# The D4/D8 model and holographic QCD

###### Abstract

As a top-down holographic approach, the D4/D8 model is expected to be the
holographic version of QCD since it almost includes all the elementary
features of QCD based on string theory. In this manuscript, we review the
fundamental properties of the D4/D8 model with respect to the D4-brane
background, embedding of flavor branes and holographic quark, gluon, meson,
baryon and glueball with various symmetries, then we also take a look at some
interesting applications and developments based on this model.

Keywords: Gauge-gravity duality; AdS/CFT correspondence; Holographic QCD

_∗_ Si-wen Li111Email<EMAIL_ADDRESS>_†_ Xiao-tong Zhang222Email:
<EMAIL_ADDRESS>
_∗ Department of Physics, College of Science,_

_Dalian Maritime University,_

_Dalian 116026, China_

_† College of Innovation and Entrepreneurship ,_

_Dalian Maritime University,_

_Dalian 116026, China_

###### Contents

  1. 1 Introduction
  2. 2 The D4/D8 model
    1. 2.1 11d supergravity and D4-brane background
    2. 2.2 Embedding the probe $\mathrm{D8/\overline{D8}}$-branes
    3. 2.3 Gluon, quark and symmetries
    4. 2.4 Mesons on the flavor brane
    5. 2.5 The wrapped D4-brane and baryon vertex
    6. 2.6 Gravitational wave as glueball
  3. 3 Developments and holographic approaches to QCD
    1. 3.1 QCD deconfinement transition
    2. 3.2 Phase diagram with chiral transition
    3. 3.3 Higgs mechanism and heavy-light meson field
    4. 3.4 Interaction of hadron and glueball
    5. 3.5 Theta dependence in QCD
  4. 4 Summary and outlook
  5. 5 Appendix A: The type II supergravity solution
  6. 6 Appendix B: Dimensional reduction for spinors
  7. 7 Appendix C: Supersymmetric meson on the flavor brane

##  1 Introduction

Although it has been about 25 years since the proposal of AdS/CFT and gauge-
gravity duality with holography [1, 2, 3], it remains to attract great
interests today. The most significant part of AdS/CFT and gauge-gravity
duality is that people can evaluate strongly coupled quantum field theory
(QFT) by analyzing quantitatively its associated gravity theory in the weak
coupling region. So it provides a holographic way to study the strongly
coupled QFT in which traditional QFT based on perturbation method is out of
reach. Accordingly, a large number of publications about strongly coupled QFT
through AdS/CFT and gauge-gravity duality have appeared, for example on Wilson
loop and quark potential [4, 5, 6], transport coefficient [7, 8, 9], fermionic
correlation function [10, 11], the Schwinger effect [12], quantum entanglement
entropy [13, 14] and quantum information on black hole [15] which have become
most remarkable works in this field.

On the other hand, QCD (quantum chromodynamics) as the fundamental theory
describing the property of strong interaction is extremely complex in the
strong coupling region, especially at finite temperature with dense matter due
to its asymptotic freedom [16, 17], hence the holographic version of QCD is
natural to become an interesting topic. While there are several models and
theories attempting to give a holographic version of QCD (e.g. bottom-up
approaches [18, 19, 20], D3/D7 approach [21], D4/D6 approach [22]), one of the
most successful achievements in holography is the D4/D8 model (also named as
Witten-Sakai-Sugimoto model) [23, 24, 25] which includes almost all the
elementary features of QCD in a very simple way, e.g. meson, baryon [26, 27,
28], glueball [29, 30, 31, 32], deconfinement transition [33, 34, 35], chiral
phase [36, 37], heavy flavor [38, 39, 40], $\theta$ term and QCD axion [41,
42, 43, 44, 45], nucleon interaction [46, 47, 48, 49, 50, 51, 52]. The D4/D8
model is based on the holographic duality between the 11-dimensional (11d)
M-theory on $\mathrm{AdS_{7}}\times S^{4}$ and the
$\mathcal{N}=\left(2,0\right)$ super conformal field theory (SCFT) on $N_{c}$
M5-branes [53]. By using the dimensional reduction in [23, 54], it reduces to
the correspondence of the pure Yang-Mills theory on $N_{c}$ D4-branes
compactified on a circle and 10d IIA supergravity (SUGRA). Flavors as $N_{f}$
pairs of D8- and anti D8-branes ($\mathrm{D8/\overline{D8}}$) can be further
introduced into the geometric background produced by the $N_{c}$ D4-branes, so
the dual theory includes flavored fundamental quarks which would be more close
to the realistic QCD. Beside, as the D4/D8 model is a T-dualized version of
D3/D9 system [55], the fundamental quark and meson in the D4/D8 model can
therefore be identified to the $4-8$333The $4-8$ string refers to the open
string connects the $N_{c}$ D4-brane and $N_{f}$ D8-branes. And it is similar
for e.g. the $8-8$ or $4-\bar{8}$ string. and $8-8$ strings respectively by
following the same discussion in D3/D9 system [55]. Moreover, the baryon
vertex is identified as a D4-brane wrapped on $S^{4}$ [26] and the glueball is
recognized to be the bulk gravitational polarization [29, 30]. The chiral
phase is determined by the embedding configuration of the
$\mathrm{D8/\overline{D8}}$-branes due to the gauge symmetry on their
worldvolume [56] while the deconfinement transition is suggested to be the
Hawking-Page transition in this model [33, 34, 35, 36]. Altogether, this model
includes all the fundamental elements of QCD thus can be treated as a
holographic version of QCD.

In this review, we will revisit the above properties of the D4/D8 model, then
take a brief look at some recently relevant developments and holographic
approaches with this model. The outline of this review is as follows. In
Section 2, we will review the relation of 11d M-theory and IIA SUGRA with
respect to the case of M5-brane and D4-brane. Afterwards, it is the embedding
configuration of the $\mathrm{D8/\overline{D8}}$-branes to the D4-brane
background, the holographic quark, gluon, meson, baryon and glueball with
various symmetries which are all the relevant objects in hadron physics. In
Section 3, we review several topics about the developments and holographic
approaches of this model which includes deconfinement transition, chiral
transition, Higgs mechanism and heavy-light meson or baryon, interaction
involving glueball and QCD $\theta$ term in holography. In the appendix, we
give the general form of the black brane solution in type II SUGRA, the
relevant dimensional reduction for spinor and discussion about supersymmetric
meson which are useful to the main content of this paper.

##  2 The D4/D8 model

In this section, we will revisit the D4-brane background and the embedding of
the probe $\mathrm{D8/\overline{D8}}$-branes. Then we will review how to
identify quarks, gluon, meson, baryon and glueball with various symmetries in
this model.

###  2.1 11d supergravity and D4-brane background

The D4-brane background of the D4/D8 model is based on the holographic duality
between the type $\mathcal{N}=\left(2,0\right)$ super conformal field theory
(SCFT) on coincident $N_{c}$ M5-branes and 11-dimensional (11d) M-theory on
$\mathrm{AdS}_{7}\times S^{4}$ [53]. In order to obtain a geometric solution,
the effective action of the M-theory is necessary which is known as the 11d
supergravity action. In the large-$N_{c}$ limit, the geometric background can
be obtained by solving its bosonic part which is consisted of metric (elfbein)
and a three-from $C_{3}$ as [57],

$S_{\mathrm{SUGRA}}^{11\mathrm{d}}=\frac{1}{2\kappa_{11}^{2}}\int
d^{11}x\sqrt{-g}\left[\mathcal{R}^{(11)}-\frac{1}{2}\left|F_{4}\right|^{2}\right]-\frac{1}{12\kappa_{11}^{2}}\int
C_{3}\wedge F_{4}\wedge F_{4},$ (2.1)

where $F_{4}=dC_{3}$. The convention in (2.1) is as follows.
$\mathcal{R}^{(11)}$ refers to the 11d scalar curvature, $\kappa_{11}$ is the
11d gravity coupling constant given by,

$2\kappa_{11}^{2}=16\pi G_{11}=\frac{1}{2\pi}\left(2\pi l_{p}\right)^{9},$
(2.2)

where $G_{11}$ is 11d Newton’s constant, $l_{p}$ is the Planck length. The
quantity $\left|F_{4}\right|^{2}$ can be obtained by a general notation of an
$n$-form $F_{n}$ as,

$\left|F_{n}\right|^{2}=\frac{1}{n!}g^{A_{1}B_{1}}g^{A_{2}B_{2}}...g^{A_{n}B_{n}}F_{A_{1}A_{2}...A_{n}}F_{B_{1}B_{2}...B_{n}},$
(2.3)

where $g_{AB}$ refers to the metric on the manifold. We note that in (2.1) the
last term is a Chern-Simons structure which is independent on the metric or
elfbein while the first term depends on the metric or elfbein through the
metric combination

$g_{AB}=e_{A}^{a}\eta_{ab}e_{B}^{b}.$ (2.4)

The solution for extremal coincident $N_{c}$ M5-branes is obtained as,

$\displaystyle ds_{\mathrm{11d}}^{2}=$ $\displaystyle
H_{5}\left(r\right)^{-\frac{1}{3}}\eta_{MN}dx^{M}dx^{N}+H_{5}\left(r\right)^{\frac{2}{3}}\left(dr^{2}+r^{2}d\Omega_{4}^{2}\right),$
$\displaystyle H_{5}\left(r\right)=$ $\displaystyle
1+\frac{r_{5}^{3}}{r^{3}},\
\left({}^{*}F_{4}\right)_{r012...5}=-\frac{3r_{5}^{3}}{r^{4}H_{5}^{2}},$ (2.5)

where $r$ denotes the radial coordinate to the M5-branes and $M,N$ run over
the M5-branes. Using the BPS condition for M5-brane,

$2\kappa_{11}^{2}T_{\mathrm{M5}}N_{c}=\int_{S^{4}}F_{4},$ (2.6)

it leads to

$r_{5}^{3}=\pi N_{c}l_{p}^{3},$ (2.7)

where $T_{\mathrm{M5}}=\frac{1}{\left(2\pi\right)^{5}l_{p}^{6}}$ refers to the
tension of the M5-brane. Taking the near horizon limit
$H_{5}\rightarrow\frac{r_{5}^{3}}{r^{3}}$ and replacing the variables as

$\left\\{r,r_{5},x^{i},\Omega_{4}\right\\}\rightarrow\left\\{r^{2},\frac{L}{2},\frac{1}{\sqrt{L}}x^{i},\frac{1}{\sqrt{L}}\Omega_{4}\right\\},$
(2.8)

the metric presented in (2.5) reduces to,

$ds_{\mathrm{11d}}^{2}=\frac{r^{2}}{L^{2}}\eta_{MN}dx^{M}dx^{N}+\frac{L^{2}}{r^{2}}\left(dr^{2}+\frac{r^{2}}{4}d\Omega_{4}^{2}\right),$
(2.9)

describing the standard form of $\mathrm{AdS}_{7}\times S^{4}$ where the
radius of $S^{4}$ is $L/2$. In addition, the action (2.1) also allows the
near-extremal M5-brane solution which, after taking near horizon limit and
replacement (2.8), is

$\displaystyle ds_{\mathrm{11d}}^{2}$
$\displaystyle=\frac{r^{2}}{L^{2}}\left[-f\left(r\right)\left(dx^{0}\right)^{2}+\sum_{i=1}^{5}dx^{i}dx^{i}\right]+\frac{L^{2}}{r^{2}}\left[\frac{dr^{2}}{f\left(r\right)}+\frac{r^{2}}{4}d\Omega_{4}^{2}\right],$
$\displaystyle f\left(r\right)$ $\displaystyle=1-\frac{r_{H}^{6}}{r^{6}},$
(2.10)

The constant $r_{H}$ refers to the location of the horizon which can be
determined by omitting the conical singularity as,

$\beta_{T}=\frac{2\pi L^{2}}{3r_{H}},$ (2.11)

where $\beta_{T}$ is the size of the compactified direction $x^{0}$.

In order to obtain a QCD-like low-energy theory in holography, Witten proposed
a scheme in [23] based on the above M5-brane solution. Specifically, the first
step is to campactify one spacial direction (denoted by $x^{5}$) of M5-branes
on a circle with periodic condition for fermions, which means the
supersymmetry remains. Accordingly the resultant theory is a supersymmetric
gauge theory above the size of the circle. Then, recall the relation between
M-theory and IIA string theory, the 11d metric presented in (2.9) or (2.10)
reduces to a 10d metric as,

$ds_{\mathrm{11d}}^{2}=e^{-\frac{2}{3}\phi}ds_{\mathrm{10d}}^{2}+e^{\frac{4}{3}\phi}\left(dx^{5}\right)^{2},$
(2.12)

with the non-trivial dilaton $e^{\phi}=\left(r/L\right)^{3/2}$. For the later
use, let us introduce another radial coordinates
$U\in\left[U_{H},\infty\right)$ by

$U=\frac{r^{2}}{2L},L=2R.$ (2.13)

So in the large $N_{c}$ limit, the 11th direction $x^{5}$ presented in (2.12)
vanishes due to (2.7) (2.8), which means the coincident $N_{c}$ M5-branes
corresponds to coincident $N_{c}$ D4-branes for $N_{c}\rightarrow\infty$. And
the remained 10d metric in (2.12) becomes the near-extremal black D4-brane
solution as444The extremal D4-brane solution can be obtained by setting
$U_{H}\rightarrow 0$.,

$\displaystyle ds_{\mathrm{10d}}^{2}$
$\displaystyle=\left(\frac{U}{R}\right)^{3/2}\left[-f_{T}\left(U\right)\left(dx^{0}\right)^{2}+\sum_{i=1}^{4}dx^{i}dx^{i}\right]+\left(\frac{R}{U}\right)^{3/2}\left[\frac{dU^{2}}{f_{T}\left(U\right)}+U^{2}d\Omega_{4}^{2}\right],$
$\displaystyle f_{T}\left(U\right)$
$\displaystyle=1-\frac{U_{H}^{3}}{U^{3}},F_{4}=3R^{3}g_{s}^{-1}\epsilon_{4},e^{\phi}=\left(\frac{U}{R}\right)^{3/4},R^{3}=\pi
g_{s}N_{c}l_{s}^{3}$ (2.14)

where $\epsilon_{4}$ refers to the volume form of a unit $S^{4}$. Once the
formula (2.12) is imposed to action (2.1), the 11d SUGRA action reduces to the
10d type IIA SUGRA action exactly which is given as,555There could be a Chern-
Simons term to the IIA SUGRA action (2.15) as
$S_{\mathrm{CS}}^{\mathrm{10d}}=-\frac{1}{4\kappa_{10}^{2}}\int B_{2}\wedge
F_{4}\wedge F_{4}$ with $B_{mn}=A_{mn5}$. For the purely black brane solution,
the $B_{2}$ field can be gauged away by setting $B_{2}=0$ which implies this
CS term could be absent in 10d action [53, 57].,

$S_{\mathrm{IIA}}^{10\mathrm{d}}=\frac{1}{2\kappa_{10}^{2}}\int
d^{10}x\sqrt{-G}e^{-2\phi}\left[\mathcal{R}^{(10)}+4\partial_{\mu}\phi\partial\phi\right]-\frac{1}{4\kappa_{10}^{2}}\int
d^{10}x\sqrt{-G}\left|F_{4}\right|^{2},$ (2.15)

where $2\kappa_{10}^{2}=16\pi G_{10}=\left(2\pi\right)^{7}l_{s}^{8}g_{s}^{2}$
is the 10d gravity coupling constant. And it would be straightforward to
verify that the solution (2.14) satisfies the equation of motion obtained by
varying action (2.15).

The next step is to perform the double Wick rotation
$\left\\{x^{0}\rightarrow-ix^{4},x^{4}\rightarrow-ix^{0}\right\\}$ on the
metric (2.14) leading to a bubble D4-brane solution as,

$\displaystyle ds_{\mathrm{10d}}^{2}$
$\displaystyle=\left(\frac{U}{R}\right)^{3/2}\left[\eta_{\mu\nu}dx^{\mu}dx^{\nu}+f\left(U\right)\left(dx^{4}\right)^{2}\right]+\left(\frac{R}{U}\right)^{3/2}\left[\frac{dU^{2}}{f\left(U\right)}+U^{2}d\Omega_{4}^{2}\right],$
$\displaystyle f\left(U\right)$ $\displaystyle=1-\frac{U_{KK}^{3}}{U^{3}},$
(2.16)

which is defined only for $U\in\left[U_{KK},\infty\right)$. We have renamed
$U_{H}$ as $U_{KK}$ in (2.16) since there is not a horizon in the bubble
solution as it is illustrated in Figure 1 .

Figure 1:  The compactified D4-brane geometry on $U-x^{4}$ plane. Left: The
black D4-brane background geometry. Right: The D4 bubble geometry.

Now the direction of $x^{4}$ is periodic as,

$x^{4}\sim x^{4}+\delta x^{4},\delta x^{4}=\frac{2\pi}{M_{KK}}=\frac{4\pi
R^{3/2}}{3U_{KK}^{1/2}},$ (2.17)

because it is identified as the time direction in the black brane solution
(2.14). $M_{KK}$ refers to the Klein-Kaluza (KK) energy scale and the
supersymmetry on the D4-branes breaks down below $M_{KK}$ by imposing the
anti-periodic condition to the supersymmetric fermions along $x^{4}$.
Accordingly, the low-energy zero modes on the D4-branes are the massless gauge
field on the D4-branes and the scalar fields as the transverse modes of the
D4-branes. While the scalar fields acquire mass via one-loop corrections, the
trace part of the scalars $\phi^{i}$ and gauge field along $x^{4}$ direction
$a_{4}$ remain to be massless. As they give irrelevant coupling terms in the
low-energy effective theory on the D4-branes, it means the dual theory below
$M_{KK}$ only contains 4d pure Yang-Mills gauge field as it is expected. Note
that the three-from $C_{3}$ in 11d SUGRA (2.1) corresponds to the Ramond-
Ramond (R-R) three-form in the type IIA string theory.

Besides, as the wrap factor $\left(U/R\right)^{3/2}$ in (2.16) never goes to
zero, the dual theory would be able to exhibit confinement according to the
behavior of the Wilson loop in this geometry. Since the solution (2.16) allows
an arbitrarily large period for $x^{0}$, it implies the dual theory on the
D4-brane could be defined at zero (or very low) temperature. Furthermore, in
order to obtain a deconfined version of holographic QCD based on (2.16) at
finite temperature, it is also possible to compactify one spacial direction
(denoted by $x^{4}$) of the D4-branes in the background (2.14) with the anti-
periodic condition for the supersymmetric fermions666There might be an issue
if we identify the black brane background (2.14) to the deconfinement phase
exactly since Wilson loop on this background may mot match to the
deconfinement QCD [58, 59]. Nevertheless, we can identify the black brane
background (2.14) to QCD phase at high temperature in which the deconfinement
will occur. as it is displayed in (2.17) and Figure 1. In this case, the
Hawking temperature $T$ in compactified background (2.14) is given by (2.11)
as,

$\beta_{T}=\frac{1}{T}=\frac{4\pi R^{3/2}}{3U_{H}^{1/2}},$ (2.18)

which can therefore be identified as the temperature in the dual theory. And
the variables in terms of the dual theory is summarized as,

$R^{3}=\frac{1}{2}\frac{g_{\mathrm{YM}}^{2}N_{c}l_{s}^{2}}{M_{KK}},\
U_{KK}=\frac{2}{3}g_{\mathrm{YM}}^{2}N_{c}M_{KK}l_{s}^{2},\
g_{s}=\frac{1}{2\pi}\frac{g_{\mathrm{YM}}^{2}}{M_{KK}l_{s}},$ (2.19)

where $g_{\mathrm{YM}}$ to the Yang-Mills (YM) coupling constant.

###  2.2 Embedding the probe $\mathrm{D8/\overline{D8}}$-branes

In the D4/D8 model, there is a stack of coincident $N_{f}$ pairs of D8- and
(anti-D8) $\overline{\mathrm{D8}}$-branes as probes embedded into the bulk
geometry illustrated in Figure 1. The relevant D-brane configuration is given
in Table 1.

| 0 | 1 | 2 | 3 | 4 | 5($U$) | 6 | 7 | 8 | 9
---|---|---|---|---|---|---|---|---|---|---
$N_{c}$ D4-branes | - | - | - | - | - |  |  |  |  |
$N_{f}$ $\mathrm{D8/\overline{D8}}$-branes | - | - | - | - |  | - | - | - | - | -
Table 1:  The D-brane configuration in the D4/D8 model. “-” denotes the
D-brane extends along this direction.

