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For a Hermitian adjoint scalar field with kinetic term $\mathrm{tr}(D_{\mu }\Phi (D^{\mu }\Phi {)}^{†})$, use $D_{\mu }\Phi ={\partial }_{\mu }\Phi +ig[A_{\mu },\Phi ]$ and generators normalized by $\mathrm{tr}(t^a{t}^b)={\delta }^{ab}/2$. Expanding at a constant vacuum expectation value ${\Phi }_0$ and matching ${\mathcal{L}}_{\mathrm{mass}}=A_{\mu }^a(m^2{)}^{ab}{A}^{b\mu }/2$ gives the displayed expression. A real combination of generators $X=x_a{t}^a$ has quadratic form $2g^2\mathrm{tr}([X,{\Phi }_0{]}^{†}[X,{\Phi }_0])\ge 0$. Thus the massless directions are exactly the generators commuting with the vacuum.

For ${\Phi }_0=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$, the two Hermitian generators associated with each matrix unit pair $E_{ij},E_{ji}$ have squared mass $g^2({\lambda }_i-{\lambda }_j{)}^2$. The surviving group is the centralizer of a Hermitian matrix in SU(n).

Let $V$ be a finite-dimensional symplectic vector space over an algebraically closed field of characteristic zero. For $a\in \mathfrak{sp}(V)$, write its Jordan–Chevalley decomposition as $a=S+N$. Applying the symplectic adjoint to this decomposition and using $a^{\ast }=-a$, uniqueness gives $S^{\ast }=-S$ and $N^{\ast }=-N$. Thus both parts belong to the symplectic Lie algebra. The operators $\mathrm{ad}S$ and $\mathrm{ad}N$ commute. The former is diagonalizable on $\mathrm{End}(V)$ in a matrix unit basis for an eigenbasis of $S$, and the latter is a nilpotent linear map, as follows fromThe restrictions to the invariant symplectic Lie algebra retain these properties. Hence these restrictions form the Jordan–Chevalley decomposition of $\mathrm{ad}a$.

If $\mathrm{ad}a$ is diagonalizable, $\mathrm{ad}N=0$. For every $v\in V$, the rank-one map $T_v(w)=\omega (v,w)v$ lies in $\mathfrak{sp}(V)$, since the two terms in $\omega (T_{v}w,u)+\omega (w,T_{v}u)$ cancel by alternation. For $v\ne 0$ its image is $kv$, so commuting with $T_v$ forces $N$ to preserve $kv$. A linear map preserving every line is scalar: compare its action on two linearly independent vectors and on their sum. Since $N$ is a nilpotent linear map, that scalar is zero. Thus $a=S$ is diagonalizable. Conversely, if $a$ is diagonalizable on $V$, its operator commutator action is diagonalizable on $\mathrm{End}(V)$ and therefore on its invariant symplectic Lie algebra. This proves both directions, including $V=0$.

For real $n\times n$ skew-symmetric matrices, each upper-triangular entry can be chosen freely, the corresponding lower-triangular entry is its negative, and diagonal entries vanish. Thus the displayed matrix unit differences span the space and are linearly independent: inspecting entry $(i,j)$ with $i<j$ recovers its coefficient. There are $n(n-1)/2$ such pairs, giving the dimension.

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 $l(2l+1)$.

In its block form, a matrix-unit basis of the unitary Lie algebra spans the anti-Hermitian block $A$, with the corresponding conjugate block below. Independent real and imaginary symmetric matrix units span $B$, with lower block $-\overline{B}$. These give $n^2+n(n+1)$ independent real generators, including diagonal symmetric entries.

A real basis of the unitary Lie algebra is $i{E}_{ii}$, $E_{ij}-E_{ji}$ and $i(E_{ij}+E_{ji})$ for $i<j$, where $E_{ij}$ are matrix units. These separately span the imaginary diagonal and real and imaginary upper off-diagonal entries.

