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For a Lie group $G$, Left translation on a Lie group and right translation on a Lie group are the diffeomorphisms $L_g(h)=gh$ and $R_g(h)=hg$. Their differentials identify the tangent space at the identity with every tangent space. Translating a basis $e_a$ of the Lie algebra gives global left-invariant vector fields $E_a(g)=(d{L}_g{)}_e{e}_a$ and right-invariant vector fields $F_a(g)=(d{R}_g{)}_e{e}_a$. Left and right translations commute. If $[e_a,e_b]=c^c{}_{ab}{e}_c$, thenThe opposite sign for the right-invariant frame is essential.

Define an affine connection ${\nabla }^{-}$ by declaring the $E_a$ parallel. Explicitly, ${\nabla }_X^{-}(f^a{E}_a)=X(f^a)E_a$. Its curvature form of a connection vanishes, since successive derivatives of the component functions satisfy the vector-field commutator identity. Its torsion tensor isSimilarly, declaring the $F_a$ parallel gives another flat affine connection ${\nabla }^{+}$, with $T^{+}(F_a,F_b)=-[F_a,F_b]$. In a left-invariant frame the latter connection obeysTo see this, express $E_a=F_{{\mathrm{Ad}}_g{e}_a}$. Along $g\exp (t{e}_b)$, differentiation of ${\mathrm{Ad}}_{g\exp (t{e}_b)}{e}_a$ gives ${\mathrm{Ad}}_g[e_b,e_a]$, producing the displayed bracket. Consequently $T^{+}(X,Y)=[X,Y]$ for left-invariant $X,Y$. These are the flat translation connections on a Lie group. They have opposite torsion; if $G$ is Abelian, both torsion tensors vanish and the two connections coincide. The phrase “with torsion” does not mean nonzero torsion for Abelian groups.

The average ${\nabla }^0=({\nabla }^{-}+{\nabla }^{+})/2$ is again an affine connection and satisfies ${\nabla }_X^{0}Y=\frac{1}{2}[X,Y]$ for left-invariant fields. Its torsion tensor is zero. With the curvature convention $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, the Jacobi identity givesDefine the Ricci tensor by $\mathrm{Ric}(X,Y)=\mathrm{Tr}(Z\mapsto R(Z,X)Y)$. Since $[[Z,X],Y]={\mathrm{ad}}_Y{\mathrm{ad}}_{X}Z$, this giveswhere $K(X,Y)=\mathrm{Tr}({\mathrm{ad}}_X{\mathrm{ad}}_Y)$ is the Killing form. Thus the canonical torsion-free connection on a Lie group is generally curved. Its exact flatness criterion is $[[\mathfrak{g},\mathfrak{g}],\mathfrak{g}]=0$: Abelian and two-step nilpotent groups are exceptions. A vanishing Ricci tensor alone does not prove flatness.

For a semisimple Lie algebra, the Killing form is nondegenerate. Its adjoint invariance gives $K([X,Y],Z)+K(Y,[X,Z])=0$, so translating it defines a bi-invariant pseudo-Riemannian metric $g_K$. For left-invariant fields,A metric-compatible torsion-free connection is uniquely the Levi-Civita connection; therefore ${\nabla }^0$ is its Levi-Civita connection. Moreover,The special feature is a canonical nondegenerate metric and an Einstein manifold structure. The metric need not be positive definite: noncompact semisimple groups generally require indefinite signature. For compact semisimple groups, using $-K$ gives a bi-invariant Riemannian metric with $\mathrm{Ric}=\frac{1}{4}{g}_{-K}$ instead.

Identify the ambient real four-dimensional vector space with $2\times 2$ real matrices byFor $A,B\in SL(2,\mathbb{R})$, the linear action $X\mapsto AX{B}^{-1}$ preserves this quadratic form. It gives a group homomorphism to the special orthogonal group of signature $(2,2)$; its determinant on matrix space is $(\det A{)}^2(\det B{)}^{-2}=1$. The group $SL(2,\mathbb{R})$ is connected: an oriented Gram–Schmidt factorization writes every element as a rotation times an upper triangular matrix with positive diagonal and determinant one, and each factor varies in a connected space. Thus the image lies in the identity component $S{O}_0(2,2)$.

