11institutetext: SISSA, Via Bonomea 265, 34136 Trieste, Italy22institutetext:
INFN, Sezione di Trieste, Via Valerio 2, 34127 Trieste, Italy

# Sequential deconfinement and self-dualities in $4d$ $\mathcal{N}\\!=\\!1$
gauge theories

Stephane Bajeot 2 Sergio Benvenuti<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

We apply the technique of sequential deconfinement to the four dimensional
$\mathcal{N}\\!=\\!1$ $Usp(2N)$ gauge theory with an antisymmetric field and
$2F$ fundamentals. The fully deconfined frame is a length-$N$ quiver. We use
this deconfined frame to prove the known self-duality of $Usp(2N)$ with an
antisymmetric field and $8$ fundamentals. Along the way we encounter a
subtlety: in certain quivers with degenerate holomorphic operators, a naive
application of Seiberg duality rules leads to an incorrect superpotential or
chiral ring.

We also consider the reduction to $3d$ $\mathcal{N}\\!=\\!2$ theories,
recovering known fully deconfined duals of $Usp(2N)$ and $U(N)$ gauge
theories, and obtaining new ones.

##  1 Introduction and summary

Dualities are among the most powerful tool to analyze quantum field theories
at strong coupling. In this paper we are interested in four dimensional
$\mathcal{N}=1$ gauge theories with rank-$2$ matter. Various recent works
derived dualities involving gauge theories with rank-$2$ matter using only
basic dualities involving gauge theories with fundamental matter, like Seiberg
Seiberg:1994pq , Intriligator-Pouliot Intriligator:1995ne  and Aharony
dualities Aharony:1997gp .

In three dimensions, Pasquetti:2019uop ; Pasquetti:2019tix  proved a $3d$
$\mathcal{N}=2$ S-confining duality for $U(N)$ with adjoint and $(1,1)$
fundamentals Aghaei:2017xqe ; Benvenuti:2018bav ; Amariti:2018wht  iterating
Aharony duality. Benvenuti:2020gvy  proved $3d$ $\mathcal{N}=2$ self-dualities
of $U(N)$ with adjoint and $(2,2)$ fundamentals and of $Usp(2N)$ with
antisymmetric and $6$ fundamentals Benvenuti:2018bav ; Amariti:2018wht ,
iterating Aharony dualities. Iterative application of confining monopole
dualities Benini:2017dud  have been proven very useful in $3d$ also in
Benvenuti:2017lle ; Benvenuti:2017kud ; Benvenuti:2017bpg ; Giacomelli:2017vgk
.

More recently, in four dimensions, Bajeot:2022kwt ; Bottini:2022vpy  proved
the S-confinement of $4d$ $\mathcal{N}=1$ $Usp(2N)$ with antisymmetric and $6$
fundamentals, and Bajeot:2022kwt  proved that all S-confining quivers with one
node (classified long time ago in Csaki:1996sm ; Csaki:1996zb ) can be
obtained from Seiberg and Intriligator-Pouliot S-confinements.

By _proving the duality_ between theory $\mathcal{T}_{1}$ and theory
$\mathcal{T}_{k}$, we mean constructing a sequence of quiver gauge theories
$\mathcal{T}_{i}$, $i=1,\ldots,k$, such that $\mathcal{T}_{i}$ is related to
$\mathcal{T}_{i+1}$ by the application of an elementary duality on a single
node. So all the theories $\mathcal{T}_{1},\ldots,\mathcal{T}_{k}$ are
infrared dual. Assuming, as is standard, that the renormalization group flows
commute with dualizing a single node, this amounts to a proof of the non-
elementary duality $\mathcal{T}_{1}\leftrightarrow\mathcal{T}_{k}$. The
strategy to construct the sequence of dual quivers requires the
_deconfinement_ of some rank-$2$ matter. The deconfinement we need in this
paper was introduced in Bajeot:2022kwt ; Bottini:2022vpy , and uplifts the
$3d$ deconfiments used in Pasquetti:2019uop ; Pasquetti:2019tix ;
Benvenuti:2020gvy .111Similar deconfinements appear in Garcia-
Etxebarria:2012ypj ; Garcia-Etxebarria:2013tba  in the study of orientifolded
dimer models, and were used in Etxebarria:2021lmq  to construct
$\mathcal{N}=1$ Lagrangians for $4d$ $\mathcal{N}=2$ SCFTs.

In a related context, Bottini:2021vms ; Hwang:2021ulb  proved the $4d$ mirrors
dualities Hwang:2020wpd  (upflit of $3d$ $\mathcal{N}=4$ mirror symmetries
Intriligator:1996ex ; Hanany:1996ie  for linear quivers) using only
Intriligator-Pouliot dualities. One difference is that, in the mirror symmetry
case, the intermediate steps are quasi-Lagrangian theories, that is they
involve gauging symmetries which only emerge in the infrared, as in
Gadde:2015xta .

In this paper we work out the _sequentially deconfined_ dual of $4d$
$\mathcal{N}=1$ $Usp(2N)$ with antisymmetric and $2F$ fundamentals, uplifting
the $3d$ $\mathcal{N}=2$ results of Benvenuti:2020gvy . This means that we
step by step prove a duality with a linear quiver gauge theories with $N$
nodes and a certain _saw_ structure, full details of the theory are in (4.5).
This fully deconfined dual frame enjoys the nice property that all the chiral
ring operators are gauge singlet fields, similarly to what happens for
Intriligator-Pouliot and Aharony dualities.222This property cannot be enjoyed
by theories with general baryonic operators in the chiral ring, such as in
$SU(N)$ SQCD or in the sequentially deconfined $3d$ $\mathcal{N}=2$ $SO$ gauge
theories BenvenutiLoMonaco .

As an application of this construction, we provide a simple proof of the
_self-duality modulo flips_ 333Flipping a gauge invariant operator
$\mathcal{O}$ means adding a gauge singlet field $\sigma$ to the theory, and
an interaction term of the form $\sigma\mathcal{O}$ to the superpotential. The
gauge singlet $\sigma$ is called a _flipper_. of $Usp(2N)$ with antisymmetric
and $8$ fundamentals, proposed long time ago in Csaki:1997cu . By _self-dual
modulo flips_ we mean that the electric and magnetic theory share the same
gauge structure, but differ by gauge singlets fields of flip type. Self-
dualities modulo flips have been discussed in Distler:1996ub ; Csaki:1997cu ;
Karch:1997jp ; Razamat:2018gbu , the simplest case is $SU(2)$ with $8$
doublets. Interestingly, given a self-duality modulo flips, it is possible to
move the singlets across the duality and construct _exactly self-dual
theories_ with enhanced infra-red global symmetry, see for instance
Razamat:2017wsk ; Razamat:2018gbu ; Sela:2019nqa ; Hwang:2020ddr ;
Hwang:2021xyw . In our case of $Usp(2N)$ with antisymmetric and $8$
fundamentals, Hwang:2020ddr  discussed various exactly self-dual theories and
discussed the associated symmetry enhancements. Sometimes these symmetry
enhancements can be understood compactifying a $6d$ $(1,0)$ SCFT on a Riemann
surface. For $Usp(2N)$ with antisymmetric and $8$ fundamentals, Hwang:2021xyw
related the self-duality and specific symmetry enhancements to a
compactification of the rank-$N$ E-string $6d$ SCFT on a $2$-sphere.

We encounter one subtlety (which is also present in the $3d$ cases but was
overlooked in Benvenuti:2020gvy ) during the process that we call _degenerate
holomorphic operator ambiguity_. As the name suggests, this phenomenon appears
when we reach a frame that contains more than one gauge invariant holomorphic
operator with the same global symmetry quantum numbers (including $U(1)_{R}$),
but only one combination is a chiral protected operator. If such an operator
is flipped by a gauge singlet, only one specific combination appears in the
superpotential. In the examples we encounter in this paper, it happens that if
we follow the rules of Seiberg duality (as is usually done) we end up with the
incorrect result. In some cases we can determine which is the precise
combination of operators appearing in the chiral ring (equivalently, the
combination that can be flipped) by going in a dual frame and using classical
F-terms relations there. Hence, in the original theory with _degenerate
holomorphic operator ambiguity_ , the ambiguity is resolved by quantum
relations. In the case of $F=4$, the precise superpotential is crucial in the
proof of the self-duality modulo flips, so this case provides a good
consistency check of our procedure.

#### Future directions

One natural question is wether there is a relation between our sequentially
deconfined dual of $Usp(2N)$ with an antisymmetric and $2F$ fundamentals and
Kutasov-Schwimmer type dualities Kutasov:1995ve ; Kutasov:1995np ;
Kutasov:1995ss , which for Usp were proposed in Brodie:1996xm . Namely one can
turn on a superpotential term $tr(A^{j})$ on the electric side, such term maps
to a singlet on the magnetic, so a Higgsing process is induced. The study of
this Higgsing might shed light on the dualities of Kutasov:1995ve ;
Kutasov:1995np ; Kutasov:1995ss ; Brodie:1996xm . We expect the _degenerate
holomorphic operator ambiguity_ encountered in this paper to play an important
role.

There are quite a few _self-dualities modulo flips_ proposed in the literature
Distler:1996ub ; Csaki:1997cu ; Karch:1997jp ; Razamat:2018gbu , such as
$SU(2N)$ with antisymmetric, conjugate antisymmetric and $(4,4)$ fundamentals,
or $SU(6)$ with $2$ antisymmetrics and $(2,6)$ fundamentals, or $SU(8)$ with
$2$ antisymmetrics and $(0,8)$ fundamentals. A natural question is if such
self-dualities can be proven using only the basic Seiberg and Intriligator-
Pouliot dualities, as done in this work for $Usp(2N)$ with antisymmetric and
$8$ fundamentals. Notice that many self-dual gauge theories have been
constructed simply ’adding one flavor’ to an S-confining gauge theory
Csaki:1997cu ; Karch:1997jp , so the fact that the S-confining dualities can
be proven Bajeot:2022kwt  is encouraging.

Related to the above point, many S-confinements and many self-dualities have
been proposed for $4d$ $\mathcal{N}=1$ $Spin(N)$ theories with spinors and
vectors Csaki:1997cu ; Karch:1997jp ; Razamat:2018gbu ; Sela:2019nqa . It
would be very interesting to find a way to deconfine spinorial matter and try
to prove such proposals.

Limits to gauge theories with orthogonal and/or symplectic gauge groups.
S-confinements for $3d$ $\mathcal{N}=2$ theories with SO/Usp gauge groups and
adjoint matter where recently proposed in Benvenuti:2021nwt , and
Amariti:2022wae  pointed out a relation to $4d$ $\mathcal{N}=1$ S-confinements
for Usp gauge group and antisymmetric matter. It might be interesting to
deform the $4d$ $\mathcal{N}=1$ sequential deconfinements as in
Amariti:2022wae . See BenvenutiLoMonaco  for the sequential deconfinement of
$3d$ $\mathcal{N}=2$ rank-$2$ matter with SO/Usp gauge groups.

It would also be interesting to study _degenerate holomorphic operator
ambiguity_ in other examples, possibly involving different kind of gauge
groups.

#### Structure of the paper

This paper is organized as follows.

In section 2 we recall our notation and the three basic dualities which will
be iteratively used in the rest of the paper.

In section 3 we discuss in detail a simple but non-trivial example, namely we
sequentially deconfine $Usp(6)$ with an antisymmetric and $8$ fundamentals.
This example is the simplest one where the issue of the _degenerate
holomorphic operator ambiguity_ appears. We construct the fully deconfined
dual, then we reconfine the quiver tail in order to prove the self-duality of
the theory.

In section 4 we present the general sequential deconfinement of $Usp(2N)$ with
an antisymmetric and $2F$ fundamentals.

In section 5 we set $2F=8$, which allows to sequentially reconfine the quiver
tail, and prove the self-duality modulo flips of the theory.

In section 6 we show how to reduce our $4d$ $\mathcal{N}=1$ $Usp(2N)$ story to
$3d$ $\mathcal{N}=2$, re-obtaining the results of for $U(N)$ and $Usp(2N)$
found in Benvenuti:2020gvy . Along the way we also derive new sequentially
deconfined duals, namely for $U(N)$ with adjoint and $(F,F)$ fundamentals with
monopole superpotentials.

##  2 Basic S-confining and dualities moves

In this section we present the basic ingredients that we use in this paper to
obtain the more complicated dualities involving rank-2 matter field. The first
basic move is the deconfinement of an antisymmetric field with a $Usp$ gauge
group as in Bajeot:2022kwt ; Bottini:2022vpy , which is a modification of the
original Berkooz deconfinement Berkooz:1995km . This form was used in
Bajeot:2022kwt ; Bottini:2022vpy  to tackle Usp(2N) with $6$ fundamentals,
which is S-confining.

Throughout the paper, blue $2N$ circles denote $Usp(2N)$ gauge nodes.

$Usp$ deconfinement:444On the r.h.s quiver in (1), we didn’t include the
flipper $\beta_{1}$ on the drawing. Sometimes, as in Bajeot:2022kwt  this
flipper is represented on the quiver by a cross, $\times$, on the
bifundamental field $b_{1}$. Moreover, the trace ”$tr$” is taken using the
$Usp$ invariant antisymmetric matrix.
Throughout the paper, a trace of an operator in the antisymmetric of a $Usp$
group is taken using the $Usp$ invariant antisymmetric matrix.

$2N$$2F$${\mathcal{W}}=0$$A$$\equiv$$2N$$2F-1$$1$$Q$$P$$A$$\longleftrightarrow$$2N$$2N-2$$2F-1$$1$$1$$v_{1}$$v_{1}$$h_{1}$${\mathcal{W}}=v_{1}b_{1}d_{1}+h_{1}d_{1}v_{2}+\beta_{1}\,\text{tr}\,(b_{1}b_{1})$$b_{1}$$d_{1}$$Q$

(1)

In order to get this deconfinement we have to use another basic move. It is
the S-confining result of Intriligator-Pouliot (IP) Intriligator:1995ne  that
involves $Usp(2N)$ gauge group and $2N+4$ fields in the fundamental
representation.