Over a field of characteristic different from two, any Lie algebra ideal in ${\mathfrak{sl}}_n$ containing a nonscalar matrix is the whole special linear Lie algebra. Indeed, if $A_{ji}\ne 0$ for $i\ne j$, then $[E_{ij},[E_{ij},A]]=-2A_{ji}{E}_{ij}$ extracts a matrix unit. For a nonscalar diagonal matrix, $[E_{ij},A]=(A_{jj}-A_{ii})E_{ij}$ extracts one instead. Its bracket with $E_{ji}$ produces $H=E_{ii}-E_{jj}$, and $[H,E_{ji}]=-2E_{ji}$ extracts the reverse unit. Brackets $[E_{ij},E_{jk}]=E_{ik}$ and $[E_{ki},E_{ij}]=E_{kj}$ spread these units to every row and column; opposite pairs span all trace-zero diagonal matrices. Thus only ideals consisting of scalar matrices can be proper. This proof works even when restrictions of distinct roots of a root system to a Cartan subalgebra coincide in positive characteristic.

Let $E_{ji}$ be the matrix unit having its only nonzero entry, one, in position $(j,i)$. Direct multiplication gives $\mathrm{tr}(A{E}_{ji})=a_{ij}$. If $A\in R$, every such pairing vanishes, forcing every entry $a_{ij}$ to be zero. Conversely, zero pairs to zero with every matrix. ThusThe matrix trace pairing is therefore a nondegenerate bilinear form, despite its failure to be positive definite when $n\ge 2$.

A noncommutative algebra may have a commutative quotient: take $A=K\times M_2(K)$, $B=K$, and $\phi (\lambda ,M)=\lambda$. This is a surjective unital homomorphism. Neither requested change of commutativity can occur for an isomorphism, since an isomorphism transfers the identity $ab=ba$ in both directions.

For the matrix classifications, write $E_{ij}$ for a matrix unit and $N=E_{12}+E_{23}$. Then $N^2=E_{13}$ and $N^3=0$. The complete square-zero sets and commutativity results areIndeed $A_1$ is diagonal. In $A_2$ and $A_5$ the two off-diagonal generators have all products zero, so multiplication has the form $(x,u)(x^{\prime },u^{\prime })=(x{x}^{\prime },x{u}^{\prime }+x^{\prime }u)$. The diagonal entries of a square force $x=0$, after which the off-diagonal part squares to zero. In $A_3=K[N]$, a square-zero element first has scalar coefficient zero and then $(yN+z{N}^2{)}^2=y^2{N}^2$, forcing $y=0$. This argument works also in characteristic two. In $A_4$, a square first forces both diagonal parameters to vanish, leaving precisely $K{E}_{12}$. Noncommutativity follows from $E_{22}{E}_{12}=0$ but $E_{12}{E}_{22}=E_{12}$.

The coordinate map $xI+y{E}_{12}+z{E}_{13}\mapsto xI+y{E}_{13}+z{E}_{23}$ is an isomorphism $A_2\to A_5$, by the common multiplication formula. Every other pair is nonisomorphic: $A_4$ is the only noncommutative member, and among the commutative members the dimensions of the square-zero sets are respectively zero, two, one, and two. These sets are vector spaces in these examples, and an algebra isomorphism must carry one onto the other. Thus the only distinct isomorphic pair is $A_2\cong A_5$.

By the Artin–Wedderburn theorem, a finite-dimensional complex simple algebra is $M_d(\mathbb{C})$, with a unique simple module up to isomorphism, namely the column module $S={\mathbb{C}}^d$. Its left regular module is $S^{\oplus d}$. The Schur lemma gives ${\mathrm{End}}_A(S)=\mathbb{C}$, soThe right action can be seen explicitly. For a row vector $r$, the map $f_r(m)=mr$ is $A$-linear, and all such maps are obtained this way, since the row space and $V_M$ both have dimension $d$. Moreover $f_r\cdot x=f_{rx}$. Any nonzero row generates the whole row space under right multiplication by matrices: if $r_j\ne 0$, suitable matrix units send it to each coordinate row. Thus $V_M$ is a simple right $A$-module.

