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The printed phrase “not reducible” is a misprint: the requested nonzero complementary submodules establish that this tensor square is reducible. The natural reason is that the flip $\tau (v\otimes w)=w\otimes v$ commutes with the Lie algebra representation. Its two eigenspaces are the symmetric square and the exterior square, givingIn the symmetric square, the four weights $\pm 2L_i$ and four weights $\pm L_1\pm L_2$ each occur once, while zero occurs twice. In the exterior square, the four weights $\pm L_1\pm L_2$ each occur once and zero occurs twice; the $\pm 2L_i$ are absent because repeated wedge factors vanish. These are the two requested complementary weight diagrams.

More precisely, the tensor-square decomposition of the defining symplectic representation identifies ${\mathrm{Sym}}^{2}V$ with the irreducible adjoint module $L(2{\omega }_1)$. The identification is an intertwiner sending a symmetric tensor $vw$ to the element of the symplectic Lie algebra $x\mapsto \omega (v,x)w+\omega (w,x)v$. For the exterior square, symplectic contraction of an exterior square has a five-dimensional kernel, the primitive exterior square, and the invariant tensor $\Omega =e_1\wedge e_3+e_2\wedge e_4$ supplies a complementary line. Thus the full irreducible decomposition isHere ${\omega }_1=L_1$ and ${\omega }_2=L_1+L_2$. The two zero weights in the exterior square split into one primitive zero weight and the invariant line.

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

Let $U$ be a four-dimensional symplectic vector space, with a symplectic basis $e_1,f_1,e_2,f_2$. Its invariant inverse-form bivector may be chosen as $\eta =e_1\wedge f_1+e_2\wedge f_2$. Put $\mathrm{vol}=e_1\wedge f_1\wedge e_2\wedge f_2$, so $\eta \wedge \eta =2\mathrm{vol}$. Define a symmetric bilinear form on the six-dimensional exterior square byIt is symmetric because both factors have degree two. It is nondegenerate: in the six-element wedge basis, every $e_a\wedge e_b$ pairs nontrivially with the wedge of the complementary pair and with no other basis vector. Since $B(\eta ,\eta )=2$, the orthogonal complementhas dimension five and inherits a nondegenerate bilinear form. It is the primitive exterior square, equivalently the kernel of the symplectic contraction of an exterior square.

The symplectic Lie algebra preserves $\eta$. It also preserves $\mathrm{vol}$, since its elements have trace zero and the induced action on ${\Lambda }^{4}U$ is multiplication by that trace. The exterior-power Lie algebra representation therefore preserves $B$ and $W$, giving a Lie algebra homomorphismTo show injectivity, suppose $A$ acts as zero on $W$. It already acts as zero on the invariant line $\mathbb{C}\eta$, so it kills all of ${\Lambda }^{2}U=W\oplus \mathbb{C}\eta$. For any basis $u_1,\dots ,u_4$, this meansIf $k\ne i$, choose $j$ distinct from $i,k$; the coefficient of $u_k\wedge u_j$ forces $A_{ki}=0$. Thus $A$ is diagonal. The same identity gives $A_{ii}+A_{jj}=0$ for every pair. Applying this to three distinct indices forces each diagonal entry to vanish. Hence $A=0$ and $\Phi$ is injective.

Finally $\dim {\mathfrak{sp}}_4=2(2\cdot 2+1)=10$ and $\dim \mathfrak{so}(W,B)=5\cdot 4/2=10$. Injectivity between these equal-dimensional vector spaces gives surjectivity. Every nondegenerate complex symmetric bilinear form on a five-dimensional space is equivalent to the standard one. We have therefore constructed the exceptional isomorphism between sp4 and so5:

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:

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.