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The Lie algebra of the compact symplectic group consists of $X^{†}=-X$ and $XJ+J{X}^T=0$. Equivalently $X=\left(\begin{array}{cc}A& B\\ -\overline{B}& \overline{A}\end{array}\right)$ with $A^{†}=-A$ and $B^T=B$. Its real dimension is $n(2n+1)$, and its complexification is the symplectic Lie algebra.

For the symplectic Lie algebra ${\mathfrak{sp}}_{2n}$, the defining crystal basis is the chain $1,2,\dots ,n,\bar{n},\dots ,\bar{1}$, with weights ${\epsilon }_1,\dots ,{\epsilon }_n,-{\epsilon }_n,\dots ,-{\epsilon }_1$. Its successive lowering-edge colors are $1,2,\dots ,n-1,n,n-1,\dots ,1$. Every step subtracts its color's simple root. Applying the crystal tensor-product rule to two copies gives a complete grid description of the tensor-square crystal. For $n\ge 2$, its only highest vertices are $1\otimes 1$, $1\otimes 2$, and $1\otimes \bar{1}$, of highest weights $2{\omega }_1$, ${\omega }_2$, and zero. For $n=1$, the middle vertex is absent and only the other two highest vertices remain.

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 ${\mathfrak{so}}_5$.

Relative to $J=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$, a symplectic Lie algebra matrix has blocks $\left(\begin{array}{cc}P& Q\\ R& -P^{\mathsf{T}}\end{array}\right)$ with $Q,R$ symmetric. Its diagonal Cartan subalgebra has basis $E_{ii}-E_{l+i,l+i}$. Write ${\epsilon }_i(h)={\lambda }_i$. A root-space decomposition is obtained fromThe matrix unit rule $[h,E_{ab}]=(h_{aa}-h_{bb})E_{ab}$ proves $[h,e_r]=r(h)e_r$. These root vectors and the diagonal basis span all the permitted blocks and are independent. There are $2l^2$ roots and $l$ diagonal basis elements, giving dimension $l(2l+1)$.

Let $V={\mathbb{R}}^4$ have symplectic basis $(e_1,f_1,e_2,f_2)$. Fix $\mathrm{vol}=e_1\wedge f_1\wedge e_2\wedge f_2$ and define a symmetric bilinear form $Q$ on ${\Lambda }^{2}V$ by $u\wedge v=Q(u,v)\mathrm{vol}$. The pairings between complementary basis bivectors split this six-dimensional space into three hyperbolic planes, so $Q$ has signature $(3,3)$.

The symplectic group preserves volume and the bivector $\Omega =e_1\wedge f_1+e_2\wedge f_2$, obtained from the inverse symplectic form, up to its fixed sign convention. Since $Q(\Omega ,\Omega )=2$, its orthogonal complement $W={\Omega }^{\perp }$ has dimension five and signature $(2,3)$. On $W$ use $-Q$, which has signature $(3,2)$. This constructs the exterior-square double cover of SO0(3,2).

If an element acts trivially on $W$, it also fixes $\Omega$, so its action on the entire exterior square is the identity. Every decomposable bivector then shows that its two-plane is preserved. A line is the intersection of two such planes; hence every line is preserved, and the matrix is scalar. Its exterior-square action forces its scalar square to equal one, giving precisely $\pm I$.

The derivative is injective by the same infinitesimal argument: zero action on ${\Lambda }^{2}V$ forces an infinitesimal scalar, whose induced scalar there is twice its value and therefore zero. The symplectic Lie algebra has dimension $2^2+2(2\cdot 3/2)=10$, from blocks $\left(\begin{array}{cc}A& B\\ C& -A^T\end{array}\right)$ with symmetric $B,C$. The orthogonal Lie algebra in five dimensions also has dimension ten. Thus the image is open. Real symplectic polar decomposition has connected compact factor $U(2)$ and a contractible positive symplectic factor, so $Sp(4,\mathbb{R})$ is connected and its image is exactly the identity component:As in the preceding noncompact cases, this does not identify the quotient with the disconnected full $SO(3,2)$.

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.

Write $T=\left(\begin{array}{cc}P& Q\\ R& S\end{array}\right)$ and $J=\left(\begin{array}{cc}0& I_l\\ -I_l& 0\end{array}\right)$. Block multiplication givesIts vanishing is equivalent toThis identifies $L$ with the symplectic Lie algebra ${\mathfrak{sp}}_{2l}(\mathbb{C})$.

