Realizing complexes with bases as cellular complexes

Question

This is a question a friend of mine asked me some time ago. I suspect the answer is "no" but can't prove it.

Every free complex of abelian groups is isomorphic to the reduced cellular complex of some finite CW-complex; in fact, one can take a wedge of balls and Moore spaces. The question is whether there is a similar result for complexes equipped with a basis.

Namely, let $C_{*}$ be a finite chain complex (i.e., the differential decreases the degree) of free abelian groups. Suppose a basis $B_i$ of each $C_i$ is chosen.

1. Is it true that $(C_{*},(B_i))$ is isomorphic as a complex with a basis to the (possibly shifted) reduced cellular complex of some CW-complex with the basis given by the cells?

2. What happens if one works with complexes of finite-dimensional vector spaces over $\mathbb{Z}/2$ instead?

Answer 1

Here is a sketch of an argument to show that all based chain complexes are realizable. (This might end up being pretty similar to Tyler's argument.)

First one gives an algebraic argument that by a change of basis the chain complex can be put in a standard "diagonal" form. Moreover, the change of basis can be achieved by a sequence of elementary operations, as in linear algebra, but now over the integers rather than a field, using the fact that the group $GL(n,Z)$ is generated by elementary matrices, including signed permutations. The most important of the elementary operations is to add plus or minus one basis element to another. Doing such an operation in $C_i$ changes the boundary maps to and from $C_i$ by multiplication by an elementary matrix and its inverse.

The "diagonalized" chain complex can easily be realized geometrically, so it remains to see that the elementary basis change operations can be realized geometrically. In the special case of top-dimensional cells, one can slide a part of one such cell over another to achieve the elementary operation of adding plus or minus one column of the outgoing boundary matrix to another. For lower-dimensional cells one wants to do the same thing and then extend the deformation over the higher cells. It should be possible to do this directly without great difficulty. The slide gives a way of attaching a product $\{cell\}\times I$, and this product deformation retracts onto either end, so one can use the deformation retraction to change how the higher-dimensional cells attach.

The argument should work for 1-cells as well as for higher-dimensional cells, so it shouldn't be necessary to assume that $C_1$ is trivial.

An alternative approach would be to temporarily thicken the cell complex into a handle structure on a smooth compact manifold-with-boundary of sufficiently large dimension, with one i-handle for each i-cell. Sliding an i-cell then corresponds to sliding an i-handle, and there is well-established machinery on how to do this sort of thing, as one sees in the proof of the h-cobordism theorem for example. Or one can use the language of morse functions and gradient-like vector fields as in Milnor's book on the h-cobordism theorem. Either way, after all the elementary basis changes have been realized by handle slides, one can collapse the handles back down to their core cells to get the desired based cellular chain complex.

There are plenty of details to fill in here in either the cell or handle approach. I don't recall seeing this result in the classical literature, but I wouldn't be too surprised if it were there somewhere, maybe in some paper or book on J.H.C.Whitehead's simple homotopy theory where elementary row and column operations play a big role.

Answer 2

It is always possible to alter an existing solution to one that has a correct basis.

Given your chain complex, use your original method to construct a CW complex X with the correct CW-chain complex, but the wrong basis. Write X⁽ⁿ⁾ for its n-skeleton. To make life easy assume that it only has one 0-simplex and no 1-simplices, perhaps by suspending a previous answer. Inductively assume that you've constructed a map Y⁽ⁿ⁾ → X⁽ⁿ⁾ of CW-complexes so that Y⁽ⁿ⁾ is n-dimensional and the induced map on CW-chains is an isomorphism, so that Y⁽ⁿ⁾ has the correct CW-chains with basis through dimension n. The Whitehead theorem implies that the map Y⁽ⁿ⁾ → X⁽ⁿ⁾ is a homotopy equivalence.

Now each X⁽ⁿ⁺¹⁾ is built by cell attachment as a (reduced) mapping cone of a map $\bigvee S^n \to X^{(n)}$, where the homology map $H_n(\bigvee S^n) \to H_n(X^{(n)})$ has a chosen isomorphism to the boundary map $C_{n+1} \to Z_n$. The Hurewicz theorem tells you that your basis (B_i) of $C_{n+1}$ lifts to a map $\bigvee S^{n} \to \bigvee S^{n}$, which induces a homology isomorphism corresponding to the inclusion of the basis (B_i) of $H_n$.

Then X⁽ⁿ⁺¹⁾ is homotopy equivalent to the mapping cone of the (homotopy equivalent) composite map $\bigvee S^n \to \bigvee S^n \to X^{(n)}$. As Y⁽ⁿ⁾ is homotopy equivalent to X⁽ⁿ⁾ we can lift this corrected-basis attaching map to a map $\bigvee S^n \to Y^{(n)}$. The mapping cone of this map, which I'll call Y⁽ⁿ⁺¹⁾, is homotopy equivalent to X⁽ⁿ⁾ and has the correct CW-chains with basis through dimension n+1.

Answer 3

A remark to Tyler's answer:

Thank you for your answer! It was my question, and I misunderstand your construction, starting from: "Now each X(n+1) is built by cell attachment...".

Consider the simplest example: X is a wedge of 2-sphere with 3-disk. We want to obtain a CW-complex Y with two 2-cells a and b and one 3-cell x, such that d(x)=a+b. We start to apply your construction: Y_2 is a wedge of two 2-spheres and homotopy equivalence X_2\to Y_2 is trivial. Then, I claim, it is impossible to extend this homotopy equivalence to a homotopy equivalence between X and Y...

Thank you again.

Answer 4

Thank you, Allen! I like your idea very much and I think it works.

It is very nice to have you comment on this, since the question was originally motivated by an idea to modify and use a technique of yours and Wagoner's for working with Cerf diagramms in contact topology.

Unfortunately, this question is a toy version of another one. I still do not understand how to modify your arguments to work out the following: is it possible to construct a CW-complex realising the given algebraic chain complex with a basis and with a linear order on basis elements (i.e. for any k the first k cells form a CW-subcomplex). Or, in Morse theory, is it possible to realise given chain complex by the complex of a strong Morse function, "sufficiently good at infinity" and with prescribed order of critical points. It would be very helpful if you could comment on this as well.

The first problem in case when the basis elements are ordered is that the corresponding group is not $GL(n,Z)$, it is the upper triangular group. So it is not clear that we can find a representative CW-complex to start sliding its cells. Moreover, I do not know what the classification of chain Z-comlpexes with ordered bases under such a group looks like.