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Mar 12, 2019 at 19:56 comment added Jim Humphreys Agreed. but my "assumption" was meant to be a reply to Venkataramana's comment.
Mar 10, 2019 at 4:40 vote accept Strongart
Mar 8, 2019 at 23:28 comment added YCor There's no reason to assume $G$ semisimple/reductive. Indeed, Borel subgroups contain the solvable radical $R$, and more precisely are the inverse images of Borel subgroups of the corresponding semisimple quotient $G/R$. Hence the question immediately reduces to the semisimple case.
Mar 8, 2019 at 23:02 history answered Jim Humphreys CC BY-SA 4.0