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A complex semisimple Lie algebra has an intrinsic Jordan–Chevalley decomposition $x=x_s+x_n$ characterized by $[x_s,x_n]=0$, a diagonalizable Adjoint representation operator $\mathrm{ad}{x}_s$, and a nilpotent Adjoint representation operator $\mathrm{ad}{x}_n$. The Adjoint representation is faithful because the centre is zero. Closure of semisimple matrix Lie algebras are closed under additive Jordan decomposition applied to this faithful representation supplies the parts and their uniqueness. For every finite-dimensional Lie algebra representation $\rho$, the operators $\rho (x_s)$ and $\rho (x_n)$ are the semisimple and nilpotent parts of $\rho (x)$. Thus the decomposition is independent of a matrix embedding; for a nonfaithful representation it passes to the semisimple quotient by its kernel.

A Lie algebra over a field $k$ is a vector space $\mathfrak{g}$ with a bilinear map $[,]:\mathfrak{g}\times \mathfrak{g}\to \mathfrak{g}$, called the Lie bracket, that is alternating and satisfies the Jacobi identity. Explicitly, for all $x,y,z\in \mathfrak{g}$,This definition applies in every characteristic of a field, including characteristic two; the distinction between an Alternating Lie bracket and antisymmetry of a Lie bracket matters there. See Pavel Etingof, Lie Groups and Lie Algebras, Definition 8.4 and Remark 8.5.

The motivating question is how to describe very small symmetries. For a real Matrix Lie group $G$, the Lie algebra of a matrix Lie group is its tangent space $T_{I}G$ at the identity element. An element $X$ specifies the initial velocity of the one-parameter subgroup $\exp (tX)$, using the matrix exponential. The Lie bracket records the first nonzero effect of changing the order of two small transformations. Even when first-order motions cancel, a second-order motion can remain: the small rotation commutator below leaves a rotation about a third axis.

This turns local symmetry questions into computations with linear maps. A basis and its structure constants of a Lie algebra determine every Lie bracket by bilinearity. For example, three matrices suffice for the sl2 matrix commutator example, while the familiar cross product describes the SO(3) Lie algebra. These descriptions help calculate rotations in three dimensions and build Lie algebra representations; the same bracket structure underlies angular momentum commutation relations. The striking feature is that a nonlinear family of motions has a local description by a vector space and one product. Global information still needs care: the Lie algebra alone does not distinguish the circle group from the additive real line.

Historically, the connection with differential equations is visible in Sophus Lie's Theorie der Transformationsgruppen, volume I (1888), prepared with Friedrich Engel. Chapter 6 interprets first-order differential operators as infinitesimal transformations, linking their action on functions to the transformations those functions admit. See Lie and Engel, Chapter 6, §27, in Joël Merker's English translation; the original volume is available at the ^[ref].

The Chevalley-Eilenberg complex consists of alternating maps from ${\mathfrak{g}}^k$ to a Lie algebra representation $V$, with differential built from the bracket and module action. For trivial real coefficients, a two-cocycle is an alternating bilinear form $\kappa$ satisfying $\kappa ([X,Y],Z)+\kappa ([Y,Z],X)+\kappa ([Z,X],Y)=0$. Coboundaries have the form $b([X,Y])$, up to the differential's sign convention. Their quotient is $H^2(\mathfrak{g},\mathbb{R})$ and classifies central extensions of a Lie algebra.

A linear map between Lie algebra representations is a representation homomorphism when it commutes with the action of every element. Its kernel and image of a linear map are invariant subspaces. This is a map of modules, distinct from a Lie algebra homomorphism between the acting algebras themselves.

A bilinear form on a Lie algebra representation is invariant when it satisfies the displayed identity. Equivalently, $T_B(v)=B(v,-)$ is an intertwining operator from the representation to its dual Lie algebra representation. For a finite-dimensional irreducible representation over an algebraically closed field, any nonzero such form is nondegenerate, and the Schur lemma makes all invariant forms proportional. In characteristic different from two, transposing twice then proves that the form is a symmetric bilinear form or an alternating bilinear form.

