This course introduces the basic structure theory of Lie algebras, which are algebraic objects designed to encode infinitesimal symmetry. The opening chapters develop the definition and first examples, then move to the standard constructions that organize the subject: ideals, quotients, subalgebras, and representations. From there, the course shifts from examples and formalism to the behavior of Lie algebras acting on vector spaces, where module-theoretic ideas become a central tool. A major theme is the distinction between solvable, nilpotent, and semisimple Lie algebras, and the way these classes are detected by structural criteria. The middle chapters build the machinery needed for this classification, culminating in Engel’s theorem, Lie’s theorem, and triangularization results. The final chapters study the Killing form and Cartan’s criterion, then use them to develop the theory of semisimple Lie algebras and prove Weyl’s theorem on complete reducibility. Each chapter adds a layer of structure that the next chapter uses, so the course moves steadily from foundational definitions to the core structural theorems of the subject. # Introduction This opening chapter sets the scope and conventions for **Lie Algebras I: Foundations**. The course studies finite-dimensional Lie algebras over a field, beginning with the bracket operation and moving toward the structural tests that separate solvable, nilpotent, and semisimple behaviour. The guiding theme is that many internal properties of a Lie algebra are best detected through its representations, especially the adjoint representation. The notes assume linear algebra, abstract algebra, basic module theory, elementary field theory, and familiarity with [Jordan normal form](/page/Jordan%20Normal%20Form). Later chapters use these prerequisites without reintroducing them, but this chapter fixes the course vocabulary and explains why the first structural theorems take the form they do. ## What Is the Course Trying to Classify? The central problem is to understand when two non-abelian algebraic systems built from a bilinear bracket should be regarded as structurally similar. Unlike groups or associative algebras, Lie algebras are governed by infinitesimal commutation: the bracket measures failure of two infinitesimal operations to commute. This makes Lie algebras both algebraic objects in their own right and linear shadows of Lie groups. The course does not attempt the full classification of semisimple Lie algebras. Instead, it builds the foundational tools needed before classification can begin: ideals, quotient Lie algebras, solvability, nilpotence, representations, trace criteria, and complete reducibility. [definition: Lie Algebra Course Object] A Lie algebra in this course is a finite-dimensional [vector space](/page/Vector%20Space) $\mathfrak g$ over a field $F$ equipped with a bilinear map \begin{align*} [-,-] : \mathfrak g \times \mathfrak g \to \mathfrak g \end{align*} satisfying skew-symmetry and the Jacobi identity. [/definition] The definition is deliberately spare. Most of the course asks how much structure is forced by these two identities once finite-dimensionality is imposed. [example: Commutators As The First Source] Let $A$ be an associative algebra over $F$, and define \begin{align*} [x,y]=xy-yx \end{align*} for $x,y\in A$. We show that this bracket is bilinear, skew-symmetric, and satisfies the Jacobi identity. For $a,b\in F$ and $x_1,x_2,y\in A$, distributivity and compatibility of scalar multiplication with multiplication in the associative algebra give \begin{align*} [ax_1+bx_2,y] &=(ax_1+bx_2)y-y(ax_1+bx_2)\\ &=a x_1y+b x_2y-a yx_1-b yx_2\\ &=a(x_1y-yx_1)+b(x_2y-yx_2)\\ &=a[x_1,y]+b[x_2,y]. \end{align*} The same calculation in the second variable gives \begin{align*} [x,ay_1+by_2] &=x(ay_1+by_2)-(ay_1+by_2)x\\ &=a xy_1+b xy_2-a y_1x-b y_2x\\ &=a[x,y_1]+b[x,y_2], \end{align*} so the bracket is bilinear. Skew-symmetry follows from \begin{align*} [y,x]=yx-xy=-(xy-yx)=-[x,y]. \end{align*} It remains to verify the Jacobi identity. Using associativity to remove parentheses in products of three elements, we compute \begin{align*} [x,[y,z]] &=x(yz-zy)-(yz-zy)x\\ &=xyz-xzy-yzx+zyx,\\ [y,[z,x]] &=y(zx-xz)-(zx-xz)y\\ &=yzx-yxz-zxy+xzy,\\ [z,[x,y]] &=z(xy-yx)-(xy-yx)z\\ &=zxy-zyx-xyz+yxz. \end{align*} Adding these three displayed identities term by term gives \begin{align*} [x,[y,z]]+[y,[z,x]]+[z,[x,y]] &=(xyz-xzy-yzx+zyx)\\ &\quad +(yzx-yxz-zxy+xzy)\\ &\quad +(zxy-zyx-xyz+yxz)\\ &=0, \end{align*} because $xyz$ cancels with $-xyz$, $xzy$ with $-xzy$, $yzx$ with $-yzx$, $zyx$ with $-zyx$, $yxz$ with $-yxz$, and $zxy$ with $-zxy$. Thus the commutator bracket makes every associative algebra into a Lie algebra. [/example] This example explains why matrix Lie algebras dominate the first lectures. The operation $XY-YX$ is concrete enough for computation, but it already contains the identities that govern abstract Lie algebras. ## Why Ideals and Quotients Come First The first structural question is how to break a Lie algebra into smaller pieces without destroying the bracket. Subalgebras are closed under the bracket, but ideals are the correct analogue of normal subgroups because they are exactly the subspaces that support quotient Lie algebras. [definition: Ideal] Let $\mathfrak g$ be a Lie algebra over $F$. An ideal of $\mathfrak g$ is a vector subspace $\mathfrak i\subset\mathfrak g$ such that \begin{align*} [x,y]\in\mathfrak i \end{align*} for all $x\in\mathfrak g$ and $y\in\mathfrak i$. [/definition] Ideals measure internal compatibility with all bracket operations from $\mathfrak g$. They make it possible to form quotients, define simple Lie algebras, and build derived and central series. The immediate test of this definition is whether it is exactly the condition needed for kernels and quotient constructions. If ideals are the correct analogue of normal subgroups, they should appear precisely as kernels of homomorphisms and precisely as the subspaces by which one can divide a Lie algebra. [quotetheorem:3751] [citeproof:3751] This is the first indication that the usual algebraic tools still work in the Lie setting, provided multiplication is replaced by the bracket. The result also explains why ideals appear before more refined structure theory: every reduction to a quotient, every discussion of simplicity, and every passage to a smaller structural object depends on this compatibility condition rather than on internal closure alone. The point is not merely that ideals are useful examples of subspaces, but that they are the exact obstruction to transferring the bracket to a quotient. If a subspace is only closed under brackets among its own elements, representatives of the same coset may produce different bracket cosets; the extra ambient stability in the definition of an ideal is what removes this ambiguity. ## How Solvability and Nilpotence Organise Complexity The next problem is to distinguish Lie algebras whose repeated commutators eventually disappear from those with persistent non-commuting structure. Two filtrations encode this: the derived series and the lower central series. They are similar in appearance but detect different degrees of non-commutativity. [definition: Derived Series] Let $\mathfrak g$ be a Lie algebra over $F$. The derived series of $\mathfrak g$ is the sequence of ideals defined by \begin{align*} \mathfrak g^{(0)} &= \mathfrak g, & \mathfrak g^{(r+1)} &= [\mathfrak g^{(r)},\mathfrak g^{(r)}] \end{align*} for $r\ge 0$. [/definition] A Lie algebra is called solvable when this series reaches $0$. Solvability records the eventual disappearance of commutators after repeatedly replacing the algebra by its derived algebra. There is a stricter way to measure how quickly commutators disappear: instead of bracketing the previous term only with itself, bracket it with the whole original algebra at every stage. This produces the lower central series, which is designed to detect nilpotence rather than mere solvability. [definition: Lower Central Series] Let $\mathfrak g$ be a Lie algebra over $F$. The lower central series of $\mathfrak g$ is the sequence of ideals defined by \begin{align*} \mathfrak g_1 &= \mathfrak g, & \mathfrak g_{r+1} &= [\mathfrak g,\mathfrak g_r] \end{align*} for $r\ge 1$. [/definition] A Lie algebra is called nilpotent when its lower central series reaches $0$. Nilpotence is stronger than solvability because each step allows commutators with the whole algebra, not only with the previous term itself. [example: The Heisenberg Algebra] Let $\mathfrak h$ have basis $x,y,z$ over $F$, with $[x,y]=z$ and with every bracket involving $z$ equal to $0$. Since the bracket is skew-symmetric, $[y,x]=-[x,y]=-z$, and $[x,x]=[y,y]=[z,z]=0$. For arbitrary elements $u=ax+by+cz$ and $v=dx+ey+fz$, bilinearity gives \begin{align*} [u,v] &=[ax+by+cz,dx+ey+fz]\\ &=ad[x,x]+ae[x,y]+af[x,z]\\ &\quad+bd[y,x]+be[y,y]+bf[y,z]\\ &\quad+cd[z,x]+ce[z,y]+cf[z,z]\\ &=0+ae z+0+bd(-z)+0+0+0+0+0\\ &=(ae-bd)z. \end{align*} Thus every commutator lies in $Fz$, and since $[x,y]=z$, we have \begin{align*} [\mathfrak h,\mathfrak h]=Fz. \end{align*} The lower central series therefore begins with \begin{align*} \mathfrak h_1&=\mathfrak h,\\ \mathfrak h_2&=[\mathfrak h,\mathfrak h]=Fz. \end{align*} The next term is \begin{align*} \mathfrak h_3 &=[\mathfrak h,\mathfrak h_2]\\ &=[\mathfrak h,Fz]. \end{align*} If $u=ax+by+cz$ and $\lambda z\in Fz$, then \begin{align*} [u,\lambda z] &=[ax+by+cz,\lambda z]\\ &=a\lambda[x,z]+b\lambda[y,z]+c\lambda[z,z]\\ &=0+0+0\\ &=0. \end{align*} Hence $\mathfrak h_3=0$, while $\mathfrak h_2=Fz\ne 0$, so $\mathfrak h$ is nilpotent of class $2$. The derived series gives the corresponding solvability calculation: \begin{align*} \mathfrak h^{(0)}&=\mathfrak h,\\ \mathfrak h^{(1)}&=[\mathfrak h,\mathfrak h]=Fz,\\ \mathfrak h^{(2)}&=[Fz,Fz]. \end{align*} For $\lambda z,\mu z\in Fz$, \begin{align*} [\lambda z,\mu z]=\lambda\mu[z,z]=0, \end{align*} so $\mathfrak h^{(2)}=0$. Thus the Heisenberg algebra is non-abelian, because $[x,y]=z\ne 0$, but its non-commutativity is entirely concentrated in the central line $Fz$. [/example] The Heisenberg algebra is the model example for the early part of the course: it is non-abelian, small enough to compute by hand, and governed by a central commutator. ## Why Representations Enter Structure Theory The next question is why linear actions reveal internal structure. A representation turns the bracket in $\mathfrak g$ into commutators of linear maps, allowing [Jordan normal form](/theorems/864), trace, eigenvalues, and invariant subspaces to enter the theory. Here $\mathfrak{gl}(V)$ denotes the Lie algebra of endomorphisms of $V$, with bracket $[A,B]=AB-BA$. Thus a representation is not just a linear action; it is a way of preserving the Lie bracket inside an algebra of linear operators. [definition: Representation] Let $\mathfrak g$ be a Lie algebra over $F$. A representation of $\mathfrak g$ on a vector space $V$ is a Lie algebra homomorphism \begin{align*} \rho : \mathfrak g \to \mathfrak{gl}(V). \end{align*} [/definition] A general representation may depend on an external choice of vector space, so the next task is to construct one representation attached to every Lie algebra without extra data. The bracket gives a natural candidate: let each element act on the underlying vector space of the algebra by bracketing with everything else. The formal definition isolates this canonical action and makes precise why the internal multiplication of $\mathfrak g$ can be studied through ordinary linear operators. [definition: Adjoint Representation] Let $\mathfrak g$ be a Lie algebra over $F$. The adjoint representation is the map \begin{align*} \operatorname{ad}:\mathfrak g\to\mathfrak{gl}(\mathfrak g), \qquad \operatorname{ad}_x(y)=[x,y]. \end{align*} [/definition] The Jacobi identity is exactly what makes $\operatorname{ad}$ a homomorphism of Lie algebras. This observation is a turning point: internal bracket identities become statements about linear operators. The first structural feature visible through the adjoint representation is its kernel. A representation can fail to see part of a Lie algebra, and for the adjoint representation this failure has a concrete source: an element disappears exactly when its bracket with every element is zero. Thus the problem is to identify the invisible directions of the canonical action, not just to name another invariant. [quotetheorem:3752] [citeproof:3752] This theorem turns the centre into the first invariant detected by a representation. It shows that the adjoint representation is faithful exactly when the centre is zero, so central elements are precisely the directions invisible to internal commutator action. Later chapters use the same philosophy with traces of products of adjoint maps, leading to the Killing form and Cartan's criterion. ## The Main Theorems Ahead The course builds toward four structural results. [Engel's theorem](/theorems/3798) says that a finite-dimensional Lie algebra whose adjoint operators are nilpotent is itself nilpotent. The common-eigenline form of [Lie's theorem](/theorems/3802), together with its triangularization form, describes representations of solvable Lie algebras over algebraically closed fields of characteristic $0$. Cartan's criterion detects solvability and semisimplicity using traces. Weyl's theorem gives complete reducibility for finite-dimensional representations of semisimple Lie algebras in characteristic $0$. Engel's theorem is a major milestone of the course and will be developed later. Its role in the roadmap is to convert a linear-algebraic condition on all adjoint operators into an intrinsic statement about the lower central series. Nilpotence is only one way repeated brackets can simplify a Lie algebra. For solvable Lie algebras, the guiding question becomes whether representations can be put into a form where the action is visibly triangular, so invariant subspaces can be read off one dimension at a time. Lie's theorem supplies the starting eigenline; its triangularization form then iterates that line to produce a full invariant flag. [remark: Cartan Criterion Roadmap] Cartan's criterion is the later structural result asserting that, for finite-dimensional Lie algebras in characteristic $0$, trace forms built from adjoint endomorphisms detect solvability of ideals and nondegeneracy of the Killing form detects semisimplicity. The exact statement requires the Killing form and several preparatory lemmas. The point for now is that trace, an elementary invariant of endomorphisms, becomes a structural test for Lie algebras. [/remark] After solvable and nilpotent behaviour has been isolated, the opposite extreme is semisimplicity. The central representation-theoretic question is whether finite-dimensional representations break into irreducible pieces without hidden extensions between them. [quotetheorem:3755] This result explains why semisimple Lie algebras are representation-theoretically rigid. The course stops at this foundational form, before the later classification theory and highest-weight methods. ## Conventions Used Throughout The base field will usually be denoted $F$. Lie algebras are denoted by fraktur letters such as $\mathfrak g$, $\mathfrak h$, and $\mathfrak{sl}_n(F)$, while vector spaces for representations are denoted by $V$ and $W$. The Lie bracket is written $[x,y]$, and the commutator of endomorphisms is written $[A,B]=AB-BA$. All Lie algebras in the core course are finite-dimensional unless stated otherwise. Hypotheses on the field, especially algebraic closedness and characteristic $0$, are part of the mathematical content rather than background decoration. Several major theorems fail or require modification when these hypotheses are removed. [remark: Characteristic Hypotheses] Skew-symmetry and alternation differ in characteristic $2$. In this course, most structural theorems are formulated under characteristic $0$ hypotheses when traces, eigenvalues, and triangularisation are used. [/remark] The course should be read as a sequence of reductions. First, construct examples and invariants. Next, learn which ideals and quotients preserve the relevant properties. Then use representations to translate bracket identities into linear algebra. Finally, apply trace and complete reducibility arguments to reach semisimplicity. The outline has now shifted from a broad roadmap to the objects that make it precise. With examples, maps, and structural tests in hand, we can begin working directly with Lie algebras as the language of infinitesimal symmetry. # 1. Lie Algebras and First Examples Lie algebras are the linear algebraic objects that record infinitesimal symmetry. Instead of a multiplication that composes elements, a Lie algebra has a bracket measuring the failure of two infinitesimal operations to commute. This opening chapter sets up the language: Lie algebras, their maps and internal subobjects, the adjoint representation, and the first matrix and derivation examples that will reappear throughout the course. ## From Associative Multiplication to Brackets What algebraic structure remains if we forget the product in an associative algebra and remember only the commutator $xy-yx$? The answer motivates the axioms of a Lie algebra: bilinearity records compatibility with the underlying vector space, skew-symmetry records antisymmetry of commutation, and the Jacobi identity replaces associativity. Without the Jacobi identity, commutators of operators would not give a stable theory of infinitesimal actions: bracketing by a fixed element would fail to behave compatibly with the bracket itself. Let $F$ be a field. Unless stated otherwise, all vector spaces in this chapter are over $F$. [definition: Lie Algebra] A Lie algebra over $F$ is an $F$-vector space $\mathfrak g$ equipped with an $F$-bilinear map \begin{align*} [-,-] : \mathfrak g \times \mathfrak g \to \mathfrak g \end{align*} such that, for all $x,y,z \in \mathfrak g$, \begin{align*} [x,x] &= 0, \\ [x,[y,z]] + [y,[z,x]] + [z,[x,y]] &= 0. \end{align*} [/definition] The first identity is alternation. If $\operatorname{char} F \ne 2$, it is equivalent to skew-symmetry, since bilinearity gives $0=[x+y,x+y]=[x,y]+[y,x]$. The second identity is the Jacobi identity; it says that bracketing with a fixed element behaves like a derivation of the bracket. [remark: Skew-Symmetry Convention] Many texts define a Lie algebra by requiring $[x,y]=-[y,x]$ together with $[x,x]=0$ in characteristic $2$, or by assuming $\operatorname{char} F \ne 2$. The alternating convention above is robust in every characteristic and implies $[x,y]=-[y,x]$ without additional hypotheses when $2$ is invertible in $F$. [/remark] The zero bracket is the boundary case of the theory. It shows that the axioms do not force non-commutativity; instead, they give a framework in which commutative and non-commutative infinitesimal structures can be compared. [example: Abelian Lie Algebra] Let $V$ be any $F$-vector space and define a bracket by $[x,y]=0$ for every $x,y\in V$. For $\lambda,\mu\in F$ and $x_1,x_2,y\in V$, \begin{align*} [\lambda x_1+\mu x_2,y] &=0 \\ &=\lambda 0+\mu 0 \\ &=\lambda[x_1,y]+\mu[x_2,y], \end{align*} and the same calculation in the second variable gives $[x,\lambda y_1+\mu y_2]=\lambda[x,y_1]+\mu[x,y_2]$. Thus the bracket is bilinear. Alternation is immediate because \begin{align*} [x,x]=0 \end{align*} for every $x\in V$. The Jacobi identity also holds term by term: \begin{align*} [x,[y,z]]+[y,[z,x]]+[z,[x,y]] &=[x,0]+[y,0]+[z,0] \\ &=0+0+0 \\ &=0. \end{align*} So $V$ is a Lie algebra. For each $x\in V$, the adjoint map satisfies \begin{align*} \operatorname{ad}_x(y)=[x,y]=0 \end{align*} for every $y\in V$, hence every $\operatorname{ad}_x$ is the zero [linear map](/page/Linear%20Map). The center is all of $V$ because every $x\in V$ satisfies $[x,y]=0$ for all $y\in V$: \begin{align*} Z(V)=V. \end{align*} The derived algebra is \begin{align*} [V,V] &=\operatorname{span}_F\{[x,y]:x,y\in V\} \\ &=\operatorname{span}_F\{0\} \\ &=0. \end{align*} Thus the abelian Lie algebra is the baseline case: no element acts nontrivially by adjoint operators, and no nonzero direction is produced by taking brackets. [/example] Associative algebras supply the most important source of first examples. The construction takes an associative product and measures its non-commutativity. This is useful because many familiar spaces already carry associative multiplication: matrices, endomorphisms, and polynomial or operator algebras all become sources of Lie brackets once the symmetric part of multiplication is discarded. The definition below isolates the operation used in all of these examples. [definition: Commutator Bracket] Let $A$ be an associative $F$-algebra. The commutator bracket on $A$ is the map \begin{align*} [-,-] : A\times A &\to A, \\ [x,y] &:= xy-yx \end{align*} for $x,y \in A$. [/definition] The point of this definition is not merely to build another operation from multiplication. If the product in $A$ is associative, the failure of two elements to commute automatically satisfies the Jacobi identity; without associativity, the cancellation in the Jacobi identity can break down. [quotetheorem:3756] [citeproof:3756] This theorem is the bridge from familiar matrix multiplication to Lie theory. The hypothesis that $A$ is associative is essential: the proof uses reassociation of degree-three monomials, and a non-associative multiplication need not make $xy-yx$ satisfy Jacobi. The theorem does not say that every Lie algebra is naturally an associative algebra with commutator bracket; it only gives a large source of examples. Its forward use is that every matrix algebra and every endomorphism algebra immediately becomes available as a Lie algebra. [example: Matrix Commutators] Let $A=M_n(F)$ with the commutator bracket $[X,Y]=XY-YX$; by *Commutator Lie Algebra*, this makes $M_n(F)$ into the Lie algebra $\mathfrak{gl}_n(F)$. For elementary matrices, the $(a,b)$ entry of $E_{ij}E_{kl}$ is \begin{align*} (E_{ij}E_{kl})_{ab} &=\sum_{r=1}^n (E_{ij})_{ar}(E_{kl})_{rb}. \end{align*} The factor $(E_{ij})_{ar}$ is nonzero exactly when $a=i$ and $r=j$, and then it equals $1$. The factor $(E_{kl})_{rb}$ is nonzero exactly when $r=k$ and $b=l$, and then it equals $1$. Hence the only possible nonzero summand occurs when $a=i$, $r=j=k$, and $b=l$, so \begin{align*} (E_{ij}E_{kl})_{ab} &=\delta_{ai}\delta_{jk}\delta_{bl}. \end{align*} This is exactly the matrix $\delta_{jk}E_{il}$, since \begin{align*} (\delta_{jk}E_{il})_{ab} &=\delta_{jk}\delta_{ai}\delta_{bl}. \end{align*} Therefore \begin{align*} E_{ij}E_{kl}=\delta_{jk}E_{il}. \end{align*} Interchanging the two elementary matrices gives \begin{align*} E_{kl}E_{ij}=\delta_{li}E_{kj}. \end{align*} Thus the commutator is \begin{align*} [E_{ij},E_{kl}] &=E_{ij}E_{kl}-E_{kl}E_{ij} \\ &=\delta_{jk}E_{il}-\delta_{li}E_{kj}. \end{align*} This formula turns matrix multiplication into explicit structure constants for $\mathfrak{gl}_n(F)$: one sparse bracket either vanishes or produces a signed elementary matrix, which is why elementary matrices are effective for later computations in $\mathfrak{sl}_2(F)$, triangular Lie algebras, and root decompositions. [/example] ## Subobjects and Structure-Preserving Maps Once Lie algebras have been defined, the next question is which linear maps and subspaces respect the bracket. The answer parallels group and ring theory: homomorphisms preserve the operation, subalgebras are closed under it, and ideals are exactly the subspaces on which quotient constructions and kernels depend. [definition: Lie Algebra Homomorphism] Let $\mathfrak g$ and $\mathfrak h$ be Lie algebras over $F$. A Lie algebra homomorphism is an $F$-linear map $\varphi: \mathfrak g \to \mathfrak h$ such that \begin{align*} \varphi([x,y]) = [\varphi(x),\varphi(y)] \end{align*} for all $x,y\in \mathfrak g$. [/definition] A homomorphism preserves all bracket expressions built from elements of $\mathfrak g$. Its kernel is not merely a subalgebra: it is stable under bracketing with arbitrary elements of the source. Before imposing that stronger kernel-like condition, it is useful to separate the weaker notion of an internal Lie algebra sitting inside a larger one. A subspace need not inherit the bracket at all: even if $x$ and $y$ lie in the subspace, the bracket $[x,y]$ may leave it. The first obstruction to treating a subspace as a Lie algebra is therefore closure under brackets of its own elements. [definition: Lie Subalgebra] A Lie subalgebra of a Lie algebra $\mathfrak g$ is a vector subspace $\mathfrak h \subset \mathfrak g$ such that \begin{align*} [x,y] \in \mathfrak h \end{align*} for all $x,y\in \mathfrak h$. [/definition] Internal closure is enough when studying $\mathfrak h$ as a Lie algebra in its own right. It is not enough for quotients: if elements outside $\mathfrak h$ can bracket into directions not controlled by $\mathfrak h$, then changing coset representatives may change the bracket of cosets. To make kernels and quotient brackets behave correctly, the subspace must be stable under bracketing with every element of the ambient algebra. This stronger stability condition is the Lie algebra analogue of normality. [definition: Ideal] An ideal of a Lie algebra $\mathfrak g$ is a vector subspace $\mathfrak a \subset \mathfrak g$ such that \begin{align*} [x,a] \in \mathfrak a \end{align*} for all $x\in \mathfrak g$ and all $a\in \mathfrak a$. [/definition] Every ideal is a subalgebra, because its stability condition includes bracketing by elements of the ideal itself. The converse fails in general: a subalgebra is internally closed, while an ideal is stable under the whole ambient algebra. The reason for isolating ideals is that two different constructions put the same pressure on a subspace. Kernels of homomorphisms must absorb brackets with arbitrary source elements, and quotient brackets are well defined only when changing representatives changes the bracket by something inside the subspace. The structural question is therefore whether these two requirements are precisely the ideal condition. [quotetheorem:3751] [citeproof:3751] This theorem is the Lie algebra analogue of normal subgroups and quotient groups. The homomorphism hypothesis is essential: the kernel of a merely linear map need not be stable under brackets, so it need not be an ideal. The ideal hypothesis in the quotient direction is also essential; the line $Fe\subset\mathfrak{sl}_2(F)$ below is a subalgebra, but the bracket of cosets modulo $Fe$ is not well-defined because $[f,e]=-h\notin Fe$. The theorem does not classify all ideals or all quotients; it identifies the precise condition under which quotient brackets make sense, which is why ideals become central in later structure theory. [example: Subalgebra That Is Not an Ideal] Work in $\mathfrak{sl}_2(F)$ with \begin{align*} e=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad f=\begin{pmatrix}0&0\\1&0\end{pmatrix},\qquad h=\begin{pmatrix}1&0\\0&-1\end{pmatrix}, \end{align*} and assume $\operatorname{char}F\ne 2$. We show that the line $Fe$ is closed under the bracket, but is not stable under bracketing by all of $\mathfrak{sl}_2(F)$. First, \begin{align*} e^2 &=\begin{pmatrix}0&1\\0&0\end{pmatrix} \begin{pmatrix}0&1\\0&0\end{pmatrix} \\ &=\begin{pmatrix} 0\cdot 0+1\cdot 0 & 0\cdot 1+1\cdot 0 \\ 0\cdot 0+0\cdot 0 & 0\cdot 1+0\cdot 0 \end{pmatrix} \\ &=\begin{pmatrix}0&0\\0&0\end{pmatrix}. \end{align*} Hence \begin{align*} [e,e] &=ee-ee \\ &=0. \end{align*} For arbitrary $\lambda,\mu\in F$, bilinearity of the commutator bracket gives \begin{align*} [\lambda e,\mu e] &=\lambda\mu[e,e] \\ &=\lambda\mu\cdot 0 \\ &=0\in Fe. \end{align*} Thus $Fe$ is a Lie subalgebra. To test the ideal condition, bracket $e$ with the element $f$ of the ambient algebra. Compute \begin{align*} fe &=\begin{pmatrix}0&0\\1&0\end{pmatrix} \begin{pmatrix}0&1\\0&0\end{pmatrix} \\ &=\begin{pmatrix} 0\cdot 0+0\cdot 0 & 0\cdot 1+0\cdot 0 \\ 1\cdot 0+0\cdot 0 & 1\cdot 1+0\cdot 0 \end{pmatrix} \\ &=\begin{pmatrix}0&0\\0&1\end{pmatrix}, \end{align*} while \begin{align*} ef &=\begin{pmatrix}0&1\\0&0\end{pmatrix} \begin{pmatrix}0&0\\1&0\end{pmatrix} \\ &=\begin{pmatrix} 0\cdot 0+1\cdot 1 & 0\cdot 0+1\cdot 0 \\ 0\cdot 0+0\cdot 1 & 0\cdot 0+0\cdot 0 \end{pmatrix} \\ &=\begin{pmatrix}1&0\\0&0\end{pmatrix}. \end{align*} Therefore \begin{align*} [f,e] &=fe-ef \\ &=\begin{pmatrix}0&0\\0&1\end{pmatrix} -\begin{pmatrix}1&0\\0&0\end{pmatrix} \\ &=\begin{pmatrix}-1&0\\0&1\end{pmatrix} \\ &=-h. \end{align*} If $-h$ belonged to $Fe$, then for some $\lambda\in F$ we would have \begin{align*} \begin{pmatrix}-1&0\\0&1\end{pmatrix} =\lambda e =\begin{pmatrix}0&\lambda\\0&0\end{pmatrix}. \end{align*} Comparing the $(2,2)$ entries gives $1=0$, impossible in a field. Hence $-h\notin Fe$. The subspace $Fe$ is therefore internally closed under brackets, but it is not an ideal because the ambient element $f$ brackets it outside the line. [/example] Several standard subspaces measure how elements commute with each other. The most global question is whether some elements have no visible commutator interaction with the algebra at all. Such elements behave like scalars in an associative algebra: they may be present in the vector space, but the bracket cannot distinguish them from zero through internal action. Isolating these elements gives a basic invariant that later controls the kernel of the adjoint representation. [definition: Center] The center of a Lie algebra $\mathfrak g$ is \begin{align*} Z(\mathfrak g) := \{x\in \mathfrak g : [x,y]=0 \text{ for all } y\in \mathfrak g\}. \end{align*} [/definition] The center detects elements that are invisible to every commutator. Often the relevant question is more local: an element may fail to commute with all of $\mathfrak g$ but still commute with a chosen subset or subspace that is being studied. The centralizer records this relative commutation condition and lets one measure the symmetries of a particular part of the algebra. [definition: Centralizer] Let $\mathfrak g$ be a Lie algebra and let $S\subset \mathfrak g$. The centralizer of $S$ in $\mathfrak g$ is \begin{align*} C_{\mathfrak g}(S) := \{x\in \mathfrak g : [x,s]=0 \text{ for all } s\in S\}. \end{align*} [/definition] After centralizers, a weaker and more flexible question becomes important: which elements preserve a chosen subspace under bracketing, even if they do not commute with it pointwise? This is the condition needed when one wants a subalgebra or ideal to remain stable under an ambient adjoint action. The normalizer records exactly this preservation condition. [definition: Normalizer] Let $\mathfrak g$ be a Lie algebra and let $\mathfrak h\subset \mathfrak g$ be a subspace. The normalizer of $\mathfrak h$ in $\mathfrak g$ is \begin{align*} N_{\mathfrak g}(\mathfrak h) := \{x\in \mathfrak g : [x,h]\in \mathfrak h \text{ for all } h\in \mathfrak h\}. \end{align*} [/definition] The center is $C_{\mathfrak g}(\mathfrak g)$, while $N_{\mathfrak g}(\mathfrak h)$ is the largest part of $\mathfrak g$ whose bracket action preserves $\mathfrak h$. If $\mathfrak h$ is an ideal, then $N_{\mathfrak g}(\mathfrak h)=\mathfrak g$. Commutation subspaces can also measure how far the whole Lie algebra is from being abelian. Instead of asking which elements commute, one collects all brackets that actually occur. This produces the part of $\mathfrak g$ generated by non-commutativity and prepares the quotient in which all brackets have been forced to vanish. [definition: Derived Algebra] The derived algebra of a Lie algebra $\mathfrak g$ is the vector subspace \begin{align*} [\mathfrak g,\mathfrak g] := \operatorname{span}_F\{[x,y] : x,y\in \mathfrak g\}. \end{align*} [/definition] The derived algebra measures the first level of non-commutativity. Without passing to the span of all brackets, the phrase "the commutator part" would not even define a subspace, and without the Jacobi identity it need not be stable under the ambient bracket. The quotient $\mathfrak g/[\mathfrak g,\mathfrak g]$ is the largest abelian quotient of $\mathfrak g$, a fact that motivates the derived series in later chapters. [quotetheorem:3758] [citeproof:3758] The Jacobi identity is the essential hypothesis behind this theorem; it is exactly what lets a bracket with a bracket be rewritten as a sum of brackets. The result does not say that $[\mathfrak g,\mathfrak g]$ is all of $\mathfrak g$ or that it is central: abelian Lie algebras have zero derived algebra, while $\mathfrak{sl}_2(F)$ has derived algebra equal to itself in the usual characteristics. The forward connection is that the derived algebra begins the derived series, which later measures solvability. ## The Adjoint Representation How does a Lie algebra act on itself? Each element $x\in\mathfrak g$ defines a linear operator