The embedding configuration of $\mathrm{D8/\overline{D8}}$-branes is
determined by solving the bosonic action for a $\mathrm{D}_{p}$-branes, which
consists of Dirac-Born-Infeld (DBI) and Wess-Zumino (WZ) terms. The action
reads [60],

$\displaystyle S_{\mathrm{D_{p}}}$
$\displaystyle=S_{\mathrm{DBI}}+S_{\mathrm{WZ}},$ $\displaystyle
S_{\mathrm{DBI}}$
$\displaystyle=-T_{\mathrm{D}_{p}}\int_{\mathrm{D}_{p}}d^{p+1}xe^{-\phi}\mathrm{STr}\left[\sqrt{-\det\left[E_{ab}+E_{aI}\left(Q^{IJ}-\delta^{IJ}\right)E_{Ib}+2\pi\alpha^{\prime}F_{ab}\right]}\sqrt{\det\left(Q_{\
J}^{I}\right)}\right],$ $\displaystyle S_{\mathrm{WZ}}$
$\displaystyle=g_{s}T_{\mathrm{D}_{p}}\int_{\mathrm{D}_{p}}\sum_{n=0,1...}C_{p-2n+1}\wedge\frac{1}{n!}\left(2\pi\alpha^{\prime}\right)^{n}\mathrm{Tr}F^{n},$
(2.20)

with the D-brane tension
$T_{\mathrm{D}_{p}}=g_{s}^{-1}\left(2\pi\right)^{-p}l_{s}^{-p-1}$ and
$e^{\Phi}=g_{s}e^{\phi}$,

$\displaystyle E_{ab}=$
$\displaystyle\left(G_{MN}+B_{MN}\right)\partial_{a}X^{M}\partial_{b}X^{N},\
E_{aI}=\left(G_{MI}+B_{MI}\right)\partial_{a}X^{M},$ $\displaystyle Q_{\
J}^{I}=$ $\displaystyle\delta_{\
J}^{I}+i2\pi\alpha^{\prime}\left[\Psi^{I},\Psi^{J}\right],\
F_{ab}=\partial_{a}A_{b}-\partial_{b}A_{a}+i\left[A_{a},A_{b}\right],$
$\displaystyle a,b=$ $\displaystyle 0,1...p,\ M,N=0,1...d,\ I,J=p+1,p+2...d.$
(2.21)

Here $G_{MN},B_{MN}$ and $\Phi$ refers respectively to metric, the
antisymmetric tensor and dilaton field in the background spacetime. $\Psi^{I}$
refers to the transverse mode of the $\mathrm{D}_{p}$-brane under the
T-duality. By choosing $p=8$, the action (2.20) leads to the action for
D8-brane on the $N_{c}$ D4-brane background as777In the D4/D8 approach, the
antisymmetric tensor $B_{MN}$ has been gauged away. ,

$S_{\mathrm{D8}}=-T_{\mathrm{D8}}\int_{\mathrm{D8}}d^{9}xe^{-\phi}\mathrm{STr}\sqrt{-\det\left[g_{ab}+\left(2\pi\alpha^{\prime}\right)F_{ab}\right]}+\frac{g_{s}}{3!}\left(2\pi\alpha^{\prime}\right)^{3}T_{\mathrm{D8}}\int_{\mathrm{D8}}C_{3}\wedge\mathrm{Tr}\left(F\wedge
F\wedge F\right).$ (2.22)

Using the induced metric on $\mathrm{D8/\overline{D8}}$-branes with respect to
the bubble D4 background (2.16),

$ds_{\mathrm{D8}}^{2}=\left(\frac{U}{R}\right)^{3/2}\eta_{\mu\nu}dx^{\mu}dx^{\nu}+\left(\frac{U}{R}\right)^{3/2}\left[f\left(U\right)\left(x^{4\prime}\right)^{2}+\left(\frac{R}{U}\right)^{3}\frac{1}{f\left(U\right)}\right]dU^{2}+R^{3/2}U^{1/2}d\Omega_{4}^{2},$
(2.23)

and the black D4-brane background (2.14) as,

$\displaystyle ds_{\mathrm{D8}}^{2}=$
$\displaystyle\left(\frac{U}{R}\right)^{3/2}\left[-f_{T}\left(U\right)\left(dx^{0}\right)^{2}+\delta_{ij}dx^{i}dx^{j}\right]$
$\displaystyle+\left(\frac{U}{R}\right)^{3/2}\left[\left(x^{4\prime}\right)^{2}+\left(\frac{R}{U}\right)^{3}\frac{1}{f_{T}\left(U\right)}\right]dU^{2}+R^{3/2}U^{1/2}d\Omega_{4}^{2},$
(2.24)

the DBI action for D8-branes becomes respectively,

$S_{\mathrm{DBI}}=-g_{s}T_{\mathrm{D8}}V_{3}\beta_{4}\Omega_{4}\int_{U_{KK}}^{\infty}dUU^{4}\left[f\left(U\right)\left(x^{4\prime}\right)^{2}+\left(\frac{R}{U}\right)^{3}\frac{1}{f\left(U\right)}\right]^{1/2},$
(2.25)

and

$S_{\mathrm{DBI}}=-g_{s}T_{\mathrm{D8}}V_{3}\beta_{T}\Omega_{4}\int_{U_{H}}^{\infty}dUU^{4}\left[\left(x^{4\prime}\right)^{2}f_{T}\left(U\right)+\left(\frac{R}{U}\right)^{3}\right]^{1/2}.$
(2.26)

Here we use $\Omega_{4}=8\pi^{2}/3$ to refer to the volume of a unit $S^{4}$.
Note that the WZ action is independent on the metric or elfbein. Vary the
action (2.25) and (2.26) with respect to $x^{4}$, the associated equation of
motion is respectively obtained as,

$\frac{d}{dU}\left[\frac{U^{4}f\left(U\right)x^{4\prime}}{\sqrt{f\left(U\right)\left(x^{4\prime}\right)^{2}+\left(\frac{R}{U}\right)^{3}\frac{1}{f\left(U\right)}}}\right]=0,$
(2.27)

and,

$\frac{d}{dU}\left[\frac{U^{4}f_{T}\left(U\right)x^{4\prime}}{\sqrt{\left(x^{4\prime}\right)^{2}f_{T}\left(U\right)+\left(\frac{R}{U}\right)^{3}}}\right]=0.$
(2.28)

As the D8- and $\overline{\mathrm{D8}}$-branes are the only probe branes, they
could be connected smoothly at the location $U=U_{0}$ which means
$x^{4}|_{U=U_{0}}\rightarrow\infty$. With this boundary condition, (2.27) and
(2.28) reduce respectively to the following solutions as,

$\left(x^{4\prime}\right)^{2}=\frac{U_{0}^{8}f\left(U_{0}\right)}{U^{3}f\left(U\right)^{2}}\frac{R^{3}}{U^{8}f\left(U\right)-U_{0}^{8}f\left(U_{0}\right)},$
(2.29)

and

$\left(x^{4\prime}\right)^{2}=\frac{U_{0}^{8}f_{T}\left(U_{0}\right)}{U^{3}f_{T}\left(U\right)}\frac{R^{3}}{U^{8}f_{T}\left(U\right)-U_{0}^{8}f_{T}\left(U_{0}\right)}.$
(2.30)

In particular, in the bubble D4-brane background, solution (2.29) implies
$x^{4}|_{U\rightarrow\infty,U_{0}\rightarrow U_{KK}}=\beta_{4}/4$ and
$x^{4\prime}|_{U_{0}=U_{KK}}=0$. Thus $x^{4}=\beta_{4}/4$ is a solution to
(2.29) representing D8- and $\overline{\mathrm{D8}}$-branes are located at the
antipodal points of $S^{1}$ while they are connected at $U=U_{KK}$, because
the size of $x^{4}$ shrinks to zero at $U=U_{KK}$. On the other hand, in the
black D4-brane background, if $U_{0}=U_{H}$, (2.30) also implies a constant
solution for $x^{4}$ while the separation of the D8- and
$\overline{\mathrm{D8}}$-branes could be arbitrary but no more than
$\beta_{4}/2$. For $U_{0}>U_{KK},U_{H}$, the solutions (2.28) (2.29) represent
D8- and $\overline{\mathrm{D8}}$-branes are joined into a single brane at
$U=U_{0}$. The configuration of the D8- and $\overline{\mathrm{D8}}$-branes in
bubble and black D4-brane background is illustrated in Figure 2 and 3.

Figure 2:  The D8-brane configuration in the bubble D4-brane background. Left:
D8- and $\overline{\mathrm{D8}}$-branes are located at the antipodal points of
$x^{4}$. Right: D8- and $\overline{\mathrm{D8}}$-branes are located at the
non-antipodal points of $x^{4}$. Figure 3:  The D8-brane configuration in the
black D4-brane background. Left: D8- and $\overline{\mathrm{D8}}$-branes are
parallel located. Right: D8- and $\overline{\mathrm{D8}}$-branes are connected
at $U=U_{0}$.

###  2.3 Gluon, quark and symmetries

As the dual theory in the D4/D8 model is expected to be QCD in the large
$N_{c}$ limit, it is natural to identify the effective theory on $N_{c}$
D4-branes below $M_{KK}$ to the color sector in QCD which implies the gauge
field $A_{\mu}^{\left(\mathrm{D4}\right)}$ on the D4-branes can be interpreted
as gluon in holography. The reason is that the low-energy theory on $N_{c}$
D4-branes is $U\left(N_{c}\right)$ pure Yang-Mills theory and it has a SUGRA
duality in the strong coupling region in the large $N_{c}$ limit as it is
discussed in Section 2.1. We note that the Lorentz symmetry of the 10d
spacetime breaks down to $SO\left(1+4\right)\times SO\left(5\right)$ when a
stack of D4-branes is introduced. However the worldvolume symmetry of the
D4-branes becomes $SO\left(1+3\right)$ since the D4-branes are compactified on
a circle in the D4/D8 model. Furthermore, when the flavors as D8- and
$\overline{\mathrm{D8}}$-branes are introduced, it is possible to create
chiral fermions in the low-energy theory which can be obtained by analyzing
the spectrum of $4-8$ or $4-\bar{8}$ string in R-sector (Ramond-sector). Both
the spectra of $4-8$ and $4-\bar{8}$ string in R-sector contain spinors with
positive and negative chirality as the representations of the Lorentz group
$SO\left(1,3\right)$. Since the GSO (Gliozzi-Scherk-Olive) projection will
remove the spinor with one of the chiralities in string theory, we can choose
the spinor with positive and negative chirality as the massless fermionic
modes (denoted by $q_{L,R}$) of $4-8$ and $4-\bar{8}$ string respectively
which accordingly can be identified as the fundamental chiral quarks in the
dual theory. We note that these chirally fermionic fields are complex spinors
since the $4-8$ and $4-\bar{8}$ strings have two orientations. And they are
also the fundamental representation of $U\left(N_{c}\right)$ and
$U\left(N_{f}\right)$. The massless modes and symmetries in the D4/D8 system
are collected in Table 2.

Fields | $U\left(N_{c}\right)$ | $SO\left(1,3\right)$ | $SO\left(5\right)$ | $U\left(N_{f}\right)_{L}\times U\left(N_{f}\right)_{R}$
---|---|---|---|---
$A_{\mu}^{\left(\mathrm{D4}\right)}$ | adj. | 4 | 1 | (1,1)
$q_{L}$ | fund. | 2+ | 1 | (fund.,1)
$q_{R}$ | fund. | 2- | 1 | (1,fund.)
$a_{4}$ | 1 | 1 | 1 | (1,1)
$\phi^{i}$ | 1 | 1 | 5 | (1,1)
Table 2:  The fields in the D4/D8 model. Here 2+, 2- denote the positive and
negative chirality spinor representations of $SO\left(1,3\right)$. $a_{4}$ and
$\phi^{i}$ are the trace parts of the gauge field along $x^{4}$ direction on
the $N_{c}$ D4-branes and the transverse modes of the $N_{c}$ D4-branes which
are decouple to the gluon and fundamental quarks in the low-energy theory.

Due to the above holographic correspondence, the chirally symmetric and broken
phase in the dual theory can be identified respectively to the disconnected
and connected configuration of the $\mathrm{D8/\overline{D8}}$-branes. It
would be clear if we employ the configuration presented in Figure 2 for
example. The effective action for the gauge fields
$A_{\mu}^{\left(\mathrm{D4}\right)}$ and fundamental fermions $q_{L,R}$ on the
$N_{c}$ D4-branes with $N_{f}$ $\mathrm{D8/\overline{D8}}$-branes can be
evaluated by expanding the DBI action which leads to,

$\displaystyle S=$
$\displaystyle\int_{\mathrm{D4}}d^{5}x\sqrt{-g}\left[\delta\left(x^{4}-X_{L}\right)q_{L}^{\dagger}\bar{\sigma}^{\mu}\left(i\nabla_{\mu}+A_{\mu}\right)q_{L}+\delta\left(x^{4}-X_{R}\right)q_{R}^{\dagger}\bar{\sigma}^{\mu}\left(i\nabla_{\mu}+A_{\mu}\right)q_{R}\right]$
$\displaystyle-\frac{1}{4g_{\mathrm{YM}}^{2}}\int_{\mathrm{D4}}d^{5}x\sqrt{-g}\mathrm{Tr}F_{MN},\
M,N=0,1...4,$ (2.31)

where $X_{L,R}$ denotes the intersection of the D4- and D8-branes, D4- and
$\overline{\mathrm{D8}}$-branes and we have omitted the notation “D4” in
$A_{\mu}$. As all the fields depend on $\left\\{x^{\mu},x^{4}\right\\}$,
$q_{L}$ is identified to be $q_{R}$ if $X_{L}=X_{R}$ which leads to an action
with single flavor symmetry $U\left(N_{f}\right)$. For the connected
configurations, we can therefore see the D8- and
$\overline{\mathrm{D8}}$-branes are separated at high energy
($U\rightarrow\infty$, $X_{L}\neq X_{R}$) according to the solutions (2.29)
and (2.30), which leads to an approximated $U\left(N_{f}\right)_{L}\times
U\left(N_{f}\right)_{R}$ chiral symmetry. However, at low energy
$(U\rightarrow U_{0},X_{L}\rightarrow X_{R})$, D8- and
$\overline{\mathrm{D8}}$-branes are joined into a single pair of D8-branes at
$U=U_{0}$ ($X_{L}=X_{R}$) which implies the $U\left(N_{f}\right)_{L}\times
U\left(N_{f}\right)_{R}$ symmetry breaks down to a single
$U\left(N_{f}\right)$. This configuration of
$\mathrm{D8/\overline{D8}}$-branes provides nicely a geometric interpretation
of chiral symmetry in this model [56].

###  2.4 Mesons on the flavor brane

As meson is the bound state in the adjoint representation of the chiral
symmetry group, it is identified as the gauge field on the flavor branes which
is the massless mode excited by $8-8$ string888Massless mode excited by
$\bar{8}-\bar{8}$ string is therefore identified as anti-meson. . The reason
is that the gauge field excited by $8-8$ (and $\bar{8}-\bar{8}$) is the
generator of $U\left(N_{f}\right)_{L}$ (and $U\left(N_{f}\right)_{R}$). Hence
we consider the gauge field on the flavor branes with non-zero components as
$A_{M}=\left\\{A_{\mu}\left(x,z\right),A_{z}\left(x,z\right)\right\\},\mu=0,1...3$
in the bubble D4-brane background (2.16). We note that, while the
supersymmetry on $N_{c}$ D4-branes breaks down by compactifying $x^{4}$ on a
circle, there is not any mechanism to break down the supersymmetry on
$\mathrm{D8/\overline{D8}}$-branes since $\mathrm{D8/\overline{D8}}$-branes
are vertical to $x^{4}$. Therefore the $8-8$ string is supersymmetric leading
to a super partner fermion $\Psi$ of the gauge field $A_{M}$ in the low-energy
theory. And we will see in Appendix C, this supersymmetric fermion is Majorana
spinor which leads to the fermionic meson (mesino) while they are absent in
the realistic QCD.

Nonetheless let us assume the supersymmetry on the flavor branes somehow
breaks down and the supersymmetric meson can be turned off in order to
continue the discussion about the QCD sector of this model. Since the
D8-branes are probes, the worldvolume gauge field is fluctuation. Thus the
effective action for $A_{M}$ can be obtained by expanding (2.22) which, for
Abelian case $N_{f}=1$, is

$\displaystyle S_{\mathrm{D8}}=$ $\displaystyle-
T_{\mathrm{D8}}\int_{\mathrm{D8}}d^{9}xe^{-\phi}\sqrt{-\det\left[g_{ab}+\left(2\pi\alpha^{\prime}\right)F_{ab}\right]}$
$\displaystyle=$ $\displaystyle-
T_{\mathrm{D8}}\int_{\mathrm{D8}}d^{9}x\sqrt{-g}e^{-\phi}\left[1+\frac{1}{4}\left(2\pi\alpha^{\prime}\right)^{2}F_{MN}F^{MN}+\mathcal{O}\left(F^{4}\right)\right],$
(2.32)

leading to the Yang-Mills (YM) action as,

$\displaystyle S_{\mathrm{YM}}=$ $\displaystyle-
T_{\mathrm{D8}}\int_{\mathrm{D8}}d^{9}x\sqrt{-g}e^{-\phi}\frac{1}{4}\left(2\pi\alpha^{\prime}\right)^{2}F_{MN}F^{MN}$
$\displaystyle=$
$\displaystyle-\frac{2}{3}R^{3/2}U_{KK}^{1/2}\left(2\pi\alpha^{\prime}\right)^{2}T_{\mathrm{D8}}\Omega_{4}\int
d^{4}xdz\left(\frac{R^{3}}{4U}\eta^{\mu\rho}\eta^{\nu\sigma}F_{\mu\nu}F_{\rho\sigma}+\frac{9U^{3}}{8U_{KK}}\eta^{\mu\nu}F_{\mu
z}F_{\nu z}\right)$ $\displaystyle=$ $\displaystyle-\kappa\int
d^{4}xdZ\left(\frac{1}{2}K^{-1/3}\eta^{\mu\rho}\eta^{\nu\sigma}F_{\mu\nu}F_{\rho\sigma}+KM_{KK}^{2}\eta^{\mu\nu}F_{\mu
Z}F_{\nu Z}\right),$ (2.33)

where we have used the Cartesian coordinates $z$ and dimensionless $Z$ defined
as,

$U^{3}=U_{KK}^{3}+U_{KK}z^{2},Z=\frac{z}{U_{KK}},K\left(Z\right)=1+Z^{2}=\frac{U^{3}}{U_{KK}^{3}},$
(2.34)

and

$\kappa=\frac{1}{3}R^{9/2}U_{KK}^{1/2}\left(2\pi\alpha^{\prime}\right)^{2}T_{\mathrm{D8}}\Omega_{4}=\frac{\lambda
N_{c}}{216\pi^{3}},\lambda=g_{\mathrm{YM}}^{2}N_{c},$ (2.35)

with the induced metric on the D8-branes as,

$ds_{\mathrm{D8}}^{2}=\left(\frac{U}{R}\right)^{3/2}\eta_{\mu\nu}dx^{\mu}dx^{\nu}+\frac{4R^{3/2}U_{KK}}{9U^{5/2}}dz^{2}+R^{3/2}U^{1/2}d\Omega_{4}^{2}.$
(2.36)

We have employed the configuration that $\mathrm{D8/\overline{D8}}$-branes are
located at the antipodal points of $S^{1}$. Then, in order to obtain a 4d
mesonic action, let us assume that
$A_{\mu}\left(x,z\right),A_{z}\left(x,z\right)$ can be expanded in terms of
complete sets $\left\\{\psi_{n}\left(z\right),\phi_{n}\left(z\right)\right\\}$
as,

$A_{\mu}\left(x,z\right)=\sum_{n}B_{\mu}^{\left(n\right)}\left(x\right)\psi_{n}\left(z\right),A_{z}\left(x,z\right)=\sum_{n}\varphi^{\left(n\right)}\left(x\right)\phi_{n}\left(z\right),$
(2.37)

where
$B_{\mu}^{\left(n\right)}\left(x\right),\varphi^{\left(n\right)}\left(x\right)$
refers to the 4d meson field. To obtain a finite action, the normalization
condition for $\psi_{n}\left(z\right)$ is chosen as,

$2\kappa\int dZK^{-1/3}\psi_{n}\psi_{m}=\delta_{mn},$ (2.38)

with the eigen equation ($n\geq 1$),

$-K^{1/3}\partial_{Z}\left(K\partial_{Z}\psi_{n}\right)=\lambda_{n}\psi_{n},$
(2.39)

where $\lambda_{n}$ is the associated eigen value. In this sense, the basic
function $\phi_{n}\left(z\right)$ can be chosen as ($n\geq 1$),

$\phi_{0}=\frac{1}{\sqrt{2\pi\kappa}}\frac{1}{U_{KK}M_{KK}}\frac{1}{K},\phi_{n}=m_{n}^{-1}\partial_{Z}\psi_{n},m_{n}=\lambda_{n}M_{KK}.$
(2.40)

Keeping these in hand, impose (2.38) - (2.40) into (2.33) then define the
vector field $V_{\mu}^{\left(n\right)}\left(x\right)$ by a gauge
transformation,

$V_{\mu}^{\left(n\right)}=B_{\mu}^{\left(n\right)}-m_{n}^{-1}\partial_{\mu}\varphi^{\left(n\right)},$
(2.41)

the Yang-Mills action (2.33) reduces to a 4d effective action for mesons as,

$S_{\mathrm{YM}}=-\kappa\int
d^{4}x\left(\frac{1}{2}\partial_{\mu}\varphi^{\left(0\right)}\partial^{\mu}\varphi^{\left(0\right)}+\sum_{n=1}^{\infty}\left[\frac{1}{4}F_{\mu\nu}^{\left(n\right)}F^{\left(n\right)\mu\nu}+\frac{1}{2}m_{n}^{2}V_{\mu}^{\left(n\right)}V^{\left(n\right)\mu}\right]\right),$
(2.42)

where
$F_{\mu\nu}^{\left(n\right)}=\partial_{\mu}V_{\nu}^{\left(n\right)}-\partial_{\nu}V_{\mu}^{\left(n\right)}$.
Accordingly, $\varphi^{\left(0\right)}$ can be interpreted as pion meson which
is the Nambu-Goldstone boson associated to the chiral symmetry breaking. By
analyzing the parity, it turns out that $\varphi^{\left(0\right)}$ is pseudo-
scalar field as it is expected.

