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Apr 5, 2010 at 20:24 comment added Jim Humphreys While the abstract Cartan is the correct gadget to introduce, the original question can be dealt with concretely if it's made clear up front what you mean by "induction". The chosen Borel subgroup itself and not just its characters must play a role in the process of induction. Any of its maximal tori, or the abstract one, will do here, but $B$ determines the positive roots and hence the dominant weights. (No assumption yet on characteristic, until you ask for irreducible representations.)
Apr 5, 2010 at 17:19 history edited Ben Webster CC BY-SA 2.5
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Feb 5, 2010 at 21:44 comment added Allen Knutson The abstract Cartan is discussed in [Chriss-Ginzburg].
Nov 11, 2009 at 17:27 vote accept Dinakar Muthiah
Nov 11, 2009 at 16:54 comment added Ben Webster None I know of. It's easy to prove; you just note that if you pick a Cartan subgroup, the action on $n$ is isomorphic (as a T-rep) to the associated graded, and the action of your chosen Cartan on the associated graded matches with the action of the abstract Cartan under the isomorphism induced by picking the Borel.
Nov 11, 2009 at 16:45 comment added Rob Harron Cool, thanks. Are there good references for this?
Nov 11, 2009 at 15:33 comment added Ben Webster I added answers to both of these above.
Nov 11, 2009 at 15:33 history edited Ben Webster CC BY-SA 2.5
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Nov 11, 2009 at 15:27 history edited Ben Webster CC BY-SA 2.5
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Nov 11, 2009 at 5:41 comment added Rob Harron So what are the roots? (You've given the simple roots)
Nov 11, 2009 at 5:25 comment added Dinakar Muthiah Ah, I see. How would you then get the simple coroots (to eventually get your fundamental weights)?
Nov 11, 2009 at 5:23 vote accept Dinakar Muthiah
Nov 11, 2009 at 5:23
Nov 11, 2009 at 5:15 history answered Ben Webster CC BY-SA 2.5