 Solution

Let $V={\mathbb{C}}^4$ with its nondegenerate alternating bilinear form. Its inverse-form bivector $\Omega \in {\Lambda }^{2}V$ is invariant under the symplectic Lie algebra, and $\Omega \wedge \Omega \ne 0$. Choose $\mathrm{vol}=\Omega \wedge \Omega /2$. The exterior product defines a symmetric bilinear form on ${\Lambda }^{2}V$ byIt is symmetric because two-forms commute under the wedge product. It is nondegenerate because each basis wedge $e_i\wedge e_j$ pairs with its complementary basis wedge. Since $B(\Omega ,\Omega )=2$, the five-dimensional orthogonal complementalso has a nondegenerate symmetric bilinear form. This is the primitive exterior square, equivalently the kernel of symplectic contraction of an exterior square.

Every element of ${\mathfrak{sp}}_4$ fixes $\Omega$ and preserves volume, since its trace is zero. The induced exterior-power Lie algebra representation consequently preserves $W$ and satisfies $B(A\xi ,\eta )+B(\xi ,A\eta )=0$. Thus it gives a Lie algebra homomorphismWe check injectivity explicitly. If $A$ acts as zero on $W$, it also kills $\Omega$, and therefore acts as zero on all of ${\Lambda }^{2}V=W\oplus \mathbb{C}\Omega$. For three distinct indices $i,j,k$, the coefficient of $e_k\wedge e_j$ in $A(e_i\wedge e_j)$ is $A_{ki}$, so all off-diagonal entries of $A$ vanish. Its action on $e_i\wedge e_j$ is then $(A_{ii}+A_{jj})e_i\wedge e_j$. All these sums vanish, which in characteristic zero forces every diagonal entry to vanish. Hence $A=0$.

Finally,The injective Lie algebra homomorphism is therefore surjective. Over $\mathbb{C}$, every nondegenerate symmetric bilinear form admits a basis in which its matrix is the identity, so this constructs the exceptional isomorphism between sp4 and so5: