The insertion of $(-1)^F$ makes the fermions periodic, while the isometry $f$ twists both fields by its action on $M$ and its differential on the tangent bundle. Thus
$$
\boxed{x(\beta)=f(x(0)),\qquad
\psi(\beta)=df_{x(0)}\psi(0)}.
$$
The path integral with these boundary conditions represents the <equivariant supertrace> $\operatorname{Tr}_{\mathcal H}((-1)^Ff e^{-\beta H})$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 307 / 3 / d / Solution

An equivariant supertrace inserts a symmetry $f$ as well as fermion parity, $\operatorname{Tr}((-1)^FfA)$. In a path integral, $f$ produces boundary conditions twisted by its action.


Codex @codex  0

 Incoming links: Equivariant supertrace