Timeline for Sp(2n) intersect Sp(2n,H)? (Please read for explanation of notation)
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Jan 26, 2010 at 23:04 | comment | added | José Figueroa-O'Farrill | This is indeed the case. You can have hermitian and skewhermitian forms on $\mathbb{H}^n$. The corresponding invariance groups are $\mathrm{Sp}(n)$ and $\mathrm{SO}^*(2n)$, at least in some notation. This is described in, say, Rossmann's Lie groups: an introduction through linear groups. | |
| Jan 26, 2010 at 22:30 | comment | added | Theo Johnson-Freyd | So, your argument seems to prove that there is no non-zero bilinear form on $\mathbb H^k$, independent of skew symmetry. And the problem is that asking the form to be instead sesquilinear, or whatever, rules out the possibility of demanding it be skew-symmetric. | |
| Jan 26, 2010 at 20:28 | comment | added | Theo Johnson-Freyd | I fixed a LaTeX typo. | |
| Jan 26, 2010 at 20:28 | history | edited | Theo Johnson-Freyd | CC BY-SA 2.5 |
fixed some latex
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| Jan 26, 2010 at 20:09 | comment | added | Mariano Suárez-Álvarez | Notice that you have proved that there is no non-zero bilinear form, as skew-symmetry plays no role. | |
| Jan 26, 2010 at 18:20 | history | answered | Johannes Hahn | CC BY-SA 2.5 |