Solution (source code)

= 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 square>
$$
\kappa:\Lambda^2W\to\mathbb C,\qquad\kappa(u\wedge v)=\Omega(u,v),
\qquad E=\ker\kappa.
$$
It is equivariant for the <symplectic Lie algebra>, and $\dim E=5$. Identify $\Lambda^4W$ 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 $\Lambda^2W$. 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 homomorphism
$$
\Phi:\mathfrak{sp}_4\longrightarrow\mathfrak{so}(E,B).
$$
We check injectivity directly. If $X$ acts trivially on $E$, it also annihilates the invariant line $\mathbb C\tau$, hence acts trivially on all of $\Lambda^2W$. In any basis $w_1,\ldots,w_4$, 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 $\left(\begin{smallmatrix}A&B\\C&-A^T\end{smallmatrix}\right)$ with $B,C$ symmetric, so $\dim\mathfrak{sp}_4=4+3+3=10$. Also $\dim\mathfrak{so}(E,B)=5\cdot4/2=10$. 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
$$
\boxed{\mathfrak{so}_5\cong\mathfrak{sp}_4.}
$$