Write $e_{ij}$ for the matrix units. Left multiplication by $a=\left(\begin{array}{cc}r& s\\ 0& t\end{array}\right)$ givesThus a composition series of a module isEach quotient is one-dimensional, hence a simple module. The required direct sum isThe first summand is simple. The second has a unique nonzero proper submodule, $\mathbb{C}{e}_{12}$: any vector $u{e}_{12}+v{e}_{22}$ with $v\ne 0$ generates $e_{12}$ by applying $e_{12}$ and then generates $e_{22}$. A decomposition into two nonzero summands would require a second one-dimensional submodule, which is impossible. Hence this summand is an indecomposable module.

In spontaneous symmetry breaking, the action respects a symmetry while a selected vacuum state is invariant only under a proper subgroup. A nonzero vacuum expectation value can identify that choice. For a local gauge theory, gauge transformations are redundancies rather than transformations between physically different states; a gauge-fixed scalar expectation value describes the Higgs mechanism and its massive excitations. It should not be interpreted as breaking an observable local redundancy.

Write the adjoint scalar field as $\Phi ={\Phi }_0+\phi$, with Hermitian gauge generators normalized by $\mathrm{tr}(t^a{t}^b)={\delta }^{ab}/2$. Since ${\Phi }_0$ is constant, the part of the covariant derivative linear in the gauge field is $D_{\mu }{\Phi }_0=ig{A}_{\mu }^a[t^a,{\Phi }_0]$. Both $t^a$ and ${\Phi }_0$ are Hermitian, so their commutator is anti-Hermitian. The scalar kinetic term therefore contributesMatching coefficients gives the adjoint scalar gauge-boson mass matrixIt is positive semidefinite: for real coefficients $x_a$, its quadratic form is $2g^2\mathrm{tr}(C^{†}C)\ge 0$, where $C=[x_a{t}^a,{\Phi }_0]$. Its null space consists of generators commuting with the vacuum, the Lie algebra of its centralizer.

For a diagonal vacuum ${\Phi }_0=\mathrm{diag}({\lambda }_1,{\lambda }_2,{\lambda }_3)$, the matrix unit $E_{ij}$ obeys $[E_{ij},{\Phi }_0]=({\lambda }_j-{\lambda }_i)E_{ij}$. The two normalized Hermitian generators built from $E_{ij},E_{ji}$ each acquireThe diagonal generators $t^3,t^8$ commute with every such vacuum and remain massless. Orthogonality of these root generators also shows that the mass matrix is diagonal in the supplied Gell-Mann matrices basis. Take $g>0$ and $v\ne 0$ below; $v=0$ restores the full SU(3) group.

For a differentiable curve $Q(t)\in SO(5)$ with $Q(0)=I_5$, differentiate $Q(t{)}^{T}Q(t)=I_5$. Its tangent matrix $X=Q^{\prime }(0)$ is real and satisfies $X^T+X=0$. Conversely the matrix exponential of any real antisymmetric matrix is orthogonal, and $\det e^{tX}=e^{t\mathrm{tr}X}=1$ because $\mathrm{tr}X=0$. Hence the real Special orthogonal Lie algebra isLet $E_{ab}$ denote a matrix unit. A simple real basis isEach independent entry above the diagonal supplies one real parameter, giving $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ parameters. The commutator of two basis matrices iswhere $M_{ba}=-M_{ab}$ and $M_{aa}=0$.

A Cartan subalgebra is spanned by the commuting matrices $M_{12},M_{34}$, which generate independent rotations in two orthogonal planes. A generic $h=a{M}_{12}+b{M}_{34}$, with $ab(a^2-b^2)\ne 0$, has distinct complex eigenvalues $ia,-ia,ib,-ib,0$. A real antisymmetric matrix commuting with it preserves those two planes and the remaining line; it is a multiple of the rotation in each plane and is zero on the line. Thus the commuting Lie subalgebra has dimension two and is maximal toral. Therefore

