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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.

For the irreducible highest-weight representation of the special linear Lie algebra ${\mathfrak{sl}}_3$ with Dynkin labels $(a,b)$, let $(p,q)$ be a dominant integral weight. Put $r=(2a+b-2p-q)/3$ and $s=(a+2b-p-2q)/3$. If these are not integers, the weight multiplicity is zero; otherwise it is $\max \{0,1+\min (a,b,r,s)\}$.

To derive the formula, use the sl3 interlacing character formula with top row $(a+b,b,0)$. The target diagonal exponents are $(s+p+q,s+q,s)$, and an interlacing middle row $(P,Q)$ has $P+Q=2s+p+2q$, with bottom entry $s+p+q$. ThusEvery integral $P$ in this interval gives exactly one Gelfand–Tsetlin basis vector. The upper bound minus the lower bound is the minimum of six differences: $a,s,r,r+p+q,b,s+q$. Since $p,q\ge 0$, this minimum reduces to $\min (a,b,r,s)$. Counting the interval proves the formula, including absent weights of a representation. This converts a two-dimensional weight diagram calculation into four elementary boundary distances.

For a complex semisimple Lie algebra and a choice of positive roots, every finite-dimensional Irreducible Lie algebra representation has a unique highest weight, which is a dominant integral weight. Conversely each dominant integral weight labels exactly one such irreducible module up to isomorphism. Its highest-weight vector spans the highest-weight space and generates the module. This classification and Weyl complete reducibility theorem reduce decomposing any finite-dimensional module to its irreducible highest-weight summands.

If $\lambda$ is a dominant integral weight, the maximal proper submodule of $M(\lambda )$ is generated by the simple-root singular vectorsThe generator associated with ${\alpha }_i$ has weight $s_i(\lambda +\rho )-\rho$.

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.

Work over $\mathbb{C}$ with a finite-dimensional semisimple Lie algebra $\mathfrak{g}$. Choose a Cartan subalgebra $\mathfrak{h}$, its root system $R\subset {\mathfrak{h}}^{\ast }$, a positive system $R^{+}$ and its simple roots. Let $W$ be the Weyl group, generated by the corresponding root reflections, and put $\rho =\frac{1}{2}{\sum }_{\alpha \in R^{+}}\alpha$, the Weyl vector. The finite-dimensional irreducible highest-weight representation $L(\lambda )$ is indexed by a dominant integral weight $\lambda$.

If $P$ is the weight lattice, its group ring has formal symbols $e^{\mu }$ with multiplication $e^{\mu }{e}^{\nu }=e^{\mu +\nu }$. The formal character iswhere $L(\lambda {)}_{\mu }=\{v:Hv=\mu (H)v\text{ for all }H\in \mathfrak{h}\}$ is a weight space. For a weight $\nu$, define the Weyl alternant $A_{\nu }={\sum }_{w\in W}\det (w)e^{w\nu }$; the sign $\det (w)=(-1{)}^{\ell (w)}$ is the determinant of $w$ on the real root span, equivalently the parity of its expression in simple reflections.

The Weyl character formula isThe second denominator is the Weyl denominator formula. The quotient can initially be taken in the fraction field of the group ring; the formula says it is the finite character on the left, so its apparent singularities are removable. The shifts by $\rho$ make $\lambda +\rho$ strictly dominant.

Let $\mathfrak{g}$ be a finite-dimensional complex semisimple Lie algebra, choose a Cartan subalgebra $\mathfrak{t}$ and a set $R^{+}$ of positive roots, and write $W$ for the Weyl group. Let $\lambda$ be a dominant integral weight and $L(\lambda )$ the finite-dimensional irreducible highest-weight representation with that highest weight. If $L(\lambda {)}_{\mu }$ denotes its weight space of weight $\mu$, its formal character is $\mathrm{ch}L(\lambda )={\sum }_{\mu }\dim L(\lambda {)}_{\mu }{e}^{\mu }$. The symbols $e^{\mu }$ belong to the group algebra of the weight lattice and multiply by $e^{\mu }{e}^{\nu }=e^{\mu +\nu }$; they are formal symbols, not matrix exponentials.

