Weights, Verma modules and the theorem of the highest weight

Written by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. Self-checked by the writing AI. Public domain (CC0).

A highest-weight vector records both a Cartan eigenvalue and a direction in which no raising operator can move it. We first construct the largest module generated by such a vector. We then impose finitely many rank-one relations and prove that the resulting irreducible quotient is finite-dimensional. This gives the classification without first choosing a matrix realization for every representation.

Throughout, $\mathfrak g$ is a finite-dimensional complex semisimple Lie algebra, $\mathfrak h$ a Cartan subalgebra, and $\Phi^+$ a choice of positive roots. Write $\mathfrak g=\mathfrak n^-\oplus\mathfrak h\oplus\mathfrak n^+, \qquad \mathfrak b=\mathfrak h\oplus\mathfrak n^+.$ We use the root decomposition and normalized root triples from The root space decomposition, the Euclidean root systems and Weyl chambers from Root systems and their Weyl groups and Cartan matrices and Dynkin diagrams, and the triangular PBW theorem from The universal enveloping algebra. The finite-dimensional rank-one theory is proved in The representations of sl(2). Explicit models used in the examples are from The simple Lie algebras.

1. The lattice and the order

Let $\alpha_1,\ldots,\alpha_r$ be the simple roots, and choose $[e_i,f_i]=h_i=\alpha_i^\vee,\quad [h_i,e_i]=2e_i,\quad [h_i,f_i]=-2f_i,\quad [e_i,f_j]=0\quad(i\ne j). \tag{1.1}$ Here a coroot is an element of $\mathfrak h$; its Euclidean representative is $2\alpha_i/(\alpha_i,\alpha_i)$. The $h_i$ form a basis. The simple-root generators and their relations, including generation of $\mathfrak n^+$ by the $e_i$, are proved in The isomorphism theorem and Serre’s theorem. In our column-coroot convention the Cartan matrix is $a_{ji}=\alpha_j(h_i)$.

The integral weight lattice, root lattice and positive root monoid are $\Lambda=\{\lambda\in\mathfrak h^*: \lambda(\alpha^\vee)\in\mathbb Z \text{ for every }\alpha\in\Phi\},\quad Q=\sum_i\mathbb Z\alpha_i,\quad Q^+=\sum_i\mathbb Z_{\ge0}\alpha_i. \tag{1.2}$ The simple coroots are a base of the coroot root system; every coroot is their integral combination. Thus the condition defining $\Lambda$ is equivalent to $\lambda(h_i)\in\mathbb Z$ for all $i$. Define the fundamental weights by $\omega_j(h_i)=\delta_{ij}.$ The dual-basis property gives $\Lambda=\bigoplus_i\mathbb Z\omega_i, \qquad \Lambda^+=\sum_i\mathbb Z_{\ge0}\omega_i, \qquad \lambda=\sum_i\lambda(h_i)\omega_i. \tag{1.3}$ The roots have integral coroot pairings, so $Q\subseteq\Lambda$. The set $\Lambda^+$ consists of the dominant integral weights; it is an additive monoid, rather than a subgroup unless $r=0$. A simple reflection acts by $s_i\lambda=\lambda-\lambda(h_i)\alpha_i$, preserving $\Lambda$ and every coset modulo $Q$.

For arbitrary complex weights we use the order $\mu\le\lambda\quad\Longleftrightarrow\quad\lambda-\mu\in Q^+. \tag{1.4}$ It is a partial order: transitivity follows from addition in $Q^+$, and antisymmetry follows from independence of the simple roots. A scalar functional positive on every simple root is useful for choosing a maximum in a finite set, but equality of its values does not imply equality of weights.

Put $\rho=\tfrac12\sum_{\alpha\in\Phi^+}\alpha$. A simple reflection permutes the positive roots other than $\alpha_i$ and changes $\alpha_i$ to its negative. Hence $s_i\rho=\rho-\alpha_i$. Comparing with the reflection formula gives $\rho(h_i)=1$, so $\rho=\sum_i\omega_i. \tag{1.5}$

For a module $V$, its weight space is $V_\mu=\{v: Hv=\mu(H)v\text{ for every }H\in\mathfrak h\}.$ A weight module means $V=\bigoplus_\mu V_\mu$. The sum is algebraic: each vector has finitely many weight components. If $x\in\mathfrak g_\alpha$, the commutator with $H$ gives $xV_\mu\subseteq V_{\mu+\alpha}. \tag{1.6}$

Lemma 1.1. Every finite-dimensional $\mathfrak g$-module is a weight module with integral weights.

Proof. On restricting to the rank-one algebra generated by $e_i,f_i,h_i$, the finite-dimensional theory decomposes the module into its $\mathfrak{sl}_2$ irreducibles. Consequently $h_i$ is diagonalizable with integer eigenvalues. The commuting diagonalizable operators $h_1,\ldots,h_r$ are simultaneously diagonalizable: decompose by the first, restrict the others to its eigenspaces, and continue. Their joint eigenvalues give weights, whose values on all $h_i$ are integers. Equations (1.2)–(1.3) give integrality. $\square$

We will also need decomposition for submodules of an infinite weight module. If $v=\sum_{j=1}^t v_{\mu_j}$ has distinct weights, choose $H\in\mathfrak h$ whose values on this finite set are distinct. The Lagrange interpolation polynomial $p_j(T)=\prod_{k\ne j}\frac{T-\mu_k(H)}{\mu_j(H)-\mu_k(H)}$ satisfies $p_j(H)v=v_{\mu_j}$. Therefore every $\mathfrak h$-stable subspace is the direct sum of its weight intersections. Its quotient is a weight module, with weight spaces the corresponding quotients. This argument uses finite support of each vector, not finite dimension of the entire module.

