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Let $\mathsf{hCW}$ denote the homotopy category of CW-complexes and $\mathsf{hCWSpec}$ the homotopy category of CW-spectra (ie. families of CW-complexes $(X_i)_{i\in\mathbb{Z}}$ with connection maps $\Sigma X_i \rightarrow X_{i+1}$ given by subcomplex inclusions).

What is the essential image of the suspension spectrum functor $\Sigma^\infty: \mathsf{hCW} \rightarrow \mathsf{hCWSpec}$?

It is folklore (see this remark on the nlab) that any CW-spectrum can be built up from cofiber sequences of the form $$\begin{array}{ccc} \bigoplus_{I_k} \Sigma^{q_k}\mathbb{S} & \rightarrow & X_k\\ \downarrow&&\downarrow\\ * & \rightarrow & X_{k+1} \end{array}$$ Since a similar fact holds for CW-complexes, it seems reasonable to expect that a CW-spectrum in the essential image of $\Sigma^\infty$ requires all stable cells to have nonnegative dimension $q_k$. But I am unsure whether this is actually sufficient...

I was told to have a look at the Hurewicz-theorem for spectra, but I don't see how this can be applied here, since it requires connective spectra. Since for negative $k$ $$\pi_k(X)=\operatorname*{colim} \limits_{n\in\Bbb N} \pi_{n+k}(X_n) \overset{def}{=}\operatorname*{colim} \limits_{n \geq \vert k\vert}\pi_{n+k}(X_n)$$ I don't see what having no negative stable cells has to do with having vanishing negative stable homotopy groups. Even in the case of finite CW-spectra I don't see a connection there. It is however very likely that I misunderstood the hint given, so I appreciate any feedback.

As always thank you very much for your attention and support.

Eric Wofsey
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Jonas Linssen
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    It's certainly not sufficient for all stable cells to have nonnegative dimension (for instance, consider $\Sigma^{\infty-1}\mathbb{RP}^2$ which has cells only in nonnegative dimension but has torsion in its $0$th homology so cannot be the suspension spectrum of a space). You seem to imply that this question was given as an exercise, which is very confusing to me--it does not have any simple answer. – Eric Wofsey Dec 09 '21 at 15:43
  • Thanks for your comment. I am currently attending a seminar on stable homotopy theory and there we often switch between different representations of (often finite) CWspectra, whatever is required in the moment. But since all of these things were stated as facts and were not shown, I lost track and thought maybe an answer to my question would shed some light on it. You say it cannot be answered in generality, which is nice to know. So I will ask more specific questions another time then :) – Jonas Linssen Dec 09 '21 at 16:26

2 Answers2

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There is not any simple characterization of when a spectrum is equivalent to a suspension spectrum. In particular, it does not suffice for the spectrum to be built out of cells of nonnegative dimension. The issue is that the attaching maps for those cells only need to exist stably (i.e., after suspending enough times), and so may not exist as unstable maps in their actual dimension.

For instance, consider the space $\mathbb{RP}^2$. It is built by attaching a $2$-cell to $S^1$ along a map $S^1\to S^1$ of degree $2$. In particular, it has only a $1$-cell and a $2$-cell (and a basepoint $0$-cell). So if we take the suspension spectrum $\Sigma^\infty\mathbb{RP}^2$ and desuspend it once, we get a spectrum $\Sigma^{\infty-1}\mathbb{RP}^2$ which has only a $0$-cell and a $1$-cell. However, there is no corresponding space, since the degree $2$ attaching map $S^1\to S^1$ cannot be desuspended to a map $S^0\to S^0$. To prove definitively that $\Sigma^{\infty-1}\mathbb{RP}^2$ cannot be equivalent to the suspension spectrum of any space, you can observe that $H_0(\Sigma^{\infty-1}\mathbb{RP}^2)\cong\mathbb{Z}/2$ but $H_0$ of a space can never have torsion.

Eric Wofsey
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    For the opposite type of example: for $A$ an abelian group $HA$ is the $\Omega$-spectrum defined by $HA(n)=K(A,n)$. $H\mathbb{Z}$ is definitely not equivalent to a (desuspended) suspension spectrum and similarly for $H\mathbb{Z}/p$. It would be reasonable to conjecture that for nontrivial $A$ it never is, but remarkably it turns out that $H\mathbb{Q}$ is. – Connor Malin Dec 09 '21 at 19:04
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A spectrum is spacelike if it is a retract of a suspension spectrum. Nick Kuhn has developed a version of homological algebra in which spacelike spectra play the role of "projective" objects. Using this framework, we see that a spectrum $E$ is spacelike iff it is a retract of $\Sigma^\infty \Omega^\infty E$ for example. Furthermore, in principle, we can use the nontriviality of various derived functors defined using these "spacelike resolutions" as obstructions to characterize whether or not a spectrum is spacelike.

From another perspective, suspension spectra inherit a coalgebraic structure from the diagonal in spaces. While not all coalgebras in spectra are suspension spectra, the nonexistence of such a structure presents another kind of obstruction to being a suspension spectrum, represented by a class in $[E, \Sigma D_2(E)]$. This has been worked out by John Klein, who also showed that this is the only obstruction in some metastable range. Furthermore, using his techniques one can define higher obstructions, the vanishing of which would establish $E$ as homotopy equivalent to a suspension spectrum.

References:

  • Kuhn, "Spacelike resolutions of spectra"
  • Kuhn, "Suspension spectra and homology equivalences"
  • Klein, "Moduli of suspension spectra"
JHF
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