 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-4/4/vi/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Let every root have squared length

L

. Since

r + s

is a root,

L = ∥ r + s ∥^{2} = 2 L + 2 (r, s)

, hence

(r, s) = - L /2

. In particular the roots are independent. The potential higher sums satisfy

∥2 r + s ∥^{2} = ∥ r + 2 s ∥^{2} = 3 L,

so neither is a root. The root-space decomposition therefore gives

[e_{r}, e_{r + s}] = [e_{s}, e_{r + s}] = 0

A = t ad e_{r}

B = u ad e_{s}

and

C = [A, B] = t u N_{r, s} ad e_{r + s}

. Part (iv) shows

[A, C] = [B, C] = 0

. To compute the sign, differentiate

F (v) = e^{- v B} A e^{v B}

F^{'} (v) = e^{- v B} [A, B] e^{v B} = C, F (0) = A .

Thus

e^{- B} A e^{B} = A + C

. Conjugating the matrix exponential, and then using the commutation of

A, C

, gives

e^{- B} e^{- A} e^{B} e^{A} = e^{- A - C} e^{A} = e^{- C} .

x_{s} (u)^{- 1} x_{r} (t)^{- 1} x_{s} (u) x_{r} (t) = x_{r + s} (- N_{r, s} t u) .

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