For finite-dimensional $V$, define its formal character in the group ring $\mathbb Z[\Lambda]$ by $\operatorname{ch}V=\sum_{\mu\in\Lambda}(\dim V_\mu)e^\mu, \qquad e^\mu e^\nu=e^{\mu+\nu}. \tag{1.7}$ It records multiplicities, so evaluating every $e^\mu$ at one gives $\dim V$. For highest-weight modules with finite weight spaces the same notation denotes a formal series supported in a translate of $-Q^+$; the completion and its coefficients will be explicit below.

2. The universal highest-weight module

A nonzero $v\in V_\lambda$ is a highest-weight vector if $\mathfrak n^+v=0$. It suffices that every $e_i$ kills it, since these generate $\mathfrak n^+$. A highest-weight module is generated by such a vector.

The one-dimensional $\mathfrak b$-module $\mathbb C_\lambda$ has $H$ acting by $\lambda(H)$ and $\mathfrak n^+$ acting by zero. This is a Lie action: $[\mathfrak b,\mathfrak b]\subseteq\mathfrak n^+$. Define the Verma module $M(\lambda)=U(\mathfrak g)\otimes_{U(\mathfrak b)}\mathbb C_\lambda, \qquad v_\lambda=1\otimes1. \tag{2.1}$ Equivalently it is the quotient by the left ideal generated by $e_i$ and $H-\lambda(H)$. Indeed, the tensor generator obeys those relations, giving a map from that quotient to (2.1). Conversely those relations make every element of $U(\mathfrak b)$ act on the quotient’s generator by its character on $\mathbb C_\lambda$; the tensor relation therefore defines an inverse map. Both compositions fix the cyclic generator. If $v$ is a highest-weight vector of weight $\lambda$ in any module, the formula $u\otimes1\longmapsto uv \tag{2.2}$ is well defined, is a module homomorphism, and is the unique map sending $v_\lambda$ to $v$. It is onto exactly when $v$ generates the target.

Theorem 2.1 (the PBW description). The map $U(\mathfrak n^-)\longrightarrow M(\lambda),\qquad u\longmapsto uv_\lambda$ is an isomorphism of left $U(\mathfrak n^-)$-modules. The weights are exactly $\lambda-Q^+$, the top weight space is $\mathbb Cv_\lambda$, and every weight space is finite-dimensional. Moreover, $\operatorname{ch}M(\lambda) =e^\lambda\prod_{\alpha\in\Phi^+}(1-e^{-\alpha})^{-1} =e^\lambda\sum_{\gamma\in Q^+}p(\gamma)e^{-\gamma}, \tag{2.3}$ where $p(\gamma)$ counts expressions of $\gamma$ as an unordered sum of positive roots, with repetitions.

Proof. Triangular PBW gives the multiplication isomorphism $U(\mathfrak n^-)\otimes U(\mathfrak b)\cong U(\mathfrak g)$, which also respects the right $U(\mathfrak b)$-actions. Tensoring over that algebra gives the asserted free module, with no remaining relation among its negative PBW monomials.

Choose and order a nonzero vector $f_\alpha\in\mathfrak g_{-\alpha}$ for every positive root. The basis vectors $\left(\prod_{\alpha\in\Phi^+}f_\alpha^{m_\alpha}\right)v_\lambda, \qquad m_\alpha\in\mathbb Z_{\ge0}, \tag{2.4}$ have weight $\lambda-\sum_\alpha m_\alpha\alpha$. Every $\gamma\in Q^+$ occurs, by using just the simple-root exponents. If $\gamma=\sum_i d_i\alpha_i$, its height is $\sum_i d_i$. Each positive root has positive integral height, so $\sum_\alpha m_\alpha\le\operatorname{ht}\gamma$; there are only finitely many such tuples. The number giving $\gamma$ is exactly $p(\gamma)$. Only the all-zero tuple has weight $\lambda$.

In the formal power-series ring $\mathbb Z[[q_1,\ldots,q_r]]$, interpret $e^{-\sum_i d_i\alpha_i}$ as $q_1^{d_1}\cdots q_r^{d_r}$. Each geometric series $(1-e^{-\alpha})^{-1}$ is legitimate there. Multiplying the finitely many series counts the tuples in (2.4), proving (2.3) coefficient by coefficient. The factor $e^\lambda$ specifies the weight translate; no analytic convergence is asserted. $\square$

For nonzero $\mathfrak g$, the vectors $f_i^m v_\lambda$ for $m\ge0$ already show that every Verma module is infinite-dimensional. A finite-dimensional complete-reducibility theorem cannot be applied to it.

3. The irreducible quotient

Theorem 3.1. The module $M(\lambda)$ has a unique maximal proper submodule $N(\lambda)$. Its quotient $L(\lambda)=M(\lambda)/N(\lambda) \tag{3.1}$ is the unique irreducible highest-weight module of weight $\lambda$. Every nonzero highest-weight module of that weight maps onto it. Furthermore $L(\lambda)\cong L(\mu)$ if and only if $\lambda=\mu$.

Proof. By the interpolation argument in Section 1, every submodule is a sum of weight spaces. A proper submodule has zero intersection with $M(\lambda)_\lambda$: containing a nonzero multiple of $v_\lambda$ would mean containing its whole cyclic module. The sum of all proper submodules still has zero top component. Thus it is proper and contains every proper submodule. Call it $N(\lambda)$. Its maximality and uniqueness follow immediately, and (3.1) is irreducible.