If $AX{B}^{-1}=X$ for every $X$, taking $X=I$ gives $A=B$. A matrix commuting with every real $2\times 2$ matrix is scalar, and its determinant-one condition then gives $A=B=\pm I$. The kernel is therefore exactly $\{(I,I),(-I,-I)\}$. Infinitesimally the action is $X\mapsto aX-Xb$, with $a,b$ traceless. If this vanishes for every $X$, then $a=b$ and it is scalar and traceless, hence zero. The differential is injective; source and target Lie algebras each have dimension six, so it is an isomorphism. The image is therefore an open subgroup of the connected target, which must be the entire target. ConsequentlyThe finite quotient is by the diagonal central subgroup, not by a sign in just one factor. The connected-component subscript is a genuine qualification of the source's displayed formula: the full $SO(2,2)$ is disconnected and cannot be isomorphic to a quotient of the connected product.

Let the anti-de Sitter radius be $\ell >0$. Its quadric is $-u^2-v^2+x^2+y^2=-{\ell }^2$, or $\det X={\ell }^2$. Thus the map $X\mapsto g=X/\ell$ identifies this quadric smoothly and globally with $SL(2,\mathbb{R})$. This is the group model of three-dimensional anti-de Sitter spacetime. The induced Lorentzian metric is bi-invariant, since the two matrix multiplication actions preserve the ambient quadratic form. At the identity, tangent matrices $a,b$ are traceless andIndeed, for a traceless matrix $a$, $\det a=-\frac{1}{2}\mathrm{Tr}(a^2)$, and polarizing $-\det (\ell a)$ gives this metric. For ${\mathfrak{sl}}_2(\mathbb{R})$ the Killing form is $K(a,b)=4\mathrm{Tr}(ab)$: in the basis $H=\mathrm{diag}(1,-1)$, $E=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and $F=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$, the brackets are $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$, giving $K(H,H)=8$, $K(E,F)=4$ and the remaining independent pairings zero. Therefore $g_{\ell }={\ell }^{2}K/8$.

The canonical torsion-free connection on a Lie group is the Levi-Civita connection of this bi-invariant pseudo-Riemannian metric. Its Ricci tensor, already calculated from the double bracket, isSubstitution into the vacuum Einstein field equations givesThus the group metric solves the three-dimensional vacuum equations with negative cosmological constant.

The antipodal map $X\mapsto -X$ is $g\mapsto -g$, multiplication by the central element $-I$. Hence the antipodal quotient is $PSL(2,\mathbb{R})=SL(2,\mathbb{R})/\{\pm I\}$, with the same local curvature and cosmological constant. To see the global time issue, useThe induced metric isOn the quadric $t$ has period $2\pi$; curves with fixed $\rho ,\phi$ are closed timelike curves. The antipodal identification is $(t,\phi )\sim (t+\pi ,\phi +\pi )$, and does not eliminate this problem. At $\rho =0$, even the half-period time shift closes after the antipodal quotient.

If global time is not identified, take the universal cover, with $t\in \mathbb{R}$. The topology of $SL(2,\mathbb{R})$ is $S^1\times {\mathbb{R}}^2$, as follows from the same rotation–triangular decomposition, so the universal cover is diffeomorphic to ${\mathbb{R}}^3$ and carries the lifted group structure $\widetilde{SL(2,\mathbb{R})}$. The metric and Einstein equation are unchanged locally. Here $t$ is a global time function with timelike gradient, excluding closed causal curves. Unwrapped anti-de Sitter spacetime is the universal covering group, rather than the original embedded quadric or its antipodal quotient; retaining the antipodal time-shift identification would wrap time again.

Let $R_a=\mathrm{ad}(T_a)$, where $\mathrm{ad}(X)Y=[X,Y]$. In the given basis, the Adjoint representation of a Lie algebra isThe Jacobi identity gives $[\mathrm{ad}X,\mathrm{ad}Y]=\mathrm{ad}[X,Y]$, so $[R_a,R_b]=f^c{}_{ab}{R}_c$, as required for a Lie algebra representation. The Killing form is the symmetric Trace form of a Lie algebra representationCyclicity of the trace proves its invariance directly:In components this is precisely $\boxed{{\kappa }_{dc}{f}^d{}_{ab}+{\kappa }_{bd}{f}^d{}_{ac}=0}$, so the Killing form is an invariant bilinear form on a Lie algebra.

