Wilson loops and defect CFT data

Which part of a complex Wilson-loop expectation value carries a physical response? A useful first experiment changes a supersymmetric mass while holding the loop and its regularization fixed. We will calculate the complete answer at rank one, recover a prescribed Bremsstrahlung coefficient, and bound the error of extracting it from two nearby mass values. This is a reproduction and a controlled training extension, not a new defect observable.

Prepare with fermionic loops and framing, source derivatives and contact terms, and the reproduction-to-research method. The calculation needs no large-rank or Fermi-gas approximation.

Fix the theory, loop, and mass path

Use three-dimensional ABJM with compact $U(N)_k\times U(N)_{-k}$, equal ranks, and no additional quotient of the center. Each node has rank $N$; the product has rank $2N$. The undeformed generic theory has twelve real Poincaré and twelve real conformal supercharges. A manifest $\mathcal N=2$ formulation displays four real Poincaré charges. We work at $N=1$ and integer $k>2$, so exceptional-level enhancement is outside this calculation.

On the round sphere of radius $\ell$, keep the positive-first-node localization phase and the supersymmetric Hopf framing of linking number $+1$. The observable is the single-winding fundamental fermionic loop, normalized by $1/(2N)$, with the graded endpoint twist already derived in the prerequisite. Its localized insertion is

$$w_F(s,t)=\frac{e^{2\pi s}+e^{2\pi t}}2, \qquad s=\ell\sigma,\quad t=\ell\widehat\sigma. \tag{D14.1}$$

At zero mass this circle preserves twelve combinations of the generic conformal theory’s charges. A nonzero mass breaks conformal symmetry; the selected sphere deformation retains the manifest rigid algebra, and the loop retains half of that smaller algebra. We use its shared localization charge, not the undeformed twelve-charge count at nonzero mass. AGG, section 2.2.

Let $m$ be the real background mass, $x=\ell m$ its dimensionless value. Choose AGG’s specific path

$$\zeta_1=2ix,\qquad\zeta_2=0, \qquad\lambda_{\rm AGG}=2\pi s,\quad\mu_{\rm AGG}=2\pi t. \tag{D14.2}$$

The last two symbols are source eigenvalues, not a ‘t Hooft coupling or chemical potential. The real $\zeta$ parameters in the paper’s initial presentation describe R-charge mixing; (D14.2) instead continues to a real physical mass near zero, as in its (7.4). The accompanying auxiliary background follows the supersymmetric mass path; it is not held at zero while $m$ changes. At nonzero $x$, we compute a sourced sphere observable. Defect conformal data will be read at $x=0$.

At rank one there is no Vandermonde. Translating the matter denominators in AGG (2.3) gives

$$Z(x)=\lim_{\epsilon\downarrow0}\int_{\mathbb R^2}ds\,dt\, e^{i\pi k(s^2-t^2)-\epsilon(s^2+t^2)}f_x(s-t), \qquad f_x(v)=\frac1{4\cosh\pi(v+x)\cosh\pi v}. \tag{D14.3}$$

The loop $\mathcal W(x)$ is the same regulated integral with insertion (D14.1), divided by the regulated partition function before taking the limit. This positive Gaussian Abel prescription defines the oscillatory integral. The source’s unnormalized character differs by a mass-independent constant, which cancels from logarithmic derivatives. The measure and positive phase agree with DT (3.16)–(3.18); its paragraph after (4.1) specifies framing one. We do not infer a conversion to another matter-theory framing from a pure Chern–Simons phase.

Recover the complex answer

Put $u=(s+t)/2$, $v=s-t$. The phase is $2\pi i k u v$ and the insertion is $e^{2\pi u}\cosh\pi v$. In the numerator, Gaussian integration over $u$ centers the remaining Gaussian at $v=i/k$. Both node exponentials contain the same factor $e^{2\pi u}$, so both require this same shift. Their different relative factors $e^{\pm\pi v}$ do not reverse it.

