Localization in a field theory

The finite model gave an exact answer because an odd symmetry converted the deformation derivative into a boundary term. A field theory adds infinitely many variables, gauge redundancy and a choice of quantum prescription. None of these may be hidden inside the phrase “a total derivative.”

The general argument below needs only that finite example and what a supercharge does. It applies to either three- or four-dimensional theories once their particular symmetry is supplied. A later, explicitly separate example uses the free chiral sphere action.

Define the integral before deforming it

Let $\Phi$ denote all integration fields, including auxiliaries and, in a gauge theory, gauge-fixing fields. Work in Euclidean signature. Choose an odd operation $Q$ with

$$Q^2=\mathcal B,$$

where $\mathcal B$ is a specified bosonic symmetry: it may combine a spacetime isometry with gauge, R-symmetry or flavor transformations. The coefficients and allowed global parameters belong to the chosen theory and background. This equation does not assign the same supersymmetry algebra to every dimension.

Let $S$ be the action, $\mathcal V$ an odd functional and $\mathcal O$ an even insertion. The algebraic requirements are

$$QS=0,\qquad \mathcal B\mathcal V=0,\qquad Q\mathcal O=0. \tag{O02.1}$$

Then $\mathcal I=Q\mathcal V$ is even and $Q\mathcal I=0$. Closure of $S$ alone does not establish that a proposed deformation is exact; its preimage $\mathcal V$ must be exhibited. The special case $\mathcal B=0$ is allowed.

Introduce a regulator labeled $\Lambda$, a bosonic integration contour $\Gamma_\Lambda$, and a measure $D_\Lambda\Phi$. The odd variables have an ordered Berezin measure. Define

$$Z_\Lambda(t)= \int_{\Gamma_\Lambda}D_\Lambda\Phi\, e^{-S_\Lambda-t\mathcal I_\Lambda}, \qquad t\geq0. \tag{O02.2}$$

Here $t$ is dimensionless; field scales and the normalization of $Q,\mathcal V$ make $\mathcal I_\Lambda$ dimensionless. The measure, contour, background sources and observable are held fixed as $t$ varies.

The regulated symmetry must obey the integrated identity

$$\int_{\Gamma_\Lambda}D_\Lambda\Phi\,QX=0 \tag{O02.3}$$

for the particular decaying functionals $X$ used below. This requires more than a classical transformation law. A regulator must preserve the chosen symmetry, and its measure must have the appropriate Jacobian. There must be no surviving boundary flux, including at large field amplitude or a singular excluded configuration. If a complex contour is used, justify the permitted change of variables or contour deformation; imposing a formal reality condition is not that justification.

In a gauge theory, gauge fixing and ghosts enter this same test. A physical supersymmetry can be combined with the odd gauge-fixing symmetry, with its ghost transformations chosen so that the combined operation closes. It is not enough to add an arbitrary ghost action and assume that it is $Q$-invariant. Pestun, version 3, sections 4.1 and 4.3 gives an explicit construction.

Derive the identity for the quantity being computed

Write $S_{\Lambda,t}=S_\Lambda+tQ\mathcal V_\Lambda$. Assuming differentiation under the integral is legitimate,

$$\begin{aligned} \frac{dZ_\Lambda}{dt} &=-\int D_\Lambda\Phi\,(Q\mathcal V_\Lambda)e^{-S_{\Lambda,t}}\\ &=-\int D_\Lambda\Phi\, Q\!\left(\mathcal V_\Lambda e^{-S_{\Lambda,t}}\right)=0. \end{aligned} \tag{O02.4}$$

The second line uses $QS_{\Lambda,t}=0$; the last uses (O02.3). The same argument with $\mathcal O\mathcal V_\Lambda$ proves independence of the numerator with a closed, parameter-independent insertion. This is the general pattern in Pestun’s introduction, pp. 5–6, with the sign here fixed by $e^{-S}$.

