= Solution
In dimension four, the two positive <roots> are $\alpha=\varepsilon_1-\varepsilon_2$ and $\beta=\varepsilon_1+\varepsilon_2$. Using the <matrix root basis of the even orthogonal Lie algebra>, define
$$
\begin{aligned}e_\alpha&=E_{12}-E_{34},&f_\alpha&=E_{21}-E_{43},&h_\alpha&=\operatorname{diag}(1,-1,1,-1),\\e_\beta&=E_{13}-E_{24},&f_\beta&=E_{31}-E_{42},&h_\beta&=\operatorname{diag}(1,1,-1,-1).
\end{aligned}
$$
The <matrix unit> identity $[E_{ij},E_{kl}]=\delta_{jk}E_{il}-\delta_{li}E_{kj}$ gives $[e_\gamma,f_\gamma]=h_\gamma$, $[h_\gamma,e_\gamma]=2e_\gamma$, and $[h_\gamma,f_\gamma]=-2f_\gamma$ for $\gamma=\alpha,\beta$. Thus each triple spans an <sl2 Lie algebra>.
The two triples commute. Their Cartan actions on one another vanish because $\alpha(h_\beta)=\beta(h_\alpha)=0$; their mixed root-vector brackets vanish because none of $\pm\alpha\pm\beta$, namely $\pm2\varepsilon_1$ and $\pm2\varepsilon_2$, is a root. This can equally be checked from the displayed matrix units. The six vectors are <linearly independent>, occupying the four distinct <root spaces> and two independent Cartan directions, so they exhaust the six-dimensional <so4 Lie algebra>. The two commuting spans are consequently ideals, and
$$
\boxed{\mathfrak{so}_4(\mathbb C)\cong\mathfrak{sl}_2(\mathbb C)\oplus\mathfrak{sl}_2(\mathbb C).}
$$
This gives an explicit <Chiral decomposition of the complexified so4 Lie algebra>, including generators for both factors.
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