S-dual Amplitude and $D_{3}$-Brane Couplings

Komeil Babaei<EMAIL_ADDRESS>and H. Babaei-
<EMAIL_ADDRESS>
1Department of Physics, University of Guilan,

P.O. Box 41335-1914, Rasht, Iran

2Department of Physics, University of Mohaghegh Ardabili,

P.O. Box 179, Ardabil, Iran

###### Abstract

Recently, it has been observed that the $IIB$ scattering amplitudes are
compatible with the standard rules of S-duality. Inspired by this observation,
we will find the tree-level S-matrix elements of one Ramond-Ramond and three
open strings by imposing this symmetry on the tree-level S-matrix elements of
one Kalb-Ramond and three open strings. We also find a $SL(2,R)$ invariant
form of the $D_{3}$-brane effective action containing four gauge fields with
derivative corrections that was derived from one-loop level four-point
amplitude. Using the expansion of the nonlinear $SL(2,R)$ invariant
structures, we find the action with derivative corrections at the level of
more gauge fields.

##  1 Introduction

It has been shown that the consistency of an electrodynamics theory under the
duality transformation can be expressed as a requirement in which the
Lagrangian must transform under duality in a particular way, defined by the
Noether-Gaillard-Zumino (NGZ) identity [1]. So, in general, a nonlinear
electrodynamics theory that saturates the NGZ identity, is a self-dual theory.
One of the familiar self-dual electrodynamics theory is the Born-Infeld theory
${\cal L}_{BI}=\sqrt{-\det(\eta_{ab}+F_{ab})}-1$, where
$F_{ab}=\partial_{a}A_{b}-\partial_{b}A_{a}$ is gauge field strength and $A$
is polarization of the gauge field.

This theory could be enjoyed the SL(2,R) S-duality (relevant to string theory
which insinuates a strong-weak coupling duality of such theories 111From the
presence of two two-form gauge fields $B_{2}$ ($NSNS$ two-form) and $C_{2}$
($RR$ two-form) in the string theory, a string can carry two types of charge.
These two-forms form a doublet of $SL(2,R)$ it follows that the string also
transform as a doublet. On the other hand, the transformation of the complex
field $\tau=C_{0}+ie^{\phi_{0}}\rightarrow-1/\tau$ evaluated at $C_{0}=0$,
changes the sign of the dilaton, which implies that the string coupling
constant maps to its inverse [5].) after taking axion and dilaton fields into
account. This action is not invariant under the S-duality, however, its
equations of motion and energy-momentum tensor are invariant under the
S-duality as it was shown for the electric-magnetic duality[2, 26].

The supersymmetric p-branes that supported by Ramond-Ramond $(RR)$ sources are
the solutions of type $II$ superstring theories. These branes have alternative
description in terms of open strings with Dirichlet boundary conditions.
$D$-branes are described by an effective Dirac-Born-Infeld $(DBI)$ action
(supplemented with extra couplings to RR fields) which is closely connected to
the Born-Infeld ($BI$) type effective action of open string theory. We
consider the $D_{3}$-brane of type $IIB$ theory which play a special role in
this theory. The action corresponding to this brane in the Einstein frame can
be interpreted as a generalization of four dimensional $BI$ action coupled to
a special background metric, dilaton, axion, etc [3].

$\displaystyle
S_{D3}=-T_{D3}\int{d^{4}x}\sqrt{-det(g_{ab}+e^{-\phi/2}B_{ab})}+T_{D3}\int[C^{(4)}+C^{(2)}B+\frac{1}{2}C^{(0)}BB],$
(1)

where all the bulk fields in the action are pull-back onto the world-volume of
$D$-brane.

It has been demonstrated that the $D_{3}$-brane action (combined with type IIB
effective action) and the corresponding equation of motion and energy-momentum
tensor are invariant under the $SL(2,R)$ symmetry of type IIB theory222This
symmetry is not shared by the full type $IIB$ superstring theory. Indeed, it
is broken by a variety of stringy and quantum effects to the infinite discrete
subgroup $SL(2,Z)$. [2]. In fact, this action satisfies the NGZ identity. As
was pointed out in [4] that the duality invariance of Hamiltonian and thus of
the corresponding energy momentum tensor should imply the invariance of the
S-matrix.

It is known that the S-matrix elements satisfy the Ward identity corresponding
to the S-duality[22]. At the level of the equations of motion, $D_{3}$-brane
effective action is invariant under nonlinear S-duality transformation [2]. So
at higher orders, one dose not expect the effective action to be consistent
with the nonlinear S-duality.

It was pointed in [6] that the $BI$ action involving on-shell gauge field is
invariant under the linear S-duality up to $F^{4}$ terms. From this point, the
disk-level scattering amplitude of four gauge fields on the world volume of a
single $D_{3}$-brane, which appears as a contact term, should be invariant
under the linear S-duality. The leading order S-matrix element of the six
gauge fields has both contact terms coming from the $F^{6}$ terms in the $BI$
action and massless poles coming from the $F^{4}$ terms in the $BI$ action. So
as expected, no part of this amplitude is not invariant under the linear
$SL(2,R)$ transformation separately, however, the combination of these two
parts is invariant[21]. It has been discussed in [6] that by imposing the
linear S-duality transformation on the S-dual invariant amplitude of four gage
fields, the pole part of the amplitude of six gage fields could be
constructed. Then, $F^{6}$-contact-terms could be found by applying the linear
S-duality again. Then using the linear S-duality, one may find the contact
terms of $F^{8}$, and so on. This result confirms the $BI$ action with abelian
gage fields as the effective action of a single D-brane.

So, in order to investigate the behavior of S-matrix elements under duality
transformation, it is convenient to separate the $\alpha^{\prime}$ expansion
of them into two parts: contact term and massless $n$-poles

$\displaystyle{\cal A}={\cal A}_{contact}+\sum_{n}A_{n-poles}.$ (2)

Neither ${\cal A}_{contact}$ nor $A_{n-poles}$ are invariant under the linear
S-duality. In fact, the combination of these two parts, ı.e. ${\cal A}$, must
be invariant. On the other hand, the pole part of amplitude may transform to
contact part under the linear S-duality transformation, so the effective
action, that is made of contact part, satisfies the Ward identity
corresponding to S-duality333On the other hand, the T-duality transformation
dose not transform the pole terms to contact terms, so the contact terms
always satisfy the Ward identity corresponding to T-duality [25].. In the case
that the S-matrix elements have no massless poles, one can find the
corresponding couplings from the consistency of them with the dualities[7].

Unlike the gauge field transformation that is carried by its field strength,
all other transformations are done by field potential. So, the pole part of
S-matrix dose not transform to contact part under the closed string $SL(2,R)$
transformation. It could be noted that if the S-matrix involves the
antisymmetric $NSNS$ ($B$-field) closed string, it could be found the
amplitude in terms of $B$-field strength $H$ by combining the pole part in
amplitude which are produced by the gauge fields and some of the contact terms
which are produced by replacement $F\rightarrow F+B$ [22].