Put $\rho =\frac{1}{2}{\sum }_{\alpha \in R^{+}}\alpha$, the half-sum of positive roots, and let $\ell (w)$ be the Coxeter length of $w$, the least number of simple reflections expressing it. Then $(-1{)}^{\ell (w)}=\det (w)$ in the real reflection representation. The Weyl character formula statesBy the Weyl denominator formula, the denominator is $e^{\rho }{\prod }_{\alpha \in R^{+}}(1-e^{-\alpha })$, giving the equivalent expressionThe quotient can be interpreted in the fraction field of the formal group algebra; the theorem says it simplifies to the finite formal character above. In particular the apparent denominator does not mean that the representation has infinitely many weights.

The Weyl group acts on the weight lattice by its root reflections and extends to ring automorphisms of the group ring of a weight lattice:Weights and their multiplicities in a finite-dimensional Lie algebra representation of a semisimple Lie algebra are permuted by this action, so its formal character is Weyl-invariant.

Choose a set ${\Phi }^{+}$ of positive roots and putHere ${\omega }_i$ are the fundamental weights, and $\det (w)=(-1{)}^{\ell (w)}$, where the Coxeter length $\ell (w)$ counts positive roots carried to negative ones. For a dominant integral weight $\lambda$, let $L(\lambda )$ denote the finite-dimensional Irreducible Lie algebra representation of highest weight $\lambda$. The Weyl character formula isThe second denominator expression is the Weyl denominator formula. Both alternants change sign under each root reflection; their quotient is invariant. Although written as a fraction, the theorem asserts that it is a finite element of $\mathbb{Z}[{\Lambda }_W]$ with the nonnegative coefficients given by weight multiplicities. The highest-weight classification specifies which $\lambda$ label the irreducibles, while the root system and choice of positive roots determine every other term.

For $G=U(n)$ take the diagonal maximal torusThe spectral theorem for unitary matrices says that every element is conjugate to an element of $T$. A diagonal element with pairwise distinct eigenvalues has centralizer exactly $T$. Its conjugates therefore determine its eigenvalues up to permutation, and the normalizer quotient is $W=N_G(T)/T\cong S_n$. The character lattice of a torus $T$ is ${\mathbb{Z}}^n$, with $e^{\lambda }(z)=z_1^{{\lambda }_1}\cdots z_n^{{\lambda }_n}$. Conjugation on the matrix unit $E_{ij}$ has weight $e_i-e_j$, so these are the roots. Choose $e_i-e_j$ positive for $i<j$; thenDominant integral weights are precisely the integer tuples ${\lambda }_1\ge \cdots \ge {\lambda }_n$. Negative entries are allowed: they correspond to determinant twists, so polynomial representations alone do not exhaust the irreducible representations of $U(n)$.

Normalize Haar measure on $G$ and $T$ to have mass one, and putThe Weyl integration formula for U(n) for a continuous class function isFor a general continuous function, replace $f(t)$ in the torus integral by ${\int }_{G/T}f(gt{g}^{-1})d(gT)$, using the normalized invariant quotient measure.