The poles of $f_x$ are at $v=i(n+1/2)$ and $v=-x+i(n+1/2)$. For real $x$ and $k>2$, moving the contour to $v=y+i/k$ crosses none. At positive $\epsilon$, the Gaussian suppresses the connecting segments at infinity. More explicitly, let $Y$ be a real centered Gaussian with variance $\epsilon/(\pi^2k^2)$. The original numerator and denominator become

$$\begin{aligned} \mathcal N_\epsilon(x)&=\frac1{4k}\mathbb E\!\left[ \frac{e^{-\epsilon(Y+i/k)^2/2}} {\cosh\pi(Y+x+i/k)}\right],\\ Z_\epsilon(x)&=\frac1{4k}\mathbb E\!\left[ \frac{e^{-\epsilon Y^2/2}} {\cosh\pi(Y+x)\cosh\pi Y}\right]. \end{aligned} \tag{D14.4}$$

This equality holds at finite regulator. The useful bound $|\cosh(a+ib)|\geq|\cos b|\cosh a$ bounds the shifted denominator away from zero. Rescale $Y$ to a fixed unit Gaussian and use dominated convergence. Thus

$$Z(x)=\frac1{4k\cosh\pi x},\qquad \boxed{\mathcal W(x)=\frac{\cosh\pi x} {\cosh(\pi x+i\pi/k)}},\qquad N=1,\ k>2. \tag{D14.5}$$

At $x=0$, this is the prerequisite’s $\sec(\pi/k)$. At nonzero real mass it is generally complex, with $\mathcal W(-x)=\mathcal W(x)^*$. Writing $a=\pi/k$, $b=\pi x$ makes the phase visible:

$$\arg\mathcal W(x)=-\arctan(\tan a\tanh b),\qquad |\mathcal W(x)|^2=\frac{\cosh^2b}{\cosh^2b-\sin^2a}. \tag{D14.6}$$

Use the continuous phase that vanishes at $x=0$. No pole or branch crossing occurs on this real path. The restriction $k>2$ is part of the contour proof, not a statement that the physical $k=1,2$ theories do not exist. Their loops require a separately justified prescription.

An independent integral tests the complex shift. Fourier transformation of the difference of two hyperbolic tangents gives, for real $x\ne0$,

$$\widehat f_x(\omega) =\int_{\mathbb R}dv\,e^{2\pi i\omega v}f_x(v) =\frac{e^{-i\pi\omega x}\sin(\pi\omega x)} {2\sinh(\pi x)\sinh(\pi\omega)}, \qquad f_x(i/k)=\int_{\mathbb R}d\omega\, e^{2\pi\omega/k}\widehat f_x(\omega). \tag{D14.7}$$

The $x=0$ limit is $\widehat f_0=\omega/[2\sinh(\pi\omega)]$. The second integral converges absolutely precisely in the same strip $1/k<1/2$. For a cutoff $T>0$, its omitted absolute tail is at most

$$\frac1{|\sinh\pi x|(1-e^{-2\pi T})} \left(\frac{e^{-c_+T}}{c_+}+\frac{e^{-c_-T}}{c_-}\right), \qquad c_\pm=\pi(1\mp2/k). \tag{D14.8}$$

This bound is deliberately for $x\ne0$; taking its singular $x\to0$ limit would be a poor error estimate. The limiting transform instead has a polynomial times exponential tail. The regulated Gaussian and the Fourier integral test different orders of integration and retain the phase.

Extract one prescribed defect datum

A line defect breaks translations transverse to itself. The displacement operator $D_i(t)$ records that breaking in the stress-tensor Ward identity. On a conformal line in three dimensions its dimension is two; the standard normalization here is $\langle D_i(t)D_j(0)\rangle_W=12B\delta_{ij}/|t|^4$. The dimensionless coefficient $B$ is the Bremsstrahlung function describing the small-acceleration response of the probe.

The nontrivial supersymmetric dictionary is a sourced Ward-identity input, not a consequence of differentiating any loop. For AGG’s fundamental fermionic loop, its mass generator and fixed source/defect prescription, AGG (7.11)–(7.12) gives

$$B=\left.\frac{i}{8\pi^2\ell}\partial_m\log\mathcal W \right|_{m=0} =\left.\frac{i}{8\pi^2}\partial_x\log\mathcal W \right|_{x=0}. \tag{D14.9}$$

The source derives this using the stress-tensor multiplet and a protected operator on a linked great circle. It explicitly excludes applying that argument to the bosonic loop, which lacks the required shared charge. We use the one-point response only; general higher mass derivatives need their complete contact and normalization terms.

Differentiating the complex answer gives

$$\left.\partial_x\log\mathcal W\right|_0=-i\pi\tan\frac\pi k, \qquad \boxed{B=\frac1{8\pi}\tan\frac\pi k} =\frac1{8k}+\frac{\pi^2}{24k^3}+O(k^{-5}). \tag{D14.10}$$

For $k=4$, $B=1/(8\pi)\simeq0.03978873577$ and the displacement coefficient is $12B=3/(2\pi)$. Replacing $\mathcal W$ by $|\mathcal W|$ makes its first derivative at zero vanish and loses this answer. The complex phase at nonzero mass is therefore essential. The expansion in (D14.10) is a large-level expansion at fixed rank one, not a planar or M-theory expansion. The locally free matter description at rank one also does not trivialize this specified globally gauged line observable.

