= Solution
The Euclidean saddle in winding sector $m$ is a constant-speed lifted path, $\phi_{\rm cl}(\tau)=\phi_0+2\pi m\tau/\mathcal T$. Write $\phi=\phi_{\rm cl}+\eta$ with Dirichlet fluctuation endpoints $\eta(0)=\eta(\mathcal T)=0$. The cross term in the kinetic action is proportional to $\int\partial_\tau\eta\,d\tau=0$, and the flux term of $\eta$ vanishes for the same reason. Therefore
$$
\frac{S_E}{\hbar}=\frac{2\pi^2I}{\hbar^2\beta}m^2-i2\pi fm+\frac I{2\hbar}\int_0^{\mathcal T}(\partial_\tau\eta)^2d\tau.
$$
Define the fluctuation prefactor by
$$
Z_0=\int_{\eta(0)=\eta(\mathcal T)=0}\mathcal D\eta\,\exp\left[-\frac I{2\hbar}\int_0^{\mathcal T}(\partial_\tau\eta)^2d\tau\right].
$$
It is the free-line diagonal Euclidean propagator for the angular coordinate with inertia $I$, or equivalently the Gaussian bridge amplitude about a given classical lifted path. It is independent of $m,\phi_0$ and flux. It is \b[not] the complete zero-flux ring <partition function>, which still includes every winding and the initial-angle integral. Translational independence supplies the factor $\int_0^{2\pi}d\phi_0=2\pi$, leaving
$$
\boxed{Z=2\pi Z_0\sum_{m\in\mathbb Z}\exp\left[-\frac{2\pi^2I}{\hbar^2\beta}m^2+i2\pi fm\right].}
$$
The <momentum-winding duality of a quantum rotor> uses a Gaussian expansion that is exact here because the kinetic action is quadratic and the flux coupling is a total derivative. It is not merely a leading semiclassical approximation.
\Image[/past-exam-of-the-quantum-condensed-matter-field-theory-course-of-the-university-of-cambridge/2016/paper-1-flux-ring-paths.png]
{title=Lifted winding sectors on a flux-threaded ring and the periodic flux dependence of quantum energy levels}
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