When $x_{f1}\neq x_{f2}$, from equation (214) we have
$\lambda_{1}\neq\lambda_{2}$. In this case, by demanding that
$\beta_{h_{i}}^{\text{1-loop}}=0$, we get

$h_{i}=\Big{(}\frac{3+\sigma_{i}\sqrt{6}}{16}\Big{)}\lambda_{i},$ (215)

where $\sigma_{1}$ and $\sigma_{2}$ can be $1$ or $-1$. Therefore,

$\begin{split}h_{p}\equiv\frac{h_{1}+h_{2}}{2}=\frac{3(\lambda_{1}+\lambda_{2})+(\sigma_{1}\lambda_{1}+\sigma_{2}\lambda_{2})\sqrt{6}}{32},\\\
h_{m}\equiv\frac{h_{1}-h_{2}}{2}=\frac{3(\lambda_{1}-\lambda_{2})+(\sigma_{1}\lambda_{1}-\sigma_{2}\lambda_{2})\sqrt{6}}{32}.\end{split}$
(216)

One can define similar linear combinations for the double trace couplings to
simplify the analyisis:

$f_{p}\equiv\frac{f_{1}+f_{2}}{2},f_{m}\equiv\frac{f_{1}-f_{2}}{2}.$ (217)

The 1-loop beta functions of these couplings (along with $\zeta$) are as
follows:

$\begin{split}\beta_{f_{p}}^{\text{1-loop}}=&f_{m}\Big{[}32h_{m}-3(\lambda_{1}-\lambda_{2})\Big{]}+8\Big{[}\zeta^{2}+(f_{p}+h_{p})(f_{p}+3h_{p})\Big{]}+8f_{m}^{2}-3f_{p}(\lambda_{1}+\lambda_{2})\\\
&+24h_{m}^{2}+\frac{9}{32}(\lambda_{1}^{2}+\lambda_{2}^{2}),\\\
\beta_{f_{m}}^{\text{1-loop}}=&f_{m}\Big{[}16f_{p}+32h_{p}-3(\lambda_{1}+\lambda_{2})\Big{]}+32f_{p}h_{m}-3f_{p}(\lambda_{1}-\lambda_{2})+48h_{p}h_{m}+\frac{9}{32}(\lambda_{1}^{2}-\lambda_{2}^{2}),\\\
\beta_{\zeta}^{\text{1-loop}}=&\zeta\Big{[}16f_{p}+32h_{p}-3(\lambda_{1}+\lambda_{2})\Big{]}.\end{split}$
(218)

By setting $\beta_{\zeta}^{\text{1-loop}}=0$ and demanding that $\zeta\neq 0$,
we get

$\begin{split}&f_{p}=-2h_{p}+\frac{3}{16}(\lambda_{1}+\lambda_{2}).\end{split}$
(219)

Similarly, setting $\beta_{f_{m}}^{\text{1-loop}}=0$, we get

$\begin{split}&f_{p}=\frac{-48h_{p}h_{m}-\frac{9}{32}(\lambda_{1}^{2}-\lambda_{2}^{2})}{32h_{m}-3(\lambda_{1}-\lambda_{2})}\\\
\end{split}$ (220)

Here, we have assumed that

$\begin{split}32h_{m}-3(\lambda_{1}-\lambda_{2})\neq 0.\end{split}$ (221)

From the value of $h_{m}$ given in equation (216), we can see that this is
equivalent to demanding

$\begin{split}\sigma_{1}\lambda_{1}\neq\sigma_{2}\lambda_{2}.\end{split}$
(222)

When $\sigma_{1}=\sigma_{2}$, this holds true trivially because $x_{f1}\neq
x_{f2}$. When $\sigma_{1}=-\sigma_{2}$, this is true because otherwise
$\lambda_{1}$ and $\lambda_{2}$ would have opposite signs. This is not
admissible because $\lambda_{i}$ is related to the gauge coupling by the
relation

$\begin{split}\lambda_{i}\equiv\frac{N_{ci}g_{i}^{2}}{(4\pi)^{2}}\
,\end{split}$ (223)

and hence both $\lambda_{1}$ and $\lambda_{2}$ must be positive. 727272Here
the reality of $g_{i}$ is the crucial assumption which is a necessary
condition for the unitarity of the theory. Therefore, we can trust equation
(220).

Now, combining (219) with (220), we get

$16h_{p}h_{m}-6h_{m}(\lambda_{1}+\lambda_{2})-6h_{p}(\lambda_{1}-\lambda_{2})+\frac{9}{32}(\lambda_{1}^{2}-\lambda_{2}^{2})=0.$
(224)

Substituting the values of $h_{p}$ and $h_{m}$ (given in (216)) into the above
equation, we get

$\begin{split}&(13+6\sqrt{6}\sigma_{1})\lambda_{1}^{2}=(13+6\sqrt{6}\sigma_{2})\lambda_{2}^{2}.\end{split}$
(225)

When $\sigma_{1}=\sigma_{2}$, we get
$\lambda_{1}^{2}=\lambda_{2}^{2}\implies\lambda_{1}=\lambda_{2}$ since we have
already shown that both $\lambda_{1}$ and $\lambda_{2}$ must be positive.
However, this cannot be true for $x_{f1}\neq x_{f2}$.

The other possibility is that $\sigma_{1}=-\sigma_{2}$. For instance, consider
the case $\sigma_{1}=-\sigma_{2}=1$. Then we get

$\begin{split}&(13+6\sqrt{6})\lambda_{1}^{2}=(13-6\sqrt{6})\lambda_{2}^{2}.\end{split}$
(226)

Now the coefficient $(13+6\sqrt{6})>0$, whereas
$(13-6\sqrt{6})\approx-1.69694<0$. Then the signs of $\lambda_{1}^{2}$ and
$\lambda_{2}^{2}$ are opposite. However, this is not consistent with the
reality of $\lambda_{1}$ and $\lambda_{2}$. Therefore, equation (226) cannot
be satisfied. Similarly, we can rule out the existence of any unitary fixed
point with $\sigma_{1}=-\sigma_{2}=-1$. Thus, from the above analysis, we can
conclude that when $x_{f1}\neq x_{f2}$, there is no unitary fixed point of the
1-loop beta functions with $\zeta\neq 0$.

###  D.2 Proof of constraint 2

When $x_{f1}=x_{f2}=x_{f}$, from equation (214), we have
$\lambda_{1}=\lambda_{2}=\lambda\equiv\frac{21-4x_{f}}{-27+13x_{f}}$. As
before the couplings for the single trace interactions are given by

$h_{i}=\Big{(}\frac{3+\sigma_{i}\sqrt{6}}{16}\Big{)}\lambda.$ (227)
$\begin{split}h_{p}=\frac{6+(\sigma_{1}+\sigma_{2})\sqrt{6}}{32}\lambda,\
h_{m}=\frac{(\sigma_{1}-\sigma_{2})\sqrt{6}}{32}\lambda.\end{split}$ (228)

The 1-loop beta functions of the couplings corresponding to the double trace
interactions simplify in this case as follows:

$\begin{split}\beta_{f_{p}}^{\text{1-loop}}=&\
32f_{m}h_{m}+8\Big{[}\zeta^{2}+(f_{p}+h_{p})(f_{p}+3h_{p})\Big{]}+8f_{m}^{2}-6f_{p}\lambda+24h_{m}^{2}+\frac{9}{16}\lambda^{2},\\\
\beta_{f_{m}}^{\text{1-loop}}=&f_{m}\Big{[}16f_{p}+32h_{p}-6\lambda\Big{]}+32f_{p}h_{m}+48h_{p}h_{m},\\\
\beta_{\zeta}^{\text{1-loop}}=&\zeta\Big{[}16f_{p}+32h_{p}-6\lambda\Big{]}.\end{split}$
(229)

Let us focus on the last two beta functions given above. Setting them equal to
zero and searching for fixed points with $\zeta\neq 0$, we get

$\begin{split}16f_{p}=-32h_{p}+6\lambda,\
32f_{p}h_{m}+48h_{p}h_{m}=0.\end{split}$ (230)

Substituting the value of $f_{p}$ obtained from the first equation into the
second one, we get

$\begin{split}(12\lambda-16h_{p})h_{m}=0\implies\Bigg{(}\frac{18-(\sigma_{1}+\sigma_{2})\sqrt{6}}{2}\Bigg{)}(\sigma_{1}-\sigma_{2})=0.\end{split}$
(231)

The only way in which the above equation can be satisfied is if
$\sigma_{1}=\sigma_{2}=\sigma$. In this case, we have

$\begin{split}h_{p}=\frac{3+\sigma\sqrt{6}}{16}\lambda,\ h_{m}=0,\
f_{p}=-\frac{\sigma\sqrt{6}}{8}\lambda.\end{split}$ (232)

Now, setting $\beta_{f_{p}}^{\text{1-loop}}=0$, we get

$\begin{split}&8(\zeta^{2}+f_{m}^{2})=-\frac{1}{32}\Bigg{[}18\sigma\sqrt{6}+39\Bigg{]}\lambda^{2}.\end{split}$
(233)

Note that the reality of the couplings $f_{m}$ and $\zeta$ leads to the LHS of
the above equation being non-negative. The RHS of the same equation is,
however, manifestly negative if $\sigma=1$. Thus, the only admissible unitary
fixed point is the one where $\sigma=-1$. Therefore, we can conclude that when
$x_{f_{1}}=x_{f_{2}}$, at any unitary fixed point where the two sectors are
coupled, the couplings corresponding to the single trace interactions have the
following values:

$h_{1}=h_{2}=\Big{(}\frac{3-\sqrt{6}}{16}\Big{)}\lambda.$ (234)

##  Appendix E Minima of the thermal effective potential

In this appendix, we will investigate the minima of the thermal effective
potential of the scalar fields in the real double bifundamental model. When
both the thermal masses (squared), $m_{\text{th},1}^{2}$ and
$m_{\text{th},2}^{2}$, are positive, one can trivially conclude that the
minimum lies at the origin of the field space. On the other hand, if either
$m_{\text{th},1}^{2}$ or $m_{\text{th},2}^{2}$ is negative, the minima would
lie away from the origin, and the baryon symmetry would be broken. To analyze
the location of the minima in such a situation, we would restrict our
attention to the fixed points discussed in section 3.2 for which
$r\equiv\frac{N_{c2}}{N_{c1}}<1$, and $m_{\text{th},1}^{2}>0$ while
$m_{\text{th},2}^{2}<0$.

First, let us employ gauge transformations to bring the matrices of the scalar
fields to the following diagonal forms:

$\displaystyle\Phi_{i}=\text{diag}\\{\phi_{i1},\cdots,\phi_{iN_{ci}}\\}.$
(235)

The thermal effective potential (up to leading order in $\lambda$) for such a
configuration is

$\displaystyle V_{\text{eff}}=$
$\displaystyle\frac{1}{2}\sum_{i=1}^{2}\sum_{a_{i}=1}^{N_{ci}}m_{\text{th},i}^{2}(\phi_{ia_{i}})^{2}+\sum_{i=1}^{2}\sum_{a_{i}=1}^{N_{ci}}\widetilde{h}_{i}(\phi_{ia_{i}})^{4}$
(236)
$\displaystyle+\sum_{i=1}^{2}\sum_{a_{i}=1}^{N_{ci}}\sum_{b_{i}=1}^{N_{ci}}\widetilde{f}_{i}(\phi_{ia_{i}})^{2}(\phi_{ib_{i}})^{2}+2\widetilde{\zeta}\sum_{a_{1}=1}^{N_{c1}}\sum_{b_{2}=1}^{N_{c2}}(\phi_{1a_{1}})^{2}(\phi_{2b_{2}})^{2}.$

One can determine the saddle points of this potential by setting its partial
derivatives with respect to all the scalar fields equal to zero as shown
below:

$\displaystyle\frac{\partial V_{\text{eff}}}{\partial\phi_{ia_{i}}}=$
$\displaystyle
2\phi_{ia_{i}}\Bigg{[}\frac{1}{2}m_{\text{th},i}^{2}+2\widetilde{h}_{i}(\phi_{ia_{i}})^{2}+2\widetilde{f}_{i}\sum_{b_{i}=1}^{N_{ci}}(\phi_{ib_{i}})^{2}+2\widetilde{\zeta}\sum_{b_{i^{\prime}}=1}^{N_{ci^{\prime}}}(\phi_{i^{\prime}b_{i^{\prime}}})^{2}\Bigg{]}=0,$
(237)

where $i^{\prime}$ denotes the complement of $i$. The above equation has the
following possible solutions:

$\displaystyle\phi_{ia_{i}}=0\ \ \text{or}\ \
\frac{1}{2}m_{\text{th},i}^{2}+2\widetilde{h}_{i}(\phi_{ia_{i}})^{2}+2\widetilde{f}_{i}\sum_{b_{i}=1}^{N_{ci}}(\phi_{ib_{i}})^{2}+2\widetilde{\zeta}\sum_{b_{i^{\prime}}=1}^{N_{ci^{\prime}}}(\phi_{i^{\prime}b_{i^{\prime}}})^{2}=0.$
(238)

Let us consider a saddle where $n_{i}$ of the $N_{ci}$ diagonal entries of
$\Phi_{i}$ are nonzero. In E.1, we will show that for $n_{1}\neq 0$, the
solution corresponds to negative values of $(\phi_{1a_{1}})^{2}$ which is in
conflict with the reality of $\phi_{1a_{1}}$. So such a saddle point cannot
correspond to a minimum of the potential. Later in E.2, we will argue that the
minima actually correspond to $n_{1}=0,\ n_{2}=N_{c2}$.