The S-matrix elements of one $RR$ two-form field and three gauge fields that
appears as a pole term and the S-matrix elements of one $B$ field and three
gauge fields that contains a contact term as well as a pole term have been
calculated in [21]. It has been shown that the combination of these two
amplitudes could be appeared in terms of a $SL(2,R)$ invariant structure. In
this paper we show that these amplitudes transform to each other under the
linear S-duality transformation

$\displaystyle{\cal A}_{1-pole}(C^{2},F^{3})\rightarrow{\cal
A}_{contact}(B,F^{3})+{\cal A}_{1-pole}(B,F^{3}).$ (3)

Inspired by the above observation that the S-matrix elements of one closed
string and three open strings should satisfy the Ward identity corresponding
to the S-duality, it could be proposed the following expression for the
massless n-pole amplitude of n $RR$ two-form.

$\displaystyle{\cal A}_{n-pole}((C^{2})^{n},F^{4-n})\rightarrow{\cal
A}_{contact}(B^{n},F^{4-n})$ $\displaystyle+$ $\displaystyle{\cal
A}_{1-pole}(B^{n},F^{4-n})+\cdots+{\cal A}_{n-pole}(B^{n},F^{4-n})$

where $n=0,1,2,3,4$. The contact part of above amplitude could be exactly
reproduced by the replacement $F\rightarrow F+B$ in the $D_{3}$-brane
effective action.

It is convenient to consider the normalization factor $2\pi\alpha^{\prime}$ in
front of the gauge field in $DBI$ action. With this normalization, the $DBI$
action is at the leading order of $\alpha^{\prime}$. The $\alpha^{\prime}$
corrections to Born-Infeld action have been studied in [9, 10, 11, 12, 13, 14]
in the $\sigma$-model approach.

The derivatives of gauge fields could be appeared in $D_{3}$-brane effective
action. The study of the behavior of four point function under the linear
S-duality could be extended to four point function with derivative
corrections[15]. It has been shown in [17] that the electromagnetic self-
duality holds for such functions with derivative corrections at the order of
$\alpha^{\prime 2}$. So, one might expect that the $SL(2,R)$ S-dual invariant
structure could be appeared for each order of $\alpha^{\prime}$ independently.
On other words, we have to answer the following question that should S-duality
be a symmetry of full $D_{3}$-brane action with all derivative corrections?
There is no exact reasoning for this, but there is still an alternative
(albeit somewhat exploratory) logic suggesting that $D_{3}$-brane action (but
not actions for other D-branes!) should be S-duality covariant444We would like
to thank A.A.Tseytlin for discussions on this point. [2] (see also [8]).

By calculating the scattering amplitude of four open string in $IIB$
superstring theory, it has been found that the leading order of $D_{3}$-brane
effective action in superstring theory contains four gauge fields and four
derivatives[12, 16, 17]

$\displaystyle L_{(\partial F{})^{4}}$ $\displaystyle=$ $\displaystyle
e^{-2\phi}\,\,\bigg{(}\frac{1}{4}\partial_{c}F^{ef}\partial^{c}F^{ab}\partial_{d}F_{ef}\partial^{d}F_{ab}-\frac{1}{2}\
\partial_{c}F_{df}\partial^{c}F^{ab}\partial_{e}F_{b}{}^{f}\partial^{e}F_{a}{}^{d}$
(4)
$\displaystyle-\partial_{c}F_{a}{}^{d}\partial^{c}F^{ab}\partial_{f}F_{de}\partial^{f}F_{b}{}^{e}+\frac{1}{8}\partial_{c}F_{ab}\partial^{c}F^{ab}\
\partial_{f}F_{de}\partial^{f}F^{de}\bigg{)}.$

This action satisfies the NGZ identity, and is consistent with electro-
magnetic self-duality[17]. We will find the form of above action that is
manifestly S-dual. In fact, we will find the above action in terms a nonlinear
S-dual structure. From the nonlinear expansion of the $SL(2,R)$ invariant
structures, we find this effective action up to level of $N$ gauge fields and
four derivatives.

The outline of the paper is as follows: We begin in section 2 by studying the
S-duality transformations of bosonic fields and find some $SL(2,R)$ invariant
structures involving these fields. In section 3, we show that the
compatibility of the S-matrix elements of one gauge field, two scalar fields
and one $RR$ field $C^{(2)}$ with the S-duality generates the S-matrix
elements of one gauge field, two scalar fields and one $NSNS$ $B$-field. We
find a S-dual invariant form of these amplitudes and then predict the axion
amplitude. In section 4, we calculate the S-matrix elements of three gauge
fields and one $NSNS$ B-field by applying the S-duality transformation on the
S-matrix elements of three gauge fields and one $RR$ field $C^{(2)}$. In
section 5, using $SL(2,R)$ invariant structures found in section 2, we write
the $D_{3}$-brane effective action involving four gauge fields and four
derivatives in S-dual form. This expression could be extended to include more
gauge fields.

##  2 Nonlinear SL(2,R) structure

The axion $C_{0}$ combines with the dilaton $\phi_{0}$ to give the complex
scalar field $\tau=C_{0}+ie^{-\phi_{0}}$. Now consider the matrix ${\cal M}$

$\displaystyle{\cal M}=e^{\phi_{0}}\pmatrix{|\tau|^{2}&C_{0}\cr C_{0}&1}.$ (5)

The nonlinear $SL(2,R)$ transformation of the complex scalar field and the
above matrix are given by[18]

$\displaystyle\tau\rightarrow\frac{p\tau+q}{r\tau+s}\,\,\,\,\,\,\,,\,\,\,\,\,\,{\cal
M}\rightarrow\Lambda{\cal
M}\Lambda^{T}\,\,\,\,\,\,;\,\,\,\,\,\,\,\,\,\,\,\,\Lambda=\left(\begin{array}[]{cc}p&q\\\
r&s\end{array}\right)\in\,{SL(2,R)}.$ (8)

For the gauge field we have the following transformation

$\displaystyle F_{ab}$ $\displaystyle\rightarrow$ $\displaystyle
sF_{ab}+r\tilde{G}_{ab},$ $\displaystyle G_{ab}$ $\displaystyle\rightarrow$
$\displaystyle pG_{ab}-q\tilde{F}_{ab},$ (9)

where ${F}_{ab}=-\frac{1}{2}\epsilon_{abcd}\tilde{F}^{cd}$, the antisymmetric
tensor $G_{ab}$ is given by $-2\partial{\cal L}(F)/\partial F^{ab}$ and ${\cal
L}$ is the Lagrangian. By considering a gauge field doublet ${\cal F}_{ab}$
containing $F_{ab}$ and $G_{ab}$ and also a background field doublet ${\cal
B}_{ab}$ containing the two forms $B^{(2)}$ and $C^{(2)}$ and using
$\tilde{(\tilde{F})}=-F$, one can write the transformation of these field
doublets as

$\displaystyle{\cal F}_{ab}\rightarrow(\Lambda^{-1})^{T}{\cal
F}_{ab}\,\,\,\,\,\,\,;\,\,\,\,\,\,\,\,\,{\cal
B}_{ab}\rightarrow(\Lambda^{-1})^{T}{\cal B}_{ab}.$ (10)

Using the above transformations, one can find that the structures ${\cal
F}^{T}{\cal M}{\cal F}$ and ${\cal B}^{T}{\cal M}{\cal B}$ are $SL(2,R)$
invariant structures.

