= Solution
The equivariant supertrace is independent of $\beta$ because positive-energy bosonic and fermionic states pair under the supercharge. Take the <short-time limit> $\beta\to0$. A finite-action path becomes constant, but the twisted boundary condition then requires $x=f(x)$; its constant saddles are exactly the <fixed-point set> $M^f$. Supersymmetry cancels the nonzero bosonic and fermionic fluctuations away from their zero modes. Consequently the path integral localizes to a tubular neighborhood of $M^f$, with the remaining Gaussian determinants giving the local fixed-point contribution to the <equivariant index theorem>.
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