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Jul 12, 2014 at 15:19 comment added user44191 Ah, sorry; I think I missed the tiny asterisk in the second subscript.
Jul 12, 2014 at 15:00 comment added Jim Humphreys My last paragraph is accurate, but I was simplifying the original formulation by replacing $\mu^*$ by $\mu$. Anyway, Victor's answer and the fuller version I got from Kumar extract the answers to (1) and (2) mainly from PRV. For (2) it's probably not strictly necessary to interpret the modules as global sections of line bundles, but for this version the best convention is to use $B^-$ rather than $B$. Algebraic group people like Andersen and Jantzen write things this way. Mixing $B, B^-$ gets confusing.
Jul 11, 2014 at 17:33 comment added user44191 I don't think that's quite accurate; it should be that $V_\lambda \otimes V_{\mu'}$ is included in $V_{\lambda + \nu} \otimes V_{\mu' + (-w_0 \nu)}$, if you are making the substitution $\mu' = -w_0 \mu$. Consider, for example, the case of $SL_3$ with $\lambda = \mu = 0, \nu = \omega$; clearly, whenever $\lambda = \mu = 0$, you get the natural inclusion of $\mathbb{C}$ into $V_\nu \otimes V_\nu^*$, but there is no inclusion of $\mathbb{C}$ into $V_\nu \otimes V_\nu$.
Apr 12, 2014 at 18:29 history edited Jim Humphreys CC BY-SA 3.0
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Apr 12, 2014 at 17:46 history answered Jim Humphreys CC BY-SA 3.0