# References for Eilenberg-Zilber shuffle product

Most of the treatments I can find in the literature for the Eilenberg-Zilber shuffle product approach it from the point of view of simplicial sets (including the original Eilenberg-MacLane paper). I was wondering if anyone knows of a reference written directly in the language of singular or PL chains. By this I mean that one should show that the (p,q)-shuffles each produce a distinct embedded p+q simplex in $\Delta^p\times \Delta^q$ and that, collectively, this produces a triangulation of $\Delta^p\times \Delta^q$. Furthermore, this should be a chain map, in the sense that the boundary of the triangulation is compatible with the same triangulation process applied to the boundaries (or in other words, the map $S_p(X)\otimes S_q(X)\to S_{p+q}(X\times Y)$ obtained by applying this process is a chain map). Dold constructs this map explicitly in his book, but it's in an exercise and he doesn't provide any proofs of properties (he does helpfully indicate with an asterisk that it is a challenging exercise to do so!)

The reason I ask is that I'm working on some lecture notes (for publication) for which I need to have an explicit construction of the homology cross product at the chain level, but the intended audience is not expected to know about simplicial sets.

If what I'm looking for does not exist, can anyone recommend the simplicial set treatment that provides the best description of the connection to the explicit geometric triangulation side of things?

• My memory may be faulty here, but I think I once saw something like this in Lefschetz's book on Algebraic Topology Oct 8, 2012 at 18:01
• @Justin. To clarify, I'm not looking for the definition of the shuffle product (I've found that in, e.g., Dold). I'm looking for a proof that the construction provides a triangulation of $\Delta^p\times \Delta^q$ that's compatible with boundaries (in the sense that the cross product induced from this point of view is a chain map). Oct 15, 2012 at 20:54
• Since MO informs me that this question has garnered a lot of traffic, I should update and mention that there is now a detailed treatment of all this in the appendix to my book Singular Intersection Homology. It should be published soon but for now it's still available on my web site: faculty.tcu.edu/gfriedman A lot of the details there come from work in Dan Ramras's undergraduate thesis. Thanks to everyone who contributed answers to the original question. Oct 1, 2018 at 5:06

Here is a very explicit way of realizing a product of simplices $\Delta^r$ and $\Delta^s$ as well as a triangulation of their product with maximal simplices naturally indexed by words obtained from shuffling $a_1\cdots a_r$ with $b_1\cdots b_s$. The main thing this uses is that the type B Coxeter arrangement is a simplicial hyperplane arrangement.

The type B Coxeter arrangement is the hyperplane arrangement given by hyperplanes $x_i \pm x_j = 0$ and $x_i = 0$. Saying it is a simplicial arrangement is saying that the natural subdivision it induces on a unit sphere centered at the origin is a simplicial subdivision. Thus, we will take as our $r$-simplex the region in ${\bf R}^r$ given by $0\le a_1 \le a_2\le \cdots \le a_r$ and $\sum a_i^2 \le 1$. Likewise, we take as our $s$-simplex the region in a different ${\bf R}^s$ given by $0\le b_1\le b_2\le \cdots \le b_s$ and $\sum b_i^2 \le 1$. The product of these simplices lives in ${\bf R}^{r+s}$ and is the set of points satisfying all of these inequalities. But this is an intersection of a convex body containing the origin (namely the set of points with $\sum a_i^2 \le 1$ and $\sum b_i^2 \le 1$) and an intersection of half-spaces given by hyperplanes in the type B Coxeter arrangement for ${\bf R}^{r+s}$. But the fact that this type B Coxeter arrangement is a simplicial arrangement means that this region is a union of simplices. This triangulation into simplices is the subdivision of our region induced by the remaining hyperplanes in the type B Coxeter arrangement for ${\bf R}^{r+s}$.

In other words, the simplices are exactly the signed permutations on alphabet $a_1,\dots ,a_r,b_1,\dots ,b_s$ satisfying $0\le a_1\le \cdots \le a_r$ and $0\le b_1\le \cdots \le b_s$. But these are given by precisely the desired shuffle words. I hope this helps with what you want.

Added later: you also ask that the boundary of the triangulation be compatible with applying the same triangulation process to the boundaries of the original simplices. I think this construction above indeed has this property, if I understand correctly what you are asking for. To see this, it is convenient first (in terms of simplifying language) to replace the balls $\sum a_i^2 \le 1$ and $\sum b_i^2 \le 1$ above by the cubes $\max |a_i| \le 1$ and $\max |b_i| \le 1$, which we may do since the type B Coxeter complex will triangulate any full-dimensional convex body having the origin in its interior.

