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Restrict a finite-dimensional representation of a complex semisimple Lie algebra to the sl2 subalgebra associated with a root $\alpha$. Its Cartan element is the coroot $h_{\alpha }$. The classification of finite-dimensional sl2 representations gives only integer eigenvalues for this element, so every weight of a representation has integral value on every coroot. This proves containment of all such representation weights in the weight lattice; it does not hold for unrestricted infinite-dimensional modules.

If a submodule $U$ of a Lie algebra representation $V$ admits an intertwining map $P:V\to U$ with $P{|}_U=I_U$, then $P^2=P$ and its kernel is an invariant complement. For a complex semisimple Lie algebra, the Hom representation on maps $V\to U$ whose restriction is a scalar identity reduces finding $P$ to splitting of a trivial quotient for a semisimple Lie algebra.

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.

For a finite-dimensional complex Lie algebra $L$, the radical $R$ of its Killing form is a Lie algebra ideal by invariance. For $x,y\in R$, their adjoint actions vanish on $L/R$, so $B_R(x,y)=B_L(x,y)=0$. The Cartan solvability criterion makes the image ${\mathrm{ad}}_R(R)$ solvable; its kernel is the abelian center, so $R$ is solvable. A semisimple Lie algebra has no nonzero solvable ideal, forcing its Killing radical to vanish.

A finite-dimensional irreducible representation of a complex semisimple Lie algebra is minuscule if its weights form one Weyl group orbit. Every weight multiplicity is then one, because multiplicities are invariant under the Weyl group and the highest weight has multiplicity one.

For a complex semisimple Lie algebra, the Cartan subalgebra is orthogonal under the Killing form to every nonzero root space. A vector in the Cartan subalgebra orthogonal to that whole subalgebra is therefore orthogonal to the entire Lie algebra and must vanish. On the Cartan subalgebra the form is $\kappa (h,h^{\prime })={\sum }_{\alpha \in \Phi }\alpha (h)\alpha (h^{\prime })$.

For a root of a root system of a complex semisimple Lie algebra, define $t_{\alpha }$ by $B(t_{\alpha },h)=\alpha (h)$ using the Killing form on its Cartan subalgebra. Then $\alpha (t_{\alpha })\ne 0$. Otherwise vectors $x\in {\mathfrak{g}}_{\alpha }$, $y\in {\mathfrak{g}}_{-\alpha }$ with $B(x,y)=1$ would generate a solvable Lie algebra with $[x,y]=t_{\alpha }$ central. The Lie theorem would make $\mathrm{ad}{t}_{\alpha }$ nilpotent, while the root-space decomposition makes it diagonalizable, forcing $t_{\alpha }$ into the zero center of a Lie algebra.

For an ideal of a Lie algebra $I$ and nondegenerate Killing form, invariance makes $I^{\perp }$ an ideal. An element of $I\cap I^{\perp }$ commutes with $I$, so the intersection is an abelian ideal. Abelian ideals lie in the radical of the Killing form, forcing the intersection to vanish. Thus $\mathfrak{g}=I\oplus I^{\perp }$ as commuting ideals. Iterating minimal nonzero ideals gives the direct-sum formulation of a semisimple Lie algebra.

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

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

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

Use the index convention $[e_a,e_c]=C_a{}^b{}_c{e}_b$. For the left- and right-invariant coframes, the Maurer-Cartan equations areTheir dual invariant vector fields satisfyFor example, evaluating $d{\lambda }^b$ on two dual fields gives $-{\lambda }^b([L_a,L_c])$, since its invariant coefficient functions are constant. The opposite right-invariant bracket gives the other sign. The mixed bracket vanishes because left and right translations commute.

For a semisimple Lie algebra, the Killing form $B(X,Y)=\mathrm{Tr}(\mathrm{ad}X\mathrm{ad}Y)$ is nondegenerate. It is adjoint invariant: using $\mathrm{ad}[Z,X]=[\mathrm{ad}Z,\mathrm{ad}X]$ and cyclicity of trace givesGroup conjugation preserves the trace as well, so this is invariant under the full group adjoint action. Translation supplies a bi-invariant pseudo-Riemannian metric.

Lower the indicated structure index by $C_{abc}=B_{be}{C}_a{}^e{}_c$. ThenThis changes sign under exchange of $a,c$ by the bracket and under exchange of $a,b$ by invariance. Hence $C_{abc}=C_{[abc]}$.

