References for Eilenberg-Zilber shuffle product - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T09:54:39Z http://mathoverflow.net/feeds/question/109159 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109159/references-for-eilenberg-zilber-shuffle-product References for Eilenberg-Zilber shuffle product Greg Friedman 2012-10-08T16:51:28Z 2012-10-17T10:21:51Z <p>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!)</p> <p>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. </p> <p>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?</p> http://mathoverflow.net/questions/109159/references-for-eilenberg-zilber-shuffle-product/109162#109162 Answer by Puff for References for Eilenberg-Zilber shuffle product Puff 2012-10-08T18:01:38Z 2012-10-08T18:01:38Z <p>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.</p> http://mathoverflow.net/questions/109159/references-for-eilenberg-zilber-shuffle-product/109163#109163 Answer by Ronnie Brown for References for Eilenberg-Zilber shuffle product Ronnie Brown 2012-10-08T18:01:43Z 2012-10-17T01:52:53Z <p>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 </p> <p><a href="http://www.sciencedirect.com/science/article/pii/S0022404902001603" rel="nofollow">Tonks, A.P. On the Eilenberg-Zilber theorem for crossed complexes. <em>J. Pure Appl. Algebra</em> 179 (2003) 199-220.</a></p> <p>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. </p> http://mathoverflow.net/questions/109159/references-for-eilenberg-zilber-shuffle-product/109665#109665 Answer by Patricia Hersh for References for Eilenberg-Zilber shuffle product Patricia Hersh 2012-10-15T01:04:50Z 2012-10-16T22:27:49Z <p>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. </p> <p>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 ${\bf R}^{r+s}$.</p> <p>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.</p> <p><strong>Added later:</strong> 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. </p> <p>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. </p> <p>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 <em>if</em> I understand correctly what you are asking for, I think you do get your desired compatibility. </p> http://mathoverflow.net/questions/109159/references-for-eilenberg-zilber-shuffle-product/109891#109891 Answer by Jake for References for Eilenberg-Zilber shuffle product Jake 2012-10-17T10:21:51Z 2012-10-17T10:21:51Z <p>A complete reference for this with full proofs can be found in Bredon's <em>Topology and Geometry</em> (Springer GTM). More specifically, this is written up in chapter IV, part 16 (p220-223) and chapter VI, part 1 (p315-318). </p>