The abelian Lie algebra $kI\subseteq \mathrm{End}(V)$, for $V\ne 0$, is a nilpotent Lie algebra but contains the nonnilpotent identity map. Thus nilpotence of the abstract Lie algebra does not imply that a particular Lie algebra representation consists of nilpotent endomorphisms. Engel theorem requires nilpotence of every represented endomorphism to obtain strict upper triangularity.

Every finite-dimensional irreducible Lie algebra representation has highest weight $\lambda =a{\Lambda }_1+b{\Lambda }_2$ for unique integers $a,b\ge 0$. The Weyl vector isso $\lambda +\rho =(a+b/2+3/2,b/2+1/2)$. In the Weyl dimension formula, the factors for the four positive roots, ordered as ${\epsilon }_1-{\epsilon }_2,{\epsilon }_2,{\epsilon }_1,{\epsilon }_1+{\epsilon }_2$, are respectivelyHenceIn particular the two fundamental representations have dimensions five and four. The highest root is $\theta ={\epsilon }_1+{\epsilon }_2=2{\Lambda }_2$, so the Adjoint representation has dimension ten. These also fix the dimensions of the crystals below. Our numbering places the long root first; interchanging a B2/C2 numbering without interchanging the associated fundamental-weight labels would give the wrong dimension formula.

Define an equivariant contraction $C:V^{\ast }\otimes {\mathrm{Sym}}^{2}V\to V$ byThis uses the natural dual pairing, so it commutes with the Lie algebra representation. Also define $s(v)={\sum }_i{e}_i^{\ast }\otimes e_{i}v$. The identity tensor ${\sum }_i{e}_i^{\ast }\otimes e_i$ is invariant, making $s$ equivariant. Direct contraction gives $C(s(v))=3v+v=4v$. Thus $s$ is injective and $C$ is surjective, andis a direct decomposition into $\ker C$ and $s(V)$. In particular $\dim \ker C=18-3=15$ and $s(V)$ is a defining irreducible module of highest weight ${\omega }_1$.

For the usual upper-triangular choice of positive roots, $T_3^{11}=z^{\ast }\otimes x^2$ lies in the kernel and is killed by both simple raising operators $E_{12},E_{23}$. Its highest weight is $2{ϵ}_1-{ϵ}_3=2{\omega }_1+{\omega }_2$, with Dynkin labels $(2,1)$. The highest-weight classification and complete reducibility give an irreducible copy $L(2,1)$ generated by this vector. The Weyl dimension formula for ${\mathfrak{sl}}_3$ isso $\dim L(2,1)=15$. It fills the entire kernel. Therefore the sl3 decomposition of the symmetric-square dual tensor product, with its tensor factors exchanged, isThe weight diagrams below record both summands. Their labels use $K_{ij}=2T_j^{ij}-T_i^{ii}$ for $j\ne i$ and $S_i={\sum }_j{T}_j^{ij}$; these are the weight bases detailed in part (iii).

For a finite-dimensional Lie algebra representation $V$, a weight of a representation relative to $\mathfrak{h}$ is a linear functional $\mu \in {\mathfrak{h}}^{\ast }$ for which the simultaneous eigenspaceis nonzero. Its elements are weight vectors, and $\dim V_{\mu }$ is the weight multiplicity. The set of weights is $\boxed{{\Lambda }_V=\{\mu :V_{\mu }\ne 0\}}$. For a complex semisimple Lie algebra, the Cartan action on a finite-dimensional module is simultaneously diagonalizable, giving $V={⨁}_{\mu \in {\Lambda }_V}{V}_{\mu }$. The set of weights and their multiplicities are distinct information: several basis vectors can occupy the same point of a weight diagram.

