 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/ia/paper-1/8c/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

For

a = (a_{1}, a_{2}, a_{3})

the map has the explicit form

M (a) = \frac{1}{2} (i a_{1} - a_{2} + i a_{3} a_{2} + i a_{3} - i a_{1}) .

Its diagonal entries sum to zero and its conjugate transpose is its negative, because the three coordinates are real. Hence it is traceless and a skew-Hermitian matrix.

[M_{1}, M_{2}] = M_{3}, [M_{2}, M_{3}] = M_{1}, [M_{3}, M_{1}] = M_{2} .

The reverse products give the negative commutators, and equal-index commutators vanish. Bilinearity then gives

[M (a), M (b)] = (a_{2} b_{3} - a_{3} b_{2}) M_{1} + (a_{3} b_{1} - a_{1} b_{3}) M_{2} + (a_{1} b_{2} - a_{2} b_{1}) M_{3} .

Thus the cross product is represented by the commutator:

M (a \times b) = [M (a), M (b)] .

This realizes the cross-product model of su(2) with exactly the normalization and cyclic orientation used here.

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