Let $\eta (X,Y,Z)=B(X,[Y,Z])$ at the identity and extend it as a left-invariant differential form. Its value is adjoint invariant, so the extended form is also right-invariant. The three-form in the question is $-\eta$ with the index-lowering convention just specified, an irrelevant overall sign for these properties. To prove closedness rather than only invariance, apply the exterior derivative on four left-invariant fields. Its six bracket terms combine by invariance intoThe final equality is the Jacobi identity. Therefore the Cartan three-formis both bi-invariant and closed.

Define the raw Killing form byFor a finite-dimensional real or complex Lie algebra in characteristic zero, it is nondegenerate exactly when the algebra is a semisimple Lie algebra. For a real compact semisimple algebra it is negative definite. Consequently the positive compact Killing metric is $g=-K$: for example on $\mathfrak{su}(2)$ with $[e_i,e_j]={\epsilon }_{ijk}{e}_k$, $K_{ij}=-2{\delta }_{ij}$ and $-K$ gives a Riemannian metric on $SU(2)$. An overall minus sign in the term “Killing metric” is common here; raw trace $K$ itself is negative definite in this example.

For a Lorentzian metric use $\mathfrak{sl}(2,\mathbb{R})$ with $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. Direct traces givewhose metric signature is $(2,1)$. For a non-Abelian degenerate example, the three-dimensional Heisenberg Lie algebra has $[x,y]=z$ central. Every product of two adjoint maps is zero, so $K=0$.

For any adjoint-invariant nondegenerate metric on a Lie group, the Koszul formula on left-invariant fields simplifies toTake $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$. The Jacobi identity givesTaking the trace of $Z\mapsto R(Z,X)Y=-\mathrm{ad}Y\mathrm{ad}X(Z)/4$ yieldsThus the metric $g=K$ is a Killing-form Einstein metric with Einstein constant $-1/4$. With $g=-K$ the same connection has constant $+1/4$. This proves the Einstein manifold assertion for every semisimple group, including products when the unscaled Killing form is used on all factors.

There is also a different, flat metric-preserving connection. Declare the global left-invariant frame parallel:Its curvature is zero because applying two derivatives to coefficient functions and subtracting their commutator gives zero. Its metric derivative vanishes since $g_{ab}$ is constant in this frame. Its torsion tensor on invariant fields isFor each fixed $c$, antisymmetry gives $T_a{}^b{}_c=C_c{}^b{}_a$, so this is the matrix of $\mathrm{ad}{e}_c$. The torsion contractions of a left-parallel Lie-group connection therefore areThe zero trace follows because a semisimple algebra is perfect: every element is a sum of brackets, and the adjoint of a bracket is a matrix commutator with zero trace. If the metric convention is $g=-K$, the second contraction is $-g_{cd}$. These flat translation connections on a Lie group are not the torsion-free Levi-Civita connection; nonzero torsion permits flatness alongside the curved Killing metric.

Here the Cartan subalgebra of a finite-dimensional complex semisimple Lie algebra is a maximal toral subalgebra: it is abelian, its adjoint operators are simultaneously diagonalizable, and its centralizer is itself. These standard properties give the root-space decompositionThe roots of a root system $\alpha$ are the nonzero linear functionals in ${\mathfrak{h}}^{\ast }$ for which ${\mathfrak{g}}_{\alpha }\ne 0$; $\Phi$ is their finite set. Each root space is one-dimensional. The zero weight space is the centralizer of $\mathfrak{h}$, hence exactly $\mathfrak{h}$. Thus the displayed sum is a direct sum of simultaneous eigenspaces of the Adjoint representation. It is the Cartan root-space decomposition meant here.

The bracket respects weight addition: $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }]\subseteq {\mathfrak{g}}_{\alpha +\beta }$, where ${\mathfrak{g}}_0=\mathfrak{h}$ and a missing weight means the zero space. This follows by applying the Jacobi identity with $H\in \mathfrak{h}$ to the bracket of two root vectors. It explains how the vector-space decomposition encodes the Lie algebra structure.

Use the finite-dimensional complex-module setting of the course. We prove the Weyl complete reducibility theorem by Haar averaging produces invariant complements, defining its constructions rather than invoking complete reducibility itself.

