Timeline for E-infinity structure on singular cochains
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 11, 2022 at 2:33 | comment | added | Hari Rau-Murthy | (Here = denotes quasiisomorphism) | |
| Apr 11, 2022 at 0:59 | comment | added | Hari Rau-Murthy | 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/… | |
| Dec 23, 2018 at 22:08 | vote | accept | Vladimir Baranovsky | ||
| Jul 2, 2015 at 6:06 | comment | added | Vladimir Baranovsky | 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 | |
| Jul 1, 2015 at 14:09 | comment | added | Tyler Lawson | 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. | |
| Jul 1, 2015 at 8:27 | comment | added | Vladimir Baranovsky | 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)? | |
| Jul 1, 2015 at 5:09 | history | answered | Tyler Lawson | CC BY-SA 3.0 |