Timeline for A computation about Whittaker functions and Eisenstein series
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 11, 2015 at 6:18 | comment | added | Jianrong Li | great! My email address is: [email protected]. If there are some parts of the book (or some other books, papers) which you would like to discuss, please do not hesitate to send me emails. | |
| Jun 11, 2015 at 0:17 | comment | added | Spencer Leslie | Excellent, these are the chapters of interest for me as well. I am currently trying to learn quantum groups and crystal graphs from Hong-Kang. | |
| Jun 10, 2015 at 11:15 | comment | added | Jianrong Li | I am trying to read Chapters 1-5, 19, 20 of the book. | |
| Jun 10, 2015 at 10:37 | comment | added | Jianrong Li | it seems that in the book they use the decomposition $G = \cup_{w \in W} Bw^{-1}B=\cup_{w \in W} Bw^{-1}U$. Then we have the representatives of $B \backslash G = \cup_{w \in W} B \backslash Bw^{-1}U$ are of the form $w^{-1}u$. We also have $B w^{-1} u = B w^{-1}$ implies that $u \in w U w^{-1}$. So we have verified the statement in the book. | |
| Jun 10, 2015 at 10:31 | comment | added | Jianrong Li | yes, I am reading the book. It is great to discuss the book with you. It seems that $Bwu=Bw$ implies that $u \in w^{-1}Uw$ (not $wUw^{-1}$): suppose that $Bwu=Bw$. Then $b w u = b' w$ for some $b, b' \in B$. By your proof, we have $b^{-1}b' \in U$. We also have $u = w^{-1} b^{-1}b' w$. Therefore $u \in w^{-1} U w$. | |
| Jun 10, 2015 at 10:25 | comment | added | Spencer Leslie | Are you currently working through this book? I ask because I am doing the same would be interested in discussing it. | |
| Jun 9, 2015 at 11:31 | comment | added | Spencer Leslie | You're right. I believe this is the correct argument now. | |
| Jun 9, 2015 at 11:28 | history | edited | Spencer Leslie | CC BY-SA 3.0 |
corrected the argument for part (1)
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| Jun 9, 2015 at 10:24 | comment | added | Jianrong Li | thank you very much. But it seems that $wu=w'u'$ and $w = w'$ imply that $u=u'$ (not $u'=wuw^{-1}$). | |
| Jun 9, 2015 at 10:22 | vote | accept | Jianrong Li | ||
| Jun 7, 2015 at 13:52 | history | answered | Spencer Leslie | CC BY-SA 3.0 |