I’ve been reading Wall’s “Finiteness conditions for CW complexes.” This paper gives necessary and sufficient conditions for a space to be homotopy equivalent to a finite cell complex. Alternatively, it gives an obstruction in ${K}$-theory for when a retract (in the homotopy category) of a finite cell complex has the homotopy type of a finite cell complex. I’d like to describe this result, and try to motivate why the existence of such an obstruction is a natural thing to expect by a simpler analogy with algebra.

There is a fruitful analogy between spaces and chain complexes. Let ${R}$ be a ring, and consider the derived category ${D(R)}$ of chain complexes of ${R}$-modules. There are various interesting subcategories of ${D(R)}$:

1. The finitely presented derived category ${D_{fp}(R)}$; this is the smallest triangulated (or stable) full subcategory of ${D(R)}$ containing ${R}$ and closed under cofiber sequences. In other words, ${D_{fp}(R)}$ consists of complexes which are quasi-isomorphic to finite complexes of finitely generated free modules.
2. The perfect derived category ${D_{pf}(R)}$: this is the category of objects ${X \in D(R)}$ such that ${\hom(X, \cdot)}$ commutes with direct sums (i.e., the compact objects). It turns out that so-called perfect complexes are those that can be represented as finite complexes of finitely generated projectives.

One should think of the finitely presented objects as analogous to the finite cell complexes in topology, and the perfect objects as analogous to the retracts of finite cell complexes. (To push the analogy: the finite cell complexes are the smallest subcategory of the ${\infty}$-category of spaces containing ${\ast}$ and closed under finite colimits. The retracts of finite cell complexes are the compact objects in this ${\infty}$-category.)

When ${R}$ is a principal ideal domain or a local ring (more generally, when every finitely generated projective is free), the two subcategories above coincide. However, in general, a perfect complex need not be representable by a finite complex of finite frees. For instance, a projective module ${P}$ over ${R}$ is finitely presented in the derived category if and only if ${P}$ is stably free: that is, it becomes free after adding a finite free summand. Conversely, if every finitely generated projective is stably free, then every finitely generated projective—and thus every perfect complex—is finitely presented. We thus find:

Proposition 1 ${D_{fp}(R) = D_{pf}(R)}$ if and only if ${\widetilde{K}^0(R) = 0}$: that is, every finitely generated projective is stably free.

Recall that the (reduced) Grothendieck group ${\widetilde{K}^0(R)}$ represents stable equivalence classes of finitely generated projective modules.

We can say a little more. Given a perfect complex ${P_\bullet \in D(R)}$, we can actually define an obstruction in ${\widetilde{K}^0(R)}$, which vanishes if and only if ${P_\bullet}$ is finitely presented.

Definition 2 The finiteness obstruction of ${P_\bullet}$ is the alternating sum ${\sum (-1)^i [P_i] \in \widetilde{K}^0(R)}$.

One checks that the above finiteness obstruction is well-defined: that is, it is invariant under quasi-isomorphism. Clearly it vanishes for a complex that can be represented as a finite complex of finite frees, and its vanishing is therefore a necessary condition for ${P_\bullet \in D_{fp}(R)}$. Conversely, if the obstruction of ${P_\bullet}$ is zero, we can assume that all but one of the terms in ${P_\bullet}$ is free (by adding shifts of chain complexes of the form ${Q \rightarrow Q}$ where ${Q}$ is projective). The vanishing of the finiteness obstruction shows that the last term must be stably free, and direct summing with an appropriate ${F \rightarrow F}$ where ${F}$ is free makes ${P_\bullet}$ a complex of frees.

We find:

Proposition 3 For ${P_\bullet \in D_{pf}(R)}$, there is an obstruction in ${\widetilde{K}^0(R)}$ which vanishes if and only if ${P_\bullet \in D_{fp}(R)}$.

This was fairly straightforward, but spaces are more complex than chain complexes, for instance because of the existence of potentially troublesome ${\pi_1}$‘s. So we’ll have to do the analysis more carefully in the next section.

1. An inductive construction

Let’s now work with spaces. Suppose given a space (without loss of generality, connected) ${X}$ which we’d like to prove is homotopy equivalent to a finite cell complex. Clearly we’d have to expect:

1. The homology of ${X}$ (in fact, of its universal cover ${\widetilde{X} }$) vanishes above a certain dimension.
2. ${\pi_1 X}$ is a finitely presented group.

These conditions are not enough to guarantee ${X}$ finite, but they will certainly be satisfied if ${X}$ is a retract of a finite cell complex. (Actually, it takes a little work to show that a retract of a finitely presented group is finitely presented in Wall’s paper.)