Consider Lagrangian in terms of two possible Lorentz invariants
$t=1/4F^{ab}F_{ab}$ and $z=1/4F^{ab}\tilde{F}_{ab}$. At the presence of axion
and dilaton couplings, it could be useful to express the Lagrangian in the
following form which the contribution of axion coupling is separated.

$\displaystyle{\cal L}={\cal L}^{{}^{\prime}}+C_{0}z$ (11)

The antisymmetric tensor $G_{ab}$ then separates as
$G_{ab}=G^{{}^{\prime}}_{ab}-C_{0}\tilde{F}_{ab}$ where $G^{{}^{\prime}}_{ab}$
could be an arbitrary nonlinear function of $F$ which is
$G^{{}^{\prime}}_{ab}=-2\frac{\partial\mathcal{L}^{{}^{\prime}}}{\partial
F^{ab}}$.

By this consideration one gets the $SL(2,R)$ invariant structure in the
following form:

$\displaystyle({\cal F}^{T})_{a}{}^{c}{\cal M}_{0}{\cal
F}_{bc}=e^{-\phi_{0}}\,\tilde{F}_{a}{}^{c}\,\tilde{F}_{bc}+e^{\phi_{0}}\,G^{{}^{\prime}}_{a}{}^{c}\,G^{{}^{\prime}}_{bc},$
(12)

where the matrix ${\cal M}_{0}$ is the matrix ${\cal M}$ in which the dilaton
and the R-R scalar are constant ($\phi_{0},C_{0}$). It is easy to check that
the above structure is invariant under the linear transformations
$G^{{}^{\prime}}\longrightarrow\tilde{F}$ ,
$\tilde{F}\longrightarrow-G^{{}^{\prime}}$ and $e^{-\phi_{0}}\longrightarrow
e^{\phi_{0}}$.

َWhen the dilaton and the R-R scalar are constant, one finds the following
S-dual multiplet[19]:

$\displaystyle(\tilde{{\cal F}}^{T})_{a}{}^{c}{\cal M}_{0}{\cal B}_{bc}$
$\displaystyle=$ $\displaystyle
e^{-\phi_{0}}F_{a}{}^{c}B_{bc}-\tilde{F}_{a}{}^{c}C_{bc}^{(2)}-C_{0}\tilde{F}_{a}{}^{c}B_{bc}.$
(13)

At the presence of two gauge fields with two derivatives, one can find the
following $SL(2,R)$ invariant structure

$\displaystyle\partial{\cal F}^{T}{\cal M}\partial{\cal F}$ $\displaystyle=$
$\displaystyle
e^{-\phi_{0}}\partial\tilde{F}\partial\tilde{F}+e^{\phi_{0}}\partial
G^{{}^{\prime}}\partial G^{{}^{\prime}}.$ (14)

By considering the nonlinear expansion of $G^{\prime}_{ab}$ (in any self dual
theory) as
$G^{{}^{\prime}}_{ab}=e^{-\phi_{0}}F+e^{-2\phi_{0}}F^{3}+\cdots+e^{-N\phi_{0}}F^{2N-1}$,
the nonlinear $SL(2,R)$ invariant structure corresponding to $2N$ gauge fields
and two derivatives could be found.

$\displaystyle\partial{\cal F}^{T}{\cal M}\partial{\cal F}$
$\displaystyle\sim$ $\displaystyle\partial F\partial F+FF\partial F\partial
F+FFFF\partial F\partial F+\cdots.$ (15)

##  3 S-dual amplitude of one gauge field, two scalar fields and one two-form

The S-matrix elements of one gauge field, two scalar fields and one two-form
vertex operator could be calculated from $D_{3}$-brane effective action. At
first, we are going to find the corresponding amplitude in which the two-form
is an antisymmetric $RR$ field $C^{(2)}$. The amplitude is given by the
following Feynman rule:

$\displaystyle{\cal
A}(C_{1}^{(2)},F_{2},\Phi_{3},\Phi_{4})=V^{a}(C_{1}^{(2)},A)G_{ab}(A)V^{b}(A,F_{2},\Phi_{3},\Phi_{4}),$
(16)

where $F_{2}$ is the polarization of the gauge boson, $\Phi_{3},\Phi_{4}$ are
the scalar fields and $A$ is the off-shell gauge field propagating between the
two vertices. The gauge boson propagator and the vertices can be read from
$D_{3}$-brane effective action

$\displaystyle G_{ab}(A)$ $\displaystyle\sim$
$\displaystyle-\frac{e^{\phi_{0}}\eta_{ab}}{k\\!\cdot\\!k}$ $\displaystyle
V^{a}(C_{1}^{(2)},A)$ $\displaystyle\sim$
$\displaystyle-k\\!\cdot\\!(\tilde{C_{1}}^{(2)})^{a}$ (17) $\displaystyle
V^{b}(A,F_{2},\Phi_{3},\Phi_{4})$ $\displaystyle\sim$ $\displaystyle
e^{-\phi_{0}}{\Phi_{3}}_{i}{\Phi_{4}}^{i}\bigg{[}k\\!\cdot\\!k_{3}k_{4}\\!\cdot\\!F_{2}^{a}+k\\!\cdot\\!F_{2}\\!\cdot\\!k_{4}k_{3}^{a}-\frac{1}{2}k_{3}\\!\cdot\\!k_{4}k\\!\cdot\\!F_{2}^{a}\bigg{]}+(3\leftrightarrow
4)$

where $(\tilde{C}^{(2)})_{ab}=\frac{1}{2}\epsilon_{abcd}(C^{(2)})^{cd}$. From
the conservation of momentum along the brane we have
$k=-k_{1}=k_{2}+k_{3}+k_{4}$. Considering the above vertices and propagator,
one finds that the S-matrix elements have not any contribution to the contact
amplitude and the amplitude (16) then appears as a pole amplitude:

$\displaystyle{\cal A}(C^{(2)}_{1},F_{2},\Phi_{3},\Phi_{4})$ $\displaystyle=$
$\displaystyle\frac{{\Phi_{3}}_{i}\Phi_{4}^{i}}{k_{1}\\!\cdot\\!k_{1}}\bigg{[}k_{1}\\!\cdot\\!\tilde{C}^{(2)}_{1}\\!\cdot\\!k_{4}\,\,k_{1}\\!\cdot\\!F_{2}\\!\cdot\\!k_{3}-k_{1}\\!\cdot\\!k_{3}\,\,k_{1}\\!\cdot\\!\tilde{C}^{(2)}_{1}\\!\cdot\\!F_{2}\\!\cdot\\!k_{4}$
(18)
$\displaystyle\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\frac{1}{2}k_{3}\\!\cdot\\!k_{4}\,\,k_{1}\\!\cdot\\!\tilde{C}^{(2)}_{1}\\!\cdot\\!F_{2}\\!\cdot\\!k_{1}\bigg{]}+(3\leftrightarrow
4).$

The amplitude dose not satisfy the Ward identity corresponding to the
S-duality transformations, however is invariant under the Ward identity
corresponding to abelian gauge symmetry and to the RR gauge symmetry.