For a normalized expectation value with $Z_\Lambda\ne0$, direct differentiation gives the more informative formula

$$\frac d{dt}\langle\mathcal O\rangle_t =-\langle\mathcal O\,Q\mathcal V\rangle_t +\langle\mathcal O\rangle_t\langle Q\mathcal V\rangle_t. \tag{O02.5}$$

Both terms vanish by the Ward identity under the stated hypotheses. A general insertion need not be protected. If $\mathcal O$ itself depends on $t$, there is an additional $\langle\partial_t\mathcal O\rangle_t$ term. Changing a physical mass or geometry can also change $Q$, the action’s closedness conditions or the contour; it is not automatically the deformation in (O02.2).

One can keep a failure visible. Define the Ward defect $\mathcal W_\Lambda[X]=\int D_\Lambda\Phi\,QX$. Then the derivative in (O02.4) is $-\mathcal W_\Lambda[\mathcal V e^{-S_{\Lambda,t}}]$, rather than zero. A boundary term or a noninvariant regulated measure must be evaluated or repaired, not discarded. Compact spacetime does not make the space of field amplitudes compact.

Finally, removing $\Lambda$ requires a renormalized measure and counterterm prescription compatible with these identities. Independence at every finite cutoff does not alone prove that a finite continuum answer exists. Do not interchange a large-deformation limit and a regulator limit without controlling them.

What remains after a large deformation

Suppose the ordinary part of $\mathcal I$ on the specified contour is nonnegative and suppresses directions away from its zero set $\mathcal M$. Determine that set for the actual deformation. When $\mathcal I_{\rm bos}$ is a sum of norms of fermion variations, its zero equations are the corresponding supersymmetry conditions. This relation needs to be checked; a list of classical solutions of the original action is not automatically the localization locus.

Let $u$ label a smooth component of $\mathcal M$. Separate its tangent zero modes from normal fluctuations and write

$$\Phi=\Phi_0(u)+t^{-1/2}\eta_\perp.$$

For a nondegenerate normal quadratic form,

$$t\mathcal I[\Phi] =\mathcal I^{(2)}_u[\eta_\perp]+O(t^{-1/2}). \tag{O02.6}$$

The original action approaches $S[\Phi_0(u)]$ along these normal fluctuations. Under the required convergence assumptions, the result takes the form

$$Z=\sum_\alpha\int_{\mathcal M_\alpha}d\mu(u)\, e^{-S[\Phi_0(u)]}\,Z_{\rm quad}(u). \tag{O02.7}$$

The sum includes every contributing component or sector of the specified contour. $d\mu$ includes the zero-mode and gauge-quotient factors; $Z_{\rm quad}$ is the Gaussian integral over the normal fluctuations and ghosts. Fermionic zero modes also need treatment: an unsaturated Berezin zero mode can make a contribution vanish. A singular or noncompact component needs more analysis than the smooth approximation just written.

For example, one complex bosonic mode and one independent fermion pair give, with the displayed measure ordering,

$$\int\frac{d^2c}{\pi}\,e^{-t\lambda|c|^2} =\frac1{t\lambda},\qquad \int d\bar\eta\,d\eta\,e^{-t\omega\bar\eta\eta} =t\omega. \tag{O02.8}$$

Thus their product is $\omega/\lambda$, not generally one. Many modes produce determinant ratios; real Gaussian variables have the corresponding real Gaussian factors. Zero eigenvalues must be separated before forming an inverse determinant. Infinite products require a regulator, a phase prescription and possible local counterterms.

The fluctuation calculation is called the one-loop contribution. Deformation independence can make its large-$t$ limit exact; it does not imply that quantum effects vanish or that every saddle is the vacuum. Willett, version 3, (3.4)–(3.10) implements this separation in a field theory.

Optional example: an exact sphere deformation

This section requires the free chiral sphere action. We now specialize to one gauge-neutral three-dimensional $\mathcal N=2$ chiral, with four real Lorentzian Poincaré supercharges before Euclidean complexification, zero central mass and no superpotential. The round background preserves that complexified rigid algebra. There is no dynamical gauge group. Take

$$A=V=C=0,\qquad H=-\frac i\ell,\qquad R=-\frac6{\ell^2},\qquad r=\frac12,\qquad \alpha=\ell^{-1}.$$

The action already derived there is

$$S_\chi=\int_{S^3}\sqrt g\,d^3x\, \left( \nabla_\mu\widetilde\phi\nabla^\mu\phi -i\widetilde\psi\gamma^\mu\nabla_\mu\psi -\widetilde F_\Phi F_\Phi +\frac{3\alpha^2}{4}\widetilde\phi\phi \right). \tag{O02.9}$$

The fields have engineering dimensions $[\phi]=1/2$, $[\psi]=1$ and $[F_\Phi]=3/2$, with opposite R-charges for their independent tilded partners. Every action term has mass dimension three. Retain the spinor contraction $uv=u^TCv$, where $C=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$.

