Reproduce a duality with its contact terms

Two partition functions can have identical absolute values and still fail a duality test. Their phases remember local source terms, and a wrong source dictionary can itself produce a pure phase. How can we distinguish these possibilities?

This is a reproduction project with a controlled training extension. Start from the SQED–XYZ mass and FI dictionary, source derivatives and contact terms, and the method for turning a reproduction into a question. We will compare two nonzero sources at once, then test a large-mass limit. The index section has its own explicit preparation.

Specify the comparison before integrating

The electric theory is three-dimensional $\mathcal N=2$ SQED: compact gauge group $U(1)$ of rank one, one pair $Q,\widetilde Q$ of gauge charges $+1,-1$, dynamical Chern–Simons level zero, and no superpotential. Its infrared XYZ description has no gauge field and has $W=XYZ$. The full and manifest supersymmetry have four real Poincaré generators in Lorentzian signature. The Euclidean round-sphere background preserves both rigid spinor families. At the undeformed fixed point there are also conformal supercharges; finite real masses do not preserve conformal symmetry.

Fix sphere radius $\ell$, orientation and background $H=-i/\ell$, $A=V=C=0$. A constant flavor mass $m$ is accompanied by its supersymmetric auxiliary source $D_B=im/\ell$, with zero flavor connection and gauginos. Define the dimensionless axial and topological-source masses $\mu=\ell m_A$ and $\eta=\ell m_J$. They are real. The level-one BF coupling to the topological source gives the electric localization factor $e^{2\pi i\eta s}$, where $s=\ell\sigma\in\mathbb R$. In the FI notation of the deformation lesson, $\zeta=m_J/(2\pi)$ and $\xi=-\zeta$.

The operator map fixes every mass; none is a fitting parameter:

Electric operator	XYZ field	Axial charge	Topological charge	R-charge	Dimensionless real mass
$Q\widetilde Q$	$X$	$2$	$0$	$2r$	$2\mu$
$V_+$	$Y$	$-1$	$+1$	$1-r$	$-\mu+\eta$
$V_-$	$Z$	$-1$	$-1$	$1-r$	$-\mu-\eta$

Here $r$ is the electric chiral R assignment. The symmetric fixed point has $r=1/3$, so the three gauge-invariant XYZ chirals have $R=\Delta=2/3$. We do not assign a physical local-primary dimension to a gauge-variant electric field. Keeping $r$ variable is useful for checking the source translation; the main benchmark uses $1/3$.

Use the determinant

$$\mathcal D_r(x)=e^{L(1-r-ix)},\qquad L'(z)=-\pi z\cot(\pi z),\qquad L(0)=0. \tag{D04.1}$$

The symbol $L$ denotes the determinant function usually written $\ell(z)$; it is unrelated to the sphere radius. Its analytic branch is real on $-1<z<1$. Thus $L(\bar z)=\overline{L(z)}$ in the domain used here. The branch and its constant are part of the observable, not an adjustable phase.

The two answers to compare are

$$\begin{aligned} Z_E(r;\mu,\eta) &=\int_{\mathbb R}ds\,e^{2\pi i\eta s} \mathcal D_r(s+\mu)\mathcal D_r(-s+\mu),\\ Z_M(r;\mu,\eta) &=\mathcal D_{2r}(2\mu) \mathcal D_{1-r}(-\mu+\eta) \mathcal D_{1-r}(-\mu-\eta). \end{aligned} \tag{D04.2}$$

There is no non-Abelian root determinant or Weyl factor. For $0<r<1$ and real $\mu,\eta$, the electric integrand decays as $e^{-2\pi(1-r)|s|}$, so its real contour is absolutely convergent. This domain includes every calculation below; no complex contour continuation is being hidden.

