 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-48/2/g/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

Vary

F_{i}

and

F_{i}^{*}

independently. The auxiliary fields have algebraic, rather than differential, Euler-Lagrange equations:

F_{i}^{*} + W_{i} = 0, F_{i} + \overline{W_{i}} = 0.

Alternatively complete the square:

L_{F} = \sum_{i} ∣ F_{i} + \overline{W_{i}} ∣^{2} - \sum_{i} ∣ W_{i} ∣^{2} .

Substitution therefore yields the on-shell auxiliary fields and nonnegative scalar potential

F_{i} = - \overline{W_{i}}, L_{F}^{on shell} = - V_{F}, V_{F} = i \sum ∣ W_{i} ∣^{2} .

The sign in the Lagrangian is minus the F-term scalar potential; changing the metric convention does not change this algebraic result. A supersymmetric vacuum requires all

W_{i} = 0

, namely F-flatness. For a noncanonical positive Kähler metric

K_{i \overset{ˉ}{j}}

, the purely bosonic elimination instead gives

V_{F} = K^{i \overset{ˉ}{j}} W_{i} \overline{W_{j}}

; fermionic connection terms must also be included in a complete nonlinear component action.

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