###  E.1 Saddle points with $n_{1}\neq 0$ correspond to imaginary field
configurations

In this subsection we will prove that all the saddle points with $n_{1}\neq 0$
correspond to imaginary values of $\phi_{1a_{1}}$. For this, let us consider
such a saddle where $\phi_{a_{i}}\neq 0$ for the first $n_{i}$ values of
$a_{i}$. The nonzero components of the scalar fields satisfy the following
equations:

$\displaystyle\frac{1}{2}m_{\text{th},1}^{2}+2\widetilde{h}_{1}(\phi_{1a_{1}})^{2}+2\widetilde{f}_{1}\sum_{b_{1}=1}^{n_{1}}(\phi_{1b_{1}})^{2}+2\widetilde{\zeta}\sum_{b_{2}=1}^{n_{2}}(\phi_{2b_{2}})^{2}=0\
\ \forall\ \ a_{1}\in\\{1,\cdots,n_{1}\\},$ (239)
$\displaystyle\frac{1}{2}m_{\text{th},2}^{2}+2\widetilde{h}_{2}(\phi_{2a_{2}})^{2}+2\widetilde{f}_{2}\sum_{b_{2}=1}^{n_{2}}(\phi_{2b_{2}})^{2}+2\widetilde{\zeta}\sum_{b_{1}=1}^{n_{1}}(\phi_{1b_{1}})^{2}=0\
\ \forall\ \ a_{2}\in\\{1,\cdots,n_{2}\\}.$

Solving these equations, we get the following values of the squares of the
fields after substituting the original couplings by the corresponding ’t Hooft
couplings:

$\displaystyle(\phi_{1a_{1}})^{2}=-\frac{N_{c1}^{2}}{64\pi^{2}}\Bigg{[}\frac{(h_{2}N_{c2}+f_{2}n_{2})m_{\text{th},1}^{2}-r\zeta
n_{2}m_{\text{th},2}^{2}}{(h_{1}N_{c1}+f_{1}n_{1})(h_{2}N_{c2}+f_{2}n_{2})-\zeta^{2}n_{1}n_{2}}\Bigg{]}\
\ \forall\ \ a_{1}\in\\{1,\cdots,n_{1}\\},$ (240)
$\displaystyle(\phi_{2a_{2}})^{2}=-\frac{N_{c2}^{2}}{64\pi^{2}}\Bigg{[}\frac{(h_{1}N_{c1}+f_{1}n_{1})m_{\text{th},2}^{2}-\frac{\zeta}{r}n_{1}m_{\text{th},1}^{2}}{(h_{1}N_{c1}+f_{1}n_{1})(h_{2}N_{c2}+f_{2}n_{2})-\zeta^{2}n_{1}n_{2}}\Bigg{]}\
\ \forall\ \ a_{2}\in\\{1,\cdots,n_{2}\\}.$

To prove that the nonzero values of $(\phi_{1a_{1}})^{2}$ are negative, we
will show that both the numerator and the denominator of the quantity within
the brackets in the first line of (240) are positive. First, let us consider
the denominator:

$\displaystyle(h_{1}N_{c1}+f_{1}n_{1})(h_{2}N_{c2}+f_{2}n_{2})-\zeta^{2}n_{1}n_{2}=$
$\displaystyle
h_{1}h_{2}N_{c1}N_{c2}+h_{1}f_{2}N_{c1}n_{2}+h_{2}f_{1}N_{c2}n_{1}$ (241)
$\displaystyle+(f_{1}f_{2}-\zeta^{2})n_{1}n_{2}.$

Note that in section 3.2.1, we have already shown that $h_{i}>0$, $f_{i}>0$
and $(f_{1}f_{2}-\zeta^{2})>0$ for all points on the fixed circle. Therefore,
all the terms in the above expression are positive-definite which ensures the
positivity of the denominator. Now, let us look at the numerator:

$\begin{split}&(h_{2}N_{c2}+f_{2}n_{2})m_{\text{th},1}^{2}-r\zeta
n_{2}m_{\text{th},2}^{2}\\\
&=h(N_{c2}-n_{2})+\frac{\pi^{2}}{48}\beta_{\text{th}}^{-2}n_{2}\Bigg{[}3(25-2\sqrt{6})\lambda^{2}-16(f_{m}+r\zeta)\Big{(}9-\sqrt{6}\Big{)}\lambda\Bigg{]}\\\
\end{split}$ (242)

Here we have substituted the thermal masses by their values given in (82).
Moreover we have used the values of the different couplings and the equation
of the fixed circle:

$\displaystyle h_{1}=h_{2}=h=\Big{(}\frac{3-\sqrt{6}}{16}\Big{)}\lambda,\
f_{p}\equiv\frac{f_{1}+f_{2}}{2}=\frac{\sqrt{6}}{8}\lambda,\
\zeta^{2}+f_{m}^{2}=\Big{(}\frac{18\sqrt{6}-39}{256}\Big{)}\lambda^{2},$ (243)

where $f_{m}\equiv\frac{f_{1}-f_{2}}{2}$. The equation of the fixed circle and
the fact that $r<1$ further impose the condition
$(f_{m}+r\zeta)<|f_{m}|+|\zeta|<\frac{\sqrt{18\sqrt{6}-39}}{8}\lambda$. This
inequality, together with the positivity of $h$ and the fact that $n_{2}\leq
N_{c2}$, ensures that the numerator is positive as shown below:

$\begin{split}(h_{2}N_{c2}+f_{2}n_{2})m_{\text{th},1}^{2}-r\zeta
n_{2}m_{\text{th},2}^{2}&>\frac{\pi^{2}}{48}\lambda^{2}\beta_{\text{th}}^{-2}n_{2}\Bigg{[}3(25-2\sqrt{6})-2\sqrt{18\sqrt{6}-39}\Big{(}9-\sqrt{6}\Big{)}\Bigg{]}\\\
&\approx 6.32\lambda^{2}\beta_{\text{th}}^{-2}n_{2}>0.\\\ \end{split}$ (244)

Therefore, we can conclude that

$\displaystyle(\phi_{1a_{1}})^{2}<0\ \ \forall\ \
a_{1}\in\\{1,\cdots,n_{1}\\},$ (245)

which implies that $\phi_{1a_{1}}$ is imaginary for these values of $a_{1}$.
This rules out the possibility of any saddle point with $n_{1}\neq 0$ being a
minimum of the potential.

###  E.2 The minima correspond to the saddle points with
$n_{1}=0,n_{2}=N_{c2}$

Let us now consider the saddle points with $n_{1}=0$. For these saddle points,
$\phi_{1a_{1}}=0$ for all $a_{1}\in\\{1,\cdots,N_{c1}\\}$, whereas the nonzero
values of $\phi_{2a_{2}}$ are as follows:

$\displaystyle(\phi_{2a_{2}})^{2}=-\frac{N_{c2}^{2}}{64\pi^{2}}\frac{m_{\text{th},2}^{2}}{h_{2}N_{c2}+f_{2}n_{2}}\
\ \forall\ \ a_{2}\in\\{1,\cdots,n_{2}\\}.$ (246)

For these saddle points to be candidates for the minima of the potential, the
above values of $(\phi_{2a_{2}})^{2}$ have to be positive. This can be
verified by noting that for the fixed points under consideration, we have
$m_{\text{th},2}^{2}<0$, $h_{2}>0$ and $f_{2}>0$. The values of potential at
these saddle points are

$\displaystyle V_{\text{eff}}\Big{|}_{n_{1}=0,n_{2}}=$
$\displaystyle-\frac{N_{c2}^{2}}{256\pi^{2}}\Bigg{[}\frac{(m_{\text{th},2}^{2})^{2}}{h_{2}(\frac{N_{c2}}{n_{2}})+f_{2}}\Bigg{]}.$
(247)

Due to the positivity of $h_{2}$ at the fixed points, we can see from the
above expression that the minima of the potential correspond to
$n_{2}=N_{c2}$. Substituting $n_{2}$ by $N_{c2}$ in (246), we get

$\displaystyle(\phi_{2a_{2}})^{2}=-\frac{N_{c2}}{64\pi^{2}}\frac{m_{\text{th},2}^{2}}{h_{2}+f_{2}}\
\ \forall\ \ a_{2}\in\\{1,\cdots,N_{c2}\\}.$ (248)

Let us note here that there are $2^{N_{c2}}$ different field configurations
that satisfy the above equation. These configurations are as follows:

$\displaystyle\Phi_{1}=0,\
\Phi_{2}=\Big{(}-\frac{N_{c2}}{64\pi^{2}}\frac{m_{\text{th},2}^{2}}{h_{2}+f_{2}}\Big{)}^{\frac{1}{2}}\text{diag}\\{\sigma_{1},\cdots,\sigma_{N_{c2}}\\},$
(249)

where $\sigma_{1},\cdots,\sigma_{N_{c2}}$ can be $1$ or $-1$. However, many of
these configurations are related to each other by $SO(N_{c2})\times
SO(N_{c2})$ gauge transformations. It can be shown that all these
configurations can be categorized into two equivalence classes. The
configurations in each class are related to each other by gauge
transformations. One can go from one class to the other by flipping the sign
of just one of the diagonal entries. Thus, we can choose one representative
from each of these classes as follows:

$\displaystyle\Phi_{1}=0,\
\Phi_{2}=\Big{(}-\frac{N_{c2}}{64\pi^{2}}\frac{m_{\text{th},2}^{2}}{h_{2}+f_{2}}\Big{)}^{\frac{1}{2}}\text{diag}\\{\pm
1,1,\cdots,1\\},$ (250)

and treat these as two distinct minima unrelated by gauge transformations.
Since we are working in a perturbative regime where the ’t Hooft couplings are
small, the thermal expectation values of the field $\Phi_{1}$ and $\Phi_{2}$
would correspond to these minima at leading order in perturbation theory.

##  Appendix F Large N scaling of planar diagrams

In this appendix we will discuss the scaling of different planar diagrams in
the double bifundamental models. For simplicity, we will restrict our
attention to the complex double bifundamental model. This would enable us to
use ’t Hooft’s large $N$ expansion in terms of oriented surfaces. The
$\mathfrak{s}\mathfrak{o}(N_{ci})$ valued gauge connections in the real double
bifundamental model generate both orientable and non-orientable diagrams.
However, the large N scaling of the planar diagrams are identical in both the
double bifundamental models.

In the following analysis, we will take the two ranks $N_{c1}$ and $N_{c2}$ to
be of comparable magnitudes, say $O(N)$, and then work in the limit
$N\rightarrow\infty$. In this limit, we will look at how different planar
diagrams scale with $N$. In the process, we will demonstrate an important
feature of these diagrams, viz., any double trace vertex in such a diagram
links two otherwise disconnected subdiagrams. As discussed in section 5, this
feature plays a crucial role in proving the survival of the fixed circles in
these models under all loop corrections at the planar limit. Throughout this
appendix, we will be using ’t Hooft’s double line notation to represent the
diagrams. As we are interested only in how different diagrams scale with $N$
where $N$ is the order of magnitude of the ranks in both the sectors, we will
not distinguish between the fields in the two sectors. Thus, unlike section 5,
we will not use different colors to show propagators and vertices belonging to
different sectors.

