Hall Droplet Sheets in Holographic QCD

Francesco Bigazzia, Aldo L. Cotronea,b, Andrea Olzib

a INFN, Sezione di Firenze; Via G. Sansone 1; I-50019 Sesto Fiorentino
(Firenze), Italy.
b Dipartimento di Fisica e Astronomia, Universitá di Firenze; Via G. Sansone
1;
I-50019 Sesto Fiorentino (Firenze), Italy.
<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
Abstract

In single-flavor QCD, the low energy description of baryons as Skyrmions is
not available. In this case, it has been proposed by Komargodski that baryons
can be viewed as kinds of charged quantum Hall droplets, or “sheets”. In this
paper we propose a string theory description of the sheets in single-flavor
holographic QCD, focusing on the Witten-Sakai-Sugimoto model. The sheets have
a “hard” gluonic core, described by D6-branes, and a “soft” mesonic shell,
dual to non-trivial D8-brane gauge field configurations. We first provide the
description of an infinitely extended sheet with massless or moderately
massive quarks. Then, we construct a semi-infinite sheet ending on a one-
dimensional boundary, a “vortex string”. The holographic description allows
for the precise calculation of sheet observables. In particular, we compute
the tension and thickness of the sheet and the vortex string, and provide
their four dimensional effective actions.

###### Contents

  1. 1 Introduction
  2. 2 The background
  3. 3 The sheet in the WSS model
    1. 3.1 Comparison with QCD
  4. 4 Infinite sheet
    1. 4.1 The core
    2. 4.2 The shell
  5. 5 Semi-infinite sheet: the vortex string
    1. 5.1 Solution in flat space
      1. 5.1.1 Step 1: monopole-like solution
      2. 5.1.2 Step 2: solution for ${A}^{c}_{z}$
      3. 5.1.3 Step 3: solution for ${A}_{\pm}$
    2. 5.2 Solution on curved space
      1. 5.2.1 The vortex string
      2. 5.2.2 The sheet
      3. 5.2.3 The charged mode
  6. 6 Summary and future developments
  7. A Including the scalar mode

##  1 Introduction

In the effective field theory of QCD, baryons can be described as solitons of
the chiral Lagrangian equipped with the Skyrme term. This picture is well
justified in the large $N$ limit, where the baryon mass, linear in $N$, scales
like the inverse squared meson coupling. However, this description does not
hold if the number of flavors reduces to $N_{f}=1$. In this case, in fact, the
group manifold of the chiral Lagrangian trivializes and it is not possible to
write down a conserved baryonic charge. Hence, one-flavored baryons, which
have spin $N/2$, require an alternative low-energy description.

An intriguing proposal has been put forward by Komargodski in [1], building on
earlier works such as [2, 3, 4, 5, 6]. In large $N$ QCD with just one quark
whose mass is parametrically lower than the QCD scale, the would-be
$\eta^{\prime}$ meson is a light particle. The $\eta^{\prime}$ is a periodic
field being the phase of the chiral condensate, and its effective theory
admits extended $\eta^{\prime}$ configurations, $(2+1)$-dimensional sheets,
that interpolate from the $\eta^{\prime}=0$ vacuum on one side to the
(equivalent) $\eta^{\prime}=2\pi$ vacuum on the other side. Around these
sheets the $\eta^{\prime}$ has a $2\pi$ monodromy. The sheets can end on a
boundary. The related $\eta^{\prime}$ configuration cannot be smooth
everywhere, because the monodromy is violated outside the boundary. Thus, the
$\eta^{\prime}$ field should be singular and the minimal singularity that must
exist is of the form of the boundary. As a consequence, a locus of
discontinuity for the $\eta^{\prime}$ field arises, which can be thought of as
a Dirac sheet, where it jumps from $0$ to $2\pi$. In [1] it has been suggested
that one-flavored baryons can be classically thought of as pancake-shaped
sheets, with a $N$-independent size and a mass proportional to $N$.

Another key element of this construction is that the effective field theory on
the sheet is a $U(1)_{N}$ topological field theory which describes the bulk
physics of a fractional quantum Hall state. Then, the sheets can be
interpreted as baryons by coupling a background baryonic gauge field to the
theory on the sheet. Indeed, this coupling is precisely like the coupling of
the electromagnetic gauge field to the emergent gauge field in the fractional
quantum Hall effect (FQHE). If the sheet has a boundary, due to the bulk-
boundary correspondence of FQHE, there can be edge excitations which carry a
conserved baryon number. The charge of these modes forbids the sheet to
collapse, stabilizing its size. Other studies of this proposal can be found
e.g. in [7, 8, 9, 10, 11, 12].

The low-energy $\eta^{\prime}$ theory is incomplete: there is a cusp
singularity in the potential which prevents us to perform explicit
calculations in this framework. The physical interpretation of the cusp is
that in correspondence of the sheet some heavy fields rearrange. These heavy
fields are likely to be glueballs, which are not taken into account in the
chiral Lagrangian.

This paper aims at exploring and testing the above proposal in top-down
holographic QCD, namely within the Witten-Sakai-Sugimoto (WSS) model [13, 14].
This is a non-supersymmetric $SU(N)$ gauge theory in $3+1$ dimensions, coupled
to $N_{f}$ fundamental flavors and a tower of massive (Kaluza-Klein) adjoint
fields. The Yang-Mills plus adjoint field sector is realized as the low energy
theory of a stack of $N$ D4-branes wrapped on a circle. At strong coupling and
in the planar limit, it is holographically described by a dual Type IIA
supergravity theory on a specific background, which remarkably accounts for
confinement and mass gap formation in the dual field theory. Fields in the
fundamental representation, i.e. quarks, are obtained by introducing $N_{f}$
pairs of D8-$\overline{\rm D}$8-branes, which, in the $N_{f}\ll N$ limit, can
be treated as probes on the above mentioned background. To minimize their
energy these branes form a unique U-shaped configuration, geometrically
realizing the spontaneous breaking of chiral symmetry. The effective action on
the flavor branes describes a $U(N_{f})$ Yang-Mills-Chern-Simons theory in
five curved dimensions. Mesons arise as excitations of the $U(N_{f})$ gauge
field, i.e. of open strings on the flavor branes, while baryons (wrapped
D4-branes in the zero-size limit, following [15]) are identified with
instantonic configurations of the $U(N_{f})$ field [16]. Remarkably, the same
problem of finding a baryon charge in the low energy description of $N_{f}=1$
large $N$ QCD arises also in the holographic model [17]. This makes even more
urgent to find a way to construct an alternative low energy description of the
corresponding baryons.

In this paper we propose a holographic realization of the sheet in the one-
flavored WSS model. This is achieved by adding an extra D6 probe brane,
wrapped on the four-sphere of the gravitational background, and extended in
$(2+1)$ Minkowski directions. The world-volume of the D6-brane, after a
dimensional reduction on the four-sphere, precisely supports the expected
$U(1)_{N}$ Chern-Simons theory realizing the bulk physics of the FQHE. In this
picture, the sheet can be viewed as a wall with a hard (gluonic) D6-brane
core, corresponding to the heavy fields associated to the singularity in the
chiral Lagrangian, and a soft shell (a mesonic cloud) coming from the non-
trivial $\eta^{\prime}$ profile. Indeed, the D6-brane can be viewed as a
source for the Maxwell gauge field living on the D8-brane, whose excitations
are identified with the $\eta^{\prime}$ meson.

The holographic construction makes it transparent that the dynamical gauge
field on the sheet is not of mesonic nature, rather it sources some mesonic
profiles through the boundary. Some discussions on this topic appear in [8,
10, 11]. Nevertheless, as we will briefly discuss in section 3.1, it is
possible that in QCD with light quarks the dynamical field is Higgsed. In this
case, one should be able to describe the sheet entirely in terms of mesonic
degrees of freedom. We will not explore this possibility in this paper,
leaving this interesting issue for a future study.

To simplify our analysis, we will focus on two configurations: the infinite
and the semi-infinite sheet with a straight boundary. In the first case, as
expected, we find a regular profile for the $\eta^{\prime}$ field both in the
massless and the massive quark case. We compute the tension and the thickness
of the sheet considering the contributions from the hard core and the mesonic
cloud. In the second case, the straight boundary can be viewed as a linear
distribution of magnetic charge in the five-dimensional Maxwell-Chern-Simons
effective theory on a curved background. The corresponding gauge field can be
found semi-analytically, through a series expansion, and numerically. From
this solution we extract the effective action for the transverse fluctuations
of the boundary, and the tension, which scales parametrically as expected in
[1]. Then, we can compute the $\eta^{\prime}$ profile which now has two
different features depending on the domain. In fact, it is continuous (in
analogy with the infinite sheet case) through the D6-brane, while it is
discontinuous outside the sheet, enforcing the monodromy of the
$\eta^{\prime}$ meson around the wall. Finally, we will show that it is
possible to turn on, with no energy cost, a chiral component of the gauge
field on the boundary of the sheet; this is possibly connected to the chiral
edge mode on the boundary of the Hall droplet.

The paper is organized as follows. After a brief review of the Witten-Sakai-
Sugimoto model in section 2, we describe our proposal for the dual picture of
the sheet in section 3. Then, we move on to the construction of the infinitely
extended sheet, both with massless and massive quarks, in section 4. Section 5
contains the main results of this paper, concerning a semi-infinite sheet
terminating on a string-like boundary. After a warm-up in flat space in
section 5.1, we provide the complete solution and its properties in section
5.2. We conclude with a summary of our results and some directions for future
investigations in section 6.

