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The complexification of a Lie algebra $L_0$ is $L=L_0{\otimes }_{\mathbb{R}}\mathbb{C}$, with the Lie bracket extended complex-bilinearly. Equivalently write $L=L_0\oplus i{L}_0$, whereComplex conjugation is an antilinear map and a Lie algebra automorphism whose fixed subalgebra is $L_0$.

If the solvable radical of $L_0$ is nonzero, its complexification is a nonzero solvable ideal of $L$. Conversely the solvable radical $R$ of $L$ is preserved by complex conjugation, because it is the unique largest solvable ideal. Consequentlyfor $z\in R$, both $(z+\overline{z})/2$ and $(z-\overline{z})/(2i)$ belong to $R_0$. If $R\ne 0$, $R_0\ne 0$ is a solvable ideal of $L_0$. This proves $L_0$ is semisimple if and only if $L_0{\otimes }_{\mathbb{R}}\mathbb{C}$ is semisimple. Consistently, the complex Killing form is just the complex-bilinear extension of the real one; its determinant in a real basis is unchanged by extending scalars.

Take the sl2R Lie algebra ${\mathfrak{sl}}_2(\mathbb{R})$ and the special unitary Lie algebra $\mathfrak{su}(2)$. Both complexify to ${\mathfrak{sl}}_2(\mathbb{C})$. This is immediate for the former; for the latter, the real basis $ih,e-f,i(e+f)$ consists of traceless matrices that are skew-Hermitian matrices and is also a complex basis of ${\mathfrak{sl}}_2(\mathbb{C})$.

They are not isomorphic as real Lie algebras. Their Killing forms are $B(X,Y)=4\mathrm{tr}(XY)$, but on $\mathfrak{su}(2)$ this is negative definite. On ${\mathfrak{sl}}_2(\mathbb{R})$ the basis $h,e+f,e-f$ has a diagonal Gram matrix with entries $8,8,-8$, so the signature of a quadratic form is $(2,1)$. A Lie algebra isomorphism preserves the Killing form and therefore its signature.

A real split semisimple Lie algebra has a Cartan subalgebra whose Adjoint representation of a Lie algebra is simultaneously diagonalizable over $\mathbb{R}$, so its root-space decomposition is defined over $\mathbb{R}$. The example ${\mathfrak{sl}}_2(\mathbb{R})$ is split: the diagonal Cartan subalgebra $\mathbb{R}h$ has eigenvalues $0,2,-2$ and real root spaces $\mathbb{R}e,\mathbb{R}f$. The example $\mathfrak{su}(2)$ is not split. Invariance makes every $\mathrm{ad}x$ skew-adjoint for the positive definite inner product $-B$, so its eigenvalues are purely imaginary. If it were diagonalizable over $\mathbb{R}$, all these eigenvalues would be zero and $\mathrm{ad}x=0$. The center is zero, so only $x=0$ has this property; no nonzero split Cartan subalgebra exists. Thus ${\mathfrak{sl}}_2(\mathbb{R})$ is split and $\mathfrak{su}(2)$ is not.

The special orthogonal group in five dimensions isIts Lie algebra consists of real skew-symmetric matrices. There are $5\cdot 4/2=10$ independent entries above the diagonal, so $\dim SO(5)=10$. Equivalently, the orthogonality equations impose fifteen independent constraints on twenty-five matrix entries; the determinant condition chooses a component without changing the dimension.

Fix the fifth coordinate. The matrices $\mathrm{diag}(S,1)$, $S\in SO(4)$, form an explicit subgroup. Under this subgroup the defining vector space splits as ${\mathbb{R}}^4\oplus \mathbb{R}{\mathbf{e}}_5$, so its branching rule is $5\downarrow SO(4)=4\oplus 1$. An element of the so5 Lie algebra can be written uniquely asConjugation by $\mathrm{diag}(S,1)$ sends $B$ to $SB{S}^{-1}$ and $w$ to $Sw$. The first summand is the six-dimensional Adjoint representation of a Lie algebra of $\mathfrak{so}(4)$, and the second is its four-dimensional vector representation. Hence the SO5 to SO4 branching gives

For the left SU(2) subgroup of SO(4), identify ${\mathbb{R}}^4$ with the quaternions. Left multiplication by a unit quaternion is a real orthogonal transformation and gives an embedded SU(2) group. More generally, $q\mapsto aq{b}^{-1}$ gives the double cover $SU(2{)}_L\times SU(2{)}_R\to SO(4)$ with kernel $\{(1,1),(-1,-1)\}$. After complexifying, the vector representation is $(2,2)$ and the Adjoint representation is $(3,1)\oplus (1,3)$. Restricting to the left factor turns the right factor into a multiplicity space. ThereforeThese are decompositions into complex irreducible representations; the real $4$ is the underlying real representation of a quaternionic doublet. Combining the branching rules gives

Write a weight as $\lambda =(x,y)$. The integrality conditions for the B2 root system give $2x\in \mathbb{Z}$ and $y-x\in \mathbb{Z}$. Hence $x=m/2$, $y=m/2+n$, with $m,n\in \mathbb{Z}$. ThusThis is the B2 weight lattice, with an integer square lattice and a second square lattice shifted by $(1/2,1/2)$. The eight roots of a root system areThe short roots lie on the coordinate axes and the long roots on the diagonals. The positive roots for the given simple-root choice are $\alpha$, $\beta$, $\alpha +\beta$, $2\alpha +\beta$.

