This course develops the representation theory of Lie algebras as a systematic study of how abstract algebraic symmetries act on vector spaces. The central goal is to classify and compute representations, starting from the algebraic structure of universal enveloping algebras and the PBW theorem, then moving to weights, characters, and the special role of $\mathfrak{sl}_2$ as the basic model that guides the general theory. Along the way, the course emphasizes the mechanisms that make representations tractable: highest weight methods, Verma modules, central characters, and the structure of semisimple and reducible modules. The chapters build in a deliberate progression. After establishing the enveloping algebra framework, the course uses $\mathfrak{sl}_2$ to motivate the language of weights and highest weight modules, then generalizes these ideas to the classification of irreducibles and the Weyl character formula. From there, it turns to tensor products, duals, invariants, and Schur functors as computational tools for building and decomposing representations, before studying branching rules and restriction to subalgebras. The final chapters place everything in the organizing framework of category $\mathcal O$ and synthesize the main results into a practical toolkit for both structural reasoning and explicit calculation. # Introduction This opening chapter fixes the viewpoint of the course and records the vocabulary that will be used throughout. The main object is a finite-dimensional complex semisimple [Lie algebra](/page/Lie%20Algebra) $\mathfrak{g}$, and the main question is how $\mathfrak{g}$ can act linearly on vector spaces. Earlier courses describe the internal structure of $\mathfrak{g}$ through Cartan subalgebras, roots, and the Weyl group; this course turns that structure into a classification and computation theory for representations. The central theme is highest weight theory. Instead of trying to list all possible actions directly, we use the triangular decomposition of $\mathfrak{g}$ to build modules from a vector killed by all positive root spaces, then determine when those modules have finite-dimensional irreducible quotients. The final destination is the Weyl character formula, which packages the dimensions of all weight spaces of an irreducible finite-dimensional representation into a single computable expression. ## The Organising Question A Lie algebra is designed to encode infinitesimal symmetry, so its representation theory asks how that infinitesimal symmetry appears inside linear algebra. For a fixed Lie algebra $\mathfrak{g}$, the raw data of a representation is a [vector space](/page/Vector%20Space) together with a compatible action of every element of $\mathfrak{g}$. [definition: Representation Of A Lie Algebra] Let $\mathfrak{g}$ be a Lie algebra over $\mathbb C$. A representation of $\mathfrak{g}$ is a pair $(V,\rho)$ consisting of a complex vector space $V$ and a Lie algebra homomorphism \begin{align*} \rho : \mathfrak{g} \to \mathfrak{gl}(V). \end{align*} [/definition] We often suppress $\rho$ and write $xv$ for $\rho(x)(v)$, where $x \in \mathfrak{g}$ and $v \in V$. With this notation, the defining compatibility condition says \begin{align*} [x,y]v = x(yv)-y(xv) \end{align*} for all $x,y \in \mathfrak{g}$ and $v \in V$. The representation definition is efficient when we want to compare Lie algebras by homomorphisms into endomorphism algebras. For the rest of the course, however, we will constantly pass to subobjects, quotients, direct sums, tensor products, and generated cyclic objects. That makes it useful to restate the same data in module language, where those constructions are native. [definition: Lie Algebra Module] Let $\mathfrak{g}$ be a Lie algebra over $\mathbb C$. A $\mathfrak{g}$-module is a complex vector space $V$ equipped with a bilinear map \begin{align*} \mathfrak{g}\times V \to V \end{align*} whose value at $(x,v)$ is denoted $xv$, such that \begin{align*} [x,y]v=x(yv)-y(xv) \end{align*} for all $x,y\in\mathfrak{g}$ and $v\in V$. [/definition] Thus representations and modules are two languages for the same object. The representation language emphasises homomorphisms into endomorphism algebras, while the module language emphasises the internal structure of a vector space carrying an action. [example: Natural Representation Of Sl Two] Let $V=\mathbb C^2$ with standard basis $v_1,v_2$. Define linear operators by \begin{align*} hv_1=v_1,\quad hv_2=-v_2,\quad ev_1=0,\quad ev_2=v_1,\quad fv_1=v_2,\quad fv_2=0. \end{align*} We check that these operators satisfy the $\mathfrak{sl}_2(\mathbb C)$ commutation relations. On $v_1$, \begin{align*} (he-eh)v_1=h(0)-e(v_1)=0=2ev_1. \end{align*} On $v_2$, \begin{align*} (he-eh)v_2=h(v_1)-e(-v_2)=v_1+v_1=2v_1=2ev_2. \end{align*} Thus $[h,e]=2e$. Similarly, \begin{align*} (hf-fh)v_1=h(v_2)-f(v_1)=-v_2-v_2=-2v_2=-2fv_1. \end{align*} and \begin{align*} (hf-fh)v_2=h(0)-f(-v_2)=0=-2fv_2. \end{align*} Thus $[h,f]=-2f$. Finally, \begin{align*} (ef-fe)v_1=e(v_2)-f(0)=v_1=hv_1. \end{align*} and \begin{align*} (ef-fe)v_2=e(0)-f(v_1)=-v_2=hv_2. \end{align*} Thus $[e,f]=h$, so the assignment $e\mapsto e$, $f\mapsto f$, $h\mapsto h$ gives the natural representation of $\mathfrak{sl}_2(\mathbb C)$ on $V$. The basis vectors are weight vectors for $h$: $v_1$ has weight $1$ and $v_2$ has weight $-1$. The operator $e$ raises $v_2$ to $v_1$ and kills $v_1$, while $f$ lowers $v_1$ to $v_2$ and kills $v_2$; hence $v_1$ is already a highest weight vector in this smallest nontrivial example. [/example] The example shows why $\mathfrak{sl}_2(\mathbb C)$ remains the running test case. Many features of semisimple representation theory are visible there in compressed form, and the general theory can be read as a systematic extension of this pattern to all root systems. ## Why The Universal Enveloping Algebra Appears A Lie algebra has no associative multiplication compatible with repeated actions on a module. Yet if $x,y\in\mathfrak{g}$ act on $V$, then the composite operator obtained by first applying $y$ and then applying $x$ is a genuine linear operator on $V$. Without an associative algebra containing these words, expressions such as $xyv$, $x^3yv$, or products of lowering operators have to be interpreted separately inside every representation. The universal enveloping algebra is introduced to keep track of all such iterated actions while imposing exactly the commutation relations required by the Lie bracket. [definition: Universal Enveloping Algebra] Let $\mathfrak{g}$ be a Lie algebra over $\mathbb C$. The universal enveloping algebra $U(\mathfrak{g})$ is the associative unital algebra generated by $\mathfrak{g}$ subject to the relations \begin{align*} xy-yx=[x,y] \end{align*} for all $x,y\in\mathfrak{g}$. [/definition] The first lecture makes this definition precise using the tensor algebra $T(\mathfrak{g})$ and a quotient by the two-sided ideal generated by $x\otimes y-y\otimes x-[x,y]$. A construction by generators and relations is only useful if it has the expected mapping property, so the next result identifies $U(\mathfrak{g})$ as the universal associative algebra through which all Lie algebra actions factor. [quotetheorem:9355] [citeproof:9355] This theorem explains why $\mathfrak{g}$-modules may be treated as $U(\mathfrak{g})$-modules. The hypothesis that $A$ is associative is essential: the conclusion is about multiplying words in a fixed order, not merely about preserving the Lie bracket. The unital condition fixes how the empty word acts, namely as the identity operator on a module. The theorem also does not say that $\mathfrak{g}$ itself has become an associative algebra, or that different words in the generators are independent; it only gives the universal place where all iterated Lie algebra actions can be interpreted consistently. Independence of words is the separate content of PBW. [example: Extending An Sl Two Action] For the natural $\mathfrak{sl}_2(\mathbb C)$-module $V=\mathbb C^2$ with basis $v_1,v_2$, the word $feh\in U(\mathfrak{sl}_2)$ acts on $V$ in the order \begin{align*} (feh)v=\rho(f)\bigl(\rho(e)(\rho(h)(v))\bigr). \end{align*} Using the defining action from the natural module, we compute its value on $v_2$ one step at a time. First \begin{align*} h v_2=-v_2. \end{align*} By linearity of the action of $e$, \begin{align*} e(hv_2)=e(-v_2)=-e(v_2)=-v_1. \end{align*} By linearity of the action of $f$, \begin{align*} f(e(hv_2))=f(-v_1)=-f(v_1)=-v_2. \end{align*} Therefore \begin{align*} (feh)v_2=-v_2. \end{align*} This shows concretely that a word in the enveloping algebra records an ordered sequence of infinitesimal actions: here $h$ acts first, then $e$, then $f$. [/example] The second key structural result is the Poincare-Birkhoff-Witt theorem. It says that although the defining relations of $U(\mathfrak{g})$ allow generators to be reordered, the resulting algebra still has the same size as a polynomial algebra on an ordered basis, in a filtered sense. [quotetheorem:8827] [citeproof:8827] PBW is the technical engine behind much of the course, but its content should be read carefully. It is a vector-space basis theorem, not a statement that the generators commute: changing the order of two generators introduces the bracket term $[x,y]$, usually in a lower filtration degree. The ordered basis also depends on first choosing an ordered basis of $\mathfrak{g}$; for a finite-dimensional Lie algebra this gives a countable monomial basis, while in broader infinite-dimensional settings one must replace this finite list by a well-ordered basis and be more careful about the formulation. What PBW prevents is the worst possible failure of the generators-and-relations construction: it says that the relations $xy-yx=[x,y]$ do not accidentally collapse the expected monomials. This gives concrete coordinates in $U(\mathfrak{g})$, supports the triangular factorisation used in highest weight theory, and lets us build modules by inducing from subalgebras. ## Highest Weights As A Classification Principle The next problem is classification. Even for a finite-dimensional semisimple Lie algebra, there are many possible modules, so we need invariants fine enough to distinguish irreducibles and computable enough to use. The root decomposition supplies such invariants through weights. [definition: Weight Space] Let $\mathfrak{g}$ be a finite-dimensional complex semisimple Lie algebra with Cartan subalgebra $\mathfrak{h}$. If $V$ is a $\mathfrak{g}$-module and $\lambda\in\mathfrak{h}^*$, the $\lambda$-weight space of $V$ is \begin{align*} V_\lambda=\{v\in V: hv=\lambda(h)v\text{ for all }h\in\mathfrak{h}\}. \end{align*} [/definition] Weights turn a representation into a finite combinatorial object when $V$ is finite-dimensional. The action of the root spaces moves vectors between weight spaces, so the geometry of the root system controls the internal layout of a module. To classify irreducibles, we need to know when this movement has a top point from which the whole module can be generated. [definition: Highest Weight Vector] Fix a triangular decomposition \begin{align*} \mathfrak{g}=\mathfrak{n}^-\oplus\mathfrak{h}\oplus\mathfrak{n}^+. \end{align*} Let $V$ be a $\mathfrak{g}$-module. A nonzero vector $v\in V$ is a highest weight vector of weight $\lambda\in\mathfrak{h}^*$ if $hv=\lambda(h)v$ for all $h\in\mathfrak{h}$ and $xv=0$ for all $x\in\mathfrak{n}^+$. [/definition] The phrase "highest" depends on the chosen positive roots. Once that choice is fixed, the positive root spaces raise weights and the negative root spaces lower them, so a highest weight vector is a starting point from which the whole irreducible module is generated by lowering operators. [quotetheorem:9356] [citeproof:9356] This theorem reduces the classification problem to identifying which highest weights occur and when two highest weight modules are isomorphic, but each hypothesis is doing work. Finite-dimensionality gives a finite set of weights, so a maximal weight exists; without it, a module may have weights extending indefinitely upward. Irreducibility is what turns the submodule generated by a highest weight vector into the whole module; a reducible module may contain several unrelated highest weight pieces. Semisimplicity and the fixed triangular decomposition provide the root-space raising and lowering operators and specify what "highest" means; changing the positive roots changes which vectors are highest. Chapters 4 and 5 answer the resulting existence and uniqueness questions using dominant integral weights and Verma modules. [example: Highest Weights In Sl Two] For $\mathfrak{sl}_2(\mathbb C)$, the finite-dimensional irreducible modules are indexed by a nonnegative integer $m$. In the standard model $L(m)$, choose a basis $v_0,\dots,v_m$ such that \begin{align*} hv_i=(m-2i)v_i. \end{align*} Thus $v_i$ is a weight vector of weight $m-2i$. In particular, \begin{align*} hv_0=mv_0. \end{align*} The raising operator satisfies $ev_0=0$, so $v_0$ is a highest weight vector of weight $m$. The lowering operator moves one step down the string: \begin{align*} fv_i=v_{i+1}\quad\text{for }0\leq i<m,\qquad fv_m=0. \end{align*} Therefore repeated lowering gives \begin{align*} v_i=f^i v_0\quad\text{for }0\leq i\leq m. \end{align*} For example, applying $h$ to $fv_i$ gives \begin{align*} h(fv_i)=hv_{i+1}=(m-2(i+1))v_{i+1}=(m-2i-2)fv_i. \end{align*} So each application of $f$ lowers the $h$-weight by $2$. The weights of $L(m)$ are exactly \begin{align*} m,\ m-2,\ m-4,\ \dots,\ -m. \end{align*} Thus the single highest weight $m$ determines the whole finite-dimensional irreducible $\mathfrak{sl}_2(\mathbb C)$-module: it fixes the length of the string, the weights appearing in it, and the fact that $v_0$ generates the module by the lowering operator $f$. [/example] The general theory keeps this $\mathfrak{sl}_2$ picture but replaces a single integer $m$ by a dominant integral weight in the weight lattice. Each simple root contributes an embedded $\mathfrak{sl}_2$ calculation, and the compatibility of all those calculations is encoded by the root system. ## Characters And Computation Classification is not the end of the story: once irreducible modules are known to exist, we still need to compute their dimensions, weight multiplicities, and [tensor product](/page/Tensor%20Product) decompositions. Listing a basis and the action matrices is usually too much data, and it is not stable under operations such as direct sums or tensor products. Characters are introduced because they remember all weight multiplicities while forgetting the choice of basis inside each weight space. This connects Lie theory with the broader use of generating functions in combinatorics: a character packages many dimensions into one algebraic expression that can be added, multiplied, and compared. [definition: Formal Character] Let $\mathfrak{g}$ be a finite-dimensional complex semisimple Lie algebra with weight lattice $P$. Let $V$ be a finite-dimensional $\mathfrak{g}$-module with weight space decomposition $V=\bigoplus_{\lambda\in P}V_\lambda$. The formal character of $V$ is the element \begin{align*} \operatorname{ch} V=\sum_{\lambda\in P} (\dim V_\lambda)e^\lambda \in \mathbb C[P], \end{align*} where the symbols $e^\lambda$ form the basis of the group algebra $\mathbb C[P]$. [/definition] The character is additive in short exact sequences and multiplicative under tensor products. This makes it a computational shadow of the representation category: many questions about modules become questions about expressions in $\mathbb C[P]$. It also records less than the full module: two non-isomorphic modules can share the same formal character if the category is not controlled by the finite-dimensional semisimple theory being developed here. Since the Weyl group already acts on the weights, the next issue is whether characters remember that symmetry. [quotetheorem:9357] [citeproof:9357] Weyl invariance is a first sign that characters are governed by the root system, not by arbitrary linear algebra. The finite-dimensional hypothesis matters because the formal character is then a finite sum in $\mathbb C[P]$, and restriction to each embedded $\mathfrak{sl}_2$ breaks into ordinary finite-dimensional $\mathfrak{sl}_2$-modules. For infinite-dimensional highest weight modules, characters may be infinite formal sums and need not have the same symmetry. Invariance alone is also far from a classification theorem: many Weyl-invariant elements of $\mathbb C[P]$ are not characters of representations, and a character records multiplicities but not chosen bases or maps between modules. The later Weyl character formula adds the missing structure by giving a closed expression for the character of the irreducible module of highest weight $\lambda$. [example: Character Of The Natural Sl Three Module] Let $\mathfrak{g}=\mathfrak{sl}_3(\mathbb C)$ and let $V=\mathbb C^3$ have coordinate basis $v_1,v_2,v_3$. The diagonal Cartan subalgebra consists of matrices \begin{align*} h=\operatorname{diag}(a_1,a_2,a_3)\quad\text{with }a_1+a_2+a_3=0. \end{align*} For such an $h$, the natural matrix action gives \begin{align*} hv_1=a_1v_1. \end{align*} Similarly, \begin{align*} hv_2=a_2v_2. \end{align*} And \begin{align*} hv_3=a_3v_3. \end{align*} Define $\varepsilon_i\in\mathfrak{h}^*$ by $\varepsilon_i(h)=a_i$. Since $a_1+a_2+a_3=0$, these functionals satisfy $\varepsilon_1+\varepsilon_2+\varepsilon_3=0$ on $\mathfrak{h}$. The three displayed equations say exactly that \begin{align*} v_i\in V_{\varepsilon_i}\quad\text{for }i=1,2,3. \end{align*} Each weight space is one-dimensional, because $V_{\varepsilon_i}=\mathbb C v_i$. Therefore the formal character is \begin{align*} \operatorname{ch}V=(\dim V_{\varepsilon_1})e^{\varepsilon_1}+(\dim V_{\varepsilon_2})e^{\varepsilon_2}+(\dim V_{\varepsilon_3})e^{\varepsilon_3}. \end{align*} Since each dimension is $1$, this becomes \begin{align*} \operatorname{ch}V=e^{\varepsilon_1}+e^{\varepsilon_2}+e^{\varepsilon_3}. \end{align*} The Weyl group of $\mathfrak{sl}_3(\mathbb C)$ is $S_3$, acting by permuting the three coordinates, so for $\sigma\in S_3$ one has $\sigma(\varepsilon_i)=\varepsilon_{\sigma(i)}$. Hence \begin{align*} \sigma(\operatorname{ch}V)=e^{\varepsilon_{\sigma(1)}}+e^{\varepsilon_{\sigma(2)}}+e^{\varepsilon_{\sigma(3)}}=e^{\varepsilon_1}+e^{\varepsilon_2}+e^{\varepsilon_3}. \end{align*} Thus the character records three one-dimensional weight spaces and is unchanged by permuting them. [/example] The computational part of the course culminates in using characters to decompose tensor products. Tensor product multiplicities are difficult at the level of bases, but character multiplication gives a controlled route to the answer. ## Roadmap Of The Course The course begins by replacing Lie algebra actions with modules over $U(\mathfrak{g})$. PBW then gives a basis theorem and the triangular factorisation \begin{align*} U(\mathfrak{g})\cong U(\mathfrak{n}^-)\otimes U(\mathfrak{h})\otimes U(\mathfrak{n}^+) \end{align*} as vector spaces, which is the algebraic mechanism behind highest weight constructions. After that, weights and characters translate representation theory into the language of the weight lattice. The Weyl group acts on this lattice, and finite-dimensional characters respect that symmetry. These lectures keep the root-system material from earlier courses in constant use. The middle part of the course constructs Verma modules. A Verma module is a universal highest weight module, usually infinite-dimensional, whose irreducible quotient is the object we want in finite-dimensional theory. This separates existence from finiteness: first build the universal object, then impose the integrality and dominance conditions that make the irreducible quotient finite-dimensional. The final part proves the Weyl character formula and applies it. The formula leads to the [Weyl dimension formula](/theorems/9385), concrete character computations, and tensor product decompositions in small rank examples. By the end, the classification of finite-dimensional irreducible representations is not just a list of labels but a working calculus. ## Standing Assumptions And Conventions Unless stated otherwise, $\mathfrak{g}$ denotes a finite-dimensional complex semisimple Lie algebra. We fix a Cartan subalgebra $\mathfrak{h}$, a root system $\Phi\subset\mathfrak{h}^*$, a set of positive roots $\Phi^+$, and the corresponding triangular decomposition \begin{align*} \mathfrak{g}=\mathfrak{n}^-\oplus\mathfrak{h}\oplus\mathfrak{n}^+. \end{align*} Modules are complex vector spaces, and finite-dimensional modules are assumed finite-dimensional over $\mathbb C$. The weight lattice is denoted $P$, the dominant integral weights are denoted $P^+$, and the Weyl group is denoted $W$. For $\lambda\in P^+$, the irreducible finite-dimensional module of highest weight $\lambda$ will be denoted $L(\lambda)$. These conventions let the course keep notation stable while moving between algebra, geometry of root systems, and explicit calculations. The reader should expect each new construction to be tested first on $\mathfrak{sl}_2(\mathbb C)$, then on $\mathfrak{sl}_3(\mathbb C)$ or another small-rank example, before being stated in full generality. # 1. Universal Enveloping Algebras and PBW After the introductory overview, this chapter sets up the algebraic machine used throughout the course. It assumes the basic language of vector spaces, associative algebras, Lie algebras, ideals, quotients, and modules from the preceding algebra courses. Representations of a Lie algebra are linear actions, but many constructions in representation theory are easier when the action is treated as a module over an associative algebra. The universal enveloping algebra $U(\mathfrak g)$ is the associative algebra that records exactly the same representation data as $\mathfrak g$, while allowing multiplication, ideals, filtrations, and module-theoretic constructions. ## From Lie Representations to Associative Algebras The first problem is to replace the Lie bracket by ordinary multiplication without losing the bracket relation. If $\rho: \mathfrak g \to \mathfrak{gl}(V)$ is a Lie algebra representation, then the operators $\rho(x)$ compose, so products such as $\rho(x_1)\rho(x_2)\cdots \rho(x_m)$ already live in an associative algebra. The construction of $U(\mathfrak g)$ asks for the most economical associative algebra containing symbols for elements of $\mathfrak g$ and imposing only the relations forced by the Lie bracket. [definition: Tensor Algebra] Let $V$ be a vector space over a field $k$. The tensor algebra of $V$ is the associative unital $k$-algebra \begin{align*} T(V) := \bigoplus_{m=0}^{\infty} V^{\otimes m}, \end{align*} where $V^{\otimes 0}=k$, and multiplication is induced by concatenation of tensors. [/definition] The tensor algebra is the free associative algebra generated by $V$. Thus it contains all formal words in elements of $V$, with no commutation relations imposed. This motivates quotienting the tensor algebra by exactly the relations saying that the commutator in the associative algebra agrees with the given Lie bracket. [definition: Universal Enveloping Algebra] Let $\mathfrak g$ be a Lie algebra over a field $k$. The universal enveloping algebra of $\mathfrak g$ is \begin{align*} U(\mathfrak g) := T(\mathfrak g) / I, \end{align*} where $I$ is the two-sided ideal generated by all elements \begin{align*} x \otimes y - y \otimes x - [x,y] \end{align*} for $x,y \in \mathfrak g$, with $[x,y]$ viewed inside $T^1(\mathfrak g)=\mathfrak g$. [/definition] We write $i: \mathfrak g \to U(\mathfrak g)$ for the [linear map](/page/Linear%20Map) obtained by composing the inclusion $\mathfrak g \subset T(\mathfrak g)$ with the quotient map. The defining relations give \begin{align*} i([x,y]) = i(x)i(y)-i(y)i(x), \end{align*} so $i$ is a Lie algebra homomorphism from $\mathfrak g$ to the Lie algebra underlying $U(\mathfrak g)$ with commutator bracket. Before proving the main mapping property, it is useful to test the construction in the case where no nonzero brackets remain. [example: Abelian Lie Algebra] Let $\mathfrak g$ be abelian with basis $x_1,\dots,x_n$, and write again $x_i$ for the image of $x_i$ in $U(\mathfrak g)$. Since $[x_i,x_j]=0$, the defining generator $x_i\otimes x_j-x_j\otimes x_i-[x_i,x_j]$ becomes $x_i\otimes x_j-x_j\otimes x_i$, so in the quotient $U(\mathfrak g)$ we have \begin{align*} x_i x_j-x_j x_i=0. \end{align*} Thus $x_i x_j=x_jx_i$ for every pair $i,j$. To identify the quotient, define a unital algebra map $T(\mathfrak g)\to k[t_1,\dots,t_n]$ by sending $x_i$ to $t_i$. For every defining generator, \begin{align*} x_i\otimes x_j-x_j\otimes x_i-[x_i,x_j]\mapsto t_i t_j-t_j t_i-0=0. \end{align*} Hence the map factors through $U(\mathfrak g)$. Conversely, because the elements $x_1,\dots,x_n$ commute in $U(\mathfrak g)$, the assignment $t_i\mapsto x_i$ defines a unital algebra map $k[t_1,\dots,t_n]\to U(\mathfrak g)$. The two maps send each generator $t_i$ to $t_i$ and each generator $x_i$ to $x_i$, so they are inverse algebra homomorphisms. Hence \begin{align*} U(\mathfrak g)\cong k[t_1,\dots,t_n], \end{align*} with $x_i$ corresponding to $t_i$. Under this identification, the ordered monomials $x_1^{a_1}\cdots x_n^{a_n}$ are exactly the usual polynomial monomials, so they form the standard basis of $U(\mathfrak g)$ in the abelian case. [/example] This example shows that $U(\mathfrak g)$ interpolates between noncommutative words and commutative polynomial algebras. The definition will be useful only if every Lie algebra map from $\mathfrak g$ into an associative algebra factors through it in a controlled way, and this motivates the universal property. [quotetheorem:9355] [citeproof:9355] The hypotheses in the universal property are exactly what makes the factorisation possible. The algebra $A$ must be associative and unital because tensor words are multiplied by concatenation and the empty word must map to $1_A$. A concrete failure occurs when $\mathfrak g=0$ and $A$ is a nonzero associative algebra without identity: then $U(0)=k$, but there is no unital algebra homomorphism $k\to A$ because $A$ has no element that can serve as the image of $1$. The requirement that $\varphi$ be a Lie algebra homomorphism is also essential: if $\varphi([x,y])\ne \varphi(x)\varphi(y)-\varphi(y)\varphi(x)$ for some $x,y$, then the defining generator $x\otimes y-y\otimes x-[x,y]$ would not lie in the kernel and no map out of the quotient could exist. The theorem does not say that every associative algebra generated by a copy of $\mathfrak g$ is equal to $U(\mathfrak g)$; it says that $U(\mathfrak g)$ is initial among such algebras, so all other compatible constructions receive a unique map from it. This is the exact comparison needed for representation theory. A Lie representation is a Lie algebra map into an endomorphism algebra, and $\operatorname{End}_k(V)$ is associative and unital under composition. The universal property should therefore turn the data of a Lie action into the data of a module over $U(\mathfrak g)$, and the next theorem records this equivalence. [quotetheorem:3777] [citeproof:3777] The Lie homomorphism condition in the representation is necessary: arbitrary linear maps $\mathfrak g\to \mathfrak{gl}(V)$ do not extend to $U(\mathfrak g)$-actions unless the bracket relation is respected. For example, if $\mathfrak g=\mathfrak{sl}_2$ and operators $E,H,F$ on $V$ fail to satisfy $[H,E]=2E$, then the relation \begin{align*} he-eh-2e=0 \end{align*} in $U(\mathfrak{sl}_2)$ would act by a nonzero operator, so there is no well-defined module action. The theorem also does not identify Lie algebra representations with modules over an arbitrary associative algebra containing $\mathfrak g$; the point is that $U(\mathfrak g)$ imposes exactly the bracket relations and no extra ones. This equivalence explains why enveloping algebras enter every later construction: submodules, quotients, cyclic modules, tensor operations, and annihilator ideals become ordinary module-theoretic notions once the action has been extended. The following example shows how the abstract extension acts on the standard ordered words that will later become a PBW basis. [example: Extending a Lie Representation] Let $\rho:\mathfrak{sl}_2\to \mathfrak{gl}(V)$ be a representation with basis $e,h,f$ and brackets $[h,e]=2e$, $[h,f]=-2f$, and $[e,f]=h$. In $U(\mathfrak{sl}_2)$ the defining relations are \begin{align*} he-eh-2e=0,\qquad hf-fh+2f=0,\qquad ef-fe-h=0. \end{align*} Because $\rho$ is a Lie algebra homomorphism, the first relation acts as zero: \begin{align*} \rho(h)\rho(e)-\rho(e)\rho(h)-2\rho(e)=\rho([h,e])-\rho(2e)=0. \end{align*} The other two relations act as zero in the same way: \begin{align*} \rho(h)\rho(f)-\rho(f)\rho(h)+2\rho(f)=\rho([h,f])-\rho(-2f)=0. \end{align*} \begin{align*} \rho(e)\rho(f)-\rho(f)\rho(e)-\rho(h)=\rho([e,f])-\rho(h)=0. \end{align*} Thus the tensor-word action $x_1\otimes\cdots\otimes x_m\mapsto \rho(x_1)\cdots\rho(x_m)$ kills the defining relations of $U(\mathfrak{sl}_2)$, so it descends to the corresponding $U(\mathfrak{sl}_2)$-module action. In particular, \begin{align*} f^a h^b e^c\cdot v=\rho(f)^a\rho(h)^b\rho(e)^c(v) \end{align*} for every $v\in V$. For example, since $ef=fe+h$ in $U(\mathfrak{sl}_2)$, the two words act equally: \begin{align*} \rho(e)\rho(f)(v)=\rho(f)\rho(e)(v)+\rho(h)(v). \end{align*} So changing a word by one of the enveloping-algebra relations does not change the resulting operator on $V$. [/example] ## Filtrations and the PBW Theorem The next problem is that $U(\mathfrak g)$ is defined as a quotient of all tensor words, so its elements initially have many representatives. To use $U(\mathfrak g)$ effectively, we need a normal form for products of elements of $\mathfrak g$. The PBW theorem supplies this normal form by comparing $U(\mathfrak g)$ with the commutative symmetric algebra on the same vector space. [definition: Filtered Algebra] A filtered algebra is an associative algebra $A$ together with subspaces \begin{align*} F_0A \subset F_1A \subset F_2A \subset \cdots \subset A \end{align*} such that $A=\bigcup_{m\ge 0}F_mA$, $1_A\in F_0A$, and $(F_mA)(F_nA)\subset F_{m+n}A$ for all $m,n\ge 0$. [/definition] A filtration records the maximum degree of an element without requiring a homogeneous decomposition inside $A$. For $U(\mathfrak g)$, the relevant measure of degree is word length in the tensor algebra, which motivates the standard filtration. [definition: Standard Filtration on the Universal Enveloping Algebra] Let $\mathfrak g$ be a Lie algebra. The standard filtration on $U(\mathfrak g)$ is \begin{align*} F_mU(\mathfrak g) := \operatorname{span}\{i(x_1)\cdots i(x_r):0\le r\le m,\ x_j\in\mathfrak g\}. \end{align*} [/definition] This filtration is compatible with multiplication because multiplying a word of length at most $m$ by a word of length at most $n$ gives a word of length at most $m+n$. To compare a filtered algebra with a genuinely graded object, we need to retain only the top filtered piece of each degree, which motivates the associated graded algebra. [definition: Associated Graded Algebra] Let $A$ be a filtered algebra. Its associated graded algebra is \begin{align*} \operatorname{gr} A := \bigoplus_{m=0}^{\infty} F_mA/F_{m-1}A, \end{align*} with $F_{-1}A:=0$, and multiplication induced from multiplication in $A$. [/definition] The key point is that commutators in $U(\mathfrak g)$ lower filtration degree. Indeed, for $x,y\in\mathfrak g$, the relation $xy-yx=[x,y]$ has the RHS in filtration degree $1$, while $xy$ and $yx$ have degree $2$. This makes $\operatorname{gr}U(\mathfrak g)$ commutative and motivates comparing it with the symmetric algebra $S(\mathfrak g)$. [quotetheorem:8827] [citeproof:8827] Each hypothesis in PBW has a visible role. The Lie bracket relations are needed because without them the tensor algebra has no reason to reorder words, while the Jacobi identity is what prevents contradictory outcomes when a word such as $x_cx_bx_a$ is reordered along different paths. For a concrete obstruction, take a three-dimensional vector space with basis $x,y,z$ and an alternating bilinear operation defined by $[x,y]=z$, $[y,z]=x$, and $[z,x]=0$. Then \begin{align*} [x,[y,z]]+[y,[z,x]]+[z,[x,y]]= [x,x]+[y,0]+[z,z]=0, \end{align*} but modifying the operation to $[x,y]=y$, $[y,z]=z$, and $[z,x]=x$ gives \begin{align*} [x,[y,z]]+[y,[z,x]]+[z,[x,y]]= [x,z]+[y,x]+[z,y]=-x-y-z\ne 0. \end{align*} For this second operation, attempting to impose the same commutator relations in a tensor quotient would make different reorderings of $zyx$ disagree by a nonzero degree-one correction, so the PBW normal-form argument has no well-defined target. The ordered basis is only a device for writing a concrete basis: changing the order changes the displayed monomials but not the intrinsic graded statement $S(\mathfrak g)\cong\operatorname{gr}U(\mathfrak g)$. PBW does not say that $U(\mathfrak g)$ is commutative or isomorphic as an algebra to $S(\mathfrak g)$; for instance $U(\mathfrak{sl}_2)$ has $he-eh=2e$, whereas $S(\mathfrak{sl}_2)$ is commutative. What it says is that the noncommutativity only appears in lower filtration degree, which is why representation theory can combine noncommutative module arguments with polynomial-style dimension and character counts. The theorem becomes concrete in rank one, where every word in the standard generators can be reordered into a fixed triangular shape. [example: Ordered Monomials In Universal Enveloping Algebra Of Special Linear Two] For $\mathfrak{sl}_2$ with ordered basis $f<h<e$, the *Poincare Birkhoff Witt Theorem* says that the ordered monomials \begin{align*} f^a h^b e^c,\qquad a,b,c\ge 0, \end{align*} form a basis of $U(\mathfrak{sl}_2)$. The defining relations are \begin{align*} he-eh-2e=0,\qquad hf-fh+2f=0,\qquad ef-fe-h=0. \end{align*} Solving these for the out-of-order adjacent pairs $eh$, $hf$, and $ef$ gives \begin{align*} eh=he-2e. \end{align*} \begin{align*} hf=fh-2f. \end{align*} \begin{align*} ef=fe+h. \end{align*} For example, the word $ehf$ is not ordered because $e$ appears before both $h$ and $f$. First rewrite the left adjacent pair $eh$: \begin{align*} ehf=(he-2e)f. \end{align*} By distributivity, \begin{align*} (he-2e)f=hef-2ef. \end{align*} Now rewrite $ef$ in each occurrence: \begin{align*} hef=h(fe+h). \end{align*} \begin{align*} 2ef=2(fe+h). \end{align*} Using distributivity again, \begin{align*} h(fe+h)-2(fe+h)=hfe+h^2-2fe-2h. \end{align*} Rewrite the remaining out-of-order pair $hf$ inside $hfe$: \begin{align*} hfe=(fh-2f)e. \end{align*} Distributivity gives \begin{align*} (fh-2f)e=fhe-2fe. \end{align*} Substituting this into the previous expression gives \begin{align*} ehf=fhe+h^2-4fe-2h. \end{align*} Each term is now a scalar multiple of an ordered monomial $f^a h^b e^c$. The same adjacent-pair rewrites move every word into ordered form, and PBW says the resulting ordered monomials are not merely spanning words but a basis. [/example] PBW also confirms that the original Lie algebra embeds into its enveloping algebra. The construction of $U(\mathfrak g)$ defines a map $i:\mathfrak g\to U(\mathfrak g)$, but injectivity is not immediate from the quotient definition; this motivates isolating the embedding consequence. [quotetheorem:9358] [citeproof:9358] The use of PBW here is essential, because quotient maps can collapse degree-one elements in general. For example, quotienting a tensor algebra by an ideal containing a nonzero vector $x\in V$ would send that generator to zero, so the existence of a quotient presentation alone gives no injectivity. PBW rules out this collapse by showing that degree-one ordered monomials remain part of a basis. The theorem does not say that $\mathfrak g$ is a subalgebra of $U(\mathfrak g)$ for the associative multiplication; it is a Lie subalgebra for the commutator bracket. From this point onward, we identify $\mathfrak g$ with its image in $U(\mathfrak g)$, and this identification