 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-7/3/b/solution

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

[C (a) C (b), C (v)] = 2 (b, v) C (a) - 2 (a, v) C (b) .

Indeed move

C (v)

past

C (b)

and then past

C (a)

using the two anticommutator identities; the three-generator products cancel.

Since all

C (w)

π (T)^{*} + π (T) = \frac{1}{4} \sum_{i} (C (e_{i}) C (T e_{i}) + C (T e_{i}) C (e_{i})) = \frac{1}{2} \sum_{i} (e_{i}, T e_{i}) I = 0.

Each scalar vanishes because

T

is skew-adjoint. This proves

π (T)^{*} = - π (T)

For the requested commutator, the first identity gives

[π (T), C (v)] = \frac{1}{2} \sum_{i} ((e_{i}, v) C (T e_{i}) - (T e_{i}, v) C (e_{i})) .

The first sum is

C (T v)

by the orthonormal basis expansion of

v

. For the second, the skew-adjoint property gives

\sum_{i} (T e_{i}, v) e_{i} = - T v

. By real linearity of

C

, the difference is therefore

[π (T), C (v)] = \frac{1}{2} (C (T v) - C (- T v)) = C (T v) .

This calculation explains the factor

1/4

V

act correctly on its Clifford generators.

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