The amplitude corresponding to antisymmetric B-field has massless pole as well
as contact term. The contact part of B-field amplitude can be found using the
Feynman rule and considering the relevant higher derivative correction of
D-brane action.

$\displaystyle{\cal
A}_{contact}\left(B_{1},F_{2},\Phi_{3},\Phi_{4}\right)=e^{-\phi_{0}}{\Phi_{3}}_{i}\Phi_{4}^{i}\bigg{[}\frac{1}{4}k_{3}\\!\cdot\\!k_{4}Tr(B_{1}\\!\cdot\\!F_{2})-k_{3}\\!\cdot\\!B_{1}\\!\cdot\\!F_{2}\\!\cdot\\!k_{4}\bigg{]}+(3\leftrightarrow
4).$ (19)

Considering the propagator and vertex $V^{b}(A,F_{2},\Phi_{3},\Phi_{4})$ in
(3) and using the fourth order in the expansion of D-brane action to find the
vertex operator $V^{a}(B_{1},A)=e^{-\phi_{0}}k_{1}\\!\cdot\\!B_{1}^{a}$, one
finds the pole part of B-field amplitude as following:

$\displaystyle{\cal A}_{pole}\left(B_{1},F_{2},\ \Phi_{3},\Phi_{4},\right)$
$\displaystyle=$ $\displaystyle
V^{a}(B_{1},A)G_{ab}(A)V^{b}(A,F_{2},\Phi_{3},\Phi_{4})$ (20) $\displaystyle=$
$\displaystyle-\frac{e^{-\Phi}{\Phi_{3}}_{i}\Phi_{4}^{i}}{k_{1}.k_{1}}\bigg{[}k_{1}\\!\cdot\\!B_{1}\\!\cdot\\!k_{4}k_{1}\\!\cdot\\!F_{2}\\!\cdot\\!k_{3}-k_{1}\\!\cdot\\!k_{3}k_{1}\\!\cdot\\!B_{1}\\!\cdot\\!F_{2}\\!\cdot\\!k_{4}$
$\displaystyle\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\frac{1}{2}k_{3}\\!\cdot\\!k_{4}k_{1}\\!\cdot\\!B_{1}\\!\cdot\\!F_{2}\\!\cdot\\!k_{1}\bigg{]}+(3\leftrightarrow
4)$

where we use the previously mentioned conservation of momentum.

Now, we are going to determine the transformation behavior of the amplitude
(18) under the S-duality transformation when the axion background dos not tack
into account $C_{0}=0$. There are two different structures in amplitude (18)
that transform under the S-duality as following:

$\displaystyle
k_{N}\\!\cdot\\!\tilde{C}^{(2)}_{1}\\!\cdot\\!F_{2}\\!\cdot\\!k_{M}$
$\displaystyle\longrightarrow$ $\displaystyle
e^{-\phi}k_{N}\\!\cdot\\!\tilde{B}_{1}\\!\cdot\\!\tilde{F}_{2}\\!\cdot\\!k_{M}$
$\displaystyle=$ $\displaystyle
e^{-\phi}\bigg{(}k_{N}\\!\cdot\\!F_{2}\\!\cdot\\!B_{1}\\!\cdot\\!k_{M}-\frac{1}{2}k_{N}\\!\cdot\\!k_{M}{\rm
Tr}(B_{1}\\!\cdot\\!F_{2})\bigg{)},$ $\displaystyle
k_{N}\\!\cdot\\!\tilde{C}^{(2)}_{1}\\!\cdot\\!k_{M}\,\,\,\,\,k_{P}\\!\cdot\\!F_{2}\\!\cdot\\!k_{Q}$
$\displaystyle\longrightarrow$ $\displaystyle
e^{-\phi}k_{N}\\!\cdot\\!\tilde{B}_{1}\\!\cdot\\!k_{M}\,\,k_{P}\\!\cdot\\!\tilde{F}_{2}\\!\cdot\\!k_{Q}$
$\displaystyle=$ $\displaystyle
e^{-\phi}\bigg{(}-\frac{1}{2}k_{N}\\!\cdot\\!k_{Q}k_{P}\\!\cdot\\!k_{M}{\rm
Tr}(B_{1}\\!\cdot\\!F_{2})+\frac{1}{2}k_{N}\\!\cdot\\!k_{P}k_{Q}\\!\cdot\\!k_{M}{\rm
Tr}(B_{1}\\!\cdot\\!F_{2})$ $\displaystyle-
k_{N}\\!\cdot\\!F_{2}\\!\cdot\\!k_{M}k_{P}\\!\cdot\\!B_{1}\\!\cdot\\!k_{Q}-k_{Q}\\!\cdot\\!k_{M}k_{N}\\!\cdot\\!F_{2}\\!\cdot\\!B_{1}\\!\cdot\\!k_{P}+k_{P}\\!\cdot\\!k_{M}k_{N}\\!\cdot\\!F_{2}\\!\cdot\\!B_{1}\\!\cdot\\!k_{Q}$
$\displaystyle+k_{N}\\!\cdot\\!k_{Q}k_{M}\\!\cdot\\!F_{2}\\!\cdot\\!B_{1}\\!\cdot\\!k_{P}-k_{N}\\!\cdot\\!k_{P}k_{M}\\!\cdot\\!F_{2}\\!\cdot\\!B_{1}\\!\cdot\\!k_{Q}\bigg{)},$

where we use the following identity:

$\displaystyle\epsilon^{abcd}\epsilon^{efgh}=-\left|\begin{array}[]{cccc}\eta^{ae}&\eta^{af}&\eta^{ag}&\eta^{ah}\cr\
\eta^{be}&\eta^{bf}&\eta^{bg}&\eta^{bh}\cr\
\eta^{ce}&\eta^{cf}&\eta^{cg}&\eta^{ch}\cr\
\eta^{de}&\eta^{df}&\eta^{dg}&\eta^{dh}\\\ \end{array}\right|.$ (25)

Applying the above relations in the amplitude (18), one obtains the amplitudes
(19) and (20). In fact the amplitude of one $RR$, two scalar and one gauge
field transform to the amplitude of one $NSNS$ B-field, two scalar and one
gauge field under the S-duality.

$\displaystyle{\cal A}\left(C^{(2)}_{1},F_{2},\Phi_{3},\Phi_{4}\right)$
$\displaystyle\longrightarrow$ $\displaystyle{\cal
A}_{contact}\left(B_{1},F_{2},\Phi_{3},\Phi_{4}\right)+{\cal
A}_{pole}\left(B_{1},F_{2},\Phi_{3},\Phi_{4}\right).$ (26)

Therefore, to get the $NSNS$ coupling to brane, one can apply the S-duality on
the pole amplitude contains $RR$ state. This is consistent with the statement
that the $NSNS$ coupling is found by replacement $F\longrightarrow F+B$ in the
D-brane effective action that has been expressed in literatures.

