 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-1/32a/d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Anti-Hermiticity makes

\overset{z}{ˉ}_{j} A_{jk} z_{k}

purely imaginary, so

I_{A}

is real. Its complex derivatives are

\frac{\partial I _{A}}{\partial z _{l}} = \frac{1}{2 i} \overset{z}{ˉ}_{j} A_{j l}, \frac{\partial I _{A}}{\partial z ˉ _{l}} = \frac{1}{2 i} A_{l k} z_{k} .

The complex-coordinate bracket from part (c) therefore gives

{I_{A}, I_{B}} = \frac{i}{2} \overset{z}{ˉ}_{j} (A_{j l} B_{l k} - B_{j l} A_{l k}) z_{k} = \frac{i}{2} \overset{z}{ˉ} [A, B] z = - \frac{1}{2 i} \overset{z}{ˉ} [A, B] z .

{I_{A}, I_{B}} = - I_{[A, B]} .

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