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By dominant root-lattice highest weights have zero weight, a minuscule weight in the root lattice has zero in its Weyl group orbit. Since every element of the Weyl group is invertible, the highest weight must itself be zero. Thus equality of the weight lattice and root lattice rules out every nontrivial minuscule representation without requiring a classification of root systems.

We first prove the useful lemma that dominant root-lattice highest weights have zero weight. Write $Q=\mathbb{Z}\Phi$ for the root lattice, and let a dominant integral weight $\lambda \in Q$ be the highest weight of a finite-dimensional irreducible representation.

First, $\lambda$ is a nonnegative integral combination of simple roots. Indeed, write $\lambda ={\lambda }_{+}-{\lambda }_{-}$, separating its positive and negative coefficients in the root basis. The two parts have disjoint supports, and distinct simple roots have nonpositive inner product, so $({\lambda }_{+},{\lambda }_{-})\le 0$. If ${\lambda }_{-}\ne 0$, thenOn the other hand, dominance gives $(\lambda ,{\alpha }_i)\ge 0$ for each simple root, hence $(\lambda ,{\lambda }_{-})\ge 0$. This contradiction proves ${\lambda }_{-}=0$.

Now suppose a nonzero weight $\mu ={\sum }_i{m}_i{\alpha }_i$ has all $m_i\in {\mathbb{Z}}_{\ge 0}$. The identityshows that some $i$ satisfies $m_i>0$ and $⟨\mu ,{\alpha }_i^{\vee }⟩>0$. We use the standard sl2 Lie algebra fact that its lowering operator is injective on any positive $h$ eigenspace in a finite-dimensional representation. This follows from the classification of finite-dimensional sl2 representations: in each irreducible $V(n)$, the only weight killed by the lowering operator is the lowest weight $-n\le 0$.

Consequently a nonzero vector of weight $\mu$ lowers to a nonzero vector of weight $\mu -{\alpha }_i$. Its simple-root coefficients remain nonnegative and their sum decreases by one. Starting at $\lambda$, repeated lowering must therefore reach the zero weight. Notice that intermediate weights need not remain dominant; positivity of the chosen coroot pairing is enough at each step.

For a minuscule representation, every weight belongs to $W\lambda$, so the zero weight just obtained lies in this orbit. Every Weyl group element is invertible, and $w\lambda =0$ forces $\lambda =0$. This proves that minuscule weights in the root lattice are zero. Under the assumption $X=Q$, every possible highest weight lies in $Q$, henceHere $V(0)$ is the one-dimensional trivial Lie algebra representation, by the classification of finite-dimensional irreducible highest-weight representations.