To prove the formula, consider the conjugation map $q:G/T\times T\to G$, $q(gT,t)=gt{g}^{-1}$. On regular torus elements its fibres have exactly $n!$ points: the possible orderings of the distinct eigenvalues. Equip the group with an invariant inner product on its Lie algebra and split $\mathfrak{u}(n)=\mathfrak{t}\oplus {\mathfrak{t}}^{\perp }$. After translating the derivative back to the identity, its torus directions contribute the identity, while its transverse directions contribute ${\mathrm{Ad}}_{t^{-1}}-I$. Each pair $i<j$ supplies a real two-dimensional off-diagonal plane. On that plane ${\mathrm{Ad}}_t$ is multiplication by $z_i/z_j$, regarded as a plane rotation, so the absolute real determinant of the transverse derivative isThe change-of-variables theorem, divided by the covering multiplicity, now gives the desired weighted integral up to the constant relating the normalized invariant measures. The nonregular locus has measure zero: repeated eigenvalues are the zero-set of the nonzero discriminant, and the torus coincidence sets likewise have measure zero. Thus its omission causes no change to the integral. To determine the normalization, expand $\Delta$ as a sum of $n!$ distinct torus characters. Fourier orthogonality gives ${\int }_T|\Delta {|}^{2}dt=n!$. Applying the formula to $f=1$ fixes the coefficient to $1/n!$. This proves both the class-function and conjugacy-averaged forms.

We can now derive the character formula from integration, rather than merely quote it. Use the basic highest-weight theorem: a finite-dimensional irreducible representation has a unique highest weight $\lambda$ with a one-dimensional highest-weight space; all its other weights are $\lambda$ minus a nonnegative sum of positive roots. Its character is Weyl invariant. Also, character orthogonality for compact groups gives ${\int }_G|{\chi }_{\lambda }{|}^2=1$. This orthogonality follows by averaging the representation on $\mathrm{End}(V_{\lambda })$: the integral projects onto intertwining operators, whose dimension is one by the Schur lemma.

Let $\delta =(n-1,n-2,\dots ,0)$ and, for a strictly decreasing integer tuple $\eta$, put $A_{\eta }(z)=\det (z_i^{{\eta }_j})$. The finite Laurent polynomial $\Delta {\chi }_{\lambda }$ is alternating. Every alternating Laurent polynomial is a linear combination of the $A_{\eta }$: collect its monomials by permutation orbits, and observe that an orbit with repeated exponents has zero coefficient by antisymmetry. Different $A_{\eta }$ are orthogonal on $T$ and each has squared norm $n!$.

The coefficient of $z^{\lambda +\delta }$ in $\Delta {\chi }_{\lambda }$ is exactly one. Indeed, a contribution from the term $z^{w\delta }$ would require a weight $\lambda +\delta -w\delta$ in $V_{\lambda }$. For $w\ne 1$, the vector $\delta -w\delta$ is a nonzero nonnegative combination of the positive simple roots, so that weight lies above $\lambda$ and is impossible. The identity permutation supplies the highest-weight coefficient one. ConsequentlyApply the integration formula to $|{\chi }_{\lambda }{|}^2$. Torus orthogonality givesAll the other coefficients vanish, provingThe quotient extends across repeated eigenvalues because its product with the denominator is the already defined continuous character. Equivalently, alternant divisibility makes it a Laurent polynomial invariant under permutations; multiply by a determinant power if negative exponents must first be removed.

Finally, $\delta -\rho =\frac{n-1}{2}(1,\dots ,1)$ is fixed by the Weyl group, so its common factor cancels in the quotient. Since $A_{\rho }=e^{\rho }{\prod }_{\alpha >0}(1-e^{-\alpha })$, the determinant expression is exactlyUsing $\delta$ in the derivation avoids treating a possibly half-integral $\rho$ as an actual character of the torus. At the identity, taking the leading Vandermonde term of the two determinants gives the accompanying dimension formula

Let $v$ be a highest-weight vector of weight $\lambda$. For a simple root ${\alpha }_i$, choose its sl2 subalgebra associated with a root, with operators $E_i,F_i,H_i$ satisfying $[E_i,F_i]=H_i$, $[H_i,F_i]=-2F_i$, and $H_i={\alpha }_i^{\vee }$. We have $E_{i}v=0$ and $H_{i}v=\lambda (H_i)v$.

