= Solution
{c}
Choose representatives $\lambda_1,\ldots,\lambda_r$ as in part (b). We use the <symplectic centralizer decomposition>
$$
\boxed{C_{\mathfrak{sp}(V)}(\mathfrak a)\cong\mathfrak{sp}(V_0)\oplus\bigoplus_{i=1}^r\mathfrak{gl}(V_{\lambda_i}).}
$$
To 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> $V_{\lambda_i}^*$. The action $A_i$ on $V_{\lambda_i}$ can be arbitrary, but its action on $V_{-\lambda_i}$ must be $-A_i^*$, where $A_i^*$ is the <dual map>. Indeed,
$$
\omega(A_iv,w)+\omega(v,B_iw)=0\quad\text{for all }v,w
$$
determines $B_i=-A_i^*$ 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 $m\leq1$: for $m\geq2$, the <matrix units> satisfy $[E_{11},E_{12}]=E_{12}\ne0$. 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> $H=\operatorname{diag}(1,-1)$ and $E=E_{12}$ with $[H,E]=2E\ne0$. 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 criterion
$$
\boxed{C_{\mathfrak{sp}(V)}(\mathfrak a)\text{ is abelian}\iff\begin{gathered}V_0=0,\\\dim V_\lambda=1\quad(0\ne\lambda\in\Psi(\mathfrak a)).\end{gathered}}
$$
Equivalently, 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 $\mathfrak a$.
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