Timeline for why the group $GL(6,V)$ has an open orbit?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Jan 14, 2013 at 17:34 | comment | added | user21574 | Dear Robert , thanks for your nice Remark | |
| Jan 14, 2013 at 12:59 | history | edited | Robert Bryant | CC BY-SA 3.0 |
added information about open $k$-form orbits
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| Jan 14, 2013 at 8:12 | comment | added | user21574 | @@Dears Kreck and Robert many thanks for your helpful comments | |
| Jan 14, 2013 at 0:41 | comment | added | user29720 | @Robert Bryant: For example, if $k = \mathbf{Q}_p$ then the number of open $G(k)$-orbits is 4 when $p$ is odd and 8 when $p = 2$, and if $k$ is a finite extension of $\mathbf{Q}_p$ then it remains 4 for odd $p$ but is a bit more complicated when $p = 2$. | |
| Jan 14, 2013 at 0:37 | comment | added | user29720 | @Robert Bryant: True, I was just illustrating a method that can be used for general situations where one is trying to analyze the relationship between $G(k)$-orbits and $G$-orbits in the algebro-geometric sense (in effect, $G(\overline{k})$-orbits, for which eigenvalue issues are not too much of a headache, etc.) where $k$ can be any field of characteristic 0 (such as $\mathbf{R}$, or $\mathbf{Q}$, etc.). For example, one might be curious about counting the number of open $G(k)$-orbits when working over other "analytic" ground fields such as $k = \mathbf{Q}_p$. | |
| Jan 14, 2013 at 0:26 | comment | added | Robert Bryant | @kreck: Thanks for outlining this sophisticated proof. However, one doesn't need such tools to classify the orbits (even the non-open ones) in $\Lambda^3(\mathbb{R}^6)$, as it can be done easily using standard exterior algebra facts. It's just not obvious (which is what I wrote). If the reader wants details on this argument, one place they are available is in the Appendix to my article "On the geometry of almost complex $6$-manifolds" (arxiv.org/pdf/math/0508428.pdf). | |
| Jan 13, 2013 at 20:23 | comment | added | Mikhail Borovoi | @Hassan: If you like the answer (and you probably like it, because you have accepted it), you can also vote it up! | |
| Jan 13, 2013 at 20:21 | comment | added | user29720 | So the open orbit as an $\mathbf{R}$-variety is $G/H$ with $H = ({\rm{SL}}_3 \times {\rm{SL}}_3) \rtimes (\mathbf{Z}/(2))$. By Cor. 1 to Prop. 36 in Ch. I of Serre's "Galois cohomology" book, $G(\mathbf{R})\backslash (G/H)(\mathbf{R})$ is identified with the kernel of the map ${\rm{H}}^1(\mathbf{R},H)\rightarrow {\rm{H}}^1(\mathbf{R},G)$. The target vanishes by Hilbert 90, and the source bijects onto ${\rm{H}}^1(\mathbf{R},H/H^0)$ of size 2 since $H^0$ has only two $\mathbf{R}$-forms (itself and "${\rm{SL}}_3(\mathbf{C})$ as an $\mathbf{R}$-group") and both have trivial degree-1 cohomology. | |
| Jan 13, 2013 at 19:57 | comment | added | user29720 | To see there are exactly 2 open orbits, we can use algebraic-group methods as follows. Let $G={\rm{GL}}_6$. By considering tangent spaces analytically (over $\mathbf{R}$ and $\mathbf{C}$) and algebraically (over $\mathbf{C}$), and dimensions of stabilizer groups analytically and algebraically, a point $\omega$ has open orbit under $G(\mathbf{R})$ in the analytic sense if and only if it has Zariski-open $G$-orbit in the algebraic sense (over $\mathbf{C}$, or use $\mathbf{R}$-schemes). By the nature of the Zariski topology, the open algebraic orbit through $\phi_0$ is the only one. (cont'd) | |
| Jan 13, 2013 at 19:44 | vote | accept | CommunityBot | ||
| Jan 13, 2013 at 17:33 | comment | added | user21574 | Dear Robert, you beautifully solved it, thanks | |
| Jan 13, 2013 at 17:30 | vote | accept | CommunityBot | ||
| Jan 13, 2013 at 19:43 | |||||
| Jan 13, 2013 at 16:26 | history | answered | Robert Bryant | CC BY-SA 3.0 |