 Solution

In dimension four, the two positive roots are $\alpha ={\epsilon }_1-{\epsilon }_2$ and $\beta ={\epsilon }_1+{\epsilon }_2$. Using the matrix root basis of the even orthogonal Lie algebra, defineThe 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 $\pm 2{\epsilon }_1$ and $\pm 2{\epsilon }_2$, 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, andThis gives an explicit Chiral decomposition of the complexified so4 Lie algebra, including generators for both factors.