Supersymmetric Wilson lines and circles

Before this lesson: Couplings and conformal invariance; What BPS and protected mean.

A Wilson loop measures the response of a gauge theory to a heavy charged probe transported around a contour. In a supersymmetric theory the probe can also couple to scalars. That extra coupling can cancel the supersymmetry variation of the gauge connection, but the cancellation has to use the same symmetry everywhere along the path.

Start with the N=4 coupling and conformal vacuum and what a protected observable means. We work on flat Euclidean four-space, with the massless $SU(N)$ theory, $N\geq2$ , and zero scalar expectation values. Its parent has sixteen real Poincaré supercharges and, in the conformal vacuum, sixteen conformal supercharges. We use their complexified spinor equations after continuation. No large- $N$ approximation or localization argument is needed below.

Specify the probe and its normalization

The fields in the action lessons have canonical kinetic terms. For a line connection it is useful to absorb the dimensionless coupling: $\mathcal A_\mu=gA_\mu$ and $\Phi_I=gX_I$ , $I=1,\ldots,6$ . These are Hermitian matrices, with ordinary fundamental generators obeying $\operatorname{Tr}_N(t^at^b)=\delta^{ab}/2$ and covariant derivative $\partial_\mu-i[\mathcal A_\mu,\cdot]$ . This rescaling changes the propagators, not the theory.

Let $x^\mu(t)$ be a smooth regular contour, $v=|\dot x|>0$ its speed, and $n^I$ a fixed real unit vector in the six scalar directions. Define

\mathcal L(t)=i\dot x^\mu\mathcal A_\mu(x(t))+v n^I\Phi_I(x(t)), \qquad W[C,n]=\frac1N\operatorname{Tr}_N\mathcal P \exp\!\left(\oint dt\,\mathcal L(t)\right). \tag{N11.1}

Here $\mathcal P$ puts later parameters on the left. Thus $U(t,t_0)$ solves $\partial_tU=\mathcal L(t)U$ , $U(t_0,t_0)=1$ . The scalar term is real rather than multiplied by $i$ ; the resulting transport is generally not unitary. Both terms have dimension zero after integration. The factor $1/N$ fixes $W=1$ at zero fields and coupling. In a different representation, specify that representation and divide by its dimension if the same normalization is desired.

Under a gauge transformation $h(x)$ , the connection along the path obeys $\mathcal L'=h\mathcal Lh^{-1}+\dot h h^{-1}$ . Differentiating $h(x(t))U(t,t_0)h(x(t_0))^{-1}$ proves the endpoint transformation law. The trace is gauge invariant for a closed contour. An open line requires endpoint states or specified behavior at infinity; a trace alone does not remove two different endpoint transformations.

An increasing reparametrization preserves both $\dot x\,dt=dx$ and $v\,dt=ds$ . Reversing the orientation changes the gauge term while the scalar coupling to length retains its sign. It is therefore not automatically the inverse of the original matrix transport.

The definition is the Hermitian form of the scalar-coupled probe in Drukker–Gross (1.4), with our $n=-\theta$ relative to their displayed scalar sign. Their fields already absorb $g$ . Pestun (1.1), (2.18) instead uses anti-Hermitian gauge-algebra fields and an explicitly continued scalar. Keep a complete convention package when comparing the factors of $i$ .

Vary the ordered exponential

First vary its defining differential equation. With fixed endpoints and $\delta U(t_0,t_0)=0$ , variation of constants gives

\delta U(t_f,t_0)=\int_{t_0}^{t_f}dt\, U(t_f,t)\,\delta\mathcal L(t)\,U(t,t_0). \tag{N11.2}

This explains why setting $\delta\mathcal L=0$ pointwise is sufficient even when the matrices at different positions do not commute. We keep the contour and $n$ fixed while varying the quantum fields. More general variations that are covariant total derivatives require their own endpoint analysis; they are not needed here.

