= Solution
{c}
We use the supplied <weight decomposition> $V=\bigoplus_{\lambda\in\Psi(\mathfrak a)}V_\lambda$. 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> $a^*$ by $\omega(av,w)=\omega(v,a^*w)$. The <symplectic Lie algebra> condition is $a^*=-a$. Write the <Jordan–Chevalley decomposition> in $\operatorname{End}(V)$ as $a=S+N$, with $S$ <diagonalizable>, $N$ a <nilpotent linear map>, and $SN=NS$. Uniqueness of the <Jordan–Chevalley decomposition>, applied to $a^*=-a$, gives $S^*=-S$ and $N^*=-N$. Thus $S,N\in\mathfrak{sp}(V)$. On $\operatorname{End}(V)$, $\operatorname{ad}S$ is <diagonalizable> by using a <matrix unit> <basis> in an <eigenbasis> of $S$, and $\operatorname{ad}N$ is a <nilpotent linear map>, since
$$
(\operatorname{ad}N)^m(T)=\sum_{j=0}^m(-1)^j\binom mjN^{m-j}TN^j
$$
vanishes for large $m$. These operators commute and preserve $\mathfrak{sp}(V)$, so their restrictions are the <Jordan–Chevalley decomposition> of $\operatorname{ad}a$. Because $\operatorname{ad}a$ is <diagonalizable>, its nilpotent part is zero: $\operatorname{ad}N=0$ 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 $v\ne0$, 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> gives
$$
0=\omega(av,w)+\omega(v,aw)=(\lambda(a)+\mu(a))\omega(v,w).
$$
If $\lambda+\mu\ne0$ as a <linear functional>, choose $a$ on which it is nonzero; then $\omega(v,w)=0$. Therefore
$$
\boxed{\omega(V_\lambda,V_\mu)=0\quad\text{unless }\mu=-\lambda.}
$$
For $v\ne0$ in $V_\lambda$, the fact that $\omega$ is a <nondegenerate bilinear form> supplies a $w\in V$ with $\omega(v,w)\ne0$. Decomposing $w$ into <weight spaces>, only its $V_{-\lambda}$ component can contribute. Thus $V_{-\lambda}\ne0$, 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 $1\leq i\leq r$. Here
$$
r=\frac{|\Psi(\mathfrak a)\setminus\{0\}|}{2}.
$$
The <linear map> $\mathfrak a\to k^r$ sending $a$ to $(\lambda_1(a),\ldots,\lambda_r(a))$ 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 $\mathfrak a\subseteq\operatorname{End}(V)$. Taking <dimensions> yields
$$
\boxed{\dim\mathfrak a\leq r=\frac{|\Psi(\mathfrak a)\setminus\{0\}|}{2}\leq n.}
$$
The 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> $\mathfrak a^*$; they need not be <linearly independent> at this stage.
Back to article page