The integrality conditions determine the weight lattice of the Lie algebra, equivalently of the simply connected Spin group $\mathrm{Spin}(5)$. For the global special orthogonal group $SO(5)$, a $2\pi$ rotation in either coordinate plane is the identity, so a genuine group representation requires integer $x,y$. Thus the half-integer coset contains spin representations that do not descend to $SO(5)$. Both representations requested here have integer weights, so their diagrams are unaffected by this distinction.

The root system of the displayed $SO(4)$ subgroup is $\{\pm (1,1),\pm (1,-1)\}$. The two orthogonal pairs give its two commuting $\mathfrak{su}(2)$ factors. Choose the left factor to have root $\delta =(1,1)$; exchanging the two diagonal pairs exchanges left and right. Its coroot pairs with a weight asThis demonstrates the diagonal-root SU(2) embedding in SO(5) directly. The short-axis root $\alpha$ would instead give $2x$, and therefore a different subgroup: on the vector representation it would produce a triplet and two singlets rather than two doublets and a singlet.

For the vector representation, simultaneously rotate the first and second coordinate planes. Over $\mathbb{C}$, the two planes give opposite pairs of weights, while the fifth coordinate is fixed. Hence its weight diagram iswith every weight multiplicity equal to one. Evaluating $x+y$ gives $+1$ twice, $-1$ twice and $0$ once, exactly two SU(2) representations of dimension two and one singlet.

For the Adjoint representation, the root-space decomposition has one one-dimensional space for each of the eight roots and a two-dimensional zero-weight Cartan subalgebra. ThusThe coroot values $x+y$ have multiplicities $1,2,4,2,1$ at $-2,-1,0,1,2$. One zero-weight state joins the $\pm 2$ states to make a triplet; the $\pm 1$ states form two doublets, leaving three zero-weight singlets. This verifies the earlier branching rule and accounts for all ten dimensions.

The positive roots generated by the two simple roots areThe Adjoint representation of a Lie algebra has one weight space for each positive and negative root and a rank-two zero-weight space. Its nonzero weights are thereforetogether with their negatives. Thus the diagram consists of a hexagon of six long roots, a hexagon of six short roots, and the origin with multiplicity two. The representation has dimension $12+2=14$.

Choose a short simple root ${\alpha }_1$ and a long simple root ${\alpha }_2$. The G2 root system has positive rootsThe two hexagons formed by the short and long roots give the usual twelve-root diagram. The fundamental weights areThe second is the highest root, so the irreducible module $L({\omega }_2)$ is the Adjoint representation of a Lie algebra.

The seven weights of $L({\omega }_1)$ are zero and the six short roots, each with multiplicity one. Its crystal, with arrows denoting the lowering operators ${\tilde{f}}_i$, is the chain

Applying the Weyl dimension formula to $\lambda =n_1{\omega }_1+n_2{\omega }_2$ gives

Differentiate ${\mathrm{Ad}}_{e^{tX}}Y=e^{tX}Y{e}^{-tX}$ at zero. This gives the Adjoint representation of a Lie algebraThe Jacobi identity is exactlyso this is a Lie-algebra representation on $\mathfrak{o}(3)$.

Requiring $D_{\mu }\varphi$ to transform as $U^{†}(D_{\mu }\varphi )U$ and using the gauge-field transformation law givesIndeed, differentiating $U^{†}\varphi U$ produces two inhomogeneous derivative terms, and those cancel against the inhomogeneous part of the transformed connection.

For $U=1+i{\alpha }^a{T}_a+O({\alpha }^2)$,If $[T_a,T_b]=i{f}_{ab}{}^c{T}_c$, thenThis is the infinitesimal Adjoint representation of a Lie algebra.

For a finite-dimensional Lie algebra representation $\varphi :\mathfrak{g}\to \mathfrak{gl}(V)$, the Trace form of a Lie algebra representation isThe Killing form is the trace form of the Adjoint representation of a Lie algebra:A bilinear form $B$ is $\mathfrak{g}$-invariant whenequivalently $B([x,y],z)=B(x,[y,z])$. For a trace form this follows from cyclicity of trace:

Differentiating ${\mathrm{Ad}}_{\exp (tX)}Y$ at zero gives the Adjoint representation of a Lie algebraThe Jacobi identity implies $[{\mathrm{ad}}_X,{\mathrm{ad}}_Y]={\mathrm{ad}}_{[X,Y]}$, proving that it is a Lie-algebra representation.

The Adjoint representation of a Lie algebra isThe Killing form is the symmetric invariant bilinear form

The Adjoint representation of a Lie algebra isIt is linear. The Jacobi identity givesso it is a Lie algebra representation.

Let $W\subseteq \mathfrak{g}$ be invariant under the Adjoint representation of a Lie algebra. Then $[\mathfrak{g},W]\subseteq W$, so $W$ is an ideal. If $\mathfrak{g}$ is a simple Lie algebra, its only ideals are $0$ and $\mathfrak{g}$. Hence the adjoint representation is irreducible. Compact type is compatible with the anti-Hermitian realization used below, although simplicity alone proves this adjoint irreducibility of a simple Lie algebra.

For a nontrivial finite-dimensional anti-Hermitian Lie algebra representation of a compact Lie algebra that is simple, its Trace form of a Lie algebra representation is a positive multiple of the Killing form. Define $T$ by $H(X,Y)=\kappa (TX,Y)$. Invariance makes $T$ commute with the Adjoint representation of a Lie algebra. Symmetry makes $T$ self-adjoint for $-\kappa$, so its eigenspaces are ideals. Simplicity implies $T=\mu I$. The kernel of the representation is an ideal and hence vanishes; therefore $H(X,X)=-\mathrm{Tr}(d(X{)}^{†}d(X))<0$ for nonzero $X$, proving $\mu >0$.

// Target: lie-theory.bigb