Bosonic supersymmetric Wilson loops

An ordinary Wilson loop follows a gauge connection around a closed curve. Supersymmetry varies that connection into a gaugino, so its holonomy is generally not invariant. The vector scalar supplies a second variation that can cancel the first. We will derive the required combination, evaluate it on the localization locus, and calculate an expectation value that retains a nontrivial complex phase.

Start from the finite-rank partition-function check and the distinction between BPS and coupling independent. We keep compact $U(N)_k\times U(N)_{-k}$, equal rank, canonical matter normalization, no extra center quotient, and the round sphere of radius $\ell$. Our main example has integer $k>2$.

Generic ABJM has twelve real Poincaré and twelve real conformal supercharges. The action and localization calculation select a three-dimensional $\mathcal N=2$ subalgebra. A bosonic circular loop of the type below is called one-sixth BPS relative to that generic full superconformal algebra. It is not the fermionic half-BPS loop, and the label does not make its expectation value independent of $k$ or $N$.

Derive the line connection

Let $a_\mu,\sigma$ be the first node’s Hermitian gauge connection and vector scalar. Our charged derivative is $\partial_\mu-i a_\mu$. Along a regular curve $x(\tau)$, write $v=|\dot x|$ and $n^\mu=\dot x^\mu/v$. Consider

$$\mathscr L_c(\tau)=i a_\mu\dot x^\mu+c\,\sigma v, \qquad W_{\mathcal R}[c]=\frac1{\dim\mathcal R} \operatorname{Tr}_{\mathcal R}\, \mathcal P\exp\!\left(\oint d\tau\,\mathscr L_c\right). \tag{AB15.1}$$

The trace is in a specified representation $\mathcal R$ of the first $U(N)$. The symbol $\mathcal P$ orders matrices along the path. The factor $1/\dim\mathcal R$ makes the operator equal to one when its connection vanishes. Both $a_\mu$ and $\sigma$ have dimension one, so both integrated terms are dimensionless.

Only two component variations are needed. With independent commuting Euclidean labels $\zeta,\widetilde\zeta$ and independent odd gauginos,

$$\delta a_\mu=-i(\zeta\gamma_\mu\widetilde\lambda+ \widetilde\zeta\gamma_\mu\lambda), \qquad \delta\sigma=-\zeta\widetilde\lambda+\widetilde\zeta\lambda. \tag{AB15.2}$$

These are the conventions of P, (6.20), applied to each color component. The two displayed variations retain this form for a non-Abelian node; no field-strength commutator is needed in this calculation. Multiplication by an odd external parameter would turn the odd differential $\delta$ into an ordinary even variation.

Substitution into (AB15.1) gives

$$\delta\mathscr L_c =v\left[\zeta(\gamma\cdot n-c)\widetilde\lambda+ \widetilde\zeta(\gamma\cdot n+c)\lambda\right]. \tag{AB15.3}$$

Because the two gauginos are independent, their coefficients must vanish separately. The Euclidean Clifford algebra gives $(\gamma\cdot n)^2=1$. Thus a nonzero preserved label requires $c=+1$ or $-1$. Choose $c=+1$ from now on.

To avoid an index-sign trap, recall that lower-column spinors contract as $\zeta\chi=\zeta^{\mathsf T}C\chi$, where $C=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$ and $\gamma_a^{\mathsf T}C=-C\gamma_a$. The row conditions in (AB15.3) are equivalent to

$$\gamma\cdot n\,\zeta=-\zeta,\qquad \gamma\cdot n\,\widetilde\zeta=+\widetilde\zeta. \tag{AB15.4}$$

These signs refer to the site’s lower-column convention. KWY’s row/column notation gives its own projectors in (2.6)–(2.7).