Choose a compact real form $\mathfrak{k}\subset \mathfrak{g}$: it is a real Lie subalgebra with $\mathfrak{g}=\mathfrak{k}\oplus i\mathfrak{k}$, and its simply connected Lie group $K$ is compact. Existence of this form and compact group is a standard fact for complex semisimple Lie algebras. Restrict the representation to $\mathfrak{k}$; integration of a Lie-algebra representation gives a group representation $R:K\to GL(V)$. Since $K$ is connected and generated by exponentials, the $\mathfrak{g}$-invariant subspace $W$ is also $K$-invariant.

Choose any positive-definite Hermitian inner product on $V$. With the Haar measure of $K$ normalized to mass one, defineThis is a Hermitian inner product, and it is positive-definite: for $v\ne 0$ every $R(k)v$ is nonzero, so the continuous positive integrand has positive integral. Right invariance of Haar measure gives $⟨R(a)v,R(a)w{⟩}_K=⟨v,w{⟩}_K$ for $a\in K$.

Let $W^{\prime }=W^{\perp }$ for this averaged inner product. Finite-dimensional inner-product linear algebra gives $V=W\oplus W^{\prime }$. If $v\in W^{\prime }$ and $w\in W$, thenbecause $R(a^{-1})w\in W$. Thus $W^{\prime }$ is $K$-invariant. Differentiating makes it $\mathfrak{k}$-invariant, and since it is a complex vector subspace, $\mathfrak{g}=\mathfrak{k}+i\mathfrak{k}$ also preserves it. ThereforeThe zero or whole-space submodule cases are included. This proves the required complement theorem without circularly assuming it.

The finite-dimensional hypothesis is essential to the statement. The nonsplit zero-weight Verma extension for sl2 is a counterexample for unrestricted modules: $M(0)$ has basis $v_n$ with $h{v}_n=-2n{v}_n$, $f{v}_n=v_{n+1}$ and $e{v}_n=n(1-n)v_{n-1}$. Its submodule spanned by $v_n$ for $n\ge 1$ has trivial one-dimensional quotient. A complement would be trivial and contained in the zero-weight line $\mathbb{C}{v}_0$, yet $f{v}_0=v_1\ne 0$. Hence no such complement exists in that infinite-dimensional case. The paper's finite-dimensional representation setting supplies the necessary hypothesis.

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 $B(X,Y)=\mathrm{tr}(\mathrm{ad}X\mathrm{ad}Y)$ be the Killing form. A real semisimple Lie algebra has nondegenerate $B$, and its invariance gives $B([X,Y],Z)=B(X,[Y,Z])$. Translating $B$ defines a bi-invariant pseudo-Riemannian metric $g=B$ on the Lie group.

For left-invariant vector fields, all derivatives of their pairwise inner products vanish. The Koszul formula reduces toNondegeneracy gives ${\nabla }_{X}Y=[X,Y]/2$. With $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, the Jacobi identity now givesTaking the trace of $X\mapsto R(X,Y)Z=-\frac{1}{4}\mathrm{ad}Z\mathrm{ad}Y(X)$ yields the Ricci tensorThus the Killing-form Einstein metric makes the group an Einstein manifold, allowing an indefinite metric. Its scalar curvature in this convention is $-\dim G/4$.

For a connected compact semisimple Lie group, $B$ is negative definite: the adjoint operators are skew-adjoint for an invariant positive inner product, so $B(X,X)=\mathrm{tr}((\mathrm{ad}X{)}^2)<0$ for $X\ne 0$. The conventional positive Riemannian metric is instead $g_{+}=-B$. Its Ricci tensor is $g_{+}/4$, and its sectional curvature isIt can vanish on commuting two-planes; positivity is not automatic in every direction. With the literal negative metric $B$, the corresponding sectional-curvature signs are reversed.

For a noncompact real semisimple group, a Cartan decomposition $\mathfrak{g}=\mathfrak{k}\oplus \mathfrak{p}$ has $B$ negative on $\mathfrak{k}$ and positive on $\mathfrak{p}$. The Killing metric is indefinite and cannot be made positive by an overall sign. Its Einstein constant remains $-1/4$ for $g=B$, but that does not assert a uniform sign for sectional curvature of all nondegenerate planes. For connected semisimple groups, negative-definite $B$ characterizes the compact case. Disconnected groups require a separate condition on the number of components before compactness follows.

