Exact Wilson loops and defect observables

An exact circular loop can determine more than the response of an almost straight line. It also fixes the first departure from a cusp whose geometric and internal angles are equal. This provides a useful research exercise: reproduce that slope, then change the smoothing used to extract it. Does an apparent change in the answer represent new defect physics, or an incorrectly normalized deformation?

Prepare with the research-entry method and the loop-to-defect calculation. This is a reproduction and controlled training extension. The finite-angle result is established; neither varying a regulator nor finding a discrepancy is itself a novelty claim.

Fix the observable before differentiating

Use massless four-dimensional N=4 SYM at the conformal origin, with gauge group $SU(N)$, $N\geq2$, rank $N-1$, and $\lambda=g_{\rm YM}^2N$. The fundamental scalar-coupled line has normalized trace $\operatorname{Tr}/N$. The bulk has sixteen real Poincaré and sixteen real conformal supercharges. The straight half-BPS line preserves half of each. A generic contour with an arbitrary scalar coupling does not retain that supersymmetry.

Keep the inherited line prescription, including its scalar coupling. Pestun’s (1.2) justifies the circular Gaussian expectation used in the finite-color lesson. Define

$$P_N(u)=\sum_{j=0}^{N-1}\binom{N}{j+1}\frac{(u/4N)^j}{j!}, \qquad W_{SU}(u)=\frac{P_N(u)}N \exp\!\left[\frac{N-1}{8N^2}u\right]. \tag{D09.1}$$

The positive variable $u$ is a coupling argument, not a new physical coupling convention. The auxiliary $U(N)$ answer instead has exponential $e^{u/(8N)}$; its central Gaussian is removed at the same $\lambda$. We do not infer a global factorization of all gauge sectors from this matrix-integral identity.

The normalization checkpoint is CHMS (2):

$$B_{SU}(u,N)=\frac{u}{2\pi^2} \left[\frac{N-1}{8N^2}+\frac{P_N'(u)}{P_N(u)}\right], \qquad C_D=12B. \tag{D09.2}$$

This is the logarithmic derivative at fixed $N$. $C_D$ is the separated displacement two-point coefficient of the straight line. A coupling-dependent multiplier of the line would alter the derivative and must not be introduced while claiming to keep the same observable. For example, $SU(2)$ gives $W=(1+u/16)e^{u/32}$ and $B(4,2)=13/(80\pi^2)$. Recover this number before proceeding; taking $\partial_uW$ instead of $\partial_u\log W$, or retaining the $U(1)$ center, fails this checkpoint.

Move the reference point along the BPS locus

Let $\varphi$ be the geometric turning angle and $\vartheta$ the angle between the two scalar-coupling directions. We use $\langle W_{\rm cusp}\rangle\sim e^{-\Gamma\log(L/\varepsilon)}$. Here $L$ and $\varepsilon$ are infrared and ultraviolet length cutoffs. Near the straight line, $\Gamma=B(\vartheta^2-\varphi^2)+\cdots$. Now choose a fixed $0<\varphi<\pi$ and vary only the internal angle:

$$\delta=\vartheta-\varphi, \qquad \Gamma(\varphi,\varphi+\delta) =\delta\,H(\varphi,\lambda,N)+O(\delta^2). \tag{D09.3}$$

The reference configuration $\vartheta=\varphi$ is BPS and has zero cusp anomalous dimension. The deformed cusp generally is not BPS. $H$ is a first derivative at this reference point, not a prediction for the entire two-angle function.

The useful supersymmetric family lives on a unit $S^2$. For a unit-speed contour $\boldsymbol x$, couple three of the six scalars through $\boldsymbol n=\boldsymbol x\times\dot{\boldsymbol x}$. The other three components vanish. The sphere-area input in CHMS (30)–(33) states that the nonintersecting loop has circular answer at $u=4\lambda A_1A_2/(A_1+A_2)^2$. This is a statement about that supersymmetric family, not arbitrary loops of equal area. We take the localization input with its source prescription; a Gaussian calculation alone is not a proof of it.