Change the measurement step, keeping the observable fixed

Suppose a computation returns $\mathcal W(\pm h)$ rather than its analytic derivative. With $h>0$ and the continuous logarithm define

$$B_h=\frac{i}{8\pi^2} \frac{\log\mathcal W(h)-\log\mathcal W(-h)}{2h} =\frac{\arctan[\tan a\tanh(\pi h)]}{8\pi^2h}. \tag{D14.11}$$

The answer is real despite coming from a phase difference. Its bias has both a leading coefficient and a useful bound:

$$B_h=B-\frac{\pi h^2}{24}\tan a\sec^2a+O(h^4), \qquad 0\leq B-B_h\leq\frac{\pi h^2}{24}\tan a\sec^2a. \tag{D14.12}$$

The bound follows from $0\leq u-\tanh u\leq u^3/3$ and $0\leq y-\arctan y\leq y^3/3$ for nonnegative arguments. It controls a step error at fixed physical integer $k$. Separately, if each computed logarithm has absolute error at most $\eta$, the induced error in $B_h$ is at most $\eta/(8\pi^2h)$. Smaller $h$ reduces the first error while amplifying the second. A claimed precision must bound both.

Keep four choices separate

Operator: the fermionic fundamental loop has a graded seam and the sum (D14.1). A bosonic node loop, a winding, or a product of traces is another insertion. Its response cannot inherit (D14.9) just from having an exponential in the matrix integral.

Framing: the prescription throughout is supersymmetric framing one. A mass-independent overall factor drops out of a logarithmic derivative. That observation is not a theorem that every change of matter-theory framing is such a factor.

Contact prescription: write $\mathcal W=Z_W/Z$. A background-only local counterterm multiplies both functionals by the same factor and cancels exactly. Defect-local renormalization and changes of the loop’s mass-dependent definition do not cancel in this way. An additional factor $C_{\rm def}(x)$ would shift the proposed response by $i\partial_x\log C_{\rm def}(0)/(8\pi^2)$. Whether such a factor is allowed is a question about the specified defect, compact charges and supersymmetric renormalization conditions; it is not permission to subtract an arbitrary phase.

Approximation: (D14.5) is exact in the proved rank-one contour domain. AGG’s large-rank Airy formulas resum a specified perturbative expansion and omit exponentially small sectors. Inserting $N=1$ into one of them would not provide an independent exact benchmark for this calculation.

A dated route beyond the reproduction

Literature checked 20 October 2026. DT constructs the fermionic loop and its shared-charge relation; AGG supplies the mass-deformed matrix integral, its perturbative large-rank analysis, and the defect Ward dictionary. A more recent framed-defect analysis, v3 checks interpolating ABJ(M) loops at two loops. It distinguishes bare and renormalized defect parameters and finds a special role for framing one. This is a perturbative result for that specified family, not an all-mass conversion formula for every loop.

A reachable report contains the full complex rank-one curve, the two integration checks, $B_h$ and its two error contributions, and the changed-source exercise below. A later project can increase the rank using the same normalized insertion and mass path, then compare a controlled finite-rank computation with the source’s large-rank approximation. That requires the larger-rank matrix or resolvent method and a separate convergence study. It is a proposed training extension; no novelty or currently open problem is claimed.

Practice

1. Change the flavor direction

Introduce two real dimensionless sources by $\zeta_1=2ix$, $\zeta_2=-2iy$. The matter factor becomes $[4\cosh\pi(v+x)\cosh\pi(v+y)]^{-1}$, with the same loop and phase. Find its normalized rank-one loop. Then follow $y=-x$ and calculate its first two logarithmic derivatives at zero. Does a zero first derivative imply that the original line has $B=0$?

Hint 1

The center Gaussian fixes the same relative shift for either matter mass. The numerator’s loop contributes $\cos(\pi/k)$ at that shift.

Hint 2

Differentiate along the entire new path, including $dy/dx=-1$. The Ward dictionary specifies a flavor generator as well as a loop.