Choose the infinitesimal gauge transformation convention $\delta \varphi =\omega \varphi$, with $\omega (x)={\omega }^a(x)t_a$, and write $A_{\mu }=A_{\mu }^a{t}_a$. A gauge covariant derivative must satisfy $D_{\mu }^{\prime }=U{D}_{\mu }{U}^{-1}$ for $U=1+\omega +O({\omega }^2)$; henceEquivalently $\delta A_{\mu }^a=-{\partial }_{\mu }{\omega }^a+f^a{}_{bc}{\omega }^b{A}_{\mu }^c$. Expanding $\delta (D_{\mu }\varphi )$ makes the cancellation transparent:Thus $\boxed{\delta (D_{\mu }\varphi )={\omega }^a{t}_a{D}_{\mu }\varphi }$: it transforms in the same representation as $\varphi$. In adjoint covariant derivative notation, $\delta A_{\mu }=-D_{\mu }^{\mathrm{ad}}\omega$.

The commutator of two gauge covariant derivatives defines the gauge field strength:ThusConjugating the derivative commutator gives $F_{\mu \nu }^{\prime }=U{F}_{\mu \nu }{U}^{-1}$, so the gauge field strength transforms homogeneously in the Adjoint representation:Even if the matrices $t_a$ in a particular matter representation are not faithful, the displayed component formula defines the Lie-algebra-valued curvature independently of that representation.

Only $g_{(ab)}$ contributes to the Yang-Mills theory Lagrangian, since $F_{\mu \nu }^a{F}^{b\mu \nu }$ is symmetric in $a,b$. Thus take $g_{ab}$ to be a constant real symmetric bilinear form. Its infinitesimal variation isArbitrary local field strengths and gauge parameters give the necessary and sufficient invariant gauge kinetic form conditionAn antisymmetric part is unconstrained but contributes nothing. A nondegenerate kinetic term additionally needs a nondegenerate bilinear form; positive energy additionally needs a positive internal metric in the signature $(+---)$ convention. Neither extra condition follows solely from gauge invariance. For a compact simple real Lie algebra, $-\kappa$ is positive definite and is the standard choice up to a positive factor.

For the intended complex-simple setting, the Killing form is nondegenerate. Consequently write any invariant bilinear form on a Lie algebra uniquely as $g(X,Y)=\kappa (SX,Y)$. Invariance of both forms impliesfor every $X,Y,Z$, so $S$ commutes with the Adjoint representation. A subspace invariant under all adjoint maps is an ideal, hence simplicity makes the adjoint representation irreducible. By Schur lemma, $S=cI$, giving the uniqueness of an invariant bilinear form on a simple Lie algebra:The same conclusion holds for compact simple real algebras by complexification, with $c$ real; nondegeneracy excludes $c=0$, and positive energy selects $c<0$ in the stated convention.

There is a genuine real-simple exception to uniqueness of the Killing form if “simple” is read as an arbitrary real simple algebra. Regard ${\mathfrak{sl}}_2(\mathbb{C})$ as real. It is real simple: its complexification consists of two complex simple factors exchanged by conjugation, so a conjugation-stable ideal is zero or the whole algebra. Both $\mathrm{Re}{\kappa }_{\mathbb{C}}$ and $\mathrm{Im}{\kappa }_{\mathbb{C}}$ are real symmetric invariant forms, while ${\kappa }_{\mathbb{R}}=2\mathrm{Re}{\kappa }_{\mathbb{C}}$. With $H=\mathrm{diag}(1,-1)$, the Killing form of the special linear Lie algebra gives ${\kappa }_{\mathbb{C}}(H,H)=8$ and ${\kappa }_{\mathbb{C}}(H,iH)=8i$. Therefore the imaginary part cannot be a real multiple of ${\kappa }_{\mathbb{R}}$. The uniqueness conclusion requires the customary complex-simple or compact-real interpretation.