Let ${\hat{T}}_a={\mathrm{ad}}_{T_a}$ on the Lie algebra itself. The Jacobi identity says, for every $X$,Therefore $[{\hat{T}}_a,{\hat{T}}_b]=c_{ab}^c{\hat{T}}_c$, proving that the displayed structure-constant matrices form the Adjoint representation of a Lie algebra.

The bilinear form ${\kappa }_{ab}=\mathrm{tr}({\hat{T}}_a{\hat{T}}_b)$ is the Killing form. It is symmetric by cyclicity of trace. Its invariance follows by expanding the two commutators:In components this is preciselyEquivalently, the adjoint action preserves the bilinear form, so it is an invariant tensor.

The invariant abelian subalgebra is an abelian ideal of a Lie algebra $\mathfrak{a}$. Assume it is nonzero, as required for the conclusion. For $X\in \mathfrak{a}$ and $Y\in \mathfrak{g}$, the map ${\mathrm{ad}}_X{\mathrm{ad}}_Y$ sends $\mathfrak{g}$ into $\mathfrak{a}$, because it is an ideal. It vanishes on $\mathfrak{a}$, because ${\mathrm{ad}}_Y$ preserves the ideal and ${\mathrm{ad}}_X$ kills it. Relative to $\mathfrak{a}$ and a vector-space complement, this map has zero diagonal blocks, so its trace is zero. ThusThis proves that abelian ideals lie in the radical of the Killing form. Choose a nonzero $X=x^a{T}_a$ in the ideal. Then ${\kappa }_{ba}{x}^a=0$ for every $b$, so the coordinate vector $x$ is a nonzero null eigenvector of the Killing matrix. The nonzero-ideal qualification cannot be dropped: the zero ideal is abelian even in a semisimple algebra with nondegenerate Killing form.

Now suppose $\kappa$ is invertible and let $C_{\kappa }={\kappa }^{bc}{t}_b{t}_c$ in any Lie algebra representation. In matrix notation invariance is $A_a^T\kappa +\kappa A_a=0$, where $(A_a{)}^d{}_b=c_{ab}^d$. Multiplying by the inverse form gives $A_a{\kappa }^{-1}+{\kappa }^{-1}{A}_a^T=0$. HenceNo commutativity of the representation matrices has been assumed. This proves the centrality of the quadratic Casimir operator formed with the inverse Killing form. Schur lemma makes it a scalar on every finite-dimensional complex irreducible representation.

For SU(2), make the normalization explicit. In a Hermitian angular-momentum basis with $[J_a,J_b]=i{ϵ}_{abc}{J}_c$, the structure-constant matrices obeyThus $C_{\kappa }=\frac{1}{2}{\sum }_a{J}_a^2$. In the compact anti-Hermitian basis $T_a=-i{J}_a$, the brackets have real structure constants ${ϵ}_{abc}$ and the Killing form of the SU(2) Lie algebra is instead $-2{\delta }_{ab}$; the inverse contraction gives the same $C_{\kappa }$. This sign change is a basis convention, not a different spectrum.

In the complex highest-weight basis $(H,E^{+},E^{-})$ used earlier, the same Killing form and its inverse areConsequentlyOn a highest vector of weight $m=2j$, the ladder calculation gives $E^{+}{E}^{-}v=mv$ and $E^{-}{E}^{+}v=0$. Its Casimir eigenvalue is thereforeCentrality gives the same value throughout the irreducible multiplet. This is the Killing-normalized SU(2) quadratic Casimir; the familiar angular-momentum operator ${\sum }_a{J}_a^2$ has eigenvalue $j(j+1)$ and is twice this inverse-Killing-form contraction.

Define a symmetric bilinear form $B$ by pairing $c$ with $d$ and $p$ with $q$, each with value one, and setting all other pairings of basis vectors to zero. Its Gram matrix in the order $(c,d,p,q)$ isThe determinant is one, so this is a nondegenerate bilinear form.