The above discussion implicitly assumes that the gauge field $A_{M}$ and its
field strength $F_{MN}$ should vanish at $\left|z\right|\rightarrow\infty$ in
order to obtain a finite 4d mesonic action. However there is an alternative
gauge choice $A_{z}=0$ which is recognized as a gauge transformation

$A_{M}\rightarrow A_{M}-\partial_{M}\Lambda,$ (2.43)

to (2.37). Here $\Lambda$ is solved as,

$\Lambda\left(x,z\right)=\varphi^{\left(0\right)}\left(x\right)\psi_{0}\left(z\right)+\sum_{n=1}^{\infty}m_{n}^{-1}\varphi^{\left(n\right)}\left(x\right)\psi_{n}\left(z\right),$
(2.44)

where

$\psi_{0}\left(z\right)=\int
dz\phi_{0}\left(z\right)=\frac{1}{\sqrt{2\pi\kappa}}\frac{1}{M_{KK}}\arctan\left(\frac{z}{U_{KK}}\right).$
(2.45)

Thus the components of $A_{M}$ under gauge condition $A_{z}=0$ becomes,

$\displaystyle A_{\mu}\left(x,z\right)=$
$\displaystyle-\partial_{\mu}\varphi^{\left(0\right)}\left(x\right)\psi_{0}\left(z\right)+\sum_{n=1}^{\infty}\left[B_{\mu}^{\left(n\right)}\left(x\right)-m_{n}^{-1}\partial_{\mu}\varphi^{\left(n\right)}\right]\psi_{n}\left(z\right)$
$\displaystyle=$
$\displaystyle-\partial_{\mu}\varphi^{\left(0\right)}\left(x\right)\psi_{0}\left(z\right)+\sum_{n=1}^{\infty}V_{\mu}^{\left(n\right)}\left(x\right)\psi_{n}\left(z\right),$
$\displaystyle A_{z}\left(x,z\right)=$ $\displaystyle 0.$ (2.46)

In the region $\left|z\right|\rightarrow\infty$, the gauge potential
$A_{\mu}\left(x,z\right)\rightarrow\pm\sqrt{\frac{\pi}{8\kappa}}\frac{1}{M_{KK}}$
which implies the gauge field strength remains to be vanished and the
effective 4d action remains to be finite.

The above setup for mesons can be generalized into multi-flavor case by taking
into account the non-Abelian version of (2.33),

$S_{\mathrm{YM}}^{\left(N_{f}\right)}=-\kappa N_{f}\int
d^{4}xdZ\left[\frac{1}{2}K^{-1/3}\eta^{\mu\rho}\eta^{\nu\sigma}\mathrm{Tr}\left(F_{\mu\nu}F_{\rho\sigma}\right)+KM_{KK}^{2}\eta^{\mu\nu}\mathrm{Tr}\left(F_{\mu
Z}F_{\nu Z}\right)\right],$ (2.47)

where $F_{MN}=\partial_{M}A_{N}-\partial_{N}A_{M}+\left[A_{M},A_{N}\right]$,
$M,N=0,1,2,3,z$ is the gauge field strength of $U\left(N_{f}\right)$. As it
has been discussed, in order to obtain a finite 4d action, the gauge field
strength must vanish in the limit $\left|z\right|\rightarrow\infty$. Under the
gauge condition $A_{z}=0$, $A_{\mu}$ must take asymptotically a pure gauge
configuration for $\left|z\right|\rightarrow\infty$ as,

$A_{\mu}\left(x,z\right)|_{z\rightarrow\pm\infty}\rightarrow\xi_{\pm}\left(x\right)\partial_{\mu}\xi_{\pm}^{-1}\left(x\right).$
(2.48)

Compare this with (2.46), the gauge potential can be expanded with boundary
condition (2.48) as,

$\displaystyle A_{\mu}\left(x,z\right)=$
$\displaystyle\xi_{+}\left(x\right)\partial_{\mu}\xi_{+}^{-1}\left(x\right)\psi_{+}\left(z\right)+\xi_{-}\left(x\right)\partial_{\mu}\xi_{-}^{-1}\left(x\right)\psi_{-}\left(z\right)+\sum_{n=1}^{\infty}V_{\mu}^{\left(n\right)}\left(x\right)\psi_{n}\left(z\right)$
$\displaystyle\equiv$
$\displaystyle\alpha_{\mu}\left(x\right)\hat{\psi}_{0}\left(z\right)+\beta_{\mu}\left(x\right)+\sum_{n=1}^{\infty}V_{\mu}^{\left(n\right)}\left(x\right)\psi_{n}\left(z\right),$
(2.49)

with

$\displaystyle\psi_{\pm}=$
$\displaystyle\frac{1}{2}\pm\hat{\psi}_{0},\hat{\psi}_{0}=\frac{1}{\pi}\arctan\left(\frac{z}{U_{KK}}\right).$
$\displaystyle\alpha_{\mu}\left(x\right)=$
$\displaystyle\xi_{+}\left(x\right)\partial_{\mu}\xi_{+}^{-1}\left(x\right)-\xi_{-}\left(x\right)\partial_{\mu}\xi_{-}^{-1}\left(x\right)$
$\displaystyle\beta_{\mu}\left(x\right)=$
$\displaystyle\frac{1}{2}\left[\xi_{+}\left(x\right)\partial_{\mu}\xi_{+}^{-1}\left(x\right)+\xi_{-}\left(x\right)\partial_{\mu}\xi_{-}^{-1}\left(x\right)\right].$
(2.50)

To obtain the chiral Lagrangian for mesons from the Yang-Mills action (2.47),
we identify the lowest vector meson field $V_{\mu}^{\left(n\right)}$ as the
$\rho$ meson $V_{\mu}^{\left(1\right)}=\rho_{\mu}$ and choose the following
gauge conditions

$\xi_{+}^{-1}\left(x\right)=U\left(x\right),\xi_{-}\left(x\right)=1,$ (2.51)

or

$\xi_{+}^{-1}\left(x\right)=\xi_{-}\left(x\right)=\exp\left[i\pi\left(x\right)/f_{\pi}\right].$
(2.52)

Inserting (2.49) into action (2.47) with the gauge condition (2.51), the 4d
Yang-Mills action (2.47) includes a part of Skyrme model [61] as,

$S_{\mathrm{YM}}^{\left(N_{f}\right)}=\int
d^{4}x\left[\frac{f_{\pi}^{2}}{4}\mathrm{Tr}\left(U^{-1}\partial_{\mu}U\right)^{2}+\frac{1}{32e^{2}}\mathrm{Tr}\left(U^{-1}\partial_{\mu}U,U^{-1}\partial_{\mu}U\right)^{2}+...\right],$
(2.53)

where the coupling constants $f_{\pi},e$ are given as,

$\displaystyle f_{\pi}^{2}$
$\displaystyle=6R^{3/2}U_{KK}^{1/2}T_{\mathrm{D8}}\Omega_{4}\left(2\pi\alpha^{\prime}\right)^{2}\int
dz\frac{U\left(z\right)^{3}}{U_{KK}}\left(\partial_{z}\psi_{+}\right)^{2}=\frac{\lambda
M_{KK}N_{c}}{54\pi^{2}},$ $\displaystyle e^{2}$
$\displaystyle=\left[\frac{32}{3}R^{3/2}U_{KK}^{1/2}T_{\mathrm{D8}}\Omega_{4}\left(2\pi\alpha^{\prime}\right)^{2}\int
dz\frac{R^{3}}{U\left(z\right)}\psi_{+}^{2}\left(\psi_{+}-1\right)^{2}\right]^{-1}=\frac{27\pi^{7}}{2b\lambda
N_{c}},$ $\displaystyle b$
$\displaystyle=\int\frac{dZ}{\left(1+Z^{2}\right)^{1/3}}\left(\arctan
Z+\frac{\pi}{2}\right)^{2}\left(\arctan Z-\frac{\pi}{2}\right)^{2}\simeq
15.25.$ (2.54)

And the interaction terms of $\pi,\rho$ mesons would be determined by the
Yang-Mills action (2.47) with the gauge condition (2.52) as,

$\displaystyle S_{\mathrm{YM}}^{\left(N_{f}\right)}=$ $\displaystyle\int
d^{4}x\mathrm{Tr}\left[-\partial_{\mu}\pi\partial^{\mu}\pi+\frac{1}{2}W_{\mu\nu}^{2}+m_{v}^{2}\rho_{\mu}^{2}+a_{\rho^{3}}\left[\rho_{\mu},\rho_{\nu}\right]W^{\mu\nu}+a_{\rho\pi^{2}}\left[\partial_{\mu}\pi,\partial_{\nu}\pi\right]W^{\mu\nu}\right]$
$\displaystyle+\mathcal{O}\left(\pi^{4},\rho_{\mu}^{4}\right),$ (2.55)

where $W_{\mu\nu}$ is the gauge field strength of $\rho_{\mu}$ and the
associated coupling constants are given by,

$\displaystyle m_{\rho}^{2}$
$\displaystyle=\Lambda_{1}M_{KK}^{2},a_{\rho^{3}}=\frac{\left(6\pi\right)^{3/2}}{\sqrt{\lambda
N_{c}}}b_{\rho^{3}},a_{\rho\pi^{2}}=\frac{\pi\left(3\pi\right)^{3/2}}{M_{KK}^{2}\sqrt{2\lambda
N_{c}}}b_{\rho\pi^{2}},$ $\displaystyle b_{\rho^{3}}$ $\displaystyle\simeq
0.45,b_{\rho\pi^{2}}\simeq 1.6,\Lambda_{1}\simeq 0.67.$ (2.56)

Therefore, we can reach to the meson tower or chiral Lagrangian starting with
the D8-brane action which provides description of meson in holography.

To close this section, let us finally take a look at the WZ term presented in
action (2.22), which can be integrated as,

$\displaystyle S_{\mathrm{WZ}}^{\mathrm{D8}}$
$\displaystyle=\frac{g_{s}}{3!}\left(2\pi\alpha^{\prime}\right)^{3}T_{\mathrm{D8}}\int_{\mathrm{D8}}C_{3}\wedge\mathrm{Tr}\left(F\wedge
F\wedge F\right)$
$\displaystyle=\frac{g_{s}}{3!}\left(2\pi\alpha^{\prime}\right)^{3}T_{\mathrm{D8}}\int_{\mathrm{D8}}F_{4}\wedge\omega_{5}$
$\displaystyle=\frac{N_{c}}{24\pi^{2}}\mathrm{Tr}\int_{M_{4}\times\mathbb{R}}\omega_{5}\left(A\right).$
(2.57)

Here $F_{4}=dC_{3}$ is the Ramond-Ramond field given in (2.14) and
$\omega_{5}\left(A\right)$ is the Chern-Simons (CS) 5-form given as,

$\omega_{5}\left(A\right)=AF^{2}-\frac{1}{2}A^{3}F+\frac{1}{10}A^{5},$ (2.58)

where $F=dA+\frac{1}{2}\left[A,A\right]$ is the gauge field strength. Under
the gauge transformation on the D8-brane,

$\delta_{\Lambda}A=d\Lambda+\left[A,\Lambda\right],\delta_{\Lambda}F=\left[F,\Lambda\right],$
(2.59)

we can compute

$\displaystyle\mathrm{Tr}\left(\delta_{\Lambda}\omega_{5}\right)$
$\displaystyle=\mathrm{Tr}\left[d\Lambda
d\left(AdA+\frac{1}{2}A^{3}\right)\right]$
$\displaystyle=d\left(\mathrm{Tr}\left[\Lambda
d\left(AdA+\frac{1}{2}A^{3}\right)\right]\right)$ $\displaystyle\equiv
d\chi_{4}\left(\Lambda,A\right).$ (2.60)

Hence by defining the boundary value of the gauge potential as,

$A_{\mu}^{L}\left(x\right)=\lim_{z\rightarrow+\infty}A_{\mu}\left(x,z\right),A_{\mu}^{R}\left(x\right)=\lim_{z\rightarrow-\infty}A_{\mu}^{R}\left(x,z\right),$
(2.61)

the WZ term reduces to the chiral anomaly of $U\left(N_{f}\right)_{L}\times
U\left(N_{f}\right)_{R}$ in QCD as,

$\displaystyle\delta_{\Lambda}S_{\mathrm{WZ}}^{\mathrm{D8}}$
$\displaystyle=\frac{N_{c}}{24\pi^{2}}\mathrm{Tr}\int_{M_{4}\times\mathbb{R}}\delta_{\Lambda}\omega_{5}\left(A\right)=\frac{N_{c}}{24\pi^{2}}\int_{M_{4}}\chi_{4}\left(\Lambda,A\right)\big{|}_{z\rightarrow-\infty}^{z\rightarrow+\infty}$
$\displaystyle=\frac{N_{c}}{24\pi^{2}}\int_{M_{4}}\left[\chi_{4}\left(\Lambda_{L},A_{L}\right)-\chi_{4}\left(\Lambda_{R},A_{R}\right)\right].$
(2.62)

And the formula for the chiral anomaly can also be expressed in the gauge
condition $A_{z}=0$ which is to perform the gauge transformation

$A^{g}=gAg^{-1}+gdg^{-1}.$ (2.63)

Then the CS 5-form reduces to

$\omega_{5}\left(A^{g}\right)=\omega_{5}\left(A\right)+\frac{1}{10}\left(gdg^{-1}\right)^{5}+d\alpha_{4},$
(2.64)

where

$\displaystyle\alpha_{4}$
$\displaystyle=-\frac{1}{2}W_{1}\left(AdA+dAA+A^{3}\right)+\frac{1}{4}W_{1}AW_{1}A-W_{1}^{3}A,$
$\displaystyle W_{1}$ $\displaystyle=dg^{-1}g.$ (2.65)

Recall the formulas (2.49) in the gauge $A_{z}=0$ and choose the gauge
condition (2.51), the WZ term (2.57) can be rewritten, after somewhat lengthy
but straightforward calculations, as

$S_{\mathrm{WZ}}^{\mathrm{D8}}=-\frac{N_{c}}{48\pi^{2}}\mathrm{Tr}\int_{M_{4}}L_{\mathrm{WZW}}-\frac{N_{c}}{240\pi^{2}}\mathrm{Tr}\int_{M_{4}\times\mathbb{R}}\left(gdg^{-1}\right)^{5},$
(2.66)

where $L_{\mathrm{WZW}}$ is the Wess-Zumino-Witten (WZW) term in [62, 63]
given as,

$\displaystyle L_{\mathrm{WZW}}=$
$\displaystyle\left[\left(A_{R}dA_{R}+dA_{R}A_{R}+A_{R}^{3}\right)\left(U^{-1}A_{L}U+U^{-1}dU\right)-\mathrm{p.c.}\right]$
$\displaystyle+\left(dA_{R}dU^{-1}A_{L}U-\mathrm{p.c.}\right)+\left[A_{R}\left(dU^{-1}U\right)^{3}-\mathrm{p.c.}\right]$
$\displaystyle+\frac{1}{2}\left[\left(A_{R}dU^{-1}U\right)^{2}-\mathrm{p.c.}\right]+\left[UA_{R}U^{-1}A_{L}dUdU^{-1}-\mathrm{p.c.}\right]$
$\displaystyle-\left[A_{R}dU^{-1}UA_{R}U^{-1}A_{L}U-\mathrm{p.c.}\right]+\frac{1}{2}\left(A_{R}U^{-1}A_{L}U\right)^{2}.$
(2.67)

Note that “$\mathrm{p.c.}$” refers to the terms by exchanging
$A_{L}\leftrightarrow A_{R},U\leftrightarrow U^{-1}$. And one can further work
out the couplings to the vector mesons by using (2.49) with $A_{z}=0$.

###  2.5 The wrapped D4-brane and baryon vertex

In the $SU\left(N_{c}\right)$ gauge theory, a baryon vertex connects to
$N_{c}$ external fundamental quarks with the color wave function combined
together by an $N_{c}$-th antisymmetric tensor of $SU\left(N_{c}\right)$
group. Accordingly the baryon vertex in gauge-gravity duality is recognized as
a D-brane wrapped on the internal sphere [26, 64]. To clarify this briefly,
let us first recall the baryon vertex in the holographic duality between
$\mathcal{N}=4$ super Yang-Mills theory and IIB string theory on
$\mathrm{AdS}_{5}\times S^{5}$. As the fundamental quarks in the super Yang-
Mills theory is created by the $N_{c}$ elementary superstrings in
$\mathrm{AdS}_{5}\times S^{5}$, it is represented by the endpoints of
elementary superstrings at the boundary of $\mathrm{AdS}_{5}$. Hence we need
$N_{c}$ elementary superstrings with same orientation somehow to terminate in
the $\mathrm{AdS}_{5}\times S^{5}$. On the other hand, since the baryon
current must be conserved, one needs to find a source to cancel the $N_{c}$
charges (baryon charge) contributed by the $N_{c}$ elementary superstrings. To
figure out these problems and work out a baryon vertex, a probe D5-brane
wrapped on $S^{5}$ provides us a good answer. The $N_{c}$ elementary
superstrings end on the D5-brane contributes $N_{c}$ to the D5-brane, however
the WZ action for such a wrapped D5-brane,

$S_{\mathrm{WZ}}^{\left(\mathrm{D5}\right)}\sim\int_{S^{5}\times\mathbb{R}}C_{4}F\sim\int_{S^{5}\times\mathbb{R}}F_{5}A\sim
N_{c}\int_{\mathbb{R}}A,$ (2.68)

(where $F_{5}=dC_{4}$ is the Ramond-Ramond field strength) nicely provides
$N_{c}$ charges to cancel the charges given by $N_{c}$ elementary
superstrings999The sign of the $N_{c}$ charge depends on the orientation of
the elementary superstrings and D5-branes.. Therefore the $A$ current would be
conserved, which implies the D5-brane is a baryon vertex.

The construction of the baryon vertex can also been employed in the D4/D8
model which is identified as a D4-brane101010To distinguish the D4-branes as
the baryon vertex from the $N_{c}$ D4-branes, we denote the baryon vertex as
$\mathrm{D4}^{\prime}$-brane in the rest of this paper. wrapped on $S^{4}$
with $N_{c}$ elementary superstrings ending on it. A remarkable point here is
that the $\mathrm{D4}^{\prime}$-branes can be described equivalently by the
instanton configuration of the gauge field on the D8-branes [65, 66]. To see
this clearly, let us consider a $\mathrm{D}_{p}$-brane with its worldvolume
gauge field strength $F$. According to (2.20), The WZ action for such a
$\mathrm{D}_{p}$-brane includes a term as a source,

$S_{\mathrm{WZ}}\sim g_{s}T_{D_{p}}\left(2\pi\alpha^{\prime}\right)^{2}\int
C_{p-3}\mathrm{Tr}F^{2}.$ (2.69)

For the single instanton configuration, the gauge field strength can be
integrated as,

$\mathrm{Tr}\int F^{2}=8\pi^{2}.$ (2.70)

Hence (2.69) can be written as,

$g_{s}T_{D_{p}}\left(2\pi\alpha^{\prime}\right)^{2}\int
C_{p-3}\mathrm{Tr}F^{2}=g_{s}T_{D_{p-4}}\int C_{p-3},$

giving rise to a same source included by a $\mathrm{D}_{p-4}$-brane.
Accordingly, we obtain a simple and interesting conclusion here, that is the
instanton in the $\mathrm{D}_{p}$-brane is the same object as a
$\mathrm{D}_{p-4}$-brane inside it.

Let us return to the D4/D8 model, so it implies the
$\mathrm{D4}^{\prime}$-branes as the baryon vertex are equivalent to the
instanton in the $\mathrm{D8/\overline{D8}}$-branes. For multiple instantons,
(2.70) is replaced by,

$\mathrm{Tr}\int F^{2}=8\pi^{2}n,$ (2.71)

where $n$ refers to the instanton number. Inserting the instanton
configuration of the gauge field denoted as $A_{\mathrm{cl}}$ with a
$U\left(N_{f}\right)_{V}$ fluctuation $\tilde{A}$ into the WZ action (2.57) of
D8-brane, it reduces to

$\frac{N_{c}}{8\pi^{2}}\int\tilde{A}\mathrm{Tr}F^{2}=nN_{c}\int\tilde{A},$
(2.72)

which implies the instantons take $U\left(1\right)_{V}$ charge $nN_{c}$. Since
the baryon number is defined as $1/N_{c}$ times the charge of the diagonal
$U\left(1\right)_{V}$ subgroup of the $U\left(N_{f}\right)_{V}$ symmetry, it
is obvious that the instanton number is equivalent to the baryon number in
this holographic system.

Moreover, when (2.71) is integrated out to be a Chern-Simons 3-form
$\omega_{3}$ as,

$\int_{M_{4}}\mathrm{Tr}F^{2}=\int_{\partial
M_{4}}\mathrm{Tr}\omega_{3}=\int\mathrm{Tr}\left(AF-\frac{1}{3}A^{3}\right),$
(2.73)

the baryon number can be obtained as,

$n=\frac{1}{8\pi^{2}}\mathrm{Tr}\int_{\mathbb{R}^{4}}F^{2}=\mathrm{Tr}\int_{\mathbb{R}^{3}}\omega_{3}|_{z\rightarrow-\infty}^{z\rightarrow+\infty}=-\frac{1}{24\pi^{2}}\mathrm{Tr}\int_{\mathbb{R}^{3}}\left(U^{-1}dU\right)^{3},$
(2.74)

where we have imposed the similar boundary condition as it is given in (2.51),

$\lim_{z\rightarrow+\infty}A_{\mu}\left(x^{\mu},z\right)=U^{-1}\left(x^{i}\right)\partial_{i}U\left(x^{i}\right),\lim_{z\rightarrow-\infty}A_{\mu}\left(x^{\mu},z\right)=0.$
(2.75)

The (2.74) gives the winding number of $U$ which means the homotopy is
$\pi_{3}\left[U\left(N_{f}\right)\right]\simeq\mathbb{Z}$. And it agrees with
the baryon number charge in the Skyrme model [62, 67].