by bracketing with $x$, and the Jacobi identity says that these operators form a representation of $\mathfrak g$. Without Jacobi, the commutator of the two operators $y\mapsto [x,y]$ and $y\mapsto [x',y]$ would not necessarily be the operator coming from $[x,x']$, so the algebra would not act on itself in a bracket-compatible way. [definition: Adjoint Map] Let $\mathfrak g$ be a Lie algebra. For $x\in\mathfrak g$, define \begin{align*} \operatorname{ad}_x : \mathfrak g &\to \mathfrak g, \\ \operatorname{ad}_x(y) &:= [x,y]. \end{align*} The adjoint map is the linear map \begin{align*} \operatorname{ad}: \mathfrak g \to \mathfrak{gl}(\mathfrak g), \qquad x\mapsto \operatorname{ad}_x. \end{align*} [/definition] Here $\mathfrak{gl}(\mathfrak g)$ means the Lie algebra of linear endomorphisms of the vector space $\mathfrak g$, with bracket $[S,T]=S\circ T-T\circ S$. The definition gives a linear operator $\operatorname{ad}_x$ for each element $x$, but it is not automatic that the assignment $x\mapsto\operatorname{ad}_x$ preserves the Lie bracket. The obstruction is exactly the compatibility between nested brackets in $\mathfrak g$ and commutators of endomorphisms. The next result verifies that the Jacobi identity supplies this compatibility, so the adjoint map is a genuine representation rather than just notation for many linear maps. [quotetheorem:3759] [citeproof:3759] The adjoint representation is the first representation attached to any Lie algebra. The Lie algebra axioms are essential here: bilinearity gives linear operators, and Jacobi gives the homomorphism identity. The theorem does not imply that $\operatorname{ad}$ is faithful; its kernel is precisely the center, so the Heisenberg algebra gives a non-faithful example. Its forward use is substantial: ideals, centers, and later semisimplicity criteria can be studied through the linear operators $\operatorname{ad}_x$. [example: Adjoint Operators in $\mathfrak{sl}_2$] Work in $\mathfrak{sl}_2(F)$ with \begin{align*} e=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad f=\begin{pmatrix}0&0\\1&0\end{pmatrix},\qquad h=\begin{pmatrix}1&0\\0&-1\end{pmatrix}, \end{align*} and assume $\operatorname{char}F\ne 2$. We compute the adjoint action from the commutator bracket $[X,Y]=XY-YX$. First, \begin{align*} he &=\begin{pmatrix}1&0\\0&-1\end{pmatrix} \begin{pmatrix}0&1\\0&0\end{pmatrix} \\ &=\begin{pmatrix} 1\cdot 0+0\cdot 0 & 1\cdot 1+0\cdot 0 \\ 0\cdot 0+(-1)\cdot 0 & 0\cdot 1+(-1)\cdot 0 \end{pmatrix} \\ &=\begin{pmatrix}0&1\\0&0\end{pmatrix}=e, \end{align*} while \begin{align*} eh &=\begin{pmatrix}0&1\\0&0\end{pmatrix} \begin{pmatrix}1&0\\0&-1\end{pmatrix} \\ &=\begin{pmatrix} 0\cdot 1+1\cdot 0 & 0\cdot 0+1\cdot(-1) \\ 0\cdot 1+0\cdot 0 & 0\cdot 0+0\cdot(-1) \end{pmatrix} \\ &=\begin{pmatrix}0&-1\\0&0\end{pmatrix}=-e. \end{align*} Therefore \begin{align*} [h,e] &=he-eh \\ &=e-(-e) \\ &=2e. \end{align*} Similarly, \begin{align*} hf &=\begin{pmatrix}1&0\\0&-1\end{pmatrix} \begin{pmatrix}0&0\\1&0\end{pmatrix} \\ &=\begin{pmatrix} 1\cdot 0+0\cdot 1 & 1\cdot 0+0\cdot 0 \\ 0\cdot 0+(-1)\cdot 1 & 0\cdot 0+(-1)\cdot 0 \end{pmatrix} \\ &=\begin{pmatrix}0&0\\-1&0\end{pmatrix}=-f, \end{align*} and \begin{align*} fh &=\begin{pmatrix}0&0\\1&0\end{pmatrix} \begin{pmatrix}1&0\\0&-1\end{pmatrix} \\ &=\begin{pmatrix} 0\cdot 1+0\cdot 0 & 0\cdot 0+0\cdot(-1) \\ 1\cdot 1+0\cdot 0 & 1\cdot 0+0\cdot(-1) \end{pmatrix} \\ &=\begin{pmatrix}0&0\\1&0\end{pmatrix}=f. \end{align*} Hence \begin{align*} [h,f] &=hf-fh \\ &=(-f)-f \\ &=-2f. \end{align*} Finally, \begin{align*} ef &=\begin{pmatrix}0&1\\0&0\end{pmatrix} \begin{pmatrix}0&0\\1&0\end{pmatrix} \\ &=\begin{pmatrix} 0\cdot 0+1\cdot 1 & 0\cdot 0+1\cdot 0 \\ 0\cdot 0+0\cdot 1 & 0\cdot 0+0\cdot 0 \end{pmatrix} \\ &=\begin{pmatrix}1&0\\0&0\end{pmatrix}, \end{align*} and \begin{align*} fe &=\begin{pmatrix}0&0\\1&0\end{pmatrix} \begin{pmatrix}0&1\\0&0\end{pmatrix} \\ &=\begin{pmatrix} 0\cdot 0+0\cdot 0 & 0\cdot 1+0\cdot 0 \\ 1\cdot 0+0\cdot 0 & 1\cdot 1+0\cdot 0 \end{pmatrix} \\ &=\begin{pmatrix}0&0\\0&1\end{pmatrix}. \end{align*} Thus \begin{align*} [e,f] &=ef-fe \\ &=\begin{pmatrix}1&0\\0&0\end{pmatrix} -\begin{pmatrix}0&0\\0&1\end{pmatrix} \\ &=\begin{pmatrix}1&0\\0&-1\end{pmatrix} \\ &=h. \end{align*} Since $[h,h]=0$ by alternation, the adjoint operator $\operatorname{ad}_h$ satisfies \begin{align*} \operatorname{ad}_h(e)&=[h,e]=2e,\\ \operatorname{ad}_h(f)&=[h,f]=-2f,\\ \operatorname{ad}_h(h)&=[h,h]=0. \end{align*} So $e,f,h$ are eigenvectors of $\operatorname{ad}_h$ with eigenvalues $2,-2,0$, respectively. Also, \begin{align*} \operatorname{ad}_e(f)&=[e,f]=h,\\ \operatorname{ad}_e(h)&=[e,h]=-[h,e]=-2e,\\ \operatorname{ad}_f(e)&=[f,e]=-[e,f]=-h,\\ \operatorname{ad}_f(h)&=[f,h]=-[h,f]=2f. \end{align*} Thus $\operatorname{ad}_h$ is diagonal in the basis $e,f,h$, while $\operatorname{ad}_e$ and $\operatorname{ad}_f$ move vectors between the three basis directions; this is the first instance of the weight-space pattern used later in the structure theory of semisimple Lie algebras. [/example] ## Standard Matrix Lie Algebras Which Lie algebras should be treated as the basic test cases? Matrix algebras provide the central supply: the full linear algebra, trace-zero matrices, triangular matrices, and nilpotent triangular matrices. [definition: General Linear Lie Algebra] The general linear Lie algebra $\mathfrak{gl}_n(F)$ is the vector space $M_n(F)$ with bracket \begin{align*} [X,Y] := XY-YX. \end{align*} [/definition] The first natural way to cut down $\mathfrak{gl}_n(F)$ is to impose a linear condition that has a chance of surviving commutators. Trace is the key invariant here: commutators have trace zero, so the trace-zero matrices form the most immediate candidate for a canonical subalgebra inside the full matrix Lie algebra. The next definition names this candidate before checking its bracket stability. [definition: Special Linear Lie Algebra] The special linear Lie algebra is \begin{align*} \mathfrak{sl}_n(F) := \{X\in \mathfrak{gl}_n(F) : \operatorname{tr}(X)=0\}. \end{align*} [/definition] The definition gives only a vector subspace of $\mathfrak{gl}_n(F)$, and vector subspaces usually have no reason to be closed under commutators. To use $\mathfrak{sl}_n(F)$ as a structural example, one must know that the trace-zero condition is preserved when bracketing with arbitrary matrices. The next result supplies that stability and upgrades $\mathfrak{sl}_n(F)$ from a convenient subspace to an ideal of the full matrix Lie algebra. [quotetheorem:3760] [citeproof:3760] The trace identity is essential: an arbitrary linear condition on matrices need not be preserved by commutators. The theorem does not say that $\mathfrak{sl}_n(F)$ is the only ideal of $\mathfrak{gl}_n(F)$, and in special characteristics extra care is needed because scalar matrices can have trace zero. The forward connection is that $\mathfrak{sl}_n(F)$ is the first recurring example of a non-abelian Lie algebra obtained as a natural ideal inside a larger algebra. [definition: Triangular Matrix Lie Algebras] Let $\mathfrak b_n(F)$ be the subspace of upper triangular matrices in $\mathfrak{gl}_n(F)$, and let $\mathfrak n_n(F)$ be the subspace of strictly upper triangular matrices. [/definition] For triangular matrices, the obstruction is again closure under commutators: preserving a flag under multiplication should force the same kind of preservation after taking differences $XY-YX$. The strictly upper triangular matrices require a second check, because their zero diagonal condition must also survive commutators with upper triangular matrices. The next result records both closure properties so these matrix spaces can serve as standard Lie algebra examples. [quotetheorem:3761] [citeproof:3761] The triangular hypotheses are doing real work: a random matrix subspace is usually not closed under commutators. The theorem does not say that $\mathfrak n_n(F)$ is an ideal of $\mathfrak{gl}_n(F)$; it is instead naturally an ideal inside $\mathfrak b_n(F)$. These examples foreshadow solvable and nilpotent Lie algebras because repeated commutators of triangular matrices become increasingly constrained. [example: Center and Derived Algebra of $\mathfrak{gl}_n$] [claim]For $n\ge 2$, the center of $\mathfrak{gl}_n(F)$ is $FI_n$, and \begin{align*} [\mathfrak{gl}_n(F),\mathfrak{gl}_n(F)]=\mathfrak{sl}_n(F). \end{align*} [/claim] [proof]Let $X=(x_{ab})\in \mathfrak{gl}_n(F)$ lie in the center. Then $[X,E_{ij}]=0$ for every elementary matrix $E_{ij}$, so $XE_{ij}=E_{ij}X$. For fixed $i,j$, the $(a,b)$ entry of $XE_{ij}$ is \begin{align*} (XE_{ij})_{ab} &=\sum_{r=1}^n x_{ar}(E_{ij})_{rb} \\ &=x_{ai}\delta_{jb}, \end{align*} because $(E_{ij})_{rb}$ is nonzero exactly when $r=i$ and $b=j$. Similarly, \begin{align*} (E_{ij}X)_{ab} &=\sum_{r=1}^n (E_{ij})_{ar}x_{rb} \\ &=\delta_{ai}x_{jb}. \end{align*} Thus \begin{align*} x_{ai}\delta_{jb}=\delta_{ai}x_{jb} \end{align*} for all $a,b,i,j$. Take $b=j$ and $a\ne i$. Then $\delta_{jb}=1$ and $\delta_{ai}=0$, so \begin{align*} x_{ai}=0. \end{align*} Since $i$ was arbitrary, every off-diagonal entry of $X$ is zero. Hence $X$ is diagonal. Now take $a=i$ and $b=j$. The equality becomes \begin{align*} x_{ii}=x_{jj}. \end{align*} Since $i$ and $j$ were arbitrary, all diagonal entries of $X$ are equal, so $X=\lambda I_n$ for some $\lambda\in F$. Conversely, for every $Y\in \mathfrak{gl}_n(F)$, \begin{align*} [\lambda I_n,Y] &=\lambda I_nY-Y\lambda I_n \\ &=\lambda Y-\lambda Y \\ &=0. \end{align*} Therefore \begin{align*} Z(\mathfrak{gl}_n(F))=FI_n. \end{align*} Next, for any $X,Y\in \mathfrak{gl}_n(F)$, \begin{align*} \operatorname{tr}([X,Y]) &=\operatorname{tr}(XY-YX) \\ &=\operatorname{tr}(XY)-\operatorname{tr}(YX) \\ &=0, \end{align*} using $\operatorname{tr}(XY)=\operatorname{tr}(YX)$. Hence \begin{align*} [\mathfrak{gl}_n(F),\mathfrak{gl}_n(F)]\subseteq \mathfrak{sl}_n(F). \end{align*} For the reverse containment, first let $i\ne j$. Since \begin{align*} E_{ii}E_{ij}=E_{ij} \end{align*} and \begin{align*} E_{ij}E_{ii}=0 \end{align*} because the column index $j$ is not equal to $i$, we get \begin{align*} [E_{ii},E_{ij}] &=E_{ii}E_{ij}-E_{ij}E_{ii} \\ &=E_{ij}-0 \\ &=E_{ij}. \end{align*} Thus every off-diagonal elementary matrix lies in the derived algebra. For diagonal trace-zero matrices, compute for $i\ne j$: \begin{align*} E_{ij}E_{ji}&=E_{ii},\\ E_{ji}E_{ij}&=E_{jj}, \end{align*} so \begin{align*} [E_{ij},E_{ji}] &=E_{ij}E_{ji}-E_{ji}E_{ij} \\ &=E_{ii}-E_{jj}. \end{align*} Thus every $E_{ii}-E_{jj}$ lies in the derived algebra. If $D=\operatorname{diag}(d_1,\dots,d_n)$ has trace zero, then \begin{align*} d_1+\cdots+d_n=0, \end{align*} so \begin{align*} d_n=-(d_1+\cdots+d_{n-1}). \end{align*} Therefore \begin{align*} D &=\sum_{i=1}^{n-1} d_i(E_{ii}-E_{nn}), \end{align*} which is a linear combination of commutators. Every trace-zero matrix is the sum of its off-diagonal part and its diagonal trace-zero part, so \begin{align*} \mathfrak{sl}_n(F)\subseteq [\mathfrak{gl}_n(F),\mathfrak{gl}_n(F)]. \end{align*} Combining the two containments gives \begin{align*} [\mathfrak{gl}_n(F),\mathfrak{gl}_n(F)]=\mathfrak{sl}_n(F). \end{align*} [/proof] The center consists exactly of the scalar matrices, while the derived algebra removes precisely the trace direction and keeps all trace-zero commutator information. [/example] The special linear example separates two phenomena that coincide in smaller or more abelian examples: the center can vanish while the derived algebra remains large. The next computation uses $\mathfrak{sl}_2$ to show this behaviour in the basic simple-looking matrix algebra. [example: Derived Algebra and Center of $\mathfrak{sl}_2$] [claim]Assume $\operatorname{char}F\ne 2$. For the standard basis $e,f,h$ of $\mathfrak{sl}_2(F)$, \begin{align*} [\mathfrak{sl}_2(F),\mathfrak{sl}_2(F)]=\mathfrak{sl}_2(F), \qquad Z(\mathfrak{sl}_2(F))=0. \end{align*} [/claim] [proof]From the computed commutator relations in $\mathfrak{sl}_2(F)$, \begin{align*} [e,f]&=h,\\ [h,e]&=2e,\\ [h,f]&=-2f. \end{align*} Since $\operatorname{char}F\ne 2$, the scalar $2$ is invertible in $F$. Hence \begin{align*} h&=[e,f]\in [\mathfrak{sl}_2(F),\mathfrak{sl}_2(F)],\\ e&=\frac{1}{2}[h,e]\in [\mathfrak{sl}_2(F),\mathfrak{sl}_2(F)],\\ f&=-\frac{1}{2}[h,f]\in [\mathfrak{sl}_2(F),\mathfrak{sl}_2(F)]. \end{align*} The derived algebra is a vector subspace of $\mathfrak{sl}_2(F)$ containing the basis $e,f,h$, so \begin{align*} [\mathfrak{sl}_2(F),\mathfrak{sl}_2(F)]=\mathfrak{sl}_2(F). \end{align*} Now let $u\in Z(\mathfrak{sl}_2(F))$. Write \begin{align*} u=ae+bf+ch \end{align*} with $a,b,c\in F$. Since $u$ is central, $[u,h]=0$. Using bilinearity and the relations above, \begin{align*} [u,h] &=[ae+bf+ch,h] \\ &=a[e,h]+b[f,h]+c[h,h]. \end{align*} Alternation gives $[h,h]=0$. Also $[e,h]=-[h,e]=-2e$ and $[f,h]=-[h,f]=2f$, because alternation and bilinearity give $[r,s]=-[s,r]$. Therefore \begin{align*} [u,h] &=a(-2e)+b(2f)+c\cdot 0 \\ &=-2ae+2bf. \end{align*} Since $[u,h]=0$ and $e,f,h$ are linearly independent, \begin{align*} -2a=0,\qquad 2b=0. \end{align*} Because $2$ is invertible in $F$, this gives \begin{align*} a=0,\qquad b=0. \end{align*} Thus $u=ch$. Since $u$ is central, $[u,e]=0$. But \begin{align*} [u,e] &=[ch,e] \\ &=c[h,e] \\ &=c(2e) \\ &=2ce. \end{align*} Again $e\ne 0$ and $2$ is invertible, so $2ce=0$ forces \begin{align*} c=0. \end{align*} Hence $u=0$, and therefore \begin{align*} Z(\mathfrak{sl}_2(F))=0. \end{align*} [/proof] The algebra $\mathfrak{sl}_2(F)$ is therefore completely generated by its own brackets, while no nonzero element commutes with all three basis directions. [/example] The trace-zero condition is not only a convenient subspace condition; it is the first example of a structural invariant arising from a homomorphism. Since $\operatorname{tr}([X,Y])=0$, the trace kills the derived algebra of $\mathfrak{gl}_n(F)$. ## The Heisenberg and Derivation Examples Not every important Lie algebra comes from taking all matrices or trace-zero matrices. The Heisenberg algebra is the smallest non-abelian example with a nonzero center, while derivation algebras explain why Lie brackets naturally encode infinitesimal symmetries of algebraic structures. [definition: Three-Dimensional Heisenberg Lie Algebra] The three-dimensional Heisenberg Lie algebra $\mathfrak h_3(F)$ is the $F$-vector space with basis $x,y,z$ and bracket determined by \begin{align*} [x,y]=z, \end{align*} together with $[x,z]=[y,z]=0$ and bilinearity and alternation. [/definition] The element $z$ is central, but the algebra is not abelian because $[x,y]=z\ne 0$. This is the prototype of a two-step nilpotent Lie algebra: brackets may be nonzero, but any bracket involving an already-created commutator vanishes. [example: Center and Derived Algebra of $\mathfrak h_3(F)$] [claim]If $z\ne 0$, then \begin{align*} Z(\mathfrak h_3(F))=Fz, \qquad [\mathfrak h_3(F),\mathfrak h_3(F)]=Fz. \end{align*} [/claim] [proof]Let $u=ax+by+cz$ and $v=a'x+b'y+c'z$ be arbitrary elements of $\mathfrak h_3(F)$. By bilinearity, \begin{align*} [u,v] &=[ax+by+cz,a'x+b'y+c'z] \\ &=aa'[x,x]+ab'[x,y]+ac'[x,z] \\ &\qquad +ba'[y,x]+bb'[y,y]+bc'[y,z] \\ &\qquad +ca'[z,x]+cb'[z,y]+cc'[z,z]. \end{align*} Alternation gives $[x,x]=[y,y]=[z,z]=0$. The defining relations give $[x,y]=z$, $[x,z]=[y,z]=0$, and bilinearity with alternation gives \begin{align*} [y,x]=-[x,y]=-z,\qquad [z,x]=-[x,z]=0,\qquad [z,y]=-[y,z]=0. \end{align*} Substituting these values, \begin{align*} [u,v] &=aa'\cdot 0+ab'z+ac'\cdot 0 +ba'(-z)+bb'\cdot 0+bc'\cdot 0 \\ &\qquad +ca'\cdot 0+cb'\cdot 0+cc'\cdot 0 \\ &=(ab'-ba')z. \end{align*} Thus every bracket lies in $Fz$, so \begin{align*} [\mathfrak h_3(F),\mathfrak h_3(F)]\subseteq Fz. \end{align*} Conversely, $z=[x,y]$ lies in the derived algebra, and therefore every scalar multiple of $z$ lies in the derived algebra because it is a vector subspace: \begin{align*} Fz\subseteq [\mathfrak h_3(F),\mathfrak h_3(F)]. \end{align*} Hence \begin{align*} [\mathfrak h_3(F),\mathfrak h_3(F)]=Fz. \end{align*} Now compute the center. Let $u=ax+by+cz$. Then \begin{align*} [u,x] &=[ax+by+cz,x] \\ &=a[x,x]+b[y,x]+c[z,x] \\ &=a\cdot 0+b(-z)+c\cdot 0 \\ &=-bz, \end{align*} and \begin{align*} [u,y] &=[ax+by+cz,y] \\ &=a[x,y]+b[y,y]+c[z,y] \\ &=az+b\cdot 0+c\cdot 0 \\ &=az. \end{align*} If $u\in Z(\mathfrak h_3(F))$, then $[u,x]=0$ and $[u,y]=0$, so \begin{align*} -bz=0,\qquad az=0. \end{align*} Since $z\ne 0$ and scalar multiplication in a vector space is faithful, this forces \begin{align*} b=0,\qquad a=0. \end{align*} Thus every central element has the form $u=cz$, so \begin{align*} Z(\mathfrak h_3(F))\subseteq Fz. \end{align*} For the reverse containment, let $cz\in Fz$. Then \begin{align*} [cz,x]&=c[z,x]=0,\\ [cz,y]&=c[z,y]=0,\\ [cz,z]&=c[z,z]=0. \end{align*} Since $cz$ brackets to zero with each basis vector $x,y,z$, bilinearity implies that $cz$ brackets to zero with every element of $\mathfrak h_3(F)$. Hence $Fz\subseteq Z(\mathfrak h_3(F))$, and therefore \begin{align*} Z(\mathfrak h_3(F))=Fz. \end{align*} [/proof] The same line $Fz$ is both the whole commutator output and the whole central direction: $\mathfrak h_3(F)$ is non-abelian, but all brackets land in a central one-dimensional subspace. [/example] The Heisenberg algebra can also be realised as strictly upper triangular $3\times 3$ matrices. With \begin{align*} x=E_{12},\qquad y=E_{23},\qquad z=E_{13}, \end{align*} one has $[x,y]=z$ and all brackets involving $z$ vanish. The Heisenberg example came from a concrete commutator table, but Lie algebras also arise from infinitesimal symmetries of multiplication. A linear operator on an associative algebra is too arbitrary to deserve this interpretation unless it interacts with products in the same way a derivative does. The Leibniz rule is the compatibility condition that singles out such operators, and the next definition names them. [definition: Derivation] Let $A$ be an associative $F$-algebra. A derivation of $A$ is an $F$-linear map $D:A\to A$ such that \begin{align*} D(ab)=D(a)b+aD(b) \end{align*} for all $a,b\in A$. [/definition] The Leibniz rule is the algebraic version of differentiating a product. Without this rule, a linear endomorphism of $A$ need not preserve any information about the multiplication, so its commutator with another such endomorphism would not describe an infinitesimal symmetry of $A$. [definition: Derivation Algebra] The derivation algebra of an associative $F$-algebra $A$ is \begin{align*} \operatorname{Der}(A) := \{D\in \mathfrak{gl}(A) : D(ab)=D(a)b+aD(b) \text{ for all } a,b\in A\}. \end{align*} [/definition] Once derivations are viewed as infinitesimal symmetries, the main obstruction is whether their commutator is still compatible with multiplication. Two derivations can be composed as linear maps, but a composition need not satisfy the Leibniz rule in the same form. The commutator is the operation where the unwanted second-order terms cancel: if $D,E\in\operatorname{Der}(A)$, a direct expansion of $DE(ab)-ED(ab)$ leaves exactly \begin{align*} [D,E](ab)=[D,E](a)b+a[D,E](b). \end{align*} Thus $\operatorname{Der}(A)$ is closed under commutators inside $\mathfrak{gl}(A)$, so it is itself a Lie algebra. This is one of the main conceptual reasons Lie algebras are tied to infinitesimal transformations: derivations are infinitesimal symmetries of multiplication, and their commutator is again such a symmetry. The Leibniz hypothesis is essential, since arbitrary endomorphisms of $A$ do not preserve the product and their commutator has no reason to do so. The statement does not say that every derivation is inner; that stronger property depends on the algebra $A$ and fails in many commutative examples. [example: Inner Derivations] [claim]Let $A$ be an associative $F$-algebra and fix $a\in A$. The map \begin{align*} \operatorname{ad}_a:A&\to A,\\ b&\mapsto ab-ba \end{align*} is a derivation of $A$. Moreover, $\operatorname{ad}_a=0$ exactly when $a$ lies in the center of the associative algebra $A$.[/claim] [proof]First $\operatorname{ad}_a$ is $F$-linear. For $\lambda,\mu\in F$ and $b_1,b_2\in A$, bilinearity of multiplication gives \begin{align*} \operatorname{ad}_a(\lambda b_1+\mu b_2) &=a(\lambda b_1+\mu b_2)-(\lambda b_1+\mu b_2)a\\ &=\lambda ab_1+\mu ab_2-\lambda b_1a-\mu b_2a\\ &=\lambda(ab_1-b_1a)+\mu(ab_2-b_2a)\\ &=\lambda\operatorname{ad}_a(b_1)+\mu\operatorname{ad}_a(b_2). \end{align*} Now let $b,c\in A$. Using associativity to remove parentheses in triple products, \begin{align*} \operatorname{ad}_a(bc) &=a(bc)-(bc)a\\ &=(ab)c-b(ca). \end{align*} Insert the term $bac$ by adding and subtracting it: \begin{align*} \operatorname{ad}_a(bc) &=(ab)c-bac+bac-b(ca). \end{align*} Since $bac=(ba)c$ and $bac=b(ac)$ by associativity, \begin{align*} \operatorname{ad}_a(bc) &=(ab)c-(ba)c+b(ac)-b(ca)\\ &=(ab-ba)c+b(ac-ca)\\ &=\operatorname{ad}_a(b)c+b\operatorname{ad}_a(c). \end{align*} Thus $\operatorname{ad}_a$ satisfies the Leibniz rule, so it is a derivation. Finally, $\operatorname{ad}_a=0$ means that for every $b\in A$, \begin{align*} 0=\operatorname{ad}_a(b)=ab-ba. \end{align*} This is equivalent to \begin{align*} ab=ba \end{align*} for every $b\in A$, which is exactly the condition that $a$ lies in the center of the associative algebra $A$.[/proof] Inner derivations therefore measure non-centrality: central elements commute with everything and produce the zero derivation, while any failure of $a$ to commute with some element of $A$ gives a nonzero infinitesimal symmetry. [/example] ## First Invariants and What They Measure The examples in this chapter are not isolated computations; they introduce the invariants used throughout the course. The center measures invisible elements for the adjoint representation, the derived algebra measures non-commutativity, and ideals identify which quotients and decompositions are compatible with the bracket. For an abelian Lie algebra $\mathfrak g$, the center is all of $\mathfrak g$ and the derived algebra is $0$. For $\mathfrak{gl}_n(F)$, the center consists of scalar matrices while the derived algebra is typically $\mathfrak{sl}_n(F)$. For $\mathfrak{sl}_2(F)$ in suitable characteristic, the derived algebra is the whole algebra and the center is zero. For $\mathfrak h_3(F)$, the center and derived algebra coincide in the line $Fz$. These four behaviours form the first mental map of the subject. Later chapters will refine them into solvable and nilpotent series, [Engel's theorem](/theorems/3798), Cartan's criterion, and the distinction between solvable radicals and semisimple Lie algebras. By the end of Chapter 1, the main examples and first structural patterns are in view. The next step is to turn those patterns into a calculus of ideals, quotients, and extensions so that Lie algebras can be built and compared systematically. # 2. Ideals, Quotients, and Constructions This chapter turns the basic objects from Chapter 1 into a working algebraic calculus. Once ideals have been identified as the Lie-algebra analogue of normal subgroups, the next questions are how to divide by them, how homomorphisms factor through them, and how larger Lie algebras can be built from smaller ones. The constructions in this chapter give the language for later structure theory: solvable and nilpotent series, extensions, radicals, and semisimple quotients all depend on quotients and exact sequences. ## Quotients and Isomorphism Theorems The first problem is to decide when a subspace can be collapsed to zero without losing the Lie bracket. For vector spaces any subspace gives a quotient vector space, but the bracket descends only when representatives can be changed without changing the resulting coset. [definition: Quotient Lie Algebra] Let $\mathfrak g$ be a Lie algebra over a field $F$, and let $I \trianglelefteq \mathfrak g$ be an ideal. The quotient Lie algebra $\mathfrak g/I$ is the quotient vector space equipped with the bracket map $(\mathfrak g/I)\times(\mathfrak g/I)\to\mathfrak g/I$ determined by \begin{align*} [x + I, y + I] := [x,y] + I \end{align*} for $x,y \in \mathfrak g$. [/definition] The quotient definition hides a real well-definedness problem: changing representatives must not change the bracket coset. Ideals are precisely the subspaces stable enough under bracketing to make that happen, whereas an arbitrary subspace can fail as soon as one brackets it with an element outside the subspace. The next result turns this obstruction into the basic criterion for when a quotient vector space carries a Lie algebra structure. [quotetheorem:3763] [citeproof:3763] This theorem is the quotient construction's diagnostic test: no ideal condition, no well-defined Lie bracket. Its hypothesis is not a technical convenience, because changing a representative by an element of $I$ must be invisible after taking brackets with every element of $\mathfrak g$. Later, when a quotient appears, the first check is therefore always whether the subspace being collapsed is stable under all adjoint actions. [example: Abelianisation] Let $[\mathfrak g,\mathfrak g]$ denote the linear span of all brackets $[x,y]$ with $x,y\in\mathfrak g$. This subspace is an ideal: for $z,x,y\in\mathfrak g$, the Jacobi identity gives \begin{align*} [z,[x,y]] &= [[z,x],y] + [x,[z,y]], \end{align*} and both summands lie in $[\mathfrak g,\mathfrak g]$ by its definition. Hence the quotient $\mathfrak g/[\mathfrak g,\mathfrak g]$ is a Lie algebra. For $x,y\in\mathfrak g$, its bracket is \begin{align*} [x+[\mathfrak g,\mathfrak g],y+[\mathfrak g,\mathfrak g]] &= [x,y]+[\mathfrak g,\mathfrak g] \\ &= 0+[\mathfrak g,\mathfrak g], \end{align*} because $[x,y]\in[\mathfrak g,\mathfrak g]$. Thus $\mathfrak g/[\mathfrak g,\mathfrak g]$ is abelian. Now let $\varphi:\mathfrak g\to\mathfrak a$ be a Lie algebra homomorphism into an abelian Lie algebra $\mathfrak a$. For every $x,y\in\mathfrak g$, \begin{align*} \varphi([x,y]) &= [\varphi(x),\varphi(y)]_{\mathfrak a} \\ &= 0, \end{align*} so by linearity $\varphi$ kills every element of $[\mathfrak g,\mathfrak g]$. Define \begin{align*} \overline{\varphi}:\mathfrak g/[\mathfrak g,\mathfrak g]\to\mathfrak a, \qquad \overline{\varphi}(x+[\mathfrak g,\mathfrak g])=\varphi(x). \end{align*} If $x+[\mathfrak g,\mathfrak g]=x'+[\mathfrak g,\mathfrak g]$, then $x-x'\in[\mathfrak g,\mathfrak g]$, so \begin{align*} \varphi(x)-\varphi(x')=\varphi(x-x')=0, \end{align*} and therefore $\overline{\varphi}$ is well defined. Also, \begin{align*} \overline{\varphi}([x+[\mathfrak g,\mathfrak g],y+[\mathfrak g,\mathfrak g]]) &=\overline{\varphi}(0+[\mathfrak g,\mathfrak g]) \\ &=0 \\ &=[\varphi(x),\varphi(y)]_{\mathfrak a} \\ &=[\overline{\varphi}(x+[\mathfrak g,\mathfrak g]),\overline{\varphi}(y+[\mathfrak g,\mathfrak g])]_{\mathfrak a}, \end{align*} so $\overline{\varphi}$ is a Lie algebra homomorphism. The quotient map $\pi:\mathfrak g\to\mathfrak g/[\mathfrak g,\mathfrak g]$ satisfies $\varphi=\overline{\varphi}\circ\pi$, which is the precise sense in which $\mathfrak g/[\mathfrak g,\mathfrak g]$ is the largest abelian quotient of $\mathfrak g$. [/example] A homomorphism loses exactly the information contained in its kernel, so the natural way to make the map injective is to collapse that kernel to zero. The obstruction is to show that no other identifications are needed: two elements should become equal after quotienting precisely when the homomorphism sends them to the same element. The first isomorphism theorem packages this principle and identifies the resulting quotient with the image. [quotetheorem:3764] [citeproof:3764] The theorem says that quotients are the exact way to remove the information lost by a homomorphism. The kernel is the necessary hypothesis: collapsing anything smaller would not make the map injective, while collapsing anything larger would identify elements with different images. This result will be used as the default mechanism for recognising quotients that arise from natural maps, such as trace, diagonal projection, and action maps. [example: Trace And Special Linear Algebra] Assume $\operatorname{char} F$ does not divide $n$, so the element $n\cdot 1_F\in F$ is nonzero and hence invertible. For $A=(a_{ij})$ and $B=(b_{ij})$ in $\mathfrak{gl}_n(F)$, \begin{align*} \operatorname{tr}(AB) &=\sum_{i=1}^n (AB)_{ii} \\ &=\sum_{i=1}^n\sum_{j=1}^n a_{ij}b_{ji} \\ &=\sum_{j=1}^n\sum_{i=1}^n b_{ji}a_{ij} \\ &=\sum_{j=1}^n (BA)_{jj} \\ &=\operatorname{tr}(BA). \end{align*} Therefore \begin{align*} \operatorname{tr}([A,B]) &=\operatorname{tr}(AB-BA) \\ &=\operatorname{tr}(AB)-\operatorname{tr}(BA) \\ &=0. \end{align*} Since $F$ is regarded as an abelian Lie algebra, its bracket is zero, so \begin{align*} \operatorname{tr}([A,B]) &=0 \\ &=[\operatorname{tr}(A),\operatorname{tr}(B)]_F. \end{align*} Thus $\operatorname{tr}:\mathfrak{gl}_n(F)\to F$ is a Lie algebra homomorphism. Its kernel is exactly \begin{align*} \ker(\operatorname{tr}) &=\{A\in\mathfrak{gl}_n(F):\operatorname{tr}(A)=0\} \\ &=\mathfrak{sl}_n(F). \end{align*} The trace map is surjective because, for every $c\in F$, \begin{align*} \operatorname{tr}\left(\frac{c}{n}I_n\right) &=\frac{c}{n}\operatorname{tr}(I_n) \\ &=\frac{c}{n}\,n \\ &=c. \end{align*} By the *[First Isomorphism Theorem For Lie Algebras](/theorems/3764)*, \begin{align*} \mathfrak{gl}_n(F)/\mathfrak{sl}_n(F) &\cong F. \end{align*} The same characteristic assumption gives an explicit splitting. For every $A\in\mathfrak{gl}_n(F)$, \begin{align*} A &=\frac{\operatorname{tr}(A)}{n}I_n+\left(A-\frac{\operatorname{tr}(A)}{n}I_n\right), \end{align*} where $\frac{\operatorname{tr}(A)}{n}I_n\in F I_n$, and \begin{align*} \operatorname{tr}\left(A-\frac{\operatorname{tr}(A)}{n}I_n\right) &=\operatorname{tr}(A)-\frac{\operatorname{tr}(A)}{n}\operatorname{tr}(I_n) \\ &=\operatorname{tr}(A)-\frac{\operatorname{tr}(A)}{n}\,n \\ &=0, \end{align*} so the second summand lies in $\mathfrak{sl}_n(F)$. The decomposition is unique because if $\lambda I_n\in F I_n\cap\mathfrak{sl}_n(F)$, then \begin{align*} 0 &=\operatorname{tr}(\lambda I_n) \\ &=\lambda n, \end{align*} and $n$ is invertible in $F$, so $\lambda=0$. Finally, $F I_n$ is central since for every $\lambda\in F$ and $A\in\mathfrak{gl}_n(F)$, \begin{align*} [\lambda I_n,A] &=\lambda I_nA-A\lambda I_n \\ &=\lambda A-\lambda A \\ &=0. \end{align*} Hence $\mathfrak{gl}_n(F)=F I_n\oplus\mathfrak{sl}_n(F)$ as a direct sum of vector spaces, and the scalar-matrix summand commutes with all of $\mathfrak{gl}_n(F)$. [/example] The next two isomorphism theorems describe how subalgebras and ideals behave under passage to quotients. They are used repeatedly when comparing a Lie algebra with a quotient by a chosen ideal. [quotetheorem:3765] [citeproof:3765] The second isomorphism theorem compares a subalgebra with its image in a quotient. The ideal hypothesis on $I$ is what permits $\mathfrak a+I$ to inherit a bracket and permits the quotient by $I$ to be formed. Conceptually, the theorem says that the only part of $\mathfrak a$ lost when passing to $\mathfrak g/I$ is precisely $\mathfrak a\cap I$. There is a separate quotient problem when two ideals are nested. If $I\subset J$, one can first collapse $I$ and then collapse the remaining image of $J$, or collapse $J$ all at once. The next theorem explains why these two-stage and one-stage quotient procedures produce the same Lie algebra. [quotetheorem:3766] [citeproof:3766] The third isomorphism theorem is the algebraic rule for quotienting in stages. The containment $I\subset J$ is essential, since $J/I$ must first make sense as a subspace of $\mathfrak g/I$. In later structure arguments this allows a chain of ideals to be shortened or shifted to a quotient without changing the final quotient algebra. [quotetheorem:3767] [citeproof:3767] This correspondence prevents quotient algebras from hiding ideal structure: every ideal downstairs has a unique ideal upstairs lying above $I$. It also shows what the theorem does not do: it does not identify arbitrary subalgebras of $\mathfrak g/I$ with ideals of $\mathfrak g$, because ideal stability must be preserved. The result is the main bookkeeping tool for transferring composition series, solvable series, and nilpotent series through quotients. The isomorphism theorems repeatedly compare images, kernels, and quotients, so it is useful to have notation that records all three at once. Exactness is the condition that whatever arrives at one Lie algebra is exactly what the next map kills. This turns a chain of homomorphisms into a compact record of how one algebra is built from a subalgebra and a quotient. [definition: Exact Sequence Of Lie Algebras] A sequence of Lie algebra homomorphisms \begin{align*} \cdots \longrightarrow \mathfrak g_{i-1} \xrightarrow{\varphi_{i-1}} \mathfrak g_i \xrightarrow{\varphi_i} \mathfrak g_{i+1} \longrightarrow \cdots \end{align*} is exact at $\mathfrak g_i$ if $\operatorname{im}\varphi_{i-1}=\ker\varphi_i$. It is exact if it is exact at every displayed term. [/definition] A short exact sequence \begin{align*} 0 \longrightarrow I \xrightarrow{\iota} \mathfrak g \xrightarrow{\pi} \mathfrak q \longrightarrow 0 \end{align*} says that $I$ is identified with an ideal of $\mathfrak g$ and that $\mathfrak q \cong \mathfrak g/I$. The point of the notation is that it remembers both the quotient and the way it is realised inside a larger algebra. ## Direct Sums, Semidirect Products, and Split Extensions The next problem is constructive: given Lie algebras already understood, how can they be combined to make new ones? Direct sums model independent systems, while semidirect products model a Lie algebra acting on another by derivations. [definition: Direct Sum Of Lie Algebras] Let $\mathfrak h$ and $\mathfrak k$ be Lie algebras over $F$. Their direct sum $\mathfrak h \oplus \mathfrak k$ is the vector space direct sum equipped with the bracket map $(\mathfrak h\oplus\mathfrak k)\times(\mathfrak h\oplus\mathfrak k)\to\mathfrak h\oplus\mathfrak k$ determined by \begin{align*} [(h_1,k_1),(h_2,k_2)] := ([h_1,h_2],[k_1,k_2]). \end{align*} [/definition] In a direct sum, the two summands commute with each other: $[(h,0),(0,k)]=(0,0)$. Thus direct sums describe extensions with no interaction between the two pieces. [example: General Linear As A Direct Sum] Assume $\operatorname{char}F$ does not divide $n$, so $n\cdot 1_F$ is invertible in $F$. For every $A\in\mathfrak{gl}_n(F)$, set \begin{align*} A_0&=\frac{\operatorname{tr}(A)}{n}I_n,\\ A_1&=A-\frac{\operatorname{tr}(A)}{n}I_n. \end{align*} Then $A_0\in F I_n$, and \begin{align*} \operatorname{tr}(A_1) &=\operatorname{tr}\left(A-\frac{\operatorname{tr}(A)}{n}I_n\right)\\ &=\operatorname{tr}(A)-\frac{\operatorname{tr}(A)}{n}\operatorname{tr}(I_n)\\ &=\operatorname{tr}(A)-\frac{\operatorname{tr}(A)}{n}\,n\\ &=0, \end{align*} so $A_1\in\mathfrak{sl}_n(F)$. Hence every $A$ has the displayed decomposition \begin{align*} A=\frac{\operatorname{tr}(A)}{n}I_n+\left(A-\frac{\operatorname{tr}(A)}{n}I_n\right) \end{align*} with one summand in $F I_n$ and the other in $\mathfrak{sl}_n(F)$. The decomposition is unique because the intersection is zero. Indeed, if $\lambda I_n\in F I_n\cap\mathfrak{sl}_n(F)$, then \begin{align*} 0 &=\operatorname{tr}(\lambda I_n)\\ &=\lambda\operatorname{tr}(I_n)\\ &=\lambda n. \end{align*} Since $n$ is invertible in $F$, multiplying by $n^{-1}$ gives $\lambda=0$, so $\lambda I_n=0$. It remains to check that both summands are ideals. For $\lambda\in F$ and $B\in\mathfrak{gl}_n(F)$, \begin{align*} [\lambda I_n,B] &=\lambda I_nB-B\lambda I_n\\ &=\lambda B-\lambda B\\ &=0, \end{align*} so $F I_n$ is central and therefore an ideal. If $S\in\mathfrak{sl}_n(F)$ and $B=(b_{ij})$, $S=(s_{ij})$, then \begin{align*} \operatorname{tr}(BS) &=\sum_{i=1}^n(BS)_{ii}\\ &=\sum_{i=1}^n\sum_{j=1}^n b_{ij}s_{ji}\\ &=\sum_{j=1}^n\sum_{i=1}^n s_{ji}b_{ij}\\ &=\sum_{j=1}^n(SB)_{jj}\\ &=\operatorname{tr}(SB), \end{align*} and therefore \begin{align*} \operatorname{tr}([B,S]) &=\operatorname{tr}(BS-SB)\\ &=\operatorname{tr}(BS)-\operatorname{tr}(SB)\\ &=0. \end{align*} Thus $[B,S]\in\mathfrak{sl}_n(F)$, so $\mathfrak{sl}_n(F)$ is an ideal of $\mathfrak{gl}_n(F)$. Consequently \begin{align*} \mathfrak{gl}_n(F)=F I_n\oplus\mathfrak{sl}_n(F) \end{align*} as a direct sum of ideals, and the scalar-matrix summand commutes with the special linear summand. [/example] Direct sums have no interaction between the two summands, so they cannot model extensions where one part genuinely acts on the other. To build such an extension, a linear action on the first summand must respect its Lie bracket; otherwise the Jacobi identity fails for mixed triples. The compatibility condition is the Lie-algebra version of the Leibniz rule, and the next definition isolates the operators that can serve as infinitesimal bracket symmetries. [definition: Derivation Of A Lie Algebra] Let $\mathfrak h$ be a Lie algebra over $F$. A derivation of $\mathfrak h$ is a linear map $D:\mathfrak h\to\mathfrak h$ such that \begin{align*} D([x,y]) = [D(x),y] + [x,D(y)] \end{align*} for all $x,y\in\mathfrak h$. [/definition] Derivations should themselves be closed under the natural commutator of endomorphisms if they are to form the target of Lie algebra actions. The point is not only that each derivation preserves the bracket, but that taking the commutator of two such bracket-preserving operators still preserves the bracket. The next result verifies this closure and identifies derivations as a Lie algebra in their own right. [quotetheorem:3762] [citeproof:3762] This theorem explains why actions by derivations are ordinary Lie algebra homomorphisms into $\operatorname{Der}(\mathfrak h)$. The inner derivations form the canonical source of such actions, but outer derivations can also occur and lead to genuinely new semidirect products. The distinction matters later because extensions are controlled not just by subalgebras, but by how one part acts on another. [definition: Action By Derivations] Let $\mathfrak k$ and $\mathfrak h$ be Lie algebras. An action of $\mathfrak k$ on $\mathfrak h$ by derivations is a Lie algebra homomorphism \begin{align*} \rho: \mathfrak k \longrightarrow \operatorname{Der}(\mathfrak h). \end{align*} [/definition] Once such an action is fixed, the natural problem is to build a Lie algebra whose underlying vector space contains both $\mathfrak h$ and $\mathfrak k$ but whose mixed bracket records the action. A direct sum would forget the action entirely, so the bracket must be modified exactly in the cross terms. The following definition gives that construction. [definition: Semidirect Product] Let $\rho:\mathfrak k\to\operatorname{Der}(\mathfrak h)$ be an action by derivations. The semidirect product $\mathfrak h\rtimes_\rho\mathfrak k$ is the vector space $\mathfrak h\oplus\mathfrak k$ equipped with the bracket map $(\mathfrak h\oplus\mathfrak k)\times(\mathfrak h\oplus\mathfrak k)\to\mathfrak h\oplus\mathfrak k$ determined by \begin{align*} [(h_1,k_1),(h_2,k_2)] := ([h_1,h_2] + \rho(k_1)(h_2) - \rho(k_2)(h_1), [k_1,k_2]). \end{align*} [/definition] When $\rho=0$, the semidirect product reduces to the direct sum. In general $\mathfrak h$ is an ideal in $\mathfrak h\rtimes_\rho\mathfrak k$, while $\mathfrak k$ is a subalgebra but need not be an ideal. [quotetheorem:3769] [citeproof:3769] The theorem is the point at which the derivation and homomorphism hypotheses are both used. If $\rho(k)$ failed to be a derivation, the Jacobi identity involving two elements of $\mathfrak h$ and one element of $\mathfrak k$ would fail. If $\rho$ failed to preserve brackets, the Jacobi identity involving one element of $\mathfrak h$ and two elements of $\mathfrak k$ would fail, so semidirect products encode exactly the permitted interaction. [example: Affine Lie Algebra Of The Line] Let $\mathfrak h=F e$ and $\mathfrak k=F d$, both with zero bracket. Define the linear map $\rho(d):\mathfrak h\to\mathfrak h$ by $\rho(d)(e)=e$, and extend linearly, so $\rho(\lambda d)(\mu e)=\lambda\mu e$. Since $\mathfrak h$ is abelian, for $a e,b e\in\mathfrak h$, \begin{align*} \rho(d)([a e,b e]) &=\rho(d)(0)\\ &=0, \end{align*} while \begin{align*} [\rho(d)(a e),b e]+[a e,\rho(d)(b e)] &=[a e,b e]+[a e,b e]\\ &=0+0\\ &=0. \end{align*} Thus $\rho(d)$ is a derivation of $\mathfrak h$. Also $\mathfrak k$ is abelian, so for $\lambda d,\mu d\in\mathfrak k$, \begin{align*} \rho([\lambda d,\mu d]) &=\rho(0)\\ &=0, \end{align*} and \begin{align*} [\rho(\lambda d),\rho(\mu d)] &=\rho(\lambda d)\rho(\mu d)-\rho(\mu d)\rho(\lambda d)\\ &=\lambda\mu\,\rho(d)^2-\mu\lambda\,\rho(d)^2\\ &=0. \end{align*} Hence $\rho:\mathfrak k\to\operatorname{Der}(\mathfrak h)$ is an action by derivations. In the semidirect product $\mathfrak h\rtimes_\rho\mathfrak k$, the underlying vector space has basis $(e,0),(0,d)$, which we write as $e,d$. The bracket is computed from the semidirect product formula: \begin{align*} [d,e] &=[(0,d),(e,0)]\\ &=\bigl([0,e]+\rho(d)(e)-\rho(0)(0),[d,0]\bigr)\\ &=(0+e-0,0)\\ &=e. \end{align*} Similarly, \begin{align*} [e,d] &=[(e,0),(0,d)]\\ &=\bigl([e,0]+\rho(0)(0)-\rho(d)(e),[0,d]\bigr)\\ &=(0+0-e,0)\\ &=-e. \end{align*} The brackets $[e,e]$ and $[d,d]$ are zero because both one-dimensional summands are abelian and the mixed action only appears when one input has a $\mathfrak k$-component and the other has a $\mathfrak h$-component. Thus this two-dimensional Lie algebra is determined by $[d,e]=e$, and it is the affine Lie algebra of the line, often written $F\rtimes F$. [/example] Semidirect products are most useful when they arise from an extension problem: a Lie algebra may contain an ideal and have a quotient, but still need extra data to identify the quotient inside the original algebra. A splitting is precisely the choice of such an internal copy of the quotient, compatible with the Lie bracket. This language turns the construction above into a recognition criterion for extensions. [definition: Split Short Exact Sequence] A short exact sequence \begin{align*} 0 \longrightarrow \mathfrak h \xrightarrow{\iota} \mathfrak g \xrightarrow{\pi} \mathfrak k \longrightarrow 0 \end{align*} is split if there is a Lie algebra homomorphism $s:\mathfrak k\to\mathfrak g$ such that $\pi\circ s=\operatorname{id}_{\mathfrak k}$. [/definition] With a splitting in hand, every element of $\mathfrak g$ can be compared to a piece from the ideal and a piece from the chosen copy of the quotient. The remaining question is whether the Lie bracket on $\mathfrak g$ is then forced by the induced action of that copy on the ideal. The next result states the precise equivalence with a semidirect product. [quotetheorem:3770] [citeproof:3770] This universal property says that the formula for the semidirect product is forced by the data of two subalgebras and their mixed bracket. The uniqueness condition is important: once the two summands and their interaction are fixed, there is no further freedom in the Lie algebra structure. This is the practical test used to recognise a semidirect product without rebuilding it from the definition. For short exact sequences, the recognition question becomes more concrete. A splitting chooses representatives of the quotient inside the middle Lie algebra, but it is not automatic that those representatives interact with the ideal by the semidirect product bracket. The obstruction is whether the chosen copy of the quotient is itself a subalgebra and whether the original bracket is fully recovered from its action on the kernel. The result below identifies exactly when this happens. [quotetheorem:3771] [citeproof:3771] The theorem also indicates what can go wrong for a nonsplit extension: there may be no subalgebra of $\mathfrak g$ mapping isomorphically onto $\mathfrak k$. In that case the quotient still exists, but the elements of $\mathfrak k$ cannot be chosen inside $\mathfrak g$ in a bracket-compatible way. Semidirect products therefore model split extensions, not all extensions. ## Simple, Indecomposable, and Representation-Theoretic Irreducibility The last problem in the chapter is to separate three notions that sound similar but measure different things. A Lie algebra may have no nonzero proper ideals, may fail to split as a direct sum of ideals, or may act on a vector space with no nonzero proper invariant subspaces. These are related only under extra hypotheses. [definition: Simple Lie Algebra] A Lie algebra $\mathfrak g$ is simple if $\mathfrak g$ is non-abelian and its only ideals are $0$ and $\mathfrak g$. [/definition] The non-abelian condition is part of the standard convention. Without it, every one-dimensional abelian Lie algebra would count as simple, which would obscure the structural role of simple Lie algebras as noncommutative building blocks. A different decomposition question ignores whether ideals exist at all and asks whether the whole algebra can be split into two independent ideal summands. This is weaker than simplicity, because an algebra may have ideals but still have no complementary ideal decomposition. The next definition names that weaker obstruction to splitting. [definition: Indecomposable Lie Algebra] A nonzero Lie algebra $\mathfrak g$ is indecomposable if there do not exist nonzero ideals $I,J\trianglelefteq\mathfrak g$ such that \begin{align*} \mathfrak g = I \oplus J \end{align*} as a direct sum of Lie algebras. [/definition] The two definitions now invite a comparison: having no nonzero proper ideals should prevent any nontrivial direct-sum decomposition into ideals. The converse is much less plausible, since an ideal can exist without having an ideal complement. The next result isolates the one direction that always holds. [quotetheorem:3772] [citeproof:3772] This theorem uses simplicity only through the absence of nonzero proper ideals. It does not say that every indecomposable algebra is simple, because failure to split as a direct sum is weaker than having no ideals at all. The next example is the standard warning: a small solvable algebra can have an ideal and still admit no complementary ideal. [example: Affine Algebra Is Indecomposable But Not Simple] Let $\mathfrak g=F e\rtimes F d$ with basis $e,d$ and bracket determined by $[d,e]=e$. Since the bracket is skew-symmetric, \begin{align*} [e,d]&=-[d,e]\\ &=-e, \end{align*} and also $[e,e]=[d,d]=0$. The line $F e$ is an ideal: for $\lambda,\mu,\nu\in F$, \begin{align*} [\mu e+\nu d,\lambda e] &=\mu\lambda[e,e]+\nu\lambda[d,e]\\ &=0+\nu\lambda e\\ &\in F e. \end{align*} Thus $\mathfrak g$ has a nonzero proper ideal, so it is not simple. We now show that $\mathfrak g$ cannot be written as a direct sum of two nonzero ideals. Since $\dim \mathfrak g=2$, such a decomposition would have the form \begin{align*} \mathfrak g=I\oplus J \end{align*} where $I$ and $J$ are one-dimensional ideals. Let $L=F(\alpha d+\beta e)$ be any one-dimensional ideal. Because $L$ is an ideal, the bracket with $e\in\mathfrak g$ must remain in $L$: \begin{align*} [e,\alpha d+\beta e] &=\alpha[e,d]+\beta[e,e]\\ &=-\alpha e+0\\ &=-\alpha e. \end{align*} If $\alpha\ne 0$, then $e\in L$, so $\alpha d+\beta e$ must be a scalar multiple of $e$, which forces $\alpha=0$, a contradiction. Hence $\alpha=0$, and therefore every one-dimensional ideal is $F e$. Consequently there are not two distinct nonzero one-dimensional ideals whose direct sum is $\mathfrak g$. The affine Lie algebra of the line is therefore indecomposable, even though it is not simple. [/example] The example shows that ideals describe internal structure of the Lie algebra itself. Representation theory shifts attention to a second vector space on which the Lie algebra acts, so the relevant subspaces are subspaces of that representation space rather than subspaces of the algebra. To keep these two notions separate, the next definition introduces representations together with their invariant subspaces. [definition: Representation And Invariant Subspace] A representation of a Lie algebra $\mathfrak g$ on a vector space $V$ is a Lie algebra homomorphism \begin{align*} \rho:\mathfrak g\to\mathfrak{gl}(V). \end{align*} A subspace $W\subset V$ is invariant under the representation if $\rho(x)(W)\subset W$ for all $x\in\mathfrak g$. [/definition] Once invariant subspaces have been identified, the analogue of a building block is a representation that has no smaller nonzero invariant part. This condition depends on the chosen action, not just on the abstract Lie algebra. The following definition gives the representation-theoretic version of irreducibility. [definition: Irreducible Representation] A representation $\rho:\mathfrak g\to\mathfrak{gl}(V)$ is irreducible if $V\ne 0$ and the only invariant subspaces of $V$ are $0$ and $V$. [/definition] Simplicity is a statement about the internal ideal structure of $\mathfrak g$. Irreducibility is a statement about a chosen action of $\mathfrak g$ on a chosen vector space $V$. There is one representation where these languages meet directly: the adjoint representation, in which $\mathfrak g$ acts on itself by brackets. In that case, asking for invariant subspaces is exactly asking for ideals, so simplicity can be translated into irreducibility of this particular action. The next theorem makes that bridge precise. [quotetheorem:3773] [citeproof:3773] The theorem is the precise bridge between the internal and representation-theoretic viewpoints. It also marks the limitation of that bridge: ideals are invariant subspaces only for the adjoint representation, not for an arbitrary representation on a different vector space. This distinction prevents a common mistake, namely trying to read simplicity of $\mathfrak g$ directly from reducibility properties of every representation. [example: A Simple Algebra May Have Reducible Representations] Let $\mathfrak g$ be any simple Lie algebra, and set $V=F\oplus F$. Define $\rho:\mathfrak g\to\mathfrak{gl}(V)$ by \begin{align*} \rho(x)=0 \end{align*} for every $x\in\mathfrak g$, where $0$ denotes the zero endomorphism of $V$. This is a Lie algebra homomorphism: for $x,y\in\mathfrak g$, \begin{align*} \rho([x,y]) &=0, \end{align*} and \begin{align*} [\rho(x),\rho(y)] &=\rho(x)\rho(y)-\rho(y)\rho(x)\\ &=0\circ 0-0\circ 0\\ &=0-0\\ &=0. \end{align*} Thus $\rho([x,y])=[\rho(x),\rho(y)]$, so $\rho$ is a representation of $\mathfrak g$ on $V$. Now take the one-dimensional subspace \begin{align*} W=F\oplus 0=\{(a,0):a\in F\}\subset V. \end{align*} It is nonzero because $(1,0)\in W$, and it is proper because $(0,1)\in V$ but $(0,1)\notin W$. For every $x\in\mathfrak g$ and every $(a,0)\in W$, \begin{align*} \rho(x)(a,0) &=0(a,0)\\ &=(0,0)\\ &\in W. \end{align*} Hence $W$ is a nonzero proper invariant subspace of $V$, so the zero representation on $F\oplus F$ is reducible. Therefore simplicity of $\mathfrak g$ does not imply that every representation of $\mathfrak g$ is irreducible. [/example] The next example returns from representation-theoretic reducibility to quotient constructions. Triangular matrix algebras provide a concrete place where an ideal removes precisely the strictly upper triangular part and leaves an abelian diagonal quotient. [example: Quotients Of Triangular Matrix Algebras] Let $\mathfrak b\subset\mathfrak{gl}_n(F)$ be the Lie algebra of upper triangular matrices, and let $\mathfrak n\subset\mathfrak b$ be the subspace of strictly upper triangular matrices. First we check that $\mathfrak n$ is an ideal of $\mathfrak b$. If $B=(b_{ij})\in\mathfrak b$ and $N=(n_{ij})\in\mathfrak n$, then for $i\geq j$, \begin{align*} (BN)_{ij} &=\sum_{k=1}^n b_{ik}n_{kj}. \end{align*} A summand $b_{ik}n_{kj}$ can be nonzero only if $i\leq k$ because $B$ is upper triangular, and only if $k<j$ because $N$ is strictly upper triangular. These inequalities would give $i\leq k<j$, contradicting $i\geq j$. Hence $(BN)_{ij}=0$ for all $i\geq j$. Similarly, \begin{align*} (NB)_{ij} &=\sum_{k=1}^n n_{ik}b_{kj}, \end{align*} and a nonzero summand would require $i<k$ and $k\leq j$, which is impossible when $i\geq j$. Thus $(NB)_{ij}=0$ for all $i\geq j$. Therefore \begin{align*} [B,N]_{ij} &=(BN-NB)_{ij}\\ &=(BN)_{ij}-(NB)_{ij}\\ &=0-0\\ &=0 \end{align*} for all $i\geq j$, so $[B,N]\in\mathfrak n$. Thus $\mathfrak n\trianglelefteq\mathfrak b$. Let $\mathfrak d$ be the Lie algebra of diagonal matrices, and define \begin{align*} \delta:\mathfrak b\to\mathfrak d, \qquad \delta(A)=\operatorname{diag}(a_{11},\ldots,a_{nn}) \end{align*} for $A=(a_{ij})$. If $A=(a_{ij})$ and $B=(b_{ij})$ are upper triangular, then \begin{align*} (AB)_{ii} &=\sum_{k=1}^n a_{ik}b_{ki}\\ &=a_{ii}b_{ii}, \end{align*} because $a_{ik}b_{ki}$ can be nonzero only when $i\leq k$ and $k\leq i$, hence only when $k=i$. Likewise, \begin{align*} (BA)_{ii} &=b_{ii}a_{ii}. \end{align*} Since entries lie in the field $F$, \begin{align*} ([A,B])_{ii} &=(AB-BA)_{ii}\\ &=a_{ii}b_{ii}-b_{ii}a_{ii}\\ &=0. \end{align*} Thus $\delta([A,B])=0$. Diagonal matrices commute with each other, since for $D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$ and $E=\operatorname{diag}(\mu_1,\ldots,\mu_n)$, \begin{align*} (DE)_{ii}&=\lambda_i\mu_i,\\ (ED)_{ii}&=\mu_i\lambda_i,\\ (DE-ED)_{ii}&=0, \end{align*} and all off-diagonal entries are already zero. Hence \begin{align*} \delta([A,B]) &=0\\ &=[\delta(A),\delta(B)], \end{align*} so $\delta$ is a Lie algebra homomorphism. The kernel of $\delta$ is exactly $\mathfrak n$: \begin{align*} \ker\delta &=\{A\in\mathfrak b:\delta(A)=0\}\\ &=\{A\in\mathfrak b:a_{11}=\cdots=a_{nn}=0\}\\ &=\mathfrak n. \end{align*} The map is surjective because every diagonal matrix belongs to $\mathfrak b$ and maps to itself under $\delta$. Therefore $\delta$ induces a bijective map \begin{align*} \overline{\delta}:\mathfrak b/\mathfrak n\to\mathfrak d, \qquad \overline{\delta}(A+\mathfrak n)=\delta(A). \end{align*} It is well defined because if $A+\mathfrak n=A'+\mathfrak n$, then $A-A'\in\mathfrak n=\ker\delta$, and hence \begin{align*} \delta(A)-\delta(A') &=\delta(A-A')\\ &=0. \end{align*} Finally, for $A,B\in\mathfrak b$, \begin{align*} \overline{\delta}([A+\mathfrak n,B+\mathfrak n]) &=\overline{\delta}([A,B]+\mathfrak n)\\ &=\delta([A,B])\\ &=0\\ &=[\delta(A),\delta(B)]\\ &=[\overline{\delta}(A+\mathfrak n),\overline{\delta}(B+\mathfrak n)]. \end{align*} Thus $\mathfrak b/\mathfrak n$ is identified with the diagonal Lie algebra $\mathfrak d$, which is abelian. This quotient keeps only the diagonal part of an upper triangular matrix and discards the strictly upper triangular part. [/example] The constructions in this chapter will be used throughout the course without further comment. Quotients isolate the part of a Lie algebra visible after an ideal has been collapsed, exact sequences describe extensions, and semidirect products give the standard model for split extensions. The distinction between ideals and invariant subspaces also sets up the later use of representations: representation theory detects structure, but its subspaces live in the module being acted on, not inside the Lie algebra unless the representation is the adjoint one. Once ideals and quotients are available, the focus naturally moves outward from the algebra itself to its actions. Representations let us study brackets through linear operators, linking the internal structure developed so far to the module-theoretic viewpoint used in the rest of the course. # 3. Representations and Modules Representations let a Lie algebra act by linear transformations, so they turn the abstract bracket into operators on vector spaces. The preceding chapters introduced Lie algebras through internal constructions such as ideals, quotients, and derived algebras; this chapter adds the external viewpoint. The guiding question is how much of a Lie algebra can be understood from the modules on which it acts. ## Actions by Linear Operators How should a Lie algebra act on a vector space without forgetting the Lie bracket? Since $\mathfrak{gl}(V)$ is itself a Lie algebra under the commutator bracket, an action should send brackets in $\mathfrak g$ to commutators of linear maps. [definition: Representation] Let $\mathfrak g$ be a Lie algebra over the course field $F$. A representation of $\mathfrak g$ on an $F$-vector space $V$ is a Lie algebra homomorphism \begin{align*} \rho: \mathfrak g \to \mathfrak{gl}(V). \end{align*} The vector space $V$ together with $\rho$ is called a $\mathfrak g$-module. [/definition] Writing $xv$ for $\rho(x)(v)$, the homomorphism condition becomes \begin{align*} [x,y]v = x(yv)-y(xv) \end{align*} for all $x,y \in \mathfrak g$ and $v \in V$. This formula is often the most efficient way to check that a proposed action is a representation. [example: Natural Representation Of Gl N] Let $V=F^n$ and define \begin{align*} \rho:\mathfrak{gl}_n(F)\to \mathfrak{gl}(F^n),\qquad \rho(A)(v)=Av. \end{align*} We show that $\rho$ is a Lie algebra homomorphism. For $A,B\in \mathfrak{gl}_n(F)$, $c\in F$, and $v\in F^n$, \begin{align*} \rho(A+cB)(v) &=(A+cB)v \\ &=Av+cBv \\ &=\rho(A)(v)+c\rho(B)(v), \end{align*} so $\rho(A+cB)=\rho(A)+c\rho(B)$ and $\rho$ is linear. The bracket in $\mathfrak{gl}_n(F)$ is the matrix commutator, so $[A,B]=AB-BA$. Hence \begin{align*} \rho([A,B])(v) &=\rho(AB-BA)(v) \\ &=(AB-BA)v \\ &=ABv-BAv \\ &=A(Bv)-B(Av) \\ &=\rho(A)(\rho(B)(v))-\rho(B)(\rho(A)(v)) \\ &=(\rho(A)\rho(B)-\rho(B)\rho(A))(v) \\ &=[\rho(A),\rho(B)](v). \end{align*} Since this holds for every $v\in F^n$, we have $\rho([A,B])=[\rho(A),\rho(B)]$. Thus ordinary matrix multiplication is a representation of $\mathfrak{gl}_n(F)$ on $F^n$. [/example] The natural representation is faithful: different matrices give different linear transformations of $k^n$. Many later structure arguments ask whether an arbitrary Lie algebra can also be realised as a concrete matrix algebra. [definition: Submodule] Let $V$ be a $\mathfrak g$-module. A subspace $W \subset V$ is a $\mathfrak g$-submodule if \begin{align*} xw \in W \end{align*} for all $x \in \mathfrak g$ and $w \in W$. [/definition] Submodules are the representation-theoretic analogue of ideals: they are exactly the subspaces on which the action restricts. After restricting an action to a stable subspace, the complementary operation is to collapse that subspace and study what action remains on the quotient vector space. The stability condition is what makes the formula $x(v+W)=xv+W$ meaningful, and the next definition records this quotient construction. [definition: Quotient Module] Let $V$ be a $\mathfrak g$-module with representation $\rho: \mathfrak g \to \mathfrak{gl}(V)$, and let $W \subset V$ be a $\mathfrak g$-submodule. The quotient module $V/W$ is the vector space quotient with representation \begin{align*} \bar{\rho}: \mathfrak g \to \mathfrak{gl}(V/W), \qquad \bar{\rho}(x)(v+W)=\rho(x)(v)+W. \end{align*} [/definition] The condition that $W$ is a submodule is exactly what makes this action independent of the representative $v$. This is the first place where module language pays for itself: many arguments become statements about invariant subspaces and their quotients. With submodules and quotient modules available, one can ask when a representation has been decomposed as far as the action allows. The obstruction to indivisibility is not an arbitrary vector subspace, but a proper nonzero subspace stable under every element of the Lie algebra. A module with no such stable piece cannot be simplified by passing to a smaller representation, so it serves as a basic representation-theoretic building block. [definition: Irreducible Module] A non-zero $\mathfrak g$-module $V$ is irreducible if its only $\mathfrak g$-submodules are $0$ and $V$. [/definition] Irreducible modules describe the pieces one hopes to find, but a module may contain irreducible submodules without splitting cleanly into them. The remaining obstruction is extension data: an invariant subspace may exist without an invariant complement, so the module is glued together rather than assembled as independent summands. Complete reducibility is the condition that this obstruction is absent. [definition: Completely Reducible Module] A $\mathfrak g$-module $V$ is completely reducible if it is a direct sum of irreducible $\mathfrak g$-submodules. [/definition] Complete reducibility is a strong condition. For finite-dimensional modules it is equivalent to saying that every submodule has a complementary submodule, but that characterisation is usually proved as a separate proposition in a module theory course. [example: Reducibility For A Triangular Action] Let $\mathfrak b$ be the Lie algebra of matrices \begin{align*} A=\begin{pmatrix}a&b\\0&d\end{pmatrix} \end{align*} acting on $F^2$ by ordinary matrix multiplication, with basis $e_1,e_2$. For $t\in F$, \begin{align*} A(te_1) &= \begin{pmatrix}a&b\\0&d\end{pmatrix} \begin{pmatrix}t\\0\end{pmatrix} = \begin{pmatrix}at\\0\end{pmatrix} =at e_1 \in Fe_1. \end{align*} Thus $Fe_1$ is a non-zero proper $\mathfrak b$-submodule of $F^2$, so the natural module is reducible. The quotient $F^2/Fe_1$ is spanned by $e_2+Fe_1$. Since \begin{align*} A e_2 &= \begin{pmatrix}a&b\\0&d\end{pmatrix} \begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}b\\d\end{pmatrix} =be_1+de_2, \end{align*} we get \begin{align*} A(e_2+Fe_1) &=Ae_2+Fe_1 \\ &=(be_1+de_2)+Fe_1 \\ &=de_2+Fe_1 \\ &=d(e_2+Fe_1). \end{align*} So the quotient is one-dimensional. There is no $\mathfrak b$-[stable complement](/theorems/2279) to $Fe_1$. Indeed, any one-dimensional complement has the form $L=F(\alpha e_1+\beta e_2)$ with $\beta\ne 0$. For \begin{align*} N=\begin{pmatrix}0&1\\0&0\end{pmatrix}\in \mathfrak b, \end{align*} we have \begin{align*} N(\alpha e_1+\beta e_2) &= \begin{pmatrix}0&1\\0&0\end{pmatrix} \begin{pmatrix}\alpha\\\beta\end{pmatrix} = \begin{pmatrix}\beta\\0\end{pmatrix} =\beta e_1. \end{align*} If $L$ were stable, then $\beta e_1\in L$, hence $e_1\in L$, forcing $L=Fe_1$, which is not a complement. Thus this module has an invariant flag $0\subset Fe_1\subset F^2$ even though the first non-zero submodule has no invariant complement. [/example] ## Standard Constructions Of Modules Which modules appear without making new choices? Several constructions are forced by linear algebra: adjoints, zero actions, duals, direct sums, tensor products, and Hom spaces. These are used throughout structure theory, especially when passing between a Lie algebra and its endomorphism spaces. [definition: Zero Action Module] Let $\mathfrak g$ be a Lie algebra and let $V$ be a $k$-vector space. The zero action module on $V$ is the representation \begin{align*} 0_V: \mathfrak g \to \mathfrak{gl}(V), \qquad 0_V(x)(v)=0 \end{align*} for all $x \in \mathfrak g$ and $v \in V$. [/definition] Modules with zero action measure invariant vectors and scalar pieces of a representation. They also appear as quotients whenever the action factors through a zero action. A different canonical construction comes from the Lie algebra acting on itself. The bracket already assigns to each element $x$ a linear operator $y\mapsto [x,y]$, and the question is whether these operators form a representation. This self-action is the basic way a Lie algebra exposes its commutator structure as module structure. [definition: Adjoint Module] Let $\mathfrak g$ be a Lie algebra. The adjoint module is the underlying vector space of $\mathfrak g$ with representation \begin{align*} \operatorname{ad}: \mathfrak g \to \mathfrak{gl}(\mathfrak g), \qquad \operatorname{ad}_x(y)=[x,y]. \end{align*} [/definition] The adjoint construction is only valid if the assignment $x\mapsto \operatorname{ad}_x$ respects the Lie bracket, rather than merely producing unrelated endomorphisms of the vector space $\mathfrak g$. This is exactly where the Jacobi identity enters: it converts the internal bracket of $\mathfrak g$ into the commutator bracket in $\mathfrak{gl}(\mathfrak g)$. Before using the adjoint action as a module, we must verify that this compatibility is not just suggestive notation. The point of the next result is to certify that every Lie algebra carries this canonical representation, so later structure theorems can study $\mathfrak g$ through its own commutator operators. [quotetheorem:3759] [citeproof:3759] The theorem has no hidden finite-dimensionality or field hypothesis: it is exactly the Jacobi identity written in operator form. Its kernel is the centre $Z(\mathfrak g)$, so the adjoint module is faithful precisely when $Z(\mathfrak g)=0$; for an abelian Lie algebra, the adjoint representation is the zero representation and therefore loses all elements. This limitation is important because the adjoint module measures commutators, not the whole Lie algebra in the presence of central elements. It will later become central to Engel's theorem, Cartan's criterion, and the Killing form. [example: Adjoint Representation Of Sl Two] Let $\mathfrak{sl}_2(k)$ have basis $e,f,h$ with \begin{align*} [h,e]=2e,\qquad [h,f]=-2f,\qquad [e,f]=h. \end{align*} For the adjoint module, $\operatorname{ad}_x(y)=[x,y]$. On the ordered basis $e,h,f$, the action of $h$ is \begin{align*} \operatorname{ad}_h(e)&=[h,e]=2e,\\ \operatorname{ad}_h(h)&=[h,h]=0,\\ \operatorname{ad}_h(f)&=[h,f]=-2f. \end{align*} Thus \begin{align*} [\operatorname{ad}_h]_{(e,h,f)} = \begin{pmatrix} 2&0&0\\ 0&0&0\\ 0&0&-2 \end{pmatrix}, \end{align*} so $e,h,f$ are eigenvectors for $\operatorname{ad}_h$ with weights $2,0,-2$ respectively. The raising and lowering operators are read from the same brackets, using antisymmetry of the Lie bracket: \begin{align*} \operatorname{ad}_e(e)&=[e,e]=0,\\ \operatorname{ad}_e(h)&=[e,h]=-[h,e]=-2e,\\ \operatorname{ad}_e(f)&=[e,f]=h, \end{align*} and \begin{align*} \operatorname{ad}_f(e)&=[f,e]=-[e,f]=-h,\\ \operatorname{ad}_f(h)&=[f,h]=-[h,f]=2f,\\ \operatorname{ad}_f(f)&=[f,f]=0. \end{align*} Therefore \begin{align*} [\operatorname{ad}_e]_{(e,h,f)} &= \begin{pmatrix} 0&-2&0\\ 0&0&1\\ 0&0&0 \end{pmatrix}, & [\operatorname{ad}_f]_{(e,h,f)} &= \begin{pmatrix} 0&0&0\\ -1&0&0\\ 0&2&0 \end{pmatrix}. \end{align*} So the adjoint module decomposes into the three displayed $h$-weight lines, while $\operatorname{ad}_e$ moves $f$ toward $h$ and $h$ toward $e$, and $\operatorname{ad}_f$ moves $e$ toward $h$ and $h$ toward $f$. [/example] The adjoint module is only one construction; the next operations build new modules functorially from modules already in hand. For a dual space, the subtle point is that an action on vectors must be transferred to linear functionals without reversing the Lie bracket. The resulting formula has to make evaluation compatible with the original action, which forces a sign convention. [definition: Dual Module] Let $V$ be a $\mathfrak g$-module with representation $\rho: \mathfrak g \to \mathfrak{gl}(V)$. The dual module $V^*$ is the dual vector space with representation \begin{align*} \rho^*: \mathfrak g \to \mathfrak{gl}(V^*), \qquad (\rho^*(x)f)(v)=-f(\rho(x)(v)) \end{align*} for all $x \in \mathfrak g$, $f \in V^*$, and $v \in V$. [/definition] The minus sign is forced by the commutator convention. It ensures that the action on $V^*$ is again a Lie algebra representation rather than an anti-representation. Representations also need a way to combine independent pieces without making them interact. If each summand already carries a $\mathfrak g$-action, the natural construction lets $\mathfrak g$ act componentwise so that stability and decomposition can be studied one summand at a time. [definition: Direct Sum Module] Let $(V_i)_{i \in I}$ be a family of $\mathfrak g$-modules with representations $\rho_i: \mathfrak g \to \mathfrak{gl}(V_i)$. The direct sum $\bigoplus_{i \in I} V_i$ is the $\mathfrak g$-module with representation \begin{align*} \rho_{\oplus}: \mathfrak g \to \mathfrak{gl}\left(\bigoplus_{i \in I} V_i\right), \qquad \rho_{\oplus}(x)((v_i)_{i \in I})=(\rho_i(x)v_i)_{i \in I}. \end{align*} [/definition] Direct sums formalise decomposing a representation into independent invariant pieces. Complete reducibility is exactly the assertion that the module is built this way from irreducibles. Some constructions should combine two representations into one whose vectors are bilinear combinations of vectors from both modules. The issue is that a Lie algebra element should differentiate both factors, since it represents an infinitesimal action. This leads to the same Leibniz rule that governs derivations. [definition: Tensor Product Module] Let $V$ and $W$ be $\mathfrak g$-modules with representations $\rho_V: \mathfrak g \to \mathfrak{gl}(V)$ and $\rho_W: \mathfrak g \to \mathfrak{gl}(W)$. The [tensor product](/page/Tensor%20Product) $V \otimes_k W$ is the $\mathfrak g$-module with representation \begin{align*} \rho_{V \otimes W}: \mathfrak g \to \mathfrak{gl}(V \otimes_k W), \qquad \rho_{V \otimes W}(x)(v \otimes w)=\rho_V(x)v \otimes w+v \otimes \rho_W(x)w. \end{align*} [/definition] This is the Leibniz rule for Lie algebra actions. It is compatible with the idea that elements of $\mathfrak g$ act as infinitesimal symmetries rather than as group elements. The same philosophy should apply not only to tensor products but also to spaces of maps. To make equivariant maps visible as invariant vectors, we need a $\mathfrak g$-action on $\operatorname{Hom}_k(V,W)$ that measures the failure of a linear map to commute with the two module structures. The natural action on a linear map compares acting after applying the map with applying the map after acting on the input. [definition: Hom Module] Let $V$ and $W$ be $\mathfrak g$-modules with representations $\rho_V: \mathfrak g \to \mathfrak{gl}(V)$ and $\rho_W: \mathfrak g \to \mathfrak{gl}(W)$. The vector space $\operatorname{Hom}_k(V,W)$ is the $\mathfrak g$-module with representation \begin{align*} \rho_{\operatorname{Hom}}: \mathfrak g \to \mathfrak{gl}(\operatorname{Hom}_k(V,W)), \qquad (\rho_{\operatorname{Hom}}(x)T)(v)=\rho_W(x)(T(v))-T(\rho_V(x)(v)). \end{align*} [/definition] The invariant vectors of this Hom module are precisely the $\mathfrak g$-module homomorphisms from $V$ to $W$. This observation is the entry point to [Schur's lemma](/theorems/2414). ## Schur's Lemma And Irreducible Modules What can a linear map between irreducible modules look like if it commutes with the Lie algebra action? Irreducibility forces any non-zero intertwining map to be as close to invertible as possible. Over an algebraically closed field, endomorphisms of a finite-dimensional irreducible module are only scalars. [definition: Module Homomorphism] Let $V$ and $W$ be $\mathfrak g$-modules. A $k$-linear map $T:V \to W$ is a $\mathfrak g$-module homomorphism if \begin{align*} T(xv)=xT(v) \end{align*} for all $x \in \mathfrak g$ and $v \in V$. [/definition] The kernel and image of a module homomorphism are submodules. For irreducible modules, that leaves very few possibilities: a nonzero map cannot have a proper nonzero kernel, and its image cannot be a proper nonzero submodule of the target. The theorem below turns this submodule obstruction into the basic rigidity statement for maps between irreducible representations. [quotetheorem:3824] [citeproof:3824] Schur's lemma is a representation-theoretic rigidity statement, and its hypotheses are doing real work. Finite-dimensionality is used to guarantee an eigenvalue once $k$ is algebraically closed; without algebraic closedness the endomorphism ring can be larger than the scalars, for example the real irreducible module $\mathbb R^2$ for the abelian Lie algebra $\mathbb R$ acting by the rotation matrix has commuting endomorphisms isomorphic to $\mathbb C$. The lemma also does not classify irreducible modules; it only describes maps after irreducibility is already known. It will later allow the course to identify central actions and control decompositions of semisimple modules. [example: Endomorphisms Of The Natural Sl Two Module] Let $V=k^2$ have basis $v_1,v_2$, with the standard action \begin{align*} hv_1=v_1,\qquad hv_2=-v_2,\qquad ev_1=0,\qquad ev_2=v_1,\qquad fv_1=v_2,\qquad fv_2=0. \end{align*} Let $T:V\to V$ be a $\mathfrak{sl}_2(k)$-module endomorphism, so $T(xv)=xT(v)$ for $x\in \mathfrak{sl}_2(k)$ and $v\in V$. Write \begin{align*} T(v_1)=av_1+bv_2,\qquad T(v_2)=cv_1+dv_2. \end{align*} Commuting with the action of $h$ on $v_1$ gives \begin{align*} hT(v_1)&=T(hv_1)\\ h(av_1+bv_2)&=T(v_1)\\ av_1-bv_2&=av_1+bv_2, \end{align*} so $-b=b$, hence $2b=0$. Since $\operatorname{char} k\ne 2$, this gives $b=0$. Commuting with $h$ on $v_2$ gives \begin{align*} hT(v_2)&=T(hv_2)\\ h(cv_1+dv_2)&=T(-v_2)\\ cv_1-dv_2&=-cv_1-dv_2, \end{align*} so $c=-c$, hence $2c=0$, and again $\operatorname{char} k\ne 2$ gives $c=0$. Therefore \begin{align*} T(v_1)=av_1,\qquad T(v_2)=dv_2. \end{align*} Now commute with $e$ on $v_2$: \begin{align*} T(ev_2)&=eT(v_2)\\ T(v_1)&=e(dv_2)\\ av_1&=dv_1. \end{align*} Thus $a=d$. Equivalently, commuting with $f$ on $v_1$ gives the same equality: \begin{align*} T(fv_1)&=fT(v_1)\\ T(v_2)&=f(av_1)\\ dv_2&=av_2. \end{align*} Hence \begin{align*} T(v_1)=av_1,\qquad T(v_2)=av_2, \end{align*} so $T=a\operatorname{id}_V$. Thus every endomorphism of the natural $\mathfrak{sl}_2(k)$-module is scalar. [/example] ## The Standard Module For Sl Two Why is $\mathfrak{sl}_2$ the test case for so much representation theory? Its two-dimensional natural module already shows the interaction between weights, raising and lowering operators, and irreducibility. This small example is the model for many later arguments in semisimple Lie theory. [definition: Standard Sl Two Module] Let $k$ be a field and let $V=k^2$ with basis $v_1,v_2$. The standard $\mathfrak{sl}_2(k)$-module is the representation \begin{align*} \rho: \mathfrak{sl}_2(k) \to \mathfrak{gl}(V), \qquad \rho(A)(v)=Av, \end{align*} where $A$ acts on the column vector $v \in k^2$ by matrix multiplication. [/definition] Using the usual basis \begin{align*} e=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad f=\begin{pmatrix}0&0\\1&0\end{pmatrix},\qquad h=\begin{pmatrix}1&0\\0&-1\end{pmatrix}, \end{align*} the action satisfies \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*} The key structural question is whether this two-dimensional module contains a stable line. A line spanned by $v_1$ is moved by $f$, while a line spanned by $v_2$ is moved by $e$, and a general line must survive all three displayed operators at once. The theorem records that no nonzero proper line satisfies these simultaneous stability conditions. [quotetheorem:3775] [citeproof:3775] This proof deliberately avoids relying on distinct $h$-weights, so it works in every characteristic. When $\operatorname{char} k \ne 2$, one can also prove the result by decomposing into the two eigenspaces of $h$ with eigenvalues $1$ and $-1$; in characteristic $2$ that weight-separation argument no longer applies because these eigenvalues coincide. The theorem therefore illustrates a limitation of a proof technique rather than a limitation of the statement itself. Later chapters develop weight-space methods much further, but this example warns that characteristic hypotheses must be attached to the exact argument being used. [example: Characteristic Two Boundary] Assume $\operatorname{char} k=2$. Since $-1=1$ in $k$, the standard action satisfies \begin{align*} hv_1&=v_1,\\ hv_2&=-v_2=v_2. \end{align*} Thus both coordinate vectors have the same $h$-eigenvalue, so the $h$-eigenspaces do not distinguish the lines $kv_1$ and $kv_2$. We now check irreducibility directly. Let $0\ne W\subset k^2$ be a submodule, and choose \begin{align*} 0\ne w=av_1+bv_2\in W. \end{align*} If $a\ne 0$, then \begin{align*} fw&=f(av_1+bv_2)\\ &=a\,fv_1+b\,fv_2\\ &=av_2+b\cdot 0\\ &=av_2\in W. \end{align*} Since $a\ne 0$, this gives $v_2\in W$, and then \begin{align*} ev_2=v_1\in W. \end{align*} Hence $W$ contains both $v_1$ and $v_2$, so $W=k^2$. If $a=0$, then $b\ne 0$ and $w=bv_2$. Therefore $v_2\in W$, and the same calculation gives \begin{align*} ev_2=v_1\in W. \end{align*} Again $W$ contains both basis vectors, so $W=k^2$. Thus every non-zero submodule is all of $k^2$, even though $h$ no longer separates the two coordinate lines. The failure is in the weight-space argument, not in irreducibility itself. [/example] ## One-Dimensional Modules And Abelianisation Which Lie algebras have non-zero one-dimensional representations? On a one-dimensional vector space, all endomorphisms commute, so the bracket must act by zero. Therefore one-dimensional representations only see the abelian quotient of the Lie algebra. [quotetheorem:3776] [citeproof:3776] This theorem has no finite-dimensionality or algebraic-closedness hypothesis; it only uses that $\mathfrak{gl}_1(k)$ is abelian. The limitation is that it classifies only one-dimensional modules, so it says nothing about higher-dimensional irreducible modules even for a perfect Lie algebra. For example, $\mathfrak{sl}_2(k)$ in characteristic $0$ has no non-zero one-dimensional action but has the standard two-dimensional module. The result connects representation theory back to the derived algebra introduced in the first chapter by showing that one-dimensional representations see exactly the abelianisation. [example: One-Dimensional Modules For Sl Two And A Solvable Algebra] For $\mathfrak{sl}_2(k)$ in characteristic $0$, use the usual basis \begin{align*} e=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad f=\begin{pmatrix}0&0\\1&0\end{pmatrix},\qquad h=\begin{pmatrix}1&0\\0&-1\end{pmatrix}, \end{align*} with \begin{align*} [h,e]=2e,\qquad [h,f]=-2f,\qquad [e,f]=h. \end{align*} Since $\operatorname{char} k=0$, the scalar $2$ is invertible in $k$. Hence \begin{align*} h&=[e,f]\in [\mathfrak{sl}_2(k),\mathfrak{sl}_2(k)],\\ e&=\frac{1}{2}[h,e]\in [\mathfrak{sl}_2(k),\mathfrak{sl}_2(k)],\\ f&=-\frac{1}{2}[h,f]\in [\mathfrak{sl}_2(k),\mathfrak{sl}_2(k)]. \end{align*} The elements $e,f,h$ span $\mathfrak{sl}_2(k)$, so \begin{align*} [\mathfrak{sl}_2(k),\mathfrak{sl}_2(k)]=\mathfrak{sl}_2(k). \end{align*} If $V$ is one-dimensional and $\rho:\mathfrak{sl}_2(k)\to \mathfrak{gl}(V)$ is a representation, then all endomorphisms of $V$ commute. Therefore, for all $u,v\in \mathfrak{sl}_2(k)$, \begin{align*} \rho([u,v])=[\rho(u),\rho(v)]=\rho(u)\rho(v)-\rho(v)\rho(u)=0. \end{align*} Since every element of $\mathfrak{sl}_2(k)$ is a linear combination of brackets, $\rho$ is the zero representation. Thus every one-dimensional $\mathfrak{sl}_2(k)$-module has zero action. Now let $\mathfrak g$ be the two-dimensional Lie algebra with basis $x,y$ and bracket $[x,y]=y$. By bilinearity and antisymmetry, \begin{align*} [ax+by,cx+dy] &=ac[x,x]+ad[x,y]+bc[y,x]+bd[y,y]\\ &=0+ad\,y+bc(-y)+0\\ &=(ad-bc)y. \end{align*} Thus every bracket lies in $ky$, and since $y=[x,y]$, we get \begin{align*} [\mathfrak g,\mathfrak g]=ky. \end{align*} For any one-dimensional representation $\rho:\mathfrak g\to \mathfrak{gl}(V)$, \begin{align*} \rho(y)=\rho([x,y])=[\rho(x),\rho(y)]=0, \end{align*} because $\mathfrak{gl}(V)$ is one-dimensional and abelian. If $\rho(x)$ is multiplication by $\lambda\in k$, then \begin{align*} \rho(ax+by)=a\rho(x)+b\rho(y)=a\lambda. \end{align*} Conversely, choosing any $\lambda\in k$ and defining $x$ to act by $\lambda$ and $y$ to act by $0$ gives \begin{align*} \rho([ax+by,cx+dy])=\rho((ad-bc)y)=0 \end{align*} and \begin{align*} [\rho(ax+by),\rho(cx+dy)]=(a\lambda)(c\lambda)-(c\lambda)(a\lambda)=0, \end{align*} so the representation identity holds. Hence one-dimensional modules for this algebra are exactly parametrised by the scalar action of $x$, while $y$ must act by zero. [/example] ## Faithful Representations And The Universal Enveloping Algebra Can every finite-dimensional Lie algebra be studied as a Lie algebra of matrices? Faithful representations answer this by asking whether the action map loses any element of the Lie algebra. The universal enveloping algebra gives a parallel answer: representations of $\mathfrak g$ are the same kind of data as modules over an associative algebra built from $\mathfrak g$. [definition: Faithful Representation] A representation $\rho:\mathfrak g \to \mathfrak{gl}(V)$ is faithful if $\ker \rho=0$. [/definition] Faithfulness means that $\mathfrak g$ embeds into $\mathfrak{gl}(V)$, so the Lie algebra can be realised concretely as linear operators. The adjoint representation is faithful exactly when $Z(\mathfrak g)=0$. [remark: Ado Theorem] Ado's theorem states that every finite-dimensional Lie algebra over a field of characteristic $0$ admits a finite-dimensional faithful representation. It is a quoted structural landmark here rather than a theorem proved in this set of notes. Its hypotheses matter: the finite-dimensional and characteristic $0$ assumptions are part of this formulation, and other versions require additional care. The result justifies the frequent use of matrix intuition, but it does not say that a naturally chosen representation, such as the adjoint representation, must be faithful. [/remark] Faithful representations place a Lie algebra inside an endomorphism algebra, but many arguments need an algebra that encodes all representations at once. The bracket relation $[x,y]$ should be remembered as a commutator relation $xy-yx$ inside an associative algebra, because associative algebra modules already have a well-developed theory. The universal enveloping algebra is designed to impose exactly those relations and no extra ones. [definition: Universal Enveloping Algebra] Let $\mathfrak g$ be a Lie algebra over $k$. The universal enveloping algebra $U(\mathfrak g)$ is the associative unital $k$-algebra generated by symbols from $\mathfrak g$ subject to the relations \begin{align*} xy-yx=[x,y] \end{align*} for all $x,y \in \mathfrak g$. [/definition] The construction forces the Lie bracket to become the commutator in an associative algebra. To justify using $U(\mathfrak g)$ as a bridge, we need more than a formal resemblance: representations of $\mathfrak g$ should be exactly the same data as modules over this associative algebra. The next result verifies that no representation is lost or added in passing through the enveloping algebra. [quotetheorem:3777] [citeproof:3777] The theorem is formal and does not require finite-dimensionality of $\mathfrak g$ or $V$. Its limitation is that it translates the problem into associative algebra language rather than making modules easier to classify; the algebra $U(\mathfrak g)$ is usually infinite-dimensional even when $\mathfrak g$ is finite-dimensional. For the zero Lie algebra, this reduces to ordinary vector spaces over $k$, while for non-abelian $\mathfrak g$ the non-commutative relation in $U(\mathfrak g)$ remembers the bracket. This bridge is why Lie algebra representations can be studied with tools from module theory over associative algebras. [remark: Role In Structure Theory] The language of modules lets structure theory use linear algebra on invariant subspaces. Engel's theorem will study the adjoint module under nilpotence hypotheses, [Lie's theorem](/theorems/3802) will analyse modules over solvable Lie algebras, and complete reducibility will return in Weyl's theorem for semisimple Lie algebras. [/remark] Representations now provide the linear-algebraic setting in which solvability can be tested and exploited. Having translated bracket identities into operator identities, we are ready to study how repeated commutators force a Lie algebra toward triangular form and, eventually, toward complete reducibility results. # 4. Solvable Lie Algebras Solvability is the first major weakening of commutativity in the course. Instead of asking that all brackets vanish, we ask whether repeated passage to commutator subalgebras eventually removes all noncommutativity. This chapter develops the derived series, explains how solvability behaves under the quotient, exact-sequence, and extension constructions from Chapter 2, and isolates the largest solvable ideal, the radical, which Chapter 9 will separate from the semisimple quotient. ## Measuring Noncommutativity by Derived Series How can we measure the failure of a Lie algebra to be abelian in a way that can be iterated? The derived algebra $[\mathfrak g,\mathfrak g]$ records all first-order commutators, so the natural next step is to take commutators inside that subalgebra again. Solvability means that this repeated commutator process eventually reaches zero. [definition: Derived Series] Let $\mathfrak g$ be a Lie algebra over a field $F$. The derived series of $\mathfrak g$ is the descending sequence of Lie subalgebras \begin{align*} \mathfrak g^{(0)} &= \mathfrak g, & \mathfrak g^{(i+1)} &= [\mathfrak g^{(i)},\mathfrak g^{(i)}] \quad (i \ge 0). \end{align*} [/definition] The first term $\mathfrak g^{(1)}$ is the derived algebra. Passing from $\mathfrak g^{(i)}$ to $\mathfrak g^{(i+1)}$ discards the abelian quotient $\mathfrak g^{(i)} / \mathfrak g^{(i+1)}$ and keeps only the commutator content of $\mathfrak g^{(i)}$. This iteration turns the informal idea of eventually becoming abelian into a precise stopping condition. The definition below names the Lie algebras for which repeated commutators exhaust all noncommutativity after finitely many derived steps. [definition: Solvable Lie Algebra] A Lie algebra $\mathfrak g$ is solvable if there exists $m \ge 0$ such that $\mathfrak g^{(m)} = 0$. [/definition] Thus an abelian Lie algebra is solvable with $\mathfrak g^{(1)}=0$, while nonabelian solvable Lie algebras may require more steps. The number of steps is not usually the main invariant in this course; the important point is that the derived series terminates. [example: Nonabelian Two-Dimensional Solvable Lie Algebra] Let $\mathfrak g=Fx\oplus Fy$ with bracket determined by bilinearity, $[x,y]=y$, and $[u,u]=0$ for every $u\in\mathfrak g$. For arbitrary elements $\alpha x+\beta y$ and $\gamma x+\delta y$, \begin{align*} [\alpha x+\beta y,\gamma x+\delta y] &=\alpha\gamma[x,x]+\alpha\delta[x,y]+\beta\gamma[y,x]+\beta\delta[y,y] \\ &=0+\alpha\delta y+\beta\gamma(-y)+0 \\ &=(\alpha\delta-\beta\gamma)y. \end{align*} Thus every bracket lies in $Fy$, and $y=[x,y]$ also lies in $[\mathfrak g,\mathfrak g]$, so \begin{align*} \mathfrak g^{(1)}=[\mathfrak g,\mathfrak g]=Fy. \end{align*} Now \begin{align*} \mathfrak g^{(2)} &=[\mathfrak g^{(1)},\mathfrak g^{(1)}] \\ &=[Fy,Fy]. \end{align*} For $ay,by\in Fy$, \begin{align*} [ay,by]=ab[y,y]=0, \end{align*} so $\mathfrak g^{(2)}=0$. Hence $\mathfrak g$ is solvable, while it is not abelian because $[x,y]=y\ne 0$. [/example] A solvable ideal is not merely a solvable subalgebra sitting inside a larger Lie algebra. The ideal condition says that bracketing with arbitrary elements of the ambient algebra cannot move it out of itself, so it is a piece of $\mathfrak g$ visible to quotient constructions. This is the right notion for separating the ``solvable part'' of a Lie algebra from the part that remains after quotienting. [definition: Solvable Ideal] Let $\mathfrak g$ be a Lie algebra. An ideal $\mathfrak a \trianglelefteq \mathfrak g$ is a solvable ideal if $\mathfrak a$ is solvable as a Lie algebra with the induced bracket. [/definition] Solvable ideals are important because they form the part of $\mathfrak g$ that can be removed before studying semisimple structure. A first wrong guess would be that maximal solvable ideals need not combine well: the sum of two well-behaved pieces might introduce new brackets between them. The ideal hypothesis prevents this obstruction, because the mixed brackets still land in both ambient ideals in a controlled way. [quotetheorem:3778] [citeproof:3778] The theorem depends essentially on the ideals being inside the same ambient algebra and on both summands being solvable. A non-solvable summand remains visible in the sum, so solvability cannot be recovered from only one side. This result is the mechanism that makes the radical possible: finite-dimensionality prevents an infinite increasing process of larger and larger solvable ideals from continuing forever. [definition: Radical] Let $\mathfrak g$ be a finite-dimensional Lie algebra over a field $F$. The radical $\operatorname{rad}(\mathfrak g)$ is the unique largest solvable ideal of $\mathfrak g$. [/definition] The definition names the object we want, but it is not automatic that all solvable ideals fit inside one largest solvable ideal. The possible obstruction is that choosing different maximal solvable ideals might produce incompatible candidates. Since sums of solvable ideals are again solvable in the finite-dimensional setting, maximal dimension can force these candidates to coincide and produce a single radical. [quotetheorem:3779] [citeproof:3779] Finite-dimensionality is the hypothesis that lets maximal dimension replace a direct infinite sum argument. Without a finiteness condition, the existence of a largest solvable ideal requires a separate chain argument and is not a consequence of the proof just given. The radical is not usually computed directly from all solvable ideals; in practice, the structural results below let us identify it by passing to quotients and using matrix examples. ## Solvability and Standard Constructions Which operations preserve solvability? Since the course repeatedly forms subalgebras, quotients, kernels, images, and extensions, solvability must be stable under these constructions if it is to be useful. The derived series behaves well with respect to homomorphisms, and this gives most of the permanence properties. [quotetheorem:3780] [citeproof:3780] This result says that solvability cannot be created by adding fewer elements: any obstruction already present in a subalgebra would also occur in the ambient algebra. The converse fails, since a non-solvable Lie algebra may contain many solvable subalgebras, including every one-dimensional subalgebra. The same comparison works for homomorphic images, except now the derived series maps onto the derived series of the image rather than sitting inside it. [quotetheorem:3781] [citeproof:3781] This theorem is one-directional: a solvable image does not force the source to be solvable, because a homomorphism may collapse the non-solvable part into its kernel. For example, the zero homomorphism from any Lie algebra has solvable image. Taking the homomorphism to be the quotient map gives the most common special case, where the collapsed part is recorded explicitly as an ideal. [quotetheorem:3782] [citeproof:3782] Quotients can only remove brackets, so solvability descends to them. The converse is false without a hypothesis on the kernel: if $\mathfrak a=\mathfrak g$, then $\mathfrak g/\mathfrak a=0$ is solvable for every Lie algebra $\mathfrak g$, including non-solvable ones. The [extension theorem](/theorems/3783) below gives the missing condition by requiring the part killed by the quotient map to be solvable as well. [quotetheorem:3783] [citeproof:3783] Both hypotheses are needed: a solvable quotient alone can hide a non-solvable ideal, while a solvable ideal alone says nothing about the remaining quotient. This theorem is often the quickest way to prove solvability, because it lets us build the algebra from a solvable ideal and a solvable quotient rather than computing every derived term. It also explains why the preceding permanence results combine into a clean iff criterion. [quotetheorem:3784] [citeproof:3784] The criterion packages the preceding results into a test across a short exact sequence. It fails if $\mathfrak a$ is only a subalgebra, because the quotient $\mathfrak g/\mathfrak a$ is not a Lie algebra unless $\mathfrak a$ is an ideal. A closely related problem arises when solvability is known only after applying a homomorphism. A solvable image does not by itself control the source, because the kernel may contain all of the non-solvable behaviour. The useful converse therefore asks for exactly the missing condition: the part killed by the homomorphism must also be solvable. [quotetheorem:3785] [citeproof:3785] The kernel hypothesis cannot be dropped: the zero map from a non-solvable Lie algebra has solvable image but non-solvable source. Thus the image criterion is a controlled converse to the homomorphic image theorem, not a general converse. The triangular examples in the next section give concrete solvable targets and also show how these abstract criteria appear inside matrix Lie algebras. ## Triangular Matrix Algebras What do solvable Lie algebras look like in the examples that motivated the definition? Upper triangular matrices provide the model: commutators of upper triangular matrices lose diagonal information, and repeated commutators move toward strictly upper triangular behaviour. These examples will later connect to [Lie's theorem](/theorems/3803), where solvable Lie algebras over algebraically closed fields act by upper triangular matrices. [definition: Upper Triangular Lie Algebra] Let $F$ be a field. The Lie algebra of upper triangular matrices in $\mathfrak{gl}_n(F)$ is \begin{align*} \mathfrak b_n(F)=\{A\in \mathfrak{gl}_n(F): A_{ij}=0 \text{ whenever } i>j\}, \end{align*} with bracket $[A,B]=AB-BA$. [/definition] The notation $\mathfrak b_n(F)$ is standard because this algebra is the matrix prototype of a Borel subalgebra. To analyse its derived series, we need to separate the diagonal entries from the entries strictly above the diagonal. The latter form the nilpotent-looking part where commutators of triangular matrices will land. [definition: Strictly Upper Triangular Lie Algebra] Let $F$ be a field. The Lie algebra of strictly upper triangular matrices is \begin{align*} \mathfrak n_n(F)=\{A\in \mathfrak{gl}_n(F): A_{ij}=0 \text{ whenever } i\ge j\}, \end{align*} with bracket $[A,B]=AB-BA$. [/definition] The diagonal entries of a commutator of two upper triangular matrices vanish. This suggests that the derived series of $\mathfrak b_n(F)$ should move away from the diagonal and eventually run out of possible nonzero entries. The theorem below turns that matrix pattern into the solvability statement needed for triangular examples. [quotetheorem:3786] [citeproof:3786] The finite matrix size is essential: there are only finitely many superdiagonals on which commutators can live. Conceptually, the result explains why upper triangular form is a strong structural conclusion in finite-dimensional representation theory: it turns repeated commutators into movement through a finite filtration. In dimension three this behaviour can be seen explicitly, with the first superdiagonal feeding the second and then disappearing. [example: Upper Triangular Matrices in Dimension Three] Let $\mathfrak b_3(F)$ be the upper triangular matrices in $\mathfrak{gl}_3(F)$, and write a general element as \begin{align*} A=\begin{pmatrix} a_{11}&a_{12}&a_{13}\\ 0&a_{22}&a_{23}\\ 0&0&a_{33} \end{pmatrix}. \end{align*} For $A,B\in \mathfrak b_3(F)$, the diagonal entries of $AB-BA$ are \begin{align*} (AB-BA)_{ii}=a_{ii}b_{ii}-b_{ii}a_{ii}=0 \qquad (i=1,2,3), \end{align*} so $[\mathfrak b_3(F),\mathfrak b_3(F)]\subset \mathfrak n_3(F)$, where $\mathfrak n_3(F)=FE_{12}\oplus FE_{13}\oplus FE_{23}$. Conversely, \begin{align*} [E_{11},E_{12}]&=E_{11}E_{12}-E_{12}E_{11}=E_{12}-0=E_{12},\\ [E_{11},E_{13}]&=E_{11}E_{13}-E_{13}E_{11}=E_{13}-0=E_{13},\\ [E_{22},E_{23}]&=E_{22}E_{23}-E_{23}E_{22}=E_{23}-0=E_{23}. \end{align*} Thus \begin{align*} \mathfrak b_3(F)^{(1)}=[\mathfrak b_3(F),\mathfrak b_3(F)]=\mathfrak n_3(F). \end{align*} Now compute the brackets among the basis elements of $\mathfrak n_3(F)$. Since $E_{ij}E_{kl}=\delta_{jk}E_{il}$, \begin{align*} [E_{12},E_{13}]&=E_{12}E_{13}-E_{13}E_{12}=0-0=0,\\ [E_{12},E_{23}]&=E_{12}E_{23}-E_{23}E_{12}=E_{13}-0=E_{13},\\ [E_{13},E_{23}]&=E_{13}E_{23}-E_{23}E_{13}=0-0=0. \end{align*} Together with skew-symmetry and $[E_{ij},E_{ij}]=0$, this gives \begin{align*} \mathfrak b_3(F)^{(2)} =[\mathfrak n_3(F),\mathfrak n_3(F)] =FE_{13}. \end{align*} Finally, for $\lambda E_{13},\mu E_{13}\in FE_{13}$, \begin{align*} [\lambda E_{13},\mu E_{13}] =\lambda\mu(E_{13}E_{13}-E_{13}E_{13}) =0, \end{align*} so \begin{align*} \mathfrak b_3(F)^{(3)}=[FE_{13},FE_{13}]=0. \end{align*} The derived series therefore terminates after three displayed terms, showing explicitly how commutators move from all upper triangular matrices to the first and second superdiagonals and then vanish. [/example] The dimension-three example shows the same upward movement of commutators in a small enough algebra to compute by hand. The smallest nonabelian example from the first section is already present inside triangular matrices. Seeing it inside $\mathfrak{sl}_2(F)$ also prepares the later contrast between solvable Borel subalgebras and the non-solvable ambient Lie algebra. [example: Borel Subalgebra of $\mathfrak{sl}_2$] Let \begin{align*} \mathfrak b=\left\{\begin{pmatrix} a & b \\ 0 & -a \end{pmatrix}: a,b\in F\right\}\subset \mathfrak{sl}_2(F), \end{align*} and set \begin{align*} h=\begin{pmatrix}1&0\\0&-1\end{pmatrix}, \qquad e=\begin{pmatrix}0&1\\0&0\end{pmatrix}. \end{align*} Then every element of $\mathfrak b$ has the form $ah+be$, since \begin{align*} ah+be &= a\begin{pmatrix}1&0\\0&-1\end{pmatrix} +b\begin{pmatrix}0&1\\0&0\end{pmatrix} \\ &= \begin{pmatrix}a&0\\0&-a\end{pmatrix} + \begin{pmatrix}0&b\\0&0\end{pmatrix} = \begin{pmatrix}a&b\\0&-a\end{pmatrix}. \end{align*} The basic brackets are \begin{align*} [h,h]&=0,\\ [e,e]&=0, \end{align*} and \begin{align*} [h,e] &=he-eh \\ &= \begin{pmatrix}1&0\\0&-1\end{pmatrix} \begin{pmatrix}0&1\\0&0\end{pmatrix} - \begin{pmatrix}0&1\\0&0\end{pmatrix} \begin{pmatrix}1&0\\0&-1\end{pmatrix} \\ &= \begin{pmatrix}0&1\\0&0\end{pmatrix} - \begin{pmatrix}0&-1\\0&0\end{pmatrix} = \begin{pmatrix}0&2\\0&0\end{pmatrix} =2e. \end{align*} Thus $[e,h]=-[h,e]=-2e$. For arbitrary elements $ah+be$ and $ch+de$ of $\mathfrak b$, bilinearity gives \begin{align*} [ah+be,ch+de] &=ac[h,h]+ad[h,e]+bc[e,h]+bd[e,e] \\ &=0+ad(2e)+bc(-2e)+0 \\ &=2(ad-bc)e. \end{align*} Therefore every bracket in $\mathfrak b$ lies in $Fe$, so $[\mathfrak b,\mathfrak b]\subset Fe$. Conversely, if $\operatorname{char}F\ne 2$, then $2\ne 0$ in $F$, so \begin{align*} e=\frac{1}{2}[h,e]\in [\mathfrak b,\mathfrak b], \end{align*} and hence $Fe\subset [\mathfrak b,\mathfrak b]$. Therefore \begin{align*} \mathfrak b^{(1)}=[\mathfrak b,\mathfrak b]=Fe. \end{align*} Finally, for $\lambda e,\mu e\in Fe$, \begin{align*} [\lambda e,\mu e] &=\lambda\mu[e,e] \\ &=\lambda\mu(ee-ee) \\ &=0, \end{align*} so \begin{align*} \mathfrak b^{(2)}=[Fe,Fe]=0. \end{align*} Thus $\mathfrak b$ is solvable, while it is not abelian because $[h,e]=2e\ne 0$ when $\operatorname{char}F\ne 2$. [/example] The triangular examples also show why solvability is weaker than nilpotence, a distinction developed in the next chapter. The algebra $\mathfrak n_n(F)$ is nilpotent under its lower central series, while $\mathfrak b_n(F)$ is solvable but generally not nilpotent because diagonal elements act nontrivially on upper triangular root spaces. [remark: Role of the Radical] For a finite-dimensional Lie algebra $\mathfrak g$, the quotient $\mathfrak g/\operatorname{rad}(\mathfrak g)$ has no nonzero solvable ideals. This quotient is the first approximation to the semisimple part of $\mathfrak g$, and later chapters will make that statement precise using the Killing form and Cartan's criterion. [/remark] The solvable case marks the first substantial structural weakening beyond commutativity, but it is not the end of the story. To understand what remains after solvable ideals are peeled away, we next sharpen the commutator viewpoint into the stronger notion of nilpotence. # 5. Nilpotent Lie Algebras This chapter studies the strongest finiteness condition on repeated commutators that appears before Engel's theorem in Chapter 6: nilpotence. It assumes the preceding material on ideals and quotient Lie algebras from Chapter 2, derived series and solvability from Chapter 4, and the Jacobi identity from Chapter 1. Solvability says that commutators eventually disappear after repeatedly passing to derived algebras; nilpotence asks for disappearance under a more rigid process, where we keep bracketing with the whole Lie algebra. The main structural point is that nilpotent Lie algebras behave like upper triangular matrices with zero diagonal: commutators move steadily toward the centre until no further motion remains. ## Lower Central Series How should we measure the extent to which a Lie algebra fails to be abelian? The derived algebra $[\mathfrak g,\mathfrak g]$ records first-order non-commutativity, but iterating the derived algebra is too coarse for nilpotence. The lower central series keeps the original algebra available at every step, so it detects whether repeated commutators with arbitrary elements eventually vanish. [definition: Lower Central Series] Let $\mathfrak g$ be a Lie algebra over a field $F$. The lower central series of $\mathfrak g$ is the descending sequence of ideals $(\gamma_i(\mathfrak g))_{i\geq 1}$ defined by \begin{align*} \gamma_1(\mathfrak g) &= \mathfrak g,\\ \gamma_{i+1}(\mathfrak g) &= [\mathfrak g,\gamma_i(\mathfrak g)] \quad (i\geq 1). \end{align*} [/definition] Thus $\gamma_2(\mathfrak g)=[\mathfrak g,\mathfrak g]$ is the derived algebra, but $\gamma_3(\mathfrak g)=[\mathfrak g,[\mathfrak g,\mathfrak g]]$ rather than $[[\mathfrak g,\mathfrak g],[\mathfrak g,\mathfrak g]]$. This distinction is the reason nilpotence is stronger than solvability. The lower central series gives a stricter test: every remaining commutator layer must be stable under bracketing with the whole algebra, not just with itself. The obstruction is that solvability can make commutators vanish only after bracketing inside already-derived pieces, while nonzero mixed commutators with the original algebra may still persist. Nilpotence is the condition that this stronger mixed-commutator process actually reaches zero. [definition: Nilpotent Lie Algebra] A Lie algebra $\mathfrak g$ is nilpotent if there exists $c\geq 0$ such that \begin{align*} \gamma_{c+1}(\mathfrak g)=0. \end{align*} The least such $c$ is called the nilpotency class of $\mathfrak g$. [/definition] An abelian Lie algebra has nilpotency class at most $1$, since $\gamma_2(\mathfrak g)=0$. A nonzero Lie algebra of nilpotency class $c$ has $\gamma_c(\mathfrak g)\neq 0$ and $[\mathfrak g,\gamma_c(\mathfrak g)]=0$, so the final nonzero term lies inside the centre. [example: Strictly Upper Triangular Matrices] Let $\mathfrak n_n(F)$ be the Lie algebra of strictly upper triangular $n\times n$ matrices over $F$, with bracket $[X,Y]=XY-YX$. For $1\leq a<b\leq n$, write $E_{ab}$ for the matrix with a $1$ in position $(a,b)$ and $0$ elsewhere. Then \begin{align*} E_{ab}E_{cd}=\delta_{bc}E_{ad}, \end{align*} so \begin{align*} [E_{ab},E_{cd}] &=E_{ab}E_{cd}-E_{cd}E_{ab}\\ &=\delta_{bc}E_{ad}-\delta_{da}E_{cb}. \end{align*} Let $V_r$ be the span of all $E_{ab}$ with $b-a\geq r$. Thus $\mathfrak n_n(F)=V_1$. If $E_{ab}\in V_r$ and $E_{cd}\in V_s$, then the first term $\delta_{bc}E_{ad}$ can be nonzero only when $b=c$, and then \begin{align*} d-a=(b-a)+(d-c)\geq r+s. \end{align*} The second term $\delta_{da}E_{cb}$ can be nonzero only when $d=a$, and then \begin{align*} b-c=(b-a)+(a-c)=(b-a)+(d-c)\geq r+s. \end{align*} Hence every basis term appearing in $[E_{ab},E_{cd}]$ lies in $V_{r+s}$, and by bilinearity, \begin{align*} [V_r,V_s]\subseteq V_{r+s}. \end{align*} We now compute the lower central series. Since $\gamma_1(\mathfrak n_n(F))=\mathfrak n_n(F)=V_1$, the claim is true for $i=1$. If $\gamma_i(\mathfrak n_n(F))\subseteq V_i$, then \begin{align*} \gamma_{i+1}(\mathfrak n_n(F)) &=[\mathfrak n_n(F),\gamma_i(\mathfrak n_n(F))]\\ &\subseteq [V_1,V_i]\\ &\subseteq V_{i+1}. \end{align*} Thus $\gamma_i(\mathfrak n_n(F))\subseteq V_i$ for all $i\geq 1$. For the reverse inclusion, fix $1\leq i\leq n-1$. We prove that every $E_{ab}$ with $b-a\geq i$ lies in $\gamma_i(\mathfrak n_n(F))$. For $i=1$ this is clear. Suppose it holds for $i-1$, and take $b-a\geq i$. Then $b-(a+1)\geq i-1$, so $E_{a+1,b}\in \gamma_{i-1}(\mathfrak n_n(F))$. Also $E_{a,a+1}\in \mathfrak n_n(F)$. Using the multiplication rule, \begin{align*} [E_{a,a+1},E_{a+1,b}] &=E_{a,a+1}E_{a+1,b}-E_{a+1,b}E_{a,a+1}\\ &=E_{ab}-0\\ &=E_{ab}. \end{align*} Therefore $E_{ab}\in [\mathfrak n_n(F),\gamma_{i-1}(\mathfrak n_n(F))]=\gamma_i(\mathfrak n_n(F))$. Hence \begin{align*} \gamma_i(\mathfrak n_n(F))=V_i \qquad (1\leq i\leq n-1). \end{align*} Finally, $V_n=0$ because no entry can lie at distance at least $n$ above the diagonal, while $V_{n-1}=F E_{1n}\neq 0$ when $n\geq 2$. Thus $\gamma_n(\mathfrak n_n(F))=0$ and $\gamma_{n-1}(\mathfrak n_n(F))\neq 0$, so $\mathfrak n_n(F)$ is nilpotent of class $n-1$. [/example] The example also shows why the lower central series is the right filtration: each commutator moves entries farther from the diagonal, so the filtration is compatible with the Lie bracket rather than being an arbitrary descending chain. For the filtration to support quotient arguments and induction, however, each term must be an ideal of the original Lie algebra, not merely a subspace or a subalgebra. The key point is that bracketing any element of $\mathfrak g$ with a term of the lower central series lands in the next term. [quotetheorem:3787] [citeproof:3787] This theorem allows the lower central series to be used in quotient arguments and induction. The ideal hypothesis is essential here: an arbitrary descending sequence of subspaces would not pass to quotients in a controlled way, nor would bracketing with elements of $\mathfrak g$ remain inside the next term. What the theorem does not say is that every useful filtration is the lower central series; later central series run in the opposite direction. The descending chain also shows why finite-dimensional nilpotence is a strong condition: it forces a finite filtration by ideals whose successive quotients are increasingly central. Before comparing nilpotence with solvability, the obstruction is that the two definitions iterate different operations. The derived series takes commutators only inside the current derived algebra, while the lower central series continues to bracket with all of $\mathfrak g$. Because this second process controls more brackets at each stage, vanishing of the lower central series should force vanishing of the derived series, but not conversely. [quotetheorem:3788] [citeproof:3788] The nilpotence hypothesis is doing real work: it controls brackets with all of $\mathfrak g$, not merely brackets inside the current derived subalgebra. The converse fails because solvability allows commutators to disappear after taking brackets inside smaller derived subalgebras, while nilpotence demands vanishing under bracketing with all of $\mathfrak g$ at each stage. This distinction will matter again when Engel's theorem converts pointwise nilpotence of adjoint maps into the stronger global condition. [example: Solvable Lie Algebra That Is Not Nilpotent] Let $\mathfrak b$ have basis $h,x$ and bracket $[h,x]=x$. Since the bracket is alternating and bilinear, we also have $[h,h]=[x,x]=0$ and $[x,h]=-[h,x]=-x$. For arbitrary elements $ah+bx$ and $ch+dx$ of $\mathfrak b$, \begin{align*} [ah+bx,ch+dx] &=ac[h,h]+ad[h,x]+bc[x,h]+bd[x,x]\\ &=0+ad\,x+bc(-x)+0\\ &=(ad-bc)x. \end{align*} Thus every bracket lies in $Fx$, so $[\mathfrak b,\mathfrak b]\subseteq Fx$. The reverse inclusion holds because $x=[h,x]\in[\mathfrak b,\mathfrak b]$, hence \begin{align*} [\mathfrak b,\mathfrak b]=Fx. \end{align*} Now \begin{align*} [Fx,Fx] &=\operatorname{span}\{[sx,tx]:s,t\in F\}\\ &=\operatorname{span}\{st[x,x]:s,t\in F\}\\ &=0. \end{align*} Therefore the derived series is \begin{align*} \mathfrak b^{(0)}&=\mathfrak b,\\ \mathfrak b^{(1)}&=[\mathfrak b,\mathfrak b]=Fx,\\ \mathfrak b^{(2)}&=[Fx,Fx]=0, \end{align*} so $\mathfrak b$ is solvable of derived length $2$. The lower central series behaves differently. Its first nontrivial term is \begin{align*} \gamma_2(\mathfrak b)=[\mathfrak b,\mathfrak b]=Fx. \end{align*} For the next term, \begin{align*} [\mathfrak b,Fx] &=\operatorname{span}\{[ah+bx,tx]:a,b,t\in F\}\\ &=\operatorname{span}\{at[h,x]+bt[x,x]:a,b,t\in F\}\\ &=\operatorname{span}\{atx:a,t\in F\}\\ &=Fx, \end{align*} because $[h,x]=x$. Hence \begin{align*} \gamma_3(\mathfrak b)=Fx. \end{align*} Repeating the same calculation gives $\gamma_i(\mathfrak b)=Fx$ for every $i\geq 2$, so the lower central series never reaches $0$. Thus $\mathfrak b$ is solvable but not nilpotent. [/example] ## Nilpotent Ideals and Central Series How can nilpotence be recognised without computing every term of the lower central series? The answer is to compare the lower central series with central filtrations, where each layer becomes central after passing to the next quotient. This viewpoint is especially useful for ideals and extensions. [definition: Central Series] Let $\mathfrak g$ be a Lie algebra. A central series for $\mathfrak g$ is a chain of ideals \begin{align*} 0=\mathfrak g_0\subseteq \mathfrak g_1\subseteq \cdots \subseteq \mathfrak g_c=\mathfrak g \end{align*} such that \begin{align*} [\mathfrak g,\mathfrak g_i]\subseteq \mathfrak g_{i-1} \end{align*} for each $1\leq i\leq c$. [/definition] The condition says that $\mathfrak g_i/\mathfrak g_{i-1}$ is contained in the centre of $\mathfrak g/\mathfrak g_{i-1}$. A central series builds the algebra upward from central layers, while the lower central series dismantles it downward by repeated commutation. The obstruction this resolves is that the lower central series is often awkward to compute directly. A central filtration gives the same information in a form suited to construction: instead of asking whether repeated commutators vanish, we ask whether the algebra can be assembled from layers that become central after quotienting. [quotetheorem:3789] [citeproof:3789] This criterion explains the phrase nilpotent by central layers: the algebra becomes abelian after quotienting by a central ideal, then again after another central quotient, and so on until nothing remains. The ideals in the chain cannot be replaced by arbitrary subspaces, because quotient Lie algebras and centrality conditions require ideals. The centrality condition is also essential: an arbitrary filtration of ideals may have abelian successive vector-space quotients without forcing any commutator to move downward. This formulation is the one used next to isolate nilpotent ideals inside a larger Lie algebra. [definition: Nilpotent Ideal] Let $\mathfrak g$ be a Lie algebra. An ideal $I\trianglelefteq \mathfrak g$ is nilpotent if $I$ is nilpotent as a Lie algebra with bracket inherited from $\mathfrak g$. [/definition] A nilpotent ideal need not be central, but it has a central structure internally. The largest such ideal, when it exists, captures the nilpotent part of $\mathfrak g$. [definition: Nilradical] Let $\mathfrak g$ be a finite-dimensional Lie algebra. The nilradical of $\mathfrak g$, denoted $\operatorname{nil}(\mathfrak g)$, is the sum of all nilpotent ideals of $\mathfrak g$. [/definition] The definition is meaningful because a finite-dimensional Lie algebra has only finite-dimensional subspaces, and the theorem below shows that the sum remains nilpotent. Thus $\operatorname{nil}(\mathfrak g)$ is the unique largest nilpotent ideal of $\mathfrak g$. [quotetheorem:3790] [citeproof:3790] [remark: Nilradical and Radical] The nilradical is contained in the solvable radical, since every nilpotent ideal is solvable. Equality need not hold: the two-dimensional algebra with basis $h,x$ and $[h,x]=x$ is solvable, so it is its own radical, but its nilradical is $Fx$. [/remark] Repeated brackets give another useful test. The lower central series is defined recursively, but computations often produce concrete left-normed brackets such as $[x_1,[x_2,[\cdots,x_m]\cdots]]$. To use nilpotence effectively, we need to know that these two ways of measuring long commutators are equivalent. [quotetheorem:3791] [citeproof:3791] This bracket-length criterion turns nilpotence into a usable vanishing test: it lets one prove nilpotence by bounding all sufficiently long iterated brackets, without computing every term of the lower central series separately. It also clarifies why nilpotence is a uniform condition on the whole algebra, not just a statement about selected pairs of elements. The criterion is especially useful because many natural computations produce nested brackets before they produce a clean description of the lower central series. It gives a way to pass between explicit calculations and structural nilpotence, which is the perspective needed when studying quotients and extensions. In the next section, this same control over long commutators explains why nilpotent algebras have central elements and why central quotients preserve the nilpotent structure. ## Centers, Quotients, and Central Extensions What does nilpotence force at the bottom of the algebra? The first decisive consequence is the existence of central elements. This is the Lie-algebra analogue of the fact that finite nilpotent groups have nontrivial centre. [quotetheorem:3792] [citeproof:3792] This theorem is the basic induction engine for finite-dimensional nilpotent Lie algebras. It lets us pass to central quotients while preserving nilpotence. The next obstruction is whether nilpotence survives when part of the algebra is collapsed. Since quotienting kills some brackets, nilpotence should pass downward, but the precise formula is needed to control the lower central series after the quotient. [quotetheorem:3793] [citeproof:3793] The converse for arbitrary ideals is false, since a quotient may hide non-nilpotent behaviour inside the ideal. For example, the two-dimensional algebra with basis $h,x$ and $[h,x]=x$ has quotient by $Fx$ abelian and ideal $Fx$ abelian, but the whole algebra is not nilpotent. This failure leaves a precise extension problem: what extra condition on the ideal prevents hidden non-nilpotent behaviour from surviving upstairs? The useful condition is centrality. If the ideal is central, then once the quotient has forced long commutators into that ideal, one more bracket with the whole algebra kills them, giving a controlled way to rebuild nilpotence from the quotient. [quotetheorem:3794] [citeproof:3794] The centrality hypothesis cannot be weakened to nilpotence of $I$: the two-dimensional non-nilpotent solvable algebra is already an extension of an abelian quotient by an abelian ideal. What centrality adds is that the lifted final lower central term has no room to keep moving. The Heisenberg algebra is the basic model where this criterion is visible in one step. [example: Heisenberg Algebra] Let $\mathfrak h_3(F)$ have basis $x,y,z$ and brackets $[x,y]=z$ and $[x,z]=[y,z]=0$. By bilinearity and alternation, also $[y,x]=-[x,y]=-z$, $[z,x]=[z,y]=0$, and $[x,x]=[y,y]=[z,z]=0$. First compute the centre. If $w=ax+by+cz$, then \begin{align*} [w,x] &=a[x,x]+b[y,x]+c[z,x]\\ &=0+b(-z)+0\\ &=-bz, \end{align*} and \begin{align*} [w,y] &=a[x,y]+b[y,y]+c[z,y]\\ &=az+0+0\\ &=az. \end{align*} Thus a central element must have $a=b=0$, so it lies in $Fz$. Conversely, $[z,x]=[z,y]=[z,z]=0$, so every scalar multiple of $z$ is central. Hence \begin{align*} Z(\mathfrak h_3(F))=Fz. \end{align*} The quotient $\mathfrak h_3(F)/Fz$ has basis $x+Fz,y+Fz$, and its only possible nonzero bracket is \begin{align*} [x+Fz,y+Fz] &=[x,y]+Fz\\ &=z+Fz\\ &=Fz. \end{align*} Therefore $\mathfrak h_3(F)/Fz$ is abelian. Now compute the lower central series. For arbitrary $u=ax+by+cz$ and $v=px+qy+rz$, \begin{align*} [u,v] &=[ax+by+cz,px+qy+rz]\\ &=ap[x,x]+aq[x,y]+ar[x,z]+bp[y,x]+bq[y,y]+br[y,z]+cp[z,x]+cq[z,y]+cr[z,z]\\ &=0+aqz+0+bp(-z)+0+0+0+0+0\\ &=(aq-bp)z. \end{align*} So every bracket lies in $Fz$, and since $z=[x,y]$, we have \begin{align*} \gamma_2(\mathfrak h_3(F))=[\mathfrak h_3(F),\mathfrak h_3(F)]=Fz. \end{align*} Finally, \begin{align*} \gamma_3(\mathfrak h_3(F)) &=[\mathfrak h_3(F),Fz]\\ &=\operatorname{span}\{[ax+by+cz,tz]:a,b,c,t\in F\}\\ &=\operatorname{span}\{at[x,z]+bt[y,z]+ct[z,z]:a,b,c,t\in F\}\\ &=0. \end{align*} Thus \begin{align*} \gamma_1(\mathfrak h_3(F))&=\mathfrak h_3(F),\\ \gamma_2(\mathfrak h_3(F))&=Fz,\\ \gamma_3(\mathfrak h_3(F))&=0. \end{align*} Since $\gamma_2(\mathfrak h_3(F))\neq 0$ but $\gamma_3(\mathfrak h_3(F))=0$, the Heisenberg algebra is nilpotent of class $2$. [/example] ## Nilpotent Adjoint Operators How does Lie-algebra nilpotence appear inside linear algebra? Every element $x\in\mathfrak g$ acts on $\mathfrak g$ by the adjoint endomorphism $\operatorname{ad}_x(y)=[x,y]$. The obstruction is that nilpotence is a global statement about all iterated brackets, while linear algebra sees one operator at a time. If long repeated brackets vanish, then repeated application of each individual $\operatorname{ad}_x$ must vanish as an endomorphism. [definition: Nilpotent Endomorphism] Let $V$ be a vector space over $F$. An endomorphism $T\in \operatorname{End}_F(V)$ is nilpotent if there exists $m\geq 1$ such that \begin{align*} T^m=0. \end{align*} [/definition] In finite dimension this means that the only eigenvalue of $T$ over an [algebraic closure](/page/Algebraic%20Closure) is $0$, and its Jordan normal form consists only of nilpotent Jordan blocks. The course uses this linear-algebraic notion as the bridge to Engel's theorem in the next chapter. [definition: Nilpotent Adjoint Operator] Let $\mathfrak g$ be a Lie algebra and let $x\in\mathfrak g$. The adjoint operator associated to $x$ is \begin{align*} \operatorname{ad}_x:\mathfrak g\to\mathfrak g,\qquad \operatorname{ad}_x(y)=[x,y]. \end{align*} It is called nilpotent if it is nilpotent as an endomorphism of the vector space $\mathfrak g$. [/definition] The next bridge is from intrinsic brackets to linear operators. Nilpotence of the Lie algebra should force each adjoint action to lose information after enough iterations, because repeated applications of $\operatorname{ad}_x$ are special cases of repeated commutators. The point to check is uniform structural: the lower central series controls these powers for every element $x$, not just for elements lying deep in the series. [quotetheorem:3795] [citeproof:3795] The theorem uses the full nilpotence hypothesis, not just solvability: in the two-dimensional algebra with $[h,x]=x$, the operator $\operatorname{ad}_h$ has eigenvalue $1$ on $Fx$, so it is not nilpotent. What the theorem does not say is that nilpotence of a few selected adjoint operators is enough; the converse requires all adjoint operators and finite-dimensional hypotheses. That converse is much deeper: for finite-dimensional Lie algebras, if every $\operatorname{ad}_x$ is nilpotent, then $\mathfrak g$ is nilpotent. That statement is Engel's theorem, and the next chapter develops the representation-theoretic argument needed to prove it. [remark: Why Engel's Theorem Is Not Formal] The implication from nilpotent Lie algebra to nilpotent adjoint operators follows directly from definitions. The reverse implication is not a manipulation of brackets alone: it requires finding a nonzero vector killed by all operators in a Lie algebra of nilpotent endomorphisms. This is why the proof belongs naturally after the present chapter, once nilpotence has been connected to linear operators. [/remark] Nilpotence gives the first setting where operator-theoretic arguments can recover intrinsic algebraic structure. That bridge is exactly what Engel's theorem makes precise, so the next chapter turns from definitions to the key representation-theoretic criterion. # 6. Engel's Theorem Engel's theorem is the first major result in the course where a condition on a representation forces a structural conclusion about a Lie algebra. The guiding question is: if every element acts nilpotently, must the algebra itself have a nilpotent shape? We first prove a representation-theoretic version for Lie algebras of endomorphisms, then apply it to the adjoint representation to obtain the abstract criterion for nilpotence. ## Nilpotent Endomorphisms and Invariant Vectors Suppose a Lie algebra $\mathfrak g \subseteq \mathfrak{gl}(V)$ consists entirely of nilpotent linear maps. A single nilpotent map has a non-zero kernel, but the theorem needs a vector killed by every element of $\mathfrak g$ at once. The key problem is to convert individual nilpotence into a common invariant vector. [definition: Nilpotent Endomorphism] Let $V$ be a vector space over a field $F$. An endomorphism $T \in \operatorname{End}_F(V)$ is nilpotent if there exists $m \in \mathbb N$ such that $T^m = 0$. [/definition] The integer $m$ may depend on $T$. In Engel's theorem the hypothesis is pointwise nilpotence: every element of the given Lie algebra is nilpotent as an endomorphism of the same finite-dimensional space. Individual nilpotent operators each have nonzero kernels, but those kernels need not meet in an evident way. Engel's argument needs a name for the stronger outcome that one nonzero vector is killed by the entire Lie algebra of operators at the same time. [definition: Common Zero Vector] Let $\mathfrak g \subseteq \mathfrak{gl}(V)$ be a Lie algebra of endomorphisms. A common zero vector for $\mathfrak g$ is a non-zero vector $v \in V$ such that $xv = 0$ for every $x \in \mathfrak g$. [/definition] Finding such a vector is stronger than finding an invariant line: the line $Fv$ is invariant and the induced action on it is zero. This is the starting point for triangularising the action. The obstruction is simultaneousness. A nilpotent endomorphism by itself supplies a nonzero kernel, but a Lie algebra of endomorphisms supplies many kernels that could a priori miss one another. The following result is the representation-theoretic heart of Engel's theorem: pointwise nilpotence forces a common zero vector. [quotetheorem:3796] [citeproof:3796] The result converts many separate nilpotence assumptions into one shared invariant vector, which is exactly the seed needed for triangularisation. Its hypotheses are deliberately global: every element of the same finite-dimensional Lie algebra of operators must be nilpotent on the same space. Without that common finite-dimensional setting, individual kernels need not provide a usable common starting line. [example: Failure of a Common Kernel Without Engel's Hypothesis] Let $V=F^2$ with standard basis $e_1,e_2$, and let \begin{align*} T=\begin{pmatrix}1&0\\0&0\end{pmatrix}. \end{align*} For a vector $v=ae_1+be_2$, matrix multiplication gives \begin{align*} Tv &= \begin{pmatrix}1&0\\0&0\end{pmatrix} \begin{pmatrix}a\\ b\end{pmatrix} = \begin{pmatrix}a\\0\end{pmatrix} = ae_1. \end{align*} Thus $Tv=0$ if and only if $a=0$, so $\ker T=Fe_2$. In particular, \begin{align*} Te_1=e_1\ne 0, \end{align*} so $T$ does not kill all of $V$. Also \begin{align*} T^m e_1=e_1 \quad \text{for every } m\ge 1, \end{align*} because $T e_1=e_1$ and repeated application leaves $e_1$ fixed. Hence $T^m\ne 0$ for every $m\ge 1$, so $T$ is not nilpotent. For the one-dimensional Lie algebra $FT$, the common zero space is \begin{align*} \{v\in V:\lambda T v=0 \text{ for every } \lambda\in F\} &=\{v\in V:Tv=0\}\\ &=Fe_2. \end{align*} This subspace is non-zero, but it is not all of $V$ since $e_1\notin Fe_2$. The example shows exactly where Engel's hypothesis enters: without nilpotence, an element may have a non-zero action on part of the space, as $T$ does on $e_1$. [/example] A common zero vector is only the first step. Once one invariant line has been found, the same question can be asked on the quotient by that line; iterating this process should produce a basis in which every operator acts by strictly upper triangular matrices. The next result packages that iteration into a simultaneous normal form. [quotetheorem:3797] [citeproof:3797] This theorem says that a Lie algebra of nilpotent operators is simultaneously put into a nilpotent normal form. The point is simultaneous triangularisation, not merely Jordan form for each separate operator. The finite-dimensional hypothesis is doing real work, since the proof constructs a finite flag and terminates only after finitely many quotient steps. Pointwise nilpotence is also stronger than saying that the Lie algebra is generated by nilpotent maps: two nilpotent endomorphisms may have a sum that is not nilpotent, so the hypothesis must apply to every linear combination lying in $\mathfrak g$. Thus Engel's theorem is a simultaneous statement about the whole Lie algebra, not a separate Jordan-form statement applied operator by operator. [example: Nilpotent Generators Are Not Enough] Let $V=F^2$ and suppose $\operatorname{char}F\ne 2$. Put \begin{align*} E &= \begin{pmatrix}0&1\\0&0\end{pmatrix}, & F_0 &= \begin{pmatrix}0&0\\1&0\end{pmatrix}. \end{align*} The two displayed generators are nilpotent because \begin{align*} E^2 &= \begin{pmatrix}0&1\\0&0\end{pmatrix} \begin{pmatrix}0&1\\0&0\end{pmatrix} = \begin{pmatrix}0\cdot 0+1\cdot 0&0\cdot 1+1\cdot 0\\0\cdot 0+0\cdot 0&0\cdot 1+0\cdot 0\end{pmatrix} = \begin{pmatrix}0&0\\0&0\end{pmatrix},\\ F_0^2 &= \begin{pmatrix}0&0\\1&0\end{pmatrix} \begin{pmatrix}0&0\\1&0\end{pmatrix} = \begin{pmatrix}0\cdot 0+0\cdot 1&0\cdot 0+0\cdot 0\\1\cdot 0+0\cdot 1&1\cdot 0+0\cdot 0\end{pmatrix} = \begin{pmatrix}0&0\\0&0\end{pmatrix}. \end{align*} However their sum is not nilpotent. Indeed, \begin{align*} E+F_0 &= \begin{pmatrix}0&1\\1&0\end{pmatrix}, \end{align*} and \begin{align*} (E+F_0)^2 &= \begin{pmatrix}0&1\\1&0\end{pmatrix} \begin{pmatrix}0&1\\1&0\end{pmatrix}\\ &= \begin{pmatrix}0\cdot 0+1\cdot 1&0\cdot 1+1\cdot 0\\1\cdot 0+0\cdot 1&1\cdot 1+0\cdot 0\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix} =I. \end{align*} It follows by induction that $(E+F_0)^{2m}=I$ for every $m\ge 1$, so no positive power of $E+F_0$ is zero. Thus any Lie subalgebra of $\mathfrak{gl}(V)$ containing both $E$ and $F_0$ also contains their linear combination $E+F_0$, and therefore does not satisfy Engel's hypothesis. The example shows why the theorem requires every element of the Lie algebra to be nilpotent, not merely a chosen generating set of operators. [/example] The next example gives the model where Engel's hypothesis does hold. Strictly upper triangular matrices are the concrete operators whose repeated action moves basis vectors steadily upward until nothing remains. [example: Strictly Upper Triangular Matrices] Let $\mathfrak n_n(F)$ be the Lie algebra of strictly upper triangular $n\times n$ matrices over $F$, with bracket $[A,B]=AB-BA$, and let $e_1,\dots,e_n$ be the standard basis of $F^n$. If $A=(a_{ij})\in \mathfrak n_n(F)$, then $a_{ij}=0$ whenever $i\ge j$. Hence, for each basis vector $e_j$, \begin{align*} Ae_j &= \begin{pmatrix} a_{1j}\\ a_{2j}\\ \vdots\\ a_{nj} \end{pmatrix} = \sum_{i=1}^{j-1} a_{ij}e_i \in \operatorname{span}(e_1,\dots,e_{j-1}), \end{align*} where the terms with $i\ge j$ vanish because $A$ is strictly upper triangular. We now show that every $A\in \mathfrak n_n(F)$ is nilpotent. From the displayed containment, $Ae_j\in \operatorname{span}(e_1,\dots,e_{j-1})$. Applying $A$ again gives \begin{align*} A^2e_j &\in A\operatorname{span}(e_1,\dots,e_{j-1})\\ &\subseteq \operatorname{span}(Ae_1,\dots,Ae_{j-1})\\ &\subseteq \operatorname{span}(e_1,\dots,e_{j-2}). \end{align*} Continuing by induction, \begin{align*} A^m e_j \in \operatorname{span}(e_1,\dots,e_{j-m}) \end{align*} for $0\le m\le j$, with the convention that $\operatorname{span}(e_1,\dots,e_0)=0$. Taking $m=j$ gives $A^j e_j=0$, and therefore $A^n e_j=0$ for every $j=1,\dots,n$. Since $A^n$ kills every vector in the basis, $A^n=0$. The standard flag is also the flag returned by the triangular form. Let \begin{align*} V_r=\operatorname{span}(e_1,\dots,e_r). \end{align*} If $v\in V_r$, write $v=\sum_{j=1}^r c_j e_j$. Then \begin{align*} Av &=A\left(\sum_{j=1}^r c_j e_j\right)\\ &=\sum_{j=1}^r c_j Ae_j\\ &\in \sum_{j=1}^r \operatorname{span}(e_1,\dots,e_{j-1})\\ &\subseteq V_r. \end{align*} Thus each $V_r$ is stable under every $A\in \mathfrak n_n(F)$, and in fact $A(V_r)\subseteq V_{r-1}$. So Engel's hypothesis holds for $\mathfrak n_n(F)$, and the simultaneous strictly upper triangular basis is the original standard basis. [/example] ## The Abstract Engel Criterion The representation-theoretic theorem becomes a theorem about Lie algebras once we let a Lie algebra act on itself. The problem is that the adjoint representation forgets the centre, so the argument has to use both the image of $\operatorname{ad}$ and the kernel $Z(\mathfrak g)$. [definition: Nilpotent Lie Algebra] Let $\mathfrak g$ be a Lie algebra over a field $F$. As in Chapter 5, define its lower central series by \begin{align*} \gamma_1(\mathfrak g) &= \mathfrak g, & \gamma_{i+1}(\mathfrak g) &= [\mathfrak g, \gamma_i(\mathfrak g)] \quad (i \ge 1). \end{align*} The Lie algebra $\mathfrak g$ is nilpotent if there exists $c \in \mathbb N$ such that $\gamma_{c+1}(\mathfrak g)=0$. [/definition] Nilpotence means repeated commutators eventually vanish when bracketing from the left by arbitrary elements. This is a Lie-algebra analogue of strictly upper triangular matrices, whose iterated commutators move entries farther above the diagonal. The remaining issue is whether this internal condition can be detected from the adjoint operators alone. Because $\operatorname{ad}_x$ records bracketing by $x$, nilpotence of all adjoint maps is the natural operator-level test; the subtle point is recovering a central series for $\mathfrak g$ from that test. [quotetheorem:3798] [citeproof:3798] The theorem is often stated as a criterion: instead of checking the lower central series directly, it suffices in finite dimension to check nilpotence of all adjoint maps. The use of the adjoint representation has one subtlety: $\operatorname{ad}$ always kills $Z(\mathfrak g)$, so adjoint information alone cannot distinguish central elements from zero. Engel's argument turns this kernel from a defect into the induction step, because a non-zero common kernel is precisely a non-zero centre. Finite-dimensionality remains essential here, since the proof repeatedly quotients by central ideals and must terminate. [quotetheorem:3799] [citeproof:3799] The forward direction explains why the condition is the right one: in a nilpotent Lie algebra, every single adjoint action must eventually push elements beyond the last non-zero commutator layer. The reverse direction is the deep part because it reconstructs a global central series from separate nilpotence statements about the maps $\operatorname{ad}x$. The criterion is special to finite-dimensional Lie algebras; it should not be read as saying that all representations by nilpotent-looking operators automatically encode nilpotence of the abstract algebra. It also distinguishes nilpotent Lie algebras from merely solvable ones, as the affine two-dimensional example below shows. [example: Heisenberg Algebra] Let $\mathfrak h_3(F)$ have basis $x,y,z$, with $[x,y]=z$ and with $z$ central, meaning \begin{align*} [z,x]=[z,y]=[z,z]=0. \end{align*} By skew-symmetry of the Lie bracket, $[y,x]=-[x,y]=-z$, and also $[x,x]=[y,y]=0$. Therefore the first non-zero commutator layer is \begin{align*} [\mathfrak h_3,\mathfrak h_3] &=\operatorname{span}\{[x,x],[x,y],[x,z],[y,x],[y,y],[y,z],[z,x],[z,y],[z,z]\}\\ &=\operatorname{span}\{0,z,0,-z,0,0,0,0,0\}\\ &=Fz. \end{align*} The next lower central term is \begin{align*} [\mathfrak h_3,Fz] &=\operatorname{span}\{[x,z],[y,z],[z,z]\}\\ &=\operatorname{span}\{0,0,0\}\\ &=0, \end{align*} because $z$ is central. Hence \begin{align*} \gamma_1(\mathfrak h_3)&=\mathfrak h_3,& \gamma_2(\mathfrak h_3)&=Fz,& \gamma_3(\mathfrak h_3)&=0. \end{align*} Since $\gamma_2(\mathfrak h_3)\ne 0$ but $\gamma_3(\mathfrak h_3)=0$, the algebra is nilpotent of class $2$. Now take an arbitrary element $u=ax+by+cz\in \mathfrak h_3(F)$. Its adjoint action on the basis is \begin{align*} \operatorname{ad}u(x) &=[u,x]\\ &=[ax+by+cz,x]\\ &=a[x,x]+b[y,x]+c[z,x]\\ &=a\cdot 0+b(-z)+c\cdot 0\\ &=-bz,\\ \operatorname{ad}u(y) &=[u,y]\\ &=[ax+by+cz,y]\\ &=a[x,y]+b[y,y]+c[z,y]\\ &=az+b\cdot 0+c\cdot 0\\ &=az,\\ \operatorname{ad}u(z) &=[u,z]\\ &=[ax+by+cz,z]\\ &=a[x,z]+b[y,z]+c[z,z]\\ &=0. \end{align*} Thus $\operatorname{ad}u$ sends each basis vector into $Fz$, and it kills $z$. Therefore, for every basis vector $v\in\{x,y,z\}$, the vector $\operatorname{ad}u(v)$ lies in $Fz$, and applying $\operatorname{ad}u$ again gives $0$. Hence \begin{align*} (\operatorname{ad}u)^2(x)&=0,& (\operatorname{ad}u)^2(y)&=0,& (\operatorname{ad}u)^2(z)&=0. \end{align*} Since $(\operatorname{ad}u)^2$ kills the basis $x,y,z$, we have $(\operatorname{ad}u)^2=0$. This shows concretely that the nilpotence of $\mathfrak h_3(F)$ and the nilpotence of all adjoint maps agree in this example. [/example] The next example marks the boundary of that agreement. The two-dimensional affine algebra is solvable, but its adjoint action contains a non-nilpotent scaling direction. [example: The Two-Dimensional Affine Lie Algebra] Let $\mathfrak a$ have basis $h,e$ with bracket $[h,e]=e$. By skew-symmetry, \begin{align*} [e,h]&=-[h,e]=-e, \end{align*} and bilinearity with skew-symmetry gives $[h,h]=[e,e]=0$. Hence \begin{align*} [\mathfrak a,\mathfrak a] &=\operatorname{span}\{[h,h],[h,e],[e,h],[e,e]\}\\ &=\operatorname{span}\{0,e,-e,0\}\\ &=Fe. \end{align*} The next lower central term is \begin{align*} [\mathfrak a,Fe] &=\operatorname{span}\{[h,\lambda e],[e,\lambda e]:\lambda\in F\}\\ &=\operatorname{span}\{\lambda[h,e],\lambda[e,e]:\lambda\in F\}\\ &=\operatorname{span}\{\lambda e,0:\lambda\in F\}\\ &=Fe. \end{align*} Thus \begin{align*} \gamma_1(\mathfrak a)&=\mathfrak a,& \gamma_2(\mathfrak a)&=Fe,& \gamma_3(\mathfrak a)&=Fe. \end{align*} If $\gamma_m(\mathfrak a)=Fe$ for some $m\ge 2$, then \begin{align*} \gamma_{m+1}(\mathfrak a) &=[\mathfrak a,\gamma_m(\mathfrak a)]\\ &=[\mathfrak a,Fe]\\ &=Fe, \end{align*} so by induction $\gamma_m(\mathfrak a)=Fe$ for every $m\ge 2$. Since $e\ne 0$, the lower central series never reaches $0$, and $\mathfrak a$ is not nilpotent. Now compute one adjoint map. On the basis $h,e$, \begin{align*} \operatorname{ad}h(h)&=[h,h]=0,\\ \operatorname{ad}h(e)&=[h,e]=e. \end{align*} Therefore \begin{align*} (\operatorname{ad}h)^m(e)=e \end{align*} for every $m\ge 1$, by induction from $\operatorname{ad}h(e)=e$. Hence $(\operatorname{ad}h)^m\ne 0$ for every $m\ge 1$, so $\operatorname{ad}h$ is not nilpotent. This is exactly the obstruction detected by *Engel Criterion*: the affine two-dimensional Lie algebra is solvable, but it is not nilpotent. [/example] ## Consequences for the Centre and the Upper Central Series Engel's theorem has a useful structural consequence: finite-dimensional nilpotent Lie algebras are built by repeatedly adjoining central layers. The natural question is how to recognise this construction intrinsically. [definition: Upper Central Series] Let $\mathfrak g$ be a Lie algebra over a field $F$. The upper central series is the increasing sequence of ideals $Z_i(\mathfrak g)$ defined by \begin{align*} Z_0(\mathfrak g) &= 0, & Z_{i+1}(\mathfrak g)/Z_i(\mathfrak g) &= Z(\mathfrak g/Z_i(\mathfrak g)) \quad (i \ge 0). \end{align*} [/definition] The first term after zero is $Z_1(\mathfrak g)=Z(\mathfrak g)$. Each new layer consists of elements that become central after quotienting out the earlier layers. This ascending construction should match nilpotence precisely when it eventually exhausts the algebra. The issue is to compare two opposite filtrations: the lower central series measures how commutators descend toward zero, while the upper central series measures how central layers build the algebra back up. [quotetheorem:3800] [citeproof:3800] This result gives a second way of seeing nilpotence: the algebra can be peeled apart by central quotients. It is especially useful when classifying low-dimensional examples, because central extensions are usually easier to control than arbitrary brackets. The converse direction also shows a limitation: having non-zero centre is only the first step, not nilpotence by itself. For example, the direct sum of a one-dimensional central Lie algebra with the two-dimensional affine Lie algebra has non-zero centre but is not nilpotent, since its affine summand still has a non-terminating lower central series. [quotetheorem:3801] [citeproof:3801] The non-zero centre is the bridge between lower central information and inductive arguments. The non-zero assumption excludes the zero Lie algebra, whose centre is zero by equality but whose nilpotence is degenerate. Finite-dimensionality is inherited from Engel's theorem in this proof; in broader infinite-dimensional settings, separate care is needed because descending or ascending central constructions need not terminate. The theorem also has only one direction: a non-zero centre does not force nilpotence, as the central direct-sum modification of the affine Lie algebra shows. Many proofs about nilpotent Lie algebras start by quotienting by a one-dimensional central ideal, but they still need the remaining quotient to retain nilpotent structure. [example: Upper Central Series of Strictly Upper Triangular Matrices] Let $\mathfrak n_n(F)$ have standard matrix units $E_{ij}$ for $1\le i<j\le n$. Their products satisfy \begin{align*} E_{ij}E_{kl} &=\delta_{jk}E_{il},& E_{kl}E_{ij} &=\delta_{li}E_{kj}, \end{align*} so their Lie bracket is \begin{align*} [E_{ij},E_{kl}] &=E_{ij}E_{kl}-E_{kl}E_{ij}\\ &=\delta_{jk}E_{il}-\delta_{li}E_{kj}. \end{align*} First compute the centre. If \begin{align*} X=\sum_{1\le i<j\le n} x_{ij}E_{ij}\in \mathfrak n_n(F), \end{align*} then for $1\le k<n$, \begin{align*} [X,E_{k,k+1}] &=\sum_{i<j}x_{ij}[E_{ij},E_{k,k+1}]\\ &=\sum_{i<j}x_{ij}\left(\delta_{jk}E_{i,k+1}-\delta_{k+1,i}E_{kj}\right)\\ &=\sum_{i<k}x_{ik}E_{i,k+1}-\sum_{k+1<j}x_{k+1,j}E_{kj}. \end{align*} If $X$ is central, this bracket is zero for every $k$. The coefficient of $E_{i,k+1}$ gives $x_{ik}=0$ whenever $i<k$, and the coefficient of $E_{kj}$ gives $x_{k+1,j}=0$ whenever $k+1<j$. As $k$ varies, every coefficient $x_{ij}$ with $(i,j)\ne (1,n)$ is forced to be zero. Conversely, \begin{align*} [E_{1n},E_{kl}] &=\delta_{nk}E_{1l}-\delta_{l1}E_{kn} =0, \end{align*} because $k<n$ and $l>1$ for every $E_{kl}\in \mathfrak n_n(F)$. Hence \begin{align*} Z(\mathfrak n_n(F))=F E_{1n}. \end{align*} For the higher layers, define \begin{align*} L_r=\operatorname{span}\{E_{ij}:1\le i<j\le n,\ j-i\ge n-r\} \end{align*} for $0\le r\le n-1$, with $L_0=0$. We show that $Z_r(\mathfrak n_n(F))=L_r$. The case $r=1$ is the centre computation above. Assume $Z_r(\mathfrak n_n(F))=L_r$. An element $E_{ij}$ lies in $Z_{r+1}(\mathfrak n_n(F))$ exactly when its image in $\mathfrak n_n(F)/L_r$ is central, which means \begin{align*} [E_{ij},E_{kl}]\in L_r \end{align*} for every $k<l$. Using the bracket formula, the only possible non-zero brackets are \begin{align*} [E_{ij},E_{j l}]&=E_{i l} && (j<l),\\ [E_{ij},E_{k i}]&=-E_{k j} && (k<i). \end{align*} Their gaps are \begin{align*} l-i&=(j-i)+(l-j)\ge (j-i)+1,\\ j-k&=(j-i)+(i-k)\ge (j-i)+1. \end{align*} Thus, if $j-i\ge n-(r+1)$, then every non-zero bracket has gap at least $n-r$, so it lies in $L_r$. Hence $L_{r+1}\subseteq Z_{r+1}(\mathfrak n_n(F))$. Conversely, if $j-i<n-(r+1)$, then $j-i+1<n-r$. If $i>1$, take $E_{i-1,i}$. Then \begin{align*} [E_{ij},E_{i-1,i}] &=-E_{i-1,j}, \end{align*} and the gap of $E_{i-1,j}$ is $j-i+1<n-r$, so this bracket is not in $L_r$. If $i=1$, then $j<n-r$, so in particular $j<n$; taking $E_{j,j+1}$ gives \begin{align*} [E_{1j},E_{j,j+1}] &=E_{1,j+1}, \end{align*} whose gap is again $j<n-r$, so it is not in $L_r$. Therefore no $E_{ij}$ with $j-i<n-(r+1)$ becomes central modulo $L_r$, and \begin{align*} Z_{r+1}(\mathfrak n_n(F))=L_{r+1}. \end{align*} By induction, \begin{align*} Z_r(\mathfrak n_n(F)) &=\operatorname{span}\{E_{ij}:j-i\ge n-r\}. \end{align*} When $r=n-1$, the condition $j-i\ge 1$ includes every strictly upper triangular matrix unit, so \begin{align*} Z_{n-1}(\mathfrak n_n(F))=\mathfrak n_n(F). \end{align*} The upper central series reaches the whole algebra exactly by adding one superdiagonal at each step, starting from the outermost entry $E_{1n}$. [/example] Engel's theorem therefore completes the transition from concrete matrix nilpotence to intrinsic Lie-algebra nilpotence. It also prepares the course for Lie's theorem, where solvable Lie algebras are studied through triangular rather than strictly triangular representations. Engel's theorem completes the passage from nilpotent operators to nilpotent Lie algebras. With that result established, we can move on to solvable algebras and ask how far their representations can be simplified by triangularization. # 7. Lie's Theorem and Triangularization This chapter explains how solvability becomes visible inside finite-dimensional representations. It assumes representations from Chapter 3, ideals and quotients from Chapter 2, derived algebras and solvability from Chapter 4, and Engel's theorem from Chapter 6. Engel's theorem converted nilpotence of operators into a structural statement about nilpotent Lie algebras; Lie's theorem is the parallel result for solvable Lie algebras over an algebraically closed field of characteristic zero. The output is a basis in which the whole representation is triangular, so the derived algebra appears on the strictly upper triangular part. ## From Solvability to a Common Eigenvector The first problem is to find a vector that every element of a solvable Lie algebra preserves up to scalar. For one operator this is the existence of an eigenvector, which uses algebraic closure. Lie's theorem says that solvability supplies enough compatibility between the operators to make a single eigenline work for the whole Lie algebra. [definition: Weight Vector] Let $L$ be a Lie algebra over a field $F$, let $V$ be an $L$-module, and let $\rho:L\to \mathfrak{gl}(V)$ be the corresponding representation. A nonzero vector $v\in V$ is a weight vector of weight $\lambda\in L^*$ if \begin{align*} \rho(x)v=\lambda(x)v \end{align*} for every $x\in L$. [/definition] A weight vector is the same thing as a one-dimensional submodule $Fv\subset V$. The terminology emphasizes that the scalars assemble into a linear functional on $L$, because $\rho(ax+by)v=(a\lambda(x)+b\lambda(y))v$. [example: Upper Triangular Matrices Have a Common Eigenvector] Let $L\subset \mathfrak{gl}_n(F)$ be the Lie algebra of upper triangular matrices, acting on $V=F^n$ by matrix multiplication. For $x=(x_{ij})\in L$, the first basis vector is the column vector \begin{align*} e_1= \begin{pmatrix} 1\\ 0\\ \vdots\\ 0 \end{pmatrix}. \end{align*} The $i$th coordinate of $xe_1$ is \begin{align*} (xe_1)_i=\sum_{j=1}^n x_{ij}(e_1)_j=x_{i1}(e_1)_1+\sum_{j=2}^n x_{ij}(e_1)_j=x_{i1}. \end{align*} Since $x$ is upper triangular, $x_{i1}=0$ for every $i>1$, while the first coordinate is $x_{11}$. Hence \begin{align*} xe_1= \begin{pmatrix} x_{11}\\ 0\\ \vdots\\ 0 \end{pmatrix} =x_{11}e_1. \end{align*} Therefore $Fe_1$ is preserved by every $x\in L$, and $e_1$ is a weight vector with weight $\lambda:L\to F$ given by $\lambda(x)=x_{11}$. The functional is linear because \begin{align*} \lambda(ax+by)=(ax+by)_{11}=ax_{11}+by_{11}=a\lambda(x)+b\lambda(y). \end{align*} This is the target form of Lie's theorem: in a triangular basis, the first coordinate line is automatically a common eigenline. [/example] The example shows why an invariant eigenline is the first visible sign of triangular form. For a general solvable Lie algebra of operators, the problem is to prove that such a line exists before any triangular basis has been chosen. Lie's theorem supplies exactly that starting line under the standard finite-dimensional hypotheses. [quotetheorem:3802] [citeproof:3802] Lie's theorem is best read as a structural threshold: solvability is strong enough to force at least one common eigenline, while a non-solvable algebra such as $\mathfrak{sl}_2(F)$ acting on its standard two-dimensional module has no invariant line. Algebraic closure is needed because even a one-dimensional abelian Lie algebra over $\mathbb R$ can act by a rotation with no real eigenvector, and characteristic zero is part of the standard setting in which the weight argument works. The theorem does not say that the representation is semisimple or diagonalizable; it only gives the first invariant line, which is the seed for the flag and triangularization results below. [remark: Why Characteristic Zero Enters] The standard proof uses characteristic zero when showing that a chosen weight space is preserved in the induction step. In positive characteristic, the binomial-coefficient and trace-type arguments behind the stability of weights can fail, and solvable Lie algebras may have finite-dimensional irreducible modules of dimension greater than one. [/remark] ## Invariant Flags and Triangular Bases Once a single invariant line has been found, the next problem is to continue the process on the quotient. Repeating the argument produces a complete flag of invariant subspaces, which is precisely the coordinate-free form of simultaneous triangularization. [definition: Invariant Flag] Let $V$ be a finite-dimensional $L$-module with $\dim V=n$. A complete invariant flag in $V$ is a chain of $L$-submodules \begin{align*} 0=V_0\subset V_1\subset \cdots \subset V_n=V \end{align*} with $\dim V_i=i$ for each $i$. [/definition] Choosing a basis $v_1,\dots,v_n$ adapted to such a flag, so that $V_i=\operatorname{span}(v_1,\dots,v_i)$, means that every representing matrix preserves each coordinate subspace. That condition is exactly upper triangularity. The invariant-line theorem becomes more useful when it can be iterated. The question is whether the first invariant line is only an isolated submodule or the beginning of a full coordinate system in which every operator has triangular form. The following result packages that iteration as the normal form attached to solvable representations. [quotetheorem:3803] [citeproof:3803] This theorem should be read as a representation-theoretic normal form. The assumptions are the same as in Lie's theorem because the proof repeatedly applies the invariant-line result: over $\mathbb R$ a rotation blocks even the first step, and in positive characteristic the common eigenvector statement can fail. [Finite dimensionality](/theorems/1534) is also essential to make the induction terminate and to interpret a complete flag as a basis. The theorem does not say that the operators commute, and it does not diagonalize them; it says that solvability is exactly strong enough to order the basis so that all actions move vectors only upward in the flag. This is the form used later when trace arguments detect solvable ideals inside larger Lie algebras. [example: The Standard Module For Upper Triangular Matrices] Let $L=\mathfrak{b}_n(F)$ be the Lie algebra of all upper triangular $n\times n$ matrices, and let $V=F^n$ have standard basis $e_1,\dots,e_n$. For $1\leq k\leq n$, set \begin{align*} V_k=\operatorname{span}(e_1,\dots,e_k), \end{align*} so the coordinate flag is \begin{align*} 0\subset V_1\subset V_2\subset\cdots\subset V_n=F^n. \end{align*} We verify that each $V_k$ is preserved by every $x=(x_{ij})\in L$. If \begin{align*} v=a_1e_1+\cdots+a_ke_k\in V_k, \end{align*} then the $i$th coordinate of $xv$ is \begin{align*} (xv)_i=\sum_{j=1}^n x_{ij}v_j=\sum_{j=1}^k x_{ij}a_j. \end{align*} For $i>k$ and $1\leq j\leq k$, we have $i>j$, so upper triangularity gives $x_{ij}=0$. Hence \begin{align*} (xv)_i=\sum_{j=1}^k 0\cdot a_j=0 \end{align*} for every $i>k$, and therefore $xv\in V_k$. The induced action on the quotient $V_k/V_{k-1}$ is determined by the class of $e_k$. Since \begin{align*} xe_k= \begin{pmatrix} x_{1k}\\ x_{2k}\\ \vdots\\ x_{kk}\\ 0\\ \vdots\\ 0 \end{pmatrix} =x_{1k}e_1+x_{2k}e_2+\cdots+x_{kk}e_k, \end{align*} we have \begin{align*} xe_k+V_{k-1}=x_{kk}e_k+V_{k-1}. \end{align*} Thus the one-dimensional quotient $V_k/V_{k-1}$ is acted on by the scalar $x_{kk}$. The standard basis is therefore already a triangular basis, with the diagonal entries recording the actions on the successive quotient lines. [/example] The next example applies triangular form to the smallest non-abelian solvable algebra. It shows how the derived element loses its diagonal action once the representation is triangularized. [example: A Solvable Non-Abelian Two-Dimensional Algebra] Let $L$ have basis $h,e$ with bracket $[h,e]=e$, and let $V$ be a finite-dimensional module over an algebraically closed field of characteristic zero. If $V=0$, there is nothing to check, so assume $\dim V=n>0$. By *Simultaneous Triangularization For Solvable Representations*, choose a basis of $V$ in which the representing matrices \begin{align*} H=\rho(h),\qquad E=\rho(e) \end{align*} are both upper triangular. Because $\rho$ is a Lie algebra representation and $[h,e]=e$, we have \begin{align*} E=\rho(e)=\rho([h,e])=[\rho(h),\rho(e)]=HE-EH. \end{align*} Fix $1\leq i\leq n$. Since $H$ and $E$ are upper triangular, \begin{align*} (HE)_{ii} &=\sum_{k=1}^n H_{ik}E_{ki}\\ &=\sum_{k<i} H_{ik}E_{ki}+H_{ii}E_{ii}+\sum_{k>i}H_{ik}E_{ki}\\ &=\sum_{k<i} H_{ik}\cdot 0+H_{ii}E_{ii}+\sum_{k>i}H_{ik}\cdot 0\\ &=H_{ii}E_{ii}, \end{align*} where $E_{ki}=0$ for $k>i$ and $H_{ik}=0$ for $k<i$. Similarly, \begin{align*} (EH)_{ii}=E_{ii}H_{ii}. \end{align*} Taking the $i$th diagonal entry of $E=HE-EH$ gives \begin{align*} E_{ii}=(HE-EH)_{ii}=H_{ii}E_{ii}-E_{ii}H_{ii}=0. \end{align*} Thus every diagonal entry of $E$ is zero. Since $E$ is already upper triangular, $E$ is strictly upper triangular, which shows concretely how the derived element $e=[h,e]$ loses its diagonal part in a triangular basis. [/example] ## The Derived Algebra in Triangular Form The next question is what part of the triangular matrices is occupied by commutators. Upper triangular matrices have an abelian diagonal quotient, so any commutator of upper triangular matrices has zero diagonal. This observation turns the abstract derived algebra into a nilpotent-looking collection of operators. [quotetheorem:3804] [citeproof:3804] The conclusion is often the most useful form of Lie's theorem in computations. Its hypotheses enter through the triangularizing basis: if algebraic closure or characteristic zero fails, the representation may not be triangularizable and the diagonal-commutator argument cannot start. Solvability is also essential; in the standard representation of $\mathfrak{sl}_2(F)$, the derived algebra is all of $\mathfrak{sl}_2(F)$, and the diagonal element does not act nilpotently. The original solvable algebra may not act nilpotently, because diagonal parts can remain; its derived algebra loses those diagonal parts and therefore acts nilpotently after triangularization. This boundary between diagonal character data and nilpotent commutator action is the bridge to Cartan-type criteria. [remark: Nilpotent Action Versus Nilpotent Lie Algebra] The statement says that each element of $[L,L]$ acts nilpotently in the given finite-dimensional representation. It does not by itself say that $[L,L]$ is a nilpotent Lie algebra unless the representation is faithful or one passes to a faithful representation such as the adjoint representation under suitable hypotheses. [/remark] The next example computes the derived algebra in the standard triangular model. It shows explicitly how diagonal information disappears from commutators and why the derived algebra becomes strictly upper triangular. [example: Derived Algebra of Upper Triangular Matrices] Let $L=\mathfrak{b}_n(F)$ be the Lie algebra of upper triangular matrices, and let $\mathfrak{n}_n(F)$ be the subspace of strictly upper triangular matrices. We show that \begin{align*} [\mathfrak{b}_n(F),\mathfrak{b}_n(F)]=\mathfrak{n}_n(F). \end{align*} If $A,B\in \mathfrak{b}_n(F)$, then the $i$th diagonal entry of $AB$ is \begin{align*} (AB)_{ii} &=\sum_{k=1}^n A_{ik}B_{ki}\\ &=\sum_{k<i} A_{ik}B_{ki}+A_{ii}B_{ii}+\sum_{k>i}A_{ik}B_{ki}\\ &=\sum_{k<i}0\cdot B_{ki}+A_{ii}B_{ii}+\sum_{k>i}A_{ik}\cdot 0\\ &=A_{ii}B_{ii}, \end{align*} because $A_{ik}=0$ for $k<i$ and $B_{ki}=0$ for $k>i$. Similarly, $(BA)_{ii}=B_{ii}A_{ii}$, so \begin{align*} ([A,B])_{ii}=(AB-BA)_{ii}=A_{ii}B_{ii}-B_{ii}A_{ii}=0. \end{align*} Since $[A,B]$ is still upper triangular, every commutator lies in $\mathfrak{n}_n(F)$, and hence \begin{align*} [\mathfrak{b}_n(F),\mathfrak{b}_n(F)]\subseteq \mathfrak{n}_n(F). \end{align*} For the reverse inclusion, fix $i<j$, and let $E_{ij}$ be the matrix unit with a $1$ in position $(i,j)$ and zeros elsewhere. Let $D_i$ be the diagonal matrix with $(D_i)_{ii}=1$ and all other diagonal entries equal to $0$. Both $D_i$ and $E_{ij}$ lie in $\mathfrak{b}_n(F)$. For any entries $(a,b)$, \begin{align*} (D_iE_{ij})_{ab} &=\sum_{k=1}^n (D_i)_{ak}(E_{ij})_{kb}\\ &=(D_i)_{aa}(E_{ij})_{ab}\\ &=(D_i)_{aa}\delta_{ai}\delta_{bj} =\delta_{ai}\delta_{bj}, \end{align*} so $D_iE_{ij}=E_{ij}$. Also, since $j\ne i$, \begin{align*} (E_{ij}D_i)_{ab} &=\sum_{k=1}^n (E_{ij})_{ak}(D_i)_{kb}\\ &=(E_{ij})_{aj}(D_i)_{jb}\\ &=\delta_{ai}(D_i)_{jb} =0. \end{align*} Therefore \begin{align*} [D_i,E_{ij}]=D_iE_{ij}-E_{ij}D_i=E_{ij}. \end{align*} Thus every standard strictly upper triangular matrix unit $E_{ij}$ with $i<j$ belongs to $[\mathfrak{b}_n(F),\mathfrak{b}_n(F)]$, and since the $E_{ij}$ form a basis of $\mathfrak{n}_n(F)$, we get \begin{align*} \mathfrak{n}_n(F)\subseteq [\mathfrak{b}_n(F),\mathfrak{b}_n(F)]. \end{align*} Hence $[\mathfrak{b}_n(F),\mathfrak{b}_n(F)]=\mathfrak{n}_n(F)$. Finally, every $N\in \mathfrak{n}_n(F)$ acts nilpotently on the standard module. Indeed, a product of $m$ strictly upper triangular matrices has zero $(i,j)$ entry whenever $j-i<m$; for $m=n$, every pair $i<j$ satisfies $j-i\leq n-1<n$, and all entries below or on the diagonal are already zero. Thus $N^n=0$. The derived algebra is therefore exactly the part of the upper triangular algebra that has lost all diagonal character data and acts nilpotently. [/example] ## Irreducible Modules for Solvable Lie Algebras The last problem is to identify what irreducibility can mean for a solvable algebra. Lie's theorem produces a nonzero proper invariant subspace unless the module already has dimension one. Thus solvable Lie algebras have no higher-dimensional irreducible modules in this setting. [definition: Irreducible Module] Let $L$ be a Lie algebra over a field $F$. A nonzero $L$-module $V$ is irreducible if its only $L$-submodules are $0$ and $V$. [/definition] This is the Lie algebra analogue of a simple module over an associative algebra. The point is that the submodules are subspaces stable under all operators coming from the Lie algebra action. With this definition in place, Lie's theorem imposes a severe restriction. If a solvable algebra acts on a module of dimension greater than one, the invariant-line result produces a proper nonzero submodule, contradicting irreducibility. The next theorem records the resulting collapse of irreducible finite-dimensional modules to the one-dimensional case under Lie's theorem hypotheses. [quotetheorem:3805] [citeproof:3805] This is a consequence of Lie's theorem rather than a statement about all Lie algebras. If $L$ is not solvable, higher-dimensional irreducible modules can occur, as with the standard two-dimensional module for $\mathfrak{sl}_2(F)$. If the field is not algebraically closed or the characteristic is positive, the invariant-line step can fail, so irreducible solvable modules need not be one-dimensional. The result identifies the representation theory of solvable Lie algebras in this setting as character theory plus extensions between characters, which is why triangular flags rather than direct-sum decompositions dominate the chapter. Once irreducible modules have been reduced to dimension one, the remaining task is to identify exactly which linear functionals can occur. A one-dimensional module is governed by a functional $\lambda\in L^*$, but the Lie bracket imposes a compatibility condition because scalar endomorphisms commute. The next result expresses that condition as vanishing on the derived algebra, so the data factors through the abelianization $L/[L,L]$. [quotetheorem:3806] [citeproof:3806] Unlike Lie's theorem, this statement has no algebraic-closure, characteristic-zero, finite-dimensionality, or solvability hypothesis on $L$; it is a direct consequence of scalar endomorphisms commuting on a one-dimensional space. Its limitation is that it classifies only one-dimensional modules, not arbitrary irreducible modules when Lie's theorem is unavailable. Combined with the previous theorem, it describes all irreducible finite-dimensional modules of a solvable Lie algebra under the hypotheses of Lie's theorem: they are characters of the abelian quotient $L/[L,L]$. This quotient viewpoint will reappear whenever representations are decomposed into weights and commutator directions are forced to act nilpotently. [example: Irreducible Modules for a Borel Algebra] Let $L=\mathfrak{b}_n(F)$, and write $x=(x_{ij})$. A one-dimensional $L$-module has the form $V=Fv$, so every $x\in L$ acts by a scalar: \begin{align*} x\cdot v=\lambda(x)v \end{align*} for some linear functional $\lambda:L\to F$. Since $[\mathfrak{b}_n(F),\mathfrak{b}_n(F)]=\mathfrak{n}_n(F)$, *Characters Vanish on the Derived Algebra* gives \begin{align*} \lambda(\mathfrak{n}_n(F))=0. \end{align*} Thus $\lambda$ depends only on the diagonal entries of $x$. If $D_i$ denotes the diagonal matrix with $1$ in the $(i,i)$ entry and $0$ elsewhere, set \begin{align*} a_i=\lambda(D_i). \end{align*} Every upper triangular matrix decomposes as \begin{align*} x=x_{11}D_1+\cdots+x_{nn}D_n+N, \end{align*} where $N\in \mathfrak{n}_n(F)$. Therefore \begin{align*} \lambda(x) &=\lambda(x_{11}D_1+\cdots+x_{nn}D_n+N)\\ &=x_{11}\lambda(D_1)+\cdots+x_{nn}\lambda(D_n)+\lambda(N)\\ &=a_1x_{11}+\cdots+a_nx_{nn}. \end{align*} Hence every one-dimensional module is determined by scalars $a_1,\dots,a_n\in F$ via \begin{align*} x\cdot v=(a_1x_{11}+\cdots+a_nx_{nn})v. \end{align*} Conversely, any choice of $a_1,\dots,a_n\in F$ defines a linear functional \begin{align*} \lambda(x)=a_1x_{11}+\cdots+a_nx_{nn}. \end{align*} For $x,y\in\mathfrak{b}_n(F)$, the diagonal entries of $xy$ and $yx$ are \begin{align*} (xy)_{ii}=x_{ii}y_{ii},\qquad (yx)_{ii}=y_{ii}x_{ii}, \end{align*} so \begin{align*} ([x,y])_{ii}=(xy-yx)_{ii}=x_{ii}y_{ii}-y_{ii}x_{ii}=0. \end{align*} Thus $[x,y]\in\mathfrak{n}_n(F)$, and \begin{align*} \lambda([x,y])=0. \end{align*} It follows that \begin{align*} [x,y]\cdot v=0 \end{align*} and also \begin{align*} x\cdot(y\cdot v)-y\cdot(x\cdot v) &=x\cdot(\lambda(y)v)-y\cdot(\lambda(x)v)\\ &=\lambda(y)\lambda(x)v-\lambda(x)\lambda(y)v\\ &=0. \end{align*} So the formula indeed defines a one-dimensional representation. The strictly upper triangular part acts by zero because its diagonal entries are all zero. Since $\mathfrak{b}_n(F)$ is solvable, *[Irreducible Modules of Solvable Lie Algebras Are One-Dimensional](/theorems/3805)* says that these one-dimensional characters are the only finite-dimensional irreducible $\mathfrak{b}_n(F)$-modules. [/example] ## The Hypotheses and Their Limits The final question is why Lie's theorem is stated with algebraic closure, finite dimension, and characteristic zero. Each hypothesis prevents a specific failure mode: missing eigenvectors, infinite descending or ascending processes, or positive-characteristic representations that evade triangularization. [example: Algebraic Closure Is Needed] Let $F=\mathbb R$, let $L=\mathbb R x$ be the one-dimensional Lie algebra with $[x,x]=0$, and let $V=\mathbb R^2$ with $x$ acting by \begin{align*} R= \begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix}. \end{align*} Since $[L,L]$ is spanned by brackets $[ax,bx]=ab[x,x]=0$, we have $[L,L]=0$, so $L$ is solvable. We show that $V$ has no one-dimensional real $L$-submodule. Such a submodule would be a line $\mathbb R v$ with $0\ne v\in \mathbb R^2$, and preservation of that line by $x$ would mean that $Rv=\lambda v$ for some $\lambda\in\mathbb R$. Write \begin{align*} v= \begin{pmatrix} a\\ b \end{pmatrix}. \end{align*} Then \begin{align*} Rv &= \begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix} \begin{pmatrix} a\\ b \end{pmatrix} = \begin{pmatrix} -b\\ a \end{pmatrix}, \end{align*} so the equation $Rv=\lambda v$ becomes \begin{align*} -b&=\lambda a,\\ a&=\lambda b. \end{align*} Substituting $a=\lambda b$ into the first equation gives \begin{align*} -b=\lambda(\lambda b)=\lambda^2 b, \end{align*} hence \begin{align*} (\lambda^2+1)b=0. \end{align*} Since $\lambda\in\mathbb R$, we have $\lambda^2+1\ne 0$, so $b=0$. Then $a=\lambda b=0$, contradicting $v\ne 0$. Thus the rotation operator has no real eigenvector, so the solvable real Lie algebra $L$ has a finite-dimensional module with no invariant line. This is exactly the algebraic-closure obstruction in Lie's theorem. [/example] This example isolates the eigenvalue obstruction, while the next one isolates the characteristic obstruction. Both failures occur before any subtle structure theory is involved: the induction behind Lie's theorem needs an eigenvector for the last operator and needs characteristic zero to control the movement of weights. [example: Positive Characteristic Can Fail] Let $F$ be an algebraically closed field of characteristic $p>0$. Let $L$ have basis $h,e$ with bracket \begin{align*} [h,e]=e. \end{align*} Then \begin{align*} [L,L]=Fe,\qquad [Fe,Fe]=0, \end{align*} so $L$ is solvable. Define a $p$-dimensional vector space $V$ with basis $v_0,v_1,\dots,v_{p-1}$, and define linear maps $H,E\in\mathfrak{gl}(V)$ by \begin{align*} Hv_i=i\,v_i,\qquad Ev_i=v_{i+1} \end{align*} for $0\leq i\leq p-2$, and \begin{align*} Hv_{p-1}=(p-1)v_{p-1},\qquad Ev_{p-1}=v_0. \end{align*} We check that these maps define a representation of $L$ by setting $\rho(h)=H$ and $\rho(e)=E$. For $0\leq i\leq p-2$, \begin{align*} (HE-EH)v_i &=H(Ev_i)-E(Hv_i)\\ &=H v_{i+1}-E(i v_i)\\ &=(i+1)v_{i+1}-i v_{i+1}\\ &=v_{i+1}\\ &=Ev_i. \end{align*} For $i=p-1$, using $p=0$ in $F$, \begin{align*} (HE-EH)v_{p-1} &=H(Ev_{p-1})-E(Hv_{p-1})\\ &=Hv_0-E((p-1)v_{p-1})\\ &=0-(p-1)v_0\\ &=v_0\\ &=Ev_{p-1}. \end{align*} Hence $[H,E]=E$, which matches the relation $[h,e]=e$. We now show that $V$ has no one-dimensional $L$-submodule. Suppose $Fv$ were such a submodule with $v\ne 0$. Since it is stable under $H$, there is some $\lambda\in F$ such that $Hv=\lambda v$. Write \begin{align*} v=a_0v_0+a_1v_1+\cdots+a_{p-1}v_{p-1}. \end{align*} Then \begin{align*} Hv=\sum_{i=0}^{p-1} i a_i v_i, \qquad \lambda v=\sum_{i=0}^{p-1}\lambda a_i v_i. \end{align*} Comparing coefficients gives \begin{align*} (i-\lambda)a_i=0 \end{align*} for every $i$. The scalars $0,1,\dots,p-1$ are distinct in characteristic $p$, so at most one coefficient $a_i$ is nonzero. Thus $v$ is a scalar multiple of some $v_i$. But $E v_i=v_{i+1}$ for $i<p-1$, and $Ev_{p-1}=v_0$, so $Ev_i$ is never a scalar multiple of $v_i$. Therefore the line $Fv_i$ is not stable under $E$, contradicting that $Fv$ is an $L$-submodule. Hence this solvable Lie algebra has a finite-dimensional module with no common eigenvector, showing exactly where the characteristic-zero hypothesis in Lie's theorem can fail. [/example] The failure in positive characteristic clarifies the roles of Engel and Lie theorem. The following remark separates their hypotheses and conclusions before the course moves on to trace criteria. [remark: Relation With Engel Theorem] Engel's theorem starts from nilpotent operators and concludes that the Lie algebra has a central vector in every nonzero module, leading to nilpotence of the algebra in the adjoint representation. Lie's theorem starts from solvability of the algebra and concludes triangular form for every finite-dimensional representation under stronger field hypotheses. The two theorems are complementary: Engel controls nilpotent action, while Lie controls solvable action by reducing it to diagonal characters plus nilpotent commutator action. [/remark] These results mark the transition from elementary definitions to structure theory. In later chapters, triangularization will feed into Cartan's criterion and the analysis of semisimple Lie algebras, where solvable ideals are detected and excluded using trace forms. Lie's theorem shows that solvability is visible inside a representation through triangular form. Having reached that point, the natural next question is how to detect the deeper semisimple structure