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The given equation is a Maurer-Cartan equation, with the sign appropriate to the chosen invariant coframe. Write $\lambda =({\lambda }_{ij})$ as an antisymmetric matrix of one-forms, so $d\lambda =\lambda \wedge \lambda$. Using the graded product rule for the exterior derivative,The two sums coincide after exchanging their dummy indices $k,\ell$. This is also associativity of matrix-valued wedge products of differential forms.

To make the relation to the Jacobi identity explicit, let ${\theta }^a$ be any independent coframe and write $d{\theta }^a=-\frac{1}{2}{c}^a{}_{bc}{\theta }^b\wedge {\theta }^c$. Its dual invariant vector fields satisfy $[E_b,E_c]=c^a{}_{bc}{E}_a$, by the exterior derivative of a one-form evaluated on vector fields. Evaluating $d^2{\theta }^a=0$ on three basis fields gives the cyclic coefficient identityThese are exactly the components of $[[E_b,E_c],E_d]+[[E_c,E_d],E_b]+[[E_d,E_b],E_c]=0$. Thus the structure constants obey the Jacobi identity.

For the real four-dimensional case, abbreviate $a={\lambda }_{12}$, $b={\lambda }_{23}$, $c={\lambda }_{31}$, $d={\lambda }_{43}$, $e={\lambda }_{41}$, $f={\lambda }_{42}$; here $d$ without an argument is a one-form, whereas $d(\cdot )$ denotes the exterior derivative. Directly expanding the given equations givesDefine ${\alpha }_1^{\pm }=a\pm d$, ${\alpha }_2^{\pm }=b\pm e$, and ${\alpha }_3^{\pm }=c\pm f$. For example,The minus combination and the other two cyclic combinations giveThe six forms are independent, since the original pairs are their half-sums and half-differences. Their dual vector fields $E_i^{+},E_i^{-}$ consequently obeyEach triple is a three-dimensional Lie algebra ideal with the SO(3) Lie algebra brackets; the two ideals commute and span the original six-dimensional algebra. This proves the real four-dimensional rotation algebra splittingThis is a Lie-algebra statement. Globally $SO(4)$ is a central quotient of $SU(2)\times SU(2)$; a direct product of the corresponding $SO(3)$ groups is not asserted.