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If V is an irreducible representation of a semi simple lie algebra having highest weight λ then what will be the highest weight of the corresponding irreducible representation V∗ (Dual of V)?

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This is a standard textbook question, which has come up previously on Math Overflow, probably more than once (see for instance my answer to question 81858). It's elementary enough to appear as an exercise when finite dimensional irreducible representations of a semisimple Lie algebra over $\mathbb{C}$ are characterized by their highest weights: e.g., Exercise 21.6 in my 1972 Lie algebra book. – Jim Humphreys Nov 1 '12 at 11:13
up vote 6 down vote accepted

To expand my short comment, the result itself (formulated by Sasha) has been around a long time and depends only on the definitions involved. Textbooks dealing with the highest weight classification of finite dimensional modules over $\mathbb{C}$ (which you assume but haven't specified) always say something about dual modules of simple modules but sometimes informally in exercises and sometimes more formally. For instance, the result appears in the book by Onishchik-Vinberg (where the tough exercise is to find it) but also much more formally in the treatise by Bourbaki: see Chapter VIII, $\S7.5$. (This statement about highest weight of the dual simple module actually carries over easily to prime characteristic as well, though the dimensions of such modules tend to shrink.)

The statement is fairly easy to visualize in terms of weight diagrams and Weyl group symmetry, since the modules are finite dimensional. An easy step (not requiring machinery and done quite generally in Bourbaki's Chapter I) is to show that $V^*$ is simple when $V$ is: this just requires duality and finite dimensionality along with the definition of the dual action, since submodules of one correspond to quotients of the other.

In the highest weight situation, the weight diagram of $V$ is symmetric under the Weyl group as is that of $V^*$. The weights (with multiplicity) of the latter are the negatives of the original weights, using just the definition of the dual action, so the "lowest" weight is $-\lambda$. To get the highest weight use $w_0$ to pass from negative to positive root action, so the answer is $-w_0 \lambda$.

P.S. Visualization aside, the algebraic step needed is to pick out a maximal vector $f^+ \in V^*$ having the indicated weight: if $v^+ \in V$ is a maximal vector of weight $\lambda$, just take $f^+$ to be the characteristic function of $w_0 v^+$ (taking value 1 there, 0 on other weight vectors in a basis of $V$). Here $w_0 v^+$ is shorthand for an arbitrary nonzero vector in the 1-dimensional weight space for $w_0 \lambda$.

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@Jim Humphreys thank you sir for your nice explanation. – Rekha Biswal Nov 2 '12 at 14:51

The highest weight is $-w_0(\lambda)$, where $w_0$ is the longest element of the Weyl group.

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@Sasha could you please explain how -w(lamda) will be the highest weight? – Rekha Biswal Nov 1 '12 at 12:42
The reason is that $w_0(\lambda)$ is an antidominant weight of $V$, hence the lowest weight, and the highest weight of $V^*$ is minus the lowest weight of $V$. – Sasha Nov 1 '12 at 13:59

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