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If $V$ is an irreducible representation of a semi simple lie algebra having highest weight $\lambda$ then what will be the highest weight of the corresponding irreducible representation $V^*$ (Dual of $V$)?

Julian Kuelshammer
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Rekha Biswal
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    It is $\lambda$ itself, see Humphrey's book "Representations of Semisimple Lie algebras in the BGG category O" section 3.2 – Aaron Nov 01 '12 at 13:24
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    Actually, it is not always $\lambda$ itself. See the answer to this question here: http://mathoverflow.net/questions/111127/finding-highest-weight-of-dual-of-a-representation-of-a-semisimple-lie-algebra – Alistair Savage May 10 '13 at 21:26

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To expand Alistair Savage's comment into an answer: As discussed in the answers to https://mathoverflow.net/q/111127/27465, as well as Irreducible Dual Representation, if $\lambda$ is the highest weight of an irreducible (finite-dimensional) representation $V$ of a (complex) semisimple Lie algebra, then the dual representation $V^*$ is irreducible of highest weight $-w_0 (\lambda)$, where $w_0$ is the longest element of the Weyl group.

Note that if the Lie algebra is simple and has root system of type $A_1, B_n, C_n, D_{2n}, E_7, E_8, F_4$ or $G_2$, then $-w_0 = id$, so then the comment by Aaron is true and the dual will have the same highest weight and hence be isomorphic to the original representation. For all other types, this is not the case in general, although it still is for some weights $\lambda$.

Torsten Schoeneberg
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