Date: Fri, 17 Jul 1998 19:16:20 -0400 (EDT) From: Michael Barr Subject: categories: duality question: alg. topology To all experts in algebraic topology: Is the duality I describe below known? It does not appear to be Poincare duality since there is no separate treatment of the torsion and torsion free parts. Let S be a simplicial complex, that is a set of faces of an N simplex closed under face operations. For what I am about to say, it is necessary that the empty face in dimension -1 be included so that the (co-)homology will be reduced. This means, for example, that there is a distinction between the empty simplicial complex (which has no homology) and the simplicial complex consisting of only the empty set (which has one cyclic homology group in degree -1). In fact, the former is dual to the N simplex and the latter to the N sphere. Let S' be defined as the set of complements of the complement of S. That is, for each simplex sigma not in S, the complementary simplex, whose vertices are those not in sigma, belongs to S' (and nothing else). Then for i = -1,...,N, the ith homology of S is the N-i-1st cohomology of S'. The algebraic topology books I have looked at do not mention Poincare duality, except for Lefschetz, written in 1940, and he uses a complex different from S'; in fact just opposite poset to S, since he is doing things in the generality of an arbitrary poset equipped with "incidence numbers" that are used in defining the boundary operator. In particular, his dual has the same number of elements as S, rather than 2^{N+1} - that number. One thing I find curious is that although this gives, essentially, the cohomology of S, there is no obvious way of making that into a contravariant functor. Date: Sat, 18 Jul 1998 15:42:04 -0400 (EDT) From: Michael Barr Subject: categories: More algebraic topology I have a real question now. Suppose S is a simplicial complex of dimension n with the property that every n-1 face of every n simplex is a face of at least two n simplexes. I want to conclude that H_n(S) is non-zero. Assume that the union of the n simplexes is connected (any connected component would inherit the hypothesis). Then it seems clear geometrically that the union of the faces would enclose one or more holes, but I don't see how to actually prove this. The space would appear to have no boundary, but it also is not a manifold since a point in one of the faces could have a branched neighborhood. Date: Sat, 18 Jul 1998 21:32:59 -0400 (EDT) From: John R Isbell Subject: categories: Re: More algebraic topology Dear Mike, You are asking for every strongly connected (finite) n-complex to have nonzero $H_n$, which I think you can find -- if your Russian suffices -- in P. S. Alexandrov Combinatorial Topology, OGIZ, 1947 (660 pp.) [MR 10, 55b]. It should be around Chapter 14 or 15. I have never tried to go that far in the book, nor to read the Russian at all. I have long owned, and have used as texts in classes, the English translations which go to Chapter 12 I think; Graylock, Rochester, vol. 1, 1956 [MR 17, 882a] and vol. 2, 1957 [MR 19, 759a]. With any luck you will strike an algebraic topologist who can give you an English reference. Yours, John Date: Mon, 20 Jul 1998 07:44:05 -0400 (EDT) From: Peter Freyd Subject: categories: 2 Barr questions I've got well over a thousand e-mails waiting to be looked at. I'm working on the pile from both ends. Two recent postings from Mike Barr: 1) The duality he describes sounds like Spanier-Whitehead duality. It's usually defined geometrically. Embed a finite complex into a higher-dimensional sphere, take the complement, contract it to a finite complex. Do this in the stable-homotopy category (obtained by forcing the suspension functor on the homotopy category to become an automorphism) and normalize the dimensions so that the dual of a space has the negative of its dimension. Show that the embeddings can be chosen in a coherent fashion to obtain a self-duality for stable homotopy. (The last step is non-trivial and the only person I know who verified it is Frank Adams, and that wasn't published. Does anyone have a citation?) 2) If every n-1 cell of an n-dimensional finite complex is the face of _exactly_ 2 n-dimensional cells then there's a fairly easy argument that the n-dimensional Z_2 cohomology is non-trivial. But if the condition is changed to "_at least_ 2 n-dimensional cells" then the space can be contractible. Start with the closed unit disk in the complex plane and glue the boundary onto the closed unit interval (which constitutes half of the intersection of the disk and the real line) by identifying a point x on the interval with e^(2(pi)xi) on the boundary. If one triangulates this space, each edge will be a face of either 2 or 3 triangles. One way of seeing that this is contractible is to consider first the space obtained by identifying just the upper half of the boundary with the unit interval by identifying a point x on the interval with e^((pi)xi) on the boundary. Recalling that the homotopy type of a space is unchanged when reasonable closed contractible subsets are collapsed to points, note that if the lower half of the boundary is collapsed to a point we obtain the first space. On the other hand, the upper half (together with the unit interval) is contractible, and when it's collapsed to a point we obtain the unit disk. By taking the suspension of this example we can get an example in every larger finite dimension. This example is a mapping cone. Take the map from the circle to the circle that wraps the first third of the circle around the circle, wraps the second third around in the opposite direction, and the last third in the original direction. This map is, of course, homotopic to the identity map. The mapping-cone construction depends only on the homotopy type of the map. The mapping cone of the identity map on the circle is, of course, the disk. Date: Mon, 20 Jul 1998 13:20:01 -0400 (EDT) From: Michael Barr Subject: categories: Answers I got the same counter-example from Peter Johnstone and Peter Freyd. PJ called it the dunce cap and described it as a triangle with the three edges identified, one in the reverse direction. PF described it in a more complicated way that made it easier to see it actually contractible. And the latter was really what I wanted to refute. When I saw PJ's example, I mistakenly thought it still had a 1-cycle. The smallest triangulation I found had 8 vertices, which is still too large to compute by hand with. The dual complex still has over 200 simplexes and that is just too many. If anyone knows a smaller triangulation, I would like to know it. Thanks to all who responded. Michael