Counterexample in cohomology for symmetric spectra? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T02:16:00Z http://mathoverflow.net/feeds/question/91954 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91954/counterexample-in-cohomology-for-symmetric-spectra Counterexample in cohomology for symmetric spectra? skupers 2012-03-22T22:12:03Z 2012-03-23T04:09:42Z <p>Symmetric spectra are a particular model for spectra, introduced by Hovey, Shipley and Smith. They have the nice property that they have a well-behaved smash product. Our interest in spectra comes from homotopy theory and hence we want to define a model category structure on it. Several model category structures are given in Hovey, Shipley and Smith. One model category structure that has the stable homotopy category as its homotopy category uses stable equivalences instead of the level equivalences, which one might naively guess are a good choice. </p> <p>These stable equivalences are supposed to be maps that induce isomorphisms on "all" generalized cohomology theories. However, they are defined using cohomology with coefficients in a particular class of spectra, the injective $\Omega$-spectra. The reason for picking this particular class is given earlier in the article:</p> <blockquote> <p>It would be nice if the 0th cohomology group of the symmetric spectrum $X$ with coefficients in the symmetric $\Omega$-spectrum $E$ could be defined as $\pi_0 Map_{Sp^\Sigma}(X,E)$, the set of homotopy classes of maps from $X$ to $E$. But, even though the contravariant functor $E^0 = \pi_0 Map_{Sp^\Sigma}(-,E)$ takes simplicial homotopy equivalences to isomorphisms, $E^0$ may not take level equivalences to isomorphisms. ... We introduce injective spectra as a class of spectra $E$ for which the functor $E^0$ behaves correctly.</p> </blockquote> <p>I'm trying to understand this problem and I think it would be helpful to see an example. So my question is: what is an example of a level equivalence $f: X \to Y$ of symmetric spectra and a symmetric $\Omega$-spectrum $E$ such that $f$ does not induce an isomorphism on $E^0$?</p> http://mathoverflow.net/questions/91954/counterexample-in-cohomology-for-symmetric-spectra/91976#91976 Answer by Tyler Lawson for Counterexample in cohomology for symmetric spectra? Tyler Lawson 2012-03-23T04:09:42Z 2012-03-23T04:09:42Z <p>The core problem is: Maps of symmetric spectra are, levelwise, maps equivariant with respect to the symmetric group, but the notion of weak equivalence ignores that.</p> <p>The injectivity notion essentially works because there is a (simplicial) model structure on symmetric spectra where</p> <ul> <li><p>weak equivalences are levelwise weak equivalences,</p></li> <li><p>cofibrations are levelwise monomorphisms, and</p></li> <li><p>fibrant objects are "injective" spectra.</p></li> </ul> <p>This is sometimes called the injective level model structure. This ensures that <code>$Map_{Sp^\Sigma}(X,E)$</code> has a sensible homotopy type, making <code>$\pi_0$</code> invariant under levelwise weak equivalence. The $\Omega$-spectrum condition is not as relevant.</p> <p>However, you've asked for an example, and so we have to come up with one. The problems with doing so are as follows.</p> <p>There is more than one way to skin this particular cat. There are at least three different level model structures on symmetric spectra (see Stefan Schwede's book project, chapter III, for coverage of them), and if we try to come up with an example we need to have one where the mapping space has an incorrect homotopy type for all of them.</p> <ul> <li><p>We need $E$ to be an $\Omega$-spectrum. Since this implicitly involves loop spaces, I want to stick to an example where $E$ is levelwise fibrant (and hence is fibrant in the projective model structure). So we need an example where the domains are not projective-cofibrant, which boils down to a condition about the symmetric groups not acting sufficiently freely.</p></li> <li><p>I am lazy, and want to construct the domain spaces using a "free" functor so that I can analyze the mapping spaces easily. There are the functors <code>$G_n$</code> described in Hovey-Shipley-Smith's paper, which are left adjoint to the forgetful functor <code>$X \mapsto X_n$</code> from symmetric spectra to <code>$\Sigma_n$</code>-spaces. Objects constructed in this way are always cofibrant in the flat level model structure, so we need the target $E$ to not be fibrant in the flat level model structure. This is a specific condition on what the fixed-point subspaces of <code>$E_n$</code> look like.</p></li> <li><p>Again in the "laziness" department, you've asked for an example where $E$ is an $\Omega$-spectrum, and these are more difficult to write down.</p></li> </ul> <hr /> <p>Thus, here is a recipe for taking some injective $\Omega$-spectrum $E$ and constructing a new one, $E'$, in a way ensuring that $E^0$ doesn't take weak equivalences to isomorphisms.</p> <p>Let $X \to Y$ be the map <code>$$G_n((E\Sigma_n)_+ \wedge S^k) \to G_n(S^k),$$</code> induced by the projection, where $S^k$ is given the trivial action of the symmetric group. Once checks by the explicit formulas for <code>$G_n$</code> that the map $X \to Y$ is a levelwise equivalence. For any symmetric spectrum $E$, the map $E^0(Y) \to E^0(X)$ is <code>$$\pi_0 Map_{\Sigma_n,*}(S^k, E_n) \to \pi_0 Map_{\Sigma_n,*}((E\Sigma_n)_+ \wedge S^k, E_n)$$</code> which is <code>$$\pi_k ({E_n}^{\Sigma_n}) \to \pi_k Map_{\Sigma_n}(E\Sigma_n, E_n).$$</code> If $E$ is injective, this is an isomorphism by assumption.</p> <p>Let <code>$$E_n' = E_n \wedge (E\Sigma_n)_+.$$</code> Here the symmetric groups act diagonally, and the structure maps are induced by the structure maps of $E$ and the functorial maps <code>$E\Sigma_n \to E\Sigma_{1+n}$</code>. Then $E^'$, levelwise, has no fixed points, and the map $(E')^0 (Y) \to (E')^0 (X)$ is <code>$$0 \to \pi_k Map_{\Sigma_n} (E\Sigma_n, (E\Sigma_n)_+ \wedge E_n) \cong \pi_k Map_{\Sigma_n} (E\Sigma_n, E_n)$$</code> by freeness. If $E$ was injective in the first place, the right-hand side is equivalent to $\pi_k E_n$. If <code>$E^*$</code> is not trivial, we can pick some $k$ and $n$ making this nonzero.</p>