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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.

Use the general definition of a Cartan subalgebra: a nilpotent Lie algebra that is a Lie subalgebra and equals its normalizer of a Lie subalgebra. In a semisimple Lie algebra, the stated structural fact says every Cartan subalgebra $\mathfrak{h}$ is a toral subalgebra and an abelian Lie algebra. Part (i) therefore givesConversely, suppose $\mathfrak{h}$ is a toral subalgebra and an abelian Lie algebra with $C_{\mathfrak{g}}(\mathfrak{h})=\mathfrak{h}$. An abelian Lie algebra is a nilpotent Lie algebra, because its lower central series of a Lie algebra vanishes at the second term. Part (i) gives $N_{\mathfrak{g}}(\mathfrak{h})=C_{\mathfrak{g}}(\mathfrak{h})=\mathfrak{h}$. Thus $\mathfrak{h}$ satisfies the defining conditions of a Cartan subalgebra, proving

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.

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.

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.