Timeline for Ring structure for $K^{-1}$?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jan 2, 2016 at 11:51 | comment | added | მამუკა ჯიბლაძე | @SebastianGoette Hands up - nothing to say :D | |
| Dec 31, 2015 at 19:21 | comment | added | Sebastian Goette | @მამუკაჯიბლაძე In your first comment you suggest to consider the ring structure coming from $\tilde K(\Sigma X)$. I just want to point out that any cup product on a suspension is trivial, no matter in which extraordinary multiplicative cohomology theory you work. On the other hand, suggestion 2 in the question seems to make sense and might actually produce a ring. | |
| Jun 3, 2014 at 9:12 | comment | added | მამუკა ჯიბლაძე | I believe a relevant question to study in connection to the last comment is - what kind of module is $K^{odd}$ over $K^{even}$ and $H^{odd}$ over $H^{even}$... (Note that they are moreover equipped with kind of scalar products $()^{odd}\otimes_{()^{even}}()^{odd}\to()^{even})$ | |
| May 29, 2014 at 15:21 | comment | added | Ho Man-Ho | @Steven Landsburg and semyon alesker: Actually my question was motivated by the Chern character. We know $ch:K^0(X)\to H^{even}(X; \mathbb{Q})$ is a ring homomorphism. However, the odd Chern character is only a group homomorphism. Even a ring structure of $K^{-1}$ exists, I don't think the odd Chern character would be a ring homomorphism unless we define/find a suitable ring structure on $H^{odd}(X; \mathbb{Q})$. I am not very satisfied that the odd Chern character is only a group homomorphism. | |
| May 28, 2014 at 18:40 | comment | added | asv | @StevenLandsburg: You are right. But I hoped that it does prove that there is no KNOWN canonical ring structure. | |
| May 28, 2014 at 18:06 | comment | added | Steven Landsburg | @semyonalesker: Your comment does not prove that there is no natural ring structure on $K^{-1}$. | |
| May 28, 2014 at 9:38 | comment | added | მამუკა ჯიბლაძე | You might try to use the suspension isomorphism $\tilde K^{-1}(X)\cong\tilde K(\Sigma X)$... | |
| May 28, 2014 at 9:36 | comment | added | asv | The product of two elements of $K^{-1}(X)$ belongs to $K^{-2}(X)$ which is identified with $K^0(X)$ via the Bott periodicity. Thus $K^{-1}(X)$ is not a ring. | |
| May 28, 2014 at 9:35 | history | edited | Ho Man-Ho |
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| May 28, 2014 at 9:30 | history | asked | Ho Man-Ho | CC BY-SA 3.0 |