Two longitude arcs separated by $\pi-\varphi$ enclose areas $A_1=2(\pi-\varphi)$ and $A_2=2(\pi+\varphi)$. Therefore

$$q=1-\frac{\varphi^2}{\pi^2},\qquad u=\lambda q, \qquad W_\varphi=W_{SU}(\lambda q). \tag{D09.4}$$

This sphere expectation varies even though the cusp logarithm on the BPS locus vanishes. The conformal map to the cylinder does not identify their finite expectation values; CHMS footnote 5 explicitly notes the conformal anomaly. To extract a cusp slope, we must vary the actual geometric and scalar sources.

Restore the source variation

Parametrize one longitude and its small deformation by

$$\boldsymbol x=(\sin\eta,\cos\eta\cos\psi,\cos\eta\sin\psi), \quad \sin\eta=\tanh\tau, \quad \psi=\epsilon\sigma(\tau). \tag{D09.5}$$

The cylinder coordinate $\tau$ runs from $-\infty$ to $+\infty$ between the sphere poles. Let $\sigma$ vanish there. Normalize $\boldsymbol x\times\partial_\eta\boldsymbol x$ by the contour speed. Its first variation at $\psi=0$ is

$$\delta\boldsymbol n =\epsilon\bigl(\operatorname{sech}\tau\,\sigma', \sigma-\tanh\tau\,\sigma',0\bigr), \qquad \delta\varphi=\epsilon\sigma. \tag{D09.6}$$

The prime here means $\partial_\tau$. In particular, the internal source is not just $\epsilon\sigma$: the term containing $\sigma'$ is essential. The speed normalization first changes at order $\epsilon^2$.

Let $d$ and $f$ denote the normalized one-point insertions of the geometric displacement and the second scalar on the undeformed pair of cylinder lines. Translation along the cylinder makes them constant. Their BPS relation is $d+f=0$; a constant simultaneous variation of both angles leaves $\Gamma$ zero. The first scalar in (D09.6) has zero one-point function by the remaining internal rotations. Thus the normalized response is

$$\delta\log W =\epsilon\int d\tau\,[(d+f)\sigma-f\tanh\tau\,\sigma'] =\epsilon f I_\sigma, \quad I_\sigma=\int d\tau\,\operatorname{sech}^2\tau\,\sigma(\tau). \tag{D09.7}$$

We integrated by parts, using the vanishing endpoint values. The same deformation changes the sphere area by $\delta A_1=-\epsilon I_\sigma$, since $dA=\cos\eta\,d\eta\,d\psi$ and $\cos\eta\,d\eta=\operatorname{sech}^2\tau\,d\tau$. At $A_1+A_2=4\pi$, the area rule gives

$$\delta\log W=-\epsilon I_\sigma\, \frac{2\varphi}{q}B_{SU}(\lambda q,N). \tag{D09.8}$$

Equating (D09.7) and (D09.8), and using $\delta\Gamma=-d\,\delta\varphi-f\,\delta\vartheta$, gives

$$\boxed{H=-f=d=\frac{2\varphi}{q}B_{SU}(\lambda q,N).} \tag{D09.9}$$

This reconstructs CHMS (43)–(53). As $\varphi\to0$, $H/(2\varphi)\to B(\lambda,N)$, so the near-BPS and near-straight responses agree. At nonzero angle $H$ is a one-point response in the cusp background; it should not be relabelled as the straight-line norm $C_D$ at the original coupling.