At the presence of the axion field, it is expected that one can find the
combination of the amplitudes (18), (19) and (20) in terms of $SL(2,R)$
invariant structures. To do this, we consider all contraction of the $SL(2,R)$
invariant structure $\tilde{{\cal F}}^{T}{\cal M}{\cal B}$ with four, two and
zero momenta as following:

$\displaystyle{({\tilde{{\cal F}}_{N}}^{T})}^{ab}{\cal M}{\cal
B}_{M}^{cd}\,\,{k_{1}}_{a}{k_{2}}_{b}{k_{3}}_{c}{k_{4}}_{d}$ $\displaystyle=$
$\displaystyle
e^{-\phi}k_{1}\\!\cdot\\!F_{N}\\!\cdot\\!k_{2}k_{4}\\!\cdot\\!B_{M}\\!\cdot\\!k_{3}+C_{0}e^{-\phi}k_{1}\\!\cdot\\!F_{N}\\!\cdot\\!k_{2}k_{4}\\!\cdot\\!\tilde{B}_{M}\\!\cdot\\!k_{3}$
(27) $\displaystyle-\frac{1}{2}k_{4}\\!\cdot\\!k_{2}k_{1}\\!\cdot\\!k_{3}{\rm
Tr}(F_{N}\\!\cdot\\!\tilde{C}_{M})+\frac{1}{2}k_{1}\\!\cdot\\!k_{4}k_{2}\\!\cdot\\!k_{3}{\rm
Tr}(F_{N}\\!\cdot\\!\tilde{C}_{M})$
$\displaystyle+k_{3}\\!\cdot\\!F_{N}\\!\cdot\\!k_{4}k_{1}\\!\cdot\\!\tilde{C}_{M}\\!\cdot\\!k_{2}-k_{2}\\!\cdot\\!k_{3}k_{4}\\!\cdot\\!F_{N}\\!\cdot\\!\tilde{C}_{M}\\!\cdot\\!k_{1}$
$\displaystyle+k_{1}\\!\cdot\\!k_{3}k_{4}\\!\cdot\\!F_{N}\\!\cdot\\!\tilde{C}_{M}\\!\cdot\\!k_{2}+k_{4}\\!\cdot\\!k_{2}k_{3}\\!\cdot\\!F_{N}\\!\cdot\\!\tilde{C}_{M}\\!\cdot\\!k_{1}$
$\displaystyle-
k_{1}\\!\cdot\\!k_{4}k_{3}\\!\cdot\\!F_{N}\\!\cdot\\!\tilde{C}_{M}\\!\cdot\\!k_{2},$
$\displaystyle k_{3}\\!\cdot\\!\tilde{{\cal F}}_{N}^{T}{\cal M}{\cal
B}_{M}\\!\cdot\\!k_{4}$ $\displaystyle=$
$\displaystyle-e^{-\phi}k_{3}\\!\cdot\\!F_{N}\\!\cdot\\!B_{M}\\!\cdot\\!k_{4}-C_{0}e^{-\phi}k_{3}\\!\cdot\\!F_{N}\\!\cdot\\!\tilde{B}_{M}\\!\cdot\\!k_{4}$
$\displaystyle+\frac{1}{2}k_{3}\\!\cdot\\!k_{4}{\rm
Tr}(F_{N}\\!\cdot\\!\tilde{C}_{M})-k_{3}\\!\cdot\\!\tilde{C}_{M}\\!\cdot\\!F_{N}\\!\cdot\\!k_{4},$
$\displaystyle{\rm Tr}(\tilde{{\cal F}}_{N}^{T}{\cal M}{\cal B}_{M})$
$\displaystyle=$ $\displaystyle-e^{-\phi}{\rm
Tr}(F_{N}\\!\cdot\\!B_{M})-C_{0}e^{-\phi}{\rm Tr}(F_{N}\\!\cdot\\!B_{M})+{\rm
Tr}(F_{N}\\!\cdot\\!\tilde{C}_{M}).$

Considering the above relations, one can find the following combination of
above $SL(2,R)$ invariant structures contains the amplitude (18) as well as
the amplitude (19) $+$ (20).

$\displaystyle\widetilde{{\cal A}}$ $\displaystyle=$
$\displaystyle-\frac{{\Phi_{3}}_{i}\Phi_{4}^{i}}{k_{1}\\!\cdot\\!k_{1}}\bigg{[}\frac{1}{2}k_{3}\\!\cdot\\!k_{4}k_{1}\\!\cdot\\!\tilde{{\cal
F}}_{2}^{T}{\cal M}{\cal
B}_{1}\\!\cdot\\!k_{1}-k_{1}\\!\cdot\\!k_{4}k_{3}\\!\cdot\\!\tilde{{\cal
F}}_{2}^{T}{\cal M}{\cal B}_{1}\\!\cdot\\!k_{1}-k_{1}\\!\cdot\\!\tilde{{\cal
F}}_{2}^{T}\\!\cdot\\!k_{3}{\cal M}k_{4}\\!\cdot\\!{\cal
B}_{1}\\!\cdot\\!k_{1}\bigg{]}$ (28)
$\displaystyle+{\Phi_{3}}_{i}\Phi_{4}^{i}\bigg{[}k_{3}\\!\cdot\\!\tilde{{\cal
F}}_{2}^{T}{\cal M}{\cal
B}_{1}\\!\cdot\\!k_{4}-\frac{1}{4}k_{3}\\!\cdot\\!k_{4}Tr(\tilde{{\cal
F}}_{2}^{T}{\cal M}{\cal B}_{1})\bigg{]}+(3\leftrightarrow 4).$

It is clear that this $SL(2,R)$ invariant amplitude (that is manifestly
invariant under the $SL(2,R)$ transformation) also contains the amplitude
${\cal A}\left(C_{0};B_{1},F_{2},\Phi_{3},\Phi_{4}\right)$. In fact, to make a
$SL(2,R)$ invariant amplitude from amplitudes (18) and (19) $+$ (20), the
amplitude corresponding to axion field should be take into account. On the
other hand, we can confirm the above $SL(2,R)$ invariant amplitude by deriving
the axion amplitude from the explicit calculation and the Feynman rule that is
resulted as following pole amplitude:

$\displaystyle{\cal A}\left(C_{0};B_{1},F_{2},\Phi_{3},\Phi_{4}\right)$
$\displaystyle=$
$\displaystyle-\frac{C_{0}e^{-\phi}{\Phi_{3}}_{i}\Phi_{4}^{i}}{k_{1}\\!\cdot\\!k_{1}}\bigg{[}k_{1}\\!\cdot\\!\tilde{B}_{1}\\!\cdot\\!k_{4}k_{1}\\!\cdot\\!F_{2}\\!\cdot\\!k_{3}-k_{1}\\!\cdot\\!k_{3}k_{1}\\!\cdot\\!\tilde{B}_{1}\\!\cdot\\!F_{2}\\!\cdot\\!k_{4}$
(29)
$\displaystyle\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\frac{1}{2}k_{3}\\!\cdot\\!k_{4}k_{1}\\!\cdot\\!\tilde{B}_{1}\\!\cdot\\!F_{2}\\!\cdot\\!k_{1}\bigg{]}+(3\leftrightarrow
4)$

This is exactly equal to the axion amplitude that lies in (28).

