How should I think of the $\infty$-category of spectra? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T00:35:37Z http://mathoverflow.net/feeds/question/74386 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74386/how-should-i-think-of-the-infty-category-of-spectra How should I think of the $\infty$-category of spectra? Akhil Mathew 2011-09-02T18:26:26Z 2011-09-03T22:26:40Z <p>I've seen a bunch of definitions of spectra in the literature, and the fanciest seems to be the $(\infty, 1)$-category of spectra obtaining by "stablizing" the higher category of spaces, as in DAG I. I don't really understand this stabilization procedure yet and would like to connect this idea to the more concrete notions that I've heard about, such as:</p> <ul> <li>The Boardman category of spectra: here a spectrum is a bunch of spaces (say, CW complexes) $E_n$ with closed cellular imbeddings $SE_n \to E_{n+1}$, and morphisms are defined via cofinal subspectra.</li> <li>Symmetric or orthogonal spectra, where one just has spaces and morphisms $SE_n \to E_{n+1}$, but there is some additional equivariance condition (and this way we get an honest symmetric monoidal category).</li> </ul> <p>Presumably from one of these other constructions one can still recover the $\infty$-category of spectra.</p> <p>For concreteness, I still like to think of a higher category as a topologically (or simplicially) enriched category, or even more concretely a set of objects together with 1-morphisms, 2-morphisms, etc. and various ways of composing them. The 1-morphisms in all these concrete categories are spectra are known (e.g. they're equivariant morphisms in the symmetric or orthogonal case). How should I think of the higher morphisms?</p> http://mathoverflow.net/questions/74386/how-should-i-think-of-the-infty-category-of-spectra/74465#74465 Answer by Omar Antolín-Camarena for How should I think of the $\infty$-category of spectra? Omar Antolín-Camarena 2011-09-03T22:26:40Z 2011-09-03T22:26:40Z <p>Basically you want to know what the space of maps between two spectra $X$ and $Y$ is. Well, each map from $X$ to $Y$ is a sequence of maps from $X_n$ to $Y_n$ and thus $\mathrm{map}(X,Y)$ is a subspace of the product of the $Y_n^{X_n}$. And you can build an entire spectrum of maps from $X$ to $Y$, whose nth space is just maps from $X$ to the nth suspension of $Y$.</p> <p>For modern constructions (S-modules, orthogonal spectra, symmetric spectra) of spectra which give symmetric monoidal categories this internal hom is adjoint to the smash product. These modern categories have model structures and between cofibrant and fibrant objects the mapping space described above is homotopically correct. (And, in Boardman's construction --which is not monoidal before passing to the homotopy category--, you still get that this mapping spectrum is the right one for maps between a CW-spectrum and an $\Omega$-spectrum). So one version the stable $(\infty,1)$-category of spectra is the topologically enriched category of fibrant-cofibrant spectra in any of these modern categories, with mapping spaces as above.</p> <p>Of course, for these model categories of spectra, any of the other ways of getting at the $(\infty,1)$-category they represent, such as Dwyer-Kan localization, will give an equivalent stable $(\infty,1)$-category of spectra.</p>