Is there a transparent explanation of why the singular cochain complex of a topological space X is an $E_\infty$ algebra. There are combinatorial proofs using, say, the surjection operad, but is there a topological picture behind those? If we wanted to restrict,say, to the level of little (2-)discs, could we describe the structure using something like Eilenberg-Zilber contractions of $C^*(X^m)$ onto $C^*(X)^{\otimes m}$, various diagonal maps and so on?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
5
If you wrote $E(n)$ for the chain complex of natural transformations $C_*(-) \to C_*(-)^{\otimes n}$, the $E(n)$ collectively form an operad, parametrizing all natural "one-to-many" transformations on chains, called the Eilenberg-Zilber operad. The defining co-action on chains turns into a natural action on cochains, and so the Eilenberg-Zilber operad acts on the singular cochain functor $C^*(-)$. (I seem to recall that there is something to be careful with here regarding the order of composition and whether this is naturally an operad or the reverse of an operad, but the details escape me.)
One can show that the homology groups $H_* E(n)$ are just $\Bbb Z$, concentrated in degree zero, for all $n$, and that the composition operations make these homology groups into the commutative operad. The method of acyclic models is a nice way to prove this (but that seems to have fallen by the wayside as a standard part of the curriculum). It's not clear that Eilenberg-Zilber operad is an $E_\infty$ operad because the symmetric groups aren't guaranteed to act nicely enough, but this is enough to show you that it accepts a weak equivalence from an $E_\infty$ operad (e.g. a cofibrant replacement) without having to know a combinatorial construction.
It seems, from your question, like you would like a close connection to the little discs operads. Unfortunately, this point of view does not provide one at all, and I don't know of a "natural" way to show this that doesn't rely on either a specific combinatorial construction.
answered Jul 1, 2015 at 5:09
-
$\begingroup$ Is it true that elements of E(n) can be written as compositions of various diagonal maps, Alexander-Whitney maps and Eilenberg-Zilber homotopies (is that done explicitly enough)? $\endgroup$ Commented Jul 1, 2015 at 8:27
-
2$\begingroup$ I'm not completely certain what qualifies as an Eilenberg-Zilber homotopy, but I feel reasonably certain that the answer to your question is "no". For all $n > 0$, including $n=1$, $E(n)$ has nonzero terms in negative filtration, does not appear to be free, and is not countable (it is an infinite product of free modules). I suspect that it does not have a nice presentation in terms of generators and relations. $\endgroup$ Commented Jul 1, 2015 at 14:09
-
$\begingroup$ By EZ homotopy I meant a homotopy contracting $C_*(X^n)$ onto $C_*(X^p) \otimes C_*(X^{n-p})$. But if the operad is not countable then it does not look right... Thank you, Tyler $\endgroup$ Commented Jul 2, 2015 at 6:06
-
$\begingroup$ There is in fact a close connection to the little discs operad. McClure and Smith show that $C_*(\text{n-th space in little discs})=E(n)$ here ams.org/journals/jams/2003-16-03/S0894-0347-03-00419-3/… $\endgroup$ Commented Apr 11, 2022 at 0:59
-
$\begingroup$ (Here = denotes quasiisomorphism) $\endgroup$ Commented Apr 11, 2022 at 2:33