By (2.2), any nonzero highest-weight module is $M(\lambda)/K$ with $K$ proper. Since $K\subseteq N(\lambda)$, its quotient maps onto (3.1). If it is itself irreducible, the nonzero surjection is an isomorphism. Finally an isomorphism $L(\lambda)\cong L(\mu)$ places each top weight among the weights of the other. Hence $\lambda\le\mu$ and $\mu\le\lambda$, so antisymmetry gives equality. Equality also plainly implies isomorphism. $\square$

For rank one write $n=\lambda(h)$, with $[e,f]=h$. The PBW basis is $f^k v_\lambda$, and a direct commutator induction gives $hf^k v_\lambda=(n-2k)f^k v_\lambda, \qquad ef^k v_\lambda=k(n-k+1)f^{k-1}v_\lambda. \tag{3.2}$ If $n\notin\mathbb Z_{\ge0}$, the raising coefficients are nonzero for every $k\ge1$. Any nonzero submodule contains a basis vector by weight projection, and repeated raising brings it to $v_\lambda$; the Verma module is irreducible. If $n\in\mathbb Z_{\ge0}$, its unique maximal submodule has basis $f^k v_\lambda$ for $k\ge n+1$, since this span is invariant and the quotient with basis $k=0,\ldots,n$ is irreducible. It is itself a Verma module of highest weight $-n-2$: send its generator to $f^{n+1}v_\lambda$, and PBW shows injectivity.

Thus $0\longrightarrow M(-n-2)\longrightarrow M(n)\longrightarrow L(n)\longrightarrow0 \tag{3.3}$ is exact and does not split. A splitting would put a highest vector of weight $n$ inside $M(n)$ generating a finite submodule. That vector must be a nonzero multiple of $v_n$, which generates the infinite Verma module. This also exhibits why complete reducibility fails in this infinite setting.

4. The finite-dimensional classification

Theorem 4.1 (highest-weight classification). The module $L(\lambda)$ is finite-dimensional exactly when $\lambda\in\Lambda^+$. Every finite-dimensional irreducible $\mathfrak g$-module is isomorphic to exactly one of these modules. Its weight multiplicities are Weyl-invariant.

We prove the assertion in stages, keeping the finiteness argument separate from Weyl symmetry.

First let $V\ne0$ be finite-dimensional. Its weights are integral by Lemma 1.1. Choose a real functional $\ell$ on the real weight space satisfying $\ell(\alpha_i)>0$, and choose a weight $\lambda$ maximizing $\ell$ on the finite weight set. No $\lambda+\alpha_i$ is a weight, so every vector in $V_\lambda$ is killed by the simple raising operators. If $V$ is irreducible, any nonzero one generates $V$, giving $V\cong L(\lambda)$. In its restriction to the $i$-th rank-one algebra, this vector is a highest-weight vector; finite-dimensional rank-one theory says $n_i=\lambda(h_i)\in\mathbb Z_{\ge0}$. Thus $\lambda\in\Lambda^+$. This also proves necessity for a finite-dimensional $L(\lambda)$.

Now fix $\lambda\in\Lambda^+$, and set $n_i=\lambda(h_i)$. In $M(\lambda)$, the nonzero vector $z_i=f_i^{n_i+1}v_\lambda \tag{4.1}$ is singular, meaning that every raising operator kills it. Formula (3.2) for the $i$-th triple gives $e_i z_i=0$; for $j\ne i$, (1.1) gives $e_j z_i=f_i^{n_i+1}e_jv_\lambda=0$. Its weight is $\lambda-(n_i+1)\alpha_i$. By the universal property, all weights of $U(\mathfrak g)z_i$ lie below that weight, so its top $\lambda$-component is zero. Consequently $T(\lambda)=M(\lambda)\Big/\sum_iU(\mathfrak g)z_i \tag{4.2}$ is nonzero and still has a one-dimensional top. It maps onto $L(\lambda)$. We next prove that (4.2) is finite-dimensional.

Lemma 4.2 (local nilpotence). Every $e_i$ and $f_i$ acts locally nilpotently on $T(\lambda)$, and hence on $L(\lambda)$.

Proof. The adjoint actions of $e_i,f_i$ on $\mathfrak g$ are nilpotent, by the finite root strings and the rank-one action on the Cartan and opposite root lines. Their extensions as derivations of $U(\mathfrak g)$ are locally nilpotent: on a word, a sufficiently high iterate distributes more derivatives than the sum of the bounds for its finitely many letters. Linearity gives the assertion for a finite sum of words.

For any element $x$, the identity $xu=[x,u]+ux$, iterated, gives $x^N(uv)=\sum_{a=0}^N\binom Na(\operatorname{ad}x)^a(u)x^{N-a}v. \tag{4.3}$ In (4.2), $f_i^{n_i+1}v_\lambda=0$. For fixed $u$, choose $R$ so that $(\operatorname{ad}f_i)^a u=0$ for $a>R$. If $N>R+n_i$, every summand in (4.3) vanishes. For $e_i$, the same proof uses $e_i v_\lambda=0$. Each vector is a finite sum of $uv_\lambda$, so the actions are locally nilpotent. Passing to a quotient preserves this property. $\square$

Lemma 4.3 (reflections on weight spaces). On a weight module where $e_i,f_i$ are locally nilpotent, the operator $S_i=\exp(e_i)\exp(-f_i)\exp(e_i) \tag{4.4}$ is invertible and maps $V_\mu$ isomorphically onto $V_{s_i\mu}$.