Willett–Yaakov, section 3.1 identifies this equality with a hyperbolic-gamma integral identity. To translate its chiral convention, set $\lambda_{\rm source}=-s$, $\mu_{\rm source}=-\mu+i(r-1/2)$ and $\eta_{\rm source}=-\eta$. For example, $1/2+i\mu_{\rm source}=1-r-i\mu$. The two source hyperbolic mass parameters coincide in this one-flavor problem, so the extra phase involving their difference is one. Together with the fixed determinant prescription this gives $Z_E=Z_M$, with no undetermined relative constant. The explicit FI factor is also displayed in Kapustin–Willett, (3.7). This is an established identity and a test of the proposed infrared duality, not a proof that every observable agrees.

Reproduce the complex number

For actual evaluation one may use

$$L(z)=-z\log(1-e^{2\pi iz}) +\frac{i}{2}\left(\pi z^2+\frac{\operatorname{Li}_2(e^{2\pi iz})}{\pi}\right) -\frac{i\pi}{12},\qquad \operatorname{Im}z>0, \tag{D04.3}$$

and conjugation in the lower half-plane. On the real interval use continuous limits, with $L(0)=0$ and $L(-z)=-L(z)$. Checking (D04.1) by an independent line integral of its derivative tests both the branch and the additive constant. The formula is given in ZD, Appendix A.1.

At $r=1/3$, $\mu=0.2$, $\eta=0.3$, direct electric quadrature and the XYZ product give

$$Z=0.26377369556745376431 -0.02129830469163532526\,i. \tag{D04.4}$$

A reproducible first assignment is to evaluate the integral on $[-60,60]$, split at $s=0,\pm\mu$ and a few intermediate points, and compare $Z_E/Z_M$ with one using 70-digit arithmetic. Retain the complex integrand. Increasing precision and changing subdivisions tests quadrature stability; a tail estimate tests a different error. For $(r,\mu,\eta)$ equal to $(1/3,0.2,0.3)$, $(0.4,-0.15,0.5)$, $(1/2,0.4,0.1)$, $(1/3,1.5,0.2)$ or $(1/3,3,0.2)$, the computations give relative complex agreement better than $10^{-50}$. This is numerical evidence at those points, separate from the sourced identity.

For a tail certificate define, for $t>0$,

$$B_a(t)=\frac{(|a|+t+1/(2\pi))e^{-2\pi t}}{1-e^{-2\pi t}}. \tag{D04.5}$$

Expanding the logarithm and dilogarithm in (D04.3) bounds the difference from its quadratic asymptote by $B_a(t)$ at $z=a+it$. For a cutoff $S>|\mu|+1$, the two electric masses outside the cutoff have absolute values summing to $2|s|$. Hence the omitted integral obeys

$$|\delta Z_E|\le \frac{\exp\{2B_{1-r}(S-|\mu|)-2\pi(1-r)S\}} {\pi(1-r)}. \tag{D04.6}$$

At the first benchmark and $S=60$, this is below $3.38\times10^{-110}$. It bounds truncation, not floating-point or quadrature error.

Test a phase before calling it a counterterm

Fix compact unit-normalized $U(1)_A$ and $U(1)_J$ source connections on spin manifolds. We compare this specified class of source bundles; it does not classify possible faithful flavor-group quotients. Hold the R-background and gravitational subtraction prescription fixed. An additional integer symmetric flavor matrix $K$ changes the sphere answer by

$$Z^{(K)}=e^{i\pi\mathbf x^T K\mathbf x}Z, \qquad \mathbf x=(\mu,\eta)^T. \tag{D04.7}$$

This follows by evaluating the supersymmetric source Chern–Simons action on $D_B=im/\ell$, as in the contact-term lesson. CT, (1.5)–(1.7) and (4.5) explains why the allowed shift is quantized. The reference (D04.2) adds no further integer terms. This does not assert that all fractional contact data, flavor–R terms or gravitational terms vanish.

As a deliberate scheme change, take $K_{AA}=2$, $K_{AJ}=-1$, $K_{JJ}=3$ on the electric side only. Then $Z_E^{(K)}/Z_M=e^{i\pi(2\mu^2-2\mu\eta+3\eta^2)}$. Its modulus is one at every real source value. Subtracting this declared polynomial restores the original comparison.

