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Over an algebraically closed field of characteristic zero, any two Cartan subalgebras of the symplectic Lie algebra are conjugate by the symplectic group. A Cartan subalgebra $\mathfrak{h}$ is an abelian Lie algebra and equals its centralizer of a Lie subalgebra. The symplectic centralizer decomposition therefore forces all its weight spaces in $V$ to be one-dimensional, nonzero, and paired by opposite weights of a representation. Choose a basis vector in each space, normalizing the perfect pairing with its opposite to one. These vectors form a symplectic basis $\mathcal{B}$. In this basis, $\mathfrak{h}$ is contained in the Lie subalgebra of diagonal matrices with entries $(t_1,\dots ,t_n,-t_1,\dots ,-t_n)$. All those diagonal matrices commute with $\mathfrak{h}$, so equality with its centralizer of a Lie subalgebra forces equality with the entire diagonal Lie subalgebra. Repeating for ${\mathfrak{h}}^{\prime }$ gives a second symplectic basis. The linear map sending one symplectic basis to the other belongs to the symplectic group and conjugates the two diagonal Lie subalgebras, proving the claim.

If $\mathfrak{a}$ is a toral subalgebra and an abelian Lie algebra, its normalizer of a Lie subalgebra equals its centralizer of a Lie subalgebra. For $x$ in the former and $a\in \mathfrak{a}$, one has $[a,x]\in \mathfrak{a}$, hence $[a,[a,x]]=0$. A diagonalizable linear map has the same kernel as its square, by expansion in an eigenbasis. Applying this to $\mathrm{ad}a$ gives $[a,x]=0$ for every $a$. The reverse containment follows directly from the definitions. No assumption that $\mathfrak{g}$ is a semisimple Lie algebra is needed.

The normalizer of a Lie subalgebra and the centralizer of a Lie subalgebra areThe second is always contained in the first. Conversely, let $x\in N_{\mathfrak{g}}(\mathfrak{a})$ and $a\in \mathfrak{a}$. Then $[a,x]\in \mathfrak{a}$, and, because $\mathfrak{a}$ is an abelian Lie algebra,By the definition of a toral subalgebra, $\mathrm{ad}a$ is diagonalizable. A diagonalizable linear map $T$ satisfies $\ker T^2=\ker T$: in an eigenbasis, both kernels consist of exactly the eigenvectors whose eigenvalue is zero. Thus $[a,x]=0$. This holds for every $a\in \mathfrak{a}$, so $x\in C_{\mathfrak{g}}(\mathfrak{a})$ andThis is the normalizer-centralizer equality for a toral abelian subalgebra; it does not require $\mathfrak{g}$ to be a semisimple Lie algebra.

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 $\mathfrak{a}$.

Let $\mathfrak{a}$ be a toral subalgebra and an abelian Lie algebra in the symplectic Lie algebra, over an algebraically closed field of characteristic zero. Its defining representation has a weight decomposition. The symplectic Lie algebra identity gives $(\lambda (a)+\mu (a))\omega (v,w)=0$ for $v\in V_{\lambda }$, $w\in V_{\mu }$, so $\omega (V_{\lambda },V_{\mu })=0$ unless $\mu =-\lambda$. Since $\omega$ is a nondegenerate bilinear form, its restrictions give a perfect pairing between $V_{\lambda }$ and $V_{-\lambda }$, and a nondegenerate bilinear form on $V_0$.

An element of the centralizer of a Lie subalgebra preserves each weight space. On $V_0$ it is any element of $\mathfrak{sp}(V_0)$. On each nonzero pair, its action $A$ on $V_{\lambda }$ is arbitrary, and the perfect pairing uniquely forces the action on $V_{-\lambda }$ to be the negative dual map $-A^{\ast }$. These independent choices preserve Lie brackets, proving the displayed direct sum of Lie algebras; the sum runs over one representative per nonzero pair.

This centralizer of a Lie subalgebra is an abelian Lie algebra precisely when $V_0=0$ and all the nonzero weight spaces are one-dimensional. Indeed, ${\mathfrak{gl}}_m$ is nonabelian for $m\ge 2$ by $[E_{11},E_{12}]=E_{12}$, while every nonzero symplectic vector space contains a two-dimensional symplectic subspace on which $[\mathrm{diag}(1,-1),E_{12}]=2E_{12}$. In the stated case the surviving summands are copies of ${\mathfrak{gl}}_1$ and commute. Repeated weights of a representation, including a zero one, are therefore exactly what obstructs this commutativity.