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Suppose that $X$ is a connected $E_{\infty}$-space, naturally $\Omega X$ is also an $E_{\infty}$-space. Can we classify all $E_{\infty}$-extensions of $X$ by $\Omega X$ (up to homotopy). I mean the following: we would like to classify of homotopy fiber sequences $A\rightarrow B\rightarrow C$ where $A\sim \Omega X$ and $C\sim X$ as $E_{\infty}$-spaces and $B\rightarrow C$ , $A\rightarrow B$ are maps of $E_{\infty}$-spaces in particular we assume that $B$ is an $E_{\infty}$-space. (space could mean a simplicial set or a topological space)

There are two obvious extensions of $E_{\infty}$-space the first one is when $B\sim X\times \Omega X$ and the second one is when $B=PX$ the path space. Is there others ? My first guess was that the set of extensions is a group and more precisely the set of homotopy classes of maps of spectra $[X,X]$ where $X$ is seen as a connective spectra. Maybe I'm wrong but the first extension corresponds to the homotopy class of the trivial map $X\rightarrow \ast\rightarrow X$ and the second extension corresponds to the identity map $id: X\rightarrow X$. But there should be other extension in bijective correspondence with $[X,X]$. Is it correct ?

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  • $\begingroup$ My guess is that it corresponds to homotopy classes of $E_{\infty}$-spaces from $X$ into the group completion of $X$. I have to sketches of an argument: First, if we also assume that $X$ is grouplike, then we can use that group-like $E_{\infty}$-spaces are the "same" as connective spectra and thus we are looking at (co)fiber sequences of spectra and can use that the homotopy category of spectra is triangulated. $\endgroup$
    – Lennart Meier
    Commented Jan 4, 2020 at 7:20
  • $\begingroup$ The second line of argument says that you are basically looking for $\Omega X$-principal bundles on $X$ and these should be classified by a map $X \to B \Omega X$. The latter is precisely the group completion of $X$ (for a modern reference see e.g. uni-muenster.de/IVV5WS/WebHop/user/nikolaus/papers/… ). For an "$E_{\infty}$-principal bundle" I expect $E_{\infty}$-maps. But this is just a guess. $\endgroup$
    – Lennart Meier
    Commented Jan 4, 2020 at 7:23
  • $\begingroup$ @LennartMeier Unless I'm mistaken $B\Omega X$ is the connected component of the identity of $X$ (the group completion is $\Omega BX$) $\endgroup$
    – Denis Nardin
    Commented Jan 4, 2020 at 7:30
  • $\begingroup$ @DenisNardin Of course, right. I was not paying attention. $\endgroup$
    – Lennart Meier
    Commented Jan 4, 2020 at 8:30
  • $\begingroup$ @LennartMeier $X$ is connected, in particular grouplike. It was also my first guess... $\endgroup$
    – Paris
    Commented Jan 4, 2020 at 9:47

1 Answer 1

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This is probably belaboring the obvious, but just take seriously the equivalence between grouplike $E_{\infty}$ spaces and connective spectra. See for example

Equivalence between $E_\infty$-spaces and connective spectra

The asumption that X is connected means that the associated spectrum is connected and not just connective. So we may as well just ignore $E_{\infty}$ spaces (much as I hate to do that!) and take $X$ to be a connected spectrum. One is asking for all fiber sequences $\Sigma^{-1}X \to Y \to X$ of spectra or equivalently all fiber sequences $Y \to X \to X$. That is, Y is equivalent to the fiber of a map $X\to X$. A quick triangulated category type argument shows that equivalence classes of such fibers correspond bijectively to maps $X\to X$, that is to elements of $[X,X]$.

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