##  2 The background

The WSS model of holographic QCD is based on the type IIA background [13]

$\displaystyle ds^{2}$ $\displaystyle=$
$\displaystyle\left(\frac{u}{R}\right)^{3/2}\left(dx^{\mu}dx_{\mu}+f(u)dx_{4}^{2}\right)+\left(\frac{R}{u}\right)^{3/2}\frac{du^{2}}{f(u)}+R^{3/2}u^{1/2}d\Omega_{4}^{2}\
,$ $\displaystyle f(u)$ $\displaystyle=$ $\displaystyle
1-\frac{u_{0}^{3}}{u^{3}}\ ,\qquad e^{\Phi}=g_{s}\frac{u^{3/4}}{R^{3/4}}\
,\qquad F_{4}=3R^{3}\omega_{4}\,,$ (2.1)

where $\mu=(0,1,2,3)$, the radial variable $u$ has dimensions of length and
ranges in $[u_{0},\infty)$, the circle $S_{x_{4}}$ has length $\beta_{4}$,
$R=(\pi g_{s}N)^{1/3}l_{s}$, $\omega_{4}$ is the volume form of $S^{4}$ (which
has volume $V(S^{4})=8\pi^{2}/3$) and absence of conical singularities gives
the relation $9u_{0}\beta_{4}^{2}=16\pi^{2}R^{3}$. $\Phi$ is the dilaton and
$F_{4}$ the Ramond-Ramond four-form field strength. Moreover
$\beta_{4}=2\pi/M_{KK}$, where $M_{KK}$ is the Kaluza-Klein (KK) and glueball
mass scale, proportional to the dynamical scale $\Lambda$

$M_{KK}\quad\leftrightarrow\quad\Lambda\,.$ (2.2)

The above background holographically accounts for confinement and mass gap
formation in the dual $SU(N)$ theory. In the holographic regime, the confining
string tension of the theory scales like $\lambda M_{KK}^{2}$, where
$\lambda\equiv g_{YM}^{2}N=2\pi g_{s}Nl_{s}M_{KK}\gg 1$. The background can be
further enriched by a Ramond-Ramond one form potential $C_{1}$ in case one is
interested in turning on a Yang-Mills $\theta$ angle [18]. In the following we
will work at $\theta=0$.

Flavor degrees of freedom are introduced by means of $N_{f}$ D8-$\overline{\rm
D}8$ branes at fixed points on the $S_{x_{4}}$ circle [14]. These branes are
usually treated in the probe approximation which is reliable if $N_{f}\ll N$.
To minimize their energy they actually form a unique U-shaped configuration on
the ($x_{4},u$) cigar-like subspace, dynamically realizing the spontaneous
breaking of chiral symmetry. If the D8-$\overline{\rm D}8$ branes are at
antipodal (resp. non antipodal) points on the $S_{x_{4}}$ circle, the tip of
the U-shaped configuration is at $u=u_{0}$ (resp. at $u_{1}>u_{0}$).

The D8-brane embedding can be described in terms of the coordinates

$y=\frac{u^{3/2}}{\sqrt{u_{0}}}\sqrt{f(u)}\cos(M_{KK}x_{4})\,,\qquad
z=\frac{u^{3/2}}{\sqrt{u_{0}}}\sqrt{f(u)}\sin(M_{KK}x_{4})\,.$ (2.3)

In these coordinates the antipodal D8-brane is localized at $y=0$ and extends
along $z\in(-\infty,\infty)$ with the tip at $z=0$. In units $M_{KK}=u_{0}=1$,
$R^{3}=9/4$ and the cigar metric in the new coordinates reads

$ds^{2}_{(y,z)}=\frac{2}{3u^{3/2}}\left[\left(1-q\,z^{2}\right)dz^{2}+\left(1-q\,y^{2}\right)dy^{2}-2zy\,q\,dzdy\right]\,,$
(2.4)

with $u=u(y,z)=(1+y^{2}+z^{2})^{1/3}$,
$q=q(y,z)=\frac{1}{y^{2}+z^{2}}\left(1-\frac{1}{u}\right)$.

The effective action for $N_{f}$ D8-branes at $y=0$, after reduction on
$S^{4}$ and to second order in derivatives reads

$S_{f}=-\kappa\int d^{4}xdz\,\left(\frac{1}{2}h(z)\,{\rm
Tr}\mathcal{F}_{\mu\nu}\mathcal{F}^{\mu\nu}+k(z){\rm Tr}\mathcal{F}_{\mu
z}\mathcal{F}^{\mu}_{\;\;z}\right)+\frac{N}{24\pi^{2}}\int\omega_{5}(\mathcal{A})\,,$
(2.5)

where four-dimensional indices $\mu,\nu$ are raised by the flat Minkowski
metric and (in units $M_{KK}=u_{0}=1$)

$\kappa=\frac{N\lambda}{216\pi^{3}}\,,\quad h(z)=(1+z^{2})^{-1/3}\,,\quad
k(z)=(1+z^{2})\,,$ (2.6)

and

$\omega_{5}(\mathcal{A})={\rm
Tr}\left(\mathcal{A}\wedge\mathcal{F}^{2}-\frac{i}{2}\mathcal{A}^{3}\wedge\mathcal{F}-\frac{1}{10}\mathcal{A}^{5}\right)\;,\quad
d\omega_{5}(\mathcal{A})={\rm Tr}\mathcal{F}^{3}\,.$ (2.7)

The gauge field $\mathcal{A}\in U(N_{f})$ can be decomposed in the Abelian and
non-Abelian parts

$\mathcal{A}=A\frac{\mathbf{1}}{\sqrt{2N_{f}}}+\hat{A}^{a}T^{a}\,,$ (2.8)

where $T^{a}$ are the $SU(N_{f})$ generators.

Using this notation the action can be rewritten as

$\displaystyle S_{f}$ $\displaystyle=-\kappa\int
d^{4}xdz\,\left(\frac{1}{2}h(z)\,{\rm
Tr}\hat{F}_{\mu\nu}\hat{F}^{\mu\nu}+k(z){\rm Tr}\hat{F}_{\mu
z}\hat{F}^{\mu}_{\;\;z}\right)+$ (2.9)
$\displaystyle\quad-\frac{\kappa}{2}\int
d^{4}xdz\,\left(\frac{1}{2}h(z)\,{F}_{\mu\nu}{F}^{\mu\nu}+k(z){F}_{\mu
z}{F}^{\mu}_{\;\;z}\right)+$
$\displaystyle\quad+\frac{N}{24\pi^{2}}\int\biggl{[}\omega_{5}^{SU(N_{f})}(\hat{A})+\frac{3}{\sqrt{2N_{f}}}{A}\,{\rm
Tr}\hat{F}^{2}+\frac{1}{2\sqrt{2N_{f}}}{A}\,{F}^{2}\biggr{]}\,.$

Here $\omega_{5}^{SU(N_{f})}$ is defined as in (2.7), written in terms of just
the non-Abelian components. It is worth noticing that it is identically zero
for $N_{f}\leq 2$.

In fact, in the following we will focus on the $N_{f}=1$ case, where the
effective action above describes a five-dimensional Maxwell-Chern-Simons
theory

$S^{N_{f}=1}_{f}=-\frac{\kappa}{4}\int d^{4}xdz\,\sqrt{-\det
g}\,g^{MN}g^{RS}F_{MR}F_{NS}+\frac{N}{24\pi^{2}}\int\frac{1}{2\sqrt{2}}A\wedge
F\wedge F$ (2.10)

on the curved background

$ds^{2}=g(z)dx^{\mu}dx_{\mu}+\frac{dz^{2}}{g(z)}\,,\qquad
g(z)=(1+z^{2})^{2/3}\,.$ (2.11)

It turns out that this provides an effective description for the
$\eta^{\prime}\sim\int{A}_{z}dz$ meson.

In fact, mesons arise as gauge field fluctuations in the above effective
action, while baryons (wrapped D4-branes in the zero size limit, following
[15]) are realized as instanton solutions [16] provided $N_{f}\geq 2$. In the
$N_{f}=1$ case we thus need an alternative description. This will be the focus
of the following sections.

##  3 The sheet in the WSS model

We propose that the holographic picture for the $N_{f}=1$ baryons considered
in [1] contains a D6-brane wrapped on $S^{4}$. When the D8-branes are
antipodally embedded, the D6-brane is localized at $z=0$, $y=0$ (i.e.
$u=u_{0}$) and can be taken at, say, $x_{3}=0$ in the geometry (2), (2.4); see
figure 1 for the displacement of the branes on the cigar of the geometry.

Figure 1: The displacement of the D-branes on the cigar of the geometry (2)
for antipodally embedded D8-branes.

The low energy theory on the D6-brane world-volume is given by a $U(1)_{N}$ CS
theory (plus the center of mass degrees of freedom) just because it is wrapped
on the $S^{4}$ [19], which supports $N$ units of flux of $F_{4}$. The DBI
action of the D6-brane gives, upon integration on the four-sphere, the three-
dimensional kinetic term for the Abelian gauge field $a$ which lives on the
D6-brane world-volume. Moreover, from the CS term of the D6-brane we get

$\frac{1}{8\pi^{2}}\int C_{3}\wedge da\wedge da=\frac{1}{8\pi^{2}}\int
F_{4}\wedge a\wedge da=\frac{N}{4\pi}\int a\wedge da\,,$ (3.1)

where we have used $\int_{S^{4}}F_{4}=2\pi N$. Remarkably the above theory
precisely agrees with Komargodski’s proposal for the theory on the bulk of the
sheet.

This picture resembles that of D6-branes as domain walls (DW) separating
different degenerate vacua in the WSS model at $\theta=\pi$ [18, 19].111The
world-volume theory in the case of QCD has been studied in field theory in
[20]. Across the domain wall there is a jump of the value of the gluon
condensate while in our case there is a monodromy of $\eta^{\prime}$. Notice
moreover that the sheet now does not separate two different vacua since the
$\eta^{\prime}=0$ and the $\eta^{\prime}=2\pi$ one are the same. The story
goes in close analogy with the axion monodromy studied in [21] where
interesting comments on the metastability of the D6-brane configuration with a
boundary on the D8-brane, can be found. In [21] in fact it is argued that the
D6 sheet is at best metastable, since it can decay by developing holes with an
exponential suppressed rate $\Gamma\sim e^{-\lambda^{2}N}$.

As already remarked, the holographic picture of the sheet does not end with
the D6-brane. The latter (which in the baryon case is expected to have a
finite, disk-like shaped extension) has to be thought of as the hard core of
the Hall droplet baryon. On top of this there will be a soft mesonic cloud due
to the non trivial configuration of the $\eta^{\prime}$ field. The latter can
be explicitly found considering the whole structure of the D8-brane action in
presence of a D6 source.

