Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 307 (Supersymmetry)

For the stated two-dimensional N=(2,2) supersymmetry conventions, define the supersymmetric covariant derivativesThe terms in which a Grassmann derivative hits the explicit Grassmann coordinate cancel the spacetime-derivative terms, while derivatives involving different signs act on independent coordinates. Therefore

The twisted chiral superfield constraints $D_{-}U={\bar{D}}_{+}U=0$ are solved by the twisted chiral coordinatesThe superfield depends only on $({\tilde{y}}^{\pm };{\theta }^{+},{\bar{\theta }}^{-})$ and has the finite Grassmann variable expansionwhere every component on the right is evaluated at $\tilde{y}$; numerical $\sqrt{2}$ factors may be absorbed into the component definitions.

For one chiral superfield $\Phi$ and one twisted chiral superfield $U$, the most general local two-derivative supersymmetric action isHere $K$ is real, $W$ is a holomorphic superpotential, and $\tilde{W}$ is a holomorphic twisted superpotential. A full superspace integral, a chiral F-term, and a twisted F-term each vary by a spacetime or Berezin total derivative, so all three are supersymmetric.

Choose the Vector R-symmetry and Axial R-symmetry conventionswith conjugate coordinates transforming oppositely, and assign compatible charges to $\Phi$ and $U$. The D-term is invariant when $K$ is neutral, up to a generalized Kähler transformation. The chiral measure has vector R-charge $-2$ and axial charge zero, whereas the twisted chiral measure has axial R-charge $-2$ and vector charge zero. Hence classical invariance requirestogether with neutrality of $K$. Equivalently, $W$ and $\tilde{W}$ must be quasi-homogeneous with these charges; absent suitable charge assignments, the corresponding superpotential breaks that R-symmetry.

The first-order kinetic term makes ${\eta }^a$ and ${\bar{\eta }}_b$ canonically conjugate Grassmann variables. Canonical quantization gives the canonical anticommutation relations

The Hilbert space is the fermionic Fock spaceIn a polarization where ${\hat{\eta }}^a$ acts by exterior multiplication and ${\hat{\bar{\eta }}}_a$ by contraction, a general state isThe fermion number operator $N={\hat{\eta }}^a{\hat{\bar{\eta }}}_a$ returns the exterior degree, and therefore

The coherent-state representation of an ordinary fermionic trace identifies the endpoints with a minus sign, producing antiperiodic fields. Inserting fermion parity $(-1{)}^N$ supplies a second minus sign. The resulting fermionic path integral therefore hasand its time-sliced coherent-state action is precisely $S[\eta ,\bar{\eta }]$. This proves the stated supertrace representation.

Diagonalize $T$ with eigenvalues ${\lambda }_a$ and use the periodic Fourier seriesEach Berezin integral contributes its quadratic coefficient, soPairing $k$ with $-k$ and using the infinite product for the hyperbolic sine gives, up to the local regularization factor,The product of the prefactors is $e^{\mathrm{tr}T/2}=1$ because $T$ is traceless. ThusThis agrees with the direct identity that the supertrace of an induced linear map on an exterior algebra is its characteristic determinant.

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.