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Analytically continue the free-particle return kernel using $M=I$ and $t=-i\hslash \beta$. This fixesSet $a=2{\pi }^{2}I/({\hslash }^2\beta )$. The Gaussian Fourier integral in Poisson resummation isApplying it to the winding sum gives ${\sum }_m{e}^{-a{m}^2}=\sqrt{\pi /a}{\sum }_n{e}^{-{\pi }^2{n}^2/a}$. The prefactors cancel exactly, $2\pi {\mathcal{Z}}_0\sqrt{\pi /a}=1$, and ${\pi }^2/a=\beta {\hslash }^2/(2I)$. Thereforein agreement with the spectral trace. This is momentum-winding duality of the quantum rotor. At low temperature the momentum ground state dominates; at high temperature the winding-zero sector gives $\mathcal{Z}\simeq \sqrt{2\pi I/(\beta {\hslash }^2)}$.

Analytically continue the supplied free propagator using $M=I$ and $t=-i\hslash \beta$. This gives $Z_0=\sqrt{I/(2\pi {\hslash }^2\beta )}$ without an arbitrary measure constant.

Apply the Poisson summation formula to $h(x)=e^{-A{x}^2}$, with $A=2{\pi }^{2}I/({\hslash }^2\beta )$. Its Gaussian integral isThe prefactors cancel exactly because $2\pi Z_0\sqrt{\pi /A}=1$. Therefore the momentum-winding duality of the quantum rotor givesin agreement with the spectrum. The thermal winding dominance of a free rotor interchanges between the two representations. At low temperature the momentum sum is dominated by n=0 and $Z\to 1$. At high temperature the winding sum is dominated by m=0 and $Z\sim \sqrt{2\pi I/(\beta {\hslash }^2)}$, the classical angular phase-space integral. Winding and momentum labels are dual descriptions; keeping only m=0 at low temperature would be incorrect.