I recently started reading up on simple homotopy theory. Here is a question I stumbled upon.

In his 1938 Paper Simplicial Spaces, Nuclei and m-Groups Whitehead introduced the notion of elementary expansions and elementary collapses of simplicial complexes. Essentially a simplicial complex $K'$ is obtained from $K$ through an elementary collapse, by removing two simplices $p$ and $q$ from $K$ such that:

  1. $p$ is a maximal simplex of $K$.
  2. $q < p$ is a (maximal, proper) free face of $p$ (i.e. not contained in any other simplex but itself and $p$).

An elementary expansion is the obvious inverse operation. There are topological realizations of these operations, with the collapse given by "pushing the free face onto the others". Lets call a map of simplicial complexes a simplicial simple homotopy equivalence, if it is homotopic to the composition of such maps.

Later on he decided that CW-complexes were a more convenient setting to work in (Simple Homotopy Types), and this was where he developed the now famous results on whitehead torsion. This is the setting that most of the following material (such as Cohens Book A course in simple homotopy theory) are presented in and probably most familiar to most topologists. By a cellular simple homotopy equivalence, I mean a map of CW-complexes as in 3. It seems to me, that it is somewhat folklore knowledge, that the second setting is a generalization of the prior in the following sense:

Let $K$,$K'$ be abstract simplicial complexes and $f:|K| \to |K'|$ a map between their realizations. Then $f$ is a simplicial simple homotopy equivalence, if and only if it is a cellular simple homotopy equivalence, with respect to the obvious $CW$-structures on $|K|$ and $|K'|$.

Is this even true? I imagined this would be a simple consequence of the simplicial approximation theorem, but couldn't figure out an easy proof. I also skimmed most of the papers from the time period I could find on the matter, but didn't really get a satisfying answer.

If yes, I would be really thankful for a reference, or a sketch of a proof.

  • 1
    $\begingroup$ Kamps-Porter wrote a book that covers the simplicial perspective on simple homotopy theory. I gave it to a friend of mine, so I don't have it at hand, but it might be helpful to look into. $\endgroup$ – Harry Gindi Mar 27 '20 at 19:49
  • 2
    $\begingroup$ I managed to get my hands on a copy. As far as I can tell, there is no direct answer to my question in there. However I find the abstract homotopy perspective on simple homotopies very enlighting, so this might prove very helpful in the long run. Thanks! $\endgroup$ – FeverTree Mar 28 '20 at 13:29

Turns out I should have read the original material Simplicial Spaces, Nuclei and m-groups a little more carefully. It was in there all along. I'm still supprised nobody ever explicitly stated this though. But I guess it was such common knowledge at the time that nobody bothered.

My question can be rephrased in the following way. For a simplicial complex $K$ (all complexes are taken to be finite) denote by $E_S(L)$ the set of inclusions of simplicial complexes $L \to K$ that are homotopy equivalences, modulo the equivalence relation generated by $L\to K \xrightarrow{s} K' = L\to K' \implies L \to K \sim L \to K'$ for $s$ a composition of (simplicial) elementary expansions (this actually has set size). This is the description Siebenmann chooses in Infinite Simple Homotopy Types. Further denote by $E_C(K)$ the obvious analogon in the CW setting. The latter of course is just the underlying set of the geometric description on the Whiteheadgroup $WH(X)$ as it is constructed in Cohens simple Homotopy Theory for example. As both equivalence relations identify homotopic maps and using the standard mapping cylinder arguments, my question rephrases to:

What is the Kernel of the obvious forgetful map $E_S(L) \to E_C(L) = WH(L) \cong WH( \pi_1(L))$ for $L$ connected. Here I mean kernel in the sense of pointed sets, as the lefthandside will only a posteriori be a group (one could of course bother with proving its a group first..).

The way Whitehead proved in Simple Homotopy Types that the last isomorphism is injective, is effectively by transforming each inclusion $L \to K$ to the form $ L \to K' = L \cup \bigcup e_i^n \bigcup e_i^{n+1}$ through cellular simple expansions and collapses $(n \geq 1)$. The corresponding element in $WH( \pi_1(L))$ is then given by $ \pi_{n+1}( K', K'^{n} \cup L) \to \pi_n(K'^n \cup L, L)$. Explicitly one checks using Hurewicz theorem and universal coverings that this is in fact an isomorphism of free $\mathbb Z (\pi_1(L))$ modules, with basis given by the cells $e_i$. One then shows, that all the elementary matrix operations making the corresponding matrix trivial on the $WH(\pi_1(L))$ side have an analogue on the simple homotopy side, proving injectivity.

What was known to me when I asked the question, was that this works in the cellular category, i.e. with cellular simple equivalences. It turns out, before passing to the cellular setting, Whitehead did the analogous proof in the simplicial world, which turns out to require a lot more technical arguments, but it is completely done in 1 and the statement is implicitly proven in the proof of theorem 20 there.

Very roughly, one first proves that subdivisions are simplicial simple equivalences, so that one can work in the p.l. category instead. Here one has the p.l. mapping cylinder see 1). The attachment of a p.l. cell along a p.l. boundary map, is then to be understood as taking the cylinder along the boundary and then gluing the cell on top of it. (I guess this circumvents problems with pushouts in the p.l. category). He then shows that the same simplicifactions as in the cellular setting are valid in the p.l. setting using such cylinders and simplicial approximation, thus, showing that $E_S(L) \to WH(\pi_1(K))$ and hence $E_S(L) \to E_C(L)$ has trivial kernel.

In particular this gives a positive answer to my original question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.