The cancellation also works inside a non-Abelian holonomy. If $U(\tau_2,\tau_1)$ denotes parallel transport, then

$$\delta U(T,0)=\int_0^T U(T,\tau)\,\delta\mathscr L(\tau)\,U(\tau,0)\,d\tau=0. \tag{AB15.5}$$

No commuting-matrix approximation was made. Gauge transformations conjugate the closed holonomy, so its trace is gauge invariant.

Choose a curve and a global preserved charge

The inherited round background has $H=-i/\ell$ and vanishing R connection, $V$ and $C_\mu$. Both label families obey $\nabla_\mu\zeta=i\gamma_\mu\zeta/(2\ell)$, with the same equation for $\widetilde\zeta$. In an oriented global sphere frame $E_a$, the spin connection contributes $i\gamma_a/(2\ell)$, so constant spinor components solve these equations.

Choose a Hopf great circle tangent to $E_1$. These are closed integral curves of a unit invariant sphere vector, with length $2\pi\ell$. In the gamma basis $\gamma_1=\operatorname{diag}(1,-1)$, take

$$n=E_1,\qquad \zeta=\begin{pmatrix}0\\1\end{pmatrix},\qquad \widetilde\zeta=\begin{pmatrix}1\\0\end{pmatrix}. \tag{AB15.6}$$

The projectors hold everywhere, not just at one point. Each rigid label family loses one of its two complex solution directions. This verifies the supersymmetry needed for localization. It does not count extra physical charges by imposing or removing Euclidean conjugacy conditions.

The full conformal ABJM statement is that the standard bosonic circle preserves four real combinations out of twenty-four, hence one sixth. On a conformally related straight line these split into two Poincaré and two conformal generators; on the circle they mix. Drukker–Trancanelli, section 3.1, distinguishes these four from those of the fermionic loop. At exceptional levels one and two the bulk algebra is enhanced; the generic one-sixth label is not a fresh count of every possible enhancement there.

What scalar does the loop couple to?

The off-shell operator uses $\sigma$ because its transformation is simple. To see its ordinary matter content, use the algebraic auxiliary equations of the physical theory. Put

$$Y^I=(A_1,A_2,B_1^\dagger,B_2^\dagger),\qquad M=\operatorname{diag}(1,1,-1,-1). \tag{AB15.7}$$

Every $Y^I$ maps the second color space to the first. The terms linear in the two auxiliary matrices are

$$\operatorname{Tr}_L D\left(\mu_L-\frac{k}{2\pi}\sigma\right) +\operatorname{Tr}_R\widehat D \left(\mu_R+\frac{k}{2\pi}\widehat\sigma\right), \tag{AB15.8}$$

where $\mu_L=\sum(A_iA_i^\dagger-B_i^\dagger B_i)$ and $\mu_R=\sum(B_iB_i^\dagger-A_i^\dagger A_i)$. Varying the two independent auxiliaries gives

$$\sigma=\frac{2\pi}{k}\sum_I M_I{}^I\,Y^IY_I^\dagger, \qquad \widehat\sigma=\frac{2\pi}{k}\sum_I M_I{}^I\,Y_I^\dagger Y^I. \tag{AB15.9}$$

This agrees with AB, Appendix (A.3). The ordering changes because the two expressions act on different color spaces. It is not a cyclic rearrangement inside an untraced matrix.

Thus the first-node loop couples to the bilinear $(2\pi/k)(A_iA_i^\dagger-B_i^\dagger B_i)$, with no extra factor of $i$ outside it in (AB15.1). The second node uses the other ordered expression in (AB15.9). This explains the two positive and two negative scalar couplings of KWY (4.1), with the coefficient restored for our canonical matter normalization.

These are equations of the undeformed auxiliary action. They must not be imposed on the localization locus after adding the deformation: doing so would incorrectly set the localized $\sigma$ to zero when the localized matter vanishes. The off-shell definition is what can be carried through that deformation.