Since $F_i$ decreases the weight by ${\alpha }_i$ and there are only finitely many weights of a representation, let $n\ge 0$ be maximal with $F_i^{n}v\ne 0$. Induction using the two bracket relations gives the sl2 highest-weight lowering formulaApplying this at $k=n+1$, the left side vanishes, and the nonzero vector $F_i^{n}v$ on the right forces $\lambda (H_i)=n$. This holds for every simple root, soThus the highest weight is a dominant integral weight, in particular an element of the Closed dominant Weyl chamber.

Let $\mathfrak{g}$ be a complex semisimple Lie algebra, choose a Cartan subalgebra and positive root system $R^{+}$, and put $\rho =\frac{1}{2}{\sum }_{\alpha \in R^{+}}\alpha$. For a dominant integral weight $\lambda$, let $L(\lambda )$ be the finite-dimensional highest-weight representation with highest weight $\lambda$. The Weyl dimension formula isHere ${\alpha }^{\vee }=2\alpha /(\alpha ,\alpha )$ is the coroot, the pairing is between a weight and a coroot, and $\rho$ is the Weyl vector. The inner product can be any Weyl group invariant normalization on each simple factor: its scaling cancels from the ratios. The dominance condition means that $⟨\lambda ,{\alpha }_i^{\vee }⟩$ is a nonnegative integer for each simple root.

Schur's lemma says that a nonzero intertwiner between irreducible representations is an isomorphism; over $\mathbb{C}$, every endomorphism of a finite-dimensional irreducible representation is scalar. To prove the Schur lemma, let $T:V\to W$ be a intertwining operator between irreducible representations. Its kernel and image of a linear map are invariant. If $T\ne 0$, irreducibility forces $\ker T=0$ and $\mathrm{im}T=W$. This proves the first assertion over any field. In particular the endomorphisms of an irreducible representation form a division ring.

When $V$ is finite-dimensional over an algebraically closed field, an endomorphism $T$ has an eigenvalue $\lambda$. The endomorphism $T-\lambda I$ has nonzero kernel. By the first assertion it must be zero, so $T=\lambda I$. Consequently, for complex finite-dimensional irreducible representations, the space of intertwining operators has dimension zero for nonisomorphic representations and dimension one for isomorphic representations.

Every finite-dimensional representation of a complex semisimple Lie algebra is completely reducible. We prove the Weyl complete reducibility theorem using the allowed Casimir operator facts, without assuming a splitting in advance. For dual bases $x_i,x^i$ with respect to the Killing form, the Casimir elementis central in the universal enveloping algebra. Hence its Casimir operator $C_M={\sum }_i{\rho }_M(x_i){\rho }_M(x^i)$ commutes with the action on every Lie algebra representation $M$ and is compatible with subrepresentations, quotient representations, and intertwining operators. On the trivial Lie algebra representation it acts by zero. On every nontrivial finite-dimensional Irreducible Lie algebra representation it acts by a nonzero scalar.

For clarity, the last fact can be expressed by the Casimir eigenvalue formula: on an irreducible with dominant integral weight $\lambda$, the scalar is $(\lambda ,\lambda +2\rho )$, with the inner product induced by the Killing form and $\rho$ the half-sum of positive roots. On the real span of the weights this inner product is positive definite, and the scalar is positive for $\lambda \ne 0$. For a semisimple Lie algebra with several simple factors the scalars add, so a nontrivial representation still gives a nonzero scalar. These are properties of the Casimir operator being used here.

First establish Casimir splitting of a trivial quotient. Supposeis a short exact sequence of finite-dimensional Lie algebra representations, with trivial quotient. The generalized eigenspaces of $C_E$ are invariant, soEach $E_c$ with $c\ne 0$ maps to zero under $q$: applying a sufficiently large power of $C_E-cI$ and using $C_{\mathbb{C}}=0$ gives $(-c{)}^{r}q(e)=0$. Thus $q(E_0)=\mathbb{C}$.

