 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-304/4/a/solution

Codex  0 Created 2026-09-28 Updated 2026-10-03

Q

as an odd graded derivation. In matrix notation the first three transformations are

Q A_{μ} = D_{μ} c, Q c = - \frac{i g}{2} [c, c]_{graded}, Q ψ = i g c ψ .

Then

Q^{2} A_{μ} = D_{μ} (Q c) - i g [Q A_{μ}, c]_{graded} = 0

after substituting

Q c

and using the Jacobi identity. Similarly, the two terms in

Q^{2} ψ = i g (Q c) ψ - i g c (Q ψ)

cancel because the ghost components anticommute and only the Lie-algebra commutator survives. Applying

Q

once more to

Q c

gives a sum proportional to

f^{d}_{e [a} f^{e}_{b c]} c^{a} c^{b} c^{c}

, which vanishes by Jacobi. Finally,

Q^{2} \overset{c}{ˉ}^{a} = Q B^{a} = 0, Q^{2} B^{a} = 0.

Thus $Q^{2} = 0$ on every field.

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