Differentiate a curve $U(t)\in SU(4)$ through $U(0)=I$, writing $X=U^{\prime }(0)$. Differentiating $U(t{)}^{†}U(t)=I$ gives $X^{†}+X=0$; differentiating $\det U(t)=1$ gives $\mathrm{tr}X=0$. Conversely, for every anti-Hermitian matrix $X$ with zero trace, the matrix exponential $e^{tX}$ is unitary and has determinant $e^{t\mathrm{tr}X}=1$. Thus the real Lie algebra isThe commutator again has zero trace and is anti-Hermitian, so this set is closed under the Lie bracket. Let $E_{jk}$ be the matrix units. A convenient real basis isThere are six independent $A_{jk}$, six independent $B_{jk}$, and three independent $H_a$. Every permitted off-diagonal complex entry is supplied by one real combination of $A_{jk},B_{jk}$; the imaginary diagonal entries have one trace constraint and are spanned by the $H_a$. ConsequentlyFor the rank of a semisimple Lie algebra, the diagonal Cartan subalgebra consists of $i\mathrm{diag}(t_1,t_2,t_3,t_4)$ with real $t_i$ and ${\sum }_i{t}_i=0$, of dimension three. Commuting anti-Hermitian matrices are simultaneously unitarily diagonalizable, so no larger commuting toral Lie subalgebra exists. In the common physics convention one uses the Hermitian generators $-i{A}_{jk},-i{B}_{jk},-i{H}_a$ and writes group elements as $e^{itT}$; these are the same generators with the factor of $i$ moved into the exponential.

The complexification of a Lie algebra gives $\mathfrak{su}(4{)}_{\mathbb{C}}={\mathfrak{sl}}_4(\mathbb{C})$, of type A3 root system. Choose simple roots ${\alpha }_1=e_1-e_2$, ${\alpha }_2=e_2-e_3$, ${\alpha }_3=e_3-e_4$, with squared lengths two. The Cartan matrix and Dynkin diagram areEach line represents a single bond. In the Dynkin label convention $w_a=⟨w,{\alpha }_a^{\vee }⟩$, the simple roots have coordinates equal to the rows of $A$.

In the defining representation, apply the simple coroots $E_{aa}-E_{a+1,a+1}$ to the four coordinate basis vectors. Their weights areEquivalently these are obtained by subtracting ${\alpha }_1,{\alpha }_2,{\alpha }_3$ successively from the highest weight. The dual representation has the negative weights, henceAll listed weight multiplicities are one, so both dimensions are four.

The highest weight $(1,0,1)$ is ${\alpha }_1+{\alpha }_2+{\alpha }_3$, the highest root. Its irreducible representation is therefore the Adjoint representation. Under the action on traceless matrices by conjugation, the off-diagonal matrix unit $E_{ij}$ has weight $v_i-v_j$, while the diagonal Cartan subalgebra has zero weight. ThusThe superscript $[3]$ indicates weight multiplicity three. The twelve nonzero weights each have weight multiplicity one, giving $12+3=15$ in total. This also follows from the tensor product representation $4\otimes \overline{4}=15\oplus 1$: the sixteen differences $v_i-v_j$ contain four zero weights, and removing the identity matrix removes one of them.

For a curve $O(t)=I+tX+O(t^2)$ in the special orthogonal group, differentiate $O(t{)}^{T}O(t)=I$ at the identity. This gives $X^T+X=0$. Conversely, the matrix exponential of any real skew-symmetric matrix is orthogonal and has determinant $\det e^{tX}=e^{t\mathrm{tr}X}=1$. HenceA basis is $L_{ij}=E_{ij}-E_{ji}$ for $1\le i<j\le N$, where $E_{ij}$ is a matrix unit. If $L_{ji}=-L_{ij}$ and $L_{ii}=0$, their Lie brackets areThere are $\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ independent off-diagonal entries. A maximal torus rotates independently in mutually orthogonal two-planes, generated by $L_{12},L_{34},\dots$. ThusThe maximality follows by simultaneously decomposing commuting real skew-symmetric matrices into rotation two-planes, with at most one remaining fixed axis.