We now supply the requested root-space decomposition, including explicit vectors. Let $E_{ab}$ be the matrix units, put $h_i=E_{ii}-E_{l+i,l+i}$, and define the linear functionals ${\epsilon }_i$ on the diagonal Cartan subalgebra by ${\epsilon }_i(h)={\lambda }_i$. Then $H={⨁}_i\mathbb{C}{h}_i$. The Cn root system and a corresponding matrix root basis of the symplectic Lie algebra areEach listed vector satisfies the block conditions just proved. For a diagonal matrix $h$, the matrix unit calculation $[h,E_{ab}]=(h_{aa}-h_{bb})E_{ab}$ gives all the requested eigenvalues explicitly:The diagonal vectors and difference-root vectors span every possible $P$ block. The sum-root and doubled-root vectors span all symmetric $Q$ and $R$ blocks. They are linearly independent, as is evident from their disjoint independent block entries. Equivalently there are $l+2l^2=l(2l+1)$ of them, exactly the dimension $l^2+2l(l+1)/2$ of $L$. The zero weight space is precisely $H$: a zero weight forces $P$ diagonal and every entry of $Q,R$ to vanish. Hence every nonzero root space is one-dimensional and

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:

Write an element of the symplectic Lie algebra in $n\times n$ blocks. Direct multiplication in $AJ+J{A}^T=0$ givesThe diagonal Cartan subalgebra consists of $H=\mathrm{diag}(h_1,\dots ,h_n,-h_1,\dots ,-h_n)$. Define ${\epsilon }_i(H)=h_i$. Under the Adjoint representation, the off-diagonal matrix units in $P$ have weights ${\epsilon }_i-{\epsilon }_j$; the symmetric entries in $Q$ have weights ${\epsilon }_i+{\epsilon }_j$, including $2{\epsilon }_i$ on the diagonal; those in $R$ have the negatives of these weights. All nonzero root spaces are one-dimensional, while the zero weight space is the diagonal Cartan subalgebra. Thus the Cn root system isChoose the positive roots and simple roots asTheir highest root and Weyl vector areFor the last identity, each pair $i<j$ contributes $2{\epsilon }_i$ before division by two, and the long positive root contributes another ${\epsilon }_i$ afterward. Successive simple roots have equal length except for ${\alpha }_n$, whose squared length is twice that of ${\alpha }_{n-1}$. The Cn Dynkin diagram is a chain with a double last bond, whose arrow points toward ${\alpha }_{n-1}$, the short root:

The displayed schematic is for $n\ge 4$; omit intermediate nodes for $n=2,3$. For $n=1$ the diagram has just the single node ${\alpha }_1=2{\epsilon }_1$.

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:

Here root systems are finite, reduced and crystallographic, as for complex semisimple Lie algebras. For a base of a root system $\Delta =\{{\alpha }_1,\dots ,{\alpha }_r\}$, the Dynkin diagram has one vertex for each simple root. Two distinct vertices have $a_{ij}{a}_{ji}$ bonds, with the arrow on a multiple bond pointing toward the shorter root. The classification of finite crystallographic Dynkin diagrams consists of the following connected diagrams; the integers describing arms count vertices away from the unique branching vertex.

This is classification up to diagram isomorphism and overall rescaling of the root system. It includes the conventional identifications $D_3\cong A_3$ and $B_1\cong A_1$ when those low-rank notations are used. The weighting here encodes root lengths and bonds; it is not the separate theory of vertex labels attached to nilpotent orbits. Dropping the crystallographic or reduced hypothesis changes the classification.

Precisely, a base of a root system is a vector-space basis $\Delta \subset \Phi$ such that each root has a unique expansion $\beta ={\sum }_i{n}_i{\alpha }_i$, with integer coefficients all nonnegative or all nonpositive. This defines the positive and negative roots. We use the column-coroot convention for the Cartan matrix:Some conventions transpose this matrix; specifying which index labels the coroot resolves that difference. The Weyl group is the subgroup of orthogonal transformations generated by the Weyl reflections

For the C3 root system, take an orthonormal basis $e_1,e_2,e_3$ of ${\mathbb{R}}^3$ andThere are six long roots of squared length $4$ and twelve short roots of squared length $2$. A base is ${\alpha }_1=e_1-e_2$, ${\alpha }_2=e_2-e_3$, ${\alpha }_3=2e_3$. The Cartan matrix convention for C3 above givesThus the double bond connects ${\alpha }_2$ to ${\alpha }_3$, with its arrow directed toward ${\alpha }_2$.

An explicit complex Lie algebra realizing this root system is the symplectic Lie algebraIt is closed under commutators, since the two identities $X^{T}J=-JX$, $Y^{T}J=-JY$ imply $[X,Y{]}^{T}J=-J[X,Y]$. Equivalently,so its dimension is $9+6+6=21$. A Cartan subalgebra isWith $E_{ab}$ denoting a matrix unit, the root spaces have generators $E_{ij}-E_{3+j,3+i}$ for $e_i-e_j$ ($i\ne j$), $E_{i,3+j}+E_{j,3+i}$ for $e_i+e_j$ ($i<j$), and $E_{i,3+i}$ for $2e_i$; the analogous lower-left generators give the negative roots. These are all eighteen root spaces, each one-dimensional, and together with $H$ they exhaust the algebra.