Proof. Each exponential is a finite sum on each vector and its inverse is the exponential with the opposite sign. Local nilpotence makes these linear operators well defined even before finite dimension has been proved.

The finite commutator expansion for conjugation gives $\exp(e_i)h_i\exp(-e_i)=h_i-2e_i, \quad \exp(-f_i)h_i\exp(f_i)=h_i-2f_i, \quad \exp(-f_i)e_i\exp(f_i)=e_i+h_i-f_i.$ These identities can also be checked by differentiating the conjugation polynomial and using (1.1); all needed iterated commutators vanish. Applying the three conjugations yields $S_i h_iS_i^{-1}=-h_i$. An element $H_0\in\ker\alpha_i$ commutes with both root operators and is fixed. Writing $H=H_0+\alpha_i(H)h_i/2$, we obtain $S_iHS_i^{-1}=H-\alpha_i(H)h_i=s_iH. \tag{4.5}$ The same reflection is its own inverse on $\mathfrak h$. Thus $H(S_iv)=S_i(s_iH)v=(s_i\mu)(H)S_iv$ for $v\in V_\mu$. Invertibility proves the isomorphism, including equality of dimensions when these are finite. $\square$

Since simple reflections generate $W$, Lemmas 4.2–4.3 prove Weyl invariance of the weights and their dimensions in (4.2) and (3.1). This fact by itself would not imply a finite weight set.

Let $w_0$ be the longest Weyl element, taking $\Phi^+$ to $-\Phi^+$; the root-system results give $w_0^2=1$. If $\mu$ is a weight of (4.2), then $\lambda-\mu\in Q^+,\qquad \lambda-w_0\mu\in Q^+.$ Applying $w_0$ to the second inclusion gives $\mu-w_0\lambda\in Q^+$. Write $\lambda-w_0\lambda=\sum_i b_i\alpha_i$ and $\lambda-\mu=\sum_i d_i\alpha_i$. The coefficients are integers, and these two cone conditions give $0\le d_i\le b_i\quad\text{for every }i. \tag{4.6}$ In particular $b_i\ge0$, since the weight $\lambda$ occurs and so does $w_0\lambda$. Only finitely many integer tuples satisfy (4.6). Each corresponding weight space is a quotient of a finite-dimensional Verma weight space. Hence $T(\lambda)$ is finite-dimensional, and so is its quotient $L(\lambda)$. The uniqueness assertion is Theorem 3.1. This completes Theorem 4.1. $\square$

For $\mathfrak g=0$, the lattice and positive cone are zero, the Verma module is $\mathbb C$, and the same classification has its single trivial module.

4.4. The finite integrable presentation

Corollary 4.4. For $\lambda\in\Lambda^+$, the quotient $T(\lambda)$ in (4.2) is already irreducible. In particular $L(\lambda)=U(\mathfrak g)\Big/\Bigl( \sum_iU(\mathfrak g)e_i+ \sum_iU(\mathfrak g)(h_i-\lambda(h_i))+ \sum_iU(\mathfrak g)f_i^{\lambda(h_i)+1}\Bigr). \tag{4.7}$ The denominator is a left ideal, so this is a presentation of a cyclic module.

Proof. We have proved that $T(\lambda)$ is finite-dimensional, is generated by its nonzero top vector $v$, and has one-dimensional top weight space. Let $N$ be a proper submodule. Weight projection shows that $N_\lambda=0$, since containing any nonzero top vector would imply $N=T(\lambda)$. Complete reducibility gives a module complement $S$, so $T(\lambda)=N\oplus S$. The projections commute with $\mathfrak h$. Therefore the projection of $v$ into $N$ belongs to $N_\lambda=0$, and $v\in S$. Its cyclicity forces $S=T(\lambda)$, hence $N=0$. Thus $T(\lambda)$ is irreducible and equals its simple quotient $L(\lambda)$. Combining the defining left-ideal relations of the Verma module with (4.2) gives (4.7). $\square$

4.5. Rational highest-weight modules

The rational Serre algebra has generators $e_i,f_i,h_i$ over $\mathbb Q$. If $n_i=\lambda(h_i)\in\mathbb Z_{\ge0}$, impose the relations in (4.7) over $\mathbb Q$, and denote the resulting module by $L_{\mathbb Q}(\lambda)$. Tensor algebras, their defining quotients, and the displayed left ideals commute with extension of a ground field. Hence $L_{\mathbb Q}(\lambda)\otimes_{\mathbb Q}\mathbb C \cong L(\lambda). \tag{4.8}$ This rational module is finite-dimensional: choose rational vectors whose tensors span the finite-dimensional right side and apply the basis argument of Corollary 6.3 in the Serre lesson. A nonzero proper rational submodule would remain nonzero and proper after tensoring, contradicting complex irreducibility. Thus it is irreducible over $\mathbb Q$.

The same presentation over any characteristic-zero extension $K/\mathbb Q$ is finite-dimensional and irreducible. Indeed the rational PBW weight decomposition extends to $K$, with the same one-dimensional top and cyclic vector; the algebra is semisimple by the rational Serre result. Complete reducibility over characteristic-zero fields, Theorem 4.1 supplies the complement argument of Corollary 4.4 over $K$ without algebraic closedness. These rational forms are available because the Cartan coefficients and the highest-weight integers in the presentation belong to $\mathbb Q$. This does not assert that an arbitrary nonsplit Lie algebra has such a presentation, nor that a characteristic-zero irreducible representation remains irreducible under every extension in general.

