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Let $U$ be a four-dimensional symplectic vector space. The symplectic contraction of an exterior square splits ${\Lambda }^{2}U$ into its five-dimensional primitive exterior square $W$ and the invariant line spanned by the inverse symplectic bivector $\Omega$. Choose $\mathrm{vol}=\Omega \wedge \Omega /2$. Exterior multiplication defines a symmetric bilinear form on ${\Lambda }^{2}U$ by $\xi \wedge \eta =B(\xi ,\eta )\mathrm{vol}$; it is symmetric because both degrees are two. In a symplectic basis, $\Omega \wedge \Omega =2\mathrm{vol}\ne 0$, and $W={\Omega }^{\perp }$, so the restriction to $W$ is nondegenerate.

The symplectic Lie algebra ${\mathfrak{sp}}_4$ preserves this form, giving a homomorphism ${\mathfrak{sp}}_4\to \mathfrak{so}(W)$. It is injective: an element acting trivially on $W$ also acts trivially on $\Omega$, hence on all of ${\Lambda }^{2}U$. In a four-vector basis, the identities $A{e}_i\wedge e_j+e_i\wedge A{e}_j=0$ first force every off-diagonal coefficient of $A$ to vanish, then force $A_{ii}+A_{jj}=0$ for every pair. Over $\mathbb{C}$ these equalities imply $A=0$.

Both Lie algebras have dimension ten: $\dim {\mathfrak{sp}}_4=4\cdot 5/2$ and $\dim {\mathfrak{so}}_5=5\cdot 4/2$. The injective map is therefore an isomorphism, proving the exceptional isomorphism between sp4 and so5:

In the defining representation of the symplectic Lie algebra ${\mathfrak{sp}}_4$, the weights are ${\epsilon }_1,{\epsilon }_2,-{\epsilon }_1,-{\epsilon }_2$. Their highest weight, for the root basis of part (b), is thereforeThe flip on the tensor square commutes with the action of ${\mathfrak{sp}}_4$, giving $V\otimes V={\mathrm{Sym}}^{2}V\oplus {\Lambda }^{2}V$ with dimensions $10$ and $6$.

If $v_1$ is a highest-weight vector of weight ${\epsilon }_1$, then $v_1\otimes v_1$ is a highest-weight vector of weight $2{\omega }_1$ in the symmetric square. By Weyl complete reducibility theorem, this ensures an irreducible summand $V(2{\omega }_1)$ occurs. Its dimension is $10$ by part (b), exhausting the symmetric square.

Let $\omega$ denote the preserved symplectic form. The symplectic contraction of an exterior square is the nonzero equivariant mapwhere the target is a trivial Lie algebra representation. Thus its kernel has dimension $5$. Choose a weight vector $v_2$ of weight ${\epsilon }_2$ in a symplectic basis, with $\omega (v_1,v_2)=0$. The vector $v_1\wedge v_2$ belongs to this kernel and has weight ${\epsilon }_1+{\epsilon }_2={\omega }_2$. It is a highest-weight vector: none of ${\omega }_2+\alpha$, for $\alpha \in {\Phi }^{+}$, is a weight of ${\Lambda }^{2}V$, whose weights are $\pm {\epsilon }_1\pm {\epsilon }_2$ and zero. Hence the kernel contains $V({\omega }_2)$; its dimension $5$ exhausts the kernel. Weyl complete reducibility theorem supplies a complementary invariant line $V(0)$.

The resulting tensor-square decomposition of the defining sp4 representation is

The primitive exterior square of a symplectic vector space is the kernel of its symplectic contraction of an exterior square. In dimension four, the invariant inverse-form bivector $\Omega$ has nonzero wedge square. Choose $\mathrm{vol}=\Omega \wedge \Omega /2$; the symmetric bilinear form $\xi \wedge \eta =B(\xi ,\eta )\mathrm{vol}$ is nondegenerate on ${\Lambda }^{2}V$. The primitive subspace is ${\Omega }^{\perp }$, and therefore inherits a nondegenerate form of dimension five. The symplectic Lie algebra acts on it by infinitesimal orthogonal transformations.

For the defining four-dimensional representation $V$ of the symplectic Lie algebra ${\mathfrak{sp}}_4$, with the short simple root numbered first, its highest weight is ${\omega }_1$. Its tensor square decomposes aswith dimensions $10,5,1$. The first summand is the symmetric square; the other two form the exterior square, split by symplectic contraction of an exterior square.