By irreducible traceless matrix Lie algebra is semisimple, let $R$ be the solvable radical. Lie's theorem triangularizes its action, with finitely many diagonal characters ${\lambda }_i:R\to \mathbb{C}$, and supplies a nonzero common eigenspace for one of them. For $x\in L$, conjugation by $e^{tx}$ preserves $R$ and gives an equivalent representation of it. Thus the multiset of characters ${\lambda }_i\circ e^{t\mathrm{ad}x}$ is constant. Each character varies continuously in $t$ and has values in that finite set of characters, so it is individually constant. Differentiating at zero gives ${\lambda }_i([x,r])=0$ for all $r\in R$.

Choose a common eigenspace $V_{\lambda }\ne 0$. For $v\in V_{\lambda }$,Thus $V_{\lambda }$ is $L$-invariant, so irreducibility makes it all of ${\mathbb{C}}^n$. Every element of $R$ acts as a scalar. Its trace is zero because $L\subseteq {\mathfrak{sl}}_n$, so that scalar is zero. Faithfulness of the matrix inclusion gives $R=0$. Hence $\boxed{L\text{ is semisimple}}$.

One sufficient diagonal-matrix condition for simplicity is that $L$ contain all traceless diagonal matrices. Indeed invariance under their adjoint actions then decomposes $L$ into that diagonal algebra and some matrix-unit spaces $\mathbb{C}{e}_{ij}$. Put a directed edge $j\to i$ when $e_{ij}\in L$. Irreducibility makes the directed graph strongly connected; otherwise a terminal component spans a proper invariant coordinate subspace. Bracketing matrix units along a simple directed path supplies the matrix unit between its endpoints. Strong connectivity therefore supplies every off-diagonal matrix unit, so $L={\mathfrak{sl}}_n$, which is simple. The mere existence of a diagonal matrix with distinct diagonal entries would not be a sufficient condition.

For the specified construction, $H$ has dimension two. Its diagonal adjoint action and the twelve distinct nonzero weights give a direct sum $L=H\oplus {⨁}_{i=1}^{12}\mathbb{C}{g}_i$, of dimension fourteen. The stated bracket condition and the weight-vector relations make it a Lie algebra. A subspace invariant under it is invariant under all the $g_i$ and hence their associative algebra, so it acts irreducibly on ${\mathbb{C}}^7$. The preceding result makes it semisimple.

Use prime-dimensional faithful irreducible module forces simplicity. To see this, decompose the algebra into simple ideals. Complete reducibility for one ideal and commutation with the others make its isotypic component invariant under the whole algebra, hence all of the representation. Writing that component as $W\otimes \mathrm{Hom}(W,V)$ and repeating produces a tensor product of irreducibles for the simple ideals. Faithfulness makes every factor nontrivial, of dimension at least two. Their dimensions multiply to seven, so only one factor is possible. Thus the constructed $L$ is simple.

The diagonal algebra $H$ is toral, and its centralizer inside $L$ is just $H$, since no $g_i$ has zero weight. A maximal toral subalgebra containing $H$ lies in that centralizer, so equals $H$; it is a Cartan subalgebra. Thus the rank is two and the twelve weights are the roots. Pair them as ${\alpha }_{i+6}=-{\alpha }_i$. The opposite-root brackets have weight zero and belong to $H$. For distinct $i,j\le 6$, the sum cannot be zero, and the supplied root-bracket result says their bracket is nonzero exactly when ${\alpha }_i+{\alpha }_j$ is a root. This is equivalent to $-{\alpha }_i-{\alpha }_j$ being a root, proving the requested equivalence for the paired brackets.

Among rank-two simple root systems, $A_2,B_2,G_2$ have respectively six, eight and twelve roots. ThereforeChoose simple roots $\alpha$ short and $\beta$ long. In $G_2$, $\beta +\alpha$ and $\beta +2\alpha$ are roots. Renumber with $g_1\in L_{\alpha }$, $g_3\in L_{\beta }$, $g_4\in L_{\beta +\alpha }$ and $g_5\in L_{\beta +2\alpha }$, retaining opposite partners in the other six positions. Root spaces are one dimensional, and the allowed bracket equality gives

We use the supplied weight decomposition $V={⨁}_{\lambda \in \Psi (\mathfrak{a})}{V}_{\lambda }$. There is a small distinction to justify: the definition of a toral subalgebra uses the Adjoint representation, while these weight spaces use the defining representation. For the symplectic Lie algebra, adjoint semisimplicity in the symplectic Lie algebra implies that each $a\in \mathfrak{a}$ is diagonalizable on $V$ as well. Here is a direct verification.

