Timeline for An attempt at an alternative calculation of the rank of $\pi_n(MO)$
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 4 at 18:16 | vote | accept | Chris | ||
| Dec 4 at 7:35 | answer | added | Chris | timeline score: 0 | |
| Dec 2 at 23:53 | comment | added | Chris | @TimCampion well, in my previous post I was confused about how the validity of Stong's argument along this line, which led me to try and take a sledgehammer to the problem with the Adams spectral sequence. Now I am just curious if there is a way to directly calculate the rank of these $\operatorname{Hom}$ groups, especially because the Adams spectral sequence is used for the harder cobordism theories. | |
| Dec 2 at 20:36 | comment | added | Tim Campion | Wait I'm confused -- if you already know that $H^\ast(MO)$ is a free module over the Steenrod algebra, then you're basically done -- let $\{x_i\}$ be a basis for $H^\ast(MO)$ over the Steenrod algebra. By representability of cohomology, this gives a map $MO \to \oplus_{i}\Sigma^{|x_i|} H\mathbb F_2$, which is obviously an isomorphism on cohomology, and hence an isomorphism... what is the point of mucking around with the Adams spectral sequence? | |
| Dec 2 at 18:01 | comment | added | Chris | @TimCampion these are great notes, thank you for the reference. However, they do not use the Adams spectral sequence as far as I can tell, which is something I am still curious about. | |
| Dec 2 at 12:36 | comment | added | Tim Campion | Haven't had a chance to really read what you're asking, but for Thom's theorem on MO, I'm very fond of Cary Malkiewich's notes people.math.binghamton.edu/malkiewich/cobordism.pdf | |
| Dec 2 at 8:18 | history | asked | Chris | CC BY-SA 4.0 |