For an explicit semisimplicity check, let $I\ne 0$ be an ideal. Simultaneous diagonalization of $\mathrm{ad}H$ makes $I$ a sum of its intersections with $H$ and these root spaces. If it contains a nonzero $h\in H$, some root has $\alpha (h)\ne 0$, and $[h,L_{\alpha }]$ supplies a root vector in $I$. Thus $I$ always contains a root vector. Bracket it with its opposite root vector to obtain a nonzero coroot $h_{\alpha }\in I$. Whenever $(\alpha ,\beta )\ne 0$, bracketing $h_{\alpha }$ with either $L_{\beta }$ or $L_{-\beta }$ puts both root spaces into $I$. The graph of the roots with edges for nonzero inner products is connected: $2e_i$ connects to $e_i+e_j$, which connects to $2e_j$, and every short root connects to an axis root. Repeating the argument yields all root spaces and then all of $H$ from their brackets. Hence $I=L$: the constructed algebra is simple, and therefore semisimple.

The reflections in $e_1-e_2$ and $e_2-e_3$ exchange adjacent coordinates, while reflection in $2e_3$ changes the sign of the third coordinate. They generate every signed permutation. Conversely reflection in any displayed root is a signed permutation. ThusThe reflecting hyperplanes are $x_i=0$ and $x_i=\pm x_j$. A fundamental open Weyl chamber is $x_1>x_2>x_3>0$. Every regular point has three nonzero coordinates of distinct absolute values; its signs and the ordering of those absolute values specify exactly one of the $2^3\cdot 3!=48$ chambers. The closure of this chamber is generated by the rays through $(1,0,0)$, $(1,1,0)$ and $(1,1,1)$. This is the Weyl chamber geometry of C3.

The left panel retains the actual long and short root lengths. The right panel radially projects a hemisphere of the unit sphere, draws every visible reflecting great-circle arc, and highlights the chamber $x_1>x_2>x_3>0$. Each spherical triangle is the section of a three-dimensional chamber by the sphere; opposite triangles on the rear hemisphere supply the remaining chambers. Roots of a root system are normal to the chamber walls, rather than generally lying along their edges.

The form attached to $V$ is $B_V(x,y)={\mathrm{tr}}_V(xy)$. More generally, for a Lie algebra representation $\rho$, the Trace form of a Lie algebra representation isThe unqualified Killing form is the special case of the Adjoint representation,The distinction matters: a Trace form of a Lie algebra representation can be degenerate even when $\mathfrak{g}$ is semisimple, for example on the trivial Lie algebra representation.

The Trace form of a Lie algebra representation is bilinear and symmetric, because $\mathrm{tr}(AB)=\mathrm{tr}(BA)$. It is an invariant bilinear form on a Lie algebra:This follows by expanding both commutators and cyclically permuting factors under the matrix trace. Equivalently,Its radical of a bilinear form is an ideal of a Lie algebra, since if $B_V(y,-)=0$, then $B_V([x,y],z)=-B_V(y,[x,z])=0$. The Killing form is also preserved by every automorphism of a Lie algebra, because the corresponding adjoint operators are conjugate. On a complex finite-dimensional Lie algebra, the Cartan criterion for semisimplicity says that the Killing form is nondegenerate exactly when the Lie algebra is semisimple. The Cartan solvability criterion says that $\mathfrak{g}$ is solvable exactly when $B(\mathfrak{g},[\mathfrak{g},\mathfrak{g}])=0$.

We next construct the sl2 subalgebra associated with a root. Use the root-space decompositionFor $u\in {\mathfrak{g}}_{\beta }$, $v\in {\mathfrak{g}}_{\gamma }$, invariance of the Killing form givesThus $B({\mathfrak{g}}_{\beta },{\mathfrak{g}}_{\gamma })=0$ unless $\beta +\gamma =0$, and $B({\mathfrak{g}}_{\beta },\mathfrak{h})=0$ for nonzero $\beta$. Nondegeneracy of $B$ on $\mathfrak{g}$ now implies that $-\alpha$ is a root and that $B$ pairs ${\mathfrak{g}}_{\alpha }$ and ${\mathfrak{g}}_{-\alpha }$ nondegenerately.

Nondegeneracy of $B{|}_{\mathfrak{h}}$ defines a unique $t_{\alpha }\in \mathfrak{h}$ byChoose $x\in {\mathfrak{g}}_{\alpha }$ and $y\in {\mathfrak{g}}_{-\alpha }$ with $B(x,y)=1$. Their Lie bracket lies in the zero root space, namely $\mathfrak{h}$, andTherefore $[x,y]=t_{\alpha }$.