Write ${\xi }_P(p)={\frac{d}{dt}|}_0\exp (t\xi )\cdot p$ for the fundamental vector field of a left Lie group action, and use ${\iota }_{X_f}\omega =df$ throughout. Such fundamental fields satisfy $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$. A moment map has linear components ${\mu }_{\xi }=⟨\mu ,\xi ⟩$ satisfyingFor the action to be Hamiltonian, these components can be chosen equivariantly, or, infinitesimally,This identifies the span of the component Hamiltonians with the Lie algebra image, rather than identifying the entire infinite-dimensional Poisson algebra $C^{\infty }(P)$ with $\mathfrak{g}$. For an actual isomorphism with $\mathfrak{g}$, injectivity is also needed; a faithful infinitesimal action supplies it.

Preservation of $\omega$ alone gives $d{\iota }_{{\xi }_P}\omega =0$, not necessarily exactness. For example, translation generated by ${\partial }_q$ on the symplectic two-torus has contraction $dp$, whose period is nonzero. That action has no globally defined component Hamiltonian. Once the contractions are exact, choose Hamiltonians linearly in $\xi$. On each connected component of $P$, the discrepancyis constant: apply $[X_f,X_g]=-X_{\{f,g\}}$ and the fundamental-field bracket. The Jacobi identity makes $\kappa$ a Lie algebra two-cocycle. Replacing ${\mu }_{\xi }$ by ${\mu }_{\xi }+b(\xi )$ changes it to $\kappa (\xi ,\eta )-b([\xi ,\eta ])$. Thus the exactness condition and vanishing of this moment-map equivariance obstruction are precisely the required infinitesimal conditions.

For a semisimple Lie algebra, both hold. First Cartan's magic formula givesSemisimplicity implies $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$, so every generator is a sum of commutators and has a globally exact contraction. This proves Hamiltonian existence for a semisimple symplectic action.

Next use the nondegenerate Killing form to define $D$ by $B(D\xi ,\eta )=\kappa (\xi ,\eta )$. Antisymmetry makes $D$ skew for $B$. The cocycle identity and invariance of $B$ imply $D[\xi ,\eta ]=[D\xi ,\eta ]+[\xi ,D\eta ]$, so $D$ is a derivation. Choose $Z$ byThen, using cyclicity of trace and $[D,\mathrm{ad}\xi ]=\mathrm{ad}(D\xi )$,Hence $D=\mathrm{ad}Z$ and $\kappa (\xi ,\eta )=B(Z,[\xi ,\eta ])$. Taking $b(\xi )=B(Z,\xi )$ removes the cocycle and proves semisimple moment-map equivariance without merely invoking a cohomology theorem. For a connected group, integrate the infinitesimal equivariance along its one-parameter subgroups to obtain group equivariance. A disconnected group can require an additional discrete-equivariance check.

The derived series of a Lie algebra is defined recursively by ${\mathfrak{g}}^{(0)}=\mathfrak{g}$ and ${\mathfrak{g}}^{(r+1)}=[{\mathfrak{g}}^{(r)},{\mathfrak{g}}^{(r)}]$, where the bracket denotes the linear span of all brackets of pairs of elements. A solvable Lie algebra is one whose derived series of a Lie algebra terminates at zero:The solvable radical $\mathrm{rad}\mathfrak{g}$ of a finite-dimensional Lie algebra is its largest solvable ideal of a Lie algebra. A semisimple Lie algebra has no nonzero solvable ideals of a Lie algebra, equivalentlyThese are definitions; identifying semisimplicity with a direct sum of simple ideals additionally uses the structure theorem in characteristic zero. The calculations and representation theorems below are over $\mathbb{C}$, as in the complex representation setting of this paper.

In the defining representation $V={\mathbb{C}}^{2n}$, the basis vectors $v_i$ and $v_{\bar{i}}$ have weights ${\epsilon }_i$ and $-{\epsilon }_i$, respectively, each with multiplicity one. Thus its formal character isThe vector $v_1$ is killed by every positive root space, since all positive-root matrices are upper triangular, and is a highest-weight vector. Any nonzero invariant subspace splits into weight spaces and therefore contains one of the natural weight vectors, since those weight spaces are one-dimensional. The root matrices connect that vector to all the others, so this representation is irreducible for $n\ge 2$. In terms of the fundamental weights of Dn, its highest weight is ${\omega }_1$ for $n\ge 3$ and ${\omega }_1+{\omega }_2$ for $n=2$.

