Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 304 3 Created 2026-10-03 Updated 2026-10-06
Set and , treating as independent coordinates for Wirtinger derivatives. Use left Grassmann derivatives, so . The given odd vector field acts by , , and zero on . Consequently its square vanishes on every coordinate. The square of an odd graded derivation is an even graded derivation, so the nilpotent operator property holds on every function:
For the action in this zero-dimensional supersymmetric field theory, and . The cubic term vanishes after multiplication by , and the other two terms cancel:Three further independent odd symmetries areFor , use ; the same cancellation proves invariance. The barred calculations exchange barred and unbarred variables and coefficients. Each graded derivation is also a nilpotent operator, by its coordinate action. Their different odd coefficients and coordinate derivatives make them linearly independent. These are odd symmetries of a zero-dimensional polynomial model.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 304 3 ii Solution Created 2026-10-03 Updated 2026-10-06
The odd symmetries of a zero-dimensional polynomial model preserve both the action and the flat integration measure: their coefficients have zero superdivergence. The confining polynomial weight removes boundary terms at infinity. Thus the integral of an odd-symmetry derivative is zero, a supersymmetric Ward identity for this finite-dimensional integral in this zero-dimensional supersymmetric field theory.
Since is a holomorphic function, the barred symmetry obeys . Applying the supersymmetric Ward identity gives the requested unnormalized insertion:One can also verify the result after Berezin integration, without invoking the symmetry terminology:The final integration by parts uses and rapid decay. The derivative on in the insertion is essential.