Timeline for determining symplecticity (if that's a word)
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| May 24, 2012 at 0:56 | comment | added | user6976 | Yes, there are no solutions over the reals (if I did not make errors). The sequence of polynomials corresponding to the first matrix has a Groebner basis (the one found by Maple had 56 polynomials, I am not sure it is optimal, perhaps I should have chosen a different order of variables). Then the standard procedure solves it. | |
| May 24, 2012 at 0:50 | comment | added | Igor Rivin | Ah, so you are saying that the obstruction is getting it over the reals. After that we are down to a Diophantine problem, which is kind of what I was more interested in, but it's a start... | |
| May 24, 2012 at 0:28 | comment | added | user6976 | The problem reduces to a system of polynomial equations. That system can be simplified and gives a quadratic equation whose discriminant is negative. You can do it yourself with Maple or any other CAS. | |
| May 24, 2012 at 0:03 | comment | added | Igor Rivin | Why is that matrix not conjugate to a symplectic matrix? Is there an obvious obstruction? | |
| May 24, 2012 at 0:01 | history | answered | user6976 | CC BY-SA 3.0 |