But if we have such a space ${X}$, let’s think about how we might try to build a finite cell complex homotopy equivalent to it and what might go wrong. As we can do in homological algebra, we’d have to keep trying to attach cells and apply the Whitehead theorem. So, to start with, we could take a suitable wedge of circles to make a finite complex ${K_1}$ (of dimension ${\leq 1}$) such that ${\pi_1 K_1 }$ surjects onto ${\pi_1 X}$. Moreover, we can arrange it so that there is a map

$\displaystyle K_1 \rightarrow X$

inducing a surjection on ${\pi_1}$.

That’s a first step, but we have to keep going. By definition, we have chosen things such that the relative homotopy group ${\pi_1(X, K_1) = 0}$. One can define the relative homotopy group as the homotopy groups of the homotopy fiber of ${K_1 \rightarrow X}$. Alternatively, ${\pi_n(X, K_1)}$ consists of homotopy classes of diagrams,

that is, homotopy classes of an element of ${\pi_{n-1}(K_1)}$ and a nullhomotopy of its image in ${X}$. (Wall uses this description, which I thought was quite nice.)

Anyway, we now want to expand ${K_1}$ to a bigger complex ${K_2}$ (of dimension ${\leq 2}$) together with a map ${K_2 \rightarrow X}$ such that ${\pi_2(X, K_2) = 0}$. In order to do this, we have to consider ${\pi_2(X, K_1)}$, which is the collection of diagrams,

We want to define ${K_1}$ by making a bunch of push-outs of such diagrams; this is analogous to the strategy in Quillen’s small object argument. However, it’s important to keep the finiteness conditions in mind here. So we need to get a handle on ${\pi_2(X, K_1)}$.

Except, because of fundamental group issues, it’s a little more convenient to first attach 2-cells to ${K_1}$ to get a new complex ${K_2'}$ with a map ${K_2' \rightarrow X}$ so that ${\pi_1 K_2' \simeq \pi_1 X}$. Then, by the Hurewicz theorem, we have an isomorphism of ${\pi_1 X}$-modules

$\displaystyle \pi_2(X, K_2') \simeq \pi_2(\widetilde{X}, \widetilde{K_2}') \simeq H_2( \widetilde{X}, \widetilde{K}_2'),$

where the tilde denotes the universal cover. Since the homology of ${\widetilde{X}}$ and the homology of ${\widetilde{K}_2'}$ is finitely generated over ${\pi_1 X}$ (see the below lemma) if ${X}$ is finitely dominated, we find that this is a finitely generated ${\pi_1 X}$-module.

So we choose generators ${\alpha_1, \dots, \alpha_n }$ for ${\pi_2(X, K_2')}$ as a ${\pi_1 X}$-module. Each such corresponds to a diagram as above, and we form the push-out of all these maps (i.e., attach 2-cells) to ${K_2'}$ to get a finite cell complex ${K_2}$ with a map ${K_2 \rightarrow X}$. More precisely, we let ${K_2}$ be the pushout

where the diagrams for ${i = 1, 2, \dots, n}$ come from the ${\alpha_i}$. The complex ${K_2}$ comes with a natural map to ${X}$. Then, the exact sequence

$\displaystyle \pi_2(K_2', K_2) \rightarrow \pi_2(X, K_2) \rightarrow \pi_2(X, K_2') \rightarrow 0$

and the fact that ${\pi_2(K_2', K_2) }$ is free on ${n}$ generators hitting the ${\alpha_i}$, shows that ${\pi_2(X, K_2')}$ becomes trivial.

We can definitely continue this process. It gets easier in higher degrees since we don’t even have to worry about the fundamental group, and we don’t need the intermediate stage ${K_2'}$. However, there’s a problem: it doesn’t obviously stop! We’ll need to figure out a strategy how to make it stop, if a certain obstruction vanishes.

This inductive and crude approach already tells us something interesting:

Theorem 4 If ${X}$ is finitely dominated, then ${X}$ is homotopy equivalent to a cell complex with finitely many cells in each dimension.

We also find:

Theorem 6 If ${X}$ is finitely dominated, then for any ${n}$, we can choose ${X}$ to be the retract of a finite cell complex ${Y}$ such that ${\pi_k Y \simeq \pi_k X}$ for ${k \leq n}$.

Proof: If ${X}$ is a retract of a finite cell complex ${Y'}$, then we apply the above procedure, starting with ${Y'}$, of attaching finitely many cells successively to make the homotopy groups of ${Y'}$ approximate those of ${X}$. $\Box$

These results relied on:

Lemma 5
Suppose ${X}$ is finitely dominated and connected. Then ${H_i(\widetilde{X})}$ is finitely generated over ${\pi_1 X}$.

