Let $\mathfrak{g}$ be a semi-simple finite dimensional Lie algebra. Denote by $L(\lambda)$ an irreducible finite-dimensional $\mathfrak{g}$-module of highest weight $\lambda$. (I.e. $\lambda$ is integral dominant weight) Suppose that the multiplicity of $L(\nu)$ in the tensor product $L(\lambda)\otimes L(\mu)$ is greater than $0$.

Does this imply that either $\lambda \geq \nu$ or $\lambda \leq \nu$? Where the comparison is the standard partial ordering on the set of weights: $\lambda \geq \nu$ iff their difference $\lambda-\nu$ is equal to the sum of positive roots with nonnegative coefficients.

  • $\begingroup$ @Anton: It's hard to upvote the question (which as Sasha points out has a negative answer) with no context or motivation given. What kind of examples have you looked at? There are lots of other examples besides Sasha's, say with $\nu=0$: take the natural module $V$ for $\mathfrak{sl}_n$, with highest weight $\varpi_1$, so the trivial module occurs as a summand of $\mathrm{End}(V) \cong V^* \otimes V$. All you need here is a highest weight not in the root lattice, so 0 isn't a subweight. $\endgroup$ – Jim Humphreys Jun 17 '13 at 18:49

What you want is not true. For example, take $\mathfrak{sl}_3$ and $\lambda = \mu = \omega_1$, the first dominant weight. Then the multiplicity of $\nu = \omega_2$ is $1$. But $$ \nu - \lambda = \omega_2 - \omega_1 = \frac13 \alpha_2 - \frac13 \alpha_1, $$ so it is neither negative, nor positive.

On the other hand, it is known that $\nu$ is in the convex hull of the weights $\lambda + w\mu$, where $w$ runs over the Weyl group. On the other hand $\mu - w\mu \ge 0$ with respect to your ordering. It follows that $$ \nu \le \lambda + \mu. $$ Analogously one can check that $\nu \ge \lambda + w_0\mu$, where $w_0$ is the longest element in the Weyl group.


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