IP S-confining duality:

$2N$$2N+4$${\mathcal{W}}=0$$Q$$\longleftrightarrow$$2N+4$${\mathcal{W}}=\text{Pfaff}\,(\mu)$$\mu$Mapping:
$\qquad\qquad tr(QQ)\longleftrightarrow\mu$ (2)

The last move concerns the same $Usp$ theory but with $2F\geq 2N+4$
fundamentals now. It is the IP duality Intriligator:1995ne . In quiver
notation it reads

IP duality:

$2N$$2F$${\mathcal{W}}=0$$Q$$\longleftrightarrow$

$2F-2N-4$

$2F$${\mathcal{W}}=tr(q\,q)\,\Phi$$q$$\Phi$Mapping: $\qquad\qquad
tr(QQ)\longleftrightarrow\Phi$ (3)

##  3 Case study: $Usp(6)$ with ${\tiny\ydiagram{1,1}}$ \+
$8\,{\tiny\ydiagram{1}}$

In this section we study the $Usp(6)$ gauge theory with matter in the
antisymmetric representation and $8$ fundamentals, that is $4$ flavors. It is
the simplest example that contains all the ingredients that we want to
exhibit. In the next sections, we will show the general case. It is known that
this theory is self-dual _modulo flips_ Csaki:1996eu . We will prove the self-
duality using the basic moves of the previous section (1), (2) and (3).

###  3.1 Sequential deconfinement

The first step is the deconfinement of the antisymmetric with (1). We get the
following two frames ${\mathcal{T}}_{0}$ and ${\mathcal{T}}_{0^{\prime}}$

${\mathcal{T}}_{0}:$$6$$7$$1$$Q,\,$ $R_{Q}$ $P,\,$ $4-4R_{A}-7R_{Q}$
${\mathcal{W}}=0$$A,\,$ $R_{A}$ $\longleftrightarrow$

${\mathcal{T}}_{0^{\prime}}:$$6$$4$$7$$1$$1$$h_{1},\,$ $3R_{A}$ $p_{1}$

$4-2R_{A}+7R_{Q}$

$p_{2}$

$4-\frac{9}{2}R_{A}+7R_{Q}$

${\mathcal{W}}=p_{1}c_{1}d_{1}+h_{1}d_{1}p_{2}+\beta_{1}\,\text{tr}\,(c_{1}c_{1})$$c_{1},\,$
$\frac{1}{2}R_{A}$ $d_{1}$

$-2+\frac{3}{2}R_{A}+7R_{Q}$

$R_{Q},\,$

$q_{1}$

The global symmetry of the original theory is $SU(8)\times U(1)_{A}\times
U(1)_{R}$. When we split the $8$ chirals into $7+1$, we split the $SU(8)$ into
$SU(7)\times U(1)_{P}$. So in the splitted form, the R-charges of the fields
should be a function of two variables (for the two global $U(1)$’s that can
mix with the $U(1)_{R}$). We choose to write the R-charges in terms of $R_{Q}$
and $R_{A}$. Then the R-charges of the other fields are determined by the
$U(1)_{R}$ ABJ anomaly555This is the same as requiring the vanishing of the
NSVZ $\beta$ function. and by the requirement that any superpotential term
should have R-charge equal to 2. We have written the R-charges of the fields
next to them.

Then we dualize the $Usp(6)$ node with (3). The fields $d_{1}$ and $\beta_{1}$
get a mass. After integrating them out and rescaling the fields to put a $+1$
in front of each term in the superpotential, we get

${\mathcal{T}}_{1}:$$2$$4$$7$$1$$1$${\mathcal{W}}=M_{1}Q_{1}Q_{1}+Q_{1}C_{1}Q_{2}+L_{1}Q_{1}V_{1}$$+\,\text{tr}\,(C_{1}\Phi
C_{1})+H_{1}V_{1}C_{1}P_{2}$$L_{1},\,$ $4-2R_{A}+6R_{Q}$ $L_{1}$$Q_{1},\,$
$1-R_{Q}$ $Q_{2}$

$\frac{1}{2}R_{A}+R_{Q}$

$V_{1}$

$-3+2R_{A}+7R_{Q}$

$P_{2}$

$4-\frac{9}{2}R_{A}-7R_{Q}$

$C_{1},\,$ $1-\frac{1}{2}R_{A}$ $H_{1},\,$ $3R_{A}$ $M_{1},\,$ $2R_{Q}$
$\Phi,\,$ $R_{A}$

(4)

In ${\mathcal{T}}_{1}$ the antisymmetric field, $\Phi$, is traceless because
the trace component has been killed by the equation of motion (E.O.M) of the
flipper $\beta_{1}$. The mapping of the chiral ring generators after this
first step is the following

$\begin{array}[]{c}{\mathcal{T}}_{0}\\\ QQ\\\ \text{tr}\,(QA^{i}Q)\\\
\text{tr}\,(QA^{j}P)\\\ \text{tr}\,(QA^{2}P)\\\ \text{tr}\,(A^{2})\\\
\text{tr}\,(A^{3})\end{array}\Longleftrightarrow\begin{array}[]{c}{\mathcal{T}}_{0^{\prime}}\\\
q_{1}q_{1}\\\ \text{tr}\,(q_{1}\,(c_{1}c_{1})^{i}\,q_{1})\\\
\text{tr}\,(q_{1}\,c_{1}(c_{1}c_{1})^{j}\,p_{2})\\\ q_{1}\,p_{1}\\\
\text{tr}\,((c_{1}c_{1})^{2})\\\
h_{1}\end{array}\Longleftrightarrow\begin{array}[]{c}{\mathcal{T}}_{1}\\\
M_{1}\\\ \text{tr}\,(Q_{2}\,\Phi^{i-1}\,Q_{2})\\\
\text{tr}\,(Q_{2}\,\Phi^{j}\,P_{2})\\\ L_{1}\\\ \text{tr}\,(\Phi^{2})\\\
H_{1}\end{array}\qquad\begin{array}[]{l}\\\ \\\ i=1,2\\\ j=0,1\\\ \\\ \\\ \\\
\end{array}$

(5)

It can be checked that the mapping (5) is consistent with the R-charges of the
operators. Now we iterate the procedure.

We now deconfine the traceless antisymmetric field $\Phi$.

${\mathcal{T}}_{1^{\prime}}:$$2$$4$$2$$7$$1$$1$$1$${\mathcal{W}}=m_{1}q_{1}q_{1}+q_{1}c_{1}q_{2}+l_{1}v_{1}q_{1}$$+\,\text{tr}\,(c_{1}c_{2}c_{2}c_{1})+h_{1}v_{1}c_{1}c_{2}p_{3}$$+\,h_{2}d_{2}p_{3}+p_{2}c_{2}d_{2}+\beta_{2}\,\text{tr}\,(c_{2}c_{2})$$l_{1},\,$
$4-2R_{A}-6R_{Q}$ $l_{1}$$d_{2}$

$-2+3R_{A}+7R_{Q}$

$1-R_{Q},\,$

$q_{1}$$q_{2},\,\,$

$\frac{1}{2}R_{A}+R_{Q}$

$v_{1}$

$-3+2R_{A}+7R_{Q}$

$p_{2}$

$-4-\frac{7}{2}R_{A}-7R_{Q}$

$p_{3}$

$-4-5R_{A}-7R_{Q}$

$c_{1}$

$1-\frac{1}{2}R_{A}$

$c_{2}$

$\frac{1}{2}R_{A}$

$h_{1},\,$ $3R_{A}$ $h_{2},\,$ $2R_{A}$ $m_{1},\,$ $2R_{Q}$

(6)

Now we dualize the $Usp(4)$ node. There will not be any antisymmetric field
for the other gauge groups because they are $Usp(2)$666The dualization creates
however singlets that correspond to the trace part of the would be
antisymmetric. We call the singlet on the left $\alpha_{1}$. The one on the
right will receive a mass with the singlet $\beta_{2}$ so we do not need to
give it a name.. The fields $q_{1}$, $d_{2}$ and $\beta_{2}$ get a mass.
Moreover, $\text{tr}\,(c_{1}c_{2}c_{2}c_{1})$ becomes a mass term therefore
there will be no link between the two $Usp(2)$ gauge groups. After integrating
these massive fields out and a rescaling we get

${\mathcal{T}}_{2}:$$2$$4$$2$$7$$1$$1$$1$${\mathcal{W}}=M_{1}\text{tr}\,(B_{1}Q_{2}Q_{2}B_{1})+M_{2}Q_{2}Q_{2}+\text{tr}\,(C_{2}\Phi
C_{2})+\alpha_{1}\,\text{tr}\,(B_{1}B_{1})+\text{tr}\,(B_{1}C_{2}C_{2}B_{1})$$+\,R_{1}B_{1}V_{2}+L_{1}V_{1}B_{1}Q_{2}+L_{2}V_{2}Q_{2}+Q_{2}C_{2}Q_{3}+H_{1}V_{1}B_{1}C_{2}P_{3}+H_{2}?$$L_{1}$$L_{1}$$L_{2}$$L_{2}$$R_{1}$

$5-4R_{A}-7R_{Q}$

$V_{1}$

$-3+2R_{A}+7R_{Q}$

$V_{2}$

$-3+\frac{7}{2}R_{A}+7R_{Q}$

$P_{3}$

$4-5R_{A}-7R_{Q}$

$B_{1},\,$ $\frac{1}{2}R_{A}$ $C_{2},\,$ $1-\frac{1}{2}R_{A}$ $H_{1},\,$
$3R_{A}$ \+ singlet $H_{2},\,$ $2R_{A}$ $M_{1},M_{2}$$Q_{2},\,$
$1-\frac{1}{2}R_{A}-R_{Q}$ $Q_{3},\,$ $R_{A}+R_{Q}$

(7)

At this step, we face a feature that we call the _degenerate holomorphic
operator ambiguity_. It arises when we ask what is the operator flipped by the
singlet $H_{2}$.

###  3.2 Degenerate holomorphic operator ambiguity

If we apply the rules of Seiberg duality locally in the quiver, as is usually
done, using the mapping (3), we would conclude that it is
${\color[rgb]{0.21,0.12,1}\definecolor[named]{pgfstrokecolor}{rgb}{0.21,0.12,1}\pgfsys@color@cmyk@stroke{0.79}{0.88}{0}{0}\pgfsys@color<EMAIL_ADDRESS>so the superpotential should contain
$H_{2}{\color[rgb]{0.21,0.12,1}\definecolor[named]{pgfstrokecolor}{rgb}{0.21,0.12,1}\pgfsys@color@cmyk@stroke{0.79}{0.88}{0}{0}\pgfsys@color<EMAIL_ADDRESS>However, for the quiver at hand, there is another candidate,
${\color[rgb]{0,0.88,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.88,0}\pgfsys@color@cmyk@stroke{0.91}{0}{0.88}{0.12}\pgfsys@color<EMAIL_ADDRESS>Both $\mathcal{O}_{1}$ and $\mathcal{O}_{2}$ are gauge singlets, singlets
under the $SU(7)$ flavor symmetry and have R-charge:
$R({\color[rgb]{0,0.88,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.88,0}\pgfsys@color@cmyk@stroke{0.91}{0}{0.88}{0.12}\pgfsys@color@cmyk@fill{0.91}{0}{0.88}{0.12}V_{1}R_{1}})=R({\color[rgb]{0.21,0.12,1}\definecolor[named]{pgfstrokecolor}{rgb}{0.21,0.12,1}\pgfsys@color@cmyk@stroke{0.79}{0.88}{0}{0}\pgfsys@color@cmyk@fill{0.79}{0.88}{0}{0}V_{2}C_{2}P_{3}})=2-2R(A)$
(which implies that the two operators have the same charges under the flavor
$U(1)^{\prime}s$). Therefore they are degenerate operators.777Let us emphasize
that there is no ambiguity with $H_{1}$ because there is only one operator
which is singlet under the gauge symmetry, the $SU(7)$ flavor symmetry and has
R-charge equal to $2-3R(A)$. This operator is $V_{1}B_{1}C_{2}P_{3}$.. So it
might be that the precise operator flipped by $H_{2}$ is not exactly
$\mathcal{O}_{1}$, but some linear combination of $\mathcal{O}_{1}$ and
$\mathcal{O}_{2}$.

We claim that the correct answer is that the operator flipped by $H_{2}$ is
${\color[rgb]{0,0.88,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.88,0}\pgfsys@color@cmyk@stroke{0.91}{0}{0.88}{0.12}\pgfsys@color<EMAIL_ADDRESS>instead of the naive $\mathcal{O}_{1}$. Our argument in favor of this
statement comes from dualizing some nodes in the quiver, as we will explain
soon. So in a sense the fact the the correct operator is not the naive one is
due to quantum relations, which become classical relation after Seiberg
duality.

Our strategy to decide the correct operator is to use dualities in order to go
to a frame where F-term equations can answer the question. In this case, we
apply IP S-confining duality (2) on the left $Usp(2)$ gauge node of theory
$\mathcal{T}_{2}$ _with the singlet $H_{2}$ removed_. We consider the same
theory with the flipping removed, the question becomes which linear
combination of the two degenerate holomorphic operators $\mathcal{O}_{1}$ and
$\mathcal{O}_{2}$ is non zero in the chiral ring. Since the answer, as we will
see, is that $\mathcal{O}_{2}$ is non zero in the chiral ring, the quantum
relation in the unflipped theory is $\mathcal{O}_{1}=0$.

