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Oct 23, 2019 at 18:31 answer added Mark Grant timeline score: 6
Oct 23, 2019 at 18:01 history edited Qayum Khan CC BY-SA 4.0
EDIT: As suggested, I renamed the base space $B$ to $T$, in order to not confuse it with the classifying space functor. Also, I added two citations.
Oct 23, 2019 at 17:30 history edited Qayum Khan CC BY-SA 4.0
EDIT: As suggested, I renamed the base space $B$ to $T$, in order to not confuse it with the classifying space functor.
Oct 23, 2019 at 16:55 comment added Denis T @QayumKhan DenisNardin is right, I mean just an application of classifying space functor to map $\Omega T \to Aut(\Omega T)$ representing left (more precisely, the side $\pi_1$ of base acts on fiber in your preferrable conventions) multiplication. Also possibly you want to use Moore loops for that to avoid some nuances with non-strictly associative actions etc.
Oct 23, 2019 at 7:48 comment added Denis Nardin @QayumKhan The tautological action is the one corresponding to the pathspace fibration $\Omega T\to P T\to T$ (morally it is the action of $\Omega T$ on itself by left(?) multiplication)
Oct 22, 2019 at 23:32 history edited Josiah Park CC BY-SA 4.0
reverting title edit (was unfamiliar with word 'delooping' initially, my apologies)
Oct 22, 2019 at 23:16 history edited Josiah Park CC BY-SA 4.0
edited title
Oct 22, 2019 at 22:49 comment added Qayum Khan @Denis T. : Любезно, what is the explicit formula for this tautological action? Does it it assume that $T$ is a topological group, say by pointwise-conjugating a loop by an element of $T$? If so, isn't this instead a map $T \longrightarrow Aut(\Omega T)$?
Oct 22, 2019 at 21:50 comment added Denis T (I rename $B$ to $T$ to avoid confusion) I think the only obstruction is factorizing action map $E \to BAut(\Omega T)$ through delooping of tautological action $T \to BAut(\Omega T)$ which is done by usual obstruction theory.
Oct 22, 2019 at 20:49 history asked Qayum Khan CC BY-SA 4.0