###  F.1 Scaling of bubble diagrams

Let us begin our analysis by considering bubble diagrams, i.e., diagrams which
have no external legs. For convenience, we will temporarily rescale all the
fields such that their propagators scale as $O(1/N)$. This would not affect
the scaling of the bubble diagrams due to the absence of any external leg in
such diagrams. Under the above rescaling of the fields, the single trace and
double trace couplings scale as $O(N)$ and $O(1)$ respectively.

Now, let us first consider a bubble diagram with only single trace vertices.
Each vertex in such a diagram contributes a factor of $N$, while each
propagator (an edge in the double line notation) contributes a factor of
$\frac{1}{N}$. Moreover, each color loop (a face in the double line notation)
in the diagram contributes a factor of $N$. Therefore, the overall scaling of
the diagram is given by

$\displaystyle N^{V-E+F}=N^{2-2\widetilde{g}}$ (251)

where $V,E$ and $F$ are the number of vertices, edges and faces in the diagram
respectively. $\widetilde{g}$ is the genus number corresponding to the diagram
737373Departing from the usual convention, we put a tilde over $g$ to avoid
any confusion with the gauge couplings.. For a planar diagram, we have
$\widetilde{g}=0$. Hence such a diagram scales as $N^{2}$.

Now let us extend our discussion to connected bubble diagrams with double
trace vertices. To be specific, let us consider a diagram with $m$ double
trace vertices. Each such double trace vertex has the structure shown in
figure 17.

Figure 17: Structure of a double trace vertex

To understand how this diagram scales with $N$, let us employ the following
trick. Let us remove all the double trace vertices from the diagram and join
the pairs of legs with identical colors attached to each such vertex. This
would leave us with a set of disconnected bubble diagrams which consist only
of single trace vertices and propagators. Let the number of such disconnected
pieces be $(m^{\prime}+1)$. Here, $m^{\prime}\leq m$ as the removal of each
double trace vertex can lead to at most two disconnected pieces. Now, as we
saw earlier, each of the disconnected pieces, comprising only of single trace
vertices and propagators, scales as $N^{2-2\widetilde{g}_{i}}$ where
$\widetilde{g}_{i}$ is the genus number corresponding to that piece.
Therefore, the overall scaling of the disconnected diagram obtained by
removing the double trace vertices is

$\displaystyle
N^{\sum_{i=1}^{m^{\prime}+1}(2-2\widetilde{g}_{i})}=N^{2(m^{\prime}+1)-2\sum_{i=1}^{m^{\prime}+1}\widetilde{g}_{i}}.$
(252)

From this we can now estimate the scaling of the original diagram with the
double trace vertices. For this, we just need to recall that in our present
convention, each double trace coupling scales as $O(1)$. Therefore, the
removal of such vertices from the diagram, by itself, does not lead to any
change in the scaling of the diagram. However, while removing a double trace
vertex, we are also joining two pairs of propagators in the diagram. This
leads to a reduction in the overall number of propagators by $2m$. Since each
such propagator scales as $1/N$ in our convention, the above reduction in
their number leads to an enhancement by a factor of $N^{2m}$. Therefore,
comparing with the overall scaling of the disconnected diagram as given in
(252), we find that the original diagram scales as

$\displaystyle
N^{2(m^{\prime}+1)-2\sum_{i=1}^{m^{\prime}+1}\widetilde{g}_{i}-2m}=N^{2-2(m-m^{\prime})-2\sum_{i=1}^{m^{\prime}+1}\widetilde{g}_{i}}.$
(253)

From the above expression, we can see that the leading contributions come from
diagrams where $m^{\prime}=m$ and $\widetilde{g}_{i}=0$ for all the
disconnected pieces. These are precisely the planar diagrams with double trace
vertices. Each double trace vertex in such a diagram connects two disconnected
planar subdiagrams. As we can see from (253), these planar diagrams scale as
$N^{2}$, which is identical to the scaling of the planar diagrams with only
single trace vertices. Thus, our choice of scaling of the different couplings
leads to a consistent large $N$ scaling of all planar bubble diagrams.

Next, we will determine the scaling of different diagrams that contribute to
the wave function and vertex renormalizations considered in section 5.
Henceforth, we will revert back to our previous convention of unrescaled
fields where the corresponding propagators are $O(1)$.

###  F.2 Scaling of diagrams with 2 external legs

Let us first consider planar connected diagrams that contribute to the wave
function renormalizations of the scalar fields. Such a diagram has 2 external
legs as show in figure 18.

Figure 18: Planar connected diagram with 2 external legs: Here, as well as in
all subsequent diagrams, a shaded blob represents a planar connected
component.

We can estimate the scaling of this diagram by first joining the two external
legs and then summing over the color indices in this propagator. This leads to
an enhancement by a factor of $N^{2}$ due to the introduction of two new color
loops. The resulting diagram is a planar bubble as shown in figure 19. As we
have already argued, such a planar bubble scales as $N^{2}$. Therefore, taking
into account the above-mentioned enhancement, we can conclude that the
original diagram in figure 18 scales as $N^{0}$.

Figure 19: Bubble diagram obtained by joining the two external legs in figure
18.

###  F.3 Scaling of diagrams with 4 external legs

Now, let us look at planar connected diagrams with 4 external legs of the
scalar fields. We can have two different types of such diagrams:

  1. 1.

Diagrams which contribute to the renormalization of the single trace vertices,

  2. 2.

Diagrams which contribute to the renormalization of the double trace vertices.

We will derive the scaling of both these types of diagrams below.

####  F.3.1 Diagrams corresponding to vertex renormalizations of single trace
couplings

Any planar diagram which contributes to the renormalization of a single trace
vertex has the structure shown in figure 20.

Figure 20: Planar connected diagram contributing to renormalization of a
single trace vertex.

Suppose such a diagram scales as $N^{x}$. We can estimate the value of $x$ by
constructing a bubble diagram via the following procedure: Take two identical
copies of the diagram in figure 20 and join the external legs with identical
colors in these two copies as shown in figure 21. While joining these legs,
sum over corresponding color indices.

Figure 21: Bubble diagram obtained by joining the corresponding external legs
of two copies of the diagram in figure 20.

This procedure introduces 4 new color loops which leads to an enhancement by a
factor of $N^{4}$. The planar bubble diagram at the end of the procedure
scales as $N^{2}$. Therefore, we have the following equation:

$\displaystyle 2x+4=2\implies x=-1.$ (254)

Therefore, all planar connected diagrams of the kind shown in figure 20 scale
as $N^{-1}$.

####  F.3.2 Diagrams corresponding to vertex renormalizations of double trace
couplings

Planar diagrams which contribute to the renormalization of a double trace
vertex are of the form shown in figure 22.747474Here let us remark that given
the characterization of a planar diagram as a cell decomposition of a sphere
(i.e., a compactified plane) in terms of a ribbon graph where the external
lines correspond to open intervals in the boundaries of the faces, the only
requirement for these external lines to represent a double trace vertex is
that for each pair of them with identical colors, the two lines must lie on
the same boundary of a common face. This implies that in addition to the
diagrams shown in figure 22, there can be diagrams where the two pairs of
external lines do not share a common face. In particular, one of the pairs of
external lines in such a diagram may not be continuously connected to the
infinity of the plane. An example of this would be a ribbon graph version of
the diagram shown in figure 5. Despite this subtle difference, the large N
scaling of such diagrams would be the same as the scaling of the diagrams
shown in figure 22 since the following arguments are equally applicable to
them.

Figure 22: Planar connected diagram contributing to renormalization of a
double trace vertex.

As in the case of the diagrams with 2 external legs, we can estimate the
scaling of the diagram in figure 22 by joining the pairs of external legs with
identical colors and then summing over the color indices. As a result, we
obtain the bubble diagram shown in figure 23.

Figure 23: Bubble diagram obtained by joining the external legs with identical
colors in figure 22.

The joining of the external legs introduces 4 new color loops leading to an
enhancement by a factor of $N^{4}$. As we have already argued, the bubble
diagram in figure 23 scales as $N^{2}$. Therefore, taking into account the
above-mentioned enhancement, we can conclude that the original diagram in
figure 22 scales as $N^{-2}$.

###  F.4 Comments

From the above analysis, we can draw the following two conclusions about
connected planar diagrams with 2 and 4 external legs:

  1. 1.

All such diagrams with a particular set of external legs scale identically
with $N$.

  2. 2.

Since we have shown that these diagrams can be augmented by definite
procedures to construct planar bubble diagrams, the result that double trace
vertices link otherwise disconnected planar subdiagrams can be extended to
these diagrams as well.

In section 5, we have repeatedly used these two features of the connected
planar diagrams to determine the structure of the planar beta functions of the
different couplings.

##  Appendix G General expressions for the planar beta functions of the double
trace couplings in the double bifundamental models

In this appendix, we will derive some general expressions for the planar beta
functions of the double trace couplings ($f_{i}$ and $\zeta$) in terms of the
corresponding wave function and vertex renormalizations. For this, we will
restrict our attention to the subspace where the planar beta functions of the
single trace couplings ($\lambda_{i}$ and $h_{i}$) vanish. As we have shown in
section 5, these beta functions ($\beta_{\lambda_{i}}$ and $\beta_{h_{i}}$)
are independent of the double trace couplings. Hence, their roots can be
determined independently. There is a discrete set of such roots out of which
only one corresponds to unitary fixed points where the two sectors are
coupled757575Here, let us remind the reader that we need to have
$x_{f1}=x_{f2}$ for the existence of this root.. At this root, the single
trace couplings in the two sectors are equal, i.e., we have

$\displaystyle\lambda_{1}=\lambda_{2}=\lambda_{0}\ ,\ h_{1}=h_{2}=h_{0}.$
(255)

We will now derive the expressions for the beta functions of the double trace
couplings on the subspace defined by the above equation.

To derive these expressions, we will work in dimensional regularization and
the $\overline{\text{MS}}$ scheme. In this scheme, the renormalized double
trace couplings are related to the corresponding bare couplings as follows:

$\displaystyle
f_{i}^{B}=\widetilde{\mu}^{\varepsilon}Z_{f_{i}}Z_{\Phi_{i}}^{-2}f_{i}\ ,\
\zeta^{B}=\widetilde{\mu}^{\varepsilon}Z_{\zeta}Z_{\Phi_{1}}^{-1}Z_{\Phi_{2}}^{-1}\zeta.$
(256)

Here $\widetilde{\mu}$ is an energy scale which is related to the
renormalization scale $\mu$ in the $\overline{\text{MS}}$ scheme by
$\widetilde{\mu}^{2}=\mu^{2}\frac{e^{\gamma_{E}}}{4\pi}$ where $\gamma_{E}$ is
the Euler-Mascheroni constant. $Z_{\Phi_{i}}$ is the wave function
renormalization of the field $\Phi_{i}$, and $Z_{f_{i}}$ and $Z_{\zeta}$ are
the vertex renormalizations of the respective couplings. The superscript $B$
indicates the bare couplings. Since these bare couplings are independent of
$\mu$, they drop out once we differentiate the above equations with respect to
$\mu$, and we get

$\begin{split}&\beta_{f_{i}}=-\varepsilon
f_{i}-f_{i}\Big{(}\sum_{j=1}^{2}\beta_{\lambda_{j}}\frac{\partial}{\partial\lambda_{j}}+\sum_{j=1}^{2}\beta_{h_{j}}\frac{\partial}{\partial
h_{j}}+\sum_{A^{\prime}}\beta_{\kappa_{A^{\prime}}}\frac{\partial}{\partial\kappa_{A^{\prime}}}\Big{)}\ln\Big{[}Z_{f_{i}}Z_{\Phi_{i}}^{-2}\Big{]},\\\
&\beta_{\zeta}=-\varepsilon\zeta-\zeta\Big{(}\sum_{j=1}^{2}\beta_{\lambda_{j}}\frac{\partial}{\partial\lambda_{j}}+\sum_{j=1}^{2}\beta_{h_{j}}\frac{\partial}{\partial
h_{j}}+\sum_{A^{\prime}}\beta_{\kappa_{A^{\prime}}}\frac{\partial}{\partial\kappa_{A^{\prime}}}\Big{)}\ln\Big{[}Z_{\zeta}Z_{\Phi_{1}}^{-1}Z_{\Phi_{2}}^{-1}\Big{]},\\\
\end{split}$ (257)

where $\kappa_{A^{\prime}}$ runs over the double trace couplings in the model.
Recall that we are working in the subspace where the planar beta functions
$\beta_{\lambda_{i}}$ and $\beta_{h_{i}}$ vanish in the
$\varepsilon\rightarrow 0$ limit. Therefore, for small nonzero values of
$\varepsilon$, these beta functions are
$\beta_{\lambda_{i}}=-\varepsilon\lambda_{0}$, $\beta_{h_{i}}=-\varepsilon
h_{0}$ on the subspace of our interest. In the neighborhood of this subspace,
it is convenient to switch to the following basis for the single trace
couplings:

