2011 December « The Unapologetic Mathematician

Nonvanishing Compactly-Supported de Rham Cohomology

Last time we saw that compactly-supported de Rham cohomology is nonvanishing in the top degree for $\mathbb{R}^n$ . I say that this is true for any oriented, connected $n$ -manifold $M$ . Specifically, if $\omega\in\Omega_c^n(M)$ , then the integral of $\omega$ over $M$ is zero if and only if $\omega=d\eta$ for some $\eta\in\Omega_c^{n-1}(M)$ . That the second statement implies the first should be obvious.

To go the other way takes more work, but it’s really nothing much new. Firstly, if $\omega$ is supported in some connected, parameterizable open subset $U\subseteq M$ then we can pull back along any parameterization and use the result from last time.

Next, we again shift from our original assertion to an equivalent one: $\omega_1$ and $\omega_2$ have the same integral over $M$ , if and only if their difference is exact. And again the only question is about proving the “only if” part. A partition of unity argument tells us that we only really need to consider the case where $\omega_i$ is supported in a connected, parameterizable open set $U_i\subseteq M$ ; if the integrals are zero we’re already done by using our previous step above, so we assume both integrals are equal to $c\neq0$ . Dividing by $c$ we may assume that each integral is $1$ .

Now, if $p_0\in M$ is any base-point then we can get from it to any other point $p\in M$ by a sequence of connected, parameterizable open subsets $W_i$ . The proof is basically the same as for the similar assertion about getting from one point to anther by rectangles from last time. We pick some such sequence taking us from $U_1$ to $U_2$ , and just like last time we pick a sequence of forms $\alpha_i$ supported in $W_{i-1}\cap W_i$ . Again, the differences between $\omega_1$ and $\alpha_1$ , between $\alpha_i$ and $\alpha_{i+1}$ , and between $\alpha_N$ and $\omega_2$ are all exact, and so their sum — the difference between $\omega_1$ and $\omega_2$ is exact as well.

And so we conclude that the map $\Omega_c^n(M)\to\mathbb{R}$ given by integration is onto, and its kernel is the image of $\Omega_c^{n-1}(M)$ under the exterior derivative. Thus, $H_c^n(M)\cong\mathbb{R}$ , just as for $\mathbb{R}^n$ .

December 8, 2011 Posted by John Armstrong | Differential Topology, Topology | 1 Comment

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 Posted by John Armstrong | Differential Topology, Topology | 5 Comments

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 Posted by John Armstrong | Differential Topology, Topology | drmathochist | 2 Comments

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 Posted by John Armstrong | Differential Topology, Topology | Leave a Comment

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 Posted by John Armstrong | Differential Topology, Topology | 2 Comments

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The Unapologetic Mathematician

Mathematics for the interested outsider

Nonvanishing Compactly-Supported de Rham Cohomology

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The Poincaré Lemma (setup)

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