# 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