= Solution
{c}
Use the standard structural fact that $\mathfrak{sp}(V)$ is a <semisimple Lie algebra> over an <algebraically closed field> of <characteristic zero>. If $\mathfrak h$ is a <Cartan subalgebra>, part (a)(ii) gives $C_{\mathfrak{sp}(V)}(\mathfrak h)=\mathfrak h$, and $\mathfrak h$ is an <abelian Lie algebra>. Part (c) therefore shows that its <weight spaces> occur as $n$ pairs of one-dimensional spaces, with no zero <weight space>.
Choose representatives $\lambda_1,\ldots,\lambda_n$ of its nonzero <weights of a representation>. Pick $p_i\ne0$ in $V_{\lambda_i}$ and $q_i\ne0$ in $V_{-\lambda_i}$, scaling $q_i$ so that $\omega(p_i,q_i)=1$. Part (b) shows that all other pairings among different pairs are zero. Each individual <weight space> is one-dimensional and the <bilinear form> is alternating, so the ordered <basis>
$$
\mathcal B=(p_1,\ldots,p_n,q_1,\ldots,q_n)
$$
is a <symplectic basis>. Every $h\in\mathfrak h$ is a <diagonal matrix> in this <basis>, with entries $\lambda_1(h),\ldots,\lambda_n(h),-\lambda_1(h),\ldots,-\lambda_n(h)$.
Let $D_{\mathcal B}$ be the <Lie subalgebra> consisting of all <diagonal matrices>
$$
D_{\mathcal B}=\{\operatorname{diag}(t_1,\ldots,t_n,-t_1,\ldots,-t_n):t_i\in k\}.
$$
Each such <matrix> lies in the <symplectic Lie algebra> and commutes with $\mathfrak h$. Hence
$$
\mathfrak h\subseteq D_{\mathcal B}\subseteq C_{\mathfrak{sp}(V)}(\mathfrak h)=\mathfrak h,
$$
and $\mathfrak h=D_{\mathcal B}$. In particular, every <Cartan subalgebra> has <dimension> $n$; the $\lambda_i$ are a <basis> of $\mathfrak h^*$.
Carry out the same construction for $\mathfrak h'$, obtaining a <symplectic basis> $\mathcal B'=(p'_1,\ldots,p'_n,q'_1,\ldots,q'_n)$ with $\mathfrak h'=D_{\mathcal B'}$. The unique <linear map> $g$ sending $p_i$ to $p'_i$ and $q_i$ to $q'_i$ is invertible and preserves all pairings of <basis vectors>. By <bilinearity>, it preserves $\omega$ on all of $V$, so $g$ belongs to the <symplectic group> $\operatorname{Sp}(V)$. The <change of basis> gives
$$
\boxed{g\mathfrak h g^{-1}=gD_{\mathcal B}g^{-1}=D_{\mathcal B'}=\mathfrak h',\qquad g\in\operatorname{Sp}(V).}
$$
Thus <conjugacy of Cartan subalgebras of the symplectic Lie algebra> follows by explicitly matching their <symplectic bases>. If $n=0$, the <Lie algebra> is zero and the same conclusion holds with $g$ the identity on the zero <vector space>.
Back to article page
