Timeline for $K$ theory and singular cohomology
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 9, 2016 at 15:35 | vote | accept | Thomas Rot | ||
| Feb 9, 2016 at 14:58 | comment | added | Tyler Lawson | @SebastianGoette Fair enough. I was aiming for a clean statement. In reduced $KO$-theory then there are isomorphisms $\widetilde{KO}^0(\Sigma^r X) \cong \widetilde{KO}^{-r}(X) \cong \widetilde{KO}^{8m-r}(X)$, and so one can deduce the other $KO$-groups of $X$ from this. | |
| Feb 9, 2016 at 14:55 | comment | added | Tyler Lawson | @ThomasRot Yes on both counts. There is a class $w \in KO^4(pt)$ such that $w^2$ is twice the generator of $KO^8(pt)$ inducing Bott periodicity. After inverting $2$, the result is a 4-periodic theory. The isomorphism is indeed implemented by the Pontryagin character. | |
| Feb 9, 2016 at 14:39 | comment | added | Thomas Rot | @SebastianGoette: thanks for your comments. | |
| Feb 9, 2016 at 14:38 | comment | added | Sebastian Goette | @ThomasRot There is also a quaternionic $K$-theory, which turns out to be real $K$-theory shifted by four. Have a look at chapter 1 in Lawson-Michelsohn to see this. Another explanation: there is a Bott-periodicity map from $KO(S^4X)$ to $KSp(X)$ and vice versa. It seems that rationally, the two theories are isomorphic, hence the answer to your question would be yes. And to your last comment: the Chern character of a real vector bundle can be expressed entirely in terms of Pontryagin classes. | |
| Feb 9, 2016 at 14:38 | comment | added | Thomas Rot | Before asking this question I thought that one might need the Pontryagin or Stiefel-Whitney character. To me this did not seem to map into the right dimensions of the cohomology, which prompted me to ask this question. Is this map realized by the Pontryagin character? | |
| Feb 9, 2016 at 14:34 | comment | added | Sebastian Goette | The original question has $H^*$ and $K^*$, not $K^0$ and $H^{4*}$. In the complex setting, this works if one regards $K^*$ as a $\mathbb Z/2$-graded theory. Real $K$-theory is $\mathbb Z/8$-graded. @Thomas Rot: I just read your comment. The solution is that you get a direct sum of two copies of $H^*$ if you take the full 8-periodic $KO$-theory. | |
| Feb 9, 2016 at 14:33 | comment | added | Thomas Rot | Thank you for your answer. I will have to read up on the Adams operations, but this seems nice. I think I will have to read more about this stuff. The point that puzzles me is the fact that $KO$ is 8 periodic. Does this mean that after tensoring with $\mathbb{Q}$ this periodicity reduces to 4 periodicity? | |
| Feb 9, 2016 at 14:03 | history | edited | Tyler Lawson | CC BY-SA 3.0 |
deleted 3 characters in body
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| Feb 9, 2016 at 13:48 | history | answered | Tyler Lawson | CC BY-SA 3.0 |