5. Which weights occur?

Call a Weyl-invariant integral weight set saturated if, whenever it contains $\mu$ with $m=\mu(\alpha^\vee)\ge0$, it also contains $\mu,\ \mu-\alpha,\ldots,\mu-m\alpha=s_\alpha\mu \tag{5.1}$ for every root $\alpha$. This asserts occurrence of intermediate weights, not equality of their multiplicities.

Proposition 5.1. For dominant integral $\lambda$, the weights of $L(\lambda)$ are saturated, have unique greatest element $\lambda$, and satisfy the more precise formula $\operatorname{Wt}L(\lambda)=\operatorname{conv}(W\lambda)\cap(\lambda+Q). \tag{5.2}$ Their multiplicities are constant on Weyl orbits, and the top multiplicity is one.

Proof of saturation. Fix a root $\alpha$. The finite-dimensional sum of weight spaces in a fixed coset $\mu+\mathbb Z\alpha$ is stable under its root triple. Restrict to that $\mathfrak{sl}_2$ and decompose into irreducibles. The Cartan subspace $\ker\alpha$ acts by the fixed weight restriction on this entire sum. In a rank-one irreducible, the $h_\alpha$-weights run through $n,n-2,\ldots,-n$. If an $h_\alpha$-weight $m\ge0$ occurs, all $m-2j$ for $0\le j\le m$ occur. Since the $\ker\alpha$ values are unchanged, these are precisely the weights in (5.1). This proves saturation. It also shows that lowering a nonzero weight vector with positive $h_\alpha$-weight by $f_\alpha$ cannot give zero: this is true on each rank-one summand. Weyl multiplicity invariance was proved in Lemma 4.3. The top assertion and greatest-weight assertion follow from the Verma quotient.

Here are the details for (5.2). First, every dominant integral $\eta\le\lambda$ occurs. Start at an occurring weight $\nu$, initially $\lambda$, with $\nu-\eta=\sum_i d_i\alpha_i\in Q^+$. If $\nu\ne\eta$, then $(\nu-\eta,\nu)=(\nu-\eta,\eta)+\|\nu-\eta\|^2>0,$ because $\eta$ is dominant. Thus some $i$ with $d_i>0$ has $(\alpha_i,\nu)>0$. The rank-one lowering assertion makes $\nu-\alpha_i$ an occurring weight, still above $\eta$. The height of its difference from $\eta$ decreases by one. Repetition terminates at $\eta$.

We use an elementary convexity fact. For a finite set $A$ in a Euclidean space, a point $z$ is in $\operatorname{conv}(A)$ if every linear functional $x$ satisfies $(x,z)\le\max_{a\in A}(x,a)$. To prove the nontrivial direction, if $z$ is outside the compact convex hull, choose its nearest point $p$. Minimizing the squared distance along the segment from $p$ to any $a$ gives $(z-p,a-p)\le0$. Then the functional $x=z-p$ has $(x,z)=(x,p)+\|x\|^2>\max_{a\in A}(x,a)$, a contradiction.

If $\nu$ occurs, then $w\nu\le\lambda$ for every $w$, by Weyl invariance. Given any $x$, choose $w$ with $wx$ in the closed dominant chamber. Pairing the cone difference with $wx$ gives $(x,\nu)=(wx,w\nu)\le(wx,\lambda)\le\max_{u\in W}(x,u\lambda).$ The convexity fact proves $\nu\in\operatorname{conv}(W\lambda)$; the Verma weights put it in $\lambda+Q$.

Conversely take $\nu$ in the right side of (5.2), and move it by $w$ to a dominant weight $\eta=w\nu$. It is integral, since $\lambda+Q\subseteq\Lambda$. The polytope and the coset are Weyl-invariant. Every $u\lambda$ occurs in $L(\lambda)$, hence $\lambda-u\lambda\in Q^+$. Expressing $\eta$ as a convex combination of these orbit points shows $\lambda-\eta$ is a nonnegative real combination of simple roots. Since it also belongs to $Q$, all these coordinates are nonnegative integers. Thus $\eta\le\lambda$, and the descent argument shows that it occurs. Weyl invariance then shows that $\nu$ occurs. This proves (5.2). $\square$

The set in (5.2) does not determine multiplicities. For example the zero weight in an adjoint module has multiplicity equal to the rank, whereas its root weights have multiplicity one. The character keeps this additional information; its general formula will be proved in Weyl's character formula.

6. Coordinates and small representations

6.1. Fundamental weights in rank two

For $A_2$, use $\alpha_1=\varepsilon_1-\varepsilon_2$, $\alpha_2=\varepsilon_2-\varepsilon_3$ in the sum-zero plane of $\mathbb R^3$. For $B_2$, use orthonormal $\varepsilon_1,\varepsilon_2$, with $\alpha_1=\varepsilon_1-\varepsilon_2$ long and $\alpha_2=\varepsilon_2$ short. For $G_2$, use the projected coordinates with $\sum_i\varepsilon_i=0$, $(\varepsilon_i,\varepsilon_j)=\delta_{ij}-1/3$, and $\alpha_1=\varepsilon_2$ short, $\alpha_2=\varepsilon_1-\varepsilon_2$ long. The fundamental weights and $\rho$ are

Type	$\omega_1$	$\omega_2$	$\rho$
$A_2$	$(2\alpha_1+\alpha_2)/3$	$(\alpha_1+2\alpha_2)/3$	$\alpha_1+\alpha_2$
$B_2$	$\alpha_1+\alpha_2=\varepsilon_1$	$\alpha_1/2+\alpha_2=(\varepsilon_1+\varepsilon_2)/2$	$3\alpha_1/2+2\alpha_2$
$G_2$	$2\alpha_1+\alpha_2=-\varepsilon_3$	$3\alpha_1+2\alpha_2=\varepsilon_1-\varepsilon_3$	$5\alpha_1+3\alpha_2$

Their derivation by solving the coroot equations is given in Exercise 7.1. In particular the $B_2$ lattice is $\mathbb Z^2\ \cup\ \big((1/2,1/2)+\mathbb Z^2\big). \tag{6.1}$ Both coordinates have the same integer or half-integer parity; allowing arbitrary half-integers separately would incorrectly include $(1/2,0)$.

