between $t_{L}$ and the insertion point of the $\psi_{-}$ in the next fermion
pair that we denote by $u$, see figure 13,

$\int\limits_{t_{L}}^{u}\frac{\mathrm{d}t_{L+1}}{(t_{L+1}-t_{0}+(L+1)\tilde{\epsilon})^{2}}=-\frac{1}{u-t_{0}+(L+1)\tilde{\epsilon}}+\frac{1}{t_{L}-t_{0}+(L+1)\tilde{\epsilon}}\,.$
(4.9)

For $L>0$ the second contribution precisely cancels the term of order
$\lambda^{L}$ in $\mathtt{F}^{(L-1)}$ (the second term in the parenthesis of
(4.8)). For $L=0$, it gives a power divergence that we can tune the counter
term to cancel, by setting $f=\lambda\,b$ in (4.6).

We remain with the contribution of the upper limit of integration. At this
upper limit, there are no longer empty regions on the line. Instead, we have
two types of points. One that corresponds to a gluon insertion

(4.10)

The other corresponds to the upper limit of a $\bar{\psi}$ integration

(4.11)

Here, $t_{0}$ is a point inside the line, were a $\psi_{-}$ is inserted. Upon
integration over $t$, each of these two contributions leads to a bulk
logarithmic divergence and corresponding bulk RG flow. To cancel these two
divergences and have an RG fixed point, we must set $b=1$ (of course $b=0$ is
also a trivial solution). The same also applies to all $t_{i}>t$, with either
$t_{0}\to t$ for (4.11) or that $t_{0}$ remands the same for (4.10).

The perturbative expansion now takes the form

$\langle\widetilde{M}^{({1\over 2},-{1\over
2})}\rangle=\sum_{L=0}^{\infty}\widetilde{\cal
M}_{L}^{f}\,,\quad\text{with}\quad{\widetilde{\cal
M}}^{f}_{L}=\int\limits_{0}^{1}\mathcal{P}\prod_{j=1}^{L}\mathrm{d}t_{j}\,\widetilde{\mathtt{F}}^{(L)}(\\{t_{j}+j\tilde{\epsilon}\\}_{j=1}^{L};t_{L+1}+(L+1)\tilde{\epsilon})\,.$
(4.12)

where the $L$ loop term is an $L$ points line integrand
$\widetilde{\mathtt{F}}_{b}^{(L)}$, similar to (2.33) and (3.22). Each
integration point can be of the type (4.10) or (4.11), and we have to sum over
all the corresponding possibilities. As before, the argument of the lines
integrand appears with a shift of $t_{j}$ by $j\tilde{\epsilon}$, so that
$t_{j}$ is integrated exactly up to $t_{j+1}$, and
$t_{L+1}=1+(L+1)\tilde{\epsilon}$, (see explanation below (2.33)).

Figure 14: The line integrand $\widetilde{F}^{(L)}$ satisfies a simple
recursion relation (14). It is derived by summing up the two vertex
possibilities at the last integration point $t_{L}$. The vertex can either be
a gluon insertion (4.10) or the $\bar{\psi}$ vertex insertion (4.11). The sum
of the two yields the one lower loop integrand $\widetilde{F}^{(L-1)}$, times
$\lambda/(t_{L+1}-t_{L})$.

By setting $b=1$ and iterating the relation

${1\over(s-t)(t-u)}+{1\over(t-u)(u-s)}+{1\over(u-s)(s-t)}=0\,,$ (4.13)

from the left as, (see figure 14),

$\displaystyle\widetilde{\mathtt{F}}^{(L)}(\\{t_{j}\\}_{j=1}^{L};t_{L+1})=$
$\displaystyle\sum_{{\rm level}\
L}\left(\frac{\lambda}{(t_{L+1}-t_{L})(t_{L}-t_{i})}-\frac{\lambda}{(t_{L+1}-t_{i})(t_{L}-t_{i})}\right)\times\dots$
$\displaystyle=$ $\displaystyle\sum_{{\rm level}\
L}\frac{\lambda}{(t_{L+1}-t_{L})(t_{L+1}-t_{i})}\times\dots$ (4.14)
$\displaystyle=$ $\displaystyle{\lambda\over
t_{L+1}-t_{L}}\times\widetilde{\mathtt{F}}^{(L-1)}(\\{t_{j}\\}_{j=1}^{L-1};t_{L+1})\,,$

we arrive at

$\widetilde{\mathtt{F}}^{(L)}(\\{t_{j}\\}_{j=1}^{L};t_{L+1})=\lambda^{L}\prod_{j=1}^{L}\frac{1}{t_{L+1}-t_{j}}\,.$
(4.15)

This is precisely of the form (2.43) that we have already encountered in the
bosonic case. We, therefore, conclude that232323It is interesting to note that
in our regularization scheme, for any value of $b$, the following relation
holds  $\langle\delta_{x_{0}^{3}}\widetilde{M}_{b}^{({1\over 2},-{1\over
2})}\rangle=-4\pi\lambda\langle M\rangle|_{\alpha=1-2b}\,.$ (4.16) This match
along the RG flow is however only a technical coincidence.

$\langle\widetilde{M}^{({1\over 2},-{1\over 2})}\rangle=4\pi x\langle
M\rangle\big{|}_{\alpha=-1}+{\cal
O}(\tilde{\epsilon})={1\over\Gamma(1+\lambda)}\Big{(}{\epsilon\,e^{\gamma_{E}}\over
x}\Big{)}^{-\lambda}+{\cal O}(\tilde{\epsilon})\,,$ (4.17)

where here $\lambda$ is the same parameter on both sides of the first
equality, not to be confused with the relation between the couplings of the
bosonic and fermionic theories (3.44).

###  4.2 The Four Towers of Operators

We denote the empty boundary operators considered above in accordance with
their tree-level spin and the number of longitudinal derivatives as

$\widetilde{\cal O}_{L}^{(0,{1\over 2})}=\text{empty
left}\,,\qquad\widetilde{\cal O}_{R}^{(0,-{1\over 2})}=\text{empty
right}\,,\qquad\text{and}\qquad\widetilde{\cal
O}^{(n,s)}=\delta_{3}^{n}\widetilde{\cal O}^{(0,s)}\,.$ (4.18)

In particular, by taking a single longitudinal path derivative we arrive at
the operators

$\widetilde{\cal O}_{L}^{(1,{1\over 2})}=-\sqrt{{4\pi\over
k}}\bar{\psi}_{+}-\frac{\lambda}{\epsilon}\,\qquad\text{and}\qquad\widetilde{\cal
O}_{R}^{(1,-{1\over 2})}=\sqrt{{4\pi\over
k}}\psi_{-}+\frac{\lambda}{\epsilon}\,.$ (4.19)

The anomalous dimension of these descendant operators is the same as that of
(4.18), equal to $-\lambda/2$. It is opposite from the anomalous dimension of
the same fermion components at the boundary of a standard Wilson line, ${\cal
O}^{(0,{1\over 2})}_{L}=\bar{\psi}_{+}$ and ${\cal O}^{(0,-{1\over
2})}_{R}=\psi_{-}$ in (3.45). Namely, the deformation of the Wilson line
action by the fermion condensation in (4.6) has the effect of flipping their
anomalous dimension. On the other hand, this deformation does not affect the
anomalous dimension of all boundary operators other than $\widetilde{\cal
O}_{L}^{(n,{1\over 2})}$ and $\widetilde{\cal O}_{R}^{(n,-{1\over 2})}$. That
is because all these operators are orthogonal to the fermions in (4.6) on a
straight line, and therefore are insensitive to them. Their anomalous
dimension and anomalous spin remain the same as if they were placed at the
boundary of the standard Wilson line and are given in (3.10), (3.36), and
(3.40). Altogether, we have four towers of boundary operators that are built
on top of the operators

$\\{\widetilde{\cal O}_{R}^{(0,-{1\over 2})},\widetilde{\cal
O}_{R}^{(0,-{3\over 2})}\\}\,\qquad\text{and}\qquad\\{\widetilde{\cal
O}_{L}^{(0,{3\over 2})},\widetilde{\cal O}_{L}^{(0,{1\over 2})}\\}\,.$ (4.20)

Operators in each tower are related to each other by path derivatives. They
have the same anomalous dimension and anomalous spin. This spectrum of
boundary operator match exactly the one of the $\alpha=-1$ line operator in
the bosonic theory, provided that the ’t Hooft couplings of the two
descriptions are related as in (3.44). In particular, the four bottom
operators (4.20) and (2.50) map to each other as

$\begin{tabular}[]{llllcc}\hline\cr\hline\cr{Fermionic}&{Tree}&{Bosonic}&{Tree}&$\mathfrak{s}$&$\Delta$\\\
\hline\cr\hline\cr$\ \widetilde{\cal O}_{R}^{(0,-{1\over 2})}\quad$&$\
\rm{Empty}\qquad$&$\ {\cal O}_{R}^{(0,0)}\quad$&$\
\phi$&$-{\lambda_{\text{bos}}\over 2}$&$\frac{1-\lambda_{\text{bos}}}{2}$\\\
\hline\cr$\ \widetilde{\cal O}_{R}^{(0,-{3\over 2})}\quad$&$\
\partial_{-}\psi_{-}\qquad$&$\ {\cal O}_{R}^{(0,-1)}\quad$&$\
\partial_{-}\phi\quad$&$-{2+\lambda_{\text{bos}}\over
2}$&$\frac{3+\lambda_{\text{bos}}}{2}$\\\ \hline\cr\hline\cr$\ \widetilde{\cal
O}_{L}^{(0,{1\over 2})}\quad$&\ \rm{Empty}&$\ {\cal O}_{L}^{(0,0)}\quad$&$\
\phi^{\dagger}\quad$&${\lambda_{\text{bos}}\over
2}$&$\frac{1-\lambda_{\text{bos}}}{2}$\\\ \hline\cr$\ \widetilde{\cal
O}_{L}^{(0,{3\over 2})}\quad$&$\ \partial_{+}\bar{\psi}_{+}\qquad$&$\ {\cal
O}_{L}^{(0,1)}\quad$&$\
\partial_{+}\phi^{\dagger}\quad$&${\lambda_{\text{bos}}+2\over
2}$&$\frac{3+\lambda_{\text{bos}}}{2}$\\\ \hline\cr\hline\cr\end{tabular}\,.$
(4.21)

###  4.3 The Evolution Equation

As before, there is a unique spin one, dimension two primary displacement
operator. It is given by

${\mathbb{D}}_{+}\propto\widetilde{\cal O}_{R}^{(0,-{1\over
2})}\,\widetilde{\cal O}_{L}^{(0,{3\over
2})}\,,\qquad{\mathbb{D}}_{-}\propto\widetilde{\cal O}_{R}^{(0,-{3\over
2})}\,\widetilde{\cal O}_{L}^{(0,{1\over 2})}\,.$ (4.22)

Using the identification in (4.21), this is precisely the form of the
displacement operator of the line operator in the bosonic theory with
$\alpha=-1$, (2.86). As in the computation of the evolution equations
appearing in previous sections 2.9 and 3.5, the dependence on the framing
vector is topological and does not contribute.

To fix the relative coefficient in (4.22) we now vary the condensed fermion
operator explicitly. As before, this is a formal functional variation of the
operator that ignores renormalization. We have four types of contributions,
coming from the functional variation of $\Gamma$, $P_{-}$, $A$, $\psi$, and
$\bar{\psi}$ in (4.7). We consider each one of these in turn. Without loss of
generality, we specify to a straight line.

##### Variation of the Topological Phase

The topological phase $\int\mathrm{d}\log n^{+}$ in the lower component of
$\widetilde{A}$ (4.3) combines with the same factor for the Wilson line in the
upper component (I.2) to an overall topological phase factor (3.10). Hence, it
does not contribute to a smooth internal variation of the path.

##### Variation of $\Gamma$

Along the empty parts of the line, we have the topological spin transport

$U(s,t)\equiv{\cal
P}\exp{\Big{[}\mathrm{i}\int\limits^{x_{s}}_{x_{t}}\mathrm{d}x^{\mu}\,\Gamma_{\mu}}\Big{]}\,,$
(4.23)

It is topological by construction. Hence, its variation is a boundary term

$\delta U(s,t)=V(s)U(s,t)-U(s,t)V(t)\,.$ (4.24)

As we show in equation (C.2), for a straight line

$V(s)\propto(\dot{v}_{s}\times e_{s})\cdot\gamma\,.$ (4.25)

This term is sandwich between $\bar{\psi}(s)P_{-}$ and the parallel transport
of $P_{-}\psi(t)$. For a straight line however, ${\left[(\dot{v}_{s}\times
e_{s})\cdot\gamma\right]_{-+}}=0$, so this term does not contribute. In other
words, the two fermion components in (4.6) have an opposite transverse spin
and therefore cannot contract with a spin one variation.

##### The Variation of the Wilson Line

Second, we have the variation of the Wilson line. As we explained before
(3.49), it amounts to inserting the field strength,

$\mathrm{i}\,\dot{x}^{\nu}v^{\mu}F_{\nu\mu}\quad\rightarrow\quad\frac{2\pi\mathrm{i}}{k}\dot{x}^{\nu}v^{\mu}\epsilon_{\nu\mu\rho}\bar{\psi}\gamma^{\rho}\psi\,.$
(4.26)

##### The Variation of $P_{-}$

The third contribution is the variation of the projector

$\delta_{v}\,\dot{x}\cdot
P_{-}=\dot{v}\cdot\left(e\mathbb{I}-\gamma\right)/2\,.$ (4.27)

For a straight line, $\dot{v}\cdot e=0$, and therefore the identity component
vanishes. We remain with

$\dot{x}\cdot\left(\begin{array}[]{cc}0&\delta_{v}P_{-}\psi(s)\\\
\bar{\psi}(s)\delta_{v}P_{-}&0\end{array}\right)=-\frac{1}{2}\sqrt{\frac{4\pi}{k}}\dot{v}_{s}^{\mu}\,\left(\begin{array}[]{cc}0&-\gamma_{\mu}\,\psi(s)\\\
-\bar{\psi}(s)\,\gamma_{\mu}&0\end{array}\right)\,.$ (4.28)

This operator is inserted on the line operator
$\mathcal{P}\exp(\mathrm{i}\int\mathrm{d}x\cdot\widetilde{A})$ and integrated
over $s$. Next, we perform integration by parts to remove the derivative from
$\dot{v}_{s}$. Using the relation (4.19) (without regularization), we arrive
at

$-v_{s}^{\mu}\,\left(\\!\\!\begin{array}[]{cc}\frac{2\pi}{k}\bar{\psi}^{a}\left[\gamma_{\mu},\dot{x}_{s}\cdot
P_{-}\right]^{b}_{a}\psi_{b}&\sqrt{\pi\over k}\gamma_{\mu}\dot{x}_{s}\cdot
D\psi\\\ \sqrt{\pi\over k}\,\dot{x}_{s}\cdot
D\bar{\psi}\gamma_{\mu}&0\end{array}\\!\\!\right)\,.$ (4.29)

The commutator term evaluates to

$\left[v^{\mu}\gamma_{\mu},\dot{x}_{\nu}P_{-}^{\nu}\right]=\mathrm{i}\dot{x}^{\nu}v^{\mu}\epsilon_{\nu\mu\rho}\gamma^{\rho}\,.$
(4.30)

It exactly cancels with the Wilson line variation (4.26). Note that for this
cancellation to happen, it is crucial that $b=1$.

##### The Variation of $\psi$ and $\bar{\psi}$

The fourth and final contribution is the variation of the fermions in (4.6),

$\left(\begin{array}[]{cc}0&P_{-}\delta_{v}\psi\\\
\delta_{v}\bar{\psi}P_{-}&0\end{array}\right)=\left(\begin{array}[]{cc}0&P_{-}v\cdot
D\psi\\\ v\cdot D\bar{\psi}P_{-}&0\end{array}\right)\,.$ (4.31)

Because of the spin of the condensated fermion, the following fermion
equations of motion are not affected by the condensation

$\displaystyle D_{3}\psi_{+}+\sqrt{2}D_{+}\psi_{-}=0\,,\quad
D_{3}\bar{\psi}_{-}+\sqrt{2}D_{-}\bar{\psi}_{+}=0\,.$ (4.32)

Using these and the relations

$\gamma^{-}P_{-}=P_{-}\gamma^{+}=0\,,\qquad\gamma^{+}P_{-}=\gamma^{+}\,,\qquad
P_{-}\gamma^{-}=\gamma^{-}\,,$ (4.33)

we see that one of the off-diagonal terms in (4.29) is zero while the other
cancels out with the corresponding off-diagonal term in (4.31). We end up with

$\displaystyle{\mathbb{D}}_{+}$ $\displaystyle=\sqrt{4\pi\over
k}\left(\begin{array}[]{cc}0&0\\\
D_{+}\bar{\psi}_{+}&0\end{array}\right)=\sqrt{4\pi\lambda}\,\widetilde{\cal
O}_{R}^{(0,-{1\over 2})}\,\widetilde{\cal O}_{L}^{(0,{3\over 2})}\,,$ (4.36)
$\displaystyle{\mathbb{D}}_{-}$ $\displaystyle=\sqrt{4\pi\over
k}\left(\begin{array}[]{cc}0&D_{-}\psi_{-}\\\
0&0\end{array}\right)=\sqrt{4\pi\lambda}\,\widetilde{\cal O}_{R}^{(0,-{3\over
2})}\,\widetilde{\cal O}_{L}^{(0,{1\over 2})}\,,$ (4.39)

in agreement with (4.22).

###  4.4 The Boundary Equation

The same logic as for other line operators (see 2.8 and 3.5 ) leads to the
following form of the boundary equation for right operators,

$\widetilde{\mathcal{O}}_{R}^{(n+2,s)}\propto\left\\{\begin{array}[]{ll}\delta_{-}\widetilde{\mathcal{O}}_{R}^{(n,s+1)}&\quad
s\geq-{1\over 2}\\\ \delta_{+}\widetilde{\mathcal{O}}_{R}^{(n,s-1)}&\quad
s\leq-{3\over 2}\end{array}\right.\,,$ (4.40)

and for left operators,242424Here, the framing is always taken perpendicular
to the line and to the direction of deformation, as explained in the footnote
above (2.48).

$\widetilde{\mathcal{O}}_{L}^{(n+2,s)}\propto\left\\{\begin{array}[]{ll}\delta_{-}\widetilde{\mathcal{O}}_{L}^{(n,s+1)}&\quad
s\geq{3\over 2}\\\ \delta_{+}\widetilde{\mathcal{O}}_{L}^{(n,s-1)}&\quad
s\leq{1\over 2}\end{array}\right.\,.$ (4.41)

Next, we derive the proportionality constant $-2$ for right operators (4.40).
The derivation for left operators follows the same steps. Recall that the
expectation value of straight mesonic line operators with $s_{R}\neq-1/2$ are
unaffected by the condensation. Hence, for these boundary equations the factor
of $-2$ carries over from the result of the previous section 3.5. It remains
to show that

$\widetilde{\cal O}_{R}^{(2,-{1\over 2})}=-2\delta_{-}{\cal O}_{R}^{(0,{1\over
2})}\,.$ (4.42)

In appendix E we verify (4.42) by an explicit computation.

###  4.5 The Two-point Function of the Displacement Operator

By plugging the displacement operator in (4.36) into (2.96), we arrive at

$\Lambda=4\pi\lambda\,x^{4}\langle\widetilde{M}^{(\frac{1}{2},-\frac{1}{2})}\rangle\langle\widetilde{M}^{(\frac{3}{2},-\frac{3}{2})}\rangle=4\pi\lambda\,x^{4}\langle\widetilde{M}^{(\frac{1}{2},-\frac{1}{2})}\rangle\langle\"{O}\textquestiondown\widetilde{M}^{(\frac{3}{2},-\frac{3}{2})}\rangle\,.$
(4.43)

Using the explicate expressions for these expectation values in (4.17) and
(3.43), we reproduce (2.100) with $\Delta=-\lambda_{\text{fer}}/2$ or
$\Delta=2+\lambda_{\text{fer}}/2$, in agreement with the duality.

##  5 Comments About Line RG Flow at Large $N$

We have found that there are two conformal line operators in the fundamental
representation. They correspond to two fixed points in a one-parameter family.
In the bosonic description, this parameter is the coefficient $\alpha$ in
front of the scalar bi-linear in the line action (2.3). In this section, we
study the RG flow which takes place in this one-parameter family of lines.

An RG flow on the line is triggered by local operators in the line action. In
the planar limit, these operators factorize into a product of right and left
boundary operators as

$\mathcal{O}_{\text{inner}}=\mathcal{O}_{R}\times\mathcal{O}_{L}\,\qquad\text{with}\qquad\Delta(\mathcal{O}_{\text{inner}})=\Delta(\mathcal{O}_{R})+\Delta(\mathcal{O}_{L})+{\cal
O}(1/N)\,.$ (5.1)

Consider first the $\alpha=1$ fixed point. The operator on the line with the
minimal conformal dimension is the bi-scalar. Its conformal dimension is

$\Delta(\mathcal{O}_{R}^{(0,0)}\times\mathcal{O}_{L}^{(0,0)})=1+\lambda\geq
1\qquad\text{for}\qquad\alpha=1\,.$ (5.2)

Hence, the $\alpha=1$ fixed point is stable.