We can package the gauge field, six scalars and gauginos using a ten-dimensional Clifford algebra. This is notation for the four-dimensional multiplet, not an extra spacetime assumption. Introduce matrices $\Gamma_\mu$ , $\mu=1,\ldots,4$ , and $\Gamma_{4+I}$ satisfying $\{\Gamma_M,\Gamma_N\}=2\delta_{MN}$ on a complex 32-component space. A chiral spinor occupies sixteen components. Every $\Gamma_M$ switches chirality. Put $S=n^I\Gamma_{4+I}$ , so $S^2=1$ and $\{S,\Gamma_\mu\}=0$ .

In the complexified Hermitian convention the bosonic variations can be written with a common spinor pairing as $\delta\mathcal A_\mu=\langle\Gamma_\mu\epsilon,\Psi\rangle$ and $\delta\Phi_I=\langle\Gamma_{4+I}\epsilon,\Psi\rangle$ . The pairing includes the charge-conjugation matrix and an irrelevant common normalization; it is nondegenerate, and $\Psi$ is the adjoint gaugino package. These are the bosonic part of the reduced transformation law, Pestun (2.3), in the Euclidean Clifford presentation. We need no on-shell fermion equation to vary this bosonic insertion.

Consequently the sufficient local condition is

K(t)\epsilon(x(t))=0, \qquad K(t)=i\dot x^\mu\Gamma_\mu+vS, \qquad K(t)^2=0. \tag{N11.3}

The final equality follows from $v^2=\dot x^2$ and $S^2=1$ . It shows why a scalar term of unit magnitude is possible: the gauge and scalar squares cancel. It does not show that a common nonzero supersymmetry solves the equation along an arbitrary curve.

Find a global symmetry on the line

The flat conformal Killing-spinor equations are $\partial_\mu\epsilon=\Gamma_\mu\epsilon_S$ and $\partial_\mu\epsilon_S=0$ . Their complete solutions are

\epsilon(x)=\epsilon_Q+x^\mu\Gamma_\mu\epsilon_S. \tag{N11.4}

The constant $\epsilon_Q$ has positive ten-dimensional chirality; $\epsilon_S$ has negative chirality and one extra inverse-length dimension. Each has sixteen complex components before the Euclidean symmetry conditions. They label Poincaré and conformal transformations respectively. This is Pestun (B.4); using full Clifford matrices makes the two chiral spaces explicit.

Take the oriented line $x^1=s$ , $x^{2,3,4}=0$ . Its condition is $(i\Gamma_1+S)(\epsilon_Q+s\Gamma_1\epsilon_S)=0$ for every real $s$ . The constant and linear coefficients must vanish separately. Define $J=i\Gamma_1S$ , with $J^2=1$ . Multiplication by $S$ in the first equation and direct expansion in the second give

J\epsilon_Q=+\epsilon_Q,\qquad J\epsilon_S=-\epsilon_S. \tag{N11.5}

Each eigenspace has dimension eight on its respective chiral space: a transverse product such as $\Gamma_2S$ preserves chirality and anticommutes with $J$ , pairing its two eigenspaces. The line retains eight Poincaré and eight conformal directions. “Half-BPS” refers here to half of the full superconformal fermionic symmetry. The complex calculation continues that count; it does not assert thirty-two real Poincaré charges or impose a Euclidean Majorana condition.

Find a global symmetry on the circle

For radius $R>0$ , choose $x=R(\cos t,\sin t,0,0)$ and $0\leq t\leq2\pi$ . Write $X(t)=\cos t\,\Gamma_1+\sin t\,\Gamma_2$ and $T(t)=-\sin t\,\Gamma_1+\cos t\,\Gamma_2$ . The speed is $R$ , and $TX=-\Gamma_{12}$ , $X\Gamma_{12}=T$ , where $\Gamma_{12}=\Gamma_1\Gamma_2$ . After dividing (N11.3) by $R$ , its left side is

(iT+S)(\epsilon_Q+RX\epsilon_S) =(S\epsilon_Q-iR\Gamma_{12}\epsilon_S) +(iT\epsilon_Q+RSX\epsilon_S). \tag{N11.6}

The first parentheses is independent of $t$ and the second has only sine and cosine terms. Thus the constant term requires

\boxed{\epsilon_S=\frac{i}{R}\Gamma_{12}S\epsilon_Q.} \tag{N11.7}

This also kills the angular term: $SX\Gamma_{12}S=-T$ . Every constant $\epsilon_Q$ then supplies a solution, and the constant equation permits no extra one. There are sixteen independent complexified directions, mixing Poincaré and conformal transformations. No nonzero pure Poincaré parameter is preserved by the whole circle with this fixed scalar direction: setting $\epsilon_S=0$ in the constant term gives $S\epsilon_Q=0$ .

