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 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.

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.

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.

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.

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.

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

For one generation, the Standard Model representation content under $SU(2{)}_L\times SU(3{)}_C$ iswith no right-handed neutrino. For $SU(3{)}_C$, three generations give $n_L=6$ from the two left-handed quark flavors and $n_R=6$ from the two right-handed quark flavors, while $n_s=0$. ThereforeFor $SU(2{)}_L$, each generation supplies three colored quark doublets and one lepton doublet, so $n_L=12$, $n_R=0$, and the single complex Higgs doublet gives $n_s=1$. Hence

Put $L=(2\pi {)}^{-1}\log (M_{\mathrm{GUT}}/m_Z)$. At the gauge coupling unification scale, write the common normalized coupling as ${\alpha }_U={\alpha }_2(M_{\mathrm{GUT}})={\alpha }_3(M_{\mathrm{GUT}})=(5/3){\alpha }_1(M_{\mathrm{GUT}})$. Running downward givesSubtracting the second equation from the third determinesEliminating $L$ from the first two equations yields

Direct integration between the two renormalization scales gives

Integrating the renormalization-group beta function and defining $\Lambda$ by ${\alpha }_3^{-1}(\Lambda )=0$ givessoThe scale $\Lambda$ is the QCD scale in this one-loop approximation.

Since ${\alpha }_i=g_i^2/(4\pi )$, the stated renormalization-group beta function impliesFor ${\beta }_i>0$, the running coupling decreases at high energy: the theory is asymptotically free and becomes strong in the infrared, as in Quantum chromodynamics. For ${\beta }_i<0$, it is infrared free but grows toward an ultraviolet Landau pole.

The result grows linearly with the squared centre-of-momentum energy, $\sigma \propto G_F^{2}s$. This growth eventually violates partial-wave unitarity and signals the breakdown of the pointlike Fermi interaction. At energies comparable to $M_W$ or $M_Z$, the full gauge-boson propagators must replace the contact interaction; the renormalizable electroweak theory then softens the high-energy behavior.

A Fierz rearrangement puts the charged-current operator in the same current ordering as the neutral-current operator. Accounting for the interchange of fermionic fields, the combined amplitude isSum over final spins and average over the initial electron spin. The fermion spin sum and gamma-matrix trace identities giveFor massless two-body scattering in the centre-of-momentum frame, $d\sigma /dt=\overline{|\mathcal{M}{|}^2}/(16\pi s^2)$ and $u=-s-t$. Integrating $-s\le t\le 0$ therefore yieldsConsequently

The first four-fermion operator is the low-energy Fermi interaction obtained by replacing the crossed $W$ propagator in the charged-current diagram by $-g_{\mu \nu }/M_W^2$. The second operator is obtained analogously from the $Z$-exchange weak neutral current; $c_V$ and $c_A$ are its electron vector and axial-vector couplings. Thus the two terms represent respectively the charged-current and neutral-current diagrams in part (v), with their interference retained when the amplitude is squared.