After ordering a <symplectic basis> in two blocks, write
$$
h=\operatorname{diag}(t_1,\ldots,t_n,-t_1,\ldots,-t_n),
\qquad \varepsilon_i(h)=t_i.
$$
Matrices in the <Symplectic Lie algebra> have block form
$$
\begin{pmatrix}A&B\\ C&-A^T\end{pmatrix},
\qquad B=B^T,quad C=C^T.
$$
The <root-space decomposition> is
$$
\mathfrak{sp}_{2n}=\mathfrak t
\oplus\bigoplus_{i\ne j}\mathfrak g_{\varepsilon_i-\varepsilon_j}
\oplus\bigoplus_{i<j}\left(\mathfrak g_{\varepsilon_i+\varepsilon_j}\oplus\mathfrak g_{-\varepsilon_i-\varepsilon_j}\right)
\oplus\bigoplus_i\left(\mathfrak g_{2\varepsilon_i}\oplus\mathfrak g_{-2\varepsilon_i}\right).
$$
For example, these one-dimensional spaces are spanned respectively by
$$
E_{ij}-E_{n+j,n+i},\quad
E_{i,n+j}+E_{j,n+i},\quad
E_{n+i,j}+E_{n+j,i},\quad
E_{i,n+i},\quad E_{n+i,i}.
$$
Thus this is the <Cn root system>
$$
R=\{\pm\varepsilon_i\pm\varepsilon_j:i<j\}\cup\{\pm2\varepsilon_i:1\leq i\leq n\}.
$$
The upper-triangular choice gives
$$
R^+=\{\varepsilon_i-\varepsilon_j:i<j\}
\cup\{\varepsilon_i+\varepsilon_j:i<j\}
\cup\{2\varepsilon_i:1\leq i\leq n\}.
$$
Its <simple root>[simple roots], <highest root>, <fundamental weight>[fundamental weights], and <half-sum of positive roots> are
$$
\begin{aligned}
\alpha_i&=\varepsilon_i-\varepsilon_{i+1} &&(1\leq i<n),&
\alpha_n&=2\varepsilon_n,\\
\theta&=2\varepsilon_1,&
\omega_k&=\varepsilon_1+\cdots+\varepsilon_k &&(1\leq k\leq n),\\
\rho&=n\varepsilon_1+(n-1)\varepsilon_2+\cdots+\varepsilon_n.
\end{aligned}
$$

Using the notation requested in the paper, the root lattice $P$ and weight lattice $Q$ are
$$
P=\left\{(m_1,\ldots,m_n)\in\mathbb Z^n:\sum_i m_i\equiv0\pmod2\right\},
\qquad Q=\mathbb Z^n,
$$
so $Q/P\cong\mathbb Z/2\mathbb Z$. This reverses the common notation in which the <root lattice> is called $Q$ and the <weight lattice> is called $P$.

Since a multiple-edge arrow in a <Dynkin diagram> points toward the shorter root, the finite and <Extended Dynkin diagram>[extended] diagrams are
$$
\alpha_1-\alpha_2-\cdots-\alpha_{n-2}-\alpha_{n-1}\Longleftarrow\alpha_n
$$
and
$$
\alpha_0\Longrightarrow\alpha_1-\alpha_2-\cdots-\alpha_{n-2}-\alpha_{n-1}\Longleftarrow\alpha_n,
\qquad \alpha_0=-2\varepsilon_1.
$$

Solved by gpt-5.6-sol high.


 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-102/3/i/solution