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!


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.

  • $\begingroup$ 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 '18 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 $\lambda^+$. $\endgroup$ – Jyrki Lahtonen Feb 26 '18 at 20:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.