##  4 S-dual amplitude of three gauge fields and one two-form

It has been shown that the amplitude of three gauge fields and one two-form in
which the two-form is $RR$ field $C^{(2)}$ or $NSNS$ B-field could be appeared
in terms of $SL(2,R)$ invariant structures in the presence of the background
dilaton and R-R scalar fields [21]. We are going to show that these two
amplitudes transform to each other under the S-duality transformation when the
axion field dose not tack into account. We strart with the amplitude of three
gage fields and one $RR$ field $C^{(2)}$ that is founded by explicit
calculation[21, 24].

$\displaystyle{\cal A}\left({C_{1}}^{(2)},F_{2},F_{3},F_{4}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{4}e^{-\phi_{0}}\frac{k_{1}\\!\cdot\\!F_{2}\\!\cdot\\!\tilde{C}^{(2)}_{1}\\!\cdot\\!k_{1}}{k_{1}\\!\cdot\\!k_{1}}Tr(F_{3}\\!\cdot\\!F_{4})$
(30)
$\displaystyle-e^{-\phi_{0}}\frac{k_{1}\\!\cdot\\!F_{2}\\!\cdot\\!F_{3}\\!\cdot\\!F_{4}\\!\cdot\\!\tilde{C}^{(2)}_{1}\\!\cdot\\!k_{1}}{k_{1}\\!\cdot\\!k_{1}}+P(2,3,4),$

where $P(2,3,4)$ stands for the other permutations of $2,3,4$. This amplitude
contains two independent structures. We apply S-duality transformation to
these two structures separately.

$\displaystyle
e^{-\phi_{0}}k_{1}\\!\cdot\\!F_{2}\\!\cdot\\!F_{3}\\!\cdot\\!F_{4}\\!\cdot\\!\tilde{C}^{(2)}\\!\cdot\\!k_{1}$
$\displaystyle\mathop{\longrightarrow}^{S}\,\,\,\,\frac{e^{-2\phi_{0}}}{16}k_{1a}\epsilon^{abkl}\epsilon_{bcnm}\epsilon^{cdgh}\epsilon_{deij}F_{2kl}{F_{3}}^{nm}F_{4gh}B^{ij}{k_{1}}^{e}=I_{234},$
$\displaystyle
e^{-\phi_{0}}k_{1}\\!\cdot\\!F_{2}\\!\cdot\\!\tilde{C}^{(2)}\\!\cdot\\!k_{1}Tr(F_{3}\\!\cdot\\!F_{4})$
$\displaystyle\mathop{\longrightarrow}^{S}\,\,\,\,\frac{e^{-2\phi_{0}}}{16}k_{1a}\epsilon^{abkl}\epsilon_{bcnm}\epsilon^{degh}\epsilon_{deij}F_{2kl}{F_{3}}_{gh}{F_{4}}^{ij}B^{nm}{k_{1}}^{e}=J_{234}.$

Because the invariance of the trace term in the original amplitude under the
gauge fields permutations, one can easily conclude $J_{234}=J_{243}$,
$J_{423}=J_{432}$ and $J_{324}=J_{342}$ in the dual amplitude. Therefore, the
gauge field permutations can produce six different terms of $I$-type and three
different terms of $J$-type that contribute in the dual amplitude.

The calculation in this case is more complicated than the case in the previous
section due to the presence of four Levi-Civita tensors. To rewrite this terms
in terms of various contractions of gauge fields, one has to use the identity
(25). Here, one encounters with different choices. In fact, there are three
ways to pairs the Levi-Civita tensors: ${\cal
E}=\big{(}\epsilon^{abkl}\epsilon_{bcnm}\big{)}\big{(}\epsilon^{degh}\epsilon_{deij}\big{)}$,
${\cal
E}^{\prime}=\big{(}\epsilon^{abkl}\epsilon_{deij}\big{)}\big{(}\epsilon^{cdgh}\epsilon_{bcnm}\big{)}$
and ${\cal
E}^{\prime\prime}=\big{(}\epsilon^{abkl}\epsilon^{cdgh}\big{)}\big{(}\epsilon_{bcnm}\epsilon_{deij}\big{)}$.
Using the standard identity (25) for the above first two parings in I-type
terms, we have:

$\displaystyle I_{NMP}({\cal E})$ $\displaystyle=$ $\displaystyle
e^{-2\phi_{0}}\bigg{[}-k_{1}\\!\cdot\\!F_{N}\\!\cdot\\!B\\!\cdot\\!F_{P}\\!\cdot\\!F_{M}\\!\cdot\\!k_{1}-k_{1}\\!\cdot\\!F_{N}\\!\cdot\\!F_{M}\\!\cdot\\!F_{P}\\!\cdot\\!B\\!\cdot\\!k_{1}-k_{1}\\!\cdot\\!F_{3}\\!\cdot\\!F_{M}\\!\cdot\\!F_{N}\\!\cdot\\!B\\!\cdot\\!k_{1}$
$\displaystyle\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\frac{1}{2}k_{1}\\!\cdot\\!F_{P}\\!\cdot\\!F_{M}\\!\cdot\\!k_{1}Tr(F_{N}\\!\cdot\\!B)+\frac{1}{2}k_{1}\\!\cdot\\!F_{N}\\!\cdot\\!B\\!\cdot\\!k_{1}Tr(F_{P}\\!\cdot\\!F_{M})$
$\displaystyle\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\frac{1}{4}k_{1}\\!\cdot\\!k_{1}Tr(F_{N}\\!\cdot\\!B)Tr(F_{P}\\!\cdot\\!F_{M})+k_{1}\\!\cdot\\!k_{1}Tr(F_{N}\\!\cdot\\!F_{M}\\!\cdot\\!F_{P}\\!\cdot\\!B)\bigg{]}$
$\displaystyle I_{NMP}({\cal E}^{\prime})$ $\displaystyle=$ $\displaystyle
e^{-2\phi_{0}}\bigg{[}k_{1}\\!\cdot\\!F_{P}\\!\cdot\\!B\\!\cdot\\!F_{N}\\!\cdot\\!F_{M}\\!\cdot\\!k_{1}-\frac{1}{2}k_{1}\\!\cdot\\!F_{P}\\!\cdot\\!B\\!\cdot\\!k_{1}Tr(F_{N}\\!\cdot\\!F_{M})$
(31)
$\displaystyle\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\frac{1}{2}k_{1}\\!\cdot\\!F_{N}\\!\cdot\\!F_{M}\\!\cdot\\!k_{1}Tr(F_{P}\\!\cdot\\!B)+\frac{1}{4}k_{1}\\!\cdot\\!k_{1}Tr(F_{N}\\!\cdot\\!F_{M})Tr(F_{P}\\!\cdot\\!B)\bigg{]}$

and for the first paring in the J-type term, we have

$\displaystyle J_{NMP}({\cal E})$ $\displaystyle=$ $\displaystyle
e^{-2\phi_{0}}\bigg{[}-k_{1}\\!\cdot\\!F_{N}\\!\cdot\\!B\\!\cdot\\!k_{1}Tr(F_{M}\\!\cdot\\!F_{P})+\frac{1}{2}k_{1}\\!\cdot\\!k_{1}Tr(F_{N}\\!\cdot\\!B)Tr(F_{M}\\!\cdot\\!F_{P})\bigg{]}$
(32)

where $N,M,P=2,3,4$ and $N\neq M\neq P$.

