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Let $\mathfrak{g}$ be a complex finite-dimensional semisimple Lie algebra with a fixed Cartan subalgebra $\mathfrak{h}$. Assume that $\lambda \in \mathfrak{h}^*$ such that $\lambda$ is a $\mathbb{Z}$-linear combination of roots of $\mathfrak{g}$.

Question: Is it true that $\lambda$ is a weight of a finite-dimensional module over $\mathfrak{g}$? Thanks!

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Yes. If $\alpha_1,\alpha_2,\ldots,\alpha_n$ are the simple roots and $\lambda=\sum_{i=1}^n m_i\alpha_i$, then $\lambda$ is a weight of the $m$-fold symmetric power $S^mL$ of the adjoint representation $L$, where $m=\sum_{i=1}^n|m_i|$.

This is because $\pm\alpha_i$ are weights of $L$, and sums of any $m$ weights of a module $V$ are weights in the symmetric power $S^mV$. The $m$-fold tensor product $V^{\otimes m}$ would do equally well.

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  • $\begingroup$ I see, thanks a lot!! Also, I was wondering if this means for a given weight $\lambda$, then $\lambda$ appears in a finite-dimensional module if and only if $\lambda$ is integral? (that is, $<\lambda, \alpha^{\vee}>$ are integers for all roots $\alpha$) $\endgroup$
    – TLSu
    Feb 26, 2018 at 19:58
  • $\begingroup$ Yes. In the orbit (under the Weyl group) of $\lambda$ there is a dominant weight $\lambda^+$. $\lambda$ then appears as a weight in the f.d. module of highest weight $\lambda^+$. $\endgroup$ Feb 26, 2018 at 20:29

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