Use matrix multiplication together with the wedge product of differential forms, suppressing wedge symbols in this calculation. Set $F=dA+A^2$ and $a=\delta A$. The gauge field strength varies as $\delta F=da+Aa+aA$, the covariant exterior derivative of $a$. The matrix trace has graded cyclicity, $\mathrm{Tr}(\alpha \beta )=(-1{)}^{pq}\mathrm{Tr}(\beta \alpha )$ for forms of degrees $p,q$. Thus the Chern-Simons 5-form can be expanded asFor example, the two cross terms in $F^{2}A$ both become $\mathrm{Tr}(A^{3}dA)$; combining the remaining two terms gives the displayed coefficients. This expansion keeps track of the noncommuting products rather than treating $A$ as a scalar one-form.

Put $B=dA$. Vary the first term and use the graded exterior derivative to move derivatives off $a$:Here $dB=0$. For the next term use $d(A^3)=B{A}^2-ABA+A^{2}B$. Graded cyclicity then givesFinally all five contributions to $\delta \mathrm{Tr}(A^5)$ are equal under the matrix trace, giving $5\mathrm{Tr}(a{A}^4)$. Combining these results proves the variation of the Chern-Simons 5-form:Consequently, on an oriented five-manifold $N$,The boundary term vanishes for a closed manifold, for variations supported in its interior, or when the tangential boundary connection is held fixed. A different boundary variational problem would require boundary conditions or a boundary action as well.

Write $A=A^{\alpha }{T}_{\alpha }$ in a basis of the Lie algebra. Independent compactly supported variations of each component yield the bulk Euler-Lagrange equationsEquivalently, with $d_{\alpha \beta \gamma }=\frac{1}{2}\mathrm{Tr}[T_{\alpha }\{T_{\beta },T_{\gamma }\}]$, these are $d_{\alpha \beta \gamma }{F}^{\beta }\wedge F^{\gamma }=0$. The symmetry of the two curvature two-forms accounts for the anticommutator. Writing simply $F\wedge F=0$ as a full matrix equation requires an additional assumption: variations must span a matrix algebra paired nondegenerately by the matrix trace. For a smaller Lie algebra, only its trace-paired projection is constrained. If the trace is literally taken in the adjoint representation of a semisimple Lie algebra, its symmetric cubic invariant vanishes, and these bulk equations are identically zero. Indeed, the adjoint generators are skew with respect to the nondegenerate invariant Killing form; transposition in this pairing reverses the sign of the symmetrized trace of three generators. The usual nontrivial five-dimensional theory therefore needs an invariant cubic trace that does not vanish; saying that the gauge field transforms in the adjoint alone does not specify such a trace.

Under a Yang-Mills gauge transformation, use $A^g=g^{-1}Ag+g^{-1}dg$, so $F^g=g^{-1}Fg$. For any Lie-algebra element $T$,Since $gT{g}^{-1}$ is still in the Lie algebra, the right-hand side vanishes whenever the original field satisfies every equation. Every gauge transform of a bulk solution remains a bulk solution. This conclusion follows from covariance of the field equations; it does not require the action itself to be exactly invariant under all large gauge transformations or under transformations changing prescribed boundary data.

Let $B$ be the given invariant bilinear form on a Lie algebra. Choose a basis $x_1,\dots ,x_d$ and its $B$-dual basis $x^1,\dots ,x^d$, so $B(x_i,x^j)={\delta }_i^j$. The Casimir element isThe tensor ${\sum }_i{x}_i\otimes x^i$ is the inverse tensor of $B$, so it is independent of the chosen basis. Multiplication in the universal enveloping algebra therefore makes ${\Omega }_B$ independent of that choice as well.

For $z\in \mathfrak{g}$, write $[z,x_i]={\sum }_j{a}_{ji}{x}_j$. Invariance means $B([z,x],y)+B(x,[z,y])=0$, whence $[z,x^i]=-{\sum }_j{a}_{ij}{x}^j$. Thereforeby interchanging $i,j$ in the second sum. Since $\mathfrak{g}$ generates its universal enveloping algebra, ${\Omega }_B$ is central. This argument does not require adding a symmetry assumption to the given bilinear form.