The D8-brane action for $N_{f}=1$ admits a CS term

$\frac{1}{2\pi}\int C_{7}\wedge\frac{F}{\sqrt{2}}\,,$ (3.2)

so that a charge for a D6-brane with the same embedding as that of the source
can be induced by a non trivial flux for the abelian gauge field $F$. As we
will see in the following, the D6-brane source actually induces a flux

$\int F_{x_{3}z}dx_{3}dz=-2\pi\sqrt{2}\,,$ (3.3)

which provides an $\overline{\rm D}6$-brane charge precisely canceling that of
the source. Hence the total D6 charge in the setup is zero. The sheet in fact
can decay and thus there cannot be a net D6 charge.

The condition (3.3) precisely accounts for a non-trivial profile of the
$\eta^{\prime}$ field. The mesonic shell of the sheet can be seen as a non-
trivial solution for the Abelian gauge field on the D8-brane satisfying (3.3).

However, this is not enough to get the action for a point-like particle with
baryonic $U(1)_{B}$ charge on the D8-branes. In fact defining
$B_{0}(t)={A}_{0}/\sqrt{2}$ and treating it as a time dependent perturbation,
we can get such a point-particle action only if the other component ${F}_{12}$
of the field strength is turned on. Indeed from the five-dimensional CS part
of the action (2.10) we get

$\frac{N}{24\pi^{2}}\int\frac{1}{2\sqrt{2}}{A}\wedge\,{F}\wedge{F}=\frac{N}{24\pi^{2}}\int
dt\,B_{0}(t)\int\frac{F}{\sqrt{2}}\wedge\frac{F}{\sqrt{2}}\,,$ (3.4)

which implies that the baryon solution has to be some kind of “Abelian
instanton” with unit instanton number

$\int({F}\wedge{F})_{x_{1}x_{2}x_{3}z}dx_{1}dx_{2}dx_{3}dz=48\pi^{2}\,.$ (3.5)

Obviously in the Maxwell-CS theory there is no solution of this kind on
$\mathbb{R}^{4}$. The only possibility is that there is some kind of non
trivial extended source. If the overlap between the sheet and the D8-branes in
the $(x_{1},x_{2})$ plane is a disk of radius $L$ then the above condition has
some hope to be satisfied. Analogously to what happens for the standard
baryons, due to the five-dimensional CS coupling the above condition induces
an electric field. We expect that the related Coulomb interaction is what
stabilizes the sheet to some finite size $L_{stable}$.222We do not have any
argument concerning the scaling of $L_{stable}$ with $\lambda$: it might be
that, as for the standard baryon in the WSS model, the size is parametrically
small, $L_{stable}\sim 1/(\lambda\Lambda)$, regardless of the large spin. This
issue can be settled only studying the full solution. In this paper we will
not consider this setup, and thus we will not impose the condition (3.5),
leaving the analysis of the actual quantum Hall droplet baryon for the future.

###  3.1 Comparison with QCD

In the WSS model, departure from the antipodal D8-brane configuration
corresponds to increasing quartic NJL-type couplings between the quarks [22].
If this departure is large, i.e. the configuration is far from the antipodal
one, the effect of large NJL-type couplings is that the critical temperature
for chiral symmetry breaking is different (larger) than that for confinement
[23]. Thus, this scenario is different from what is expected to happen in QCD.
Instead, if the departure from the antipodal configuration is not large, the
chiral symmetry breaking and confinement temperatures are still coincident and
there is no clear contradiction with QCD expectations.

Thus, the configurations closer to QCD are the antipodal one and the
moderately non-antipodal ones. In the antipodal case, the holographic
description of the sheet is a D6-brane lying on top of the D8-branes at the
tip of the cigar (as in figure 1). In flat space, such a configuration is
unstable towards the formation of a D8-D6 bound state, where the D6 is melt in
the D8 world-volume. This happens because the spectrum of D8-D6 strings
contains a scalar. The scalar is tachyonic if the branes are on top of each
other (and there are no fluxes turned on). The melting of the D6 corresponds
to the condensation of this scalar.333It is not immediately evident how to
describe the three-dimensional scalar field on the world-volume of the sheet
in terms of the elementary fields of the four-dimensional theory. This
corresponds to a complete Higgsing of the $U(1)$ gauge group of the three-
dimensional theory living on the world-volume of the sheet (the D6-brane). In
this scenario it should be again possible to describe finite energy baryons. A
very similar situation is found in the case of the domain-wall at $\theta=\pi$
[19].

A small departure from the antipodal configuration is sufficient to give a
positive mass to the scalar, avoiding the Higgsing mechanism and so the
melting of the D6-brane. Thus, there is a family of moderately non-antipodal
configurations whose phase diagram is consistent with the QCD one and for
which the sheet is realized with an actual D6-brane which is not melt.444We
thank Ofer Aharony for exchange of ideas on this issue. The D6 melting could
be avoided also by introducing sufficiently large masses for the quarks [3,
19].

The condensed scalar scenario is not the one explored in [1]. While it would
be of obvious interest to explore the melted-D6 configuration, in this paper
we aim at studying the original proposal in [1], corresponding to an actual
D6-brane. In order to simplify the computations we will consider the antipodal
configuration without large quark masses, aiming at revealing some basic
features of the sheet. This is obviously an oversimplification, which
eventually must be lifted in future studies.

##  4 Infinite sheet

In this section we study the infinitely extended wall corresponding to a sheet
without boundaries. The sheet has a “hard”, gluonic core corresponding to the
D6-brane, and a “soft” shell from the $\eta^{\prime}$ profile, i.e. a meson
cloud, as advocated e.g. in [3].

###  4.1 The core

As we have already pointed out, the “hard”, gluonic core of the sheet is
holographically given by the wrapped D6-brane. Assuming the sheet to be
extended along the $x_{1},x_{2}$ directions, its tension $T_{hard}$ can be
extracted from the DBI term of the wrapped D6-brane action

$S^{DBI}_{D6}=-T_{6}\int d^{7}xe^{-\Phi}\sqrt{-g_{7}}\equiv-T_{hard}\int
dtdx_{1}dx_{2}\,,$ (4.1)

where $T_{6}=(2\pi)^{-6}\,l_{s}^{-7}$ and

$T_{hard}=\frac{1}{\pi^{3}3^{6}}\lambda^{2}NM_{KK}^{3}\,.$ (4.2)

As expected, the tension scales as $N$, but is enhanced by the $\lambda^{2}$
factor, as usual in holographic theories.

Considering the thickness $\delta$ of a D-brane sitting at the tip of the
background cigar to be given by the effective local string length $l_{s,eff}$,
one gets for the thickness of the hard core of the wall (up to a numerical
factor)

$\delta_{hard}\sim l_{s,eff}\sim\frac{1}{\lambda^{1/2}M_{KK}}\,,$ (4.3)

which comes from the scaling $T_{s}\sim l_{s,eff}^{-2}\sim\lambda M_{KK}^{2}$
of the Yang-Mills string tension in the model.

###  4.2 The shell

A localized D6-brane is a source for the RR $C_{7}$ field. This in turn will
generate a non-trivial profile for the D8-brane mode corresponding to the
$\eta^{\prime}$. The profile of the $\eta^{\prime}$, the meson cloud,
constitutes the “soft shell” of the sheet.

In order to study the D6-D8 setup we have in mind we should better consider
the total action $S_{f}^{N_{f}=1}+S_{C_{7}}$ in the presence of D6-brane
sources where $S_{f}^{N_{f}=1}$ is given in (2.10) and

$S_{C_{7}}=-\frac{1}{4\pi}(2\pi l_{s})^{6}\int
dC_{7}\wedge^{\star}dC_{7}+\frac{1}{2\pi}\int
C_{7}\wedge\frac{F}{\sqrt{2}}\wedge\omega_{1}+\int C_{7}\wedge\omega_{3}\,,$
(4.4)

is the ten-dimensional action for the Ramond-Ramond potential $C_{7}$. The
kinetic term comes from the type IIA supergravity action; the second term is
due to the flavor D8-brane, whose embedding, in the antipodal case, is
described by a delta-like one form $\omega_{1}=\delta(y)dy$; the third term is
due to the D6-brane source, whose embedding is described by the three form
$\omega_{3}$.

Let us consider the equation of motion for $C_{7}$. Setting
${}^{\star}F_{8}=(2\pi l_{s})^{-6}\widetilde{F}_{2}$ we have, for D8-brane at
$y=0$ and D6-brane at $x_{3}=z=y=0$,

$d\widetilde{F}_{2}=\frac{F}{\sqrt{2}}\wedge\delta(y)dy+2\pi\delta(z)\delta(y)\delta(x_{3})dz\wedge
dy\wedge dx_{3}\,,$ (4.5)

where $F=dA$ is the $U(1)$ field strength on the D8-brane.

We can formally solve the above equation (with $dC_{1}=0$) as

$\widetilde{F}_{2}=\frac{A}{\sqrt{2}}\wedge\delta(y)dy+2\pi\Theta(x_{3})\delta(z)\delta(y)dz\wedge
dy\,.$ (4.6)

Now, the full action for $C_{7}$ turns out to be equivalent to555See e.g.
Appendix B in [24].

