Timeline for Maps from mod-$p$ Eilenberg-MacLane spectrum to connective $K$-theory spectrum
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 17, 2019 at 22:38 | comment | added | Dustin Clausen | Dear user438991. No, I'm not saying that. I'm saying \pi_1 KU = 0, so there's no lim^1 contribution from the Milnor sequence, and if you work out the lim^0 contribution (which involves \pi_0 KU=Z) you find it amounts to Hom(Q,Z) which is 0 anyway. | |
| Apr 17, 2019 at 9:26 | comment | added | Dustin Clausen | Dear user438991, you're very welcome! If you follow the same line of argument with HZ, you end up with [HZ,ku] = [KU smash HZ, KU]_{KU-modules} = [KU_Q, KU]_{KU-modules}. Here KU_Q is the rationalization of KU. This is the filtered colimit over the sequence where all the objects are KU and the n^{th} map is multiplication by n. Thus the calculation can be gotten from that of [\Sigma^i KU,KU]_{KU-modules} = \pi_i KU for i=0,1 by the Milnor sequence. You should see that actually [HZ,ku]=0 after all. | |
| Apr 15, 2019 at 11:20 | history | edited | Dustin Clausen | CC BY-SA 4.0 |
added 132 characters in body
|
| Apr 15, 2019 at 10:12 | comment | added | Dustin Clausen | You can probably shortcut the computational proof by using Snaith's presentation KU = \Sigma^\infty_+ BU(1)[bott^{-1}]. Then it's enough to invert Bott on the homology of the space BU(1). That this dies with finite coefficients follows from the fact that the pontryagin ring structure on H_*(BU(1)) is a divided power algebra. | |
| Apr 15, 2019 at 10:11 | comment | added | Dustin Clausen | I guess there are also more computational proofs. You could brute force calculate HZ/p smash KU: by Bott perodicity, it is (with shifts) a certain N-indexed colimit of Z/p-homologies of the space BU ( = colim n BU(n) ). The latter is a well-known calculation, and then it's a matter of identifying the effect of the Bott map \Sigma^2 BU --> BU on homology to be able to pass to the colimit. I guess the key for the vanishing should be the result (of Bott) that the Hurewicz image of the n^{th} power Bott element in \pi_{2n}BU is divisible by (n-1)!, hence dies (mod p) for large n. | |
| S Apr 15, 2019 at 9:16 | history | suggested | user43326 | CC BY-SA 4.0 |
Improved visual presentation
|
| Apr 15, 2019 at 8:26 | comment | added | Dustin Clausen | Another proof is using Adams' self-map v_1: \Sigma^d S/p --> S/p. You can pin down the minimal d (2p-2 for odd p, 8 for p=2), but the only important thing here is that it's positive. When you smash with KU this v_1 becomes a power of the Bott element and hence is invertible, but when you smash with HZ it becomes null for degree reasons. Hence HZ/p smash KU ( = S/p smash HZ smash KU) carries a self-map which is both invertible and null, therefore it must vanish. | |
| Apr 15, 2019 at 8:23 | comment | added | Dustin Clausen | Dear user438991, I think this fact has come up a few times on MathOverflow. Anyway, here are two proofs. One is using the theory of complex oriented cohomology theories: HZ/p gives the additive formal group and KU gives the multiplicative formal group, so on HZ/p smash KU you must have a formal group which is isomorphic to both the additive and multiplicative one. But this is impossible over a ring of characteristic p, because the heights are different. | |
| Apr 15, 2019 at 7:43 | review | Suggested edits | |||
| S Apr 15, 2019 at 9:16 | |||||
| Apr 15, 2019 at 6:52 | comment | added | user438991 | Dear @DustinClausen. I can't see why is it true that $H\mathbb{Z}/p$ smash with $KU$ is 0. | |
| Apr 11, 2019 at 20:27 | history | answered | Dustin Clausen | CC BY-SA 4.0 |