Isomorphic Lie Algebra $so(4)$ and $su(2) \times su(2)$

Question

Show that the Lie algebras of $SO(4)$ and $SU(2) \times SU(2)$ are isomorphic.

I managed to show that $so(4) = \{A \in M_4(\mathbb{R}) \mid A^T = -A\}$ and $su(2) = \{A \in GL_2(\mathbb{C}) \mid A^\dagger = -A, \text{tr}(A) = 0\}$ with $su(2) \times su(2)$ being the Lie algebra of $SU(2)\times SU(2)$. However I am not able to find any mapping that preserves the Lie bracket structure. I am aware of the Pauli matrices being the generators of $su(2)$ but I can't find similar generators of $so(4)$ that preserve the bracket.

Answer 1

See $SU(2)$ as the group of the quaternions with norm $1$. Let $\operatorname{Im}\mathbb H$ be the space of pure quaternions, that is, the space $\{\alpha i+\beta j+\gamma k\mid \alpha,\beta,\gamma\in\mathbb R\}$. Then$$\begin{array}{rccc}\Psi\colon&SU(2)\times SU(2)&\longrightarrow&\operatorname{Aut}(\operatorname{Im}\mathbb H)\\&(q_1,q_2)&\mapsto&\left(\begin{array}{ccc}\operatorname{Im}\mathbb H&\longrightarrow&\operatorname{Im}\mathbb H\\r&\mapsto&q_1.r.q_2^*\end{array}\right)\end{array}$$induces an isomorphism from $(SU(2)\times SU(2))/\{\pm(\operatorname{Id},\operatorname{Id})\}$ onto the group of all isometries of $\operatorname{Im}\mathbb H$ with determinant $1$. This last group is basically $SO(3,\mathbb R)$. So, the derivative of $\Psi$ at the identity element of $SU(2)\times SU(2)$ is the Lie algebra isomorphism that you're after.