Timeline for Generating the symplectic group
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| May 27, 2011 at 13:54 | comment | added | Igor Rivin | @Artie: ah, I guessed the Calabi-Yau connection. I have been looking a little at the monodromy connection, but I guess this is really quite different. I am curious why you need to know finite index -- alg. geometers are usually happy with Zariski-dense... | |
| May 27, 2011 at 11:56 | comment | added | user5117 | Igor: they come from certain autoequivalences of Calabi--Yau varieties acting on the cohomology. | |
| May 26, 2011 at 14:41 | comment | added | Igor Rivin | Out of curiosity, where do your matrices come from? | |
| May 26, 2011 at 13:36 | vote | accept | CommunityBot | moved from User.Id=5117 by developer User.Id=36770 | |
| May 26, 2011 at 13:36 | history | edited | user5117 | CC BY-SA 3.0 |
update with derek holt's solution
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| May 26, 2011 at 12:28 | answer | added | Derek Holt | timeline score: 18 | |
| May 25, 2011 at 20:54 | answer | added | Igor Rivin | timeline score: 14 | |
| May 25, 2011 at 15:33 | comment | added | Ian Agol | Here's a related question: mathoverflow.net/questions/9628/… | |
| May 25, 2011 at 14:58 | comment | added | user5117 | I see. Thanks a lot for the explanation! | |
| May 25, 2011 at 14:42 | comment | added | Tom De Medts | The matrix $B$ is of the form $B = \begin{pmatrix} 0 & X \\ -X^{-1} & B \end{pmatrix}$, and the essential observation is that $\begin{pmatrix} I & X \\ 0 & I \end{pmatrix} \begin{pmatrix} I & 0 \\ -X^{-1} & I \end{pmatrix} = \begin{pmatrix} 0 & X \\ -X^{-1} & I \end{pmatrix}$. The fact that $\begin{pmatrix} 0 & X \\ -X^{-1} & I \end{pmatrix} \begin{pmatrix} I & Y \\ 0 & I \end{pmatrix} = \begin{pmatrix} 0 & X \\ -X^{-1} & I - X^{-1}Y \end{pmatrix}$ takes care of the last factor. | |
| May 25, 2011 at 14:18 | comment | added | user5117 | That's interesting. May I ask how you found this expression for B? | |
| May 25, 2011 at 13:54 | comment | added | Tom De Medts | The matrix $B$ can be written as $B = \left(\begin{array}{rrrr} 1 & 0 & 0 & -1 \\ 0 & 1 & -1 & 5 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) \cdot J^{-1} \cdot \left(\begin{array}{rrrr} 1 & 0 & -5 & -1 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) \cdot J \cdot \left(\begin{array}{rrrr} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)$. | |
| May 25, 2011 at 12:29 | comment | added | user5117 | Dear Tom, thanks for your suggestion. Could I ask you to elaborate a little? Those relations are all between matrices of specific types (rotations, translations, semi-involutions); I don't know, for instance, how B decomposes as a product of such. | |
| May 25, 2011 at 11:03 | answer | added | Charles Matthews | timeline score: 5 | |
| May 25, 2011 at 10:35 | comment | added | Tom De Medts | Have you already tried to use the relations in Stanek's (freely available) paper ams.org/journals/proc/1963-014-05/S0002-9939-1963-0153748-8/… ? | |
| May 25, 2011 at 9:38 | history | asked | user5117 | CC BY-SA 3.0 |