In contrast, when $\alpha=-1$, the scalar bi-linear with dimension
$\Delta=1-\lambda\leq 1$ is relevant. It is the only relevant deformation of
the $\alpha=-1$ fixed point, whose coefficient grows along the RG flow. The
sign of the relevant coupling determines the direction of that flow. Given
that the $\alpha=1$ is the only stable fixed point in this one-parameter
family, there is a definite flow toward this point. If the sign of the
relevant coupling is reversed, the magnitude of the bi-scalar coefficient will
continue to grow. To understand where this flow ends, we turn to the fermionic
description.

Figure 15: The coefficient of the scalar bi-linear $\alpha$ in (2.3) undergoes
an RG flow on the line. The fixed point at $\alpha=1$ is stable while the
fixed point at $\alpha=-1$ is unstable. Starting from the unstable fixed
point, two flows exist. One moves in the direction of the stable fixed point,
while the other moves in the other direction toward a topological spin
transport that is essentially trivial (4.2).

The fermionic dual of the $\alpha=-1$ operator is the condensed fermion
operator (4.6) with $b=1$. When expended in perturbation theory, the
corresponding line is decomposed from “empty” segments with the topological
spin connection (4.2) and “filled” segments with a Wilson line between a pair
of fundamental and anti-fundamental fermions, see figure 12. The duals of the
fundamental scalar boundary operators are the empty endpoints. Deforming the
bosonic line action by the scalar bi-linear is consequently equivalent, to
leading order in the deformation parameter, to modifying the coefficient in
front of $P_{-}$ in the bottom right component of the connection
$\widetilde{A}$ in (4.6). This modification changes the weight of the empty
segments. It thus generates an RG flow on the line in which the empty segments
grow or shrink accordingly. This flow ends when the line is either completely
filled or completely empty. The wholly filled line is the Wilson line
operator, dual to the $\alpha=1$ operator in the bosonic description. The
completely empty line is an almost trivial line, with only the topological
spin connection (4.2). Figure 15 shows the RG flow picture that emerges.
Another support for this RG picture comes from the beta function for the
scalar bi-linear. In section 2.6 we have found that it is proportional to
$\beta_{\alpha}\propto(1-\alpha^{2})$.

Curiously, the $\alpha=-1$ unstable line operator is the endpoint of a
different RG flow. To describe it, we work in the fermionic description and
start with the direct sum of the Wilson line and an anyonic parallel
transport, defined by the connection $\tilde{A}_{0}$ in (4.3). As a direct sum
of conformal lines, it is conformal as well. The line operator

${\cal
O}_{\text{inner}}=-{\mathrm{i}\over\sqrt{k}}\left(\begin{array}[]{cc}0&P_{-}\psi\\\
\bar{\psi}P_{-}&0\end{array}\right)\,,$ (5.3)

is relevant and therefore can be used to trigger an RG flow on the line. It
ends when the coefficient of this operator reaches the value $\sqrt{4\pi}$,
which is the condensed fermion conformal line operator (4.6).252525Note that
the boundary entropy $g$ of a direct sum of two lines is $g_{1\oplus
2}=\log\left(e^{g_{1}}+e^{g_{2}}\right)$. Hence, the direct sum of the Wilson
line operator with the trivial parallel transport gives an operator with an
higher $g$-function, in agreement with the $g$-theorem Cuomo:2021rkm .

Finally, the same RG flow picture 15 has appeared in Aharony:2022ntz  when
studying an RG flow on a line of the four-dimensional QED in the large charge
limit. It also appeared in the study of spin impurities in Cuomo:2022xgw . It
would be interesting to understand how these cases are related.

##  6 Perturbative Bootstrap of the Line Integrand

In this section, we demonstrate how the evolution and boundary equations of a
mesonic line operator can be used to recursively reconstruct its perturbative
expansion. The drawback of the approach presented here is that the
manipulations are formal. This is because we do not perform the line
integrations, which are the only place where divergences, regularization, and
renormalization of the line come about. In particular, the framing
regularization of the line is not taken into account. Hence, these approach
can be applied to any line operator, conformal or not. For the sake of
simplicity of the presentation, we simply set $\alpha=0$ in (2.3) and adjust
the form of the displacement operator accordingly. We will show explicitly how
the resulting formal solution reproduces the result of perturbation theory at
one loop order.

To reconstruct the perturbative expansion of the line, we introduce a set of
functions $\mathtt{B}^{(L)}_{\mu_{1}\ldots\mu_{m}}$. They are generalizations
of the line integrand in (2.37) that also includes the gluon-gluon exchange on
the line. We bootstrap them recursively from the corresponding evolution and
boundary equations. In spirit, the manipulations in this section are in line
with the original Makeenko-Migdal solution to the four-dimensional loop
equation Makeenko:1979pb .

While working formally, the main point of this section is that the evolution
and boundary equations are very restrictive constraints. Motivated by this, in
bootstrap  we present a physical non-perturbative bootstrap of the
renormalized mesonic line operators. Indeed, we find there that the evolution
and boundary equations combined with the spectrum of the boundary operators
determine the expectation values uniquely.

###  6.1 Setup

We consider the mesonic line operator in the bosonic theory that is formally
defined as

$M[{\cal
C}_{10}]\equiv\langle\phi^{\dagger}(x_{1})\,W[\mathcal{C}_{10}]\phi(x_{0})\rangle=\sum_{L=0}^{\infty}\lambda^{L}M^{(L)}[{\cal
C}_{10}]\,.$ (6.1)

Here, ${\cal C}_{10}$ is some arbitrary smooth path between $x_{0}$ and
$x_{1}$ and $W$ is the standard Wilson line operator in (3.2). We start by
writing an ansatz for the perturbative solution of the form

$M^{(L)}[\mathcal{C}_{10}]\equiv{1\over
4\pi}\sum_{m=0}^{2L}\int_{s_{1}>s_{2}>\dots>s_{m}}\mathrm{d}s_{1}\dots\mathrm{d}s_{m}\
\dot{x}_{s_{1}}^{\mu_{1}}\cdots\dot{x}_{s_{m}}^{\mu_{m}}\,\mathtt{B}^{(L)}_{\mu_{1}\ldots\mu_{m}}(x_{s_{1}},\dots,x_{s_{m}})\
,$ (6.2)

where the $\mathtt{B}^{(n)}_{\mu_{1}\ldots\mu_{m}}(s_{1},\dots,s_{m})$ is a
dimension $m+1$ function of $(x_{s_{1}},\dots,x_{s_{m}})$ as well as $x_{0}$
and $x_{1}$, but is independent of the path $\mathcal{C}$. We think of these
points as the $L$-loop contribution from $m$ gluons inserted on the line.
However, we will not use Feynman diagrams directly. The relation between
$\mathtt{B}^{(L)}_{\mu_{1}\ldots\mu_{m}}$ and the integrand
$\mathtt{B}_{\alpha=0}^{(L)}$ in (2.33) will not be relevant for us.262626It
involves fixing a light cone gauge and performing some of the line
integrations, which correspond to the gluon exchanges on the line.

The gauge symmetry of the CS theory manifests itself in an ambiguity in the
form of the ansatz (6.2). Namely, (6.2) is invariant under simultaneous
transformations

$\displaystyle\mathtt{B}^{(L)}_{\mu_{1}\ldots\mu_{m}}\rightarrow\mathtt{B}^{(L)}_{\mu_{1}\ldots\mu_{m}}+{\partial\over\partial
x^{\mu_{j}}_{s_{j}}}\Omega_{\mu_{1},\dots\mu_{j-1}\mu_{j+1}\dots\mu_{m}}(x_{s_{j}}|x_{s_{1}},\dots,x_{s_{j-1}},x_{s_{j+1}},\dots,x_{s_{m}})\
,$ (6.3)
$\displaystyle\mathtt{B}^{(L)}_{\mu_{1}\ldots\mu_{m-1}}\rightarrow\mathtt{B}^{(L)}_{\mu_{1}\ldots\mu_{m-1}}+\Omega_{\mu_{1},\dots\mu_{j-1}\mu_{j+1}\dots\mu_{m}}(x_{s_{j-1}}|x_{s_{1}},\dots,x_{s_{j-1}},x_{s_{j+1}},\dots,x_{s_{m}})$
$\displaystyle\hskip
108.0pt-\Omega_{\mu_{1},\dots\mu_{j-1}\mu_{j+1}\dots\mu_{m}}(x_{s_{j+1}}|x_{s_{1}},\dots,x_{s_{j-1}},x_{s_{j+1}},\dots,x_{s_{m}})\
,$

with $s_{0}=0$ and $s_{m+1}=1$. Note that this ambiguity is more than the
original CS gauge symmetry because we can use different functions, $\Omega$,
for different segments of the line.

Our aim is to demonstrate that up to this gauge ambiguity, $\mathtt{B}^{(L)}$
is uniquely fixed by the boundary and evolution equations. In other words,
$\mathtt{B}^{(L)}$ in (6.2) is fixed by the scalar and gauge field equations
of motion restricted to the line. For this, it is crucial to include all
contact terms, which are sources when the line has a self-crossing point or
when the two endpoints coincide.

Consider the boundary equation first. This equation takes the form272727It
originated from the following gauge non-invariant Schwinger-Dyson equation:
$0=\int[\mathcal{D}A][\mathcal{D}\phi]\,\frac{\delta}{\delta\phi^{i}(x_{1})}\Big{[}e^{-S}\,(\mathcal{P}e^{\mathrm{i}\int_{{\cal
C}_{10}}A\cdot\mathrm{d}x})^{i}_{\ j}\phi(x_{0})^{j}\Big{]}\,,$ where repeated
color indices $i,j$ are being summed over.

$\Box_{x_{0}}M[\mathcal{C}_{10}]=\Box_{x_{1}}M[\mathcal{C}_{10}]=-\delta^{3}(x_{0}-x_{1})\times{1\over
N}{\rm tr~{}}W[\mathcal{C}_{10}]\ ,$ (6.4)

where $\Box=\\{\delta_{+},\delta_{-}\\}+\delta_{3}^{2}$ is the boundary
Laplacian. The right-hand side in (6.4) is a contact term that is sourced when
the two endpoints coincide. It is proportional to the closed Wilson loop that
is obtained by joining the endpoints together. This term was not relevant
before when we classified the boundary operators at finite coupling.

Second, we consider the evolution equation. It is obtained by setting
$\alpha=0$ in (2.90)-(2.9.2) and reads

Figure 16: The evolution equation for a mesonic line operator with a Wilson
line along the path ${\cal C}_{10}$, is given in (6.5). Local functional
variation of the path result in the insertion of the field strength operator.
When substituting the equation of motion $F=J$, an additional contact term
contributes at self-crossing points. In the planar limit, both terms factorize
into a product of two lines. In particular, the self crossing term factorize
into a product of a closed Wilson loop $W[\widetilde{\mathcal{C}}_{ss}]$ times
a mesonic line along the path $\widetilde{{\cal C}}_{10}$ in the figure.

$\displaystyle{\delta M[\mathcal{C}_{10}]\over\delta
x^{\mu}_{s}}=2\pi\mathrm{i}\lambda\,\epsilon_{\mu\nu\rho}\dot{x}^{\nu}_{s}\Big{[}$
$\displaystyle\Big{(}{\delta\over\delta
x^{\rho}_{s}}M[\mathcal{C}_{1s}]\Big{)}M[\mathcal{C}_{s0}]-M[\mathcal{C}_{1s}]\Big{(}{\delta\over\delta
x^{\rho}_{s}}M[\mathcal{C}_{s0}]\Big{)}$ (6.5)
$\displaystyle-\int\limits_{\mathcal{C}_{10}}\mathrm{d}x_{t}^{\rho}\,\delta^{3}(x_{s}-x_{t})M[\widetilde{\mathcal{C}}_{10}]\times{1\over
N}{\rm tr~{}}W[\widetilde{\mathcal{C}}_{ss}]\Big{]}+{\cal O}(1/N)\,.$

Here the term in the second line is the contact term that is sourced when the
contour has a self-crossing point, similar to (6.6). At such point, the
operator factorizes into the product of a closed Wilson loop times a mesonic
line, see figure 16.

The closed Wilson loop on the right-hand side of (6.4) and (6.5) satisfies a
loop equation that is the same as the one satisfied by the mesonic line
operator in (6.5). The only difference is that instead of an open line between
fundamental and anti-fundamental scalars, we have a closed loop with a color
trace. As a result, the first term in (6.5) becomes $1/N$ suppressed and we
remained with

$\frac{\delta}{\delta x_{s}^{\mu}}{\rm
tr~{}}W[\mathcal{C}_{10}]=-{2\pi\mathrm{i}\over
k}\int\limits_{\mathcal{C}_{10}}\mathrm{d}x_{t}^{\rho}\,\delta^{3}(x_{s}-x_{t})\,{\rm
tr~{}}W[\widetilde{\mathcal{C}}_{st}]\times{\rm
tr~{}}W[\widetilde{\mathcal{C}}_{ts}]+{\cal O}(1/N)\,.$ (6.6)

Here, the loop at any of the self-crossing point $x_{s}$ splits into two
disjoint closed loops, denoted by $\widetilde{\mathcal{C}}_{ts}$ and
$\widetilde{\mathcal{C}}_{st}$, respectively. This equation can also be
derived by plugging (6.4) into (6.5).

###  6.2 Tree Level

At tree level, (6.2) is independent of the shape of the path, so the boundary
Laplacian $\Box_{x}$ reduces to the normal Laplacian $\partial^{2}_{x}$. The
boundary equation (6.4) then implies that

$\mathtt{B}^{(0)}={1\over|x_{0}-x_{1}|}\,,$ (6.7)

where we used the three dimensional Green’s function
$\partial^{2}_{x}\frac{1}{|x|}=-4\pi\delta^{3}(x)$. One may worry about
additional constant and linear solutions to (6.7). These however are ruled out
by translation symmetry and the fact that $\mathtt{B}^{(0)}$ has dimension
one.

###  6.3 One Loop

At loop level, the boundary Laplacian, which is decomposed from path
derivatives $\Box=\\{\delta_{+},\delta_{-}\\}+\delta_{3}^{2}$, also act in the
line integrations. To map the boundary equation (6.4) into a constraint on
$\mathtt{B}^{(L)}$, it is convenient to use the following identities

$\displaystyle\Box_{x}\int\limits_{x}^{x_{1}}\mathrm{d}x^{\mu}_{s}\,\Gamma_{\mu}(x,x_{s})=\int\limits_{x}^{x_{1}}\mathrm{d}x^{\mu}_{s}\,\partial^{2}_{x}\Gamma_{\mu}(x,x^{\mu}_{s})-2\partial_{u}^{\mu}\Gamma_{\mu}(u,x)\Big{|}_{u=x}\,,$
$\displaystyle\Box_{x}\int\limits_{x}^{x_{1}}\mathrm{d}x_{s}^{\mu}\int\limits_{x}^{x_{s}}\mathrm{d}x^{\nu}_{t}\,\Gamma_{\mu\nu}(x,x_{s},x_{t})=\int\limits_{x}^{x_{1}}\mathrm{d}x_{s}^{\mu}\Big{(}\int\limits_{x}^{x_{s}}\mathrm{d}x^{\nu}_{t}\,\partial^{2}_{x}\Gamma_{\mu\nu}(x,x_{s},x_{t})-2\partial^{\sigma}_{y}\Gamma_{\mu\sigma}(y,x_{s},x)\Big{|}_{y=x}\Big{)}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad+\Gamma_{\mu}^{\
\mu}(x,x,x)\,,$ (6.8)
$\displaystyle\Box_{x}\int\limits_{x}^{x_{1}}\mathrm{d}x_{s}^{\mu}\int\limits_{x}^{x_{s}}\mathrm{d}x^{\nu}_{t}\int\limits_{x}^{x_{t}}\mathrm{d}x^{\rho}_{u}\,\Gamma_{\mu\nu\rho}(x,x_{s},x_{t},x_{u})=\int\limits_{x}^{x_{1}}\mathrm{d}x_{s}^{\mu}\int\limits_{x}^{x_{s}}\mathrm{d}x^{\nu}_{t}\int\limits_{x}^{x_{t}}\mathrm{d}x^{\rho}_{u}\,\partial^{2}_{x}\Gamma_{\mu\nu\rho}(x,x_{s},x_{t},x_{u})$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad+\int\limits_{x}^{x_{1}}\mathrm{d}x_{s}^{\mu}\Big{(}\Gamma_{\mu\nu}^{\
\
\nu}(x,x_{s},x,x)-2\int\limits_{x}^{x_{s}}\mathrm{d}x^{\nu}_{t}\,\partial^{\sigma}_{y}\Gamma_{\mu\nu\sigma}(y,x_{s},x_{t},x)\Big{|}_{y=x}\Big{)}\,,$

where we recall that $\partial^{2}_{x}$ denotes the three dimensional
Laplacian. The last equation also applies to multiple line integration which
depends on four or more integration variables.

At one loop order we have to solve for the three functions $\mathtt{B}^{(1)}$,
$\mathtt{B}^{(1)}(x)$, and $\mathtt{B}^{(1)}(x_{s},x_{t})$, that also depend
on $x_{0}$ and $x_{1}$ but not on the path ${\cal C}$. We plug the ansatz
(6.2) into the box equation (6.4) and evaluate both sides at order $\lambda$.
Using (6.3) with $x=x_{0}$ we get

$\displaystyle\Box_{x_{0}}M^{(1)}[\mathcal{C}_{10}]=$
$\displaystyle\frac{\lambda}{4\pi}\Big{[}\partial^{2}_{x_{0}}\mathtt{B}^{(1)}-2\partial_{x_{0}}^{\mu}\mathtt{B}_{\mu}^{(1)}(x_{0})+\int\limits_{0}^{1}\mathrm{d}s\,\dot{x}_{s}^{\mu}\,\partial^{2}_{x_{0}}\mathtt{B}_{\mu}^{(1)}(x_{s})$
(6.9)
$\displaystyle\qquad+\eta^{\mu\nu}\mathtt{B}^{(1)}_{\mu\nu}(x_{0},x_{0})-2\int\limits_{0}^{1}\mathrm{d}s\,\dot{x}_{s}^{\nu}\partial_{x_{0}}^{\mu}\mathtt{B}^{(1)}_{\nu\mu}(s,0)+\int\limits_{s>t}\\!\mathrm{d}s\,\mathrm{d}t\,\dot{x}_{s}^{\mu}\dot{x}_{t}^{\nu}\,\partial^{2}_{x_{0}}\mathtt{B}^{(1)}_{\mu\nu}(x_{s},x_{t})\Big{]}$
$\displaystyle=$ $\displaystyle-\delta^{3}(x_{0}-x_{1})\times{1\over N}{\rm
tr~{}}W[\mathcal{C}_{10}]\Big{|}_{\mathcal{O}(\lambda)}\,.$

This equation should hold for an arbitrary shape of the path
$\mathcal{C}_{10}$. At the level we work at, we must treat the expectation
value of the closed Wilson loop in the last line on the same footing as we
treat the mesonic line operator. Namely, we make the following ansatz for it

${1\over N}{\rm
tr~{}}W[\mathcal{C}_{10}]\Big{|}_{\mathcal{O}(\lambda)}=\frac{\lambda}{2}\int\limits_{s>t}\\!\mathrm{d}s\,\mathrm{d}t\,\dot{x}_{s}^{\mu}\dot{x}_{t}^{\nu}\,\mathtt{G}_{\mu\nu}(x_{s},x_{t})+\mathcal{O}(\lambda^{2})\
,\ $ (6.10)

where terms with three or more integration points are suppressed by powers of
the coupling. Plugging the ansatz into (6.6), to leading order in $\lambda$,
the equation that determines $\mathtt{G}_{\mu\nu}$ is simply

${\partial\over\partial
x_{s}^{\mu}}\mathtt{G}_{\nu\rho}(x_{s},x_{t})-{\partial\over\partial
x_{s}^{\nu}}\mathtt{G}_{\mu\rho}(x_{s},x_{t})=-2\pi\mathrm{i}\lambda\,\epsilon_{\mu\nu\rho}\,\delta^{3}(x_{s}-x_{t})\,.$
(6.11)

where the equation is understood to hold up to total derivatives.282828This
equation originates from the integral equation
$\oint_{\mathcal{C}_{10}}\mathrm{d}x^{\rho}_{t}(\mathtt{LHS}-\mathtt{RHS})=0$,
so the solution is up to a total derivative term that vanishes for a closed
loop. We can identify $\mathtt{G}_{\mu_{1}\mu_{2}}$ with the gauge dependent
gluon propagator in position space

$\langle A_{\mu}^{I}(x)A_{\nu}^{J}(y)\rangle={1\over
k}\mathtt{G}_{\mu\nu}(x-y)\delta^{IJ}\,.$ (6.12)

By comparing the double line integration in (6.9), we conclude that up to
total derivatives, $\mathtt{B}^{(1)}_{\mu\nu}(x_{s},x_{t})$ is the gluon
propagator multiplies a free scalar propagator $1/|x_{01}|$,

$\mathtt{B}^{(1)}_{\mu\nu}(x_{s},x_{t})=\frac{1}{2|x_{01}|}\mathtt{G}_{\mu\nu}(x_{s},x_{t})+{\partial\over\partial
x_{s}^{\mu}}F_{\nu}(x_{s},x_{t})+{\partial\over\partial
x_{t}^{\nu}}\tilde{F}_{\mu}(x_{s},x_{t})\,.$ (6.13)

We can always use the redundancy (LABEL:gaugered) to set $F=\tilde{F}=0$. This
is equivalent to redefining $\mathtt{B}^{(1)}_{\mu}$.

