Timeline for Fixed space of the square of a symmetric matrix over $\mathbb{F}_2$
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Mar 11, 2015 at 20:10 | comment | added | Geoff Robinson | The condition that $M$ be in ${\rm Sp}(2n,2)$ is that $M^{t}\left(\begin{array}{clcr} 0 & I_{n}\\I_{n} & 0 \end{array}\right)M = \left(\begin{array}{clcr} 0 & I_{n}\\I_{n} & 0 \end{array}\right).$ | |
| Mar 11, 2015 at 18:46 | comment | added | Lisa S. | @NoamD.Elkies: Yes, you are right. I thought that the antisymmetry condition in characteristic $2$ simply translates to the matrix being symmetric. Did I make a mistake by forgetting that alternating $\neq$ antisymmetric in characteristic $2$? What is the concrete condition for $M$ to be in $\mathrm{Sp}_{2n}$? If you could answer my question in the comment, I would be most grateful. | |
| Mar 11, 2015 at 18:44 | vote | accept | Lisa S. | ||
| Mar 11, 2015 at 10:48 | answer | added | Stefan Kohl♦ | timeline score: 5 | |
| Mar 11, 2015 at 9:06 | review | Close votes | |||
| Mar 11, 2015 at 9:56 | |||||
| Mar 11, 2015 at 8:01 | comment | added | Geoff Robinson | @NoamD.Elkies (and darij). I just address the original question about symmetric matrices. | |
| Mar 11, 2015 at 7:59 | comment | added | Geoff Robinson | @DaveWitteMorris :Done. (Lisa- you are welcome). | |
| Mar 11, 2015 at 7:56 | answer | added | Geoff Robinson | timeline score: 8 | |
| Mar 11, 2015 at 5:54 | comment | added | darij grinberg | What if $M$ is alternating? | |
| Mar 11, 2015 at 5:44 | comment | added | Dave Witte Morris | @GeoffRobinson, please post your comment as an answer (so the question no longer shows up as unanswered). Thanks (and well done)! | |
| Mar 11, 2015 at 2:16 | comment | added | Noam D. Elkies | In this context $A$ must be in ${\rm Sp}_{2n}$, yes? | |
| Mar 11, 2015 at 1:12 | comment | added | Geoff Robinson | $\left(\begin{array}{clcr}1&1&0&0\\1&1&1&0\\0&1&1&0\\0&0&0&1 \end{array} \right).$ | |
| Mar 11, 2015 at 1:09 | comment | added | Lisa S. | @GeoffRobinson: Thanks. Which one? | |
| Mar 11, 2015 at 1:05 | comment | added | Geoff Robinson | There is a $4 \times 4$-matrix for which the dimension is odd. | |
| Mar 11, 2015 at 0:55 | history | edited | Lisa S. |
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| Mar 11, 2015 at 0:53 | comment | added | Lisa S. | It's not homework. If $A$ is a principally polarized abelian variety over a finite field $\mathbb{F}_q$ of odd characteristic, is $A(\mathbb{F}_{q^2})[2]$ even dimensional? | |
| Mar 11, 2015 at 0:50 | comment | added | Igor Rivin | Where does this come from? Looks a lot like homework. | |
| Mar 10, 2015 at 23:51 | history | asked | Lisa S. | CC BY-SA 3.0 |