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When we consider the weights of an irrep of a simple Lie algebra, are they always in a single orbit under the Weyl group of the Lie algebra, or do they form a set of disjoint orbits?

If they form multiple orbits, is there an efficient way of figuring out the orbits without actually going through the humongous amount of job of acting each Weyl reflection onto each weight?

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  • $\begingroup$ They form multiple orbits. Each orbit has a unique dominant weight, enumerating the orbits is the same as enumerating the dominant weights. $\endgroup$ – Nate Nov 6 '14 at 16:58
  • $\begingroup$ Thank you! That's a succint answer about what I wanted to know. $\endgroup$ – Chan Y. Park Nov 6 '14 at 17:05
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The weights are not always in a single orbit---this happens already the first time it can, for the irreducible $3$-dimensional representation of $\mathfrak{sl}_2(\mathbb{C})$ (which also happens to be the adjoint representation). The weights of this representation are $-\alpha,0$, and $\alpha$, where $\alpha$ is the unique positive root.

For finite dimensional irreducible modules, the set of weights is Weyl group stable, but this fails for arbitrary irreducibles (so, given what you write I assume you are interested in finite dimensional irreducibles). If $L(\lambda)$ is the finite dimensional irreducible with highest weight $\lambda$, then it follows from the Weyl character formula that the dominant weights appearing in the weight space decomposition of $L(\lambda)$ are precisely those dominant weights that are at most $\lambda$ in dominance order. As @Nate wrote, these give you orbit representatives.

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  • $\begingroup$ Thank you for your detailed answer! $\endgroup$ – Chan Y. Park Nov 6 '14 at 17:05

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