At this point, it is convenient to resolve the ambiguity (LABEL:gaugered) by
fixing a gauge. A choice that is equivalent to Lorentz gauge in the CS theory
is to pick

$\partial_{x_{s_{j}}}^{\mu_{j}}\mathtt{B}^{(n)}_{\mu_{1}\ldots\mu_{m}}(x_{s_{1}},\dots,x_{s_{m}})=0\,.$
(6.14)

From (6.13) it follows that $\mathtt{G}_{\mu\nu}$ has to satisfy the same
gauge constraint. This gauge condition still allows a linear gauge
transformation such as $\Omega(x)=v_{\mu}x^{\mu}$. Demanding that the
$\mathtt{B}^{(L)}$ decays at infinity require $v^{\mu}=0$, and no gauge
redundancy is left, up to a constant $\Omega$ that decouples from $M$. In this
gauge, we have that

$\mathtt{B}^{(1)}_{\mu\nu}(x_{s},x_{t})=\frac{1}{2|x_{01}|}\mathtt{G}_{\mu\nu}(x_{s},x_{t})=\frac{\mathrm{i}}{2|x_{01}|}\epsilon_{\mu\nu\rho}\partial^{\rho}_{x_{s}}\frac{1}{|x_{s}-x_{t}|}\,.$
(6.15)

After plugging (6.15) into the boundary equation (6.9), it reduces to

$\partial^{2}_{x_{0}}\mathtt{B}^{(1)}_{\mu}(x_{s})=2\partial_{x_{0}}^{\nu}\mathtt{B}^{(1)}_{\mu\nu}(x_{s},x_{t})\Big{|}_{t=0}\
,\qquad\partial^{2}_{x_{0}}\mathtt{B}^{(1)}=2\partial_{x_{0}}^{\mu}\mathtt{B}^{(1)}_{\mu}(x_{t})\Big{|}_{t=0}\,.$
(6.16)

Next, we move to the evolution equation (6.5). By plugging (6.2) into it and
imposing the Lorentz-type gauge (6.14) we get for the left-hand side

$\displaystyle{\delta M[\mathcal{C}_{10}]\over\delta x^{\mu}_{s}}=$
$\displaystyle{\lambda\over
4\pi}\dot{x}_{s}^{\nu}\Big{[}{\partial\over\partial
x_{s}^{\mu}}\mathtt{B}^{(1)}_{\nu}(x_{s})-{\partial\over\partial
x_{s}^{\nu}}\mathtt{B}^{(1)}_{\mu}(x_{s})$ (6.17) $\displaystyle+$
$\displaystyle\int\limits_{0}^{1}\mathrm{d}t\,\dot{x}^{\rho}_{t}\Big{(}{\partial\over\partial
x_{s}^{\mu}}\mathtt{B}^{(1)}_{\nu\rho}(x_{s},x_{t})-{\partial\over\partial
x_{s}^{\nu}}\mathtt{B}^{(1)}_{\mu\rho}(x_{s},x_{t})\Big{)}\Big{]}+{\cal
O}(\lambda^{2})\,,$

where we used that
$\mathtt{B}^{(1)}_{\mu\nu}(x,y)=\mathtt{B}^{(1)}_{\nu\mu}(y,x)$ and our
Lorentz gauge choice (6.14). The term $\mathtt{B}^{(1)}_{\mu\nu}(x_{s},x_{t})$
in (6.15) satisfies the Bianchi identity up to a contact term when
$x_{s},x_{t}$ coincides,

${\partial\over\partial
x_{s}^{\mu}}\mathtt{B}^{(1)}_{\nu\rho}(x_{s},x_{t})+{\partial\over\partial
x_{s}^{\nu}}\mathtt{B}^{(1)}_{\rho\mu}(x_{s},x_{t})+{\partial\over\partial
x_{s}^{\rho}}\mathtt{B}^{(1)}_{\mu\nu}(x_{s},x_{t})=-\frac{2\pi\mathrm{i}}{|x_{01}|}\epsilon_{\mu\nu\rho}\,\delta^{(3)}(x_{s}-x_{t})\,.$
(6.18)

By plugging (6.18) into (6.17) and using that
$\partial_{x_{s}}^{\rho}\mathtt{B}^{(1)}_{\mu\nu}(x_{s},x_{t})=-\partial_{x_{t}}^{\rho}\mathtt{B}^{(1)}_{\mu\nu}(x_{s},x_{t})$
and
$\mathtt{B}^{(1)}_{\mu\nu}(x_{s},x_{t})=-\mathtt{B}^{(1)}_{\nu\mu}(x_{s},x_{t})$,
we express the second line of (6.17) as a sum of contact and a boundary terms

$\displaystyle\int\limits_{0}^{1}\mathrm{d}t\,\dot{x}^{\rho}_{t}\Big{(}{\partial\over\partial
x_{s}^{\mu}}\mathtt{B}^{(1)}_{\nu\rho}(x_{s},x_{t})-{\partial\over\partial
x_{s}^{\nu}}\mathtt{B}^{(1)}_{\mu\rho}(x_{s},x_{t})\Big{)}$ (6.19)
$\displaystyle=$
$\displaystyle\,\mathtt{B}_{\mu\nu}^{(1)}(x_{s},x_{1})-\mathtt{B}_{\mu\nu}^{(1)}(x_{s},x_{0})-\frac{2\pi\mathrm{i}}{|x_{01}|}\epsilon_{\mu\nu\rho}\int\limits_{0}^{1}\mathrm{d}t\,\dot{x}^{\rho}_{t}\delta^{(3)}(x_{s}-x_{t})\,.$

The right-hand side of the evolution equation (6.5) is already of order
$\lambda$. Hence, we only need to plug in the tree-level expressions (6.7)
there. We get

$\mathtt{RHS}=\lambda\left(\frac{\mathrm{i}}{2}\epsilon_{\mu\nu\rho}\dot{x}_{s}^{\nu}V^{\rho}(x_{s})-2\pi\mathrm{i}\,\epsilon_{\mu\nu\rho}\dot{x}_{s}^{\nu}\int\mathrm{d}x_{t}^{\rho}\delta^{3}(x_{s}-x_{t})\frac{1}{4\pi|x_{10}|}\right)+{\cal
O}(\lambda^{2})\,,$ (6.20)

where

$V^{\rho}(x_{s})\equiv\frac{1}{|x_{0s}|}\partial_{x_{s}}^{\rho}\frac{1}{|x_{1s}|}-\frac{1}{|x_{1s}|}\partial_{x_{s}}^{\rho}\frac{1}{|x_{0s}|}\,.$
(6.21)

The contact terms term in (6.19) and (6.20) cancel against each other. The
evolution equation then reduces to the following relation between functions of
points, with no dependence on the path $\mathcal{C}_{10}$

${\partial\over\partial
x_{s}^{\mu}}\mathtt{B}^{(1)}_{\nu}(x_{s})-{\partial\over\partial
x_{s}^{\nu}}\mathtt{B}^{(1)}_{\mu}(x_{s})+\mathtt{B}_{\mu\nu}^{(1)}(x_{s},x_{1})-\mathtt{B}_{\mu\nu}^{(1)}(x_{s},x_{0})=\frac{\mathrm{i}}{2}\epsilon_{\mu\nu\rho}V^{\rho}(x_{s})\,.$
(6.22)

Acting on both sides with $\partial^{\mu}_{x_{s}}$ and using the gauge
condition (6.14), this equation reduces to

$\partial^{2}_{x}\mathtt{B}^{(1)}_{\nu}(x)=\frac{\mathrm{i}}{2}\epsilon_{\mu\nu\rho}\partial^{\mu}_{x}V^{\rho}(x)\,.$
(6.23)

The rotational covariant solution to this Laplace’s equation that decays at
infinity is

$\mathtt{B}^{(1)}_{\mu}(x)=\frac{\mathrm{i}}{2}\int\frac{\mathrm{d}^{3}y}{4\pi|x-y|}\epsilon_{\mu\nu\rho}\partial^{\nu}_{y}V^{\rho}(y)\,.$
(6.24)

It reminds to solve for $\mathtt{B}^{(1)}$, which is a function of $x_{0}$ and
$x_{1}$ only. The remaining non-trivial constraint from the boundary equation
(6.9) is

$\partial^{2}_{x_{0}}\mathtt{B}^{(1)}=2\partial_{x_{0}}^{\mu}\mathtt{B}^{(1)}_{\mu}(x_{t})\Big{|}_{t=0}\,.$
(6.25)

From the explicit form of $\mathtt{B}^{(1)}_{\mu}(x)$ in (6.24) one can check
that $\partial_{x_{0}}^{\mu}\mathtt{B}^{(1)}_{\mu}(x_{t})\Big{|}_{t=0}=0$.
Therefore, we have to solve $\partial^{2}_{x_{0}}\mathtt{B}^{(1)}=0$. The only
dimension one translation and rotation symmetric solution is
$\mathtt{B}^{(1)}=0$.

We conclude that at one-loop order, the combination of the evolution equation
and the box equation determine the solution up to the gauge freedom
(LABEL:gaugered). The result agrees with the explicit 1-loop calculation in
the Lorentz gauge, presented in the appendix A. We can proceed systematically
like this and reconstruct the full formal perturbative expansion of the
mesonic line operator (6.1). Turning on a non-zero value for $\alpha$ adds
terms to the equations, but does not lead to any ambiguity in their solution.

##  7 Lift to Superconformal Line Operators in the ${\cal N}=2$ CS-matter
Theory

For all of the conformal line operators we have considered, the anomalous
dimension and anomalous spin of the boundary operators were the same up to a
sign. This section shows that this relation follows from supersymmetry, even
though the theories we have considered are not supersymmetric. In short, the
reason for this is the following. Every line operator we have considered has a
lift into a different supersymmetric line operator in the ${\cal N}=2$ CS-
matter theory. The lift has the property that the expectation value of the
corresponding mesonic line operator in the supersymmetric and non-
supersymmetric CS-matter theories is identical to the leading order in $1/N$.
At the same time, the primary boundary operators are BPS operators of the
supersymmetry that is preserved by the line. The three-dimensional transverse
spin becomes an R-symmetry of the one-dimensional super-conformal algebra.
Hence, the BPS condition of the boundary operators relates their conformal
dimension to their transverse spin.

One implication of our study for the ${\cal N}=2$ CS-matter theory is that the
lift of the $\alpha=-1$ operator is very different from the lift of the
$\alpha=1$ one. To our surprise, only the latter was explicitly considered
before.

It is important to note that the analysis in this section is formal because it
is done at the level of the classical action, and we do not renormalize the
operators. Classically, the line operators we consider preserve a one-
dimensional $\mathfrak{osp}(2|2)$ superconformal symmetry. However, the line
supersymmetry is allowed to be anomalous, and we can have a non-trivial RG
flow on the line.292929See Bianchi:2016vvm  for an example of a line operator
with anomalous supersymmetry. The number of broken supersymmetries is related
to the number of dimensions $3/2$ fermionic operator on the line Agmon:2020pde
.303030We thank Yifan Wang and Nathan Agmon for illuminating discussions on
this point. Our analysis in sections 2-4 shows that the lines studied in
section 7.1 are not anomalous. At the same time, the family of lines described
in section 7.3 has only two isolated superconformal fixed points.

For each of these superconformal line operators, we consider the BPS boundary
operators on which the line can end. At the formal level, these boundary
operators preserve one-half of the line supersymmetry. Provided that the line
is superconformal at the quantum level, the BPS boundary operators are
guaranteed to remain BPS also at the quantum level. This is because they do
not have other boundary operators to combine into long multiples. In summary,
most of the formal analysis in this section holds true also at the quantum
level.

###  7.1 The $\alpha=\pm 1$ Superconformal Line Operators

We consider the ${\cal N}=2$ CS-matter theory of a single chiral superfield in
the fundamental representation, Kapustin:2009kz ; Hama:2010av ; Jain:2012qi .
After integrating out all auxiliary fields, the action of the theory takes the
form $S_{{\cal N}=2}=S_{CS}+S_{m}$ where the CS action is given in (2.2) and

$S_{m}\\!=\\!\int\\!\mathrm{d}^{3}x\left[D_{\mu}\phi^{\dagger}D^{\mu}\phi-\mathrm{i}\bar{\psi}\gamma^{\mu}D_{\mu}\psi-\frac{4\pi\mathrm{i}}{k}(\phi\phi)(\bar{\psi}\psi)-\frac{2\pi\mathrm{i}}{k}(\bar{\psi}\phi)(\phi^{\dagger}\psi)+\frac{4\pi^{2}}{k^{2}}(\phi^{\dagger}\phi)^{3}\right]\,.$
(7.1)

Here, in this section, we use the standard SUSY convention for the kinetic
term of the fermions. It differs from the normalization we have used so far
(3.1) by a factor of $-\mathrm{i}$. The dictionary between the two conventions
is

$\psi^{1}_{\text{here}}=\sqrt{\mathrm{i}}\,\psi_{+}\,,\quad\psi^{2}_{\text{here}}=\sqrt{\mathrm{i}}\,\psi_{-}\,,\quad\bar{\psi}^{1}_{\text{here}}=-\sqrt{\mathrm{i}}\,\bar{\psi}_{+}\,,\quad\bar{\psi}^{2}_{\text{here}}=\sqrt{\mathrm{i}}\,\bar{\psi}_{-}\,.$
(7.2)

The theory (7.1) is expected to be self-dual Giveon:2008zn ; Benini:2011mf ;
Kapustin:2011gh ; Willett:2011gp . In the planar limit, the CS level and the
’t Hooft coupling are related to the dual ones as $k\to-k$ and $\lambda-{\rm
sign}(k)\to\lambda$. The theory (7.1) posses a three dimensional ${\cal N}=2$
superconformal symmetry. In appendix F we uniformize our notations for the
algebra and its realization with the existing literature cited there. Readers
who are familiar with it can skip this appendix, which otherwise is needed for
reading this section.

Consider the modified Wilson line operator (2.3) or equivalently (F.18), along
a straight line in the third direction. A general supersymmetry transformation
can be parameterized using two constant Grassmannian spinors, $\epsilon$ and
$\bar{\epsilon}$, as

$\delta=\epsilon\,Q+\bar{\epsilon}\,\bar{Q}\,.$ (7.3)

The condition that the line stays invariant under this transformation reads
313131The same equation holds for superconformal transformations. Taking
$\epsilon=x_{3}\gamma^{3}\epsilon_{c}$ and using (F.5), one finds that the
constraint equation for $\epsilon_{c}$ reads
$(\alpha\mathbb{I}+\gamma_{3})\epsilon_{c}=0$. Since $\alpha^{2}=1$, this is
identical to (7.4). One can also verify that the solution is compatible with
the reality condition (F.34).

$\bar{\epsilon}^{a}\varepsilon_{ab}(\gamma_{3}+\alpha\mathbb{I})^{b}_{\
c}=0\qquad\text{and}\qquad(\gamma_{3}+\alpha\mathbb{I})^{b}_{\
c}\epsilon^{c}=0\,.$ (7.4)

Since $\det(\gamma_{3}+\alpha\mathbb{I})=1-\alpha^{2}$, for the line to be
invariant we must have $\alpha=\pm 1$ and half of the SUSY variations must
vanish. We have

$\delta=\epsilon^{2}Q_{2}+\bar{\epsilon}^{1}\bar{Q}_{1}\qquad\text{for}\quad\alpha=+1\,,$
(7.5)

and

$\delta=\epsilon^{1}Q_{1}+\bar{\epsilon}^{2}\bar{Q}_{2}\qquad\text{for}\quad\alpha=-1\,.$
(7.6)

In either of these cases, the line brakes half of the supersymmetry, leaving
two supercharges unbroken. The line also preserves their conjugate
superconformal charges.

The bosonic symmetries of the straight line are generated by the conformal
$\mathfrak{sl}(2,\mathbb{R})$ algebra and the
$\mathfrak{so}(2)\simeq\mathfrak{u}(1)$ transverse rotations. Together with
the two supercharges and the two superconformal charges that preserve the
line, they generate the line superconformal group $\mathfrak{osp}(2|2)$. The
non-trivial commutation relations that we will need below are the ones between
the $Q$’s and the $S$’s. Using (F.31), we find

$\begin{aligned} \\{Q_{2},\bar{S}^{2}\\}&=-\mathrm{i}D+(\mathcal{R}+M_{12})\\\
\\{S^{1},\bar{Q}_{1}\\}&=-\mathrm{i}D-(\mathcal{R}+M_{12})\end{aligned}\qquad\text{for
}\alpha=+1\,,$ (7.7)

and

$\begin{aligned} \\{Q_{1},\bar{S}^{1}\\}&=-\mathrm{i}D+(\mathcal{R}-M_{12})\\\
\\{S^{2},\bar{Q}_{2}\\}&=-\mathrm{i}D-(\mathcal{R}-M_{12})\end{aligned}\qquad\text{for
}\alpha=-1\,.$ (7.8)

Here, $D$ is the dilatation generator, ${\cal R}$ is the three-dimensional
R-charge generator, and $M_{12}$ generates rotations in the transverse plane
to the line, see appendix F for details.

Next, we study the actions of this superconformal symmetry of the line on the
boundary operators.

###  7.2 Boundary Operators of the $\alpha=\pm 1$ Superconformal Lines

The two superconformal line operators, (2.3) with $\alpha=\pm 1$, can end on
operators in the fundamental/anti-fundamental representation. Among these, we
would like to understand what are the boundary operators that are super-
conformal primaries, preserving one-half of the straight line super and
superconformal charges. Depending on the sign of $\alpha$, these boundary
operators are annihilated by one supercharge and one superconformal charge in
(7.7) or (7.8) correspondingly. The right hand side of these equations then
relates their conformal dimension $\Delta$ (eigenvalue of $\mathrm{i}D$) to
their transverse spin $\mathfrak{s}$ (eigenvalue of $-M_{12}$).

It is sufficient to consider the four bottom operators of the lowest dimension
and transverse spin. All the other operators in the towers are related to
these by path derivatives. Correspondingly, their dimension and transverse
spin are related to the bottom operator in the tower by an integer shift.

At weak coupling, the boundary operators of the lowest dimension are the
scalar and fermion fields. We now study their superconformal transformation.
For convenience, we place them at the origin. A general superconformal
transformation can be parameterized by
$\epsilon=\epsilon_{s}+x^{\mu}\gamma_{\mu}\epsilon_{c}$ (see (F.9)), with
$\epsilon_{s},\epsilon_{c}$ being constant spinors. The corresponding
variation of the fundamental fields takes the form

$\delta_{\epsilon}(\cdot)=[\epsilon_{s}Q,\cdot]+\mathrm{i}[\epsilon_{c}S,\cdot]\,,$
(7.9)

with a similar equation for the conjugate transformation.

####  7.2.1 Boundary Operators of the $\alpha=1$ Line

The $\alpha=1$ line is preserved by the supercharges $Q_{2},\bar{Q}_{1}$ and
their conjugate superconformal charges $\bar{S}^{2},S^{1}$. These generate the
following transformation of the fields at the origin, (see (F.13))

$\displaystyle\delta\phi$
$\displaystyle=-\bar{\epsilon}^{1}_{s}\psi^{2}\,,\quad$
$\displaystyle\delta\phi^{\dagger}$
$\displaystyle=\epsilon^{2}_{s}\bar{\psi}^{1}\,,$ (7.10)
$\displaystyle\delta\psi^{1}$
$\displaystyle=\mathrm{i}\sqrt{2}\epsilon^{2}_{s}D_{+}\phi\,,\quad$
$\displaystyle\delta\bar{\psi}^{1}$
$\displaystyle=\mathrm{i}\bar{\epsilon}^{1}_{s}\mathbb{D}_{3}\phi^{\dagger}+\mathrm{i}\bar{\epsilon}_{c}^{1}\phi^{\dagger}\,,$
$\displaystyle\delta\psi^{2}$
$\displaystyle=-\mathrm{i}\epsilon^{2}_{s}\mathbb{D}_{3}\phi+\mathrm{i}\epsilon_{c}^{2}\phi\,,\quad$
$\displaystyle\delta\bar{\psi}^{2}$
$\displaystyle=\mathrm{i}\sqrt{2}\bar{\epsilon}^{1}_{s}D_{-}\phi^{\dagger}\,,$

where ${\mathbb{D}}_{3}\phi=(\partial_{3}-\mathrm{i}A-\sigma)\phi$ is the full
line covariant derivative in the third direction. Here, the auxiliary field
$\sigma\simeq\phi\phi^{\dagger}$ and the auxiliary field $F$ has been
integrated out and set to zero.

From this transformation rule, we learn that
$\\{\phi,\phi^{\dagger},\psi^{1},\bar{\psi}^{2}\\}$ are superconformal
primaries, annihilated by $\bar{S}^{2}$ and $S^{1}$, because their
transformations are independent of $\epsilon_{c}$ and $\bar{\epsilon}_{c}$. In
addition, each of those primary fields is annihilated by one supercharge, so
from the line perspective they are all $1/2$ BPS operators. Concretely, $\phi$
and $\bar{\psi}^{2}$ are annihilated by $Q_{2}$, because their transformations
are independent of $\epsilon_{s}^{2}$. Similarly, $\phi^{\dagger}$ and
$\psi^{1}$ are annihilated by $\bar{Q}_{1}$. We can therefore use (7.7) to
relate their dimension to the transverse spin.

