Subject: Re: two postings Date: Fri, 06 Aug 2004 10:56:35 -0400 From: "Nicholas J. Kuhn" (It is odd to answer to find myself answering my colleague's query via email, but Greg is in Finland and I am not ...) Let F(n,k) be ordered k-tuples of points in R^n, i.e. R^{nk} - (fat diagonal). Fox and Neuwirth in `The Braid Groups', Math Scand 10 (1962), give a very explicit Sigma_k equivariant cell decomposition of R^{2k), with the (fat diagonal) as a subcomplex, and from this (using duality) it is clear that B(2,k) is equivalent to a cell complex of dimension k-1, with a cell for each k-1 tuple (i_0,..,i_{k-1}) with the entries 0 and 1, of dimension the number of 1's. As I have known since I was a graduate student in the late 1970's, this story generalizes to all n without any significant change: thus B(n,k) is equivalent to a complex of dimension (n-1)(k-1) with a cell of dimension (i_1 + .. + i_{k-1}) for every k-1 tuple (i_1,..,i_{k-1}) with entries between 0 and n-1. I don't know if anyone else had noticed this before. So I referenced Fox and Neuwirth for this fact when I mentioned it in a Barcelona conference proceeding paper (in Prog. Math. 196, 2001, proof of Lemma 3.6). This cell structure, I think, is not well adapted for studying the operad structure on little cubes: this is why I didn't pursue this. In Jeff Smith's 1981 thesis (eventually published), he constructed a simplicial little cubes operad. Maybe he, or someone else, can comment on whether his model gives the (n-1)(k-1) dimension number. Nick Kuhn > > Subject: Question for the mailing list > Date: Fri, 06 Aug 2004 06:06:42 -0400 > From: "Gregory Arone" > > Here is a reference question for the list: > > Let B(n,k) be the space of unordered k-tuples of distinct > points in R^n. I need to use the following fact: B(n,k) is > homotopy equivalent to an (n-1)(k-1) - dimensional CW > complex. I am not asking for a proof or a recent > reference. My question is: what is the earliest reference > for this, and who does the credit properly belong to? I > would like to get this right. > > Thanks, Greg