$S_{eff}=-\frac{1}{4\pi(2\pi
l_{s})^{6}}\int\widetilde{F}_{2}\wedge^{\star}\widetilde{F}_{2}\,,$ (4.7)

just as in the case with no explicit D6-brane source. With the same arguments
as in [14, 24] we can thus obtain the four-dimensional effective Lagrangian
for the $\eta^{\prime}$ field as

${\cal
L}=-\frac{\kappa}{\pi}(\partial_{\mu}\varphi)(\partial^{\mu}\varphi)+2cm\cos\varphi-\frac{\chi_{g}}{2}\,\underset{k\in\mathbb{Z}}{\text{Min}}\,(\varphi-2\pi\Theta(x_{3})+2\pi
k)^{2}\,,$ (4.8)

where we have also included the contribution of the flavor mass term ($m$
being the quark mass and $c$ a constant parameter) introduced in the string
theory setup as in [25]. We have introduced $\chi_{g}$, that is the
topological susceptibility of the unflavored gauge theory, which in units of
$M_{KK}$ reads

$\chi_{g}=\frac{\lambda^{3}}{4(3\pi)^{6}}\,.$ (4.9)

The Lagrangian above has been obtained by using the following ansatz for the
gauge field

$A_{z}(z,x_{3})=-\frac{\sqrt{2}}{\pi(1+z^{2})}\varphi(x_{3})\,,$ (4.10)

and integrating over $z$. Hence

$\int A_{z}dz\equiv-\sqrt{2}\varphi\equiv
2\frac{\eta^{\prime}}{f_{\pi}}\,,\quad f_{\pi}=2\sqrt{\frac{\kappa}{\pi}}\,.$
(4.11)

From the Lagrangian (4.8), we can derive the equation of motion for the field
$\varphi=\varphi(x_{3})$. We are going to search for a configuration
interpolating from 0 for $x_{3}\rightarrow-\infty$ to $2\pi$ for
$x_{3}\rightarrow+\infty$ and such that $\varphi(0)=\pi$. In this case the
potential is minimized for $k=0$ and the equation of motion reads

$\partial^{2}\varphi(x_{3})-m_{WV}^{2}\left(\varphi(x_{3})-2\pi\Theta(x_{3})\right)-m_{\pi}^{2}\sin(\varphi(x_{3}))=0,$
(4.12)

where

$m_{WV}^{2}\equiv\frac{\chi_{g}\pi}{2\kappa}=\frac{\lambda^{2}}{27\pi^{2}N}$
(4.13)

is the Witten-Veneziano mass. Treating the D8-brane in the probe approximation
amounts on taking $m_{WV}\ll 1$. The other parameter in the equation of motion
above is

$m_{\pi}^{2}=\dfrac{4cm}{f_{\pi}^{2}},$ (4.14)

which represents the quark mass contribution to the $\eta^{\prime}$ mass. When
$N_{f}>1$ these become the standard parameters related to pions, as shown in
[25]. As observed in [19], the parameter $m_{\pi}$ has to be small,
$m_{\pi}\ll 1$, in order to neglect higher order corrections to the mass term
of the WSS model.

#### Massless quark

In the massless case $m=0$ the equation of motion for $\varphi$ (4.12) has
solution

$\varphi(x_{3})=\pi e^{m_{WV}\,x_{3}}-\pi
e^{-m_{WV}\,x_{3}}\left(e^{m_{WV}\,x_{3}}-1\right)^{2}\Theta(x_{3})\,.$ (4.15)

This solution has been derived in the presence of an explicit gluonic source.
Similar solutions without explicit sources, with regular or singular
potentials, have a very long history, see e.g. [26, 3, 4].

The thickness of the “soft” part of the wall (the one given by the
$\eta^{\prime}$ profile) is clearly linear in $1/m_{WV}$ (from (4.15)), so up
to a numerical factor,

$\delta_{soft}\sim\frac{1}{m_{WV}}\sim\frac{\sqrt{N}}{\lambda M_{KK}}\,.$
(4.16)

The $\varphi\sim\eta^{\prime}$ solution (4.15) interpolates between zero (for
$x_{3}\rightarrow-\infty$) and $2\pi$ ($x_{3}\rightarrow\infty$) and it takes
the value $\varphi(0)=\pi$ in $x_{3}=0$. See figure 2 (left panel).

Figure 2: Left: the profile of $\varphi=-\sqrt{2}\eta^{\prime}/f_{\pi}$ for
$m_{WV}/M_{KK}=0.1,0.3,0.5$ (green, yellow and blue lines). Right: the profile
of $\varphi$ for $m_{WV}/M_{KK}=0.3$ and
$m_{\pi}^{2}/M_{KK}^{2}=0,0.05,0.1,0.2$ (blue, green, yellow and red lines).

Correspondingly, the Abelian field strength on the D8-brane reads

$F_{x_{3}z}=-\frac{\sqrt{2}m_{WV}}{1+z^{2}}\left(e^{m_{WV}\,x_{3}}-2\Theta(x_{3})\sinh(m_{WV}\,x_{3})\right)\,,$
(4.17)

where we have reinserted the $z$-dependent factor. This satisfies

$\int dx_{3}dzF_{x_{3}z}=-2\sqrt{2}\pi\,,$ (4.18)

hence, as anticipated in eq. (3.3) the $\eta^{\prime}$ profile effectively
induces a unit of $\overline{\rm D}6$-brane charge.

Due to the field strength (4.17), the D8-brane has an excess of energy,
providing the tension of the soft shell of the sheet. This can be deduced from
the on-shell value of the action (2.10)

$S_{f}^{N_{f}=1}=-\frac{\kappa}{2}\int dt\,dx_{1}\,dx_{2}\int
dz\,dx_{3}(1+z^{2})F_{3z}^{2}\,.$ (4.19)

Inserting the solution (4.17) and performing the integration in $z,x_{3}$ we
get the tension

$T_{soft}=\frac{1}{3^{3}2^{3}\pi^{2}}\lambda
Nm_{WV}M_{KK}^{2}=\frac{1}{3^{9/2}2^{3}\pi^{3}}\lambda^{2}N^{1/2}M_{KK}^{3}\,.$
(4.20)

Note that considering the expression of $f_{\pi}$ in (4.11), one has
$T_{soft}\sim f_{\pi}^{2}m_{\eta^{\prime}}$.

#### Massive quark

Figure 3: The function $\Psi(m_{\pi}/m_{WV})$ providing the quark-mass
dependence of the soft part of the tension of the sheet.

In the massive-quark case one can integrate numerically the equation derived
from (4.8) with $k=0$, with a similar behavior as in the massless case, see
figure 2 (right panel). Also in this case the solution corresponds to a unit
$\overline{\rm D}6$-brane charge.

The thickness of the soft part of the wall now depends on the quark mass $m$
(and hence on $m_{\pi}$) as well and roughly behaves as (at small $\varphi$
this comes from (4.8))

$\delta_{soft}\sim\frac{1}{\sqrt{m_{WV}^{2}+m_{\pi}^{2}}}\,.$ (4.21)

In order to estimate the tension, we have to insert the numerical solution for
$F_{x_{3}z}$ in formula (4.19), which gives

$T_{soft}=\frac{1}{3^{3}2^{3}\pi^{2}}\lambda
Nm_{WV}\Psi(m_{\pi}/m_{WV})M_{KK}^{2}\,.$ (4.22)

The result is exactly as in (4.20) but for the substitution

$m_{WV}\longrightarrow m_{WV}\Psi\left(\dfrac{m_{\pi}}{m_{WV}}\right),$ (4.23)

where $\Psi(q)$ is a function, going as $\sqrt{1+q^{2}}$ at small $q$, which
is reported in figure 3.

#### Resume

Table 1 contains the parametric dependence of tension and thickness of the
core and shell parts of the sheet discussed above.

| Massless quark | Massive quark
---|---|---
Core tension $T_{hard}$ | $\lambda^{2}N\Lambda^{3}$ | $\lambda^{2}N\Lambda^{3}$
Core thickness $\delta_{hard}$ | $\lambda^{-1/2}\Lambda^{-1}$ | $\lambda^{-1/2}\Lambda^{-1}$
Shell tension $T_{soft}$ | $\lambda Nm_{WV}\Lambda^{2}\sim\lambda^{2}N^{1/2}\Lambda^{3}$ | $\lambda Nm_{WV}\Psi(m_{\pi}/m_{WV})\Lambda^{2}$
Shell thickness $\delta_{soft}$ | $m_{WV}^{-1}\sim\lambda^{-1}N^{1/2}\Lambda^{-1}$ | $(m_{WV}^{2}+m_{\pi}^{2})^{-1/2}$
Table 1: Summary of parametric dependence of tension and thickness of the core
and shell parts of the sheet. For the reader’s convenience, we used the
dynamical scale notation $\Lambda$ instead of $M_{KK}$.

##  5 Semi-infinite sheet: the vortex string

In this section we describe the semi-infinite sheet, whose boundary is a
straight infinite vortex string. The sheet is dual to a D6-brane which for
convenience is placed at $x_{3}=z=y=0$ and extended along $x_{2}>0$. The
string is its boundary, that we take at $x_{2}=0$, attached to the D8-brane,
see figure 4.

Figure 4: D6-brane dual to the semi-infinite sheet. It is taken to be extended
along the $x_{1}$ axis and along the positive values of the $x_{2}$ axis. The
directions $x_{3}$ and $z$ are orthogonal to the sheet. Every point is also
extended in time and along other four dimensions which are wrapped on $S^{4}$.

The boundary is a magnetic source on the D8-brane worldvolume. In this case
the action terms (4.4) take the form

$\displaystyle S_{C_{7}}$ $\displaystyle=$ $\displaystyle-\frac{1}{4\pi}(2\pi
l_{s})^{6}\int dC_{7}\wedge^{\star}dC_{7}$ (5.1)
$\displaystyle+\frac{1}{2\pi}\int
C_{7}\wedge\left[\frac{F}{\sqrt{2}}\wedge\delta(y)dy+2\pi\Theta(x_{2})\delta(x_{3})dx_{3}\wedge\delta(z)dz\wedge\delta(y)dy\right]\,.$

The corresponding equation for ${}^{\star}dC_{7}$,

$d\widetilde{F}_{2}=\frac{F}{\sqrt{2}}\wedge\delta(y)dy+2\pi\Theta(x_{2})\delta(x_{3})dx_{3}\wedge\delta(z)dz\wedge\delta(y)dy\,,$
(5.2)

provides the violation of the Bianchi identity due to the magnetic source (the
boundary of the D6)

$dF=-2\sqrt{2}\pi\delta(x_{2})\delta(x_{3})\delta(z)dx_{2}\wedge dx_{3}\wedge
dz\,.$ (5.3)

Thus, we are led to consider the problem of finding the fields generated by a
uniform linear distribution of magnetic charge in five-dimensional
electromagnetism with CS and $\widetilde{F}_{2}$ source terms on the curved
background (2.11). We have not found such type of solution, or part of it, in
the literature, so we derive it in this section. The strategy will be the
following.

  * 1.

We will first find a configuration ${F}^{m}$ satisfying the Bianchi equation
(5.3) and $d^{\star}{F}^{m}=0$. This corresponds to a kind of magnetic
monopole-like solution for a source extended along the $x_{1}$ axis.

  * 2.

