What are the algebras over $\Omega^k\Sigma^k$ ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T09:28:16Z http://mathoverflow.net/feeds/question/57151 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57151/what-are-the-algebras-over-omegak-sigmak What are the algebras over $\Omega^k\Sigma^k$ ? Toribio Smith 2011-03-02T18:28:25Z 2011-03-02T19:42:35Z <p>Let $Ho(Spc)$ be the homotopy category of spaces. There is an adjoint pair $$\Sigma^k \colon Ho(Spc) \leftrightarrows Ho(Spc)\colon \Omega^k,$$ where $\Sigma^k$ is the $k$-th supension functor and $\Omega^k$ is the $k$-fold loop space functor ($k\ge 1)$. This adjunction has as associated monad the functor $\Omega^k\Sigma^k$. My question is what are the algebras over this monad (a precise reference to this fact would be welcome).</p> <p>If we denote by $Ho^s$ the homotopy category of spectra, then there is another adjunction $$\Sigma^{\infty} \colon Ho(Spc) \leftrightarrows Ho^s\colon \Omega^{\infty},$$ where $\Sigma^{\infty}$ is the suspension spectrum functor. My question is again what are the algebras over the monad $\Omega^{\infty}\Sigma^{\infty}$.</p> http://mathoverflow.net/questions/57151/what-are-the-algebras-over-omegak-sigmak/57152#57152 Answer by Tyler Lawson for What are the algebras over $\Omega^k\Sigma^k$ ? Tyler Lawson 2011-03-02T18:41:14Z 2011-03-02T19:09:26Z <p>(Everything I say here is up to homotopy equivalence.)</p> <p>Algebras over $\Omega^k \Sigma^k$ are spaces $X$ equivalent to a $k$-fold loop space $\Omega^k Y$. Algebras over $\Omega^\infty \Sigma^\infty$ are infinite loop spaces; this is a little harder to say, but it is essentially that there is a sequence of spaces <code>$Y_n$</code> with <code>$Y_0 = X$</code> and equivalences <code>$Y_n \simeq \Omega Y_{n+1}$</code>.</p> <p>The original and still canonical reference, which covers all of this in detail, is J.P. May's book "The geometry of iterated loop spaces," Lectures Notes in Mathematics 271.</p> <p>EDIT: As Neil points out, I've misread the question. The statements above are for spaces, not for objects in the homotopy category of spaces.</p> http://mathoverflow.net/questions/57151/what-are-the-algebras-over-omegak-sigmak/57161#57161 Answer by Neil Strickland for What are the algebras over $\Omega^k\Sigma^k$ ? Neil Strickland 2011-03-02T19:42:35Z 2011-03-02T19:42:35Z <p>I know interesting answers to two questions that are not the same as the one asked, but are related.</p> <ol> <li><p>Consider the monad $T=\Omega^\infty L_{K(n)}\Sigma^\infty$ on the homotopy category of spaces. It is straightforward to construct a functor $$\Omega^\infty:Ho(\{K(n)-\text{local spectra}\})\to \{T-\text{algebras}\}.$$ One can show using the Bousfield-Kuhn functor and related ideas that this is actually an equivalence.</p></li> <li><p>Consider the monad $Q=\Omega^\infty\Sigma^\infty$ on the category of based spaces (not up to homotopy). If we use spectra in the sense of Lewis and May, there is an evident functor $\Omega^\infty:\{\text{spectra}\}\to\{Q-\text{algebras}\}$. This is actually full and faithful (even on spectra whose homotopy groups are concentrated in negative degrees), which means that the point-set level $Q$-action carries a lot more information than you might naively guess. The key point in the proof is that we can use a trick with the Hopf map and space-filling curves to express $S^2$ as the coequaliser of two based maps from $S^3$ to $S^3$. This gives a natural way to express $\Omega^2X$ as the equaliser of two maps from $\Omega^3X$ to $\Omega^3X$, which allows us to do a bunch of things by induction.</p></li> </ol>