Timeline for How to inject PGL (n, k) in PGL (n +1, k)
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 21, 2012 at 22:41 | vote | accept | Rajkarov | ||
| Mar 16, 2012 at 15:16 | comment | added | Derek Holt | Rajkarov: No. For example ${\rm PGL}(5,3)$ has no subgroup isomorphic to ${\rm PGL}(4,3)$. The 2- to 3-dimensional embedding is exceptional. | |
| Mar 16, 2012 at 11:56 | comment | added | François Brunault | I think the embedding $\mathrm{PGL}(2,K) \to \mathrm{PGL}(3,K)$ can be seen as follows : let $\mathrm{GL}_(2,K)$ act naturally on the space of binary quadratic forms over $K$. If $g \in \mathrm{GL}_(2,K)$ is such that $Q \circ g =\lambda Q$ for every binary quadratic form $Q$, then by taking squares of linear forms we see that $g$ fixes every elements of $\mathbf{P}^1(K)$, so that $g$ is scalar. | |
| Mar 16, 2012 at 11:49 | comment | added | Rajkarov | Thank you Mr Holt for precision. I'm interested in this injection for $k$ non archimedean local field. You say that $PGL(2,K)$ embeds in $PGL(3,K)$ but is it true that $PGL(2n,K)$ embeds in $PGL(2n+1,K)$ ? | |
| Mar 16, 2012 at 10:07 | history | answered | Derek Holt | CC BY-SA 3.0 |