= Solution
A <Cartan subalgebra> $\mathfrak h$ of a finite-dimensional complex <Semisimple Lie algebra> is a maximal abelian subalgebra consisting of semisimple elements. The <root-space decomposition> is
$$
\mathfrak g=\mathfrak h\oplus\bigoplus_{\alpha\in\Phi}\mathfrak g_\alpha,
\qquad
\mathfrak g_\alpha=\{X:[H,X]=\alpha(H)X\text{ for all }H\in\mathfrak h\},
$$
and the nonzero functionals $\alpha$ are the roots. A <Cartan-Weyl basis> consists of a basis $H_i$ of $\mathfrak h$ and root vectors $E_\alpha\in\mathfrak g_\alpha$. Its brackets have the form
$$
[H_i,H_j]=0,qquad [H_i,E_\alpha]=\alpha(H_i)E_\alpha,qquad
[E_\alpha,E_{-\alpha}]=H_\alpha,
$$
and $[E_\alpha,E_\beta]=N_{\alpha\beta}E_{\alpha+\beta}$ when $\alpha+\beta$ is a root, and zero when $\alpha+\beta$ is neither a root nor zero.
For the complexified <so4 Lie algebra>, take $H^1=T^{(12)}$ and $H^2=T^{(34)}$. Write
$$
A_1^\pm=T^{(13)}\pm T^{(24)},
\qquad
A_2^\pm=T^{(14)}\pm T^{(23)}.
$$
Direct use of the stated commutation relations gives
$$
\begin{array}{c|rrrr}
&A_1^+&A_1^-&A_2^+&A_2^-\\ \hline
\operatorname{ad}H^1&A_2^-&-A_2^+&A_1^-&-A_1^+\\
\operatorname{ad}H^2&-A_2^-&-A_2^+&A_1^-&A_1^+
\end{array}.
$$
The simultaneous <eigenvectors>, hence the step generators, may be chosen as
$$
\begin{aligned}
E_{++}&=A_2^+-iA_1^-,& \alpha_{++}&=(i,i),\\
E_{+-}&=A_2^-+iA_1^+,& \alpha_{+-}&=(i,-i),\\
E_{-+}&=A_2^--iA_1^+,& \alpha_{-+}&=(-i,i),\\
E_{--}&=A_2^++iA_1^-,& \alpha_{--}&=(-i,-i).
\end{aligned}
$$
Thus the roots relative to $(H^1,H^2)$ are $(\pm i,\pm i)$. Replacing $H^a$ by $-iH^a$ gives the usual real coordinates $(\pm1,\pm1)$. The only nonzero brackets between step generators, apart from those obtained by antisymmetry, are
$$
\boxed{[E_{++},E_{--}]=4i(H^1+H^2),
\qquad [E_{+-},E_{-+}]=4i(H^1-H^2).}
$$
An <isomorphism> of Lie algebras is a bijective <linear map> preserving the Lie bracket. Define
$$
\begin{aligned}
J_1^\pm&=-\tfrac12(T^{(23)}\pm T^{(14)}),\\
J_2^\pm&=-\tfrac12(T^{(31)}\pm T^{(24)}),\\
J_3^\pm&=-\tfrac12(T^{(12)}\pm T^{(34)}).
\end{aligned}
$$
Then
$$
[J_i^\pm,J_j^\pm]=\epsilon_{ijk}J_k^\pm,
\qquad [J_i^+,J_j^-]=0.
$$
The two spans are commuting copies of the complexified $\mathfrak{su}_2$, and together contain all six basis elements of $\mathfrak{so}_4$. <Chiral decomposition of the complexified so4 Lie algebra> therefore gives
$$
\boxed{\mathfrak{so}_4(\mathbb C)\cong
\mathfrak{su}_2(\mathbb C)\oplus\mathfrak{su}_2(\mathbb C).}
$$
Under this isomorphism the <Adjoint representation> is the direct sum of the adjoint representations of the two factors:
$$
\boxed{\mathbf6=(\mathbf3,\mathbf1)\oplus(\mathbf1,\mathbf3).}
$$
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