Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 102 / 3 / i

Solution

After ordering a symplectic basis in two blocks, write

h = diag (t_{1}, \dots, t_{n}, - t_{1}, \dots, - t_{n}), ε_{i} (h) = t_{i} .

(A C B - A^{T}), B = B^{T}, q u a d C = C^{T} .

sp_{2 n} = t \oplus ⨁_{i \neq = j} g_{ε_{i} - ε_{j}} \oplus ⨁_{i < j} (g_{ε_{i} + ε_{j}} \oplus g_{- ε_{i} - ε_{j}}) \oplus ⨁_{i} (g_{2 ε_{i}} \oplus g_{- 2 ε_{i}}) .

For example, these one-dimensional spaces are spanned respectively by

E_{ij} - E_{n + j, n + i}, E_{i, n + j} + E_{j, n + i}, E_{n + i, j} + E_{n + j, i}, E_{i, n + i}, E_{n + i, i} .

Thus this is the Cn root system

R = {\pm ε_{i} \pm ε_{j} : i < j} \cup {\pm 2 ε_{i} : 1 \leq i \leq n} .

The upper-triangular choice gives

R^{+} = {ε_{i} - ε_{j} : i < j} \cup {ε_{i} + ε_{j} : i < j} \cup {2 ε_{i} : 1 \leq i \leq n} .

α_{i} θ ρ = ε_{i} - ε_{i + 1} = 2 ε_{1}, = n ε_{1} + (n - 1) ε_{2} + \dots + ε_{n} . ω_{k} (1 \leq i < n), = ε_{1} + \dots + ε_{k} α_{n} = 2 ε_{n}, (1 \leq k \leq n),

Using the notation requested in the paper, the root lattice

P

Q

are

P = {(m_{1}, \dots, m_{n}) \in Z^{n} : \sum_{i} m_{i} \equiv 0 (mod 2)}, Q = Z^{n},

Q / P ≅ Z /2 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 diagrams are

α_{1} - α_{2} - \dots - α_{n - 2} - α_{n - 1} ⟸ α_{n}

and

α_{0} ⟹ α_{1} - α_{2} - \dots - α_{n - 2} - α_{n - 1} ⟸ α_{n}, α_{0} = - 2 ε_{1} .

Solved by gpt-5.6-sol high.

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