6.2. The three and eight dimensions of sl(3)

In the standard module $\mathbb C^3$, the three coordinate vectors have weights $\varepsilon_1,\varepsilon_2,\varepsilon_3$, understood as restrictions to the trace-zero diagonal. Taking $e_1=E_{12},e_2=E_{23}$ and opposite $f_i$, its lowering chain is $v_1\xrightarrow{f_1}v_2\xrightarrow{f_2}v_3.$ Its raising arrows reverse this chain. Any nonzero invariant subspace has a nonzero weight component by projection, hence one of these basis vectors, and the arrows give all three. The module is irreducible. Its highest weight is $\varepsilon_1=\omega_1$, so it is $L(\omega_1)$ and $\operatorname{ch}L(\omega_1)=e^{\varepsilon_1}+e^{\varepsilon_2}+e^{\varepsilon_3}. \tag{6.2}$ The dual is irreducible as well: the annihilator of a proper invariant subspace would be an invariant subspace in the original. Its highest weight is $-\varepsilon_3=\omega_2$.

The adjoint module of the simple algebra $\mathfrak{sl}_3$ is irreducible, since its submodules are ideals. Its root vectors $E_{ij}$ have the six weights $\varepsilon_i-\varepsilon_j$, each of multiplicity one, and its two-dimensional Cartan space has weight zero. The highest vector $E_{13}$ is killed by both simple raisings and has weight $\alpha_1+\alpha_2=\omega_1+\omega_2$. Thus $\operatorname{ch}L(\omega_1+\omega_2)=2+\sum_{i\ne j}e^{\varepsilon_i-\varepsilon_j}, \qquad \dim L(\omega_1+\omega_2)=8. \tag{6.3}$ Permutation of the three coordinates gives the Weyl symmetry, with the zero weight fixed and still of multiplicity two.

6.3. The four and five dimensions of sp(4)

Here use the $C_2$ base $\alpha_1=\varepsilon_1-\varepsilon_2$ short, $\alpha_2=2\varepsilon_2$ long. Its fundamental weights are $\omega_1=\varepsilon_1$, $\omega_2=\varepsilon_1+\varepsilon_2$. This numbering differs from the $B_2$ coordinates in Section 6.1.

Let the standard basis be $v_1,v_2,v_{-2},v_{-1}$, with alternating form $B(v_i,v_{-j})=\delta_{ij}$, $B(v_{-j},v_i)=-\delta_{ij}$. Use $e_1=E_{1,2}-E_{-2,-1},\quad f_1=E_{2,1}-E_{-1,-2}, \qquad e_2=E_{2,-2},\quad f_2=E_{-2,2}. \tag{6.4}$ The four distinct weights are $\varepsilon_1,\varepsilon_2,-\varepsilon_2,-\varepsilon_1$. The lowering chain is $v_1\mapsto v_2\mapsto v_{-2}\mapsto-v_{-1}$, using $f_1,f_2,f_1$. Every reverse arrow under the corresponding $e_i$ is nonzero. Weight projection and these arrows prove irreducibility, so the module is $L(\omega_1)$ of dimension four, with character $\sum_i(e^{\varepsilon_i}+e^{-\varepsilon_i})$.

In $\Lambda^2V$, the vector $\omega=v_1\wedge v_{-1}+v_2\wedge v_{-2}$ is invariant, as proved from preservation of the alternating form in the classical-model lesson. Set $W=\Lambda^2V/\mathbb C\omega$. Put $a=v_1\wedge v_2,\quad b=v_1\wedge v_{-2},\quad c=v_1\wedge v_{-1}-v_2\wedge v_{-2},\quad d=v_2\wedge v_{-1},\quad t=v_{-1}\wedge v_{-2}. \tag{6.5}$ Their classes are a basis of the quotient: the four nonzero-weight wedge monomials are independent, and the zero-weight plane has independent vectors $c,\omega$. Their weights are respectively $\varepsilon_1+\varepsilon_2,\varepsilon_1-\varepsilon_2,0,-\varepsilon_1+\varepsilon_2,-\varepsilon_1-\varepsilon_2$. Directly applying (6.4) on each wedge factor gives $f_2a=b,\quad f_1b=-c,\quad f_1c=2d,\quad f_2d=-t; \qquad e_2b=a,\quad e_1c=-2b,\quad e_1d=c,\quad e_2t=-d. \tag{6.6}$ Thus every basis line reaches every other through nonzero arrows. Weight projection proves irreducibility. Both $e_i$ kill $a$, whose weight is $\omega_2$. Consequently $W=L(\omega_2),\quad \dim W=5,\quad \operatorname{ch}W=1+\sum_{\sigma,\tau\in\{1,-1\}}e^{\sigma\varepsilon_1+\tau\varepsilon_2}. \tag{6.7}$ The full exterior square has dimension six and contains the extra trivial summand; the five-dimensional quotient is the fundamental module.