Define the symplectic adjoint $a^{\ast }$ by $\omega (av,w)=\omega (v,a^{\ast }w)$. The symplectic Lie algebra condition is $a^{\ast }=-a$. Write the Jordan–Chevalley decomposition in $\mathrm{End}(V)$ as $a=S+N$, with $S$ diagonalizable, $N$ a nilpotent linear map, and $SN=NS$. Uniqueness of the Jordan–Chevalley decomposition, applied to $a^{\ast }=-a$, gives $S^{\ast }=-S$ and $N^{\ast }=-N$. Thus $S,N\in \mathfrak{sp}(V)$. On $\mathrm{End}(V)$, $\mathrm{ad}S$ is diagonalizable by using a matrix unit basis in an eigenbasis of $S$, and $\mathrm{ad}N$ is a nilpotent linear map, sincevanishes for large $m$. These operators commute and preserve $\mathfrak{sp}(V)$, so their restrictions are the Jordan–Chevalley decomposition of $\mathrm{ad}a$. Because $\mathrm{ad}a$ is diagonalizable, its nilpotent part is zero: $\mathrm{ad}N=0$ on $\mathfrak{sp}(V)$.

To conclude that $N=0$ without an assumption on the center of a Lie algebra, observe that each map $T_v(w)=\omega (v,w)v$ lies in $\mathfrak{sp}(V)$. If $v\ne 0$, its image is $kv$, because $\omega$ is a nondegenerate bilinear form. Since $N$ commutes with $T_v$, it preserves $kv$. Hence $Nv\in kv$ for every $v$, forcing $N$ to be a scalar linear map: compare its action on two linearly independent vectors and their sum. A scalar nilpotent linear map is zero. For $V=0$ the assertion is immediate. This proves the needed diagonalizability. Commuting diagonalizable linear maps admit simultaneous diagonalization, giving the displayed weight decomposition.

Now let $v\in V_{\lambda }$ and $w\in V_{\mu }$. For every $a\in \mathfrak{a}$, preservation of the alternating bilinear form givesIf $\lambda +\mu \ne 0$ as a linear functional, choose $a$ on which it is nonzero; then $\omega (v,w)=0$. ThereforeFor $v\ne 0$ in $V_{\lambda }$, the fact that $\omega$ is a nondegenerate bilinear form supplies a $w\in V$ with $\omega (v,w)\ne 0$. Decomposing $w$ into weight spaces, only its $V_{-\lambda }$ component can contribute. Thus $V_{-\lambda }\ne 0$, and $\Psi (\mathfrak{a})=-\Psi (\mathfrak{a})$. The same argument on both sides shows that $\omega :V_{\lambda }\times V_{-\lambda }\to k$ is a perfect pairing, so $\dim V_{\lambda }=\dim V_{-\lambda }$. On $V_0$, the restriction of $\omega$ is a nondegenerate bilinear form that is an alternating bilinear form, so $\dim V_0$ is even.

Choose one representative ${\lambda }_i$ from each pair of nonzero weights of a representation, for $1\le i\le r$. HereThe linear map $\mathfrak{a}\to k^r$ sending $a$ to $({\lambda }_1(a),\dots ,{\lambda }_r(a))$ is injective: an element in its kernel acts by zero on every nonzero weight space, and also on $V_0$, so it is the zero element of $\mathfrak{a}\subseteq \mathrm{End}(V)$. Taking dimensions yieldsThe final inequality follows because each pair contributes at least two to $\dim V=2n$. Equivalently, the nonzero weights of a representation span the dual space ${\mathfrak{a}}^{\ast }$; they need not be linearly independent at this stage.

