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Oct 13, 2016 at 7:59 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
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Mar 17, 2016 at 2:14 history edited Jeanne Scott CC BY-SA 3.0
order of the cross product
Mar 16, 2016 at 20:03 answer added Jeanne Scott timeline score: 1
Mar 15, 2016 at 19:13 comment added Jeanne Scott I realise this is more narrow. However, once orientations are assigned to the type $B_2$ and type $G_2$ edges the corresponding unipotent radical is well defined --- in fact it can be constructed directly by generators and relations alone without recourse to either the Lie algebra or the enveloping Kac-Moody group. best, A. Leverkühn
Mar 15, 2016 at 18:55 comment added Jim Humphreys Also, your notion of "Coxeter diagram" is much more limited than the usual definition of "Coxeter graph" (or "Coxeter matrix") and creates ambiguity in passing to Lie algebras or groups: non-isomorphic ones may have isomorphic Coxeter groups attached.
Mar 15, 2016 at 17:59 history edited Jeanne Scott CC BY-SA 3.0
orientations added
Mar 11, 2016 at 2:03 history edited Jeanne Scott
tags change
Mar 10, 2016 at 15:59 comment added Jim Humphreys The tag 'lie-groups' here seems inappropriate, since Lie groups don't in general have an intrinsic Jordan decomposition (in particular, "unipotent radical" isn't generally defined). Maybe 'kac-moody-algebras' or 'algebraic-groups'? So far there isn't a tag for Kac-Moody groups.
Mar 10, 2016 at 3:27 history asked Jeanne Scott CC BY-SA 3.0