The baryon mass $m_{B}$ can be roughly obtained by evaluating the energy
carried by the $\mathrm{D4}^{\prime}$-branes which can be read from its DBI
action as,

$\displaystyle S_{\mathrm{D4}^{\prime}}=$ $\displaystyle-T_{\mathrm{D4}}\int
dx^{0}d\Omega_{4}e^{-\phi}\sqrt{-g_{00}g_{S^{4}}}$ $\displaystyle=$
$\displaystyle-\frac{1}{27}M_{KK}\lambda N_{c}^{2}\int_{\mathbb{R}}dx^{0},$
$\displaystyle=$ $\displaystyle-m_{B}\int_{\mathbb{R}}dx^{0}$ (2.76)

where the bubble D4-brane background has been chosen for the confined property
of baryon and $g_{S^{4}}$ refers to the metric on $S^{4}$ presented in (2.16).
This formula illustrates a stable position of the baryonic
$\mathrm{D4}^{\prime}$-brane by minimizing its energy, which is $U=U_{KK}$
since the bubble geometry shrinks at $U=U_{KK}$. In the black D4-brane, one
can follow the same formula (2.76) to evaluate the baryon mass. However, if
the baryonic $\mathrm{D4}^{\prime}$-brane is the only probe brane, it can not
stay at $U=U_{H}$ stably in the black D4-brane background since gravity will
pull it into the horizon. In this sense, the baryon vertex exists in the
bubble D4-brane background only and it is consistent with its property of
confinement. When the probe $\mathrm{D8/\overline{D8}}$-branes are embedded
into the bulk geometry, due to the balance condition, the baryonic
$\mathrm{D4}^{\prime}$-brane can be restricted inside the D8-branes if
$\mathrm{D8/\overline{D8}}$-branes are connected, as it is displayed in Figure
4111111Authors of [68] claim that, according to the numerical calculation,
there is not a wrapped configuration for the baryonic
$\mathrm{D4}^{\prime}$-brane in the black D4-brane background thus this
background may correspond to the deconfinement phase of QCD. We note that this
issue is not figured out even if the baryon vertex is introduced into the
black D4-brane background..

Figure 4:  Various D-brane configurations in the D4/D8 model. Purple refers to
the baryon vertex ($\mathrm{D4}^{\prime}$-brane) locates at the stable
position. When the $\mathrm{D8/\overline{D8}}$-branes are connected as (a),
(b), (c), the baryon vertex can locate stably at the connected position inside
the $\mathrm{D8/\overline{D8}}$-branes. There is not a stable position for
baryon vertex if $\mathrm{D8/\overline{D8}}$-branes are parallel as (d).

Therefore it can be described equivalently by the instanton configuration on
the D8-branes.

To obtain the baryon mass or baryon spectrum in this model, it is worth
searching for an exact instanton solution for the gauge field on the
D8-branes. As baryon lives in the low-energy region of QCD, we may find an
approximated solution for the instanton configuration in the strong coupling
limit i.e. $\lambda\rightarrow\infty$. To this goal, let us take a look at the
gauge field on the D8-branes whose dynamic is described by the Yang-Mills
action (2.33) plus the Chern-Simons action presented in (2.57) 121212As the
size of instanton takes order of $\lambda^{-1/2}$, it may lead to a puzzle if
the Yang-Mills action is taken into account only because the high order
derivatives in the DBI action contributes more importantly. However, according
to the holographic duality, taking the near-horizon limit requires
$\alpha^{\prime}\rightarrow 0$ which implies the Yang-Mills action dominates
the dynamics in the DBI action. This puzzle is not figured out completely in
[27, 28] and we may additionally set $\alpha^{\prime}\lambda\rightarrow 0$
when Yang-Mills action is taken into account only in this setup.. Since the
size of instanton is of order $\lambda^{-1/2}$, it would be convenient to
rescale the coordinate $\left\\{x^{0},x^{i},z\right\\}$ and the gauge
potential $A$ as,

$\displaystyle x^{M}\rightarrow\lambda^{-1/2}x^{M},\ $ $\displaystyle
x^{0}\rightarrow x^{0}$ $\displaystyle A_{M}\rightarrow\lambda^{1/2}A_{M},\ $
$\displaystyle A_{0}\rightarrow A_{0}$ $\displaystyle F_{MN}\rightarrow\lambda
F_{MN},\ $ $\displaystyle F_{0M}\rightarrow\lambda^{1/2}F_{0M},$ (2.77)

where $M,N=1,2,3,z$. In the large $\lambda$ limit, the Yang-Mills action
(2.33) can be expanded as,

$\displaystyle S_{\mathrm{YM}}=$ $\displaystyle-\frac{\kappa}{\lambda}\int
d^{4}xdz\mathrm{Tr}\left[\frac{\lambda}{2}F_{MN}^{2}+\left(-\frac{z^{2}}{6}F_{ij}^{2}+z^{2}F_{iz}^{2}-F_{0M}^{2}\right)+\mathcal{O}\left(\lambda^{-1}\right)\right]$
$\displaystyle-\frac{\kappa}{2\lambda}\int
d^{4}xdz\left[\frac{\lambda}{2}\hat{F}_{MN}^{2}+\left(-\frac{z^{2}}{6}\hat{F}_{ij}^{2}+z^{2}\hat{F}_{iz}^{2}-\hat{F}_{0M}^{2}\right)+\mathcal{O}\left(\lambda^{-1}\right)\right],$
(2.78)

while the Chern-Simons action (2.57) remains under the rescaling (2.77). We
have employed the D-brane configuration presented in (a) of Figure 4. The
$U\left(N_{f}\right)$ group is decomposed as $U\left(N_{f}\right)\simeq
U\left(1\right)\times SU\left(N_{f}\right)$ and correspondingly its generator
is decomposed as,

$A+\frac{1}{\sqrt{2N_{f}}}\hat{A}=A^{a}t^{a}+\frac{1}{\sqrt{2N_{f}}}\hat{A},$
(2.79)

where $\hat{A},A$ refers to the generators of $U\left(1\right)$,
$SU\left(N_{f}\right)$ respectively and $t^{a}$ ($a=1,2...N_{f}^{2}-1$) are
the normalized bases satisfying

$\mathrm{Tr}\left(t^{a}t^{b}\right)=\frac{1}{2}\delta^{ab}.$ (2.80)

So the Chern-Simons action (2.57) can be derived as,

$\displaystyle S_{\mathrm{CS}}=$
$\displaystyle\frac{N_{c}}{24\pi^{2}}\int\omega_{5}^{SU\left(N_{f}\right)}\left(A\right)+\frac{N_{c}}{24\pi^{2}}\sqrt{\frac{2}{N_{f}}}\epsilon_{MNPQ}\int
d^{4}xdz\bigg{[}\frac{3}{8}\hat{A}_{0}\mathrm{Tr}\left(F_{MN}F_{PQ}\right)$
$\displaystyle-\frac{3}{2}\hat{A}_{M}\mathrm{Tr}\left(\partial_{0}A_{N}F_{PQ}\right)+\frac{3}{4}\hat{F}_{MN}\mathrm{Tr}\left(A_{0}F_{PQ}\right)+\frac{1}{16}\hat{A}_{0}\hat{F}_{MN}\hat{F}_{PQ}-\frac{1}{4}\hat{A}_{M}\hat{F}_{0N}\hat{F}_{PQ}$
$\displaystyle+\left(\mathrm{total\ derivatives}\right)\bigg{]}.$ (2.81)

And the equation of motion for $\hat{A},A$ can be derived by varying action
(2.78) plus (2.81) which allows an instanton solution as,

$A_{M}^{\mathrm{cl}}=-if\left(\xi\right)g\left(x\right)\partial_{M}g^{-1},$
(2.82)

where

$\displaystyle f\left(\xi\right)$
$\displaystyle=\frac{\xi^{2}}{\xi^{2}+\rho^{2}},\xi=\sqrt{\left(x^{M}-X^{M}\right)^{2}},$
$\displaystyle g\left(x\right)$
$\displaystyle=\left(\begin{array}[]{cc}g^{SU\left(2\right)}\left(x\right)&0\\\
0&\mathbf{1}_{N_{f}-2}\end{array}\right),g^{SU\left(2\right)}\left(x\right)=\frac{1}{\xi}\left[\left(z-Z\right)\mathbf{1}_{2}-i\left(x^{i}-X^{i}\right)\tau^{i}\right].$
(2.85)

Here $\mathbf{1}_{N}$ is an $N\times N$ identity matrix and $\tau^{i}$’s are
the Pauli matrices. The position and the size of the instanton are denoted by
the constants $X^{M}=\left\\{X^{i},Z\right\\}$ and $\rho$ respectively to
which have been rescaled as (2.77). The configuration (2.82) (2.85) is the
Belavin–Polyakov–Schwarz–Tyupkin (BPST) solution embedding into
$SU\left(N_{f}\right)$ which represents the $SU\left(2\right)$ Euclidean
instanton, and one may verify this solution satisfies (2.70). Then the
$U\left(1\right)$ part of the gauge field is solved as,

$\hat{A}_{0}^{\mathrm{cl}}=\sqrt{\frac{2}{N_{f}}}\frac{1}{8\pi^{2}a}\frac{1}{\xi^{2}}\left[1-\frac{\rho^{4}}{\left(\xi^{2}+\rho^{2}\right)^{2}}\right],\
\hat{A}_{M}^{\mathrm{cl}}=0.$ (2.86)

which leads to a non-zero $A_{0}$ as,

$A_{0}^{\mathrm{cl}}=\frac{1}{16\pi^{2}a}\frac{1}{\xi^{2}}\left[1-\frac{\rho^{4}}{\left(\xi^{2}+\rho^{2}\right)^{2}}\right]\left(\mathcal{P}_{2}-\frac{2}{N_{f}}\mathbf{1}_{N_{f}}\right),$
(2.87)

where $\mathcal{P}_{2}$ is an $N_{f}\times N_{f}$ matrix
$\mathcal{P}_{2}=\mathrm{diag}\left(1,1,0,...0\right)$.

Keeping these in hand, it is possible to evaluate the classical baryon mass
through the soliton mass $M$ with respect to the D8-brane action as
$S\left[A^{\mathrm{cl}}\right]=-\int Mdt$ which is obtained as,

$M=8\pi^{2}\kappa+\frac{8\pi^{2}\kappa}{\lambda}\left(\frac{\rho^{2}}{6}+\frac{Z^{2}}{3}+\frac{1}{320\pi^{4}a^{2}\rho^{2}}\right),$
(2.88)

by inserting (2.82) - (2.87) into action (2.78) plus (2.81). On the other
hand, since the low-energy effective theory on the D8-branes can reduce to
Skyrme model, we can further employ the idea in the Skyrme model of baryon,
which is identified as the excitation of the collective modes, in order to
search for the baryon spectrum. The classically effective Lagrangian for
baryon describing the dynamics of the collective coordinates
$\mathcal{X}^{\alpha}$ in the moduli space by the one instanton solution.
which refers to the world line element with a baryonic potential
$U\left(\mathcal{X}^{\alpha}\right)$ in the moduli space

$L\left(\mathcal{X}^{\alpha}\right)=\frac{m_{X}}{2}\mathcal{G}_{\alpha\beta}\dot{\mathcal{X}}^{\alpha}\dot{\mathcal{X}}^{\beta}-U\left(\mathcal{X}^{\alpha}\right)+\mathcal{O}\left(\lambda^{-1}\right),$
(2.89)

where “$\cdot$” refers to the derivative respected to time, the collective
coordinates $\mathcal{X}^{\alpha}$ denotes to
$\left\\{X^{M},\rho,y^{a}\right\\}$ and $W=y^{a}t^{a}$ is the
$SU\left(N_{f}\right)$ orientation of the instanton. The potential
$U\left(\mathcal{X}^{\alpha}\right)$ is the classical soliton mass given by
$S\left[A^{\mathrm{cl}}\right]=-\int dtU\left(\mathcal{X}^{\alpha}\right)$.
The basic idea to quantize the classical Lagrangian (2.89) is to move slowly
the classical soliton so that the collective coordinates
$\mathcal{X}^{\alpha}$ are promoted to be time-dependent [69]. Approximately,
the $SU\left(N_{f}\right)$ gauge field potential is becomes time-dependent by
a gauge transformation,

$\displaystyle A_{M}\left(t,x\right)$
$\displaystyle=W\left(t\right)A_{M}^{\mathrm{cl}}\left(x,\mathcal{X}^{\alpha}\right)W\left(t\right)^{-1}-iW\left(t\right)\partial_{M}W\left(t\right)^{-1},$
$\displaystyle A_{0}\left(t,x\right)$
$\displaystyle=W\left(t\right)A_{0}^{\mathrm{cl}}\left(x,\mathcal{X}^{\alpha}\right)W\left(t\right)^{-1}+\Delta
A_{0},$ $\displaystyle\hat{A}_{M}\left(t,x\right)$ $\displaystyle=0,\
\hat{A}_{0}\left(t,x\right)=\hat{A}_{0}^{\mathrm{cl}}\left(t,x\right),$ (2.90)

and the associated field strength becomes,

$\displaystyle F_{MN}=$ $\displaystyle
W\left(t\right)F_{MN}^{\mathrm{cl}}W\left(t\right)^{-1},$ $\displaystyle
F_{0M}=$ $\displaystyle
W\left(t\right)\left(\dot{\mathcal{X}}^{\alpha}\frac{\partial}{\partial\mathcal{X}^{\alpha}}A_{M}^{\mathrm{cl}}-D_{M}^{\mathrm{cl}}\Sigma-
D_{M}^{\mathrm{cl}}A_{0}^{\mathrm{cl}}\right)W\left(t\right)^{-1},$
$\displaystyle\hat{F}_{0M}=$ $\displaystyle\hat{F}_{0M}^{\mathrm{cl}},\
\hat{F}_{MN}=\hat{F}_{MN}^{\mathrm{cl}},$ (2.91)

where

$\displaystyle D_{M}^{\mathrm{cl}}A_{0}$
$\displaystyle=\partial_{M}A_{0}+i\left[A_{M}^{\mathrm{cl}},A_{0}\right],$
$\displaystyle\Sigma$ $\displaystyle=W\left(t\right)^{-1}\Delta
A_{0}W\left(t\right)-i\dot{W}\left(t\right)^{-1}W\left(t\right).$ (2.92)

$\Delta A_{0}$ must be determined by its equation of motion as,

$D_{M}^{\mathrm{cl}}\left(\dot{X}^{N}\frac{\partial}{\partial
X^{N}}A_{M}^{\mathrm{cl}}+\dot{\rho}\frac{\partial}{\partial\rho}A_{M}^{\mathrm{cl}}-D_{M}^{\mathrm{cl}}\Sigma\right)=0.$
(2.93)

While for generic $N_{f}$, the exact solution for $\Sigma$ may be out of
reach, the solution with $N_{f}=2,3$ is collected respectively in [27, 28].
Accordingly the Lagrangian of the collective modes is given by

$\displaystyle
S\left[\mathcal{A}\right]-S\left[\mathcal{A}^{\mathrm{cl}}\right]$
$\displaystyle=\int
dt\left[L_{\mathrm{YM}}\left(\mathcal{X}^{\alpha}\right)+L_{\mathrm{CS}}\left(\mathcal{X}^{\alpha}\right)\right]=\int
dtL\left(\mathcal{X}^{\alpha}\right)$ $\displaystyle
S_{\mathrm{YM}}\left[\mathcal{A}\right]-S_{\mathrm{YM}}\left[\mathcal{A}^{\mathrm{cl}}\right]$
$\displaystyle=\int dtL_{\mathrm{YM}}\left(\mathcal{X}^{\alpha}\right),$
$\displaystyle
S_{\mathrm{CS}}\left[\mathcal{A}\right]-S_{\mathrm{CS}}\left[\mathcal{A}^{\mathrm{cl}}\right]$
$\displaystyle=\int dtL_{\mathrm{CS}}\left(\mathcal{X}^{\alpha}\right),$
(2.94)

which leads to,

$\displaystyle L\left(\mathcal{X}^{\alpha}\right)$
$\displaystyle=-M+aN_{c}\mathrm{Tr}\int
d^{3}xdz\left(\dot{X}^{N}F_{MN}^{\mathrm{cl}}+\dot{\rho}\frac{\partial}{\partial\rho}A_{M}-\dot{X}^{N}D_{M}^{\mathrm{cl}}A_{N}^{\mathrm{cl}}-D_{M}^{\mathrm{cl}}\Sigma\right)^{2}+\mathcal{O}\left(\lambda^{-1}\right)$
$\displaystyle=-M_{0}+\frac{m_{X}}{2}\delta_{ij}\dot{X}^{i}\dot{X}^{j}+L_{Z}+L_{\rho}+L_{\rho
W}+\mathcal{O}\left(\lambda^{-1}\right),$ (2.95)

where

$\displaystyle L_{Z}=$
$\displaystyle\frac{m_{Z}}{2}\left(\dot{Z}^{2}-\omega_{Z}^{2}Z^{2}\right),L_{\rho}=\frac{m_{\rho}}{2}\left(\dot{\rho}-\omega_{\rho}^{2}\rho^{2}\right)-\frac{K}{m_{\rho}\rho^{2}},$
$\displaystyle L_{\rho W}=$
$\displaystyle\frac{m_{\rho}\rho^{2}}{2}\sum_{a}C_{a}\left[\mathrm{Tr}\left(-iW^{-1}\dot{W}t^{a}\right)\right]^{2},a=1,2...N_{f}^{2}-1$
(2.96)

and

$M_{0}=8\pi^{2}\kappa,m_{X}=m_{Z}=\frac{m_{\rho}}{2}=8\pi^{2}\kappa\lambda^{-1},K=\frac{2}{5}N_{c}^{2},\omega_{Z}^{2}=4\omega_{\rho}^{2}=\frac{2}{3}.$
(2.97)

Here we note that the formulas are in the unit of $M_{KK}=1$, $C_{a}$’s are
constants dependent on the $SU\left(N_{f}\right)$ instanton solution and the
metric of the moduli space can be further obtained by comparing (2.95) with
(2.89). For example, we have $C_{1,2,3}=1$ for $N_{f}=2$, and
$C_{1,2,3}=1,C_{4,5,6,7}=1/2,C_{8}=0$ for $N_{f}=3$. Afterwards, the baryon
states can be obtained by quantizing the Lagrangian (2.95) that is to replace
the derivative term by
$\dot{X}^{\alpha}\rightarrow-\frac{i}{m_{X}}\partial_{\alpha}$
straightforwardly. Hence the quantized Hamiltonian associated to (2.95) is
collected as131313We note that for generic $N_{f}$, the baryonic Hamiltonian
must be supported by additional constraint according to [70] although it may
not change the baryon spectrum. ,

$\displaystyle H=$ $\displaystyle M_{0}+H_{Z}+H_{\rho}+H_{\rho W},$
$\displaystyle H_{Z}=$
$\displaystyle-\frac{1}{2m_{Z}}\partial_{Z}^{2}+\frac{1}{2}m_{Z}\omega_{Z}^{2}Z^{2},$
$\displaystyle H_{\rho}=$
$\displaystyle-\frac{1}{2m_{\rho}}\frac{1}{\rho^{\eta}}\partial_{\rho}\left(\rho^{\eta}\partial_{\rho}\right)+\frac{1}{2}m_{\rho}\omega_{\rho}^{2}\rho^{2}+\frac{K}{m_{\rho}\rho^{2}},$
$\displaystyle H_{\rho W}=$
$\displaystyle\frac{m_{\rho}\rho^{2}}{2}\sum_{a}C_{a}\left[\mathrm{Tr}\left(-iW^{-1}\dot{W}t^{a}\right)\right]^{2}=\frac{2}{m_{\rho}\rho^{2}}\sum_{a}C_{a}\left(J^{a}\right)^{2},$
(2.98)

where $\eta=N_{f}^{2}-1$ and $J^{a}$’s are the operators of the angular
momentum of $SU\left(N_{f}\right)$. The baryon spectrum can be finally
obtained by evaluating the eigen values of the Hamiltonian (2.98) which takes
a fortunately analytical formula [27, 28].

###  2.6 Gravitational wave as glueball

According to AdS/CFT and gauge-gravity duality [29, 30, 31, 32], the glueball
operator $\hat{O}$ can be identified as the source of gravitational
fluctuation in bulk since it is included in energy-momentum tensor of Yang-
Mills theory in the dual theory as $\hat{O}\sim F_{\mu\nu}F^{\mu\nu}\sim
T_{\mu\nu}$ coupling to metric. Thus due to the confined property of glueball,
we can choose the bubble D4-brane background (2.16) compactified on a circle
with gravitational fluctuation in order to investigate glueball in holography.