The essential nonisotropic root lemma is that $\alpha (t_{\alpha })\ne 0$. Suppose instead that it vanished. Then $[t_{\alpha },x]=[t_{\alpha },y]=0$, so $⟨x,y,t_{\alpha }⟩$ would be a solvable Lie algebra with derived algebra $\mathbb{C}{t}_{\alpha }$. Apply the Lie theorem to its action on $\mathfrak{g}$ by the Adjoint representation. The commutator $\mathrm{ad}{t}_{\alpha }=[\mathrm{ad}x,\mathrm{ad}y]$ is strictly upper triangular in a suitable basis, hence nilpotent. But $t_{\alpha }\in \mathfrak{h}$, so the root-space decomposition makes $\mathrm{ad}{t}_{\alpha }$ diagonalizable. A diagonalizable nilpotent linear map is zero. Thus $t_{\alpha }$ is central in $\mathfrak{g}$. The center of a Lie algebra of a semisimple Lie algebra is zero; equivalently a central element lies in the radical of the Killing form. This forces $t_{\alpha }=0$, contradicting $\alpha \ne 0$.

Writing $c=\alpha (t_{\alpha })$, defineThe root-space decomposition and $[x,y]=t_{\alpha }$ giveThe three vectors are linearly independent because they lie in the distinct summands ${\mathfrak{g}}_{\alpha }$, $\mathfrak{h}$, and ${\mathfrak{g}}_{-\alpha }$. Their span is therefore a copy of the sl2 Lie algebra.

The weight lattice consists of the functionals integral on all coroots. With $H_{\alpha }$ the coroot above, the weight lattice iswhere the fundamental weights satisfy ${\omega }_i(H_{{\alpha }_j})={\delta }_{ij}$ for the simple roots ${\alpha }_j$. Here $P$ lies in the real span of the roots, viewed inside ${\mathfrak{h}}^{\ast }$.

The classification of finite-dimensional sl2 representations says that every finite-dimensional complex sl2 Lie algebra representation is a direct sum of irreducibles $L_m$, $m\in {\mathbb{Z}}_{\ge 0}$, on which the standard $H$ has eigenvalues $m,m-2,\dots ,-m$. Restrict any finite-dimensional Lie algebra representation of $\mathfrak{g}$ to each sl2 subalgebra associated with a root. If $v$ has weight $\lambda$, then $H_{\alpha }v=\lambda (H_{\alpha })v$, so $\lambda (H_{\alpha })$ is an integer. Thus every weight lies in $P$. The same restrictions show that the commuting simple coroots act diagonalizably, justifying the simultaneous weight-space decomposition.

For $\mathfrak{sp}(4)$, the roots are $\pm 2{\epsilon }_1$, $\pm 2{\epsilon }_2$, and $\pm {\epsilon }_1\pm {\epsilon }_2$. Work on ${\mathbb{C}}^4$ with the alternating bilinear form having matrixThe symplectic Lie algebra isUsing the matrix units $E_{ij}$, take the Cartan subalgebraDefine ${\epsilon }_i(a_1{h}_1+a_2{h}_2)=a_i$. A regular diagonal element of $\mathfrak{h}$ has centralizer precisely $\mathfrak{h}$, and every element of $\mathfrak{h}$ acts diagonalizably. Thus it is a Cartan subalgebra. The requested Cartan decomposition is the root-space decompositionChoose positive roots ${\epsilon }_1-{\epsilon }_2$, $2{\epsilon }_2$, ${\epsilon }_1+{\epsilon }_2$, $2{\epsilon }_1$. The symplectic root sl2 triple are given explicitly byFor the negative root spaces, use the corresponding $Y_{\alpha }$. These eight root vectors, together with $h_1,h_2$, form a basis: the block description above has dimension $4+3+3=10$, and the ten listed vectors are independent. Finally, the matrix unit identityverifies $[X_{\alpha },Y_{\alpha }]=H_{\alpha }$ for every row. The diagonal differences verify $[H_{\alpha },X_{\alpha }]=2X_{\alpha }$ and $[H_{\alpha },Y_{\alpha }]=-2Y_{\alpha }$. Thus each row supplies a basis of the required sl2 subalgebra associated with a root.

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

Write an element of the diagonal Cartan subalgebra asand define the linear functionals ${\epsilon }_i(t)=t_i$. The root-space decomposition of the symplectic Lie algebra ${\mathfrak{sp}}_6(\mathbb{C})$ then has the C3 root system

After ordering a symplectic basis in two blocks, writeMatrices in the symplectic Lie algebra have block formThe root-space decomposition isFor example, these one-dimensional spaces are spanned respectively byThus this is the Cn root systemThe upper-triangular choice givesIts simple roots, highest root, fundamental weights, and half-sum of positive roots are