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Oct 7, 2014 at 8:46 comment added Henrik Winther I edited in "in characteristic 0". My thought was that people interested in algebraic groups would read the answer by Dietrich, where this is explained, but I see now that it was confusing.
Oct 7, 2014 at 8:43 history edited Henrik Winther CC BY-SA 3.0
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Oct 6, 2014 at 17:19 comment added Jim Humphreys Actually, there need not exist a Levi decomposition for an algebraic group in prime characteristic. But in characteristic 0 Chevalley's old framework is applicable. (In either case, the Jordan decomposition for algebraic groups makes it somewhat more natural to deal with the unipotent radical and a possible reductive Levi factor.)
Oct 6, 2014 at 12:01 history wiki removed S. Carnahan
Oct 6, 2014 at 8:55 history answered Henrik Winther CC BY-SA 3.0