Explicitly, the scalar $\phi$ is invariant under $Q_{2}$ and
$\bar{S}^{2}=(Q_{2})^{\dagger}$. The vanishing of the
$\\{Q_{2},\bar{S}^{2}\\}$ anticommutator (7.7) gives,323232 Note that when
acting on fields, $M_{12}$ gives minus the spin. Explicitly, we have
$[\mathrm{i}D,\mathcal{O}]=\Delta_{O}\mathcal{O}$,
$[\mathcal{R},\mathcal{O}]=r_{\mathcal{O}}\mathcal{O}$, and
$[M_{12},\mathcal{O}]=-M_{12}^{R}\mathcal{O}$, where $M_{12}^{R}$ is the
representation matrix of the Lorentz representation $R$. For spinors,
$M_{12}^{R}=-\frac{\mathrm{i}}{4}[\gamma_{1},\gamma_{2}]=\frac{1}{2}\gamma^{3}$,
see Minwalla:1997ka  for more details. Note also that the fermionic spin is
obtained by acting with the spin generators from the left, for both $\psi$ and
$\bar{\psi}$.

$\Delta_{\phi}=\frac{1}{2}-\mathfrak{s}(\phi)\,.$ (7.11)

Similar relations are found for the rest of the $1/2$ BPS primaries, they are
given by

$\Delta_{\phi^{\dagger}}=\frac{1}{2}+\mathfrak{s}(\phi^{\dagger})\,,\qquad\Delta_{\psi^{1}}=\frac{1}{2}+\mathfrak{s}(\psi^{1})\,,\qquad\Delta_{\bar{\psi}^{2}}=\frac{1}{2}-\mathfrak{s}(\bar{\psi}^{2})\,.$
(7.12)

The fields $\psi^{2}$ and $\bar{\psi}^{1}$ on the contrary are descendent of
$\phi$ and $\phi^{\dagger}$ respectively, (7.10). The scaling dimension of the
SUSY parameters $\bar{\epsilon}_{s}^{1}$, $\epsilon_{s}^{2}$ is $-1/2$ and
their transverse spin is $+1/2$ and $-1/2$ correspondingly. Hence, it follows
from (7.10) that

$\displaystyle\Delta_{\psi^{2}}$ $\displaystyle=\Delta_{\phi}+{1\over 2}\,,$
$\displaystyle\qquad\mathfrak{s}(\psi^{2})$
$\displaystyle=\mathfrak{s}(\phi)-{1\over 2}\,,$ (7.13)
$\displaystyle\Delta_{\bar{\psi}^{1}}$
$\displaystyle=\Delta_{\phi^{\dagger}}+{1\over 2}\,,$
$\displaystyle\qquad\mathfrak{s}(\bar{\psi}^{1})$
$\displaystyle=\mathfrak{s}(\phi^{\dagger})+{1\over 2}\,.$

Once combined with (7.11), (7.12), we find that

$\Delta_{\psi^{2}}={1\over
2}-\mathfrak{s}(\psi^{2})\,,\qquad\Delta_{\bar{\psi}^{1}}={1\over
2}+\mathfrak{s}(\bar{\psi}^{1})\,.$ (7.14)

In addition to those relations, the two bottom bosonic operators with spin,
$\mathcal{O}_{L}^{(0,-1)}=D_{-}\phi^{\dagger}$ and
$\mathcal{O}_{R}^{(0,1)}=D_{+}\phi$ are descendants of the primary fermion
$\bar{\psi}^{2}$ and $\psi^{1}$ respectively, (7.10). Therefore, their
dimension and spin are also related as

$\Delta_{D_{+}\phi}=\frac{1}{2}+\mathfrak{s}(D_{+}\phi),\qquad\Delta_{D_{-}\phi^{\dagger}}=\frac{1}{2}-\mathfrak{s}(D_{-}\phi^{\dagger})\,.$
(7.15)

####  7.2.2 Boundary Operators of the $\alpha=-1$ Line

The $\alpha=-1$ line is preserved by the supercharges $Q_{1}$, $\bar{Q}_{2}$
and their conjugate superconformal charges $\bar{S}^{1}$, $S^{2}$. The
corresponding field transformations at the origin read

$\displaystyle\delta\phi$
$\displaystyle=\bar{\epsilon}^{2}_{s}\psi^{1}\,,\quad$
$\displaystyle\delta\phi^{\dagger}$
$\displaystyle=-\epsilon^{1}_{s}\bar{\psi}^{2}\,,$ (7.16)
$\displaystyle\delta\psi^{1}$
$\displaystyle=\mathrm{i}\epsilon^{1}_{s}\mathbb{D}_{3}\phi+\mathrm{i}\epsilon_{c}^{1}\phi\,,\quad$
$\displaystyle\delta\bar{\psi}^{1}$
$\displaystyle=\mathrm{i}\sqrt{2}\bar{\epsilon}^{2}_{s}D_{+}\phi^{\dagger}\,,$
$\displaystyle\delta\psi^{2}$
$\displaystyle=\mathrm{i}\sqrt{2}\epsilon^{1}_{s}D_{-}\phi\,,\quad$
$\displaystyle\delta\bar{\psi}^{2}$
$\displaystyle=-\mathrm{i}\bar{\epsilon}^{2}_{s}\mathbb{D}_{3}\phi^{\dagger}+\mathrm{i}\bar{\epsilon}^{2}_{c}\phi^{\dagger}\,,$

where the covariant derivative now takes the form
$\mathbb{D}_{3}=\partial_{3}-\mathrm{i}(A_{3}+\mathrm{i}\sigma)$.

One can verify that $\phi,\phi^{\dagger}$ and $\psi^{2},\bar{\psi}^{1}$ are
superconformal primaries and are $1/2$-BPS operators. The relevant $1/2$-BPS
conditions (7.8) now read

$\Delta_{\phi}=\frac{1}{2}+\mathfrak{s}(\phi)\,,\quad\Delta_{\phi^{\dagger}}=\frac{1}{2}-\mathfrak{s}(\phi^{\dagger})\,,\quad\Delta_{\psi^{2}}=\frac{1}{2}-\mathfrak{s}(\psi^{2})\,,\quad\Delta_{\bar{\psi}^{1}}=\frac{1}{2}+\mathfrak{s}(\bar{\psi}^{1})\,.$
(7.17)

Comparing with the $\alpha=1$ case, see (7.11) and (7.12), the sign of the
spin term for bosons flips.

The fermion components $\psi^{1}$ and $\bar{\psi}^{2}$ on the other hand, are
SUSY descendants of $\phi$ and $\phi^{\dagger}$ respectively. Their conformal
dimensions and transverse spins are related to those of the primaries as

$\displaystyle\Delta_{\psi^{1}}$ $\displaystyle=\Delta_{\phi}+{1\over 2}\,,$
$\displaystyle\qquad\mathfrak{s}(\psi^{1})$
$\displaystyle=\mathfrak{s}(\phi)+{1\over 2}\,,$ (7.18)
$\displaystyle\Delta_{\bar{\psi}^{2}}$
$\displaystyle=\Delta_{\phi^{\dagger}}+{1\over 2}\,,$
$\displaystyle\qquad\mathfrak{s}(\bar{\psi}^{2})$
$\displaystyle=\mathfrak{s}(\phi^{\dagger})-{1\over 2}\,.$

Once combined with (7.8), we find that

$\Delta_{\psi^{1}}={1\over
2}+\mathfrak{s}(\psi^{1})\,,\qquad\Delta_{\bar{\psi}^{2}}={1\over
2}-\mathfrak{s}(\bar{\psi}^{2})\,.$ (7.19)

In addition to those relations, the two bottom bosonic operators with spin,
$\mathcal{O}_{L}^{(0,1)}=D_{+}\phi^{\dagger}$ and
$\mathcal{O}_{R}^{(0,-1)}=D_{-}\phi$ are descendants of the primary fermion
$\bar{\psi}^{1}$ and $\psi^{2}$ respectively, (7.16). Therefore, their
dimension and spin are also related as

$\Delta_{D_{-}\phi}=\frac{1}{2}-\mathfrak{s}(D_{-}\phi)\,,\qquad\Delta_{D_{+}\phi^{\dagger}}=\frac{1}{2}+\mathfrak{s}(D_{+}\phi^{\dagger})\,.$
(7.20)

####  7.2.3 Summary

In total, we find that the sign of $\alpha$ only enters the scalar relations

$\Delta_{\phi}=\frac{1}{2}-\alpha\,\mathfrak{s}(\phi)\,,\qquad\Delta_{\phi^{\dagger}}=\frac{1}{2}+\alpha\,\mathfrak{s}(\phi^{\dagger})\,,$
(7.21)

while the fermionic relations always take that form (7.14), (7.19). It follows
that the anomalous dimension and the anomalous spin of all these boundary
operators are the same (up to a sign), and that only the bosons are sensitive
to the sign of $\alpha$.

These results are in agreement with our explicit computations of the previous
sections. However, as discussed above, due to potential anomalies it is not a
priori guaranteed that the classical analysis in this section also holds in
the quantum theory. Using our results from sections 2 and 3 we can show that
such anomalies do not occur. To show this we note that for any spacetime
supersymmetry that is broken by the line, there is a protected fermionic
operator on the line of dimension $3/2$ and transverse spin $1/2$,
Agmon:2020pde . Similarly to the protected operators discussed in section
2.11, this operator is responsible for infinitesimal deformations of the line
in the direction of the broken symmetry. In the case of the operators (2.3),
we have two such operators

$\Sigma_{-}=\phi\times\bar{\psi}_{-}\,,\qquad\Sigma_{+}=\psi_{+}\times\phi^{\dagger}\qquad\text{for}\qquad\alpha=1\,,$
(7.22)

and

$\widetilde{\Sigma}_{+}=\phi\times\bar{\psi}_{+}\,,\qquad\widetilde{\Sigma}_{-}=\psi_{-}\times\phi^{\dagger}\qquad\text{for}\qquad\alpha=-1\,.$
(7.23)

These two superconformal primary operators on the line (which are also BPS)
correspond to the two broken supersymmetries. As the displacement operator,
they are constructed such that the right and left anomalous dimensions cancel
each other. The fact that there are only two such operators at finite coupling
(and not four or more) implies that the conformal line operator only breaks
two out of the four spacetime supercharges (at least at large $N$).

From the spectrum of the boundary operators, we expect that the $\alpha=1$ be
mapped back to itself under the Fermion-Boson duality. Flipping the sign of
$\alpha$ has the effect of flipping the sign of the anomalous dimension for
the boundary bosons. It however does not affect the anomalous dimension of the
boundary fermions. Hence, under Fermion-Boson duality, the $\alpha=-1$ line
operator cannot be mapped to itself.

Like the line operators, (7.22), (7.23), and the displacement operator, we
also have protected bosonic operators at any nonzero integer spin and
protected fermionic operators at any half integer spin. Their existence
follows from the existence of high spin conserved charges and super charges
that are only broken at order $1/N$, see section 2.11.

###  7.3 The Supersymmetric Condensed Fermion Line

The condensed fermionic line operator considered in section 4 also has a lift
into a $1/2$-BPS line operator in the ${\cal N}=2$ CS-matter theory. This
supersymmetric line operator was considered before in the context of quiver
gauge theories, with the fermion in the bi-fundamental representation
Drukker:2009hy ; Ouyang:2015iza ; Lee:2010hk ; Mauri:2018fsf ; Drukker:2019bev
.333333The construction of such fermionic type BPS line operators was first
considered in the $\mathcal{N}=6$ ABJM theory Drukker:2009hy . The lift of the
condensed fermionic line operator is obtained by taking the rank of one of the
gauge groups to one. This limit was discussed previously in Drukker:2020opf .

The starting point of the construction, following Drukker:2020opf , is to
embed the $\alpha=1$ BPS Wilson line discussed in the previous sections into
the upper diagonal of a $(2|2)$ super-connection,

$\mathcal{L}_{0}=\begin{pmatrix}A_{\mu}\dot{x}^{\mu}-\mathrm{i}\sigma|\dot{x}|&\quad
0\\\ 0&\quad 0\\\ \end{pmatrix}\,.$ (7.24)

This connection is preserved by $Q_{2}$ and $\bar{Q}_{1}$. Next, one
constructs a family of (formally) SUSY line operators that are parameterized
by a complex number $u$ as343434 Note that the only physical parameter is the
product of the coefficients in front of the fermion and the conjugate fermion
in $\mathcal{L}_{u}$. Here, we have chosen them to be equal.

${\cal L}_{0}\quad\rightarrow\quad{\cal
L}_{u}\equiv\mathcal{L}_{0}-\begin{pmatrix}u^{2}\,\phi\phi^{\dagger}&\mathrm{i}u\,\psi^{2}\\\
\mathrm{i}u\,\bar{\psi}^{1}&u^{2}\,\phi^{\dagger}\phi\end{pmatrix}|\dot{x}|\,.$
(7.25)

To see that ${\cal L}_{u}$ indeed leads to an $1/2$-BPS line operator, one
note that

${\cal L}_{u}={\cal
L}_{0}-\left(\mathrm{i}\mathcal{Q}_{-}\mathcal{G}_{u}+\mathcal{G}_{u}^{2}\right)|\dot{x}|\qquad\text{with}\qquad\mathcal{G}_{u}=\begin{pmatrix}0&u\,\phi\\\
u\,\phi^{\dagger}&0\end{pmatrix}\,,$ (7.26)

where $\mathcal{Q}_{\pm}=Q_{2}\pm\bar{Q}_{1}$. Under a SUSY variation, ${\cal
L}_{u}$ transforms into a total derivative

$\mathcal{Q}_{-}\mathcal{L}_{u}=-\mathcal{D}_{\mathcal{L}_{u}}\mathcal{G}_{u}\,,\qquad\mathcal{Q}_{+}\mathcal{L}_{u}=-\mathcal{D}_{\mathcal{L}_{u}}\begin{pmatrix}0&u\phi\\\
-u\phi^{\dagger}&0\end{pmatrix}\,,$ (7.27)

where $\mathcal{D}_{\mathcal{L}_{u}}$ is the covariant derivative with respect
to the deformed connection ${\cal L}_{u}$.

To see that ${\cal L}_{u}$ is also superconformal we consider the conjugate
charges $\mathcal{S}^{\pm}=(\mathcal{Q}_{\pm})^{\dagger}=\bar{S}^{2}\pm
S^{1}$. Similarly to (7.27), their action on the deformed connection
reads353535More explicitly, notice the action of $\mathcal{S}$ on the deformed
connection reads,  $\displaystyle\mathcal{S}^{\pm}\mathcal{L}_{u}$
$\displaystyle=\mathcal{S}^{\pm}\mathcal{L}_{0}-\mathrm{i}\mathcal{S}^{\pm}\mathcal{Q}_{-}\mathcal{G}_{u}-\\{\mathcal{S}^{\pm}\mathcal{G}_{u},\mathcal{G}_{u}\\}\,.$
(7.28) The first term vanishes because the original $\alpha=1$ line is
superconformal. For the second term, using the field transformation rules
(F.13), one finds
$-\mathrm{i}\,\mathcal{S}^{\pm}\mathcal{Q}_{-}\mathcal{G}_{u}=-\mathrm{i}\begin{pmatrix}0&\pm
u(x_{s}^{3}\mathbb{D}_{3}\phi+\phi)\\\
-u(x_{s}^{3}\mathbb{D}_{3}\phi^{\dagger}+\phi^{\dagger})&0\\\
\end{pmatrix}\,.$ (7.29) The terms in the bracket nicely combine into a total
derivative. For the last term, one notices that
$\mathcal{S}^{\pm}\mathcal{G}_{u}=\pm\mathrm{i}x_{s}^{3}\mathcal{Q}_{\pm}\mathcal{G}_{u}\,.$
(7.30) Summing over all the contributions leads to (7.31).

$\mathcal{S}^{-}\mathcal{L}_{u}=\mathrm{i}\,\mathcal{D}_{\mathcal{L}_{u}}\Big{[}x_{s}^{3}\mathcal{G}_{u}\Big{]},\qquad\mathcal{S}^{+}\mathcal{L}_{u}=-\mathrm{i}\,\mathcal{D}_{\mathcal{L}_{u}}\Bigg{[}x_{s}^{3}\begin{pmatrix}0&u\phi\\\
-u\phi^{\dagger}&0\\\ \end{pmatrix}\Bigg{]}\,.$ (7.31)

The family of line operators $\\{W_{u}={\cal P}\,\exp\int{\cal
L}_{u}\mathrm{d}s\,,\ u\in{\mathbb{C}}\\}$ are all formally superconformal.
Changing $u$, however, corresponds to a deformation of ${\cal L}_{u}$ by a
classically marginal operator that preserves supersymmetry. We should
therefore ask which, if any, of these operators is also superconformal in the
quantum theory. To answer this question, we note that at leading order in
$1/N$, the fermions do not contribute to the expectation value of the mesonic
line operator

$M=(\phi^{\dagger}\ 0)\cdot{\cal P}e^{\mathrm{i}\int{\cal
L}_{u}\mathrm{d}s}\cdot\begin{pmatrix}\phi\\\ 0\end{pmatrix}\,.$ (7.32)

It, therefore, takes the same value as the expectation value of the operator
(2.4) in the bosonic CS-matter theory. This picks up the points

$u^{2}={4\pi\mathrm{i}\over k}\,,\qquad\text{and}\qquad u^{2}=0\,,$ (7.33)

as the two quantum conformal operators, in correspondence with the $\alpha=-1$
and $\alpha=1$ operators respectively.

As discussed above, to check whether the $u=\sqrt{4\pi\mathrm{i}/k}$ line is
superconformal at the quantum level, we have to count the protected fermionic
operators on the line of dimension $3/2$ and transverse spin $\pm 1/2$. At the
tree level, there are six such operators. Two of them consist of the empty
endpoints

$\begin{pmatrix}D_{-}\phi\\\ 0\\\ \end{pmatrix}\times(0\
1)\,,\qquad\text{and}\qquad\begin{pmatrix}0\\\ 1\\\
\end{pmatrix}\times(D_{+}\phi^{\dagger}\ 0)\,.$ (7.34)

As discussed in the next section, the anomalous dimensions and spins of these
two operators cancel between their left and right factors. Hence, they remain
of dimension $3/2$ and transverse spin $\pm 1/2$ at the quantum level. The
other four take the form

$\begin{pmatrix}\widetilde{{\cal O}}_{R}\\\ 0\\\
\end{pmatrix}\times(\widetilde{{\cal O}}_{L}\
0)\,,\qquad\text{with}\qquad(\widetilde{{\cal O}}_{R},\widetilde{{\cal
O}}_{L})\in\\{(\phi,\bar{\psi}_{\pm}),\,(\psi_{\pm},\phi^{\dagger})\\}\,.$
(7.35)

The anomalous dimensions of the right and left factors of these operators sum
up, so they will obtain non-zero anomalous dimensions. We therefore conclude
that the $u=\sqrt{4\pi\mathrm{i}/k}$ line is a superconformal $1/2$-BPS line
at the quantum level.

The reason for the anomalous dimensions of the operators in (7.35) can be
understood as follows. The condensation of the fermions has the effect of
flipping the anomalous dimension of the two fermion components that are being
condensed while leaving the other two components the same as for the normal
Wilson line operator of section 3. Namely, the anomalous dimensions of
operators $\psi_{-}(\psi^{2})$ and $\bar{\psi}_{+}(\bar{\psi}^{1})$ flips the
sign, while the anomalous dimensions of $\psi_{+}(\psi^{1})$ and
$\bar{\psi}_{-}(\bar{\psi}^{2})$ remains the same. After this flip, all
fermion components on the condensed line have the same anomalous dimension,
equal to $-\lambda/2$. Similarly, the condensation flips the sign of the
scalar’s anomalous dimension. This is because the upper-diagonal component of
${\cal L}_{u}-{\cal L}_{0}$ in (7.25) flips the sign of the coefficient in
front of the bi-scalar, from $\alpha=1$ in ${\cal L}_{0}$ to $\alpha=-1$ in
${\cal L}_{u}$ with $u={\sqrt{4\pi\mathrm{i}/k}}$. After this flip, the
scalar’s anomalous dimension is equal to $-\lambda/2$.

####  7.3.1 Boundary Operators of the Condensed Fermion Line

The lift of the tree-level boundary operators in table (4.21) to
superconformal line construct from ${\cal L}_{u}$ is

$\widetilde{\cal O}_{R}^{(0,-{1\over 2})}\to\begin{pmatrix}0\\\
1\end{pmatrix}\,,\quad\widetilde{\cal O}_{R}^{(0,-{3\over
2})}\to\begin{pmatrix}\partial_{-}\psi_{-}\\\
0\end{pmatrix}\,,\quad\widetilde{\cal O}_{L}^{(0,{1\over 2})}\to\left(0\
1\right)\,,\quad\widetilde{\cal O}_{L}^{(0,{3\over
2})}\to\left(\partial_{+}\bar{\psi}_{+}\ 0\right)\,.$ (7.36)

In appendix G, we analyze the superconformal properties of these operators. We
find that the operators $\widetilde{\cal O}_{R}^{(0,-{1\over 2})}$ and
$\widetilde{\cal O}_{L}^{(0,{1\over 2})}$ are $1/2$-BPS superconformal
primaries while the operators $\widetilde{\cal O}_{R}^{(0,-{3\over 2})}$ and
$\widetilde{\cal O}_{L}^{(0,{3\over 2})}$ are superconformal descendants. The
corresponding relations between their conformal dimensions and transverse
spins are derived in the appendix. They read

$\Delta(\widetilde{{\cal
O}}_{R}^{(0,-\frac{1}{2})})=\frac{1}{2}+\mathfrak{s}(\widetilde{{\cal
O}}_{R}^{(0,-\frac{1}{2})})\,,\qquad\Delta(\widetilde{{\cal
O}}_{L}^{(0,\frac{1}{2})})=\frac{1}{2}-\mathfrak{s}(\widetilde{{\cal
O}}_{L}^{(0,-\frac{1}{2})})\,,$ (7.37)

and

$\Delta(\widetilde{\cal O}_{R}^{(0,-{3\over
2})})=\frac{1}{2}-\mathfrak{s}(\widetilde{\cal O}_{R}^{(0,-{3\over
2})})\,,\qquad\Delta(\widetilde{\cal O}_{L}^{(0,{3\over
2})})=\frac{1}{2}+\mathfrak{s}(\widetilde{\cal O}_{L}^{(0,{3\over 2})})\,.$
(7.38)

##  8 Mesonic Line Operators in the Mass Deformed CS-matter Theories

The recursion techniques used in this paper can also be applied to the mass-
deformed version of the CS-matter theories studied in this paper. These
theories are defined in the same way as the ones studied in the other
sections, but with the physical mass not tuned to zero. This mass, $m$, is
defined as the location of the pole in the exact propagator. Because it is the
only dimensionful parameter in the theory, all non-zero masses are equivalent.