Our calculations show that one can get to the amplitude of ${\cal
A}\left({B_{1}}^{(2)},F_{2},F_{3},F_{4}\right)$ only by selecting the specific
parings of Levi-Civita tensors as following:

$\displaystyle{\cal
A}\left({C_{1}}^{(2)},F_{2},F_{3},F_{4}\right)\mathop{\longrightarrow}^{S}\frac{I_{234}({\cal
E}^{\prime})+I_{324}({\cal E}^{\prime})+I_{423}({\cal
E}^{\prime})+I_{243}({\cal E})+I_{342}({\cal E})+I_{432}({\cal
E})}{k_{1}\\!\cdot\\!k_{1}}$
$\displaystyle\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\frac{1}{2}\frac{J_{234}({\cal
E})+J_{324}({\cal E})+J_{423}({\cal E})}{k_{1}\\!\cdot\\!k_{1}}.$ (33)

Replacing (31) and (32) in the above relation and using the conservation of
momentum, one finds the following expression, after a straightforward
calculation.

$\displaystyle\frac{1}{2}e^{-2\phi_{0}}\left({\rm
Tr}(B_{1}\\!\cdot\\!F_{2}\\!\cdot\\!F_{3}\\!\cdot\\!F_{4})-\frac{1}{4}{\rm
Tr}(B_{1}\\!\cdot\\!F_{2}){\rm Tr}(F_{3}\\!\cdot\\!F_{4})\right)$
$\displaystyle-\frac{1}{4}e^{-2\phi_{0}}\frac{k_{1}\\!\cdot\\!F_{2}\\!\cdot\\!B_{1}\\!\cdot\\!k_{1}}{k_{1}\\!\cdot\\!k_{1}}Tr(F_{3}\\!\cdot\\!F_{4})+e^{-2\phi_{0}}\frac{k_{1}\\!\cdot\\!F_{2}\\!\cdot\\!F_{3}\\!\cdot\\!F_{4}\\!\cdot\\!B_{1}\\!\cdot\\!k_{1}}{k_{1}\\!\cdot\\!k_{1}}+P(2,3,4).$
(34)

As we have mentioned in (3), this is exactly equal to the amplitude of one
$B$-field and three gauge fields containing pole term as well as contact term
that calculated in [21].

##  5 S-dual effective action with four derivative corrections

It has been shown that the one-loop correction to the $D_{3}$-brane Abelian
Born-Infeld action (4) containing four gauge fields with four derivative
proportional to $(s^{2}+t^{2}+u^{2})F^{4}$ where $s,t$ and $u$ are the
Mandelstam variables [16]. This result has been found from supersymmetry
fixing [23] and string Disk-level scattering amplitude calculation[20]. The
$D_{3}$-brane effective action at the order of $(\partial F)^{4}$ is invariant
under the $SL(2,R)$ duality. From the Lagrangian viewpoint, this action
saturates the NGZ identity[17].

We are going to find this action in terms of some $SL(2,R)$ invariant
structures. To do this, one should consider all contractions of two nonlinear
structures $W^{abcdefg}W_{ghnmpq}$, which any of them contains two gauge field
derivative terms as following

$\displaystyle W^{abcdef}=\partial^{a}{({\cal F}^{T})}^{bc}{\cal
M}\,\,\,\,\partial^{d}{{\cal F}}^{ef}.$ (35)

Using ”xAct” [27], one finds the following 33 different $WW$ contractions:

$\displaystyle w_{1}\
W_{c}{}^{efcab}W_{def}{}^{d}{}_{ab}+w_{2}W_{a}{}^{ab}{}_{c}{}^{ef}\
W_{def}{}^{d}{}_{b}{}^{c}+w_{3}W_{bef}{}^{cab}W_{dc}{}^{fe}{}_{a}{}^{d}+w_{4}W_{bc}{}^{fcab}\
W_{def}{}^{e}{}_{a}{}^{d}$ $\displaystyle+w_{5}W_{cdf}{}^{cab}\
W_{eb}{}^{fe}{}_{a}{}^{d}+w_{6}\
W_{a}{}^{ab}{}_{cb}{}^{c}W_{d}{}^{de}{}_{fe}{}^{f}+w_{7}W_{a}{}^{ab}{}_{cd}{}^{e}\
W^{d}{}_{b}{}^{c}{}_{fe}{}^{f}+w_{8}W_{a}{}^{ab}{}_{dc}{}^{e}\
W^{d}{}_{b}{}^{c}{}_{fe}{}^{f}$ $\displaystyle+w_{9}W_{a}{}^{ab}{}_{b}{}^{cd}\
W^{e}{}_{cdfe}{}^{f}+w_{10}W_{a}{}^{abd}{}_{b}{}^{c}\
W^{e}{}_{cdfe}{}^{f}+w_{11}W^{cabe}{}_{a}{}^{d}W_{fce}{}^{f}{}_{bd}+w_{12}W_{cdf}{}^{cab}\
W^{e}{}_{a}{}^{df}{}_{be}$ $\displaystyle+w_{13}W^{cab}{}_{dcf}\
W^{e}{}_{a}{}^{df}{}_{be}+w_{14}W^{cabe}{}_{a}{}^{d}W_{fcd}{}^{f}{}_{be}\
+w_{15}W^{cab}{}_{def}\ W^{d}{}_{ac}{}^{f}{}_{b}{}^{e}+w_{16}W_{ca}{}^{dcab}\
W_{edf}{}^{f}{}_{b}{}^{e}$ $\displaystyle+w_{17}W^{cabd}{}_{ac}\
W_{edf}{}^{f}{}_{b}{}^{e}+w_{18}W_{ca}{}^{dcab}\
W_{fde}{}^{f}{}_{b}{}^{e}+w_{19}W^{cabd}{}_{ac}\
W_{fde}{}^{f}{}_{b}{}^{e}+w_{20}W_{a}{}^{ab}{}_{b}{}^{cd}\
W_{def}{}^{f}{}_{c}{}^{e}$ $\displaystyle+w_{21}W^{cab}{}_{def}\
W^{d}{}_{ab}{}^{f}{}_{c}{}^{e}+w_{22}W_{a}{}^{ab}{}_{def}\
W^{d}{}_{b}{}^{cf}{}_{c}{}^{e}+w_{23}W^{cabd}{}_{ab}\
W_{edf}{}^{f}{}_{c}{}^{e}+w_{24}W_{a}{}^{abd}{}_{b}{}^{c}\
W_{edf}{}^{f}{}_{c}{}^{e}$ $\displaystyle+w_{25}W^{cabd}{}_{ab}\
W_{fde}{}^{f}{}_{c}{}^{e}+w_{26}W_{a}{}^{abd}{}_{b}{}^{c}\
W_{fde}{}^{f}{}_{c}{}^{e}+w_{27}W_{bef}{}^{cab}\
W^{d}{}_{ac}{}^{f}{}_{d}{}^{e}+w_{28}W_{a}{}^{ab}{}_{cef}\
W^{d}{}_{b}{}^{cf}{}_{d}{}^{e}$
$\displaystyle+w_{29}W_{a}{}^{ab}{}_{cb}{}^{c}\
W_{edf}{}^{fde}+w_{30}W_{bac}{}^{cab}W_{edf}{}^{fde}+w_{31}\
W_{cab}{}^{cab}W_{edf}{}^{fde}+w_{32}W_{a}{}^{ab}{}_{cb}{}^{c}\
W_{fde}{}^{fde}$ $\displaystyle+w_{33}W_{cab}{}^{cab}W_{fde}{}^{fde},$ (36)