For the sl2 Lie algebra, takeThe cyclic identity for the trace gives $B([Z,X],Y)+B(X,[Z,Y])=0$. The form is nondegenerate, since $B(e,f)=1$, $B(h,h)=2$, and all the other pairings except $B(f,e)$ are zero. The dual of $(e,f,h)$ is $(f,e,h/2)$, soBy the classification of finite-dimensional sl2 representations, an irreducible of dimension $n$ has a highest-weight vector $v$ with $ev=0$ and $hv=mv$, where $m=n-1$. Since $ef=fe+h$, one obtainsCentrality and the Schur lemma then give the answer on the entire representation:This Casimir eigenvalue for sl2 depends on normalization. Choosing the Killing form instead, $K=4B$, scales the dual basis and the Casimir element by $1/4$, giving $(n^2-1)I/8$.

A finite-dimensional complex Lie algebra is a semisimple Lie algebra when its solvable radical is zero, equivalently when it has no nonzero solvable ideals. Its Killing form is the symmetric bilinear formThe cyclic trace identity and ${\mathrm{ad}}_{[z,x]}=[{\mathrm{ad}}_z,{\mathrm{ad}}_x]$ giveThus the Killing form is an invariant bilinear form on a Lie algebra. In particular its radical $R=\{x:B_L(x,L)=0\}$ is an ideal.

We supply the trace argument needed for nondegeneracy rather than assuming the Cartan criterion for semisimplicity. The matrix form of the Cartan solvability criterion says: if $A\subseteq {\mathrm{End}}_{\mathbb{C}}(V)$ and $\mathrm{tr}(uv)=0$ for every $u\in [A,A]$, $v\in A$, then $A$ is solvable. To prove this direction, fix $u\in [A,A]$. Let $V={⨁}_{\lambda }{V}_{\lambda }$ be its generalized eigenspace decomposition, and define $s$ to act on $V_{\lambda }$ by the scalar $\overline{\lambda }$. On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, the semisimple part of ${\mathrm{ad}}_u$ has eigenvalue $\lambda -\mu$, whereas ${\mathrm{ad}}_s$ acts by $\overline{\lambda }-\overline{\mu }=\overline{\lambda -\mu }$. Polynomial interpolation on the finitely many eigenvalues, with all derivatives through the sizes of the nilpotent blocks set to zero, therefore givesHere the same difference always has the same conjugate, so the interpolation is consistent; the prescribed zero derivatives remove every nilpotent block. Since ${\mathrm{ad}}_u(A)\subseteq [A,A]$, it follows that $[s,A]\subseteq [A,A]$.

Write $u={\sum }_j[a_j,b_j]$. By cyclicity and the trace hypothesis,On the other hand, $\mathrm{tr}(us)={\sum }_{\lambda }\dim (V_{\lambda })|\lambda {|}^2$. Thus all eigenvalues of $u$ vanish, and every element of $[A,A]$ is nilpotent. The Engel theorem makes $[A,A]$ nilpotent as a Lie algebra, hence solvable; $A/[A,A]$ is abelian, so $A$ is solvable. This is the Conjugate-spectrum proof of Cartan solvability.

Apply this to the Killing radical $R$. For $x,y\in R$, the adjoint actions preserve $R$ and act as zero on $L/R$, because $R$ is an ideal. Computing traces in a basis adapted to $R$ givesThe matrix Lie algebra ${\mathrm{ad}}_R(R)$ consequently satisfies the trace criterion and is solvable. Its kernel is $Z(R)$, an abelian ideal; a central extension of a solvable algebra is solvable. Thus $R$ is solvable. This proves the reusable assertion that the Killing radical is a solvable ideal. Semisimplicity forces $R=0$, and hence $B_L$ is nondegenerate.

A Cartan subalgebra is a nilpotent subalgebra $H$ which is self-normalizing:For a complex semisimple algebra this is equivalently a maximal toral subalgebra. The nilpotent, self-normalizing definition permits a proof of the restricted nondegeneracy without first assuming the toral characterization.