The insertion is a normalized character

For the chosen preserved charge, use the same gauge-fixed localization calculation and contour as for the partition function. On its locus,

$$a=\widehat a=0,\qquad \sigma=\ell^{-1}\operatorname{diag}(s_1,\ldots,s_N),\qquad \widehat\sigma=\ell^{-1}\operatorname{diag}(t_1,\ldots,t_N). \tag{AB15.10}$$

The auxiliary values remain $D=i\sigma/\ell$ and $\widehat D=i\widehat\sigma/\ell$. The $s_i,t_i$ are real. Path ordering becomes trivial on this locus; the integral of $\sigma$ around the great circle is $2\pi\ell\sigma$.

$$w_{\mathcal R}(s)= \frac{\chi_{\mathcal R}(e^{2\pi s})}{\dim\mathcal R}, \qquad w_L(s)=\frac1N\sum_i e^{2\pi s_i},\qquad w_R(t)=\frac1N\sum_i e^{2\pi t_i}. \tag{AB15.11}$$

Here $\chi_{\mathcal R}$ means the trace of the displayed matrix in representation $\mathcal R$; $w_L,w_R$ are fundamental representations of their respective nodes. A trace over both nodes or a product of their traces is a different operator. The radius cancels between the circumference and the definition of $s$.

Let $d\mu_k(s,t)$ denote the complete previously derived matrix measure, including $(N!)^{-2}$, the two root products, all matter determinant factors and phase $e^{i\pi k(\sum s_i^2-\sum t_i^2)}$. Then

$$\langle W_L\rangle_{\rm loc} =\lim_{\epsilon\downarrow0} \frac{\int d\mu_k(s,t)\, e^{-\epsilon\sum(s_i^2+t_i^2)}\,w_L(s)} {\int d\mu_k(s,t)\, e^{-\epsilon\sum(s_i^2+t_i^2)}}. \tag{AB15.12}$$

The limit is part of the prescription and needs checking. The general insertion follows KWY (3.69); the positive first-node phase is the one inherited here and also displayed in DT (3.16). KWY (4.4) displays the opposite phase, requiring a level-sign translation or node exchange when comparing complex results.

Supersymmetric point splitting displaces the loop onto a neighboring Hopf fiber. This specifies a framing, the ribbon used to regulate coincident points. We retain that localization prescription, with no finite line redefinition. In our boundary orientation on $S^3\subset\mathbb C^2$, equally oriented Hopf fibers link $+1$. The first-node Abelian pure-Chern–Simons calibration follows directly from a Gaussian:

$$\lim_{\epsilon\downarrow0} \frac{\int_{\mathbb R}ds\,e^{-(\epsilon-i\pi k)s^2+2\pi s}} {\int_{\mathbb R}ds\,e^{-(\epsilon-i\pi k)s^2}} =\lim_{\epsilon\downarrow0} e^{\pi^2/(\epsilon-i\pi k)}=e^{i\pi/k}.$$

KWY’s discussion after (3.33) calls its prescription framing $-1$ in its conventions. The actual oriented prescription and matrix phase above fix our meaning; importing a framing label alone does not fix a complex answer. The pure-theory calibration does not determine a general framing conversion in the matter theory.

A finite-rank calculation with an unbounded insertion

At $N=1$, $w_L=e^{2\pi s}$. This grows on the real contour, so the bounded-test-function argument for the partition function cannot simply be reused. Let $U_\epsilon$ be the numerator of (AB15.12), set $x=s-t$, $y=(s+t)/2$, and write $f(x)=1/[4\cosh^2(\pi x)]$. Integrating the ordinary convergent $y$ Gaussian first gives

$$U_\epsilon= \sqrt{\frac{\pi}{2\epsilon}}\int_{\mathbb R}dx\,f(x) \exp\!\left[-\frac{\epsilon x^2}{2}+\pi x+ \frac{\pi^2(1+ikx)^2}{2\epsilon}\right]. \tag{AB15.13}$$

The narrow Gaussian is now centered at a complex location. The matter factor has poles at $x=i(n+\tfrac12)$, $n\in\mathbb Z$. For $k>2$, shifting the contour to $x=z+i/k$, $z\in\mathbb R$, crosses none of them. At each fixed $\epsilon>0$, the quadratic decay makes the connecting vertical segments vanish.