6.4. The seven and fourteen dimensions of G₂

Use the proved model $\mathfrak g_2=\operatorname{Der}(\mathbb O)$ from Section 4 of the classical and exceptional models lesson. In its Zorn coordinates write $e_+,e_-$ for the diagonal idempotents, $h=e_+-e_-$, and $u(x),v(y)$ for the two off-diagonal three-dimensional spaces. Its derivations are parametrized by $A\in\mathfrak{sl}_3$, $p\in\mathbb C^3$, $q\in(\mathbb C^3)^*$, with $D_{A,p,q}h=2u(p)+2v(q),\quad D_{A,p,q}u(x)=-q(x)h+u(Ax)+v(p\times x),$ $D_{A,p,q}v(y)=-y(p)h+u(q\times y)-v(A^ty). \tag{6.8}$ These formulas preserve the trace-zero space $\mathbb O_0=\mathbb Ch\oplus u(\mathbb C^3)\oplus v((\mathbb C^3)^*), \tag{6.9}$ of dimension seven. Write $K_A=D_{A,0,0}$, $P_p=D_{0,p,0}$, $Q_q=D_{0,0,q}$, and let $x_i,x_i^*$ be dual coordinate bases. The diagonal $K_A$ give the distinct weights $0,\varepsilon_i,-\varepsilon_i$, with respective lines $\mathbb Ch,\mathbb Cu(x_i),\mathbb Cv(x_i^*)$.

A nonzero invariant subspace contains a nonzero weight component. If it contains $u(x_i)$, apply $Q_{x_i^*}$ to get $-h$; if it contains $v(x_i^*)$, apply $P_{x_i}$ to get $-h$. Once it contains $h$, (6.8) supplies all six remaining lines. Thus (6.9) is irreducible.

The normalized simple raisings are $e_1=P_{x_2}$, $e_2=K_{E_{12}}$, as proved in that model. They both kill $v(x_3^*)$, by (6.8), and this vector has weight $-\varepsilon_3=\omega_1$ for the $G_2$ base of Section 6.1. We have proved directly that $\mathbb O_0=L(\omega_1),\quad \operatorname{ch}L(\omega_1)=1+\sum_{i=1}^3(e^{\varepsilon_i}+e^{-\varepsilon_i}), \quad \dim L(\omega_1)=7. \tag{6.10}$

The adjoint module is irreducible because $\mathfrak g_2$ is simple. Its twelve roots are the six $\pm\varepsilon_i$ and the six $\varepsilon_i-\varepsilon_j$; the zero weight has multiplicity two. The vector $K_{E_{13}}$ is killed by $e_1,e_2$: the brackets are zero since $E_{13}x_2=0$ and $[E_{12},E_{13}]=0$. Its weight is $\varepsilon_1-\varepsilon_3=\omega_2$. Therefore $\operatorname{ch}L(\omega_2)=2+\sum_{\alpha\in\Phi(G_2)}e^\alpha, \qquad \dim L(\omega_2)=14. \tag{6.11}$ These dimensions come from explicit irreducible modules, independently of a general dimension formula.

7. Exercises with complete solutions

Exercise 7.1 — solve the coroot equations (easy)

Compute the fundamental weights and $\rho$ for the three bases in Section 6.1, and determine the $B_2$ weight lattice in orthonormal coordinates.

Solution. If $\omega=x\alpha_1+y\alpha_2$, its simple-coroot values are $(x,y)A$. The three column-coroot matrices are $A_{A_2}=\begin{pmatrix}2&-1\\-1&2\end{pmatrix},\quad A_{B_2}=\begin{pmatrix}2&-2\\-1&2\end{pmatrix},\quad A_{G_2}=\begin{pmatrix}2&-1\\-3&2\end{pmatrix}.$ For $A_2$, the equations $2x-y=1,-x+2y=0$ give $(x,y)=(2/3,1/3)$; replacing the right side by $(0,1)$ gives $(1/3,2/3)$. For $B_2$, $2x-y=1,-2x+2y=0$ give $(1,1)$, and right side $(0,1)$ gives $(1/2,1)$. For $G_2$, $2x-3y=1,-x+2y=0$ give $(2,1)$, and right side $(0,1)$ gives $(3,2)$. Adding the two fundamental weights gives every $\rho$ in the table. As a separate check, the positive roots are $\alpha_1,\alpha_2,\alpha_1+\alpha_2$ in $A_2$; add $\alpha_1+2\alpha_2$ in $B_2$; in $G_2$ the list is $\alpha_1,\alpha_2,\alpha_1+\alpha_2,2\alpha_1+\alpha_2,3\alpha_1+\alpha_2,3\alpha_1+2\alpha_2$. Their half-sums give the same answers.

For $B_2$, a vector $m_1\varepsilon_1+m_2\varepsilon_2$ is integral exactly when $m_1-m_2\in\mathbb Z$ and $2m_2\in\mathbb Z$, its simple-coroot conditions. Thus both coordinates are integers, or both are half-integers with nonintegral halves. Conversely either condition ensures both coroot values are integral. This proves (6.1).

Exercise 7.2 — find a singular vector (medium)

If $n=\lambda(h_i)\in\mathbb Z_{\ge0}$, prove that $f_i^{n+1}v_\lambda$ is nonzero and singular in $M(\lambda)$, and that it generates a proper submodule.

