1. Consider the co-monad $M:=\Sigma^\infty \Omega^\infty$ on the category of spectra. It is clear that given a pointed space $X$, $M\Sigma^\infty X=\Sigma^\infty E(X)$,where $E(X)$ is the free unital $E_\infty$-algebra on $X$. As $\Sigma^\infty$ commutes with colimit, $M\Sigma^\infty X$ is equivalent to the free $E_\infty$-algebra on the spectrum $\Sigma^\infty X$, that is, to: $$ \lor_n (\wedge^n (\Sigma^\infty X)_{hS_n}).\tag{1} $$ Is this right?

  2. It looks like given an arbitrary spectrum $A$, there is no equivalence between $\Sigma^\infty\Omega^\infty A$ and $ \lor_n (\wedge^n (A)_{hS_n}) $ Is there any general condition for having such an equivalence?

  • 2
    $\begingroup$ I think for (1) you should assume that $X$ is a pointed connected space, right? $\endgroup$
    – Tim Campion
    Jan 3, 2022 at 4:26

1 Answer 1


For connected spectra $A$, there is an equivalence $$\Sigma^\infty \Omega^\infty A \simeq \bigvee_{n=1}^\infty A^{\wedge n}_{h\Sigma_n}$$ if and only if $A$ is a wedge summand of a suspension spectrum.

This is Theorem 1.2 in N. Kuhn's paper Suspension Spectra and Homology Equivalences (TAMS, 1983). Kuhn calls spectra that are wedge summands of suspension spectra spacelike.

I am guessing that if $A$ is not connected then such an equivalence can not exist. It is easy to see that there can not be an equivalence of ring spectra.


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