The inverse radius in the relation is required by dimensions. The matrix $\Gamma_{12}S$ switches chirality, as needed. In the reduced timelike-Clifford notation, replacing $iS$ by the chosen matrix $\Gamma^0$ translates the relation into Pestun (2.19). That source subsequently selects a smaller subalgebra for its four-sphere localization construction; we have counted the full flat conformal theory’s loop symmetry.

A circle can be supersymmetric and still have a nontrivial answer

Use Feynman gauge for the first weak-coupling term. In the absorbed normalization,

\langle\mathcal A_\mu^a(x)\mathcal A_\nu^b(y)\rangle_0 =\frac{g^2\delta^{ab}\delta_{\mu\nu}}{4\pi^2|x-y|^2},\qquad \langle\Phi_I^a(x)\Phi_J^b(y)\rangle_0 =\frac{g^2\delta^{ab}\delta_{IJ}}{4\pi^2|x-y|^2}. \tag{N11.8}

The two factors of $i$ in the gauge insertion give a minus sign. At separated points the combined contour contraction is

\langle\mathcal L^a(t)\mathcal L^b(s)\rangle_0 =\frac{g^2\delta^{ab}}{4\pi^2} \frac{|\dot x(t)||\dot x(s)|-\dot x(t)\cdot\dot x(s)}{|x(t)-x(s)|^2}. \tag{N11.9}

On a straight line with the same orientation at both points, the numerator vanishes. On the circle it is $R^2[1-\cos(t-s)]$ and the denominator is $2R^2[1-\cos(t-s)]$ . Their ratio is $1/2$ , with a smooth coincident limit in this combined kernel. Keep this common point-splitting prescription: separate gauge and scalar self-contractions are singular.

Define $C_F$ by $\sum_a t^at^a=C_F1$ . The ordered second term of (N11.1) now gives

\begin{aligned} \langle W\rangle &=1+\int_0^{2\pi}dt\int_0^t ds\, \frac{g^2 C_F}{8\pi^2}+O(g^4)\\ &=1+\frac{g^2 C_F}{4}+O(g^4). \end{aligned} \tag{N11.10}

There is no additional $1/2!$ on an already ordered triangular domain. For $SU(N)$ , taking the trace of the defining Casimir equation gives $NC_F=(N^2-1)/2$ , hence

\langle W\rangle_{SU(N)} =1+\frac{\lambda}{8}\left(1-\frac1{N^2}\right)+O(\lambda^2), \qquad\lambda=g^2N. \tag{N11.11}

For the explicitly different $U(N)$ theory, adjoining $t^0=1/\sqrt{2N}$ adds $1/(2N)$ to $C_F$ and gives $C_F=N/2$ . Its first correction is $\lambda/8$ , including at $N=1$ . The finite-color distinction matters before any planar limit. These are weak-coupling expansions at fixed $N$ ; nothing in this calculation sums higher-loop diagrams.

This reproduces the contour mechanism in Drukker–Gross (2.21)–(2.23) while retaining the traceless color correction. The source also explains why a conformal map taking the line through infinity cannot be used to equate its normalized expectation value with the circle’s. Symmetry of the insertion does not set its expectation value to one. The exact circle answer requires the separately constructed sphere theory and localization argument.

Exercises

1. Detune the scalar coupling

Replace $vS$ by $\alpha vS$ , with real constant $\alpha$ , and keep a regular contour. Which $\alpha$ permit a nonzero pointwise spinor kernel? For a straight line, calculate the separated-point contraction and identify the short-distance problem when $\alpha=0$ .

Hint 1

Square $i\dot x\cdot\Gamma+\alpha vS$ using its anticommuting pieces.

Hint 2

The scalar exchange acquires $\alpha^2$ , while the two gauge factors of $i$ retain their sign.

