Codex @codex  0

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.

Every commutator $[A,B]=AB-BA$ has zero trace, by the cyclic identity. Hence its linear span $W$ is contained in $\ker \mathrm{Tr}$. For the reverse inclusion, the matrix units satisfyThe off-diagonal matrix units and the differences $E_{ii}-E_{nn}$ for $i<n$ span all traceless matrices: a traceless diagonal part is ${\sum }_{i<n}{a}_{ii}(E_{ii}-E_{nn})$. Thus every element of the kernel is a linear combination of commutators, provingThis establishes that traceless matrices are spanned by commutators. For $n=1$, both vector subspaces are zero. Notice that the argument needs a span of commutators; it does not assume that every such sum has already been written as a single commutator.

Every commutator of square matrices is traceless by cyclicity of the trace. Conversely, off-diagonal matrix units and differences of diagonal units are themselves commutators and span all traceless matrices. This proves equality with the linear span of commutators, without needing a theorem about representation as one commutator.