What localization shares across three and four dimensions

The four-sphere SYM calculation and the three-sphere ABJM calculation both replace an interacting path integral by an ordinary integral. What makes them instances of the same method, and what prevents us from transferring one answer to the other? We will compare their actual inputs and calculate a response that remains nontrivial after localization.

Begin with the finite-rank circular loop and the first ABJM partition-function checks. Their prerequisite chains contain the two derivations. Here we keep those derivations in their teaching homes and examine the choices they depend on.

The identity shared by both calculations

Write the Euclidean action as $S$ and choose an odd symmetry $Q$. For an odd functional $V$, consider

$$Z_t[\mathcal O]=\int_{\mathcal C}\mathcal D\Phi\, \mathcal O\,e^{-S-tQV},\qquad t\geq0. \tag{X01.1}$$

$\Phi$ denotes all fields, including auxiliary and gauge-fixing fields when required; $\mathcal C$ is the chosen integration cycle. The observable obeys $Q\mathcal O=0$. The deformation is invariant when $Q^2V=0$, whether or not $Q^2$ vanishes on every field. For an even observable, the graded product rule then gives

$$\frac{dZ_t[\mathcal O]}{dt} =-\int_{\mathcal C}\mathcal D\Phi\, Q\!\left(\mathcal O V e^{-S-tQV}\right)=0. \tag{X01.2}$$

The last equality is a Ward identity: a symmetry variation integrates to zero. It needs $QS=0$, an invariant measure and cycle, and control of boundary terms or limiting operations. These conditions are the substance of the argument in Pestun’s introduction, not consequences of the word “exact.” If both $Z_t[\mathcal O]$ and $Z_t[1]$ satisfy it and the denominator is nonzero, their ratio is also independent of $t$.

One chooses $QV$ so its bosonic part has a nonnegative real part on the selected contour. Large $t$ suppresses configurations away from its zero set. Fluctuations transverse to that set still contribute a determinant; gauge identifications still contribute a measure. The surviving variables are integrated, not automatically set to zero. Noncompact directions and an oscillatory final integral can require additional care even though spacetime is compact.

Compare the charges actually being used

For the three-dimensional calculation use the round $S^3$ of radius $\ell$, with the rigid sources of the ABJM background: $A_\mu=V_\mu=C_\mu=0$ and $H=-i/\ell$. The source paper’s scalar curvature is $R_P=-6/\ell^2$; the GR convention has $R_{\rm GR}=+6/\ell^2$. The two-form potential underlying $H$ is patched, rather than a global potential with nonzero integrated exterior derivative.

A definite-R-charge choice is $Q_3=\delta_\zeta$, with $\widetilde\zeta=0$ and $\nabla_\mu\zeta=i\gamma_\mu\zeta/(2\ell)$. In the global frame used by the sphere lessons, take the constant lower-index column $\zeta=(0,1)^{\mathsf T}$. Same-family anticommutators in P (6.1) vanish:

$$Q_3^2=0. \tag{X01.3}$$

This is the algebra on the off-shell covariant multiplets before any change in notation from gauge fixing. A single charge also appears in KWY §3.1 and in Willett’s deformation construction. The conjugate spinor used to build a positive deformation does not by itself add a second symmetry generator to $Q_3$.

There is nevertheless a Hopf direction in this geometry. Add the independent label $\widetilde\zeta=(1,0)^{\mathsf T}$ and define $Q_{\rm pair}=\delta_\zeta+\delta_{\widetilde\zeta}$. With $uv=u^{\mathsf T}Cv$, $C=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$, the normalized bilinears are $\zeta\widetilde\zeta=1$ and $K=\zeta\gamma^\mu\widetilde\zeta\,\partial_\mu=E_1$, where $E_1$ is a unit Hopf tangent. For a gauge-invariant scalar $\mathcal O_r$ of R-charge $r$ and zero central charge,

$$Q_{\rm pair}^2\mathcal O_r =-2iE_1^\mu\partial_\mu\mathcal O_r +\frac{2r}{\ell}\mathcal O_r. \tag{X01.4}$$

Gauge-charged components require the additional gauge transformation. Equations (X01.3) and (X01.4) describe different choices. In particular, the geometric Hopf direction associated with a spinor does not imply that the single-charge square is a Hopf rotation. Euclidean $\zeta$ and $\widetilde\zeta$ have not been constrained to be complex conjugates.

