 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-302/4/v/solution

Substituting

J_{m} = ϵ_{mij} M^{ij} /2

and

K_{i} = M^{0 i}

yields

[K_{i}, K_{j}] = - ϵ_{ijk} J_{k}, [J_{i}, K_{j}] = ϵ_{ijk} K_{k}, [J_{i}, J_{j}] = ϵ_{ijk} J_{k} .

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