The dual field to the glueball operator is the gravitational fluctuation
coupling the energy-momentum which therefore refers to the gravitational
polarization. By employing the relation of 11d M-theory and 10d type IIA
string theory in Section 2.1, it would be convenient to find the gravitational
polarization in 11d theory. For example, the lowest exotic scalar glueball
with quantum number $J^{CP}=0^{++}$ corresponds to the exotic polarizations of
the bulk gravitational polarization and its 11d components are given as
($\mu,\nu=0,1,2,3$),

$\displaystyle\delta g_{44}$ $\displaystyle=$
$\displaystyle-\frac{r^{2}}{L^{2}}f\left(r\right)H_{E}\left(r\right)G_{E}\left(x\right),$
$\displaystyle\delta g_{\mu\nu}$ $\displaystyle=$
$\displaystyle\frac{r^{2}}{L^{2}}H_{E}\left(r\right)\left[\frac{1}{4}\eta_{\mu\nu}-\left(\frac{1}{4}+\frac{3r_{KK}^{6}}{5r^{6}-2r_{KK}^{6}}\right)\frac{\partial_{\mu}\partial_{\nu}}{M_{E}^{2}}\right]G_{E}\left(x\right),$
$\displaystyle\delta g_{55}$ $\displaystyle=$
$\displaystyle\frac{r^{2}}{4L^{2}}H_{E}\left(r\right)G_{E}\left(x\right),$
$\displaystyle\delta g_{rr}$ $\displaystyle=$
$\displaystyle-\frac{L^{2}}{r^{2}}\frac{1}{f\left(r\right)}\frac{3r_{KK}^{6}}{5r^{6}-2r_{KK}^{6}}H_{E}\left(r\right)G_{E}\left(x\right),$
$\displaystyle\delta g_{r\mu}$ $\displaystyle=$
$\displaystyle\frac{90r^{7}r_{KK}^{6}}{M_{E}^{2}L^{2}\left(5r^{6}-2r_{KK}^{6}\right)^{2}}H_{E}\left(r\right)\partial_{\mu}G_{E}\left(x\right),$
(2.99)

where $H_{E}\left(r\right)$ must be determined by its eigen equation given by,

$\frac{1}{r^{3}}\frac{d}{dr}\left[r\left(r^{6}-r_{KK}^{6}\right)\frac{d}{dr}H_{E}\left(r\right)\right]+\left[\frac{432r^{2}r_{KK}^{12}}{\left(5r^{6}-2r_{KK}^{6}\right)^{2}}+L^{4}M_{E}^{2}\right]H_{E}\left(r\right)=0,$
(2.100)

and $G_{E}\left(x\right)$ refers to the 4d glueball field. By imposing metric
with gravitational fluctuation (2.99) as
$g_{MN}=g_{MN}^{\left(0\right)}+\delta g_{MN}$ and solution of $C_{3}$ into
11d SUGRA action (2.1), we can obtain

$S_{\mathrm{SUGRA}}^{11\mathrm{d}}=-\frac{1}{2}\mathcal{C}_{E}\beta_{4}\beta_{5}\int
d^{4}x\left[\left(\partial_{\mu}G_{E}\right)^{2}+M_{E}^{2}G_{E}^{2}\right],$
(2.101)

representing the standard kinetic action for scalar glueball field $G_{E}$,
where $\beta_{5}$, $g_{MN}^{\left(0\right)}$ refers respectively to the size
of $x^{5}$ and the bubble version of (2.10). $\mathcal{C}_{E}$ is a numerical
constant given by,

$\displaystyle\mathcal{C}_{E}$
$\displaystyle=0.057396\left[H_{E}\left(r_{KK}\right)\right]^{2}\frac{r_{KK}^{4}}{L^{3}},$
$\displaystyle\left[H_{E}\left(r_{KK}\right)\right]^{-1}$
$\displaystyle=0.0069183\lambda^{1/2}N_{c}M_{KK}.$ (2.102)

Hence the eigen value of (2.100) determines the mass spectrum of exotic scalar
glueball. The mass spectrum for various glueballs can be obtained by taking
into account different gravitational polarizations e.g. dilatonic scalar
glueball with $J^{CP}=0^{++}$,

$\displaystyle\delta G_{11,11}$
$\displaystyle=-3\frac{r^{2}}{L^{2}}H_{D}\left(r\right)G_{D}\left(x\right),$
$\displaystyle\delta G_{\mu\nu}$
$\displaystyle=\frac{r^{2}}{L^{2}}H_{D}\left(r\right)\left[\eta^{\mu\nu}-\frac{\partial^{\mu}\partial^{\nu}}{M_{D}^{2}}\right]G_{D}\left(x\right),$
(2.103)

and tensor glueball with $J^{CP}=2^{++}$,

$\delta
G_{\mu\nu}=-\frac{r^{2}}{L^{2}}H_{T}\left(r\right)T_{\mu\nu}\left(x\right).$
(2.104)

The eigen equations for $H_{D}\left(r\right)$ and $H_{T}\left(r\right)$ are
given as,

$\frac{1}{r^{3}}\frac{d}{dr}\left[r\left(r^{6}-r_{KK}^{6}\right)\frac{d}{dr}H_{D,T}\left(r\right)\right]+L^{4}M_{D,T}^{2}H_{D,T}\left(r\right)=0,$
(2.105)

which determines the mass spectrum of dilatonic scalar and tensor glueball.
The mass spectrum of various glueballs are collected in Table 3 for the
reader’s convenience. The labels $\mathrm{S,T,V,N,M,L}$ refer to the solutions
for six independent wave equations for various scalar, vector tensor modes of
glueballs.

Mode | $\mathrm{S}_{4}$ | $\mathrm{T}_{4}$ | $\mathrm{V}_{4}$ | $\mathrm{N}_{4}$ | $\mathrm{M}_{4}$ | $\mathrm{L}_{4}$
---|---|---|---|---|---|---
$J^{PC}$ | $0^{++}$ | $0^{++}/2^{++}$ | $0^{-+}$ | $1^{+-}$ | $1^{--}$ | $0^{++}$
$n=0$ | 7.30835 | 22.0966 | 31.9853 | 53.3758 | 83.0449 | 115.002
$n=1$ | 46.9855 | 55.5833 | 72.4793 | 109.446 | 143.581 | 189.632
$n=2$ | 94.4816 | 102.452 | 126.144 | 177.231 | 217.397 | 227.283
$n=3$ | 154.963 | 162.699 | 193.133 | 257.959 | 304.531 | 378.099
$n=4$ | 228.709 | 236.328 | 273.482 | 351.895 | 405.011 | 492.171
Table 3:  Glueball mass spectrum $m_{n}^{2}$ of $\mathrm{AdS}_{7}$ in the
units of $r_{KK}^{2}/L^{4}=M_{KK}^{2}/9$ in [32].

And this model predicts the properties of glueball in a very simple and
powerful way.

##  3 Developments and holographic approaches to QCD

In this section, we will review some holographic approaches to QCD by using
the D4/D8 model and some developments of this model in recent years which
includes basically the topics about phase transition, heavy flavor, hadron
interaction and theta angle in QCD.

###  3.1 QCD deconfinement transition

While the confinement phase of QCD corresponds to the bubble D4-brane geometry
given in (2.16), it is less clear whether the black D4-brane background (2.14)
corresponds to the deconfinement phase in holography exactly [58, 59]. This
issue is recognized by investigating the associated Wilson loop in the bubble
(2.16) and black brane background (2.14) respectively. Nonetheless, it would
be interesting to compare the deconfinement transition in QCD with the
Hawking-Page transition in D4-brane system through the gauge-gravity duality
to find a holographic description of the deconfinement transition exactly. To
this goal, let us first recall the holographic relation between the partition
functions $Z$ of the bulk gravity and its dual field theory,

$Z=e^{-F}=e^{-S_{\mathrm{SUGRA}}^{ren}},$ (3.1)

which implies the classical (onshell) renormalized SUGRA action
$S_{\mathrm{SUGRA}}^{ren}$ is equivalent to the free energy $F$ of the dual
theory (in the Euclidean version). The classical SUGRA action
$S_{\mathrm{SUGRA}}^{ren}$ can be collected by

$S_{\mathrm{SUGRA}}^{ren}=S_{\mathrm{IIA}}^{E}+S_{\mathrm{GH}}+S_{\mathrm{CT}}^{\mathrm{bulk}},$
(3.2)

where $S_{\mathrm{IIA}}^{E}$ refers to the Euclidean version of IIA SUGRA
action given in (2.15). And $S_{\mathrm{GH}}$ refers to the standard Gibbons-
Hawking term given by [71],

$S_{\mathrm{GH}}=\frac{1}{\kappa_{10}^{2}}\int_{\partial\mathcal{M}}e^{-2\Phi}\sqrt{h}K,$
(3.3)

where $h_{MN}$ refers to the metric on the holographic boundary
$\partial\mathcal{M}$ with
$\partial\mathcal{M}=\left\\{r=\varepsilon\right\\}$ for
$\varepsilon\rightarrow 0$, and

$K=h^{MN}\nabla_{M}n_{N}=-\frac{1}{\sqrt{\left|g\right|}}\partial_{r}\left(\frac{\sqrt{\left|g\right|}}{\sqrt{g_{rr}}}\right)\big{|}_{r=\varepsilon},$
(3.4)

is trace of the extrinsic curvature. $S_{\mathrm{CT}}^{\mathrm{bulk}}$ refers
to the counterterm of the bulk fields presented in action (2.15) given as
[72],

$S_{\mathrm{CT}}^{\mathrm{bulk}}=-\frac{5}{2\kappa_{10}^{2}}\frac{g_{s}^{1/3}}{R}\int
d^{9}xe^{-7\Phi/3}\sqrt{h}.$ (3.5)

Using (3.2) - (3.5) by picking up the bubble (2.16) and black brane background
(2.14) solution respectively, we can obtain the free energy of the dual theory
by a simple formula as,

$\displaystyle F_{\mathrm{conf.}}$
$\displaystyle=S_{\mathrm{SUGRA}}^{ren,\mathrm{bub}}=-\frac{2N_{c}^{2}M_{KK}^{4}\lambda
V_{4}}{3^{7}\pi^{2}},$ $\displaystyle F_{\mathrm{deconf}.}$
$\displaystyle=S_{\mathrm{SUGRA}}^{ren,\mathrm{black}}=-\frac{2^{7}N_{c}^{2}T^{6}\pi^{4}\lambda
V_{4}}{3^{7}M_{KK}^{2}},$ (3.6)

where $V_{4}$ refers to the volume of $\mathbb{R}^{4}$, $T$ is the Hawking
temperature in the black D4-brane solution (2.16). Comparing the free energies
given in (3.6), we obtain the critical temperature by
$F_{\mathrm{conf.}}=F_{\mathrm{deconf}.}\left(T=T_{c}\right)$ for the Hawking-
Page transition as,

$T_{c}=\frac{M_{KK}}{2\pi},$ (3.7)

which is expected to be the deconfinement transition in QCD in the large
$N_{c}$ limit. While this may be a trivial result for QCD, it is theoretically
expected in gravity side since the bubble solution (2.14) is obtained by a
double Wick rotation to the black brane solution (2.16) i.e. (3.7) means
exactly $\beta_{4}=\beta_{T}$. However, this does not mean that Hawking-Page
transition has nothing to do with the QCD deconfinement transition because the
fundamental flavored matter has not been taken into account.

In order to obtain a critical temperature close to the realistic QCD with the
D4/D8 model, the flavored matter on the D8-branes must somehow contribute to
the free energy. It means in the gravity side, flavor branes have to back
react to the bulk geometry thus they would not be probes. For such a
holographic setup, we have to require $N_{f}/N_{c}$ is fixed in the large
$N_{c}$ limit in order to go beyond the probe approximation for the flavor
branes. Besides we further need $N_{f}/N_{c}\ll 1$ otherwise the dynamics of
the dual theory is determined by flavors instead of colors, and in the gravity
side $N_{f}/N_{c}\ll 1$ is also necessary since $N_{c}$ D4-branes must
dominate the bulk geometry otherwise the holographic duality given in the
previous sections would not be valid141414See similar setups in [73, 74, 75]
for the D3/D7 system.. Then the next step is to confirm the embedding
configuration of the $\mathrm{D8/\overline{D8}}$-branes. Since the
configuration of the $\mathrm{D8/\overline{D8}}$-branes relates to the chiral
symmetry discussed in Section 2.2 - 2.3, we can identify respectively the
bubble D4-brane background where $\mathrm{D8/\overline{D8}}$-branes are
located at the antipodal points of $x^{4}$ (the left one in Figure 2) as the
confined phase with broken chiral symmetry and the black D4-brane background
where $\mathrm{D8/\overline{D8}}$-branes are parallel (the left one in Figure
3) as the deconfined phase with the restored chiral symmetry. While this
identification does not distinguish exactly the chiral transition from
deconfinement transition and is not unique, it is the most simple setup to
include the elementary features in the QCD deconfinement transition. However,
keep the above requirements in hand, it is not enough to give a holographic
setup quantitatively, because when the flavored backreaction is considered, it
would be extremely challenging to search for a SUGRA solution technologically
with respect to the D-brane configuration in the D4/D8 model as in Table 1. To
simplify the calculation and keep the fundamental features of QCD, authors of
[34, 35] suggest to consider the case that the
$\mathrm{D8/\overline{D8}}$-branes are smeared on the $x^{4}$ direction
homogeneously so that the harmonic function for the D8-branes are identified
uniquely thus it is possible to search for a geometric solution in this setup.

Altogether, let us write down the IIA SUGRA action plus the dynamics of
$N_{f}$ D8-branes smeared on $x^{4}$ as the total action $S$,

$\displaystyle S=$ $\displaystyle\frac{1}{2\kappa_{10}^{2}}\int
d^{10}x\sqrt{-G}e^{-2\phi}\left[\mathcal{R}^{(10)}+4\partial_{\mu}\phi\partial\phi\right]-\frac{1}{4\kappa_{10}^{2}}\int
d^{10}x\sqrt{-G}\left|F_{4}\right|^{2}$
$\displaystyle-\frac{g_{s}N_{f}T_{\mathrm{D8}}M_{KK}}{\pi}\int
d^{10}x\frac{\sqrt{-\det\left[g+\left(2\pi\alpha^{\prime}\right)F\right]}}{\sqrt{g_{44}}}e^{-\phi}.$
(3.8)

To search for an approximate solution under the condition $N_{f}/N_{c}\ll 1$,
the solution with $N_{f}=0$ is therefore the zero-th order solution to the
equations of motion from (3.8) which is nothing but the bubble and black
D4-brane solution given in (2.16) and (2.14). Then let us attempt to find a
solution of $\mathcal{O}\left(N_{f}/N_{c}\right)$ to the bubble D4-brane
(2.16) first. For a homogeneous solution, the ansatz of the metric to solve
the action (3.8) can be chosen as [33, 34, 35],

$\displaystyle ds^{2}$
$\displaystyle=e^{2\hat{\lambda}}\left(-dt^{2}+dx_{i}dx^{i}\right)+e^{2\tilde{\lambda}}\left(dx^{4}\right)^{2}+l_{s}^{2}e^{-2\varphi}d\rho^{2}+l_{s}^{2}e^{2\nu}d\Omega_{4}^{2},$
$\displaystyle\varphi$ $\displaystyle=2\Phi-4\lambda-\tilde{\lambda}-4\nu,$
(3.9)

where $\Phi$ refers to the dilaton field,
$\hat{\lambda},\tilde{\lambda},\varphi,\nu$ are unknown functions depending on
the holographic coordinate $\rho$ only. And $\rho$ is the logarithmic
coordinate defined as,

$e^{-3a\rho}=1-\frac{U_{KK}^{3}}{U^{3}},a=\frac{\sqrt{2}Q_{c}U_{KK}^{3}}{3R^{3}g_{s}}=\frac{U_{KK}^{3}}{l_{s}^{3}g_{s}^{2}}.$
(3.10)

Impose the ansatz (3.9) into (3.8), it reduces to a 1d effective action,

$\displaystyle S$ $\displaystyle=\mathcal{V}\int
d\rho\left[-4\dot{\hat{\lambda}}^{2}-\dot{\tilde{\lambda}}^{2}-4\dot{\nu}^{2}+\dot{\varphi}^{2}+V\right],$
$\displaystyle V$
$\displaystyle=12e^{-2\nu-2\varphi}-Q_{c}^{2}e^{4\hat{\lambda}+\tilde{\lambda}-4\nu-4\varphi}-Q_{f}e^{2\hat{\lambda}-\frac{\tilde{\lambda}}{2}+2\nu-\frac{3}{2}\varphi},$
(3.11)

which has to be supported by the zero-energy constraint (dot refers to the
derivative respected to $\rho$)

$-4\dot{\hat{\lambda}}^{2}-\dot{\tilde{\lambda}}^{2}-4\dot{\nu}^{2}+\dot{\varphi}^{2}-V=0,$
(3.12)

with

$Q_{c}=\frac{3\pi
N_{c}}{\sqrt{2}}=\frac{3}{\sqrt{2g_{s}}}\frac{R^{3}}{l_{s}^{3}},Q_{f}=\frac{2\kappa_{10}^{2}N_{f}g_{s}T_{\mathrm{D8}}M_{KK}l_{s}^{2}}{\pi},\mathcal{V}=\frac{1}{2\kappa_{10}^{2}}V_{3}\Omega_{4}\beta_{T}\frac{2\pi}{M_{KK}}l_{s}^{3}.$
(3.13)

Here $\beta_{T}$ is the size of $x^{0}$ representing the temperature in the
dual theory $\beta_{T}=1/T$ and the only nonzero component of the gauge field
potential on the D8-branes is a constant $A_{0}$ representing the chemical
potential in the dual theory. Next, we expand the all the relevant functions
$\hat{\lambda},\tilde{\lambda},\varphi,\nu$ up to $N_{f}/N_{c}$ as,

$\Psi=\Psi_{0}+\epsilon_{f}\Psi_{1}+\mathcal{O}\left(\epsilon_{f}^{2}\right),$
(3.14)

where

$\epsilon_{f}=\frac{R^{3/2}u_{0}^{1/2}g_{s}}{l_{s}}Q_{f}=\frac{1}{12\pi^{3}}\lambda^{2}\frac{N_{f}}{N_{c}}\ll
1.$ (3.15)

So the zero-th order solution of $\hat{\lambda},\tilde{\lambda},\varphi,\nu$
reads by comparing the metric ansatz (3.9) with the bubble D4-brane solution
(2.16) as,

$\displaystyle\hat{\lambda}_{0}$
$\displaystyle=f_{0}+\frac{3}{4}\log\frac{U_{KK}}{R},$
$\displaystyle\tilde{\lambda}_{0}$
$\displaystyle=f_{0}-\frac{3}{2}a\rho+\frac{3}{4}\log\frac{U_{KK}}{R},$
$\displaystyle\Phi_{0}$
$\displaystyle=f_{0}+\frac{3}{4}\log\frac{U_{KK}}{R}+\log g_{s},$
$\displaystyle\nu_{0}$
$\displaystyle=\frac{1}{3}f_{0}+\frac{1}{4}\log\frac{U_{KK}}{R}+\log\frac{R}{l_{s}},$
$\displaystyle f_{0}$
$\displaystyle=-\frac{1}{4}\log\left(1-e^{-3a\rho}\right),$ (3.16)

Put (3.14) - (3.16) back into the equation of motion varied from action
(3.11), we can obtain a series of equations for
$\hat{\lambda}_{1},\tilde{\lambda}_{1},\varphi_{1},\nu_{1}$ as