Use the independent commuting spinor labels $\zeta,\widetilde\zeta$ of that lesson, chosen with nonzero constant contraction $s=\zeta\widetilde\zeta$. Their components are constant in the global sphere frame because its spin connection gives $\nabla_\mu\zeta=i\alpha\gamma_\mu\zeta/2$, and likewise for $\widetilde\zeta$. A covariant derivative of a label is therefore not zero.

Choose the odd operation $Q=\delta_\zeta$, and write $\widetilde Q=\delta_{\widetilde\zeta}$ for the independent second operation, not a Hermitian adjoint. Here $Q^2=0$ off shell: this is the nilpotent specialization of the general bosonic square. For example, $Q^2\phi=2(\zeta\zeta)F_\Phi=0$, since a commuting spinor has vanishing antisymmetric self-contraction. Keeping the auxiliaries makes the same statement hold on the other fields.

Set $K^\mu=\zeta\gamma^\mu\widetilde\zeta$. The Killing equations give $\nabla_\mu K_\nu=\alpha\epsilon_{\mu\nu\rho}K^\rho$ and $\nabla_\mu K^\mu=0$. Consider the neutral even expression

$$G=\widetilde\psi\psi-\alpha\widetilde\phi\phi, \qquad \mathcal V=\frac1{2s}\widetilde Q \int_{S^3}\sqrt g\,d^3x\,G . \tag{O02.10}$$

Its first variation, with the odd product rule, is

$$\frac{\widetilde QG}{\sqrt2} =\widetilde F_\Phi(\widetilde\zeta\psi) +\frac{3\alpha}{2}\phi(\widetilde\zeta\widetilde\psi) +i(\nabla_\mu\phi) (\widetilde\zeta\gamma^\mu\widetilde\psi). \tag{O02.11}$$

Applying $Q$ gives the action plus a definite current:

$$\frac12 Q\widetilde QG=s\mathcal L_\chi+\nabla_\mu J^\mu, \tag{O02.12}$$ $$J^\mu= \frac{3i\alpha}{2}K^\mu\widetilde\phi\phi -i\epsilon^{\nu\mu\rho}K_\rho (\nabla_\nu\phi)\widetilde\phi -i(\zeta\gamma^\mu\widetilde\psi) (\widetilde\zeta\psi).$$

Here are the cancellations behind this identity. The auxiliary product is $-s\widetilde F_\Phi F_\Phi$. The two scalar derivatives give $s\nabla\widetilde\phi\nabla\phi -i\epsilon^{\nu\mu\rho}K_\rho(\nabla_\nu\phi)(\nabla_\mu\widetilde\phi)$. Integrating the second term by parts differentiates $K$ and produces $2i\alpha K^\mu(\nabla_\mu\phi)\widetilde\phi$. It combines with the remaining first derivatives into the first two terms of $J^\mu$. The scalar algebraic coefficient is $(\alpha/2)(3\alpha/2)=3\alpha^2/4$.

