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Dec 14, 2009 at 15:38 vote accept Jean Delinez
Dec 12, 2009 at 5:44 comment added David Bar Moshe As given in an earlier comment, a contragredient representation is the representation on the dual vector space. Let pi(g) be a representation, then its contragredient is pi(g^-1)^t. In the Wikipedia article on the dual representation, this terminology is mentioned. For SU(n), the contragredient of the fundamental representation is also the antisymmetric (n-1) power representation. In the following article by A. Borel there is a nice explanation: hkumath.hku.hk/~imr/records0001/borel.pdf
Dec 11, 2009 at 19:39 comment added Jean Delinez What does "contragradient" in "contragredient fundamental n-1 dimensional representation" mean?
Dec 11, 2009 at 19:18 history edited David Bar Moshe CC BY-SA 2.5
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Dec 11, 2009 at 15:07 comment added David Bar Moshe I corrected the error in my answer concerning the action of the abelian factor in the denominator on the holomorphic tanget bundle, thanks to Greg Kuperberg's remark.
Dec 11, 2009 at 15:04 history edited David Bar Moshe CC BY-SA 2.5
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Dec 11, 2009 at 9:55 comment added David Bar Moshe The U(1) subgroup that I referred to is the U(1) subgroup of U(n) which intersects the SU(n-1) in the unit element only. According to Borel's theorem on Kaehlerian coset spaces of semisimple Lie groups (which include CPn), (A. Borel, Proc. Natl. Acad. Sci. USA 40, 1147 (1954)) The U(n-1) group in the denominator is a centralizer of a torus in U(n). The U(1) subgroup that I referred to is the torus itself, and its abelian and semisimple components commute.
Dec 11, 2009 at 7:18 comment added Greg Kuperberg If it's the defining representation on SU(n-1) or its dual, then it can't be trivial on the U(1) part. The U(1) part intersects SU(n-1) in a cyclic group that acts non-trivially. Surely you should use the defining representation on all of U(n-1).
Dec 11, 2009 at 7:11 history answered David Bar Moshe CC BY-SA 2.5