A finite-window experiment

Instead of taking an infinitely long plateau, choose the smooth profile

$$\sigma_T(\tau)= \frac{\tanh(\tau+T)-\tanh(\tau-T)}{2\tanh T}, \qquad T>0. \tag{D09.10}$$

It equals one at $\tau=0$ and decays at both ends. With $t=\tanh T$, its sphere form is $\sigma_T=\cos^2\eta/(1-t^2\sin^2\eta)$. The denominator stays positive for finite $T$, and the quadratic vanishing makes the contour approach the original longitude smoothly at each pole. Substitute $v=\tanh\tau$ to find

$$I_T=\int_{-1}^{1}\frac{1-v^2}{1-t^2v^2}\,dv =\frac{2[t-T(1-t^2)]}{t^3}. \tag{D09.11}$$

It tends to 2 as $T\to\infty$, but is smaller at every finite $T$. If one uses the actual source response and incorrectly divides it by 2, the inferred slope is $H_{\rm naive}=I_T H/2$. For $T=1,2,4$, the relative underestimate is respectively $0.22665685$, $0.08169530$, and $0.00403186$. More arithmetic precision does not cure this bias. Dividing by the measured $I_T$, equivalently retaining the actual area change, recovers $H$ without that large-$T$ approximation.

For $SU(2)$ the full target is particularly simple:

$$H_2(\varphi,\lambda) =\frac{\varphi\lambda}{\pi^2} \left[\frac1{32}+\frac1{16+\lambda q}\right]. \tag{D09.12}$$

At $\lambda=4$, $\varphi=\pi/2$, it is $51/(304\pi)=0.0534006716953\ldots$. Here is a finite-color comparison, all at the same physical $\lambda=4$:

Color number	$B_{SU}(4,N)$	$H(\pi/2,4,N)$
2	0.016464692342	0.053400671695
3	0.019508568844	0.063281992983
8	0.021600284924	0.070073991744
Planar limit	0.021942491772	0.071185300066

The last row uses $B_0(u)=\sqrt u\,I_2(\sqrt u)/[4\pi^2I_1(\sqrt u)]$, where $I_n$ is the modified Bessel function. It is a limit, not an alternative exact answer at $N=2$.

For an independent numerical route at $N=2$, write the traceless Hermitian matrix as $M=\boldsymbol r\cdot\boldsymbol\sigma$, with eigenvalues $\pm r$. Its radial measure is proportional to $r^2e^{-8r^2/u}dr$ and its insertion is $\cosh r$. Differentiating the normalized measure gives

$$u\partial_u\log W =\frac8u\left[ \frac{\langle r^2\cosh r\rangle}{\langle\cosh r\rangle} -\langle r^2\rangle\right], \qquad \langle r^2\rangle=\frac{3u}{16}. \tag{D09.13}$$

Direct convergent radial quadratures therefore test the coupling response without numerically differentiating the polynomial. Separately integrate the embedded contour area and (D09.7). Agreement tests the matrix normalization, the source derivative and the finite window through different calculations. When using a finite deformation step, halve that step and check its bias separately from quadrature precision.

What is the next research step?

Literature checked 20 October 2026. Read Pestun (1.1)–(1.2) for the circle’s gauge-theory origin; CHMS section 3 for the source deformation above; then Gromov–Levkovich-Maslyuk’s near-BPS expansion (3.1)–(3.2). The last paper uses the planar quantum spectral curve: functional equations with specified complex analyticity and asymptotic conditions. Its $g=\sqrt\lambda/(4\pi)$ is not $g_{\rm YM}$. Section 3.2 supplies the next near-BPS coefficient; that problem is already studied, not an open question created by this dossier.

A bounded later reproduction is to recover one weak-coupling coefficient of that next response at a specified angle, starting from its stated expression and matching expansion variables before comparing numbers. This requires learning the paper’s contour kernels and branch conventions; running a general spectral-curve solver is a further project. It does not follow from another derivative of the single circular answer. A second variation introduces connected defect correlations, source-variation terms and contact prescriptions that the one-point calculation did not determine.

Keep two limits under control. At fixed finite $\lambda$, $\varphi\to\pi^-$ makes $u=\lambda q$ small, giving $H\to\lambda(1-N^{-2})/(8\pi)$. In the planar theory at $\lambda q\gg1$, however, $H_0\sim\varphi\sqrt\lambda/(2\pi^2\sqrt q)$. The latter approximation cannot be extended to $q=0$ at fixed coupling. Nor does an infinitesimal slope bound the error at finite $\delta$, especially near coincident endpoints. Take the $\delta$ derivative first at fixed interior angle. A failed comparison initially calls for checking these limits, the gauge-group center, the deformation coordinate and the finite window—not announcing a new defect observable.

