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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:

  • 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.
  • 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).

Presumably from one of these other constructions one can still recover the $\infty$-category of spectra.

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?

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As it happens, I think this method of "stabilizing" ends up looking a lot like the Boardman construction or Symmetric spectra construction: that is, you basically look at sequences of objects in your category such that the loop of each object is the preceding one. i.e. the stabilization of an infinity category "is" the category of "spectrum objects" inside the thing you started with. –  Dylan Wilson Sep 3 '11 at 0:57
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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 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 to the nth suspension of Y. –  Omar Antolín-Camarena Sep 3 '11 at 4:14
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(cont.) 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 in my previous comment is homotopically correct. –  Omar Antolín-Camarena Sep 3 '11 at 4:16
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Yes, this is analogous to how chain complexes form a dg-category. As I keep telling you, just pretend spectra are chain complexes. :) –  Omar Antolín-Camarena Sep 3 '11 at 22:28
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Oh, and the use of $N(\mathbb{Z} \times \mathbb{Z})$ is not for extra coherence data, it's a clever way to specify in a single diagram that you don't want maps $X_n \to X_{n+1}$ but rather maps $\Sigma X_n \to X_{n+1}$ (a spectrum is not a diagram of shape $\mathbb{Z}$). –  Omar Antolín-Camarena Sep 3 '11 at 22:33
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1 Answer

up vote 5 down vote accepted

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 $Y$.

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.

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.

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Accepted, as promised! Thanks. –  Akhil Mathew Sep 4 '11 at 0:27
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