Timeline for symplectic matrices
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Mar 5, 2011 at 21:06 | comment | added | Elizabeth S. Q. Goodman | Retagged from "symplectic", but I'm not sure I should have. A "symplectic-matrices" or "symplectic-group" tag would be better than the ambiguous "symplectic" tag, though, so I hope that one dies. | |
| Mar 5, 2011 at 20:35 | history | edited | Elizabeth S. Q. Goodman |
edited tags
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| Oct 1, 2010 at 3:50 | comment | added | Theo Johnson-Freyd | (and maybe, looking at wikipedia, you don't want it to fix the form infinitesimally, but rather honestly). | |
| Oct 1, 2010 at 3:46 | comment | added | Theo Johnson-Freyd | (and, I guess, since you are over $\mathbb F_2$, $1=-1$ is fine in the matrix.) | |
| Oct 1, 2010 at 3:45 | comment | added | Theo Johnson-Freyd |
I don't understand the question. I assume that $(AB|CD)=\begin{pmatrix}A&B\\C&D\end{pmatrix}$ is a block matrix over $\mathbb F_2$. Your notation suggests moreover that $A,B,C,D$ are all squares. "symplectic" presumably means that it fixes (infinitesimally) some particular symplectic form, which (if I had to guess) you are taking to be $(AB|CD)=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$, where by "$1$" I mean the identity square matrix of whatever size you're working with. If this is all correct, fine, but you should edit the question to make precise the notation.
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| Sep 30, 2010 at 22:19 | comment | added | Steven Sam | Does $(AB|CD)$ mean $\begin{pmatrix} A & B \\ C & D \end{pmatrix}$? | |
| Sep 30, 2010 at 20:03 | comment | added | Willie Wong | I fixed the math display to look like the one you put into the duplicate post. I hope it is what you want. | |
| Sep 30, 2010 at 20:01 | history | edited | Willie Wong | CC BY-SA 2.5 |
Fix the math to look like the duplicate. As the OP intended it.
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| Sep 30, 2010 at 19:23 | history | asked | Robert Gunning | CC BY-SA 2.5 |