is what allows later formulas to write $xv$ for the action of a Lie algebra element on a module without repeatedly mentioning the canonical map $i$. It also lets PBW be applied directly to structural decompositions of $\mathfrak g$, which is the step needed next for semisimple Lie algebras and their triangular decompositions. ## Triangular Factorisation for Semisimple Lie Algebras The final problem in this chapter is to adapt PBW to the structure theory of a complex semisimple Lie algebra. Highest weight theory depends on separating lowering operators, Cartan elements, and raising operators. The triangular decomposition of $U(\mathfrak g)$ is the algebraic form of that separation. Let $\mathfrak g$ be a finite-dimensional complex semisimple Lie algebra with Cartan subalgebra $\mathfrak h$, root system $\Phi\subset\mathfrak h^*$, choice of positive roots $\Phi^+$, and nilpotent subalgebras \begin{align*} \mathfrak n^+ := \bigoplus_{\alpha\in\Phi^+}\mathfrak g_\alpha, \qquad \mathfrak n^- := \bigoplus_{\alpha\in\Phi^+}\mathfrak g_{-\alpha}. \end{align*} Then the vector space decomposition is \begin{align*} \mathfrak g = \mathfrak n^- \oplus \mathfrak h \oplus \mathfrak n^+. \end{align*} To make PBW reflect this decomposition, the basis order must put the three summands in the same sequence that highest weight theory uses. This motivates the triangular order. [definition: Triangular Order] Choose ordered bases of $\mathfrak n^-$, $\mathfrak h$, and $\mathfrak n^+$. The triangular order on a basis of $\mathfrak g$ places all basis elements of $\mathfrak n^-$ first, then all basis elements of $\mathfrak h$, then all basis elements of $\mathfrak n^+$. [/definition] With this order, PBW monomials first use negative root vectors, then Cartan elements, then positive root vectors. This motivates the triangular PBW factorisation: the vector space decomposition of $\mathfrak g$ should become a factorisation statement for $U(\mathfrak g)$, in the form needed for Verma modules and highest weight vectors. [quotetheorem:9359] [citeproof:9359] The triangular hypotheses are doing more than naming three convenient subspaces. The decomposition $\mathfrak g=\mathfrak n^-\oplus\mathfrak h\oplus\mathfrak n^+$ ensures that a triangularly ordered basis is actually a basis of all of $\mathfrak g$; without a direct-sum decomposition, the proposed tensor product basis would either miss directions or count them twice. The multiplication order is also essential: in $U(\mathfrak{sl}_2)$, the product $ef$ is not already triangular, and the relation $ef=fe+h$ shows exactly how commuting factors creates lower triangular terms. The factorisation is a vector space statement, not a claim that $U(\mathfrak g)$ is the direct product or tensor product algebra of the three enveloping algebras, since the subalgebras do not commute with one another. Its forward role is to separate lowering, weight-measuring, and raising operators, which is the algebraic mechanism behind Verma modules, highest weight vectors, and character computations. In the smallest semisimple example, the statement recovers the ordered monomials already seen above. [example: Triangular Factorisation For Special Linear Two] For $\mathfrak{sl}_2$, take \begin{align*} \mathfrak n^- = \mathbb C f, \qquad \mathfrak h = \mathbb C h, \qquad \mathfrak n^+ = \mathbb C e. \end{align*} Choose the triangular order $f<h<e$. Since each of $\mathfrak n^-$, $\mathfrak h$, and $\mathfrak n^+$ is one-dimensional and abelian, their enveloping algebras have bases \begin{align*} 1,f,f^2,\ldots \quad;\quad 1,h,h^2,\ldots \quad;\quad 1,e,e^2,\ldots . \end{align*} Under the multiplication map \begin{align*} U(\mathfrak n^-)\otimes U(\mathfrak h)\otimes U(\mathfrak n^+) \longrightarrow U(\mathfrak{sl}_2), \end{align*} the elementary tensor $f^a\otimes h^b\otimes e^c$ is sent to \begin{align*} f^a h^b e^c. \end{align*} By the *Poincare Birkhoff Witt Theorem* with the order $f<h<e$, the monomials \begin{align*} f^a h^b e^c,\qquad a,b,c\ge 0, \end{align*} form a basis of $U(\mathfrak{sl}_2)$. Therefore the multiplication map sends the tensor product basis of $U(\mathfrak n^-)\otimes U(\mathfrak h)\otimes U(\mathfrak n^+)$ bijectively onto a basis of $U(\mathfrak{sl}_2)$, so every element of $U(\mathfrak{sl}_2)$ has a unique finite expression \begin{align*} \sum_{a,b,c} \lambda_{a,b,c} f^a h^b e^c,\qquad \lambda_{a,b,c}\in \mathbb C. \end{align*} This is the rank-one triangular factorisation: in highest weight modules, the factors $e$, $h$, and $f$ appear in the algebraic order that separates raising operators, weight-measuring operators, and lowering operators. [/example] This chapter leaves us with the main algebraic language for the course. From now on, a representation of $\mathfrak g$ may be treated as a $U(\mathfrak g)$-module, PBW gives normal forms inside $U(\mathfrak g)$, and the triangular factorisation isolates the three pieces used in highest weight theory. The next chapter begins by using $\mathfrak h$ to decompose representations into weight spaces and to package them into formal characters. # 2. Weights and Characters This chapter introduces the bookkeeping language that turns representations of a semisimple Lie algebra into finite combinatorial objects. Using the PBW theorem and triangular factorisation from Chapter 1, the Cartan subalgebra $\mathfrak h$ becomes the part of $\mathfrak g$ that can be diagonalised, while the root spaces describe how the remaining generators move between eigenspaces. The central questions are: which linear functionals occur as weights, how are they constrained by the roots, and why do characters remember enough structure to guide classification? ## Weight Spaces and Formal Characters A representation of $\mathfrak g$ is usually too large to understand generator by generator. The first simplification is to restrict the action to the abelian Cartan subalgebra $\mathfrak h$, where simultaneous eigenspaces can be used to decompose the module. [definition: Weight Space] Let $\mathfrak g$ be a finite-dimensional complex semisimple Lie algebra with Cartan subalgebra $\mathfrak h$, and let $V$ be a $\mathfrak g$-module. For $\lambda \in \mathfrak h^*$, the $\lambda$-weight space of $V$ is \begin{align*} V_\lambda := \{v \in V : h\cdot v = \lambda(h)v \text{ for all } h \in \mathfrak h\}. \end{align*} A weight of $V$ is a functional $\lambda \in \mathfrak h^*$ such that $V_\lambda \ne 0$. [/definition] The definition isolates simultaneous eigenvectors, but by itself it does not say whether enough of them exist to recover the whole module. Since later arguments read a representation from its weight spaces, we need a decomposition theorem that rules out missing generalized eigenspace pieces in the finite-dimensional semisimple setting. [quotetheorem:9360] [citeproof:9360] This theorem is the first point at which representation theory becomes discrete: a module is replaced by a finite multiset of weights, each counted with its multiplicity. The finite-dimensional hypothesis matters; for example, an infinite-dimensional highest-weight module for $\mathfrak{sl}_2(\mathbb C)$ may have infinitely many nonzero weight spaces, so its weight data is no longer a finite multiset. The semisimple Lie-algebra setting matters as well, because the proof uses the triangular decomposition and the PBW spanning argument from a highest vector. The theorem does not classify which finite multisets can occur; it only says that finite-dimensional modules can be read through their weight multiplicities, which is the data encoded next by formal characters. [definition: Formal Character] Let $\mathcal C_{\mathrm{fd}}(\mathfrak g,\mathfrak h)$ be the class of finite-dimensional $\mathfrak g$-modules admitting a weight space decomposition with respect to $\mathfrak h$, and let $L_V\subset\mathfrak h^*$ be the additive subgroup generated by the weights of $V$. The formal character construction is the map \begin{align*} \operatorname{ch}:\mathcal C_{\mathrm{fd}}(\mathfrak g,\mathfrak h)\longrightarrow \bigcup_V \mathbb Z[L_V], \end{align*} defined on a module $V=\bigoplus_\lambda V_\lambda$ by \begin{align*} \operatorname{ch} V := \sum_{\lambda \in \mathfrak h^*} (\dim V_\lambda)e^\lambda, \end{align*} where $e^\lambda$ denotes the basis element of the group algebra $\mathbb Z[L_V]$ corresponding to $\lambda$. [/definition] The notation $e^\lambda$ is formal: it is designed so that multiplication satisfies $e^\lambda e^\mu=e^{\lambda+\mu}$. This makes characters compatible with tensor products. [example: Characters and Tensor Products] Let $V=\bigoplus_\lambda V_\lambda$ and $W=\bigoplus_\mu W_\mu$. For $v\in V_\lambda$, $w\in W_\mu$, and $h\in\mathfrak h$, the tensor product action gives \begin{align*} h\cdot(v\otimes w)=(h\cdot v)\otimes w+v\otimes(h\cdot w)=\lambda(h)(v\otimes w)+\mu(h)(v\otimes w)=(\lambda+\mu)(h)(v\otimes w). \end{align*} Thus $V_\lambda\otimes W_\mu\subset (V\otimes W)_{\lambda+\mu}$. Since $V$ and $W$ are direct sums of their weight spaces, linear algebra gives \begin{align*} V\otimes W=\bigoplus_{\lambda,\mu} V_\lambda\otimes W_\mu. \end{align*} The summands with $\lambda+\mu=\nu$ are exactly the summands on which every $h\in\mathfrak h$ acts by the scalar $\nu(h)$, so \begin{align*} (V\otimes W)_\nu=\bigoplus_{\lambda+\mu=\nu} V_\lambda\otimes W_\mu. \end{align*} Taking dimensions gives \begin{align*} \dim (V\otimes W)_\nu=\sum_{\lambda+\mu=\nu} \dim(V_\lambda\otimes W_\mu)=\sum_{\lambda+\mu=\nu}(\dim V_\lambda)(\dim W_\mu). \end{align*} Therefore \begin{align*} \operatorname{ch}(V\otimes W)=\sum_\nu\left(\sum_{\lambda+\mu=\nu}(\dim V_\lambda)(\dim W_\mu)\right)e^\nu. \end{align*} On the other hand, multiplication in the group algebra satisfies $e^\lambda e^\mu=e^{\lambda+\mu}$, so \begin{align*} (\operatorname{ch}V)(\operatorname{ch}W)=\left(\sum_\lambda(\dim V_\lambda)e^\lambda\right)\left(\sum_\mu(\dim W_\mu)e^\mu\right)=\sum_{\lambda,\mu}(\dim V_\lambda)(\dim W_\mu)e^{\lambda+\mu}. \end{align*} Grouping the last sum by $\nu=\lambda+\mu$ gives the same coefficient of every $e^\nu$, hence $\operatorname{ch}(V\otimes W)=(\operatorname{ch}V)(\operatorname{ch}W)$. This is why formal characters turn tensor products of modules into products in the group algebra. [/example] This example explains why formal characters are more useful than lists of weights: they convert representation-theoretic operations into algebraic operations in a group algebra. The next obstruction is that the formal symbols $e^\lambda$ are only useful when the possible exponents live in a controlled lattice. The adjoint action supplies the directions that control this lattice, because non-Cartan elements of $\mathfrak g$ do not preserve a weight space; they move it along root directions. ## Roots and Integral Weights The next question is which linear functionals on $\mathfrak h$ are allowed to occur in finite-dimensional representation theory. The roots of $\mathfrak g$ already live in $\mathfrak h^*$ through the adjoint action, and they impose arithmetic restrictions on all finite-dimensional weights. [definition: Root] Let $\mathfrak g$ be a finite-dimensional complex semisimple Lie algebra with Cartan subalgebra $\mathfrak h$. For $\alpha\in\mathfrak h^*$, the corresponding root space is \begin{align*} \mathfrak g_\alpha := \{x\in \mathfrak g : [h,x]=\alpha(h)x \text{ for all } h\in\mathfrak h\}. \end{align*} A root is a nonzero $\alpha\in\mathfrak h^*$ such that $\mathfrak g_\alpha\ne 0$. [/definition] The adjoint representation is the model example: it shows that roots are weights, but with the zero-weight space $\mathfrak h$ also present. [example: Adjoint Representation Weights] For the adjoint representation, the underlying vector space is $V=\mathfrak g$ and the action is $h\cdot x=[h,x]$. If $\alpha\ne 0$, then the $\alpha$-weight space is \begin{align*} V_\alpha=\{x\in\mathfrak g:[h,x]=\alpha(h)x\text{ for all }h\in\mathfrak h\}. \end{align*} By the definition of the root space, the right-hand side is exactly $\mathfrak g_\alpha$. Hence the nonzero weights of the adjoint representation are precisely the roots $\alpha\in\Phi$, with multiplicity $\dim\mathfrak g_\alpha$. For the zero weight, the weight-space condition is \begin{align*} V_0=\{x\in\mathfrak g:[h,x]=0\text{ for all }h\in\mathfrak h\}. \end{align*} Since $\mathfrak h$ is abelian, every $x\in\mathfrak h$ satisfies $[h,x]=0$ for all $h\in\mathfrak h$, so $\mathfrak h\subseteq V_0$. For a Cartan subalgebra in a complex semisimple Lie algebra, the zero eigenspace of the adjoint action of $\mathfrak h$ is $\mathfrak h$ itself, so $V_0=\mathfrak h$. Therefore the formal character is \begin{align*} \operatorname{ch}\mathfrak g=(\dim\mathfrak h)e^0+\sum_{\alpha\in\Phi}(\dim\mathfrak g_\alpha)e^\alpha. \end{align*} For complex semisimple $\mathfrak g$, each root space $\mathfrak g_\alpha$ is one-dimensional, so every nonzero adjoint weight occurs with multiplicity $1$. [/example] The example shows that roots are not external decoration: they are weights in the adjoint module and directions along which other modules will move. To compare a general weight with a root direction, we need the corresponding Cartan element inside the embedded $\mathfrak{sl}_2$ subalgebra. [definition: Coroot] The coroot assignment is the map \begin{align*} \Phi \longrightarrow \mathfrak h,\qquad \alpha\longmapsto \alpha^\vee. \end{align*} For a root $\alpha$ of $\mathfrak g$, choose a standard $\mathfrak{sl}_2$-triple $(e_\alpha,h_\alpha,f_\alpha)$ with $e_\alpha\in\mathfrak g_\alpha$, $f_\alpha\in\mathfrak g_{-\alpha}$, and $[e_\alpha,f_\alpha]=h_\alpha$. The coroot is $\alpha^\vee:=h_\alpha$, normalised relative to this triple by \begin{align*} \alpha(\alpha^\vee)=2. \end{align*} [/definition] The coroot packages the rank-one Cartan direction attached to $\alpha$, so evaluating a weight on $\alpha^\vee$ gives the integer-like coordinate seen by that $\mathfrak{sl}_2$ subalgebra. This motivates separating the linear functionals that pass all these rank-one arithmetic tests from those that cannot occur in finite-dimensional modules. [definition: Integral Weight] A functional $\lambda\in\mathfrak h^*$ is an integral weight if \begin{align*} \lambda(\alpha^\vee)\in\mathbb Z \end{align*} for every root $\alpha$. [/definition] The definition gives a condition on a single functional, but representation theory needs a common domain for all characters and highest weights. Without such a domain, two characters might be written in different group algebras even when they belong to modules for the same Lie algebra, making addition, tensor products, and Weyl group actions ambiguous. We therefore collect all integral weights into one lattice, which will be the coefficient lattice for the formal exponentials appearing in finite-dimensional characters. [definition: Weight Lattice] The weight lattice of $\mathfrak g$ is \begin{align*} P := \{\lambda\in\mathfrak h^* : \lambda(\alpha^\vee)\in\mathbb Z \text{ for every root } \alpha\}. \end{align*} [/definition] The lattice $P$ is still too large to classify irreducible finite-dimensional modules directly, because it contains weights pointing in many Weyl-equivalent directions. After choosing positive roots, we need a preferred chamber inside $P$; this is obtained by imposing nonnegative pairings with the simple coroots. [definition: Dominant Weight] Fix a set of simple roots $\Delta\subset\Phi$. An integral weight $\lambda\in P$ is dominant if \begin{align*} \lambda(\alpha^\vee)\ge 0 \end{align*} for every $\alpha\in\Delta$. [/definition] Dominance singles out the chamber that will later contain highest weights, while integrality is a constraint on every weight appearing anywhere in a finite-dimensional module. We now justify that this lattice condition is not optional: it is forced locally by restricting to the $\mathfrak{sl}_2$ subalgebra attached to each root. [quotetheorem:9361] [citeproof:9361] This result turns the collection of possible weights from a complex vector space into a lattice. The hypothesis that $V$ is finite-dimensional is essential: Verma modules for $\mathfrak{sl}_2$ may have arbitrary complex highest weight, so the Cartan eigenvalues need not be integral. The theorem also does not say that every integral weight occurs in a given module, or even that every integral weight is dominant; it only gives a necessary arithmetic condition on weights that already occur. This necessary condition is the first filter on possible characters, and the examples already familiar from matrices show the lattice in concrete coordinates. [example: Natural Representation of Special Linear N] Let $\mathfrak g=\mathfrak{sl}_n(\mathbb C)$ act on $V=\mathbb C^n$ by matrix multiplication, and let $\mathfrak h$ be the diagonal trace-zero subalgebra. For \begin{align*} h=\operatorname{diag}(a_1,\dots,a_n)\in\mathfrak h, \end{align*} we have $\sum_i a_i=0$, and the standard basis vector $e_i$ satisfies \begin{align*} h\cdot e_i=\operatorname{diag}(a_1,\dots,a_n)e_i=a_i e_i. \end{align*} Define $\varepsilon_i\in\mathfrak h^*$ by \begin{align*} \varepsilon_i(\operatorname{diag}(a_1,\dots,a_n))=a_i. \end{align*} Then $h\cdot e_i=\varepsilon_i(h)e_i$ for every $h\in\mathfrak h$, so $e_i\in V_{\varepsilon_i}$. Since $e_1,\dots,e_n$ form a basis of $V$, the weight decomposition is \begin{align*} V=V_{\varepsilon_1}\oplus\cdots\oplus V_{\varepsilon_n}, \end{align*} with $V_{\varepsilon_i}=\mathbb C e_i$ and $\dim V_{\varepsilon_i}=1$. Therefore \begin{align*} \operatorname{ch}V=e^{\varepsilon_1}+\cdots+e^{\varepsilon_n}. \end{align*} The roots are read from the adjoint action on elementary matrices. For $i\ne j$, let $E_{ij}$ be the matrix with a $1$ in position $(i,j)$ and zeros elsewhere. Multiplying diagonal matrices with $E_{ij}$ gives \begin{align*} hE_{ij}=a_iE_{ij}. \end{align*} Similarly, \begin{align*} E_{ij}h=a_jE_{ij}. \end{align*} Hence \begin{align*} [h,E_{ij}]=hE_{ij}-E_{ij}h=(a_i-a_j)E_{ij}. \end{align*} Because $(\varepsilon_i-\varepsilon_j)(h)=a_i-a_j$, this says \begin{align*} [h,E_{ij}]=(\varepsilon_i-\varepsilon_j)(h)E_{ij}. \end{align*} Thus $E_{ij}$ lies in the root space for $\varepsilon_i-\varepsilon_j$, and the nonzero roots of $\mathfrak{sl}_n(\mathbb C)$ are precisely the functionals $\varepsilon_i-\varepsilon_j$ with $i\ne j$. [/example] This coordinate picture shows individual weights and also records a useful computational warning: the symbols $\varepsilon_i$ are restricted to the trace-zero hyperplane, so $\varepsilon_1+\cdots+\varepsilon_n=0$ as a functional on $\mathfrak h$. Thus weights for $\mathfrak{sl}_n$ are often computed by first using diagonal entries and then remembering the trace-zero relation. It also raises a dynamic question: what happens when we apply a Lie algebra element outside $\mathfrak h$ to a weight vector? Since the nonzero root spaces are the remaining pieces of $\mathfrak g$, the answer should say that root vectors move weight spaces in root directions. [quotetheorem:9362] [citeproof:9362] This theorem explains why the weight diagram of a module is organised along root directions. Each hypothesis has a role. The vector must start in an actual weight space: if $v=v_++v_-$ in the two-dimensional $\mathfrak{sl}_2(\mathbb C)$-module with weights $1$ and $-1$, then $v$ is not a weight vector, and applying $e$ gives a vector in only one of the possible shifted spaces rather than a statement attached to a single starting weight. The operator must lie in one root space: the element $e+f$ is not in a single root space and sends $v_++v_-$ to a sum of components in different weight spaces, so there is no fixed shift $\alpha$. The Cartan weight-space condition on $V_\lambda$ is also essential; for a module whose $\mathfrak h$-action has a non-diagonal Jordan block, generalized eigenvectors do not satisfy the displayed calculation as eigenvectors. Finally, the conclusion is only an inclusion because a root operator may kill a weight vector, as the raising operator does on a highest-weight vector in an $\mathfrak{sl}_2$ representation. This local shifting rule is the mechanism behind root strings, highest weights, and later character formulas. The simplest place to see the shift theorem at work is rank one. In that case there is only one root direction, so a weight diagram becomes a single string and the raising and lowering operators move along it until they reach an endpoint. [example: Special Linear Two Weight Strings] Let $\mathfrak g=\mathfrak{sl}_2(\mathbb C)$ with basis $e,h,f$ and relations $[h,e]=2e$, $[h,f]=-2f$, and $[e,f]=h$. Realise the irreducible module of highest weight $m\in\mathbb Z_{\ge 0}$ as the space $V_m$ of homogeneous polynomials of degree $m$ in $x,y$, with basis \begin{align*} v_k=x^{m-k}y^k\qquad 0\le k\le m. \end{align*} Use the standard action \begin{align*} h=x\frac{\partial}{\partial x}-y\frac{\partial}{\partial y},\qquad e=x\frac{\partial}{\partial y},\qquad f=y\frac{\partial}{\partial x}. \end{align*} For $v_k=x^{m-k}y^k$, the Cartan element acts by \begin{align*} h\cdot v_k=x\frac{\partial}{\partial x}(x^{m-k}y^k)-y\frac{\partial}{\partial y}(x^{m-k}y^k)=(m-k)x^{m-k}y^k-kx^{m-k}y^k=(m-2k)v_k. \end{align*} Thus the weights are \begin{align*} m,\ m-2,\ m-4,\ \dots,\ -m, \end{align*} and each has multiplicity $1$, because each weight space is spanned by exactly one basis vector $v_k$. The raising and lowering operators move between adjacent basis vectors: \begin{align*} e\cdot v_k=x\frac{\partial}{\partial y}(x^{m-k}y^k)=kx^{m-k+1}y^{k-1}=k v_{k-1}. \end{align*} \begin{align*} f\cdot v_k=y\frac{\partial}{\partial x}(x^{m-k}y^k)=(m-k)x^{m-k-1}y^{k+1}=(m-k)v_{k+1}. \end{align*} Since $v_{k-1}$ has weight $m-2(k-1)=(m-2k)+2$ and $v_{k+1}$ has weight $m-2(k+1)=(m-2k)-2$, the operator $e$ raises the weight by $2$ and $f$ lowers the weight by $2$. Therefore \begin{align*} \operatorname{ch}V_m=e^m+e^{m-2}+e^{m-4}+\cdots+e^{-m}. \end{align*} The weight diagram is a single finite string, with endpoints killed by the missing adjacent basis vectors: $e\cdot v_0=0$ and $f\cdot v_m=0$. [/example] Rank-one strings remain the local model in higher rank: every root gives an $\mathfrak{sl}_2$ direction, and the global weight set is assembled from all these directions at once. The obstruction is that different root strings overlap, so there is no preferred linear ordering of all weights in higher rank. The Weyl group is the device that records the reflections forced by every rank-one string without asking the full diagram to be a single string. ## Weyl Group Symmetry of Characters The final question in this chapter is why characters have symmetries that are not visible from the definition alone. The Weyl group acts on $\mathfrak h^*$ by reflecting across root hyperplanes, and finite-dimensional representation theory forces the multiset of weights to be invariant under these reflections. [definition: Weyl Group Action on Weights] For a root $\alpha$, define the reflection $s_\alpha:\mathfrak h^*\to\mathfrak h^*$ by \begin{align*} s_\alpha(\lambda)=\lambda-\lambda(\alpha^\vee)\alpha. \end{align*} The Weyl group $W$ is the subgroup of $GL(\mathfrak h^*)$ generated by the reflections $s_\alpha$ for $\alpha\in\Phi$. [/definition] This action preserves the weight lattice because integral coroot pairings are stable under root reflections. It also preserves the root system itself, so it acts on the same combinatorial data used to build characters. [example: Weyl Group of Special Linear N] For $\mathfrak{sl}_n(\mathbb C)$, write $\mathfrak h$ for the diagonal trace-zero matrices and define $\varepsilon_i(\operatorname{diag}(a_1,\dots,a_n))=a_i$. For $i\ne j$, the root $\alpha_{ij}=\varepsilon_i-\varepsilon_j$ has coroot $E_{ii}-E_{jj}$, so \begin{align*} \varepsilon_k(\alpha_{ij}^\vee)=\varepsilon_k(E_{ii}-E_{jj})=\delta_{ki}-\delta_{kj}. \end{align*} The reflection attached to $\alpha_{ij}$ therefore sends $\varepsilon_k$ to \begin{align*} s_{\alpha_{ij}}(\varepsilon_k)=\varepsilon_k-\varepsilon_k(\alpha_{ij}^\vee)(\varepsilon_i-\varepsilon_j). \end{align*} If $k=i$, this gives $s_{\alpha_{ij}}(\varepsilon_i)=\varepsilon_i-(\varepsilon_i-\varepsilon_j)=\varepsilon_j$. If $k=j$, this gives $s_{\alpha_{ij}}(\varepsilon_j)=\varepsilon_j+\varepsilon_i-\varepsilon_j=\varepsilon_i$. If $k\ne i,j$, then $\varepsilon_k(\alpha_{ij}^\vee)=0$, so $s_{\alpha_{ij}}(\varepsilon_k)=\varepsilon_k$. Thus the root reflection $s_{\alpha_{ij}}$ swaps $\varepsilon_i$ and $\varepsilon_j$ while fixing every $\varepsilon_k$ with $k\ne i,j$. These transpositions generate $S_n$, so the Weyl group acts by permuting $\varepsilon_1,\dots,\varepsilon_n$. For the natural representation, the character is \begin{align*} \operatorname{ch}\mathbb C^n=e^{\varepsilon_1}+\cdots+e^{\varepsilon_n}. \end{align*} If $w\in W$ corresponds to a permutation $\sigma\in S_n$, then \begin{align*} w(\operatorname{ch}\mathbb C^n)=e^{\varepsilon_{\sigma(1)}}+\cdots+e^{\varepsilon_{\sigma(n)}}. \end{align*} The last sum contains the same formal basis terms as $e^{\varepsilon_1}+\cdots+e^{\varepsilon_n}$, only in a different order, so $w(\operatorname{ch}\mathbb C^n)=\operatorname{ch}\mathbb C^n$. This gives a quick obstruction to proposed characters. Suppose $n\ge 3$ and apply the transposition swapping $\varepsilon_2$ and $\varepsilon_3$ to $e^{\varepsilon_1}+e^{\varepsilon_2}$. The result is \begin{align*} e^{\varepsilon_1}+e^{\varepsilon_3}. \end{align*} The functionals $\varepsilon_2$ and $\varepsilon_3$ are distinct on $\mathfrak h$, for example on $h=\operatorname{diag}(0,1,-1,0,\dots,0)$ we have $\varepsilon_2(h)=1$ and $\varepsilon_3(h)=-1$. Hence $e^{\varepsilon_2}$ and $e^{\varepsilon_3}$ are distinct basis elements in the group algebra, so \begin{align*} e^{\varepsilon_1}+e^{\varepsilon_2}\ne e^{\varepsilon_1}+e^{\varepsilon_3}. \end{align*} Thus $e^{\varepsilon_1}+e^{\varepsilon_2}$ fails Weyl invariance, which rules it out as the character of a finite-dimensional $\mathfrak{sl}_n(\mathbb C)$-module in the Weyl-invariant setting. [/example] The example shows Weyl symmetry in the easiest character, but the classification program needs this symmetry for every finite-dimensional semisimple module. By the weight-space decomposition theorem, such a module has a formal character, and the same rank-one string symmetry used for $\mathfrak{sl}_2$ weights applies along each root direction. Thus each simple reflection should preserve the full formal character. [quotetheorem:9357] [citeproof:9357] The theorem is a structural constraint on any proposed character, and its hypotheses are not cosmetic. [Finite dimensionality](/theorems/1534) is essential because infinite highest-weight modules need not have characters invariant under reflection; an $\mathfrak{sl}_2$ Verma module has weights descending from its highest weight without the reflected upward half. The complex semisimple setting supplies the root space decomposition, the embedded rank-one subalgebras $\mathfrak s_\alpha\cong\mathfrak{sl}_2$, and the Weyl group generated by their reflections; for a general Lie algebra with no such root datum, the statement has no comparable meaning. The Cartan weight decomposition is also required: the expression $\operatorname{ch}V$ records ordinary eigenspaces, so modules with non-diagonalisable Cartan action are not measured by this character without replacing the setup. The semisimplicity assumption matches the chapter's decomposition framework and guarantees that $V$ is assembled from finite-dimensional irreducibles whose weight spaces behave as in the proof; over complex semisimple $\mathfrak g$, Weyl's theorem means this assumption is automatic for finite-dimensional modules, while outside that setting extensions can carry data invisible to a character. The theorem also does not say that every Weyl-invariant finite sum with nonnegative coefficients is a character of a module; dominance and highest-weight theory still have to identify which invariant sums come from irreducibles and their direct sums. Conceptually, this is the bridge from linear representation theory to the geometry of Weyl chambers and, later, to character formulas where symmetric functions encode the same reflection constraints. In type $A$, for example, characters of polynomial representations become symmetric polynomials, so the Weyl group symmetry seen here is the representation-theoretic source of the symmetry familiar from Schur functions. [remark: Characters Are Not Complete in General] For finite-dimensional semisimple $\mathfrak g$-modules, the character determines the isomorphism class because the module is a [direct sum](/page/Direct%20Sum) of irreducibles and highest weights can be read from the character recursively. For arbitrary modules, especially infinite-dimensional modules, the character may fail to exist as a finite formal sum or may not determine extension data. [/remark] The chapter has therefore translated modules into weight-lattice combinatorics. The next stage of the course uses a choice of positive roots to single out highest weights, constructs universal highest weight modules, and then isolates the finite-dimensional irreducibles among them. # 3. The $\mathfrak{sl}_2$ Template The previous chapter introduced weights and characters as bookkeeping devices for representations of semisimple Lie algebras. This chapter assumes the basic language of Lie algebra modules, eigenspaces of a linear operator, tensor products of modules, and finite-dimensional linear algebra. It studies the smallest non-abelian semisimple example, $\mathfrak{sl}_2(\mathbb C)$, in enough detail that it becomes a template for the general theory. The main questions are classification, computation, and tensor products: which finite-dimensional irreducible modules occur, what are their characters, and how do tensor products split? ## Highest Weights and Irreducible Modules The first problem is to recognize an irreducible finite-dimensional module from a single vector. For $\mathfrak{sl}_2(\mathbb C)$, the Cartan subalgebra is one-dimensional, so a highest weight is just a complex number; finite-dimensionality will force this number to be a non-negative integer. [definition: Standard Basis of Sl Two] Let $\mathfrak{sl}_2(\mathbb C)$ be the Lie algebra of $2 \times 2$ complex matrices with trace $0$. Let $E_{ij}$ denote the matrix with $1$ in entry $(i,j)$ and $0$ elsewhere. Its standard basis is \begin{align*} e = E_{12}, \qquad f = E_{21}, \qquad h = E_{11}-E_{22}. \end{align*} [/definition] For this basis, direct matrix multiplication gives the commutation relations \begin{align*} [h,e] = 2e, \qquad [h,f] = -2f, \qquad [e,f] = h. \end{align*} These relations say that $e$ raises $h$-weights by $2$, while $f$ lowers them by $2$. This is the first appearance of the root string picture that later becomes the general root-string structure for semisimple Lie algebras. [definition: Weight Vector] Let $V$ be a $\mathfrak{sl}_2(\mathbb C)$-module. A non-zero vector $v \in V$ is a weight vector of weight $\lambda \in \mathbb C$ if \begin{align*} hv = \lambda v. \end{align*} The corresponding weight space is \begin{align*} V_\lambda = \{v \in V : hv = \lambda v\}. \end{align*} [/definition] Once weights are present, the finite-dimensional condition gives endpoints: a raising process cannot continue forever inside a finite-dimensional vector space. This makes the following notion the natural starting point for constructing modules. [definition: Highest Weight Vector] Let $V$ be a $\mathfrak{sl}_2(\mathbb C)$-module. A non-zero vector $v \in V$ is a highest weight vector of highest weight $m \in \mathbb C$ if \begin{align*} hv = mv, \qquad ev = 0. \end{align*} [/definition] A highest weight vector generates the whole cyclic module obtained by applying powers of $f$, but it is not yet known which highest weights can occur in finite dimension. Without a finite-dimensional endpoint, the same highest-weight construction produces the infinite family of vectors $(f^i v_0)_{i\ge 0}$ rather than an irreducible finite-dimensional module. The next theorem isolates the precise obstruction: the lowering string must stop at the moment when the coefficient in the raising formula vanishes. [quotetheorem:9363] [citeproof:9363] The theorem turns the classification problem into a single integer parameter. Finite-dimensionality is essential: if $m\in \mathbb C$ is arbitrary, the Verma module generated by $v_0$ with $ev_0=0$ and $hv_0=mv_0$ has basis $\{f^i v_0:i\ge 0\}$ unless a quotient is imposed, so the conclusion $m\in\mathbb Z_{\ge0}$ fails outside the finite-dimensional setting. Irreducibility is also essential, since $L(1)\oplus L(1)$ has highest weight $1$ but is not a single copy of $L(1)$. The theorem does not classify arbitrary cyclic highest-weight modules or infinite-dimensional representations; it classifies the finite-dimensional irreducible building blocks that the rest of the chapter will assemble into larger modules. [example: Symmetric Powers of the Natural Module] Let $m\in \mathbb Z_{\ge 0}$, and let the $\mathfrak{sl}_2(\mathbb C)$-action on $\operatorname{Sym}^m(W)$ be the induced derivation action, so $X(uv)=(Xu)v+u(Xv)$. Since $hx=x$ and $hy=-y$, each monomial $x^{m-i}y^i$ satisfies \begin{align*}h(x^{m-i}y^i)=(m-i)x^{m-i}y^i-i x^{m-i}y^i=(m-2i)x^{m-i}y^i.\end{align*} Also $ex=0$ and $ey=x$, so \begin{align*}e(x^m)=m x^{m-1}ex=0.\end{align*} Thus $x^m$ is a highest weight vector of weight $m$. The lowering and raising actions on the monomial basis are explicit: \begin{align*}f(x^{m-i}y^i)=(m-i)x^{m-i-1}y^{i+1}+i x^{m-i}y^{i-1}fy=(m-i)x^{m-i-1}y^{i+1}.\end{align*} \begin{align*}e(x^{m-i}y^i)=(m-i)x^{m-i-1}ex\cdot y^i+i x^{m-i}y^{i-1}ey=i x^{m-i+1}y^{i-1}.\end{align*} Define \begin{align*}v_i=\frac{m!}{(m-i)!}x^{m-i}y^i \qquad 0\le i\le m.\end{align*} Then \begin{align*}hv_i=(m-2i)v_i.\end{align*} For $0\le i<m$, \begin{align*}fv_i=\frac{m!}{(m-i)!}(m-i)x^{m-i-1}y^{i+1}=\frac{m!}{(m-i-1)!}x^{m-i-1}y^{i+1}=v_{i+1},\end{align*} and $fv_m=0$. For $1\le i\le m$, \begin{align*}ev_i=\frac{m!}{(m-i)!}i x^{m-i+1}y^{i-1}=i(m-i+1)\frac{m!}{(m-i+1)!}x^{m-i+1}y^{i-1}=i(m-i+1)v_{i-1},\end{align*} and $ev_0=0$. These are exactly the basis formulas in *Sl Two Highest Weight Classification*, so $\operatorname{Sym}^m(\mathbb C^2)$ realizes $L(m)$. [/example] This example is more than a model: it gives a concrete realisation of every finite-dimensional irreducible module. In general semisimple Lie theory, the analogue is that dominant integral highest weights classify irreducibles, but the modules are no longer all symmetric powers of one natural representation. ## Raising Operators, Lowering Operators, and Weight Strings The next question is how much of a finite-dimensional module can be read directly from the operators $e$, $f$, and $h$. The answer is that finite-dimensional $\mathfrak{sl}_2$-modules break into strings whose weights differ by $2$, and each string is governed by the same coefficients as in $L(m)$. [quotetheorem:9364] [citeproof:9364] This result explains why the diagrams for $\mathfrak{sl}_2$ modules are arranged in chains of weights. The module hypothesis is needed because the expressions $ev$, $fv$, and $hv$ only make sense after $e$, $f$, and $h$ act linearly on the same vector space. The weight-vector hypothesis is also doing real work: if $v$ is not an $h$-eigenvector, then $hv$ is not a scalar multiple of $v$, so the calculation cannot assign a single shifted weight to $ev$ or $fv$. The statement is only an inclusion statement: it does not say that a weight space has dimension one, and it does not say that a module is a direct sum of chains. For example, $L(1)\oplus L(1)$ has two-dimensional weight spaces at weights $1$ and $-1$, while the theorem only records where $e$ and $f$ can send those spaces. The next definition packages the special case where a single highest weight vector is followed downward until the lowering process terminates. [definition: Weight String] Let $V$ be a $\mathfrak{sl}_2(\mathbb C)$-module. A weight string generated by a highest weight vector $v_0$ of weight $m$ is a finite indexed family \begin{align*} \{v_i : 0\le i\le r\} \end{align*} where $v_i=f^i v_0$ is non-zero for $0\le i\le r$ and $f^{r+1}v_0=0$. [/definition] The classification theorem says that irreducible strings have length $m+1$ and weights $m-2i$ for $0\le i\le m$. Reducible modules may contain several such strings, but a list of strings does not by itself rule out extension data between them. The obstruction to a clean decomposition is that a submodule might have a quotient isomorphic to another irreducible without admitting a complementary submodule; complete reducibility is the assertion that this obstruction disappears in the finite-dimensional $\mathfrak{sl}_2(\mathbb C)$ setting. [quotetheorem:3817] [citeproof:3817] This is the concrete shadow of [Weyl's complete reducibility theorem](/theorems/3755). Finite-dimensionality is essential: Verma modules for $\mathfrak{sl}_2(\mathbb C)$ are infinite-dimensional highest-weight modules and have proper submodules in special integral cases, so they need not split as direct sums of finite strings. The theorem also does not say that an arbitrary Lie algebra has semisimple finite-dimensional representation theory; for a one-dimensional abelian Lie algebra, a single Jordan block gives a finite-dimensional indecomposable module that is not a direct sum of one-dimensional eigenspaces. The result supplies the decomposition principle needed for characters to become a complete computational tool in the next section. [example: Decomposing a Four-Dimensional Module by Strings] Let $v_0$ be a highest weight vector of weight $3$, so $hv_0=3v_0$ and $ev_0=0$. We compute the lowering string generated by $v_0$. Using $[h,f]=-2f$, \begin{align*}h(fv_0)=fhv_0+[h,f]v_0=3fv_0-2fv_0=fv_0.