that remains once solvable behavior has been excluded, and the Killing form provides that test. # 8. The Killing Form and Cartan's Criterion This chapter explains how trace computations inside the adjoint representation detect structure inside a finite-dimensional Lie algebra. The central object is the Killing form, a symmetric [bilinear form](/page/Bilinear%20Form) built from $\operatorname{ad}$, and the central question is how much of the Lie bracket can be read from its orthogonality relation. The chapter continues the study of solvable ideals from Chapter 4 and nilpotent ideals from Chapter 5 by giving Cartan's criterion, which turns trace identities into solvability statements. The obstruction running through the chapter is that the bracket itself is not a scalar invariant: it changes form under a change of basis and is difficult to compare across ideals, quotients, and representations. Trace forms solve this by extracting numbers from endomorphisms in a basis-independent way. The restrictions in the hypotheses are not cosmetic: finite-dimensionality is needed for ordinary traces, characteristic $0$ prevents trace identities from collapsing for arithmetic reasons, and algebraic closedness is used when triangularisation and Jordan decomposition enter the proof of Cartan's criterion. ## Trace Forms and Invariance How can a Lie algebra produce bilinear forms without first choosing a basis or an inner product? The main source is a representation: endomorphisms have traces, and traces are unchanged under cyclic permutation. Applied to the adjoint representation, this gives a canonical form attached to the Lie algebra itself. Let $F$ be a field and let $\mathfrak g$ be a finite-dimensional Lie algebra over $F$. In this chapter we assume, unless stated otherwise, that $F$ has characteristic $0$. [definition: Trace Form Of A Representation] Let $\rho: \mathfrak g \to \mathfrak{gl}(V)$ be a finite-dimensional representation of $\mathfrak g$ over $F$. The trace form associated to $\rho$ is the bilinear form $B_\rho: \mathfrak g \times \mathfrak g \to F$ defined by \begin{align*} B_\rho(x,y) = \operatorname{tr}(\rho(x)\rho(y)). \end{align*} [/definition] The trace form is symmetric because $\operatorname{tr}(AB)=\operatorname{tr}(BA)$ for endomorphisms of a finite-dimensional vector space. It is also compatible with the Lie bracket in a way that makes it useful for ideals and quotients. [definition: Invariant Bilinear Form] Let $\mathfrak g$ be a Lie algebra over $F$. A bilinear form $B: \mathfrak g \times \mathfrak g \to F$ is invariant if \begin{align*} B([x,y],z)=B(x,[y,z]) \end{align*} for all $x,y,z \in \mathfrak g$. [/definition] Invariant forms are the Lie-algebra analogue of bilinear forms preserved by a [group action](/page/Group%20Action). The condition says that each operator $\operatorname{ad}y$ is skew-adjoint with respect to $B$, in the sense that $B([y,x],z)+B(x,[y,z])=0$. The trace form attached to a representation is the main source of such bilinear forms. To use trace forms structurally, one must know that their cyclic symmetry is compatible with the Lie bracket, not merely that the form is symmetric. The following theorem verifies this compatibility for every finite-dimensional representation. [quotetheorem:3807] [citeproof:3807] Trace forms turn representations into invariant bilinear forms, so they provide a way to convert operator data into intrinsic-looking algebraic data. This is useful because invariance lets brackets be moved from one input of the form to the other, making ideals and orthogonal complements interact with the Lie algebra structure rather than with the underlying vector space alone. The construction also explains why trace identities become structure theorems later: a vanishing trace condition can detect which directions act nilpotently, solvably, or by zero in a representation. The most important representation for this purpose is the adjoint representation, where no external module has to be chosen. Since $\operatorname{ad}x$ records how $x$ brackets with every element of $\mathfrak g$, its trace form is the canonical form attached to the Lie algebra itself. This motivates naming the adjoint trace form separately. [definition: Killing Form] Let $\mathfrak g$ be a finite-dimensional Lie algebra over $F$. For each $x \in \mathfrak g$, let \begin{align*} \operatorname{ad}x: \mathfrak g \to \mathfrak g, \qquad (\operatorname{ad}x)(y)=[x,y], \end{align*} and let $\operatorname{ad}: \mathfrak g \to \mathfrak{gl}(\mathfrak g)$ denote the adjoint representation $x \mapsto \operatorname{ad}x$. The Killing form of $\mathfrak g$ is the bilinear form $\kappa: \mathfrak g \times \mathfrak g \to F$ defined by \begin{align*} \kappa(x,y)=\operatorname{tr}(\operatorname{ad}x\,\operatorname{ad}y). \end{align*} [/definition] Thus the Killing form is the trace form of the adjoint representation. The point of using $\operatorname{ad}$ is that no external representation has to be chosen: the Lie algebra acts on itself by bracketing. If this action has a large kernel, as happens for central elements, the Killing form cannot see those directions; if the adjoint action is rich enough, the form records substantial information about the bracket. It is symmetric and invariant by the preceding theorem. [quotetheorem:3808] [citeproof:3808] After the theorem, invariance should be read as a constraint on how the Killing form behaves with respect to all inner symmetries of the Lie algebra. A bilinear form chosen arbitrarily on the underlying vector space would not usually make ideals interact well with orthogonality, and would therefore be poor at detecting structure. The theorem does not say that the Killing form is nondegenerate; the next example is the basic warning that canonical forms can still lose information. [example: Killing Form Of An Abelian Lie Algebra] Let $\mathfrak g$ be an abelian finite-dimensional Lie algebra, so $[x,y]=0$ for every $x,y \in \mathfrak g$. We show that its Killing form is identically zero. For any fixed $x \in \mathfrak g$, the adjoint map is \begin{align*} \operatorname{ad}x:\mathfrak g &\to \mathfrak g,\\ z &\mapsto [x,z]. \end{align*} Since $\mathfrak g$ is abelian, $[x,z]=0$ for every $z \in \mathfrak g$, hence $\operatorname{ad}x=0$ as an endomorphism of $\mathfrak g$. Therefore, for any $x,y \in \mathfrak g$, \begin{align*} \operatorname{ad}x\,\operatorname{ad}y &=0 \circ 0\\ &=0. \end{align*} By the definition of the Killing form, \begin{align*} \kappa(x,y) &=\operatorname{tr}(\operatorname{ad}x\,\operatorname{ad}y)\\ &=\operatorname{tr}(0)\\ &=0. \end{align*} Thus $\kappa$ vanishes on every pair $(x,y)$, even if $\mathfrak g$ is nonzero. The calculation shows that degeneracy of the Killing form measures failure of the adjoint action to see directions in the Lie algebra, not failure of the Lie algebra itself to have nonzero elements. [/example] ## Orthogonality and Ideals What does invariance buy us beyond a formal identity? Its main consequence is that orthogonality interacts well with ideals. In particular, the [orthogonal complement of an ideal](/theorems/3809) is again stable under the adjoint action. [definition: Orthogonal Complement] Let $B: \mathfrak g \times \mathfrak g \to F$ be a bilinear form and let $U \subset \mathfrak g$ be a subspace. The orthogonal complement of $U$ with respect to $B$ is \begin{align*} U^\perp = \{x \in \mathfrak g : B(x,u)=0 \text{ for all } u \in U\}. \end{align*} [/definition] When $B=\kappa$, the notation $U^\perp$ will mean orthogonality with respect to the Killing form unless another form is named. Orthogonal complements are useful only if they respect the Lie algebra operations one is studying. Without invariance, $U^\perp$ is merely a linear-algebraic subspace and need not be closed under bracketing with elements of $\mathfrak g$; invariance is exactly the hypothesis that moves a bracket from one argument of $B$ to the other. [quotetheorem:3809] [citeproof:3809] This small result is often the first place where invariant forms behave like structure-theoretic objects rather than numerical gadgets. The ideal hypothesis is necessary: for an arbitrary subspace $U$, the bracket $[i,g]$ appearing in the proof need not return to $U$, so the orthogonality argument breaks. In later chapters, nondegeneracy of the Killing form will turn decompositions by ideals into orthogonal decompositions, and this is the mechanism that makes those decompositions compatible with the bracket. [quotetheorem:3810] [citeproof:3810] Cartan's criterion connects the solvable radical to the degeneracy of the Killing form on commutator directions. The theorem should be read as a necessary vanishing test: if an element lies in the radical, then pairing it with every commutator by the Killing form gives zero. This explains why the radical is difficult for the adjoint trace form to see and why nondegeneracy of the Killing form will later be tied to the absence of solvable ideals. The converse direction, combined with semisimplicity, will become the main bridge between the Killing form and structural decomposition. The present chapter uses the criterion first as a diagnostic, then turns to the linear version where trace vanishing for matrix Lie algebras forces solvability. ## The Trace Criterion for Linear Lie Algebras How can solvability be detected from matrices? Lie theorem says that representations of solvable Lie algebras over an algebraically closed field can be triangularised. Cartan criterion is a converse in trace form: enough vanishing of traces forces solvability. [quotetheorem:3811] [citeproof:3811] This result should be read as a converse to the triangular examples: commutator subalgebras of triangular Lie algebras consist of strictly upper triangular matrices, and those have zero trace after multiplication by an upper triangular matrix. The hypothesis is stronger than asking merely that every element of $[\mathfrak h,\mathfrak h]$ have trace zero; it requires vanishing after multiplication by every element of $\mathfrak h$, which is what detects the semisimple part. Algebraic closedness enters because the proof discusses eigenvalues and triangular forms, while characteristic $0$ keeps the trace pairing from losing information through modular cancellation. The theorem does not identify the derived algebra explicitly; it gives the structural conclusion that the whole Lie algebra is solvable. [example: Upper Triangular Matrices] Let $\mathfrak b \le \mathfrak{gl}_n(F)$ be the upper triangular matrices, and let $\mathfrak n \le \mathfrak b$ be the strictly upper triangular matrices. We show that $[\mathfrak b,\mathfrak b]\subseteq \mathfrak n$ and that every $x\in[\mathfrak b,\mathfrak b]$ satisfies $\operatorname{tr}(xy)=0$ for all $y\in\mathfrak b$. Take $A=(a_{ij})$ and $B=(b_{ij})$ in $\mathfrak b$. For each diagonal position $i$, \begin{align*} (AB)_{ii} &=\sum_{k=1}^n a_{ik}b_{ki}. \end{align*} Since $A$ is upper triangular, $a_{ik}=0$ when $i>k$. Since $B$ is upper triangular, $b_{ki}=0$ when $k>i$. Thus the only index $k$ that can contribute is $k=i$, so \begin{align*} (AB)_{ii}=a_{ii}b_{ii}. \end{align*} The same argument gives \begin{align*} (BA)_{ii}=b_{ii}a_{ii}. \end{align*} Because entries lie in the field $F$, $a_{ii}b_{ii}=b_{ii}a_{ii}$, and hence \begin{align*} ([A,B])_{ii} &=(AB-BA)_{ii}\\ &=(AB)_{ii}-(BA)_{ii}\\ &=a_{ii}b_{ii}-b_{ii}a_{ii}\\ &=0. \end{align*} The product and difference of upper triangular matrices are upper triangular, so $[A,B]$ is upper triangular with zero diagonal; therefore $[A,B]\in\mathfrak n$. Since $[\mathfrak b,\mathfrak b]$ is the linear span of such commutators and $\mathfrak n$ is a subspace, it follows that \begin{align*} [\mathfrak b,\mathfrak b]\subseteq \mathfrak n. \end{align*} Now let $x=(x_{ij})\in[\mathfrak b,\mathfrak b]$ and $y=(y_{ij})\in\mathfrak b$. From the containment just proved, $x$ is strictly upper triangular. For each $i$, \begin{align*} (xy)_{ii} &=\sum_{k=1}^n x_{ik}y_{ki}. \end{align*} If $x_{ik}\neq 0$, then strict upper triangularity gives $i<k$. If $y_{ki}\neq 0$, then upper triangularity of $y$ gives $k\le i$. These inequalities cannot both hold, so every summand is zero and \begin{align*} (xy)_{ii}=0. \end{align*} Therefore \begin{align*} \operatorname{tr}(xy) &=\sum_{i=1}^n (xy)_{ii}\\ &=\sum_{i=1}^n 0\\ &=0. \end{align*} Thus $\mathfrak b$ satisfies the trace-vanishing hypothesis of *Cartan Trace Criterion For Linear Lie Algebras*; over an algebraically closed field of characteristic $0$, the criterion recovers the solvability of the upper triangular Lie algebra. [/example] The previous example concerns a solvable algebra whose derived algebra is already nilpotent. The next example isolates the nilpotent part itself, where the adjoint action is so strongly triangular that the Killing form vanishes completely. This distinction between solvable and nilpotent behaviour is one of the reasons the trace criterion is formulated using $[\mathfrak h,\mathfrak h]$ rather than all of $\mathfrak h$. [example: Strictly Upper Triangular Matrices] Let $\mathfrak n \le \mathfrak{gl}_n(F)$ be the Lie algebra of strictly upper triangular matrices. For $r\geq 1$, let \begin{align*} \mathfrak n^r=\{A=(a_{ij})\in \mathfrak{gl}_n(F): a_{ij}=0 \text{ whenever } j-i<r\}. \end{align*} Thus $\mathfrak n^1=\mathfrak n$, and $\mathfrak n^r=0$ for $r\geq n$. We first check how brackets move entries upward. If $A\in \mathfrak n^r$ and $B\in \mathfrak n^s$, then \begin{align*} (AB)_{ij} &=\sum_{k=1}^n a_{ik}b_{kj}. \end{align*} A summand $a_{ik}b_{kj}$ can be nonzero only if \begin{align*} k-i\geq r \qquad\text{and}\qquad j-k\geq s. \end{align*} Adding these inequalities gives \begin{align*} j-i=(k-i)+(j-k)\geq r+s. \end{align*} Hence $(AB)_{ij}=0$ whenever $j-i<r+s$, so $AB\in \mathfrak n^{r+s}$. The same argument gives $BA\in \mathfrak n^{r+s}$, and therefore \begin{align*} [A,B]=AB-BA\in \mathfrak n^{r+s}. \end{align*} In particular, \begin{align*} [\mathfrak n,\mathfrak n^r]\subseteq \mathfrak n^{r+1}. \end{align*} The lower central series therefore satisfies \begin{align*} \gamma_1(\mathfrak n)&=\mathfrak n=\mathfrak n^1,\\ \gamma_2(\mathfrak n)&=[\mathfrak n,\mathfrak n]\subseteq \mathfrak n^2,\\ \gamma_3(\mathfrak n)&=[\mathfrak n,\gamma_2(\mathfrak n)]\subseteq [\mathfrak n,\mathfrak n^2]\subseteq \mathfrak n^3, \end{align*} and inductively $\gamma_m(\mathfrak n)\subseteq \mathfrak n^m$ for every $m\geq 1$. Since $\mathfrak n^n=0$, we get $\gamma_n(\mathfrak n)=0$, so $\mathfrak n$ is nilpotent. Now fix $x\in \mathfrak n$. The containment \begin{align*} [x,\mathfrak n^r]\subseteq \mathfrak n^{r+1} \end{align*} shows that $\operatorname{ad}x$ sends each layer $\mathfrak n^r$ into the next smaller layer $\mathfrak n^{r+1}$. Choose a basis of $\mathfrak n$ adapted to the filtration \begin{align*} \mathfrak n=\mathfrak n^1\supseteq \mathfrak n^2\supseteq \cdots \supseteq \mathfrak n^{n-1}\supseteq \mathfrak n^n=0. \end{align*} With respect to this basis, the matrix of $\operatorname{ad}x$ has zero diagonal, because every basis vector belonging to the quotient layer $\mathfrak n^r/\mathfrak n^{r+1}$ is sent into $\mathfrak n^{r+1}$ and therefore has no component in its own quotient layer. For $x,y\in \mathfrak n$, both $\operatorname{ad}x$ and $\operatorname{ad}y$ have strictly upper triangular matrices in this adapted basis. Their product is again strictly upper triangular: if $P=(p_{ij})$ and $Q=(q_{ij})$ are strictly upper triangular, then \begin{align*} (PQ)_{ii} &=\sum_{k=1}^{\dim \mathfrak n} p_{ik}q_{ki}. \end{align*} A nonzero factor $p_{ik}$ requires $i<k$, while a nonzero factor $q_{ki}$ requires $k<i$; these inequalities cannot both hold. Hence every summand is zero and \begin{align*} (PQ)_{ii}=0 \end{align*} for every $i$. Applying this to $P=\operatorname{ad}x$ and $Q=\operatorname{ad}y$ gives \begin{align*} \kappa(x,y) &=\operatorname{tr}(\operatorname{ad}x\,\operatorname{ad}y)\\ &=\sum_i(\operatorname{ad}x\,\operatorname{ad}y)_{ii}\\ &=\sum_i 0\\ &=0. \end{align*} Thus the Killing form of $\mathfrak n$ is identically zero. In this example the bracket is nontrivial, but the adjoint action is filtered so strongly toward higher superdiagonals that all adjoint trace pairings vanish. [/example] ## Cartan Criterion for Solvability What does the trace criterion become when the representation is the adjoint representation? It becomes a criterion entirely internal to the Lie algebra. The hypothesis says that the derived algebra is orthogonal to the whole Lie algebra under the Killing form. [quotetheorem:3812] [citeproof:3812] The criterion is frequently stated as $[\mathfrak g,\mathfrak g] \subseteq \mathfrak g^\perp$. It turns solvability into a single orthogonality condition. The condition is not that the whole Killing form vanishes: a solvable nonabelian Lie algebra may have nonzero values involving elements outside the derived algebra. Conversely, vanishing on the derived algebra is strong because it says that every commutator is invisible to every adjoint trace, and Cartan's trace criterion converts that invisibility into triangular structure. The next major use of this criterion is the semisimple case, where failure of solvability is reflected by nondegeneracy rather than by trace vanishing. [remark: Field Extensions] The algebraic closedness assumption is mainly a convenience for applying Lie theorem and Jordan decomposition. Over many fields of characteristic $0$, the result is proved after extending scalars to an algebraic closure and then descending solvability back to the original field. The trace identities are preserved by scalar extension. [/remark] ## Computing the Killing Form of $\mathfrak{sl}_2$ Can the Killing form distinguish the first semisimple example from the solvable and nilpotent examples above? The calculation for $\mathfrak{sl}_2$ gives the model for later semisimple theory: the form is nondegenerate and pairs root directions with opposite root directions. In representation-theoretic terms, the adjoint representation has enough trace data to see the whole Lie algebra; in root-space terms, the form pairs positive and negative root directions and gives a nonzero length to the Cartan direction. Let $F$ be a field of characteristic $0$, and let $\mathfrak{sl}_2(F)$ have standard basis \begin{align*} e=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad f=\begin{pmatrix}0&0\\1&0\end{pmatrix},\qquad h=\begin{pmatrix}1&0\\0&-1\end{pmatrix}. \end{align*} The bracket relations are \begin{align*} [h,e]=2e,\qquad [h,f]=-2f,\qquad [e,f]=h. \end{align*} [example: Killing Form Of $\mathfrak{sl}_2$] Work in the ordered basis $(e,f,h)$ of $\mathfrak{sl}_2(F)$. From \begin{align*} [h,e]=2e,\qquad [h,f]=-2f,\qquad [e,f]=h, \end{align*} we also have \begin{align*} [e,h]=-2e,\qquad [f,e]=-h,\qquad [f,h]=2f. \end{align*} Thus the columns of each adjoint matrix are the coordinates of the images of $e,f,h$: \begin{align*} \operatorname{ad}e &= \begin{pmatrix} 0&0&-2\\ 0&0&0\\ 0&1&0 \end{pmatrix},& \operatorname{ad}f &= \begin{pmatrix} 0&0&0\\ 0&0&2\\ -1&0&0 \end{pmatrix},& \operatorname{ad}h &= \begin{pmatrix} 2&0&0\\ 0&-2&0\\ 0&0&0 \end{pmatrix}. \end{align*} Using $\kappa(x,y)=\operatorname{tr}(\operatorname{ad}x\,\operatorname{ad}y)$, first compute \begin{align*} (\operatorname{ad}e)^2 &= \begin{pmatrix} 0&0&-2\\ 0&0&0\\ 0&1&0 \end{pmatrix} \begin{pmatrix} 0&0&-2\\ 0&0&0\\ 0&1&0 \end{pmatrix} \\ &= \begin{pmatrix} 0&-2&0\\ 0&0&0\\ 0&0&0 \end{pmatrix}, \end{align*} so \begin{align*} \kappa(e,e) &=\operatorname{tr}((\operatorname{ad}e)^2)\\ &=0+0+0\\ &=0. \end{align*} Similarly, \begin{align*} (\operatorname{ad}f)^2 &= \begin{pmatrix} 0&0&0\\ 0&0&2\\ -1&0&0 \end{pmatrix} \begin{pmatrix} 0&0&0\\ 0&0&2\\ -1&0&0 \end{pmatrix}\\ &= \begin{pmatrix} 0&0&0\\ -2&0&0\\ 0&0&0 \end{pmatrix}, \end{align*} and therefore \begin{align*} \kappa(f,f) &=0. \end{align*} For the mixed pair $e,f$, \begin{align*} \operatorname{ad}e\,\operatorname{ad}f &= \begin{pmatrix} 0&0&-2\\ 0&0&0\\ 0&1&0 \end{pmatrix} \begin{pmatrix} 0&0&0\\ 0&0&2\\ -1&0&0 \end{pmatrix}\\ &= \begin{pmatrix} 2&0&0\\ 0&0&0\\ 0&0&2 \end{pmatrix}. \end{align*} Hence \begin{align*} \kappa(e,f) &=\operatorname{tr}(\operatorname{ad}e\,\operatorname{ad}f)\\ &=2+0+2\\ &=4. \end{align*} By symmetry of trace, \begin{align*} \kappa(f,e) &=\operatorname{tr}(\operatorname{ad}f\,\operatorname{ad}e)\\ &=\operatorname{tr}(\operatorname{ad}e\,\operatorname{ad}f)\\ &=4. \end{align*} Next, \begin{align*} \operatorname{ad}e\,\operatorname{ad}h &= \begin{pmatrix} 0&0&-2\\ 0&0&0\\ 0&1&0 \end{pmatrix} \begin{pmatrix} 2&0&0\\ 0&-2&0\\ 0&0&0 \end{pmatrix}\\ &= \begin{pmatrix} 0&0&0\\ 0&0&0\\ 0&-2&0 \end{pmatrix}, \end{align*} so \begin{align*} \kappa(e,h) &=0. \end{align*} Also, \begin{align*} \operatorname{ad}f\,\operatorname{ad}h &= \begin{pmatrix} 0&0&0\\ 0&0&2\\ -1&0&0 \end{pmatrix} \begin{pmatrix} 2&0&0\\ 0&-2&0\\ 0&0&0 \end{pmatrix}\\ &= \begin{pmatrix} 0&0&0\\ 0&0&0\\ -2&0&0 \end{pmatrix}, \end{align*} so \begin{align*} \kappa(f,h) &=0. \end{align*} Finally, \begin{align*} (\operatorname{ad}h)^2 &= \begin{pmatrix} 2&0&0\\ 0&-2&0\\ 0&0&0 \end{pmatrix} \begin{pmatrix} 2&0&0\\ 0&-2&0\\ 0&0&0 \end{pmatrix}\\ &= \begin{pmatrix} 4&0&0\\ 0&4&0\\ 0&0&0 \end{pmatrix}, \end{align*} and hence \begin{align*} \kappa(h,h) &=\operatorname{tr}((\operatorname{ad}h)^2)\\ &=4+4+0\\ &=8. \end{align*} Therefore the Gram matrix of $\kappa$ in the basis $(e,f,h)$ is \begin{align*} \begin{pmatrix} 0&4&0\\ 4&0&0\\ 0&0&8 \end{pmatrix}. \end{align*} Its determinant is \begin{align*} \det \begin{pmatrix} 0&4&0\\ 4&0&0\\ 0&0&8 \end{pmatrix} &= 8\det \begin{pmatrix} 0&4\\ 4&0 \end{pmatrix}\\ &=8(0\cdot 0-4\cdot 4)\\ &=-128. \end{align*} Since $F$ has characteristic $0$, $-128\neq 0$ in $F$, so the Killing form on $\mathfrak{sl}_2(F)$ is nondegenerate. This contrasts with the abelian and strictly upper triangular examples: here the adjoint action has enough trace data to detect every basis direction. [/example] This computation foreshadows the next chapter. Solvable Lie algebras have Killing form vanishing on their derived algebra, whereas $\mathfrak{sl}_2$ has a nondegenerate Killing form. Cartan criterion explains why this contrast is structural rather than accidental. Cartan's criterion and the Killing form now separate the semisimple case from the solvable one in a structural way. With that dichotomy in place, the course can pass from recognition of semisimplicity to the representation theory that semisimple Lie algebras control. # 9. Semisimple Lie Algebras Semisimple Lie algebras are the Lie algebras whose internal structure has no solvable part left to peel away. Chapters 4 through 8 developed solvability, nilpotence, Engel's theorem, Lie's theorem, and Cartan's criterion; here those tools are assembled into the first structural picture of the non-solvable case. The central questions are how to recognize semisimplicity, how the Killing form detects ideals, and why every semisimple Lie algebra splits into simple building blocks without invoking the classification. ## Removing the Solvable Part What obstruction prevents a finite-dimensional Lie algebra from being built out of simple pieces? The obstruction is the largest solvable ideal, called the radical. Semisimplicity means that this obstruction is absent. [definition: Simple Lie Algebra] Let $F$ be a field and let $\mathfrak g$ be a Lie algebra over $F$. The Lie algebra $\mathfrak g$ is simple if $\mathfrak g$ is non-abelian and its only ideals are $0$ and $\mathfrak g$. [/definition] The non-abelian condition excludes the one-dimensional abelian Lie algebra, whose ideal theory is small but whose bracket carries no semisimple structure. Simple Lie algebras are the intended analogues of simple groups and simple modules: they have no proper quotient or subobject visible to the category of Lie algebras. [example: The Lie Algebra sl2] Let $F$ be a field of characteristic $0$, and let $\mathfrak{sl}_2(F)$ have basis \begin{align*} e = \begin{pmatrix}0&1\\0&0\end{pmatrix}, \qquad f = \begin{pmatrix}0&0\\1&0\end{pmatrix}, \qquad h = \begin{pmatrix}1&0\\0&-1\end{pmatrix}. \end{align*} Multiplying matrices and subtracting in the Lie bracket $[A,B]=AB-BA$ gives \begin{align*} [h,e] &= \begin{pmatrix}1&0\\0&-1\end{pmatrix} \begin{pmatrix}0&1\\0&0\end{pmatrix} - \begin{pmatrix}0&1\\0&0\end{pmatrix} \begin{pmatrix}1&0\\0&-1\end{pmatrix} \\ &= \begin{pmatrix}0&1\\0&0\end{pmatrix} - \begin{pmatrix}0&-1\\0&0\end{pmatrix} = \begin{pmatrix}0&2\\0&0\end{pmatrix} =2e, \end{align*} and similarly \begin{align*} [h,f] &= \begin{pmatrix}1&0\\0&-1\end{pmatrix} \begin{pmatrix}0&0\\1&0\end{pmatrix} - \begin{pmatrix}0&0\\1&0\end{pmatrix} \begin{pmatrix}1&0\\0&-1\end{pmatrix} \\ &= \begin{pmatrix}0&0\\-1&0\end{pmatrix} - \begin{pmatrix}0&0\\1&0\end{pmatrix} = \begin{pmatrix}0&0\\-2&0\end{pmatrix} =-2f, \end{align*} while \begin{align*} [e,f] &= \begin{pmatrix}0&1\\0&0\end{pmatrix} \begin{pmatrix}0&0\\1&0\end{pmatrix} - \begin{pmatrix}0&0\\1&0\end{pmatrix} \begin{pmatrix}0&1\\0&0\end{pmatrix} \\ &= \begin{pmatrix}1&0\\0&0\end{pmatrix} - \begin{pmatrix}0&0\\0&1\end{pmatrix} = \begin{pmatrix}1&0\\0&-1\end{pmatrix} =h. \end{align*} We show that every non-zero ideal $I\trianglelefteq \mathfrak{sl}_2(F)$ equals $\mathfrak{sl}_2(F)$. Choose \begin{align*} 0\ne x=ae+bf+ch\in I. \end{align*} Because $I$ is an ideal, $[h,x]\in I$ and $[h,[h,x]]\in I$. Using bilinearity and the three bracket relations, \begin{align*} [h,x] &=a[h,e]+b[h,f]+c[h,h] \\ &=2ae-2bf, \end{align*} and therefore \begin{align*} [h,[h,x]] &=[h,2ae-2bf] \\ &=2a[h,e]-2b[h,f] \\ &=4ae+4bf. \end{align*} If $c\ne 0$, then \begin{align*} x-\frac{1}{4}[h,[h,x]] &=(ae+bf+ch)-(ae+bf) \\ &=ch, \end{align*} so $h\in I$. Then \begin{align*} [h,e]=2e\in I,\qquad [h,f]=-2f\in I, \end{align*} and since $2$ is invertible in characteristic $0$, $e,f,h\in I$. It remains to handle $c=0$, so $x=ae+bf$ with $(a,b)\ne (0,0)$. If $a\ne 0$ and $b=0$, then $e\in I$; if $a=0$ and $b\ne 0$, then $f\in I$. If $a\ne 0$ and $b\ne 0$, then \begin{align*} [x,[h,x]] &=[ae+bf,2ae-2bf] \\ &=-2ab[e,f]+2ab[f,e] \\ &=-2ab\,h+2ab(-h) \\ &=-4ab\,h, \end{align*} so $h\in I$. Thus in every case $I$ contains one of $e,f,h$, and the bracket relations force it to contain all three. Hence $I=\mathfrak{sl}_2(F)$, so $\mathfrak{sl}_2(F)$ is simple, and therefore semisimple in the sense introduced below. [/example] For a general Lie algebra, there may be many solvable ideals. Their sum remains solvable in finite dimension, so there is a largest one. [definition: Radical] Let $F$ be a field and let $\mathfrak g$ be a finite-dimensional Lie algebra over $F$. The radical of $\mathfrak g$, denoted $\operatorname{rad}(\mathfrak g)$, is the largest solvable ideal of $\mathfrak g$. [/definition] The existence of the radical uses finite dimensionality: the sum of two solvable ideals is solvable, and an ascending chain of ideals stabilizes. The radical measures how much of $\mathfrak g$ is controlled by the solvable theory of the previous chapters. After isolating the largest solvable ideal, the complementary structural question is what it means for there to be no solvable part left at all. This condition is not the same as simplicity: a direct sum of simple Lie algebras is usually not simple, but it still has no nonzero solvable ideal. The next definition names this radical-free situation. [definition: Semisimple Lie Algebra] Let $F$ be a field and let $\mathfrak g$ be a finite-dimensional Lie algebra over $F$. The Lie algebra $\mathfrak g$ is semisimple if $\operatorname{rad}(\mathfrak g)=0$. [/definition] This definition is deliberately negative: it says that no non-zero solvable ideal survives. Later characterisations replace this condition by computable tests involving traces and the Killing form. [definition: Levi Factor] Let $F$ be a field and let $\mathfrak g$ be a finite-dimensional Lie algebra over $F$. A Levi factor of $\mathfrak g$ is a semisimple subalgebra $\mathfrak s \le \mathfrak g$ such that \begin{align*} \mathfrak g = \mathfrak s \ltimes \operatorname{rad}(\mathfrak g). \end{align*} [/definition] The full Levi decomposition theorem asserts the existence of such a factor over fields of characteristic $0$ under standard hypotheses. In this course it is used as context rather than proved; the present chapter only needs the radical and the case in which the radical vanishes. [example: Upper Triangular Matrices] Let $E_{ij}$ denote the matrix unit with $1$ in position $(i,j)$ and $0$ elsewhere, and let \begin{align*} \mathfrak n_r=\operatorname{span}\{E_{ij}: j-i\ge r\} \end{align*} inside the upper triangular matrices. Thus $\mathfrak n_1$ is the strictly upper triangular Lie algebra and $\mathfrak n_r=0$ for $r\ge n$. We first compute the first derived algebra. If $A=(a_{ij})$ and $B=(b_{ij})$ are upper triangular, then the $i$th diagonal entry of $AB$ is \begin{align*} (AB)_{ii}=\sum_{k=1}^n a_{ik}b_{ki}=a_{ii}b_{ii}, \end{align*} because $a_{ik}=0$ for $k<i$ and $b_{ki}=0$ for $k>i$. Similarly $(BA)_{ii}=b_{ii}a_{ii}$, so \begin{align*} [A,B]_{ii}=(AB-BA)_{ii}=a_{ii}b_{ii}-b_{ii}a_{ii}=0. \end{align*} Hence $[\mathfrak b_n(F),\mathfrak b_n(F)]\subseteq \mathfrak n_1$. Conversely, for every $i<j$, \begin{align*} [E_{ii},E_{ij}] &=E_{ii}E_{ij}-E_{ij}E_{ii} \\ &=E_{ij}-0 \\ &=E_{ij}. \end{align*} Thus each basis element $E_{ij}$ with $i<j$ lies in $[\mathfrak b_n(F),\mathfrak b_n(F)]$, so \begin{align*} [\mathfrak b_n(F),\mathfrak b_n(F)]=\mathfrak n_1. \end{align*} Now compute how brackets affect superdiagonals. For matrix units, \begin{align*} E_{ij}E_{kl} = \begin{cases} E_{il}, & j=k,\\ 0, & j\ne k, \end{cases} \end{align*} so \begin{align*} [E_{ij},E_{kl}] = \delta_{jk}E_{il}-\delta_{li}E_{kj}. \end{align*} If $E_{ij}\in \mathfrak n_r$ and $E_{kl}\in \mathfrak n_s$, then $j-i\ge r$ and $l-k\ge s$. In the first non-zero term we have $j=k$, hence \begin{align*} l-i=(l-k)+(j-i)\ge s+r. \end{align*} In the second non-zero term we have $l=i$, hence \begin{align*} j-k=(j-i)+(l-k)\ge r+s. \end{align*} Therefore \begin{align*} [\mathfrak n_r,\mathfrak n_s]\subseteq \mathfrak n_{r+s}. \end{align*} Let $\mathfrak b_n^{(0)}=\mathfrak b_n$ and $\mathfrak b_n^{(m+1)}=[\mathfrak b_n^{(m)},\mathfrak b_n^{(m)}]$ be the derived series. We have $\mathfrak b_n^{(1)}=\mathfrak n_1$, and the containment above gives \begin{align*} \mathfrak b_n^{(2)}=[\mathfrak n_1,\mathfrak n_1]\subseteq \mathfrak n_2, \end{align*} then \begin{align*} \mathfrak b_n^{(3)} \subseteq [\mathfrak n_2,\mathfrak n_2] \subseteq \mathfrak n_4, \end{align*} and by induction \begin{align*} \mathfrak b_n^{(m)}\subseteq \mathfrak n_{2^{m-1}} \qquad \text{for } m\ge 1. \end{align*} Choose $m$ with $2^{m-1}\ge n$. Then $\mathfrak n_{2^{m-1}}=0$, so $\mathfrak b_n^{(m)}=0$. Hence $\mathfrak b_n(F)$ is solvable. Since $\mathfrak b_n(F)$ is an ideal of itself and is solvable, its radical is the whole algebra: \begin{align*} \operatorname{rad}(\mathfrak b_n(F))=\mathfrak b_n(F). \end{align*} [/example] This example shows why semisimplicity is a restriction on ideals rather than on elements. Upper triangular algebras contain diagonal elements that are far from nilpotent, but the Lie algebra is still governed by a solvable ideal structure. ## Detecting Semisimplicity with the Killing Form How can one decide whether the radical is zero without listing every ideal? The answer in characteristic $0$ is that the adjoint representation leaves a trace pairing on $\mathfrak g$, and its degeneracy detects solvable ideals. [definition: Killing Form] Let $F$ be a field and let $\mathfrak g$ be a finite-dimensional Lie algebra over $F$. The Killing form of $\mathfrak g$ is the symmetric bilinear form $\kappa_{\mathfrak g}:\mathfrak g \times \mathfrak g \to F$ defined by \begin{align*} \kappa_{\mathfrak g}(x,y)=\operatorname{tr}(\operatorname{ad}_x \operatorname{ad}_y). \end{align*} [/definition] The invariance identity \begin{align*} \kappa_{\mathfrak g}([x,y],z)=\kappa_{\mathfrak g}(x,[y,z]) \end{align*} is the key formal property. It follows from the cyclic property of trace and makes orthogonal complements of ideals into ideals. The remaining question is whether this bilinear form detects semisimplicity rather than merely reflecting it after the fact. A criterion in terms of nondegeneracy would turn the structural condition of having no solvable ideals into a concrete computation with traces of adjoint maps. [quotetheorem:3813] [citeproof:3813] This is the main recognition theorem of the chapter. It converts an ideal-theoretic condition into a finite-dimensional linear algebra computation. Finite dimensionality is essential because the proof uses traces of endomorphisms and chain-stopping arguments for ideals. The characteristic $0$ hypothesis is also doing real work: Cartan's criterion can fail in small positive characteristic, where trace