Put $C_\epsilon=\pi^2k^2+\epsilon^2$. Completing the exponent after this shift yields

$$U_\epsilon=\frac{\pi}{\sqrt{C_\epsilon}}\, e^{i\pi/k+\epsilon/(2k^2)} \mathbb E_\epsilon\!\left[ e^{(\pi-i\epsilon/k)z}f(z+i/k)\right], \qquad z\sim\mathcal N\!\left(0,\frac{\epsilon}{C_\epsilon}\right). \tag{AB15.14}$$

This is a real Gaussian expectation of a complex function. Moreover, $|f(z+i/k)|\leq[4\cos^2(\pi/k)]^{-1}$. Writing $z=\sqrt{\epsilon/C_\epsilon}\,u$ gives a fixed Gaussian weight; its product with $e^{\pi\sqrt{\epsilon/C_\epsilon}|u|}$ is uniformly integrable for sufficiently small $\epsilon$. Dominated convergence therefore gives $U_\epsilon\to e^{i\pi/k}f(i/k)/k$. Divide by the already proved $Z_\epsilon\to1/(4k)$:

$$\boxed{\langle W_L\rangle_{\rm loc} =\frac{e^{i\pi/k}}{\cos^2(\pi/k)},\qquad \langle W_R\rangle_{\rm loc} =\frac{e^{-i\pi/k}}{\cos^2(\pi/k)}, \quad N=1,\ k>2.} \tag{AB15.15}$$

The second equality follows by exchanging the nodes and complex conjugating the regulated measure. At $k=4$, the first answer is $\sqrt2(1+i)$. Its modulus is two, not one: the supersymmetric loop includes a real scalar exponential and is not a unitary parallel transporter.

This proves the stated Abel limit in a pole-free domain. It does not prove that substituting $k=1$ or $2$ into an analytically continued formula evaluates the original regulated integral. Nor does the restriction make those physical theories meaningless. A different domain requires its own contour or continuation prescription.

As a separate computational check, the matter Fourier transform is $\widehat f(\omega)=\omega/[2\sinh(\pi\omega)]$, obtained by a beta integral. Inserting its Fourier representation into the original two-variable Gaussian gives a real-frequency integral that agrees with (AB15.14). Its limiting decay condition is again $2/k<1$. This tests the phase and the convergence restriction by a different order of calculation.

Exercises

1. Change the scalar coefficient

For the same oriented circle and labels (AB15.6), which constant $c$ makes $W_{\mathcal R}[c]$ invariant? Can $c=-1$ preserve other labels? What happens if the tangent is reversed while retaining $c=1$?

Hint 1

The chosen labels fix eigenspaces, not merely the eigenvalues available to some unspecified spinor.

Hint 2

For a unit tangent, the determinant of $\gamma\cdot n-c$ is $c^2-1$. Reversing $n$ exchanges its two eigenspaces.

Solution

For the displayed labels, use $\zeta^{\mathsf T}C=(1,0)$ and $\widetilde\zeta^{\mathsf T}C=(0,-1)$. Equation (AB15.3) becomes $v(1-c)(\widetilde\lambda_1+\lambda_2)$, so these labels require $c=1$. The purely gauge loop, $c=0$, is not invariant.

For $c=-1$, choose instead $\zeta=(1,0)^{\mathsf T}$ and $\widetilde\zeta=(0,1)^{\mathsf T}$. These solve the opposite projectors. No other constant $c$ admits a nonzero label because $c^2-1$ would be nonzero.