$\displaystyle\lambda_{p,m}\equiv\frac{\lambda_{1}\pm\lambda_{2}}{2},\
h_{p,m}\equiv\frac{h_{1}\pm h_{2}}{2}.$ (258)

In terms of this basis, the operator
$\Big{(}\sum\limits_{j=1}^{2}\beta_{\lambda_{j}}\frac{\partial}{\partial\lambda_{j}}+\sum\limits_{j=1}^{2}\beta_{h_{j}}\frac{\partial}{\partial
h_{j}}\Big{)}$ on the above-mentioned subspace is as follows

$\begin{split}\sum_{j=1}^{2}\beta_{\lambda_{j}}\frac{\partial}{\partial\lambda_{j}}+\sum_{j=1}^{2}\beta_{h_{j}}\frac{\partial}{\partial
h_{j}}&=-\varepsilon\lambda_{0}\frac{\partial}{\partial\lambda_{p}}-\varepsilon
h_{0}\frac{\partial}{\partial h_{p}}.\end{split}$ (259)

Thus, the equations in (257) can be rewritten as

$\begin{split}&\beta_{f_{i}}=-\varepsilon
f_{i}-f_{i}\Big{(}-\varepsilon\lambda_{0}\frac{\partial}{\partial\lambda_{p}}-\varepsilon
h_{0}\frac{\partial}{\partial
h_{p}}+\sum_{A^{\prime}}\beta_{\kappa_{A^{\prime}}}\frac{\partial}{\partial\kappa_{A^{\prime}}}\Big{)}\ln\Big{[}Z_{f_{i}}Z_{\Phi_{i}}^{-2}\Big{]},\\\
&\beta_{\zeta}=-\varepsilon\zeta-\zeta\Big{(}-\varepsilon\lambda_{0}\frac{\partial}{\partial\lambda_{p}}-\varepsilon
h_{0}\frac{\partial}{\partial
h_{p}}+\sum_{A^{\prime}}\beta_{\kappa_{A^{\prime}}}\frac{\partial}{\partial\kappa_{A^{\prime}}}\Big{)}\ln\Big{[}Z_{\zeta}Z_{\Phi_{1}}^{-1}Z_{\Phi_{2}}^{-1}\Big{]}.\\\
\end{split}$ (260)

Now, we can expand the wave function and vertex renormalizations in powers of
$\frac{1}{\varepsilon}$ as follows:

$\displaystyle
Z_{f_{i}}=1+\sum_{n=1}^{\infty}\frac{Z_{f_{i}}^{(n)}}{\varepsilon^{n}},\
Z_{\zeta}=1+\sum_{n=1}^{\infty}\frac{Z_{\zeta}^{(n)}}{\varepsilon^{n}},\
Z_{\Phi_{i}}=1+\sum_{n=1}^{\infty}\frac{Z_{\Phi_{i}}^{(n)}}{\varepsilon^{n}}.$
(261)

To satisfy the equations in (260) at each order in the
$\frac{1}{\varepsilon}$-expansion, we need the beta functions to have the
following forms:

$\displaystyle\beta_{f_{i}}=\varepsilon\beta_{f_{i}}^{(1)}+\beta_{f_{i}}^{(0)},\
\beta_{\zeta}=\varepsilon\beta_{\zeta}^{(1)}+\beta_{\zeta}^{(0)}.$ (262)

Solving the $O(\varepsilon)$ and $O(1)$ terms in (260) we get

$\begin{split}\beta_{f_{i}}^{(1)}=-f_{i},\
\beta_{\zeta}^{(1)}=-\zeta,\end{split}$ (263)
$\begin{split}&\beta_{f_{i}}^{(0)}=f_{i}\Big{(}\lambda_{0}\frac{\partial}{\partial\lambda_{p}}+h_{0}\frac{\partial}{\partial
h_{p}}+\sum_{A^{\prime}}\kappa_{A^{\prime}}\frac{\partial}{\partial\kappa_{A^{\prime}}}\Big{)}\Bigg{[}Z_{f_{i}}^{(1)}-2Z_{\Phi_{i}}^{(1)}\Bigg{]},\\\
&\beta_{\zeta}^{(0)}=\zeta\Big{(}\lambda_{0}\frac{\partial}{\partial\lambda_{p}}+h_{0}\frac{\partial}{\partial
h_{p}}+\sum_{A^{\prime}}\kappa_{A^{\prime}}\frac{\partial}{\partial\kappa_{A^{\prime}}}\Big{)}\Bigg{[}Z_{\zeta}^{(1)}-Z_{\Phi_{1}}^{(1)}-Z_{\Phi_{2}}^{(1)}\Bigg{]}.\\\
\end{split}$ (264)

Finally, taking the $\varepsilon\rightarrow 0$ limit, we get the following
expressions of the planar beta functions of the double trace couplings on the
subspace $\lambda_{1}=\lambda_{2}=\lambda_{0},\ h_{1}=h_{2}=h_{0}$:

$\begin{split}&\beta_{f_{i}}=f_{i}\sum_{A}\kappa_{A}\frac{\partial}{\partial\kappa_{A}}\Bigg{[}Z_{f_{i}}^{(1)}-2Z_{\Phi_{i}}^{(1)}\Bigg{]},\\\
&\beta_{\zeta}=\zeta\sum_{A}\kappa_{A}\frac{\partial}{\partial\kappa_{A}}\Bigg{[}Z_{\zeta}^{(1)}-Z_{\Phi_{1}}^{(1)}-Z_{\Phi_{2}}^{(1)}\Bigg{]},\\\
\end{split}$ (265)

where $\kappa_{A}$ runs over the values $\lambda_{p},h_{p},f_{1},f_{2}$ and
$\zeta$. Note that the expressions in (265) do not involve differentiation
with respect to $\lambda_{m}$ and $h_{m}$. Therefore, while evaluating these
beta functions, we can restrict our attention to the wave function and vertex
renormalizations in a region where $\lambda_{m}=h_{m}=0$, or equivalently,
$\lambda_{1}=\lambda_{2}=\lambda$ and $h_{1}=h_{2}=h$ with $(\lambda,h)$ lying
in the neighborhood of $(\lambda_{0},h_{0})$. As discussed in section 5.2,
this plays an important role in proving the survival of the fixed circle under
all loop corrections in the planar limit.

##  Appendix H Finite $N$ corrections

In this appendix we will explore the fate of the fixed circle in the real
double bifundamental model when the corrections due to finiteness of $N_{ci}$
are taken into account. We will assume that $N_{c1}$ and $N_{c2}$ are of
comparable magnitudes (say, $O(N)$, where $N$ is a large number). As we have
already seen in appendix D, the values of $x_{f1}$ and $x_{f2}$ must be the
same (say, $x_{0f}$) at the leading order in $\frac{1}{N}$ for the existence
of unitary fixed points where the two sectors are coupled to each other. While
considering finite N corrections to such a fixed point, we will allow for a
difference between $x_{f1}$ and $x_{f2}$ at subleading orders in the
$\frac{1}{N}$ expansion. So, we will take

$\begin{split}x_{fi}=\sum_{n=0}^{\infty}x_{fi}^{(n)}.\end{split}$ (266)

where $x_{fi}^{(n)}$ is the $O(\frac{1}{N^{n}})$ term in the expansion of
$x_{fi}$. Here, $x_{f1}^{(0)}=x_{f2}^{(0)}=x_{0f}=\frac{21}{4}-\epsilon$ with
$0<\epsilon\ll 1$. The common value ($\lambda_{0}$) of the gauge couplings in
the two sectors for the fixed points at the planar limit is determined by this
small parameter $\epsilon$. This relation can be used to express $x_{0f}$ as a
perturbative expansion in $\lambda_{0}$.

Therefore, we see that there are two small parameters in the problem:
$\frac{1}{N}$ and $\lambda_{0}$. We will work in a regime where
$\frac{1}{N}\ll\lambda_{0}$. We will denote the $n^{\text{th}}$ order term in
the $1/N$-expansion of a coupling $g$ which lies at a fixed point of the beta
functions by $g^{(n)}$. For each value of $n$, all the couplings have
perturbative expansions in $\lambda_{0}$. We will determine the first few
terms in such perturbative expansions for the $O(1/N)$ and $O(1/N^{2})$
corrections to the couplings.

For later convenience, we find it useful to define the following quantities:

$\begin{split}&\lambda_{p}^{(n)}=\frac{\lambda_{1}^{(n)}+\lambda_{2}^{(n)}}{2},\
\lambda_{m}^{(n)}=\frac{\lambda_{1}^{(n)}-\lambda_{2}^{(n)}}{2},\\\
h_{p}^{(n)}=\frac{h_{1}^{(n)}+h_{2}^{(n)}}{2},&\
h_{m}^{(n)}=\frac{h_{1}^{(n)}-h_{2}^{(n)}}{2},\
f_{p}^{(n)}=\frac{f_{1}^{(n)}+f_{2}^{(n)}}{2},\
f_{m}^{(n)}=\frac{f_{1}^{(n)}-f_{2}^{(n)}}{2}.\\\ \end{split}$ (267)

The $\lambda_{0}$-expansion of these quantities for $n=0$ are as follows:

$\begin{split}&\lambda_{p}^{(0)}\equiv\lambda_{0},\ \lambda_{m}^{(0)}=0,\\\ &\
h_{m}^{(0)}=0,\ h_{p}^{(0)}\equiv
h_{0}=\Big{(}\frac{3-\sqrt{6}}{16}\Big{)}\lambda_{0}+\Bigg{[}\frac{93\sqrt{6}-201+8(7-5\sqrt{6})x_{0f}}{768\sqrt{6}}\Bigg{]}\lambda_{0}^{2}+O(\lambda_{0}^{3})\\\
&\qquad\qquad\qquad\qquad\quad=\Big{(}\frac{3-\sqrt{6}}{16}\Big{)}\lambda_{0}-\Bigg{(}\frac{234-31\sqrt{6}}{1536}\Bigg{)}\lambda_{0}^{2}+O(\lambda_{0}^{3}),\\\
&f_{m}^{(0)}\equiv f_{0m},\ \zeta^{(0)}\equiv\zeta_{0},\ f_{p}^{(0)}\equiv
f_{0p}=\frac{\sqrt{6}}{8}\lambda_{0}-\Bigg{[}\frac{2160+339\sqrt{6}+56\sqrt{6}x_{0f}}{2304}\Bigg{]}\lambda_{0}^{2}+O(\lambda_{0}^{3})\\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad=\frac{\sqrt{6}}{8}\lambda_{0}-\Bigg{(}\frac{720+211\sqrt{6}}{768}\Bigg{)}\lambda_{0}^{2}+O(\lambda_{0}^{3}).\end{split}$
(268)

Here we have substituted $x_{0f}$ in the expression of $h_{0}$ and $f_{0p}$ by
its leading order value in the $\lambda_{0}$ expansion:

$\displaystyle x_{0f}=\frac{21}{4}+O(\lambda_{0}).$ (269)

Note that the quantities $f_{0m}$, $\zeta_{0}$ are constrained to lie on a
circle which is given by

$\begin{split}(\zeta_{0}^{2}+f_{0m}^{2})&=\Big{(}\frac{18\sqrt{6}-39}{256}\Big{)}\lambda_{0}^{2}-\Bigg{[}\frac{-5784+5607\sqrt{6}+8(51\sqrt{6}-172)x_{0f}}{12288}\Bigg{]}\lambda_{0}^{3}+O(\lambda_{0}^{4})\\\
&=\Big{(}\frac{18\sqrt{6}-39}{256}\Big{)}\lambda_{0}^{2}-\Bigg{[}\frac{2583\sqrt{6}-4336}{4096}\Bigg{]}\lambda_{0}^{3}+O(\lambda_{0}^{4}).\end{split}$
(270)

We will now check whether the circle of fixed points satisfying the above
constraints survives when the finite N corrections are taken into account. In
particular, we will focus on possible fixed points where the two sectors are
coupled, i.e., $\zeta\neq 0$.

