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The formal character of a weight module for a minuscule representation of highest weight $\lambda$ is $\mathrm{ch}V(\lambda )={\sum }_{\mu \in W\lambda }{e}^{\mu }$. The sum is over distinct points of the Weyl group orbit, so $\dim V(\lambda )=|W\lambda |$.

A dominant integral weight $\lambda$ is minuscule when its irreducible highest-weight representation is a minuscule representation. Equivalently, $⟨\lambda ,{\alpha }^{\vee }⟩\in \{0,1\}$ for every positive root $\alpha$. This convention includes the zero weight and its trivial Lie algebra representation.

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.

The triple bond in the original Dynkin diagram gives the Cartan integerswhere $\alpha$ is the long root and $\beta$ the short root. Their product is $4{\cos }^2\theta =3$, and the inner product of distinct simple roots is nonpositive. The ratio of the two integers gives the squared-length ratio. ThereforeIt is convenient to normalize $(\alpha ,\alpha )=6$, $(\beta ,\beta )=2$ and $(\alpha ,\beta )=-3$. Scaling the inner product does not affect the root system or the fundamental-weight relations.

For nonproportional roots $\gamma ,\delta$, the root-string theorem states that the root string is consecutive:The endpoints are maximal, and reflection in $\gamma$ reverses the string. The Cartan integer determines $p-q$, not in general the total $p+q+1$ by itself. This result follows by restricting the Adjoint representation to the sl2 subalgebra associated with a root. If the roots are distinct simple roots, $\delta -\gamma$ cannot be a root: its simple-root coefficients have opposite signs. Thus $p=0$ andHere length means number of roots; the number of intervals between them is one less. Distinctness matters. If $\gamma =\delta$, a reduced root system gives the set $\{\gamma ,-\gamma \}$, with a missing zero between them, so the consecutive-string theorem and the displayed simple-root formula do not apply.

In the G2 root system, the initial strings areFor $\delta =\alpha +3\beta$, $⟨\delta ,{\alpha }^{\vee }⟩=-1$. Since $\delta -\alpha =3\beta$ is not a root in a reduced root system, $p=0$ and $q=1$: this generates $2\alpha +3\beta$. The remaining strings explain why the construction stops. The $\alpha$-strings through $\beta$ and $\alpha +\beta$ are the same two-element string; the one through $\alpha +2\beta$ is a singleton since subtracting $\alpha$ gives $2\beta$, and its Cartan integer is zero; the one through $2\alpha +3\beta$ is the string $\{\alpha +3\beta ,2\alpha +3\beta \}$. The $\beta$-strings through $\alpha$, $\alpha +\beta$, $\alpha +2\beta$ and $\alpha +3\beta$ are the initial four-element string. Finally, $2\alpha +3\beta$ is orthogonal to $\beta$; its $\beta$-string is a singleton because $2\alpha +2\beta =2(\alpha +\beta )$ is not a root. Apply the same reasoning to negatives. Using the permitted completeness of this procedure givesThe short positive roots are $\beta ,\alpha +\beta ,\alpha +2\beta$, of squared length two; the other three are long, of squared length six. Each root space is one-dimensional and the Cartan subalgebra has dimension two, soHere the dimension refers to the Lie algebra, with one Cartan generator per rank, not just the number of roots.

Write a prospective weight as $w=a\alpha +b\beta$. The pairings with simple coroots areThe fundamental weights are dual to those coroots. Solving the two linear systems gives, in the long-root-first numbering of this paper,The representation with Dynkin labels $(0,1)$ has highest weight $\lambda ={\omega }_2$, a short root. Numbering the short root first, as some references do, would call this the $(1,0)$ representation instead; the representation itself is unchanged.

The weight set of a finite-dimensional irreducible highest-weight representation is invariant under the Weyl group and lies in the convex hull of the orbit of its highest weight. All weights also differ from the highest weight by an element of the root lattice. The orbit of $\lambda$ comprises the six short roots, so all six are weights. The lowering operators give the chainIn particular zero occurs: at weight $\beta$, its pairing with ${\beta }^{\vee }$ is two, so the lowering operator is nonzero by the finite-dimensional sl2 Lie algebra representation theory. The $\beta$-string through it is the usual three-weight string $\beta ,0,-\beta$; higher weight $2\beta$ would lie outside the highest-weight convex hull.

There can be no further weights. Every point of that convex hull has squared norm at most $\Vert \lambda {\Vert }^2=2$, while a root-lattice point hasThe integer solutions of the bound $\Vert w{\Vert }^2\le 2$ are precisely zero and the six short roots: $a=0$ gives $b=-1,0,1$; $a=1$ gives $b=1,2$; and $a=-1$ gives $b=-1,-2$. ThusThe final dimension uses the stipulated nondegeneracy of the weights. It also agrees with the Weyl dimension formula without that stipulation. The zero weight is not in the Weyl orbit of the nonzero weights, so this is not a minuscule representation.

In the diagram above, a coordinate label $(a,b)$ means $a\alpha +b\beta$. The left panel contains all twelve roots of a root system; the right panel contains the six short-root weights and the zero weight. The long-root-first convention is the same as in the calculations.

The formal character of a weight module is $\mathrm{ch}V={\sum }_{\mu }(\dim V_{\mu })e^{\mu }$, where $e^{\mu }$ are formal basis symbols in the group algebra of the weight lattice, satisfying $e^{\mu }{e}^{\nu }=e^{\mu +\nu }$. For a finite-dimensional irreducible highest-weight representation, the highest weight has weight multiplicity one. Another standard result is that weight multiplicities are invariant under the Weyl group: $\dim V_{w\mu }=\dim V_{\mu }$.

A minuscule representation has all its weights in a single Weyl group orbit. Since the highest weight $\lambda$ is present, that orbit is $W\lambda$, and the invariance just stated shows every point of it is present with multiplicity one. Consequently the formal character of a minuscule representation isThe sum is over distinct weights, rather than over all elements of $W$: summing $e^{w\lambda }$ over $w\in W$ would count each weight $|{\mathrm{Stab}}_W(\lambda )|$ times.

Use the coroot ${\alpha }^{\vee }$ in the pairing. The original PDF retains this mark, which is lost in the local TeX transcription. Fix any positive root $\alpha$, and put $n=⟨\lambda ,{\alpha }^{\vee }⟩$. For a dominant integral weight, $n$ is a nonnegative integer; positive coroots are nonnegative integral combinations of simple coroots.

Let $v_{\lambda }$ be a highest-weight vector, and restrict to the sl2 subalgebra associated with a root $\alpha$. It is killed by its raising operator $e_{\alpha }$ and has $h_{\alpha }$ eigenvalue $n$. If $n\ge 1$, its lowering operator produces a nonzero vector: otherwise $[e_{\alpha },f_{\alpha }]v_{\lambda }=n{v}_{\lambda }$ would vanish. Therefore $\lambda -\alpha$ is a weight.

Choose a Weyl group invariant positive-definite inner product on the real weight space. All weights of a minuscule representation have the same norm, since they belong to $W\lambda$. ButIf $n\ge 2$, this is negative, a contradiction. Thus for every positive root,When $n=1$, the lower weight is precisely the Weyl reflection $s_{\alpha }\lambda =\lambda -\alpha$, consistent with the orbit condition.

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.