Reversing the tangent with $c=1$ also exchanges the selected eigenspaces. The gauge term changes sign, but $\sigma|\dot x|$ does not. Thus reversing the parametrization is not the same operation as negating the entire supersymmetric line connection. Keep the curve, scalar sign and representation separate.

2. Distinguish three traces

For one $U(2)$ node, let $e^{2\pi s}$ have eigenvalues $u_1,u_2$. Find the normalized fundamental insertion, a twice-wound fundamental insertion, and the insertion in the symmetric square of the fundamental. Compare them with the square of the fundamental insertion at $u_1=2,u_2=3$.

Hint 1

The symmetric-square states have weights $u_1^2,u_1u_2,u_2^2$, each once. Its dimension is three.

Hint 2

Winding twice squares the holonomy before taking one trace. Squaring a trace gives two independently summed color indices.

Solution

The four expressions are

$$\begin{aligned} w_{\square}&=\frac{u_1+u_2}{2},& w_{\square}^{(2)}&=\frac{u_1^2+u_2^2}{2},\\ w_{\operatorname{Sym}^2}&=\frac{u_1^2+u_1u_2+u_2^2}{3},& (w_\square)^2&=\frac{(u_1+u_2)^2}{4}. \end{aligned} \tag{AB15.16}$$

Their respective values are $5/2$, $13/2$, $19/3$ and $25/4$. All become one at $u_1=u_2=1$, so the trivial-connection check alone would not distinguish them. More generally, $\chi_{\operatorname{Sym}^2}(U)=\tfrac12[(\operatorname{Tr}U)^2+\operatorname{Tr}U^2]$, but its dimension is $N(N+1)/2$ and must also be included.

This is a comparison of insertions before integration. It does not assert factorization of two loop expectation values.

3. Wind the rank-one loop

For positive integer winding $p$, the localized insertion is $e^{2\pi p s}$. Extend the contour proof to a domain where no matter pole is crossed. Find the result and its phase, and state why the fundamental bound on $k$ is no longer enough.

Hint 1

Replace $1+ikx$ in (AB15.13) by $p+ikx$ and $\pi x$ by $\pi p x$.

Hint 2

The useful contour shift is $x=z+ip/k$. Check its location against the nearest pole at $i/2$ before taking the limit.

Solution

The pole-free condition is $k>2p$. After the shift, the finite-regulator numerator becomes

$$U_{\epsilon,p}=\frac{\pi}{\sqrt{C_\epsilon}}\, e^{i\pi p^2/k+\epsilon p^2/(2k^2)} \mathbb E_\epsilon\!\left[ e^{(\pi p-i\epsilon p/k)z} f(z+ip/k)\right]. \tag{AB15.17}$$

The bound on the shifted matter factor is now $|f(z+ip/k)|\leq[4\cos^2(\pi p/k)]^{-1}$, finite in that domain. The same fixed-width Gaussian argument gives

$$\boxed{\langle W_L^{(p)}\rangle_{\rm loc} =e^{i\pi p^2/k}\sec^2(\pi p/k),\qquad k>2p.} \tag{AB15.18}$$

The phase depends on $p^2$, not $p$. The answer is also not the $p$th power of (AB15.15): a wound trace is a different insertion from a product of traces. For example, $k=5,p=2$ lies in the proved domain, whereas $k=3,p=2$ does not. An analytic continuation to the latter must not be advertised as the contour argument just given.

The bosonic operator now has a fixed node, representation, scalar coupling and regularization prescription. Adding fermionic couplings changes the operator and its supersymmetry variation; relating it to a bosonic loop requires an additional argument, not just a replacement of the trace in (AB15.11).

Fermionic loops, superconnections, and framing extends this operator by odd off-diagonal couplings. Its endpoint condition determines the correct trace before the two loop averages can be compared.

The localization paper companion checks the source prescription on a linked pair of loops and explains why shared supersymmetry does not imply factorization.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading three-dimensional localization with matter