$\displaystyle\frac{\ddot{\hat{\lambda}}_{1}}{a^{2}}-\frac{9}{2}\frac{e^{-3a\rho}}{\left(1-e^{-3a\rho}\right)^{2}}\left(4\hat{\lambda}_{1}+\tilde{\lambda}_{1}-\Phi_{1}\right)$
$\displaystyle=\frac{1}{4}\frac{e^{-\frac{3}{2}a\rho}}{\left(1-e^{-3a\rho}\right)^{13/6}},$
$\displaystyle\frac{\ddot{\tilde{\lambda}}_{1}}{a^{2}}-\frac{9}{2}\frac{e^{-3a\rho}}{\left(1-e^{-3a\rho}\right)^{2}}\left(4\hat{\lambda}_{1}+\tilde{\lambda}_{1}-\Phi_{1}\right)$
$\displaystyle=-\frac{1}{4}\frac{e^{-\frac{3}{2}a\rho}}{\left(1-e^{-3a\rho}\right)^{13/6}},$
$\displaystyle\frac{\ddot{\Phi}_{1}}{a^{2}}-\frac{9}{2}\frac{e^{-3a\rho}}{\left(1-e^{-3a\rho}\right)^{2}}\left(4\hat{\lambda}_{1}+\tilde{\lambda}_{1}-\Phi_{1}\right)$
$\displaystyle=\frac{5}{4}\frac{e^{-\frac{3}{2}a\rho}}{\left(1-e^{-3a\rho}\right)^{13/6}},$
$\displaystyle\frac{\ddot{\nu}_{1}}{a^{2}}-\frac{3}{2}\frac{e^{-3a\rho}}{\left(1-e^{-3a\rho}\right)^{2}}\left(4\hat{\lambda}_{1}+\tilde{\lambda}_{1}-5\Phi_{1}+12\nu_{1}\right)$
$\displaystyle=\frac{1}{4}\frac{e^{-\frac{3}{2}a\rho}}{\left(1-e^{-3a\rho}\right)^{13/6}},$
(3.17)

which can be solved analytically by

$\displaystyle\hat{\lambda}_{1}$
$\displaystyle=\frac{3}{8}f+y-\frac{1}{4}\left(A_{2}-A_{1}\right)-\frac{1}{4}\left(B_{2}-B_{1}\right)a\rho,$
$\displaystyle\tilde{\lambda}_{1}$
$\displaystyle=-\frac{1}{8}f+y-\frac{1}{4}\left(A_{2}+B_{2}a\rho\right)-\frac{3}{4}\left(A_{1}+B_{1}a\rho\right),$
$\displaystyle\Phi_{1}$
$\displaystyle=\frac{11}{8}f+y+\frac{1}{4}\left(A_{1}+B_{1}a\rho\right)-\frac{5}{4}\left(A_{2}+B_{2}a\rho\right),$
$\displaystyle\nu_{1}$ $\displaystyle=\frac{11}{24}f+q,$ (3.18)

with hyper geometrical functions

$\displaystyle f=$ $\displaystyle\frac{4}{9}e^{-3a\rho/2}\
_{3}F_{2}\left(\frac{1}{2},\frac{1}{2},\frac{13}{6};\frac{3}{2},\frac{3}{2};e^{-3a\rho}\right),$
$\displaystyle y=$ $\displaystyle
C_{2}-\coth\left(\frac{3a\rho}{2}\right)\left[C_{1}+C_{2}\left(\frac{3}{2}a\rho+1\right)\right]+z,$
$\displaystyle q=$
$\displaystyle\frac{1}{12}\left[A_{1}-5A_{2}+a\rho\left(B_{1}-5B_{2}\right)\right]+\frac{5}{3}z-\coth\left(\frac{3a\rho}{2}\right)\left[M_{1}+M_{2}\left(3a\rho+2\right)\right]+2M_{2},$
$\displaystyle z=$
$\displaystyle-\frac{e^{-9a\rho/2}\left(e^{-3a\rho}+1\right)\left[9e^{3a\rho}\
_{3}F_{2}\left(\frac{1}{2},\frac{1}{2},\frac{19}{6};\frac{3}{2},\frac{3}{2};e^{-3a\rho}\right)+\
_{3}F_{2}\left(\frac{3}{2},\frac{3}{2},\frac{19}{6};\frac{5}{2},\frac{5}{2};e^{-3a\rho}\right)\right]}{162\left(1-e^{-3a\rho}\right)}$
$\displaystyle-\frac{8e^{-3a\rho/2}\left(10e^{-3a\rho}+3\right)}{819\left(1-e^{-3a\rho}\right)}\
_{2}F_{1}\left(\frac{1}{6},\frac{1}{2};\frac{3}{2};e^{-3a\rho}\right)+-\frac{e^{-15a\rho/2}\left(38e^{3a\rho}+8e^{6a\rho}-40\right)}{273\left(1-e^{-3a\rho}\right)^{13/6}},$
(3.19)

where $A_{1,2},B_{1,2},C_{1,2},M_{1,2}$ are integration constants. The
integration constants can be determined by analyzing the asymptotics and using
the zero-energy constraint (3.12) which leads to

$\displaystyle A_{2}$
$\displaystyle=-2A_{1},A_{1}=\frac{81\sqrt{3}\pi^{2}\left(-9+\sqrt{3}\pi-12\log
2+9\log
3\right)}{43120\left(2\right)^{2/3}\Gamma\left(-14/3\right)\Gamma\left(-2/3\right)^{2}},$
$\displaystyle k$
$\displaystyle=C_{1}+C_{2}=\frac{\pi^{3/2}\left(3+\sqrt{3}\pi-12\log 2+9\log
3\right)}{78\Gamma\left(-2/3\right)\Gamma\left(1/6\right)},$
$\displaystyle\frac{5k}{3}$ $\displaystyle=M_{1}+M_{2},$ (3.20)

while the other constants must be confirmed by imposing additional physical
condition. Nevertheless, one may find the phase transition depends only on the
integration constants given in (3.20).

Follow the same step, it is also possible to obtain a solution of order
$N_{f}/N_{c}$ to the black D4-brane solution by using the metric ansatz,

$\displaystyle ds^{2}$
$\displaystyle=-e^{2\tilde{\lambda}}dt^{2}+e^{2\hat{\lambda}}dx_{i}dx^{i}+e^{2\lambda_{s}}\left(dx^{4}\right)^{2}+l_{s}^{2}e^{-2\varphi}d\rho^{2}+l_{s}^{2}e^{2\nu}d\Omega_{4}^{2},$
$\displaystyle\varphi$
$\displaystyle=2\Phi-3\hat{\lambda}-\tilde{\lambda}-\lambda_{s}-4\nu,$ (3.21)

with a non-zero dynamical chemical potential

$2\pi\alpha^{\prime}A=A_{t}dt.$ (3.22)

We note that $U_{KK}$ is replaced by $U_{H}$ in the black brane case. Put
(3.21) and (3.22) into action (3.8), it leads to a 1d action as,

$\displaystyle S$ $\displaystyle=\mathcal{V}\int
d\rho\left[-3\dot{\hat{\lambda}}^{2}-\dot{\tilde{\lambda}}^{2}-\dot{\lambda}_{s}^{2}-4\dot{\nu}^{2}+\dot{\varphi}^{2}+V\right],$
$\displaystyle V$
$\displaystyle=12e^{-2\nu-2\varphi}-Q_{c}^{2}e^{3\hat{\lambda}+\tilde{\lambda}+\lambda_{s}-4\nu-4\varphi}-Q_{f}e^{\frac{3}{2}\hat{\lambda}-\frac{1}{2}\lambda_{s}+\frac{\tilde{\lambda}}{2}+2\nu-\frac{3}{2}\varphi}\sqrt{1-\frac{1}{l_{s}^{2}}e^{-2\tilde{\lambda}+2\varphi}\dot{A}_{t}^{2}}.$
(3.23)

Taking into account the near-horizon geometry, the DBI action presented in
(3.23) can be expanded with respect to small gauge field potential. Then keep
the quadratic action for $A_{t}$, we can obtain an analytical leading order
solution by the equations of motion derived from (3.23) as,

$\displaystyle A_{t}$
$\displaystyle=\frac{2}{3}qU_{H}\left(1-\sqrt{1-e^{-3a\rho}}\right),q=\frac{g_{s}l_{s}^{4}}{R^{3/2}U_{H}^{5/2}}n$
$\displaystyle\hat{\lambda}_{1}$
$\displaystyle=\frac{f}{28}-\frac{3}{16}q^{2}g+y-\frac{1}{4}\left(a_{2}-a_{1}-a_{3}\right)-\frac{1}{4}\left(b_{2}-b_{1}-b_{3}\right)a\rho,$
$\displaystyle\lambda_{s1}$
$\displaystyle=\hat{\lambda}_{1}-\frac{f}{21}+\frac{q}{4}g-a_{1}-b_{1}a\rho,$
$\displaystyle\tilde{\lambda}_{1}$
$\displaystyle=\hat{\lambda}_{1}+\frac{q^{2}}{2}g-a_{3}-b_{3}a\rho,$
$\displaystyle\Phi_{1}$
$\displaystyle=\hat{\lambda}_{1}+\frac{2}{21}f-a_{2}-b_{2}a\rho,$
$\displaystyle\nu_{1}$ $\displaystyle=\frac{11}{252}f-\frac{q^{2}}{16}g+w,$
(3.24)

with

$\displaystyle f=$
$\displaystyle\frac{6}{\left(1-e^{-3a\rho}\right)^{1/6}}+\sqrt{3}\tan^{-1}\left[\frac{2\left(1-e^{-3a\rho}\right)^{1/6}-1}{\sqrt{3}}\right]+\sqrt{3}\tan^{-1}\left[\frac{2\left(1-e^{-3a\rho}\right)^{1/6}+1}{\sqrt{3}}\right]$
$\displaystyle-2\tanh^{-1}\left[\left(1-e^{-3a\rho}\right)^{1/6}\right]-\coth^{-1}\left[\left(1-e^{-3a\rho}\right)^{1/6}+\frac{1}{\left(1-e^{-3a\rho}\right)^{1/6}}\right],$
$\displaystyle g=$
$\displaystyle\frac{4}{9}\log\left(e^{-3a\rho/2}\sqrt{e^{3a\rho}-1}+1\right)-\frac{4}{9}e^{-3a\rho/2}\sqrt{e^{3a\rho}-1},$
$\displaystyle y=$ $\displaystyle
c_{2}-\left[c_{1}+\left(1+\frac{3}{2}a\rho\right)c_{2}\right]\coth\left(\frac{3a\rho}{2}\right)+q^{2}j+z,$
$\displaystyle w=$ $\displaystyle
2m_{2}-\left[m_{1}+\left(2+3r\right)m_{2}\right]\coth\left(\frac{3a\rho}{2}\right)+\frac{1}{12}\left(a_{1}-5a_{2}+a_{3}+b_{1}a\rho-5b_{2}a\rho+b_{3}a\rho\right)$
$\displaystyle+\frac{5}{3}z-q^{2}j,$ $\displaystyle j=$
$\displaystyle\frac{1}{72}\left[4\sqrt{1-e^{-3a\rho}}+\left(-9a\rho+6\sqrt{1-e^{-3a\rho}}-6\log\left(\sqrt{1-e^{-3a\rho}}+1\right)\right)\coth\left(\frac{3a\rho}{2}\right)\right],$
$\displaystyle z=$
$\displaystyle\frac{3e^{3a\rho}\left(1-e^{-3a\rho}\right)^{5/6}-\frac{\sqrt{3}}{2}\left(e^{3a\rho}+1\right)\left[\tan^{-1}\left(\frac{2\sqrt[6]{1-e^{-3a\rho}}-1}{\sqrt{3}}\right)+\tan^{-1}\left(\frac{2\sqrt[6]{1-e^{-3a\rho}}+1}{\sqrt{3}}\right)\right]}{546\left(e^{3a\rho}-1\right)}$
$\displaystyle+\frac{1}{2}\left(e^{3a\rho}+1\right)\frac{2\tan^{-1}\left(\frac{2\sqrt[6]{1-e^{-3a\rho}}-1}{\sqrt{3}}\right)+\coth^{-1}\left[\left(1-e^{-3a\rho}\right)^{1/6}+\frac{1}{\left(1-e^{-3a\rho}\right)^{1/6}}\right]}{546\left(e^{3a\rho}-1\right)},$
(3.25)

where $a_{1,2,3},b_{1,2,3},c_{1,2},m_{1,2}$ are integration constants and $n$
refers to the $U\left(1\right)$ charge density. And the zero-energy constraint
is given by

$\displaystyle-3\dot{\hat{\lambda}}^{2}-\dot{\lambda}_{s}^{2}-\dot{\tilde{\lambda}}^{2}-4\dot{\nu}^{2}+\dot{\varphi}^{2}+\frac{Q_{f}}{2l_{s}}e^{\frac{3}{2}\hat{\lambda}-\frac{1}{2}\lambda_{s}+\frac{3\tilde{\lambda}}{2}+2\nu+\frac{1}{2}\varphi}\dot{A}_{t}^{2}-\mathcal{P}$
$\displaystyle=0,$ (3.26)

with

$\mathcal{P}=12e^{-2\nu-2\varphi}-Q_{c}^{2}e^{3\hat{\lambda}+\tilde{\lambda}+\lambda_{s}-4\nu-4\varphi}-Q_{f}e^{\frac{3}{2}\hat{\lambda}-\frac{1}{2}\lambda_{s}+\frac{\tilde{\lambda}}{2}+2\nu-\frac{3}{2}\varphi}.$
(3.27)

Now it is possible to obtain the free energy of the dual theory involving the
flavored matters by imposing the above leading order solutions into the action
given in (3.8) after holographic renormalization. Before this, we need to add
an additional holographic counterterm to (3.2) in order to cancel the
divergence in the DBI action presented in (3.8) which is turned out to be [34,
76, 77, 78]

$S_{\mathrm{CT}}^{\mathrm{D8}}=\frac{Q_{f}}{\kappa_{10}^{2}l_{s}^{2}}\int_{\partial\mathcal{M}}d^{9}x\frac{\sqrt{h}}{\sqrt{h_{44}}}\left[\frac{R}{g_{s}^{1/3}}\chi_{1}e^{-2\Phi/3}-\frac{2R^{2}}{g_{s}^{2/3}}\chi_{2}e^{-\Phi/3}\left(K-\frac{8}{3}\bm{n}\cdot\nabla\Phi-\bm{n}\cdot\frac{\nabla\left(\sqrt{g_{44}}\right)}{\sqrt{g_{44}}}\right)\right].$
(3.28)

Here $\bm{n}$ refers to the normal vector of $\partial\mathcal{M}$ and
$\chi_{1,2}$ are renormalized constants. For example, with the back reaction
from D8-branes to the bubble background, it leads to,

$\chi_{1}=-\frac{631}{5005},\chi_{2}=-\frac{2}{2145}.$ (3.29)

For the black brane background, it leads to

$\chi_{1}=-\frac{607}{5005},\chi_{2}=-\frac{4}{15015}.$ (3.30)

Hence we finally obtained the renormalized SUGRA action as,

$F=S_{\mathrm{SUGRA}}^{ren}=S_{\mathrm{IIA}}^{E}+S_{\mathrm{GH}}+S_{\mathrm{CT}}^{\mathrm{bulk}}+S_{\mathrm{CT}}^{\mathrm{D8}},$
(3.31)

with suitable choice of $\chi_{1,2}$. Respectively, the confined and
deconfined free energy with flavors can be computed straightforwardly by
plugging the solutions of order $N_{f}/N_{c}$ into (3.31), as

Figure 5:  The critical temperature of Hawking-Page transition as the
temperature of QCD deconfinement transition in the D4/D8 model with chemical
potential $\mu$ or magnetic field $B$.

$\displaystyle F_{\mathrm{conf.}}$
$\displaystyle=-\frac{2N_{c}^{2}M_{KK}^{4}\lambda
V_{4}}{3^{7}\pi^{2}}\left[1-\frac{48}{7}\epsilon_{f}\frac{\pi^{3/2}}{\Gamma\left(-2/3\right)\Gamma\left(1/6\right)}\right],$
$\displaystyle F_{\mathrm{deconf.}}$
$\displaystyle=-\frac{2^{7}N_{c}^{2}T^{6}\pi^{4}\lambda
V_{4}}{3^{7}M_{KK}^{2}}\left[1+\frac{4}{7}\epsilon_{fT}\left(1+\frac{7}{6}q^{2}\right)\right],$
(3.32)

where

$\epsilon_{fT}=\frac{R^{3/2}U_{H}^{1/2}g_{s}Q_{f}}{l_{s}^{2}}=\frac{\lambda^{2}}{12\pi^{3}}\frac{2\pi
T}{M_{KK}}\frac{N_{f}}{N_{c}}\ll 1,$ (3.33)

and we have used the choice of the relevant constants given in (3.20).
Therefore, compare the free energy given in (3.32), we can obtain the critical
temperature with flavors as,

$\frac{2\pi
T_{c}}{M_{KK}}=1-\frac{1}{126\pi^{3}}\lambda^{2}\frac{N_{f}}{N_{c}}\left[1+\frac{12\pi^{3/2}}{\Gamma\left(-2/3\right)\Gamma\left(1/6\right)}\right]-\frac{27}{16\pi}\frac{N_{f}}{N_{c}}\frac{\mu^{2}}{M_{KK}^{2}},$
(3.34)

where $\mu$ is the chemical potential in the dual theory given by
$A_{t}\big{|}_{U\rightarrow\infty}=\mu$. And the behavior of the Hawking-Page
transition given in (3.34) agrees qualitatively with the QCD deconfinement
transition [79, 80, 81, 82].

Moreover, when the backreaction to the bulk geometry of the flavor branes is
picked up, it is also possible to evaluate QCD deconfinement transition under
an external magnetic field. Because extremely strong magnetic field may also
give rise to deconfinement transition in QCD [83, 84, 85, 86]. The setup
follows mostly the same discussion given above while we need to turn on a
constant magnetic field in the DBI action presented in (3.8), as the only non-
zero component of the gauge field strength. Then we can derive the effective
1d action by using the metric ansatz (3.9) and (3.21) with respect to the
bubble and black D4-brane background. Fortunately, it is possible to find an
analytical solution, which leads to critical temperature $T_{c}$ as

$\frac{2\pi
T_{c}}{M_{KK}}=1+\frac{N_{f}}{N_{c}}\left(X\lambda^{2}+\frac{B^{2}}{M_{KK}^{4}}Y\right),$
(3.35)

by comparing free energy in the same way. Here $B$ refers to the external
magnetic field, and $X,Y$ are numerical numbers. For the probe approximation
limit of the D8-brane, $X,Y$ are calculated as,

$\displaystyle X$
$\displaystyle=-\frac{1}{126\pi^{3}}\left[1-\frac{8\pi^{3/2}}{3\Gamma\left(1/6\right)\Gamma\left(4/3\right)}\right]\simeq
5\times 10^{-4},$ $\displaystyle Y$
$\displaystyle=-\frac{81}{16\pi}\left[1-\frac{2\pi^{3/2}}{3\Gamma\left(1/6\right)\Gamma\left(4/3\right)}\right]\simeq-0.408.$
(3.36)

By considering the backreaction of the D8-branes, $X,Y$ are calculated as,

$X\simeq 5\times 10^{-4},Y\simeq-2.44.$ (3.37)

And the behavior of the critical temperature illustrated in (3.35) also
coincides qualitatively with the QCD deconfinement transition under external
magnetic field predicted by lattice QCD [83]. We plot out the behavior of the
critical temperature given in (3.34) and (3.35) in Figure 5. In this sense, we
could conclude at least, investigating the Hawking-Page transition in the
D4/D8 model is very suggestive to study the QCD deconfinement transition in
holography which also covers partly the discussion in some bottom-up
approaches [87, 88].

###  3.2 Phase diagram with chiral transition

As we have reviewed the deconfinement transition in the D4/D8 model which can
not be distinguished from the chiral transition, let us focus on the chiral
transition in the D4/D8 model since QCD has various phases with chiral
symmetry.

Recall the relation of $\mathrm{D8/\overline{D8}}$-brane configuration and
chiral symmetry, the chiral transition is identified as the transition from
connected to disconnected configuration of the
$\mathrm{D8/\overline{D8}}$-branes. Hence we need to choose the black D4-brane
background in order to include both the connected and disconnected
$\mathrm{D8/\overline{D8}}$-brane configuration. The main idea to evaluate the
phase transition follows the Section 3.1, which is to compute the free energy
in holography. As we will work with respect to the black D4-brane background
only, the contribution from the bulk geometry would be irrelevant, because the
difference of the free energy determines the phase transition. Keeping this in
mind, we can quickly write down the D8-brane action for mesonic (broken chiral
symmetry) and quark matter phase (restored chiral symmetry) with a
$U\left(1\right)$ chemical potential $\hat{A}_{0}$, which corresponds to the
connected and disconnected D8-brane configuration respectively in Figure 3
as151515We note that in this setup, the Chern-Simons action vanishes.,

$S_{\mathrm{DBI}}^{\mathrm{D8}}=\mathcal{N}\frac{V_{3}}{T}\int_{u_{0}}^{\infty}duu^{5/2}\sqrt{1+u^{3}f_{T}\left(x_{4}^{\prime}\right)^{2}-\hat{a}_{0}^{\prime
2}},$ (3.38)

where the variables in (3.38) are dimensionless as,

$x_{4}=x^{4}M_{KK},u=\frac{U}{R}\frac{1}{\left(M_{KK}R\right)^{2}},\hat{a}_{0}=\hat{A}_{0}\frac{4\pi}{\lambda
M_{KK}},\mathcal{N}=2T_{\mathrm{D8}}\Omega_{4}R^{5}\left(M_{KK}R\right)^{7}.$
(3.39)

The equation of motion can be obtained by varying (3.38) respected to $x_{4}$
and $\hat{a}_{0}$ which are,

$\displaystyle\frac{u^{5/2}\hat{a}_{0}^{\prime}}{\sqrt{1+u^{3}f_{T}x_{4}^{\prime
2}-\hat{a}_{0}^{\prime 2}}}$ $\displaystyle=n_{I},$
$\displaystyle\frac{u^{11/2}f_{T}x_{4}^{\prime}}{\sqrt{1+u^{3}f_{T}x_{4}^{\prime
2}-\hat{a}_{0}^{\prime 2}}}$
$\displaystyle=f_{T}^{1/2}\left(u_{0}\right)u_{0}^{4},$ (3.40)

where “$\prime$” refers to the derivative with respect to $u$. The constant
$n_{I}$ corresponds to the $U\left(1\right)$ charge density which is therefore
the baryon number in this setup. In the mesonic phase, the baryon number is
zero i.e. $n_{I}=0$, and the equations of motion in (3.40) can be solved by
the following boundary condition according to the connected configuration in
Figure 3, as

$\displaystyle\hat{a}_{0}^{\prime}\left(u_{0}\right)$ $\displaystyle=0,\
\hat{a}_{0}\left(\infty\right)=\mu,$ $\displaystyle
x_{4}^{\prime}\left(u_{0}\right)$ $\displaystyle=\infty,\
\frac{l}{2}=\int_{u_{0}}^{\infty}dux_{4}^{\prime},$ (3.41)

where constant $\mu$ refers to the chemical potential in the dual theory,
constant $l$ refers to the separation of the
$\mathrm{D8/\overline{D8}}$-branes at boundary $u\rightarrow\infty$. Thus the
solution is

$\hat{a}_{0}=\mu,x_{4}^{\prime}=\frac{f_{T}^{1/2}\left(u_{0}\right)u_{0}^{4}}{f_{T}^{1/2}\left(u\right)u^{3/2}\sqrt{u^{8}f_{T}\left(u\right)-u_{0}^{8}f_{T}\left(u_{0}\right)}}.$
(3.42)

Put the solution (3.42) back into action (3.38), we can obtain the free energy
as,

$F_{\mathrm{mesonic}}=\mathcal{N}\int_{u_{0}}^{\infty}duu^{5/2}\frac{u^{4}f_{T}^{1/2}\left(u\right)}{\sqrt{u^{8}f_{T}\left(u\right)-u_{0}^{8}f_{T}\left(u_{0}\right)}}.$
(3.43)

For the quark matter phase, the boundary condition reads from the disconnected
configuration in Figure 3 as,

$\hat{a}_{0}\left(u_{H}\right)=0,\hat{a}_{0}\left(\infty\right)=\mu,x_{4}^{\prime}=0,$
(3.44)

which leads to a solution with hyper geometrical functions as,

$\hat{a}_{0}\left(u\right)=\mu-\frac{n_{I}^{2/5}\Gamma\left(3/10\right)\Gamma\left(6/5\right)}{\sqrt{\pi}}+u\
_{2}F_{1}\left(\frac{1}{5},\frac{1}{2},\frac{6}{5},-\frac{u^{5}}{n_{I}^{2}}\right).$
(3.45)

Therefore the free energy is computed as,

$F_{\mathrm{quarkmatter}}=\mathcal{N}\int_{u_{H}}^{\infty}du\frac{u^{5}}{\sqrt{u^{5}+n_{I}^{5}}}.$
(3.46)

We note that the condition $\hat{a}_{0}\left(u_{H}\right)=0$ implies $n_{I}$
is a function of $\mu$ as,

$0=\mu-\frac{n_{I}^{2/5}\Gamma\left(3/10\right)\Gamma\left(6/5\right)}{\sqrt{\pi}}+u_{H}\
{}_{2}F_{1}\left(\frac{1}{5},\frac{1}{2},\frac{6}{5},-\frac{u_{H}^{5}}{n_{I}^{2}}\right).$
(3.47)

Then the phase diagram can be obtained by compare the free energy given in
(3.43) and (3.47) with constraint (3.46). Notice that while the free energy
given in (3.43) and (3.47) is divergent, their difference which determines the
phase diagram remains to be finite. Thus it is not necessary to do the
holographic renormalization in this case.