For invariance, put $T(x,y,z)=B([x,y],z)$. The nonzero values on distinct basis triples are generated byand the reversed triples have value minus one. Every value involving $c$, or a repeated basis vector, vanishes. Thus $T$ is an alternating trilinear form, explicitly $d^{\ast }\wedge p^{\ast }\wedge q^{\ast }$. In particular $T(x,y,z)=T(y,z,x)$. By symmetry of $B$, this reads $B([x,y],z)=B(x,[y,z])$, the defining identity for an invariant bilinear form on a Lie algebra.

Suppose a finite-dimensional Lie algebra representation $\rho$ realized this form as $B(x,y)=\mathrm{tr}(\rho (x)\rho (y))$. The Lie theorem says that every finite-dimensional representation of a complex solvable Lie algebra admits a basis in which all representing matrices are upper triangular. A commutator of upper triangular matrices has zero diagonal, hence is strictly upper triangular. Therefore $\rho ([\mathfrak{g},\mathfrak{g}])$ is strictly upper triangular, and its product with any representing matrix still has zero diagonal. It follows thatThis is the general fact that trace forms of solvable Lie algebras annihilate the derived algebra. Here $c\in [\mathfrak{g},\mathfrak{g}]$, so such a trace form would have $B(c,d)=0$, contradicting the constructed value one. Thus the nondegenerate invariant form exists, but no finite-dimensional representation has it as its trace form. Finite dimensionality is implicit in the ordinary representation trace; no trace on arbitrary infinite-dimensional endomorphisms is being assumed.

The proof of the linearity of a uniform pro-p group has three distinct steps: pass to a finite-dimensional Lie algebra, integrate a faithful representation on a sufficiently small open subgroup, and extend it to the entire group. Working first on a small subgroup is essential because a faithful Lie algebra representation need not send the original integral lattice into the convergence domain of the matrix exponential.

Let $G$ be a uniform pro-p group and let $L_G$ be its intrinsic powerful Lie lattice. Thenis a finite-dimensional Lie algebra over a field of characteristic zero. The Ado theorem, the classical Lie-theoretic ingredient in this proof, supplies a faithful finite-dimensional Lie algebra representationOne must use a faithful representation rather than just the Adjoint representation, which annihilates the center of a Lie algebra.

Choose a p-adic lattice $\Lambda \subset V$. Since $L_G$ is finitely generated as a module, the entries of the representing matrices of a module basis have bounded denominators. Hence for sufficiently large $k$,For odd $p$, the matrix exponential and p-adic logarithm converge and are inverse between $p\mathrm{End}(\Lambda )$ and $1+p\mathrm{End}(\Lambda )$. Put $H=G^{p^k}$; under the uniform pro-p group and powerful Lie lattice correspondence it is the BCH group on $p^k{L}_G$. DefineHere ${\exp }_G$ denotes the inverse of the intrinsic logarithm identification, not an assumed matrix representation of $G$. The Lie algebra homomorphism $\rho$ respects the convergent Baker--Campbell--Hausdorff formula, soThus $\tau$ is a continuous group homomorphism. If $\tau ({\exp }_{G}X)=1$, taking the matrix logarithm gives $\rho (X)=0$, whence $X=0$. Therefore $\tau$ is faithful and all its matrices preserve $\Lambda$. This proves linearity of the open normal subgroup $H$.

To finish, use the extension of a faithful representation from an open normal subgroup. Choose representatives $t_1=1,t_2,\dots ,t_r$ of the finite left cosets $G/H$, where $r=[G:H]$. On the finite free module $W={⨁}_{j=1}^r\Lambda$, define the action of $g\in G$ by writing uniquelyand mapping the $j$-th block to the $i$-th block by $\tau (h_{ij})$. If $g^{\prime }{t}_j=t_{\ell }{h}^{\prime }$ and $g{t}_{\ell }=t_{i}h$, then $g{g}^{\prime }{t}_j=t_{i}h{h}^{\prime }$; this proves the representation identity because $\tau (h{h}^{\prime })=\tau (h)\tau (h^{\prime })$. Each block matrix is invertible over ${\mathbb{Z}}_p$, and the map is continuous because the finite coset permutation is locally constant and the remaining matrix entries are obtained continuously from $\tau$.

