The essential image of the suspension spectrum functor $\Sigma^\infty$

algebraic-topologyspectrastable-homotopy-theory

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.

Best Answer

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.

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