\end{align*} Also, using $[e,f]=h$, \begin{align*}e(fv_0)=fev_0+hv_0=0+3v_0=3v_0.\end{align*} Thus $fv_0\ne 0$, and it has weight $1$. Applying the same commutator relations again, \begin{align*}h(f^2v_0)=f(hfv_0)+[h,f]fv_0=f(fv_0)-2f^2v_0=-f^2v_0.\end{align*} Moreover, \begin{align*}e(f^2v_0)=e(f(fv_0))=f(e(fv_0))+h(fv_0)=f(3v_0)+fv_0=4fv_0.\end{align*} So $f^2v_0\ne 0$, and it has weight $-1$. One more step gives \begin{align*}h(f^3v_0)=f(hf^2v_0)+[h,f]f^2v_0=f(-f^2v_0)-2f^3v_0=-3f^3v_0.\end{align*} Also, \begin{align*}e(f^3v_0)=e(f(f^2v_0))=f(e(f^2v_0))+h(f^2v_0)=f(4fv_0)-f^2v_0=3f^2v_0.\end{align*} Hence $f^3v_0\ne 0$, and it has weight $-3$. Finally, $f^4v_0$ would have weight $-5$ by the same lowering calculation, but $V$ has no weight space of weight $-5$, so $f^4v_0=0$. The four non-zero vectors \begin{align*}v_0,\quad fv_0,\quad f^2v_0,\quad f^3v_0\end{align*} lie in the distinct weight spaces $V_3,V_1,V_{-1},V_{-3}$, so they are linearly independent. Since these four weight spaces each have multiplicity one, they already account for all of $V$. The submodule generated by $v_0$ is therefore all of $V$, and its basis has the same lowering string as $L(3)$; by *Sl Two Highest Weight Classification*, $V\cong L(3)$. This shows that, in this multiplicity-one situation, the highest weight vector determines the whole module. [/example] The same reasoning becomes more delicate when weight multiplicities exceed one. Then the problem is not only to find the possible strings, but also to determine how many copies of each irreducible string occur. ## Characters and Tensor Product Decompositions The final computational problem in this chapter is to turn direct-sum decompositions into arithmetic. Weight multiplicities alone can be misleading before complete reducibility: the same formal list of weights may come from a direct sum or from an indecomposable extension in representation theories where splitting fails. For finite-dimensional $\mathfrak{sl}_2(\mathbb C)$-modules, complete reducibility removes this obstruction, so characters become reliable bookkeeping devices for decompositions. [definition: Sl Two Character] Let $V$ be a finite-dimensional $\mathfrak{sl}_2(\mathbb C)$-module with weight decomposition \begin{align*} V = \bigoplus_{\lambda \in \mathbb C} V_\lambda. \end{align*} The character of $V$ is the element of the free abelian group on formal symbols $q^\lambda$ defined by \begin{align*} \operatorname{ch} V = \sum_{\lambda \in \mathbb C} (\dim V_\lambda)q^\lambda. \end{align*} [/definition] For finite-dimensional completely reducible $\mathfrak{sl}_2(\mathbb C)$-modules, the classification theorem implies that only integral powers occur, so the character lies in $\mathbb Z[q,q^{-1}]$. Characters add over direct sums and multiply over tensor products, so they should turn structural decompositions into polynomial identities. The computation still needs the finite-dimensional classification theorem: without it, an infinite highest-weight module would contribute an infinite formal series rather than a Laurent polynomial. The next theorem computes the irreducible characters explicitly, giving the rank-one version of the Weyl character formula. [quotetheorem:9365] [citeproof:9365] The quotient form is the rank-one prototype of the Weyl character formula. Finite-dimensionality is required here: the Verma module of highest weight $m$ has formal character $\sum_{i=0}^{\infty}q^{m-2i}$, so the finite symmetric string is a special feature of $L(m)$. The formula by itself does not prove a decomposition theorem; without complete reducibility, equalities of characters would not force a module to split into irreducible summands. In a representation theory with non-split extensions, two modules can have the same composition factors and hence the same character while only one is a direct sum of irreducibles. The finite-sum form is usually better for tensor product calculations, because products of characters can be decomposed into sums of these strings after complete reducibility is available. The remaining obstruction is combinatorial rather than structural: after multiplying two strings, we must identify which irreducible strings account for the resulting multiplicities. [quotetheorem:2479] [citeproof:2479] This rule is the first tensor product rule in the course. Complete reducibility is essential to the argument: the character product identifies which irreducible summands must occur only because the tensor product is already known to split. The assumption $a\ge b$ is only a convention for writing the list without negative indexing; if $b>a$, the same formula holds after interchanging the two factors. The rule is also special to rank one, since in higher rank a highest weight can occur with multiplicity greater than one and character subtraction no longer follows a single chain. Later, the same strategy reappears with dominant weights, Weyl characters, and multiplicity formulas replacing the rank-one string subtraction. The same decomposition is the algebraic form of angular-momentum addition in quantum mechanics. Irreducible $\mathfrak{sl}_2$-modules correspond, after passing to the compact real form $\mathfrak{su}(2)$, to spin representations; the summands above record which total spin values can occur when two systems are coupled. This interpretation gives a concrete model for the rank-one calculation: the highest possible weight appears first, and the remaining allowed weights descend in steps of two. [example: Tensor Product of Symmetric Powers] By the symmetric-power realization of the irreducible module $L(m)$, we may replace each $L(m)$ in the *Clebsch-Gordan Rule* by $\operatorname{Sym}^m(\mathbb C^2)$. Thus, for $a\ge b$, \begin{align*}\operatorname{Sym}^a(\mathbb C^2)\otimes \operatorname{Sym}^b(\mathbb C^2)\cong \bigoplus_{j=0}^{b}\operatorname{Sym}^{a+b-2j}(\mathbb C^2).\end{align*} Taking $a=2$ and $b=1$, the possible values of $j$ are $0$ and $1$. The summand for $j=0$ is $\operatorname{Sym}^{2+1-2\cdot 0}(\mathbb C^2)=\operatorname{Sym}^3(\mathbb C^2)$, and the summand for $j=1$ is $\operatorname{Sym}^{2+1-2\cdot 1}(\mathbb C^2)=\operatorname{Sym}^1(\mathbb C^2)$. Hence \begin{align*}\operatorname{Sym}^2(\mathbb C^2)\otimes \operatorname{Sym}^1(\mathbb C^2)\cong \operatorname{Sym}^3(\mathbb C^2)\oplus \operatorname{Sym}^1(\mathbb C^2).\end{align*} The dimensions match: $\dim \operatorname{Sym}^2(\mathbb C^2)=3$ and $\dim \operatorname{Sym}^1(\mathbb C^2)=2$, so the left-hand side has dimension $3\cdot 2=6$. On the right, $\dim \operatorname{Sym}^3(\mathbb C^2)=4$ and $\dim \operatorname{Sym}^1(\mathbb C^2)=2$, so the right-hand side has dimension $4+2=6$. The same decomposition is visible on characters. By the *Sl Two Character Formula*, \begin{align*}\operatorname{ch} L(2)=q^2+q^0+q^{-2}=q^2+1+q^{-2}.\end{align*} Also, \begin{align*}\operatorname{ch} L(1)=q+q^{-1}.\end{align*} Multiplying these two characters gives \begin{align*}(q^2+1+q^{-2})(q+q^{-1})=q^2q+q^2q^{-1}+1q+1q^{-1}+q^{-2}q+q^{-2}q^{-1}.\end{align*} Combining exponents term by term gives \begin{align*}q^2q+q^2q^{-1}+1q+1q^{-1}+q^{-2}q+q^{-2}q^{-1}=q^3+q+q+q^{-1}+q^{-1}+q^{-3}.\end{align*} Thus \begin{align*}(q^2+1+q^{-2})(q+q^{-1})=q^3+2q+2q^{-1}+q^{-3}.\end{align*} On the other hand, \begin{align*}\operatorname{ch} L(3)=q^3+q+q^{-1}+q^{-3}.\end{align*} Adding $\operatorname{ch} L(1)=q+q^{-1}$ gives \begin{align*}\operatorname{ch} L(3)+\operatorname{ch} L(1)=(q^3+q+q^{-1}+q^{-3})+(q+q^{-1})=q^3+2q+2q^{-1}+q^{-3}.\end{align*} So the dimension count and the character calculation both confirm the decomposition $\operatorname{Sym}^2(\mathbb C^2)\otimes \operatorname{Sym}^1(\mathbb C^2)\cong \operatorname{Sym}^3(\mathbb C^2)\oplus \operatorname{Sym}^1(\mathbb C^2)$. [/example] The chapter's template is now complete. Highest weights classify irreducibles, lowering operators build their bases, characters encode their weights, and tensor products are decomposed by subtracting irreducible characters from the top weight downward. For a general complex semisimple Lie algebra, each part remains present, but weights become lattice-valued, lowering operators come from many negative root spaces, and tensor product multiplicities are no longer forced by a single chain. # 4. Highest Weight Modules Highest weight theory turns the structure results from weights, PBW, and the rank-one $\mathfrak{sl}_2$ template into a classification machine. Chapter 2 explained weight decompositions and root operators, while Chapter 3 showed in rank one how a highest vector controls an irreducible module. We now ask which weights can occur at the top of an irreducible finite-dimensional module, how a module is generated from such a top vector, and why the answer is controlled by the simple roots. Throughout this chapter $\mathfrak g$ is a finite-dimensional complex semisimple Lie algebra with Cartan subalgebra $\mathfrak h$, root system $\Phi \subset \mathfrak h^*$, chosen positive roots $\Phi^+$, simple roots $\Delta=\{\alpha_1,\dots,\alpha_r\}$, and triangular decomposition \begin{align*} \mathfrak g = \mathfrak n^- \oplus \mathfrak h \oplus \mathfrak n^+. \end{align*} For each root $\alpha$ choose root vectors $e_\alpha \in \mathfrak g_\alpha$ and $f_\alpha \in \mathfrak g_{-\alpha}$, and for a simple root write $e_i=e_{\alpha_i}$, $f_i=f_{\alpha_i}$, and $h_i=\alpha_i^\vee$. ## The Search for a Top Weight A finite-dimensional module has only finitely many weights, and the positive roots impose a partial order on them. The first question is whether every irreducible finite-dimensional module has a weight from which no further raising is possible. Such a vector is the representation-theoretic analogue of the highest vector in the irreducible $\mathfrak{sl}_2$-module of dimension $m+1$. [definition: Root Order] Let $\lambda,\mu \in \mathfrak h^*$. We write $\mu \le \lambda$ if \begin{align*} \lambda-\mu \in \sum_{i=1}^{r}\mathbb N_0\alpha_i. \end{align*} We write $\mu < \lambda$ if $\mu \le \lambda$ and $\mu \ne \lambda$. [/definition] The order records how weights change under the negative root spaces. Acting by $\mathfrak g_{-\alpha}$ lowers a weight by $\alpha$, while acting by $\mathfrak g_\alpha$ raises it by $\alpha$. To turn a maximal weight into a useful generator, we need a name for a vector that is both a weight vector and immune to every positive-root raising operator. [definition: Highest Weight Vector] Let $V$ be a $\mathfrak g$-module. A non-zero vector $v \in V$ is a highest weight vector of weight $\lambda \in \mathfrak h^*$ if $hv=\lambda(h)v$ for all $h\in \mathfrak h$ and $\mathfrak n^+v=0$. [/definition] The first condition places $v$ in the weight space $V_\lambda$, and the second says that $v$ has no weight above it. By itself this only identifies a possible top vector inside a larger module; it does not say that the rest of the module is controlled by that vector. To make highest weight methods apply to an entire representation rather than to a single distinguished vector, we need a module-level condition: the chosen top vector must generate everything by the $\mathfrak g$-action. This is the point of the next definition, which packages the highest vector together with the cyclicity needed for PBW and weight arguments. [definition: Highest Weight Module] A $\mathfrak g$-module $V$ is a highest weight module of highest weight $\lambda$ if there exists a highest weight vector $v_\lambda \in V_\lambda$ such that \begin{align*} V=U(\mathfrak g)v_\lambda. \end{align*} [/definition] The definition is deliberately phrased in terms of generation. A module may contain a vector killed by $\mathfrak n^+$ without being generated by that vector, but highest weight theory focuses on the cyclic case because PBW then gives a strong spanning description. The next theorem makes this precise by showing that all raising and Cartan factors disappear once they reach the highest vector. [quotetheorem:9366] [citeproof:9366] This theorem is the bridge from algebra to combinatorics: the possible weights sit below $\lambda$ in the root order. The cyclicity hypothesis is essential here. A direct sum of two highest weight modules may contain a vector killed by $\mathfrak n^+$ in one summand, but that vector cannot generate the other summand, so PBW only describes the cyclic submodule it generates. The theorem also says nothing by itself about irreducibility or finite-dimensionality: $U(\mathfrak n^-)v_\lambda$ may still have infinitely many non-zero weight spaces, as happens for Verma modules introduced below. It remains to justify why finite-dimensional irreducible modules have such a top vector in the first place, rather than merely admitting a weight decomposition. The PBW spanning theorem starts from a cyclic highest vector, but a general finite-dimensional module initially presents only a finite set of weights and no distinguished generator. The next result uses maximality in the root order to find a vector killed by all raising operators, and then uses irreducibility to promote the submodule generated by that vector to the whole representation. [quotetheorem:9367] [citeproof:9367] The hypotheses here are doing real work. Finite dimensionality gives a finite set of weights, so maximal weights exist in the root order; irreducibility then prevents the highest-weight submodule from being only a proper piece of the representation. Without finite dimensionality, highest-weight modules such as Verma modules can still be generated by a highest vector but need not be finite-dimensional. Without irreducibility, a module may contain several highest-weight summands rather than one distinguished cyclic top. The result therefore justifies treating finite-dimensional irreducible modules as objects controlled by a single highest weight, which is exactly what the standard representation of $\mathfrak{sl}_n(\mathbb C)$ illustrates next. [example: Standard Module of Special Linear n] Let $\mathfrak g=\mathfrak{sl}_n(\mathbb C)$ act on $V=\mathbb C^n$ with basis $v_1,\dots,v_n$, and let $E_{ij}$ be the matrix unit satisfying $E_{ij}v_j=v_i$ and $E_{ij}v_\ell=0$ for $\ell\ne j$. For $h=\operatorname{diag}(t_1,\dots,t_n)$ with $\sum_i t_i=0$, define $\varepsilon_i(h)=t_i$. Then \begin{align*} hv_1=\operatorname{diag}(t_1,\dots,t_n)v_1=t_1v_1=\varepsilon_1(h)v_1. \end{align*} Thus $v_1$ is a weight vector of weight $\varepsilon_1$ on the trace-zero Cartan subalgebra. For the standard choice of positive roots, $\mathfrak n^+$ consists of the span of the $E_{ij}$ with $i<j$. If $i<j$, then $j>1$, so $E_{ij}v_1=0$ because $E_{ij}$ only acts non-trivially on $v_j$. Hence $\mathfrak n^+v_1=0$, and $v_1$ is a highest weight vector. The negative root operators in the first column recover the other basis vectors: \begin{align*} E_{j1}v_1=v_j \quad \text{for } j=2,\dots,n. \end{align*} Together with $v_1$, these vectors form the basis $v_1,\dots,v_n$, so $U(\mathfrak g)v_1=V$. Finally, for the simple coroots $\alpha_i^\vee=E_{ii}-E_{i+1,i+1}$, we have \begin{align*} \varepsilon_1(\alpha_1^\vee)=\varepsilon_1(E_{11}-E_{22})=1. \end{align*} For $i>1$, \begin{align*} \varepsilon_1(\alpha_i^\vee)=\varepsilon_1(E_{ii}-E_{i+1,i+1})=0. \end{align*} Therefore $\varepsilon_1$ has the defining values of the first fundamental weight $\omega_1$ on the simple coroots, so the standard module $\mathbb C^n$ is a highest weight module with highest weight $\omega_1$. [/example] This example is the model for the rest of the chapter: the highest vector is visible at the top of the chosen ordering, while the entire module is recovered by applying lowering operators. It also shows why the highest-weight condition is stronger than merely naming a large weight: the root spaces determine concrete paths through the whole representation. Once these paths are restricted to a single simple-root direction, they become ordinary $\mathfrak{sl}_2$ strings, so their possible lengths are governed by the rank-one theory. The next section extracts the numerical restrictions forced by these paths in each simple-root direction. ## Dominance Forced by Root Subalgebras The classification cannot allow arbitrary $\lambda\in\mathfrak h^*$. Each simple root determines a copy of $\mathfrak{sl}_2$, and any finite-dimensional $\mathfrak g$-module restricts to a finite-dimensional module for that subalgebra. The familiar $\mathfrak{sl}_2$ string calculation therefore forces integrality and non-negativity conditions on the highest weight. [definition: Integral and Dominant Weights] A weight $\lambda\in\mathfrak h^*$ is integral if \begin{align*} \lambda(\alpha_i^\vee)\in \mathbb Z \quad \text{for all } i=1,\dots,r. \end{align*} It is dominant if \begin{align*} \lambda(\alpha_i^\vee)\in \mathbb N_0 \quad \text{for all } i=1,\dots,r. \end{align*} The set of dominant integral weights is denoted $P^+$. [/definition] The non-negative integers $\lambda(\alpha_i^\vee)$ are the coordinates that will label irreducible finite-dimensional modules. To express those coordinates intrinsically, we introduce the basis dual to the simple coroots. [definition: Fundamental Weights] The fundamental weights $\omega_1,\dots,\omega_r\in\mathfrak h^*$ are defined by \begin{align*} \omega_i(\alpha_j^\vee)=\delta_{ij} \end{align*} for $1\le i,j\le r$. [/definition] After this definition, every dominant integral weight has a unique expression $\lambda=\sum_i m_i\omega_i$ with $m_i\in\mathbb N_0$. This notation is useful because the $\mathfrak{sl}_2$ restriction attached to $\alpha_i$ reads off the coefficient $m_i$ directly. We now prove that finite-dimensional highest weight modules can only produce such non-negative integer coefficients, which is the first half of the classification theorem. [quotetheorem:9368] [citeproof:9368] This argument is local in the Dynkin diagram: each simple root direction contributes one integer constraint. The finite-dimensional hypothesis is doing real work. For example, the Verma module $M(\lambda)$ exists for any $\lambda\in\mathfrak h^*$, including weights with $\lambda(\alpha_i^\vee)\notin\mathbb N_0$, but such a module cannot be finite-dimensional when the simple-root $\mathfrak{sl}_2$ string has no allowed finite top length. The theorem is therefore a necessary condition for finite-dimensional highest weight modules, not an existence theorem. The full semisimple algebra couples these directions, but the first arithmetic obstruction is already visible inside the rank-one subalgebras. [example: Special Linear Three Dominant Weights] For $\mathfrak{sl}_3(\mathbb C)$, take the Cartan subalgebra of diagonal trace-zero matrices and the standard positive roots $\alpha_1=\varepsilon_1-\varepsilon_2$ and $\alpha_2=\varepsilon_2-\varepsilon_3$. The simple coroots are $\alpha_1^\vee=E_{11}-E_{22}$ and $\alpha_2^\vee=E_{22}-E_{33}$, so the fundamental weights satisfy $\omega_1(\alpha_1^\vee)=1$, $\omega_1(\alpha_2^\vee)=0$, $\omega_2(\alpha_1^\vee)=0$, and $\omega_2(\alpha_2^\vee)=1$. Hence a weight $a\omega_1+b\omega_2$ is dominant integral exactly when \begin{align*} (a\omega_1+b\omega_2)(\alpha_1^\vee)=a \in \mathbb N_0 \end{align*} and \begin{align*} (a\omega_1+b\omega_2)(\alpha_2^\vee)=b \in \mathbb N_0. \end{align*} In the standard representation, the vector $v_1$ has weight $\varepsilon_1$ and is killed by the upper-triangular root vectors $E_{12},E_{13},E_{23}$. On simple coroots, \begin{align*} \varepsilon_1(\alpha_1^\vee)=\varepsilon_1(E_{11}-E_{22})=1 \end{align*} and \begin{align*} \varepsilon_1(\alpha_2^\vee)=\varepsilon_1(E_{22}-E_{33})=0. \end{align*} Thus $\varepsilon_1=\omega_1$ on the trace-zero Cartan subalgebra, so the standard representation has highest weight $\omega_1$. For the dual representation, let $v_1^*,v_2^*,v_3^*$ be the [dual basis](/theorems/414), with action $(X\varphi)(v)=-\varphi(Xv)$. If $h=\operatorname{diag}(t_1,t_2,t_3)$ and $t_1+t_2+t_3=0$, then \begin{align*} (hv_3^*)(v_\ell)=-v_3^*(hv_\ell)=-t_\ell v_3^*(v_\ell). \end{align*} This is $0$ for $\ell\ne 3$ and $-t_3$ for $\ell=3$, so $hv_3^*=-\varepsilon_3(h)v_3^*$. The vectors $E_{12},E_{13},E_{23}$ kill $v_3^*$ because each sends basis vectors into the span of $v_1,v_2$ before $v_3^*$ is evaluated. Also \begin{align*} (-\varepsilon_3)(\alpha_1^\vee)=-\varepsilon_3(E_{11}-E_{22})=0 \end{align*} and \begin{align*} (-\varepsilon_3)(\alpha_2^\vee)=-\varepsilon_3(E_{22}-E_{33})=1. \end{align*} Thus the dual standard representation has highest weight $\omega_2$. For the adjoint representation, the root vector $E_{13}$ has weight $\varepsilon_1-\varepsilon_3$ because \begin{align*} [h,E_{13}]=hE_{13}-E_{13}h=t_1E_{13}-t_3E_{13}=(\varepsilon_1-\varepsilon_3)(h)E_{13}. \end{align*} Moreover $\varepsilon_1-\varepsilon_3=(\varepsilon_1-\varepsilon_2)+(\varepsilon_2-\varepsilon_3)=\alpha_1+\alpha_2$. The upper root vectors kill $E_{13}$ under the bracket: $[E_{12},E_{13}]=0$, $[E_{23},E_{13}]=0$, and $[E_{13},E_{13}]=0$. Finally, \begin{align*} (\alpha_1+\alpha_2)(\alpha_1^\vee)=(\varepsilon_1-\varepsilon_3)(E_{11}-E_{22})=1 \end{align*} and \begin{align*} (\alpha_1+\alpha_2)(\alpha_2^\vee)=(\varepsilon_1-\varepsilon_3)(E_{22}-E_{33})=1. \end{align*} Therefore $\alpha_1+\alpha_2=\omega_1+\omega_2$, so the adjoint representation has highest weight $\omega_1+\omega_2$. [/example] The $\mathfrak{sl}_3$ picture is small enough to visualise as a two-dimensional weight lattice. Moving down by $\alpha_1$ and $\alpha_2$ gives the weight diagram, while the two coordinates $a$ and $b$ record the allowed lengths of the top strings. [remark: Why All Positive Roots Are Not Needed] Dominance is tested on simple coroots rather than on every positive coroot. For a finite-dimensional representation, the Weyl group and root-string structure imply non-negativity on all positive coroots once the simple inequalities hold. The simple-root formulation is the minimal set of inequalities defining the closed dominant chamber. [/remark] This observation keeps the classification finite: a highest weight is specified by $r=\operatorname{rank}\mathfrak g$ non-negative integers rather than by a condition for every vector in the root system. It also explains why the simple roots, rather than all positive roots, are the correct data for constructing modules. The existence problem can therefore be phrased in a finite form: start with the $r$ integers attached to a dominant weight, impose the lowering relations predicted by those integers, and ask whether the resulting irreducible quotient is finite-dimensional. ## Generated Modules and Irreducible Quotients Having found necessary conditions, the next problem is existence. Given a dominant integral weight $\lambda$, we need a module whose top vector has weight $\lambda$, and then we need to isolate its irreducible finite-dimensional quotient. The universal construction is the Verma module, which packages the relations required of a highest weight vector and imposes no additional lowering relations. [definition: Verma Module] Let $\lambda\in\mathfrak h^*$. Let $\mathbb C_\lambda$ be the one-dimensional $(\mathfrak b=\mathfrak h\oplus\mathfrak n^+)$-module on which $h\in\mathfrak h$ acts by $\lambda(h)$ and $\mathfrak n^+$ acts by $0$. The Verma module of highest weight $\lambda$ is \begin{align*} M(\lambda)=U(\mathfrak g)\otimes_{U(\mathfrak b)}\mathbb C_\lambda. \end{align*} [/definition] The Verma module is almost never finite-dimensional, but it is the correct universal source. The point is that formal lowering operations overgenerate: if no relations are imposed beyond $\mathfrak n^+v_\lambda=0$ and $hv_\lambda=\lambda(h)v_\lambda$, then all PBW monomials in negative root vectors survive until a quotient kills them. In finite-dimensional modules extra relations such as powers of simple lowering operators must appear, but the universal construction deliberately postpones those relations. Any highest weight module of highest weight $\lambda$ should therefore be obtained by sending the canonical highest vector to the chosen highest vector and then quotienting by the relations that hold in the target. The next result states this universal mapping property, which is what makes Verma modules control all cyclic highest weight modules. The construction should be read as a replacement for writing down every possible highest-weight module separately: the canonical vector in $M(\lambda)$ carries exactly the relations shared by all highest vectors of weight $\lambda$. Once a target module contains such a vector, the map out of $M(\lambda)$ is forced by where the canonical highest vector goes. [quotetheorem:9369] [citeproof:9369] The universal property reduces uniqueness questions to submodules of $M(\lambda)$, since every highest weight module is a quotient of it. The hypothesis that the chosen vector is a highest weight vector is indispensable. For example, in the standard $\mathfrak{sl}_2$-module with basis $v_+,v_-$, the lower vector $v_-$ has weight lower than the highest weight but is not killed by the raising operator $e$; sending the canonical highest vector of a Verma module to $v_-$ would violate the defining tensor relation $e(1\otimes 1)=0$. The theorem also does not say that the quotient is unique, finite-dimensional, or irreducible. Imposing different lowering relations on the same universal highest vector can produce different highest weight quotients before one passes to the irreducible quotient. The next obstruction is that a universal module can have many quotients; to obtain a canonical irreducible object, we must know that all proper quotients factor through a single largest proper submodule. [quotetheorem:9370] [citeproof:9370] This theorem defines $L(\lambda)$ for every weight, but it does not make $L(\lambda)$ finite-dimensional. In rank one, if $\lambda(h)$ is not a non-negative integer, the irreducible highest weight quotient has an infinite string of weights descending by the simple root. The same obstruction appears in each simple-root direction for general $\mathfrak g$. Thus the classification of finite-dimensional modules still needs a finiteness criterion. Dominance has already been shown necessary; we now need the converse direction, namely that the relations predicted by the simple-root $\mathfrak{sl}_2$ strings are enough to produce a finite-dimensional quotient. [quotetheorem:9371] [citeproof:9371] This is the constructive half of the classification. Its guiding principle is the same as in $\mathfrak{sl}_2$: dominance tells us exactly which powers of the simple lowering operators must eventually vanish. The dominance assumption cannot be dropped. In rank one, if $\mathfrak g=\mathfrak{sl}_2$ and $\lambda(h)\notin\mathbb N_0$, then there is no relation of the form $f^{\lambda(h)+1}v_\lambda=0$ with a non-negative integer exponent, and the irreducible highest weight quotient has an infinite descending weight string. The theorem does not claim that the Verma module itself is finite-dimensional; the finite-dimensional object is obtained only after quotienting by the integrability relations. This distinction matters in later character theory, where the infinite Verma module and its finite irreducible quotient have different characters but share the same highest weight. [example: Exterior Powers of the Standard sln Module] Let $V=\mathbb C^n$ have basis $v_1,\dots,v_n$, and let $1\le k\le n-1$. On $\Lambda^k V$, the induced action is \begin{align*} X(u_1\wedge\cdots\wedge u_k)=\sum_{a=1}^k u_1\wedge\cdots\wedge Xu_a\wedge\cdots\wedge u_k. \end{align*} For $h=\operatorname{diag}(t_1,\dots,t_n)$ with $\sum_i t_i=0$, this gives \begin{align*} h(v_1\wedge\cdots\wedge v_k)=(t_1+\cdots+t_k)(v_1\wedge\cdots\wedge v_k). \end{align*} Thus $v_1\wedge\cdots\wedge v_k$ has weight $\varepsilon_1+\cdots+\varepsilon_k$. Now let $E_{ij}$ be strictly upper triangular, so $i<j$. If $j>k$, then $E_{ij}$ kills each factor $v_1,\dots,v_k$. If $j\le k$, the only non-zero term replaces $v_j$ by $v_i$, and since $i<j$ the vector $v_i$ already appears among $v_1,\dots,v_{j-1}$. Hence that wedge has a repeated factor and is $0$. Therefore every strictly upper-triangular matrix kills $v_1\wedge\cdots\wedge v_k$, so this vector is a highest weight vector. For the simple coroots $\alpha_\ell^\vee=E_{\ell\ell}-E_{\ell+1,\ell+1}$, we compute \begin{align*} (\varepsilon_1+\cdots+\varepsilon_k)(\alpha_\ell^\vee)=\sum_{a=1}^k \varepsilon_a(E_{\ell\ell}-E_{\ell+1,\ell+1}). \end{align*} If $\ell<k$, the two non-zero contributions are $1$ and $-1$, so the value is $0$. If $\ell=k$, only $\varepsilon_k(E_{kk})$ contributes, so the value is $1$. If $\ell>k$, neither diagonal entry occurs among $\varepsilon_1,\dots,\varepsilon_k$, so the value is $0$. Thus \begin{align*} (\varepsilon_1+\cdots+\varepsilon_k)(\alpha_\ell^\vee)=\delta_{k\ell}, \end{align*} which is exactly the defining property of $\omega_k$. The lowering operators generate every wedge basis vector from the top one. Indeed, if $I=\{i_1<\cdots<i_k\}\ne\{1,\dots,k\}$, choose an index $s$ with $i_s>s$ and $i_s-1\notin I$; replacing $i_s$ by $i_s-1$ gives a lexicographically smaller $k$-subset $I'$. The operator $E_{i_s,i_s-1}$ sends the basis wedge for $I'$ to $\pm v_{i_1}\wedge\cdots\wedge v_{i_k}$, because it replaces the single factor $v_{i_s-1}$ by $v_{i_s}$ and kills the other factors. Induction on $\sum_s(i_s-s)$ shows that $U(\mathfrak{sl}_n)(v_1\wedge\cdots\wedge v_k)=\Lambda^k V$. The same basis check also shows irreducibility. If $I\ne\{1,\dots,k\}$ and $s$ is the first index with $i_s>s$, then $s\notin I$ and the positive root operator $E_{s,i_s}$ sends the wedge for $I$ to a non-zero wedge. Hence the only weight vector killed by all strictly upper-triangular matrices is a scalar multiple of $v_1\wedge\cdots\wedge v_k$. Any non-zero submodule has a maximal weight vector, hence contains this highest vector, and therefore equals all of $\Lambda^k V$. By *Highest Weight Classification*, $\Lambda^k V\cong L(\omega_k)$. When $k=n-1$, this recovers the dual standard representation. If $\Omega=v_1\wedge\cdots\wedge v_n$ and $\Phi:\Lambda^{n-1}V\to V^*$ is defined by $u\wedge w=\Phi(w)(u)\Omega$, then for $X\in\mathfrak{sl}_n$ we have $X\Omega=0$, so \begin{align*} u\wedge Xw=-(Xu)\wedge w=-(\Phi(w)(Xu))\Omega. \end{align*} This is precisely the dual action, so $\Lambda^{n-1}V\cong V^*$ and its highest weight is $\omega_{n-1}$. [/example] These examples give the first family of fundamental representations. More complicated irreducibles arise from dominant weights such as $2\omega_1+\omega_3$, but the classification says that they are still uniquely determined by the coefficients of the fundamental weights. ## Classification by Highest Weight The final question is whether the label $\lambda$ distinguishes irreducible finite-dimensional modules. The answer is yes: a non-zero homomorphism between irreducible highest weight modules must respect the top weight, and the top weight space in an irreducible highest weight module is one-dimensional. [quotetheorem:9372] [citeproof:9372] We now have the three pieces needed for classification: every finite-dimensional irreducible has a dominant highest weight, every dominant integral weight produces a finite-dimensional irreducible module, and the highest weight determines that module up to isomorphism. Each hypothesis in the uniqueness statement matters. If irreducibility is dropped, then $L(\lambda)\oplus L(\lambda)$ and $L(\lambda)$ have the same highest weight label available but are not isomorphic; if finite-dimensionality is dropped, Verma modules and their irreducible quotients share a highest weight while having different structure. Requiring the same highest weight is also essential, since $L(\lambda)$ and $L(\mu)$ are distinguished by their maximal weights when $\lambda\ne\mu$. The following theorem records the resulting bijection and becomes the reference point for character formulas and tensor product decompositions later in the course. [quotetheorem:9373] [citeproof:9373] The theorem turns representation theory into a problem about the dominant chamber in the weight lattice, but its hypotheses are essential. It classifies finite-dimensional irreducible modules for complex semisimple Lie algebras; it does not classify reducible modules, infinite-dimensional highest weight modules, or representations of non-semisimple Lie algebras. Reducible finite-dimensional modules are then handled by complete reducibility, while infinite-dimensional categories require additional tools such as Verma modules and category $\mathcal O$. Later chapters refine this classification by computing the character of $L(\lambda)$, describing its weight multiplicities, and analysing how tensor products decompose. [example: Classification for Special Linear Two] For $\mathfrak{sl}_2(\mathbb C)$ take the standard basis $e,f,h$ with $[h,e]=2e$, $[h,f]=-2f$, and $\alpha^\vee=h$. There is one fundamental weight $\omega$, defined by \begin{align*} \omega(\alpha^\vee)=\omega(h)=1. \end{align*} Thus a weight $\lambda=a\omega$ satisfies \begin{align*} \lambda(\alpha^\vee)=(a\omega)(h)=a\omega(h)=a, \end{align*} so $\lambda$ is dominant integral exactly when $a=m\in\mathbb N_0$. For $\lambda=m\omega$, the irreducible highest weight module $L(m\omega)$ is generated by a highest vector $v_0$ with \begin{align*} ev_0=0,\qquad hv_0=(m\omega)(h)v_0=mv_0. \end{align*} Set $v_j=f^jv_0$. Since $[h,f]=-2f$, we have $hf=fh-2f$, and the induction step is \begin{align*} h(f^{j+1}v_0)=(hf)f^jv_0=(fh-2f)f^jv_0=f(hf^jv_0)-2f^{j+1}v_0=(m-2j-2)f^{j+1}v_0. \end{align*} Starting from $hv_0=mv_0$, this gives \begin{align*} hv_j=(m-2j)v_j. \end{align*} The finite-dimensional rank-one $\mathfrak{sl}_2$ calculation gives $v_0,v_1,\dots,v_m\ne0$ and $v_{m+1}=f^{m+1}v_0=0$, so these $m+1$ weight vectors form a basis. Hence $L(m\omega)$ has dimension $m+1$, and its weights in the standard $h$-normalisation are \begin{align*} m,\ m-2,\ m-4,\ \dots,\ -m. \end{align*} This is exactly the rank-one version of highest weight classification: one non-negative integer $m$ determines the irreducible module. [/example] This final example confirms that the general theorem recovers the classification already known in rank one. The higher-rank theory keeps the same top-weight philosophy, with several simple root directions replacing the single lowering