On $S^4_\ell$, use the precise conformal Killing spinor and seven-auxiliary-field completion constructed in the four-sphere lesson. Call that chosen operation $Q_4$. In its stereographic coordinates $x^\mu$, its spacetime vector is

$$v=\frac1\ell\left(-x^2\partial_1+x^1\partial_2 -x^4\partial_3+x^3\partial_4\right). \tag{X01.5}$$

In the source’s anti-Hermitian field convention its square is a rotation, a gauge transformation with parameter $\Lambda=v^\mu A_\mu^P+\phi_0^P-\cos\theta\,\phi_9^P$, and an internal rotation. Here $\theta$ is the polar angle and $\phi_A^P$ are the source’s six scalar components. The internal coefficients appearing in the square are $\mathcal J_{56}=1/\ell$, $\mathcal J_{78}=-1/\ell$, with antisymmetric partners. Thus on a gauge-invariant internal-singlet scalar, $Q_4^2=-\mathcal L_v$. There is no dilation in this selected square. The vector vanishes at the poles, even though the spinor itself does not vanish there. This selected off-shell completion does not furnish a finite auxiliary completion of every generator of full $\mathcal N=4$ SYM.

The notation $\mathcal N=2$ also counts different things in different dimensions. The manifest flat subalgebra has four real Poincaré charges in three dimensions and eight in four. Generic ABJM has twelve Poincaré plus twelve conformal charges; four-dimensional $\mathcal N=4$ SYM has sixteen plus sixteen. Sphere charges mix the flat Poincaré and conformal generators. Selecting one operation for localization neither changes the theory’s full supersymmetry nor proves that an arbitrary deformation preserves it. ABJM’s exceptional-level enhancement is additional information.

Curvature and the two meanings of a fixed set

Even the scalar curvature terms differ. For a conformal scalar in $d>2$ dimensions, the GR-sign coupling is $\xi_dR_{\rm GR}|\phi|^2$, with $\xi_d=(d-2)/[4(d-1)]$. On the round sphere,

$$m_{\rm curv}^2=\xi_d\frac{d(d-1)}{\ell^2} =\frac{d(d-2)}{4\ell^2}. \tag{X01.6}$$

This gives $3/(4\ell^2)$ for the ABJM matter scalar and $2/\ell^2$ for a SYM scalar. It is a background curvature coupling, not a chosen flat-space mass deformation. The source-sign translation in three dimensions is essential to obtaining the positive first value.

The localization locus is a set of field configurations where the deformation vanishes. A spacetime fixed point is a point where a geometric transformation vanishes. These are different spaces:

Input	Round ABJM calculation	Round massless SYM calculation
Gauge data	Compact $U(N)_k\times U(N)_{-k}$, integer $k>0$, no extra quotient	$SU(N)$, or a separately specified $U(N)$ theory; $\theta=0$ here
Smooth locus	Flat gauge fields; two constant commuting scalars; matter zero	A constant Cartan scalar with the contour and auxiliary fields fixed in the sphere-locus lesson
Remaining variables	$s_i=\ell\sigma_i$, $t_i=\ell\widehat\sigma_i$, both real	Dimensionless Hermitian $M=2\pi\ell a$, traceless for $SU(N)$
Additional sectors	No smooth magnetic-flux sum on $S^3$ in this problem	Polar instanton factors before taking the correctly identified massless limit
Observable used in the first calculation	The absolute sphere partition function	A normalized fundamental circular Wilson loop

For ABJM the auxiliary contour gives $D=i\sigma/\ell$ at the locus, and similarly at the other node. Flat connections are gauge-trivial because $S^3$ is simply connected. This does not exclude monopole defects from other problems; none has been inserted here. The Hopf vector has no zeros, yet the field-space locus is nonempty.

For SYM the polar sectors are part of the localization construction. Their disappearance in the final massless Gaussian answer uses the determinant and equivariant-mass analysis of the Gaussian-model lesson. It does not follow from ignoring the poles. Other four-sphere theories need not have the same cancellation.

The remaining integral is part of the answer

For the normalized $SU(N)$ loop, write $\alpha=2N/\lambda$, where $\lambda=g_{\rm YM}^2N>0$ in the site’s ordinary-trace convention. The Gaussian result is

$$\langle W\rangle= \frac{\int_{\operatorname{Tr}M=0}dM\, e^{-\alpha\operatorname{Tr}M^2}\,N^{-1}\operatorname{Tr}e^M} {\int_{\operatorname{Tr}M=0}dM\, e^{-\alpha\operatorname{Tr}M^2}}. \tag{X01.7}$$

Diagonalizing $M$ contributes $\prod_{i<j}(m_i-m_j)^2$, the Vandermonde square. The overall constant measure cancels in this ratio; the eigenvalue-dependent factor does not.