In this section, we compute the expectation values of the mesonic line
operators along a straight line of length $x$ for the massive case. We find
that even though the $3d$ theory is gapped, for $\alpha=\pm 1$, there is no RG
flow on the line. Moreover, the mesonic line operators in the fermionic and
bosonic theories are consistent with the bosonization duality, provided that
we identify the dimensionless parameter $m^{2}x^{2}$ between the two theories.

###  8.1 Bosonic Case

##### Massive Boson Propagator

For massive case, we chose the regularization scheme

$\langle\phi^{\dagger}_{i}(p)\phi^{j}(q)\rangle^{\text{tree}}_{\epsilon}\equiv(2\pi)^{3}\delta^{3}(p+q)\,{\delta_{i}^{j}\over
p^{2}+m^{2}}\times e^{\mathrm{i}\epsilon\,p_{3}}\,.$ (8.1)

Thus, the exact propagator reads

$\langle\phi^{\dagger}_{i}(p)\phi^{j}(q)\rangle_{\epsilon}=(2\pi)^{3}\delta^{3}(p+q)\,{\delta_{i}^{j}\over
p^{2}+m^{2}+e^{\mathrm{i}\epsilon\,p_{3}}(m_{c.t.}^{2}-\Sigma(p))}\,e^{\mathrm{i}\epsilon\,p_{3}}\,,$
(8.2)

where $m$ is the physical mass. The Schwinger-Dyson equation determining
$\Sigma$ is still given in figure 2. One can also include the $\lambda_{4}$
and $\lambda_{6}$ type vertex. Diagrammatically, they are given by the
standard $\lambda\phi^{4}$ and $\lambda\phi^{6}$ bubble-type vertices.
Importantly, in the planar limit those contributions are independent of the
momenta of the self-energy. Therefore, in any scheme, the effect of
$\lambda_{4}$ and $\lambda_{6}$ on the self-energy can be absorbed into a
shift of the mass counter-term. Hence, we shall not take them into account
explicitly in the following.

We can now apply the differential equation techniques described in the
previous section (H). Notice that the rotational symmetry in the transverse
plane determines $e^{\mathrm{i}\epsilon p_{3}}(m_{c.t}^{2}-\Sigma(p))\equiv
f(p_{\perp};\lambda,m)$ to be a function of $|p_{\perp}|$ only. Taking
differentials with respect to $p^{-}$, one finds,

$\frac{\partial}{\partial p^{-}}\Sigma(p)=\frac{\partial}{\partial
p^{-}}\Big{(}(a)+(b)+(c)\Big{)}=0\,.$ (8.3)

This implies $\Sigma(p)$ is not a function of $p^{-}$. As the $p^{-}$
dependence of $\Sigma(p)$ is only trough $p_{\perp}^{2}=2p^{+}p^{-}$, we
conclude that $\Sigma(p)$, and thus $f$, are independent of $|p_{\perp}|$.

The computation then simplifies drastically, as one can take $p^{\pm}=0$ in
the Schwinger-Dyson equation. In this limit, diagrams (b) and (c) vanish
because the integrands of these diagrams are antisymmetric with respect to
exchanging of the two loop momenta. Diagram (a) then evaluates to

$(a)=-\frac{\lambda^{2}}{4\epsilon^{2}}e^{-\epsilon\sqrt{m^{2}-f(\lambda,m)}}\,.$
(8.4)

The Schwinger-Dyson equation now reads,

$\Sigma(p)=e^{-\mathrm{i}\epsilon
p_{3}}f(\lambda,m)+m_{c.t.}^{2}=(a)+(b)+(c)=-\frac{\lambda^{2}}{4\epsilon^{2}}e^{-2\epsilon\sqrt{m^{2}-f(\lambda,m)}}\
,$ (8.5)

As there is no momenta dependence on the RHS, one has to take

$f(\lambda,m)=0,\qquad m^{2}_{c.t.}=-{\lambda^{2}\over
4\epsilon^{2}}e^{-2\epsilon|m|}\,.$ (8.6)

One conclude that the exact massive propagator is given by

$\langle\phi^{\dagger}_{i}(p)\phi^{j}(q)\rangle_{\epsilon}=(2\pi)^{3}\delta^{3}(p+q)\,{\delta_{i}^{j}\over
p^{2}+m^{2}}\,e^{\mathrm{i}\epsilon\,p_{3}}\,,$ (8.7)

and the tree-level mass $m$ becomes the physical pole mass.

##### Massive Line Integrand.

The generalization of the analysis in section 2.4 is straightforward. Equation
(2.29) becomes,

$\displaystyle{\cal M}_{L}=$
$\displaystyle\,(-2\pi\lambda)^{L}\int\limits_{0}^{1}\mathrm{d}t_{L}\int\limits_{0}^{t_{L}}\mathrm{d}t_{L-1}\cdots\int\limits_{0}^{t_{2}}\mathrm{d}t_{1}\int\prod_{j=0}^{L}\frac{\mathrm{d}^{3}k_{j}}{(2\pi)^{3}}\times$
(8.8)
$\displaystyle\frac{e^{\mathrm{i}k_{0}^{3}x(t_{1}+\tilde{\epsilon})}}{k_{0}^{2}+m^{2}}\mathtt{V}_{\alpha}(k_{0},k_{1})\frac{e^{\mathrm{i}k_{1}^{3}x(t_{21}+\tilde{\epsilon})}}{k_{1}^{2}+m^{2}}\
\cdots\
\frac{e^{\mathrm{i}k_{L-1}^{3}x(t_{LL-1}+\tilde{\epsilon})}}{k_{L-1}^{2}+m^{2}}\mathtt{V}_{\alpha}(k_{L-1},k_{L})\frac{e^{\mathrm{i}k_{L}^{3}x(t_{L+1L}+\tilde{\epsilon})}}{k_{L}^{2}+m^{2}}\,.$

This leads to a simple modification of the integral necessary for the
recursion (2),

$\int\frac{\mathrm{d}^{3}k_{1}}{(2\pi)^{3}}\frac{k_{1}^{+}\,e^{\mathrm{i}k_{1}^{3}x\,t_{21}}}{(k_{1}^{2}+m^{2})(k_{0}-k_{1})^{+}(k_{1}-k_{2})^{+}}\\\
=\frac{1}{4\pi x\
t_{21}}\frac{1}{(k_{0}-k_{2})^{+}}\left(e^{-\sqrt{k^{2}_{2\perp}+m^{2}}x\,t_{21}}-e^{-\sqrt{k^{2}_{0\perp}+m^{2}}x\,t_{21}}\right)\,,$
(8.9)

The identification of $k_{j}^{3}$ is with
$\mathrm{i}\sqrt{k_{j\perp}^{2}+m^{2}}$. Under this identification, the
exponential term in (8.9) combines with
$e^{\mathrm{i}k_{2}^{3}x\,t_{32}}\simeq
e^{-\sqrt{k_{2\perp}^{2}+m^{2}}x\,t_{32}}$ or
$e^{\mathrm{i}k_{0}^{3}x\,t_{10}}\simeq
e^{-\sqrt{k_{2\perp}^{2}+m^{2}}x\,t_{10}}$ in (8.8), to induce exactly the
same recursion relation as in the massless case (2.41) and (2.42). This also
leads to the same solution to the recursion, up to a modification of the seed
term,

$\mathtt{B}_{m^{2}>0}^{(0)}(t)=\int\frac{\mathrm{d}^{3}k}{(2\pi)^{3}}\frac{e^{\mathrm{i}k^{3}xt}}{k^{2}+m^{2}}=\frac{e^{-|m|x\,t}}{4\pi\,x\,t}\
.$ (8.10)

Hence, similarly to (2.46), we find that

$\langle M_{10}\rangle_{m^{2}>0}={e^{-|m|x}\over 4\pi
x}\times\begin{dcases}{1\over\Gamma(1-\lambda)}\Big{(}{\epsilon\,e^{\gamma_{E}}\over
x}\Big{)}^{\lambda}\,,\quad\ \ \alpha=+1,\\\\[5.0pt]
{1\over\Gamma(1+\lambda)}\Big{(}{\epsilon\,e^{\gamma_{E}}\over
x}\Big{)}^{-\lambda}\,,\quad\alpha=-1\,.\\\ \end{dcases}$ (8.11)

We see that there is no RG flow taking place on the line.

##### Derivative Operator.

Following the same logic, the recursion for the derivative operator can also
be generalized to the massive case, and the only modification is the recursion
seed.

To verify the duality, we present the result for the recursion seed here.
Explicitly, the recursion seed of the operator $\langle
M_{10}^{(1,-1)}\rangle$ can be obtained by replacing all propagators in figure
6 with the massive propagator. After some algebra, one finds,

$\mathtt{B}^{(0,-1)}(t)=\frac{e^{-|m|tx}}{4\pi(x\,t)^{3}}\times\frac{(\lambda+1)(\lambda+2+2|m|t\,x)}{2}+\mathcal{O}(\tilde{\epsilon})\,.$
(8.12)

The other seed term $\mathtt{B}^{(0,1)}(t)$ is obtained by flipping the sign
of $\lambda$. Hence, similar to (2.72), to leading order in
$\tilde{\epsilon}$, we find

$\displaystyle\langle M_{10}^{(1,-1)}\rangle$
$\displaystyle=\frac{e^{-|m|x}}{4\pi
x^{3}}\times\left(\frac{\epsilon\,e^{\gamma_{E}}}{x}\right)^{-\lambda}\times\frac{(1+\lambda)(2+\lambda+2|m|x)}{2\Gamma(1+\lambda)}\,,$
(8.13) $\displaystyle\langle M_{10}^{(-1,1)}\rangle$
$\displaystyle=\frac{e^{-|m|x}}{4\pi
x^{3}}\times\left(\frac{\epsilon\,e^{\gamma_{E}}}{x}\right)^{+\lambda}\times\frac{(1-\lambda)(2-\lambda+2|m|x)}{2\Gamma(1-\lambda)}\,.$

###  8.2 Fermionic Case

The recursion relation can also be applied to the massive fermionic case.

##### Massive Fermion Propagator

The symmetry argument that leads to the Schwinger-Dyson equation (3.15) and
(3.16) still applies to the massive case, so the equations to be solved are
still given by these two. Assuming the exact propagator has a mass pole at
$m$, we can generalize the condition (3.13) as363636Note that the mass pole
condition only determines $|m|$. However, we can reproduce the sign by
agreement with the free mass when $\lambda\to 0$. The duality map implies that
$m_{\mathrm{bos}}=-m_{\mathrm{fer}}$ Minwalla:2015sca .

$\left(p_{\mu}+\Sigma_{\mu}(p)\right)^{2}+\Sigma_{0}(p)^{2}=p^{2}+m^{2}\,.$
(8.14)

The non-vanishing components of the fermionic self-energy then read

$\displaystyle\Sigma_{0}$
$\displaystyle=|m|-\frac{\lambda}{\epsilon}\Big{(}e^{-\epsilon|m|}-e^{-\epsilon\sqrt{|p_{\perp}|^{2}+m^{2}}}\Big{)}\,,$
(8.15)
$\displaystyle\simeq|m|-\lambda\Big{(}\sqrt{|p_{\perp}|^{2}+m^{2}}-|m|\Big{)}+{\cal
O}(\epsilon^{2})\,,$ $\displaystyle\Sigma_{+}$
$\displaystyle=-\frac{\lambda}{2\epsilon^{2}}\Big{(}e^{-\sqrt{p_{\perp}^{2}+m^{2}}}-e^{-|m|\epsilon}\Big{)}\Big{(}2|m|\epsilon+\lambda
e^{-\sqrt{p_{\perp}^{2}+m^{2}}}-\lambda e^{-|m|\epsilon}\Big{)}$
$\displaystyle\simeq\frac{p^{-}}{p_{\perp}^{2}}\Big{[}|m|\lambda\big{(}\sqrt{p_{\perp}^{2}+m^{2}}-|m|\big{)}+\lambda^{2}\Big{(}|m|\big{(}\sqrt{p_{\perp}^{2}+m^{2}}-|m|\big{)}-\frac{p_{\perp}^{2}}{2}\Big{)}\Big{]}+{\cal
O}(\epsilon^{2})\,.$

##### Massive Line Integrand.

The fermionic recursion relation stays the same form as for the massless case,
for the same reason as for the bosonic case. The first two recursion seed
terms now reads

$\displaystyle\big{(}\mathtt{F}^{(0)}(t)\big{)}_{-+}$
$\displaystyle=-\frac{e^{-|m|xt}}{4\pi x^{2}\,t^{2}}\left(1+\frac{\lambda
t}{t+\tilde{\epsilon}}e^{-|m|\epsilon}\right)\,,$ (8.16)
$\displaystyle\big{(}\mathtt{F}^{(0)}(t)\big{)}_{+-}$
$\displaystyle=+\frac{e^{-|m|xt}}{4\pi
x^{2}\,t^{2}}\left(1+2|m|tx-\frac{\lambda
t}{t+\tilde{\epsilon}}e^{-|m|\epsilon}\right)\,.$

When inserting the initial value $t=1+L\epsilon$, to leading order, one finds,

$\displaystyle\big{(}\mathtt{F}^{(0)}(t)\big{)}_{-+}$
$\displaystyle\simeq-\frac{e^{-|m|x}}{4\pi x^{2}}(1+\lambda)+{\cal
O}(\epsilon)\,,$ (8.17) $\displaystyle\big{(}\mathtt{F}^{(0)}(t)\big{)}_{+-}$
$\displaystyle\simeq+\frac{e^{-|m|x}}{4\pi x^{2}}(1-\lambda+2|m|x)+{\cal
O}(\epsilon)\,.$

As a consequence, the expectation values of the massive mesonic lines
read373737The new $|m|x$ dependence of $\langle M^{(-{1\over 2},{1\over
2})}\rangle$ might look confusing, but it is needed for the duality. Recall
from the dictionary (3.45) that this operator is dual to the bosonic operator
with derivatives. At the tree-level, the massive bosonic operator with
derivatives is proportional to
$\propto\int\frac{\mathrm{d}^{3}p}{(2\pi)^{3}}e^{\mathrm{i}p_{3}x}\frac{p^{+}p^{-}}{p^{2}+m^{2}}\propto\frac{e^{-|m|x}}{x^{3}}(1+|m|x)\,,$
therefore the dual fermionic operator must involve the $|m|x$ dependence.

$\begin{split}\langle M^{({1\over 2},-{1\over 2})}\rangle&=-{e^{-|m|x}\over
4\pi x^{2}}\Big{(}{\epsilon\,e^{\gamma_{E}}\over x}\Big{)}^{+\lambda}\
\frac{1+\lambda}{\Gamma(1-\lambda)}+{\cal O}(\tilde{\epsilon})\,,\\\ \langle
M^{(-{1\over 2},{1\over 2})}\rangle&=+{e^{-|m|x}\over 4\pi
x^{2}}\Big{(}{\epsilon\,e^{\gamma_{E}}\over x}\Big{)}^{-\lambda}\
\frac{1-\lambda+2|m|x}{\Gamma(1+\lambda)}+{\cal
O}(\tilde{\epsilon})\,.\end{split}$ (8.18)

##### Derivative Operators.

The recursion relation also applies to the derivative cases. The only
modification is the recursion seed.

For instance, one can modify the propagator in
$\mathtt{Figure}\,\ref{fig:ferderseed}.a$ and
$\mathtt{Figure}\,\ref{fig:ferderseed}.b$ correspondingly. After some algebra,
one obtains

$\mathtt{F}^{(0,-{3\over
2})}_{-+}(t)=-\frac{e^{-|m|tx}}{4\pi(tx)^{4}}\times\frac{(1+\lambda)(2+\lambda)(3+\lambda+2|m|xt)}{2}+{\cal
O}(\tilde{\epsilon})\,.$ (8.19)

Therefore,

$\langle M^{({3\over 2},-{3\over 2})}\rangle=-{e^{-|m|x}\over 4\pi
x^{4}}\left(\frac{\epsilon\,e^{\gamma_{E}}}{x}\right)^{\lambda}{1\over\Gamma(1-\lambda)}\frac{(1+\lambda)(2+\lambda)(3+\lambda+2|m|x)}{2}\,.$
(8.20)

##### Condensed Fermion Operator.

The derivation of the condensed fermionic operator can also be repeated.
First, we observe that in the presence of mass, the definition of the
condensed line operator must be modified. Repeating the derivation of the
displacement operator (4.36) is one way to see this. All the steps go through
as before, with one exception: the fermion equation of motion that are
unaffected by the condensation (4.32) must now include the mass term
contribution on the right-hand side, leading to

${\mathbb{D}}_{+}\propto\left(\begin{array}[]{cc}0&m\psi_{+}\\\
D_{+}\bar{\psi}_{+}&0\end{array}\right)\,,\qquad{\mathbb{D}}_{-}\propto\left(\begin{array}[]{cc}0&D_{-}\psi_{-}\\\
m\bar{\psi}_{-}&0\end{array}\right)\,.$ (8.21)

The additional off-diagonal term violates the condition that
$\mathbb{D}_{\mu}$ has a quantum dimension two in the UV, provided that $m$
has dimension one.383838Note that $\psi_{+}$ has the opposite anomalous
dimension than that of $D_{+}\bar{\psi}_{+}$. Therefore, the naive definition
of the line does not result in a good operator, and $m$-dependent terms are
expected to be added to the definition of the condensed line to cancel out
these unwanted contributions.

The massive modification of operator (4.6) is simple to implement: merely
scale the topological spin connection in the empty part by the total length as

$U_{\mathcal{C}}(s,t)\rightarrow
U_{\mathcal{C}}(s,t)\times\exp\Big{(}-\int\limits_{\mathcal{C}}|m||\mathrm{d}x|\Big{)}\,.$
(8.22)

Under this modification, each filled or empty component of the condensed
fermionic line operator contributes a factor proportional to the length of the
corresponding segment. These factors add up to a factor $\exp(-|m|x)$
proportional to the total length of the condensed line.

The argument in section 4 now can be generalized to the massive case. After
factorizing out the $\exp(-mt)$ factor, the $L$-ladder integrand now reads,
393939Recall that the $t_{j}$ terms are shifted by $j+\tilde{\epsilon}$; after
shifting back and factoring out the total length dependence, there will be a
$L$-dependent exponential remainder factor.

$\big{(}\mathtt{F}^{(L)}(t_{L+1})\big{)}_{-+}=-\frac{e^{-|m|x(L+1)\epsilon}}{4\pi
x^{2}}\Big{(}\frac{1}{(t_{L+1}+(L+1)\tilde{\epsilon})^{2}}\\\ +\frac{\lambda
e^{-|m|\epsilon}}{(t_{L+1}+(L+1)\tilde{\epsilon})(t_{L+1}+(L+2)\tilde{\epsilon})}\Big{)}\prod_{j=1}^{L}\frac{-\lambda}{t_{j}+j\tilde{\epsilon}}\,.$
(8.23)

Integrating the first term yields two terms, with the term proportional to
$t_{L}$ canceling out the second term in the previous order. The argument in
section 4 can now be repeated verbatim. The value of the contact term
coefficient $f$ and the coefficient $b$ now becomes,

$f=\lambda be^{-|m|\epsilon},\qquad b=1\,.$ (8.24)

The expectation value of the condensed line picks up an additional exponential
factor compared with the massless case,

$\langle\widetilde{M}^{({1\over 2},-{1\over
2})}\rangle={e^{-|m|x}\over\Gamma(1+\lambda)}\Big{(}{\epsilon\,e^{\gamma_{E}}\over
x}\Big{)}^{-\lambda}+{\cal O}(\tilde{\epsilon})\,.$ (8.25)

The form of the displacement operator remains the same as (4.36). The only
change in the derivation is that the unaffected fermion equation of motion
(4.32) now depends on the mass term, which was then canceled by the
contribution from the deformed parallel transport.

###  8.3 Evolution and Boundary Equations

The mass term does not affect the gluon equations of motion. Therefore, the
bosonic and fermionic evolution equations remain the same as in the conformal
case and are given by (2.84) and (3.50) respectively. The boundary equation,
however, is modified in a simple way, as we derive below.

