Timeline for Non existence of commutative singular cochains vs quasi iso between cochains on BT and its cohomology
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 19, 2018 at 14:03 | comment | added | Sean Tilson | I think his chapter 4 is a good, more modern, take on what May did in his 1970s paper. | |
| Feb 16, 2018 at 22:24 | comment | added | J Cameron | @SeanTilson Yes, I was confused. Thanks for the pointer to Bob Bruner's adams spectral sequence book! | |
| Feb 16, 2018 at 15:06 | comment | added | Sean Tilson | I think it is the frobenous in degree 0,because it is the cup square. Do you know about Dyer-Lashof operations? Those might be a better perspective... | |
| Feb 15, 2018 at 14:09 | comment | added | J Cameron | That helps, thanks. @SeanTilson yes, it is nonzero on degree zero elements, I guess it sends them to the identity. | |
| Feb 15, 2018 at 11:12 | comment | added | Sean Tilson | The $\cup_i$ products encode (a lot of) the $E_{\infty}$ structure. In particular, $sq^i(x):=x\cup_{\vert x\vert -i} x$. So if a map doesn't respect $\cup_i$ products then it can not respect Steenrod operations. Also, $sq^0$ is nonzero on degree $0$ elements of a genuinely commutative DGA, right? | |
| Feb 15, 2018 at 9:16 | comment | added | Fernando Muro | From a highly categorical point of view, being commutative is not a property but a structure. Cochains form an $E_\infty$-algebra, and also cohomology in a rather trivial way, but the Guggenheim-May quasi-isomorphism is only an $A_\infty$-map, not compatible with the $E_\infty$ structures. Both $E_\infty$-algebras are not quasi-isomorphic as such, the Steenrod operations argument works. Concerning your second question, what you loose is all that higher structure. | |
| Feb 15, 2018 at 5:43 | history | asked | J Cameron | CC BY-SA 3.0 |