Proof: Without loss of generality, ${X}$ is a finite CW complex. Then we can prove below, more generally, that for any finite complex ${X}$ equipped with a map ${X \rightarrow BG}$ (for ${G}$ a group), then ${H_j(X \times_{BG} EG)}$ is finitely generated over ${G}$. Applying this when ${G = \pi_1 X}$ gives the result. (Below means a later post.) $\Box$

2. Stopping the algorithm

The previous section gave an algorithm for, given a finitely dominated space, expressing it as a countable CW complex with finitely many cells in each dimension. That is, given ${X}$, we produced a sequence of finite cell complexes over ${X}$,

$\displaystyle K_1 \subset K_2 \subset \dots \subset K_n \subset$

with ${\dim K_i \leq i}$, such that ${\pi_j(X, K_i) =0 }$ for ${0 \leq j \leq i}$. Unfortunately, this algorithm does not terminate: that is, it does not produce a finite cell complex in general. We can try to refine it using the following fact: there exists an ${N \geq 3}$ with the following property:

1. ${H_i(\widetilde{X} , G) =0 }$ for ${i > N}$ and for all groups ${G}$.
2. ${H^{N+1}(X, G) = 0}$ for all coefficient systems on ${X}$.

Motivated by this, let’s see if we can try to stop the algorithm at ${N}$. That is, let’s say that we have constructed ${\phi: K_{N-1} \rightarrow X}$, an ${N-1}$-dimensional complex, such that ${\pi_i(X, K_{N-1}) =0}$ for ${i \leq N-1}$. Can we construct ${K_N}$ such that ${K_N \rightarrow X}$ is a homotopy equivalence?

By construction, we are going to want to take ${K_N}$ to be ${K_{N-1} \cup_{\bigsqcup S^{N-1}} \bigsqcup D^N}$ for some maps ${S^{N-1} \rightarrow K_{N-1}}$. These should form generators of the finitely generated ${\mathbb{Z}[\pi_1X]}$-module

$\displaystyle \pi_N(X, K_{N-1}) \simeq H_{N}( \widetilde{X}, \widetilde{K}_{N-1}) .$

Suppose this module is free. Then let ${\alpha_1, \dots, \alpha_m}$ be generators for the module and use these maps (diagrams) to build ${K_N}$ from ${K_{N-1}}$. We have then an exact sequence

$\displaystyle \pi_N(K_N, K_{N-1}) \rightarrow \pi_N(X, K_{N-1}) \rightarrow \pi_N(X, K_N) \rightarrow 0,$

which shows that the attaching makes ${\pi_N(X, K_N) =0}$, as desired. However, the first map is this time an isomorphism, which is even better than it was before: we disturb the situation less in higher dimensions.

In fact, let’s look in homology; the long exact sequence for the triple ${(\widetilde{X}, \widetilde{K}_N, \widetilde{K}_{N-1})}$ has the following properties:

1. ${H_*(\widetilde{K}_N), H_*(\widetilde{X})}$ vanish in degrees ${>N}$. This is by construction of ${K_N}$ and by choice of ${N}$.
2. ${H_*(\widetilde{K}_N) \rightarrow H_*(\widetilde{X})}$ is an isomorphism in degrees ${\ast \leq N}$ by ${N}$-connectedness.
3. However, the map for ${\ast = N+1}$ is not only a surjection, but an isomorphism: the map has higher connectivity than expected! This is because of the exact sequence

$\displaystyle H_{N+1}(\widetilde{X}, \widetilde{K}_N) \rightarrow H_N( \widetilde{K}_N, \widetilde{K}_{N-1}) \stackrel{\simeq}{\rightarrow} H _N(\widetilde{X}, \widetilde{K}_{N-1}) \rightarrow H_N(\widetilde{X}, \widetilde{K}_N) ,$

which (together with vanishing of ${H_{N+1}(\widetilde{X}, \widetilde{K}_{N-1})}$) implies that ${H_{N+1}(\widetilde{X}, \widetilde{K}_N) = 0}$.

4. From this, we find that the map ${\widetilde{K_N} \rightarrow \widetilde{X}}$, and thus the map ${K_N \rightarrow X}$, is a homology (hence homotopy) equivalence.

In particular:

Proposition 7 If for some ${N}$ (large enough to satisfy (1) above), we can choose ${K_{N-1}}$ such that ${H_N(\widetilde{X}, \widetilde{K}_{N-1})}$ is finite free over ${\mathbb{Z}[\pi_1 X]}$, then ${X}$ is homotopy equivalent to an ${N}$-dimensional finite complex.

Here again we find the finite freeness of something necessary to build a construction. And it is here that the finiteness obstruction enters: it’s going to be the class in ${K}$-theory of this group precisely.

Proposition 8 If ${N}$ is above, the group ${H_N(\widetilde{X}, \widetilde{K}_N)}$ is finite projective over ${\mathbb{Z}[\pi_1 X]}$, and its class in ${\widetilde{K}^0(\pi_1 X)}$ is independent of the choice of ${K_N}$.

Wall proves this result by obstruction theory, and this class is called the Wall finiteness obstruction: sure enough, its vanishing is the necessary and sufficient condition for ${X}$ to be finite. In the next post, I’d like to describe this argument, but in such a way that makes clearer the parallels with algebra.