 Solution

We construct the exceptional isomorphism between sp4 and so5. Let $W={\mathbb{C}}^4$ have a nondegenerate alternating form $\Omega$ and symplectic basis $a_1,b_1,a_2,b_2$. Define the symplectic contraction of an exterior squareIt is equivariant for the symplectic Lie algebra, and $\dim E=5$. Identify ${\Lambda }^{4}W$ with $\mathbb{C}$ using $a_1\wedge b_1\wedge a_2\wedge b_2$. Wedge product then gives a symmetric bilinear form $B(\xi ,\eta )=\xi \wedge \eta$ on ${\Lambda }^{2}W$. It is nondegenerate, as the wedge-basis vectors pair with their complementary pairs.

The bivector $\tau =a_1\wedge b_1+a_2\wedge b_2$ is invariant under $\mathfrak{sp}(W)$, satisfies $B(\tau ,\tau )=2$, and has $B(\xi ,\tau )=\kappa (\xi )$. Consequently $E={\tau }^{\perp }$ and $B{|}_E$ is nondegenerate. The exterior-power Lie algebra representation preserves wedge product and the volume form, so restriction gives a homomorphismWe check injectivity directly. If $X$ acts trivially on $E$, it also annihilates the invariant line $\mathbb{C}\tau$, hence acts trivially on all of ${\Lambda }^{2}W$. In any basis $w_1,\dots ,w_4$, the equation $X(w_i\wedge w_j)=0$ forces every off-diagonal coefficient $X_{ki}$ to vanish: choose $j$ distinct from $i,k$ and inspect the coefficient of $w_k\wedge w_j$. Thus $X$ is diagonal. Its diagonal entries $x_i$ satisfy $x_i+x_j=0$ for every pair, which forces them all to be zero. Therefore $\Phi$ is injective.

A symplectic matrix has infinitesimal block form $\left(\begin{array}{cc}A& B\\ C& -A^T\end{array}\right)$ with $B,C$ symmetric, so $\dim {\mathfrak{sp}}_4=4+3+3=10$. Also $\dim \mathfrak{so}(E,B)=5\cdot 4/2=10$. Injectivity and equal dimensions make $\Phi$ an isomorphism. Every nondegenerate complex symmetric form in dimension five is equivalent by a change of basis to the given antidiagonal form, so