Localization Loci, Zero Modes, and One-Loop Determinants

The localized answer is a sum or integral over every component of the zero set of the bosonic deformation, in every allowed topological sector. Nonzero fluctuations form a regulated superdeterminant of the gauge-fixed deformation complex; true zero modes instead become collective coordinates, stabilizer factors, or unsaturated fermionic insertions. This separation is the core of a correct one-loop formula.

Required background. Gauge fixing and the localization deformation complex supplies the graded operator whose cohomology and determinant are required. Heat kernels, zeta functions, and spectral determinants supplies spectral regularization.

Helpful background. Generalized Killing spinors and global spin-R bundles supplies flux sectors and fixed-point bundle data.

Deriving the localization locus

Choose a cycle $\Gamma$ and deformation for which the bosonic part can be expressed as a sum of nonnegative norms,

$$(\widehat QV)_{\rm bos} =\sum_a\lVert\widehat Q\Psi_a\rVert_\Gamma^2.$$

The localization equations are therefore

$$\widehat Q\Psi_a=0$$

for every fermion $\Psi_a$, together with gauge fixing and the chosen reality slice. This compact statement must be expanded before solving: auxiliaries may impose algebraic shifts, curvature changes the equations, and a complex contour can replace a real norm by a Morse function.

The full locus is a disjoint union

$$\mathcal M_Q=\bigsqcup_{\nu}\mathcal M_{Q,\nu},$$

where $\nu$ labels allowed bundle topology, magnetic flux, instanton number, holonomy sector, or defect charge. A perturbative solution in the trivial bundle does not prove that the other components vanish. On $S^4$, for example, the smooth bulk locus is accompanied by pointlike instanton and anti-instanton sectors at the fixed poles Pestun 2012, §§3.4 and 4.3.

Quadratic fluctuations and the superdeterminant

Near a saddle $\Phi_0\in\mathcal M_{Q,\nu}$, split fields into collective coordinates $m$, nonzero bosonic fluctuations $x$, and nonzero fermionic fluctuations $\psi$. The large-$t$ action has the schematic form

$$S_t=S_{\rm cl}(\Phi_0) +\frac t2\langle x,K_Bx\rangle +t\langle\bar\psi,K_F\psi\rangle+O(t^{-1/2}).$$

Gaussian integration gives, before regularization,

$$Z_{\rm 1-loop} =\frac{\det'K_F}{\sqrt{\det'K_B}},$$

or a Pfaffian when the fermionic quadratic form is antisymmetric. The prime removes every kernel mode. The square root and Pfaffian require an orientation and phase prescription; they are not fixed by the absolute eigenvalues.

Supersymmetry pairs most nonzero modes. If a first-order operator $D:E_0\to E_1$ controls the complex, then $D^\dagger D$ and $DD^\dagger$ have identical nonzero singular values. Their formal determinant quotient cancels, while

$$\ker D \quad\text{and}\quad \ker D^\dagger$$

remain. In an equivariant problem the unpaired kernels carry weights under $\mathcal H=\widehat Q^2$, and the one-loop answer is reconstructed from the equivariant index

$$\operatorname{ind}_{\mathcal H}D =\operatorname{Tr}_{\ker D}e^{\mathcal H} -\operatorname{Tr}_{\ker D^\dagger}e^{\mathcal H}.$$

Turning this character into an infinite product still requires a choice of expansion chamber and spectral regulator.

Zero modes are integrations, not determinants

Different kernels have different meanings:

• tangent zero modes to $\mathcal M_{Q,\nu}$ become collective coordinates with the induced measure $d\mu_\nu(m)$;
• gauge zero modes are divided by the volume of the actual stabilizer;
• bosonic noncompact moduli require a contour and convergence prescription;
• fermionic zero modes must be saturated by insertions or interactions, otherwise that sector vanishes;
• obstruction modes can change the local measure to an Euler class rather than a smooth volume form.

Consequently the general structure is

$$Z=\sum_\nu \int_{\Gamma_\nu\subset\mathcal M_{Q,\nu}} d\mu_\nu(m)\; e^{-S_{\rm cl}(m)} Z_{{\rm 1-loop},\nu}(m) Z_{{\rm nonpert},\nu}(m).$$

$Z_{\rm nonpert}$ records saddle structure not captured by Gaussian fluctuations, such as pointlike instantons. Its separation from $Z_{\rm 1-loop}$ is conventional only after the moduli-space compactification is specified.

Regulating the product

For a positive elliptic operator $K$ with zero modes removed, zeta regularization defines

$$\zeta_K(s)=\sum_{\lambda\ne0}\lambda^{-s}, \qquad \log\det'K=-\zeta_K'(0).$$

If eigenvalues are complex or the operator is first order, a spectral cut fixes $\lambda^{-s}$ and hence the phase. Heat-kernel, zeta, Pauli–Villars, and index-character prescriptions can differ by local counterterms when they preserve the same symmetries. Agreement of absolute values is not agreement of the quantum observable.

Several checks are especially powerful:

1. dimensions and dependence on the renormalization scale;
2. cancellation in a free supersymmetric multiplet with known spectrum;
3. behavior when a mass crosses zero and an eigenmode joins the kernel;
4. agreement between explicit harmonics and an index calculation;
5. decoupling of a very massive multiplet up to allowed local terms;
6. invariance under large gauge transformations and the global gauge group.

A concrete five-dimensional treatment develops the cohomological complex, its fixed locus, gauge fixing, and the determinant in Qiu and Zabzine 2017, §§3–5.

Common pitfalls

A primed determinant without a zero-mode measure. Removing a zero eigenvalue is only half the operation; state what replaces it.

The perturbative locus is called complete. Fluxes, reducible connections, pointlike instantons, and singular strata must be excluded by an argument or included in the sector sum.

The index is called the determinant. An index gives the virtual unpaired representation. Converting it to a product requires weights, a chamber, a regulator, and a phase.

Exercises

1. Singular-value cancellation. Let $D$ have nonzero singular values $s_n$. Show that

$$\frac{\det'^{1/2}(DD^\dagger)}{\det'^{1/2}(D^\dagger D)}=1$$

before phase effects.

Solution

Both positive operators have the same nonzero eigenvalues $s_n^2$ with the same multiplicities. Their regulated products therefore cancel when the same regulator is used. The statement says nothing about the unequal kernels or about phases of an underlying first-order fermion operator.

2. Fermion zero modes. Why does a saddle with one unsaturated Grassmann zero mode contribute zero to the partition function?

Solution

The zero-mode integral contains $\int d\eta\,1=0$. A contribution requires an insertion or interaction supplying a factor of $\eta$; then $\int d\eta\,\eta=1$.

References

• Pestun, Vasily. “Localization of Gauge Theory on a Four-Sphere and Supersymmetric Wilson Loops.” Communications in Mathematical Physics 313 (2012): 71–129. doi:10.1007/s00220-012-1485-0. Open preprint.
• Qiu, Jian, and Maxim Zabzine. “Review of Localization for 5D Supersymmetric Gauge Theories.” Journal of Physics A: Mathematical and Theoretical 50 (2017): 443014. doi:10.1088/1751-8121/aa8e4a. Open preprint.

Further reading

• Pestun, Vasily, and Maxim Zabzine, eds. “Localization Techniques in Quantum Field Theories.” Journal of Physics A: Mathematical and Theoretical 50 (2017), special issue. Foreword and chapter guide.

Next step

The formal finite-dimensional answer becomes a defined observable only after choosing complex contours, zero-mode prescriptions, and regularization.