Timeline for irreducible subgroup of SL(n,R)
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 19, 2010 at 16:57 | answer | added | Victor Protsak | timeline score: 4 | |
| Aug 19, 2010 at 10:09 | comment | added | damiano | Let $G$ be any subgroup of $SO(2,R)$. This group has a natural real representation of dimension two that is irreducible with only a couple of exceptions. The same representation is not irreducible over the complex numbers. Note that among the various choices for $G$ there are finitely generated infinite groups with any finite number of generators. | |
| Aug 19, 2010 at 9:47 | comment | added | user8617 | If we assume the subgroup of SL(n,R) is finitely generated with more than 2 generators and infinite, is there a counter example? | |
| Aug 19, 2010 at 9:20 | comment | added | damiano | The integers act on $R^2$ via $1 \mapsto \begin{pmatrix}\cos(n) & \sin(n) \cr -\sin(n) & \cos(n) \end{pmatrix}$; this action has no non-trivial invariant subspaces over the real numbers, but decomposes into a sum of two one-dimensional representations over the complex numbers. | |
| Aug 19, 2010 at 9:08 | comment | added | user8617 | Thanks for your answer. If we assume the subgroup of SL(n,R) is finitey generated infinite group, is there a counter example? | |
| Aug 19, 2010 at 8:57 | comment | added | damiano | This is false: the cyclic group with three elements acts on $R^3$, permuting the coordinates and fixing the subspace $V$ whose coordinates sum to zero. The representation $V$ is irreducible over the reals, but not over the complex numbers. | |
| Aug 19, 2010 at 8:43 | history | asked | user8617 | CC BY-SA 2.5 |