Timeline for Generators K-theory of Cuntz algebras
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 20, 2016 at 17:01 | vote | accept | John N. | ||
| Feb 20, 2016 at 16:38 | answer | added | Nik Weaver | timeline score: 4 | |
| Feb 20, 2016 at 13:47 | comment | added | John N. | Thank you very much for the help. I'll read Abrams' paper. | |
| Feb 20, 2016 at 13:36 | comment | added | Benjamin Steinberg | I am not sure where to find a reference for the operator case. You can find this for Leavitt path algebras in the survey of Abrams on the first decade of Leavitt path algebras. I am sure the operator theory proof is essentially the same but as an algebraist I wouldn't know how to write it in a technically correct way. | |
| Feb 20, 2016 at 13:28 | comment | added | John N. | Ok. So we can choose 1, and say that it is a generator of the K_0 group. Could you give me a reference or expand your comment in an answer that proves that the class of 1 is not zero? Thank you for the help | |
| Feb 20, 2016 at 12:25 | comment | added | Benjamin Steinberg | Yes you are right it is equivalent to 1. But isn't the class of 1 the generator not 0 for n>2? All n of those idempotents are equivalent to 1 so the cuntz relation gives n[1]=[1]. | |
| Feb 20, 2016 at 12:19 | comment | added | John N. | @ Benjamin Steinberg: I am not sure. Probably I am wrong, but S_S_1^* is equivalent to 1 (S_1 is an isometry). However, the class of [1] is 0. | |
| Feb 20, 2016 at 12:16 | comment | added | Benjamin Steinberg | I think it should correspond to the idempotent s_1s_1* | |
| Feb 20, 2016 at 9:46 | history | asked | John N. | CC BY-SA 3.0 |