The $Usp(2)$ we dualize is coupled to $6$ fundamentals and so it confines,
producing a traceless antisymmetric field $B$ (the trace part is killed by the
flipper $\alpha_{1}$). We get

$4$$2$$7$$1$$1$$1$${\mathcal{W}}=M_{1}\text{tr}\,(Q_{2}BQ_{2})+M_{2}Q_{2}Q_{2}+\text{tr}\,(C_{2}BC_{2})+XV_{2}+L_{1}PQ_{2}+L_{2}V_{2}Q_{2}+Q_{2}C_{2}Q_{3}$$+\,H_{1}PC_{2}P_{3}+H_{2}?+\text{Pfaff}\,$
$\begin{pmatrix}B&\vdots&P&\vdots&X\\\ \dots&\dots&\dots&\dots&\dots\\\
&\vdots&0&\vdots&s\\\ &&\dots&\dots&\dots\\\ &&&\vdots&0\end{pmatrix}$
$B$$L_{1}$$L_{1}$$L_{2}$$L_{2}$$P$

$-3+\frac{5}{2}R_{A}+7R_{Q}$

$X$$V_{2}$$P_{3}$$C_{2}$$H_{1}$\+ singlet
$H_{2}$$M_{1},M_{2}$$Q_{2}$$Q_{3}$$s,\,$ $2-2R_{A}$

(8)

The Pfaffian term gives: Pfaff $\begin{pmatrix}B&\vdots&P&\vdots&X\\\
\dots&\dots&\dots&\dots&\dots\\\ &\vdots&0&\vdots&s\\\ &&\dots&\dots&\dots\\\
&&&\vdots&0\end{pmatrix}$
$\sim\varepsilon_{4}\,(B^{2})s+\varepsilon_{4}\,(BPX)$

The fields $X$ and $V_{2}$ are massive. The E.O.M of $X$ gives: $V_{2}+BP=0$.

In addition, the F-term equation for the singlet $H_{1}$ gives:
$PC_{2}P_{3}=0$.

Combining these two informations, we can resolve the ambiguity about the
operators
${\color[rgb]{0.21,0.12,1}\definecolor[named]{pgfstrokecolor}{rgb}{0.21,0.12,1}\pgfsys@color@cmyk@stroke{0.79}{0.88}{0}{0}\pgfsys@color<EMAIL_ADDRESS>and
${\color[rgb]{0,0.88,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.88,0}\pgfsys@color@cmyk@stroke{0.91}{0}{0.88}{0.12}\pgfsys@color<EMAIL_ADDRESS>Indeed in this frame these operators become

$\displaystyle{\color[rgb]{0.21,0.12,1}\definecolor[named]{pgfstrokecolor}{rgb}{0.21,0.12,1}\pgfsys@color@cmyk@stroke{0.79}{0.88}{0}{0}\pgfsys@color<EMAIL_ADDRESS>$\displaystyle{\color[rgb]{0.21,0.12,1}\definecolor[named]{pgfstrokecolor}{rgb}{0.21,0.12,1}\pgfsys@color@cmyk@stroke{0.79}{0.88}{0}{0}\pgfsys@color<EMAIL_ADDRESS>V_{2}C_{2}P_{3}\,\,}\overset{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\text{{\scriptsize
E.O.M
$X$}}}}{{\color[rgb]{0.21,0.12,1}\definecolor[named]{pgfstrokecolor}{rgb}{0.21,0.12,1}\pgfsys@color@cmyk@stroke{0.79}{0.88}{0}{0}\pgfsys@color@cmyk@fill{0.79}{0.88}{0}{0}=}}{\color[rgb]{0.21,0.12,1}\definecolor[named]{pgfstrokecolor}{rgb}{0.21,0.12,1}\pgfsys@color@cmyk@stroke{0.79}{0.88}{0}{0}\pgfsys@color@cmyk@fill{0.79}{0.88}{0}{0}\,\,BPC_{2}P_{3}\,\,}\overset{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\text{{\scriptsize
F-term
$H_{1}$}}}}{{\color[rgb]{0.21,0.12,1}\definecolor[named]{pgfstrokecolor}{rgb}{0.21,0.12,1}\pgfsys@color@cmyk@stroke{0.79}{0.88}{0}{0}\pgfsys@color@cmyk@fill{0.79}{0.88}{0}{0}\simeq}}{\color[rgb]{0.21,0.12,1}\definecolor[named]{pgfstrokecolor}{rgb}{0.21,0.12,1}\pgfsys@color@cmyk@stroke{0.79}{0.88}{0}{0}\pgfsys@color@cmyk@fill{0.79}{0.88}{0}{0}\,\,0}$
(9)
$\displaystyle{\color[rgb]{0,0.88,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.88,0}\pgfsys@color@cmyk@stroke{0.91}{0}{0.88}{0.12}\pgfsys@color<EMAIL_ADDRESS>$\displaystyle{\color[rgb]{0,0.88,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.88,0}\pgfsys@color@cmyk@stroke{0.91}{0}{0.88}{0.12}\pgfsys@color<EMAIL_ADDRESS>s}$ (10)

The symbol $\simeq$ in the last of equality in (9) means an equivalence in the
chiral ring.

Therefore we conclude that the non-zero operator in the chiral ring is
${\color[rgb]{0,0.88,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.88,0}\pgfsys@color@cmyk@stroke{0.91}{0}{0.88}{0.12}\pgfsys@color<EMAIL_ADDRESS>and so it should be this one that enters in the superpotential with $H_{2}$.

###  3.3 Fully deconfined frame

We now go back to our deconfining procedure, and dualize the right $Usp(2)$
node in ${\mathcal{T}}_{2}$ using (3). We reach a frame that we call fully
deconfined as in Benvenuti:2020gvy .

${\mathcal{T}}_{Dec}:$$2$$4$$6$$7$$1$$1$$1$$\vdots$${\mathcal{W}}=m_{1}\text{tr}\,(b_{1}b_{2}q_{3}q_{3}b_{2}b_{1})+m_{2}\text{tr}\,(b_{2}q_{3}q_{3}b_{2})+m_{3}q_{3}q_{3}+\alpha_{1}\,\text{tr}\,(b_{1}b_{1})+\alpha_{2}\,\text{tr}\,(b_{2}b_{2})$$+\,\text{tr}\,(b_{1}a_{2}b_{1})+\text{tr}\,(b_{2}a_{2}b_{2})+r_{1}b_{1}v_{2}+r_{2}b_{2}v_{3}+l_{1}v_{1}b_{1}b_{2}q_{3}+l_{2}v_{2}b_{2}q_{3}+l_{3}v_{3}q_{3}+h_{1}v_{1}b_{1}r_{2}+h_{2}v_{1}r_{1}$$l_{1}$$l_{1}$$\dots$$l_{3}$$l_{2}$$l_{3}$$r_{1}$$r_{2}$$a_{2}$

$5-4R_{A}-7R_{Q}$

$v_{1}$

$-3+2R_{A}+7R_{Q}$

$v_{2}$

$-3+\frac{7}{2}R_{A}+7R_{Q}$

$v_{3}$

$4-5R_{A}-7R_{Q}$

$b_{1},\,$ $\frac{1}{2}R_{A}$ $b_{2},\,$ $\frac{1}{2}R_{A}$ $h_{1},\,$
$3R_{A}$ $h_{2},\,$ $2R_{A}$ $m_{1},\dots,m_{3}$$q_{3},\,$ $R_{A}+R_{Q}$

(11)

The antisymmetric field $a_{2}$ is traceless, as all the antisymmetric field
of $Usp$ that will appear in this paper. Once again there is the question of
the operator flipped by $h_{2}$ because $v_{2}\,r_{2}$ has the same quantum
numbers as $v_{1}\,r_{1}$. Using the same procedure of confining from the
left, we would obtain that the operator $v_{2}r_{2}$ is $0$ on the chiral
ring. Therefore we claim that the correct final superpotential is the one with
this switching procedure and not the one that we would have got using naive
iteration of IP dualities. The final mapping of the chiral ring generators is

$\begin{array}[]{c}{\mathcal{T}}_{1}\\\ M_{1}\\\ Q_{2}\,Q_{2}\\\
\text{tr}\,(Q_{2}\,\Phi\,Q_{2})\\\ Q_{2}\,P_{2}\\\
\text{tr}\,(Q_{2}\,\Phi\,P_{2})\\\ L_{1}\\\ \text{tr}\,(\Phi^{2})\\\
H_{1}\end{array}\Longleftrightarrow\begin{array}[]{c}{\mathcal{T}}_{1^{\prime}}\\\
m_{1}\\\ q_{2}\,q_{2}\\\ \text{tr}\,(q_{2}\,(c_{2}c_{2})\,q_{2})\\\
q_{2}\,c_{2}\,p_{3}\\\ q_{2}\,p_{2}\\\ l_{1}\\\ h_{2}\\\
h_{1}\end{array}\Longleftrightarrow\begin{array}[]{c}{\mathcal{T}}_{2}\\\
M_{1}\\\ M_{2}\\\ Q_{3}\,Q_{3}\\\ Q_{3}\,P_{3}\\\ L_{2}\\\ L_{1}\\\ H_{2}\\\
H_{1}\end{array}\Longleftrightarrow\begin{array}[]{c}{\mathcal{T}}_{Dec}\\\
m_{1}\\\ m_{2}\\\ m_{3}\\\ l_{3}\\\ l_{2}\\\ l_{1}\\\ h_{2}\\\
h_{1}\end{array}$

(12)

Combining the two mappings (5) and (12), we get the mapping between
${\mathcal{T}}_{0}$ and ${\mathcal{T}}_{Dec}$

$\begin{array}[]{c}{\mathcal{T}}_{1}\\\ \text{tr}\,(Q\,A^{i}\,Q)\\\
\text{tr}\,(Q\,A^{j}\,P)\\\
\text{tr}\,(A^{k})\end{array}\quad\Longleftrightarrow\quad\begin{array}[]{c}{\mathcal{T}}_{DEC}\\\
m_{i+1}\\\ l_{3-j}\\\ h_{N+1-k}\\\ \end{array}\qquad\begin{array}[]{l}\\\
i=0,1,2\\\ j=0,1,2\\\ k=2,3\end{array}$

(13)

###  3.4 Self-duality

We already said that this theory is self-dual Csaki:1996eu . Let us see now
how we can use our ${\mathcal{T}}_{Dec}$ frame to prove the self-duality. The
strategy is to _reconfine_ the quiver tail. We notice that the left $Usp(2)$
has $6$ fundamentals, so we start confining from the left. The effect of this
confinement is to kill the antisymmetric field $a_{2}$. In addition, the
fields $\alpha_{1}$, $v_{2}$ and $h_{2}$ get a mass and we produce a Pfaffian
superpotential as in (8). We get

$\mathcal{R}_{1}:$$4$$6$$7$$1$$1$$\vdots$${\mathcal{W}}=m_{1}\text{tr}\,(b_{2}b_{2}b_{2}q_{3}q_{3}b_{2})+m_{2}\text{tr}\,(b_{2}q_{3}q_{3}b_{2})+m_{3}q_{3}q_{3}+\alpha_{2}\,\text{tr}\,(b_{2}b_{2})+r_{2}b_{2}v_{3}$$+\,l_{1}p_{1}b_{2}q_{3}+l_{3}v_{3}q_{3}+h_{1}p_{1}r_{2}+l_{2}p_{1}b_{2}b_{2}b_{2}q_{3}$$l_{1}$$l_{1}$$\dots$$l_{3}$$l_{2}$$l_{3}$$r_{2}$

$5-\frac{11}{2}R_{A}-7R_{Q}$

$p_{1}$

$-3+\frac{5}{2}R_{A}+7R_{Q}$

$v_{3}$

$4-5R_{A}-7R_{Q}$

$b_{2},\,$ $\frac{1}{2}R_{A}$ $h_{1},\,$ $3R_{A}$
$m_{1},\dots,m_{3}$$q_{3},\,$ $R_{A}+R_{Q}$

(14)

Then we can confine the $Usp(4)$ node. We will reach the self-dual frame of
the original theory. Indeed, we produce a traceless antisymmetric field, $B$,
for the $Usp(6)$ and the fields $h_{1}$, $\alpha_{2}$ and $v_{3}$ get a mass.
The final quiver reads

$\mathcal{R}_{2}\equiv\mathcal{R}_{final}:$$6$$7$$1$$\dots$$\dots$$\vdots$${\mathcal{W}}=\,\sum_{j=1}^{3}m_{j}\,\text{tr}\,(q_{3}\,B^{3-j}\,q_{3})$$\,+\sum_{j=1}^{3}\,l_{j}\,\text{tr}\,(p_{2}\,B^{j-1}\,q_{3})$$l_{1}$$l_{1}$$l_{3}$$l_{3}$$m_{1},\dots,m_{3}$$p_{2}$$q_{3}$$B$

(15)

We can repackage the final result into a manifestly $SU(8)$ invariant way

$\mathcal{R}_{final}:$$6$$8$$\vdots$${\mathcal{W}}=\,\sum_{j=1}^{3}\mu_{j}\,\text{tr}\,(\tilde{Q}\,B^{3-j}\,\tilde{Q})$$\mu_{1},\dots,\mu_{3}$$\tilde{Q}$$B$

(16)

Where we define

$\mu_{j}=\begin{pmatrix}\makebox[0.0pt][l]{$\smash{\overbrace{\phantom{\begin{matrix}\phantom{12}m_{j}\phantom{12}\end{matrix}}}^{\text{7}}}$}\phantom{12}m_{j}\phantom{12}&\vdots&\makebox[0.0pt][l]{$\smash{\overbrace{\phantom{\begin{matrix}l_{4-j}\end{matrix}}}^{\text{1}}}$}l_{4-j}\\\
\dotfill&\vdots\dotfill&\dotfill\\\ &\vdots&0\end{pmatrix}\begin{aligned}
&\left.\begin{matrix}\vphantom{O_{N}}\\\ \\\ \end{matrix}\\!\\!\right\\}7\\\
&\left.\begin{matrix}\\\ \end{matrix}\\!\\!\right\\}1\\\
\end{aligned},\qquad\tilde{Q}=\begin{pmatrix}\makebox[0.0pt][l]{$\smash{\overbrace{\phantom{\begin{matrix}\phantom{12}q_{N}\phantom{12}\end{matrix}}}^{\text{7}}}$}\phantom{12}q_{N}\phantom{12}&\vdots&\makebox[0.0pt][l]{$\smash{\overbrace{\phantom{\begin{matrix}p_{3-1}\end{matrix}}}^{\text{1}}}$}p_{3-1}\end{pmatrix}$
(17)

During this sequential reconfinement, from ${\mathcal{T}}_{Dec}$ and
$\mathcal{R}_{final}$, the mapping is

$\begin{array}[]{c}{\mathcal{T}}_{Dec}\\\ m_{i}\\\ l_{i}\\\ h_{2}\\\
h_{1}\end{array}\Longleftrightarrow\begin{array}[]{c}\mathcal{R}_{1}\\\
m_{i}\\\ l_{i}\\\ \text{tr}\,(b_{2}b_{2})^{2}\\\
h_{1}\end{array}\Longleftrightarrow\begin{array}[]{c}\mathcal{R}_{final}\\\
m_{i}\\\ l_{i}\\\ \text{tr}\,(B^{2})\\\ \text{tr}\,(B^{3})\end{array}$

(18)

Comparing with the mapping (13), we read the mapping for the self-duality

$\begin{array}[]{c}{\mathcal{T}}_{1}\\\ \text{tr}\,(Q\,A^{i}\,Q)\\\
\text{tr}\,(Q\,A^{j}\,P)\\\
\text{tr}\,(A^{k})\end{array}\quad\Longleftrightarrow\quad\begin{array}[]{c}\mathcal{R}_{final}\\\
m_{i+1}\\\ l_{3-j}\\\ \text{tr}\,(B^{k})\end{array}\qquad\begin{array}[]{l}\\\
i=0,\dots,2\\\ j=0,\dots,2\\\ k=2,\dots,3\end{array}$ (19)

This is precisely the mapping given in Csaki:1996eu .

