why does a fundamental representation of SL(n) have only one dominant weight? I am asking this because /\i(v) is a fundamental representation , but that can happen only if theres only one dominant weight and all others are conjugates under the weyl group action to it.
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$\begingroup$ Every irreducible representation of SL(n) has only one dominant weight. This should be proven in any book on representations of Lie groups, as long as it is not too abstract and is not Fulton-Harris (which probably has it, too, but is unreadable). $\endgroup$– darij grinbergCommented May 14, 2011 at 8:02
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3$\begingroup$ darij, I think you may be confusing "dominant" with "highest". $\endgroup$– S. Carnahan ♦Commented May 14, 2011 at 8:15
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$\begingroup$ Oh, thanks Scott. Somehow I mistook "dominant" for "highest wrt dominance order". $\endgroup$– darij grinbergCommented May 14, 2011 at 10:21
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2 Answers
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The question is not well-formulated, as Bugs Bunny observes. In the case of the special linear group or Lie algebra (say over the field $\mathbb{C}$), it happens to be true that the fundamental dominant weights are precisely the minuscule ones (defined by the condition that the weight is nonzero while all pairings with arbitrary positive coroots are 0 or 1), equivalent to being minimal (and nonzero) in the usual partial ordering of dominant weights. In particular, for an irreducible representation having such a highest weight all its weights are conjugate under the Weyl group, with only the highest weight therefore being dominant.
For irreducible root systems of other than type $A$, the minuscule weights are also fundamental but there are very few of them. For example, the 0 weight always occurs in irreducible representations for type $G_2$ and rules out existence of minuscule weights. Bourbaki is the most standard reference (Groupes et algebres de Lie, VIII, 7.3), but in textbooks the details are usually left as an exercise.
In type A, it's a standard exercise to compute the highest weights occurring in the various exterior powers of the standard representation (whose highest weight is the first fundamental weight) and see directly that these are minuscule and in particular fundamental.
answered May 14, 2011 at 15:30
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$\begingroup$ I just want to point out a very pretty paper arxiv.org/pdf/0908.1091v1 -- I'm not sure it answers the question, but it seems to give a nice way of thinking of minuscule and dominant weights... $\endgroup$– JGordonCommented May 14, 2011 at 17:23
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What do you mean by "why"? Do you want a complete proof - it is rather straightforward and an easy exercise.
If you want a moral reason then this is because it is rather small or miniscule, to be precise, i.e. its weights form a Weyl group orbit.
answered May 14, 2011 at 10:38
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2$\begingroup$ The (peculiar) spelling is actually minuscule ("minus", not "mini"). Anyway, the Wikipedia entry chooses an odd reference and gives a nonstandard definition allowing the 0 weight to be called minuscule. (My preferred term for that variant is "minimal".) This is usually forbidden. $\endgroup$ Commented May 14, 2011 at 15:35