M\times[0,1])\to\Omega^{k-1}(M)$. That is, for any differential form $\omega$ we will have the equation

$\displaystyle\iota_1^*\omega-\iota_0^*\omega=d(I\omega)+I(d\omega)$

Given this, we can write

\displaystyle\begin{aligned}f_1^*\omega-f_0^*\omega&=\iota_1^*H^*\omega-\iota_1^*H^*\omega\\&=d(I(H^*\omega))+I(d(H^*\omega))\\&=d([I\circ H^*]\omega)+[I\circ H^*](d\omega)\end{aligned}

which shows that $I\circ H^*$ is then a chain homotopy from $f_0$ to $f_1$.

Sometimes the existence of the chain homotopy $I$ is referred to as the Poincaré lemma; sometimes it’s the general fact that a homotopy $H$ induces the chain homotopy $I\circ H^*$; sometimes it’s a certain corollary of this fact, which we will get to later. Given my categorical bent, I take it to be the general assertion that we have a 2-functor between the homotopy 2-category and that of chain complexes, chain maps, and chain homotopies.

As a side note: now we can finally understand what the name “chain homotopy” means.

December 2, 2011

## 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

## (Pseudo-)Riemannian Metrics

Ironically, in order to tie what we’ve been doing back to more familiar material, we actually have to introduce more structure. It’s sort of astonishing in retrospect how much structure comes along with the most basic, intuitive cases, or how much we can do before even using that structure.

In particular, we need to introduce something called a “Riemannian metric”, which will move us into the realm of differential geometry instead of just topology. Everything up until this point has been concerned with manifolds as “shapes”, but we haven’t really had any sense of “size” or “angle” or anything else we could measure. Having these notions — and asking that they be preserved — is the difference between geometry and topology.

Anyway, a Riemannian metric on a manifold $M$ is nothing more than a certain kind of tensor field $g$ of type $(0,2)$ on $M$. At each point $p\in M$, the field $g$ gives us a tensor:

$\displaystyle g_p\in\mathcal{T}_p^*M\otimes\mathcal{T}_p^*M\cong\left(\mathcal{T}_pM\otimes\mathcal{T}_pM\right)^*$

We can interpret this as a bilinear function which takes in two vectors $v_p,w_p\in\mathcal{T}_pM$ and spits out a number $g_p(v_p,w_p)$. That is, $g_p$ is a bilinear form on the space $\mathcal{T}_pM$ of tangent vectors at $p$.

So, what makes $g$ into a Riemannian metric? We now add the assumption that $g_p$ is not just a bilinear form, but that it’s an inner product. That is, $g_p$ is symmetric, nondegenerate, and positive-definite. We can let the last condition slip a bit, in which case we call $g$ a “pseudo-Riemannian metric”. When equipped with a metric, we call $M$ a “(pseudo-)Riemannian manifold”.

It’s common to also say “Riemannian” in the case of negative-definite metrics, since there’s little difference between the cases of signature $(n,0)$ and $(0,n)$. Another common special case is that of a “Lorentzian” metric, which is signature $(n-1,1)$ or $(1,n-1)$.

As we might expect, $g$ is called a metric because it lets us measure things. Specifically, since $g_p$ is an inner product it gives us notions of the length and angle for tangent vectors at $p$. We must be careful here; we do not yet have a way of measuring distances between points on the manifold $M$ itself. The metric only tells us about the lengths of tangent vectors; it is not a metric in the sense of metric spaces. However, if two curves cross at a point $p$ we can use their tangent vectors to define the angle between the curves, so that’s something.

September 20, 2011

## Stokes’ Theorem on Manifolds

Now we come back to Stokes’ theorem, but in the context of manifolds with boundary.

If $M$ is such a manifold of dimension $n$, and if $\omega$ is a compactly-supported $n$-form, then as usual we can use a partition of unity to break up the form into pieces, each of which is supported within the image of an orientation-preserving singular $n$-cube. For each singular cube $c$, either the image $c([0,1]^n)$ is contained totally within the interior of $M$, or it runs up against the boundary. In the latter case, without loss of generality, we can assume that $c([0,1]^n)\cap M$ is exactly the face $c_{n,0}([0,1]^{n-1})$ of $c$ where the $n$th coordinate is zero.

In the first case, our work is easy:

$\displaystyle\int\limits_Md\omega=\int\limits_cd\omega=\int\limits_{\partial d}\omega=\int\limits_{\partial M}\omega$

since $\omega$ is zero everywhere along the image of $\partial c$, and along $\partial M$.

In the other case, the vector fields $\frac{\partial}{\partial u^i}$ — in order — give positively-oriented basss of the tangent spaces of the standard $n$-cube. As $c$ is orientation, preserving, the ordered collection $\left(c_*\frac{\partial}{\partial u^1},\dots,c_*\frac{\partial}{\partial u^n}\right)$ gives positively-oriented bases of the tangent spaces of the image of $c$. The basis $\left(c_*\left(-\frac{\partial}{\partial u^n}\right),c_*\frac{\partial}{\partial u^1},\dots,c_*\frac{\partial}{\partial u^{n-1}}\right)$ is positively-oriented if and only if $n$ is even, since we have to pull the $n$th vector past $n-1$ others, picking up a negative sign for each one. But for a point $(a,0)$ with $a\in[0,1]^{-1}$, we see that

$\displaystyle c_{*(a,0)}\left(\frac{\partial}{\partial u^i}\right)=(c_{n,0})_{*a}\left(\frac{\partial}{\partial u^i}\right)$

for all $1\leq i\leq n-1$. That is, these image vectors are all within the tangent space of the boundary, and in this order. And since $c_*\left(-\frac{\partial}{\partial u^n}\right)$ is outward-pointing, this means that $c_{n,0}:[0,1]^{n-1}\to\partial M$ is orientation-preserving if and only if $n$ is even.

Now we can calculate

\displaystyle\begin{aligned}\int\limits_Md\omega&=\int\limits_cd\omega\\&=\int\limits_{\partial c}\omega\\&=\int\limits_{(-1)^nc_{n,0}}\omega\\&=(-1)^n\int\limits_{c_{n,0}}\omega\\&=(-1)^n(-1)^n\int\limits_{\partial M}\omega\\&=\int\limits_{\partial M}\omega\end{aligned}

where we use the fact that integrals over orientation-reversing singular cubes pick up negative signs, along with the sign that comes attached to the $(n,0)$ face of a singular $n$-cube to cancel each other off.

So in general we find

\displaystyle\begin{aligned}\int\limits_{\partial M}\omega&=\sum\limits_{\phi\in\Phi}\int\limits_{\partial M}\phi\omega\\&=\sum\limits_{\phi\in\Phi}\int\limits_Md(\phi\omega)\\&=\sum\limits_{\phi\in\Phi}\int\limits_M\left(d\phi\wedge\omega+\phi d\omega\right)\\&=\sum\limits_{\phi\in\Phi}\int\limits_Md\phi\wedge\omega+\int\limits_Md\omega\end{aligned}

The last sum is finite, since on of the support of $\omega$ all but finitely many of the $\phi$ are constantly zero, meaning that their differentials are zero as well. Since the sum is (locally) finite, we have no problem pulling it all the way inside:

$\displaystyle\sum\limits_{\phi\in\Phi}\int\limits_Md\phi\wedge\omega=\int\limits_Md\left(\sum\limits_{\phi\in\Phi}\phi\right)\wedge\omega=\int\limits_Md\left(1\right)\wedge\omega=0$

so the sum cancels off, leaving just the integral, as we’d expect. That is, under these circumstances,

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

which is Stokes’ theorem on manifolds.

September 16, 2011