For ABJM, the inherited absolute prescription is instead

$$\begin{aligned} Z(N,k)&=\lim_{\epsilon\downarrow0}\frac1{(N!)^2} \int_{\mathbb R^{2N}}d^Ns\,d^Nt\, e^{i\pi k\sum_i(s_i^2-t_i^2)-\epsilon\sum_i(s_i^2+t_i^2)} \,\mathcal H(s,t),\\ \mathcal H(s,t)&= \frac{\prod_{i<j}[2\sinh\pi(s_i-s_j)]^2 [2\sinh\pi(t_i-t_j)]^2} {\prod_{i,j}[2\cosh\pi(s_i-t_j)]^2}. \end{aligned} \tag{X01.8}$$

The opposite phases come from the two Chern–Simons levels, the hyperbolic numerators from vector multiplets, and the denominator from bifundamental matter. The Weyl factors and factors of two are fixed by the integral’s derivation. Its positive first-node phase is translated explicitly from the source convention. The regulator is retained during manipulations; compact spacetime has not made this ordinary integral absolutely convergent at $\epsilon=0$.

Protected line insertions can become matrix characters in both settings, but this does not identify the lines, their scalar couplings, or their framing prescriptions. For example, ABJM’s bosonic circle inserts $N^{-1}\sum_i e^{2\pi s_i}$ into (X01.8), with its own charge and framing analysis. The two integrals do not acquire the same measure because both contain a character.

A physical coupling still changes the answer

The auxiliary coefficient $t$ in (X01.1) is not the physical coupling $\lambda$. This can be checked without referring to a formal path integral. Put $X=\operatorname{Tr}M^2$ and $w=N^{-1}\operatorname{Tr}e^M$. Differentiating the convergent Gaussian numerator and denominator gives

$$\frac{d\langle w\rangle_\alpha}{d\alpha} =-\langle wX\rangle_\alpha +\langle w\rangle_\alpha\langle X\rangle_\alpha =-\operatorname{Cov}_\alpha(w,X). \tag{X01.9}$$

There is no reason for this connected correlation to vanish. For $SU(2)$, $M=r\widehat n\cdot\boldsymbol\sigma$ gives $X=2r^2$, $w=\cosh r$, and radial weight $r^2e^{-2\alpha r^2}dr$. The completed finite-rank calculation gives

$$\langle w\rangle_\alpha= \left(1+\frac1{4\alpha}\right)e^{1/(8\alpha)},\qquad \langle X\rangle_\alpha=\frac3{2\alpha}. \tag{X01.10}$$

Differentiating yields $-e^{1/(8\alpha)}[3/(8\alpha^2)+1/(32\alpha^3)]$, which is strictly negative. Since $\alpha=4/\lambda$, the loop increases with $\lambda$. At $\lambda=4$ it is $\tfrac54e^{1/8}$; at $\lambda=16$ it is $2e^{1/2}$. Localization gives this physical dependence exactly; it does not remove it. The ABJM level is discrete, so varying integer $k$ is yet another operation: the already-derived $Z(1,k)=1/(4k)$ changes between theories without contradicting (X01.2).

The reusable method is therefore a symmetry identity with controlled boundary conditions. Its implementation requires a specified charge, background, integration cycle, locus, measure and observable each time. The following exercises make three of those distinctions quantitative.

Exercises

1. Test the square before testing the charge

Along a Hopf fiber of length $2\pi\ell$, use arc length $\tau$ and a periodic scalar mode $f_m=e^{im\tau/\ell}$, $m\in\mathbb Z$, of R-charge $r$. Find its eigenvalues under $Q_3^2$ and $Q_{\rm pair}^2$. In the stereographic patch of $S^4$, take $g_{mn}=(x^1+ix^2)^m(x^3+ix^4)^n$, with $m,n\geq0$, as a neutral internal-singlet scalar test function. Find its $Q_4^2$ eigenvalue. Does annihilation by a square establish that a field-theory operator is $Q$-closed?

Hint 1

Use $E_1=\partial_\tau$ on the fiber. For the four-dimensional test, apply the two plane rotations separately.

Hint 2

Both plane rotations in (X01.5) have the same orientation. Distinguish the action of $v$ from the minus sign in $Q_4^2=-\mathcal L_v$.

Solution

Equation (X01.3) annihilates every mode. Equation (X01.4) gives $Q_{\rm pair}^2f_m=2(m+r)f_m/\ell$. Its zero-eigenvalue condition is $m=-r$; if the chosen charge is not an integer, no such periodic scalar mode exists in this simple trivialization.

