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Oct 12, 2014 at 13:33 comment added Jim Humphreys @Allen: No, it's a general fact (if "antidominant" is defined in the natural way: see 3.5 for discussion of terminology). However, there seems to be no simple unified proof, so I wound up treating integral weights in 4.4 but waited until 4.8 to deal with non-integral ones. Here as elsewhere in the subject it's annoying to make detours into more elaborate arguments for arbitary $\lambda \in \mathfrak{h}^*$. And sometimes the integral case is all one wants.
Oct 11, 2014 at 18:12 comment added Allen Knutson The "simple iff antidominant" statement is only in the presence of the $\lambda$ integral assumption, right?
Oct 10, 2014 at 18:58 history answered Jim Humphreys CC BY-SA 3.0