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Jan 8, 2020 at 2:27 vote accept Paris
Jan 4, 2020 at 15:33 answer added Peter May timeline score: 19
Jan 4, 2020 at 9:47 comment added Paris @LennartMeier $X$ is connected, in particular grouplike. It was also my first guess...
Jan 4, 2020 at 8:30 comment added Lennart Meier @DenisNardin Of course, right. I was not paying attention.
Jan 4, 2020 at 7:30 comment added Denis Nardin @LennartMeier Unless I'm mistaken $B\Omega X$ is the connected component of the identity of $X$ (the group completion is $\Omega BX$)
Jan 4, 2020 at 7:23 comment added Lennart Meier 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.
Jan 4, 2020 at 7:20 comment added Lennart Meier 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.
Jan 4, 2020 at 7:16 history edited Lennart Meier CC BY-SA 4.0
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Jan 3, 2020 at 23:46 history edited Paris CC BY-SA 4.0
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Jan 3, 2020 at 22:57 history edited Paris CC BY-SA 4.0
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Jan 3, 2020 at 22:51 history asked Paris CC BY-SA 4.0