Timeline for Root in positive Weyl chamber
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Apr 3, 2015 at 17:23 | answer | added | Daniel Juteau | timeline score: 3 | |
| Mar 28, 2015 at 23:04 | answer | added | Allen Knutson | timeline score: 8 | |
| Mar 28, 2015 at 19:05 | vote | accept | shu | ||
| Mar 28, 2015 at 16:31 | answer | added | Jim Humphreys | timeline score: 4 | |
| Mar 28, 2015 at 14:12 | history | edited | shu | CC BY-SA 3.0 |
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| Mar 28, 2015 at 12:19 | comment | added | Sasha | @shu: This is a standard argument which does not depend on the classification. I would advise you to read any book on root systems. | |
| Mar 28, 2015 at 11:19 | comment | added | shu | @JimHumphreys, I rewrite the question. I hope it is more clear, and it make sense. | |
| Mar 28, 2015 at 11:18 | history | edited | shu | CC BY-SA 3.0 |
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| Mar 28, 2015 at 11:12 | history | edited | shu | CC BY-SA 3.0 |
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| Mar 28, 2015 at 11:07 | comment | added | shu | @Sasha, I totally agree. I rewrite the question. But I do not know how to get you first comment. Why the set $R\cap \overline{K}$ has only one or two elements? Is this can be obtained only by reading the Dykin diagram without long calculation. | |
| Mar 28, 2015 at 11:00 | history | edited | shu | CC BY-SA 3.0 |
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| Mar 28, 2015 at 5:24 | comment | added | Sasha | I guess there are only two cases when there is a root in $K$. One is $A_2$ with the highest root being $\rho$. The other is $A_1$ with the highest root being $2\rho$. | |
| Mar 27, 2015 at 22:10 | comment | added | Jim Humphreys | @shu: This example is correct, since the highest root happens to coincide with $\rho$. But it is apparently the only such case. (I echoed Sasha's comment too quickly, since the highest root can lie inside the chamber $K$. But it doesn't usually coincide with $\rho$. So I don't quite understand what you asking.) | |
| Mar 27, 2015 at 21:07 | comment | added | shu | Dear @JimHumphreys, In the case $A_2$, $\Pi=\{e_1-e_2,e_2-e_3\}$ and $R^+=\{e_1-e_2,e_1-e_3,e_2-e_3\}$. Then $\rho=e_1-e_3$ is a Root and in the open Weyl chamber. Am I right? | |
| Mar 27, 2015 at 20:52 | comment | added | Jim Humphreys | @shu: Your question isn't clear to me. The set $R \cap K$ is empty to begin with (as Sasha comments, only one or two roots can be dominant weights, and they don't lie in the interior of $K$). On the other hand, $\rho$ isn't a root but does lie in the (dominant) open Weyl chamber $K$. What are you trying to find a proof for without using classification? | |
| Mar 27, 2015 at 20:07 | comment | added | shu | @DietrichBurde, to learn and try to use it for calculating some explicit example or to construct some contraexample (like symmetric space.) This question comes from the Rigidity of Witten genus for homogeneous space. I do not know why you ask this question? | |
| Mar 27, 2015 at 19:49 | comment | added | Dietrich Burde | If Lie algebras are far away your area, why do you need it ? | |
| Mar 27, 2015 at 19:40 | comment | added | shu | Dear @Sasha, thank you for the comment. Is there some reference? What means simple-lanced? I am sorry, Lie algebra is far away my area... | |
| Mar 27, 2015 at 19:35 | comment | added | Sasha | It is easy to classify $R \cap \overline{K}$. In the simply-laced case this set consists just of one root (the highest root). In the non-simply-laced case it consists of two roots (the highest long root and the highest short root). However, typically both lie on a wall of $K$. | |
| Mar 27, 2015 at 19:21 | history | asked | shu | CC BY-SA 3.0 |