If $g\notin H$, its permutation of $G/H$ moves the identity coset, so its block representation is not the identity. If $g\in H$ and its block representation is the identity, the block corresponding to $t_1=1$ is $\tau (g)=1$, and faithfulness of $\tau$ gives $g=1$. Thus this induced representation is faithful. Choosing a basis of $W$ provesThe continuous injection is a topological embedding, since $G$ is compact and the target is Hausdorff. The trivial group is immediate. For uniform groups at $p=2$, the same proof uses the usual stronger powerfulness condition and the convergence ball $4\mathrm{End}(\Lambda )$ instead; the preceding questions concern odd $p$.

Realize the defining sl2 Lie algebra Lie algebra representation as $\mathbb{C}x\oplus \mathbb{C}y$, with $H=x{\partial }_x-y{\partial }_y$, $E=x{\partial }_y$, $F=y{\partial }_x$. A basis of the symmetric power is $e_i=x^{4-i}{y}^i$ for $0\le i\le 4$, with $H{e}_i=(4-2i)e_i$. Hence the exterior square basis $w_{ij}=e_i\wedge e_j$, $i<j$, has weight $8-2(i+j)$.

Collecting equal weights gives the complete weight-space decomposition:Every one of the ten exterior basis vectors appears exactly once. Thus $U={⨁}_{m=6,4,2,0,-2,-4,-6}{U}_m$, with the displayed bases and dimensions.

The printed phrase “not reducible” is a misprint: the requested nonzero complementary submodules establish that this tensor square is reducible. The natural reason is that the flip $\tau (v\otimes w)=w\otimes v$ commutes with the Lie algebra representation. Its two eigenspaces are the symmetric square and the exterior square, givingIn the symmetric square, the four weights $\pm 2L_i$ and four weights $\pm L_1\pm L_2$ each occur once, while zero occurs twice. In the exterior square, the four weights $\pm L_1\pm L_2$ each occur once and zero occurs twice; the $\pm 2L_i$ are absent because repeated wedge factors vanish. These are the two requested complementary weight diagrams.

More precisely, the tensor-square decomposition of the defining symplectic representation identifies ${\mathrm{Sym}}^{2}V$ with the irreducible adjoint module $L(2{\omega }_1)$. The identification is an intertwiner sending a symmetric tensor $vw$ to the element of the symplectic Lie algebra $x\mapsto \omega (v,x)w+\omega (w,x)v$. For the exterior square, symplectic contraction of an exterior square has a five-dimensional kernel, the primitive exterior square, and the invariant tensor $\Omega =e_1\wedge e_3+e_2\wedge e_4$ supplies a complementary line. Thus the full irreducible decomposition isHere ${\omega }_1=L_1$ and ${\omega }_2=L_1+L_2$. The two zero weights in the exterior square split into one primitive zero weight and the invariant line.

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

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

For $v\in V_{\lambda }$ and $w\in W_{\mu }$, the tensor product of Lie algebra representations acts byThus its weight space of weight $\nu$ isSubstituting this into the formal character and using $e(\lambda +\mu )=e(\lambda )e(\mu )$ provesEvery sum is finite for the Lie algebra representations under consideration.

The ambient group ring of a weight lattice isAddition is coefficientwise, and multiplication uses addition in the weight lattice. For a finite-dimensional Lie algebra representation $V={⨁}_{\lambda }{V}_{\lambda }$, its formal character isThe formal character records the dimensions of all weight spaces. The ambient lattice ring should be distinguished from the representation ring of a semisimple Lie algebra, which identifies with its Weyl-invariant subring.