Use the generalized-weight decomposition for a nilpotent Lie algebra for the adjoint action of $H$:For completeness, the stability underlying this decomposition follows directly from nilpotence of $H$. For fixed $h$, every ${\mathrm{ad}}_h$ on $H$ is nilpotent. If $T=\rho (h)$ and $S=\rho (y)$ in a finite-dimensional representation, then $({\mathrm{ad}}_T{)}^{r}S=0$ for sufficiently large $r$. The identityshows that $S$ preserves each generalized eigenspace of $T$. Starting with a basis of $H$, refine these primary decompositions successively; all summands remain $H$-invariant. Each resulting summand has only one eigenvalue for each basis element. The Lie theorem triangularizes the action on that summand, so those eigenvalues extend to a single linear character $\lambda$ on all of $H$. This proves the displayed decomposition and nilpotence of all shifted operators there.

Since $H$ is nilpotent, $H\subseteq L^0$. The space $L^0$ is a subalgebra: repeated use of the derivation rule for ${\mathrm{ad}}_h$ shows that the bracket of two generalized zero-eigenvectors is another such vector. If $L^0/H\ne 0$, the adjoint action of $H$ on this quotient consists entirely of nilpotent maps. The Engel theorem supplies a nonzero coset $x+H$ with $[H,x]\subseteq H$, contradicting self-normalization. Therefore the zero generalized weight space of a Cartan subalgebra is exactly $H$.

Finally, $B_L(H,L^{\lambda })=0$ when $\lambda \ne 0$. Choose $h\in H$ with $\lambda (h)\ne 0$ and put $T={\mathrm{ad}}_h$. On $H$, a power $T^m$ vanishes. On $L^{\lambda }$, $T$ is invertible. For $u\in H$ and $v\in L^{\lambda }$, write $v=T^{m}w$; invariance givesIf $u\in H$ is orthogonal to $H$, it is now orthogonal to every summand of $L$, so nondegeneracy of $B_L$ implies $u=0$. The restriction $B_L{|}_{H\times H}$ is nondegenerate. This establishes the nondegeneracy of the Killing form on a Cartan subalgebra for every Cartan subalgebra, without requiring a chosen root basis.

A Lie algebra representation $V$ is completely reducible when it is a direct sum of irreducible submodules. In finite dimension this is equivalent to saying that every submodule $W\subseteq V$ has an invariant complementary subspace. Indeed, one may split off a minimal nonzero submodule and induct on dimension; conversely, if $V$ is a finite sum of simple submodules, choose a maximal sum $U$ of these with $U\cap W=0$. If $W+U\ne V$, some simple summand $S$ is not contained in $W+U$. Then $S\cap (W+U)=0$ by simplicity, so $U+S$ is a larger sum disjoint from $W$, a contradiction. Thus $V=W\oplus U$.

We prove the required splitting by Haar averaging produces invariant complements. The structural input is the compact real form of a complex semisimple Lie algebra: there is a real semisimple algebra $\mathfrak{k}$ with $L=\mathfrak{k}\oplus i\mathfrak{k}$, whose simply connected integrating Lie group $K$ is compact. This structural fact is independent of the complete-reducibility assertion. One standard construction uses root vectors normalized so that $[e_{\alpha },e_{-\alpha }]=h_{\alpha }$, with real structure constants compatible with the involution $e_{\alpha }\mapsto -e_{-\alpha }$ and $h_{\alpha }\mapsto -h_{\alpha }$. The real spanis closed under brackets, complexifies to $L$, and has negative-definite Killing form. The compactness criterion for real semisimple Lie algebras then gives a compact simply connected $K$. Thus this route uses root structure and compact integration, not the theorem being proved.

Restrict $\rho :L\to {\mathrm{End}}_{\mathbb{C}}(V)$ to $\mathfrak{k}$. The integration of a Lie-algebra representation gives a representation $R:K\to {\mathrm{GL}}_{\mathbb{C}}(V)$, since $K$ is simply connected. Choose any positive-definite Hermitian inner product $⟨,{⟩}_0$ and average using normalized Haar measure:The integral exists by compactness. It is positive-definite because $R(g)v\ne 0$ for $v\ne 0$, and it is $K$-invariant by translation invariance of Haar measure. Equivalently, every $\rho (a)$ with $a\in \mathfrak{k}$ is skew-Hermitian for this inner product.