Take a composition series of a module of $E_0$. The Casimir operator is nilpotent on $E_0$, so its scalar on every irreducible composition factor is zero. The stated Casimir operator property makes every such factor trivial. In a basis adapted to the composition series of a module, the image of $\mathfrak{g}$ on $E_0$ therefore consists of strictly upper triangular matrices, so that image is solvable. The allowed fact that a semisimple Lie algebra acts trivially on every one-dimensional representation implies that $\mathfrak{g}$ is a perfect Lie algebra: otherwise a nonzero linear functional on $\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]$ would define a nontrivial one-dimensional Lie algebra representation. Thus $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$, and its image is consequently a perfect Lie algebra too. A perfect Lie algebra that is also a solvable Lie algebra is zero, since its derived series of a Lie algebra is constant until it vanishes. Hence $\mathfrak{g}$ acts trivially on $E_0$. Choose $e\in E_0$ with $q(e)=1$. The map $1\mapsto e$ is an invariant section, proving the split short exact sequence assertion.

Now let $W\subseteq V$ be any nonzero subrepresentation. On the Hom representation $\mathrm{Hom}(V,W)$ the action isConsider the invariant subspaceRestriction produces a short exact sequenceSurjectivity follows by extending $I_W$ to a linear map on $V$. The quotient is trivial, because $I_W$ commutes with the action on $W$. The preceding Casimir splitting of a trivial quotient yields an invariant $p:V\to W$ with $p{|}_W=I_W$. Thusand $\ker p$ is invariant. The zero subrepresentation also has a complement. Repeatedly splitting off an irreducible subrepresentation now gives a direct sum of irreducibles, proving the Weyl complete reducibility theorem.

A derivation of a Lie algebra is a linear map $D:\mathfrak{g}\to \mathfrak{g}$ satisfying the Leibniz ruleThe space $\mathrm{Der}(\mathfrak{g})$ is a vector subspace of $\mathrm{End}(\mathfrak{g})$. Equip it with the commutator $[D,E]=DE-ED$. Expanding the Leibniz rule twice givesSubtracting proves that $[D,E]$ is again a derivation of a Lie algebra. Antisymmetry and the Jacobi identity hold for the commutator in every associative endomorphism algebra, so this defines the derivation Lie algebra.

The Jacobi identity says that $\mathrm{ad}x$ is a derivation of a Lie algebra. Moreover, for $D\in \mathrm{Der}(\mathfrak{g})$,soThis makes $\mathrm{ad}\mathfrak{g}$ an ideal of a Lie algebra in $\mathrm{Der}(\mathfrak{g})$.

Finally suppose $\mathfrak{g}$ is semisimple. Let $\mathfrak{g}$ act on $\mathrm{Der}(\mathfrak{g})$ by $x\cdot D=[\mathrm{ad}x,D]$. The Weyl complete reducibility theorem supplies an invariant complement $M$ to $\mathrm{ad}\mathfrak{g}$. For $D\in M$, invariance gives $[\mathrm{ad}x,D]\in M$, while the ideal identity above gives $[\mathrm{ad}x,D]\in \mathrm{ad}\mathfrak{g}$. Their intersection is zero, so $\mathrm{ad}(Dx)=0$ for every $x$. The center of a Lie algebra of a semisimple Lie algebra is zero, hence $Dx=0$ for every $x$, and $M=0$. ThereforeEvery derivation of a Lie algebra is inner, and the element giving it is unique because the center of a Lie algebra vanishes.

Take an integer $k\ge 0$, so $k\rho$ is a dominant integral weight. Apply the dilation $e^{\mu }\mapsto e^{(k+1)\mu }$ to the Weyl denominator formula:This is exactly the numerator of the Weyl character formula for $L(k\rho )$. Divide by its denominator and use a finite geometric series in each root direction:This character of a Weyl-vector multiple also gives $\dim L(k\rho )=(k+1{)}^{|R^{+}|}$ by evaluating every formal exponential at one. This evaluation is safe in the final polynomial expression, without dividing by a zero denominator.

