Diagonalize $T$ with eigenvalues $\lambda_a$ and use the periodic <Fourier series>
$$
\eta^a(t)=\sum_{k\in\mathbb Z}\eta_k^ae^{2\pi ikt},
\qquad
\bar\eta^a(t)=\sum_{k\in\mathbb Z}\bar\eta_k^ae^{-2\pi ikt}.
$$
Each <Berezin integral> contributes its quadratic coefficient, so
$$
Z=\prod_{a=1}^n\prod_{k\in\mathbb Z}(2\pi ik+\lambda_a).
$$
Pairing $k$ with $-k$ and using the infinite product for the hyperbolic sine gives, up to the local regularization factor,
$$
\prod_{k\in\mathbb Z}(2\pi ik+\lambda)
=2\sinh\frac\lambda2
=e^{\lambda/2}(1-e^{-\lambda}).
$$
The product of the prefactors is $e^{\operatorname{tr}T/2}=1$ because $T$ is traceless. Thus
$$
\boxed{Z=\prod_a(1-e^{-\lambda_a})=\det(1-e^{-T})}.
$$
This agrees with the direct identity that the <supertrace> of an induced linear map on an exterior algebra is its characteristic determinant.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 307 / 2 / d / Solution

{title2=$\operatorname{Str}A=\operatorname{Tr}((-1)^FA)$}

A supertrace is the trace weighted by fermion parity. Supersymmetry makes paired positive-energy states cancel in a supertrace.


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