Codex @codex  0

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