identities no longer control solvability in the same way. The theorem does not classify semisimple Lie algebras; it gives the nondegenerate bilinear form that will make ideals split orthogonally in the next section. [example: Killing Form of sl2] With respect to the ordered basis $(e,f,h)$, the bracket relations \begin{align*} [e,f]=h,\qquad [h,e]=2e,\qquad [h,f]=-2f \end{align*} give \begin{align*} \operatorname{ad}_e(e)&=[e,e]=0,& \operatorname{ad}_e(f)&=[e,f]=h,& \operatorname{ad}_e(h)&=[e,h]=-2e,\\ \operatorname{ad}_f(e)&=[f,e]=-h,& \operatorname{ad}_f(f)&=[f,f]=0,& \operatorname{ad}_f(h)&=[f,h]=2f,\\ \operatorname{ad}_h(e)&=[h,e]=2e,& \operatorname{ad}_h(f)&=[h,f]=-2f,& \operatorname{ad}_h(h)&=[h,h]=0. \end{align*} Therefore the matrices of the three adjoint maps are \begin{align*} [\operatorname{ad}_e]_{(e,f,h)} &= \begin{pmatrix} 0&0&-2\\ 0&0&0\\ 0&1&0 \end{pmatrix},& [\operatorname{ad}_f]_{(e,f,h)} &= \begin{pmatrix} 0&0&0\\ 0&0&2\\ -1&0&0 \end{pmatrix},& [\operatorname{ad}_h]_{(e,f,h)} &= \begin{pmatrix} 2&0&0\\ 0&-2&0\\ 0&0&0 \end{pmatrix}. \end{align*} Using $\kappa_{\mathfrak{sl}_2}(x,y)=\operatorname{tr}(\operatorname{ad}_x\operatorname{ad}_y)$, we compute the non-zero entries. First, \begin{align*} [\operatorname{ad}_e][\operatorname{ad}_f] &= \begin{pmatrix} 0&0&-2\\ 0&0&0\\ 0&1&0 \end{pmatrix} \begin{pmatrix} 0&0&0\\ 0&0&2\\ -1&0&0 \end{pmatrix} = \begin{pmatrix} 2&0&0\\ 0&0&0\\ 0&0&2 \end{pmatrix}, \end{align*} so $\kappa(e,f)=2+0+2=4$. Also \begin{align*} [\operatorname{ad}_h]^2 &= \begin{pmatrix} 2&0&0\\ 0&-2&0\\ 0&0&0 \end{pmatrix} \begin{pmatrix} 2&0&0\\ 0&-2&0\\ 0&0&0 \end{pmatrix} = \begin{pmatrix} 4&0&0\\ 0&4&0\\ 0&0&0 \end{pmatrix}, \end{align*} so $\kappa(h,h)=4+4+0=8$. The remaining mixed products with $h$ have zero trace: \begin{align*} [\operatorname{ad}_e][\operatorname{ad}_h] &= \begin{pmatrix} 0&0&0\\ 0&0&0\\ 0&-2&0 \end{pmatrix}, & [\operatorname{ad}_f][\operatorname{ad}_h] &= \begin{pmatrix} 0&0&0\\ 0&0&0\\ -2&0&0 \end{pmatrix}. \end{align*} Finally, $[\operatorname{ad}_e]^2$ and $[\operatorname{ad}_f]^2$ are strictly off-diagonal in this basis, so their traces are $0$. Hence the Gram matrix of the Killing form in the basis $(e,f,h)$ is \begin{align*} \begin{pmatrix} \kappa(e,e)&\kappa(e,f)&\kappa(e,h)\\ \kappa(f,e)&\kappa(f,f)&\kappa(f,h)\\ \kappa(h,e)&\kappa(h,f)&\kappa(h,h) \end{pmatrix} = \begin{pmatrix} 0&4&0\\ 4&0&0\\ 0&0&8 \end{pmatrix}. \end{align*} Its determinant is \begin{align*} 0\cdot \begin{vmatrix} 0&0\\ 0&8 \end{vmatrix} - 4\cdot \begin{vmatrix} 4&0\\ 0&8 \end{vmatrix} + 0\cdot \begin{vmatrix} 4&0\\ 0&0 \end{vmatrix} = -4(32) = -128. \end{align*} Since $F$ has characteristic $0$, the scalar $-128$ is non-zero in $F$. Thus $\kappa_{\mathfrak{sl}_2}$ is nondegenerate, and *[Cartan Semisimplicity Criterion](/theorems/3813)* shows that $\mathfrak{sl}_2(F)$ is semisimple, matching the earlier direct proof that it is simple. [/example] The criterion is also useful for seeing what semisimplicity is not. A Lie algebra may have a large semisimple-looking quotient while retaining a central or solvable part that the Killing form detects as degenerate. [example: gln Is Reductive but Not Semisimple] Let $F$ be a field of characteristic $0$ and let $n\ge 1$. For every $X\in \mathfrak{gl}_n(F)$, set \begin{align*} t=\frac{\operatorname{tr}(X)}{n}, \qquad Y=X-tI_n. \end{align*} Since characteristic $0$ implies $n\ne 0$ in $F$, the scalar $1/n$ is defined, and \begin{align*} \operatorname{tr}(Y) &=\operatorname{tr}(X)-t\operatorname{tr}(I_n) \\ &=\operatorname{tr}(X)-\frac{\operatorname{tr}(X)}{n}\cdot n \\ &=0. \end{align*} Thus $Y\in \mathfrak{sl}_n(F)$ and \begin{align*} X=Y+tI_n\in \mathfrak{sl}_n(F)+FI_n. \end{align*} The intersection is zero: if $\lambda I_n\in \mathfrak{sl}_n(F)$, then \begin{align*} 0=\operatorname{tr}(\lambda I_n)=\lambda \operatorname{tr}(I_n)=\lambda n, \end{align*} so $\lambda=0$. Hence \begin{align*} \mathfrak{gl}_n(F)=\mathfrak{sl}_n(F)\oplus FI_n \end{align*} as a direct sum of vector spaces. Both summands are ideals. If $A\in \mathfrak{gl}_n(F)$ and $Y\in \mathfrak{sl}_n(F)$, then \begin{align*} \operatorname{tr}([A,Y]) &=\operatorname{tr}(AY-YA) \\ &=\operatorname{tr}(AY)-\operatorname{tr}(YA) \\ &=0, \end{align*} using the cyclic identity $\operatorname{tr}(AY)=\operatorname{tr}(YA)$ for matrices. Therefore $[A,Y]\in \mathfrak{sl}_n(F)$. Also, for $\lambda\in F$, \begin{align*} [A,\lambda I_n] &=A(\lambda I_n)-(\lambda I_n)A \\ &=\lambda A-\lambda A \\ &=0, \end{align*} so $FI_n$ is central. The ideal $FI_n$ is abelian because \begin{align*} [\lambda I_n,\mu I_n] &=(\lambda I_n)(\mu I_n)-(\mu I_n)(\lambda I_n) \\ &=\lambda\mu I_n-\mu\lambda I_n \\ &=0. \end{align*} Hence its derived algebra is $[FI_n,FI_n]=0$, so $FI_n$ is solvable. Since $FI_n$ is a non-zero solvable ideal, it is contained in $\operatorname{rad}(\mathfrak{gl}_n(F))$, and therefore \begin{align*} \operatorname{rad}(\mathfrak{gl}_n(F))\ne 0. \end{align*} Thus $\mathfrak{gl}_n(F)$ is not semisimple. When $\mathfrak{sl}_n(F)$ is semisimple, the quotient by the centre satisfies \begin{align*} \mathfrak{gl}_n(F)/FI_n \cong \mathfrak{sl}_n(F), \end{align*} so $\mathfrak{gl}_n(F)$ consists of a semisimple part together with an abelian central part; this is the basic reductive, but not semisimple, example. [/example] ## Ideals and Orthogonal Complements Once the Killing form is nondegenerate, what does it say about ideals inside a semisimple Lie algebra? The answer is that ideals behave like orthogonal summands: each ideal has a canonical complementary ideal given by its Killing-orthogonal complement. [definition: Orthogonal Complement of an Ideal] Let $\mathfrak g$ be a finite-dimensional Lie algebra with Killing form $\kappa_{\mathfrak g}$, and let $\mathfrak a \trianglelefteq \mathfrak g$ be an ideal. The orthogonal complement of $\mathfrak a$ is \begin{align*} \mathfrak a^{\perp}=\{x\in \mathfrak g : \kappa_{\mathfrak g}(x,a)=0 \text{ for all } a\in \mathfrak a\}. \end{align*} [/definition] The definition uses the Killing form on the ambient algebra. [Invariance of the Killing form](/theorems/3808) ensures that $\mathfrak a^{\perp}$ is stable under brackets with all elements of $\mathfrak g$. To make this complement useful for decomposition, one must also know that the ideal itself has no radical part invisible to the form. The next issue is therefore nondegeneracy on the ideal, because only then can $\mathfrak a$ meet its orthogonal complement only at zero. [quotetheorem:3814] [citeproof:3814] This result is the mechanism behind direct sum decompositions. It says that no ideal of a semisimple Lie algebra is hidden inside its own orthogonal complement. The hypothesis that $\mathfrak g$ is semisimple is necessary: in a solvable algebra, the Killing form may be degenerate on a large ideal, and the intersection with its orthogonal complement can be non-zero. The statement concerns the restriction of the ambient Killing form $\kappa_{\mathfrak g}$, not an arbitrary bilinear form placed on $\mathfrak a$ and not merely the Killing form of $\mathfrak a$ before comparison. [quotetheorem:3815] [citeproof:3815] Thus ideal complements in the semisimple case are canonical once the Killing form is fixed. This is much stronger than the existence of a vector-space complement. For nonsemisimple algebras, the Killing form can be degenerate, so an orthogonal complement may have the wrong dimension or fail to give a direct sum. The assumption that $\mathfrak a$ is an ideal is also essential: arbitrary subalgebras are not preserved by bracketing with all of $\mathfrak g$, so invariance does not make their orthogonal complements into ideals. This splitting theorem is the engine for the decomposition into simple ideals, because every proper ideal can be split off and the process repeated. [example: Splitting a Product] Let $L=\mathfrak{sl}_2(F)$ and write \begin{align*} \mathfrak g=L\oplus L,\qquad \mathfrak a=L\oplus 0. \end{align*} The bracket on the direct product is componentwise: \begin{align*} [(x_1,x_2),(y_1,y_2)]=([x_1,y_1],[x_2,y_2]). \end{align*} For $u=(u_1,u_2)$ and $v=(v_1,v_2)$, the adjoint map on $L\oplus L$ is \begin{align*} \operatorname{ad}_{u}(z_1,z_2) &=[(u_1,u_2),(z_1,z_2)] \\ &=([u_1,z_1],[u_2,z_2]), \end{align*} so, with respect to any ordered basis of the first copy of $L$ followed by the same ordered basis of the second copy, its matrix has block form \begin{align*} \operatorname{ad}_{u} = \begin{pmatrix} \operatorname{ad}_{u_1} & 0\\ 0 & \operatorname{ad}_{u_2} \end{pmatrix}. \end{align*} Therefore \begin{align*} \operatorname{ad}_{u}\operatorname{ad}_{v} &= \begin{pmatrix} \operatorname{ad}_{u_1} & 0\\ 0 & \operatorname{ad}_{u_2} \end{pmatrix} \begin{pmatrix} \operatorname{ad}_{v_1} & 0\\ 0 & \operatorname{ad}_{v_2} \end{pmatrix} \\ &= \begin{pmatrix} \operatorname{ad}_{u_1}\operatorname{ad}_{v_1} & 0\\ 0 & \operatorname{ad}_{u_2}\operatorname{ad}_{v_2} \end{pmatrix}, \end{align*} and hence \begin{align*} \kappa_{\mathfrak g}(u,v) &=\operatorname{tr}(\operatorname{ad}_{u}\operatorname{ad}_{v}) \\ &=\operatorname{tr}(\operatorname{ad}_{u_1}\operatorname{ad}_{v_1}) +\operatorname{tr}(\operatorname{ad}_{u_2}\operatorname{ad}_{v_2}) \\ &=\kappa_L(u_1,v_1)+\kappa_L(u_2,v_2). \end{align*} Now let $(x_1,x_2)\in \mathfrak g$. This element lies in $\mathfrak a^{\perp}$ exactly when \begin{align*} 0 &=\kappa_{\mathfrak g}((x_1,x_2),(y,0)) \\ &=\kappa_L(x_1,y)+\kappa_L(x_2,0) \\ &=\kappa_L(x_1,y) \end{align*} for every $y\in L$. The Killing form on $\mathfrak{sl}_2(F)$ was computed above to have Gram matrix \begin{align*} \begin{pmatrix} 0&4&0\\ 4&0&0\\ 0&0&8 \end{pmatrix}, \end{align*} whose determinant is $-128\ne 0$ in characteristic $0$. Thus $\kappa_L$ is nondegenerate, so $\kappa_L(x_1,y)=0$ for every $y\in L$ forces $x_1=0$. Conversely, if $x_1=0$, then the displayed equality gives $\kappa_{\mathfrak g}((0,x_2),(y,0))=0$ for every $y\in L$. Therefore \begin{align*} \mathfrak a^{\perp}=0\oplus L. \end{align*} The decomposition is consequently \begin{align*} \mathfrak g=(L\oplus 0)\oplus (0\oplus L)=\mathfrak a\oplus \mathfrak a^{\perp}, \end{align*} and the two summands commute because \begin{align*} [(x,0),(0,y)]=([x,0],[0,y])=(0,0) \end{align*} for all $x,y\in L$. This makes the abstract orthogonal-complement splitting concrete: in a product of two copies of $\mathfrak{sl}_2(F)$, the Killing-orthogonal complement of one factor is exactly the other factor. [/example] ## Decomposition into Simple Ideals If every ideal in a semisimple Lie algebra has a complementary ideal, how far can the splitting process continue? Finite dimensionality forces the process to stop, and the terminal pieces are simple ideals. [quotetheorem:3816] [citeproof:3816] This theorem is a structural decomposition, not a classification. It says that semisimple Lie algebras are assembled from simple ideals, but it does not identify the possible simple ideals. Finite dimensionality is needed so that the repeated splitting process terminates; without it there can be infinite descending or ascending decompositions that are not captured by a finite direct sum. Characteristic $0$ enters through Cartan's criterion and the nondegeneracy of the Killing form, and modular Lie algebras in positive characteristic require separate structure theory. The theorem also prepares the representation theory of the next chapter: once $\mathfrak g$ is a direct sum of commuting simple ideals, representations can be analysed factor by factor, and Weyl's theorem will show that finite-dimensional representations are completely reducible under the same characteristic $0$ hypotheses. [remark: Uniqueness of the Simple Summands] The decomposition is unique up to reordering at the level of minimal non-zero ideals. Every simple ideal in a semisimple Lie algebra is one of the simple summands in the decomposition, because its bracket with the other summands is zero and its projection to the direct sum has only one non-zero component. [/remark] The uniqueness statement separates semisimple structure from reductive structure: simple ideals are forced, while central abelian summands belong to the radical unless the definition is broadened. [example: Semisimple Versus Reductive Decomposition] Let \begin{align*} \mathfrak g=\mathfrak{sl}_2(F)\oplus \mathfrak{sl}_3(F) \end{align*} with componentwise bracket. The two summands are ideals, because for $x,a\in \mathfrak{sl}_2(F)$ and $y,b\in \mathfrak{sl}_3(F)$, \begin{align*} [(x,0),(a,b)]=([x,a],0)\in \mathfrak{sl}_2(F)\oplus 0, \qquad [(0,y),(a,b)]=(0,[y,b])\in 0\oplus \mathfrak{sl}_3(F). \end{align*} The algebra $\mathfrak{sl}_2(F)$ is simple by the calculation above, and $\mathfrak{sl}_3(F)$ is simple by the standard simplicity theorem for $\mathfrak{sl}_n(F)$ in characteristic $0$. Since \begin{align*} [\mathfrak{sl}_2(F)\oplus 0,\;0\oplus \mathfrak{sl}_3(F)] = 0, \end{align*} this is a direct sum of commuting simple ideals, so it is semisimple. Now let \begin{align*} \mathfrak h=\mathfrak{sl}_2(F)\oplus Fz \end{align*} where $[z,x]=0$ for every $x\in \mathfrak h$. The subspace $Fz$ is an ideal because, for $\lambda\in F$ and $x\in \mathfrak h$, \begin{align*} [x,\lambda z]=\lambda[x,z]=-\lambda[z,x]=0\in Fz. \end{align*} It is abelian because, for $\lambda,\mu\in F$, \begin{align*} [\lambda z,\mu z] = \lambda\mu [z,z] = 0. \end{align*} Therefore its derived algebra is \begin{align*} (Fz)^{(1)}=[Fz,Fz]=0, \end{align*} so $Fz$ is solvable. Since $Fz\ne 0$ is a solvable ideal of $\mathfrak h$, it is contained in $\operatorname{rad}(\mathfrak h)$, and hence \begin{align*} \operatorname{rad}(\mathfrak h)\ne 0. \end{align*} Thus $\mathfrak h$ is not semisimple. The decomposition \begin{align*} \mathfrak h=\mathfrak{sl}_2(F)\oplus Fz \end{align*} nevertheless separates a semisimple ideal from an abelian central ideal, which is the basic reductive but non-semisimple pattern. [/example] The chapter therefore closes the first structural arc of the course. Solvable Lie algebras are controlled by triangularisation and derived series; semisimple Lie algebras are controlled by the Killing form and by direct sums of simple ideals. The next step is to study representations of semisimple Lie algebras, where Weyl's theorem shows that the same absence of solvable radical forces complete reducibility. The structure theory developed so far explains why semisimple Lie algebras behave rigidly under decomposition. Weyl's theorem is the representation-theoretic payoff: it turns that rigidity into complete reducibility for finite-dimensional modules. # 10. Weyl's Theorem on Complete Reducibility Weyl's theorem is the representation-theoretic culmination of the structure theory developed so far. Earlier chapters used Engel's theorem, Lie's theorem, Cartan's criterion, and the Killing form to distinguish nilpotent, solvable, and semisimple Lie algebras. This chapter explains the decisive feature of semisimple Lie algebras over algebraically closed fields of characteristic zero: finite-dimensional representations break into simple pieces. The result is not a classification theorem, but it gives the structural foundation on which later highest-weight theory is built. ## Why Complete Reducibility Is the Right Target A representation may contain an invariant subspace without having a compatible invariant complement. The central question is therefore not whether submodules exist, but whether every submodule can be split off so that the module is a direct sum of smaller representations. [definition: Module] Let $F$ be a field and let $\mathfrak g$ be a Lie algebra over $F$. A $\mathfrak g$-module is an $F$-vector space $V$ equipped with an $F$-linear map $\rho: \mathfrak g \to \mathfrak{gl}(V)$ such that \begin{align*} \rho([x,y]) = \rho(x)\rho(y)-\rho(y)\rho(x) \end{align*} for all $x,y\in \mathfrak g$. [/definition] We usually write $xv$ for $\rho(x)(v)$. With this convention the module identity becomes $[x,y]v=x(yv)-y(xv)$. The analogue of an invariant subspace is the part of $V$ that is closed under every operator coming from $\mathfrak g$. We need this notion because such subspaces are the only candidates for smaller representations sitting inside $V$, so they are the objects that can obstruct or enable decomposition. [definition: Submodule] Let $V$ be a $\mathfrak g$-module. A subspace $W\subset V$ is a $\mathfrak g$-submodule if $xw\in W$ for all $x\in\mathfrak g$ and all $w\in W$. [/definition] The obstruction to decomposing a representation is exactly the possible failure of invariant complements. A submodule may exist without behaving like a direct summand, and then the representation contains extension data rather than just independent pieces. We need a separate condition for the stronger situation in which this obstruction never occurs. [definition: Completely Reducible Module] A finite-dimensional $\mathfrak g$-module $V$ is completely reducible if for every $\mathfrak g$-submodule $W\subset V$ there exists a $\mathfrak g$-submodule $U\subset V$ such that $V = W \oplus U$. [/definition] Equivalently, in the finite-dimensional setting, $V$ is completely reducible precisely when it is a direct sum of simple submodules. To make this reformulation precise, we need a name for the pieces that cannot be decomposed any further inside the module category. These are the representation-theoretic atoms: once a module is written as a sum of such pieces, no smaller invariant subspaces remain to split within each summand. [definition: Simple Module] A nonzero $\mathfrak g$-module $V$ is simple if its only $\mathfrak g$-submodules are $0$ and $V$. [/definition] This definition isolates the irreducible building blocks, but Weyl theorem says more: for semisimple Lie algebras, these building blocks assemble without hidden extension data. [example: Reducibility Versus Complete Reducibility] Let $\mathfrak g=F x$ be the one-dimensional abelian Lie algebra and let $V=F^2$ have basis $e_1,e_2$, with $x$ acting by \begin{align*} N=\begin{pmatrix}0&1\\0&0\end{pmatrix}. \end{align*} With respect to this basis, \begin{align*} x e_1=N e_1=0,\qquad x e_2=N e_2=e_1. \end{align*} Hence the line $W=F e_1$ is stable under $x$, because for every $c\in F$, \begin{align*} x(c e_1)=c\,x e_1=c\cdot 0=0\in W. \end{align*} Thus $W$ is a nonzero proper submodule of $V$, so $V$ is reducible. We now show that $W$ has no $x$-stable complement. Any one-dimensional complement $L$ to $W$ is spanned by a vector $v=a e_1+b e_2$ with $b\ne 0$, since $b=0$ would put $v$ in $W$. Its image under $x$ is \begin{align*} xv &=x(ae_1+be_2)\\ &=a\,x e_1+b\,x e_2\\ &=a\cdot 0+b e_1\\ &=b e_1. \end{align*} If $L$ were $x$-stable, then $b e_1$ would belong to $L$. Since $b\ne 0$, this would imply $e_1\in L$, so $L\cap W$ would contain $F e_1\ne 0$, contradicting that $V=W\oplus L$. Therefore no complementary line is a submodule, and reducibility need not imply complete reducibility for solvable Lie algebras. [/example] This example is the pattern Weyl theorem rules out in the semisimple case. The nilpotent Jordan block is an extension of the one-dimensional module by itself that does not split. ## Weyl Theorem and Splitting Short Exact Sequences The useful form of complete reducibility is often phrased in terms of exact sequences. The problem is to decide when a short exact sequence of modules carries an invariant section. [definition: Short Exact Sequence of Modules] A short exact sequence of finite-dimensional $\mathfrak g$-modules is a diagram \begin{align*} 0 \to U \xrightarrow{i} V \xrightarrow{p} W \to 0 \end{align*} where $i$ and $p$ are $\mathfrak g$-module homomorphisms, $i$ is injective, $p$ is surjective, and $\operatorname{im} i=\ker p$. [/definition] A splitting is a module map $s:W\to V$ with $p\circ s=\operatorname{id}_W$. Its image is then a $\mathfrak g$-submodule complementary to $i(U)$. The point of translating complements into exact sequences is that it separates the obstruction from the chosen coordinates on $V$. If every finite-dimensional sequence of this kind splits for modules over a semisimple Lie algebra, then invariant subspaces always come with invariant complements. [quotetheorem:3817] [citeproof:3817] This splitting statement turns an invariant subspace from a possible complication into a decomposition tool. Once a submodule is found, it can be split off and the argument can be repeated, so the theorem supplies the categorical form of complete reducibility rather than just one chosen decomposition. The converse direction is also important: complete reducibility should not depend on whether we phrase the problem as a submodule complement or as a short exact sequence. Establishing the equivalence lets Weyl's theorem be used interchangeably in either language. [quotetheorem:3818] [citeproof:3818] Conversely, the splitting statement implies complete reducibility by applying it to $0\to W\to V\to V/W\to 0$. Thus Weyl theorem can be read as the assertion that the finite-dimensional module category of a semisimple Lie algebra is semisimple. [example: Finite-Dimensional Modules for Sl Two] Let $F$ be algebraically closed of characteristic zero and let $\mathfrak g=\mathfrak{sl}_2(F)$. The Lie algebra $\mathfrak{sl}_2(F)$ is semisimple, so by *Weyl complete reducibility*, every finite-dimensional $\mathfrak g$-module is completely reducible. Thus, if $W\subset V$ is a $\mathfrak g$-submodule, there is a $\mathfrak g$-submodule $U\subset V$ such that \begin{align*} V=W\oplus U. \end{align*} Starting with a finite-dimensional $\mathfrak g$-module $V$, if $V=0$ there is nothing to decompose, and if $V$ is simple then $V$ is already one simple summand. If $V$ is not simple, choose a nonzero proper submodule $W\subset V$. Complete reducibility gives \begin{align*} V=W\oplus U. \end{align*} Since $W\ne 0$ and $W\ne V$, both summands have smaller dimension than $V$: \begin{align*} 0<\dim W<\dim V,\qquad 0\le \dim U<\dim V. \end{align*} Repeating the same splitting process on any nonsimple nonzero summand must stop, because the dimensions strictly decrease at each step. Therefore \begin{align*} V=S_1\oplus\cdots\oplus S_r \end{align*} for simple $\mathfrak{sl}_2(F)$-modules $S_1,\dots,S_r$. This conclusion does not use the classification of the simple $\mathfrak{sl}_2(F)$-modules. It says that, once the simple constituents appear, they assemble by direct sum: for example, any short exact sequence \begin{align*} 0\to S\to E\to T\to 0 \end{align*} with $S$ and $T$ simple must split by complete reducibility applied to the submodule $S\subset E$, so there are no nonsplit finite-dimensional extensions among the simple constituents. [/example] This is why the representation theory of $\mathfrak{sl}_2(F)$ can proceed in two stages: first prove complete reducibility, then classify the simple summands. ## The Casimir Element and the Splitting Mechanism The proof of Weyl theorem needs a replacement for averaging over a finite group. For semisimple Lie algebras, the Killing form supplies dual bases and hence a canonical operator that commutes with the action. [definition: Killing Form] Let $\mathfrak g$ be a finite-dimensional Lie algebra over $F$. The Killing form of $\mathfrak g$ is the symmetric bilinear form $\kappa(x,y)=\operatorname{tr}(\operatorname{ad}x\,\operatorname{ad}y)$ for $x,y\in\mathfrak g$. [/definition] In the previous chapter, Cartan criterion identified semisimplicity with nondegeneracy of this form in characteristic zero. Nondegeneracy allows us to choose dual bases. Dual bases let the Lie algebra act twice and then sum the result in a way that is independent of coordinates. This construction is the infinitesimal substitute for averaging: it produces a canonical endomorphism of every finite-dimensional module that can be used to control equivariance. [definition: Casimir Operator] Let $\mathfrak g$ be a finite-dimensional semisimple Lie algebra over $F$, and let $\kappa$ be its Killing form. Choose a basis $x_1,\dots,x_n$ of $\mathfrak g$ and the $\kappa$-[dual basis](/theorems/414) $y_1,\dots,y_n$, so that $\kappa(x_i,y_j)=\delta_{ij}$. For a finite-dimensional $\mathfrak g$-module $V$ with representation $\rho:\mathfrak g\to\mathfrak{gl}(V)$, the Casimir operator on $V$ is $C_V=\sum_{i=1}^n \rho(x_i)\rho(y_i)\in\mathfrak{gl}(V)$. [/definition] Although the formula uses a basis, the operator does not depend on the chosen dual bases. It is the representation-level image of the Casimir element associated with the Killing form. For the Casimir operator to help with splitting, coordinate independence is not enough. We need it to respect the module structure, meaning that it should commute with the action of every element of $\mathfrak g$ and therefore define a natural endomorphism in the representation category. [quotetheorem:3819] [citeproof:3819] Centrality is the point at which Schur-type arguments enter. On a simple module over an algebraically closed field, any endomorphism commuting with the action is scalar. Thus the Casimir operator converts structural information about $\mathfrak g$ into a numerical invariant on each irreducible summand, which is exactly the kind of control needed in reducibility arguments. The next question is what this central operator looks like on modules that cannot be decomposed further. Determining its scalar action on simple modules supplies the representation-theoretic input used later to build invariant splittings. [quotetheorem:3820] [citeproof:3820] In the splitting proof, the same operator is applied not only to the original module but also to Hom-spaces. Its centrality lets the correction procedure preserve equivariance. [explanation: How the Casimir Splits an Extension] Given an exact sequence $0\to U\to V\to W\to 0$, choose a linear section $t:W\to V$. The section need not be a module map, and the defect is the family of maps \begin{align*} d_t(x)(w)=x\,t(w)-t(xw),\qquad x\in\mathfrak g,\ w\in W. \end{align*} This defect lands in $U$ and satisfies the cocycle identity coming from the Lie bracket. The Casimir construction acts on the finite-dimensional module $\operatorname{Hom}_F(W,U)$. Semisimplicity gives enough nondegeneracy to solve the correction equation for a linear map $a:W\to U$ so that $t-a$ has zero defect. Thus $t-a$ is a $\mathfrak g$-module section. The proof is a Lie-algebra analogue of averaging a projection over a finite group, with the Casimir operator taking the role of the averaging device. [/explanation] ## Consequences for Ideals and the Adjoint Representation Weyl theorem applies to every finite-dimensional module, including the adjoint module. This turns representation-theoretic decompositions into structural decompositions of the Lie algebra itself. [definition: Adjoint Module] Let $\mathfrak g$ be a Lie algebra over $F$. The adjoint module of $\mathfrak g$ is the $\mathfrak g$-module whose underlying vector space is $\mathfrak g$ and whose action is $x\cdot y=[x,y]$. [/definition] The submodules of the adjoint module are precisely the ideals of $\mathfrak g$. To translate complete reducibility into a structural decomposition of the Lie algebra, we need invariant complements in the adjoint module, because those complements are exactly complements by ideals. [quotetheorem:3821] [citeproof:3821] This recovers the standard structure statement for semisimple Lie algebras: semisimple means a direct sum of simple ideals, not merely absence of solvable ideals. [example: Adjoint Decomposition in a Product] Let $\mathfrak s=\mathfrak{sl}_2(F)$ and write \begin{align*} \mathfrak g=\mathfrak s\oplus \mathfrak s,\qquad \mathfrak g_1=\mathfrak s\oplus 0,\qquad \mathfrak g_2=0\oplus \mathfrak s. \end{align*} For $(a,b),(u,v)\in \mathfrak g$, the bracket in the direct product is \begin{align*} [(a,b),(u,v)]=([a,u],[b,v]). \end{align*} Hence \begin{align*} [(a,b),(u,0)]&=([a,u],[b,0])=([a,u],0)\in \mathfrak g_1,\\ [(a,b),(0,v)]&=([a,0],[b,v])=(0,[b,v])\in \mathfrak g_2. \end{align*} Thus $\mathfrak g_1$ and $\mathfrak g_2$ are stable under the adjoint action of $\mathfrak g$, so they are submodules of the adjoint module. The action separates by factors. For $a,b,u,v\in\mathfrak s$, \begin{align*} (a,0)\cdot(u,v)&=[(a,0),(u,v)]=([a,u],0),\\ (0,b)\cdot(u,v)&=[(0,b),(u,v)]=(0,[b,v]). \end{align*} So the first copy of $\mathfrak{sl}_2(F)$ acts only on $\mathfrak g_1$, and the second copy acts only on $\mathfrak g_2$. Since $\mathfrak{sl}_2(F)$ is simple by *Simplicity of $\mathfrak{sl}_2$*, the only $\mathfrak g$-submodules of $\mathfrak g_1$ are $0$ and $\mathfrak g_1$, and the same holds for $\mathfrak g_2$. Finally, \begin{align*} \mathfrak g_1+\mathfrak g_2&=\{(u,0)+(0,v):u,v\in\mathfrak s\}\\ &=\{(u,v):u,v\in\mathfrak s\}\\ &=\mathfrak g, \end{align*} and \begin{align*} \mathfrak g_1\cap\mathfrak g_2 &=\{(u,0):u\in\mathfrak s\}\cap\{(0,v):v\in\mathfrak s\}\\ &=\{(0,0)\}. \end{align*} Therefore the adjoint module decomposes as \begin{align*} \mathfrak g=\mathfrak g_1\oplus\mathfrak g_2 =(\mathfrak{sl}_2(F)\oplus 0)\oplus(0\oplus\mathfrak{sl}_2(F)), \end{align*} a direct sum of two simple ideals. [/example] The same reasoning explains why invariant subspaces of the adjoint module carry algebraic meaning: they are exactly the ideals whose presence decomposes the Lie algebra. The next question is whether every semisimple Lie algebra admits such a decomposition into simple ideal summands, not merely in this product example. [quotetheorem:3822] [citeproof:3822] This is stronger than the existence of a vector-space complement. The complement is stable under the adjoint action, so it is compatible with the Lie bracket and is itself an ideal. The theorem therefore turns the representation-theoretic complete reducibility of the adjoint module into an internal decomposition of the Lie algebra. It is the finite-dimensional structural endpoint of the semisimple theory developed in these notes. ## Contrast with Solvable Lie Algebras The final lesson is that complete reducibility is a semisimple phenomenon. Solvable Lie algebras have many invariant flags by Lie theorem, but those flags need not split. [quotetheorem:3823] [citeproof:3823] This result points in the opposite direction from Weyl theorem. Solvable Lie algebras produce invariant lines, while semisimple Lie algebras force invariant complements. [example: A Non-Split Solvable Module] Let $\mathfrak g=F x$ be one-dimensional abelian, and let $V=F^2$ have basis $e_1,e_2$. Suppose $x$ acts by \begin{align*} A=\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}, \end{align*} so \begin{align*} x e_1=Ae_1=\lambda e_1,\qquad x e_2=Ae_2=e_1+\lambda e_2. \end{align*} The line $W=F e_1$ is stable under $x$, because for every $c\in F$, \begin{align*} x(c e_1)=c\,x e_1=c\lambda e_1\in F e_1. \end{align*} Thus $W$ is a one-dimensional submodule. In the quotient $V/W$, the class $\overline{e_2}=e_2+W$ satisfies \begin{align*} x\overline{e_2} &=\overline{x e_2}\\ &=\overline{e_1+\lambda e_2}\\ &=\overline{e_1}+\lambda\overline{e_2}\\ &=0+\lambda\overline{e_2}\\ &=\lambda\overline{e_2}, \end{align*} so $V/W$ is the one-dimensional module on which $x$ acts by $\lambda$. We now show that $W$ has no stable complement. Any one-dimensional complement $L$ to $W$ is spanned by a vector $v=a e_1+b e_2$ with $b\ne 0$, since $b=0$ would put $v$ in $W$. If $L$ were stable under $x$, then $xv$ would belong to $L$, so for some $\mu\in F$ we would have $xv=\mu v$. But \begin{align*} xv &=x(ae_1+be_2)\\ &=a\,x e_1+b\,x e_2\\ &=a\lambda e_1+b(e_1+\lambda e_2)\\ &=(a\lambda+b)e_1+b\lambda e_2, \end{align*} while \begin{align*} \mu v=\mu(ae_1+be_2)=a\mu e_1+b\mu e_2. \end{align*} Comparing the $e_2$-coefficients gives $b\lambda=b\mu$, and since $b\ne 0$, this gives $\mu=\lambda$. Comparing the $e_1$-coefficients then gives \begin{align*} a\lambda+b=a\mu=a\lambda, \end{align*} hence $b=0$, contradicting $b\ne 0$. Therefore no complementary line is a submodule, so $V$ is a nonsplit extension of the one-dimensional $\lambda$-module by itself. [/example] The example shows why Lie theorem is not a complete reducibility theorem. A flag can exist without decomposing the module into the direct sum of its successive quotients. [remark: Role of the Hypotheses] The hypotheses in Weyl theorem are structural. Semisimplicity supplies the nondegenerate Killing form and removes solvable extension phenomena. Algebraic closedness supports the scalar form of Schur lemma used in the Casimir argument. Characteristic zero is used in Cartan criterion, trace arguments, and the splitting mechanism. [/remark] Together these results complete the foundational arc of the course. The preceding chapters identified semisimple Lie algebras through the Killing form; Weyl theorem now shows that their finite-dimensional representation theory is governed by direct sums of simple modules. Later theory can therefore focus on understanding the simple modules themselves. ## References