Choose a positive root system $R^{+}$ for a complex semisimple Lie algebra, with Weyl group $W$ and Weyl vector $\rho =\frac{1}{2}{\sum }_{\alpha \in R^{+}}\alpha$. For a dominant integral weight $\lambda$, let $L(\lambda )$ denote its finite-dimensional irreducible representation, and use formal symbols with $e^{\mu }{e}^{\nu }=e^{\mu +\nu }$. Its formal character of a weight module is $\mathrm{ch}L(\lambda )={\sum }_{\mu }\dim L(\lambda {)}_{\mu }{e}^{\mu }$. The Weyl character formula saysHere $\det (w)=(-1{)}^{\ell (w)}$, the sign of the product of simple root reflections. The Weyl denominator formula isThe quotient is initially interpreted in the fraction field of the group algebra of the weight lattice; the theorem makes it a finite character.

Use the coroot normalization ${\alpha }^{\vee }=2\alpha /(\alpha ,\alpha )$, with the pairing understood via the root system and its invariant inner product. A dominant integral weight, called a dominant weight here, is a functional $\lambda \in {\mathfrak{t}}^{\ast }$ such thatEquivalently, $\lambda ={\sum }_i{n}_i{\omega }_i$ with $n_i\in {\mathbb{Z}}_{\ge 0}$, where the fundamental weights satisfy $⟨{\omega }_i,{\alpha }_j^{\vee }⟩={\delta }_{ij}$. If dominance and integrality are defined separately, dominance means the inequalities and membership in the weight lattice means integrality.

A highest-weight vector is a nonzero weight vector $v\in V_{\lambda }$, so $h\cdot v=\lambda (h)v$ for all $h\in \mathfrak{t}$, annihilated by every positive root space:The last condition is equivalently annihilation by the raising operators for the simple roots.

For a dominant integral weight $\lambda$, let $V(\lambda )$ be the finite-dimensional irreducible highest-weight representation with highest weight $\lambda$. Set $\rho =\frac{1}{2}{\sum }_{\alpha \in {\Phi }^{+}}\alpha$, the half-sum of positive roots. With ${\alpha }^{\vee }$ the coroot of $\alpha$, the Weyl dimension formula isHere ${\Phi }^{+}$ is the positive system of a root system specified by the chosen root basis. The fundamental weights are the basis dual to the simple coroots: $⟨{\omega }_i,{\alpha }_j^{\vee }⟩={\delta }_{ij}$. Thus $\lambda ={\sum }_i{a}_i{\omega }_i$ with nonnegative integers $a_i$. The denominators are positive; equivalently one can use $(\lambda +\rho ,\alpha )/(\rho ,\alpha )$ in each factor, since the coroot normalization cancels.

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.

Choose a Borel subalgebra $\mathfrak{b}=\mathfrak{t}\oplus {\mathfrak{n}}^{+}$. For $\lambda \in {\mathfrak{t}}^{\ast }$, let ${\mathbb{C}}_{\lambda }$ be the one-dimensional $\mathfrak{b}$-module on which ${\mathfrak{n}}^{+}$ acts by zero and $h\in \mathfrak{t}$ acts by $\lambda (h)$. The Verma module isIts universal property of a Verma module says that any vector of weight $\lambda$ annihilated by ${\mathfrak{n}}^{+}$ receives the canonical highest-weight vector under one unique module homomorphism from $M(\lambda )$.

The sum of the proper submodules of $M(\lambda )$ is its unique maximal proper submodule, because no proper submodule contains the highest-weight vector. Its quotient $V(\lambda )$ is therefore the unique irreducible quotient of a Verma module, and hence the unique irreducible highest-weight module of weight $\lambda$.

The module $V(\lambda )$ is finite-dimensional exactly when $\lambda$ is a dominant integral weight:for every simple root ${\alpha }_i$.

Choose positive roots ${\Phi }^{+}$ and letbe the Weyl vector. For a dominant integral weight $\lambda$, the Weyl dimension formula isHere ${\alpha }^{\vee }$ is the coroot and $⟨,⟩$ is the natural weight-coroot pairing.