The flip of the two tensor factors is equivariant, henceUse the invariant symmetric bilinear form $Q(u,v)=u^{T}Jv$. Contraction $C(u\otimes v)=Q(u,v)$ is an equivariant map to the trivial Lie algebra representation. Its inverse-form tensoris invariant and satisfies $C(\tau )=2n$. ConsequentlyThe traceless summand contains the highest-weight vector $v_1\otimes v_1$ of weight $2{\epsilon }_1$. The Weyl complete reducibility theorem states that every finite-dimensional complex semisimple Lie algebra representation is a direct sum of irreducible representations, so the submodule generated by this highest vector is $L(2{\epsilon }_1)$. The Weyl dimension formula verifies that it is the entire traceless space: only positive roots involving ${\epsilon }_1$ change their factors, andThis equals $\dim {\mathrm{Sym}}^{2}V-1=n(2n+1)-1$. Thus the traceless symmetric square of the defining even orthogonal representation is irreducible, including $n=2$.

For the other summand, the exterior square of the defining orthogonal representation has the equivariant identificationThe image is skew relative to $Q$, and in a basis making $Q$ the identity the images are the independent elementary skew matrices. The map is therefore an isomorphism onto the Adjoint representation of a Lie algebra. For $n\ge 3$, the connected Dn Dynkin diagram gives a simple Lie algebra (with $D_3=A_3$); its adjoint module is irreducible because its invariant subspaces are exactly its ideals. Its highest weight is ${\epsilon }_1+{\epsilon }_2$, and $v_1\wedge v_2$ is a highest vector. The tensor-square decomposition of the defining even orthogonal representation is consequentlyThe three dimensions are $(n+1)(2n-1)$, $n(2n-1)$ and one, adding to $4n^2$. The highest weights are $2{\omega }_1$, ${\omega }_2$ and zero for $n\ge 4$; for $n=3$ the middle weight is ${\omega }_2+{\omega }_3$.

For $n=2$, the Chiral decomposition of the complexified so4 Lie algebra splits the adjoint module into its two three-dimensional ideals. ThusThe weights are $2{\omega }_1+2{\omega }_2$, $2{\omega }_1$, $2{\omega }_2$ and zero. Equivalently, labelling highest weights by the two sl2 Lie algebra factors, this is $(2,2)\oplus (2,0)\oplus (0,2)\oplus (0,0)$. If $n=1$ is allowed, the abelian algebra has $V=L({\epsilon }_1)\oplus L(-{\epsilon }_1)$ and $V\otimes V=L(2{\epsilon }_1)\oplus 2L(0)\oplus L(-2{\epsilon }_1)$; there is then no single highest weight for an irreducible natural module.

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

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

In the finite-dimensional complex semisimple Lie algebra setting, the rank of a semisimple Lie algebra is the dimension of a Cartan subalgebra $\mathfrak{h}$, equivalently the number of independent commuting Cartan generators. For a compact real algebra, use its complexification and the corresponding maximal torus. Simultaneously diagonalizing the commuting adjoint actions of $\mathfrak{h}$ gives the root-space decompositionThe nonzero linear functionals $\alpha$ for which this space is nonzero are the roots of a root system. An invariant inner product identifies them with vectors in a real Euclidean root space. These are the usual semisimple assumptions behind the rank and root language here.

The displayed generators $H,E^{+},E^{-}$ form an sl2 triple. Start with a nonzero lowest-weight element $X_0$ and define $X_{n+1}=[E^{+},X_n]$. The Jacobi identity givesInduction starting with $[H,X_0]=-\lambda X_0$ consequently gives the finite sl2 lowest-weight ladder weightsNow commute the lowering equation with $E^{+}$ and again use the Jacobi identity:The left side is $q_n{X}_n$. Whenever $X_n\ne 0$, coefficient comparison givesThe initial condition is $q_0=0$, meaning $[E^{-},X_0]=0$; directly $[E^{-},X_1]=[-H,X_0]=\lambda X_0$, so $q_1=\lambda$. Summing the recurrence from $0$ to $n-1$ yieldsIn fact the vector identity $[E^{-},X_n]=n(\lambda -n+1)X_{n-1}$ follows inductively from the same Jacobi identity without needing a separately assumed lowering coefficient.

If $X_{n_0}$ is the last nonzero element and $[E^{+},X_{n_0}]=0$, apply $E^{-}$ to this zero next element. The lowering identity givesso $\lambda =n_0$, a nonnegative integer. The nonzero endpoint is necessary. Literally allowing $X_{n_0}=0$ would make the printed conclusion false: take $X_0=E^{-}$ in the adjoint representation of ${\mathfrak{sl}}_2$. Then $\lambda =2$, $X_1=H$, $X_2=-2E^{+}$, $X_3=0$. Choosing $n_0=3$ gives a vanishing next bracket but not $\lambda =3$. The intended endpoint is the first terminating bracket after the nonzero ladder.