The complete field can be written as

${F}={F}^{m}+d{A}^{c}\,,$ (5.4)

being $A^{c}$ the gauge connection associated to a closed field strength. If
we consider a potential ${A}^{m}$ for ${F}^{m}$, we have

$d{A}^{m}={F}^{m}+{\cal S}\,,$ (5.5)

where ${\cal S}$ is a Dirac sheet (a Dirac string for an extended source). A
particular gauge choice of the Dirac sheet allows to solve equation (5.2) as

$\widetilde{F}_{2}=\left(\frac{{A}^{c}+{A}^{m}}{\sqrt{2}}\right)\wedge\delta(y)dy+2\pi\Theta(x_{3})\delta(z)dz\wedge\delta(y)dy\,.$
(5.6)

As in [14] and Section 4, the integral of this field along the $(z,y)$ cigar

$\int_{cig}\widetilde{F}_{2}=\int\left(\frac{{A}^{c}_{z}+{A}_{z}^{m}}{\sqrt{2}}\right)dz+2\pi\Theta(x_{3})\,,$
(5.7)

enters, through the supergravity term (4.7), the equation for ${A}^{c}_{z}$.
If there are no CS terms (as it will be the case), and considering that
$d^{\star}{F}^{m}=0$, the equation reads schematically

$d^{\star}d{A}^{c}_{z}=\int\left({A}^{c}_{z}+{A}_{z}^{m}\right)dz+2\pi\sqrt{2}\Theta(x_{3})\,.$
(5.8)

Given that ${A}_{z}^{m}$ is known from step 1, we can solve this equation for
${A}^{c}_{z}$.

  * 3.

We can consider the charged case by switching on the components
${A}_{\pm}(x_{\pm})$ of the gauge field ($x^{\pm}=(x_{1}\pm t)/2$). One can
easily realize that ${A}_{\pm}$ do not enter the previous equations, but on
the contrary the field ${A}_{z}={A}^{c}_{z}+{A}^{m}_{z}$ generates a CS term
for their equations, which schematically read

$d^{\star}d{A}_{\pm}={F}_{\pm x_{2}}\wedge{F}_{x_{3}z}+...\,.$ (5.9)

The latter can be solved independently after steps 1 and 2 are completed,
since the only unknowns are ${A}_{\pm}$.

In order to illustrate this procedure, in the next section we consider as a
warm-up the flat space case. Then, in section 5.2 we will provide the solution
on the curved space (2.11). The procedure does not depend on the explicit form
of the metric function $g(z)$ and can in principle be applied to a generic
metric of the form (2.11).

###  5.1 Solution in flat space

In this section we consider the case of five-dimensional flat spacetime, i.e.
$g(z)=1$. In this case the configuration has spherical symmetry in the
directions $x_{2},x_{3},z$ transverse to $t,x_{1}$ where the line source is
placed, so we employ spherical coordinates

$x_{2}=\rho\cos\theta\,,\quad x_{3}=\rho\sin\theta\cos\phi\,,\quad
z=\rho\sin\theta\sin\phi\,.$ (5.10)

It is also convenient to parameterize the $(t,x_{1})$-part of spacetime with
light-cone coordinates

$x_{\pm}=\frac{x_{1}\pm t}{2}\,,$ (5.11)

such that

$ds^{2}=4dx_{+}dx_{-}+d\rho^{2}+\rho^{2}d\Omega_{2}\,,$ (5.12)

with the two-sphere parameterized by angles $\theta,\phi$.

We consider an ansatz where:

  * •

all the fields just depend on the radial direction $\rho$;

  * •

the magnetic charge from equation (5.3) is

$\displaystyle-2\sqrt{2}\pi\delta^{3}(\rho)\rho^{2}d\rho\wedge\omega_{2}\,,$
(5.13)

with $\omega_{2}=\sin\theta d\theta\wedge d\phi$ being the two-sphere volume
form;

  * •

the field strength has just two components

$F_{\theta\phi}\omega_{2}\,,\qquad F_{\pm\rho}dx_{\pm}\wedge d\rho\,,$ (5.14)

where either $A_{+}$ or $A_{-}$ are turned on. This implies that we have
(only) one non-vanishing term in $F\wedge F$, entering the equation for
$A_{\pm}$.

It is readily verified that the ansatz is consistent with the equations of
motion and Bianchi identity.

####  5.1.1 Step 1: monopole-like solution

We start by searching for a solution ${F}^{m}$ to the Bianchi equation
$d{F}^{m}=-2\sqrt{2}\pi\delta^{3}(\rho)\rho^{2}d\rho\wedge\omega_{2}\ $ and
the equation of motion (without CS term, as explained above)
$d^{\star}{F}^{m}=0$.

Considering translation invariance along $x_{1}$ and spherical symmetry, the
problem is very similar to that of finding the field strength generated by the
standard Dirac monopole in four dimensions. Thus, the solution is quite
similar

${F}^{m}={F}^{m}_{\theta\phi}\omega_{2}=-\frac{1}{2\rho^{2}}\rho^{2}\sqrt{2}\omega_{2}\,,$
(5.15)

where $\omega_{2}=\sin{\theta}d\theta\wedge d\phi$ and we have made explicit
the singular behavior at the origin $\rho=0$. It is straightforward to check
that (5.15) solves the equations of motion and that

$\int_{M_{3}}d{F}^{m}=\int_{M_{2}}{F}^{m}=-2\sqrt{2}\pi\,,$ (5.16)

where $M_{3},M_{2}$ are three- and two-dimensional closed surfaces containing
the point $\rho=0$, $M_{2}$ being the boundary of $M_{3}$.

A standard way of writing a potential ${A}^{m}$ for this field is

${A}^{m}={A}^{m}_{\phi}d\phi=-\sqrt{2}\frac{(1-\cos\theta)}{2\rho\sin\theta}\rho\sin\theta
d\phi\,.$ (5.17)

In this form, one has

$d{A}^{m}={F}^{m}+{\cal S}\,,$ (5.18)

where ${\cal S}$ is a Dirac sheet along $\theta=\pi$, i.e. extended along
$x_{1}$ and the negative branch of $x_{2}$

${\cal S}=-2\sqrt{2}\pi\Theta(-x_{2})\delta(x_{3})\delta(z)dx_{3}\wedge dz\,.$
(5.19)

As usual, one could choose the position of the Dirac sheet to be along any
other direction, and glue the solutions up to gauge transformations.

The energy associated to the field strength (5.15) is readily checked to be
divergent, due to the small-$\rho$ behavior. This can be interpreted as the
usual divergence close to the charges (electric or magnetic) in standard
electromagnetism.

The same field (5.15) with sign changed represents the solution corresponding
to an opposite sign of the charge distribution,
$d{F}^{m}=2\sqrt{2}\pi\,\delta^{3}(\rho)\rho^{2}d\rho\wedge\omega_{2}$. This
corresponds to choosing an $\overline{\rm D}6$-brane instead of a D6-brane in
the stringy setup.

####  5.1.2 Step 2: solution for ${A}^{c}_{z}$

In flat space this step is trivial, since there is no $(z,y)$-cigar over which
we can integrate $\widetilde{F}_{2}$ with a finite result. But our five-
dimensional electromagnetic problem is well-defined without the
$\widetilde{F}_{2}$ source term. Without source, and since the CS term does
not enter the equations for ${A}_{z},{A}_{x_{2}},{A}_{x_{3}}$, we can just
consider (5.17) as a solution of the equations, without the need to turn on
further components in directions $x_{2},x_{3},z$.

####  5.1.3 Step 3: solution for ${A}_{\pm}$

Even if it is not forced upon us by the equations, we can switch on the
components ${A}_{\pm}(x_{\pm},\rho)$ of the gauge field. The equations read

$\frac{1}{g_{5}^{2}}d^{\star}{F}=N{F}\wedge{F}\,,$ (5.20)

where the five-dimensional coupling constant is in this case
$g_{5}^{2}=1/(8\pi^{2}\kappa)$. Considering the solution for
${F}_{\theta\phi}$ (5.15) in the CS term, the equations reduce to

$\partial_{\rho}(\rho^{2}\partial_{\rho}{A}_{\pm})=\mp
g_{5}^{2}N\partial_{\rho}{A}_{\pm}\,,$ (5.21)

with the solutions

$\partial_{\rho}{A}_{\pm}=c(x_{\pm})\frac{e^{\pm\frac{g_{5}^{2}N}{\rho}}}{\rho^{2}}\,.$
(5.22)

Here we have included a possible arbitrary function $c(x_{\pm})$.666Note that
the component ${F}_{+-}$ of the gauge field strength would mix the equations
among each other, so we are keeping it off in this paper. Since
$g_{5}^{2}N\sim 1/(\lambda M_{KK})$, the exponential behavior of the solution,
due to the CS term, sets in for radii smaller than $\rho\sim 1/(\lambda
M_{KK})$, that is very close to the source (since $\lambda\gg 1$).

Now we can note two interesting facts. First, only one of the two modes,
${A}_{-}$,777The fact that $A_{-}$ is normalizable depends on the sign choice
of the magnetic source, i.e. on whether we consider a D6 or an $\overline{\rm
D}6$-brane in the string setup. is normalizable. Second, if we just keep the
normalizable mode, i.e. as long as we do not switch on both components
${F}_{\rho-}$ and ${F}_{\rho+}$ in (5.22), it has zero energy: it is tempting
to read this as an indication that the mode $c(x_{\pm})$ has zero potential.
These properties seem to suggest that the functions $c(x_{\pm})$ could be
related to the chiral edge modes expected to be present on the boundary of the
sheet: they are only left or right moving and have zero Hamiltonian.

###  5.2 Solution on curved space

In this section we report on the main result of this paper, that is the
solution for the semi-infinite sheet on the curved background (2.11). For
simplicity, we just consider the massless-quark case.

The boundary of the sheet extends along $t,x_{1}$, while the sheet itself is
placed at $z=x_{3}=0$ and extends along $x_{2}\geq 0$ up to the boundary at
$x_{2}=0$, see figure 4. The configuration without the sheet has translational
invariance along $x_{1}$ and axial symmetry around the boundary, so we can use
polar coordinates888The curvature along $z$ breaks the spherical symmetry we
had in flat space. The difference between the angular variable $\varphi$ and
the normalized $\eta^{\prime}/f_{\pi}$ field should be hopefully clear from
the context.