Notice that in the reconfinement the precise operator flipped by the singlet
$H_{2}$ was crucial to obtain the duality with the correct amount of gauge
singlets.

In the next section we generalize our discussion to arbitrary $N$ and $F$.

##  4 Sequential deconfinement of $Usp(2N)$ with ${\tiny\ydiagram{1,1}}$ \+
$2F\,{\tiny\ydiagram{1}}$

In this section we study the general case. The $Usp(2N)$ gauge theory with a
traceless antisymmetric field $A$ and $2F$ complex chiral fields (the number
of fundamentals should be even to avoid the global anomaly).

We proceed as in Benvenuti:2020gvy  to derive a chain of $2N$ dual frames
consisting of quiver theories, with number of gauge nodes ranging from $1$ to
$N$. Let us start with the first $\mathcal{T}_{0}$ quiver

${\mathcal{T}}_{0}:$$2N$$2F$${\mathcal{W}}=0$$A$$\equiv$$2N$$2F-1$$1$$Q$$P$${\mathcal{W}}=0$$A$

(20)

###  4.1 Deconfine and dualize: first step

We start by using the deconfinement (1), we obtain

${\mathcal{T}}_{0^{\prime}}:$$2N$$2N-2$$2F-1$$1$$1$$p_{1}$$p_{2}$$h_{1}$${\mathcal{W}}=p_{1}c_{1}d_{1}+h_{1}d_{1}p_{2}+\beta_{1}\,\text{tr}\,(c_{1}c_{1})$$c_{1}$$d_{1}$$q_{1}$

(21)

Then we dualize the $Usp(2N)$ node with (3). This step is the same as in (4).

${\mathcal{T}}_{1}:$$2F-6$$2N-2$$2F-1$$1$$1$${\mathcal{W}}=M_{1}Q_{1}Q_{1}+Q_{1}C_{1}Q_{2}+L_{1}Q_{1}V_{1}$$+\,\text{tr}\,(C_{1}\Phi
C_{1})+H_{1}V_{1}C_{1}P_{2}$$L_{1}$$L_{1}$$Q_{1}$$Q_{2}$$V_{1}$$P_{2}$$C_{1}$$H_{1}$$M_{1}$$\Phi$

(22)

The mapping of the chiral ring generators after this first step is the
following

$\begin{array}[]{c}{\mathcal{T}}_{0}\\\ QQ\\\ \text{tr}\,(QA^{i}Q)\\\
\text{tr}\,(QA^{j}P)\\\ \text{tr}\,(QA^{N-1}P)\\\ \text{tr}\,(A^{m})\\\
\text{tr}\,(A^{N})\end{array}\Longleftrightarrow\begin{array}[]{c}{\mathcal{T}}_{0^{\prime}}\\\
q_{1}q_{1}\\\ \text{tr}\,(q_{1}\,(c_{1}c_{1})^{i}\,q_{1})\\\
\text{tr}\,(q_{1}\,c_{1}(c_{1}c_{1})^{j}\,p_{2})\\\ q_{1}\,p_{1}\\\
\text{tr}\,((c_{1}c_{1})^{m})\\\
h_{1}\end{array}\Longleftrightarrow\begin{array}[]{c}{\mathcal{T}}_{1}\\\
M_{1}\\\ \text{tr}\,(Q_{2}\,\Phi^{i-1}\,Q_{2})\\\
\text{tr}\,(Q_{2}\,\Phi^{j}\,P_{2})\\\ L_{1}\\\ \text{tr}\,(\Phi^{m})\\\
H_{1}\end{array}\qquad\begin{array}[]{l}\\\ \\\ i=1,\ldots,N-1\\\
j=0,\ldots,N-2\\\ \\\ m=2,\ldots,N-1\\\ \\\ \end{array}$

(23)

Now we iterate the procedure. We deconfine the traceless antisymmetric field,
$\Phi$ and then we dualize. Let us write explicitly another step and then it
will be enough to obtain the general story.

###  4.2 Second step

After the deconfinement we get

${\mathcal{T}}_{1^{\prime}}:$$2F-6$$2N-2$$2N-4$$2F-1$$1$$1$$1$${\mathcal{W}}=m_{1}q_{1}q_{1}+q_{1}c_{1}q_{2}+l_{1}v_{1}q_{1}$$+\,\text{tr}\,(c_{1}c_{2}c_{2}c_{1})+h_{1}v_{1}c_{1}c_{2}p_{3}$$+\,h_{2}d_{2}p_{3}+p_{2}c_{2}d_{2}+\beta_{2}\,\text{tr}\,(c_{2}c_{2})$$l_{1}$$l_{1}$$d_{2}$$q_{1}$$q_{2}$$v_{1}$$p_{2}$$p_{3}$$c_{1}$$c_{2}$$h_{1}$$h_{2}$$m_{1}$

(24)

Now we dualize the $Usp(2N-2)$ node. The fields $q_{1}$, $d_{2}$ and
$\beta_{2}$ get a mass. In addition $\text{tr}\,(c_{1}c_{2}c_{2}c_{1})$
becomes a mass term therefore there will be no link between the $Usp(2F-6)$
and $Usp(2N-4)$ gauge group. After the integration of these massive fields and
a rescaling we get

${\mathcal{T}}_{2}:$$2F-6$$2(2F-6)$$2N-4$$2F-1$$1$$1$$1$${\mathcal{W}}=M_{1}\text{tr}\,(B_{1}Q_{2}Q_{2}B_{1})+M_{2}Q_{2}Q_{2}+\text{tr}\,(C_{2}\Phi
C_{2})+\text{tr}\,(B_{1}A_{1}B_{1})+\alpha_{1}\text{tr}\,(B_{1}B_{1})$$+\,\text{tr}\,(B_{1}C_{2}C_{2}B_{1})+R_{1}B_{1}V_{2}+L_{1}V_{1}B_{1}Q_{2}+L_{2}V_{2}Q_{2}+Q_{2}C_{2}Q_{3}+H_{1}V_{1}B_{1}C_{2}P_{3}+H_{2}V_{2}C_{2}P_{3}$$L_{1}$$L_{1}$$L_{2}$$L_{2}$$R_{1}$$V_{1}$$V_{2}$$P_{3}$$B_{1}$$C_{2}$$H_{1}$$H_{2}$$M_{1},M_{2}$$Q_{2}$$Q_{3}$$\Phi$$A_{1}$

(25)

We recall that the antisymmetric field $A_{1}$ is traceles, as well as $\Phi$.

The mapping after this second step is given by888Using the table of R-charges
of the fields given in appendix A in Table 1, it is easy to check that the
mapping is consitent.

$\begin{array}[]{c}{\mathcal{T}}_{1}\\\ M_{1}\\\ Q_{2}\,Q_{2}\\\
\text{tr}\,(Q_{2}\,\Phi^{i-1}\,Q_{2})\\\
\text{tr}\,(Q_{2}\,\Phi^{j}\,P_{2})\\\
\text{tr}\,(Q_{2}\,\Phi^{N-2}\,P_{2})\\\ L_{1}\\\ \text{tr}\,(\Phi^{m})\\\
\text{tr}\,(\Phi^{N-1})\\\ H_{1}\\\
\end{array}\Longleftrightarrow\begin{array}[]{c}{\mathcal{T}}_{1^{\prime}}\\\
m_{1}\\\ q_{2}\,q_{2}\\\ \text{tr}\,(q_{2}\,(c_{2}c_{2})^{i-1}\,q_{2})\\\
\text{tr}\,(q_{2}\,c_{2}(c_{2}c_{2})^{j}\,p_{3})\\\ q_{2}\,p_{2}\\\ l_{1}\\\
\text{tr}\,((b_{2}b_{2})^{m})\\\ h_{2}\\\ h_{1}\\\
\end{array}\Longleftrightarrow\begin{array}[]{c}{\mathcal{T}}_{2}\\\ M_{1}\\\
M_{2}\\\ \text{tr}\,(Q_{3}\,\Phi^{i-2}\,Q_{3})\\\
\text{tr}\,(Q_{3}\,\Phi^{j}\,P_{3})\\\ L_{2}\\\ L_{1}\\\
\text{tr}\,(\Phi^{m})\\\ H_{2}\\\ H_{1}\\\
\end{array}\qquad\begin{array}[]{l}\\\ \\\ i=2,\dots,N-1\\\ j=0,\ldots,N-3\\\
\\\ \\\ m=2,\ldots,N-2\\\ \\\ \end{array}$

(26)

###  4.3 After k steps

After the iteration of $k$ steps, we get the following quiver

${\mathcal{T}}_{k}:$$2F-6$$2(2F-6)$$\dots$$k(2F-6)$$2N-2k$$2F-1$$1$$1$$1$$\dots$$1$$1$${\mathcal{W}}=\,\sum_{i=1}^{k}M_{i}\,\text{tr}\,(B_{i}B_{i+1}\dots
B_{k-1}\,Q_{k}\,Q_{k}\,B_{k-1}\dots
B_{i+1}B_{i})+\text{tr}\,(C_{k}\,\Phi\,C_{k})+\sum_{i=1}^{k-1}\alpha_{i}\text{tr}\,(B_{i}B_{i})$$+\,\sum_{i=1}^{k-1}\text{tr}\,(B_{i}\,A_{i}\,B_{i})+\sum_{i=1}^{k-2}\text{tr}\,(B_{i}\,A_{i+1}\,B_{i})+\text{tr}\,(B_{k-1}C_{k}C_{k}B_{k-1})+\sum_{i=1}^{k-1}R_{i}B_{i}V_{i+1}$$+\,\sum_{i=1}^{k}L_{i}V_{i}B_{i}B_{i+1}\dots
B_{k-1}Q_{k}+Q_{k}C_{k}Q_{k+1}+\sum_{i=1}^{k}H_{i}V_{i}B_{i}B_{i+1}\dots
B_{k-1}C_{k}P_{k+1}$$\dots$$\vdots$$L_{1}$$L_{1}$$L_{2}$$L_{k}$$L_{k}$$V_{1}$$V_{2}$$V_{k}$$P_{k+1}$$R_{1}$$R_{2}$$R_{k-1}$$B_{1}$$B_{2}$$B_{k-1}$$C_{k}$$H_{1}$$H_{2}$$H_{3}$$H_{k}$$M_{1},\dots,M_{k}$$Q_{k}$$Q_{k+1}$$A_{1}$$A_{2}$$\Phi$

(27)

The last term in the superpotential should be taken with a grain of salt.
Indeed as explained in the appendix A, when $k$ is great enough some operators
become degenerate with $V_{i}B_{i}B_{i+1}\dots B_{k-1}C_{k}P_{k+1}$ and then
the superpotential should be modified. This is the _degenerate holomorphic
operator ambiguity_ that we described in (7). Since it is a $k$-dependent
statement, we decided to be cavalier when writing this term in (27) and write
the modified version in ${\mathcal{T}}_{N-1}$.