Now make a different mistake: assign $Y,Z$ axial mass $+\mu$ while leaving $X$ at $2\mu$. The erroneous product $\widetilde Z_M$ has $|\widetilde Z_M/Z_M|=1$ too, by conjugation and interchange of $Y,Z$. At (D04.4) its ratio is approximately $0.66734393836+0.74474966797i$. A phase alone has not identified a scheme change.

At $\eta=0$, choose the logarithm continuously from $\mu=0$. The wrong ratio has

$$\log\frac{\widetilde Z_M}{Z_M} =2\{L(r-i\mu)-L(r+i\mu)\},\qquad \left.\partial_\mu^3\log\frac{\widetilde Z_M}{Z_M}\right|_0 =4iL'''(r). \tag{D04.8}$$

Differentiating (D04.1) gives $L'''(r)=2\pi^2\csc^2(\pi r)[1-\pi r\cot(\pi r)]$, which is nonzero at $r=1/3$. A quadratic flavor counterterm has zero third derivative. Even allowing an affine phase would not repair this error. Recheck the charges before fitting counterterms. This diagnostic concerns the fixed conformal/source prescription; it is not permission to add arbitrary higher-order local functions of the masses.

Change one input with an error bound

Increase positive $\mu$ at fixed real $\eta$ and $r=1/3$. The three XYZ masses have signs $+,-,-$ once $\mu>|\eta|$. The leading terms of (D04.3) and its conjugate give

$$Z_{\rm as}(\mu,\eta) =\exp\left[-\frac{4\pi\mu}{3} +i\pi\left(\mu^2-\eta^2-\frac1{36}\right)\right]. \tag{D04.9}$$

The constant phase matters. Its origin can be checked from the electric description. In the region where both electric masses are positive, their leading product is $\exp[-4\pi\mu/3+i\pi(\mu^2+s^2-5/18)]$. It induces the level-$+1$ gauge Gaussian. With a positive Abel regulator,

$$\lim_{\varepsilon\downarrow0}\int_{\mathbb R} e^{-(\varepsilon-i\pi)s^2+2\pi i\eta s}\,ds =e^{i\pi/4-i\pi\eta^2}. \tag{D04.10}$$

Thus $-5/18+1/4=-1/36$, reproducing the complete phase. This checks the effective Gaussian coefficient and orientation. The intermediate approximation is not uniform over all $s$, so it alone is not a proof of the full integral’s asymptotics.

The exact dual product supplies that control. Expanding (D04.3) explicitly yields the remainder $\sum_{n\ge1}e^{2\pi inz}(z/n+i/(2\pi n^2))$. Taking absolute values proves (D04.5). Consequently

$$\left|\frac{Z}{Z_{\rm as}}-1\right|\le e^{E}-1, \quad E=B_{1/3}(2\mu)+B_{1/3}(\mu-\eta)+B_{1/3}(\mu+\eta). \tag{D04.11}$$

For $\eta=0.2$ the actual relative errors at $\mu=1.5,2,3$ are approximately $4.6762\times10^{-4}$, $2.6714\times10^{-5}$ and $7.4396\times10^{-8}$; the respective bounds are $5.5894\times10^{-4}$, $3.0764\times10^{-5}$ and $8.2183\times10^{-8}$. The decreasing discrepancy is a finite-mass correction, not a numerical failure of duality. This estimate is nonuniform when $\eta$ approaches $\mu$ because one magnetic chiral remains light.

A separately prepared index comparison

First complete gauge projection and the SQED–XYZ index. The sphere integral cannot supply the index’s flux sum or Hilbert-space grading. Return to the undeformed conformal theory, use $r=1/3$, and write $t=x^{1/3}$ for the index expansion variable; the axial and topological fugacities are $a,v$. The chosen trace uses $\{q,q^\dagger\}=D-R-J_3$ and exponent $D+J_3$.

Turn on fixed topological background flux $N=2$ and zero axial background flux. In the convention of the index lesson the BF insertion is $u^2v^m$, where $u$ is the compact gauge holonomy and $m\in\mathbb Z$ is dynamical flux. The $u$ integral now extracts matter charge $-2$. It does not sum over $N$.

