Subject: Re: 2 responses on symmetric products Date: Fri, 9 Feb 2001 14:23:13 -0500 (Eastern Standard Time) From: "Nicholas J. Kuhn" To: Donald Davis CC: nick Since we have the old guys weighing in on this classic question, let me add a couple of comments and a question: (1) As Jean Pierre suggested, Math Sci Net is a good place to start. E.G. typing in Anywhere "homology" and Anywhere "symmetric products" brings up 74 papers, about half of which are relevant. (2) Early on, Dold, then Puppe, had results of the following form: There is a functor F_n such that F_nH_*(X) = H_*(SP^nX), where additive structure is a functor of additive structure, cup products depend on cup products (I've switched to cohomology, where I should probably assume some finite type hypotheses), A-module stucture depends on A--module structure (now I am assuming mod p coefs). So they are saying that there are functors F_n: K --> K where K = unstable algebras over A, with F_n(H^*(X)) = H^*(SP^n(X)). However these papers don't actually say was F_n is. (Sneaky of them!) Thus we move on to... (3) The calculational literature. Of course as n --> infinity, one gets the infinite symmetric product, and we will assume known a functor F: K --> K such that FH^*(X) = H^(SP^infty(X)). Nakaoka, and others, including Milgram show that F_n is a quotient of F. However, they tend to focus on Ker(F_n --> F_n-1). It seems to me that they are identifying the graded functor associated to the filtered functor F, which is slightly weaker than computing the filtration itself. So, is the following explicitly given in the literature? (equivalent dual homology versions would also be fine.) QUESTION What is the kernel of H^(SP^infty(X);Z/p) --> H^*(SP^n(X);Z/p), as a functor of H^*(X;Z/p), regarded as an object in K? I have always assumed the answer is that this kernel is spanned as a vector space by the "obvious" subset of the standard elements in H^*(SP^n(X);Z/p) (cup products of iterated Steenrod operations done to guys in H^*(X)), but I have always felt queasy about whether or not anyone has actually proved this. Nick Kuhn