Solution

The square is $(\alpha^2-1)v^2$ . For $\alpha^2\ne1$ this is a nonzero multiple of the identity, so the matrix is invertible and has no kernel. Both $\alpha=+1$ and $-1$ allow the local cancellation; the negative choice reverses the scalar direction and changes which symmetries survive.

With $x^1=t$ , the contraction is $g^2\delta^{ab}(\alpha^2-1)/[4\pi^2(t-s)^2]$ . At $\alpha=0$ the uncanceled inverse-square singularity produces a regulator-dependent local line divergence. It is not a finite negative prediction for an unrenormalized Wilson line. Unit scalar coupling cancels this particular singularity and permits supersymmetry; it alone does not prove global supersymmetry for a general contour.

2. Stretch the circle into an ellipse

Keep $n$ fixed, but take $x=(a\cos t,b\sin t,0,0)$ with positive unequal $a,b$ . At every point (N11.3) still has a half-dimensional kernel. Show that no nonzero global parameter of the form (N11.4) solves this condition. It suffices to compare the opposite endpoints of the two axes.

Hint 1

At $t=0,\pi$ , divide by the local speed $b$ and add the two equations. At $t=\pi/2,3\pi/2$ , divide by $a$ and do the same.

Hint 2

Both sums relate $S\epsilon_Q$ to $i\Gamma_{12}\epsilon_S$ , but their coefficients are different semiaxes.

Solution

The first pair yields $S\epsilon_Q-ia\Gamma_{12}\epsilon_S=0$ ; the second yields $S\epsilon_Q-ib\Gamma_{12}\epsilon_S=0$ . Subtracting gives $i(a-b)\Gamma_{12}\epsilon_S=0$ . Since $\Gamma_{12}^2=-1$ and $a\ne b$ , it follows that $\epsilon_S=0$ and then $\epsilon_Q=0$ .

For $a=b$ the two conditions coincide and become the circle relation; thus the test has the correct limit. The failure concerns the fixed scalar-direction probe we defined. It does not classify loops with a position-dependent scalar coupling, different background fields, or a different line operator.

3. Reparametrize and wind the same circle

Let $x=R(\cos f(t),\sin f(t),0,0)$ with $f(t)=m(t+\eta\sin t)$ , $0\leq t\leq2\pi$ , positive integer $m$ and $|\eta|<1$ . Keep the normalized fundamental trace and fixed $n$ . Find the preserved-spinor relation and the order- $g^2$ correction. Evaluate it for $SU(3)$ at $m=2$ . Explain why this is not the product of $m$ independent loop expectations.

Hint 1

$f'>0$ , so the local spinor equation after dividing by the speed depends on the geometric angle $f$ , not on how quickly it is traversed.

Hint 2

The combined contour ratio is $f'(t)f'(s)/2$ . Integrate over the ordered triangle and use $f(2\pi)-f(0)=2\pi m$ .

Solution

The same relation $\epsilon_S=i\Gamma_{12}S\epsilon_Q/R$ works at every angle, including repeated visits. The ordered contraction is

\frac{g^2C_F}{8\pi^2}\int_0^{2\pi}dt\,f'(t) \int_0^t ds\,f'(s) =\frac{g^2C_Fm^2}{4}.

It is independent of $R$ and $\eta$ . We use the same combined-kernel limiting prescription also at repeated contour points; the numerator and denominator cancel before that limit. This calculation establishes the stated order only, not all higher-order coincident-contour properties.

For $SU(3)$ , $C_F=4/3$ , so $m=2$ gives $\langle W_2\rangle=1+4g^2/3+O(g^4)=1+4\lambda/9+O(\lambda^2)$ . The single insertion is the trace of the holonomy traversed twice. Contractions can join different traversals, producing the quadratic $m^2$ dependence. Multiplying independently averaged single-loop answers would instead give a linear $m$ correction and describes a different observable.

The line and circle now have a defined representation, scalar coupling, preserved symmetry and first quantitative check. Next, put the theory on the four-sphere and construct the spinor and curvature coupling needed to turn this observable into a finite integral while retaining these normalizations.