In the $m=0$ sector, the lowest contribution is two charge-$-1$ scalar letters, giving $a^2t^2$. In a sector with $|m|=p$, the vacuum contributes $a^{-p}t^{2p}$ and each lowest scalar costs $t^{3p+1}$. Neutralizing the insertion requires two such letters. Thus the first $m=\pm1$ terms occur at $t^{10}$, with coefficient $a(v+v^{-1})$; $p\ge2$ cannot contribute below $t^{18}$. On the XYZ side, $Y,Z$ have background fluxes $+2,-2$ and their combined vacuum factor is $a^2t^2$. Exercise 3 completes the comparison through $t^{10}$.

This is within the zero-axial-flux identity proved in IDX3, section 3.3. Its half-integer source labels obey $m_{\rm site}=2s_{\rm source}$ and $N=2n_{\rm source}$; its reference R assignment is translated by $\alpha_{\rm source}=a x^{-1/6}$ and $w_{\rm source}=v$. An extra mixed counterterm is defined operationally here by $a^{\kappa_{\rm idx}N}$, with integer $\kappa_{\rm idx}$ and reference zero. We do not equate its sign to a sphere coefficient without translating the background orientation and BF conventions. The zero-background-flux index would miss this factor entirely.

Literature status and the next experiment

Literature review: 20 October 2026. The sphere identity, its source deformations and the zero-axial-flux index identity are established results, not open questions. Read ZD section 3.1 for the integral identity, CT sections 1 and 4 for what counterterms may change, then IDX3 section 3.3 for the separate index proof. A later use is Cheng–Sułkowski, section 4.1, (4.13)–(4.15), which reuses the SQED–XYZ identity before gauging flavor symmetries in section 4.2. This is a selected literature route, not an assertion that the latest work ends in 2023.

The reachable deliverable is a source table, the complex ratio over a small two-dimensional grid, a declared counterterm matrix, and separate integration and asymptotic errors. If the ratio fails, first test the determinant branch, BF sign and mass map. If it is a phase, test its derivatives and quantization before interpreting it. Agreement supports this observable in this source family; it does not establish equivalence on arbitrary bundles, equality of unprotected correlators, or a new duality.

For a next training experiment, approach $\eta=\mu$ while keeping the nearly massless determinant exact and expanding only the other two. Derive a new uniform remainder estimate before plotting an apparent breakdown. Gauging a flavor source is a more substantial change: it introduces a new contour and flux sum, so it requires the explicit construction and convergence conditions in the later literature rather than simply relabeling this answer.

Exercises

1. Recover a mixed contact term without unwrapping a whole grid

Suppose two calculations differ by the declared quadratic phase in (D04.7), with no additional affine term. Form the diagonal ratios $R(h,0)R(-h,0)/R(0,0)^2$ and $R(0,h)R(0,-h)/R(0,0)^2$, and the four-corner ratio $R(h,h)R(-h,-h)/[R(h,-h)R(-h,h)]$, where $R=Z_E^{(K)}/Z_M$. At $h=1/4$, evaluate them for the stated $K$. Can one isolated measurement determine arbitrary integer levels?

Hint 1

Multiply the exponentials before choosing an argument. The four-corner combination cancels both diagonal terms.

Hint 2

The mixed result is $e^{8\pi iK_{AJ}h^2}$. A principal argument contains only information modulo $2\pi$.

Solution

The diagonal ratios are $e^{2\pi iK_{AA}h^2}$ and $e^{2\pi iK_{JJ}h^2}$; the corner ratio is $e^{8\pi iK_{AJ}h^2}$. At $h=1/4$ they are $e^{i\pi/4}$, $e^{3i\pi/8}$ and $e^{-i\pi/2}$. On a continuously chosen small-source branch they recover $(2,3,-1)$.

One grid spacing does not recover unbounded integer levels: at this spacing the diagonal phases identify a level only modulo $16$, and the mixed phase only modulo $4$. Halve $h$ and follow a continuous branch toward zero, or combine data with a justified level bound. The ratio itself remains well-defined; only the logarithm has this ambiguity. A nonzero third derivative such as (D04.8) is an independent reason to reject the quadratic-counterterm hypothesis.