For the ${\mathfrak{sl}}_3$ module ${\Gamma }_{1,2}$, $L_1+L_2+L_3=0$ and the highest weight is $L_1-2L_3={\omega }_1+2{\omega }_2$. The decomposition $V\otimes {\mathrm{Sym}}^2{V}^{\ast }={\Gamma }_{1,2}\oplus V^{\ast }$ givesCollecting equal weights yieldsThe six first weights and three second weights have multiplicity one; each of the three weights $-L_i$ has multiplicity two. Their total is $6+3+6=15$, as required.

Work over $\mathbb{C}$ with finite-dimensional Lie algebras and Lie algebra representation spaces. The adjoint form of Engel theorem isHere a nilpotent Lie algebra has terminating lower central series, ${\gamma }_1\mathfrak{g}=\mathfrak{g}$, ${\gamma }_{j+1}\mathfrak{g}=[\mathfrak{g},{\gamma }_j\mathfrak{g}]$. We prove the stronger linear statement: if $V\ne 0$, $L\subseteq \mathfrak{gl}(V)$ and every member of $L$ is a nilpotent endomorphism, then there is a nonzero $v\in V$ killed by all of $L$, and $L$ can be simultaneously represented by strictly upper triangular matrices.

For the proof of Engel theorem by induction and normalizers, first establish a useful nilpotence fact. If $x^m=0$, left and right multiplication by $x$ on $\mathrm{End}(V)$ commute. Expanding their difference givesFor $N=2m-1$, every summand contains a power of $x$ at least $m$, so $(\mathrm{ad}x{)}^{2m-1}=0$. Restrictions and induced quotient maps remain nilpotent.

Now prove the common-kernel statement, the Engel lemma, by induction on $\dim L$. The zero algebra is immediate. Choose a maximal proper Lie subalgebra $M$. Its action on $L/M$ by commutators consists of nilpotent maps by the preceding fact. Applying the lower-dimensional induction hypothesis to its image gives a nonzero coset $y+M$ with $[M,y]\subseteq M$. HenceThe normalizer of a Lie subalgebra is a Lie subalgebra by the Jacobi identity. Maximality implies $N_L(M)=L$, so $M$ is an ideal of a Lie algebra. Moreover $L/M$ has no proper nonzero Lie subalgebra, and every one-dimensional subspace is a Lie subalgebra. Thus $\dim L/M=1$.

Induction also provides a nonzero common annihilatorIt is stable under $L$: for $m\in M$, $x\in L$, $v\in W$,because $[m,x]\in M$. Choose $x$ representing a basis of $L/M$. The restriction of the nilpotent endomorphism $x$ to the nonzero space $W$ has nonzero kernel. A vector in this kernel is killed by both $M$ and $x$, hence by all of $L$. This finishes the induction. When an action is not faithful, induction is applied to its image, whose dimension is no larger than $\dim M$; no faithfulness assumption is hidden.

Starting with the common-kernel line, apply the same statement to each successive quotient Lie algebra representation space. It constructs a complete flag of invariant subspacesAn adapted basis represents every member of $L$ by a strictly upper triangular matrix. Every product of $d$ such matrices is zero, so every iterated commutator with $d$ entries is zero as well.

Finally, if every $\mathrm{ad}x$ on $\mathfrak{g}$ is nilpotent, apply the linear statement to the Adjoint representation. With $d=\dim \mathfrak{g}$, every product of $d$ adjoint operators vanishes, so ${\gamma }_{d+1}\mathfrak{g}=0$. Conversely, if ${\gamma }_{c+1}\mathfrak{g}=0$, then $(\mathrm{ad}x{)}^{c}y=0$ for all $x,y$, proving the other implication. This proves Engel's theorem and its simultaneous-triangularization form. The hypothesis concerns every element of the algebra, not merely a chosen set of generators.