For the plus-sign nested commutators, use $H=H_1$, $E^{\pm }=E_1^{\pm }$ and $X_0=E_2^{+}$. The cross relation gives $[E_1^{-},E_2^{+}]=0$, while $[H_1,E_2^{+}]=K_{21}{E}_2^{+}$. Thus the lowest-weight label is $\lambda =-K_{21}$. If the $n$-fold raised element is nonzero and the $(n+1)$-fold element vanishes, the preceding endpoint argument gives $\lambda =n$ and henceFor the minus-sign ladder, use the sl2 triple $H^{\prime }=-H_1$, $E^{\prime +}=E_1^{-}$, $E^{\prime -}=E_1^{+}$ and start at $X_0=E_2^{-}$. Its lowest-weight equation is again $[H^{\prime },X_0]=K_{21}{X}_0$, and its lowering bracket vanishes by the cross relation. The same argument proves exactly the same result, including the endpoint assumption that is explicit in the nested-commutator question.

For arbitrary roots of a root system $\alpha ,\beta$, choose the normalized sl2 subalgebra associated with a root $\beta$, whose generators obey $[H_{\beta },E_{\beta }^{\pm }]=\pm 2E_{\beta }^{\pm }$ and $[E_{\beta }^{+},E_{\beta }^{-}]=H_{\beta }$. The coroot normalization givesThe normalization has a concrete construction. Use the nondegenerate Killing form $B$, and define $t_{\beta }\in \mathfrak{h}$ by $B(t_{\beta },H)=\beta (H)$. Invariance makes root spaces orthogonal unless their roots sum to zero, so there exist $e\in {\mathcal{L}}_{\beta }$ and $f\in {\mathcal{L}}_{-\beta }$ with $B(e,f)=1$. Invariance then gives $B([e,f],H)=B(e,[f,H])=\beta (H)$, hence $[e,f]=t_{\beta }$. With $(\beta ,\beta )=\beta (t_{\beta })$, take $E_{\beta }^{+}=e$, $E_{\beta }^{-}=2f/(\beta ,\beta )$ and $H_{\beta }=2t_{\beta }/(\beta ,\beta )$. Their brackets have the required normalization and $\alpha (H_{\beta })=2(\alpha ,\beta )/(\beta ,\beta )$. Let $v$ be a nonzero $\alpha$ root vector with $H_{\beta }$ weight $w=\alpha (H_{\beta })$. Repeatedly apply $\mathrm{ad}{E}_{\beta }^{-}$ until reaching a last nonzero vector $v_0=(\mathrm{ad}{E}_{\beta }^{-}{)}^{p}v$. Its weight is $w-2p$. This process terminates because different weights are linearly independent and the Lie algebra is finite-dimensional. Starting at $v_0$, repeatedly raise; this also terminates for the same reason. The proven lowest-weight endpoint formula says $w-2p=-N$ for a nonnegative integer $N$. ThereforeThis proves integrality of Cartan integers directly from the finite ladder, rather than assuming the crystallographic axiom.

Finally use the indices actually printed in the PDF: $E_1^{+}=R^1{}_2$, $E_2^{+}=R^2{}_3$, $E_1^{-}=R^2{}_1$, $E_2^{-}=R^3{}_2$. The converted TeX has corrupted these indices. The prescribed bracket is that of trace-free matrix units. The sum of the diagonal generators is zero, leaving two independent diagonal elements, and their pair brackets giveFor example $[R^1{}_2,R^2{}_1]=R^1{}_1-R^2{}_2$, and the analogous second pair gives $H_2$. The cross raising-lowering brackets vanish. A diagonal $H$ with entries $h_r$ has $[H,R^r{}_s]=(h_r-h_s)R^r{}_s$. Thus $H_1$ acts on $E_1^{+},E_2^{+}$ with weights $2,-1$, and $H_2$ acts on them with weights $-1,2$. The negative generators have the opposite weights. In the question's index ordering,This is the Cartan matrix of the A2 root system, and the Lie algebra generated by these matrix units has rank two. As an additional bracket check, $[E_1^{+},E_2^{+}]=R^1{}_3\ne 0$ whereas $[E_1^{+},R^1{}_3]=0$, so the nested ladder has $n=1$ and indeed $K_{21}=-1$. The negative ladder likewise terminates after one nonzero bracket.