Choose representatives ${\lambda }_1,\dots ,{\lambda }_r$ as in part (b). We use the symplectic centralizer decompositionTo prove it, a linear map commuting with every $a\in \mathfrak{a}$ must preserve every weight space: if $v\in V_{\lambda }$, then $a(Tv)=T(av)=\lambda (a)Tv$. Conversely, preserving the weight spaces makes a linear map commute with the scalar action of every $a$ on each space.

On $V_0$, the symplectic Lie algebra condition says exactly that the restriction belongs to $\mathfrak{sp}(V_0)$. On $V_{{\lambda }_i}\oplus V_{-{\lambda }_i}$, use the perfect pairing from part (b) to identify $V_{-{\lambda }_i}$ with the dual space $V_{{\lambda }_i}^{\ast }$. The action $A_i$ on $V_{{\lambda }_i}$ can be arbitrary, but its action on $V_{-{\lambda }_i}$ must be $-A_i^{\ast }$, where $A_i^{\ast }$ is the dual map. Indeed,determines $B_i=-A_i^{\ast }$ uniquely. There are no additional conditions between different pairs because $\omega$ vanishes between different pairs. Every choice of these blocks therefore gives an element of the centralizer of a Lie subalgebra. The block correspondence preserves Lie brackets, proving the claimed direct sum of Lie algebras.

The general linear Lie algebra ${\mathfrak{gl}}_m(k)$ is an abelian Lie algebra exactly when $m\le 1$: for $m\ge 2$, the matrix units satisfy $[E_{11},E_{12}]=E_{12}\ne 0$. A nonzero symplectic vector space has a symplectic basis and contains a two-dimensional symplectic subspace. The symplectic Lie algebra acting on that plane, extended by zero on its symplectic orthogonal complement, contains the matrices $H=\mathrm{diag}(1,-1)$ and $E=E_{12}$ with $[H,E]=2E\ne 0$. Thus $\mathfrak{sp}(V_0)$ is an abelian Lie algebra exactly when $V_0=0$.

Since different summands in a direct sum of Lie algebras commute, we obtain the precise criterionEquivalently, there are $2n$ distinct nonzero weights of a representation, each with a one-dimensional weight space. The condition concerns multiplicities in the defining representation; it does not assert that $\dim \mathfrak{a}=n$ for every such $\mathfrak{a}$.

For distinct coordinates $i,j$, let $E_{ij}(t)=I+t{e}_{ij}$, where $e_{ij}$ is the matrix unit. For $t\ne 0$ this is a transvection: it fixes the hyperplane with $j$th coordinate zero pointwise, has determinant one and inverse $E_{ij}(-t)$. We show these elementary transvection matrices generate $SL(n,F)$ over any field.

Let $M\in SL(n,F)$. Row addition is left multiplication by an $E_{ij}(t)$. If the current diagonal pivot is zero, a nonzero entry occurs below it in the same remaining column, since the remaining square block is invertible; adding that row makes the pivot nonzero. Row additions then clear the entries below the pivot. Repeating produces an upper-triangular matrix with nonzero diagonal. Clearing the entries above the diagonal, starting from the last column, gives a diagonal matrix $D=\mathrm{diag}(a_1,\dots ,a_n)$ with ${\prod }_i{a}_i=1$. Thus it remains to realize determinant-one diagonal matrices using elementary transvections.

In a two-coordinate block putDirect multiplication givesThe diagonal matrix $D$ is the product, for $i=1,\dots ,n-1$, of these blocks on coordinates $i,n$, with entries $a_i,a_i^{-1}$. Their final $n$th entry is ${\prod }_{i<n}{a}_i^{-1}=a_n$. Therefore $D$, and hence $M$, is a product of elementary transvections. All row operations used are invertible transvections, and no division except by a nonzero pivot or $a_i$ has been made. This provesThe identity element $E_{ij}(0)$ can be omitted from the generating set. This is transvections generate the special linear group; ordinary row scaling alone would not prove it, because such a scaling can leave $SL(n,F)$.