We are using a shuffled word such as $a_1a_2b_1a_3b_2b_3b_4$ to describe the simplex $0\le a_1 \le a_2 \le b_1 \le a_3\le b_2 \le b_3 \le b_4\le 1$. Now applying the boundary to this simplex means taking an alternating sum over all ways of turning one of these weak inequalities into an equality. If this equality is between some $a_i$ and some $b_j$, then the resulting face is an interior face of the triangulation -- because we can swap $a_i$ with $b_j$ to obtain a different shuffled word indexing a different simplex also having this face in its boundary. On the other hand, when a face results from an equality $a_i = a_{i+1}$ or $b_i = b_{i+1}$ or one from setting the first letter equal to 0 or the last letter equal to 1, then this is necessarily giving a boundary face in our triangulation of the product of simplices.

For comparison, when we take the boundary of our original simplex $\Delta^r$ (resp. $\Delta^s$), we take an alternating sum over ways of setting inequalities to equalities $a_i = a_{i+1}$ or $a_1=0$ or $a_r=1$ (resp. $b_j = b_{j+1}$ or $b_1=1$ or $b_s=1$). Applying our triangulation procedure after doing this, thus gives shuffled words possessing exactly one such equality of the type that puts you at the boundary of the original triangulation. So if I understand correctly what you are asking for, I think you do get your desired compatibility.

• I should have also probably said that the signs do work out so the map to the alternating sum over shuffles commutes with the boundary map -- this follows from the fact that crossing a hyperplane swaps two adjacent letters giving a new shuffled word with opposite sign. Oct 17, 2012 at 0:40
• Thanks! This is a very interesting approach. Is it hard to verify that the type B Coxeter arrangement is a simplicial hyperplane arrangement. Can you recommend a reference for that? Oct 17, 2012 at 20:50
• Sure, this is exactly the type B Coxeter complex. One reference is p. 26 in James Humphreys' book "Reflection groups and Coxeter groups". This is a special case of a more general result: given a finite collection of hyperplanes through the origin such that the reflection group generated by the reflections across these hyperplanes is a finite group not containing any additional reflections, then the group is also a Coxeter group and the subdivision of the unit sphere is a simplicial subdivision yielding the Coxeter complex associated to that Coxeter group. Oct 17, 2012 at 21:11
• To answer your other question, this isn't difficult to verify. One way is to use the theory of Coxeter groups (or even buildings) to think of faces as cosets of parabolic subgroups, as discussed in Humphreys or in Ken Brown's book on Buildings, among others. Another way is by observing that the face poset for the closure of a maximal region in the triangulation is a Boolean algebra (with elements given by which inequalities are turned to equalities) and that each closed cell is a regular CW complex, implying together that the closed cell must be a simplex. Oct 17, 2012 at 21:45

This is very late, I know, but I just stumbled across this question as I happen to be thinking about these things myself. There is a nice discussion of the simplicial cross product in Hatcher's "Algebraic Topology" text, on pp. 277-278. There Hatcher refers to Eilenberg and Steenrod's "Foundations of Algebraic Topology" textbook, p. 68, for a proof of precisely the fact you mention. I have not yet waded through their proof, but this would seem to be the canonical reference.

In fact they seem to be proving something more general, that the shuffles induce a natural $\Delta$-complex structure on a product of (finite) $\Delta$-complexes.

Have you looked at Hilton and Wylie "Homological Algebra"? As far as I recall they deal with the case of singular chains and also singular cubical homology.

• Thanks. I had a look at Hilton and Wylie, and unfortunately, they handle the Eilenberg-Zilber theorem in much the same way other texts do - abstractly by the method of acyclic models. Oct 9, 2012 at 18:43

The explicit form of the theorem due to Eilenberg and Mac Lane is in the paper you are referring to "On the groups $H(\Pi,n): I$". A stronger and more precise form is in

Tonks, A.P. On the Eilenberg-Zilber theorem for crossed complexes. J. Pure Appl. Algebra 179 (2003) 199-220.

Also Tonks verifies the important fact that the tensor product is a strong deformation retract of the crossed complex of the product, as is well known for the chain complex case and used in Homological Perturbation Theory. You may find his references to the chain complex case helpful.

• Thanks, Ronnie. I had a look, but Tonks's paper also seems to be mostly about simplicial sets. Perhaps I can reverse engineer what I want from a proof that, for simplicial sets, $|K|\times |L|=|K\times L|$. I suppose what I really want is a proof of this for semi-simplicial sets (Delta sets). Oct 15, 2012 at 20:56

A complete reference for this with full proofs can be found in Bredon's Topology and Geometry (Springer GTM). More specifically, this is written up in chapter IV, part 16 (p220-223) and chapter VI, part 1 (p315-318).

• I just had a look and it seems that Bredon also uses acyclic models - he does not provide an explicit triangulation of the product $\Delta^p\times \Delta^q$. Oct 17, 2012 at 20:45