###  H.1 Finite $N$ corrections to the gauge couplings

First, let us look at the finite N corrections to the values of the gauge
couplings. From the form of the planar beta functions of the gauge couplings,
we see that $\lambda_{0}$ satisfies an equation of the following form at any
fixed point:

$\begin{split}-\lambda_{0}^{2}\Bigg{(}\frac{21-4x_{0f}}{6}\Bigg{)}-\lambda_{0}^{3}\Bigg{(}\frac{27-13x_{0f}}{6}\Bigg{)}+A_{0}\lambda_{0}^{4}+O(\lambda_{0}^{5})=0.\end{split}$
(271)

Here, we have included a contribution of the 3-loop diagrams to the planar
beta functions of the gauge couplings since it is necessary for determining
the $O(1/N)$ and $O(1/N^{2})$ terms in the values of the different couplings
at the fixed points.767676There may be terms involving the coupling $h_{0}$ in
such 3-loop contributions to the planar beta functions. As we have already
seen, the coupling $h_{0}=(\frac{3-\sqrt{6}}{16})\lambda_{0}+O(\lambda_{0})$
at any fixed point. One can substitute this relation between $h_{0}$ and
$\lambda_{0}$ to express the quartic terms in the planar beta function of
$\lambda_{0}$ as $A_{0}\lambda_{0}^{4}$. Although we have not evaluated the
coefficient $A_{0}$, we will see that most of the important conclusions in
this appendix do not rely on its exact value.

The value of $\lambda_{0}$ in the above expression is determined by $x_{0f}$.
One can invert this relation to express $x_{0f}$ as follows:

$\begin{split}x_{0f}=\frac{21}{4}-\frac{165}{16}\lambda_{0}+\Big{(}\frac{2145-96A_{0}}{64}\Big{)}\lambda_{0}^{2}+O(\lambda_{0}^{3}).\end{split}$
(272)

Now, we will determine the values of the $O(1/N)$ and $O(1/N^{2})$ corrections
to the gauge couplings. As we will see, these corrections take the form of
perturbative expansions in $\lambda_{0}$ where the leading order terms are
independent of $\lambda_{0}$. For the $O(1/N^{2})$ contributions to the
couplings, we will discuss only these leading order terms. However, for the
$O(1/N)$ contributions, we will need to take into account the first subleading
terms in the $\lambda_{0}$-expansion as they play a role in determining the
leading order terms in the $O(1/N^{2})$ corrections. Moreover, we will allow
the quantities $x_{fi}^{(n)}$ to have a perturbative expansion in
$\lambda_{0}$ as shown below:

$\displaystyle
x_{fi}^{(n)}=(x_{fi}^{(n)})_{0}+\lambda_{0}(x_{fi}^{(n)})_{1}+O(\lambda_{0}^{2}).$
(273)

We will check if the coefficients in this expansion can be fine-tuned such
that the degeneracy in the fixed points survives at a given order.

Having provided a general outline of our strategy, let us now turn to the
analysis of the $O(1/N)$ and $O(1/N^{2})$ corrections to the gauge couplings
at the fixed points.

####  H.1.1 $O(1/N)$ terms in the gauge couplings

To study the $O(1/N)$ corrections to the gauge couplings, we will assume that
the leading order term in $\lambda_{i}^{(1)}$ is independent of $\lambda_{0}$.
We will soon provide a consistency check for this assumption. Keeping this
assumption in mind, one can get the following perturbative expansion for the
beta function of $\lambda_{i}^{(1)}$:

$\begin{split}\beta_{\lambda_{i}^{(1)}}=&-\lambda_{0}\lambda_{i}^{(1)}\Bigg{(}\frac{21-4x_{0f}}{3}\Bigg{)}+\lambda_{0}^{2}\Bigg{(}\frac{2x_{fi}^{(1)}}{3}+\frac{22}{3N_{ci}}\Bigg{)}-\lambda_{0}^{2}\lambda_{i}^{(1)}\Bigg{(}\frac{27-13x_{0f}}{2}\Bigg{)}\\\
&+\lambda_{0}^{3}\Bigg{(}\frac{13x_{fi}^{(1)}}{6}+\frac{29}{24N_{ci}}\Bigg{)}+4A_{0}\lambda_{0}^{3}\lambda_{i}^{(1)}+O(\lambda_{0}^{4}/N).\end{split}$
(274)

While writing the second last term in the above expansion, we have substituted
$x_{0f}$ by $\frac{21}{4}$ as all higher order terms in $x_{0f}$ lead to
contributions at $O(\lambda_{0}^{4}/N)$ or a higher order in
$\beta_{\lambda_{i}^{(1)}}$.

Now, setting $\beta_{\lambda_{i}^{(1)}}=0$, we get the following value of
$\lambda_{i}^{(1)}$ at the fixed point:

$\begin{split}\lambda_{i}^{(1)}=&-\frac{16\Big{(}11+N_{ci}(x_{fi}^{(1)})_{0}\Big{)}}{165N_{ci}}\\\
&+\Bigg{[}\frac{-55\Bigg{(}601+8N_{ci}\Big{(}13(x_{fi}^{(1)})_{0}+2(x_{fi}^{(1)})_{1}\Big{)}\Bigg{)}+256A_{0}\Big{(}11+N_{ci}(x_{fi}^{(1)})_{0}\Big{)}}{9075N_{ci}}\Bigg{]}\lambda_{0}\\\
&+O(\lambda_{0}^{2}/N).\end{split}$ (275)

Note that the leading order term in $\lambda_{i}^{(1)}$ is independent of
$\lambda_{0}$ as mentioned earlier. We will see that this feature is shared by
the leading order terms in the $O(1/N)$ and $O(1/N^{2})$ corrections to all
the couplings.

####  H.1.2 $O(1/N^{2})$ terms in the gauge couplings

The beta functions of the $O(1/N^{2})$ correction to the gauge couplings can
be similarly expanded as follows:

$\begin{split}\beta_{\lambda_{i}^{(2)}}=&-\Big{(}2\lambda_{0}\lambda_{i}^{(2)}+\lambda_{i}^{(1)2}\Big{)}\Bigg{(}\frac{21-4x_{0f}}{6}\Bigg{)}+\frac{4}{3}\lambda_{0}\lambda_{i}^{(1)}\Bigg{(}x_{fi}^{(1)}+\frac{11}{N_{ci}}\Bigg{)}+\frac{2}{3}\lambda_{0}^{2}x_{fi}^{(2)}\\\
&-\Big{(}\lambda_{0}^{2}\lambda_{i}^{(2)}+\lambda_{0}\lambda_{i}^{(1)2}\Big{)}\Bigg{(}\frac{27-13x_{0f}}{2}\Bigg{)}+\lambda_{0}^{2}\lambda_{i}^{(1)}\Bigg{(}\frac{13x_{fi}^{(1)}}{2}+\frac{29}{8N_{ci}}\Bigg{)}\\\
&+6A_{0}\lambda_{0}^{2}\lambda_{i}^{(1)2}+O(\lambda_{0}^{3}/N^{2}).\end{split}$
(276)

At the fixed point, we have

$\begin{split}\lambda_{i}^{(2)}=&-\frac{16}{1497375N_{ci}^{2}}\Bigg{[}128A_{0}\Big{(}11+N_{ci}(x_{fi}^{(1)})_{0}\Big{)}^{2}\\\
&\qquad\qquad\qquad\qquad-55\Bigg{(}319+601N_{ci}(x_{fi}^{(1)})_{0}+N_{ci}^{2}\Big{(}52(x_{fi}^{(1)})_{0}^{2}-165x_{fi}^{(2)}\Big{)}\Bigg{)}\Bigg{]}\\\
&+O(\lambda_{0}/N^{2}).\end{split}$ (277)

From the expressions of $\lambda_{i}^{(1)}$ and $\lambda_{i}^{(2)}$ one can
extract the expressions of
$\lambda_{p,m}^{(1)}\equiv\frac{\lambda_{1}^{(1)}\pm\lambda_{2}^{(1)}}{2}$ and
$\lambda_{p,m}^{(2)}\equiv\frac{\lambda_{1}^{(2)}\pm\lambda_{2}^{(2)}}{2}$.
These will be useful in determining the finite $N$ corrections to the quartic
couplings as we will discuss below.

###  H.2 Finite $N$ corrections to the quartic couplings

The analysis of the finite $N$ corrections to the quartic couplings at the
fixed points is quite similar to the one for the gauge couplings presented
above. We will first discuss the $O(1/N)$ corrections and show that by tuning
$x_{fi}^{(1)}$ and $x_{f2}^{(1)}$ appropriately, the degeneracy in the fixed
points can be preserved up to the first subleading order in $\lambda_{0}$. We
will then look at the $O(1/N^{2})$ corrections. There we will again see that
the degeneracy in the fixed points can be preserved up to the leading order in
$\lambda_{0}$ by appropriately tuning the values of $x_{f1}^{(2)}$ and
$x_{f2}^{(2)}$.

####  H.2.1 $O(1/N)$ terms in the quartic couplings

The beta functions of the $O(1/N)$ corrections to the quartic couplings have
the following forms:

$\begin{split}\beta_{h_{p}^{(1)}}=&(32h_{0}-6\lambda_{0})h_{p}^{(1)}-\Bigg{(}(6h_{0}-\frac{3}{8}\lambda_{0})\lambda_{p}^{(1)}-(16h_{0}^{2}+3h_{0}\lambda_{0}-\frac{3\lambda_{0}^{2}}{8})\Big{(}\frac{1}{N_{c1}}+\frac{1}{N_{c2}}\Big{)}\Bigg{)}\\\
&+\Big{(}-288h_{0}^{2}+80\lambda_{0}h_{0}+\frac{9}{4}\lambda_{0}^{2}\Big{)}h_{p}^{(1)}+\Big{(}40h_{0}^{2}+\frac{9}{2}h_{0}\lambda_{0}-\frac{15}{8}\lambda_{0}^{2}\Big{)}\lambda_{p}^{(1)}+O(\lambda_{0}^{3}/N),\end{split}$
(278)

$\begin{split}\beta_{h_{m}^{(1)}}=&(32h_{0}-6\lambda_{0})h_{m}^{(1)}-\Bigg{(}(6h_{0}-\frac{3}{8}\lambda_{0})\lambda_{m}^{(1)}-(16h_{0}^{2}+3h_{0}\lambda_{0}-\frac{3\lambda_{0}^{2}}{8})\Big{(}\frac{1}{N_{c1}}-\frac{1}{N_{c2}}\Big{)}\Bigg{)}\\\
&+\Big{(}-288h_{0}^{2}+80\lambda_{0}h_{0}+\frac{9}{4}\lambda_{0}^{2}\Big{)}h_{m}^{(1)}+\Big{(}40h_{0}^{2}+\frac{9}{2}h_{0}\lambda_{0}-\frac{15}{8}\lambda_{0}^{2}\Big{)}\lambda_{m}^{(1)}+O(\lambda_{0}^{3}/N),\end{split}$
(279)

$\begin{split}\beta_{\zeta^{(1)}}=&\zeta_{0}\Bigg{[}16f_{p}^{(1)}+32h_{p}^{(1)}-6\lambda_{p}^{(1)}+(8h_{0}+3\lambda_{0})\Big{(}\frac{1}{N_{c1}}+\frac{1}{N_{c2}}\Big{)}\\\
&\quad+\Big{(}128\lambda_{0}-320h_{0}\Big{)}h_{p}^{(1)}+\Big{(}64f_{0p}+128h_{0}+\frac{21}{2}\lambda_{0}\Big{)}\lambda_{p}^{(1)}+64\lambda_{0}f_{p}^{(1)}+64f_{0m}\lambda_{m}^{(1)}\Bigg{]}\\\
&+O(\lambda_{0}^{3}/N),\end{split}$ (280)