Solution

The pole locations move only horizontally for real $x,y$, so $k>2$ still permits the shift. With $a=\pi/k$,

$$\mathcal W(x,y)= \frac{\cos a\cosh\pi x\cosh\pi y} {\cosh(\pi x+ia)\cosh(\pi y+ia)}.$$

At $y=-x$, the product in the denominator is real and positive: writing $b=\pi x$, $\mathcal W(x,-x)=\cos a\cosh^2b/(\cosh^2b-\sin^2a)$. It has the same zero-mass value $\sec a$, but its first logarithmic derivative is zero and its second is $-2\pi^2\tan^2a$. The original path had $y=0$, with nonzero first derivative $-i\pi\tan a$. The difference is the flavor combination being probed. Equation (D14.9) cannot be transferred to the new direction without a new Ward-identity translation. The undeformed line and its original Bremsstrahlung coefficient have not changed.

2. Restore the bulk normalization and the seagull

On the original one-source path, calculate $\partial_x^2\log Z_W$, $\partial_x^2\log Z$, and $\partial_x^2\log\mathcal W$ at zero. Then derive the general second response when a source-independent insertion $w$ is averaged with $e^{-S(x)}$. Explain why it is not just a connected product of two first-derivative insertions.

Hint 1

Use $Z_W=Z\mathcal W$ before taking derivatives.

Hint 2

Write $A=\partial_xS$ and $C=\partial_x^2S$; differentiating $\langle A\rangle$ also differentiates $A$ itself.

Solution

Here $Z_W=[4k\cosh(\pi x+ia)]^{-1}$, while $Z=[4k\cosh\pi x]^{-1}$. Consequently

$$(\log Z_W)''_0=-\pi^2\sec^2a,\qquad (\log Z)''_0=-\pi^2,\qquad (\log\mathcal W)''_0=-\pi^2\tan^2a.$$

Let $\langle\cdot\rangle_w$ mean normalization by $Z_W$. Direct differentiation gives

$$(\log\mathcal W)''= \langle A A\rangle_{w,c}-\langle C\rangle_w -\langle A A\rangle_c+\langle C\rangle.$$

The explicit $C$ term is often called a seagull or local source term. For this localized integral, $A=\pi\tanh\pi(v+x)$ and $C=\pi^2\operatorname{sech}^2\pi(v+x)$. In the Abel limit the normalized insertion evaluates analytic functions at $v=i/k$, whereas the vacuum evaluates them at $v=0$. The corresponding localized connected $AA$ terms vanish; the two explicit $C$ values reproduce the displayed answer. This statement about the localized source response does not say that arbitrary physical separated-point correlators vanish. The measure is complex, and no positive probability-variance interpretation was assumed.

A common background counterterm cancels from $\log Z_W-\log Z$. Dropping $Z$ or the $C$ terms changes the observable. The first derivative alone conceals the bulk subtraction here because $Z'(0)=0$.

3. Resolve a derivative rather than a phase plot

At $k=4$, compare $B_h$ for $h=0.1,0.05,0.025$ with $B$. Check the bound (D14.12), derive the leading step bias, and form a combination of $B_h$ and $B_{h/2}$ that removes it. If each logarithm has absolute error at most $\eta$, why can taking arbitrarily small $h$ fail?

Hint 1

Expand $\tanh(\pi h)$ and then the arctangent to cubic order before dividing by $h$.

Hint 2

A quadratic error changes by a factor four under $h\mapsto h/2$. Track the two input-log errors by the triangle inequality.

Solution

Writing $t=\tan a$, $\arctan[t\tanh(\pi h)]=\pi th-\pi^3t(1+t^2)h^3/3+O(h^5)$. This gives the coefficient in (D14.12). The bounds quoted there follow without assuming a small step: integrate $1-\operatorname{sech}^2u=\tanh^2u\leq u^2$ and $1-(1+y^2)^{-1}\leq y^2$.

Using (D14.11) and $B=0.03978873577\ldots$ gives

Step $h$	$B_h$	Error $B-B_h$	Upper bound $\pi h^2/12$
$0.1$	$0.03740276038$	$0.00238597540$	$0.00261799388$
$0.05$	$0.03914993738$	$0.00063879839$	$0.00065449847$
$0.025$	$0.03962611330$	$0.00016262248$	$0.00016362462$

The positive errors decrease toward a factor four when $h$ is halved. The extrapolation $B^{\rm ext}_h=(4B_{h/2}-B_h)/3$ cancels the $h^2$ term and has $O(h^4)$ step bias at fixed $k$.

Two logarithmic errors of size at most $\eta$ give derivative error at most $\eta/(8\pi^2h)$; the extrapolated error bound is $3\eta/(8\pi^2h)$ if all four input logarithms obey that same bound. Thus smaller steps demand more accurate complex loop values. The absolute-value-only computation returns zero for every $h$ and fails this test even if its numerical precision is arbitrarily high.