Let $W$ be an $L$-submodule. It is invariant under $\mathfrak{k}$, hence under the exponentials generating the connected group $K$. For $v\in W^{\perp }$, $w\in W$ and $g\in K$,Consequently the orthogonal complement $W^{\perp }$ is $K$-invariant, and differentiation makes it $\mathfrak{k}$-invariant. It is a complex vector subspace, so it is also invariant under $i\mathfrak{k}$ and hence under $L$. Therefore $V=W\oplus W^{\perp }$ as $L$-modules.

Induction on $\dim V$ now expresses $V$ as a finite direct sum of irreducible submodules. Every finite-dimensional representation of a finite-dimensional complex semisimple Lie algebra is completely reducible. The compact real form need not be the real form originally used to present $L$; in particular the averaging argument does not require the given presentation to be unitary.

Use a real Lie algebra convention $[T_a,T_b]=f_{ab}{}^c{T}_c$. Its Adjoint representation of a Lie algebra acts on $\mathfrak{g}$ itself:Linearity is immediate. The Jacobi identity givesso $[{\mathrm{ad}}_X,{\mathrm{ad}}_Y]={\mathrm{ad}}_{[X,Y]}$, the defining Lie algebra representation condition. In the basis $T_b$, its matrix entries areThus the adjoint generators are the structure constants arranged as matrices. The representation has kernel equal to the center of a Lie algebra; it need not be faithful for an arbitrary algebra.

The Adjoint representation of a Lie group is ${\mathrm{Ad}}_g(Y)=gY{g}^{-1}$. It satisfies ${\mathrm{Ad}}_{gh}={\mathrm{Ad}}_g{\mathrm{Ad}}_h$, preserves the identity, and sends inverses to inverse matrices. Differentiating $e^{tX}Y{e}^{-tX}$ gives $[X,e^{tX}Y{e}^{-tX}]$, soConsequently the positive adjoint exponentials furnish the representation on elements $g=e^X$ and their products, and the intrinsic conjugation action defines it globally.

The inverse-conjugation formula in the PDF needs a minus adjoint exponent with this standard definition. Its first-order term is $Y-[X,Y]$, whereas $e^{+{\mathrm{ad}}_X}Y$ has first-order term $Y+[X,Y]$. For example $X=T_1,Y=T_2$ in a nonabelian algebra with $[T_1,T_2]\ne 0$ already distinguishes the two. The inverse conjugation and adjoint antirepresentations identity explains the group-order issue as well: $F(g)={\mathrm{Ad}}_{g^{-1}}$ satisfies $F(gh)=F(h)F(g)$. Retaining the printed inverse conjugation as a left action without this order reversal would not give an ordinary group representation.

The Killing form is the symmetric bilinear formTo prove degeneracy for a non-semisimple Lie algebra, take its nonzero solvable radical and the last nonzero member $\mathfrak{a}$ of its derived series of a Lie algebra. This is a nonzero abelian ideal of a Lie algebra. For $x\in \mathfrak{a}$, ${\mathrm{ad}}_x$ maps $\mathfrak{g}$ into $\mathfrak{a}$ and kills $\mathfrak{a}$. Every ${\mathrm{ad}}_y$ preserves $\mathfrak{a}$. Hence ${\mathrm{ad}}_x{\mathrm{ad}}_y$ maps the full space into $\mathfrak{a}$ and has zero restriction there, so its trace is zero. Thus $B(x,y)=0$ for every $y$. This is the abelian ideals lie in the radical of the Killing form argument, and establishes a nonzero kernel and vanishing determinant.

For a compact real semisimple Lie algebra, the adjoint action is unitary in an invariant positive inner product. Its infinitesimal generators are skew-Hermitian, givingStrictness follows because the adjoint kernel is the zero center. This explains the compactness criterion from the Killing form: “strictly negative” means negative-definite, not that every entry of its matrix is negative.

In the first three-generator example, take columns to be images of the basis $(X,Y,H)$. Direct use of the brackets givesTaking traces of products yields the Killing form for cyclic three-generator bracketsIt is nondegenerate, hence semisimple by the preceding degeneracy result, but it has mixed signature and is not compact. An explicit realization iswhich span the real traceless two-by-two matrices and satisfy exactly these brackets.