The mapping of the chiral ring generators is the following

$\begin{array}[]{c}{\mathcal{T}}_{(k-1)^{\prime}}\\\ m_{1,\dots,k-1}\\\
l_{1,\dots,k-1}\\\ h_{1,\dots,k}\\\ q_{k}\,q_{k}\\\
\text{tr}\,(q_{k}\,(c_{k}\,c_{k})^{i-k+1}\,q_{k})\\\
\text{tr}\,(q_{k}\,c_{k}\,(c_{k}\,c_{k})^{j}\,p_{k+1})\\\ q_{k}\,p_{k}\\\
\text{tr}\,((c_{k}c_{k})^{m})\end{array}\quad\Longleftrightarrow\quad\begin{array}[]{c}{\mathcal{T}}_{k}\\\
M_{1,\dots,k-1}\\\ L_{1,\dots,k-1}\\\ H_{1,\dots,k}\\\ M_{k}\\\
\text{tr}\,(Q_{k+1}\,\Phi^{i-k}\,Q_{k+1})\\\
\text{tr}\,(Q_{k+1}\,\Phi^{j}\,P_{k+1})\\\ L_{k}\\\
\text{tr}\,(\Phi^{m})\end{array}\qquad\begin{array}[]{l}\\\ \\\ \\\ \\\ \\\
i=k,\dots,N-1\\\ j=0,\dots,N-k-1\\\ \\\ m=2,\dots,N-k\end{array}$

(28)

###  4.4 After N-1 steps

${\mathcal{T}}_{N-1}:$$2F-6$$2(2F-6)$$\dots$

$(N-1)(2F-6)$

$2$$2F-1$$1$$1$$1$$\dots$$1$$1$${\mathcal{W}}=\,\sum_{i=1}^{N-1}M_{i}\,\text{tr}\,(B_{i}B_{i+1}\dots
B_{N-2}\,Q_{N-1}\,Q_{N-1}\,B_{N-2}\dots
B_{i+1}B_{i})+\sum_{i=1}^{N-2}\alpha_{i}\text{tr}\,(B_{i}B_{i})$$+\,\sum_{i=1}^{N-2}\text{tr}\,(B_{i}\,A_{i}\,B_{i})+\sum_{i=1}^{N-3}\text{tr}\,(B_{i}\,A_{i+1}\,B_{i})+\text{tr}\,(B_{N-2}C_{N-1}C_{N-1}B_{N-2})$$+\,\sum_{i=1}^{N-2}R_{i}B_{i}V_{i+1}+\,\sum_{i=1}^{N-1}L_{i}V_{i}B_{i}B_{i+1}\dots
B_{N-2}Q_{N-1}+Q_{N-1}C_{N-1}Q_{N}$$+\,H_{1}V_{1}B_{1}\dots
B_{N-2}C_{N-1}P_{N}+\sum_{i=2}^{N-1}H_{i}V_{1}B_{1}\dots
B_{N-1-i}R_{N-i}$$\dots$$\vdots$$L_{1}$$L_{1}$$L_{2}$$L_{N-1}$$L_{N-1}$$V_{1}$$V_{2}$$V_{N-1}$$P_{N}$$R_{1}$$R_{2}$$R_{N-2}$$B_{1}$$B_{2}$$B_{N-2}$$C_{N-1}$$H_{N-1}$$H_{N-2}$$H_{2}$$H_{1}$$M_{1},\dots,M_{N-1}$$Q_{N-1}$$Q_{N}$$A_{1}$$A_{2}$

(29)

The last term in the superpotential is the one after switching, see Appendix
A. Since the last node is $Usp(2)$, there is no antisymmetric and we can
directly use (3).

###  4.5 Final step and fully deconfined frame

${\mathcal{T}}_{Dec}:$$2F-6$

$2(2F-6)$

$\dots$

$(N-1)(2F-6)$

$N(2F-6)$

$2F-1$$1$$1$$1$$\dots$$1$$1$${\mathcal{W}}=\,\sum_{i=1}^{N}m_{i}\,\text{tr}\,(b_{i}b_{i+1}\dots
b_{N-1}\,q_{N}\,q_{N}\,b_{N-1}\dots
b_{i}b_{i+1})+\sum_{i=1}^{N-1}\alpha_{i}\text{tr}\,(b_{i}b_{i})$$+\,\sum_{i=1}^{N-1}\text{tr}\,(b_{i}\,a_{i}\,b_{i})+\sum_{i=1}^{N-2}\text{tr}\,(b_{i}\,a_{i+1}\,b_{i})+\sum_{i=1}^{N-1}r_{i}b_{i}v_{i+1}$$+\,\sum_{i=1}^{N}l_{i}v_{i}b_{i}b_{i+1}\dots
b_{N-1}q_{N}+\sum_{i=1}^{N-1}h_{i}v_{1}b_{1}\dots
b_{N-1-i}r_{N-i}$$\dots$$\vdots$$l_{1}$$l_{1}$$l_{2}$$l_{N-1}$$l_{N}$$l_{N}$$v_{1}$$v_{2}$$v_{N-1}$$v_{N}$$r_{1}$$r_{2}$$r_{N-2}$$r_{N-1}$$b_{1}$$b_{2}$$b_{N-2}$$b_{N-1}$$h_{N-1}$$h_{N-2}$$h_{2}$$h_{1}$$m_{1},\dots,m_{N}$$q_{N}$$a_{1}$$a_{2}$$a_{N-1}$

(30)

We have obtained the fully deconfined frame. The mapping is

$\begin{array}[]{c}{\mathcal{T}}_{N-1}\\\ M_{1,\dots,N-1}\\\
L_{1,\dots,N-1}\\\ H_{1,\dots,N-1}\\\ Q_{N}\,Q_{N}\\\
Q_{N}\,P_{N}\end{array}\quad\Longleftrightarrow\quad\begin{array}[]{c}{\mathcal{T}}_{DEC}\\\
m_{1,\dots,N-1}\\\ l_{1,\dots,N-1}\\\ h_{1,\dots,N-1}\\\ m_{N}\\\
l_{N}\end{array}$

(31)

Collecting all the mappings, we write the mapping from the ${\mathcal{T}}_{0}$
frame to ${\mathcal{T}}_{Dec}$999The mapping is consitent with the R-charges
given in Table 2.

$\begin{array}[]{c}{\mathcal{T}}_{0}\\\ \text{tr}\,(Q\,A^{i}\,Q)\\\
\text{tr}\,(Q\,A^{j}\,P)\\\
\text{tr}\,(A^{k})\end{array}\quad\Longleftrightarrow\quad\begin{array}[]{c}{\mathcal{T}}_{DEC}\\\
m_{i+1}\\\ l_{N-j}\\\ h_{N+1-k}\end{array}\qquad\begin{array}[]{l}\\\
i=0,\dots,N-1\\\ j=0,\dots,N-1\\\ k=2,\dots,N\end{array}$

(32)

One interesting property of the deconfined frame is that all chiral ring
generators are elementary gauge singlets.

##  5 Reconfinement and self-duality for $Usp(2N)$ with
${\tiny\ydiagram{1,1}}$ \+ $8\,{\tiny\ydiagram{1}}$

It was proposed in Csaki:1996eu  that in the special case of $F=4$ the theory
studied in the previous section, $Usp(2N)$ with antisymmetric and $2F$
fundamentals, is self-dual modulo flips. In this section we use our
${\mathcal{T}}_{Dec}$ frame (30) to prove this result. Let us rewrite it,
specifying $F=4$:

$\mathcal{R}_{0}:$$2$$4$$\dots$

$2(N-1)$

$2N$$7\,\,\,\,\,\,$$1$$1$$1$$\dots$$1$$1$${\mathcal{W}}=\,\sum_{i=1}^{N}m_{i}\,\text{tr}\,(b_{i}b_{i+1}\dots
b_{N-1}\,q_{N}\,q_{N}\,b_{N-1}\dots
b_{i+1}b_{i})+\sum_{i=1}^{N-1}\alpha_{i}\text{tr}\,(b_{i}b_{i})$$+\,\sum_{i=2}^{N-1}\text{tr}\,(b_{i}\,a_{i}\,b_{i})+\sum_{i=1}^{N-2}\text{tr}\,(b_{i}\,a_{i+1}\,b_{i})+\sum_{i=1}^{N-1}r_{i}b_{i}v_{i+1}$$+\,\sum_{i=1}^{N}l_{i}v_{i}b_{i}b_{i+1}\dots
b_{N-1}q_{N}+\sum_{i=1}^{N-1}h_{i}v_{1}b_{1}\dots
b_{N-1-i}r_{N-i}$$\dots$$\vdots$$l_{1}$$l_{1}$$l_{2}$$l_{N-1}$$l_{N}$$l_{N}$$v_{1}$$v_{2}$$v_{N-1}$$v_{N}$$r_{1}$$r_{2}$$r_{N-2}$$r_{N-1}$$b_{1}$$b_{2}$$b_{N-2}$$b_{N-1}$$h_{N-1}$$h_{N-2}$$h_{2}$$h_{1}$$m_{1},\dots,m_{N}$$q_{N}$$a_{2}$$a_{N-1}$

(33)

$\begin{array}[]{c}{\mathcal{T}}_{0}\\\ \text{tr}\,(Q\,A^{j}\,Q)\\\
\text{tr}\,(Q\,A^{j}\,P)\\\
\text{tr}\,(A^{i})\end{array}\quad\Longleftrightarrow\quad\begin{array}[]{c}R_{0}\\\
m_{j+1}\\\ l_{N-j}\\\ h_{N+1-i}\end{array}\qquad\begin{array}[]{l}\\\
j=0,\dots,N-1\\\ j=0,\dots,N-1\\\ i=2,\dots,N\end{array}$

(34)

We see that the $Usp(2)$ gauge group is coupled to $4+1+1$ chiral fields in
the fundamental representation and so it confines (2). This step is similar as
in (14). The confinement will give a mass to the traceless antisymmetric field
$a_{2}$ as well as $v_{2},\,\alpha_{1}$ and $h_{N-1}$. After integrating them
out and computing the Pfaffian superpotential (See discussion below (8)) we
get

$\mathcal{R}_{1}:$$4$$\dots$

$2(N-1)$

$2N$$7\,\,\,\,\,\,$$1$$1$$\dots$$1$$1$${\mathcal{W}}=\,m_{1}\,\text{tr}\,(b_{2}b_{2}b_{2}\dots
b_{N-1}\,q_{N}\,q_{N}\,b_{N-1}\dots
b_{2})+\sum_{i=2}^{N}m_{i}\,\text{tr}\,(b_{i}b_{i+1}\dots
b_{N-1}\,q_{N}\,q_{N}\,b_{N-1}\dots
b_{i+1}b_{i})$$+\,\sum_{i=2}^{N-1}\alpha_{i}\text{tr}\,(b_{i}b_{i})+\sum_{i=3}^{N-1}\text{tr}\,(b_{i}\,a_{i}\,b_{i})+\sum_{i=2}^{N-2}\text{tr}\,(b_{i}\,a_{i+1}\,b_{i})+\sum_{i=2}^{N-1}r_{i}b_{i}v_{i+1}$$+\,l_{1}p_{1}b_{2}\dots
b_{N-1}q_{N}+l_{2}p_{1}b_{2}b_{2}b_{2}\dots
b_{N-1}q_{N}+\sum_{i=3}^{N}l_{i}v_{i}b_{i}b_{i+1}\dots
b_{N-1}q_{N}+\sum_{i=1}^{N-2}h_{i}p_{1}b_{2}\dots
b_{N-1-i}r_{N-i}$$\dots$$\vdots$$l_{1}$$l_{1}$$l_{2}$$l_{3}$$l_{N-1}$$l_{N}$$l_{N}$$v_{N-1}$$v_{N}$$r_{2}$$r_{N-2}$$r_{N-1}$$b_{2}$$b_{N-2}$$b_{N-1}$$h_{N-2}$$h_{2}$$h_{1}$$m_{1},\dots,m_{N}$$p_{1}$$q_{N}$$a_{N-1}$

(35)

The mapping of the chiral ring generators is

$\begin{array}[]{c}R_{0}\\\ m_{j},\,l_{j}\\\ h_{i}\\\
h_{N-1}\end{array}\quad\Longleftrightarrow\quad\begin{array}[]{c}R_{1}\\\
m_{j},\,l_{j}\\\ h_{i}\\\
\text{tr}\,(b_{2}b_{2})^{2}\end{array}\quad\begin{array}[]{c}\text{it will
stay like that}\\\ i=1,\dots,N-2\\\ \end{array}$

(36)

We can now iterate. Indeed the $Usp(4)$ group is coupled to $6+1+1$
fundamentals and so it also confines

$\mathcal{R}_{2}:$$6$$\dots$

$2(N-1)$

$2N$$7\,\,\,\,\,\,$$1$$1$$\dots$$1$$1$

${\mathcal{W}}=\,m_{1}\,\text{tr}\,([b_{3}b_{3}]^{3}b_{4}\dots
b_{N-1}\,q_{N}\,q_{N}\,b_{N-1}\dots
b_{4})+m_{2}\,\text{tr}\,([b_{3}b_{3}]^{2}b_{4}\dots
b_{N-1}\,q_{N}\,q_{N}\,b_{N-1}\dots b_{4})$

$+\,\sum_{i=3}^{N}m_{i}\,\text{tr}\,(b_{i}b_{i+1}\dots
b_{N-1}\,q_{N}\,q_{N}\,b_{N-1}\dots
b_{i+1}b_{i})+\sum_{i=3}^{N-1}\alpha_{i}\text{tr}\,(b_{i}b_{i})+\sum_{i=4}^{N-1}\text{tr}\,(b_{i}\,a_{i}\,b_{i})$$+\,\sum_{i=3}^{N-2}\text{tr}\,(b_{i}\,a_{i+1}\,b_{i})+\sum_{i=3}^{N-1}r_{i}b_{i}v_{i+1}+l_{1}p_{2}b_{3}\dots
b_{N-1}q_{N}+l_{2}p_{2}[b_{3}b_{3}]b_{3}\dots
b_{N-1}q_{N}$$+\,l_{3}p_{2}[b_{3}b_{3}]^{2}b_{3}\dots
b_{N-1}q_{N}+\sum_{i=4}^{N}l_{i}v_{i}b_{i}b_{i+1}\dots
b_{N-1}q_{N}+\sum_{i=1}^{N-3}h_{i}p_{2}b_{3}\dots
b_{N-1-i}r_{N-i}$$\vdots$$\dots$$\vdots$$l_{1}$$l_{1}$$l_{3}$$l_{4}$$l_{N-1}$$l_{N}$$l_{N}$$v_{N-1}$$v_{N}$$r_{3}$$r_{N-2}$$r_{N-1}$$b_{3}$$b_{N-2}$$b_{N-1}$$h_{N-3}$$h_{2}$$h_{1}$$m_{1},\dots,m_{N}$$p_{2}$$q_{N}$$a_{N-1}$

(37)

