 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-1/26i/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Let

Skew_{2 n}

be the skew-symmetric

2 n

2 n

F : M_{2 n} (R) \to Skew_{2 n}, F (A) = A^{T} J A .

Then

S p_{n} (R) = F^{- 1} (J)

, and

D F_{A} (H) = H^{T} J A + A^{T} J H .

At a symplectic

A

, put

H = A X

. The derivative becomes

X^{T} J + J X

. Given skew

K

, choosing

X = - \frac{1}{2} J K

gives

X^{T} J + J X = K

, so the derivative is surjective.

dim M_{2 n} = 4 n^{2}, dim Skew_{2 n} = \frac{( 2 n ) ( 2 n - 1 )}{2} = 2 n^{2} - n,

2 n^{2} + n

 New to topics? Read the docs here!