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Is there a model category $C$ on an additive category such that its homotopy category $Ho(C)$ is the stable homotopy category of spectra and the additive structure on $Ho(C)$ is induced from that on $C$.

Basically I want to add and subtract maps in $C$ without going to its homotopy category.

I'm not asking for $C$ to be a derived category or anything like that. Just that it should have an additive structure.

As John Palmieri pointed out I should really say what structure I want the equivalence (between $Ho(C)$ and the stable homotopy category) to preserve. Since I do want it to be a triangulated equivalence, Cisinski indicates why this is not possible.

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I am a little confused by your question. Are you looking for any example of such a model category C? In that case I think taking C to be the model category of (bounded) chain complexes on a suitable abelian category gives you such an example. Or do you have a fix particular C in mind? –  Chris Schommer-Pries Mar 3 '10 at 14:08
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The question asks that the homotopy category Ho(C) be equivalent to the ordinary stable homotopy category of topological spaces. In particular, chain complexes are excluded because they have an enrichment in topological abelian groups (per Tilman's answer). –  Tyler Lawson Mar 3 '10 at 15:09
    
I want $Ho(C)$ to be the stable homotopy category, I mean the category of spectra. I hope the way I rewrote the question helps. –  Don Stanley Mar 3 '10 at 15:22
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So like Tyler said ... –  Don Stanley Mar 3 '10 at 15:22
    
Thanks for adding that. I think it is much clearer now. –  Chris Schommer-Pries Mar 3 '10 at 17:09

3 Answers 3

up vote 25 down vote accepted

The answer is: no there isn't such a thing. Here is a rough argument (a full proof would deserve a little more care).

Using the main result of

S. Schwede, The stable homotopy category is rigid, Annals of Mathematics 166 (2007), 837-863

your question is equivalent to the following: does there exist a model category $C$, which is additive, and such that $C$ is Quillen equivalent to the usual model category of spectra?

In particular, we might ask: does there exist an additive category $C$, endowed with a Quillen stable model category structure, such that the corresponding stable $(\infty,1)$-category is equivalent to the stable $(\infty,1)$-category of spectra?

Replacing $C$ by its full subcategory of cofibrant objects, your question might be reformulated as: does there exist a category of cofibrant objects $C$ (in the sense of Ken Brown), with small sums (and such that weak equivalences are closed under small sums), and such that the corresponding $(\infty,1)$-category (obtained by inverting weak equivalence of $C$ in the sense of $(\infty,1)$-categories) is equivalent to the stable $(\infty,1)$-category of spectra? If the answer is no, then there will be no additive model category $C$ such that $Ho(C)$ is (equivalent to) the category of spectra (as a triangulated category).

So, assume there is an additive category of cofibrant objects $C$, with small sums, such that $Ho(C)$ is (equivalent to) the category $S$ of spectra (as a triangulated category). Let $C_f$ be the full subcategory of $C$ spanned by the objects which correspond to finite spectra in $S$. Then $Ho(C_f)\simeq S_f$, where, by abuse of notations, $Ho(C_f)$ is the $(\infty,1)$-category obtained from $C_f$ by inverting weak equivalences, while $S_f$ stands for the stable $(\infty,1)$-category of finite spectra (essentially the Spanier-Whitehead category of finite CW-complexes). Given any (essentially) small additive category $A$ denote by $K(A)$ the "derived $(\infty,1)$-category of $A$" (that is the $(\infty,1)$-category obtained from the category of bounded complexes of $A$, by inverting the chain homotopy equivalences). Then, the canonical functor $A\to K(A)$ (which sends an object $X$ to itself, seen as a complex concentrated in degree $0$), has the following universal property: given a stable $(\infty,1)$-category $T$, any functor $A\to T$ which sends split short exact sequences of $A$ to distinguished triangles (aka homotopy cofiber sequences) in $T$ extends uniquely into a finite colimit preserving functor $K(A)\to T$. In particular, the functor $C_f\to Ho(C_f)\simeq S_f$ extends uniquely to a finite colimit preserving functor $F:K(C_f)\to S_f$. Let $Ker(F)$ be the full $(\infty,1)$-subcategory of $K(C_f)$ spanned by objects which are sent to zero in $S_f$. Then the induced functor $$K(C_f)/Ker(F)\to S_f$$ is an equivalence of (stable) $(\infty,1)$-categories (to see this, you may use the universal property of $S_f$: given a stable $(\infty,1)$-category $T$, a finite colimit preserving functor $S_f\to T$ is the same as an object of $T$; see Corollary 10.16 in DAG I). This implies that, for any object $X$ of $S_f$, if $X/n$ denotes the cone of the map $n:X\to X$ (multiplication by an integer $n$), then $n.X/n\simeq 0$ (see Proposition 1 in Schwede's paper Algebraic versus topological triangulated categories). But such a property is known to fail whenever $X$ is a finite spectrum for $n=2$ (see Proposition 2 in loc. cit.). Hence there isn't such a $C$...