Each complex coordinate in the four-dimensional problem has $v$ eigenvalue $i/\ell$. The product therefore obeys $Q_4^2g_{mn}=-i(m+n)g_{mn}/\ell$, and only $m=n=0$ in this holomorphic polynomial family has zero eigenvalue. These are local test functions, not a claim about the complete globally smooth sphere spectrum.

If $Q\mathcal O=0$, applying $Q$ again necessarily gives $Q^2\mathcal O=0$. The converse fails: every vector is killed by a nilpotent square, while its first action can be nonzero. The nonnilpotent squares likewise impose a necessary invariance condition, not the full supersymmetry variation. One must still test $Q\mathcal O$.

2. Measure a susceptibility after localization

For the $SU(2)$ Gaussian distribution above, calculate $\operatorname{Var}_\alpha(X)$ directly from radial moments and verify $d\langle X\rangle_\alpha/d\alpha=-\operatorname{Var}_\alpha(X)$. Explain why this nonzero susceptibility is compatible with independence of the deformation coefficient $t$.

Hint 1

The radial normalization is $I_0=\int_0^\infty r^2e^{-2\alpha r^2}dr$. Use $u=2\alpha r^2$ to express $I_j=\int_0^\infty r^{2j+2}e^{-2\alpha r^2}dr$ through a gamma integral.

Hint 2

$\langle X\rangle=2I_1/I_0$ and $\langle X^2\rangle=4I_2/I_0$. The connected quantity subtracts the square of the mean.

Solution

The substitution gives $I_j=\Gamma(j+3/2)/[2(2\alpha)^{j+3/2}]$. Hence $\langle X\rangle=3/(2\alpha)$ and $\langle X^2\rangle=15/(4\alpha^2)$. Their combination is $\operatorname{Var}(X)=3/(2\alpha^2)>0$, equal to $-d\langle X\rangle/d\alpha$ as required. It has the inverse-square scaling expected from rescaling the three Gaussian components.

Changing $\alpha$ changes the physical Gaussian action inherited from the SYM coupling. Changing $t$ adds the chosen $Q$-exact deformation to a fixed theory. Equation (X01.2) controls the latter operation under its assumptions. It does not identify either insertion $X$ or the physical coupling derivative with that operation.

3. A central variable need not be a removable volume

Compare two already-derived center integrations. For $U(N)$ SYM, write $M=M_0+c\mathbf1$, $\operatorname{Tr}M_0=0$, with central weight $e^{-\alpha Nc^2}dc$. Find the central contribution to the loop $N^{-1}\operatorname{Tr}e^M$.

For rank-one ABJM, set $x=s-t$, $y=(s+t)/2$ and insert the bounded diagnostic $h_\xi(x)=\cos(2\pi\xi x)$ into the regulated integral, with fixed real $\xi$. Using the normalized approximate delta function obtained in the rank-one lesson, find its normalized $\epsilon\downarrow0$ expectation. Why would discarding either central integration as a gauge volume give an unjustified result?

Hint 1

The first observable factors as $e^cN^{-1}\operatorname{Tr}e^{M_0}$. Complete the real Gaussian square. For the second, $y$ multiplies the relative variable through $s^2-t^2=2xy$.

Hint 2

After the $y$ integration, the remaining positive kernel is proportional to $e^{-A_\epsilon x^2}/[4\cosh^2(\pi x)]$, where $A_\epsilon=\pi^2k^2/(2\epsilon)+\epsilon/2$ tends to infinity.

Solution

The normalized central factor is $\langle e^c\rangle=\exp[1/(4\alpha N)]=\exp[\lambda/(8N^2)]$. It multiplies the traceless answer. Dropping $c$ would change the loop even though the same Gaussian normalization occurs in numerator and denominator.

For ABJM, let $f(x)=1/[4\cosh^2(\pi x)]$. The normalized diagnostic is the ratio of $\int e^{-A_\epsilon x^2}f(x)h_\xi(x)dx$ to $\int e^{-A_\epsilon x^2}f(x)dx$. Rescale $x=u/\sqrt{A_\epsilon}$. The factors $f$ and $h_\xi$ are bounded and continuous at zero, so dominated convergence against $e^{-u^2}$ gives $h_\xi(0)=1$. This is an integral diagnostic, not an assertion that every such insertion is a protected physical line.

The common variable $y$ enforces the relative constraint and contributes the finite normalization that yields $Z(1,k)=1/(4k)$. It is not an unused infinite volume. The four-dimensional central scalar instead has its own nonzero Gaussian variance. A common-looking center decomposition does not make these two integrations interchangeable.