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Are non-equivalent irreps with dimension n of a lie algebra share the same weight set? E.g. in su(2), given a dimension of the irrep, the weight set is the same no matter what exactly the irrep is. Is it generally true for any lie algebra?

Shadumu
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$\mathfrak{su}(2)$ only has a single irrep in each dimension. Anyway, the answer is no: the (multi)set of weights of an irreducible representation of a semisimple (otherwise I don't know what you mean by "weights") Lie algebra completely determines it up to isomorphism.

Qiaochu Yuan
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  • Could you give me a example of a lie algebra with multiple weight sets, Thx – Shadumu Nov 11 '15 at 21:18
  • @user3229471: I don't know what you mean by "multiple weight sets." Can you elaborate? Do you mean multiple irreps of the same dimension? Almost every semisimple Lie algebra has these; try, for example, $\mathfrak{su}(2) \times \mathfrak{su}(2) \cong \mathfrak{so}(4)$. – Qiaochu Yuan Nov 11 '15 at 21:20
  • So I mean for different irreps of the same dimension, their weight sets are always different? – Shadumu Nov 11 '15 at 21:47
  • Yes, that's what I'm saying. (You need to say "multiset," though; the multiplicities matter. If you're really asking about the set of weights and not the multiset, maybe there are counterexamples.) – Qiaochu Yuan Nov 11 '15 at 21:51
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    Actually an irrep is completely determined by its highest weight... So if you are given a representation which you know to be irreducible, then just the highest weight (which is of multiplicity one) is enough. On the other hand, if you compare two general representations, then you need all weights with multiplicities. – Daniel Juteau Nov 12 '15 at 08:52