# Connective spectra versus simplicial abelian groups - very basic question

Hello,

I have very general , "introductory" questions (It is quite hard for me to seek for specific things in the algebraic topology literature).

I guess that connective spectra have a model structure. So do simplicial abelian groups. Are these Quillen equivalent?

Secondly, I think of a simplicial abelian group as a space with strictly associative and commutative operation, while I think of a connective spectrum as a space with an operation which is associative and commutative up to all higher coherences (i.e. some words like $E_{\infty}$). So these are similar. How do I see what extra richness is encoded in a spectrum? For example, what mental pictures do I lose when I think of a connective spectrum as a right-bounded chain complex?

I think that the last is the most important for me, to have some small mental picture which I should have for spectra but not for simplicial abelian groups / chain complexes.

Thank you, Sasha

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I believe that the model category/$\infty$-category of simplicial abelian groups should be equivalent to connective $H\mathbb{Z}$-module spectra, but not all connective spectra are $Hmathbb{Z}$-module spectra. For example, the sphere spectrum should not correspond to a simplicial abelian group. If it did, you should be able to compute all its homotopy groups. –  Chris Brav Apr 24 '12 at 16:39
by connective spectrum do you really mean connective ring spectrum? Regardless, you should look at some of the work of Shipley. She has shown that the category of HR-module spectra is the same as the category of DG-R-modules, which seems to be what you are asking for. There is a lot of structure that you will not be able to capture with just the DG-category that is in the category of connective spectra. –  Sean Tilson Apr 25 '12 at 3:10
I don't think you mean ring spectrum. If you change the word "multiplication" to "addition" in the first second of your second paragraph, I think your question will be clearer. –  Tom Goodwillie Apr 25 '12 at 3:52
@Tom: OK, Thank you, I changed "multiplication" to "an operation". –  Sasha Apr 26 '12 at 16:55

One key point is that if you let $S/2$ denote the cofibre of twice the identity on the sphere spectrum (also known as the mod $2$ Moore spectrum, or $\Sigma^{-1}\mathbb{R}P^2$), then twice the identity map on $S/2$ is nonzero. This cannot happen in the homotopy category of simplicial abelian groups, or (roughly speaking) any other triangulated category arising from algebra rather than topology.

UPDATE: as a bridge between Dustin's answer and mine, you can note that the category of differential graded $MU_*MU$-comodules is a purely algebraic category that supports most of the same Steenrod/chromatic phenomena as the category of spectra, but still the cofibre of multiplication by two has exponent two.

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