$x_{2}=r\cos{\varphi}\,,\qquad x_{3}=r\sin{\varphi}\,,$ (5.23)

in terms of which the background metric reads

$ds^{2}=g(z)(4dx_{+}dx_{-}+dr^{2}+r^{2}d\varphi^{2})+\frac{dz^{2}}{g(z)}\,.$
(5.24)

The ansatz we put forward is such that:

  * •

all the fields are independent on $x_{\pm}$ (eventually apart from the field
${A}_{\pm}$ which can depend on $x_{\pm}$);

  * •

the field strength has the components

$F_{x_{2}z}\,,\quad F_{x_{3}z}\,,\quad F_{x_{2}x_{3}}\,,\quad F_{\pm
z}\,,\quad F_{\pm x_{2}}\,,\quad F_{\pm x_{3}}\,.$ (5.25)

Thus, the non-vanishing CS terms have components

$F_{x_{2}z}F_{\pm x_{3}}\,,\quad F_{x_{3}z}F_{\pm x_{2}}\,,\quad
F_{x_{2}x_{3}}F_{\pm z}\,,$ (5.26)

and all enter just the equation for $A_{\pm}$.

The ansatz is consistent with the equations of motion and Bianchi identity
(5.3). We proceed to solve them in three steps as explained above.

####  5.2.1 The vortex string

The goal of this section is to find a solution $F^{m}$ of the Bianchi equation
(5.3) such that $d^{\star}F^{m}=0$ on the background (5.24). In this sub-
section the only non-trivial fields will be
$F_{x_{2}z},F_{x_{3}z},F_{x_{2}x_{3}}$.

The configuration is less symmetric than in flat space, so we are not able to
write a compact solution. But we can expand the fields in a basis of
eigenfunctions depending only on $z$ and finding the corresponding modes.

We start by writing an ansatz which solves automatically the equation of
motion $d^{\star}F^{m}=0$, which reads

$F^{m}_{x_{2}z}=-g(z)^{-3/2}\partial_{x_{3}}H\,,\quad
F^{m}_{x_{3}z}=g(z)^{-3/2}\partial_{x_{2}}H\,,\quad
F^{m}_{x_{2}x_{3}}=g(z)^{1/2}\partial_{z}H\,,$ (5.27)

with the function $g(z)$ given in (2.11). This form can also be written as

$F^{m}=d\varphi\wedge[g(z)^{-3/2}r\partial_{r}Hdz-g(z)^{1/2}r\partial_{z}Hdr]\,.$
(5.28)

The function $H=H(x_{2},x_{3},z)$ is determined by the Bianchi equation (5.3),
which reads

$g(z)^{-3/2}(\partial_{x_{2}}^{2}+\partial_{x_{3}}^{2})H+\partial_{z}(g^{1/2}\partial_{z}H)=-2\sqrt{2}\pi\delta(x_{2})\delta(x_{3})\delta(z-Z)\,,$
(5.29)

where $Z$ is a (possibly pseudo)-modulus for the position of the string in the
$z$ direction. Here we consider just the classical configuration taking $Z=0$
at the end, but we should remind that, eventually, $Z$ should become a time
dependent operator at the quantum level. Taking inspiration from similar
equations in [16], we search for a solution in the form

$H(x_{2},x_{3},z)=\sum_{n=0}^{\infty}Y_{n}(r)\zeta_{n}(z)\zeta_{n}(Z)\,,$
(5.30)

where we have set to zero, without loss of generality, the
modulus999$R=\sqrt{X_{2}^{2}+X_{3}^{2}}$, where the moduli $X_{2}$ and $X_{3}$
together with the pseudo-modulus $Z$ represent the center of mass collective
coordinates of the vortex string. $R$ in the argument of the $Y_{n}(r-R)$.
Equation (5.29) is equivalent to the infinite set of equations

$\displaystyle(\partial_{x_{2}}^{2}+\partial_{x_{3}}^{2}-\lambda_{n})Y_{n}=-2\sqrt{2}\pi\delta(x_{2})\delta(x_{3})\,,$
(5.31) $\displaystyle
g(z)^{3/2}\partial_{z}(g^{1/2}\partial_{z}\zeta_{n})+\lambda_{n}\zeta_{n}=0\,,$
(5.32)

supported by the completeness condition

$\sum_{n=0}^{\infty}g(z)^{-3/2}\zeta_{n}(z)\zeta_{n}(Z)=\delta(z-Z)\,,$ (5.33)

satisfied by the functions $\zeta_{n}(z)$ with orthonormality condition given
by

$\int dzg(z)^{-3/2}\zeta_{n}(z)\zeta_{m}(z)=\delta_{nm}\,.$ (5.34)

As long as $\lambda_{n}\neq 0$, the first set of equations (5.31) has
solutions in terms of the modified Bessel function of order zero of the second
kind101010Remember that we are working in units where $M_{KK}=1$. One can
obtain the correct dimensionalities by reinserting the appropriate powers of
$M_{KK}$ in these formulas.

$Y_{n}(r)=\sqrt{2}K_{0}(\sqrt{\lambda_{n}}r)\,,\qquad n>0\,.$ (5.35)

They behave as logarithms at small $r$, while fall off exponentially fast at
large $r$. For $\lambda_{0}\equiv 0$ the solution of (5.31) is just the
logarithm

$Y_{0}(r)=-\sqrt{2}\log{(r)}\,,\qquad n=0\,.$ (5.36)

We have chosen to put to zero the constant solution, since the ansatz (5.27)
only contains derivatives of $H$.

We have numerically solved the second set of equations (5.32), determining the
eigenvalues $\lambda_{n}$, by requiring that the derivatives of the functions
$\zeta_{n}(z)$ vanish at $z\rightarrow\pm\infty$ for $n>0$, so that they are
orthogonal to the even part of the zero-mode, which is just a constant,
$\zeta_{0}=1/\sqrt{\pi}$. The masses of the first even modes computed from
(5.32) are then

$\lambda_{n}=0,1.6,4.6,9.2,15.2,22.8,31.8,\dots$ (5.37)

We have checked that the numerical solutions form a complete set according to
(5.33).

Notice that since the functions $\zeta_{n}$ have definite parity, so that
$\zeta_{n}(Z=0)=0$ for odd parity, classically the function $H$ is even in
$z$. A plot of the resulting solution for $H$ is reported in figure 5 (left
panel).

Figure 5: Left: the function $H(r,z)$ obtained from the first six (even) modes
in (5.30). Right: The profile of $-\frac{\sqrt{2}}{f_{\pi}}\eta^{\prime}$ for
$m_{WV}/M_{KK}=0.3$.

It can be checked numerically that with such solution in (5.28) one has the
correct magnetic charge in (5.3)

$\int_{M_{3}}dF^{m}=\int_{M_{2}}F^{m}=-2\sqrt{2}\pi\,,$ (5.38)

where $M_{3},M_{2}$ are three- and two-dimensional closed surfaces containing
the point $r=z=0$, $M_{2}$ being the boundary of $M_{3}$.111111We used spheres
of different radii to perform the check.

We can choose to express the solution in terms of the potentials

$\displaystyle A^{m}_{z}=g(z)^{-3/2}r\varphi\partial_{r}H\,,\qquad
A^{m}_{r}=-g(z)^{1/2}r\varphi\partial_{z}H\,.$ (5.39)

This corresponds to choosing the Dirac sheet ${\cal S}$, defined by

$dA^{m}=F^{m}+{\cal S}\,,$ (5.40)

to be placed along the negative $x_{2}$ direction

${\cal S}=-2\sqrt{2}\pi\Theta(-x_{2})\delta(x_{3})\delta(z)dx_{3}\wedge dz\,.$
(5.41)

We can see explicitly the stringy nature of the solution by plugging formulas
(5.28), (5.30) in the action (2.10) and integrating in $z$, obtaining the
four-dimensional effective action for the vortex string

$S^{N_{f}=1}_{f}=-\frac{1}{2}\sum_{n=0}^{\infty}\int
d^{4}x\left[(\partial_{r}\widetilde{Y}_{n}(r,\varphi))^{2}+\lambda_{n}(\widetilde{Y}_{n}(r,\varphi))^{2}\right]\,,$
(5.42)

where

$\widetilde{Y}_{n}\equiv\sqrt{\kappa}\zeta_{n}(0)Y_{n}\,.$ (5.43)

As we can see, we have an infinite tower of massive modes depending only on
the two coordinates $x_{2},x_{3}$ transverse to the string (with dynamics only
in the radial direction), plus a massless zero-mode.

Putting the Lagrangian density in (5.42) on-shell on the solutions (5.35) and
integrating in $r,\varphi$, one can derive the tension of the vortex string
(re-inserting the $M_{KK}$ factors)

$T_{v}=M_{KK}^{2}\frac{\lambda
N}{108\pi^{3}}\int_{0}^{\infty}dr\Biggl{\\{}\frac{1}{r}+\pi
r\sum_{n=1}^{\infty}\zeta_{n}(0)^{2}\lambda_{n}\left[K_{0}^{2}(\sqrt{\lambda_{n}}r)+K_{1}^{2}(\sqrt{\lambda_{n}}r)\right]\Biggr{\\}}\,.$
(5.44)

As expected [3, 4, 6, 1], this scales linearly with $N$ and is divergent at
the location $r=0$ of the string. One has to remember that the solution is
derived under the hypothesis that the field strength is small in order to
consider only the first term in the expansion of the square-root of the DBI
action. Thus, the solution we found cannot be trusted close to the source,
where the divergence is expected to be avoided in the full DBI setup.

On the other hand, we do know that the core of the string is made by the
boundary of the D6 on the D8-brane, and that it has the width (4.3)

$\delta_{v,hard}\sim\frac{1}{\lambda^{1/2}\Lambda}\,.$ (5.45)

This can thus be reasonably used to cut-off the integral (5.44) at small $r$.

