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If a dominant integral weight $\lambda$ lies in the root lattice, its finite-dimensional irreducible highest-weight representation contains zero as a weight. First $\lambda$ is a nonnegative integral combination of simple roots: a negative part would have negative inner product with $\lambda$, contradicting dominance. For any nonzero weight $\mu =\sum m_i{\alpha }_i$ with $m_i\ge 0$, the identity $(\mu ,\mu )=\sum m_i(\mu ,{\alpha }_i)>0$ supplies an index with $m_i>0$ and $⟨\mu ,{\alpha }_i^{\vee }⟩>0$. By Injectivity of sl2 lowering above weight zero, applying the corresponding lowering operator produces a nonzero vector of weight $\mu -{\alpha }_i$. Repetition reaches zero.

The dominance order isSuppose $\lambda \ne \mu$. Put $\delta =\lambda -\mu$. Since $\mu$ is dominant,Writing $\delta ={\sum }_i{n}_i{\alpha }_i$, some index with $n_i>0$ therefore satisfies $(\lambda ,{\alpha }_i)>0$, equivalently $⟨\lambda ,{\alpha }_i^{\vee }⟩>0$. The ${\mathfrak{sl}}_2$ lowering operatoris injective by the Injectivity of sl2 lowering above weight zero. HenceThe new weight still dominates $\mu$ in the partial order. Iterating until reaching $\mu$ givesThis is the weight multiplicity decreases away from a dominant weight property.