We can iterate. After $k$ confinement, we get

$\mathcal{R}_{k}:$

$2k+2$

$\dots$

$2(N-1)$

$2N$$7\,\,\,\,\,\,$$1$$1$$\dots$$1$$1$${\mathcal{W}}=\,\sum_{j=1}^{k}m_{j}\,\text{tr}\,([b_{k+1}b_{k+1}]^{k+2-j}\,b_{k+2}\dots
b_{N-1}\,q_{N}\,q_{N}\,b_{N-1}\dots
b_{k+2})$$+\,\sum_{i=k+1}^{N}m_{i}\,\text{tr}\,(b_{i}b_{i+1}\dots
b_{N-1}\,q_{N}\,q_{N}\,b_{N-1}\dots
b_{i+1}b_{i})+\sum_{i=k+1}^{N-1}\alpha_{i}\text{tr}\,(b_{i}b_{i})$$+\,\sum_{i=k+2}^{N-1}\text{tr}\,(b_{i}\,a_{i}\,b_{i})+\sum_{i=k+1}^{N-2}\text{tr}\,(b_{i}\,a_{i+1}\,b_{i})+\sum_{i=k+1}^{N-1}r_{i}b_{i}v_{i+1}$$+\,\sum_{j=1}^{k+1}l_{j}\,p_{k}[b_{k+1}b_{k+1}]^{j-1}\,b_{k+1}\dots
b_{N-1}q_{N}+\sum_{i=k+2}^{N}l_{i}v_{i}b_{i}b_{i+1}\dots
b_{N-1}q_{N}$$+\,\sum_{i=1}^{N-k-1}h_{i}p_{k}b_{k+1}\dots
b_{N-1-i}r_{N-i}$$\vdots$$\dots$$\vdots$$l_{1}$$l_{1}$$l_{k+1}$$l_{k+2}$$l_{N-1}$$l_{N}$$l_{N}$$v_{N-1}$$v_{N}$$r_{k+1}$$r_{N-2}$$r_{N-1}$$b_{k+1}$$b_{N-2}$$b_{N-1}$$h_{N-k-1}$$h_{2}$$h_{1}$$m_{1},\dots,m_{N}$$p_{k}$$q_{N}$$a_{N-1}$

(38)

The mapping between two successive reconfinement is

$\begin{array}[]{c}R_{k-1}\\\ h_{i}\\\ h_{N-k}\\\
\text{tr}\,(b_{k}b_{k})^{k}\\\ \vdots\\\
\text{tr}\,(b_{k}b_{k})^{2}\end{array}\quad\Longleftrightarrow\quad\begin{array}[]{c}R_{k}\\\
h_{i}\\\ \text{tr}\,(b_{k+1}b_{k+1})^{k+1}\\\
\text{tr}\,(b_{k+1}b_{k+1})^{k}\\\ \vdots\\\
\text{tr}\,(b_{k+1}b_{k+1})^{2}\end{array}\qquad\begin{array}[]{c}i=1,\dots,N-k-1\\\
\\\ \\\ \\\ \end{array}$

(39)

For $k=N-2$, there won’t be any antisymmetric field left. We get

$\mathcal{R}_{N-2}:$

$2N-2$

$2N$$7$$1$$1$${\mathcal{W}}=\,\sum_{j=1}^{N-2}m_{j}\,\text{tr}\,([b_{N-1}b_{N-1}]^{N-j}\,q_{N}\,q_{N})+m_{N-1}\,\text{tr}\,(b_{N-1}\,q_{N}\,q_{N}\,b_{N-1})+m_{N}q_{N}q_{N}$$+\,\alpha_{N-1}\text{tr}\,(b_{N-1}b_{N-1})+r_{N-1}b_{N-1}v_{N}+\sum_{j=1}^{N-1}\,l_{j}\,p_{N-2}\,[b_{N-1}b_{N-1}]^{j-1}\,b_{N-1}\,q_{N}+l_{N}v_{N}q_{N}+h_{1}p_{N-2}r_{N-1}$$\dots$$\dots$$\vdots$$l_{1}$$l_{1}$$l_{N-1}$$l_{N}$$l_{N}$$v_{N}$$r_{N-1}$$b_{N-1}$$h_{1}$$m_{1},\dots,m_{N}$$p_{N-2}$$q_{N}$

(40)

Now we can do the last confinement with the $Usp(2N-2)$ group. It will produce
the traceless antisymmetric field, $B$ for $Usp(2N)$ (the trace part is killed
by the flipper $\alpha_{N-1}$). We get

$\mathcal{R}_{N-1}:$$2N$$7$$1$${\mathcal{W}}=\,\sum_{j=1}^{N}m_{j}\,\text{tr}\,(q_{N}\,B^{N-j}\,q_{N})$$\,+\sum_{j=1}^{N}\,l_{j}\,\text{tr}\,(p_{N-1}\,B^{j-1}\,q_{N})$$\dots$$\dots$$\vdots$$l_{1}$$l_{1}$$l_{N}$$l_{N}$$m_{1},\dots,m_{N}$$p_{N-1}$$q_{N}$$B$

(41)

The mapping for this last reconfinement is given by

$\begin{array}[]{c}R_{N-2}\\\ h_{1}\\\ \text{tr}\,(b_{N-1}b_{N-1})^{N-1}\\\
\text{tr}\,(b_{N-1}b_{N-1})^{N-2}\\\ \vdots\\\
\text{tr}\,(b_{N-1}b_{N-1})^{2}\end{array}\quad\Longleftrightarrow\quad\begin{array}[]{c}R_{N-1}\\\
\text{tr}\,(B^{N})\\\ \text{tr}\,(B^{N-1})\\\ \text{tr}\,(B^{N-2})\\\
\vdots\\\ \text{tr}\,(B^{2})\end{array}$ (42)

We can repackage the last frame into a manifestly $SU(8)$ invariant way to
obtain the final frame

$\mathcal{R}_{final}:$$2N$$8$${\mathcal{W}}=\,\sum_{j=1}^{N}\mu_{j}\,\text{tr}\,(\tilde{Q}\,B^{N-j}\,\tilde{Q})$$\vdots$$\mu_{1},\dots,\mu_{N}$$\tilde{Q}$$B$

(43)

Where we define

$\mu_{j}=\begin{pmatrix}\makebox[0.0pt][l]{$\smash{\overbrace{\phantom{\begin{matrix}\phantom{12}m_{j}\phantom{12}\end{matrix}}}^{\text{7}}}$}\phantom{12}m_{j}\phantom{12}&\vdots&\makebox[0.0pt][l]{$\smash{\overbrace{\phantom{\begin{matrix}l_{N+1-j}\end{matrix}}}^{\text{1}}}$}l_{N+1-j}\\\
\dotfill&\vdots\dotfill&\dotfill\\\ &\vdots&0\end{pmatrix}\begin{aligned}
&\left.\begin{matrix}\vphantom{O_{N}}\\\ \\\ \end{matrix}\right\\}7\\\
&\left.\begin{matrix}\\\ \end{matrix}\right\\}1\\\
\end{aligned},\qquad\tilde{Q}=\begin{pmatrix}\makebox[0.0pt][l]{$\smash{\overbrace{\phantom{\begin{matrix}\phantom{12}q_{N}\phantom{12}\end{matrix}}}^{\text{7}}}$}\phantom{12}q_{N}\phantom{12}&\vdots&\makebox[0.0pt][l]{$\smash{\overbrace{\phantom{\begin{matrix}p_{N-1}\end{matrix}}}^{\text{1}}}$}p_{N-1}\end{pmatrix}$
(44)

Now combining the mappings, we see that the reconfinement procedure gives

$\begin{array}[]{c}R_{0}\\\
h_{i}\end{array}\quad\Longleftrightarrow\quad\begin{array}[]{c}R_{final}\\\
\text{tr}\,(B^{N+1-i})\end{array}\qquad\begin{array}[]{c}\\\
i=1,\dots,N-1\end{array}$ (45)

Now if we compare the original frame ${\mathcal{T}}_{0}$ and the last frame
after reconfinement $\mathcal{R}_{final}$ we see the self-duality and we
obtain the following mapping

$\begin{array}[]{c}T_{0}\\\ \text{tr}\,(Q\,A^{j}\,Q)\\\
\text{tr}\,(Q\,A^{j}\,P)\\\
\text{tr}\,(A^{i})\end{array}\quad\Longleftrightarrow\quad\begin{array}[]{c}R_{final}\\\
m_{j+1}\\\ l_{N-j}\\\ \text{tr}\,(B^{i})\end{array}\qquad\begin{array}[]{c}\\\
j=0,\dots,N-1\\\ j=0,\dots,N-1\\\ i=2,\dots,N\end{array}$ (46)

Which is precisely the mapping proposed in Csaki:1996eu .

##  6 Reduction to $3d$ $\mathcal{N}=2$ sequential deconfinement

It is possible to reduce $4d$ $\mathcal{N}=1$ theories on a circle, obtaining
$3d$ $\mathcal{N}=2$ theories. Generically, this steps introduces a
superpotential term linear in the basic monopole operator (exceptions are, for
instance, theories with $8$ supercharges). Once in $3d$, it is possible to
turn on real mass deformations, that do not exist in $4d$. Starting from a
$4d$ $Usp(2N)$ gauge theory, $3d$ real masses allow to flow to $Usp(2N)$ or
$U(N)$ gauge groups with or without various types of monopole superpotentials.
This process has been discussed in detail in the case without rank-$2$ matter
Benini:2017dud , and for the case of $2F=8$ Amariti:2018wht ;
Benvenuti:2018bav . A brane interpretation has been found in Amariti:2017gsm .
Examples of $4d$ $\mathcal{N}=1$ simplectic quivers reduced and deformed to
$3d$ $\mathcal{N}=2$ or $3d$ $\mathcal{N}=4$ unitary quivers have been
discussed in Pasquetti:2019tix ; Pasquetti:2019hxf , mostly from the
superconformal index perspective.

On the electric side, the story is as follows:

$Usp(2N)$$2F+2$${\mathcal{W}}_{4d}=0$reduction on
$S^{1}$$Usp(2N)$$2F+2$${\mathcal{W}}_{3d}={\mathfrak{M}}$real masses
$(+^{F+1},-^{F+1})$real masses
$(0^{2F},+,-)$$Usp(2N)$$2F$${\mathcal{W}}=0$$U(N)$$(F+1,F+1)$${\mathcal{W}}={\mathfrak{M}}^{+}+{\mathfrak{M}}^{-}$real
masses $(0^{F},+;0^{F},-)$$U(N)$$(F,F)$${\mathcal{W}}={\mathfrak{M}}^{+}$real
masses $(0^{F-1},+;0^{F-1},-)$$U(N)$$(F-1,F-1)$${\mathcal{W}}=0$

(47)

Where the rank-$2$ field is a traceless antisymmetric for $Usp(2N)$ and a
traceless adjoint for $U(N)$. The monopoles ${\mathfrak{M}}$,
${\mathfrak{M}}^{\pm}$ are the monopoles with minimal GNO charges. See
Benini:2017dud  for more details.

We could also turn on different real masses, possibly leading to non-zero
Chern-Simons terms as in Amariti:2018wht ; Benvenuti:2018bav ;
Benvenuti:2020gvy , but we refrain to do this in the present paper.

In the remaining of this section we perform the reduction and deformation of
the fully deconfined dual, recovering the results found in Benvenuti:2020gvy
working in $3d$.

#### Reduction to the deconfined dual of $Usp(2N)$ with antisymmetric and
$2F+2$ fundamentals, $\mathcal{W}={\mathfrak{M}}$

We put the $4d$ duality on a circle. On the electric side we get $3d$
$\mathcal{N}=2$ $Usp(2N)$ with antisymmetric and $(2F+1)_{Q}+1_{P}$
fundamentals, $\mathcal{W}={\mathfrak{M}}$, with global symmetry
$SU(2F+2)\times U(1)$. On the magnetic side we obtain the same quiver as in
$4d$, (30), with $F\rightarrow F+1$. The difference is that the
superpontential now includes $N$ additional terms, linear in the monopole
operators with GNO charges for a single gauge group,
$\sum_{i=1}^{N}{\mathfrak{M}}^{0^{i-1},\bullet,0^{N-i}}$:

$Usp(2F\\!-\\!4)$

$Usp(2(2F\\!-\\!4))$

$\dots$

$Usp((N\\!-\\!1)(2F\\!-\\!4))$

$Usp(N(2F\\!-\\!4))$

$2F+1$$1$$1$$1$$\dots$$1$$1$$\dots$$\vdots$$a_{1}$$a_{2}$$a_{N-1}$$l_{1}$$l_{1}$$l_{2}$$l_{N-1}$$l_{N}$$l_{N}$$v_{1}$$v_{2}$$v_{N-1}$$v_{N}$$r_{1}$$r_{2}$$r_{N-2}$$r_{N-1}$$b_{1}$$b_{2}$$h_{N-1}$$h_{N-2}$$h_{2}$$h_{1}$$m_{1},\dots,m_{N}$$q_{N}$${\mathcal{W}}=\,\sum_{i=1}^{N}m_{i}\,\text{tr}\,(b_{i}b_{i+1}\dots
b_{N-1}\,q_{N}\,q_{N}\,b_{N-1}\dots
b_{i}b_{i+1})+\sum_{i=1}^{N-1}\alpha_{i}\text{tr}\,(b_{i}\,b_{i})+\sum_{i=1}^{N-1}\text{tr}\,(b_{i}\,a_{i}\,b_{i})$$+\,\sum_{i=1}^{N-2}\text{tr}\,(b_{i}\,a_{i+1}\,b_{i})+\sum_{i=1}^{N-1}r_{i}b_{i}v_{i+1}+\,\sum_{i=1}^{N}l_{i}v_{i}b_{i}b_{i+1}\dots
b_{N-1}q_{N}+\,\sum_{i=1}^{N-1}h_{i}v_{1}b_{1}\dots
b_{N-1-i}r_{N-i}$$+\,\sum_{i=1}^{N}{\mathfrak{M}}^{0^{i-1},\bullet,0^{N-i}}$

(48)

For convenience we reproduce also the mapping of the chiral ring generators:

$\begin{array}[]{c}\text{tr}\,(Q\,A^{i}\,Q)\\\ \text{tr}\,(Q\,A^{j}\,P)\\\
\text{tr}\,(A^{k})\end{array}\quad\Longleftrightarrow\quad\begin{array}[]{c}m_{i+1}\\\
l_{N-j}\\\ h_{N+1-k}\end{array}\qquad\begin{array}[]{l}i=0,\dots,N-1\\\
j=0,\dots,N-1\\\ k=2,\dots,N\end{array}$

(49)

This is the same mapping as in $4d$, at this level there are no monopoles in
the chiral ring, due to the presence of linear monopole terms in the
superpotential.