direction of $\mathfrak{sl}_2$. # 5. Verma Modules Verma modules answer the construction problem left by highest weight theory. Chapter 4 identified dominant highest weights as the labels for finite-dimensional irreducibles, but it also left open the universal construction of a module generated by a vector of a prescribed highest weight. The idea of this chapter is to build the largest possible highest weight module first, then recover irreducible modules by quotienting. Throughout, let $\mathfrak g$ be a finite-dimensional complex semisimple Lie algebra with triangular decomposition \begin{align*} \mathfrak g = \mathfrak n^- \oplus \mathfrak h \oplus \mathfrak n^+, \end{align*} and let $\mathfrak b = \mathfrak h \oplus \mathfrak n^+$ be the chosen Borel subalgebra. We write $U(\mathfrak a)$ for the universal enveloping algebra of a Lie algebra $\mathfrak a$. ## Induction from the Borel Subalgebra The first question is how to impose the conditions for a highest weight vector without imposing any further relations. A highest weight vector of weight $\lambda \in \mathfrak h^*$ should be killed by $\mathfrak n^+$, and $\mathfrak h$ should act on it by the scalar $\lambda$. Induction from $\mathfrak b$ to $\mathfrak g$ packages exactly these requirements. For a fixed weight $\lambda$, we first turn $\mathbb C$ into a one-dimensional $\mathfrak b$-module. This object records the desired highest weight behaviour before the lowering operators have been allowed to act. [definition: Highest Weight Line] Let $\lambda \in \mathfrak h^*$. The highest weight line $\mathbb C_\lambda$ is the one-dimensional $\mathfrak b$-module spanned by a vector $v_\lambda$ such that $h v_\lambda = \lambda(h)v_\lambda$ for all $h \in \mathfrak h$ and $x v_\lambda = 0$ for all $x \in \mathfrak n^+$. [/definition] The highest weight line remembers only the desired behaviour at the top, so it is not yet a representation of all of $\mathfrak g$. The problem now is to extend this $\mathfrak b$-module to a $\mathfrak g$-module while adding all lowering directions and no extra relations; this motivates defining the induced module below. [definition: Verma Module] Let $\lambda \in \mathfrak h^*$. The Verma module of highest weight $\lambda$ is \begin{align*} M(\lambda) := U(\mathfrak g) \otimes_{U(\mathfrak b)} \mathbb C_\lambda. \end{align*} The element $1 \otimes v_\lambda$ is denoted again by $v_\lambda$. [/definition] This construction is designed so that $v_\lambda$ is a highest weight vector and every other vector is obtained from it by the $U(\mathfrak g)$-action. To make the module usable, we need to know whether the induction has introduced hidden vector-space relations; the PBW theorem gives the answer. [quotetheorem:9374] [citeproof:9374] The PBW model says that a Verma module is obtained by freely applying all lowering operators to the top vector. The triangular decomposition is essential here: without separating $\mathfrak n^-$ from $\mathfrak b$, the tensor product would not identify a clean free set of lowering directions. The theorem is only a vector-space description, not a statement that the $\mathfrak g$-action is diagonal or finite-dimensional; even in $\mathfrak{sl}_2$, it produces infinitely many lowering powers. The next issue is functorial: if some other module already contains a vector satisfying the same highest weight relations, the free construction should map to it in a unique way. [quotetheorem:9375] [citeproof:9375] The theorem explains why Verma modules are the correct universal objects for highest weight theory. The hypotheses on $v$ are exactly the highest weight relations; if $\mathfrak n^+v\ne 0$ or if $\mathfrak h$ does not act through the character $\lambda$, the proposed formula $u v_\lambda \mapsto uv$ would not respect the tensor relations defining $M(\lambda)$. The theorem does not say that the resulting map is injective: kernels are precisely where singular vectors and proper quotients enter. Any highest weight module of highest weight $\lambda$ is a quotient of $M(\lambda)$, so classification can be attacked by studying submodules of Verma modules. [example: Verma Module for sl2] Let $\mathfrak g=\mathfrak{sl}_2(\mathbb C)$ with basis $e,h,f$ and relations $[h,e]=2e$, $[h,f]=-2f$, and $[e,f]=h$. Since $\mathfrak n^-=\mathbb C f$, the PBW model gives a basis of $M(\lambda)$: \begin{align*} \{f^k v_\lambda : k\in \mathbb Z_{\ge 0}\}. \end{align*} The relation $[h,f]=-2f$ gives $h f=f h-2f$, and induction on $k$ gives \begin{align*} h f^k=f^k h-2k f^k. \end{align*} Therefore \begin{align*} h(f^k v_\lambda)=(f^k h-2k f^k)v_\lambda=(\lambda-2k)f^k v_\lambda. \end{align*} So $f^k v_\lambda$ has weight $\lambda-2k$. To compute the action of $e$, use $e v_\lambda=0$ and expand the commutator: \begin{align*} e f^k v_\lambda=(f^k e+[e,f^k])v_\lambda=[e,f^k]v_\lambda. \end{align*} Using $[e,f]=h$ and the Leibniz rule for commutators, \begin{align*} [e,f^k]=\sum_{i=0}^{k-1} f^i h f^{k-1-i}. \end{align*} For each summand, the identity $h f^m=f^m h-2m f^m$ gives \begin{align*} f^i h f^{k-1-i}v_\lambda=f^i(f^{k-1-i}h-2(k-1-i)f^{k-1-i})v_\lambda=(\lambda-2(k-1-i))f^{k-1}v_\lambda. \end{align*} Hence \begin{align*} e f^k v_\lambda=\sum_{i=0}^{k-1}(\lambda-2(k-1-i))f^{k-1}v_\lambda. \end{align*} The scalar is \begin{align*} \sum_{i=0}^{k-1}(\lambda-2(k-1-i))=k\lambda-2\sum_{j=0}^{k-1}j=k\lambda-k(k-1)=k(\lambda-k+1). \end{align*} Thus \begin{align*} e f^k v_\lambda=k(\lambda-k+1)f^{k-1}v_\lambda. \end{align*} The infinitely many basis vectors show that $M(\lambda)$ is infinite-dimensional for every $\lambda$, while the displayed formula shows exactly when a lower basis vector can be killed by $e$ and become a new highest weight vector. [/example] ## Weights and Singular Vectors The next question is how much of a Verma module is already determined by the root system. Because $M(\lambda)$ is built from $U(\mathfrak n^-)$, its weights are obtained from $\lambda$ by subtracting non-negative sums of positive roots. This gives a combinatorial skeleton for every Verma module. To state the weight decomposition, let $\Phi^+$ be the set of positive roots and write \begin{align*} Q_+ := \sum_{\alpha \in \Phi^+} \mathbb Z_{\ge 0}\alpha. \end{align*} The PBW monomials in negative root vectors record exactly how far below the top weight a vector lies, so they also count the dimensions of the weight spaces. [quotetheorem:9376] [citeproof:9376] The formula separates the universal combinatorics from the dependence on $\lambda$. The PBW ordering and the choice of positive roots matter: changing the Borel changes which weights are regarded as lying below the top, while changing the order of root vectors changes the displayed basis but not the dimensions. The theorem also does not decide irreducibility, since identical weight multiplicities can occur for Verma modules with very different submodule structures. The set of weights and their multiplicities are obtained by translating the same positive-root partition data, while the value of $\lambda$ controls whether a lower weight vector can start a new highest weight submodule. That possibility motivates the terminology of singular vectors. [definition: Singular Vector] Let $M$ be a $\mathfrak g$-module. A nonzero vector $w \in M$ is a singular vector of weight $\mu \in \mathfrak h^*$ if $w \in M_\mu$ and $\mathfrak n^+ w = 0$. [/definition] A singular vector behaves like a new highest weight vector inside an existing module. In a Verma module, such a vector generates a proper highest weight submodule unless it is the original top vector, so the next theorem explains why every nonzero submodule must be detected in this way. [quotetheorem:9377] [citeproof:9377] This result is the reason submodules of Verma modules are visible through singular vectors. The weight-module hypothesis is doing real work: for a general module without a usable weight decomposition, a submodule need not contain a vector that can be isolated as a highest component. The theorem also does not classify all submodules; it only guarantees that each nonzero submodule begins with at least one singular vector. The next examples show how this mechanism appears first in rank one and then in rank two. [example: Singular Vectors in sl2 Verma Modules] In $M(\lambda)$ for $\mathfrak{sl}_2(\mathbb C)$, the positive nilpotent subalgebra is $\mathfrak n^+=\mathbb C e$, so a nonzero weight vector is singular exactly when it is killed by $e$. For $k\geq 1$, the action formula computed above gives \begin{align*} e f^k v_\lambda = k(\lambda-k+1)f^{k-1}v_\lambda . \end{align*} Since $f^{k-1}v_\lambda\ne 0$ in the PBW basis and $k\ne 0$, this vector is killed by $e$ exactly when \begin{align*} k(\lambda-k+1)=0 . \end{align*} Thus, for $k\geq 1$, \begin{align*} e f^k v_\lambda=0 \end{align*} holds exactly when \begin{align*} \lambda-k+1=0, \end{align*} equivalently \begin{align*} k=\lambda+1 . \end{align*} Therefore a lower PBW basis vector $f^k v_\lambda$ can be singular only if $\lambda+1$ is a positive integer, which is the same as $\lambda\in\mathbb Z_{\geq 0}$. In that case $f^{\lambda+1}v_\lambda$ is singular. Its weight is computed from $h(f^k v_\lambda)=(\lambda-2k)f^k v_\lambda$ with $k=\lambda+1$: \begin{align*} h(f^{\lambda+1}v_\lambda)=(\lambda-2(\lambda+1))f^{\lambda+1}v_\lambda . \end{align*} The scalar is \begin{align*} \lambda-2(\lambda+1)=\lambda-2\lambda-2=-\lambda-2 . \end{align*} So $f^{\lambda+1}v_\lambda$ is a highest weight vector of weight $-\lambda-2$. The submodule it generates has basis \begin{align*} \{f^r f^{\lambda+1}v_\lambda : r\in\mathbb Z_{\geq 0}\}=\{f^{r+\lambda+1}v_\lambda : r\in\mathbb Z_{\geq 0}\}, \end{align*} and these vectors are linearly independent because they are distinct PBW basis vectors of $M(\lambda)$. Hence this submodule is a copy of $M(-\lambda-2)$, with its highest vector identified with $f^{\lambda+1}v_\lambda$. If $\lambda\notin\mathbb Z_{\geq 0}$, then no integer $k\geq 1$ satisfies $k=\lambda+1$, so no basis vector below $v_\lambda$ is singular; in rank one, this means there is no visible proper highest weight submodule generated by a lower PBW basis vector. [/example] ## The Maximal Submodule and the Irreducible Quotient The final structural question is how to extract the irreducible representation controlled by the highest weight $\lambda$. Since every proper submodule of $M(\lambda)$ misses the top weight line, their sum is still proper. This produces a canonical maximal submodule. Before stating the result, recall that $M(\lambda)_\lambda = \mathbb C v_\lambda$. A proper submodule cannot contain $v_\lambda$, because $v_\lambda$ generates all of $M(\lambda)$; this observation makes it possible to add all proper submodules at once. [quotetheorem:9378] [citeproof:9378] The unique maximal submodule isolates exactly the part of the Verma module that must be removed to obtain a simple module. The word "proper" is essential: summing all submodules would include $M(\lambda)$ itself, while summing only selected proper submodules would give a quotient depending on choices. The theorem is special to this highest weight situation; it does not assert that arbitrary modules have a canonical maximal submodule. This quotient retains the highest weight line while killing every obstruction coming from a proper submodule, so it deserves its own notation. [definition: Irreducible Highest Weight Quotient] Let $\lambda \in \mathfrak h^*$. Define \begin{align*} L(\lambda) := M(\lambda)/J(\lambda), \end{align*} where $J(\lambda)$ is the unique maximal proper submodule of $M(\lambda)$. [/definition] The quotient has been defined to have no proper submodule inherited from $M(\lambda)$. We still need to record its universal classification role: every simple highest weight module with the same top weight must arise from this quotient and no other one. [quotetheorem:9379] [citeproof:9379] This theorem is the bridge from universal modules to classification. The simplicity of $V$ in the uniqueness statement is necessary: a nonsimple highest weight quotient of $M(\lambda)$ can still have highest weight $\lambda$ without being equal to $L(\lambda)$. The theorem also classifies only simple highest weight modules for the chosen Borel, not arbitrary simple $\mathfrak g$-modules without a highest weight vector. Verma modules are usually too large, but their simple quotients are the irreducible highest weight modules. [example: Finite-Dimensional sl2 Quotients] Let $\mathfrak{sl}_2(\mathbb C)$ have basis $e,h,f$ as above, and take $\lambda=n\in\mathbb Z_{\ge 0}$. The rank-one formula \begin{align*} e f^k v_n=k(n-k+1)f^{k-1}v_n \end{align*} gives \begin{align*} e f^{n+1}v_n=(n+1)(n-(n+1)+1)f^n v_n \end{align*} and the scalar is \begin{align*} (n+1)(n-(n+1)+1)=(n+1)(n-n-1+1)=(n+1)\cdot 0=0. \end{align*} Thus $f^{n+1}v_n$ is singular. Its weight is \begin{align*} h(f^{n+1}v_n)=(n-2(n+1))f^{n+1}v_n=(-n-2)f^{n+1}v_n. \end{align*} Let $N=U(\mathfrak{sl}_2)f^{n+1}v_n$. Since $f$ raises the exponent of $f$ by one, $h$ preserves each weight line, and \begin{align*} e f^k v_n=k(n-k+1)f^{k-1}v_n, \end{align*} the span of the tail vectors \begin{align*} \{f^k v_n:k\ge n+1\} \end{align*} is a submodule, and it is exactly the submodule generated by $f^{n+1}v_n$. Therefore the quotient by this submodule has representatives \begin{align*} \overline{v_n},\overline{fv_n},\overline{f^2v_n},\ldots,\overline{f^n v_n}. \end{align*} They are linearly independent in the quotient because the PBW basis of $M(n)$ splits into the disjoint sets $\{f^k v_n:0\le k\le n\}$ and $\{f^k v_n:k\ge n+1\}$. Hence \begin{align*} \dim L(n)=n+1. \end{align*} The action on these representatives is \begin{align*} f\overline{f^k v_n}=\overline{f^{k+1}v_n} \end{align*} for $0\le k<n$, while $f\overline{f^n v_n}=\overline{f^{n+1}v_n}=0$. Also \begin{align*} e\overline{f^k v_n}=k(n-k+1)\overline{f^{k-1}v_n}. \end{align*} For $1\le k\le n$, the two factors $k$ and $n-k+1$ are both nonzero, so the only basis representative killed by $e$ is $\overline{v_n}$. Thus the quotient is the finite-dimensional irreducible highest weight module of highest weight $n$, while the original Verma module $M(n)$ still has the infinite PBW basis $\{f^k v_n:k\in\mathbb Z_{\ge 0}\}$. [/example] The comparison with finite-dimensional representation theory is now precise: finite-dimensional irreducibles arise among the $L(\lambda)$, not among the Verma modules themselves. For a semisimple Lie algebra, dominant integral highest weights produce the finite-dimensional simple modules studied earlier, while arbitrary weights still produce simple highest weight modules that may be infinite-dimensional. [example: First Verma Modules for sl3] Let $\mathfrak g=\mathfrak{sl}_3(\mathbb C)$ with simple roots $\alpha_1,\alpha_2$ and positive roots $\alpha_1,\alpha_2,\alpha_1+\alpha_2$. Choose negative root vectors $f_{\alpha_1}$, $f_{\alpha_1+\alpha_2}$, and $f_{\alpha_2}$, in that PBW order. By *[PBW Model of a Verma Module](/theorems/9374)*, the vectors \begin{align*} f_{\alpha_1}^{a}f_{\alpha_1+\alpha_2}^{b}f_{\alpha_2}^{c}v_\lambda \qquad a,b,c\in\mathbb Z_{\geq 0} \end{align*} form a basis of $M(\lambda)$. Each factor lowers the weight by its root. Hence \begin{align*} \operatorname{wt}\bigl(f_{\alpha_1}^{a}f_{\alpha_1+\alpha_2}^{b}f_{\alpha_2}^{c}v_\lambda\bigr)=\lambda-a\alpha_1-b(\alpha_1+\alpha_2)-c\alpha_2. \end{align*} Combining the coefficients of $\alpha_1$ and $\alpha_2$ gives \begin{align*} \lambda-a\alpha_1-b(\alpha_1+\alpha_2)-c\alpha_2=\lambda-(a+b)\alpha_1-(b+c)\alpha_2. \end{align*} For total root height $1$, the possibilities are $(a,b,c)=(1,0,0)$ and $(0,0,1)$, giving the weights $\lambda-\alpha_1$ and $\lambda-\alpha_2$. For total root height $2$, the first possibilities include \begin{align*} (a,b,c)=(0,1,0),\ (1,1,0),\ (0,0,2),\ (2,0,0),\ (1,0,1). \end{align*} These give, among others, \begin{align*} \lambda-(\alpha_1+\alpha_2),\quad \lambda-2\alpha_1,\quad \lambda-2\alpha_2. \end{align*} Now compute the multiplicity of the weight $\lambda-(\alpha_1+\alpha_2)$. We need \begin{align*} (a+b)\alpha_1+(b+c)\alpha_2=\alpha_1+\alpha_2. \end{align*} Equating the coefficients of the simple roots gives \begin{align*} a+b=1,\qquad b+c=1. \end{align*} If $b=1$, then $a=0$ and $c=0$, giving the basis vector $f_{\alpha_1+\alpha_2}v_\lambda$. If $b=0$, then $a=1$ and $c=1$, giving the basis vector $f_{\alpha_1}f_{\alpha_2}v_\lambda$ in the chosen PBW order. Therefore \begin{align*} \dim M(\lambda)_{\lambda-(\alpha_1+\alpha_2)}=2. \end{align*} This is the first rank-two place where the multiplicity comes from partitions into positive roots: $\alpha_1+\alpha_2$ can occur either as one positive root or as the sum $\alpha_1+\alpha_2$ of two simple positive roots. [/example] Verma modules therefore supply a uniform construction for highest weight representations, a combinatorial weight formula, and a canonical route to irreducible modules. The next stage of the course studies when these quotients are finite-dimensional and how their characters are computed. # 6. Complete Reducibility and Central Tools Chapters 4 and 5 built simple highest weight modules from Verma modules, while Chapter 2 introduced characters to track their weight multiplicities. This chapter explains why finite-dimensional representation theory of a complex semisimple Lie algebra behaves like linear algebra over a semisimple algebra: every finite-dimensional module splits into irreducibles, and the centre of the enveloping algebra detects the irreducible constituents. The bridge is the Casimir element, a canonical central element of $U(\mathfrak g)$ whose scalar action on highest weight modules gives a powerful splitting tool. Throughout this chapter, $\mathfrak g$ is a finite-dimensional complex semisimple Lie algebra, $\mathfrak h \subset \mathfrak g$ is a Cartan subalgebra, $\Phi$ is the root system, and $\Phi^+$ is a choice of positive roots. We write \begin{align*} \rho=\frac{1}{2}\sum_{\alpha \in \Phi^+}\alpha. \end{align*} We use the invariant [bilinear form](/page/Bilinear%20Form) $(\cdot,\cdot)$ on $\mathfrak h^*$ induced by the Killing form, and write $L(\lambda)$ for the irreducible highest weight module of highest weight $\lambda$ when it is finite-dimensional. ## The Casimir Element as a Central Operator A representation of $\mathfrak g$ may contain many weights and many submodules, so a useful operator should be intrinsic rather than tied to a chosen basis of the module. The question is whether the Lie algebra itself produces canonical endomorphisms of every module. The Casimir element answers this by converting the Killing form into a central element of $U(\mathfrak g)$. [definition: Dual Bases for the Killing Form] Let $\mathfrak g$ be a finite-dimensional semisimple Lie algebra with Killing form $\kappa$. Two ordered bases $(x_i)_{i=1}^n$ and $(x^i)_{i=1}^n$ of $\mathfrak g$ are dual bases for $\kappa$ if \begin{align*} \kappa(x_i,x^j)=\delta_{ij} \end{align*} for all $1 \le i,j \le n$. [/definition] Dual bases are not unique, so the coordinate expression must be used in a way that does not depend on a particular basis. The inverse Killing form gives a canonical tensor, and multiplying its two tensor factors inside $U(\mathfrak g)$ produces the intrinsic element whose action can be compared across all modules in the chapter. The obstruction is that the formula $\sum_i x_i x^i$ appears to depend on coordinates, while representation-theoretic arguments need an operator that is attached to $\mathfrak g$ itself. Naming this basis-independent element separates the construction from the later facts about it: first one produces a well-defined element of the enveloping algebra, and only afterward can one ask whether it is central and how it acts on irreducible modules. [definition: Casimir Element] Let $(x_i)_{i=1}^n$ and $(x^i)_{i=1}^n$ be dual bases of $\mathfrak g$ for the Killing form. The Casimir element is \begin{align*} \Omega = \sum_{i=1}^n x_i x^i \in U(\mathfrak g). \end{align*} [/definition] The definition lies in the associative algebra $U(\mathfrak g)$, so $x_i x^i$ denotes multiplication in $U(\mathfrak g)$ rather than a product in $\mathfrak g$. Although the displayed formula uses a chosen pair of dual bases, the resulting element is independent of that choice: it is the tensor corresponding to the inverse Killing form, transported through the multiplication map $\mathfrak g\otimes \mathfrak g\to U(\mathfrak g)$. This well-definedness is what lets us speak of the Casimir element rather than a Casimir element attached to coordinates. The next result explains why this element is useful on every representation at once. [quotetheorem:8828] [citeproof:8828] Centrality makes $\Omega$ a natural endomorphism of any $\mathfrak g$-module: if $V$ is a $\mathfrak g$-module, then the action of $\Omega$ commutes with every $x\in\mathfrak g$. This conclusion uses semisimplicity through the nondegenerate Killing form; for a non-semisimple Lie algebra the Killing form may be degenerate, so the same construction need not even produce dual bases on the whole algebra. Centrality by itself also does not imply that $\Omega$ acts by a scalar on every module: on a reducible or indecomposable module it can have several eigenspaces or even nontrivial Jordan behaviour, and scalar action is forced only after imposing irreducibility together with [Schur's lemma](/theorems/2414). Splitting arguments therefore need more than centrality: they need numerical information about the eigenvalues. For highest weight modules, the eigenvalue can be read directly from the highest weight, which makes the Casimir usable in computations. [quotetheorem:9380] [citeproof:9380] This formula turns a structural object in $U(\mathfrak g)$ into a computable expression on the weight lattice, but its hypotheses are doing real work. The dominant integral condition is what places $L(\lambda)$ in the finite-dimensional highest weight classification; for a non-dominant highest weight, the corresponding Verma module is usually infinite-dimensional, and the same scalar does not imply finite-dimensional irreducibility. The formula also gives only one central scalar, so it cannot classify simple modules by itself: different highest weights in higher rank may have the same value of the quadratic expression. The point is therefore computational rather than classificatory; the Casimir supplies a useful spectral invariant, while the full centre is needed for complete separation. The simplest case already shows how the abstract normalisation matches familiar $\mathfrak{sl}_2$ calculations. [example: Casimir Calculation for Sl Two] Let $\mathfrak g=\mathfrak{sl}_2(\mathbb C)$ with standard basis $e,f,h$ satisfying $[h,e]=2e$, $[h,f]=-2f$, and $[e,f]=h$. The irreducible module $L(m)$ has highest weight $m\omega$, where $m\in\mathbb N\cup\{0\}$ and $\omega$ is the fundamental weight. For $\mathfrak{sl}_2$, the unique positive root is $\alpha=2\omega$, so \begin{align*} \rho=\frac{1}{2}\alpha=\omega. \end{align*} By the *Casimir Eigenvalue Formula*, the Casimir scalar on $L(m)$ is \begin{align*} (m\omega,m\omega+2\omega)=(m\omega,(m+2)\omega). \end{align*} Using bilinearity of $(\cdot,\cdot)$, \begin{align*} (m\omega,(m+2)\omega)=m(m+2)(\omega,\omega). \end{align*} In the rescaled normalisation \begin{align*} C=ef+fe+\frac{1}{2}h^2, \end{align*} let $v_m$ be a highest weight vector in $L(m)$, so $ev_m=0$ and $hv_m=mv_m$. Then \begin{align*} efv_m=([e,f]+fe)v_m=hv_m+f(ev_m)=mv_m+0=mv_m. \end{align*} Also, \begin{align*} fev_m=f(ev_m)=0. \end{align*} Finally, \begin{align*} h^2v_m=h(hv_m)=h(mv_m)=m(hv_m)=m^2v_m. \end{align*} Therefore \begin{align*} Cv_m=mv_m+0+\frac{1}{2}m^2v_m=\frac{1}{2}m(m+2)v_m. \end{align*} Thus the rescaled quadratic Casimir acts on $L(m)$ by the scalar $\frac{1}{2}m(m+2)$, matching the highest-weight formula after the corresponding change of normalisation. [/example] The scalar separates many irreducibles, but not every pair in every root system. The broader principle is that the whole centre $Z(U(\mathfrak g))$, rather than just $\Omega$, controls central characters. ## Weyl Complete Reducibility The main structural problem is whether a finite-dimensional representation can contain a submodule that does not have an invariant complement. For arbitrary Lie algebras this happens frequently, even in dimension two. Semisimplicity of $\mathfrak g$ rules out such extensions, and the proof uses the Casimir element to split modules. [definition: Completely Reducible Module] Let $\mathfrak g$ be a Lie algebra and let $V$ be a finite-dimensional $\mathfrak g$-module. The module $V$ is completely reducible if there exist simple $\mathfrak g$-submodules $V_1,\dots,V_r \subset V$ such that \begin{align*} V = V_1 \oplus \cdots \oplus V_r. \end{align*} [/definition] Complete reducibility says more than the existence of composition factors: it says the composition factors can be realised simultaneously as submodules. The obstruction is the possible presence of nonsplit extensions, where a submodule exists but every candidate complement fails to be stable under the Lie algebra action. For finite-dimensional modules over a semisimple Lie algebra, Weyl's theorem removes exactly this obstruction. [quotetheorem:3817] [citeproof:3817] The theorem means that finite-dimensional representation theory of semisimple Lie algebras is governed by simple modules and their multiplicities, but both adjectives in the theorem are essential. If $\mathfrak g$ is not semisimple, extensions need not split: for the one-dimensional abelian Lie algebra with basis $x$, the two-dimensional module on which $x$ acts by a single nilpotent Jordan block has a one-dimensional submodule with no invariant complement. If finite-dimensionality is dropped, semisimple Lie algebras also have nonsplit modules; Verma modules have proper submodules whose quotient is simple, but those submodules usually do not split off. Thus Weyl complete reducibility is a finite-dimensional theorem for semisimple Lie algebras, not a general property of all Lie algebra representations. It also justifies using characters additively, because every finite-dimensional character is a finite sum of irreducible characters. [example: Complete Reducibility for Sl Two Tensor Representations] Let $V=L(1)$ have basis $v_+,v_-$ with \begin{align*} ev_+=0,\quad fv_+=v_-,\quad hv_+=v_+,\quad ev_-=v_+,\quad fv_-=0,\quad hv_-=-v_-. \end{align*} On $V\otimes V$ we use the diagonal action $x(u\otimes v)=(xu)\otimes v+u\otimes (xv)$. The symmetric square has basis \begin{align*} v_+\otimes v_+,\quad s=v_+\otimes v_-+v_-\otimes v_+,\quad v_-\otimes v_-, \end{align*} and the action preserves this span because \begin{align*} e(v_+\otimes v_+)=0,\quad f(v_+\otimes v_+)=v_-\otimes v_+ + v_+\otimes v_-=s,\quad h(v_+\otimes v_+)=2(v_+\otimes v_+). \end{align*} Also \begin{align*} es=e(v_+\otimes v_-)+e(v_-\otimes v_+)=v_+\otimes v_+ + v_+\otimes v_+=2(v_+\otimes v_+), \end{align*} \begin{align*} fs=f(v_+\otimes v_-)+f(v_-\otimes v_+)=v_-\otimes v_-+v_-\otimes v_-=2(v_-\otimes v_-), \end{align*} and \begin{align*} hs=h(v_+\otimes v_-)+h(v_-\otimes v_+)=(v_+\otimes v_- - v_+\otimes v_-)+(-v_-\otimes v_+ + v_-\otimes v_+)=0. \end{align*} Finally, \begin{align*} e(v_-\otimes v_-)=v_+\otimes v_-+v_-\otimes v_+=s,\quad f(v_-\otimes v_-)=0,\quad h(v_-\otimes v_-)=-2(v_-\otimes v_-). \end{align*} Thus $S^2V$ is a three-dimensional highest weight module with highest weight $2$, so $S^2V\cong L(2)$. The alternating square is spanned by \begin{align*} a=v_+\otimes v_- - v_-\otimes v_+. \end{align*} Its action is \begin{align*} ea=e(v_+\otimes v_-)-e(v_-\otimes v_+)=v_+\otimes v_+ - v_+\otimes v_+=0, \end{align*} \begin{align*} fa=f(v_+\otimes v_-)-f(v_-\otimes v_+)=v_-\otimes v_- - v_-\otimes v_-=0, \end{align*} and \begin{align*} ha=h(v_+\otimes v_-)-h(v_-\otimes v_+)=(v_+\otimes v_- - v_+\otimes v_-)-(-v_-\otimes v_+ + v_-\otimes v_+)=0. \end{align*} So $\Lambda^2V$ is the one-dimensional trivial module $L(0)$. Since every tensor $u\otimes v$ is the sum of its symmetric and alternating parts, \begin{align*} u\otimes v=\frac{1}{2}(u\otimes v+v\otimes u)+\frac{1}{2}(u\otimes v-v\otimes u), \end{align*} and a tensor that is both symmetric and alternating must equal its negative, the intersection is zero. Therefore \begin{align*} V\otimes V=S^2V\oplus \Lambda^2V\cong L(2)\oplus L(0), \end{align*} giving an explicit decomposition into stable irreducible summands. [/example] Tensor products for larger Lie algebras behave the same way structurally, although computing the summands requires more weight combinatorics. The natural representation of $\mathfrak{sl}_3$ gives the first example where several fundamental weights appear. [example: Complete Reducibility for Sl Three Tensor Representations] Let $V=\mathbb C^3$ have basis $v_1,v_2,v_3$, and let $E_{ij}$ act by $E_{ij}v_k=\delta_{jk}v_i$. We take the positive root vectors to be $E_{12},E_{23},E_{13}$, so $v_1$ is a highest weight vector of weight $\omega_1$. The flip map $\tau(u\otimes v)=v\otimes u$ commutes with the diagonal action because \begin{align*} \tau(xu\otimes v+u\otimes xv)=v\otimes xu+xv\otimes u=x(v\otimes u). \end{align*} Thus the $+1$ and $-1$ eigenspaces of $\tau$, namely $S^2V$ and $\Lambda^2V$, are $\mathfrak{sl}_3$-submodules. For the symmetric square, $v_1\otimes v_1$ has weight $2\omega_1$, and \begin{align*} E_{12}(v_1\otimes v_1)=(E_{12}v_1)\otimes v_1+v_1\otimes(E_{12}v_1)=0. \end{align*} Similarly $E_{23}(v_1\otimes v_1)=0$ and $E_{13}(v_1\otimes v_1)=0$, so $v_1\otimes v_1$ is a highest weight vector. Applying lowering operators gives \begin{align*} E_{21}(v_1\otimes v_1)=v_2\otimes v_1+v_1\otimes v_2. \end{align*} Then \begin{align*} E_{21}(v_2\otimes v_1+v_1\otimes v_2)=2(v_2\otimes v_2). \end{align*} Also, \begin{align*} E_{32}(v_2\otimes v_1+v_1\otimes v_2)=v_3\otimes v_1+v_1\otimes v_3. \end{align*} Applying $E_{32}$ once more to $v_3\otimes v_1+v_1\otimes v_3$ gives $0$, while \begin{align*} E_{32}(v_2\otimes v_2)=v_3\otimes v_2+v_2\otimes v_3. \end{align*} Finally, \begin{align*} E_{32}(v_3\otimes v_2+v_2\otimes v_3)=2(v_3\otimes v_3). \end{align*} These six symmetric tensors form a basis of $S^2V$, so the submodule generated by $v_1\otimes v_1$ is all of $S^2V$. Hence $S^2V$ is the irreducible highest weight module $L(2\omega_1)$. For the alternating square, set \begin{align*} a_{12}=v_1\otimes v_2-v_2\otimes v_1. \end{align*} Its weight is $\omega_2$, since it is the sum of the weights of $v_1$ and $v_2$. The positive root vectors kill it: \begin{align*} E_{12}a_{12}=v_1\otimes v_1-v_1\otimes v_1=0. \end{align*} Also, \begin{align*} E_{23}a_{12}=0\otimes v_2+v_1\otimes 0-0\otimes v_1-v_2\otimes 0=0. \end{align*} And \begin{align*} E_{13}a_{12}=0. \end{align*} Thus $a_{12}$ is a highest weight vector of weight $\omega_2$. Lowering gives \begin{align*} E_{32}a_{12}=v_1\otimes v_3-v_3\otimes v_1. \end{align*} Also, \begin{align*} E_{21}(v_1\otimes v_3-v_3\otimes v_1)=v_2\otimes v_3-v_3\otimes v_2. \end{align*} The three tensors $a_{12}$, $v_1\otimes v_3-v_3\otimes v_1$, and $v_2\otimes v_3-v_3\otimes v_2$ form a basis of $\Lambda^2V$, so $\Lambda^2V\cong L(\omega_2)$. The map $\Lambda^2V\to V^*$ sending $v_i\wedge v_j$ to the functional $v_k^*$ with sign determined by $v_i\wedge v_j\wedge v_k=v_1\wedge v_2\wedge v_3$ is $\mathfrak{sl}_3$-equivariant, so this is the usual identification $\Lambda^2V\cong V^*$. Every tensor decomposes as \begin{align*} u\otimes v=\frac{1}{2}(u\otimes v+v\otimes u)+\frac{1}{2}(u\otimes v-v\otimes u). \end{align*} If a tensor is both symmetric and alternating, then $\tau(t)=t$ and $\tau(t)=-t$, so $2t=0$ and hence $t=0$ over $\mathbb C$. Therefore \begin{align*} V\otimes V=S^2V\oplus \Lambda^2V\cong L(2\omega_1)\oplus L(\omega_2). \end{align*} This explicit decomposition shows that the tensor product splits into stable highest weight summands rather than remaining an undifferentiated $9$-dimensional module. [/example] Weyl's theorem is special to finite-dimensional modules. Infinite-dimensional highest weight modules, such as Verma modules, have submodules that need not split, and this failure is part of why category $\mathcal O$ has richer homological structure. [remark: Finite Dimensional Hypothesis] The finite-dimensional hypothesis in Weyl's theorem is essential. A Verma module $M(\lambda)$ has a unique simple quotient $L(\lambda)$, but the kernel of the quotient map is usually not a direct summand. Thus complete reducibility belongs to the finite-dimensional theory developed in this part of the course. [/remark] This distinction explains why finite-dimensional character theory is so effective: extensions disappear in the category under consideration. To identify the simple summands, however, we still need tools that tell different irreducibles apart. ## Schur's Lemma and Central Characters Once every module splits into simples, the next problem is to distinguish those simple modules by intrinsic operators. Central elements of $U(\mathfrak g)$ commute with the whole $\mathfrak g$-action, so Schur's lemma forces them to act by scalars on irreducibles. Collecting these scalars gives the central character. [quotetheorem:3824] [citeproof:3824] Schur's lemma applies to central elements because their action commutes with the action of $\mathfrak g$, but the proof also shows where the hypotheses enter. The scalar endomorphism conclusion uses that the base field is algebraically closed, since the argument chooses an eigenvalue of $T$; over a non-[algebraically closed field](/page/Algebraically%20Closed%20Field), a simple module can have a larger division algebra of endomorphisms. Finite-dimensionality is used for the same eigenvalue step, and for infinite-dimensional simple modules an endomorphism need not have an eigenvalue. The lemma also says nothing scalar about a reducible module as a whole: central elements may act by different scalars on different simple summands, or by more complicated operators before complete reducibility is known. This motivates packaging their scalar actions on each simple module into a homomorphism from the centre. [definition: Central Character] Let $V$ be a finite-dimensional simple $\mathfrak g$-module. The central character of $V$ is the algebra homomorphism \begin{align*} \chi_V:Z(U(\mathfrak g))\to \mathbb C \end{align*} defined by $zv=\chi_V(z)v$ for all $z\in Z(U(\mathfrak g))$ and all $v\in V$. [/definition] For a highest weight simple module $L(\lambda)$, we also write $\chi_\lambda$ for $\chi_{L(\lambda)}$. The Casimir eigenvalue formula gives one value of this character, but classifying simple modules requires knowing whether the full central character identifies the highest weight. [quotetheorem:9381] [citeproof:9381] [example: Casimir Does Not Replace the Full Centre] Take $\mathfrak g=\mathfrak{sl}_3(\mathbb C)$, and write the fundamental weights as $\omega_1,\omega_2$. With the usual normalisation for type $A_2$, \begin{align*} (\omega_1,\omega_1)=\frac{2}{3},\quad (\omega_2,\omega_2)=\frac{2}{3},\quad (\omega_1,\omega_2)=\frac{1}{3}. \end{align*} Also $\rho=\omega_1+\omega_2$. For $\lambda=\omega_1$, the Casimir scalar is \begin{align*} (\omega_1,\omega_1+2\rho)=(\omega_1,3\omega_1+2\omega_2). \end{align*} Using bilinearity, \begin{align*} (\omega_1,3\omega_1+2\omega_2)=3(\omega_1,\omega_1)+2(\omega_1,\omega_2)=3\cdot\frac{2}{3}+2\cdot\frac{1}{3}=\frac{8}{3}. \end{align*} For $\mu=\omega_2$, the Casimir scalar is \begin{align*} (\omega_2,\omega_2+2\rho)=(\omega_2,2\omega_1+3\omega_2), \end{align*} and bilinearity gives \begin{align*} (\omega_2,2\omega_1+3\omega_2)=2(\omega_2,\omega_1)+3(\omega_2,\omega_2)=2\cdot\frac{1}{3}+3\cdot\frac{2}{3}=\frac{8}{3}. \end{align*} Thus $L(\omega_1)$ and $L(\omega_2)$ have the same Casimir eigenvalue. They are nevertheless different simple modules. In the standard realisation of $A_2$, the shifted weights $\omega_1+\rho=2\omega_1+\omega_2$ and $\omega_2+\rho=\omega_1+2\omega_2$ correspond to the triples \begin{align*} \left(\frac{5}{3},-\frac{1}{3},-\frac{4}{3}\right)\quad\text{and}\quad\left(\frac{4}{3},\frac{1}{3},-\frac{5}{3}\right) \end{align*} respectively. These triples are not permutations of one another, so the shifted weights are not in the same Weyl-group orbit. Since the full centre separates simple highest weight modules by the Weyl orbit of the shifted weight, the full central character separates $L(\omega_1)$ from $L(\omega_2)$. The example shows that the quadratic Casimir records only one central scalar, while the full centre records enough scalar data to distinguish the two simple modules. [/example] Central characters also organise arbitrary finite-dimensional modules after Weyl complete reducibility. If $V\cong \bigoplus_i L(\lambda_i)^{\oplus m_i}$, then the action of $Z(U(\mathfrak g))$ on $V$ is diagonal by central character, and the multiplicity spaces are the simultaneous eigenspaces for the centre. This is the Lie-theoretic analogue of decomposing a semisimple associative algebra module by simultaneous spectral data from a commutative algebra of operators. In later infinite-dimensional settings, especially category $\mathcal O$, the same central-character blocks remain important even though the modules inside a block need not split into simples. [remark: Role in the Rest of the Course] Complete reducibility lets us reduce finite-dimensional questions to irreducible highest weight modules. Central characters then provide intrinsic labels for those irreducibles, while the Casimir element gives a computable central operator used in splitting arguments and examples. These tools prepare for the later character formulae, where the remaining task is to compute the multiplicities inside each $L(\lambda)$ and inside tensor products. [/remark] # 7. Weyl Character Formula This chapter turns the classification of finite-dimensional highest weight modules into computable formulas. It assumes the earlier construction of highest weight modules $M(\lambda)$ and $L(\lambda)$, the root-space decomposition of a complex semisimple Lie algebra, the weight lattice $P$, dominant integral weights $P^+$, the Weyl group $W$, and the Weyl chamber picture governed by a choice of positive roots. Chapters 4 through 6 classified finite-dimensional irreducibles by highest weight and justified reducing to them; the actual character still contains hidden multiplicities. The Weyl character formula resolves this by comparing a symmetric finite-dimensional character with an alternating sum over the Weyl group. The denominator identity is the algebraic device that makes the cancellation visible. ## Alternating Sums and the Weyl Denominator The first problem is to package the Weyl [group action](/page/Group%20Action) in a way that remembers signs. Ordinary Weyl-invariant sums are too symmetric to isolate a highest weight, while alternating sums vanish on weights fixed by a reflection and therefore detect walls of Weyl chambers. Let $\mathfrak g$ be a finite-dimensional complex semisimple Lie algebra, with Cartan subalgebra $\mathfrak h$, root system $\Phi \subset \mathfrak h^*$, chosen positive roots $\Phi^+$, Weyl group $W$, and weight lattice $P$. The choice of $\Phi^+$ determines the triangular decomposition \begin{align*} \mathfrak g=\mathfrak n^-\oplus\mathfrak h\oplus\mathfrak n^+, \end{align*} where $\mathfrak n^+=\bigoplus_{\alpha\in\Phi^+}\mathfrak g_\alpha$ and $\mathfrak n^-=\bigoplus_{\alpha\in\Phi^+}\mathfrak g_{-\alpha}$. Write \begin{align*} \rho = \frac{1}{2}\sum_{\alpha \in \Phi^+}\alpha. \end{align*} For a weight $\mu \in P$, let $e^\mu$ denote the corresponding basis element in the group algebra $\mathbb Z[P]$. The sign representation is the part of the Weyl group action that records whether an element preserves or reverses orientation on the real span of the roots. [definition: Weyl Alternating Sum] The Weyl alternating-sum map is the function \begin{align*} A:P&\longrightarrow \mathbb Z[P],& \mu&\longmapsto A_\mu:=\sum_{w \in W} \det(w)e^{w\mu}. \end{align*} For each $\mu\in P$, the element $A_\mu$ is called the Weyl alternating sum of $\mu$. [/definition] The defining feature of $A_\mu$ is anti-invariance: applying $v \in W$ sends $A_\mu$ to $\det(v)A_\mu$. This already explains why $A_\mu=0$ whenever $\mu$ lies on a reflecting hyperplane, since the corresponding reflection fixes $\mu$ but changes the sign of the alternating sum. [example: Alternating Sum For Type A One] For $\mathfrak{sl}_2$, the Weyl group is $W=\{1,s\}$, with $\det(1)=1$ and $\det(s)=-1$. If the fundamental weight is $\omega$, then the simple reflection acts by \begin{align*} s(m\omega)=-m\omega. \end{align*} Using the definition $A_\mu=\sum_{w\in W}\det(w)e^{w\mu}$ with $\mu=m\omega$, the two Weyl group terms are \begin{align*} \det(1)e^{1(m\omega)}=e^{m\omega}. \end{align*} The reflection term is \begin{align*} \det(s)e^{s(m\omega)}=-e^{-m\omega}. \end{align*} Therefore \begin{align*} A_{m\omega}=e^{m\omega}-e^{-m\omega}. \end{align*} When $m=0$, this becomes \begin{align*} A_0=e^0-e^0=0. \end{align*} When $m>0$, the two weights $m\omega$ and $-m\omega$ are distinct, so the alternating sum records the two points in the Weyl orbit with opposite signs. [/example] The type $A_1$ computation shows that alternating sums have a built-in factor measuring distance from the wall. In higher rank, the corresponding wall factors should be contributed by all positive roots. The next theorem identifies the alternating sum $A_\rho$ with exactly that product, giving the denominator that will later divide the highest-weight numerator. [quotetheorem:9382] [citeproof:9382] The product form is useful because it resembles the character of a Verma module denominator. The alternating-sum form is useful because it transforms predictably under $W$. The regularity of $\rho$ is essential here: if a weight lies on a reflecting hyperplane, its alternating sum vanishes, so it cannot serve as a nonzero denominator. For example, in type $A_1$ the weight $0$ is fixed by the simple reflection, and \begin{align*} A_0=e^0-e^0=0. \end{align*} The formula also depends on the chosen positive system, but changing the positive roots changes both sides compatibly through the Weyl group action. The Weyl character formula is obtained by putting a highest weight numerator over this denominator. [remark: Regularity Of Rho] The weight $\rho$ lies strictly inside the dominant Weyl chamber: for every simple coroot $\alpha_i^\vee$, one has $\langle \rho,\alpha_i^\vee\rangle=1$. Therefore no non-identity Weyl group element fixes $\rho$, so $A_\rho$ is nonzero. [/remark] ## The Character Formula from Cancellation The next question is how to recover the character of $L(\lambda)$ from the universal highest weight object with highest weight $\lambda$. Verma modules have simple characters, but they are infinite-dimensional; finite-dimensional irreducibles arise after quotienting out submodules generated by reflected singular vectors. The Weyl character formula says that the finite quotient is obtained by an alternating cancellation over the Weyl group. [definition: Formal Character] Let $\mathcal C_{\mathrm{wt}}$ be the class of weight $\mathfrak g$-modules $M=\bigoplus_{\mu\in P}M_\mu$ with $\dim M_\mu<\infty$ for every $\mu$, and such that the set of weights is contained in a finite union of sets of the form $\lambda-Q^+$, where $Q^+$ is the nonnegative integer span of $\Phi^+$. Let $\widehat{\mathbb Z[P]}$ denote the corresponding completion of the group algebra, allowing infinite sums supported in such downward cones. The formal character is the map \begin{align*} \operatorname{ch}:\mathcal C_{\mathrm{wt}}\longrightarrow \widehat{\mathbb Z[P]} \end{align*} defined by $\operatorname{ch}(M)=\sum_{\mu\in P}\dim(M_\mu)e^\mu$. [/definition] For finite-dimensional modules the image lies in the ordinary group algebra $\mathbb Z[P]$. For Verma modules it is interpreted in the completed group algebra because infinitely many weights occur in negative root directions. The finite-dimensionality of each weight space is part of the definition, since otherwise the coefficient of $e^\mu$ would not be an integer multiplicity. Since the later cancellation formula is built from Verma modules, we first need their character in a product form compatible with the Weyl denominator. [quotetheorem:9383] [citeproof:9383] This formula turns the problem of characters into the problem of deciding which Verma module characters survive in the irreducible quotient. Its hypotheses matter in two separate ways: finite-dimensional weight spaces are needed so that each coefficient is a finite multiplicity, and the completed group algebra is needed because the product expands into infinitely many terms below $\lambda$. In type $A_1$, for instance, \begin{align*} \operatorname{ch}M(\lambda)=e^\lambda(1+e^{-\alpha}+e^{-2\alpha}+\cdots), \end{align*} so even the simplest Verma module character is not an element of the ordinary group algebra $\mathbb Z[P]$. If a weight module had an infinite-dimensional highest weight space, the coefficient of $e^\lambda$ would be infinite and the formal character would not land in $\widehat{\mathbb Z[P]}$ with integer coefficients. The theorem also has a strict limitation: it does not determine the composition factors or submodule lattice of $M(\lambda)$; it records only the sizes of the weight spaces before quotienting. The cancellation pattern is governed by the shifted, or dot, Weyl action. [definition: Dot Action] The dot action is the map \begin{align*} W\times \mathfrak h^*&\longrightarrow \mathfrak h^*,& (w,\lambda)&\longmapsto w\cdot\lambda:=w(\lambda+\rho)-\rho. \end{align*} For each $w\in W$, the corresponding affine map is \begin{align*} \mathfrak h^*&\longrightarrow \mathfrak h^*,& \lambda&\longmapsto w(\lambda+\rho)-\rho. \end{align*} [/definition] The definition records the shift by $\rho$ that already appeared in the denominator formula. This is needed because reducible Verma modules are linked across shifted chamber walls, so the cancellation terms have highest weights $w\cdot\lambda$ rather than $w\lambda$. The character formula below uses the course's BGG-style cancellation theorem for finite-dimensional highest weight modules as an input: for dominant integral $\lambda$, the irreducible $L(\lambda)$ has an alternating Verma-module character expansion indexed by $W$. The theorem packages that cancellation into a quotient whose numerator is the alternating sum through $\lambda+\rho$ and whose denominator is the Weyl denominator. [quotetheorem:9384] [citeproof:9384] The formula has two important built-in consistency checks. First, the quotient is $W$-invariant, because both numerator and denominator are anti-invariant. Second, the highest term is $e^\lambda$ with coefficient $1$, since the identity term in $A_{\lambda+\rho}$ divided by the identity term in $A_\rho$ gives the highest weight contribution. The hypothesis $\lambda\in P^+$ is not cosmetic: if $\lambda$ is integral but not dominant, then a singular shifted weight $\lambda+\rho$ gives zero numerator, while a regular shifted weight can be moved by the dot action to the dominant chamber at the cost of a sign. In neither case is the expression the character of a highest-weight finite-dimensional module with highest weight $\lambda$. If $\lambda$ is not integral, the expression need not lie in the integral group algebra of the weight lattice, so it is not a character of a representation with integral weights. The quotient gives all multiplicities in principle, but it does not display them term-by-term without further expansion; this is why the dimension and Kostant multiplicity formulas are needed next. [example: Characters Of The Natural And Dual Modules For Sl Three] For $\mathfrak{sl}_3$, write the weights of the natural module as $\varepsilon_1,\varepsilon_2,\varepsilon_3$ with simple roots $\alpha_1=\varepsilon_1-\varepsilon_2$ and $\alpha_2=\varepsilon_2-\varepsilon_3$. The highest weight is $\varepsilon_1=\omega_1$, and the remaining two weights are obtained by subtracting simple roots: \begin{align*} \varepsilon_2=\varepsilon_1-\alpha_1=\omega_1-\alpha_1. \end{align*} \begin{align*} \varepsilon_3=\varepsilon_2-\alpha_2=\omega_1-\alpha_1-\alpha_2. \end{align*} Each weight space is one-dimensional, so \begin{align*} \operatorname{ch}L(\omega_1)=e^{\omega_1}+e^{\omega_1-\alpha_1}+e^{\omega_1-\alpha_1-\alpha_2}. \end{align*} The dual module has weights $-\varepsilon_3,-\varepsilon_2,-\varepsilon_1$, with highest weight $-\varepsilon_3=\omega_2$. Since $\alpha_2=\varepsilon_2-\varepsilon_3$ and $\alpha_1+\alpha_2=\varepsilon_1-\varepsilon_3$, we have \begin{align*} -\varepsilon_2=-\varepsilon_3-\alpha_2=\omega_2-\alpha_2. \end{align*} \begin{align*} -\varepsilon_1=-\varepsilon_3-(\alpha_1+\alpha_2)=\omega_2-\alpha_1-\alpha_2. \end{align*} Thus \begin{align*} \operatorname{ch}L(\omega_2)=e^{\omega_2}+e^{\omega_2-\alpha_2}+e^{\omega_2-\alpha_1-\alpha_2}. \end{align*} These formulas match the Weyl character quotient in type $A_2$. For $L(\omega_1)$, the proposed character is $e^{\omega_1}(1+e^{-\alpha_1}+e^{-\alpha_1-\alpha_2})$, and the denominator product is $e^\rho(1-e^{-\alpha_1})(1-e^{-\alpha_2})(1-e^{-\alpha_1-\alpha_2})$. Put $x=e^{-\alpha_1}$ and $y=e^{-\alpha_2}$. Then the product of the proposed character with the denominator factor is \begin{align*} e^{\omega_1+\rho}(1+x+xy)(1-x)(1-y)(1-xy). \end{align*} Expanding one factor at a time gives \begin{align*} (1+x+xy)(1-x)=1+xy-x^2-x^2y. \end{align*} \begin{align*} (1+xy-x^2-x^2y)(1-y)=1-y+xy-xy^2-x^2+x^2y^2. \end{align*} \begin{align*} (1-y+xy-xy^2-x^2+x^2y^2)(1-xy)=1-y-x^2+x^2y^3+x^3y-x^3y^3. \end{align*} Therefore the numerator is the six-term alternating orbit sum \begin{align*} e^{\omega_1+\rho}(1-y-x^2+x^2y^3+x^3y-x^3y^3)=A_{\omega_1+\rho}. \end{align*} The same computation with $\alpha_1$ and $\alpha_2$ interchanged gives the numerator for $L(\omega_2)$, so the two displayed characters are exactly the Weyl character formula specializations for the natural and dual representations. [/example] The first non-minuscule example in type $A_2$ shows that multiplicities are not always $1$. The character formula encodes this extra information through cancellations rather than by listing weights directly. [example: The Adjoint Character For Sl Three] For $\mathfrak{sl}_3$, the roots are \begin{align*} \Phi=\{\alpha_1,\alpha_2,\alpha_1+\alpha_2,-\alpha_1,-\alpha_2,-\alpha_1-\alpha_2\}. \end{align*} Since $\alpha_1=2\omega_1-\omega_2$ and $\alpha_2=-\omega_1+2\omega_2$, the highest root is \begin{align*} \alpha_1+\alpha_2=(2\omega_1-\omega_2)+(-\omega_1+2\omega_2)=\omega_1+\omega_2. \end{align*} Thus the adjoint representation has highest weight $\omega_1+\omega_2$. Using the root-space decomposition, \begin{align*} \mathfrak{sl}_3=\mathfrak h\oplus\bigoplus_{\alpha\in\Phi}\mathfrak g_\alpha. \end{align*} The Cartan subalgebra $\mathfrak h$ has weight $0$ and dimension $2$, while each root space $\mathfrak g_\alpha$ is one-dimensional with weight $\alpha$. Hence the character is \begin{align*} \operatorname{ch}L(\omega_1+\omega_2)=2e^0+e^{\alpha_1}+e^{\alpha_2}+e^{\alpha_1+\alpha_2}+e^{-\alpha_1}+e^{-\alpha_2}+e^{-\alpha_1-\alpha_2}. \end{align*} Equivalently, \begin{align*} \operatorname{ch}L(\omega_1+\omega_2)=2e^0+\sum_{\alpha\in\Phi}e^\alpha. \end{align*} The repeated zero weight records the two-dimensional Cartan subalgebra, while the six remaining weight spaces are exactly the six root spaces. [/example] ## Dimension and Multiplicity Consequences Once the character is known as a quotient of alternating sums, the next question is what numerical information can be extracted without expanding the whole character. Setting all exponentials equal to $1$ should give the dimension, but both numerator and denominator vanish at that point. The Weyl dimension formula evaluates the limiting quotient. [quotetheorem:9385] [citeproof:9385] The formula reduces a representation-theoretic dimension to a finite product of integers and rational numbers determined by the root system. Dominance is again necessary for the representation-theoretic interpretation: outside $P^+$, the product may still be a formal signed expression, but it is not the dimension of a finite-dimensional highest weight module with that highest weight. In type $A_1$, taking $\lambda=-2\omega$ gives $\lambda+\rho=-\omega$, and the product equals $-1$; this signed value cannot be the dimension of any finite-dimensional representation. Taking $\lambda=-\omega$ gives a zero numerator factor, reflecting that $\lambda+\rho=0$ lies on the wall rather than producing a highest-weight module of dimension $0$. The denominator factors never vanish for positive roots because $\rho$ is regular, while the numerator records how far $\lambda+\rho$ lies from each wall. The formula also gives only the total dimension, so it cannot distinguish different characters with the same dimension. In simply laced type $A$, the coroots are the roots, and the expression becomes especially concrete. [example: Fundamental Dimensions For Sl N] For $\mathfrak{sl}_n$, the irreducible module $L(\omega_k)$ is the $k$th exterior power of the natural module. We compute its dimension from the *Weyl Dimension Formula*. The positive roots are $\varepsilon_i-\varepsilon_j$ with $1\le i<j\le n$, and in type $A_{n-1}$ one has \begin{align*} \langle \rho,\varepsilon_i-\varepsilon_j\rangle=j-i. \end{align*} The fundamental weight $\omega_k$ pairs with $\varepsilon_i-\varepsilon_j$ by \begin{align*} \langle \omega_k,\varepsilon_i-\varepsilon_j\rangle=1 \text{ if } i\le k<j, \text{ and } \langle \omega_k,\varepsilon_i-\varepsilon_j\rangle=0 \text{ otherwise.} \end{align*} Therefore every positive root not crossing from an index $i\le k$ to an index $j>k$ contributes the factor $1$, and the dimension product becomes \begin{align*} \dim L(\omega_k)=\prod_{i=1}^{k}\prod_{j=k+1}^{n}\frac{j-i+1}{j-i}. \end{align*} For a fixed $i$, the [inner product](/page/Inner%20Product) is \begin{align*} \prod_{j=k+1}^{n}\frac{j-i+1}{j-i}=\frac{k-i+2}{k-i+1}\cdot\frac{k-i+3}{k-i+2}\cdots\frac{n-i+1}{n-i}. \end{align*} All intermediate factors cancel, so \begin{align*} \prod_{j=k+1}^{n}\frac{j-i+1}{j-i}=\frac{n-i+1}{k-i+1}. \end{align*} Substituting this into the outer product gives \begin{align*} \dim L(\omega_k)=\prod_{i=1}^{k}\frac{n-i+1}{k-i+1}. \end{align*} The numerator is $n(n-1)\cdots(n-k+1)=n!/(n-k)!$, and the denominator is $k(k-1)\cdots 1=k!$, hence \begin{align*} \dim L(\omega_k)=\frac{n!}{k!(n-k)!}=\binom{n}{k}. \end{align*} This agrees with the realization $L(\omega_k)\cong \bigwedge^k\mathbb C^n$, whose standard basis vectors are indexed by the $k$-element subsets of $\{1,\dots,n\}$. [/example] Characters also determine multiplicities of individual weights, and the adjoint example shows why this matters: dimension alone cannot distinguish a repeated zero weight from many different nonzero weights. The Weyl formula gives a direct route by expanding a rational expression. The next theorem isolates the coefficient of a chosen weight in terms of the number of ways to subtract positive roots. [quotetheorem:9386] [citeproof:9386] This statement separates two effects that are merged in the character formula: the partition function counts all ways of lowering a weight inside a Verma module, while the alternating Weyl sum subtracts the contributions belonging to submodules. The dominant integral hypothesis is needed because the finite-dimensional irreducible $L(\lambda)$ is the object whose multiplicities are being computed; for a general weight, the same coefficient extraction describes a formal cancellation rather than actual finite-dimensional weight spaces. A sharp failure already appears in type $A_1$: with $\lambda=-2\omega$, the dimension formula gives $-1$, and Kostant's coefficient extraction is attached to a signed dot-action cancellation rather than to an irreducible finite-dimensional highest-weight module. With $\lambda=-\omega$, the shifted weight $\lambda+\rho=0$ is singular, so the Weyl numerator vanishes and the expression cannot be a character with highest term $e^{-\omega}$. There is also a support condition hidden in the partition function: if $w(\lambda+\rho)-(\mu+\rho)$ is not in $Q^+$, then its contribution is $0$. For instance, taking $\mu$ outside the convex hull of the weights of $L(\lambda)$ makes every signed term vanish, so the formula returns no weight space rather than a negative multiplicity. The formula is exact, but it is not always efficient, because computing $p(\nu)$ can itself require enumerating many decompositions into positive roots. This [multiplicity formula](/theorems/2420) is most useful when only a small part of the character is needed. It also explains why multiplicities can be larger than $1$: different decompositions into positive roots can contribute to the same weight, while Weyl group signs remove the parts forced by submodules. [example: Zero Weight Multiplicity In The Sl Three Adjoint Module] For type $A_2$, take the positive roots to be $\alpha_1,\alpha_2,\alpha_1+\alpha_2$. In the adjoint representation the highest weight is the highest root \begin{align*} \lambda=\omega_1+\omega_2=\alpha_1+\alpha_2. \end{align*} Also $\rho=\omega_1+\omega_2=\alpha_1+\alpha_2$, so for $\mu=0$ the *[Kostant Multiplicity Formula](/theorems/9386)* gives \begin{align*} \dim L(\lambda)_0=\sum_{w\in W}\det(w)p\bigl(w(2\alpha_1+2\alpha_2)-(\alpha_1+\alpha_2)\bigr). \end{align*} Use $s_1(\alpha_1)=-\alpha_1$, $s_1(\alpha_2)=\alpha_1+\alpha_2$, $s_2(\alpha_1)=\alpha_1+\alpha_2$, and $s_2(\alpha_2)=-\alpha_2$. The identity contribution has argument \begin{align*} 2\alpha_1+2\alpha_2-(\alpha_1+\alpha_2)=\alpha_1+\alpha_2. \end{align*} There are exactly two decompositions of $\alpha_1+\alpha_2$ into nonnegative sums of positive roots: \begin{align*} \alpha_1+\alpha_2=(\alpha_1+\alpha_2). \end{align*} \begin{align*} \alpha_1+\alpha_2=\alpha_1+\alpha_2. \end{align*} Thus $p(\alpha_1+\alpha_2)=2$. For the simple reflections, the arguments are \begin{align*} s_1(2\alpha_1+2\alpha_2)-(\alpha_1+\alpha_2)=2(-\alpha_1)+2(\alpha_1+\alpha_2)-(\alpha_1+\alpha_2)=-\alpha_1+\alpha_2. \end{align*} \begin{align*} s_2(2\alpha_1+2\alpha_2)-(\alpha_1+\alpha_2)=2(\alpha_1+\alpha_2)+2(-\alpha_2)-(\alpha_1+\alpha_2)=\alpha_1-\alpha_2. \end{align*} For the two length-$2$ elements, \begin{align*} s_1s_2(2\alpha_1+2\alpha_2)-(\alpha_1+\alpha_2)=s_1(2\alpha_1)-(\alpha_1+\alpha_2)=-3\alpha_1-\alpha_2. \end{align*} \begin{align*} s_2s_1(2\alpha_1+2\alpha_2)-(\alpha_1+\alpha_2)=s_2(2\alpha_2)-(\alpha_1+\alpha_2)=-\alpha_1-3\alpha_2. \end{align*} Finally the longest element sends both positive simple-root coefficients here to negative ones: \begin{align*} w_0(2\alpha_1+2\alpha_2)-(\alpha_1+\alpha_2)=-2\alpha_1-2\alpha_2-(\alpha_1+\alpha_2)=-3\alpha_1-3\alpha_2. \end{align*} Each non-identity argument has a negative coefficient in the simple-root basis, so it is not in $Q^+$ and its partition count is $0$. Hence \begin{align*} \dim L(\omega_1+\omega_2)_0=1\cdot p(\alpha_1+\alpha_2)=2. \end{align*} The multiplicity $2$ is exactly the dimension of the Cartan subalgebra, which is the zero-weight space in the adjoint representation of $\mathfrak{sl}_3$. [/example] The same identity also sits at the meeting point of several other subjects. Geometrically, the Weyl numerator is an alternating orbit sum attached to chamber walls. In invariant theory, the quotient by the denominator is a symmetric object obtained from an anti-invariant numerator divided by the basic anti-invariant factor. In combinatorics, the type $A$ specializations recover Schur characters, hook-length products, and partition-counting formulas. The chapter ends with a practical viewpoint. The Weyl character formula is not only a closed expression for characters; it is a computational method. The denominator records all lowering by positive roots, the numerator enforces the finite-dimensional quotient by signed Weyl cancellation, and the dimension and multiplicity formulas are the two main numerical shadows of the same identity. # 8. Tensor Products and Decomposition Tensor products are the first place where the classification of irreducible highest weight modules becomes genuinely computational. Earlier chapters named the irreducibles as $L(\lambda)$ for dominant integral weights $\lambda$, and the Weyl character formula gave a compact expression for their characters. This chapter asks how to break $V(\lambda) \otimes V(\mu)$ into irreducible summands, using characters, highest weight constraints, and the combinatorics of the [Littlewood-Richardson rule](/theorems/5183) for type $A$. The guiding point is that tensor products preserve finite-dimensionality but rarely preserve irreducibility. The highest possible weight is easy to predict, while the lower summands encode subtle multiplicity information. Characters turn this problem into algebra in the group ring of the weight lattice, and for $\mathfrak{sl}_n$ the same algebra is modelled by Young diagrams and tableaux. ## Characters as a Computational Tool The decomposition problem starts with a basic question: if two representations have known weight-space decompositions, what can be read from the weights of their tensor product? The tensor product weight spaces are built by adding weights, so the formal character behaves multiplicatively. This turns tensor product decomposition into the problem of expanding a product of characters in the basis of irreducible characters. [definition: Tensor Product Representation] Let $\mathfrak{g}$ be a complex Lie algebra, and let $V,W$ be $\mathfrak{g}$-modules with representation maps $\rho_V:\mathfrak{g}\to \operatorname{End}(V)$ and $\rho_W:\mathfrak{g}\to \operatorname{End}(W)$. The tensor product representation is the representation map \begin{align*} \rho_{V\otimes W}:\mathfrak{g}\to \operatorname{End}(V\otimes W) \end{align*} defined on pure tensors by \begin{align*} \rho_{V\otimes W}(x)(v \otimes w) = \rho_V(x)v \otimes w + v \otimes \rho_W(x)w \end{align*} for all $x \in \mathfrak{g}$, $v \in V$, and $w \in W$. [/definition] This is the infinitesimal version of the diagonal action on a product: the Lie algebra element acts on both tensor factors and the two contributions are added. To make this action useful for decomposition, we next need to identify its weight spaces in terms of the weight spaces of the two factors. [quotetheorem:9387] [citeproof:9387] The theorem reduces tensor products to an arithmetic operation on characters, but its hypotheses are doing real work. The weight-module assumption is what lets us speak of a direct sum of weight spaces; for a one-dimensional abelian Lie algebra acting on a two-dimensional module by a single Jordan block, there is an eigenvalue but no decomposition into eigenspaces, so the displayed formula for weight spaces has no analogue. Finite-dimensionality keeps the character as a finite formal sum in this chapter and ensures that each coefficient is an ordinary finite dimension. Even under these hypotheses, the theorem gives only weight multiplicities in the tensor product, not irreducible summand multiplicities: several larger irreducibles can contribute to the same lower weight. To turn character multiplication into a decomposition statement, we need notation for the coefficients that appear when the product character is written as a sum of irreducible characters. [definition: Tensor Product Multiplicity] Let $\mathfrak{g}$ be a complex semisimple Lie algebra and let $\lambda,\mu,\nu \in P^+$. The tensor product multiplicity $c_{\lambda\mu}^{\nu}$ is the multiplicity of the irreducible module $L(\nu)$ in a decomposition of $L(\lambda)\otimes L(\mu)$: \begin{align*} L(\lambda)\otimes L(\mu) \cong \bigoplus_{\nu\in P^+} L(\nu)^{\oplus c_{\lambda\mu}^{\nu}}. \end{align*} [/definition] Complete reducibility makes this number well-defined, independent of the chosen decomposition. A character identity is initially only a necessary check on a proposed decomposition, since it records weights rather than chosen submodules. The missing step is to know that no two different decompositions into irreducibles can have the same character. In the finite-dimensional semisimple setting, irreducible characters are separated by their highest weights, so the next result turns character multiplication from a consistency test into an actual multiplicity calculation. [quotetheorem:9388] [citeproof:9388] The character method depends on complete reducibility and on the highest-weight classification of finite-dimensional semisimple modules. Outside this setting, a character can record the same weights while missing extension data: for the one-dimensional abelian Lie algebra, a two-dimensional Jordan block with eigenvalue $\lambda$ has the same formal character as a direct sum of two one-dimensional weight modules, but it is not a direct sum of irreducibles. Thus character multiplication is a decomposition method here because finite-dimensional modules over complex semisimple Lie algebras are completely reducible, not because characters detect arbitrary module structure. The method becomes concrete when a representation has a visible symmetry commuting with the Lie algebra action. The tensor square of the natural representation is the basic example, since the flip map already exhibits the two summands that the character calculation would find. [example: Natural Tensor Square] Let $\mathfrak{g}=\mathfrak{sl}_n$ and let $V=\mathbb C^n$ with the natural action. For $x\in \mathfrak{sl}_n$ and pure tensors $v\otimes w$, the tensor product action is \begin{align*} x\cdot(v\otimes w)=(xv)\otimes w+v\otimes(xw). \end{align*} If $\tau(v\otimes w)=w\otimes v$, then \begin{align*} \tau(x\cdot(v\otimes w))=\tau((xv)\otimes w+v\otimes(xw))=w\otimes(xv)+(xw)\otimes v. \end{align*} On the other hand, \begin{align*} x\cdot\tau(v\otimes w)=x\cdot(w\otimes v)=(xw)\otimes v+w\otimes(xv). \end{align*} Thus $\tau(x\cdot(v\otimes w))=x\cdot\tau(v\otimes w)$, so the $+1$ and $-1$ eigenspaces of $\tau$ are $\mathfrak{sl}_n$-submodules. Every tensor $v\otimes w$ splits as \begin{align*} v\otimes w=\frac{1}{2}(v\otimes w+w\otimes v)+\frac{1}{2}(v\otimes w-w\otimes v). \end{align*} The first summand is fixed by $\tau$ and the second is negated by $\tau$, so \begin{align*} \mathbb C^n\otimes \mathbb C^n \cong \operatorname{Sym}^2(\mathbb C^n)\oplus \bigwedge^2(\mathbb C^n). \end{align*} If $e_1,\dots,e_n$ is the standard basis, then $e_1\otimes e_1$ has weight $\omega_1+\omega_1=2\omega_1$, giving the symmetric summand highest weight $2\omega_1$. For $n\ge 3$, the vector $e_1\wedge e_2=e_1\otimes e_2-e_2\otimes e_1$ is a highest weight vector of weight $\omega_2$, so the exterior summand has highest weight $\omega_2$. Hence \begin{align*} V(\omega_1)\otimes V(\omega_1) \cong V(2\omega_1)\oplus V(\omega_2) \end{align*} for $n\ge 3$, by the highest-weight classification of finite-dimensional $\mathfrak{sl}_n$-modules. For $n=2$, $\bigwedge^2(\mathbb C^2)$ is spanned by $e_1\wedge e_2$, and \begin{align*} x\cdot(e_1\wedge e_2)=(xe_1)\wedge e_2+e_1\wedge(xe_2)=(\operatorname{tr}x)(e_1\wedge e_2)=0. \end{align*} Since $x\in\mathfrak{sl}_2$ has trace $0$, the exterior square is the one-dimensional trivial representation in the $n=2$ case. [/example] This example shows how character multiplication, symmetry, and highest weights reinforce the same answer. The summand with highest weight $2\omega_1$ comes from symmetrising the highest weight vector with itself, while the exterior square records the next dominant weight available in the tensor square. ## Highest Weights in Tensor Products The next problem is to bound which irreducibles can occur before doing a full character calculation. Since every weight in $V(\lambda)$ is obtained from $\lambda$ by subtracting non-negative integer combinations of positive roots, every weight in the tensor product lies below $\lambda+\mu$. This gives a strong first constraint and identifies the top summand. [quotetheorem:9389] [citeproof:9389] The bound is often the fastest way to remove impossible summands, but it is only a first constraint. The highest-weight hypotheses ensure that $\lambda$ and $\mu$ are genuine maximal weights and that every other weight is obtained by subtracting positive roots; for a general weight module with no maximal weight, there need not be a top candidate $\lambda+\mu$ at all. Complete reducibility is also being used when the proof speaks of irreducible summands rather than only submodules generated by highest weight vectors. The theorem does not determine the lower multiplicities: for example, in $\mathfrak{sl}_2$ the bound identifies the top term in $V(2\omega_1)\otimes V(2\omega_1)$ as $V(4\omega_1)$, but it does not by itself decide which lower summands occur or with what multiplicity. This motivates the following definition: the unique top summand should be named separately because it is the first term removed in character calculations and the fixed reference point for all lower constituents. [definition: Cartan Component] Let $\mathfrak{g}$ be a complex semisimple Lie algebra and let $\lambda,\mu\in P^+$. The Cartan component of $L(\lambda)\otimes L(\mu)$ is the unique irreducible summand isomorphic to $L(\lambda+\mu)$. [/definition] The Cartan component is the tensor product analogue of adding leading terms. The remaining summands measure the failure of a tensor product of irreducibles to stay irreducible, and finding them requires care because weight multiplicities are not the same as summand multiplicities. [remark: Multiplicity Versus Weight Multiplicity] The multiplicity of a weight $\nu$ inside $V(\lambda)\otimes V(\mu)$ is not the same as the multiplicity of the irreducible summand $V(\nu)$. A vector of weight $\nu$ contributes to a copy of $V(\nu)$ only when it is killed by all positive root spaces, and lower weight spaces also receive contributions from irreducibles with larger highest weights. [/remark] This distinction explains why character subtraction is necessary. A raw weight count sees all summands at once; to find irreducible constituents, one repeatedly removes the full character of the largest remaining highest weight module. [example: The Adjoint Summand in Type A Two By Dual] For $\mathfrak{sl}_n$, let $V=\mathbb C^n$ with standard basis $e_1,\dots,e_n$, and let $V^*=(\mathbb C^n)^*$. The dual representation has highest weight $\omega_{n-1}$, so we compute $V(\omega_1)\otimes V(\omega_{n-1})$ by identifying $V\otimes V^*$ with $\operatorname{End}(V)$ through \begin{align*} v\otimes f \longmapsto T_{v,f}, \qquad T_{v,f}(u)=f(u)v. \end{align*} For $x\in \mathfrak{sl}_n$, the dual action is $(x\cdot f)(u)=-f(xu)$, hence \begin{align*} T_{xv,f}(u)+T_{v,x\cdot f}(u)=f(u)xv-f(xu)v. \end{align*} Since $T_{v,f}(u)=f(u)v$, this is \begin{align*} f(u)xv-f(xu)v=xT_{v,f}(u)-T_{v,f}(xu)=(xT_{v,f}-T_{v,f}x)(u). \end{align*} Thus the tensor product action corresponds to the commutator action \begin{align*} x\cdot T=xT-Tx \end{align*} on $\operatorname{End}(V)$. The scalar line $\mathbb C I$ is a submodule because \begin{align*} x\cdot I=xI-Ix=0. \end{align*} The trace-zero subspace is also a submodule, since for $T\in \mathfrak{sl}_n\subset \operatorname{End}(V)$, \begin{align*} \operatorname{tr}(xT-Tx)=\operatorname{tr}(xT)-\operatorname{tr}(Tx)=0 \end{align*} by cyclicity of trace. Every endomorphism splits uniquely into its scalar and trace-zero parts: \begin{align*} T=\frac{\operatorname{tr}(T)}{n}I+\left(T-\frac{\operatorname{tr}(T)}{n}I\right). \end{align*} The second summand has trace \begin{align*} \operatorname{tr}\left(T-\frac{\operatorname{tr}(T)}{n}I\right)=\operatorname{tr}(T)-\frac{\operatorname{tr}(T)}{n}\operatorname{tr}(I)=\operatorname{tr}(T)-\operatorname{tr}(T)=0. \end{align*} Therefore \begin{align*} \operatorname{End}(\mathbb C^n)=\mathbb C I\oplus \mathfrak{sl}_n \end{align*} as $\mathfrak{sl}_n$-modules. The first summand is the trivial representation $V(0)$. For the second summand, the matrix unit $E_{1n}$ has weight $\varepsilon_1-\varepsilon_n=\omega_1+\omega_{n-1}$. If $E_{ij}$ is a positive root vector with $i<j$, then \begin{align*} [E_{ij},E_{1n}]=E_{ij}E_{1n}-E_{1n}E_{ij}=\delta_{j1}E_{in}-\delta_{ni}E_{1j}=0, \end{align*} because $i<j$ makes $j=1$ impossible and $i=n$ impossible. Thus $E_{1n}$ is a highest weight vector of weight $\omega_1+\omega_{n-1}$, so the trace-zero summand is $V(\omega_1+\omega_{n-1})$. Hence \begin{align*} V(\omega_1)\otimes V(\omega_{n-1})\cong V(0)\oplus V(\omega_1+\omega_{n-1}). \end{align*} For $n=3$, this reads \begin{align*} V(\omega_1)\otimes V(\omega_2)\cong V(0)\oplus V(\omega_1+\omega_2). \end{align*} The one-dimensional summand is the invariant trace direction, while the Cartan component is the adjoint representation. [/example] The example also illustrates the top-weight theorem: the adjoint representation is the Cartan component, while the one-dimensional summand appears lower down through the invariant trace pairing. ## Littlewood-Richardson Rule for Type A For general semisimple Lie algebras, tensor product multiplicities are governed by deep combinatorics. The model case is $\mathfrak{sl}_n$, where polynomial representations are encoded by partitions and tensor products are controlled by the Littlewood-Richardson rule. The problem becomes: given two Young diagrams, which diagrams can be formed by adding boxes in a way compatible with semistandard tableaux? [definition: Partition and Young Diagram] A partition is a finite non-increasing sequence $\lambda=(\lambda_1,\dots,\lambda_r)$ of non-negative integers. Its Young diagram is the set of boxes with row $i$ containing $\lambda_i$ boxes, aligned on the left. [/definition] Partitions provide coordinates for polynomial representations of $GL_n(\mathbb C)$, and after restricting to $\mathfrak{sl}_n$ they encode highest weights up to the determinant shift. To count tensor product multiplicities rather than merely label representations, we need skew fillings that record how one diagram is added to another. [definition: Littlewood-Richardson Tableau] Let $\lambda\subset \nu$ be Young diagrams. A Littlewood-Richardson tableau of skew shape $\nu/\lambda$ and content $\mu=(\mu_1,\mu_2,\dots)$ is a filling of the boxes of $\nu/\lambda$ with positive integers such that rows are weakly increasing, columns are strictly increasing, the integer $i$ appears $\mu_i$ times, and the reverse row-reading word is a lattice word. [/definition] The lattice-word condition is the extra representation-theoretic constraint. Without it, skew fillings count too many ways of adding boxes, including fillings that do not correspond to highest weight vectors in the tensor product. Imposing the lattice condition isolates the admissible highest weight data, turning the tensor product question into a finite tableau count. [quotetheorem:5183] [citeproof:5183] The theorem is the type $A$ replacement for repeated character subtraction, but it is stated for polynomial $GL_n(\mathbb C)$-representations rather than arbitrary representations. The condition that the partitions have at most $n$ parts is necessary because a Young diagram with more than $n$ rows gives no non-zero Schur functor on $\mathbb C^n$; for instance, $\bigwedge^{n+1}\mathbb C^n=0$. After restriction to $\mathfrak{sl}_n$, one must also remember the determinant shift: adding a full column of height $n$ tensors the corresponding $GL_n(\mathbb C)$-representation by a power of the determinant, which is invisible to $\mathfrak{sl}_n$. Thus the rule packages tensor product multiplicities into a finite counting problem only after the correct type $A$ dictionary has been fixed, and it is especially efficient for small diagrams. [example: The Product of Two Natural Representations for Type A] The natural representation of $GL_n(\mathbb C)$ is $S_{(1)}(\mathbb C^n)$, so its tensor square is computed by the Littlewood-Richardson coefficients $c_{(1),(1)}^\nu$. Since the content $(1)$ has one box, the skew diagram $\nu/(1)$ must consist of exactly one added box. The only partitions of $2$ containing $(1)$ are $(2)$ and $(1,1)$. For $\nu=(2)$, the skew shape $(2)/(1)$ is one box in the first row. Filling that box with $1$ gives a row-weakly-increasing and column-strict tableau, and its reverse row-reading word is $1$, which is a lattice word because after reading the only letter there is one entry equal to $1$ and no entries equal to $2$ or $3$ or any larger label. Hence \begin{align*} c_{(1),(1)}^{(2)}=1. \end{align*} For $\nu=(1,1)$, the skew shape $(1,1)/(1)$ is one box in the second row. Filling that box with $1$ again gives a row-weakly-increasing and column-strict tableau, and the reverse row-reading word is again $1$, so \begin{align*} c_{(1),(1)}^{(1,1)}=1. \end{align*} There are no other possible $\nu$, so the *Littlewood-Richardson Rule* gives \begin{align*} S_{(1)}(\mathbb C^n)\otimes S_{(1)}(\mathbb C^n)\cong S_{(2)}(\mathbb C^n)\oplus S_{(1,1)}(\mathbb C^n). \end{align*} Here $S_{(2)}(\mathbb C^n)=\operatorname{Sym}^2(\mathbb C^n)$ and $S_{(1,1)}(\mathbb C^n)=\bigwedge^2(\mathbb C^n)$. After restriction to $\mathfrak{sl}_n$, the vector $e_1\otimes e_1$ has weight $\omega_1+\omega_1=2\omega_1$, so the symmetric summand has highest weight $2\omega_1$. For $n\ge 3$, the vector $e_1\wedge e_2$ has weight $\varepsilon_1+\varepsilon_2=\omega_2$, so the exterior summand has highest weight $\omega_2$. Therefore \begin{align*} V(\omega_1)\otimes V(\omega_1)\cong V(2\omega_1)\oplus V(\omega_2) \end{align*} for $n\ge 3$. This is the same decomposition as the symmetric/exterior splitting of the tensor square, now recovered by counting the two admissible one-box Littlewood-Richardson tableaux. [/example] This recovers the symmetric/exterior decomposition from a counting rule. The two possible diagrams are the two ways to place the second box: in the first row or in the second row. [example: Two Fundamental Decompositions for $\mathfrak{sl}_3$] For $\mathfrak{sl}_3$, the natural representation is $V(\omega_1)=\mathbb C^3$, and its dual is $V(\omega_2)$. In the $GL_3$ partition notation, the Littlewood-Richardson rule gives \begin{align*} S_{(1)}(\mathbb C^3)\otimes S_{(1)}(\mathbb C^3)\cong S_{(2)}(\mathbb C^3)\oplus S_{(1,1)}(\mathbb C^3). \end{align*} Here $S_{(2)}(\mathbb C^3)=\operatorname{Sym}^2(\mathbb C^3)$ and $S_{(1,1)}(\mathbb C^3)=\bigwedge^2\mathbb C^3$. After restriction to $\mathfrak{sl}_3$, the vector $e_1\otimes e_1$ has weight $\omega_1+\omega_1=2\omega_1$, so the symmetric summand has highest weight $2\omega_1$. The vector $e_1\wedge e_2=e_1\otimes e_2-e_2\otimes e_1$ has weight $\varepsilon_1+\varepsilon_2=\omega_2$, so the exterior summand has highest weight $\omega_2$. Thus \begin{align*} V(\omega_1)\otimes V(\omega_1)\cong V(2\omega_1)\oplus V(\omega_2). \end{align*} For the mixed tensor product, identify $\mathbb C^3\otimes(\mathbb C^3)^*$ with $\operatorname{End}(\mathbb C^3)$ by $v\otimes f\mapsto T_{v,f}$, where $T_{v,f}(u)=f(u)v$. The tensor product action becomes the commutator action: \begin{align*} x\cdot T=xT-Tx. \end{align*} The scalar line is fixed because \begin{align*} x\cdot I=xI-Ix=0. \end{align*} The trace-zero subspace is stable because \begin{align*} \operatorname{tr}(xT-Tx)=\operatorname{tr}(xT)-\operatorname{tr}(Tx)=0 \end{align*} by cyclicity of trace. Every endomorphism splits as \begin{align*} T=\frac{\operatorname{tr}(T)}{3}I+\left(T-\frac{\operatorname{tr}(T)}{3}I\right), \end{align*} and the second summand is trace-zero since \begin{align*} \operatorname{tr}\left(T-\frac{\operatorname{tr}(T)}{3}I\right)=\operatorname{tr}(T)-\frac{\operatorname{tr}(T)}{3}\operatorname{tr}(I)=\operatorname{tr}(T)-\operatorname{tr}(T)=0. \end{align*} Therefore \begin{align*} \mathbb C^3\otimes(\mathbb C^3)^*\cong \mathbb C I\oplus \mathfrak{sl}_3. \end{align*} The scalar summand is $V(0)$. In the trace-zero summand, the matrix unit $E_{13}$ has weight $\varepsilon_1-\varepsilon_3=\omega_1+\omega_2$, and for a positive root vector $E_{ij}$ with $i<j$, \begin{align*} [E_{ij},E_{13}]=E_{ij}E_{13}-E_{13}E_{ij}=\delta_{j1}E_{i3}-\delta_{3i}E_{1j}=0, \end{align*} because $j=1$ and $i=3$ are both impossible when $i<j$ in $\{1,2,3\}$. Hence $E_{13}$ is a highest weight vector of weight $\omega_1+\omega_2$, and \begin{align*} V(\omega_1)\otimes V(\omega_2)\cong V(\omega_1+\omega_2)\oplus V(0). \end{align*} The dimensions match the decompositions: $\dim \operatorname{Sym}^2(\mathbb C^3)=3\cdot 4/2=6$, $\dim \bigwedge^2\mathbb C^3=3\cdot 2/2=3$, and $3\cdot 3=6+3$; also $\dim \mathfrak{sl}_3=3^2-1=8$, so $3\cdot 3=8+1$. The first decomposition is the symmetric/exterior split, while the second is the decomposition of endomorphisms into trace-zero and scalar parts. [/example] These two decompositions are the basic computational patterns for type $A_2$. The first shows how adding boxes creates new highest weights, and the second shows how duality introduces invariant pairings and hence the representation $V(0)$. ## Character Algorithms and Multiplicity Constraints The final question is how the different methods fit together in actual computations. The practical answer is to combine the highest weight bound with character subtraction: locate the largest remaining dominant weight, remove the corresponding irreducible character, and repeat until the product character is exhausted. [definition: Dominance-Ordered Character Subtraction] Let $\mathfrak{g}$ be a complex semisimple Lie algebra. Dominance-ordered character subtraction is the procedure that starts from $\operatorname{ch}V(\lambda)\operatorname{ch}V(\mu)$, chooses a maximal dominant weight $\nu$ still appearing with non-zero coefficient, subtracts the appropriate multiple of $\operatorname{ch}V(\nu)$, and iterates until no weights remain. [/definition] The procedure works because an irreducible character has a unique highest term. When checking a proposed decomposition, however, character subtraction is easy to misread unless there are independent numerical constraints. The next result supplies those constraints in a form usable during computations: the largest possible highest weight is bounded by the sum of the input highest weights, and the total dimensions must still add up. These tests give quick ways to rule out impossible decompositions before doing a full character calculation. [quotetheorem:9390] [citeproof:9390] These constraints again rely on the finite-dimensional semisimple setting: complete reducibility supplies the summand decomposition, and finite-dimensionality makes the dimension identity a finite numerical check. The dimension equation is not enough to determine the decomposition, because different lists of irreducibles can have the same total dimension while satisfying the same top-weight bound. For example, in $\mathfrak{sl}_2$ the product $V(\omega_1)\otimes V(\omega_1)$ has dimension $4$ and top summand $V(2\omega_1)$; the dimension count then permits one additional one-dimensional summand, but dimension alone would not explain why that summand is the trivial module rather than some other candidate in a different category. In higher rank the ambiguity is more serious, since several non-isomorphic lower highest-weight modules may have compatible dimensions and lie below the same top weight. Together with the top-weight constraint and character multiplication, the dimension equation gives a reliable workflow for small-rank examples. [example: A Character-Subtraction Workflow] Let \begin{align*} P=\operatorname{ch}V(\lambda)\operatorname{ch}V(\mu) \end{align*} be the tensor product character, written as a finite sum of monomials $e^\alpha$ with integer coefficients. If its maximal dominant weight is $\lambda+\mu$, then the Cartan component occurs once by *[Tensor Product Highest Weight Bound](/theorems/9389)*, so the first step replaces $P$ by \begin{align*} P_1=P-\operatorname{ch}V(\lambda+\mu). \end{align*} This removes the top monomial because $\operatorname{ch}V(\lambda+\mu)$ contains $e^{\lambda+\mu}$ with coefficient $1$, and no irreducible character with lower highest weight contains a weight above its own highest weight. Now suppose the remaining character $P_1$ has maximal dominant weight $\nu$, and the coefficient of $e^\nu$ in $P_1$ is $m$. Since $\operatorname{ch}V(\nu)$ contains $e^\nu$ with coefficient $1$ and contains no weight larger than $\nu$, the next subtraction is \begin{align*} P_2=P_1-m\operatorname{ch}V(\nu). \end{align*} The coefficient of $e^\nu$ in $P_2$ is \begin{align*} m-m\cdot 1=0. \end{align*} No larger weight is changed, because $P_1$ has no dominant weight larger than $\nu$ still present and $\operatorname{ch}V(\nu)$ has no weight larger than $\nu$. Repeating the same step gives a sequence \begin{align*} P_{k+1}=P_k-m_k\operatorname{ch}V(\nu_k), \end{align*} where $\nu_k$ is a maximal dominant weight still appearing in $P_k$ and $m_k$ is the coefficient of $e^{\nu_k}$. The process terminates because the original tensor product character has only finitely many weights, and each step removes at least one maximal dominant weight from the remaining support without creating any new higher weights. The output is the expansion of the product character in irreducible characters, hence the tensor product decomposition. [/example] This algorithmic view is the bridge to later topics. Weyl characters make the computation uniform across all types, while type $A$ tableaux give a concrete combinatorial model in the most important family of examples. The theme to carry forward is that tensor products are controlled simultaneously by the top weight, the full character, and the representation-theoretic meaning of highest weight vectors inside a tensor product. # 9. Duals, Invariants, and Schur Functors Duals and invariants are two ways of turning representation-theoretic questions into computable algebra. The guiding problem in this chapter is how to recognise representations inside duals, tensor products, and tensor powers without decomposing everything from scratch. We assume the highest weight classification for finite-dimensional semisimple Lie algebra modules, the Weyl group action on weights, and complete reducibility. The chapter first explains how duals transform highest weights, then uses invariant vectors to encode Hom spaces and tensor product multiplicities, and finally introduces Schur functors as the partition-indexed construction behind polynomial representations of $\mathfrak{gl}_n$ and their restrictions to $\mathfrak{sl}_n$. ## Dual Representations and the Longest Weyl Group Element Given a representation $V$ of a Lie algebra $\mathfrak g$, the ordinary vector-space dual $V^*$ should again carry a representation, but the action must be chosen so that the natural pairing $V^* \times V \to \mathbb C$ is compatible with the Lie algebra action. The resulting module is contragredient: the element $x \in \mathfrak g$ acts on a functional by differentiating the argument with a minus sign. [definition: Dual Representation] Let $\mathfrak g$ be a complex Lie algebra and let $V$ be a $\mathfrak g$-module with action $\rho: \mathfrak g \to \mathfrak{gl}(V)$. The dual representation on $V^*$ is the action map \begin{align*} \mathfrak g \times V^* \to V^* \end{align*} defined by \begin{align*} (x \cdot \varphi)(v) = -\varphi(x \cdot v) \end{align*} for all $x \in \mathfrak g$, $\varphi \in V^*$, and $v \in V$. [/definition] The minus sign is forced by the requirement that the evaluation pairing be invariant under the diagonal action. This convention also agrees with the antipode on the Hopf algebra $U(\mathfrak g)$, where $x$ acts on the dual by precomposition with $-x$. [example: Invariant Evaluation Pairing] Let $V$ be finite-dimensional, choose a basis $(v_i)$, and let $(v_i^*)$ be the dual basis. Identify $V\otimes V^*$ with $\operatorname{End}(V)$ by sending $v\otimes \varphi$ to the rank-one operator $w\mapsto \varphi(w)v$. Under this identification, \begin{align*} \left(\sum_i v_i\otimes v_i^*\right)(w)=\sum_i v_i^*(w)v_i=w \end{align*} for every $w\in V$, so $\sum_i v_i\otimes v_i^*$ corresponds to $\operatorname{id}_V$. Now compute the diagonal action on a rank-one tensor. For $x\in\mathfrak g$, \begin{align*} x\cdot(v\otimes\varphi)=(xv)\otimes\varphi+v\otimes(x\cdot\varphi). \end{align*} Evaluating the corresponding endomorphism at $w\in V$ and using the dual action $(x\cdot\varphi)(w)=-\varphi(xw)$ gives \begin{align*} \bigl(x\cdot(v\otimes\varphi)\bigr)(w)=\varphi(w)xv-\varphi(xw)v. \end{align*} If $T(w)=\varphi(w)v$, this is exactly \begin{align*} (\rho(x)T-T\rho(x))(w)=\rho(x)(\varphi(w)v)-T(\rho(x)w)=\varphi(w)xv-\varphi(xw)v. \end{align*} Therefore the diagonal action on $V\otimes V^*$ corresponds to the commutator action $T\mapsto \rho(x)T-T\rho(x)$ on $\operatorname{End}(V)$. For $T=\operatorname{id}_V$, \begin{align*} \rho(x)\operatorname{id}_V-\operatorname{id}_V\rho(x)=\rho(x)-\rho(x)=0, \end{align*} so the coevaluation tensor $\sum_i v_i\otimes v_i^*$ is fixed by every $x\in\mathfrak g$. The evaluation map is invariant by the same sign cancellation: for $\varphi\in V^*$ and $v\in V$, \begin{align*} \operatorname{ev}\bigl(x\cdot(\varphi\otimes v)\bigr)=\operatorname{ev}\bigl((x\cdot\varphi)\otimes v+\varphi\otimes xv\bigr)=-\varphi(xv)+\varphi(xv)=0. \end{align*} Thus the coevaluation tensor spans a copy of the zero-action one-dimensional representation inside $V\otimes V^*$ whenever $V\ne 0$, and evaluation is the corresponding invariant contraction. [/example] The example shows that duality is tied to invariant pairings, but it does not yet answer which irreducible module the dual is. For highest weight theory, this question matters because all finite-dimensional irreducibles are labelled by dominant weights, so the next task is to compute the label of $L(\lambda)^*$ from the label of $L(\lambda)$. [quotetheorem:9391] [citeproof:9391] This result packages a useful rule: duality reverses the weight diagram and then reorients it so that the highest point is measured with respect to the original positive system. The hypotheses are doing real work. Finite-dimensionality ensures that the module has both highest and lowest weights, semisimplicity supplies the Weyl group symmetry needed to identify the lowest weight as $w_0\lambda$, and the chosen positive roots determine what "highest" means after taking negatives. For a concrete failure outside this setting, take a Verma module $M(\lambda)$ for $\mathfrak{sl}_2$: its weights run downward without a lowest weight, and the full algebraic dual $M(\lambda)^*$ contains functionals on all weight spaces at once, so it is not an ordinary Verma-type highest weight module generated by a single highest weight vector. It also uses irreducibility: for a reducible module, one applies the rule to irreducible summands after a decomposition is known. [example: Duals of Fundamental Representations for Special Linear] Let $\mathfrak g=\mathfrak{sl}_n(\mathbb C)$, let $V=\mathbb C^n$, and fix $1\le k\le n-1$. The fundamental representation $L(\omega_k)$ is $\Lambda^k V$. In type $A_{n-1}$, the longest Weyl group element satisfies $w_0(\omega_k)=-\omega_{n-k}$, so *Highest Weight of the Dual Module* gives \begin{align*} (\Lambda^k V)^* \cong L(-w_0\omega_k). \end{align*} Since $-w_0\omega_k=\omega_{n-k}$, this becomes \begin{align*} (\Lambda^k V)^* \cong L(\omega_{n-k}). \end{align*} Using again that $L(\omega_j)\cong \Lambda^j V$ for $1\le j\le n-1$, we obtain \begin{align*} (\Lambda^k V)^* \cong \Lambda^{n-k}V. \end{align*} The isomorphism is the determinant pairing. Choose the volume form identifying $\Lambda^n V$ with $\mathbb C$, and define \begin{align*} \Psi(u)(a)=a\wedge u \end{align*} for $a\in \Lambda^k V$ and $u\in \Lambda^{n-k}V$, with $a\wedge u$ viewed as a complex number through $\Lambda^n V\cong \mathbb C$. If $e_1,\dots,e_n$ is the standard basis and $e_I=e_{i_1}\wedge\cdots\wedge e_{i_k}$ for $I=\{i_1<\cdots<i_k\}$, then $e_I\wedge e_J=0$ unless $J=I^c$, while \begin{align*} e_I\wedge e_{I^c}=\pm e_1\wedge\cdots\wedge e_n. \end{align*} Thus the pairing matches each basis vector $e_I$ with the complementary basis vector $e_{I^c}$ up to sign, so it is nondegenerate and $\Psi:\Lambda^{n-k}V\to(\Lambda^kV)^*$ is a vector-space isomorphism. It remains to see that $\Psi$ respects the $\mathfrak{sl}_n$-action. For $x\in\mathfrak{sl}_n$, the induced action on $\Lambda^n V$ is multiplication by $\operatorname{tr}(x)=0$, so \begin{align*} 0=x\cdot(a\wedge u)=(x\cdot a)\wedge u+a\wedge(x\cdot u). \end{align*} Therefore \begin{align*} -\bigl((x\cdot a)\wedge u\bigr)=a\wedge(x\cdot u). \end{align*} Using the dual action on $(\Lambda^kV)^*$, we get \begin{align*} (x\cdot\Psi(u))(a)=-\Psi(u)(x\cdot a). \end{align*} Substituting the definition of $\Psi$ gives \begin{align*} (x\cdot\Psi(u))(a)=-\bigl((x\cdot a)\wedge u\bigr). \end{align*} By the previous identity, \begin{align*} (x\cdot\Psi(u))(a)=a\wedge(x\cdot u)=\Psi(x\cdot u)(a). \end{align*} Hence $\Psi$ is an $\mathfrak{sl}_n$-module isomorphism, and the dual of $\Lambda^kV$ is the complementary exterior power $\Lambda^{n-k}V$. [/example] Self-duality is therefore controlled by the action of $-w_0$ on dominant weights. In type $A_{n-1}$, this exchanges $\omega_k$ and $\omega_{n-k}$, while in several other types many fundamental representations are self-dual. ## Invariant Vectors, Hom Spaces, and Tensor Product Multiplicities A central computational problem is to determine how often one representation appears inside another construction, especially a tensor product. Invariants turn this into a problem about maps: a vector fixed by $\mathfrak g$ is the same data as a homomorphism from the one-dimensional zero-action module. [definition: Invariant Subspace] Let $\mathfrak g$ be a complex Lie algebra and let $V$ be a $\mathfrak g$-module. The invariant subspace is \begin{align*} V^{\mathfrak g} = \{v \in V : x \cdot v = 0 \text{ for all } x \in \mathfrak g\}. \end{align*} [/definition] The notation $V^{\mathfrak g}$ records the vectors on which the entire Lie algebra acts as zero. This is the multiplicity space of the one-dimensional zero-action representation when $\mathfrak g$ is semisimple and $V$ is finite-dimensional. [example: Invariants in V Tensor V Star] Let $V$ be an irreducible finite-dimensional $\mathfrak g$-module, and identify $V\otimes V^*$ with $\operatorname{End}(V)$ by sending $v\otimes \varphi$ to the operator $T_{v,\varphi}$ defined by $T_{v,\varphi}(w)=\varphi(w)v$. For $x\in \mathfrak g$, the diagonal action gives \begin{align*} x\cdot(v\otimes\varphi)=(xv)\otimes\varphi+v\otimes(x\cdot\varphi). \end{align*} Evaluating the corresponding endomorphism at $w\in V$ gives \begin{align*} T_{xv,\varphi}(w)+T_{v,x\cdot\varphi}(w)=\varphi(w)xv+(x\cdot\varphi)(w)v. \end{align*} By the definition of the dual action, $(x\cdot\varphi)(w)=-\varphi(xw)$, so this becomes \begin{align*} \varphi(w)xv-\varphi(xw)v. \end{align*} On the other hand, \begin{align*} (\rho(x)T_{v,\varphi}-T_{v,\varphi}\rho(x))(w)=\rho(x)(\varphi(w)v)-T_{v,\varphi}(xw). \end{align*} Since $\varphi(w)$ is a scalar and $T_{v,\varphi}(xw)=\varphi(xw)v$, this is \begin{align*} \varphi(w)xv-\varphi(xw)v. \end{align*} Thus the diagonal action on $V\otimes V^*$ corresponds exactly to the commutator action $T\mapsto \rho(x)T-T\rho(x)$ on $\operatorname{End}(V)$. Therefore a tensor is $\mathfrak g$-invariant exactly when the corresponding endomorphism $T$ satisfies \begin{align*} \rho(x)T-T\rho(x)=0 \end{align*} for every $x\in\mathfrak g$, equivalently $T\in \operatorname{End}_{\mathfrak g}(V)$. Since $V$ is irreducible, *Schur's lemma* gives \begin{align*} \operatorname{End}_{\mathfrak g}(V)=\mathbb C\operatorname{id}_V. \end{align*} Hence \begin{align*} (V\otimes V^*)^{\mathfrak g}\cong \mathbb C\operatorname{id}_V \end{align*} is one-dimensional. If $(v_i)$ is a basis of $V$ with dual basis $(v_i^*)$, then the tensor $\sum_i v_i\otimes v_i^*$ corresponds to $\operatorname{id}_V$, because \begin{align*} \left(\sum_i v_i\otimes v_i^*\right)(w)=\sum_i v_i^*(w)v_i=w. \end{align*} Thus the invariant line in $V\otimes V^*$ is precisely the line spanned by $\sum_i v_i\otimes v_i^*$. [/example] The preceding example identifies invariants by translating tensors into endomorphisms. This raises the question of whether the same translation is available for maps from any module $V$ to any module $W$. The Hom-tensor adjunction is the theorem below that supplies this conversion and makes invariant-space calculations usable for decomposition problems. [quotetheorem:9392] [citeproof:9392] The adjunction is often used with $W$ a tensor product. It converts a multiplicity question into an invariant-space computation, which can then be attacked using weights, characters, or explicit contractions. The finite-dimensional hypothesis is what lets us identify $V^* \otimes W$ with the full vector space $\operatorname{Hom}(V,W)$ using a basis of $V$. For infinite-dimensional modules, the algebraic dual is usually too large and the tensor-Hom map may only describe finite-rank maps unless extra topological or restricted-dual structures are imposed. Thus the theorem is a reliable tool in the finite-dimensional setting of the course, but it should not be imported unchanged into category $\mathcal O$ or topological representation theory. [quotetheorem:9393] [citeproof:9393] This formula is a useful bridge between abstract decomposition and concrete tensor calculus. The semisimplicity and finite-dimensionality assumptions are essential because the word "multiplicity" is being interpreted through complete reducibility: a summand $L(\nu)$ contributes one copy to $\operatorname{Hom}_{\mathfrak g}(L(\nu),M)$, and no extension data is present. Outside this setting, a module may have subquotients isomorphic to $L(\nu)$ without splitting as a direct sum, so Hom dimensions need not record composition multiplicities. For instance, a nonsplit short exact sequence $0 \to L \to E \to L \to 0$ has two composition factors isomorphic to $L$, but $\dim \operatorname{Hom}_{\mathfrak g}(L,E)$ may be $1$ when only the embedded copy of $L$ gives a map into $E$. In classical types, invariant theory often supplies explicit generators for the invariant spaces in the formula, turning the abstract multiplicity into a calculation with pairings, contractions, and symmetrisers. [example: Detecting the Zero-Action Summand in a Tensor Product] Let $\mathfrak g=\mathfrak{sl}_2(\mathbb C)$, and let $a,b\ge 0$ with $L(m)$ denoting the irreducible highest weight module of highest weight $m$. The zero-action summand in $L(a)\otimes L(b)$ has multiplicity \begin{align*} \dim \operatorname{Hom}_{\mathfrak g}(\mathbb C,L(a)\otimes L(b))=\dim (L(a)\otimes L(b))^{\mathfrak g}. \end{align*} By *Hom Tensor Adjunction for Lie Algebra Modules*, applied with $V=L(a)^*$ and $W=L(b)$, \begin{align*} \operatorname{Hom}_{\mathfrak g}(L(a)^*,L(b))\cong ((L(a)^*)^*\otimes L(b))^{\mathfrak g}. \end{align*} Since $L(a)$ is finite-dimensional, the canonical double-dual map gives $((L(a)^*)^*)\cong L(a)$ as $\mathfrak g$-modules, so \begin{align*} \dim \operatorname{Hom}_{\mathfrak g}(L(a)^*,L(b))=\dim (L(a)\otimes L(b))^{\mathfrak g}. \end{align*} For $\mathfrak{sl}_2$, the longest Weyl group element sends the highest weight $a$ to $-a$, hence *Highest Weight of the Dual Module* gives \begin{align*} L(a)^*\cong L(-(-a))=L(a). \end{align*} Therefore \begin{align*} \operatorname{Hom}_{\mathfrak g}(L(a)^*,L(b))\cong \operatorname{Hom}_{\mathfrak g}(L(a),L(b)). \end{align*} If $a=b$, then $L(a)=L(b)$ and *Schur's lemma* gives \begin{align*} \operatorname{Hom}_{\mathfrak g}(L(a),L(a))=\mathbb C\operatorname{id}_{L(a)}. \end{align*} Thus $\dim (L(a)\otimes L(a))^{\mathfrak g}=1$. If $a\ne b$, then $L(a)$ and $L(b)$ have different highest weights and are non-isomorphic irreducibles, so *Schur's lemma* gives \begin{align*} \operatorname{Hom}_{\mathfrak g}(L(a),L(b))=0. \end{align*} Thus $\dim (L(a)\otimes L(b))^{\mathfrak g}=0$ when $a\ne b$. The invariant vector exists exactly when the two tensor factors are dual to each other, and for irreducible $\mathfrak{sl}_2$-modules this means the two highest weights are equal. [/example] ## Schur Functors and Partitions for $\mathfrak{gl}_n$ and $\mathfrak{sl}_n$ Representations Tensor powers of the natural representation of $\mathfrak{gl}_n$ contain many familiar constructions: symmetric powers, exterior powers, and their mixed analogues. The obstacle is that symmetric and exterior powers alone only impose a single kind of symmetry, while most irreducible polynomial representations require different symmetries in different tensor positions. Schur functors solve this by combining row symmetrisation with column antisymmetrisation, so partitions become both combinatorial diagrams and highest weight labels. This connects the representation theory of $\mathfrak{gl}_n$ with the [symmetric group](/page/Symmetric%20Group) action on tensor space. [definition: Partition] A partition is a finite non-increasing sequence of nonnegative integers \begin{align*} \lambda = (\lambda_1,\lambda_2,\dots,\lambda_r) \end{align*} with $\lambda_1 \ge \lambda_2 \ge \cdots \ge \lambda_r \ge 0$. Its size is $|\lambda|=\lambda_1+\cdots+\lambda_r$, and its length $\ell(\lambda)$ is the number of nonzero parts. [/definition] Partitions serve as highest weights for polynomial representations of $\mathfrak{gl}_n$ when their length is at most $n$. To construct the corresponding representation rather than just name its highest weight, we need an operation that imposes the row symmetries and column antisymmetries encoded by the Young diagram. Applying only a symmetric power would forget the column restrictions, while applying only an exterior power would forget the row restrictions; the Young symmetriser is the operator that enforces both at once. The tensor power $E^{\otimes |\lambda|}$ is far too large for this purpose: it contains tensors with every possible symmetry type, while the partition $\lambda$ prescribes one particular row-column symmetry pattern. The next definition packages the required projection operation. It starts with the tensor power determined by the number of boxes of $\lambda$, applies a Young symmetriser for the diagram, and names the resulting image as the representation attached to the partition. [definition: Schur Functor] Let $\lambda$ be a partition of $d$ and let $E$ be a finite-dimensional complex vector space. Let \begin{align*} c_\lambda : E^{\otimes d} \to E^{\otimes d} \end{align*} be a Young symmetriser associated to $\lambda$, viewed as an endomorphism induced by the place-permutation action of $S_d$ on $E^{\otimes d}$. The Schur functor $S_\lambda(E)$ is the subspace \begin{align*} S_\lambda(E)=\operatorname{im}(c_\lambda) \subseteq E^{\otimes d}. \end{align*} [/definition] This definition depends on a choice of Young symmetriser, but different standard choices give isomorphic irreducible polynomial $GL(E)$-modules. For representation theory of Lie algebras, we differentiate the $GL(E)$-action to obtain a $\mathfrak{gl}(E)$-module. [example: Symmetric and Exterior Powers as Schur Functors] For the one-row partition $(d)$, the Young diagram has one row and no column containing more than one box. Thus the Young symmetriser only symmetrises the $d$ tensor positions. Concretely, for $v_1,\dots,v_d\in E$, the symmetrised tensor is represented by \begin{align*} \sum_{\sigma\in S_d} v_{\sigma(1)}\otimes v_{\sigma(2)}\otimes\cdots\otimes v_{\sigma(d)}. \end{align*} This is unchanged when the tensor positions are permuted, so its image is exactly the symmetric tensors in $E^{\otimes d}$. Hence \begin{align*} S_{(d)}(E)\cong \operatorname{Sym}^d(E). \end{align*} For the one-column partition $(1^d)$, the Young diagram has one column and no row containing more than one box. Thus the Young symmetriser only antisymmetrises the $d$ tensor positions. On a pure tensor it gives the alternating sum \begin{align*} \sum_{\sigma\in S_d} \operatorname{sgn}(\sigma)\, v_{\sigma(1)}\otimes v_{\sigma(2)}\otimes\cdots\otimes v_{\sigma(d)}. \end{align*} Under the quotient map from alternating tensors to exterior products, this tensor corresponds to $d! \, v_1\wedge\cdots\wedge v_d$, so the image is the exterior power: \begin{align*} S_{(1^d)}(E)\cong \Lambda^d E. \end{align*} Now suppose $\dim E=n$. If $d>n$, then any $d$ vectors $v_1,\dots,v_d\in E$ are linearly dependent. Choose scalars $a_1,\dots,a_d$, not all zero, with \begin{align*} a_1v_1+\cdots+a_dv_d=0. \end{align*} If $a_j\ne 0$, then \begin{align*} v_j=-a_j^{-1}\sum_{i\ne j} a_i v_i. \end{align*} Substituting this into $v_1\wedge\cdots\wedge v_d$ and using multilinearity gives a sum of terms in which two wedge factors are equal, and each such term is zero by alternation. Therefore \begin{align*} \Lambda^dE=0, \end{align*} so \begin{align*} S_{(1^d)}(E)=0. \end{align*} In contrast, $\operatorname{Sym}^d(E)$ is nonzero whenever $E\ne 0$: if $0\ne v\in E$, then $v^d$ is a nonzero element of $\operatorname{Sym}^d(E)$. Thus the same row-column rule recovers symmetric and exterior powers, while also showing why a partition with more than $n$ rows vanishes on $\mathbb C^n$. [/example] The example recovers the first two families of polynomial representations, but it also exposes the dimension obstruction: column antisymmetry vanishes once a column asks for more independent vectors than $\mathbb C^n$ can supply. The classification question is therefore which partitions actually give nonzero polynomial $\mathfrak{gl}_n$-modules, and what highest weight those modules carry. [quotetheorem:9394] [citeproof:9394] The condition $\ell(\lambda)\le n$ is the representation-theoretic shadow of a linear algebra obstruction: antisymmetrising more than $n$ vectors in $\mathbb C^n$ gives zero. For example, the column partition $(1^{n+1})$ gives $S_{(1^{n+1})}(\mathbb C^n)=\Lambda^{n+1}\mathbb C^n=0$, so not every formal partition can produce a nonzero module in fixed dimension. The theorem classifies irreducible polynomial $\mathfrak{gl}_n$-modules arising from tensor powers of the natural representation; it does not classify all finite-dimensional $\mathfrak{gl}_n$-modules, since determinant twists and duals introduce weights that need not be partitions. For $\mathfrak{sl}_n$, two $\mathfrak{gl}_n$ highest weights that differ by a multiple of $\varepsilon_1+\cdots+\varepsilon_n$ restrict to the same $\mathfrak{sl}_n$ highest weight. This is why determinant twists disappear after restriction from $GL_n$ to $SL_n$ or from $\mathfrak{gl}_n$ to $\mathfrak{sl}_n$. [example: Schur Functors on Complex N Space] Let $E=\mathbb C^n$ with standard basis $e_1,\dots,e_n$, and assume $n\ge 3$. For the partition $\lambda=(2,1)$, label the two boxes in the first row by tensor positions $1,2$ and the lower box in the first column by tensor position $3$. The row symmetriser averages positions $1,2$, and the column antisymmetriser alternates positions $1,3$. Thus, on a pure tensor $a\otimes b\otimes c\in E^{\otimes 3}$, one Young symmetriser sends it to the sum of the row-symmetrised tensor minus the same tensor with positions $1$ and $3$ exchanged: \begin{align*} a\otimes b\otimes c+b\otimes a\otimes c-c\otimes b\otimes a-b\otimes c\otimes a. \end{align*} The image of this operator is $S_{(2,1)}(E)\subseteq E^{\otimes 3}$. By the Schur-functor highest-weight rule for polynomial $\mathfrak{gl}_n$-modules, the partition $(2,1)$ gives highest weight \begin{align*} 2\varepsilon_1+\varepsilon_2+0\varepsilon_3+\cdots+0\varepsilon_n=2\varepsilon_1+\varepsilon_2. \end{align*} To restrict this weight to $\mathfrak{sl}_n$, let $\delta=\varepsilon_1+\cdots+\varepsilon_n$. The fundamental weights of $\mathfrak{sl}_n$ are represented by \begin{align*} \omega_k=\varepsilon_1+\cdots+\varepsilon_k-\frac{k}{n}\delta. \end{align*} Therefore \begin{align*} \omega_1+\omega_2=\left(\varepsilon_1-\frac{1}{n}\delta\right)+\left(\varepsilon_1+\varepsilon_2-\frac{2}{n}\delta\right)=2\varepsilon_1+\varepsilon_2-\frac{3}{n}\delta. \end{align*} On the Cartan subalgebra of $\mathfrak{sl}_n$, an element has diagonal entries summing to $0$, so $\delta$ restricts to $0$. Hence $2\varepsilon_1+\varepsilon_2$ and $\omega_1+\omega_2$ define the same $\mathfrak{sl}_n$ weight. Thus $S_{(2,1)}(\mathbb C^n)$ has $\mathfrak{gl}_n$ highest weight $2\varepsilon_1+\varepsilon_2$, and its restriction to $\mathfrak{sl}_n$ has highest weight $\omega_1+\omega_2$. [/example] Schur functors are not merely a classification device; they give a practical computational bridge. Decomposing tensor powers can be approached by symmetric group combinatorics, while highest weight theory translates the output back into irreducible Lie algebra modules. [remark: Schur Weyl Viewpoint] In Schur-Weyl duality, $GL_n$ acts on $(\mathbb C^n)^{\otimes d}$ diagonally and $S_d$ acts by permuting tensor positions. These two actions commute, and the simultaneous decomposition explains why partitions index both Specht modules and polynomial $\mathfrak{gl}_n$-modules. This viewpoint turns tensor product calculations into combinatorics of partitions, tableaux, and Littlewood-Richardson coefficients. [/remark] # 10. Branching and Restriction Branching asks what information remains when a representation is viewed by a smaller Lie algebra. Earlier chapters built irreducible modules from highest weights and computed their characters inside a fixed semisimple Lie algebra. This chapter studies the functor of restriction, especially the standard $\mathfrak{gl}_n$ chain whose interlacing rule also guides the corresponding special-linear calculations after central data are handled. The guiding question is how a highest weight for the large algebra breaks into highest weights for the subalgebra. ## Restriction Along Lie Subalgebras A representation of a Lie algebra often carries more symmetry than the problem currently needs. If $\mathfrak{k}\subset \mathfrak{g}$ is a Lie subalgebra, then every $\mathfrak{g}$-module can be viewed as a $\mathfrak{k}$-module by forgetting part of the action. The first question is what this operation preserves and what new decomposition problem it creates. [definition: Restricted Representation] Let $\mathfrak{k}\subset \mathfrak{g}$ be a Lie subalgebra and let $V$ be a $\mathfrak{g}$-module with action $\rho:\mathfrak{g}\to \mathfrak{gl}(V)$. The restriction of $V$ to $\mathfrak{k}$ is the $\mathfrak{k}$-module $\operatorname{Res}^{\mathfrak{g}}_{\mathfrak{k}}V$ with action $\rho|_{\mathfrak{k}}:\mathfrak{k}\to \mathfrak{gl}(V)$. [/definition] Restriction does not change the underlying vector space, but it changes the algebra used to decompose it. An irreducible $\mathfrak{g}$-module may become reducible over $\mathfrak{k}$ because fewer operators are available to connect its vectors. [example: Restricting The Natural Module] Let $\mathfrak{sl}_{n-1}\subset \mathfrak{sl}_n$ be embedded in the upper-left block, and let $V=\mathbb{C}^n$ have basis $e_1,\dots,e_n$. If $X=(x_{ij})\in\mathfrak{sl}_{n-1}$, then its embedded action on $V$ is determined by \begin{align*} X\cdot e_j=\sum_{i=1}^{n-1}x_{ij}e_i\quad\text{for }1\le j\le n-1,\qquad X\cdot e_n=0. \end{align*} Thus $W=\operatorname{span}(e_1,\dots,e_{n-1})$ is stable under $\mathfrak{sl}_{n-1}$, and the displayed formula is exactly the natural action of $\mathfrak{sl}_{n-1}$ on $\mathbb{C}^{n-1}$. The remaining line $\mathbb{C}e_n$ is fixed pointwise, since every $X\in\mathfrak{sl}_{n-1}$ satisfies $X\cdot e_n=0$. Every vector $v\in\mathbb{C}^n$ decomposes uniquely as \begin{align*} v=(v_1e_1+\cdots+v_{n-1}e_{n-1})+v_ne_n. \end{align*} Both summands lie in $\mathfrak{sl}_{n-1}$-stable subspaces, so as $\mathfrak{sl}_{n-1}$-modules, \begin{align*} \operatorname{Res}^{\mathfrak{sl}_n}_{\mathfrak{sl}_{n-1}}\mathbb{C}^n \cong \mathbb{C}^{n-1}\oplus \mathbb{C}. \end{align*} Restriction has kept the same vector space, but the smaller algebra no longer connects the last basis direction to the first $n-1$ directions, so a fixed one-dimensional summand appears. [/example] The basic operation is formal, but the main issue is decomposition. Since the course works over $\mathbb{C}$ and with finite-dimensional semisimple Lie algebras, complete reducibility turns restriction into a finite multiplicity problem. [quotetheorem:3817] [citeproof:3817] For restriction, Weyl complete reducibility changes the problem from existence of a decomposition to identification of its pieces. The same underlying vector space may break into several irreducible $\mathfrak{k}$-modules, and different embeddings $\mathfrak{k}\subset \mathfrak g$ can produce different answers. Thus the theorem supplies the structural permission to decompose, but it does not identify the summands or count how many times each one occurs. The useful invariant is the multiplicity with which a specified irreducible $\mathfrak{k}$-module appears after restriction. [definition: Branching Multiplicity] Let $\mathfrak{k}\subset \mathfrak{g}$ be semisimple Lie algebras. If $L_{\mathfrak{g}}(\lambda)$ is an irreducible finite-dimensional $\mathfrak{g}$-module and $L_{\mathfrak{k}}(\mu)$ is an irreducible finite-dimensional $\mathfrak{k}$-module, the branching multiplicity is \begin{align*} \left[\operatorname{Res}^{\mathfrak{g}}_{\mathfrak{k}}L_{\mathfrak{g}}(\lambda):L_{\mathfrak{k}}(\mu)\right]. \end{align*} [/definition] These multiplicities depend on the embedding $\mathfrak{k}\subset \mathfrak{g}$, not only on the abstract isomorphism type of $\mathfrak{k}$. For semisimple subalgebras coming from a standard subsystem of roots, the restriction has a close relationship with weights. [remark: Restriction Of Weights] Let $\mathfrak{h}_{\mathfrak{k}}\subset \mathfrak{h}_{\mathfrak{g}}$ be compatible Cartan subalgebras. If $V=\bigoplus_\nu V_\nu$ is a $\mathfrak{g}$-weight decomposition, then each $\mathfrak{g}$-weight $\nu\in \mathfrak{h}_{\mathfrak{g}}^*$ restricts to a linear functional $\nu|_{\mathfrak{h}_{\mathfrak{k}}}\in \mathfrak{h}_{\mathfrak{k}}^*$. The $\mathfrak{k}$-weight space of weight $\eta$ is the sum of all $V_\nu$ with $\nu|_{\mathfrak{h}_{\mathfrak{k}}}=\eta$. [/remark] This remark gives a necessary numerical constraint: a $\mathfrak{k}$-irreducible of highest weight $\mu$ can occur only if its weights are present among the restricted weights of the original module. It is not yet a full rule, because one must identify which restricted weights are highest weights for $\mathfrak{k}$ and how often they generate new summands. ## Branching From $\mathfrak{gl}_n$ To $\mathfrak{gl}_{n-1}$ The standard upper-left inclusion $\mathfrak{gl}_{n-1}\subset \mathfrak{gl}_n$ is the first branching problem where a complete and usable answer appears. The natural question is: given an irreducible polynomial $\mathfrak{gl}_n$-module of highest weight $\lambda$, which irreducible $\mathfrak{gl}_{n-1}$-modules appear after restriction? The corresponding $\mathfrak{sl}$ question is obtained by forgetting the scalar direction, but that forgetting can combine several $\mathfrak{gl}_{n-1}$ rows into the same $\mathfrak{sl}_{n-1}$ highest weight. For this section it is convenient to use partition notation for dominant integral highest weights. A dominant highest weight for polynomial representations of $\mathfrak{gl}_n$ is written as a weakly decreasing sequence $\lambda=(\lambda_1,\dots,\lambda_n)$ of integers. For $\mathfrak{sl}_n$, adding the same integer to all entries does not change the restricted highest weight, so the sequence records a representative of the corresponding weight class. [definition: Interlacing Weights] Let $\lambda=(\lambda_1,\dots,\lambda_n)$ be weakly decreasing and let $\mu=(\mu_1,\dots,\mu_{n-1})$ be weakly decreasing. We say that $\mu$ interlaces $\lambda$ if \begin{align*} \lambda_1\ge \mu_1\ge \lambda_2\ge \mu_2\ge \cdots \ge \mu_{n-1}\ge \lambda_n. \end{align*} [/definition] Interlacing is the numerical shadow of highest-weight restriction: each entry of the smaller highest weight must sit between two adjacent entries of the larger one. The central theorem of this section says that this necessary-looking inequality is also sufficient, and that no hidden multiplicity remains to be computed. [quotetheorem:9395] [citeproof:9395] The multiplicity-one feature is special and powerful at the $\mathfrak{gl}$ level: the restricted module is read as a set of interlacing rows rather than as a weighted multiset. The hypotheses should not be blurred. The theorem uses the standard upper-left inclusion and polynomial highest-weight modules; for other embeddings, such as a root $\mathfrak{sl}_2\subset\mathfrak{sl}_3$, multiplicities can occur. Even for $\mathfrak{sl}_{n-1}$ obtained from the standard inclusion, passing from $\mathfrak{gl}_{n-1}$ to $\mathfrak{sl}_{n-1}$ can identify several interlacing rows with the same $\mathfrak{sl}_{n-1}$ highest weight, so the safest form of the rule is the $\mathfrak{gl}$ statement above followed by this quotienting step. [example: Exterior Powers Of The Natural Module] Let $V=\mathbb{C}^n$ with basis $e_1,\dots,e_n$, and write $W=\operatorname{span}(e_1,\dots,e_{n-1})$. The exterior power $\Lambda^r V$ has highest weight represented by \begin{align*} \lambda=(\underbrace{1,\dots,1}_{r},0,\dots,0). \end{align*} For a $\mathfrak{gl}_{n-1}$ interlacing row $\mu=(\mu_1,\dots,\mu_{n-1})$, the inequalities are \begin{align*} \lambda_i\ge \mu_i\ge \lambda_{i+1}\quad\text{for }1\le i\le n-1. \end{align*} If $i<r$, then $\lambda_i=\lambda_{i+1}=1$, so $1\ge \mu_i\ge 1$ and hence $\mu_i=1$. If $i>r$, then $\lambda_i=\lambda_{i+1}=0$, so $0\ge \mu_i\ge 0$ and hence $\mu_i=0$. For $i=r$, when $1\le r\le n-1$, the inequality is $1\ge \mu_r\ge 0$, so $\mu_r$ is either $1$ or $0$. Thus the only possible rows are \begin{align*} (\underbrace{1,\dots,1}_{r},0,\dots,0) \quad\text{and}\quad (\underbrace{1,\dots,1}_{r-1},0,\dots,0), \end{align*} with the first absent when $r=n$ and the second absent when $r=0$. By *Basic Branching Rule For General Linear Lie Algebras*, each allowed interlacing row occurs once, so after passing to $\mathfrak{sl}_{n-1}$ we get \begin{align*} \operatorname{Res}^{\mathfrak{sl}_n}_{\mathfrak{sl}_{n-1}}\Lambda^r\mathbb{C}^n\cong \Lambda^r\mathbb{C}^{n-1}\oplus \Lambda^{r-1}\mathbb{C}^{n-1}. \end{align*} The same decomposition is visible on wedge bases. Every basis wedge $e_{i_1}\wedge\cdots\wedge e_{i_r}$ with $i_1<\cdots<i_r$ either has $i_r<n$, in which case it lies in $\Lambda^r W$, or has $i_r=n$, in which case it has the form \begin{align*} e_{i_1}\wedge\cdots\wedge e_{i_{r-1}}\wedge e_n \quad\text{with}\quad e_{i_1}\wedge\cdots\wedge e_{i_{r-1}}\in \Lambda^{r-1}W. \end{align*} These two kinds of basis vectors are disjoint and exhaust the wedge basis, so \begin{align*} \Lambda^r V=\Lambda^r W\oplus\left(\Lambda^{r-1}W\wedge e_n\right). \end{align*} For $X\in\mathfrak{sl}_{n-1}$, the embedded action satisfies $X\cdot W\subseteq W$ and $X\cdot e_n=0$, so both summands are $\mathfrak{sl}_{n-1}$-stable. Thus the interlacing calculation matches the concrete separation of wedges that avoid $e_n$ from wedges that contain $e_n$. [/example] The exterior-power example shows how interlacing recovers a familiar multilinear decomposition. For general highest weights, the same inequalities replace direct basis inspection. [example: A Three-Dimensional Branching Diagram] For $\mathfrak{sl}_3$, take the corresponding polynomial $\mathfrak{gl}_3$ highest weight $(a,b,0)$ with $a\ge b\ge 0$, and restrict first to the upper-left $\mathfrak{gl}_2$. By *Basic Branching Rule For General Linear Lie Algebras*, the $\mathfrak{gl}_2$ summands are indexed by rows $(\mu_1,\mu_2)$ satisfying \begin{align*}a\ge \mu_1\ge b\ge \mu_2\ge 0.\end{align*} For the derived algebra $\mathfrak{sl}_2\subset\mathfrak{gl}_2$, the Cartan element may be taken as $h=\operatorname{diag}(1,-1)$. A $\mathfrak{gl}_2$ highest-weight vector of weight $(\mu_1,\mu_2)$ has $h$-weight \begin{align*}\mu_1\cdot 1+\mu_2\cdot(-1)=\mu_1-\mu_2.\end{align*} Therefore this $\mathfrak{gl}_2$ summand becomes an $\mathfrak{sl}_2$ irreducible with highest weight $\mu_1-\mu_2$. Thus the multiplicity of $L_{\mathfrak{sl}_2}(r)$ is the number of integer pairs $(\mu_1,\mu_2)$ such that \begin{align*}a\ge \mu_1\ge b\ge \mu_2\ge 0\quad\text{and}\quad \mu_1-\mu_2=r.\end{align*} Equivalently, writing $\mu_1=\mu_2+r$, the inequalities become \begin{align*}a\ge \mu_2+r\ge b\quad\text{and}\quad b\ge \mu_2\ge 0.\end{align*} So $\mu_2$ must lie in the integer interval \begin{align*}\max(0,b-r)\le \mu_2\le \min(b,a-r).\end{align*} Hence \begin{align*}\left[\operatorname{Res}^{\mathfrak{sl}_3}_{\mathfrak{sl}_2}L_{\mathfrak{sl}_3}(a,b):L_{\mathfrak{sl}_2}(r)\right]=\max\left(0,\min(b,a-r)-\max(0,b-r)+1\right).\end{align*} This makes the role of the central $\mathfrak{gl}_2$ weight explicit: two different rows can have the same difference $\mu_1-\mu_2$, so discarding the centre before counting can merge distinct $\mathfrak{gl}_2$ branching summands into one $\mathfrak{sl}_2$ multiplicity. [/example] The previous example depends on the standard embedded copy of $\mathfrak{sl}_2$. Other copies of $\mathfrak{sl}_2$ inside $\mathfrak{sl}_3$, such as root subalgebras, are better analysed directly from weights and root strings. ## Gelfand-Tsetlin Patterns Iterating the previous restriction suggests a finer question. If $L^{\mathfrak{gl}_n}(\lambda)$ restricts to many possible $\mathfrak{gl}_{n-1}$ highest weights, and each of those restricts to many possible $\mathfrak{gl}_{n-2}$ highest weights, can the whole process be encoded in one combinatorial object? [definition: Gelfand-Tsetlin Pattern] A Gelfand-Tsetlin pattern of top row $\lambda=(\lambda_{n,1},\dots,\lambda_{n,n})$ is a collection of integers $\lambda_{k,i}$ for $1\le i\le k\le n$, arranged in rows of lengths $n,n-1,\dots,1$, with top row $\lambda_{n,i}=\lambda_i$, such that every adjacent pair of rows interlaces: \begin{align*} \lambda_{k+1,i}\ge \lambda_{k,i}\ge \lambda_{k+1,i+1} \end{align*} for $1\le i\le k$ and $1\le k<n$. [/definition] Each row is a possible highest weight after restricting one step further down the chain \begin{align*} \mathfrak{gl}_n\supset \mathfrak{gl}_{n-1}\supset \cdots \supset \mathfrak{gl}_1. \end{align*} The definition turns a sequence of branching choices into one triangular array, and the next theorem identifies these arrays with the full set of iterated branching paths. [quotetheorem:9396] [citeproof:9396] This theorem explains why Gelfand-Tsetlin patterns are more than bookkeeping. They provide a concrete combinatorial model for a basis compatible with the whole $\mathfrak{gl}$ chain of subalgebras. The standard chain, the retained central data, and multiplicity-free $\mathfrak{gl}$ branching are essential to this indexing. If one passes at each stage to only the derived algebra $\mathfrak{sl}_k$, different $\mathfrak{gl}_k$ rows may collapse to the same $\mathfrak{sl}_k$ highest weight, so a path of special-linear highest weights may not identify a basis vector without extra labels. The theorem is also a counting and indexing statement rather than a complete construction of the representation: it does not by itself give the explicit formulas for the action of the Chevalley generators on the Gelfand-Tsetlin basis. Those formulas are additional structure built on top of the branching pattern. [example: Patterns For The Natural Representation] For the natural representation of $\mathfrak{gl}_3$, take the polynomial highest weight $(1,0,0)$. A Gelfand-Tsetlin pattern with this top row is determined by a middle row $(\mu_1,\mu_2)$ and a bottom entry $\nu$, with interlacing inequalities \begin{align*} 1\ge \mu_1\ge 0,\qquad 0\ge \mu_2\ge 0,\qquad \mu_1\ge \nu\ge \mu_2. \end{align*} Since all entries are integers, $1\ge \mu_1\ge 0$ gives $\mu_1\in\{0,1\}$, while $0\ge \mu_2\ge 0$ gives $\mu_2=0$. Thus the only possible middle rows are $(1,0)$ and $(0,0)$. If the middle row is $(1,0)$, then the bottom inequality is $1\ge \nu\ge 0$, so $\nu=1$ or $\nu=0$. If the middle row is $(0,0)$, then the bottom inequality is $0\ge \nu\ge 0$, so $\nu=0$. Therefore the three patterns are represented by the row data \begin{align*} (1,0,0)\supset (1,0)\supset (1), \end{align*} \begin{align*} (1,0,0)\supset (1,0)\supset (0), \end{align*} and \begin{align*} (1,0,0)\supset (0,0)\supset (0). \end{align*} There are exactly $3$ Gelfand-Tsetlin patterns with top row $(1,0,0)$, matching the dimension of the natural module $\mathbb{C}^3$. [/example] The same count for larger top rows rapidly becomes nontrivial. The pattern description packages all intermediate restrictions into an array whose inequalities can be checked by inspection. ## Weight Diagrams As A Branching Tool The interlacing rule is the clean formula for the standard chain, but in practice one often restricts to a subalgebra not written as the previous rank in the chain. Weight diagrams give a practical method: restrict the weights, identify candidate highest weights, and subtract the characters of irreducible summands. [explanation: The Weight-Diagram Algorithm] Suppose $\mathfrak{k}\subset \mathfrak{g}$ is a semisimple subalgebra with compatible Cartan subalgebras. Start with the multiset of $\mathfrak{g}$-weights of $V$, including multiplicities. Restrict each weight to $\mathfrak{h}_{\mathfrak{k}}$, obtaining a multiset of $\mathfrak{k}$-weights. Choose a dominant maximal restricted weight $\mu$, record one copy of $L_{\mathfrak{k}}(\mu)$, and subtract the full weight multiset of $L_{\mathfrak{k}}(\mu)$ from the restricted multiset. Repeating this process terminates because finite-dimensional weight sets are finite and dominance gives a descending procedure. [/explanation] The method is a character calculation written geometrically. Its accuracy depends on subtracting whole irreducible weight diagrams, not just removing highest weights. [example: Restricting The Adjoint Module Of $\mathfrak{sl}_3$ To A Root $\mathfrak{sl}_2$] Let $\mathfrak{k}\cong\mathfrak{sl}_2$ be the root subalgebra generated by $e_{\alpha_1},f_{\alpha_1},h_{\alpha_1}$ inside $\mathfrak{sl}_3$. The adjoint module of $\mathfrak{sl}_3$ has root weights $\alpha_1,-\alpha_1,\alpha_2,-\alpha_2,\alpha_1+\alpha_2,-\alpha_1-\alpha_2$ together with the zero weight twice, coming from the Cartan subalgebra. To restrict these weights to $\mathfrak{k}$, evaluate them on $h_{\alpha_1}$. The type $A_2$ Cartan pairings are \begin{align*} \alpha_1(h_{\alpha_1})=2,\qquad \alpha_2(h_{\alpha_1})=-1. \end{align*} Therefore \begin{align*} (-\alpha_1)(h_{\alpha_1})=-2,\qquad (-\alpha_2)(h_{\alpha_1})=1. \end{align*} Also \begin{align*} (\alpha_1+\alpha_2)(h_{\alpha_1})=\alpha_1(h_{\alpha_1})+\alpha_2(h_{\alpha_1})=2+(-1)=1. \end{align*} Similarly, \begin{align*} (-\alpha_1-\alpha_2)(h_{\alpha_1})=-\alpha_1(h_{\alpha_1})-\alpha_2(h_{\alpha_1})=-2-(-1)=-1. \end{align*} The two zero weights restrict to $0$ and $0$, so the restricted $\mathfrak{sl}_2$ weight multiset is \begin{align*} \{2,-2,-1,1,1,-1,0,0\}=\{2,1,1,0,0,-1,-1,-2\}. \end{align*} For $\mathfrak{sl}_2$, the irreducible module $L_{\mathfrak{sl}_2}(m)$ has weights $m,m-2,\dots,-m$, each with multiplicity $1$. The largest remaining weight is $2$, so one summand is $L_{\mathfrak{sl}_2}(2)$, whose weight multiset is \begin{align*} \{2,0,-2\}. \end{align*} Removing these weights from the restricted multiset leaves \begin{align*} \{2,1,1,0,0,-1,-1,-2\}\setminus\{2,0,-2\}=\{1,1,0,-1,-1\}. \end{align*} The largest remaining weight is $1$, so remove one copy of $L_{\mathfrak{sl}_2}(1)$, whose weights are $\{1,-1\}$: \begin{align*} \{1,1,0,-1,-1\}\setminus\{1,-1\}=\{1,0,-1\}. \end{align*} The largest remaining weight is again $1$, so remove a second copy: \begin{align*} \{1,0,-1\}\setminus\{1,-1\}=\{0\}. \end{align*} The remaining weight multiset $\{0\}$ is exactly the trivial module $L_{\mathfrak{sl}_2}(0)$. Hence \begin{align*} \operatorname{Res}^{\mathfrak{sl}_3}_{\mathfrak{k}}\mathfrak{sl}_3 \cong L_{\mathfrak{sl}_2}(2)\oplus L_{\mathfrak{sl}_2}(1)\oplus L_{\mathfrak{sl}_2}(1)\oplus L_{\mathfrak{sl}_2}(0). \end{align*} The two copies of $L_{\mathfrak{sl}_2}(1)$ show that multiplicities can appear for root-subalgebra restrictions, even though the standard $\mathfrak{gl}_n$ to $\mathfrak{gl}_{n-1}$ branching rule is multiplicity-free. [/example] Weight diagrams also clarify why Levi subalgebras require a slightly richer record than semisimple root subalgebras. A Levi subalgebra keeps a subset of simple-root directions together with the Cartan directions centralising them, so the next definition fixes the object whose restriction remembers both the semisimple and central parts. [definition: Levi Subalgebra From Simple Roots] Let $\mathfrak{g}$ be a complex semisimple Lie algebra with Cartan subalgebra $\mathfrak{h}$ and simple roots $\Delta$. For a subset $I\subset \Delta$, the standard Levi subalgebra $\mathfrak{l}_I$ is generated by $\mathfrak{h}$ together with the root spaces $\mathfrak{g}_{\alpha}$ for all roots $\alpha$ in the root subsystem generated by $I$. [/definition] Unlike semisimple subalgebras, a Levi subalgebra has a centre. Its irreducible modules are controlled by the semisimple derived algebra together with central characters. [remark: Central Weights In Levi Restriction] When restricting to a Levi subalgebra $\mathfrak{l}_I$, weights must be retained on the whole Cartan subalgebra $\mathfrak{h}$, not only on the derived algebra $[\mathfrak{l}_I,\mathfrak{l}_I]$. Two summands with the same derived-algebra highest weight may differ by their action of the centre of $\mathfrak{l}_I$. This is why Levi branching is often recorded using the full projected weight data. [/remark] The conceptual pattern of the chapter is now complete. Restriction is the functor, branching multiplicities are the data to compute, interlacing solves the standard general-linear chain, Gelfand-Tsetlin patterns encode iterated branching with full Cartan labels, and weight diagrams provide the practical fallback for embedded subalgebras where no closed formula is being used. This is also where the subject connects outward: the same interlacing arrays appear as semistandard-tableau data in symmetric function theory, while closely related branching questions occur in physics when decomposing angular-momentum states after reducing the symmetry algebra. # 11. Category $\mathcal O$ as an Organizing Framework The preceding chapters constructed highest weight modules one at a time: Verma modules gave universal objects, irreducible highest weight modules appeared as their quotients, and finite-dimensional representations were singled out by dominance and integrality. This chapter packages those constructions into a single abelian category, the BGG category $\mathcal O$. The gain is organizational: subquotients, extensions, central characters, and Weyl group symmetry can be discussed inside one controlled setting rather than separately for each module. The guiding question is which class of $U(\mathfrak g)$-modules is large enough to contain Verma modules and finite-dimensional modules, but small enough that weight theory, central characters, and finite length arguments still work. ## The BGG Category and Verma Modules Highest weight theory uses modules that decompose into weights and on which positive root operators eventually stop moving vectors upward. The definition of category $\mathcal O$ isolates exactly these two features, together with finite generation over $U(\mathfrak g)$. [definition: BGG Category O] Let $\mathfrak g$ be a finite-dimensional complex semisimple Lie algebra with triangular decomposition \begin{align*} \mathfrak g = \mathfrak n^- \oplus \mathfrak h \oplus \mathfrak n^+. \end{align*} The BGG category $\mathcal O$ is the full subcategory of $U(\mathfrak g)$-modules $M$ satisfying: 1. $M$ is finitely generated as a $U(\mathfrak g)$-module. 2. $M$ is $\mathfrak h$-semisimple, so \begin{align*} M = \bigoplus_{\lambda \in \mathfrak h^*} M_\lambda, \qquad M_\lambda = \{m \in M : h m = \lambda(h)m \text{ for all } h \in \mathfrak h\}. \end{align*} 3. $M$ is locally $U(\mathfrak n_+)$-finite, meaning $U(\mathfrak n_+)m$ is finite-dimensional for every $m \in M$. [/definition] The finite generation condition prevents uncontrolled direct sums, the weight decomposition keeps the Cartan action visible, and local $\mathfrak n_+$-finiteness formalizes the highest-weight direction. The definition is asymmetric in $\mathfrak n_+$ and $\mathfrak n_-$ because category $\mathcal O$ is built to study highest weight modules rather than lowest weight modules. [example: Category O Contains Verma Modules] Let $\lambda \in \mathfrak h^*$, and let $M(\lambda)=U(\mathfrak g)v_\lambda$ be the Verma module with $\mathfrak n^+v_\lambda=0$ and $hv_\lambda=\lambda(h)v_\lambda$ for all $h\in\mathfrak h$. It is generated by the single vector $v_\lambda$, so it is finitely generated over $U(\mathfrak g)$. By the *PBW theorem*, multiplication gives a vector space spanning set \begin{align*} f_{\alpha_1}^{a_1}\cdots f_{\alpha_r}^{a_r}v_\lambda \end{align*} where $\alpha_1,\dots,\alpha_r$ are the positive roots, $f_{\alpha_i}\in\mathfrak g_{-\alpha_i}$, and $a_i\in\mathbb Z_{\ge 0}$. For such a monomial, the commutator relation $[h,f_{\alpha_i}]=-\alpha_i(h)f_{\alpha_i}$ gives \begin{align*} h f_{\alpha_1}^{a_1}\cdots f_{\alpha_r}^{a_r}v_\lambda = \left(\lambda-\sum_{i=1}^r a_i\alpha_i\right)(h) f_{\alpha_1}^{a_1}\cdots f_{\alpha_r}^{a_r}v_\lambda. \end{align*} Thus $M(\lambda)$ is a weight module, and its weights all have the form $\lambda-\beta$ with $\beta\in\mathbb Z_{\ge 0}\Phi^+$. It remains to check local $U(\mathfrak n^+)$-finiteness. Fix a weight vector $m\in M(\lambda)_{\lambda-\beta}$. If $x\in U(\mathfrak n^+)$ has weight $\gamma\in\mathbb Z_{\ge 0}\Phi^+$, then $xm$ has weight $\lambda-\beta+\gamma=\lambda-(\beta-\gamma)$. Since no weights occur above $\lambda$, this vector is zero unless $\beta-\gamma\in\mathbb Z_{\ge 0}\Phi^+$. There are only finitely many such $\gamma$, and for each remaining weight $\lambda-\delta$ there are only finitely many PBW monomials with total root weight $\delta$. Hence \begin{align*} U(\mathfrak n^+)m \subseteq \bigoplus_{0\le \delta\le \beta} M(\lambda)_{\lambda-\delta} \end{align*} is finite-dimensional. Therefore every Verma module satisfies all three defining conditions of category $\mathcal O$. [/example] This example is the reason category $\mathcal O$ is the natural home for Verma modules. The restrictions in the definition are also essential: if finite generation is dropped, arbitrary direct sums of Verma modules enter and finite-length statements fail; if weight decomposability is dropped, generalized weight modules can hide the Cartan action; if local $U(\mathfrak n_+)$-finiteness is dropped, positive root operators can create infinite upward strings. To use $\mathcal O$ as a framework rather than a container, we need closure under the operations used to form kernels, quotients, and composition series, together with a finiteness theorem that excludes these pathologies. [quotetheorem:9397] [citeproof:9397] Finite length means that category $\mathcal O$ turns questions about arbitrary objects into questions about simple objects and how they fit together. Each hypothesis is doing work: without finite generation, an infinite direct sum $\bigoplus_{k\ge 1} M(\lambda_k)$ satisfies the weight and local $\mathfrak n_+$-finite conditions but need not have finite length; without finite-dimensional weight spaces, repeated composition arguments can fail to terminate inside a single weight. The theorem does not classify the simple objects or compute extension groups; it only guarantees that such questions make sense in an abelian finite-length setting. The next problem is to divide this finite-length category into smaller pieces, because composition factors linked by extensions cannot be studied independently from the central action. ## Blocks and Central Characters A module in $\mathcal O$ can have many weights, but the center $Z(U(\mathfrak g))$ gives a coarser invariant that is stable under extensions. Scalar central characters are not enough for nonsimple modules: in a short exact sequence of two modules with the same scalar central character, a central element may act by a Jordan block rather than by a scalar on the middle term. The first step is therefore to name the generalized scalar data by which central elements act. [definition: Central Character] Let $M$ be a $U(\mathfrak g)$-module. A central character of $M$ is an algebra homomorphism \begin{align*} \chi : Z(U(\mathfrak g)) \to \mathbb C \end{align*} such that for every $z \in Z(U(\mathfrak g))$, the endomorphism \begin{align*} z-\chi(z)\operatorname{id}_M : M \to M \end{align*} acts locally nilpotently on every $m \in M$. [/definition] For a simple module, Schur's lemma turns local nilpotence into scalar action: each central element acts by a scalar. For a nonsimple object of $\mathcal O$, generalized eigenspaces are needed because extensions may replace scalar action by Jordan blocks; this motivates grouping modules according to generalized central character. [definition: Block of Category O] For a central character $\chi$, the block $\mathcal O_\chi$ is the full subcategory of $\mathcal O$ consisting of modules on which every $z-\chi(z)$, with $z \in Z(U(\mathfrak g))$, acts locally nilpotently. [/definition] Blocks separate modules according to generalized central character. To turn this definition into a computable classification of blocks, we need to identify when two highest weights give the same central character. [quotetheorem:9398] [citeproof:9398] This theorem explains how Weyl group orbits enter block theory, and the $\rho$-shift is essential: the center sees the shifted, or dot, action rather than the ordinary Weyl action on highest weights. For $\mathfrak{sl}_2$, the ordinary reflection sends a weight $n$ to $-n$, while the dot action sends it to $-n-2$; the pair $L(n)$ and $L(-n-2)$ has the same central character, not the pair predicted by the unshifted reflection. The highest-weight and semisimplicity hypotheses are also essential, since an indecomposable extension of two simples with the same central character may have the center acting by a nontrivial Jordan block rather than by scalars. Thus the theorem classifies scalar central characters of simple highest weight modules, not arbitrary indecomposable objects in $\mathcal O$. It also does not say that two simples with the same central character occur in the same Verma module, nor does it compute multiplicities. The next question is more representation-theoretic: if a simple module appears inside a Verma module, how much freedom is there in its highest weight? [quotetheorem:9399] [citeproof:9399] The converse is subtler: lying in the same dot orbit is necessary for occurrence as a composition factor, but it is not sufficient in this form, and the theorem gives no multiplicity formula. The Verma-module hypothesis is essential because the argument uses the central character inherited from the highest weight vector. For example, a nonsplit extension between two simples in the same block, or a projective cover in that block, is an object of $\mathcal O$ whose composition factors are not controlled by the submodule lattice of a single Verma module. Such objects still belong to the same central-character block, but their extension structure requires more than the linkage condition. This is the first point where the chapter connects to the broader structure of category $\mathcal O$: actual multiplicities are governed in general by Kazhdan-Lusztig theory, translation functors, and the geometry of flag varieties. We therefore test the linkage statement in the rank-one case, where singular vectors can be written down explicitly. ## The Rank-One Model: Category O for $\mathfrak{sl}_2$ In general, linkage gives an abstract constraint before it gives a calculation: it says that composition factors must lie in a dot orbit, but it does not exhibit the singular vectors that produce submodules. Rank one is the place where the obstruction can be computed directly. For $\mathfrak{sl}_2$, the Weyl group has two elements, the weight strings are one-dimensional, and the possible singular vector is detected by a single coefficient. [example: Weights in an $\mathfrak{sl}_2$ Verma Module] Let $\mathfrak{sl}_2$ have basis $e,h,f$ with $[h,e]=2e$, $[h,f]=-2f$, and $[e,f]=h$. For $\lambda\in\mathbb C$, the Verma module $M(\lambda)$ is generated by a vector $v_\lambda$ satisfying $ev_\lambda=0$ and $hv_\lambda=\lambda v_\lambda$, and it has basis $v_\lambda,fv_\lambda,f^2v_\lambda,\dots$. Since $[h,f]=-2f$, we have $hf=fh-2f$, and induction gives \begin{align*} h f^k = f^k h - 2k f^k. \end{align*} Applying this to $v_\lambda$ yields \begin{align*} h f^k v_\lambda = (f^k h - 2k f^k)v_\lambda = (\lambda-2k)f^k v_\lambda. \end{align*} Thus $f^k v_\lambda$ has weight $\lambda-2k$. We now compute when $f^k v_\lambda$ is killed by $e$. From $[e,f]=h$, we have $ef=fe+h$. For $k\ge 1$, \begin{align*} [e,f^k]=\sum_{i=0}^{k-1} f^i[e,f]f^{k-1-i}=\sum_{i=0}^{k-1} f^i h f^{k-1-i}. \end{align*} Using $h f^m=f^m h-2m f^m$ with $m=k-1-i$, each summand becomes \begin{align*} f^i h f^{k-1-i}=f^i(f^{k-1-i}h-2(k-1-i)f^{k-1-i})=f^{k-1}h-2(k-1-i)f^{k-1}. \end{align*} Therefore \begin{align*} [e,f^k]=k f^{k-1}h-2\left(\sum_{i=0}^{k-1}(k-1-i)\right)f^{k-1}=k f^{k-1}h-k(k-1)f^{k-1}. \end{align*} Since $ev_\lambda=0$, we get \begin{align*} e f^k v_\lambda = [e,f^k]v_\lambda = k\lambda f^{k-1}v_\lambda-k(k-1)f^{k-1}v_\lambda = k(\lambda-k+1)f^{k-1}v_\lambda. \end{align*} Because $f^{k-1}v_\lambda$ is a basis vector, the vector $f^k v_\lambda$ with $k\ge 1$ is killed by $e$ exactly when $k(\lambda-k+1)=0$, equivalently when $k=\lambda+1$. Hence a new highest weight vector appears precisely when $\lambda+1$ is a positive integer. [/example] The formula detects reducibility by locating possible singular vectors. Once such a vector appears, we need to identify the submodule it generates and list the resulting composition factors. [quotetheorem:9400] [citeproof:9400] This theorem is the first complete block calculation in category $\mathcal O$. The condition $\lambda\in\mathbb Z_{\ge 0}$ is exactly the condition that the possible singular vector occurs at an actual positive PBW degree $k=\lambda+1$; for nonintegral $\lambda$ there is no such integer $k$, and for negative integral weights the candidate degree is not positive. The theorem is special to rank one in its explicit two-factor form: in higher rank, several simple reflections can produce several possible singular directions, and the Bruhat order replaces this single string. The two relevant highest weights $n$ and $-n-2$ are related by the dot action of the nonidentity Weyl group element, since for $\mathfrak{sl}_2$ we have $\rho=1$ and \begin{align*} s \cdot n = s(n+1)-1 = -n-2. \end{align*} [example: The Principal Block for $\mathfrak{sl}_2$] For $\mathfrak{sl}_2$, the Weyl group is $W=\{1,s\}$, and with the normalization $\rho=1$ the dot action is \begin{align*} s\cdot \lambda=s(\lambda+1)-1=-(\lambda+1)-1=-\lambda-2. \end{align*} Thus the principal dot orbit, the orbit of $0$, is \begin{align*} 1\cdot 0=0, \end{align*} and \begin{align*} s\cdot 0=-(0+1)-1=-2. \end{align*} So the possible simple highest weights in the principal central-character block are $0$ and $-2$, by *[Harish-Chandra Parametrization of Central Characters](/theorems/9398)*. In $M(0)$, the rank-one formula \begin{align*} e f^k v_0=k(0-k+1)f^{k-1}v_0=k(1-k)f^{k-1}v_0 \end{align*} shows that \begin{align*} e f v_0=1(1-1)v_0=0. \end{align*} Also \begin{align*} h f v_0=(0-2)f v_0=-2f v_0, \end{align*} so $fv_0$ is a highest weight vector of weight $-2$. Hence it generates a Verma submodule isomorphic to $M(-2)$. Since $-2\notin \mathbb Z_{\ge 0}$, the rank-one composition-[factor theorem](/theorems/3235) gives that $M(-2)$ is simple, so $M(-2)=L(-2)$. The submodule generated by $fv_0$ contains $fv_0,f^2v_0,f^3v_0,\dots$, so the quotient $M(0)/M(-2)$ is spanned by the image of $v_0$ alone. On this quotient, \begin{align*} e\overline{v_0}=0,\qquad f\overline{v_0}=0,\qquad h\overline{v_0}=0, \end{align*} because $fv_0$ lies in the submodule and $v_0$ has weight $0$. Therefore $M(0)/M(-2)\cong L(0)$, and $M(0)$ realizes the nontrivial extension with simple submodule $L(-2)$ and simple quotient $L(0)$ inside the principal block. [/example] The picture suggests the general small-rank pattern: Verma modules are arranged along Weyl group orbits, and homomorphisms between them are controlled by reflected weights. In higher rank, several simple reflections can appear and Verma filtrations become partially ordered by the Bruhat order. ## Finite-Dimensional Modules Inside Category O Finite-dimensional irreducible modules were classified earlier by dominant integral highest weights. Category $\mathcal O$ reinterprets them as special quotients of Verma modules sitting in blocks with many infinite-dimensional companions. [quotetheorem:9401] [citeproof:9401] This result places the finite-dimensional theory inside the larger infinite-dimensional category. Finite-dimensionality supplies all three defining hypotheses of $\mathcal O$ at once; the simplicity hypothesis is needed only for the dominant-integral highest weight classification and the uniqueness of the simple quotient. The theorem does not say that the whole block is finite-dimensional, and it does not describe the other infinite-dimensional simples sharing the same central character. The finite-dimensional module is often the visible quotient at the top of a Verma module, while the remaining submodules encode the reflections required by linkage. [example: The Finite-Dimensional Quotient for $\mathfrak{sl}_2$] For $n\in \mathbb Z_{\ge 0}$, let $Q=M(n)/M(-n-2)$ and write $\overline{v_n}$ for the image of $v_n$ in $Q$. The submodule $M(-n-2)$ is generated by the singular vector $f^{n+1}v_n$, so in $Q$ we have \begin{align*} f^{n+1}\overline{v_n}=0. \end{align*} Thus $Q$ is spanned by \begin{align*} \overline{v_n},f\overline{v_n},\dots,f^n\overline{v_n}. \end{align*} These vectors are nonzero and have distinct weights, because \begin{align*} h f^k\overline{v_n}=(n-2k)f^k\overline{v_n} \end{align*} for $0\le k\le n$. Hence they are linearly independent, and they form a basis of $Q$. The actions stay inside this basis. For $0\le k<n$, \begin{align*} f(f^k\overline{v_n})=f^{k+1}\overline{v_n}, \end{align*} while \begin{align*} f(f^n\overline{v_n})=f^{n+1}\overline{v_n}=0. \end{align*} Also $e\overline{v_n}=0$, and the rank-one formula from the preceding computation gives, for $1\le k\le n$, \begin{align*} e f^k\overline{v_n}=k(n-k+1)f^{k-1}\overline{v_n}. \end{align*} It remains to see that $Q$ is simple. Let $0\ne N\subseteq Q$ be a submodule. Since the weights $n,n-2,\dots,-n$ are distinct, $N$ contains a nonzero weight vector, say $f^k\overline{v_n}$ for some $0\le k\le n$. Applying $e$ repeatedly gives \begin{align*} e^k f^k\overline{v_n}=\left(\prod_{j=1}^k j(n-j+1)\right)\overline{v_n}. \end{align*} Each factor $j(n-j+1)$ is nonzero for $1\le j\le k\le n$, so $\overline{v_n}\in N$. Applying $f$ then gives every basis vector $f^r\overline{v_n}$ with $0\le r\le n$, so $N=Q$. Therefore \begin{align*} M(n)/M(-n-2)\cong L(n), \end{align*} the irreducible $(n+1)$-dimensional $\mathfrak{sl}_2$-module. [/example] The organizing lesson is that finite-dimensional representation theory is not separate from category $\mathcal O$; it is the dominant integral part of the highest weight story. Blocks remember the central character, linkage explains which simple modules can occur together, and rank-one calculations show how singular vectors produce the submodule structure that the general theory must manage. # 12. Synthesis and Computational Toolkit This final chapter collects the computational consequences of the highest weight theory developed earlier in the course. The previous chapters built the classification theorem, the Weyl character and dimension formulae, duality, and semisimplicity; here those results are assembled into a practical toolkit for doing examples. The emphasis is not on new structure, but on knowing which theorem answers which concrete representation-theoretic question. ## From Highest Weight Data to Representation-Theoretic Output The basic computational problem is the following. Given a complex semisimple Lie algebra $\mathfrak{g}$, a choice of Cartan subalgebra $\mathfrak{h}$ and positive roots, and a weight $\lambda$, decide whether $\lambda$ defines a finite-dimensional irreducible module and, if it does, compute the useful invariants of $L(\lambda)$. There are two separate checks hidden in this question. For instance, in type $A_2$ the weight $\omega_1-\omega_2$ has integral fundamental-weight coordinates but fails dominance, while $\frac{1}{2}\omega_1$ has nonnegative coefficient but is not integral. The first useful definition packages exactly the condition that survives both tests. [definition: Dominant Integral Input] Let $\mathfrak{g}$ be a finite-dimensional complex semisimple Lie algebra with simple roots $\alpha_1,\dots,\alpha_r$ and fundamental weights $\omega_1,\dots,\omega_r$. A weight \begin{align*} \lambda = a_1\omega_1+\cdots+a_r\omega_r \end{align*} is a dominant integral input if $a_i\in \mathbb{Z}$ and $a_i\ge 0$ for all $1\le i\le r$. [/definition] The definition isolates the input condition that makes the finite-dimensional classification usable. The next theorem is needed because dominance alone is only the first check in a calculation: after passing it, we still need a systematic route to character, dimension, duality, and tensor decompositions. [explanation: Workflow For Finite-Dimensional Irreducibles] Let $\mathfrak{g}$ be a finite-dimensional complex semisimple Lie algebra with chosen positive roots $\Phi^+$, Weyl group $W$, half-sum of positive roots \begin{align*} \rho=\frac{1}{2}\sum_{\alpha\in\Phi^+}\alpha, \end{align*} and longest Weyl group element $w_0$. The following workflow computes the finite-dimensional irreducible representation attached to a weight $\lambda$ whenever it exists. 1. Express $\lambda$ in the fundamental weight basis and test whether $\lambda$ is dominant integral. 2. If $\lambda$ is dominant integral, form the irreducible highest weight module $L(\lambda)$; if not, no finite-dimensional irreducible highest weight module with highest weight $\lambda$ exists for the chosen positive system. 3. Compute the character from the Weyl character formula \begin{align*} \operatorname{ch} L(\lambda) = \frac{\sum_{w\in W} \det(w)e^{w(\lambda+\rho)}}{\sum_{w\in W}\det(w)e^{w\rho}}. \end{align*} 4. Compute the dimension from the Weyl dimension formula, using the invariant bilinear form on the real span of the roots: \begin{align*} \dim L(\lambda) = \prod_{\alpha\in \Phi^+}\frac{(\lambda+\rho,\alpha)}{(\rho,\alpha)}. \end{align*} 5. Compute the dual highest weight by \begin{align*} L(\lambda)^* \cong L(-w_0\lambda). \end{align*} 6. Compute tensor product decompositions by multiplying characters and expanding the product as a nonnegative integer combination of irreducible characters. [/explanation] This is a checklist, not a new classification theorem. Its hypotheses matter at each step: if $\lambda$ is not dominant integral, the Weyl character and dimension formulae do not define a finite-dimensional irreducible module with highest weight $\lambda$; if the positive roots are changed, the dominance chamber and the element $w_0$ used in the duality formula are interpreted relative to that new choice. The point is that the earlier theorems are compatible: the same highest weight controls construction, character, dimension, duality, and tensor product bookkeeping. [example: Full Computation For L Two Omega One Plus Omega Two In Sl Three] Let $\mathfrak{g}=\mathfrak{sl}_3(\mathbb{C})$ and let $\lambda=2\omega_1+\omega_2$. Since the fundamental-weight coordinates of $\lambda$ are $(2,1)$, both coordinates are nonnegative integers, so $\lambda$ is dominant integral and labels a finite-dimensional irreducible module. For type $A_2$, the positive roots are $\alpha_1$, $\alpha_2$, and $\alpha_1+\alpha_2$, and $\rho=\omega_1+\omega_2$. Normalize the invariant form so that $(\alpha_i,\alpha_i)=2$; then $(\omega_i,\alpha_j)=\delta_{ij}$. If $\mu=a\omega_1+b\omega_2$, then \begin{align*} \mu+\rho=(a+1)\omega_1+(b+1)\omega_2. \end{align*} Thus the three numerator factors in the Weyl dimension formula are \begin{align*} (\mu+\rho,\alpha_1)=a+1. \end{align*} \begin{align*} (\mu+\rho,\alpha_2)=b+1. \end{align*} \begin{align*} (\mu+\rho,\alpha_1+\alpha_2)=a+b+2. \end{align*} For $\rho=\omega_1+\omega_2$, the corresponding denominator factors are \begin{align*} (\rho,\alpha_1)=1,\qquad (\rho,\alpha_2)=1,\qquad (\rho,\alpha_1+\alpha_2)=2. \end{align*} Therefore \begin{align*} \dim L(a\omega_1+b\omega_2)=\frac{(a+1)(b+1)(a+b+2)}{2}. \end{align*} Substituting $(a,b)=(2,1)$ gives \begin{align*} \dim L(2\omega_1+\omega_2)=\frac{(2+1)(1+1)(2+1+2)}{2}. \end{align*} \begin{align*} \dim L(2\omega_1+\omega_2)=\frac{3\cdot 2\cdot 5}{2}=15. \end{align*} The Weyl character formula gives \begin{align*} \operatorname{ch}L(2\omega_1+\omega_2)=\frac{\sum_{w\in S_3}\det(w)e^{w(3\omega_1+2\omega_2)}}{\sum_{w\in S_3}\det(w)e^{w(\omega_1+\omega_2)}}. \end{align*} Expanding this type $A_2$ character gives the following weights and multiplicities: \begin{align*} \operatorname{ch}L(2\omega_1+\omega_2)=e^{2\omega_1+\omega_2}+e^{2\omega_2}+2e^{\omega_1}+e^{-2\omega_1+3\omega_2}+2e^{-\omega_1+\omega_2}+2e^{-\omega_2}+e^{3\omega_1-\omega_2}+e^{2\omega_1-2\omega_2}+e^{\omega_1-3\omega_2}+e^{-3\omega_1+2\omega_2}+e^{-2\omega_1}+e^{-\omega_1-2\omega_2}. \end{align*} The sum of the displayed multiplicities is \begin{align*} 1+1+2+1+2+2+1+1+1+1+1+1=15, \end{align*} so the character expansion has the same total dimension as the dimension-formula computation. For the dual, the longest Weyl group element in type $A_2$ satisfies $w_0\omega_1=-\omega_2$ and $w_0\omega_2=-\omega_1$. Hence \begin{align*} -w_0(2\omega_1+\omega_2)=-(2w_0\omega_1+w_0\omega_2). \end{align*} \begin{align*} -w_0(2\omega_1+\omega_2)=-(-2\omega_2-\omega_1)=\omega_1+2\omega_2. \end{align*} Therefore \begin{align*} L(2\omega_1+\omega_2)^*\cong L(\omega_1+2\omega_2). \end{align*} Finally, tensoring the standard representation $L(\omega_1)$ with the adjoint representation $L(\omega_1+\omega_2)$ can be computed in type $A_2$ by the Pieri rule. The highest weight $\omega_1+\omega_2$ corresponds to the partition $(2,1,0)$, and tensoring with $L(\omega_1)$ adds one box. The possible partitions are \begin{align*} (3,1,0),\qquad (2,2,0),\qquad (2,1,1). \end{align*} Converting these back to $\mathfrak{sl}_3$ highest weights gives \begin{align*} (3,1,0)\mapsto 2\omega_1+\omega_2. \end{align*} \begin{align*} (2,2,0)\mapsto 2\omega_2. \end{align*} \begin{align*} (2,1,1)\mapsto \omega_1, \end{align*} where the last partition differs from $(1,0,0)$ by the determinant weight $(1,1,1)$, which is trivial on $\mathfrak{sl}_3$. Thus \begin{align*} L(\omega_1)\otimes L(\omega_1+\omega_2)\cong L(2\omega_1+\omega_2)\oplus L(2\omega_2)\oplus L(\omega_1). \end{align*} The dimensions match because \begin{align*} \dim L(\omega_1)\dim L(\omega_1+\omega_2)=3\cdot 8=24. \end{align*} \begin{align*} \dim L(2\omega_1+\omega_2)+\dim L(2\omega_2)+\dim L(\omega_1)=15+6+3=24. \end{align*} The same module $L(2\omega_1+\omega_2)$ therefore appears consistently through the dominant-weight test, the dimension formula, the character expansion, duality, and tensor-product bookkeeping. [/example] The example shows why the character is more than a formal decoration. It stores the decomposition data needed for tensor products, while the dimension formula gives a rapid consistency check on the final answer. ## Rank Two Tables As Computation Laboratories Rank two Lie algebras are small enough to compute by hand and rich enough to display the main phenomena. The guiding question in this section is how much representation theory can be recovered from a short list of low highest weights. [explanation: Reading Rank Two Tables] For a rank two root system, write a dominant integral weight as $a\omega_1+b\omega_2$ and abbreviate it by $(a,b)$. A dimension table records $\dim L(a,b)$ for low values of $a$ and $b$. A tensor table records decompositions generated by the fundamental representations, and it should always be checked against both highest weights and total dimensions. [/explanation] The type $A_2$ case is the most familiar because it is the representation theory of $\mathfrak{sl}_3(\mathbb{C})$. Its two fundamental representations are dual to each other. [example: Low Weights For Sl Three] Write $L(a,b)$ for $L(a\omega_1+b\omega_2)$. For type $A_2$, the dimension formula from the preceding computation is \begin{align*}\dim L(a,b)=\frac{(a+1)(b+1)(a+b+2)}{2}.\end{align*} Substituting the low weights gives \begin{align*}\dim L(0,0)=\frac{1\cdot 1\cdot 2}{2}=1,\quad \dim L(1,0)=\frac{2\cdot 1\cdot 3}{2}=3,\quad \dim L(0,1)=\frac{1\cdot 2\cdot 3}{2}=3.\end{align*} Also, \begin{align*}\dim L(2,0)=\frac{3\cdot 1\cdot 4}{2}=6,\quad \dim L(1,1)=\frac{2\cdot 2\cdot 4}{2}=8,\quad \dim L(0,2)=\frac{1\cdot 3\cdot 4}{2}=6.\end{align*} The next useful row is \begin{align*}\dim L(2,1)=\frac{3\cdot 2\cdot 5}{2}=15,\quad \dim L(1,2)=\frac{2\cdot 3\cdot 5}{2}=15,\quad \dim L(2,2)=\frac{3\cdot 3\cdot 6}{2}=27.\end{align*} Let $V=L(1,0)$ be the standard $3$-dimensional representation. Then $V^*=L(0,1)$, and \begin{align*}V\otimes V=\operatorname{Sym}^2 V\oplus \bigwedge^2 V.\end{align*} Here $\operatorname{Sym}^2 V$ has highest weight $2\omega_1$, so $\operatorname{Sym}^2 V\cong L(2,0)$, while $\bigwedge^2 V\cong V^*\cong L(0,1)$. Hence \begin{align*}L(1,0)\otimes L(1,0)\cong L(2,0)\oplus L(0,1).\end{align*} The dimension check is \begin{align*}3\cdot 3=9=6+3.\end{align*} Next, \begin{align*}V\otimes V^*\cong \operatorname{End}(V).\end{align*} The trace decomposition is \begin{align*}\operatorname{End}(V)=\mathfrak{sl}_3(\mathbb{C})\oplus \mathbb{C}\operatorname{id}_V.\end{align*} As an $\mathfrak{sl}_3(\mathbb{C})$-module, the adjoint representation has highest weight $\omega_1+\omega_2$, and the scalar endomorphisms are the trivial module. Therefore \begin{align*}L(1,0)\otimes L(0,1)\cong L(1,1)\oplus L(0,0).\end{align*} The dimensions match because \begin{align*}3\cdot 3=9=8+1.\end{align*} For the last product, identify type $A_2$ highest weights with $\operatorname{GL}_3$ partitions modulo adding the determinant weight $(1,1,1)$. The weight $(1,1)$ corresponds to the partition $(2,1,0)$. Tensoring with $L(1,0)$ adds one box, so the possible partitions are \begin{align*}(3,1,0),\quad (2,2,0),\quad (2,1,1).\end{align*} Converting back by taking adjacent differences gives \begin{align*}(3,1,0)\mapsto (3-1)\omega_1+(1-0)\omega_2=2\omega_1+\omega_2.\end{align*} Also, \begin{align*}(2,2,0)\mapsto (2-2)\omega_1+(2-0)\omega_2=2\omega_2.\end{align*} Finally, subtracting the determinant weight gives $(2,1,1)-(1,1,1)=(1,0,0)$, and \begin{align*}(1,0,0)\mapsto (1-0)\omega_1+(0-0)\omega_2=\omega_1.\end{align*} Thus \begin{align*}L(1,0)\otimes L(1,1)\cong L(2,1)\oplus L(0,2)\oplus L(1,0).\end{align*} The dimension check is \begin{align*}3\cdot 8=24=15+6+3.\end{align*} These three decompositions show the standard symmetric/alternating split, the decomposition of endomorphisms into traceless and scalar parts, and the tensor-product calculation used for $L(2\omega_1+\omega_2)$ above. [/example] The type $B_2$ case, realised by $\mathfrak{so}_5(\mathbb{C})$, introduces a spin representation alongside the vector representation. This makes it a useful test case for not assuming that every fundamental representation is obtained from exterior powers of the same standard module. [example: Low Weights For So Five] For $\mathfrak{so}_5(\mathbb{C})$, take simple roots $\alpha_1=e_1-e_2$ and $\alpha_2=e_2$, so $\omega_1=e_1$ is the vector highest weight and $\omega_2=\frac{1}{2}(e_1+e_2)$ is the spin highest weight. The positive roots are \begin{align*} \alpha_1=e_1-e_2,\quad \alpha_2=e_2,\quad \alpha_1+\alpha_2=e_1,\quad \alpha_1+2\alpha_2=e_1+e_2. \end{align*} Their half-sum is \begin{align*} \rho=\frac{1}{2}\bigl((e_1-e_2)+e_2+e_1+(e_1+e_2)\bigr)=\frac{3}{2}e_1+\frac{1}{2}e_2=\omega_1+\omega_2. \end{align*} For $\lambda=a\omega_1+b\omega_2$, we have \begin{align*} \lambda+\rho=\left(a+\frac{b}{2}+\frac{3}{2}\right)e_1+\left(\frac{b}{2}+\frac{1}{2}\right)e_2. \end{align*} The four Weyl dimension factors are \begin{align*} \frac{(\lambda+\rho,\alpha_1)}{(\rho,\alpha_1)}=\frac{\left(a+\frac{b}{2}+\frac{3}{2}\right)-\left(\frac{b}{2}+\frac{1}{2}\right)}{\frac{3}{2}-\frac{1}{2}}=a+1. \end{align*} Also, \begin{align*} \frac{(\lambda+\rho,\alpha_2)}{(\rho,\alpha_2)}=\frac{\frac{b}{2}+\frac{1}{2}}{\frac{1}{2}}=b+1. \end{align*} For the root $e_1$, \begin{align*} \frac{(\lambda+\rho,\alpha_1+\alpha_2)}{(\rho,\alpha_1+\alpha_2)}=\frac{a+\frac{b}{2}+\frac{3}{2}}{\frac{3}{2}}=\frac{2a+b+3}{3}. \end{align*} For the root $e_1+e_2$, \begin{align*} \frac{(\lambda+\rho,\alpha_1+2\alpha_2)}{(\rho,\alpha_1+2\alpha_2)}=\frac{\left(a+\frac{b}{2}+\frac{3}{2}\right)+\left(\frac{b}{2}+\frac{1}{2}\right)}{\frac{3}{2}+\frac{1}{2}}=\frac{a+b+2}{2}. \end{align*} Therefore the Weyl dimension formula gives \begin{align*} \dim L(a,b)=\frac{(a+1)(b+1)(2a+b+3)(a+b+2)}{6}. \end{align*} Substituting the first few dominant weights gives \begin{align*} \dim L(0,0)=\frac{1\cdot 1\cdot 3\cdot 2}{6}=1,\quad \dim L(1,0)=\frac{2\cdot 1\cdot 5\cdot 3}{6}=5,\quad \dim L(0,1)=\frac{1\cdot 2\cdot 4\cdot 3}{6}=4. \end{align*} The next dimensions are \begin{align*} \dim L(2,0)=\frac{3\cdot 1\cdot 7\cdot 4}{6}=14,\quad \dim L(1,1)=\frac{2\cdot 2\cdot 6\cdot 4}{6}=16,\quad \dim L(0,2)=\frac{1\cdot 3\cdot 5\cdot 4}{6}=10. \end{align*} Let $V=L(1,0)$ be the $5$-dimensional vector representation. Its tensor square splits into symmetric and alternating tensors: \begin{align*} V\otimes V=\operatorname{Sym}^2 V\oplus \bigwedge^2 V. \end{align*} The invariant quadratic form gives one one-dimensional zero-action summand inside $\operatorname{Sym}^2 V$, and the traceless symmetric tensors have highest weight $2\omega_1$, so \begin{align*} \operatorname{Sym}^2 V\cong L(2,0)\oplus L(0,0). \end{align*} The exterior square is the adjoint representation of $\mathfrak{so}_5(\mathbb{C})$, whose highest weight is $2\omega_2$, so \begin{align*} \bigwedge^2 V\cong L(0,2). \end{align*} Hence \begin{align*} L(1,0)\otimes L(1,0)\cong L(2,0)\oplus L(0,2)\oplus L(0,0). \end{align*} The dimensions match: \begin{align*} 5\cdot 5=25=14+10+1. \end{align*} Let $S=L(0,1)$ be the $4$-dimensional spin representation. The standard spin tensor identity in type $B_2$ gives \begin{align*} S\otimes S\cong \bigwedge^0 V\oplus \bigwedge^1 V\oplus \bigwedge^2 V. \end{align*} Using $\bigwedge^0 V\cong L(0,0)$, $\bigwedge^1 V\cong L(1,0)$, and $\bigwedge^2 V\cong L(0,2)$, this becomes \begin{align*} L(0,1)\otimes L(0,1)\cong L(1,0)\oplus L(0,2)\oplus L(0,0). \end{align*} The numerical check is \begin{align*} 4\cdot 4=16=5+10+1. \end{align*} Finally, tensoring the vector representation with the spin representation produces two dominant highest weights: the Cartan product has highest weight $\omega_1+\omega_2$, and Clifford contraction leaves a copy of the spin representation. Thus \begin{align*} L(1,0)\otimes L(0,1)\cong L(1,1)\oplus L(0,1). \end{align*} The dimensions are \begin{align*} 5\cdot 4=20=16+4. \end{align*} These computations show that in type $B_2$ the vector and spin fundamental representations play different roles: the vector square sees symmetric tensors, exterior tensors, and the quadratic form, while the spin square recovers exterior powers of the vector representation. [/example] The exceptional type $G_2$ is the smallest place where the root lengths and Weyl group geometry have a strong visible effect on the tables. It is also a good reminder that the notation $(a,b)$ is meaningful only after the convention for which simple root is short has been fixed. [example: Low Weights For G Two] For type $G_2$, use the convention in which $\omega_1$ is the highest weight of the $7$-dimensional fundamental representation and $\omega_2$ is the highest weight of the adjoint representation. With $\alpha_1$ short and $\alpha_2$ long, the positive roots are \begin{align*} \alpha_1,\quad \alpha_2,\quad \alpha_1+\alpha_2,\quad 2\alpha_1+\alpha_2,\quad 3\alpha_1+\alpha_2,\quad 3\alpha_1+2\alpha_2. \end{align*} The corresponding positive coroots, written in the simple coroot basis, are \begin{align*} \alpha_1^\vee,\quad \alpha_2^\vee,\quad \alpha_1^\vee+3\alpha_2^\vee,\quad 2\alpha_1^\vee+3\alpha_2^\vee,\quad \alpha_1^\vee+\alpha_2^\vee,\quad \alpha_1^\vee+2\alpha_2^\vee. \end{align*} For $\lambda=a\omega_1+b\omega_2$, we have $\lambda+\rho=(a+1)\omega_1+(b+1)\omega_2$, so pairing with these six positive coroots gives the numerator factors \begin{align*} a+1,\quad b+1,\quad a+3b+4,\quad 2a+3b+5,\quad a+b+2,\quad a+2b+3. \end{align*} For $\rho=\omega_1+\omega_2$, the corresponding denominator factors are \begin{align*} 1,\quad 1,\quad 4,\quad 5,\quad 2,\quad 3. \end{align*} Thus the Weyl dimension formula gives \begin{align*} \dim L(a,b)=\frac{(a+1)(b+1)(a+b+2)(a+2b+3)(a+3b+4)(2a+3b+5)}{120}. \end{align*} Substituting the lowest dominant weights gives \begin{align*} \dim L(0,0)=\frac{1\cdot 1\cdot 2\cdot 3\cdot 4\cdot 5}{120}=1. \end{align*} Also, \begin{align*} \dim L(1,0)=\frac{2\cdot 1\cdot 3\cdot 4\cdot 5\cdot 7}{120}=7. \end{align*} And \begin{align*} \dim L(0,1)=\frac{1\cdot 2\cdot 3\cdot 5\cdot 7\cdot 8}{120}=14. \end{align*} The next dimensions are \begin{align*} \dim L(2,0)=\frac{3\cdot 1\cdot 4\cdot 5\cdot 6\cdot 9}{120}=27. \end{align*} Similarly, \begin{align*} \dim L(1,1)=\frac{2\cdot 2\cdot 4\cdot 6\cdot 8\cdot 10}{120}=64. \end{align*} Finally, \begin{align*} \dim L(0,2)=\frac{1\cdot 3\cdot 4\cdot 7\cdot 10\cdot 11}{120}=77. \end{align*} Let $V=L(1,0)$ and $A=L(0,1)$. The $G_2$-invariant bilinear form on $V$ gives a one-dimensional zero-action summand in $\operatorname{Sym}^2 V$, and the traceless symmetric tensors have highest weight $2\omega_1$, so \begin{align*} \operatorname{Sym}^2 V\cong L(2,0)\oplus L(0,0). \end{align*} The alternating square splits into the adjoint representation and a copy of $V$: \begin{align*} \bigwedge^2 V\cong L(0,1)\oplus L(1,0). \end{align*} Since $V\otimes V=\operatorname{Sym}^2 V\oplus \bigwedge^2 V$, this gives \begin{align*} L(1,0)\otimes L(1,0)\cong L(2,0)\oplus L(0,1)\oplus L(1,0)\oplus L(0,0). \end{align*} The dimensions check term by term: \begin{align*} 7\cdot 7=49=27+14+7+1. \end{align*} Multiplying the formal characters of $L(1,0)$ and $L(0,1)$ and then subtracting irreducible characters by highest dominant weight gives the dominant summands $(1,1)$, $(2,0)$, and $(1,0)$. Hence \begin{align*} L(1,0)\otimes L(0,1)\cong L(1,1)\oplus L(2,0)\oplus L(1,0). \end{align*} The dimension check is \begin{align*} 7\cdot 14=98=64+27+7. \end{align*} For the adjoint square, the same character-subtraction procedure gives the dominant summands $(0,2)$, $(3,0)$, $(2,0)$, $(0,1)$, and $(0,0)$. The remaining dimension needed for $L(3,0)$ is \begin{align*} \dim L(3,0)=\frac{4\cdot 1\cdot 5\cdot 6\cdot 7\cdot 11}{120}=77. \end{align*} Therefore \begin{align*} L(0,1)\otimes L(0,1)\cong L(0,2)\oplus L(3,0)\oplus L(2,0)\oplus L(0,1)\oplus L(0,0). \end{align*} Its total dimension is \begin{align*} 14\cdot 14=196=77+77+27+14+1. \end{align*} These computations show how the $7$-dimensional fundamental representation and the $14$-dimensional adjoint representation generate the first useful part of the type $G_2$ tensor table. [/example] These tables are not substitutes for the Weyl character formula. Their purpose is to make repeated calculations faster and to give a way of detecting mistakes before a long expansion is trusted. ## Common Pitfalls In Highest Weight Computations Most wrong answers in this part of the course arise from applying a valid theorem outside its hypotheses. The next problem is diagnostic: when a calculation produces a plausible-looking module, which hypothesis should be checked first? [remark: Non-Dominant Highest Weights] A weight such as $\omega_1-\omega_2$ for $\mathfrak{sl}_3(\mathbb{C})$ is integral but not dominant. It can still be used as a highest weight for a Verma module, but it does not label a finite-dimensional irreducible module for the chosen positive roots. Changing the positive root system changes the dominance chamber, so the phrase "dominant" is always relative to the chosen simple roots. [/remark] The dominance condition is separate from integrality. A weight may lie in the weight lattice but fail the inequalities required for finite-dimensional highest weight theory. [example: Integral Does Not Mean Dominant] In type $A_2$, write weights in the fundamental-weight basis. Let \begin{align*} \lambda=2\omega_1-\omega_2. \end{align*} The coefficients of $\lambda$ are $2$ and $-1$, so both coefficients are integers and $\lambda$ lies in the weight lattice \begin{align*} \mathbb{Z}\omega_1+\mathbb{Z}\omega_2. \end{align*} However, the dominant integral condition requires every fundamental-weight coefficient to be a nonnegative integer. Since \begin{align*} -1<0, \end{align*} the weight $2\omega_1-\omega_2$ is integral but not dominant. The Verma module $M(\lambda)$ is still defined for this weight: by definition, \begin{align*} M(\lambda)=U(\mathfrak{g})\otimes_{U(\mathfrak{b})}\mathbb{C}_\lambda, \end{align*} and the vector $1\otimes 1$ has weight $\lambda$ and is killed by $\mathfrak{n}^+$. Thus $M(2\omega_1-\omega_2)$ has a highest weight vector of weight $2\omega_1-\omega_2$. What fails is the finite-dimensional classification input: because the second coordinate is negative, this $\lambda$ is not a dominant integral input, so it does not label a finite-dimensional irreducible highest weight module for the chosen positive roots. For example, reflecting may move the weight to a different chamber, but it does not change the fact that the original coordinates were \begin{align*} (2,-1). \end{align*} Dominance is tested relative to the fixed simple roots and fixed positive chamber, so Weyl-group motion changes the chamber in which the weight is viewed rather than repairing the dominance inequality in the original chamber. [/example] The example separates the weight-lattice condition from the dominance condition. Non-dominant weights are excluded from the finite-dimensional classification, but they are not meaningless. To study arbitrary highest weights, we need a construction that does not assume dominance or finite-dimensionality at the outset. The triangular decomposition provides one: prescribe the highest weight on the Borel subalgebra, let $\mathfrak n^+$ act by zero, and induce up to $\mathfrak g$. The resulting object is the universal highest weight module defined next. [definition: Verma Module] Let $\mathfrak{g}$ be a complex semisimple Lie algebra with triangular decomposition \begin{align*} \mathfrak{g}=\mathfrak{n}^-\oplus\mathfrak{h}\oplus\mathfrak{n}^+. \end{align*} For $\lambda\in\mathfrak{h}^*$, let $\mathbb{C}_\lambda$ be the one-dimensional $\mathfrak{b}=\mathfrak{h}\oplus\mathfrak{n}^+$-module on which $h\in\mathfrak{h}$ acts by $\lambda(h)$ and $\mathfrak{n}^+$ acts by $0$. The Verma module of highest weight $\lambda$ is \begin{align*} M(\lambda)=U(\mathfrak{g})\otimes_{U(\mathfrak{b})}\mathbb{C}_\lambda. \end{align*} [/definition] The definition gives a universal highest weight module before finite-dimensional restrictions are imposed. Because $M(\lambda)$ is usually too large and often reducible, the remaining problem is to extract the simple highest weight object it controls. The connection with the irreducible modules used in classification comes from passing to the unique simple quotient rather than from imposing dominance at the start. [quotetheorem:9402] [citeproof:9402] The theorem explains why $L(\lambda)$ exists for any highest weight in the category of highest weight modules, but finite dimensionality still requires dominance and integrality. The quotient operation removes submodules; it does not by itself impose the finite-dimensional classification. For example, if $\lambda=2\omega_1-\omega_2$ in type $A_2$, the simple quotient $L(\lambda)$ exists as a highest weight module for the chosen triangular decomposition, but it is not one of the finite-dimensional irreducibles classified by dominant integral weights. The construction also depends on the chosen Borel subalgebra: changing the triangular decomposition changes which vector is regarded as highest. [remark: Root Lattice Versus Weight Lattice] The root lattice is generated by the roots, while the weight lattice is generated by the fundamental weights. Characters of finite-dimensional representations use weights, and the highest weights in the classification lie in the dominant part of the weight lattice. In non-simply connected group-level questions the distinction becomes sharper, but even for Lie algebras it matters for coordinates and dominance tests. [/remark] A final source of mistakes is confusing the formal character ring with a [polynomial ring](/page/Polynomial%20Ring) in ordinary variables. The symbols $e^\mu$ are basis elements of a group algebra indexed by weights, not exponentials of numbers. [example: Formal Exponentials In Character Multiplication] Let $V=\bigoplus_\mu V_\mu$ and $W=\bigoplus_\nu W_\nu$, with $\dim V_\mu=m_\mu$ and $\dim W_\nu=n_\nu$. In the tensor product, a vector $v\otimes w$ with $v\in V_\mu$ and $w\in W_\nu$ has weight $\mu+\nu$, because for $h\in\mathfrak{h}$, \begin{align*} h\cdot(v\otimes w)=(h\cdot v)\otimes w+v\otimes(h\cdot w)=\mu(h)v\otimes w+\nu(h)v\otimes w=(\mu+\nu)(h)(v\otimes w). \end{align*} Thus the $\gamma$-weight space of $V\otimes W$ is obtained by collecting exactly those pairs of weights whose sum is $\gamma$: \begin{align*} (V\otimes W)_\gamma=\bigoplus_{\mu+\nu=\gamma} V_\mu\otimes W_\nu. \end{align*} Taking dimensions gives \begin{align*} \dim (V\otimes W)_\gamma=\sum_{\mu+\nu=\gamma}\dim(V_\mu\otimes W_\nu). \end{align*} Since $\dim(V_\mu\otimes W_\nu)=\dim V_\mu\dim W_\nu=m_\mu n_\nu$, this becomes \begin{align*} \dim (V\otimes W)_\gamma=\sum_{\mu+\nu=\gamma}m_\mu n_\nu. \end{align*} Therefore \begin{align*} \operatorname{ch}(V\otimes W)=\sum_\gamma \dim (V\otimes W)_\gamma e^\gamma. \end{align*} Substituting the preceding formula for $\dim (V\otimes W)_\gamma$ gives \begin{align*} \operatorname{ch}(V\otimes W)=\sum_\gamma\left(\sum_{\mu+\nu=\gamma}m_\mu n_\nu\right)e^\gamma. \end{align*} This is the same rule as multiplying the formal sums \begin{align*} \left(\sum_\mu m_\mu e^\mu\right)\left(\sum_\nu n_\nu e^\nu\right)=\sum_{\mu,\nu}m_\mu n_\nu e^{\mu+\nu}. \end{align*} Grouping the terms with the same exponent $\gamma=\mu+\nu$ gives \begin{align*} \sum_{\mu,\nu}m_\mu n_\nu e^{\mu+\nu}=\sum_\gamma\left(\sum_{\mu+\nu=\gamma}m_\mu n_\nu\right)e^\gamma. \end{align*} Thus formal exponentials multiply by adding weights, not by evaluating numerical exponentials. Once this product character is known, tensor product decomposition is a separate step: one expands the resulting character as a nonnegative integer sum of irreducible characters, usually by removing the irreducible character with the largest remaining dominant highest weight and repeating until no terms remain. [/example] ## Summary Of Classification And Computation Methods The course ends with a consolidation question: which theorem should be used for which kind of representation-theoretic task? The answer is that the methods form a chain, and each link has a different role. The point of the final classification theorem is to collect these roles into one usable test: dominant integral highest weights label the finite-dimensional simple modules, Weyl characters compute them, and character arithmetic converts tensor product questions into a finite decomposition problem. [quotetheorem:9403] [citeproof:9403] This summary is the conceptual endpoint of the course. The universal enveloping algebra provides the construction language, highest weights provide the classification labels, Weyl symmetry provides the character formula, and semisimplicity turns character arithmetic into tensor product decomposition. The theorem is not a classification of all highest weight modules: non-dominant weights still give Verma modules and simple quotients, but these modules are generally infinite-dimensional. It is also not a statement about arbitrary Lie algebras; for the one-dimensional abelian Lie algebra $\mathbb{C}x$, every scalar $\lambda\in\mathbb{C}$ gives a one-dimensional module on which $x$ acts by $\lambda$, so there is no finite root system or dominant integral cone controlling the classification. Finite dimensionality is also essential for the character toolkit: infinite-dimensional Verma modules have formal characters with infinite support, and their tensor products are not governed by a finite irreducible character basis in the way used above. [explanation: Practical Checklist] For a concrete calculation, first fix the root datum and the convention for the simple roots. Next express the proposed highest weight in the fundamental weight basis and test dominance. Then compute the Weyl character or, for rank two and other familiar cases, use a trusted table checked against the dimension formula. For duals, apply $-w_0$ to the highest weight. For tensor products, multiply characters, identify the largest remaining dominant weight, subtract its irreducible character, and repeat until no terms remain. [/explanation] The techniques in this chapter are intentionally redundant: dimensions check characters, characters check tensor decompositions, and highest weight constraints check whether a proposed summand can occur. This redundancy is what makes the toolkit reliable in examples beyond the first few tables. ## Beyond This Course The highest-weight classification developed here is the finite-dimensional starting point for several larger parts of representation theory. Verma modules and category $\mathcal O$ keep the same highest-weight language but drop finite dimensionality, so the role of singular vectors and composition factors becomes central. The Borel-Weil and Borel-Weil-Bott theorems connect dominant weights to line bundles on flag varieties, turning the algebraic classification into geometric representation theory. For compact Lie groups, the same weights describe irreducible representations through their complexified Lie algebras, linking this course to harmonic analysis and character theory. Tensor product questions continue into crystals, canonical bases, and Littlewood-Richardson theory, where combinatorial models make character multiplication more explicit. Within Androma, natural continuations include [Lie Algebras I: Foundations](/page/Lie%20Algebras%20I%3A%20Foundations), [Lie Algebras II: Structure and Classification](/page/Lie%20Algebras%20II%3A%20Structure%20and%20Classification), [Algebraic Combinatorics I: Symmetric Functions](/page/Algebraic%20Combinatorics%20I%3A%20Symmetric%20Functions), and [Algebraic Combinatorics III: Combinatorial Representation Theory](/page/Algebraic%20Combinatorics%20III%3A%20Combinatorial%20Representation%20Theory). Those notes reuse the same structural pattern seen here: algebraic structure produces weights or combinatorial labels, those labels organize modules or symmetric functions, and further geometry or combinatorics explains how objects decompose. ## References - [Lie Algebras I: Foundations](/page/Lie%20Algebras%20I%3A%20Foundations) - [Lie Algebras II: Structure and Classification](/page/Lie%20Algebras%20II%3A%20Structure%20and%20Classification) - [Algebraic Combinatorics I: Symmetric Functions](/page/Algebraic%20Combinatorics%20I%3A%20Symmetric%20Functions) - [Algebraic Combinatorics III: Combinatorial Representation Theory](/page/Algebraic%20Combinatorics%20III%3A%20Combinatorial%20Representation%20Theory)