There is also a divergence of (5.44) at large distances, due to the zero mode.
It is similar to the one of the classical global $U(1)$ string. In that
context, it is expected to be cut-off by the presence of other strings or
other physical effects.121212For example it can be regulated by the curvature
of spacetime in a cosmological context. Calling $R_{c}$ the long distance cut-
off, one obtains the estimate for the tension of the “soft” shell of the
string as

$T_{v,soft}\sim\Lambda^{2}\lambda
N\log{(R_{c}/\delta_{v,hard})}\sim\Lambda^{2}\lambda
N\log{(\lambda^{1/2}R_{c}\Lambda)}\,,$ (5.46)

while the “hard” core of the strings, coming from the D6-brane, has tension

$T_{v,hard}\sim\Lambda^{2}\lambda^{3/2}N\,.$ (5.47)

This follows by considering the boundary of the sheet to have radius
$\delta_{v,hard}$ (5.45), which is a reasonable definition in the antipodal
D8-brane configuration we are considering. In the non-antipodal cases, the
power of $\lambda$ is expected to be equal to two.

Finally, the logarithmic behavior of the zero mode means that the soft shell
of the string is very spread in space, so that it is not meaningful to
identify a thickness different from the above-mentioned large-distance cut-off
and

$\delta_{v,soft}\sim R_{c}\,.$ (5.48)

The above estimates are summarized in table 2.

Core tension $T_{v,hard}$ | $\lambda^{3/2}N\Lambda^{2}$
---|---
Core thickness $\delta_{v,hard}$ | $\lambda^{-1/2}\Lambda^{-1}$
Shell tension $T_{v,soft}$ | $\lambda\log{(\lambda^{1/2}R_{c}\Lambda)}N\Lambda^{2}$
Shell thickness $\delta_{v,soft}$ | $R_{c}$
Table 2: Summary of parametric dependence of tension and thickness of the core
and shell parts of the vortex loop (string). For the reader’s convenience, we
used the dynamical scale notation $\Lambda$ instead of $M_{KK}$. $R_{c}$ is a
large-distance cut-off.

####  5.2.2 The sheet

In this section we work out the solution of the equation of motion for the
closed part $A^{c}_{z}$ of $A_{z}$ including the source from
$\widetilde{F}_{2}$.

Let us begin by computing the term in $\widetilde{F}_{2}$. Consider equation
(5.2) for $d\widetilde{F}_{2}$ with the gauge field

$F^{m}+dA^{c}=dA^{m}+{\cal S}+dA^{c}\,.$ (5.49)

With the choice of Dirac sheet as in (5.41) we can solve (5.2) as

$\widetilde{F}_{2}=\left[\frac{1}{\sqrt{2}}(A^{m}+A^{c})+2\pi\Theta(x_{3})\delta(z)dz\right]\wedge\delta(y)dy\,.$
(5.50)

Its integral on the cigar gives

$\int_{cig}\widetilde{F}_{2}=\frac{1}{\sqrt{2}}\int(A^{m}_{z}+A^{c}_{z})dz+2\pi\Theta(x_{3})\,.$
(5.51)

As before, this gives a source term in the equation for $A^{c}_{z}$ of the
form

$-\frac{\chi_{g}}{2}\left[\frac{1}{\sqrt{2}}\int(A^{m}_{z}+A^{c}_{z})dz+2\pi\Theta(x_{3})\right]^{2}\,.$
(5.52)

Note that we know the explicit form of $A^{m}_{z}$, and so its integral, from
(5.39).

The $z$-dependence of the field $A^{c}_{z}$ can be dealt with the ansatz
(4.10)

$A^{c}_{z}=-\frac{\sqrt{2}}{\pi g(z)^{3/2}}\phi(x_{2},x_{3})\,.$ (5.53)

Its equation of motion reduces to an equation for $\phi(x_{2},x_{3})$

$(\partial_{x_{2}}^{2}+\partial_{x_{3}}^{2})\phi-
m_{WV}^{2}\left(\phi-\frac{1}{\sqrt{2}}\int
A^{m}_{z}dz-2\pi\Theta(x_{3})\right)=0\,.$ (5.54)

It is straightforward to integrate numerically this equation. The sum of the
solution and the integral of $A^{m}_{z}$ gives the profile of the
$\eta^{\prime}$ field

$-\frac{\sqrt{2}}{f_{\pi}}\eta^{\prime}=\left(\phi(x_{2},x_{3})-\frac{1}{\sqrt{2}}\int
A^{m}_{z}dz\right)\,.$ (5.55)

We impose the boundary conditions $\eta^{\prime}(x_{2},-\infty)=0$ (trivial
vacuum on one side of the sheet),
$\eta^{\prime}(\infty,x_{3})=\eta^{\prime}_{\infty}(x_{3})$, where
$\eta^{\prime}_{\infty}(x_{3})$ is the profile found in section 4.2 for the
infinite sheet. We plot this field in figure 5 (right panel). While for
$x_{2}>0$ the profile resembles the one of the infinite sheet by construction,
for $x_{2}<0$ it is visible the discontinuity due to the Dirac sheet,
enforcing the monodromy of the $\eta^{\prime}$ field around the D6-brane wall
(whose boundary is at $x_{2}=x_{3}=0$ in Minkowski).

As usual, the solution induces a D6-brane charge on the D8-brane, which we can
calculate in terms of the flux of $F_{x_{3}z}$, obtaining

$\int dx_{3}dzF_{x_{3}z}=\int dzA_{z}|^{x_{3}=+\infty}_{x_{3}=-\infty}-\int
dx_{3}dz{\cal S}_{x_{3}z}=-2\sqrt{2}\pi\Theta(x_{2})\,,$ (5.56)

where ${\cal S}$ is the Dirac sheet in formula (5.41). The result is that
there is a unit $\overline{\rm D}6$-brane charge induced only for $x_{2}>0$,
where the sheet is located. Once again, the condition (3.3) accounting for the
cancelation of the total D6-brane charge in the setup is satisfied.

The parametric dependence of the tension and thickness of the sheet are the
same as in the infinite sheet case, see table 1.

The effective action for the system of the vortex string and the sheet can be
derived from the actions (2.10), (4.7) considering that the complete field
strength reads

$\displaystyle F_{x_{2}x_{3}}=g(z)^{1/2}\partial_{z}H\,,$ (5.57)
$\displaystyle
F_{x_{2}z}=-g(z)^{-3/2}\left[\partial_{x_{3}}H+\frac{\sqrt{2}}{\pi}\partial_{x_{2}}\phi\right]\,,$
(5.58) $\displaystyle
F_{x_{3}z}=g(z)^{-3/2}\left[\partial_{x_{2}}H-\frac{\sqrt{2}}{\pi}\partial_{x_{3}}\phi\right]\,.$
(5.59)

By means of formula (5.30) and integrating in $z$ the result can be derived to
be

$\displaystyle S$ $\displaystyle=$ $\displaystyle-\frac{1}{2}\int
d^{4}x\Biggl{\\{}\sum_{n=0}^{\infty}\left[(\partial_{r}\widetilde{Y}_{n})^{2}+\lambda_{n}(\widetilde{Y}_{n})^{2}\right]+\left[(\partial_{x_{2}}\widetilde{\phi})^{2}+(\partial_{x_{3}}\widetilde{\phi})^{2}\right]$
(5.60)
$\displaystyle\qquad\qquad\qquad+\chi_{g}\left[\sqrt{\frac{\pi}{2\kappa}}\widetilde{\phi}-\frac{1}{\sqrt{2}}\int
A^{m}_{z}dz-2\pi\Theta(x_{3})\right]^{2}\Biggr{\\}}\,,$

where

$\widetilde{Y}_{n}\equiv\sqrt{\kappa}\zeta_{n}(0)Y_{n}\,,\qquad\widetilde{\phi}\equiv\sqrt{\frac{2\kappa}{\pi}}\phi\,,$
(5.61)

and $\int A_{z}^{m}dz$ can be given as a series of the $\widetilde{Y}_{n}$
using formulas (5.39), (5.30). The action (5.60) is that for a string
configuration (the part in $\widetilde{Y}_{n}$) and a membrane (the part in
$\widetilde{\phi}$) with a non-trivial potential.

This action must be used with caution. In fact, while it gives the correct
equation of motion for $\widetilde{\phi}$, it does not provide the correct
equations for the $\widetilde{Y}_{n}$. This is because of the coupling term
between the $\widetilde{Y}_{n}$ in $\int A_{z}^{m}$ and the $\widetilde{\phi}$
in the second line of (5.60), which must be dropped by hands from the
equations for the $\widetilde{Y}_{n}$. Ultimately, the reason for this resides
in the fact that the $\widetilde{Y}_{n}$ and $\widetilde{\phi}$ are both modes
of the same gauge field in five dimensions. We chose to solve the equation of
motion of the latter by enforcing the extra constraint $d^{\star}F^{m}=0$ on
the monopole-like part of the solution (which is written in terms of the
$\widetilde{Y}_{n}$). The latter constraint is not included in (5.60) and
amounts to dropping the coupling term between the $\widetilde{Y}_{n}$ and the
$\widetilde{\phi}$ in the equations of motion of the $\widetilde{Y}_{n}$, as
stated above.

In any case, we made an extra check that the configuration we found in two-
steps, first solving for $F^{m}$ enforcing the constraint $d^{\star}F^{m}=0$,
and then solving for the rest of $F$, satisfies the complete equation of
motion for the five-dimensional gauge field.

####  5.2.3 The charged mode

In this section we solve for the components $A_{\pm}$ of the D8-brane gauge
field. Their equations of motion reduce to

$\displaystyle\partial_{z}\left(g^{3/2}(z)\partial_{z}{A}_{\pm}\right)+\dfrac{1}{\sqrt{g(z)}}\left(\partial_{x_{3}}^{2}+\partial_{x_{2}}^{2}\right){A}_{\pm}\pm\dfrac{N}{16\pi^{2}\kappa}\Bigl{[}\sqrt{g(z)}\partial_{z}{A}_{\pm}\partial_{z}H+\,\,$
$\displaystyle+\dfrac{\partial_{x_{2}}{A}_{\pm}}{g^{3/2}(z)}\left(\partial_{x_{2}}H-\frac{\sqrt{2}}{\pi}\partial_{x_{3}}\phi\right)+\dfrac{\partial_{x_{3}}{A}_{\pm}}{g^{3/2}(z)}\left(\partial_{x_{3}}H+\frac{\sqrt{2}}{\pi}\partial_{x_{2}}\phi\right)\Biggr{]}=0\,.$
(5.62)

Since the only unknown in these equations are $A_{\pm}$, one can find them
either numerically or employing the basis $\zeta_{n}(z)$ to reduce them to a
set of equations in just the $x_{2},x_{3}$ variables. In any case, the
solutions can be multiplied by a generic function $c(x_{\pm})$, as in the flat
space case. The associated energy is zero as long as there is just one of the
two components $F_{\pm M}$ turned on ($M=x_{\mp},x_{2},x_{3},z$). Again, this
mode could be related to the chiral edge mode expected on the border of the
sheet.