$\begin{split}\beta_{f_{p}^{(1)}}=&16(1+4\lambda_{0})(\zeta_{0}\zeta^{(1)}+f_{0m}f_{m}^{(1)})+(48h_{0}+32f_{0p})h_{p}^{(1)}+\Big{(}\frac{9\lambda_{0}}{8}-6f_{0p}\Big{)}\lambda_{p}^{(1)}\\\
&+(16f_{0p}+32h_{0}-6\lambda_{0})f_{p}^{(1)}+f_{0m}(32h_{m}^{(1)}-6\lambda_{m}^{(1)})\\\
&+(8h_{0}+3\lambda_{0})\Big{(}\frac{f_{0p}+f_{0m}}{N_{c1}}+\frac{f_{0p}-f_{0m}}{N_{c2}}\Big{)}+32(f_{0m}^{2}+\zeta_{0}^{2})\lambda_{p}^{(1)}\\\
&+\Big{(}-1152h_{0}^{2}-320h_{0}f_{0p}+192\lambda_{0}h_{0}+9\lambda_{0}^{2}+128\lambda_{0}f_{0p}\Big{)}h_{p}^{(1)}\\\
&+\Big{(}96h_{0}^{2}+18h_{0}\lambda_{0}-\frac{45}{8}\lambda_{0}^{2}+128h_{0}f_{0p}+\frac{21}{2}\lambda_{0}f_{0p}+32f_{0p}^{2}\Big{)}\lambda_{p}^{(1)}\\\
&+\Big{(}-160h_{0}^{2}+128\lambda_{0}h_{0}+\frac{21}{4}\lambda_{0}^{2}+64\lambda_{0}f_{0p}\Big{)}f_{p}^{(1)}\\\
&+f_{0m}\Bigg{(}-320h_{0}h_{m}^{(1)}+128\lambda_{0}h_{m}^{(1)}+128h_{0}\lambda_{m}^{(1)}+64f_{0p}\lambda_{m}^{(1)}+\frac{21}{2}\lambda_{0}\lambda_{m}^{(1)}\Bigg{)}+O(\lambda_{0}^{3}/N),\end{split}$
(281)

$\begin{split}\beta_{f_{m}^{(1)}}=&(48h_{0}+32f_{0p})h_{m}^{(1)}+\Big{(}\frac{9\lambda_{0}}{8}-6f_{0p}\Big{)}\lambda_{m}^{(1)}+(16f_{0p}+32h_{0}-6\lambda_{0})f_{m}^{(1)}\\\
&+f_{0m}(16f_{p}^{(1)}+32h_{p}^{(1)}-6\lambda_{p}^{(1)})+(8h_{0}+3\lambda_{0})\Big{(}\frac{f_{0p}+f_{0m}}{N_{c1}}-\frac{f_{0p}-f_{0m}}{N_{c2}}\Big{)}\\\
&+\Big{(}-1152h_{0}^{2}-320h_{0}f_{0p}+192\lambda_{0}h_{0}+128\lambda_{0}f_{0p}+9\lambda_{0}^{2}\Big{)}h_{m}^{(1)}\\\
&+\Bigg{(}96h_{0}^{2}+128h_{0}f_{0p}+18h_{0}\lambda_{0}+\frac{21}{2}\lambda_{0}f_{0p}-\frac{45}{8}\lambda_{0}^{2}+32\Big{(}f_{0p}^{2}+f_{0m}^{2}-\zeta_{0}^{2}\Big{)}\Bigg{)}\lambda_{m}^{(1)}\\\
&+f_{0m}\Big{(}-320h_{0}h_{p}^{(1)}+128\lambda_{0}h_{p}^{(1)}+128h_{0}\lambda_{p}^{(1)}+64f_{0p}\lambda_{p}^{(1)}+64\lambda_{0}f_{p}^{(1)}+\frac{21}{2}\lambda_{0}\lambda_{p}^{(1)}\Big{)}\\\
&+\Big{(}-160h_{0}^{2}+128\lambda_{0}h_{0}+64\lambda_{0}f_{0p}+\frac{21}{4}\lambda_{0}^{2}\Big{)}f_{m}^{(1)}+O(\lambda_{0}^{3}/N).\end{split}$
(282)

In the above expression of $\beta_{\zeta}^{(1)}$, we have ignored terms of the
form $\zeta^{(1)}\Big{[}\cdots\Big{]}$ where the dots stand for a coefficient
which is not suppressed by 1/N. This coefficient is identical to the
coefficient of $\zeta_{0}$ in $\beta_{\zeta_{0}}$, and hence is set equal to
zero at the fixed point.

Now, setting $\beta_{h_{p}^{(1)}}$ and $\beta_{h_{m}^{(1)}}$ to zero, we get
the following expressions for the $O(1/N)$ corrections to $h_{p}$ and $h_{m}$:

$\begin{split}h_{p}^{(1)}=&\frac{(-3+\sqrt{6})}{330N_{c1}N_{c2}}\Bigg{[}11(N_{c1}+N_{c2})+N_{c1}N_{c2}\Big{(}(x_{f1}^{(1)})_{0}+(x_{f2}^{(1)})_{0}\Big{)}\Bigg{]}\\\
&+\frac{\lambda_{0}}{871200N_{c1}N_{c2}}\Bigg{[}\Big{(}55(-7290+2947\sqrt{6})-8448(-3+\sqrt{6})A_{0}\Big{)}(N_{c1}+N_{c2})\\\
&\qquad\qquad\quad+N_{c1}N_{c2}\Bigg{\\{}\Big{(}55(-702+281\sqrt{6})-768(-3+\sqrt{6})A_{0}\Big{)}\Big{(}(x_{f1}^{(1)})_{0}+(x_{f2}^{(1)})_{0}\Big{)}\\\
&\qquad\qquad\qquad\qquad\qquad+2640(-3+\sqrt{6})\Big{(}(x_{f1}^{(1)})_{1}+(x_{f2}^{(1)})_{1}\Big{)}\Bigg{\\}}\Bigg{]}+O(\lambda_{0}^{2}/N),\end{split}$
(283)

$\begin{split}h_{m}^{(1)}=&\frac{(-3+\sqrt{6})}{330N_{c1}N_{c2}}\Bigg{[}-11(N_{c1}-N_{c2})+N_{c1}N_{c2}\Big{(}(x_{f1}^{(1)})_{0}-(x_{f2}^{(1)})_{0}\Big{)}\Bigg{]}\\\
&+\frac{\lambda_{0}}{871200N_{c1}N_{c2}}\Bigg{[}-\Big{(}55(-7290+2947\sqrt{6})-8448(-3+\sqrt{6})A_{0}\Big{)}(N_{c1}-N_{c2})\\\
&\qquad\qquad\quad+N_{c1}N_{c2}\Bigg{\\{}\Big{(}55(-702+281\sqrt{6})-768(-3+\sqrt{6})A_{0}\Big{)}\Big{(}(x_{f1}^{(1)})_{0}-(x_{f2}^{(1)})_{0}\Big{)}\\\
&\qquad\qquad\qquad\qquad\qquad+2640(-3+\sqrt{6})\Big{(}(x_{f1}^{(1)})_{1}-(x_{f2}^{(1)})_{1}\Big{)}\Bigg{\\}}\Bigg{]}+O(\lambda_{0}^{2}/N),\end{split}$
(284)

Substituting these values of $h_{p}^{(1)}$ and $h_{m}^{(1)}$ in the expression
of $\beta_{\zeta^{(1)}}$ and then setting $\beta_{\zeta^{(1)}}=0$, we get the
following value of $f_{p}^{(1)}$:

$\begin{split}f_{p}^{(1)}=&-\frac{2}{55\sqrt{6}N_{c1}N_{c2}}\Bigg{[}11(N_{c1}+N_{c2})+N_{c1}N_{c2}\Big{(}(x_{f1}^{(1)})_{0}+(x_{f2}^{(1)})_{0}\Big{)}\Bigg{]}\\\
&+\frac{\lambda_{0}}{871200N_{c1}N_{c2}}\Bigg{[}11(101475-7195\sqrt{6}+1536\sqrt{6}A_{0})(N_{c1}+N_{c2})\\\
&\qquad\qquad+N_{c1}N_{c2}\Bigg{\\{}\Big{(}110(720-101\sqrt{6})+1536\sqrt{6}A_{0}\Big{)}\Big{(}(x_{f1}^{(1)})_{0}+(x_{f2}^{(1)})_{0}\Big{)}\\\
&\qquad\qquad\qquad\qquad-5280\sqrt{6}\Big{(}(x_{f1}^{(1)})_{1}+(x_{f2}^{(1)})_{1}\Big{)}\Bigg{\\}}\Bigg{]}\\\
&+\frac{32f_{0m}}{165N_{c1}N_{c2}}\Bigg{[}-11(N_{c1}-N_{c2})+N_{c1}N_{c2}\Big{(}(x_{f1}^{(1)})_{0}-(x_{f2}^{(1)})_{0}\Big{)}\Bigg{]}+O(\lambda_{0}^{2}/N).\end{split}$
(285)

Now substituting the values of $h_{p}^{(1)}$, $h_{m}^{(1)}$ and $f_{p}^{(1)}$
in the expression of $\beta_{f_{p}^{(1)}}$, and then setting
$\beta_{f_{p}^{(1)}}=0$, we get

$\begin{split}&2(1+4\lambda_{0})\Big{(}\zeta_{0}\zeta^{(1)}+f_{0m}f_{0m}^{(1)}\Big{)}\\\
&=\frac{(13-6\sqrt{6})\lambda_{0}}{880N_{c1}N_{c2}}\Bigg{[}11(N_{c1}+N_{c2})+N_{c1}N_{c2}\Big{(}(x_{f1}^{(1)})_{0}+(x_{f2}^{(1)})_{0}\Big{)}\Bigg{]}\\\
&\quad+\frac{4f_{0m}}{55\sqrt{6}N_{c1}N_{c2}}\Bigg{[}11(N_{c1}-N_{c2})-N_{c1}N_{c2}\Big{(}(x_{f1}^{(1)})_{0}-(x_{f2}^{(1)})_{0}\Big{)}\Bigg{]}\\\
&\quad+\frac{\lambda_{0}^{2}}{1548800N_{c1}N_{c2}}\Bigg{[}11\Big{(}5(8793\sqrt{6}-5516)+512(6\sqrt{6}-13)A_{0}\Big{)}(N_{c1}+N_{c2})\\\
&\qquad\qquad\quad+N_{c1}N_{c2}\Bigg{\\{}\Big{(}55(32+567\sqrt{6})+512(6\sqrt{6}-13)A_{0}\Big{)}\Big{(}(x_{f1}^{(1)})_{0}+(x_{f2}^{(1)})_{0}\Big{)}\\\
&\qquad\qquad\qquad\qquad\qquad+55(416-192\sqrt{6})\Bigg{\\}}\Big{(}(x_{f1}^{(1)})_{1}+(x_{f2}^{(1)})_{1}\Big{)}\Bigg{]}\\\
&\quad+\frac{f_{0m}\lambda_{0}}{435600N_{c1}N_{c2}}\Bigg{[}11(101475-28315\sqrt{6}+1536\sqrt{6}A_{0})(N_{c2}-N_{c1})\\\
&\qquad\qquad\qquad\qquad+N_{c1}N_{c2}\Bigg{\\{}(79200-32230\sqrt{6}+1536\sqrt{6}A_{0})\Big{(}(x_{f1}^{(1)})_{0}-(x_{f2}^{(1)})_{0}\Big{)}\\\
&\qquad\qquad\qquad\qquad\qquad\qquad\quad-5280\sqrt{6}\Big{(}(x_{f1}^{(1)})_{1}-(x_{f2}^{(1)})_{1}\Big{)}\Bigg{\\}}\Bigg{]}+O(\lambda_{0}^{3}/N).\end{split}$
(286)

The above expression gives the $O(1/N)$ correction to the location of the
fixed points along the radial direction of fixed circle in the
$N\rightarrow\infty$ limit.