For a more ambitious approach, let us include the baryonic phase in the black
D4-brane background, that is to take into account the configuration (c) in
Figure 4 which has broken chiral symmetry with baryon vertex. Since baryon
vertex is identified as $\mathrm{D4}^{\prime}$-brane described equivalently by
instantons on the D8-branes, we employ the BPST instanton solution given in
(2.85) to represent baryon on the flavor brane. For multiple baryons we can
summarize the instanton field strength as what we have discussed in Section
2.5, in this sense baryons are treated as instanton gas on the flavor brane.
On the other hand, as the instanton size takes order of $\lambda^{-1/2}$ and
the DBI action (2.20) does not define how to treat it with non-Abelian gauge
field161616The symmetrized trace in DBI action is usually used for all terms
of $\mathcal{O}\left(F^{4}\right)$ and higher, however it is known to be
incomplete starting from $\mathcal{O}\left(F^{6}\right)$ [89]., we may
generalize the DBI action by taking all order of gauge field strength into
non-Abelian case through the identity for Abelian gauge field strength
$\hat{F}$ as,

$\displaystyle\sqrt{\det\left(g+2\pi\alpha^{\prime}\hat{F}\right)}=$
$\displaystyle
U^{4}\left(\frac{R}{U}\right)^{3/4}\big{\\{}\left(2\pi\alpha^{\prime}\right)^{2}f_{T}\hat{F}_{iU}^{2}+\left(1+u^{3}f_{T}x_{4}^{\prime
2}-\hat{a}_{0}^{\prime
2}\right)\left[1+\left(\frac{R}{U}\right)^{3}\frac{\left(2\pi\alpha^{\prime}\right)^{2}}{2}\hat{F}_{ij}^{2}\right]$
$\displaystyle+\left(\frac{R}{U}\right)^{3}\frac{\left(2\pi\alpha^{\prime}\right)^{4}f_{T}\left(\hat{F}_{ij}\hat{F}_{kU}\epsilon_{ijk}\right)^{2}}{4}\big{\\}}^{1/2}.$
(3.48)

For non-Abelian generalization, we follow [90] to replace the quadratic terms
of $\hat{F}$ by its non-Abelian version $F$ then take trace of each term
separately as,

$\hat{F}_{iU}^{2}\rightarrow\mathrm{Tr}F_{iU}^{2},\hat{F}_{ij}^{2}\rightarrow\mathrm{Tr}F_{ij}^{2},\hat{F}_{ij}\hat{F}_{kU}\rightarrow\mathrm{Tr}F_{ij}F_{kU}.$
(3.49)

Afterwards we impose the BPST instanton solution (2.85) with multiple number
$n_{I}$ to $F$ in order to represent baryons 171717[27, 28] illustrate that in
the case of $N_{f}=2$ the non-Abelian part of $A_{0}$ presented in (2.87)
vanishes. And we do not attempt to consider baryon with $N_{f}>2$ in this
section. . Altogether, we reach to a generalized version of action for
baryonic phase as,

$F_{\mathrm{baryonic}}=S^{\mathrm{D8}}=\mathcal{N}\frac{V_{3}}{T}\int_{u_{c}}^{\infty}\left[\sqrt{\left(1+g_{1}+u^{3}f_{T}x_{4}^{\prime
2}-\hat{a}_{0}^{\prime
2}\right)\left(1+g_{2}\right)}-n_{I}\hat{a}_{0}\left(u\right)q\left(u\right)\right],$
(3.50)

where

$\displaystyle g_{1}\left(u\right)$
$\displaystyle=\frac{f_{T}\left(u\right)u^{1/2}}{3u_{c}^{2}\sqrt{f_{c}\left(u\right)}}n_{I}q\left(u\right),f_{c}\left(u\right)=1-\frac{u_{c}^{3}}{u^{3}},$
$\displaystyle g_{2}\left(u\right)$
$\displaystyle=\frac{u_{c}^{2}\sqrt{f_{c}\left(u\right)}}{3u^{7/2}}n_{I}q\left(u\right),q\left(Z\right)=\frac{3\rho^{4}}{4\left(Z^{2}+\rho^{2}\right)^{5/2}}.$
(3.51)

We note that the last term in (3.50) is the Chern-Simons action and
$q\left(Z\right)$ is the average instanton field strength defined by the
summary of the BPST instanton (2.85) as [91]

$\frac{1}{V_{3}}\sum_{n=1}^{N_{I}}\int
d^{3}X\frac{4\left(\rho/\gamma\right)^{4}}{\left[\left(\vec{X}-\vec{X}_{0n}\right)^{2}+\left(Z/\gamma\right)^{2}+\left(\rho/\gamma\right)^{2}\right]^{4}}=\frac{2}{3}\pi^{2}\gamma
q\left(Z\right)n_{I},$ (3.52)

with normalization condition

$\int_{-\infty}^{+\infty}q\left(Z\right)dZ=1.$ (3.53)

$\vec{X}_{0n}$ refers to the center of the $n$-th instanton. $Z$ is the
Cartesian coordinate, in the case of (c) in Figure 4, it is defined as,

$U^{3}=U_{c}^{3}+U_{c}Z^{2},$ (3.54)

where we use $U_{c}$ to denote the connected position of
$\mathrm{D8/\overline{D8}}$-branes with instantons to distinguish it from
$U_{0}$ in which instantons are absent. $N_{I}$ is the instanton number which
relates to its number density $n_{I}$ as
$n_{I}=N_{I}96\pi^{4}/\left(\lambda^{2}M_{KK}^{3}\right)$. With the boundary
condition

$\frac{l}{2}=\int_{u_{c}}^{\infty}dux_{4}^{\prime},\hat{a}_{0}\left(\infty\right)=\mu,$
(3.55)

the equations of motion obtained by varying (3.50),

$\displaystyle\frac{u^{5/2}\hat{a}_{0}^{\prime}\sqrt{1+g_{2}}}{\sqrt{1+g_{1}+u^{3}f_{T}x_{4}^{\prime
2}-\hat{a}_{0}^{\prime 2}}}$ $\displaystyle=n_{I}Q$
$\displaystyle\frac{u^{11/2}f_{T}x_{4}^{\prime}\sqrt{1+g_{2}}}{\sqrt{1+g_{1}+u^{3}f_{T}x_{4}^{\prime
2}-\hat{a}_{0}^{\prime 2}}}$ $\displaystyle=k,$ (3.56)

where $k$ is an integration constant to be determined and

$Q\left(u\right)=\int_{u_{c}}^{u}q\left(v\right)dv=\frac{u^{3/2}\sqrt{f_{c}}}{2}\frac{3\rho^{2}u_{c}+2\left(u^{3}-u_{c}^{3}\right)}{\left(u^{3}-u_{c}^{3}+\rho^{2}u_{c}\right)^{3/2}},$
(3.57)

can be solved as,

$\displaystyle\hat{a}_{0}^{\prime 2}$
$\displaystyle=\frac{\left(n_{I}Q\right)^{2}}{u^{5}}\frac{1+g_{1}}{1+g_{2}-\frac{k^{2}}{u^{8}f_{T}}+\frac{\left(n_{I}Q\right)^{2}}{u^{5}}},$
$\displaystyle x_{4}^{\prime 2}$
$\displaystyle=\frac{k^{2}}{u^{11}f_{T}^{2}}\frac{1+g_{1}}{1+g_{2}-\frac{k^{2}}{u^{8}f_{T}}+\frac{\left(n_{I}Q\right)^{2}}{u^{5}}}.$
(3.58)

Plugging solution (3.58) back into (3.50), we can obtain the free energy of
the baryonic phase. In order to obtain the phase transition, we need to
further minimize the free energy given in (3.50) with respect to
$n_{I},\rho,u_{c}$ as the parameters which leads to three constraints as,

$\displaystyle 0=$
$\displaystyle\int_{u_{c}}^{\infty}\left[\frac{u^{5/2}}{2}\left(\frac{\partial
g_{1}}{\partial n_{I}}\zeta^{-1}+\frac{\partial g_{2}}{\partial
n_{I}}\zeta\right)+\hat{a}_{0}^{\prime}Q\right]-\mu,$ $\displaystyle 0=$
$\displaystyle\int_{u_{c}}^{\infty}\left[\frac{u^{5/2}}{2}\left(\frac{\partial
g_{1}}{\partial\rho}\zeta^{-1}+\frac{\partial
g_{2}}{\partial\rho}\zeta\right)+n_{I}\hat{a}_{0}^{\prime}\frac{\partial
Q}{\partial\rho}\right],$ $\displaystyle
u_{c}^{5/2}\sqrt{\left[1+g_{1}\left(u_{c}\right)\right]\left[1+g_{2}\left(u_{c}\right)-\frac{k^{2}}{u_{c}^{8}f_{T}\left(u_{c}\right)}\right]}=$
$\displaystyle\int_{u_{c}}^{\infty}\left[\frac{u^{5/2}}{2}\left(\frac{\partial
g_{1}}{\partial u_{c}}\zeta^{-1}+\frac{\partial g_{2}}{\partial
u_{c}}\zeta\right)+n_{I}\hat{a}_{0}^{\prime}\frac{\partial Q}{\partial
u_{c}}\right],$ (3.59)

where

$\zeta=\frac{\sqrt{1+g_{1}}}{\sqrt{1+g_{2}-\frac{k^{2}}{u^{8}f_{T}}+\frac{\left(n_{I}Q\right)^{2}}{u^{5}}}}.$
(3.60)

With all above in hand, we can obtain numerically a holographic diagram
included mesonic, baryonic and quark matter phase of QCD by comparing the
associated free energies given in (3.43) (3.46) and (3.50). The resultant
phase diagram is given in Figure 6.

Figure 6:  Holographic QCD phase diagram vs realistic QCD phase diagram in the
$T-\mu$ plane.

and we can see the holographic diagram includes all the elementary phases in
realistic QCD although the confined geometry is not included in the current
discussion.

We note that it is very difficult to work out a reasonable model describing
QCD matter over a very wide density regime with traditional models or theories
of QCD. For example, the quark-meson model (e.g. [92, 93]) and the Nambu-Jona-
Lasinio (NJL) model [94, 95, 96, 97, 98] are very useful to get some insight
into the chiral and deconfinement phase transitions and quark matter phases,
however the nuclear matter is usually not included in these models. In
addition, nucleon-meson models e.g. [99, 100, 101, 102] are based on the
properties of nuclear matter and may be able to describe moderately dense
nuclear matter realistically, while they give a poor description of quark
matter with restored chiral symmetry. In this sense, this holographic model
provides a very powerful way to study QCD phase diagram in a very wide density
regime based on string theory.

###  3.3 Higgs mechanism and heavy-light meson field

One of the interesting developments of the D4/D8 model is to include heavy
flavor by using the Higgs mechanism in D-brane system. Recall that the
fundamental quarks in the D4/D8 model are identified to be the $4-8$ and
$4-\bar{8}$ strings. Since $N_{c}$ D4-branes and $N_{f}$ D8-branes are
coincident, we find that the $4-8$ and $4-\bar{8}$ strings have a vanishing
vacuum expectation value (VEV). Therefore the fundamental quarks created by
$4-8$ and $4-\bar{8}$ strings are massless which implies this model can
describe the mesons with light flavors only. Hence it is naturally motivated
to include the massive heavy flavor in this model. To this goal, in this
section, let us review the Higgs mechanism in D-brane system and see how to
use it to introduce heavy flavor.

First, we take a look at the Higgs mechanism in D-brane system by considering
the configuration of an open string connecting two stacks of the separated
D-branes as it is illustrated in (a) of Figure 7.

Figure 7:  The Higgs mechanism in string theory (a) and the configuration of
heavy flavor in D4/D8 model (b). (a) Higgs mechanism in string theory: The
gauge symmetry on a stack of coincident $N_{1}+N_{2}$ D-branes could be
$U\left(N_{1}+N_{2}\right)$, while it breaks down to
$U\left(N_{1}\right)\times U\left(N_{2}\right)$ if the D-branes somehow move
apart to become two stacks of coincident $N_{1}$ and $N_{2}$ D-branes. The
open string connecting the two stacks of the D-branes has a non-zero VEV hence
its ground states acquire non-zero mass which corresponds to the separation of
the D-branes. (b) Configuration of heavy flavor in D4/D8 model: Red line
refers to the stack of $N_{f}$ $\mathrm{D8/\overline{D8}}$-branes in the
original model which now is identified as light flavor. Blue line refers to
another one pair of $\mathrm{D8/\overline{D8}}$-branes separated from $N_{f}$
$\mathrm{D8/\overline{D8}}$-branes which is identified as heavy flavor. The
green line refers to the open string connecting the light and heavy brane
which is the heavy-light string, and it acquires non-zero VEV to create
massive ground states.

In this D-brane configuration, the worldvolume symmetry
$U\left(N_{1}+N_{2}\right)$ breaks down to $U\left(N_{1}\right)\times
U\left(N_{2}\right)$ when the D-branes move separately, where we use
$N_{1},N_{2}$ to refer to the D-branes number in each stack. Accordingly the
transverse modes of the D-brane acquire a non-zero VEV due to the separation
of the D-branes. Hence the multiplets created by the open string connecting
the separated D-branes become massive just like the Higgs mechanism in the
standard model of the particle physics [53, 60]. Let us investigate this
mechanism quantitatively by recalling the D-brane action in (2.20). In the
holographic approach, we need the near-horizon limit i.e.
$\alpha^{\prime}\rightarrow 0$, thus the D-brane action can be expanded to be
Yang-Mills plus Wess-Zumino action as it is in Section 2.4. Now we pick up the
transverse modes in the DBI action, it reduces to an additional quadratic
action for the transverse modes $\Psi^{I}$ with $\alpha^{\prime}\rightarrow 0$
as,

$\displaystyle S\left[\Psi^{I}\right]$
$\displaystyle=-T_{\mathrm{D}_{p}}\frac{\left(2\pi\alpha^{\prime}\right)^{2}}{4}\int_{\mathrm{D}_{p}}d^{p+1}x\sqrt{-\det
g}e^{-\phi}\mathrm{Tr}\left\\{2D_{a}\Psi^{I}D_{a}\Psi^{I}+\left[\Psi^{I},\Psi^{J}\right]^{2}\right\\},$
$\displaystyle a$ $\displaystyle=0,1...p,\ I,J=p+1...d,$ (3.61)

where the covariant derivative is

$D_{a}\Psi^{I}=\partial_{a}\Psi^{I}+i\left[\mathbf{A}_{a},\Psi^{I}\right],$
(3.62)

and $\mathbf{A}_{a}$ is the gauge field potential on the D-brane. Consider a
stack of coincident $N_{1}+N_{2}$ D-branes, $\mathbf{A}_{a}$ could be the
$U\left(N_{1}+N_{2}\right)$ generator as an
$\left(N_{1}+N_{2}\right)\times\left(N_{1}+N_{2}\right)$ matrix. However, if
the coincident $N_{1}+N_{2}$ D-branes move apart to become two stacks of
$N_{1}$ and $N_{2}$ D-branes, the gauge potential $\mathbf{A}_{a}$ becomes,

$\mathbf{A}_{a}=\left(\begin{array}[]{cc}A_{a}&\Phi_{a}\\\
\Phi_{a}^{\dagger}&0\end{array}\right),$ (3.63)

where $A_{a}$ is the $U\left(N_{1}\right)$ gauge potential as an $N_{1}\times
N_{1}$ matrix. $\Phi_{a}$ is the multiplet created by the open string
connecting to the two stacks of the D-brane which is an $N_{1}\times N_{2}$
matrix-valued field and the last element can be gauged away by the residual
symmetry. On the other hand, when the D-branes are separated, the transverse
mode $\Psi^{I}$ will have a non-zero VEV since the open string connecting the
separated D-branes can not shrink to zero. Therefore we can write down
$\Psi^{I}$ with a VEV as,

$\Psi^{I}\rightarrow\Psi^{I}+V^{I},$ (3.64)

where

$V^{I}\sim\left(\begin{array}[]{cc}V&0\\\ 0&v\end{array}\right),$ (3.65)

to represent $U\left(N_{1}+N_{2}\right)\rightarrow U\left(N_{1}\right)\times
U\left(N_{2}\right)$. Thus plugging (3.62) - (3.65) into (3.61), one obtains a
mass term in the action as

$\mathrm{Tr}\left(V^{2}\Phi_{a}^{\dagger}\Phi_{a}+v^{2}\Phi_{a}\Phi_{a}^{\dagger}\right),$
(3.66)

and $\Phi_{a}$ can be interpreted as the heavy-light field acquiring a mass
through the VEV of the transverse mode $\Psi^{I}$.

With this Higgs mechanism in string theory, let us employ it in the D4/D8
model by considering (b) of Figure 7. In this configuration, there is one pair
of $\mathrm{D8/\overline{D8}}$-branes separated from $N_{f}$
$\mathrm{D8/\overline{D8}}$-branes which is identified as heavy flavor brane
with an open string (the heavy-light string) connecting them. We note that the
configuration in (b) of Figure 7 is generalized version of (a) of Figure 7 in
the curved spacetime. Then we can write down the D8-brane action with heavy
flavor by imposing the following replacement,

$A_{a}\rightarrow\mathbf{A}_{a}=\left(\begin{array}[]{cc}A_{a}&\Phi_{a}\\\
\Phi_{a}^{\dagger}&0\end{array}\right),F_{ab}\rightarrow\bm{\mathrm{F}}_{ab}=\left(\begin{array}[]{cc}F_{ab}+i\alpha_{ab}&f_{ab}\\\
f_{ab}^{\dagger}&i\beta_{ab}\end{array}\right),$ (3.67)

to (2.33), where $A_{a},F_{ab}$ are $N_{f}\times N_{f}$ matrix-valued fields
as we have specified Section 2. $\Phi_{a}$ is an $N_{f}\times 1$ matrix-valued
multiplet created by the heavy-light string which is interpreted as the heavy-
light meson field and181818The index in the square brackets is ranked as
$T_{\left[ab\right]}=\frac{1}{2!}\left(T_{ab}-T_{ba}\right)$ and we have
chosen the gauge field as Hermitian field
$\bm{\mathrm{A}}_{a}^{\dagger}=\bm{\mathrm{A}}_{a}$.

