Timeline for Finite spectrum annihilated by multiplication by two
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 15, 2013 at 13:53 | comment | added | Charles Rezk | @AkhilMathew I don't think so. After all, you just need a unital map $T_*(M)\to R$, not a ring map. | |
| Nov 14, 2013 at 17:10 | comment | added | Akhil Mathew | Is associativity necessary? It seems to me that this argument shows that any ring spectrum (up to homotopy, and not necessarily associative) with $2 = 0$ cannot be a finite spectrum (which is a more general version of the statement that the mod $2$ Moore spectrum fails to be a ring spectrum). | |
| Nov 14, 2013 at 15:49 | comment | added | Saul Glasman | I asked @TylerLawson on chat to explain the easy argument referred to in the last paragraph, and here's what he said (edited for length): "the cohomology of T_*(M) inherits a coalgebra structure from the ring structure, so there's a map H^*(T_*M) -> H^*(T_*M) ⊗ H^*(T_*M)... however, the Cartan formula tells you that Sq^n(u) maps to Sq^n(u⊗u) = sum Sq^p(u)⊗Sq^q(u) in particular, by induction we know Sq^{n-1}(u)⊗Sq^1(u) is nonzero thus Sq^n(u) must have been nonzero in the first place." | |
| Nov 14, 2013 at 15:19 | comment | added | Charles Rezk | @TylerLawson It does seem easy. I guess the same argument shows that if $R$ is an associative ring of "characteristic $\alpha$", then you have $\xi_1$. | |
| Nov 14, 2013 at 14:30 | vote | accept | Akhil Mathew | ||
| Nov 14, 2013 at 5:36 | comment | added | Tyler Lawson | At odd primes, how many relations to you have to impose in an associative algebra beyond $p=0$ to detect $\xi_1$? | |
| Nov 14, 2013 at 5:31 | comment | added | Tyler Lawson | Huh. That's... much easier than I expected a proof to be. | |
| Nov 14, 2013 at 4:26 | history | answered | Charles Rezk | CC BY-SA 3.0 |