##### The Bosonic Boundary Equation.

Since the tree-level scalar equation of motion is deformed by the mass term,
the boundary equations must also be modified. The derivation of the quantum
boundary equation generalizes to the massive case in a straightforward way.
First, we notice that the identity (2.57) generalizes to the massive case as

$\int\frac{\mathrm{d}^{3}k_{1}}{(2\pi)^{3}}\frac{(k_{1}^{+})^{a}}{(k_{1}^{2}+m^{2})(k_{0}-k_{1})^{+}}e^{\mathrm{i}k_{1}^{3}xt}=-\frac{(k_{0}^{+})^{a-1}}{4\pi
xt}e^{-\sqrt{k_{0\perp}^{2}+m^{2}}|x\,t}\,.$ (8.26)

Therefore, all the identifications of the plus component of loop momenta made
in sec (2.83) still applies to the massive case.

Replacing the scalar propagators in (2.81) and (2.82) by the massive
propagator and repeating the steps, one finds that the ${\cal O}(\lambda)$ and
${\cal O}(\lambda^{2})$ contributions cancels exactly, while the tree-level
part is cancelled by adding a vertex proportional to the mass,

$\mathbf{V}_{m^{2}{\cal O}_{L}^{(0,-s)}}=-m^{2}(\mathrm{i}q^{+})^{s}\,.$
(8.27)

The quantum equations of motion becomes,

$2\mathbf{V}_{\delta_{+}\delta_{-}{\cal
O}_{L}^{(0,-s)}}(k_{L})+\mathbf{V}_{\delta_{3}^{2}{\cal
O}_{L}^{(0,-s)}}(k_{L})+\mathbf{V}_{m^{2}{\cal O}_{L}^{(0,-s)}}=0\,,$ (8.28)

where the equality is understood to hold under the integration. This implies
the quantum equation of motion holds

${\cal O}_{L}^{(2,-s)}+2\delta_{+}{\cal O}_{L}^{(0,-s-1)}-m^{2}{\cal
O}_{L}^{(0,-s)}=0\,.$ (8.29)

The form stays the same as for the tree-level equation of motion. The other
boundary equation is modified in the same way.

##### The Fermionic Boundary Equation.

The fermionic boundary equation is also modified as for the tree-level
equation of motion. All the arguments that leads to (3.59) can be applied to
the massive case, with one minor modification: the identification
$q_{3}\rightarrow\mathrm{i}|q_{\perp}|$ under the integral sign now becomes
$q_{3}\rightarrow\mathrm{i}\sqrt{q_{\perp}^{2}+m^{2}}$. One therefore conclude
that

${\cal O}_{L}^{(2,-s)}+2\delta_{+}{\cal O}_{L}^{(0,-s-1)}-m^{2}{\cal
O}_{L}^{(0,-s)}=0\,.$ (8.30)

###  8.4 Duality Check

For the massive case, the ratio (2.96)

${\langle\mathcal{O}_{L}\,{\cal W}_{1s}\,{\mathbb{D}}_{\mu}(s)\,{\cal
W}_{st}\,{\mathbb{D}}^{\mu}(t)\,{\cal
W}_{t0}\,\mathcal{O}_{R}\rangle\over\langle\mathcal{O}_{L}\,{\cal
W}_{10}\,\mathcal{O}_{R}\rangle}\,$

is still a normalization-independent physical quantity that one could match.
However, since we do not have conformal invariance, this ratio does not take a
simple form as it does in (2.96).

#####  $\alpha=1$ Bosonic Case.

For the same reason as for the massless case, for $\alpha=1$, a non-zero
expectation value is obtained for $s_{R}=-s_{L}=0$ or $s_{R}=-s_{L}=1$ only,
being the bottom operators in (2.49). In the former case, only the
$(\mu,\nu)=(-,+)$ term in (2.95) contributes and the ratio reads
($\Delta=(1+\lambda)/2$)

${\langle\mathcal{O}_{L}\,{\cal W}_{1s}\,{\mathbb{D}}_{\mu}(s)\,{\cal
W}_{st}\,{\mathbb{D}}^{\mu}(t)\,{\cal
W}_{t0}\,\mathcal{O}_{R}\rangle\over\langle\mathcal{O}_{L}\,{\cal
W}_{10}\,\mathcal{O}_{R}\rangle}\\\ ={1\over
x_{st}^{4}}\left({x_{10}x_{st}\over
x_{1s}x_{t0}}\right)^{\Delta}{\sin(\pi\lambda)\over\pi}(1-\lambda)\lambda(2+2|m|x_{st}-\lambda)\,.$
(8.31)

In the latter case, only the $(\mu,\nu)=(+,-)$ term contributes. The result is
($\Delta=(3-\lambda)/2$)

${\langle\mathcal{O}_{L}\,{\cal W}_{1s}\,{\mathbb{D}}_{\mu}(s)\,{\cal
W}_{st}\,{\mathbb{D}}^{\mu}(t)\,{\cal
W}_{t0}\,\mathcal{O}_{R}\rangle\over\langle\mathcal{O}_{L}\,{\cal
W}_{10}\,\mathcal{O}_{R}\rangle}\\\ ={1\over
x_{st}^{4}}\left({x_{10}x_{st}\over
x_{1s}x_{t0}}\right)^{\Delta}{\sin(\pi\lambda)\over\pi}(1-\lambda)\lambda\frac{(2+2|m|x_{1s}-\lambda)(2+2|m|x_{t0}-\lambda)}{(2+2|m|x_{10}-\lambda)}\,.$
(8.32)

##### Fermionic Case.

One can also compute the same ratio in fermionic theories. There are two
nontrivial contributions. For $\mathcal{O}_{L}=\bar{\psi}_{+}$ and
$\mathcal{O}_{R}=\psi_{-}$, we find

${\langle\mathcal{O}_{L}\,{\cal W}_{1s}\,{\mathbb{D}}_{\mu}(s)\,{\cal
W}_{st}\,{\mathbb{D}}^{\mu}(t)\,{\cal
W}_{t0}\,\mathcal{O}_{R}\rangle\over\langle\mathcal{O}_{L}\,{\cal
W}_{10}\,\mathcal{O}_{R}\rangle}\\\ ={1\over
x_{st}^{4}}\left({x_{10}x_{st}\over
x_{1s}x_{t0}}\right)^{\Delta}{\sin(\pi\lambda)\over\pi}\lambda(\lambda+1)(1+2|m||x_{st}|-\lambda)\,,$
(8.33)

which matches with the (8.31), provided that we identify $m^{2}x^{2}_{st}$
between the two theories.

The other one ratio is obtained by taking $\mathcal{O}_{L}=\bar{\psi}_{-}$ and
$\mathcal{O}_{R}=\psi_{+}$, we find

${\langle\mathcal{O}_{L}\,{\cal W}_{1s}\,{\mathbb{D}}_{\mu}(s)\,{\cal
W}_{st}\,{\mathbb{D}}^{\mu}(t)\,{\cal
W}_{t0}\,\mathcal{O}_{R}\rangle\over\langle\mathcal{O}_{L}\,{\cal
W}_{10}\,\mathcal{O}_{R}\rangle}\\\ ={1\over
x_{st}^{4}}\left({x_{10}x_{st}\over
x_{1s}x_{t0}}\right)^{\Delta}{\sin(\pi\lambda)\over\pi}(1+\lambda)\lambda\frac{(1+2|m|x_{1s}-\lambda)(1+2|m|x_{t0}-\lambda)}{(1+2|m|x_{10}-\lambda)}\,,$
(8.34)

which matches with the (8.32), provided that we identify $m^{2}x^{2}$ between
the two theories.

##### Infinite Mass Limit.

Interestingly, at $x|m|\to\infty$ limit, the ratio (8.34) becomes,

$\lim_{x|m|\to\infty}{\langle\mathcal{O}_{L}\,{\cal
W}_{1s}\,{\mathbb{D}}_{\mu}(s)\,{\cal W}_{st}\,{\mathbb{D}}^{\mu}(t)\,{\cal
W}_{t0}\,\mathcal{O}_{R}\rangle\over\langle\mathcal{O}_{L}\,{\cal
W}_{10}\,\mathcal{O}_{R}\rangle}={|m|\over x_{st}^{3}}\left({x_{10}x_{st}\over
x_{1s}x_{t0}}\right)^{\Delta-1}{\sin(\pi\lambda)\over\pi}(1+\lambda)\lambda\,.$
(8.35)

This result resembles a non-trivial, $\lambda$-dependent conformal result,
wherein the dimensions are shifted by half a unit. It suggests that in the IR
limit and after dressing out the trivial $\exp(-|m|x)$ factor, we remained
with a non-trivial CFT on the line. Further investigation of this limit would
be of great interest. In addition, it would be intriguing to determine if the
unconventional form of the planar S-matrix, as studied in Jain:2014nza ;
Mehta:2022lgq ; Gabai:2022snc , is related to the Yukawa-like potential in
(8.11), (8.18), (8.20), and (8.25).

###### Acknowledgements.

We thank Yunfeng Jiang for his collaboration during the initial stages of this
work. We are grateful to Nathan Agmon, Ofer Aharony, Nissan Itzhaki, Zohar
Komargodski, Yifan Wang, and Shimon Yankielowicz for useful discussions. We
thank Zohar Komargodski for comments on the manuscript. AS and DlZ are
supported by the Israel Science Foundation (grant number 1197/20). BG is
supported by the Simons Collaboration Grant on the Non-Perturbative Bootstrap
and by the DOE grant DE-SC0007870. BG would like to thank the Galileo Galilei
Institute for its hospitality. DlZ would like to thank the support of the
Simons Foundation for the Varna Workshop organized by the International Center
for Mathematical Sciences in Sofia.

##  Appendix A The Mesonic Wilson Line in the Bosonic Theory at 1-loop Order

In this section, we present the one-loop expectation value of mesonic line
operator (2.4) in the bosonic theory along an arbitrary smooth path ${\cal C}$
between $x_{0}$ and $x_{1}$,

${\cal M}[{\cal C},n]\equiv\langle\phi^{\dagger}(x_{1})\Big{[}{\cal
P}e^{\mathrm{i}\int\limits_{\cal
C}A\cdot\mathrm{d}x+\frac{2\pi\mathrm{i}\alpha}{k}\phi\phi^{\dagger}|dx|}\Big{]}_{n}\phi(x_{0})\rangle\,.$
(A.1)

For non-self-intersecting paths and for $\alpha=0$, the one-loop result is
finite. For $\alpha\neq 0$, it only diverges near the endpoints of the line.
Accordingly, the RG flow on the line only starts at two-loop order. The one-
loop result, of course, depends on the path. Here, we express it as single and
double integrals along the path, with no bulk integration. For $\alpha=0$, we
find that the result coincides with the one bootstrapped in section 6.

At tree level, the expectation value of the line operator is given by the free
scalar propagator,

${\cal
M}[\mathcal{C},n]_{\text{tree}}=\frac{1}{4\pi}\frac{1}{|x_{0}-x_{1}|}\,.$
(A.2)

It is thus independent of the path.

At one-loop, regardless of the choice of the gauge, there are three non-
trivial gluon exchange diagrams – gluon exchange on the Wilson line, gluon
exchange on the scalar propagator, gluon exchange between the Wilson line and
the scalar propagator, see figure 4. The fourth contribution, 4(d), comes from
the bi-scalar deformation and is equal to an integrated product of two free
scalar propagators. In the following, we analyze them one by one in Lorentz
gauge.

###  A.1 Feynman Rules in Lorentz Gauge

We work in Lorentz gauge ($\partial^{\mu}A_{\mu}=0$) and evaluate the diagrams
in position space directly. The standard gauge-fixing procedure leads to the
following gauge-fixing and Faddeev-Popov ghost terms Guadagnini:1989kr ,

$\displaystyle S_{\text{GF}}$
$\displaystyle=\frac{\mathrm{i}k}{2\pi}\int\mathrm{d}^{3}x\,\mathrm{tr}A_{\mu}\partial^{\mu}B,\qquad
S_{\text{FP}}=\int\mathrm{d}^{3}x\,{\rm
tr~{}}{\partial_{\mu}c^{\dagger}D^{\mu}c}\,.$ (A.3)

Here the auxiliary scalar field $B$ is the Lagrange multiplier imposing the
gauge condition, and $c$, $c^{\dagger}$ are complex anti-commuting ghost
fields.

The corresponding free propagators are

$\displaystyle\langle\phi^{\dagger}_{j}(x)\phi^{i}(y)\rangle$
$\displaystyle=+\frac{1}{4\pi}\delta^{i}_{j}\frac{1}{|x-y|}\,,$ (A.4)
$\displaystyle\langle A_{\mu}^{I}(x)A_{\nu}^{J}(y)\rangle$
$\displaystyle=-\frac{\mathrm{i}}{k}\delta^{IJ}\epsilon_{\mu\nu\rho}\frac{(x-y)^{\rho}}{|x-y|^{3}}=\frac{\mathrm{i}}{k}\delta^{IJ}\epsilon_{\mu\nu\rho}\partial^{\rho}_{x}\frac{1}{|x-y|}\,,$
$\displaystyle\langle A_{\mu}^{I}(x)B^{J}(y)\rangle$
$\displaystyle=-\frac{\mathrm{i}}{k}\delta^{IJ}\frac{(x-y)^{\mu}}{|x-y|^{3}}\,.$

Detailed perturbative analysis shows that there are no loop corrections to the
gluon propagator up to two-loops Chen:1992ee .

###  A.2 Gluon Exchange on the Wilson line

The gluon exchange diagram on the Wilson line is plotted in figure 4(a). We
parametrize the smooth non-self-intersecting Wilson line
$x_{s}^{\mu}=x^{\mu}(s)$ with $\dot{x}_{s}^{3}>0$. For definiteness, we set
$0<s<1$ unless otherwise stated. Using the Feynman rules (A.4), we obtain the
expression404040The factor of one quarter in (A.5) has resulted from the
normalization ${\rm tr~{}}(T^{I}T^{J})=\frac{1}{2}\delta^{IJ}$ and from
converting ordered integrals to full integrals.

$\displaystyle\mathcal{M}_{a}=$ $\displaystyle{\cal
M}[\gamma]_{\text{tree}}\frac{\mathrm{i}\lambda}{4}\int\limits_{0}^{1}\mathrm{d}s\int\limits_{0}^{1}\mathrm{d}t\,\epsilon_{\mu\nu\rho}\frac{(x_{s}-y_{t})^{\rho}}{|x_{s}-y_{t}|^{3}}\dot{x}_{s}^{\mu}\dot{y}_{t}^{\nu}.$
(A.5)

As discussed around equation (I.2), the Wilson line is defined using framing
regularization prescription. At one loop order, it amounts to deforming one of
the two integration points in (A.5) as $y_{t}=x_{t}+\epsilon\,n_{t}$, where
$n$ is the framing vector and $\epsilon\to 0$, see Witten:1988hf  for details.

For convenience, we shall now review some basic properties of this integral.
We divide the integral above to two distinct finite pieces, the Writhing
number and the Twisting number.

##### Writhing Number.

The unregularized integral in (A.5) is called the “Writhing Number”. This
expression is finite but not topological.

##### Twisting Number.

Although the region of integration in (A.5) where $x_{t}\rightarrow x_{s}$ is
not divergent, it contains a well-defined finite piece. This so-called
“Twisting Number” is extracted from the framed integral by integrating only
the infinitesimal neighborhood of the collision point. For computational
convenience, let us switch to the arc-length parameterization for the moment
by setting $\mathrm{d}s=\sqrt{\mathrm{d}x^{\mu}\mathrm{d}x_{\mu}}$. The upper
limit of integration is the length of the curve $L_{\gamma}$. In the vicinity
of $t\rightarrow s$, we can approximate
$x_{t}=x_{s}+(t-s)\dot{x}_{s}+\mathcal{O}((t-s)^{2})$,

$\displaystyle\mathcal{T}_{a}=$ $\displaystyle\lim_{\epsilon,\delta\to
0}\int\limits_{0}^{L_{\gamma}}\mathrm{d}s\int\limits_{s-\delta}^{s+\delta}\mathrm{d}t\,\epsilon_{\mu\nu\rho}\frac{(x_{t}-x_{s}+\epsilon
n_{s})^{\rho}}{|x_{s}-x_{t}-\epsilon
n_{s}|^{3}}\dot{x}_{s}^{\mu}(\dot{x}_{t}^{\nu}+\epsilon\dot{n}_{t}^{\nu})$
(A.6) $\displaystyle=$ $\displaystyle\lim_{\epsilon,\delta\to
0}\int\limits_{0}^{L_{\gamma}}\mathrm{d}s\int\limits_{s-\delta}^{s+\delta}\mathrm{d}t\epsilon_{\mu\nu\rho}\,\frac{\epsilon^{2}n_{s}^{\rho}}{|(t-s)\dot{x}_{s}+\epsilon
n_{s}|^{3}}\dot{x}_{s}^{\mu}\dot{n}_{s}^{\nu}$ $\displaystyle=$
$\displaystyle\lim_{\epsilon,\delta\to
0}\int\limits_{0}^{L_{\gamma}}\mathrm{d}s\,\epsilon_{\mu\nu\rho}\dot{x}_{s}^{\mu}\dot{n}_{s}^{\nu}n_{s}^{\rho}\int\limits_{s-\delta}^{s+\delta}\mathrm{d}t\frac{\epsilon^{2}}{[(t-s)^{2}+\epsilon^{2}]^{\frac{3}{2}}}=2\int\limits_{0}^{L_{\gamma}}\mathrm{d}s\,\epsilon_{\mu\nu\rho}\dot{x}_{s}^{\mu}\dot{n}_{s}^{\nu}n_{s}^{\rho}\
,$

where the $\epsilon\to 0$ is taken first. Here, we have used that in arc-
length parameterization $|\dot{x}_{s}|=|n_{s}|\equiv 1$. We see that
$\mathcal{T}_{a}$ depends on the framing regularization.

The quantity $\mathcal{T}_{a}$ is also called the relative torsion of the
curve. Its geometric meaning is discussed in appendix C.

###  A.3 Scalar Self-energy

The second diagram is a gluon exchange on the scalar propagator (self-energy);
see figure 4(b). This 1-loop diagram does not depends on the shape of the
Wilson loop. It reads

$\displaystyle\mathcal{M}_{b}=$
$\displaystyle\frac{\mathrm{i}^{3}2\lambda}{(4\pi)^{3}}\int\mathrm{d}^{3}y_{1}\mathrm{d}^{3}y_{2}\,\epsilon_{\mu\nu\rho}\partial^{\rho}_{y_{1}}{1\over|y_{1}-y_{2}|}\times$
$\displaystyle\left[\frac{1}{|x_{0}-y_{1}|}(\partial_{y_{1}}^{\mu}\frac{1}{|y_{1}-y_{2}|})(\partial_{y_{2}}^{\nu}\frac{1}{|y_{2}-x_{1}|})-(\partial_{y_{1}}^{\mu}\frac{1}{|x_{0}-y_{1}|})\frac{1}{|y_{1}-y_{2}|}(\partial_{y_{2}}^{\nu}\frac{1}{|y_{2}-x_{1}|})\right.$
$\displaystyle\left.-\frac{1}{|x_{0}-y_{1}|}(\partial_{y_{1}}^{\mu}\partial_{y_{2}}^{\nu}\frac{1}{|y_{1}-y_{2}|})\frac{1}{|y_{2}-x_{1}|}+(\partial_{y_{1}}^{\mu}\frac{1}{|x_{0}-y_{1}|})(\partial_{y_{2}}^{\nu}\frac{1}{|y_{1}-y_{2}|})\frac{1}{|y_{2}-x_{1}|}\right]$
$\displaystyle=$
$\displaystyle-\frac{\mathrm{i}4\lambda}{(4\pi)^{3}}\int\mathrm{d}^{3}y_{1}\mathrm{d}^{3}y_{2}\,\epsilon_{\mu\nu\rho}\partial^{\rho}_{y_{1}}{1\over(y_{1}-y_{2})^{2}}(\partial_{y_{1}}^{\mu}\frac{1}{|x_{0}-y_{1}|})(\partial_{y_{2}}^{\nu}\frac{1}{|y_{2}-x_{1}|})$
$\displaystyle=$
$\displaystyle-\frac{\mathrm{i}4\lambda}{(4\pi)^{3}}\int\mathrm{d}^{3}y_{1}\mathrm{d}^{3}y_{2}\,\epsilon_{\mu\nu\rho}\partial^{\rho}_{y_{1}}\left[{1\over(y_{1}-y_{2})^{2}}(\partial_{y_{1}}^{\mu}\frac{1}{|x_{0}-y_{1}|})(\partial_{y_{2}}^{\nu}\frac{1}{|y_{2}-x_{1}|})\right]=0\,,$

where in the first step, we observe that only one term is non-zero, and in the
last step, we have an integral over a total derivative.

The vanishing of the diagram can be seen more easily from momentum space. If
we denote the incoming/outgoing momenta by $p$ and the scalar loop momenta by
$q$, the two derivative vertices in (4)(b) contribute a factor of
$(p+q)^{\mu}(p+q)^{\nu}$, which is contracted with the gluon propagator
$\langle A_{\mu}A_{\nu}\rangle$. Since the gluon propagator is antisymmetric
in $\mu\nu$, the contraction vanishes automatically.