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This is one of the best answers I've ever seen on MO. –  Harry Gindi Mar 4 '10 at 0:57
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A nice answer. One question: doesn't Schwede's rigidity result require Ho(C) to be triangulated equivalent to the stable homotopy category, rather than merely an equivalent category? (Perhaps this is something the OP is willing to assume.) –  Tyler Lawson Mar 4 '10 at 3:14
    
Yes, you are right. We have to require the equivalence Ho(C)=S to be compatible with the triangulated structures. –  Denis-Charles Cisinski Mar 4 '10 at 11:59

Let's extend the question somewhat: given a commutative ring spectrum $R$, is there an additive category $C$ such that $Ho(C)$ is the homotopy category of $R$-module spectra?

I don't know the answer to your question, but I think that if we assume a little more, then this can only happen if $R$ is a product of Eilenberg-Mac Lane spectra. Here's a sketch. Let's assume that $C$ is a simplicial model category, i.e. the hom-functors $Hom_C(X,Y)$ actually take values in topological spaces and $\pi_0(Hom_C(X,Y)) = [X,Y]_R$ if $X$ is cofibrant and $Y$ is fibrant.

Let's assume $R$ is cofibrant-fibrant. Then $Hom_C(R,R) \simeq \Omega^\infty R$. But the left hand side is an abelian group, so $\Omega^\infty R$ is equivalent to a product of Eilenberg-Mac Lane spaces.

The sphere, by the way, is of course not a product of Eilenberg-Mac Lane spectra, and $\Omega^\infty S$ is not a product of Eilenberg-Mac Lane spaces.

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You should have a look at Irakli Patchkoria's paper arxiv.org/abs/1108.6309 . Your question in the first sense (where you don't require compatibility with enrichments) has several interesting non-Eilenberg-MacLane examples. –  Lennart Meier Dec 14 '11 at 12:48

In the category $Ho(C)$, say that every map is a fibration and a cofibration, and define the weak equivalences to be the isomorphisms. This makes $Ho(C)$ into a model category; its homotopy category is $Ho(C)$, and it is certainly additive.

It's not what you want, though; can you make your question more precise?

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Is this category complete and cocomplete? You also need those for a model category. I guess I also want $C$ to be monoidal. Thanks I'll add that. –  Don Stanley Mar 3 '10 at 21:28
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If $Ho(C)$ is triangulated then it isn't complete and cocomplete. At best, in the direction of cocompleteness for instance, it can have filtered colimits but this would imply it was phantomless. –  Greg Stevenson Mar 3 '10 at 21:48
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Of course you're right, $Ho(c)$ is not complete or cocomplete, it only has weak limits and colimits. It is, however, an additive symmetric monoidal category with a class of morphisms which you can formally invert (because they're already invertible) to get the stable homotopy category. –  John Palmieri Mar 3 '10 at 23:45

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