In fact, one can argue without solving equation (5.2.3) that only one of the
two modes $A_{\pm}$ is normalizable. The reason is that the equation
simplifies drastically close to the string at $z\sim x_{2}\sim x_{3}\sim 0$.
It can be checked that it reduces to the flat space one (5.21), with the two
solutions (5.22), only one of which ($A_{-}$) is normalizable at the origin.
Thus, there is only a left moving or a right moving normalizable mode.

One can derive an explicit numeric solution for this mode given particular
boundary conditions. We leave to a future study the search for a more general
solution. The analysis of the equation in the asymptotic regions of large
$z,x_{2},x_{3}$ shows that one can impose the solution to go to constant
values. It is natural to choose a zero constant at large $|z|$. Another
boundary condition can be imposed at $z=0$, where equation (5.2.3) can be
integrated numerically in $x_{2},x_{3}$. In this case one can impose that
close to the origin the solution is the flat space one (a primitive of
equation (5.22) at $z=0$), and that at large $|x_{2}|,|x_{3}|$ it goes to a
constant.131313This condition is the same as in flat space as well. The
difference between the flat space case and the curved one (5.2.3) at $z=0$ is
encoded in the two terms $\partial_{x_{2},x_{3}}\phi$. These are limited and
multiplied by derivatives $\partial_{x_{2},x_{3}}A_{\pm}$, so a constant
boundary condition is still the most natural one.

With these boundary conditions, one can integrate numerically equation
(5.2.3). Plots representing two slices of the solution at $z=0$ (the boundary
condition) and at a fixed $z>0$ are reported in figure 6.

Figure 6: Slices of the solution for $A_{-}$ at $z=0$ (left) and $z=5$
(right). For these plots we fixed $N/16\pi^{2}\kappa=1$ and
$A_{-}(x_{2},x_{3},0)=1$ at large $|x_{2}|,|x_{3}|$. The $z=0$ slice is almost
the same as in flat space. As $z$ increases, the solution reduces in amplitude
and eventually goes to zero at large $z$ (as dictated by the boundary
conditions).

##  6 Summary and future developments

In this paper we have put forward a proposal for the string dual, in the
context of the WSS model of holographic QCD, of the quantum Hall droplet, or
sheet, that should describe one-flavor baryons at low energies [1]. This
object consists of a D6-brane wrapped on the four-sphere of the ten-
dimensional background (much like the dual of the domain wall at
$\theta=\pi$), together with the non-trivial D8-brane gauge field sourced by
the D6-brane. The D6-brane constitutes the hard, gluonic core of the sheet,
responsible for the non-analyticity of the $\eta^{\prime}$ low-energy action.
The D8-brane gauge field provides the soft, mesonic (in particular
$\eta^{\prime}$) part of the sheet.

Concerning QCD, the holographic construction highlights the possibility of the
scenario where the sheet is realized in a Higgs phase. The latter corresponds
to the melting of the D6 in the D8 world-volume. In this paper we have instead
considered the scenario where the D6 are not melted, employing anyway the
simplified setting of antipodal D8 embedding.

After providing a few evidences for the proposal, e.g. the emergence of the
expected CS theory on the world-volume of the sheet, we have constructed two
explicit configurations. The first one corresponds to an infinitely extended
sheet without boundaries. The profile of the $\eta^{\prime}$, interpolating
from 0 to $2\pi$ on the two sides of the sheet, has been derived. The
holographic construction allows to calculate precisely the tension and
thickness of the sheet, both in the massless and massive-quark cases.

The second configuration we have considered is semi-infinite, with a straight
infinite boundary (where the D6 terminate on the D8). The boundary constitutes
a continuous distribution of magnetic charge for the five-dimensional Maxwell-
CS theory (on curved space) on the D8-brane. We solved for this interesting
problem both in flat space, where an analytic form for the gauge field can be
provided, and on the curved Witten background. In the latter case, we have
provided a semi-analytic solution. We have derived a four-dimensional
effective action for the fluctuating modes and calculated the tension and
thickness of the hard and soft part of the string-like boundary of the sheet.

Finally, we have observed that one can switch on a chiral component of the
gauge field on the boundary having zero energy, which is possibly connected to
the chiral edge mode one could consider on the boundary of the Hall droplet.

We list a few directions for possible developments of the work presented in
this paper.

  * •

A few issues that we left open include a better understanding of the
$c_{\pm}(x_{\pm})$ mode in relation to the chiral edge mode; the exploration
of other choices of the Dirac sheet and the explicit construction of a Wu-Yang
monopole-like solution without Dirac sheet; the possibility of the existence
of a more general solution relaxing the constraint $d^{\star}F^{m}=0$.

  * •

It would be interesting to study the D6 system as a BIon-like configuration on
the world-volume of the D8, following the idea presented in [27] and briefly
explored in the appendix.

  * •

As already remarked, one should really study the system for the non-antipodal
D8-brane embedding, to be sure that the D6-brane is not dissolved on the D8.
Moreover, one could try and figure out how the system of D8 plus dissolved D6
looks like.

  * •

Eventually the main goal is to find a solution for the baryon, with disk-
shaped sheet and the chiral edge mode turned on. This configuration could be
quantized to derive the baryon spectrum.

  * •

It would be interesting to work out a solution in the deconfined, chiral
symmetry breaking phase [23], using the black brane background.

Acknowledgments

We are indebted with Ofer Aharony, Zohar Komargodski and Cobi Sonnenschein for
their collaboration at the beginning of this project. We also thank Andrea
Cappelli and Domenico Seminara for comments and very helpful discussions.

##  Appendix A Including the scalar mode

In section 5 we have ignored the scalar mode on the D8 world-volume
parameterizing its transverse direction. Here we consider its contribution at
the lowest (quadratic) order (as we have done with the field strength). Let us
call this mode $\phi$. At leading order its effect is to shift the Lagrangian
density as

${\cal L}_{0}\rightarrow{\cal
L}_{0}+\frac{1}{2}\sqrt{g}\partial_{M}\phi\partial^{M}\phi\,,$ (A.1)

with $M,N=+,-,2,3,z$. We consider the ansatz

$\phi=\phi(x_{2},x_{3},z,x_{\pm})\,.$ (A.2)

The equation for the scalar mode

$\partial_{M}\left(\sqrt{-g}\,g^{MN}\partial_{N}\phi\right)=0,$ (A.3)

gives

$\partial_{z}\left(g(z)^{5/2}\partial_{z}\phi\right)+g(z)^{1/2}\left(\partial_{x_{3}}^{2}+\partial_{x_{2}}^{2}\right)\phi=0.$

We can try to solve this equation through a series expansion of $\phi$, as we
did for the magnetic field strength. The ansatz for the solution reads

$\phi=a(x_{\pm})\sum\limits_{n=0}^{\infty}\xi_{n}(z)\xi_{n}(0)J_{n}(r),$ (A.4)

with a generic function $a(x_{\pm})$ as found for the field strength of the
charged mode. Then, the equation of motion becomes

$\displaystyle\sum\limits_{n=0}^{\infty}\left(\partial_{z}(g(z)^{5/2}\partial_{z}\xi_{n}(z))\xi_{n}(0)J_{n}(r)+g(z)^{1/2}\xi_{n}(z)\xi_{n}(0)\left(\partial_{x_{3}}^{2}+\partial_{x_{2}}^{2}\right)J_{n}(r)\right)=0.$
(A.5)

We can define $\widetilde{J}_{n}(r)\equiv\xi_{n}(0)J_{n}(r)$, in order to
absorb the $n^{th}$ factor $\xi_{n}(0)$. The above equation can be solved
requiring that $\xi_{n}(z)$ and $\widetilde{J}_{n}$ are solutions to the
following equations

$\displaystyle\left(\left(\partial_{x_{3}}^{2}+\partial_{x_{2}}^{2}\right)-\tau_{n}\right)\widetilde{J}_{n}(r)=0,$
(A.6)
$\displaystyle-g^{-1/2}(z)\partial_{z}(g(z)^{5/2}\partial_{z}\xi_{n}(z))=\tau_{n}\xi_{n}(z).$
(A.7)

The first one has a solution analogous to the $Y_{n}$:

$\widetilde{J}_{n}(r)=\mathcal{N}K_{0}\left(\sqrt{\tau_{n}}r\right)\,\,\,\text{if}\,\,\,n>0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\widetilde{J}_{0}(r)=-\mathcal{N}\ln(r)\,\,\,\text{if}\,\,\,n=0,$
(A.8)

with $\mathcal{N}$ a normalization constant determined by requiring the on-
shell action to be an effective action for the modes $\widetilde{J}_{n}$
canonically normalized. The eigenvalue equation for the functions $\xi_{n}(z)$
has to be addressed with the following orthonormality condition

$\displaystyle\int dz\,g(z)^{1/2}\xi_{n}(z)\xi_{m}(z)=\delta_{nm}.$ (A.9)

The profile of the scalar grows big close to the line source, consistently
with a BIon-like behavior. Of course, as for the gauge field, we cannot thrust
the solution close to the source.

Then, we can compute the on-shell action for the scalar mode

$S=-\kappa\sum\limits_{n=0}^{\infty}\int\,d^{4}x\,a(x_{\pm})\left(\partial_{i}\widetilde{J}_{n}(r)\partial_{i}\widetilde{J}_{n}(r)+\tau_{n}\widetilde{J}_{n}(r)\widetilde{J}_{n}(r)\right).$
(A.10)

This action looks just the same as (5.42),141414The energy associated to this
action has the same divergence as the tension of the vortex string. but with
different profiles $\widetilde{J}_{n}$. However, we have also the non trivial
function $a(x_{\pm})$. The BPS case, which is described in the paper [27] by
Callan and Maldacena, is realized by taking $a(x_{\pm})$ constant. We leave
this direction to a future study.

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