Timeline for reflection LIE groups
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| May 14, 2013 at 15:12 | comment | added | Harm Derksen | There are no simple reflection Lie groups. If $G$ is a simple reflection Lie group, then it is generated by commutators and commutators lie in $SU(n)$. So G is contained in $SU(n)$ but $SU(n)$ does not contain any (generalized) reflections. But perhaps one may consider the case for which $G/[G,G]$ is finite. | |
| May 14, 2013 at 4:16 | comment | added | Theo Johnson-Freyd | I don't have a useful answer to your question, but I vaguely recall at least one conference talk in which $O(n)$ was compared, with quite some success, to a finite reflection group, so this question is not entirely out of left field. | |
| May 14, 2013 at 3:53 | comment | added | Allen Knutson | Indeed, if we compose $G \to U(V) \to U(V \otimes {\mathbb C}^2)$ then $G$ is not such a group. Hi Harm, welcome to MO! | |
| May 14, 2013 at 3:26 | comment | added | Misha | Two trivial remarks: 1. If the subgroup G is simple as an abstract group, this is just the question on whether G contains any (complex) reflections at all. 2. One then asks which irreducible representations $G\to U(n)$ sends any involution to a reflection. The latter question probably has the answer "hardly any", but making this precise would require doing some computations. | |
| May 14, 2013 at 2:38 | history | asked | Harm Derksen | CC BY-SA 3.0 |