Changing the sign of $[X,H]$ changes the first and third adjoint matrices towhile ${\mathrm{ad}}_Y$ is unchanged. Nowthe negative-definite compact case. The basis $X=-i{\sigma }_1$, $Y=-i{\sigma }_2$, $H=-i{\sigma }_3$ realizes it as the real special unitary Lie algebra $\mathfrak{su}(2)$.

A finite-dimensional Lie algebra over the complex numbers $L$ is a semisimple Lie algebra when its solvable radical is zero, equivalently when it has no nonzero solvable ideals. Its Killing form isThe cyclic property of the trace makes this bilinear form symmetric and gives its invariance of a bilinear form on a Lie algebra:It follows that $R=\{x:B_L(x,L)=0\}$ is an ideal of a Lie algebra. For $x\in R$, $\mathrm{ad}x$ induces the zero map on $L/R$. Therefore, for $x,y\in R$, the matrix trace splits over the invariant subspace $R$ and the quotient to give $B_R(x,y)=B_L(x,y)=0$. In particular $B_R([R,R],R)=0$. The Cartan solvability criterion implies that $R$ is solvable. Since $L$ is semisimple, $R=0$: the Killing form is nondegenerate.

For completeness, the trace step in the Cartan solvability criterion is precisely the mechanism of the previous solution. For a complex matrix Lie algebra $\mathfrak{a}$ with $\mathrm{tr}(xy)=0$ for $x\in [\mathfrak{a},\mathfrak{a}]$, $y\in \mathfrak{a}$, set $U=[\mathfrak{a},\mathfrak{a}]$ and $W=\mathfrak{a}$. If $\beta \in M$ and $x=[a,b]$, then $\mathrm{tr}(x\beta )=\mathrm{tr}(a[b,\beta ])=0$, since $[b,\beta ]\in U$. Linearity and the trace orthogonality nilpotence lemma show that every member of $U$ is nilpotent. The Engel theorem makes $U$ nilpotent and hence $\mathfrak{a}$ solvable. Apply this to $\mathfrak{a}=\mathrm{ad}R$; the kernel of this Adjoint representation of a Lie algebra is the abelian center of $R$, so $R$ is solvable as claimed.

For an arbitrary complex Lie algebra, a Cartan subalgebra $H$ means a nilpotent Lie algebra that is self-normalizing: $N_L(H)=\{x:[x,H]\subseteq H\}=H$. This definition does not assume that $H$ is abelian. We prove that it is abelian when $L$ is semisimple.

Use the generalized-weight decomposition for a nilpotent Lie algebra for the action of $H$ on $L$. Its zero generalized weight space isWe have $H\subseteq L^0$, since $H$ is nilpotent. If $L^0/H\ne 0$, the Engel theorem gives a nonzero coset annihilated by every $h\in H$. Its representative $x$ satisfies $[H,x]\subseteq H$, contradicting $N_L(H)=H$. Thus $L^0=H$.

For a nonzero generalized weight $\lambda$, choose $h_0\in H$ with $\lambda (h_0)\ne 0$. The operator $\mathrm{ad}{h}_0$ is invertible on $L^{\lambda }$ and nilpotent on $H$. For $h\in H$, $z\in L^{\lambda }$, write $z=(\mathrm{ad}{h}_0{)}^{m}w$ with $m$ large enough that $(\mathrm{ad}{h}_0{)}^{m}h=0$. Invariance of the Killing form givesOn the other hand, $H$ is solvable, so the Lie theorem triangularizes its action on $L$. For $a,b,h\in H$, the matrix $\mathrm{ad}[a,b]$ is strictly upper triangular, while $\mathrm{ad}h$ is upper triangular. Thus $B_L([H,H],H)=0$. Together with $L=H\oplus {⨁}_{\lambda \ne 0}{L}^{\lambda }$, this yields $B_L([H,H],L)=0$. Nondegeneracy gives $[H,H]=0$.

Finally, if $x$ commutes with $H$, it normalizes $H$, hence lies in $H$. Any abelian subalgebra containing $H$ consists of such elements. Thus $H$ is a maximal abelian subalgebra, indeed $C_L(H)=H$.