Timeline for Maximal torus of Chevalley group $Sp(4)$
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Aug 25, 2022 at 14:09 | comment | added | Martin Sleziak | The link in the post is dead - but the text can be accessed in various Wayback Machine snapshots, such as here and here. | |
| Dec 5, 2015 at 18:37 | vote | accept | MathStudent | ||
| Dec 5, 2015 at 10:33 | history | edited | MathStudent | CC BY-SA 3.0 |
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| Dec 5, 2015 at 10:21 | history | edited | MathStudent | CC BY-SA 3.0 |
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| Dec 4, 2015 at 17:30 | history | edited | MathStudent | CC BY-SA 3.0 |
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| Dec 4, 2015 at 17:28 | comment | added | Alex B. | Who is Nill? $ $ | |
| Dec 4, 2015 at 17:12 | answer | added | Jim Humphreys | timeline score: 5 | |
| Dec 4, 2015 at 17:01 | comment | added | MathStudent | @ paul. what is your $x_{\alpha}(t)$? It will be helpful if you please elaborate your comment when you have time. | |
| Dec 4, 2015 at 16:20 | comment | added | Paul Broussous | I did the computation and did find a diagonal matrix: $h_\alpha (t) ={\rm Diag}(t,t^{-1}, t^{-1},t)$. | |
| Dec 4, 2015 at 15:33 | history | edited | MathStudent | CC BY-SA 3.0 |
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| Dec 4, 2015 at 15:29 | comment | added | MathStudent | I think that the element you mentioned is not an element of $Sp(4)$. So my new question is : Does there exists a chevalley basis of $Sp(2n)$ such that the $h_{\alpha}(t)$ are diagonal matrices? | |
| Dec 4, 2015 at 12:11 | comment | added | Nick Gill | I don't quite follow your calculations, but perhaps the problem is that a split torus in $Sp_4(K)$ need not consist of diagonal matrices, it just needs to be conjugate to a set of diagonal matrices. Perhaps you've just chosen a basis for which the $h_\alpha(t)$ are not diagonal? (My guess for how to fix this would be to have $x_\alpha(t)=1+t(E_{12}-E_{34})$ -- i.e. swap the last two vectors in your basis -- but I don't have the wherewithal to do the calculation right now.) | |
| Dec 4, 2015 at 11:06 | history | asked | MathStudent | CC BY-SA 3.0 |