Finally, setting $\beta_{f_{m}^{(1)}}=0$, we get the following two constraints
on the values of $(x_{f_{i}}^{(1)})_{0}$ and $(x_{f_{i}}^{(1)})_{1}$:

$\begin{split}&(x_{f_{1}}^{(1)})_{0}-(x_{f_{2}}^{(1)})_{0}=11\Big{(}\frac{1}{N_{c2}}-\frac{1}{N_{c1}}\Big{)},\\\
&(x_{f_{1}}^{(1)})_{1}-(x_{f_{2}}^{(1)})_{1}=-\frac{9}{376}(437-55\sqrt{6})\Big{(}\frac{1}{N_{c2}}-\frac{1}{N_{c1}}\Big{)}.\end{split}$
(287)

The important point to note here is that after substituting the expressions of
all the quantities evaluated earlier, the coefficients of $f_{0m}$ and
$f_{m}^{(1)}$ in $\beta_{f_{m}^{(1)}}$ vanish and one is left with a quantity
that vanishes under the above constraints. This means that when the values of
$(x_{f_{i}}^{(1)})_{0}$ and $(x_{f_{i}}^{(1)})_{1}$ are tuned to satisfy these
constraints, no additional condition is imposed on the coupling $f_{m}$ and
the degeneracy in the fixed points survives up to $O(\lambda_{0}/N)$
corrections to the couplings. Now, substituting these relations between
$x_{f_{1}}^{(1)}$ and $x_{f_{2}}^{(1)}$ in the expressions of $f_{p}^{(1)}$
and $2(1+4\lambda_{0})\Big{(}\zeta_{0}\zeta^{(1)}+f_{0m}f_{m}^{(1)}\Big{)}$,
we get

$\begin{split}f_{p}^{(1)}=&-\frac{4}{55\sqrt{6}N_{c2}}\Big{(}11+N_{c2}(x_{f2}^{(1)})_{0}\Big{)}\\\
&+\frac{\lambda_{0}}{40946400N_{c1}N_{c2}}\\\
&\qquad\Bigg{[}81675(165-7\sqrt{6})N_{c2}+11\Big{(}5(1662705-124871\sqrt{6})+144384\sqrt{6}A_{0}\Big{)}N_{c1}\\\
&\qquad+188N_{c1}N_{c2}\Bigg{\\{}(39600-5555\sqrt{6}+768\sqrt{6}A_{0})(x_{f2}^{(1)})_{0}-2640\sqrt{6}(x_{f2}^{(1)})_{1}\Bigg{\\}}\Bigg{]}\\\
&+O(\lambda_{0}^{2}/N),\end{split}$ (288)

$\begin{split}&2(1+4\lambda_{0})\Big{(}\zeta_{0}\zeta^{(1)}+f_{0m}f_{0m}^{(1)}\Big{)}\\\
&=\frac{(13-6\sqrt{6})\lambda_{0}}{440N_{c2}}\Big{(}11+N_{c2}(x_{f2}^{(1)})_{0}\Big{)}\\\
&\quad+\frac{\lambda_{0}^{2}}{774400N_{c2}}\Bigg{[}11\Big{(}5(8793\sqrt{6}-5516)+512(6\sqrt{6}-13)A_{0}\Big{)}\\\
&\qquad\qquad\qquad\quad+\Big{(}55(567\sqrt{6}+32)+512(6\sqrt{6}-13)A_{0}\Big{)}N_{c2}(x_{f2}^{(1)})_{0}\\\
&\qquad\qquad\qquad\quad-1760(6\sqrt{6}-13)N_{c2}(x_{f2}^{(1)})_{1}\Bigg{]}-\frac{3(165-7\sqrt{6})f_{0m}\lambda_{0}}{752N_{c1}N_{c2}}(N_{c1}-N_{c2})\\\
&\quad+O(\lambda_{0}^{3}/N).\end{split}$ (289)

The fact that $h_{p}^{(1)}$, $h_{m}^{(1)}$ and $f_{p}^{(1)}$ are independent
of $f_{0m}$ implies that even after taking the $O(1/N)$ corrections into
account, the closed curve of fixed points still lies on a plane. When
$N_{c1}\neq N_{c2}$, the dependence of the quantity
$2(1+4\lambda_{0})(f_{0m}f_{m}^{(1)}+\zeta_{0}\zeta^{(1)})$ on $f_{0m}$
indicates a deformation of the shape of the closed curve away from its
circular form in the planar limit. When the two ranks are equal, i.e.,
$N_{c1}=N_{c2}$, the quantity
$2(1+4\lambda_{0})(f_{0m}f_{m}^{(1)}+\zeta_{0}\zeta^{(1)})$ is independent of
$f_{0m}$ which means that the closed curve of fixed points remains a circle up
to this order.

To explore whether the above features survive at higher orders in
$\lambda_{0}$, one needs to evaluate the contributions of the higher loop
diagrams to the beta functions. For now, we will assume that the degeneracy in
the fixed points survives at $O(1/N)$ up to all orders in $\lambda_{0}$ and
check whether the leading order contributions at $O(1/N^{2})$ lift the
degeneracy.

####  H.2.2 $O(1/N^{2})$ terms in the quartic couplings

The beta functions of the $O(1/N^{2})$ corrections to the quartic couplings
have the following forms:

$\begin{split}\beta_{h_{p}^{(2)}}=&(32h_{0}-6\lambda_{0})h_{p}^{(2)}+\Big{(}\frac{3}{8}\lambda_{0}-6h_{0}\Big{)}\lambda_{p}^{(2)}+16(h_{p}^{(1)2}+h_{m}^{(1)2})+\frac{3}{16}(\lambda_{p}^{(1)2}+\lambda_{m}^{(1)2})\\\
&-6(\lambda_{p}^{(1)}h_{p}^{(1)}+\lambda_{m}^{(1)}h_{m}^{(1)})+(3\lambda_{0}+32h_{0})\Big{(}\frac{h_{p}^{(1)}+h_{m}^{(1)}}{N_{c1}}+\frac{h_{p}^{(1)}-h_{m}^{(1)}}{N_{c2}}\Big{)}\\\
&+(3h_{0}-\frac{3\lambda_{0}}{4})\Big{(}\frac{\lambda_{p}^{(1)}+\lambda_{m}^{(1)}}{N_{c1}}+\frac{\lambda_{p}^{(1)}-\lambda_{m}^{(1)}}{N_{c2}}\Big{)}+(40\lambda_{0}-288h_{0})\Big{(}h_{p}^{(1)2}+h_{m}^{(1)2}\Big{)}\\\
&+\Big{(}\frac{9}{4}h_{0}-\frac{15}{8}\lambda_{0}\Big{)}\Big{(}\lambda_{p}^{(1)2}+\lambda_{m}^{(1)2}\Big{)}+\Big{(}\frac{9}{2}\lambda_{0}+80h_{0}\Big{)}\Big{(}\lambda_{p}^{(1)}h_{p}^{(1)}+\lambda_{m}^{(1)}h_{m}^{(1)}\Big{)}+O(\lambda_{0}^{2}/N^{2}),\end{split}$
(290)

$\begin{split}\beta_{h_{m}^{(2)}}=&(32h_{0}-6\lambda_{0})h_{m}^{(2)}+\Big{(}\frac{3}{8}\lambda_{0}-6h_{0}\Big{)}\lambda_{m}^{(2)}+32h_{p}^{(1)}h_{m}^{(1)}+\frac{3}{8}\lambda_{p}^{(1)}\lambda_{m}^{(1)}\\\
&-6\Big{(}\lambda_{p}^{(1)}h_{m}^{(1)}+\lambda_{m}^{(1)}h_{p}^{(1)}\Big{)}+(3\lambda_{0}+32h_{0})\Big{(}\frac{h_{p}^{(1)}+h_{m}^{(1)}}{N_{c1}}-\frac{h_{p}^{(1)}-h_{m}^{(1)}}{N_{c2}}\Big{)}\\\
&+(3h_{0}-\frac{3\lambda_{0}}{4})\Big{(}\frac{\lambda_{p}^{(1)}+\lambda_{m}^{(1)}}{N_{c1}}-\frac{\lambda_{p}^{(1)}-\lambda_{m}^{(1)}}{N_{c2}}\Big{)}+(80\lambda_{0}-576h_{0})h_{p}^{(1)}h_{m}^{(1)}\\\
&+\Big{(}\frac{9}{2}h_{0}-\frac{15}{4}\lambda_{0}\Big{)}\lambda_{p}^{(1)}\lambda_{m}^{(1)}+\Big{(}\frac{9}{2}\lambda_{0}+80h_{0}\Big{)}\Big{(}\lambda_{p}^{(1)}h_{m}^{(1)}+\lambda_{m}^{(1)}h_{p}^{(1)}\Big{)}+O(\lambda_{0}^{2}/N^{2}),\end{split}$
(291)

$\begin{split}\beta_{\zeta^{(2)}}=&\zeta_{0}\Bigg{[}16f_{p}^{(2)}+32h_{p}^{(2)}-6\lambda_{p}^{(2)}\\\
&\quad+\Big{(}\frac{8h_{p}^{(1)}+3\lambda_{p}^{(1)}+8h_{m}^{(1)}+3\lambda_{m}^{(1)}}{N_{c1}}+\frac{8h_{p}^{(1)}+3\lambda_{p}^{(1)}-8h_{m}^{(1)}-3\lambda_{m}^{(1)}}{N_{c2}}\Big{)}\\\
&\quad-160\Big{(}h_{p}^{(1)2}+h_{m}^{(1)2}\Big{)}+\frac{21}{4}\Big{(}\lambda_{p}^{(1)2}+\lambda_{m}^{(1)2}\Big{)}+128\Big{(}\lambda_{p}^{(1)}h_{p}^{(1)}+\lambda_{m}^{(1)}h_{m}^{(1)}\Big{)}\\\
&\quad+64\Big{(}\lambda_{p}^{(1)}f_{p}^{(1)}+\lambda_{m}^{(1)}f_{m}^{(1)}\Big{)}\Bigg{]}+O(\lambda_{0}^{2}/N^{2}),\\\
\end{split}$ (292)

$\begin{split}\beta_{f_{p}^{(2)}}=&16\Big{(}\zeta_{0}\zeta^{(2)}+f_{0m}f_{m}^{(2)}\Big{)}+8(1+4\lambda_{0})\Big{(}f_{m}^{(1)2}+\zeta^{(1)2}\Big{)}+64\lambda_{p}^{(1)}\Big{(}\zeta_{0}\zeta^{(1)}+f_{0m}f_{m}^{(1)}\Big{)}\\\
&+\Big{(}16f_{0p}+32h_{0}-6\lambda_{0}\Big{)}f_{p}^{(2)}+\Big{(}32f_{0p}+48h_{0}\Big{)}h_{p}^{(2)}+\Big{(}\frac{9}{8}\lambda_{0}-6f_{0p}\Big{)}\lambda_{p}^{(2)}\\\
&+24\Big{(}h_{p}^{(1)2}+h_{m}^{(1)2}\Big{)}+\frac{9}{16}\Big{(}\lambda_{p}^{(1)2}+\lambda_{m}^{(1)2}\Big{)}+f_{0m}\Big{(}32h_{m}^{(2)}-6\lambda_{m}^{(2)}\Big{)}\\\
&+\Big{(}32h_{p}^{(1)}-6\lambda_{p}^{(1)}\Big{)}f_{p}^{(1)}+\Big{(}32h_{m}^{(1)}-6\lambda_{m}^{(1)}\Big{)}f_{m}^{(1)}+8f_{p}^{(1)2}\\\
&+\Big{(}\frac{(8h_{0}+3\lambda_{0})(f_{p}^{(1)}+f_{m}^{(1)})+(f_{0p}+f_{0m})(8h_{p}^{(1)}+3\lambda_{p}^{(1)}+8h_{m}^{(1)}+3\lambda_{m}^{(1)})}{N_{c1}}\\\
&\qquad+\frac{(8h_{0}+3\lambda_{0})(f_{p}^{(1)}-f_{m}^{(1)})+(f_{0p}-f_{0m})(8h_{p}^{(1)}+3\lambda_{p}^{(1)}-8h_{m}^{(1)}-3\lambda_{m}^{(1)})}{N_{c2}}\Big{)}\\\
&+\Big{(}9h_{0}-\frac{45}{8}\lambda_{0}+\frac{21}{4}f_{0p}\Big{)}\Big{(}\lambda_{p}^{(1)2}+\lambda_{m}^{(1)2}\Big{)}+(96\lambda_{0}-1152h_{0}-160f_{0p})\Big{(}h_{p}^{(1)2}+h_{m}^{(1)2}\Big{)}\\\
&+2(96h_{0}+9\lambda_{0}+64f_{0p})\Big{(}\lambda_{p}^{(1)}h_{p}^{(1)}+\lambda_{m}^{(1)}h_{m}^{(1)}\Big{)}+64(2\lambda_{0}-5h_{0})\Big{(}h_{p}^{(1)}f_{p}^{(1)}+h_{m}^{(1)}f_{m}^{(1)}\Big{)}\\\
&+2\Bigg{(}64h_{0}+\frac{21}{4}\lambda_{0}+32f_{0p}\Bigg{)}\Big{(}\lambda_{p}^{(1)}f_{p}^{(1)}+\lambda_{m}^{(1)}f_{m}^{(1)}\Big{)}+32\lambda_{0}f_{p}^{(1)2}\\\
&+f_{0m}\Big{(}-320h_{p}^{(1)}h_{m}^{(1)}+64\lambda_{m}^{(1)}f_{p}^{(1)}+128\lambda_{p}^{(1)}h_{m}^{(1)}+128\lambda_{m}^{(1)}h_{p}^{(1)}+\frac{21}{2}\lambda_{p}^{(1)}\lambda_{m}^{(1)}\Big{)}\\\
&+O(\lambda_{0}^{2}/N^{2}),\end{split}$ (293)

