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

Expand $X^a=x^a+\theta {\psi }^a$, $D{X}^a={\psi }^a-i\theta {\dot{x}}^a$, and $g_{ab}(X)=g_{ab}(x)+\theta {\psi }^c{\partial }_c{g}_{ab}$. Performing the Berezin integral givesAfter a fermionic integration by parts, this is the manifestly covariant expressionUnder a diffeomorphism, $x$ is a point of $M$, $\psi \in T_{x}M$ is a tangent vector, $D_t\psi$ is its covariant derivative along a curve, and every index is contracted with the Riemannian metric; the action is therefore invariant.

Substitution of $\delta x^a=ϵ{\psi }^a$ and $\delta {\psi }^a=-iϵ{\dot{x}}^a$ into the covariant component action leaves a total time derivative; the connection-dependent terms cancel by metric compatibility and the symmetry of the Levi-Civita connection. The boundary term vanishes for the stated decay. The Noether charge isup to the overall convention inherited from the supersymmetry parameter. Its graded square gives the Hamiltonian, $\{Q,Q\}=2H$ after canonical quantization.

The quantized fermions obey a Clifford algebra, $\{{\hat{\psi }}^a,{\hat{\psi }}^b\}=g^{ab}$. When $M$ is a spin manifold, they act on the spinor bundle andthe Hilbert space of square-integrable spinor fields. The supercharge becomes the Dirac operator, the Hamiltonian is one half of its square, and because $M$ is even dimensional, $(-1{)}^F$ is the spinor chirality operator.

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. ThusThe path integral with these boundary conditions represents the equivariant supertrace ${\mathrm{Tr}}_{\mathcal{H}}((-1{)}^{F}f{e}^{-\beta H})$.

The equivariant supertrace is independent of $\beta$ because positive-energy bosonic and fermionic states pair under the supercharge. Take the short-time limit $\beta \to 0$. 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.