Exercises

1. Extract a response from a shorter profile

Replace the plateau by $\sigma(\tau)=\operatorname{sech}^2\tau$. Compute its area weight and the slope inferred by someone who still divides $-\delta\log W/\epsilon$ by 2. Recover the correct answer for $SU(2)$ at $\lambda=4$, $\varphi=\pi/2$.

Hint 1

The source response contains $I_\sigma$, not the height of the profile at its center.

Hint 2

Set $v=\tanh\tau$ in the integral of $\operatorname{sech}^4\tau$.

Solution

The weight is $I_\sigma=\int_{-1}^{1}(1-v^2)dv=4/3$. Thus $-\delta\log W/\epsilon=4H/3$, and dividing by 2 gives $2H/3$, a one-third underestimate. Dividing by $4/3$ returns the true $H=51/(304\pi)$; the naive value is $17/(152\pi)$. No change of coupling or physical defect norm has occurred. Only the source profile and its normalization changed. This profile also arises as the smooth $T\to0$ limit of (D09.10), and (D09.11) tends to $4/3$, providing a second check.

2. Test a nonuniform strong-coupling limit

Take the planar limit first. Let $\lambda\to\infty$ while choosing $q=\lambda^{-2}$, so $\varphi=\pi\sqrt{1-\lambda^{-2}}$. Determine the leading slope from the correct effective-coupling regime and compare with the result obtained by inserting this path into the large-$u$ approximation. State the near-BPS qualification.

Hint 1

The Bessel argument is $\sqrt{u}$ with $u=\lambda q$, not $\sqrt\lambda$.

Hint 2

Use $B_0(u)=u/(16\pi^2)+O(u^2)$ when $u\to0$.

Solution

Here $u=1/\lambda\to0$. Hence $H_0=\varphi\lambda/(8\pi^2)[1+O(\lambda^{-1})]\sim\lambda/(8\pi)$. The misapplied strong-$u$ formula gives $\varphi\lambda^{3/2}/(2\pi^2)$, larger by a leading factor $4\sqrt\lambda$. The physical coupling is large, but the source-specific effective coupling is small. This calculation concerns a derivative at $\delta=0$ for each fixed $\varphi<\pi$. A finite internal-angle displacement must remain inside the angle domain and small enough for the local expansion; the calculation supplies no uniform finite-$\delta$ error bound as the endpoints approach one another.

3. Translate the next coefficient before comparing it

The planar spectral-curve paper expands in $z=(\cos\varphi-\cos\vartheta)/\sin\varphi$ as $\Gamma=z\Delta^{(1)}+z^2\Delta^{(2)}+O(z^3)$. At fixed $0<\varphi<\pi$, translate to $\Gamma=\delta H+\delta^2 C+O(\delta^3)$ with $\delta=\vartheta-\varphi$. If $\Delta^{(2)}$ vanishes at one perturbative order, must $C$ vanish?

Hint 1

Expand the cosine through second order in $\delta$ before identifying coefficients.

Hint 2

The linear term in $z$ itself contributes to the quadratic term in $\delta$.

Solution

Since $z=\delta+\tfrac12\cot\varphi\,\delta^2+O(\delta^3)$, $H=\Delta^{(1)}$ and $C=\Delta^{(2)}+\tfrac12\cot\varphi\,H$. Thus a vanishing $\Delta^{(2)}$ need not imply zero $C$. For instance, the paper’s one-loop expression (5.11) is $\Gamma=2g^2\varphi z+O(g^4)$, so $H=\lambda\varphi/(8\pi^2)$ and $C=\lambda\varphi\cot\varphi/(16\pi^2)$ at that order. At $\varphi=\pi/3$, the latter is $\lambda/(48\sqrt3\pi)$. This is a change of expansion coordinate, not a new dynamical contribution. The coordinate conversion is general, but the spectral-curve coefficient invoked here is planar; it cannot be treated as a finite-$N$ prediction.