$\begin{split}\beta_{f_{m}^{(2)}}=&\Big{(}16f_{0p}+32h_{0}-6\lambda_{0}\Big{)}f_{m}^{(2)}+f_{0m}\Big{(}16f_{p}^{(2)}+32h_{p}^{(2)}-6\lambda_{p}^{(2)}\Big{)}\\\
&+\Big{(}32f_{0p}+48h_{0}\Big{)}h_{m}^{(2)}+\Big{(}\frac{9}{8}\lambda_{0}-6f_{0p}\Big{)}\lambda_{m}^{(2)}+\Big{(}16f_{p}^{(1)}+32h_{p}^{(1)}-6\lambda_{p}^{(1)}\Big{)}f_{m}^{(1)}\\\
&+\Big{(}32f_{p}^{(1)}+48h_{p}^{(1)}\Big{)}h_{m}^{(1)}+\Big{(}\frac{9}{8}\lambda_{p}^{(1)}-6f_{p}^{(1)}\Big{)}\lambda_{m}^{(1)}\\\
&+\Big{(}\frac{(8h_{0}+3\lambda_{0})(f_{p}^{(1)}+f_{m}^{(1)})+(f_{0p}+f_{0m})(8h_{p}^{(1)}+3\lambda_{p}^{(1)}+8h_{m}^{(1)}+3\lambda_{m}^{(1)})}{N_{c1}}\\\
&\qquad-\frac{(8h_{0}+3\lambda_{0})(f_{p}^{(1)}-f_{m}^{(1)})+(f_{0p}-f_{0m})(8h_{p}^{(1)}+3\lambda_{p}^{(1)}-8h_{m}^{(1)}-3\lambda_{m}^{(1)})}{N_{c2}}\Big{)}\\\
&+4(48\lambda_{0}-576h_{0}-80f_{0p})h_{p}^{(1)}h_{m}^{(1)}+\Big{(}18h_{0}-\frac{45}{4}\lambda_{0}+\frac{21}{2}f_{0p}\Big{)}\lambda_{p}^{(1)}\lambda_{m}^{(1)}\\\
&+2(64\lambda_{0}-160h_{0})\Big{(}h_{p}^{(1)}f_{m}^{(1)}+h_{m}^{(1)}f_{p}^{(1)}\Big{)}\\\
&+2(96h_{0}+9\lambda_{0}+64f_{0p})\Big{(}\lambda_{p}^{(1)}h_{m}^{(1)}+\lambda_{m}^{(1)}h_{p}^{(1)}\Big{)}\\\
&+2\Bigg{(}64h_{0}+\frac{21}{4}\lambda_{0}+32f_{0p}\Bigg{)}\Big{(}\lambda_{p}^{(1)}f_{m}^{(1)}+\lambda_{m}^{(1)}f_{p}^{(1)}\Big{)}\\\
&+64\lambda_{0}f_{p}^{(1)}f_{m}^{(1)}+64f_{0m}\Big{(}f_{p}^{(1)}\lambda_{p}^{(1)}+f_{m}^{(1)}\lambda_{m}^{(1)}\Big{)}-64\zeta_{0}\zeta^{(1)}\lambda_{m}^{(1)}\\\
&+f_{0m}\Bigg{[}128\Big{(}\lambda_{p}^{(1)}h_{p}^{(1)}+\lambda_{m}^{(1)}h_{m}^{(1)}\Big{)}-160\Big{(}h_{p}^{(1)2}+h_{m}^{(1)2}\Big{)}+\frac{21}{4}\Big{(}\lambda_{p}^{(1)2}+\lambda_{m}^{(1)2}\Big{)}\Bigg{]}\\\
&+O(\lambda_{0}^{2}/N^{2}).\end{split}$ (294)

In the above expression of $\beta_{\zeta^{(2)}}$, we have ignored terms of the
forms $\zeta^{(2)}\Big{[}\cdots\Big{]}$ and $\zeta^{(1)}\Big{[}\cdots\Big{]}$
because the dots correspond to coefficients of $\zeta_{0}$ in
$\beta_{\zeta_{0}}$ and $\beta_{\zeta^{(1)}}$ respectively, and these
coefficients have already been set to zero at the fixed point.

As earlier, by setting $\beta_{h_{p}^{(2)}}$, $\beta_{h_{m}^{(2)}}$ and
$\beta_{\zeta^{(2)}}$ to zero, one can determine the values of $h_{p}^{(2)}$,
$h_{m}^{(2)}$ and $f_{p}^{(2)}$. Then by setting $\beta_{f_{p}^{(2)}}$ to
zero, one can find the $O(1/N^{2})$ correction to the quantity
$(1+4\lambda_{p})(\zeta^{2}+f_{m}^{2})$. We will give the forms of these
corrections shortly. But first let us discuss the constraint following from
setting $\beta_{f_{m}^{(2)}}$ to zero. Just like the $O(1/N)$ corrections, the
coefficients of $f_{0m}$, $f_{m}^{(1)}$ and $f_{m}^{(2)}$ in
$\beta_{f_{m}^{(2)}}$ vanish, and no additional condition is imposed on
$f_{m}$ up to the leading order term in the $\lambda_{0}$-expansion when
$\beta_{f_{m}^{(2)}}=0$. In fact, this equation leads to a relation between
the quantities $x_{f1}^{(2)}$ and $x_{f2}^{(2)}$ which is as follows:

$\displaystyle
x_{f1}^{(2)}=x_{f2}^{(2)}-\frac{6(71\sqrt{6}-163)(N_{c2}-N_{c1})}{55(6\sqrt{6}-13)N_{c1}N_{c2}^{2}}\Big{(}11+N_{c2}(x_{f2}^{(1)})_{0}\Big{)}+O(\lambda_{0}^{2}/N^{2}).$
(295)

Thus, when this relation is satisfied, the degeneracy in the fixed points
survives at this order. Now, imposing the above relation as well as the
relation between $x_{f1}^{(1)}$ and $x_{f2}^{(1)}$ obtained earlier, we get

$\displaystyle h_{p}^{(2)}=$
$\displaystyle\frac{1}{1497375(36-13\sqrt{6})N_{c1}N_{c2}^{2}}$ (296)
$\displaystyle\Bigg{[}\Big{(}4235(8821-3606\sqrt{6})+46464(25\sqrt{6}-62)A_{0}\Big{)}N_{c1}$
$\displaystyle+27225(32-13\sqrt{6})N_{c2}\Big{(}11+N_{c2}(x_{f2}^{(1)})_{0}\Big{)}$
$\displaystyle+\Big{(}55(148262-60519\sqrt{6})+8448(25\sqrt{6}-62)A_{0}\Big{)}N_{c1}N_{c2}(x_{f2}^{(1)})_{0}$
$\displaystyle+N_{c1}N_{c2}^{2}\Bigg{\\{}\Big{(}55(7865-3207\sqrt{6})+384(25\sqrt{6}-62)A_{0}\Big{)}(x_{f2}^{(1)})_{0}^{2}$
$\displaystyle\qquad\qquad\quad+27225(25\sqrt{6}-62)x_{f2}^{(2)}\Bigg{\\}}\Bigg{]}+O(\lambda_{0}/N^{2}),$

$\displaystyle h_{m}^{(2)}=$
$\displaystyle\frac{(13\sqrt{6}-32)(N_{c1}-N_{c2})}{55(36-13\sqrt{6})N_{c1}N_{c2}^{2}}\Bigg{[}11+N_{c2}(x_{f2}^{(1)})_{0}\Bigg{]}+O(\lambda_{0}/N^{2}),$
(297)

$\begin{split}f_{p}^{(2)}&=\frac{(165-7\sqrt{6})(N_{c1}-N_{c2})}{5170N_{c1}N_{c2}^{2}}\Bigg{[}11+N_{c2}(x_{f2}^{(1)})_{0}\Bigg{]}\\\
&\quad-\frac{\sqrt{6}}{8984250N_{c2}^{2}}\Bigg{[}\Bigg{(}11+N_{c2}(x_{f2}^{(1)})_{0}\Bigg{)}\Bigg{\\{}1536A_{0}\Big{(}11+N_{c2}(x_{f2}^{(1)})_{0}\Big{)}\\\
&\qquad\qquad\qquad\qquad+55\Big{(}-656+4125\sqrt{6}+2(-101+120\sqrt{6})N_{c2}(x_{f2}^{(1)})_{0}\Big{)}\Bigg{\\}}\\\
&\qquad\qquad\qquad\qquad+108900N_{c2}^{2}x_{f2}^{(2)}\Bigg{]}+O(\lambda_{0}/N^{2}),\end{split}$
(298)

$\begin{split}&2\Big{(}\zeta_{0}\zeta^{(2)}+f_{0m}f_{m}^{(2)}\Big{)}+(1+4\lambda_{0})\Big{(}f_{m}^{(1)2}+\zeta^{(1)2}\Big{)}+8\lambda_{p}^{(1)}\Big{(}\zeta_{0}\zeta^{(1)}+f_{0m}f_{m}^{(1)}\Big{)}\\\
&=\frac{6\sqrt{6}-13}{9075N_{c2}^{2}}\Big{(}11+N_{c2}(x_{f2}^{(1)})_{0}\Big{)}^{2}\\\
&\quad+\frac{\lambda_{0}}{145200N_{c2}^{2}}\Bigg{[}\Big{(}(-2216+441\sqrt{6})+\frac{768}{55}(13-6\sqrt{6})A_{0}\Big{)}\Big{(}11+N_{c2}(x_{f2}^{(1)})_{0}\Big{)}^{2}\\\
&\qquad\qquad\qquad\qquad+\Big{(}72(163-71\sqrt{6})-32(13-6\sqrt{6})N_{c2}(x_{f2}^{(1)})_{1}\Big{)}\Big{(}11+N_{c2}(x_{f2}^{(1)})_{0}\Big{)}\\\
&\qquad\qquad\qquad\qquad+330(13-6\sqrt{6})N_{c2}^{2}x_{f2}^{(2)}\Bigg{]}-\Bigg{[}\frac{3(165-7\sqrt{6})(N_{c1}-N_{c2})\lambda_{0}}{752N_{c1}N_{c2}}\Bigg{]}f_{m}^{(1)}\\\
&\quad+\Bigg{[}\frac{(165-7\sqrt{6})(N_{c1}-N_{c2})\Big{(}11+N_{c2}(x_{f2}^{(1)})_{0}\Big{)}}{2585N_{c1}N_{c2}^{2}}\Bigg{]}f_{0m}+O(\lambda_{0}^{2}/N^{2}).\end{split}$
(299)

The last quantity is the $O(1/N^{2})$ correction to
$(1+4\lambda_{p})(\zeta^{2}+f_{m}^{2})$ up to the first subleading order in
the $\lambda_{0}$-expansion. When $N_{c1}\neq N_{c2}$, this quantity has terms
which are dependent on $f_{0m}$ and $f_{m}^{(1)}$. These terms provide the
correction to the shape of the closed curve of fixed points at $O(1/N^{2})$.
Notice that when $N_{c1}=N_{c2}$, this quantity becomes independent of
$f_{0m}$ and $f_{m}^{(1)}$ which indicates that the circular form of the
closed curve is preserved up to this order.

To determine whether the degeneracy in the fixed points survives at higher
orders in $\lambda_{0}$ as well as to evaluate the higher order corrections to
the couplings at the fixed points, one needs to compute the contributions of
the higher loop diagrams to the beta functions. This lies beyond the scope of
this paper.

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