2. Reverse the mass and test a nonuniform limit

Find the leading expression for large negative $\mu$ at fixed $\eta$. Explain what happens to the induced electric Gaussian. Does the divergence of the bound as $\mu\downarrow|\eta|$ imply a singularity of the exact sphere partition function?

Hint 1

Use conjugation of $L$ and the symmetry under exchanging $Y,Z$.

Hint 2

The negative-level Fresnel integral is the conjugate of (D04.10). At the mass wall one determinant is evaluated at a finite real argument $1/3$.

Solution

For real sources, $Z(-\mu,\eta)=\overline{Z(\mu,\eta)}$, since conjugation reverses both masses and the answer is even in $\eta$. Therefore, writing $M=|\mu|$,

$$Z\sim\exp\left[-\frac{4\pi M}{3} -i\pi\left(M^2-\eta^2-\frac1{36}\right)\right].$$

Both heavy electric fermion masses now have negative sign; the gauge Gaussian is $e^{-i\pi s^2}$. Its Abel limit is $e^{-i\pi/4+i\pi\eta^2}$, which reproduces the conjugated constant as well as the topological-source term. The bound (D04.11) applies with $M$ in place of $\mu$.

When $M-|\eta|$ tends to zero, the corresponding large-imaginary-argument expansion is invalid. The exact factor is $e^{L(1/3)}$, finite and nonzero, so this divergent bound is not an exact singularity. Keep that determinant unexpanded. This distinguishes failure of a controlled approximation from failure of the original identity.

3. Compare a background-flux sector

After the index preparation, calculate the $N=2$ index through $t^{10}$ with formal fugacities $a,v$. Then add the declared mixed index counterterm $\kappa_{\rm idx}=3$ on one side. Explain why the leading term is not one.

Hint 1

In $m=0$, extract $u^{-2}$ from the matter product. Through this order it is enough to retain scalar weights $a u^{\pm1}t$ and $a u^{\pm1}t^7$, and fermion weights $a^{-1}u^{\pm1}t^5$.

Hint 2

Add the first $m=\pm1$ terms derived above. On the magnetic side use the prefactor $a^2t^2$ and expand the flux-shifted $Y,Z$ factors.

Solution

The zero-flux coefficient is obtained from

$$[u^{-2}]\, \frac{(1-t^5/(au))(1-ut^5/a)} {(1-aut)(1-at/u)(1-aut^7)(1-at^7/u)}.$$

At $t^2,t^4$ it gives $a^2,a^4$. At $t^6$ the bosonic $a^6$ receives the fermionic subtraction $-1$. At $t^8$ the bosonic coefficient $a^8+a^2$ and fermionic coefficient $-2a^2$ leave $a^8-a^2$. At $t^{10}$ the corresponding coefficients are $a^{10}+2a^4$ and $-2a^4$, leaving $a^{10}$. Adding the two minimal monopole sectors gives

$$\begin{aligned} I_{N=2}={}&a^2t^2+a^4t^4+(a^6-1)t^6+(a^8-a^2)t^8\\ &+[a^{10}+a(v+v^{-1})]t^{10}+O(t^{12}). \end{aligned}$$

Independently, the XYZ product to this order is the expansion of

$$a^2t^2\, \frac{(1-t^4/a^2)(1-t^{10}/a^2)} {(1-a^2t^2)(1-a^2t^8)(1-vt^8/a)(1-t^8/(av))},$$

which gives the same coefficients. The minus terms are graded contributions, not negative ordinary state multiplicities. The added counterterm multiplies the whole answer by $a^{3N}=a^6$. This is invisible at $N=0$ but visible here. The nontrivial background changes the Hilbert problem and its vacuum charges; it is not the ordinary vacuum trace normalized to begin with the identity operator. Its leading $a^2t^2$ agrees with both the electric gauge projection and the magnetic zero-point factor.