Solution. Induction from $[e_i,f_i]=h_i$ and $[h_i,f_i]=-2f_i$ gives $[e_i,f_i^m]=m f_i^{m-1}(h_i-m+1).$ On $v_\lambda$, this has value $m(n-m+1)f_i^{m-1}v_\lambda$. Setting $m=n+1$ makes it zero. For every $j\ne i$, $[e_j,f_i]=0$, so $e_j f_i^{n+1}v_\lambda=0$. The simple $e_j$ generate all positive root spaces, proving singularity. PBW makes $f_i^{n+1}v_\lambda\ne0$. Its weight is $\lambda-(n+1)\alpha_i$, and its generated module is a quotient of the Verma module with that highest weight. All its weights are strictly below $\lambda$, so it cannot contain $v_\lambda$ and is proper. The argument includes $n=0$, where the singular vector is $f_i v_\lambda$.

Exercise 7.3 — retain the zero-weight multiplicity (medium)

Compute all weights and multiplicities of the adjoint $\mathfrak{sl}_3$-module, identify its highest weight, and check Weyl invariance.

Solution. For $H=\operatorname{diag}(t_1,t_2,t_3)$ with zero trace, $[H,E_{ij}]=(t_i-t_j)E_{ij}$. The six off-diagonal lines have weights $\pm\alpha_1,\pm\alpha_2,\pm(\alpha_1+\alpha_2)$, each once. The diagonal trace-zero matrices commute with every $H$ and form a two-dimensional zero-weight space. These eight independent vectors account for the whole algebra. Both $[E_{12},E_{13}]$ and $[E_{23},E_{13}]$ are zero, so the highest vector has weight $\alpha_1+\alpha_2=\omega_1+\omega_2$. Simplicity of $\mathfrak{sl}_3$ makes the adjoint module irreducible. The Weyl group is $S_3$, permuting the $\varepsilon_i$. It permutes the six root lines, all of multiplicity one, and fixes weight zero, of multiplicity two. Thus the character is (6.3), and the dimension is $6+2=8$. Counting zero as a single occurrence would incorrectly give seven.

Exercise 7.4 — complete the converse for every type (hard)

For arbitrary dominant integral $\lambda$, prove that $L(\lambda)$ is finite-dimensional. Explain exactly where finite weight spaces and a bounded weight set enter the proof.

Solution. Set $n_i=\lambda(h_i)$. Exercise 7.2 produces singular vectors $z_i=f_i^{n_i+1}v_\lambda$. Each generated submodule has zero $\lambda$-component, so their sum is proper. The quotient $T$ by this sum is nonzero and maps onto $L(\lambda)$.

The adjoint actions of the simple root vectors are nilpotent on $\mathfrak g$, and repeated Leibniz expansion makes them locally nilpotent on $U(\mathfrak g)$. For fixed $u$, let $R$ bound its nonzero adjoint iterates. Formula (4.3) shows $f_i^N uv_\lambda=0$ once $N>R+n_i$, and $e_i^N uv_\lambda=0$ once $N>R$, using the respective top relations. These statements apply to every finite sum of cyclic vectors. Hence the finite exponentials (4.4) act invertibly on $T$. The rank-one conjugations (4.5) identify its $\mu$- and $s_i\mu$-spaces; composing gives Weyl invariance, including multiplicities.

Every weight satisfies $\mu\le\lambda$, because $T$ is a Verma quotient. Weyl invariance also gives $w_0\mu\le\lambda$; applying $w_0$, which reverses the positive cone, gives $w_0\lambda\le\mu$. With $\lambda-w_0\lambda=\sum_i b_i\alpha_i$ and $\lambda-\mu=\sum_i d_i\alpha_i$, this gives the finite set of possibilities $0\le d_i\le b_i$. Each weight space has dimension at most the finite partition number $p(\sum_i d_i\alpha_i)$. In particular $\dim T\le\sum_{0\le d_i\le b_i}p\!\left(\sum_i d_i\alpha_i\right)<\infty.$ The quotient $L(\lambda)$ is therefore finite-dimensional. Weyl symmetry supplies the second cone bound; it alone would allow an infinite invariant set. Finite PBW weight spaces supply the dimension bound once the finite set is known. Every step works for a product of simple types, so there is no exceptional-type omission.

8. What this lesson does not prove, and references

All the assertions about Verma modules, their irreducible quotients, finite-dimensional classification, Weyl multiplicities and saturation have been proved here. The root, rank-one, PBW and explicit-model results named at the start are imports from their complete earlier proofs. No additional theorem is assumed without proof. The general Weyl character and dimension formulas belong to Lesson 16; the small characters here were computed from explicit bases.

For comparison and further reading, see Milne, Algebraic Groups, corrected 2021 text, §22a, for the algebraic-group analogue, whose character lattice has its own isogeny restrictions; and Etingof, Lie Groups and Lie Algebras I, §25, especially Theorem 25.17 (MIT OpenCourseWare, 2020), for a direct Lie-algebra treatment. Our order, explicit finite bound, convexity argument and module computations are given in full above.

Type	\(\omega_1\)	\(\omega_2\)	\(\rho\)
\(A_2\)	\((2\alpha_1+\alpha_2)/3\)	\((\alpha_1+2\alpha_2)/3\)	\(\alpha_1+\alpha_2\)
\(B_2\)	\(\alpha_1+\alpha_2=\varepsilon_1\)	\(\alpha_1/2+\alpha_2=(\varepsilon_1+\varepsilon_2)/2\)	\(3\alpha_1/2+2\alpha_2\)
\(G_2\)	\(2\alpha_1+\alpha_2=-\varepsilon_3\)	\(3\alpha_1+2\alpha_2=\varepsilon_1-\varepsilon_3\)	\(5\alpha_1+3\alpha_2\)