$\displaystyle\alpha_{ab}=$ $\displaystyle 2\Phi_{[a}\Phi_{b]}^{\dagger},\
\beta_{ab}=2\Phi_{[a}^{\dagger}\Phi_{b]},$ $\displaystyle f_{ab}=$
$\displaystyle 2\partial_{[a}\Phi_{b]}+2iA_{[a}\Phi_{b]}\equiv
2D_{[a}\Phi_{b]},$ (3.68)

we obtain the action (3.61) for $\Phi_{a}$ as

$\displaystyle S\left[\Phi_{\mu,z}\right]=$
$\displaystyle-2\kappa\mathrm{Tr}\int
dzdx^{4}f\left(z\right)\left(\partial_{\mu}\Phi_{\nu}^{\dagger}-\partial_{\nu}\Phi_{\mu}^{\dagger}\right)\left(\partial^{\mu}\Phi^{\nu}-\partial^{\nu}\Phi^{\mu}\right)$
$\displaystyle-\kappa\mathrm{Tr}\int
dzdx^{4}g\left(z\right)\left(\partial_{\mu}\Phi_{z}^{\dagger}-\partial_{z}\Phi_{\mu}^{\dagger}\right)\left(\partial^{\mu}\Phi_{z}-\partial_{z}\Phi^{\mu}\right)$
$\displaystyle-\tilde{v}\mathrm{Tr}\int
dzdx^{4}a\left(z\right)\left[2f\left(z\right)\Phi^{\dagger\mu}\Phi_{\mu}+g\left(z\right)\Phi_{z}^{\dagger}\Phi_{z}\right],$
(3.69)

where $z$ is the Cartesian coordinates given in (2.34) and the VEV of
T-dualitied $\Psi^{4}=2\pi\alpha^{\prime}x^{4}$ is chosen as [103],

$\Psi^{4}=\left(\begin{array}[]{cc}-\frac{v}{N_{f}-1}\mathbf{1}_{N_{f}-1}&0\\\
0&v\end{array}\right),$ (3.70)

with

$\tilde{v}=v\kappa/\left(2\pi\alpha^{\prime}\right)^{2},f\left(z\right)=\frac{R^{3}}{4U\left(z\right)},g\left(z\right)=\frac{9}{8}\frac{U\left(z\right)^{3}}{U_{KK}},a\left(z\right)=\left[\frac{U\left(z\right)}{R}\right]^{3/2}.$
(3.71)

Note that $\Psi^{4}$ is the only transverse mode of D8-brane. The heavy-light
meson tower can be obtained by expanding $\Phi_{\mu,z}$ as what we have
specified in Section 2.4. For example, the transverse modes of heavy-light
meson field is suggested as [38, 39],

$\Phi_{\mu}=\sum_{n}\phi_{\left(n\right)}^{H}\left(z\right)B_{\mu}^{\left(n\right)}\left(x\right),\Phi_{z}=0,$
(3.72)

which leads to

$S\left[\Phi_{\mu,z}\right]=\sum_{n}\int
d^{4}x\left[\frac{1}{2}\partial_{[\mu}B_{\nu]}^{\left(n\right)\dagger}\partial^{[\mu}B^{\left(n\right)\nu]}+m_{n}^{2}B_{\mu}^{\left(n\right)\dagger}B^{\left(n\right)\mu}\right],$
(3.73)

with the normalization

$4\kappa\int
dzf\left(z\right)\phi_{\left(n\right)}^{H}\phi_{\left(m\right)}^{H}=\delta_{mn},$
(3.74)

and eigenvalue equation,

$-\frac{d}{dz}\left(g\frac{d\phi_{\left(n\right)}^{H}}{dz}\right)+2f\left(z\right)\left[-m_{n}^{2}+\tilde{v}^{2}a\left(z\right)\right]\phi_{n}^{H}=0.$
(3.75)

For the transverse modes longitudinal modes, the expansion is suggested as,

$\displaystyle\Phi_{\mu}=$
$\displaystyle-\sum_{n}\frac{1}{2a\left(z\right)f\left(z\right)m_{n}^{2}}\frac{d}{dz}\left[a\left(z\right)g\left(z\right)\epsilon_{n}\left(z\right)\right]\partial_{\mu}D_{\nu}\left(x\right),$
$\displaystyle\Phi_{z}=$
$\displaystyle\sum_{n}\epsilon_{n}\left(z\right)D_{n}\left(x\right),$ (3.76)

leading to

$S\left[\Phi_{\mu,z}\right]=\sum_{n}\int
d^{4}x\left(\partial_{\mu}D_{n}^{\dagger}\partial^{\mu}D_{n}+m_{n}^{2}D_{n}^{\dagger}D_{n}\right),$
(3.77)

with the normalization

$2\kappa=\int
dza\left(z\right)g\left(z\right)\epsilon_{n}\left(z\right)\epsilon_{m}\left(z\right)=\frac{m_{n}^{2}}{\tilde{v}^{2}}\delta_{mn}.$
(3.78)

We note that, with the replacement (3.67), the Chern-Simons action (2.57) for
the D8-branes reduces to additional terms as,

$\displaystyle\mathcal{L}_{\mathrm{CS}}\left[\Phi_{\mu,z}\right]=$
$\displaystyle-\frac{N_{c}}{24\pi^{2}}\left(d\Phi^{\dagger}Ad\Phi+d\Phi^{\dagger}dA\Phi+\Phi^{\dagger}dAd\Phi\right)$
$\displaystyle+\frac{iN_{c}}{16\pi^{2}}\left(d\Phi^{\dagger}A^{2}\Phi+\Phi^{\dagger}A^{2}d\Phi+\Phi^{\dagger}AdA\Phi+\Phi^{\dagger}dAA\Phi\right)$
$\displaystyle+\frac{5N_{c}}{48\pi^{2}}\Phi^{\dagger}A^{3}\Phi+\mathcal{O}\left(\Phi^{4},A\right).$
(3.79)

Using the expansion (3.72) (3.76) and (2.46), the DBI and CS term includes the
interaction between light and heavy-light mesons.

It is also possible to obtain the baryon spectrum with heavy flavor by
considering baryon vertex as the instantons on the flavor brane [104, 105,
106]. Follow the steps in Section 2.5, the equations of motion for the heavy-
light meson field is derived as,

$\displaystyle
D_{M}D_{M}\Phi_{N}-D_{N}D_{M}\Phi_{M}+2iF_{MN}\Phi_{M}+\mathcal{O}\left(\lambda^{-1}\right)$
$\displaystyle=0,$ $\displaystyle
D_{M}\left(D_{0}\Phi_{M}-D_{M}\Phi_{0}\right)-iF^{0M}\Phi_{M}$
$\displaystyle-\frac{1}{64\pi^{2}a}\epsilon_{MNPQ}\mathcal{K}_{MNPQ}+\mathcal{O}\left(\lambda^{-1}\right)$
$\displaystyle=0,$ (3.80)

with $M,N=1,2,3,z$ and

$\mathcal{K}_{MNPQ}=i\partial_{M}A_{N}\partial_{P}\Phi_{Q}-A_{M}A_{N}\partial_{P}\Phi_{Q}-\partial_{M}A_{N}A_{P}\Phi_{Q}-\frac{5i}{6}A_{M}A_{N}A_{P}\Phi_{Q},$
(3.81)

where $F=dA+\frac{i}{2}\left[A,A\right]$ is computed by the BPST instanton
solution given in (2.82) - (2.87). Thus $\Phi_{\mu,z}$ can be solved as
$\Phi_{\mu,z}=e^{\pm im_{H}t}\phi_{\mu,z}\left(x\right)$ by

$\displaystyle\phi_{0}$
$\displaystyle=-\frac{1}{1024a\pi^{2}}\left[\frac{25\rho}{2\left(x^{2}+\rho^{2}\right)^{5/2}}+\frac{7}{\rho\left(x^{2}+\rho^{2}\right)^{3/2}}\right]\chi,$
$\displaystyle\phi_{M}$
$\displaystyle=\frac{\rho}{\left(x^{2}+\rho^{2}\right)^{3/2}}\sigma_{M}\chi,$
(3.82)

where $\sigma_{M}$ is the embedded Pauli matrices as
$\sigma_{M}/2=\left(t_{i},-\mathbf{1}_{N_{f}}\right)$ and $\chi$ is the
$SU\left(N_{f}\right)$ spinor independent on $x,z$. The soliton mass as the
baryonic potential can be evaluated by inserting (3.82) and the BPST solution
(2.82) - (2.87) to the full action for the D8-branes. Afterwards one reaches
to an effective Hamiltonian for baryon state by following Section 2.5 as it is
given in [105, 106]. We note that [40] also gives another generalization with
heavy flavor into black D4-brane background.

###  3.4 Interaction of hadron and glueball

The interaction of hadrons relates to many significant topics in QCD and
nuclear physics, and its holographic description by the D4/D8 model has been
reviewed briefly in Section 2 and [24, 25]. In this section, we will take a
look at the interaction in hadron physics involving glueballs since the D4/D8
model provides explicit definition of meson, baryon and glueball.

The main idea to include the interaction of meson and glueball is to consider
the D8-brane action with a gravitational fluctuation. Recall the discussion in
Section 2.4 and 2.6, since meson is identified as the gauge field on the
D8-branes (created by $8-8$ string) and glueball is identified as the
gravitational polarization (close string), the interaction of meson and
glueball is nothing but the interaction of open and close string which can be
therefore included into the D8-brane action when the metric fluctuation is
picked up. For example, when we put the gravitational polarization (2.99) into
the D8-brane action (2.32) with the meson tower given in (2.37), by
integrating out the $z$ dependence (the holographic coordinate) it reduces to
interaction action involving $\pi,\rho$ meson and exotic glueball $G_{E}$
after some straightforward but messy calculations as,

$\displaystyle S_{G_{E}}^{\pi-\rho}=$ $\displaystyle-\mathrm{Tr}\int
d^{4}x\bigg{\\{}c_{1}\left[\frac{1}{2}\partial_{\mu}\pi\partial_{\nu}\pi\frac{\partial^{\mu}\partial^{\nu}}{M_{E}^{2}}G_{E}+\frac{1}{4}\left(\partial_{\mu}\pi\right)^{2}\left(1-\frac{\partial^{2}}{M_{E}^{2}}\right)G_{E}\right]$
$\displaystyle+c_{2}M_{KK}^{2}\left[\frac{1}{2}\rho_{\mu}\rho_{\nu}\frac{\partial^{\mu}\partial^{\nu}}{M_{E}^{2}}G_{E}+\frac{1}{4}\left(\rho_{\mu}\right)^{2}\left(1-\frac{\partial^{2}}{M_{E}^{2}}\right)G_{E}\right]$
$\displaystyle+c_{3}\left[\frac{1}{2}\eta^{\sigma\lambda}\partial_{[\mu}\rho_{\sigma]}\partial_{[\nu}\rho_{\lambda]}\frac{\partial^{\mu}\partial^{\nu}}{M_{E}^{2}}G_{E}-\frac{1}{8}\partial_{[\mu}\rho_{\nu]}\partial^{[\mu}\rho^{\nu]}\left(1+\frac{\partial^{2}}{M_{E}^{2}}\right)G_{E}\right]+c_{4}\frac{3}{2M_{E}^{2}}\rho_{\mu}\partial^{[\mu}\rho^{\nu]}\partial_{\nu}G_{E}$
$\displaystyle+c_{5}\left[\partial_{\mu}\pi\left[\pi,\rho_{\nu}\right]\frac{\partial^{\mu}\partial^{\nu}}{M_{E}^{2}}G_{E}+\frac{1}{2}\partial_{\mu}\pi\left[\pi,\rho^{\mu}\right]\left(1-\frac{\partial^{2}}{M_{E}^{2}}\right)G_{E}\right]$
$\displaystyle+\left[\frac{1}{2}\tilde{c}_{1}\left(\partial_{\mu}\pi\right)^{2}+\frac{1}{2}\tilde{c}_{2}M_{KK}^{2}\left(\rho_{\mu}\right)^{2}+\frac{1}{4}\tilde{c}_{3}\partial_{[\mu}\rho_{\nu]}\partial^{[\mu}\rho^{\nu]}+\tilde{c}_{5}\partial_{\mu}\pi\left[\pi,\rho^{\mu}\right]\right]G_{E}\bigg{\\}},$
(3.83)

where the coefficients $c_{i}$’s and $\tilde{c}_{i}$’s are coupling constants
and numerically computed as (in the unit of
$\lambda^{-1/2}N_{c}^{-1}M_{KK}^{-1}$),

$\displaystyle c_{1}$ $\displaystyle=\int dZ\frac{\bar{H}_{E}}{\pi K}=62.6,\ \
\ \ c_{2}=2\kappa\int dZK\left(\psi_{1}^{\prime}\right)\bar{H}_{E}=7.1,$
$\displaystyle c_{3}$ $\displaystyle=2\kappa\int
dZK^{-1/3}\left(\psi_{1}\right)^{2}\bar{H}_{E}=69.7,$ $\displaystyle c_{4}$
$\displaystyle=2\kappa M_{KK}^{2}\int
dZ\frac{20ZK}{\left(5K-2\right)^{2}}\psi_{1}\psi_{1}^{\prime}H_{E}=10.6M_{KK}^{2},$
$\displaystyle c_{5}$ $\displaystyle=\int dZ\frac{\psi_{1}\bar{H}_{E}}{\pi
K}=2019.6N_{c}^{-1/2},$ $\displaystyle\tilde{c}_{1}$ $\displaystyle=\int
dZ\frac{H_{E}}{4\pi K}=16.4,\ \ \ \ \tilde{c}_{2}=\frac{1}{2}\kappa\int
dZK\left(\psi_{1}^{\prime}\right)^{2}H_{E}=3.0,$ $\displaystyle\tilde{c}_{3}$
$\displaystyle=\frac{1}{2}\kappa\int
dZK^{-1/3}\left(\psi_{1}\right)^{2}H_{E}=18.1,\ \ \ \ \tilde{c}_{5}=\int
dZ\frac{\psi_{1}H_{E}}{4\pi K}=508.2N_{c}^{-1/2}.$ (3.84)

Then the associated amplitude of glueball decay can be further evaluated by
using the effective action $S_{G_{E}}^{\pi-\rho}$ with the coupling constants.
And one can also compute the effective action of meson involving other types
of glueball by changing the formulas of the bulk gravitational polarization as
it is discussed in [29, 30, 31, 32].

The current setup to obtain an effective action of meson and glueball
interaction can also be generalized by including heavy flavor [107] which is
to take into account the configuration (b) in Figure 7 and the heavy-light
meson field. The main idea is to pick up the gravitational polarization in
bulk metric when we write down the D8-brane action with heavy flavor brane
(i.e. with the replacement given in (3.67)). For example, by considering the
gravitational polarization for the exotic glueball $G_{E}$ in (2.99), the
effective action of heavy-light meson and glueball provides terms as,

$\displaystyle\partial^{2}G_{E}\partial_{\mu}Q_{\nu}^{\dagger}\partial^{\mu}Q^{\nu},$
$\displaystyle\partial^{2}G_{E}Q_{\mu}^{\dagger}Q^{\mu},\partial^{2}G_{E}\partial_{\mu}Q^{\dagger}\partial^{\mu}Q,\partial^{2}G_{E}Q^{\dagger}Q,$
$\displaystyle\partial^{\mu}\partial^{\rho}G_{E}\partial_{\mu}Q_{\nu}^{\dagger}\partial_{\rho}Q^{\nu},$
$\displaystyle\partial^{\mu}\partial^{\nu}G_{E}\partial_{\mu}Q^{\dagger}\partial_{\nu}Q,\partial^{\mu}\partial^{\nu}G_{E}Q_{\mu,}^{\dagger}Q_{\nu,}$
$\displaystyle G_{E}\partial_{\mu}Q_{\nu}^{\dagger}\partial^{\mu}Q^{\nu},$
$\displaystyle G_{E}Q_{\mu}^{\dagger}Q^{\nu},G_{E}Q^{\dagger}Q,$
$\displaystyle\partial_{\sigma}G_{E}\partial^{\rho}Q^{\sigma\dagger}\partial_{\rho}Q,$
$\displaystyle\partial_{\sigma}G_{E}\partial^{\rho}Q^{\sigma\dagger}Q_{\rho},\partial^{\mu}G_{E}Q_{\mu}^{\dagger}Q,$
(3.85)

which are the same types as the interaction given in (3.83). Here we have used
$Q_{\mu},Q$ to denote the vector and scalar heavy-light meson field, and the
lowest heavy-light meson is identified to be D-meson with a charm quark.
Accordingly, the effective action with heavy-flavor and glueball may be useful
to study the oscillation of D-meson pairs ($D-\bar{D}$) or B-meson pairs
($B-\bar{B}$) [108, 109]. Note that since the heavy-light multiplet is created
by the heavy-light string, even if the heavy flavor is taken into account, the
interaction of heavy-light meson and glueball remains to be the open/close
string interaction through holography. Besides, in the presence of the heavy-
light meson and glueball, the effective action also mixes the interaction
terms of glueball, light and heavy-light meson which may describe the various
interaction in hadron physics.

It is also possible to include the interaction of baryon (or baryonic meson)
and glueball in a parallel way, that is to consider the interaction of
baryonic $\mathrm{D4}^{\prime}$-branes and bulk close string [110].
Specifically, one can derive the Yang-Mills action presented in (2.32) with
the gravitational polarization (2.99), then insert the BPST instanton
configuration (2.82) - (2.87) as baryon under the large $\lambda$ rescaling
(2.77). Afterwards, by following the discussion in Section 2.5, we can obtain
additional terms to the collective Hamiltonian (2.98) as,

$\Delta H\left(t\right)\simeq\cos\left(\omega
t\right)\left\\{-\frac{27}{2}\pi^{2}+\left[-\frac{59049\pi^{4}}{20\rho^{2}}+\frac{27M_{E}^{2}+468}{16}\pi^{2}\left(2Z^{2}+\rho^{2}\right)\right]\lambda^{-1}+\mathcal{O}\left(\lambda^{-2}\right)\right\\}.$
(3.86)

Using the standard technique in quantum mechanics for the time-dependent
perturbed Hamiltonian, it is possible to work out the decay rate of baryon
involving glueball and its associated select rule. We note that, when the
heavy flavor is included as in Section 2.3, moreover the decay rate of heavy-
light baryon or baryonic meson involving glueball is able to be achieved. For
example consider the exotic gravitational polarization (2.99), we can reach to
the time-dependent perturbed Hamiltonian as [111],

$\Delta
H\left(t\right)\simeq\lambda^{-1/2}M_{KK}^{-1}\left(\frac{5}{216\pi}m_{H}^{2}+\frac{15m_{H}}{32\rho^{2}}\right)G_{E}\chi^{\dagger}\chi,$
(3.87)

where we have taken the limit $m_{H}\rightarrow\infty$ followed
$\lambda\rightarrow\infty$ to simplify the formula and
$\chi^{\dagger}\chi=N_{Q}$ refers to the number of heavy-flavor quark in the
heavy-light meson. Then the decay of heavy-light baryonic matter involving
glueball can be evaluated by using (3.87) to the quantum mechanical system
(2.98) with heavy flavors. To close this section, we summarize the strings as
various hadrons in the D4/D8 model in Figure 8, and we can see the various
interactions of hadrons are interaction of strings through gauge-gravity
duality in this model.

Figure 8:  Strings as various hadrons in the D4/D8 model.

###  3.5 Theta dependence in QCD

In Yang-Mills theory, there could be an topological term proportional to the
$\theta$ angle [112]. In the large $N_{c}$ limit, the full Lagrangian takes
the form as,

$S=\frac{N_{c}}{2\lambda}\mathrm{Tr}\int\ ^{\star}F\wedge
F-i\frac{\lambda}{8\pi^{2}}\frac{\theta}{N_{c}}\mathrm{Tr}\int F\wedge F.$
(3.88)

While the value of $\theta$ angle may be experimentally small, it leads to
many interesting effects e.g. glueball spectrum [113], deconfinement
transition [114, 115], chiral magnetic effect [116, 117], especially its large
$N_{c}$ limit [118]. Since the D4/D8 model is a holographic version of QCD, it
is possible to introduce the $\theta$ term to the dual theory through the
gauge-gravity duality.

The main idea to include an Yang-Mills $\theta$ term in holography is to
introduce coincident $N_{0}$ D0-branes acting to the $N_{c}$ D4-brane
background (2.14). In this sense, we have to require $N_{0}/N_{c}$ is fixed
when $N_{c}\rightarrow\infty$. And for the SUGRA approach, the dynamics of the
Ramond-Ramond 1-form $C_{1}$ must be picked up into the IIA SUGRA action
(2.15) in order to include the charge of the D0-branes as,

$S_{\mathrm{IIA}}^{10\mathrm{d}}=\frac{1}{2\kappa_{10}^{2}}\int
d^{10}x\sqrt{-G}e^{-2\phi}\left[\mathcal{R}^{(10)}+4\partial_{\mu}\phi\partial\phi\right]-\frac{1}{4\kappa_{10}^{2}}\int
d^{10}x\sqrt{-G}\left(\left|F_{4}\right|^{2}+\left|F_{2}\right|^{2}\right),$
(3.89)

where $F_{2}=dC_{1}$. To obtain an analytical solution, we assume the
D0-branes are smeared homogeneous along $x^{4}$, hence the associated
equations of motion to (3.89) can be solved as,

$\displaystyle ds^{2}$
$\displaystyle=H_{4}^{-1/2}\left[-H_{0}^{-1/2}f_{T}\left(U\right)\left(dx^{0}\right)^{2}+H_{0}^{1/2}dx_{i}dx^{i}\right]+H_{4}^{1/2}H_{0}^{1/2}\left[\frac{dU^{2}}{f_{T}\left(U\right)}+U^{2}d\Omega_{4}^{2}\right],$
$\displaystyle H_{4}$
$\displaystyle=1+\frac{R^{3}}{U^{3}},H_{0}=1-\frac{U_{H}^{3}}{U^{3}}\frac{\Theta^{2}}{1+\Theta^{2}},e^{\phi}=H_{4}^{-1/4}H_{0}^{3/4},C_{1}=\frac{\Theta}{g_{s}}\frac{f_{T}}{H_{0}}dx^{4},F_{4}=3R^{3}g_{s}^{-1},$
(3.90)

where $\Theta$ is a constant parameter. Take the near horizon limit so that
$H_{4}\rightarrow R^{3}/U^{3}$ and impose the double Wick rotation as it is