For the fermions, move the derivative on $\widetilde\psi$ by parts, retaining the derivatives of both labels. The completeness identity $\sum_a(\gamma_a)_i{}^j(\gamma_a)_k{}^l =2\delta_i{}^l\delta_k{}^j-\delta_i{}^j\delta_k{}^l$ combines the derivative terms into $-is\widetilde\psi\gamma^\mu\nabla_\mu\psi$. To see the algebraic cancellation, write $X=(\zeta\widetilde\psi)(\widetilde\zeta\psi)$, $Y=(\zeta\psi)(\widetilde\zeta\widetilde\psi)$ and $k=i\alpha/2$. The same completeness identity, with the odd interchange signs, gives

$$\sum_a\left[ (\zeta\gamma_a\widetilde\psi)(\widetilde\zeta\gamma_a\psi) +(\zeta\gamma_a\psi)(\widetilde\zeta\gamma_a\widetilde\psi) \right]=-3(X+Y).$$

The residual algebraic expression is therefore $\tfrac{3\alpha}{2}Y-3ikX+3ik(X+Y)=0$. The integration by parts leaves the last term of $J^\mu$. These are identities of Grassmann polynomials, without a matter field equation.

All currents are global and smooth on the closed sphere. Consequently

$$Q\mathcal V=S_\chi. \tag{O02.13}$$

This proves exactness in the adopted action convention. Willett’s sphere example uses the opposite sign of the Killing-spinor connection and $H$; its component signs cannot be copied into (O02.9). Here they follow from the stated sphere transformations.

Check the contour and a complete finite cutoff

On the bosonic contour $\widetilde\phi=\phi^*$, $\widetilde F_\Phi=-F_\Phi^*$,

$$(S_\chi)_{\rm bos} =\int\sqrt g\,d^3x\, \left(|\nabla\phi|^2+\frac{3}{4\ell^2}|\phi|^2+|F_\Phi|^2\right). \tag{O02.14}$$

It is strictly positive away from $\phi=F_\Phi=0$. This identifies the sole bosonic zero of this deformation; exactness came from (O02.10)–(O02.13), not from positivity.

A concrete free-theory regulator retains scalar harmonics through a fixed degree in every global-frame component of the multiplet. Frame derivatives act within each scalar-harmonic degree. The constant label and gamma matrices, together with the constant spin connection, therefore preserve this finite space under $Q$. Keeping arbitrary unrelated scalar and spinor cutoffs would not have this property.

The regulated transformation is linear and off diagonal between ordinary and odd variables, so its infinitesimal measure supertrace vanishes. On this contour a complex derivative is an ordinary combination of real derivatives: for $c=u+iv$, $\partial_c=\tfrac12(\partial_u-i\partial_v)$. Polynomial times Gaussian boundary terms vanish at infinity; Berezin derivatives integrate to zero. Thus the finite-dimensional Ward step is justified directly.

We have not imposed a conjugation relation on fermions or claimed that the complex odd transformation respects the bosonic conjugation equations term by term. Its action on the contour integral is justified by these complex-linear integration-by-parts identities. This suffices for the free finite cutoff; a general interacting contour requires its own analysis.

In particular $S_t=(1+t)S_\chi$. If $N$ scalar harmonic functions are retained, $\phi,F_\Phi$ supply $2N$ complex ordinary coefficients and $\psi,\widetilde\psi$ supply $4N$ odd coefficients. Rescaling every coefficient by $(1+t)^{-1/2}$ gives measure factors $(1+t)^{-2N}$ and $(1+t)^{2N}$, which cancel.

For an explicit value, retain only the normalized constant harmonic. Write its scalar and auxiliary coefficients as $c,f$; set $\eta=\widetilde\psi^TC$ for the two independent row coefficients. The spin connection gives $\gamma^\mu\nabla_\mu\psi=3i\psi/(2\ell)$ even for constant components. Hence

$$S_{t,0}=a\left(\lambda|c|^2+|f|^2+ B\sum_{i=1}^2\eta_i\psi_i\right), \quad a=1+t,\quad \lambda=\frac3{4\ell^2},\quad B=\frac3{2\ell}.$$

With $d^2c/\pi$, $d^2f/\pi$ and ordered odd measure $d\eta_2\,d\psi_2\,d\eta_1\,d\psi_1$,

$$Z_0(a)=\frac1{a\lambda}\frac1a(aB)^2=3, \qquad \langle|c|^2\rangle=\frac1{a\lambda}. \tag{O02.15}$$

The full constant multiplet has an invariant partition integral and a changing unprotected scalar moment. Its answer is not the full sphere partition function. Determining the regulated product of all harmonic contributions, including its phase and normalization, is a separate task.

