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Over $\mathbb{C}$, the symplectic Lie algebra on a four-dimensional symplectic vector space acts faithfully on its five-dimensional primitive exterior square, preserving its symmetric bilinear form. Faithfulness follows because an operator killing all wedges $e_i\wedge e_j$ has all off-diagonal coefficients zero and satisfies $A_{ii}+A_{jj}=0$ for every pair, hence is zero. The source and target Lie algebras both have dimension ten, so this faithful map is an isomorphism to the Special orthogonal Lie algebra ${\mathfrak{so}}_5$.

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.

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:

Use the ordered alphabetwith weights $\mathrm{wt}(i)={\epsilon }_i$ and $\mathrm{wt}(\bar{i})=-{\epsilon }_i$. The crystal of the defining symplectic representation is the chainThe label of an arrow is its simple root color; a Kashiwara operator ${\tilde{f}}_i$ follows an arrow of color $i$ and lowers the weight by ${\alpha }_i$.

Here is a grid description of the entire tensor product of crystals, valid for every $n$. Put one vertex $a\otimes b$ at row $a$ and column $b$, in the alphabet order above. Let $P_i$ be the set of chain positions with outgoing color $i$, and $Q_i$ the positions with incoming color $i$:Their indicators are ${\phi }_i$ and ${\epsilon }_i$. Use the crystal tensor-product ruleAn arrow is omitted if the selected factor has no outgoing arrow of that color. Thus the complete grid has a downward color-$i$ arrow when $a\in P_i$ and $b\notin Q_i$, and a rightward color-$i$ arrow when $b\in P_i$ and $a\notin P_i$. There are no other arrows. These two coordinate rules specify every vertex and every edge of the requested general-rank drawing. The following figure displays the full grid at $n=3$, including all components, together with the standard chain; increasing $n$ uses exactly the same grid construction.

For completeness, highest vertices can be read from the raising rule, which selects the first factor when ${\phi }_i(a)\ge {\epsilon }_i(b)$. If $a$ has an incoming edge of color $i$, then ${\phi }_i(a)=0$; either the raising operator acts nontrivially on $a$, or ${\epsilon }_i(b)=1$ and it acts nontrivially on $b$. Thus a highest vertex must have $a=1$. At $a=1$, all ${\phi }_i(a)$ vanish except ${\phi }_1(a)=1$. To be highest, the incoming color of $b$, if any, must consequently be $1$. For $n\ge 2$, the possibilities are exactlyTheir weights are $2{\epsilon }_1=2{\omega }_1$, ${\epsilon }_1+{\epsilon }_2={\omega }_2$, and zero. Each occurs once. Connected highest-weight components of the crystal basis describe the irreducible representations in the tensor product of Lie algebra representations, soThis is the tensor-square decomposition of the defining symplectic representation. The symmetric square is $L(2{\omega }_1)$; the exterior square is the direct sum of its invariant inverse-form line and the primitive exterior square $L({\omega }_2)$. Their dimensions are respectively $n(2n+1)$, $n(2n-1)-1$, and one, summing to $4n^2=\dim (V\otimes V)$. At $n=1$, the chain has two vertices and the highest tensor vertices are only $1\otimes 1$ and $1\otimes \bar{1}$: the decomposition is $L(2{\omega }_1)\oplus L(0)$, of dimensions three and one.

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:

For $n\ge 2$, the defining Lie algebra representation $V$ of ${\mathfrak{sp}}_{2n}$ has the displayed tensor product decomposition. The crystal of the defining symplectic representation and the crystal tensor-product rule give exactly these three highest vertices, each occurring once. The symmetric square is $L(2{\omega }_1)$ of dimension $n(2n+1)$. The exterior square splits into its invariant inverse-form line and the primitive exterior square $L({\omega }_2)$ of dimension $n(2n-1)-1$. Their dimensions sum to $4n^2$. At $n=1$, the exterior square is only the invariant line, giving $L(2{\omega }_1)\oplus L(0)$.