#### Flow to the deconfined dual of $Usp(2N)$ with antisymmetric and $2F$
fundamentals, $\mathcal{W}=0$

We now discuss what happens on the fully deconfined quiver 48 upon turning on
real masses. We first turn on a real mass of the form $(0^{2F},+,-)$ (that is
we are moving to the left in the diagram 47), on the electric side $Q_{2F+1}$
and $P$ become massive. Notice that the rank of the global symmetry decrease
by one unit. Accordingly the mesons $\text{tr}\,(Q_{I}\,A^{i}\,Q_{2F+1})$ and
$\text{tr}\,(Q_{I}\,A^{i}\,P)$ have non-zero real mass, for $I=1,\ldots,2F$.
Notice that $\text{tr}\,(Q_{2F+1}\,A^{i}\,P)$ has zero total real mass. The
electric theory becomes $3d$ $\mathcal{N}=2$ $Usp(2N)$ with antisymmetric and
$2F$ fundamentals, $\mathcal{W}=0$, with global symmetry $SU(2F)\times
U(1)\times U(1)$.

It follows from the mapping 49 that the singlets $(l_{i})_{I}$ and
$(m_{i})_{I,2F+1}$ become massive for $I=1,\ldots,2F$, while $(l_{i})_{2F+1}$
remain massless. The $l_{j}$’s and $(m_{i})_{I,2F+1}$’s become massive imply
that also the elementary gauge variant fields $v_{i},r_{i}$ and
$(q_{N})_{2F+1}$ become massive. In other words the saw structure in the fully
deconfined quiver disappears, and we are left with

$\\!Usp(2F\\!-\\!4)\\!$

$\\!Usp(2(2F\\!-\\!4))\\!$

$\\!\dots\\!$

$\\!Usp((N\\!-\\!1)(2F\\!-\\!4))\\!$

$\\!Usp(N(2F\\!-\\!4))\\!$

$\,\,2\,F\,\,$${\mathcal{W}}=\,\sum_{i=1}^{N}m_{i}\,\text{tr}\,(b_{i}b_{i+1}\dots
b_{N-1}\,q\,q\,b_{N-1}\dots
b_{i}b_{i+1})+\sum_{i=1}^{N-1}\alpha_{i}\text{tr}\,(b_{i}\,b_{i})+\sum_{i=1}^{N-1}\text{tr}\,(b_{i}\,a_{i}\,b_{i})$$+\,\sum_{i=1}^{N-2}\text{tr}\,(b_{i}\,a_{i+1}\,b_{i})+\,\sum_{i=1}^{N-1}h_{i}{\mathfrak{M}}^{\bullet^{N-i},0^{i}}+\,\sum_{i=1}^{N}(l_{i})_{2F+1}{\mathfrak{M}}^{0^{i-1},\bullet^{N-i+1}}$$\vdots$$a_{1}$$a_{2}$$a_{N-1}$$b_{1}$$b_{2}$$m_{1},\dots,m_{N}$$q$

(50)

Notice that the linear monopole superpotential
$\sum_{i=1}^{N}{\mathfrak{M}}^{0^{i-1},\bullet,0^{N-i}}$ is lifted, while the
massless gauge singlets $h_{i}$ and $(l_{i})_{2F+1}$ now flip monopole
operators. (this is similar to the dimensional reduction of Seiberg and
Intriligator-Pouliot dualities Aharony:2013dha ). These interactions are
generated dynamically, one way to understand them is that such interactions
are allowed by all global symmetries, and if they are not generated the gauge
singlets would be free fields, which cannot be correct.

Equation (50) agrees with the results of section $2.4$ of Benvenuti:2020gvy
(modulo renaming $h_{i}\rightarrow\gamma_{i}$ and
$(l_{i})_{2F+1}\rightarrow\sigma_{i}$), obtained by sequentually decofining in
$3d$, using the deconfining duality $antisymm_{2N\times 2N}\leftrightarrow
Usp(2N-2)-[2N],\mathcal{W}=\gamma{\mathfrak{M}}$. The only difference is the
precise extended monopole flipped by $h_{i}$, the subtlety related to the
degenerate holomorphic operators which can in principle be flipped by $h_{i}$
was not appreciated in Benvenuti:2020gvy . One can check that with the
superpotential above, setting $F=3$, the tail reconfines appropriately and it
is possible to derive the self-duality modulo flips of $3d$ $\mathcal{N}=2$
$Usp(2N)$ with antisymmetric and $6$ fundamentals, $\mathcal{W}=0$, at each
step one $h_{i}$ singlet is eaten, while the extended monopoles flipped by
$(l_{i})_{2F+1}$ ’shorten’ according to the rules of Benvenuti:2020wpc .

#### Flow to the deconfined dual of $U(N)$ with adjoint and $F+1$ flavors,
$\mathcal{W}={\mathfrak{M}}^{+}+{\mathfrak{M}}^{-}$

We now start from the $3d$ duality $Usp(2N)$ with $2F+2$,
$\mathcal{W}={\mathfrak{M}}$ $\leftrightarrow$ 48, and turn on a real mass of
the form $(+^{F+1},-^{F+1})$ (that is we are moving to the right in the
diagram 47). This type of real mass induces a Higgsing of the form
$Usp(2N_{i})\rightarrow U(N_{i})$ on both sides of the duality, the
antisymmetric fields are replaced by adjoints and a pair of fundamentals is
replaced by a fundamental plus an antifundamental. The Higgsing is induced by
a vev of ’Coulomb branch type’, that is, on both sides of the duality, we are
going to a specific sublocus of the moduli space of vacua (inside the so
called $\mathcal{N}=2$ Coulomb branch) where there is the maximum amount of
massless fields. Moreover, the monopole superpotentials ${\mathfrak{M}}$’s are
replaced by $({\mathfrak{M}}^{+}+{\mathfrak{M}}^{-})$’s. See Benini:2017dud ;
Pasquetti:2019tix ; Pasquetti:2019hxf  for more details.

On the electric side we flow to $U(N)$ with adjoint and
$(F_{Q}+1_{P},(F+1)_{\tilde{Q}})$ flavors and
$\mathcal{W}={\mathfrak{M}}^{+}+{\mathfrak{M}}^{-}$, with global symmetry
$SU(F+1)\times SU(F+1)\times U(1)$101010We split the global symmetry
artificially into $SU(F+1)_{\tilde{Q}}\times SU(F)_{Q}\times U(1)_{P}\times
U(1)_{A}$ to match the visible symmetries in the magnetic side.. The rank of
the global symmetry decreases by one unit, as it should.

On the magnetic side, we end up with a fully deconfined quiver

$U(F\\!-\\!2)$

$U(2(F\\!-\\!2))$

$\dots$

$U((N\\!-\\!1)(F\\!-\\!2))$

$U(N(F\\!-\\!2))$

$\,F\,$$\\!F\\!+\\!1\\!$$1$$1$$1$$1$$q$$\tilde{q}$$M_{i}$$\ldots$$\dots$$l_{1}$$l_{N}$$l_{1}$$l_{2}$$l_{N-1}$$l_{N}$$v_{1}$$v_{2}$$v_{N-1}$$v_{N}$$r_{1}$$r_{N-2}$$r_{N-1}$$b_{1},\tilde{b}_{1}$$b_{2},\tilde{b}_{2}$$a_{1}$$a_{2}$$a_{N-1}$${\mathcal{W}}=\,\sum_{i=1}^{N}M_{i}\,\text{tr}\,(b_{i}b_{i+1}\dots
b_{N-1}\,q\,\tilde{q}\,\tilde{b}_{N-1}\dots\tilde{b}_{i}\tilde{b}_{i+1})+\sum_{i=1}^{N-1}\alpha_{i}\text{tr}\,(b_{i}\,\tilde{b}_{i})+\sum_{i=1}^{N-1}\text{tr}\,(b_{i}\,a_{i}\,\tilde{b}_{i})$$+\,\sum_{i=1}^{N-2}\text{tr}\,(b_{i}\,a_{i+1}\,\tilde{b}_{i})+\sum_{i=1}^{N-1}tr(r_{i}b_{i}v_{i+1})+\,\sum_{i=1}^{N}l_{i}tr(\tilde{q}\tilde{b}_{N-1}\dots\tilde{b}_{i+1}\tilde{b}_{i}v_{i})$$+\,\sum_{i=1}^{N-1}h_{i}tr(r_{N-i}\tilde{b}_{N-1-i}\dots\tilde{b}_{1}v_{1})+\sum_{i=1}^{N}\left({\mathfrak{M}}^{0^{i-1},+,0^{N-i}}+{\mathfrak{M}}^{0^{i-1},-,0^{N-i}}\right)$

(51)

Where we did not draw the $h_{i}$ singlets111111The double arrows
$U(N_{1})\leftrightarrow U(N_{2})$ stand for a pair of bifundamentals with
opposite orientation.. The global symmetry of (51) is $SU(F+1)\times
SU(F)\times U(1)\times U(1)$.

The mapping is121212We have checked that the mapping is consistent by
computing the R-charges of the operators (as a function of two variables
coressponding to the two $U(1)$ symmetries) on both sides.

$\begin{array}[]{c}\text{tr}\,(\tilde{Q}_{I}\,A^{i}\,Q^{J})\\\
\text{tr}\,(\tilde{Q}_{I}\,A^{i}\,P)\\\
\text{tr}\,(A^{k})\end{array}\quad\Longleftrightarrow\quad\begin{array}[]{c}(M_{i+1})_{I}^{J}\\\
(l_{N-i})_{I}\\\ h_{N+1-k}\end{array}\qquad\begin{array}[]{l}i=0,\dots,N-1\\\
i=0,\dots,N-1\\\ k=2,\dots,N\end{array}$ (52)

In the special case $F=3$, we can use the confining duality for $U(N)$ with
$(N+2,N+2)$ flavors and $\mathcal{W}={\mathfrak{M}}^{-}+{\mathfrak{M}}^{+}$
Benini:2017dud  and reconfine the tail in 51, deriving the self-duality modulo
flips of $3d$ $\mathcal{N}=2$ $U(N)$ with adjoint and $(4,4)$ fundamentals,
$\mathcal{W}={\mathfrak{M}}^{+}+{\mathfrak{M}}^{-}$, discussed in
Benvenuti:2018bav .

#### Flow to the deconfined dual of $U(N)$ with adjoint and $F$ flavors,
$\mathcal{W}={\mathfrak{M}}^{+}$

We now turn on real masses $(0^{F},+;0^{F},-)$ in the previous duality.

On the electric side one monopole superpotential is lifted and we flow to
$U(N)$ with adjoint and $(F_{Q},F_{\tilde{Q}})$ flavors and
$\mathcal{W}={\mathfrak{M}}^{+}$, with global symmetry $SU(F)\times
SU(F)\times U(1)\times U(1)$. The rank of the global symmetry decreases by one
unit, as expected.

On the magnetic side, using 52, the singlets $(l_{j})_{I}$ and
$(M_{i})_{F+1}^{J}$ become massive. This in turn implies that
$\tilde{q}_{F+1}$ gets a mass, together with the $v_{i}$’s and the $r_{i}$’s.
Hence the saw disappears and we end up with

$U(F\\!-\\!2)$

$U(2(F\\!-\\!2))$

$\dots$

$U((N\\!-\\!1)(F\\!-\\!2))$

$U(N(F\\!-\\!2))$

$\,F\,$$\,F\,$$q$$\tilde{q}$$M_{i}$$\ldots$$b_{1},\tilde{b}_{1}$$b_{2},\tilde{b}_{2}$$a_{1}$$a_{2}$$a_{N-1}$${\mathcal{W}}=\,\sum_{i=1}^{N}M_{i}\,\text{tr}\,(b_{i}b_{i+1}\dots
b_{N-1}\,q\,\tilde{q}\,\tilde{b}_{N-1}\dots\tilde{b}_{i}\tilde{b}_{i+1})+\sum_{i=1}^{N-1}\alpha_{i}\text{tr}\,(b_{i}\,\tilde{b}_{i})+\sum_{i=1}^{N-1}\text{tr}\,(b_{i}\,a_{i}\,\tilde{b}_{i})$$+\,\sum_{i=1}^{N-2}\text{tr}\,(b_{i}\,a_{i+1}\,\tilde{b}_{i})+\sum_{i=1}^{N}(l_{i})_{2F+1}\,{\mathfrak{M}}^{0^{i-1},+^{N-i+1}}+\,\sum_{i=1}^{N-1}h_{i}{\mathfrak{M}}^{+^{N-i},0^{i}}+\,\sum_{i=1}^{N}{\mathfrak{M}}^{0^{i-1},-,0^{N-i}}$

(53)

Notice that half of the linear monopole superpotential disappeared and that
the massless gauge singlets $(l_{i})_{F}$ and $h_{i}$ now flip monopole
operators instead of mesons constructed with the saw.

In the special case $F=3$, we can use the confining duality for $U(N)$ with
$(N+1,N+1)$ flavors and $\mathcal{W}={\mathfrak{M}}^{-}+h{\mathfrak{M}}^{+}$
Benini:2017dud  to reconfine the tail in 53, deriving the self-duality modulo
flips of $3d$ $\mathcal{N}=2$ $U(N)$ with adjoint and $(3,3)$ fundamentals,
$\mathcal{W}={\mathfrak{M}}^{+}$ Benvenuti:2018bav .