Practice

1. A constant answer for a changing insertion

In the finite model at $g=0$, take the explicitly parameter-dependent insertion $\mathcal O_t=a\rho$, where $a=1+t$. Calculate its expectation value and both terms in $d\langle\mathcal O_t\rangle/dt =\langle\partial_t\mathcal O_t\rangle -\operatorname{Cov}(\mathcal O_t,QV)$. Does the result establish that a fixed Q-closed observable was being measured?

Hint 1

Use $QV=\rho-2\psi_1\psi_2$ and the ordered measure of the finite lesson. The odd part contributes to the covariance.

Hint 2

The needed moments are $\langle\rho\rangle=1/a$, $\langle\rho^2\rangle=2/a^2$ and $\langle\rho\psi_1\psi_2\rangle=1/(2a^2)$.

Solution

The expectation is $\langle a\rho\rangle=1$. Also $\langle QV\rangle=0$, while

$$\langle\mathcal O_tQV\rangle =a\left(\frac2{a^2}-\frac1{a^2}\right)=\frac1a.$$

The explicit derivative contributes $\langle\partial_t\mathcal O_t\rangle=\langle\rho\rangle=1/a$. It cancels the covariance, giving zero. But $Q\mathcal O_t=2a(x\psi_1+y\psi_2)\ne0$, and the insertion changes with $t$. A constant expectation value can occur without the fixed-observable protection in (O02.1). Omitting either the explicit derivative or the odd part of $QV$ gives the wrong explanation.

2. A continuous localization locus

Add an angle $\theta\sim\theta+2\pi$ to the finite model, with $Q\theta=0$ and measure $d\theta/(2\pi)$. Let the action be $aQV$, independent of $\theta$, with the original normalized transverse measure. Find the localization locus, $Z$, and $\langle\cos^2\theta\rangle$. What changes if $\theta$ instead ranges over $\mathbb R$ with ordinary Lebesgue measure?

Hint 1

The deformation still suppresses $x,y$, but it has no quadratic term in $\theta$.

Hint 2

Evaluate the transverse integral first, and retain the integral over the entire zero set.

Solution

The locus is $x=y=0$ with arbitrary $\theta$: a circle. The transverse integral is one, leaving

$$Z=\int_0^{2\pi}\frac{d\theta}{2\pi}=1,\qquad \langle\cos^2\theta\rangle =\int_0^{2\pi}\frac{d\theta}{2\pi}\cos^2\theta=\frac12.$$

Choosing only $\theta=0$ would give the incorrect insertion value one. The zero mode belongs to the remaining integral, not to an inverse Gaussian determinant. On the noncompact line, the partition integral has an infinite factor $\int_{\mathbb R}d\theta$. The transverse cancellation does not remove it; a finite normalized expectation then requires additional infrared data or a prescription.

3. An auxiliary saddle and its measure

Use the optional sphere example. Someone sets the constant auxiliary coefficient $f=0$ using its classical equation, discards its integral, and keeps the two fermion pairs. Compute the resulting $a$ dependence. How does actually integrating out $f$ repair it?

Hint 1

Keep the scalar and fermion factors in (O02.15), and remove only $1/a$.

Hint 2

The saddle location is zero, but the Gaussian around it has a width and a normalization.

Solution

The proposed expression is

$$Z_{\rm omitted}(a)=\frac1{a\lambda}(aB)^2=3a.$$

Its derivative is three. Exact auxiliary integration instead gives $\int d^2f\,e^{-a|f|^2}/\pi=1/a$, restoring $Z_0=3$. Deleting a variable at its saddle is not integrating it out. An on-shell formulation must also translate the transformation algebra and inherited measure; the off-shell Ward proof cannot simply be carried over after this deletion.

Before using localization in another theory, identify the actual $Q$, its square, the preimage of the deformation, the integration prescription, the full zero set and the remaining Gaussian problem. Continue the sphere example with the free chiral determinant: include every harmonic block and define its ultraviolet subtraction. Another theory may also require gauge zero modes or additional saddle sectors. The general identity tells us why a justified simplification preserves the chosen observable, and leaves those concrete computations to be done.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading three-dimensional localization with matter