###  A.4 Scalar-Wilson Line Gluon Exchange

The diagram in figure 4(c) is the exchange of a gluon between the scalar
propagator and the Wilson line. It is given by

$\displaystyle\mathcal{M}_{c}=$
$\displaystyle\frac{\mathrm{i}\lambda}{32\pi^{2}}\int\limits_{0}^{1}\mathrm{d}s\
\dot{x}_{s}^{\nu}\int\mathrm{d}^{3}y\Big{[}\frac{\partial}{\partial
y_{\mu}}(\frac{1}{|x_{0}-y|})\frac{1}{|y-x_{1}|}-\frac{\partial}{\partial
y_{\mu}}(\frac{1}{|x_{1}-y|})\frac{1}{|y-x_{0}|}\Big{]}\epsilon_{\mu\nu\rho}\frac{y^{\rho}-x_{s}^{\rho}}{|y-x_{s}|^{3}},$
(A.7) $\displaystyle\equiv$
$\displaystyle-\frac{\mathrm{i}\lambda}{32\pi^{2}}\int\limits_{0}^{1}\mathrm{d}s\,{\cal
M}_{c}(s)\,.$

The spacetime integral over $y$ in $\mathcal{M}_{c}(s)$ can be evaluated
explicitly,

$\displaystyle{\cal M}_{c}(s)=$
$\displaystyle\dot{x}_{s}^{\nu}(\frac{\partial}{\partial
x_{0\mu}}-\frac{\partial}{\partial
x_{1\mu}})\epsilon_{\mu\nu\rho}\int\mathrm{d}^{3}y\frac{1}{|x_{0}-y|}\frac{1}{|y-x_{1}|}\frac{y^{\rho}-x_{s}^{\rho}}{|y-x_{s}|^{3}}$
(A.8) $\displaystyle=$
$\displaystyle\dot{x}_{s}^{\nu}(\frac{\partial}{\partial
x_{0\mu}}-\frac{\partial}{\partial
x_{1\mu}})\epsilon_{\mu\nu\rho}\int\mathrm{d}^{3}y\frac{1}{|y-x_{0s}|}\frac{1}{|y-x_{1s}|}\frac{y^{\rho}}{|y|^{3}}$
$\displaystyle=$ $\displaystyle
2\pi\dot{x}_{s}^{\nu}\epsilon_{\mu\nu\rho}(\frac{\partial}{\partial
x_{0\mu}}-\frac{\partial}{\partial
x_{1\mu}})\frac{|x_{0s}|+|x_{1s}|-|x_{01}|}{|x_{0s}||x_{1s}|+x_{0s}\cdot
x_{1s}}\Big{[}\frac{x_{0s}^{\rho}}{|x_{0s}|}+\frac{x_{1s}^{\rho}}{|x_{1s}|}\Big{]}$
$\displaystyle=$
$\displaystyle\frac{8\pi}{|x_{01}||x_{0s}||x_{1s}|}\frac{1}{|x_{0s}|+|x_{1s}|+|x_{01}|}\epsilon_{\mu\nu\rho}x_{01}^{\mu}x_{1s}^{\nu}\dot{x}_{s}^{\rho}\,.$

Here, we have performed the $y$ integration using the following formula
Guadagnini:1989am ,

$\int\mathrm{d}^{3}y\frac{y^{\mu}}{|y|^{3}|y-a||y-b|}=2\pi\frac{|a|+|b|-|a-b|}{|a||b|+a\cdot
b}\left(\frac{a^{\mu}}{|a|}+\frac{b^{\mu}}{|b|}\right)\,.$ (A.9)

We now show that the integral in (A.8) is finite for a smooth path. The
integrand is finite for $x_{s}\neq x_{0},x_{1}$; therefore, the only potential
divergences are at the boundaries. Consider the limit where $x_{s}\rightarrow
x_{0}$. In arc-length parametrization, we can approximate $x_{s}$ by

$x_{s}=x_{0}+(s-\frac{s^{3}\kappa^{2}}{6})\mathbf{T}+(\frac{s^{2}\kappa}{2}+\frac{s^{3}\kappa^{\prime}}{6})\mathbf{N}+\frac{s^{3}\kappa\tau}{6}\mathbf{B}+\mathcal{O}(s^{4})\,,$
(A.10)

where $\mathbf{T}=\dot{x}_{s}$, $\mathbf{N}=\ddot{x}_{s}$,
$\mathbf{B}=\mathbf{T}\times\mathbf{N}$ are the vierbein (tetrad) of the curve
at $x_{0}$. Here, $\kappa$, $\tau$ are the curvature and torsion at $x_{0}$,
respectively.

Near the boundary, the $\epsilon$-tensor factor in (A.8) can be approximated
as

$\epsilon_{\mu\nu\rho}x_{01}^{\mu}x_{1s}^{\nu}\dot{x}_{s}^{\rho}\simeq(x_{01})\cdot\Big{(}(s\mathbf{T}+s^{2}\mathbf{N})\times(\mathbf{T}+s\mathbf{N})\Big{)}=\mathcal{O}(s^{2})\,,$
(A.11)

while the denominator behaves like $|x_{0s}|=\mathcal{O}(s)$. Hence, the
integrand vanishes near the $x_{0}$ endpoint. A similar analysis applies at
$x_{1}$.

###  A.5 Bi-Scalar Contribution

The last diagram, figure 4(d), is given by products of two tree-level
propagators,

$\mathcal{M}_{d}=-\frac{\alpha\lambda}{8\pi}\int\limits_{0}^{1}\mathrm{d}s\frac{|\dot{x}_{s}|}{|x_{1s}|\,|x_{s0}|}=-\frac{\alpha\lambda|x_{01}|}{8\pi}\int\limits_{0}^{1}\mathrm{d}s\frac{1}{|x_{1s}|\,|x_{s0}|}\,.$
(A.12)

Near the endpoints, the terms in the denominator behave linearly,
$|x_{0s}|\simeq{\cal O}(s),\ |x_{1s}|\simeq{\cal O}(1-s)$, which lead to
logarithmic divergences at the boundaries and the corresponding anomalous
dimension $\gamma=\alpha\lambda/2$, see section 2.3.

##  Appendix B The Mesonic Wilson Line in the Fermionic Theory at 1-loop Order

In this section, we present the one-loop expectation value of mesonic Wilson
line operator (3.21) in the fermionic theory along an arbitrary smooth path
${\cal C}$ between $x_{0}$ and $x_{1}$,

${\cal M}^{f}[{\cal C},n]\equiv\langle\bar{\psi}(x_{1})\Big{[}{\cal
P}e^{\mathrm{i}\int\limits_{\cal
C}A\cdot\mathrm{d}x}\Big{]}_{n}\psi(x_{0})\rangle\,.$ (B.1)

Now, ${\cal M}^{f}[{\cal C},n]$ is a $2\times 2$ matrix in the fermionic
indices. For non-self-intersecting paths, the one-loop result only diverges
near the endpoints of the line.

At tree level, the expectation value of the mesonic line operator is given by
the free fermion propagator,

$\displaystyle{\cal M}[\mathcal{C}]_{\text{tree}}=$
$\displaystyle\frac{1}{4\pi}\gamma_{\mu}\frac{x^{\mu}_{1}-x_{0}^{\mu}}{|x_{1}-x_{0}|^{3}}\,.$
(B.2)

It is thus independent of the path.

At one loop, there are three non-trivial diagrams, regardless of the choice of
the gauge.414141It is because diagrams with cubic gluon vertex $A^{3}$ start
at two-loop. They are the gluon exchange on the Wilson line in figure 4(a),
the gluon exchange on the scalar propagator in figure 4(b), and the gluon
exchange between the Wilson line and the scalar propagator in figure 4(c). We
now analyze them in turn.

###  B.1 Gluon Exchange on the Wilson line

The first diagram 4(a) does not depend on the matter content of the theory. It
is therefore the same as in (A.5), with the tree level scalar propagator
replaced by the fermionic one, (B.2).

###  B.2 Fermion Self-energy

The second diagram is a gluon exchange on the fermion propagator (self-
energy); see figure 4(b). It has been computed in momentum space in
Giombi:2011kc . Here, we reproduce it directly in position space. The
expression is

$\displaystyle\mathcal{M}_{b}^{f}=$
$\displaystyle\frac{\mathrm{i}^{2}N}{(4\pi)^{3}}{\rm
tr~{}}(T^{I}T^{J})\,\gamma^{\"{O}\textquestiondown\alpha}\gamma^{\mu}\gamma^{\beta}\gamma^{\nu}\gamma^{\delta}\\!\int\mathrm{d}^{3}y_{1}\mathrm{d}^{3}y_{2}\frac{(x_{1}-y_{1})_{\delta}}{|x_{1}-y_{1}|^{3}}\frac{(y_{12})_{\beta}}{|y_{12}|^{3}}\frac{(y_{2}-x_{0})_{\alpha}}{|x_{0}-y_{2}|^{3}}\langle
A_{\nu}^{I}(y_{1})A_{\mu}^{J}(y_{2})\rangle$ (B.3)

To proceed, we can project $\mathcal{M}_{b}$ to the complete matrix basis
$\\{\mathbb{I},\gamma^{\mu}\\}$. For instance,

$\displaystyle{\rm tr~{}}\mathcal{M}_{b}^{f}=$
$\displaystyle\,-\frac{\lambda}{32\pi^{3}}\int\mathrm{d}^{3}y_{1}\mathrm{d}^{3}y_{2}\,\frac{x_{1}^{\sigma}-y_{1}^{\sigma}}{|x_{1}-y_{1}|^{3}}\frac{1}{|y_{12}|^{4}}\frac{(y_{2}-x_{0})_{\sigma}}{|x_{0}-y_{2}|^{3}}$
(B.4) $\displaystyle=$
$\displaystyle\,+\frac{\lambda}{32\pi^{3}}\frac{\partial}{\partial
x_{1}^{\sigma}}\frac{\partial}{\partial
x_{0\sigma}}\int\mathrm{d}^{3}y_{1}\mathrm{d}^{3}y_{2}\,\frac{1}{|x_{1}-y_{1}|}\frac{1}{|y_{12}|^{4}}\frac{1}{|x_{0}-y_{2}|}$
$\displaystyle=$
$\displaystyle\,-\frac{\lambda}{32\pi^{3}}\int\mathrm{d}^{3}y_{1}\mathrm{d}^{3}y_{2}\,\frac{1}{|y_{1}|}\frac{1}{|y_{12}|^{4}}\Box_{y_{2}}\frac{1}{|x_{01}-y_{2}|}$
$\displaystyle=$
$\displaystyle\,+\frac{\lambda}{8\pi^{2}}\int\mathrm{d}^{3}y_{1}\,\frac{1}{|y_{1}|}\frac{1}{|y_{1}-x_{01}|^{4}}\,.$

Using Feynman parametrization to collect the denominators, and dimensional
regularization with $d=3+\varepsilon$, we get424242In fact, we have to do
analytical continuation for $d>4$, which amounts to dropping a linearly
divergent term.

${\rm
tr~{}}\mathcal{M}_{b}^{f}=\frac{\lambda}{8\pi^{2}}\frac{3}{4}\mu^{3-d}\int\limits_{0}^{1}\mathrm{d}u\frac{u}{\sqrt{1-u}}\frac{(2\pi)^{d}}{(4\pi)^{\frac{d}{2}}}\frac{\Gamma(\frac{5}{2}-\frac{d}{2})}{\Gamma(\frac{5}{2})}\Big{(}\frac{1}{u(1-u)x_{10}^{2}}\Big{)}^{\frac{5}{2}-\frac{d}{2}}=-\frac{\lambda}{4\pi
x_{01}^{2}}+{\cal O}(\varepsilon)\,.$ (B.5)

Similarly, we can project $\mathcal{M}_{b}^{f}$ to the $\gamma$ matrices,

$\displaystyle{\rm tr~{}}\mathcal{M}_{b}^{f}\gamma^{\alpha}$
$\displaystyle\propto\int\mathrm{d}^{3}y_{1}\mathrm{d}^{3}y_{2}\,\epsilon^{\alpha\lambda_{1}\lambda_{3}}\frac{(x_{1}-y_{1})_{\lambda_{3}}}{|x_{1}-y_{1}|^{3}}\frac{1}{|y_{12}|^{4}}\frac{(y_{2}-x_{0})_{\lambda_{1}}}{|x_{0}-y_{2}|^{3}}$
(B.6)
$\displaystyle\propto\epsilon^{\alpha\lambda_{1}\lambda_{3}}\partial^{x_{1}}_{\lambda_{3}}\partial^{x_{0}}_{\lambda_{1}}\int\mathrm{d}^{3}y_{1}\mathrm{d}^{3}y_{2}\frac{1}{|y_{1}-x_{1}||y_{12}|^{4}|y_{2}-x_{0}|}$
$\displaystyle\propto\epsilon^{\alpha\lambda_{1}\lambda_{3}}\partial^{x_{1}}_{\lambda_{3}}\partial^{x_{0}}_{\lambda_{1}}\int\mathrm{d}^{3}y_{1}\frac{1}{|y_{1}-x_{1}||y_{1}-x_{0}|^{2}}$
$\displaystyle\propto\epsilon^{\alpha\lambda_{1}\lambda_{3}}\partial^{x_{1}}_{\lambda_{3}}\partial^{x_{0}}_{\lambda_{1}}\left(\log
x_{10}^{2}+\mathrm{regulator}+\mathrm{finite}\right)$ $\displaystyle=0\,.$

To summarize, the 1-loop self-energy is given by

$\mathcal{M}_{b}^{f}=-\frac{\lambda}{8\pi x_{01}^{2}}\mathbb{I}\,.$ (B.7)

This result suggests that the one-loop correction to the fermionic propagator
in momentum space is proportional to $1/|p|$, or the 1PI part receives finite
correction $\Sigma(p)\propto|p|$, in agreement with Giombi:2011kc .

###  B.3 Fermion-Wilson Line Gluon Exchange

The last diagram is the gluon exchange between the fermion propagator and the
Wilson line; see figure 4(c). It is given by

$\mathcal{M}_{c}^{f}=-\frac{\mathrm{i}^{2}N}{2(4\pi)^{2}}\int\limits_{0}^{1}\mathrm{d}s\,\dot{x}_{s}^{\nu}\int\mathrm{d}^{3}y\,\gamma^{\sigma}\gamma^{\mu}\gamma^{\rho}\frac{(x_{1}-y)_{\rho}}{|x_{1}-y|^{3}}\frac{(y-x_{0})_{\sigma}}{|y-x_{0}|^{3}}\langle
A_{\mu}^{I}(y)A_{\nu}^{I}(x_{s})\rangle\,.$ (B.8)

We can compute it by projecting on the complete matrix basis as before. The
results are

${\rm
tr~{}}\mathcal{M}_{c}^{f}=-\frac{\lambda}{8\pi}\int\limits_{0}^{1}\mathrm{d}s\,\dot{x}_{s}^{\nu}(\delta^{\sigma}_{\lambda}\delta^{\rho}_{\nu}-\delta^{\sigma}_{\nu}\delta^{\rho}_{\lambda})\partial^{\rho}_{x_{1}}\partial^{\sigma}_{x_{0}}\frac{|x_{0s}|+|x_{1s}|-|x_{01}|}{|x_{0s}||x_{1s}|+x_{0s}\cdot
x_{1s}}\Big{[}\frac{x_{0s}^{\lambda}}{|x_{0s}|}+\frac{x_{1s}^{\lambda}}{|x_{1s}|}\Big{]},$
(B.9)

and

${\rm
tr~{}}\gamma^{\alpha}\mathcal{M}_{c}^{f}=\frac{\mathrm{i}\lambda}{8\pi}\int\limits_{0}^{1}\mathrm{d}s\,\dot{x}_{s}^{\nu}(\delta^{\sigma\mu}\delta^{\rho\alpha}-\delta^{\sigma\rho}\delta^{\mu\alpha}+\delta^{\sigma\alpha}\delta^{\mu\rho})\epsilon_{\mu\nu\lambda}\partial^{\rho}_{x_{1}}\partial^{\sigma}_{x_{0}}\frac{|x_{0s}|+|x_{1s}|-|x_{01}|}{|x_{0s}||x_{1s}|+x_{0s}\cdot
x_{1s}}\Big{[}\frac{x_{0s}^{\lambda}}{|x_{0s}|}+\frac{x_{1s}^{\lambda}}{|x_{1s}|}\Big{]}\,.$
(B.10)

For a straightline, $x_{s}^{\mu}=x_{0}^{\mu}+sx_{10}^{\mu}$, we find

$\mathcal{M}_{c}^{f}\Big{|}_{x_{s}=x_{0}+sx_{10}}=\frac{\lambda}{8\pi
x_{01}^{2}}\mathbb{I}\int\limits_{0}^{1}\frac{\mathrm{d}s}{s(1-s)}\,.$ (B.11)

Based on the preceding expressions, the one-loop integral can only be
logarithmic divergent near the endpoint. Any smooth line is straight near its
endpoints, so the divergent behavior is identical to the expression given
above.

##  Appendix C The Topological Parallel Transport

In this appendix, we discuss the construction of the topological parallel
transport.

###  C.1 Construction

##### Definition.

The parallel transport $U_{\mathcal{C}}(s,t)$ preserves the local $\pm$
component of the spinor indices along the segment of a generic curve
$\mathcal{C}$. When it does not lead to confusion, we shall omit the subscript
$\mathcal{C}$. For definiteness, we choose a parameterization of the curve
s.t. the start/end-points are labeled by $t$ and $s$, respectively.

Recall that the local spinor $\pm$ components at any point $x_{s}$ are defined
as the eigenvectors of $e_{s}\cdot\gamma$, where
$e_{s}=\dot{x}_{s}/|\dot{x}_{s}|$. As a result, the parallel transport
$U_{\mathcal{C}}(s,t)$ preserves eigenvectors of $\gamma\cdot e$,

$(\gamma\cdot e_{s})U_{\mathcal{C}}(s,t)=U_{\mathcal{C}}(s,t)(\gamma\cdot
e_{t})\,.$ (C.1)

We also demand that if we break the path into two as ${\cal C}={\cal
C}_{1}\cup{\cal C}_{2}$ then $U_{{\cal C}_{1}\cup{\cal C}_{2}}(s,t)=U_{{\cal
C}_{1}}(s,u)U_{{\cal C}_{2}}(u,t)$. Hence, we are looking for a solution of
the following form,

$U_{\mathcal{C}}(s,t)=\mathcal{P}e^{\mathrm{i}\int\limits_{t}^{s}\mathrm{d}\tau\,\Gamma(\tau)}\equiv\mathcal{P}e^{\mathrm{i}\int\limits_{t}^{s}\mathrm{d}\tau\,\Gamma_{\mu}\dot{x}^{\mu}_{\tau}}\,.$
(C.2)

Without loss of generality, we take the initial point to be $t=0$. Taking
derivatives of (C.1) w.r.t $s$, we find the following differential equation
for $U_{\mathcal{C}}(s)\equiv U_{\mathcal{C}}(s,0)$,

$(\gamma\cdot\dot{e}_{s})U_{\mathcal{C}}(s)+(\gamma\cdot
e_{s})\frac{\mathrm{d}U_{\mathcal{C}}(s)}{\mathrm{d}s}=\frac{\mathrm{d}U_{\mathcal{C}}(s)}{\mathrm{d}s}(\gamma\cdot
e_{0})=\frac{\mathrm{d}U_{\mathcal{C}}(s)}{\mathrm{d}s}U_{\mathcal{C}}(s)^{-1}(\gamma\cdot
e_{s})U_{\mathcal{C}}(s),$ (C.3)

where in the last step we have used the (C.1) to translate $\gamma\cdot e_{0}$
to the point $x_{s}$. In terms of $\Gamma$, the equation above reads,

$\gamma\cdot\dot{e}_{s}+\mathrm{i}(\gamma\cdot
e_{s})\Gamma_{s}=\mathrm{i}\Gamma_{s}(\gamma\cdot e_{s})\,.$ (C.4)

##### Solving for the Exponent.

Let us first discuss the possible structures of $\Gamma_{s}$. It is a scalar
matrix function acting on the spinor space. It can therefore be expressed in
terms of a complete matrix basis $\\{\mathbb{I},\vec{\gamma}\\}$. To absorb
the Lorentz indices of $\vec{\gamma}$, the only way is to contract them with
the local orthonormal frame
$\\{e,\frac{\dot{e}}{|\dot{e}|},e\times\frac{\dot{e}}{|\dot{e}|}\\}$. Remember
that this local orthonormal frame is a complete basis, so there are no longer
any independent vector fields at a given point.

As a consequence, $\Gamma$ has to take the following general form,434343Note
that the first term is proportional to $\ddot{e}\cdot\gamma$.

$\Gamma_{s}=\frac{\alpha}{2}\epsilon_{\mu\nu\rho}e_{s}^{\mu}\dot{e}_{s}^{\nu}\gamma^{\rho}+\tilde{\beta}\dot{e}\cdot\gamma+\beta|\dot{e}|e\cdot\gamma+\delta\,\mathbb{I}\,,$
(C.5)

where four unknown coefficients $\alpha,\beta,\tilde{\beta},\delta$ were
introduced. They can depend on local scalar functions of the curve, such as
the curvature and the torsion.444444As well as the relative angle between
$\dot{e}$ and the framing vector, if such exists.