#### Flow to the deconfined dual of $U(N)$ with adjoint and $F-1$ flavors,
$\mathcal{W}=0$

We now turn on real masses $(0^{F-1},+;0^{F-1},-)$ in the previous duality.

On the electric side one monopole superpotential is lifted and we flow to
$U(N)$ with adjoint and $F-1,F-1$ flavors and $\mathcal{W}=0$, with global
symmetry $SU(F-1)\times SU(F-1)\times U(1)^{3}$. Again, the rank of the global
symmetry decreases by one unit.

On the magnetic side,

$U(F\\!-\\!2)$

$U(2(F\\!-\\!2))$

$\dots$

$U((N\\!-\\!1)(F\\!-\\!2))$

$U(N(F\\!-\\!2))$

$\,F\\!-\\!1\,$$\,F\\!-\\!1\,$$q$$\tilde{q}$$M_{i}$$\ldots$$b_{1},\tilde{b}_{1}$$b_{2},\tilde{b}_{2}$$a_{1}$$a_{2}$$a_{N-1}$${\mathcal{W}}=\,\sum_{i=1}^{N}M_{i}\,\text{tr}\,(b_{i}b_{i+1}\dots
b_{N-1}\,q\,\tilde{q}\,\tilde{b}_{N-1}\dots\tilde{b}_{i}\tilde{b}_{i+1})+\sum_{i=1}^{N-1}\alpha_{i}\text{tr}\,(b_{i}\,\tilde{b}_{i})+\sum_{i=1}^{N-1}\text{tr}\,(b_{i}\,a_{i}\,\tilde{b}_{i})$$+\,\sum_{i=1}^{N-2}\text{tr}\,(b_{i}\,a_{i+1}\,\tilde{b}_{i})+\sum_{i=1}^{N}\left((l_{i})_{2F+1}\,{\mathfrak{M}}^{0^{i-1},+^{N-i+1}}+(M_{i})_{F,F}\,{\mathfrak{M}}^{0^{i-1},-^{N-i+1}}\right)$$+\,\sum_{i=1}^{N-1}h_{i}\,{\mathfrak{M}}^{+^{i},0^{N-i}}+\sum_{i=1}^{N-1}{\mathfrak{M}}^{0^{i-1},-,0^{N-i}}$

(54)

The result 54 agrees with section $3.2$ of Benvenuti:2020gvy  (modulo renaming
$h_{i}\rightarrow\gamma_{i}$, $(l_{i})_{2F+1}\rightarrow\sigma^{+}_{i}$,
$(M_{i})_{F,F}\rightarrow\sigma_{i}^{-}$ and $F\rightarrow F+1$). Also here,
the difference is the precise extended monopole flipped by $h_{i}$, the
subtlety related to the degenerate holomorphic operators which can in
principle be flipped by $h_{i}$ was not appreciated in Benvenuti:2020gvy . In
the special case $F=3$, we can use the confining duality for $U(N)$ with
$(N+1,N+1)$ flavors and $\mathcal{W}={\mathfrak{M}}^{-}+h{\mathfrak{M}}^{+}$
to reconfine the tail in 54 and derive the self-duality modulo flips of $3d$
$\mathcal{N}=2$ $U(N)$ with adjoint and $(2,2)$ fundamentals, $\mathcal{W}=0$,
discussed in Benvenuti:2018bav .131313At each reconfining step, one linear
monopole term disappears, one $h_{i}$ is eaten, while the $(M_{i})_{I}^{J}$’s
($I,J=1,2$) and the two towers $(l_{i})_{2F+1}$ and $(M_{i})_{F,F}$ survive.
The self-duality reads
$U(N)+\Phi+(2_{Q},2_{\tilde{Q}}),\mathcal{W}=0.\Leftrightarrow
U(N)+\phi+(2_{q},2_{\tilde{q}}),\mathcal{W}=\sum_{i}\left(M_{i}tr(\tilde{q}\phi^{i}q)+(l_{i})_{2F+1}{\mathfrak{M}}^{+}_{\phi^{i}}+(M_{i})_{F,F}{\mathfrak{M}}^{-}_{\phi^{i}}\right).$
(55)

###### Acknowledgements.

We are grateful to Simone Giacomelli, Sara Pasquetti and Shlomo Razamat for
useful conversations.
Stephane Bajeot is partially supported by the INFN Research Project STEFI.
Sergio Benvenuti is partially supported by the INFN Research Project GAST.

##  Appendix A R-charges and degenerate holomorphic operator ambiguity in
generic frame ${\mathcal{T}}_{k}$

We start by writing the R-charges of the fields in the generic frame
${\mathcal{T}}_{k}$ (27).

Fields | R-charges in ${\mathcal{T}}_{k}$
---|---
$V_{i}$ | $R_{V_{i}}=5-2F+(N-\frac{5}{2}+\frac{3i}{2})R_{A}+(2F-1)R_{Q}$
$L_{i}$ | $R_{L_{i}}=2F-4-(N-2+i)R_{A}-(2F-2)R_{Q}$
$M_{i}$ | $R_{M_{i}}=2R_{Q}+(i-1)R_{A}$
$H_{i}$ | $(N-i+1)R_{A}$
$R_{j}$ | $R_{R_{j}}=2F-3-(N+\frac{3j}{2}-\frac{1}{2})R_{A}-(2F-1)R_{Q}$
$B_{j}$ | $R_{B_{j}}=\frac{1}{2}R_{A}$
$A_{j}$ | $R_{A_{j}}=2-R_{A}$
$C_{k}$ | $R_{C_{k}}=1-\frac{1}{2}R_{A}$
$Q_{k}$ | $R_{Q_{k}}=1+\frac{1}{2}(1-k)R_{A}$
$Q_{k+1}$ | $R_{Q_{k+1}}=R_{Q}+\frac{k}{2}R_{A}$
$P_{k+1}$ | $R_{F_{k+1}}=2F-4-(2N-2+\frac{k}{2})R_{A}-(2F-1)R_{Q}$
$\Phi$ | $R_{\Phi}=R_{A}$
Table 1: R-charges in the frame ${\mathcal{T}}_{k}$ with $k=1,\dots,N-1$,
$i=1,\dots,k$ and $j=1,\dots,k-1$.

Now we want to find the degenerate operators that can couple to $H_{i}$, as we
did in (7) in the case of $Usp(6)$. In order to so, we look at the R-charges
of the operator $V_{i}B_{i}\dots B_{k-1}C_{k}P_{k+1}$ which is the natural
candidate to be coupled to $H_{i}$. We find

$R(V_{i}B_{i}\dots B_{k-1}C_{k}P_{k+1})=2-(N-i+1)R_{A}$ (56)

The potential degenerate operators should be a singlet under the non-abelian
global symmetry and should have the same R-charges (56). We can build the
degenerate operators from the fields $V_{a}$, $B_{b}$ and $R_{c}$. Indeed, we
start from $V_{m}$, then we put some $B_{b}$ and end with $R_{n}$ ($n\geq
m$)141414We cannot end with $V_{j+1}$ because the F-term equation of $R_{j}$
sets the combination $B_{j}V_{j+1}$ to $0$.. The form of these operators is
then: $V_{m}B_{m}\dots B_{n-1}R_{n}$. The number of fields $B$ is $n-m$. The
R-charge is

$R_{V_{m}}+(n-m)R_{B_{b}}+R_{R_{n}}=2-(2+(n-m))R_{A}$ (57)

If we compare with (56), we find the following condition

$n-m=N-i-1$ (58)

with $1\leq m\leq n\leq k-1$, $i=1,\dots,k$ and $k=1,\dots,N-1$.

We can already make two remarks:

  * •

For $i=1$, the constraint becomes $n-m=N-2$ but the maximal value of $n-m$ is
$k-2$ and $k$ satisfies $k\leq N-1$. We conclude that there is never a
solution for $i=1$. Therefore there is never a degenerate operator associated
to $H_{1}$.

  * •

If $k=1$ then $n$ and $m$ don’t exist. Conclusion, in order to get degenerate
operators we should have $N\geq 3$ which means that degenerate operators will
pop up in frames with at least $3$ gauge groups (which correspond to $k=2$).

We can ask the more precise question: What is the first frame,
${\mathcal{T}}_{k_{min}}$, when some operators degenerate? In order to answer
that we have to try to maximize the l.h.s of (58) and minimize the r.h.s.
Therefore it is enough to look at $n=k_{min}-1$, $m=1$ and $i=k_{min}$ to
determine $k_{min}$ (it could also have degenerate operators in
${\mathcal{T}}_{k_{min}}$ not associated to $n=k_{min}-1$, $m=1$ and
$i=k_{min}$). We obtain

$k_{min}-2=N-k_{min}-1\quad\Rightarrow\quad
k_{min}=\left\lceil\frac{N+1}{2}\right\rceil$ (59)

So when we reach ${\mathcal{T}}_{k_{min}}$ we start having this issue of
degenerate operators.

Now let us solve (58) in the case $k=N-1$. In this case, $1\leq m\leq n\leq
N-2$

$i=2:$ | $n=N+m-3$
---|---
|  $\exists$ solution for $m=1$
$i=3:$ | $n=N+m-4$
|  $\exists$ solution for $m=1,2$
$\vdots$ |
$i=N-1$: | $n=m$
|  $\exists$ solution for $m=1,\dots,N-2$

Conclusion, the degenerate operators (with respect to $V_{i}B_{i}\dots
B_{N-2}C_{N-1}F_{N}$) that potentially coupled to $H_{i}$ in
${\mathcal{T}}_{N-1}$ are:

$H_{2}:$ | $V_{1}B_{1}\dots B_{N-3}R_{N-2}$ | 1 operator
---|---|---
$H_{3}:$ | $V_{1}B_{1}\dots B_{N-4}R_{N-3},\,V_{2}B_{2}\dots B_{N-3}R_{N-2}$ | 2 operators
$\quad\vdots$ |  |
$H_{N-1}:$ | $V_{1}R_{1},\,V_{2}R_{2},\,\dots,\,V_{N-2}R_{N-2}$ | N-2 operators

Finally, we can study the final frame ${\mathcal{T}}_{Dec}$. The R-charges are
the following

Fields | R-charges in ${\mathcal{T}}_{Dec}$
---|---
$v_{i}$ | $5-2F+(N-\frac{5}{2}+\frac{3i}{2})R_{A}+(2F-1)R_{Q}$
$l_{i}$ | $2F-4-(N-2+i)R_{A}-(2F-2)R_{Q}$
$m_{i}$ | $2R_{Q}+(i-1)R_{A}$
$h_{j}$ | $(N-i+1)R_{A}$
$r_{j}$ | $2F-3-(N+\frac{3j}{2}-\frac{1}{2})R_{A}-(2F-1)R_{Q}$
$b_{j}$ | $\frac{1}{2}R_{A}$
$a_{j}$ | $2-R_{A}$
$q_{N}$ | $1-R_{Q}-\frac{1}{2}(N-1)R_{A}$
Table 2: R-charges in ${\mathcal{T}}_{Dec}$ with $i=1,\dots,N$ and
$j=1,\dots,N-1$.

Now using the previous result for ${\mathcal{T}}_{N-1}$, we can summarize all
the candidates to be coupled to $h_{i}$ in the final frame
${\mathcal{T}}_{Dec}$

$h_{1}:$ | $v_{1}b_{1}\dots b_{N-2}r_{N-1}$ | No degenerate operator
---|---|---
$h_{2}:$ | $v_{1}b_{1}\dots b_{N-3}r_{N-2},\,v_{2}b_{2}\dots b_{N-2}r_{N-1}$ | 2 operators
$h_{3}:$ | $v_{1}b_{1}\dots b_{N-4}r_{N-3},\,v_{2}b_{2}\dots b_{N-3}r_{N-2},\,v_{3}b_{3}\dots B_{N-2}r_{N-1}$ | 3 operators
$\quad\vdots$ |  |
$h_{N-1}:$ | $v_{1}r_{1},\,v_{2}r_{2},\,\dots,\,v_{N-2}r_{N-2},\,v_{N-1}r_{N-1}$ | N-1 operators

Now the question is obvious, which operator is the correct one?

In the special case of $F=4$, we could use the same argument that we used in
Section 3. It goes as follows. When we reach the frame
${\mathcal{T}}_{k_{min}}$ some operators become degenerate. In order to decide
the correct operator, we start confining from the left (it is possible because
in the case of $F=4$ the gauge group becomes $Usp(2)$). Then at some point we
will discover that using the F-term equation for $H_{1}$ (which is never
associated to a degenerate operator as we saw) we can select the correct
operator associated to some $H_{a}$ (as in (9)). Then, the procedure is
iterative meaning that we should re-use our previous results for $H_{a}$ and
do more and more reconfinement to select all the correct operators associated
to the other $H_{b}$. All in all, we end up in the frame ${\mathcal{T}}_{N-1}$
with the following superpotential term

$\delta\mathcal{W}_{N-1}=H_{1}V_{1}B_{1}\dots
B_{N-2}C_{N-1}P_{N}+\sum_{i=2}^{N-1}H_{i}V_{1}B_{1}\dots B_{N-1-i}R_{N-i}$
(60)

which becomes in the final frame ${\mathcal{T}}_{Dec}$

$\delta\mathcal{W}_{Dec}=\sum_{i=1}^{N-1}h_{i}v_{1}b_{1}\dots
b_{N-1-i}r_{N-i}$ (61)

Unfortunately, in the case of $F>4$ the previous argument fails because we
cannot reconfine from the left. In this case we can deconfine the
antisymmetric field $A_{1}$ but we didn’t manage to find constraints and
remove the degeneracy. Therefore, in this case the superpotential that we
wrote in (29) and (30) are ambiguous. We wrote them with the results (60),
(61) obtain in the case $F=4$ but it is logically possible that they are wrong
for $F>4$.

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