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The Dynkin diagram in part (b) is a three-node chain, hence $D_3=A_3$. By the classification of complex semisimple Lie algebras by their root systems, this gives the Isomorphism between so6 and sl4:There is also a concrete realization. On $U={\mathbb{C}}^4$, choose a nonzero alternating four-form $\mathrm{vol}$. The six-dimensional exterior square ${\Lambda }^{2}U$ carries the nondegenerate symmetric bilinear form $B$ specified byIt is symmetric because exchanging two pairs entails four transpositions, and it is nondegenerate because each basis bivector pairs nontrivially with its complementary pair. The special linear Lie algebra acts by $A(u\wedge v)=Au\wedge v+u\wedge Av$. Its action on ${\Lambda }^{4}U$ is multiplication by $\mathrm{tr}A=0$, so it preserves $B$. We obtain a nonzero Lie algebra homomorphism ${\mathfrak{sl}}_4\to \mathfrak{so}({\Lambda }^{2}U,B)$. Its kernel is an ideal of a Lie algebra, hence zero because ${\mathfrak{sl}}_4$ is a simple Lie algebra. Both sides have dimension $15$, so this homomorphism is a Lie algebra isomorphism.

Let $V={\mathbb{C}}^4$. The special linear Lie algebra $\mathfrak{sl}(V)$ acts on ${\Lambda }^{2}V$. The wedge productis a nondegenerate symmetric bilinear form, and the action preserves it because $\mathfrak{sl}(V)$ acts trivially on ${\Lambda }^{4}V$. This gives an injective homomorphismBoth Lie algebras have dimension $15$, so the map is an isomorphism. This realizes the Isomorphism between so6 and sl4.

The preceding Dynkin diagram identifies the root system $D_3$ with $A_3$. The classification of finite-dimensional complex simple Lie algebras by connected Dynkin diagrams therefore givesThis is the Isomorphism between so6 and sl4. Both algebras have dimension $15$, consistently with the classification.