= Solution
Use $\epsilon_{xyz}=+1$ and the supplied <commutators>. Antisymmetry also gives $[K_a,L_b]=\epsilon_{abc}K_c$. Expanding the combinations,
$$
[L_a^\pm,L_b^\pm]
=\frac14\left([L_a,L_b]+[K_a,K_b]
\pm[L_a,K_b]\pm[K_a,L_b]\right)
=\epsilon_{abc}L_c^\pm.
$$
The mixed <commutator> cancels:
$$
[L_a^+,L_b^-]
=\frac14\left([L_a,L_b]-[L_a,K_b]+[K_a,L_b]-[K_a,K_b]\right)=0.
$$
Thus
$$
\boxed{[L_a^+,L_b^+]=\epsilon_{abc}L_c^+,\qquad
[L_a^-,L_b^-]=\epsilon_{abc}L_c^-,\qquad
[L_a^+,L_b^-]=0.}
$$
There are two mutually commuting three-generator rotation algebras. As real compact <Lie algebras>,
$$
\boxed{\mathfrak{so}(4)\cong\mathfrak{su}(2)\oplus\mathfrak{su}(2).}
$$
For example, the <anti-Hermitian> basis $-i\sigma_a/2$ obeys the same commutation convention in each $\mathfrak{su}(2)$ factor. This is an algebra statement. Globally, the simply connected group is $\operatorname{Spin}(4)\cong SU(2)\times SU(2)$, and
$$
SO(4)\cong\frac{SU(2)\times SU(2)}{\{(I,I),(-I,-I)\}}.
$$
Dropping that central quotient would incorrectly identify the global groups.
Back to article page
