= Solution
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>:
$$
\boxed{\mathfrak{so}_6\cong\mathfrak{sl}_4.}
$$
There is also a concrete realization. On $U=\mathbb C^4$, choose a nonzero alternating four-form $\mathrm{vol}$. The six-dimensional <exterior square> $\Lambda^2U$ carries the <nondegenerate> <symmetric bilinear form> $B$ specified by
$$
B(u\wedge v,w\wedge z)=\mathrm{vol}(u,v,w,z).
$$
It 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 $\Lambda^4U$ is multiplication by $\operatorname{tr}A=0$, so it preserves $B$. We obtain a nonzero <Lie algebra homomorphism> $\mathfrak{sl}_4\to\mathfrak{so}(\Lambda^2U,B)$. 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>.
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