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Feb 14, 2013 at 19:33 comment added Amin It may be that you're right, I admit that this part in this book (actually some other as well) is written so loosely that I didn't try to double check everything. Having said this, to quote it, it's written "The representation of G in $V^\omega$ obtained in this way can be extended to $V$" (p. 155). Now for my question, it's modest I think : given a $U(\mathcal(G))$ -module, why can we extend it to a $G$-module ? Perhaps it's a trick that I'm missing.
Feb 14, 2013 at 19:00 history answered Faisal CC BY-SA 3.0