Using the following commutation relations,

$\displaystyle(\gamma\cdot e_{s})(\gamma\cdot e_{s})$
$\displaystyle=(\gamma\cdot\dot{e}_{s})(\gamma\cdot\dot{e}_{s})=0,$ (C.6)
$\displaystyle(\gamma\cdot e_{s})(\gamma\cdot\dot{e}_{s})$
$\displaystyle=\mathrm{i}\epsilon_{\mu\nu\rho}e^{\mu}\dot{e}^{\nu}\gamma^{\rho},$
$\displaystyle(\gamma\cdot
e_{s})(\mathrm{i}\epsilon_{\mu\nu\rho}e^{\mu}\dot{e}^{\nu}\gamma^{\rho})$
$\displaystyle=\dot{e}_{s}\cdot\gamma,$

one can verify that

$[\Gamma_{s},\gamma\cdot
e_{s}]=\mathrm{i}\alpha\gamma\cdot\dot{e}_{s}-2\tilde{\beta}\epsilon_{\mu\nu\rho}e^{\mu}\dot{e}^{\nu}\sigma^{\rho}.$
(C.7)

Therefore, we must take $\alpha=-1,\tilde{\beta}=0$. We conclude that the
parallel transport with the desired properties is given by

$\displaystyle U_{\mathcal{C}}(s,0)$
$\displaystyle=\mathcal{P}\exp\int\limits_{0}^{s}\mathrm{d}\tau\left(-\frac{\mathrm{i}}{2}\epsilon_{\mu\nu\rho}e_{\tau}^{\mu}\dot{e}_{\tau}^{\nu}\gamma^{\rho}+\beta|\dot{e}|e\cdot\gamma+\delta\mathbb{I}\right)\,.$
(C.8)

The coefficients $\beta$ and $\delta$ are not constrained. As we now explain,
demanding that the path be topological specifies their form.

##### General Topological Transport.

A topological spin transport is, by definition, invariant under smooth
deformation of the path with fixed boundaries. The first term in (C.8) is,
however, not invariant under such a deformation. Instead, it gives a
contribution that cannot be canceled by setting $\beta$ to be any local
function of the path. That is, a function of the curvature and
torsion.454545See also discussions in Witten:1988hf  about the non-topological
nature of Polyakov’s spin factor Polyakov:1988md . To overcome this
obstruction, one must introduce a framing vector $n$, that we take to be a
unit orthogonal vector. In the presence of a framing vector, the form of the
topological spin connection is unique, up to the freedom in the choice of
$\delta$ in (C.8). In the next section we show that this topological
connection is given by

$\Gamma_{s}=-\frac{1}{2}\Big{(}(e\times\dot{e})-\tau_{n}(s)e\Big{)}\cdot\gamma+\delta\,\mathbb{I}\,,$
(C.9)

where $\times$ stands for the cross product for two vectors and

$\tau_{n}(s)=\frac{\left(\dot{n}\times n\right)\cdot e}{|e\times n|^{2}}\,,$
(C.10)

is the so-called relative torsion. In the arc-length parameterization, where
$|\dot{x}_{s}|=|n_{s}|=1$, it reduces to (A.6). The constant $\delta$ in (C.9)
leads to an overall phase factor. For the rest of this appendix, we set
$\delta=0$. In the main text, we include such an overall topological phase
factor that we separate from $\Gamma$, see the term
$\frac{\lambda}{2}\partial_{\mu}\log n^{+}$ in $(\widetilde{A}_{0})_{\mu}$,
(4.3).

###  C.2 Functional Variation

The topological constraint means that the spin transport is invariant under
smooth variations of the path. We now show that $\Gamma_{s}$ in (C.9) indeed
has this property.

##### Variation of Vector Fields.

We shall use the arc-length parameterization throughout the section unless
otherwise stated.

We deform the curve $x_{s}^{\mu}$ by a vector $v^{\mu}$, $x_{s}^{\mu}\mapsto
x_{s}^{\mu}+\epsilon\,v^{\mu}$. To linear order in $\epsilon$ we have

$\displaystyle e^{\mu}$ $\displaystyle\quad\rightarrow\quad
e^{\mu}+\epsilon(\dot{v}^{\mu}-(e\cdot\dot{v})e^{\mu})+{\cal
O}(\epsilon^{2})\,,$ (C.11) $\displaystyle\dot{e}^{\mu}$
$\displaystyle\quad\rightarrow\quad\dot{e}^{\mu}+\epsilon\Big{(}\ddot{v}^{\mu}-e^{\mu}[(\dot{e}\cdot\dot{v}+e\cdot\ddot{v})]-(e\cdot\dot{v})\dot{e}^{\mu}\Big{)}+{\cal
O}(\epsilon^{2})\,.$

Using the linear order result, we find the linear order variation of the first
term in (C.9) is

$e\times\dot{e}\quad\rightarrow\quad
e\times\dot{e}+\epsilon\Big{(}\dot{v}\times\dot{e}-2(e\cdot\dot{v})e\times\dot{e}+e\times\ddot{v}\Big{)}+\mathcal{O}(\epsilon^{2})\,.$
(C.12)

##### Variation of the Transport.

The functional variation of $U(1,0)$ is given by,

$\delta
U(1,0)=\frac{\mathrm{i}}{2}\int\limits_{0}^{1}\mathrm{d}s\,U(1,s)\Big{(}\delta\tau_{n}(s)e+\tau_{n}(s)\delta
e-\delta(e\times\dot{e})\Big{)}\cdot\gamma\,U(s,0)\,.$ (C.13)

Under the constraint $e\cdot v=0$, we can trade all derivatives on $v$
appearing in the inner products by derivatives on $e$. For example, using

$0=\frac{\mathrm{d}}{\mathrm{d}s}(e\cdot v)\quad\Rightarrow\quad
e\cdot\dot{v}=-\dot{e}\cdot v\,,$ (C.14)

we can simplify the second term of the variation of $e\times\dot{e}$,

$\delta(e\times\dot{e})=\dot{v}\times\dot{e}+2(v\cdot\dot{e})e\times\dot{e}+e\times\ddot{v}\,.$
(C.15)

Now, performing IBP for the $e\times\ddot{v}$ term once, we find,

$\displaystyle\delta U(1,0)$
$\displaystyle=\frac{\mathrm{i}}{2}\int\limits_{0}^{1}\mathrm{d}s\,U(1,s)\Big{(}\delta\tau_{n}e-\dot{v}\times\dot{e}-2(v\cdot\dot{e})e\times\dot{e}+\dot{e}\times\dot{v}-[(e\times\dot{e})\cdot\dot{v}]e\Big{)}\cdot\gamma\,U(s,0)$
$\displaystyle-\frac{\mathrm{i}}{2}\int\mathrm{d}s\,\frac{\mathrm{d}}{\mathrm{d}s}\Big{(}U(1,s)(e\times\dot{v})\cdot\gamma\,U(s,0)\Big{)}\,.$
(C.16)

Next, we will show that the expression in the first line of (C.2) vanishes,

$\dot{v}\times\dot{e}+2(v\cdot\dot{e})e\times\dot{e}-\dot{e}\times\dot{v}+[(e\times\dot{e})\cdot\dot{v}]e-\delta\tau_{n}e=0\,.$
(C.17)

As a result, the variation is the boundary term in the second line. The
parallel transport described above is, therefore, topological.

##### Frenet Frame.

A convenient method to deal with geometric quantities is to use the local
orthonormal Frenet frame for the curve,
$\\{\mathbf{T},\mathbf{N},\mathbf{B}\\}$. Here $\mathbf{T}=e$ is the unit
tangent, $\mathbf{N}=\dot{e}/|\dot{e}|$ is the unit normal,
$\mathbf{B}=\mathbf{T}\times\mathbf{N}$ is the so-called bi-normal. As we move
along the path, they rotate as

$\frac{\mathrm{d}}{\mathrm{d}s}\begin{pmatrix}\mathbf{T}\\\ \mathbf{N}\\\
\mathbf{B}\\\ \end{pmatrix}=|\dot{x}_{s}|\begin{pmatrix}0&\kappa&0\\\
-\kappa&0&\tau\\\ 0&-\tau&0\\\ \end{pmatrix}\begin{pmatrix}\mathbf{T}\\\
\mathbf{N}\\\ \mathbf{B}\\\ \end{pmatrix}\,,$ (C.18)

where the two scalar function $\kappa,\tau$ are the curvature and the torsion
of the curve, respectively. Our strategy is to simplify the first four terms
in (C.17), then show that the last term cancels the sum of them.

#####  $v$ Dependent Terms.

The first four terms can be expressed in a compact way. Explicitly, we will
show that

$\dot{v}\times\dot{e}+2(v\cdot\dot{e})e\times\dot{e}-\dot{e}\times\dot{v}+[(e\times\dot{e})\cdot\dot{v}]e=-[(e\times\dot{e})\cdot\dot{v}]e\,.$
(C.19)

The normal deformation $v$ can be projected on the Frenet-Serret frame,

$v=a\,\mathbf{N}+c\,\mathbf{B}\,.$ (C.20)

The derivative of $v$ can then be computed using (C.18),

$\dot{v}=-a\kappa\mathbf{T}+(\dot{a}-c\tau)\mathbf{N}+(\dot{c}+a\tau)\mathbf{B}\equiv-a\kappa\mathbf{T}+\mathfrak{a}\mathbf{N}+\mathfrak{c}\mathbf{B}.$
(C.21)

For the $v$-dependence terms in (C.17), we can now compute them term by term,

$\displaystyle\dot{v}\times\dot{e}$
$\displaystyle=\kappa\dot{v}\times\mathbf{N}=-a\kappa^{2}\mathbf{B}-\mathfrak{c}\mathbf{T},$
(C.22) $\displaystyle(v\cdot\dot{e})e\times\dot{e}$
$\displaystyle=a\kappa^{2}\mathbf{T}\times\mathbf{N}=a\kappa^{2}\mathbf{B},$
$\displaystyle(e\times\dot{e})\cdot\dot{v}$
$\displaystyle=\kappa\mathbf{B}\cdot\dot{v}=\mathfrak{c}\,.$

Hence, we see that

$-\kappa\mathfrak{c}\mathbf{T}\cdot\sigma=-[(e\times\dot{e})\cdot\dot{v}](e\cdot\sigma)\,.$
(C.23)

These expressions result in the relation (C.19).

For plane curves, $v$ only has normal component $c=\dot{c}=\tau=0$ so the sum
(C.23) vanishes (and so does the fifth term of (C.17)). Generally, for non-
planar curves, they will not vanish. Below, we will show that they cancel.

##### Geometry of the Relative Torsion.

Let us first study the geometric meaning of $\tau_{n}$. The framing vector $n$
is a unit normal vector, so we can expand it onto the Frenet-Serret frame,

$n=\cos\theta_{s}\mathbf{N}+\sin\theta_{s}\mathbf{B}.$ (C.24)

After some simple algebra, one can verify that the relative torsion differs
from the geometric torsion by a total derivative of the angle
$\theta$,464646Recall that the geometric torsion is defined in $\eqref{eqn-
apd-PT-Frenet}$. Explicitly, it is defined by taking the normal $n$ in
$\tau_{n}$ to be the geometric normal $\mathbf{N}$, $\tau=\tau_{\mathbf{N}}$.

$\tau_{n}=\frac{\left(\dot{n}\times n\right)\cdot e}{|e\times
n|^{2}}=\tau+\dot{\theta}_{s}\,.$ (C.25)

From this expression, we learn that the meaning of the “relative” is the
relative angle between the geometric normal $\mathbf{N}$ and the framing
normal $n$.

With this geometric meaning in mind, computing the functional variation of
$\tau_{n}$ reduces to computing the variation of the geometric torsion and the
relative angle $\dot{\theta}_{s}$.

##### Variation of the Torsion.

The deformation by a vector $v$ induces the deformation of the unit tangent,
which is given by (see equation (C.21))

$\displaystyle\delta\mathbf{T}$
$\displaystyle=\mathfrak{a}\mathbf{N}+\mathfrak{c}\mathbf{B}\,.$ (C.26)

Consider the functional variation of the second Frenet-Serret equation. Taking
inner product with binormal $\mathbf{B}$, we get

$\mathbf{B}\cdot\delta\Big{(}\frac{\mathrm{d}\mathbf{N}}{ds}\Big{)}=\delta(|\dot{x}|\tau)-\kappa\mathbf{B}\cdot\delta\mathbf{T}+\tau\underbrace{\mathbf{B}\cdot\delta\mathbf{B}}_{=\delta(\mathbf{B}\cdot\mathbf{B})=0}\Rightarrow\delta(|\dot{x}|\tau)=\kappa\mathbf{B}\cdot\delta\mathbf{T}+\mathbf{B}\cdot\delta\partial_{s}\mathbf{N}\,.$
(C.27)

Notice that the second term is actually a total derivative

$\mathbf{B}\cdot\delta(\partial_{s}\mathbf{N})=\frac{\mathrm{d}}{\mathrm{d}s}(\mathbf{B}\cdot\delta\mathbf{N})-\underbrace{\frac{\mathrm{d}\mathbf{B}}{\mathrm{d}s}\cdot\delta\mathbf{N}}_{=\tau\mathbf{N}\cdot\delta\mathbf{N}=\tau\delta(\mathbf{N}\cdot\mathbf{N})=0}\,.$
(C.28)

Therefore, the final expression for the functional variation of
$\tau|\dot{x}|$ is given by474747The functional variation of $\tau$ should
correspond to the variation of the combination $\tau|\dot{x}|$ in a generic
parametrization of the line. In arc-length parameterization, $|\dot{x}|=1$, so
the variation of the combination is equal to the variation of $\tau$. Refer to
the section’s concluding paragraph for clarification.

$\delta(|\dot{x}|\tau)=\kappa\mathfrak{c}+\frac{\mathrm{d}}{\mathrm{d}s}\Big{(}\mathbf{B}\cdot\delta\mathbf{N}\Big{)}\,.$
(C.29)

##### Variation of the Relative Torsion.

Since the relative torsion $\tau_{n}$ differs from the geometric torsion
$\tau$ only by $\dot{\theta}$, it suffices to study the variation of $\theta$.
Recall $n$ is kept fixed,

$\cos\theta_{s}=n\cdot\mathbf{N}\quad\Rightarrow\quad-\sin\theta_{s}\delta\theta_{s}=n\cdot\delta\mathbf{N}=\sin\theta_{s}\mathbf{B}\cdot\delta\mathbf{N}\,,$
(C.30)

or in other words,

$\delta\theta_{s}=-\mathbf{B}\cdot\delta\mathbf{N},\quad\Rightarrow\quad\delta\dot{\theta}_{s}=-\frac{\mathrm{d}}{\mathrm{d}s}\Big{(}\mathbf{B}\cdot\delta\mathbf{N}\Big{)}\,.$
(C.31)

Combining (C.29) and (C.31), we find a simple result,

$\delta\tau_{n}=\kappa\mathfrak{c}\,.$ (C.32)

It cancels the remaining bulk term (C.23), so we conclude that (C.18) indeed
holds true.

##### Remark on Functional Variation.

We would like to show that functional variation commutes with derivatives. For
that aim, consider two smooth curves related by a small variation. The first
curve ${\cal C}$ parameterized by $x_{s}$, and the other
$\mathcal{C}^{\epsilon}$ parameterized by $y_{s}=x_{s}+\epsilon v_{s}$. Here,
we have chosen the same parametrization for the two curves. Provided that we
use the arc length parametrization of the path ${\cal C}$, the resulting
parametrization of ${\cal C}^{\epsilon}$ is not its arc-length one.
Consequently, one must carefully keep track of the functional variations of
the geometric quantities. For example, consider the functional variation of
the derivative of the tangent vector $\mathbf{T}$

$\delta\frac{\mathrm{d}\mathbf{T}}{\mathrm{d}s}\equiv\frac{\mathrm{d}\mathbf{T}^{\epsilon}_{s}}{\mathrm{d}s}-\frac{\mathrm{d}\mathbf{T}_{s}}{\mathrm{d}s}=|\dot{x}^{\epsilon}_{s}|\kappa^{\epsilon}_{s}\mathbf{N}^{\epsilon}_{s}-\kappa|\dot{x}_{s}|\mathbf{N}=\delta(\kappa|\dot{x}_{s}|)\mathbf{N}+\kappa|\dot{x}_{s}|\delta\mathbf{N}\,.$
(C.33)

For the deformed unit tangent, one has to use the generic Frenet-Serret
equation, with $|\dot{x}|\neq 1$.

To summarize, when taking functional variations of the Frenet-Serret
equations, the functional variation of $\kappa,\tau$ is the variation of the
combination $|\dot{x}|\kappa,|\dot{x}|\tau$. With this in mind, we conclude
that the functional variation commutes with derivatives.

###  C.3 The Dependence on the Framing Vector

Recall that the dependence of a mesonic line $M^{({1\over 2},-{1\over 2})}$ on
the framing vector $n$, given in (3.10), is
$\left(2\,n_{L}^{+}n_{R}^{-}\right)^{\frac{\lambda+1}{2}}$. We now show that
the same framing factor is also generated by the topological spin connection
in (4.23) and the term $\frac{\lambda}{2}\partial_{\mu}\log n^{+}$ in
$(\widetilde{A}_{0})_{\mu}$, (4.3). Hence, this dependence continues to hold
for the condensed fermion operator (4.7), which is decomposed from segments
with Wilson line and segments with the topological parallel transport, see
figure 12.

The topological spin connection (4.2) is decomposed from two pieces, the
geometric transport $(e\times\dot{e})\cdot\gamma$, and the relative torsion
$\tau_{n}\,e\cdot\gamma$. In section C.1, we have seen that the geometric
transport, as a $\beta=\delta=0$ case of the generic parallel transport (C.8),
satisfies (C.1). Namely, it allows one to transport any terms of the form
$e\cdot\gamma$ to the endpoints of the line. Since the relative torsion term
is of this form, it can be transported to the end of the line as

$\mathcal{P}e^{\mathrm{i}\int\limits_{0}^{1}\mathrm{d}s\,\Gamma_{s}}=\big{[}e^{\frac{\mathrm{i}}{2}e_{s}\cdot\gamma\int\limits_{0}^{1}\mathrm{d}s\,\tau_{n}}\big{]}\cdot\big{[}\mathcal{P}e^{-\frac{\mathrm{i}}{2}\int\limits_{0}^{1}\mathrm{d}s\,(e_{s}\times\dot{e}_{s})\cdot\gamma}\big{]}=\big{[}\mathcal{P}e^{-\frac{\mathrm{i}}{2}\int\limits_{0}^{1}\mathrm{d}s\,(e_{s}\times\dot{e}_{s})\cdot\gamma}\big{]}\cdot\big{[}e^{\frac{\mathrm{i}}{2}e_{t}\cdot\gamma\int\limits_{0}^{1}\mathrm{d}s\,\tau_{n}}\big{]}\,.$
(C.34)

After this, the relative torsion term becomes a pure phase acting on the plus
and minus spin components,

$P_{\pm}(s)\cdot
e^{\frac{\mathrm{i}}{2}e_{s}\cdot\gamma\int\limits_{0}^{1}\mathrm{d}s\,\tau_{n}}=P_{\pm}(s)e^{\mp{\mathrm{i}\over
2}\int\limits_{t}^{s}\tau_{n}},\qquad
e^{\frac{\mathrm{i}}{2}e_{t}\cdot\gamma\int\limits_{0}^{1}\mathrm{d}s\,\tau_{n}}\cdot
P_{\pm}(t)=e^{\mp{\mathrm{i}\over 2}\int\limits_{t}^{s}\tau_{n}}P_{\pm}(t)\,.$
(C.35)

Consider next a straight line along the third direction. We then parameterize
the components of the framing vector using polar coordinates as
$n^{\pm}_{\tau}=e^{\pm\mathrm{i}\theta_{\tau}}/\sqrt{2}$. We find

$\frac{\mathrm{i}}{2}\tau_{n}=\frac{\mathrm{i}}{2}\epsilon_{\mu\nu\rho}\dot{n}^{\mu}n^{\nu}e^{\rho}=\frac{\mathrm{i}}{2}\epsilon_{+-3}\left(\dot{n}^{+}n^{-}-\dot{n}^{-}n^{+}\right)=-\frac{\mathrm{i}}{2}\dot{\theta}_{\tau}=-\frac{1}{2}\mathrm{d}\log
n^{+}_{\tau}\,,$ (C.36)

and therefore

$e^{\mp{\mathrm{i}\over
2}\int\limits_{0}^{1}\mathrm{d}s\,\tau_{n}(s)}=\Big{(}\frac{n_{0}^{+}}{n_{1}^{+}}\Big{)}^{\pm\frac{1}{2}}=\Big{(}2\,n_{0}^{+}n_{1}^{-}\Big{)}^{\pm\frac{1}{2}}\,.$
(C.37)

Because the spin transport (C.34) is topological, the same result holds for a
path that is not straight.

The spin factor (C.37) transports the classical $\pm 1/2$ spin of the fermions
to the endpoints. The anomalous part of the spin is taken care of by the term
$\frac{\lambda}{2}\partial_{\mu}\log n^{+}$ in $(\widetilde{A}_{0})_{\mu}$,
(4.3). Together, they result in the framing factor in (4.5).

##  Appendix D Computation of the Perturbative Integral

In this appendix, we show how to re-sum (2.44) in the small $\tilde{\epsilon}$
limit. Namely, we wish to compute the wave function renormalization factor in

$\langle M_{01}\rangle={\tilde{c}_{0}(-\lambda)\over 4\pi
x}\left({\epsilon\over
x}\right)^{\lambda}\,,\qquad\text{for}\quad\alpha=1\quad\text{and}\quad\lambda\geq
0\,.$ (D.1)

The result we derive in this appendix is