where $w_{i}$ are some unknown coefficients that would be fiexed by imposing
some physical identity.

Using the $SL(2,R)$ invariant structure (14) at the level of four gauge fields
and four derivatives, one finds that the $WW$ structures appear in terms of
the gauge field contents $(\partial F)^{4},(\partial
F)^{2}(\partial\tilde{F})^{2}$ and $(\partial\tilde{F})^{4}$. In order to make
the action (4) in terms of the above $WW$ structures, we should find the
contribution of dual gauge field $\tilde{F}$ in the last two field contents in
terms of gauge field $F$ by using the identity (25). For the third field
content, we encounter with four Levi-Civita tensors and with different choices
to pair these tensors as was mentioned in the previous section: ${\cal
E},{\cal E}^{\prime}$ and ${\cal E}^{\prime\prime}$. Any of these parings of
Levi-Civita tensors led to different results in which the gauge fields appear
in different kinds of contraction terms $(\partial F)^{4}$. But one can show
that these results are equal when the identities presenting in the appendix
have been tack into account.

After finding the above $WW$ structures in terms of gauge field contraction
terms $(\partial F)^{4}$ and using the identities in the appendix properly, we
find the following 9 equations between the constants:

$\displaystyle w_{15}$ $\displaystyle=$
$\displaystyle-\frac{3}{2}+2(w_{10}+w_{17}-w_{20}-w_{24})-3(w_{29}-w_{30}-w_{6})+w_{22}+w_{8},$
$\displaystyle w_{13}$ $\displaystyle=$
$\displaystyle\frac{2}{3}(w_{10}+w_{15}-w_{20}-w_{22}-w_{8})+\frac{1}{3}(-w_{17}+w_{24}),$
$\displaystyle w_{21}$ $\displaystyle=$
$\displaystyle\frac{2}{3}(w_{10}+w_{17}-w_{20}-w_{22}-w_{24})+\frac{1}{2}(w_{12}-w_{27}-w_{7})+\frac{1}{6}w_{15}-\frac{7}{6}w_{8}+w_{16}+w_{2},$
$\displaystyle w_{26}$ $\displaystyle=$
$\displaystyle-\frac{1}{3}(w_{8}-w_{15})+\frac{2}{3}(-w_{10}+w_{20}+w_{22})+\frac{8}{3}(-w_{17}+w_{24})+6(w_{29}-w_{30}-w_{6})$
$\displaystyle\,\,+w_{14}-w_{16}+w_{19}+w_{27}-w_{7},$ $\displaystyle w_{28}$
$\displaystyle=$
$\displaystyle\frac{4}{3}(-w_{10}-w_{17}+w_{20}+w_{24})+\frac{1}{3}(w_{8}-w_{15}+w_{22})+4(w_{29}-w_{30}-w_{6})+w_{27}-w_{7},$
$\displaystyle w_{3}$ $\displaystyle=$
$\displaystyle\frac{4}{3}(w_{10}-w_{20})+\frac{1}{3}(w_{15}+w_{17}-w_{22}-w_{24}-w_{8})-w_{29}+w_{30}+w_{6},$
$\displaystyle w_{4}$ $\displaystyle=$ $\displaystyle
w_{17}-w_{24}-w_{29}+w_{30}+w_{6},$ $\displaystyle w_{5}$ $\displaystyle=$
$\displaystyle\frac{2}{3}(w_{10}-w_{20}-w_{8})+\frac{1}{2}(w_{14}+w_{29}-w_{30}-w_{6}-w_{7})+\frac{1}{6}(w_{15}+w_{17}-w_{22}-w_{24})$
$\displaystyle\,-2w_{1}-w_{18}+w_{2}$ $\displaystyle w_{9}$ $\displaystyle=$
$\displaystyle\frac{1}{6}(-w_{10}-w_{15}-w_{17}+w_{20}+w_{22}+w_{24}+w_{8}).$

The nonzero coefficients could be fixed as follows by comparing the result
with the action (4) :

$\displaystyle\
w_{3}=w_{4}=w_{5}=w_{6}=\frac{1}{2},\,\,\,\,\,\,\,\,\,\,\,\,w_{7}=-\frac{3}{2},\,\,\,\,\,\,\,\,\,\,\,\,\,w_{27}=\frac{3}{2},\,\,\,\,\,\,\,\,\,\,\,\,\,\,w_{28}=1.$

By substituting these nonzero coefficients in (36), we can find the $SL(2,R)$
invariant form of the action (4).

$\displaystyle L_{(\partial F{})^{4}}^{S}$ $\displaystyle=$
$\displaystyle\frac{1}{2}W_{bef}{}^{cab}W_{dc}{}^{fe}{}_{a}{}^{d}+\frac{1}{2}W_{a}{}^{ab}{}_{cb}{}^{c}W_{d}{}^{de}{}_{fe}{}^{f}+\frac{1}{2}W_{bc}{}^{fcab}W_{def}{}^{e}{}_{a}{}^{d}+\frac{1}{2}W_{cdf}{}^{cab}W_{eb}{}^{fe}{}_{a}{}^{d}$
(37)
$\displaystyle-\frac{3}{2}W_{a}{}^{ab}{}_{cd}{}^{e}W^{d}{}_{b}{}^{c}{}_{fe}{}^{f}+\frac{3}{2}W_{bef}{}^{cab}W^{d}{}_{ac}{}^{f}{}_{d}{}^{e}+W_{a}{}^{ab}{}_{cef}W^{d}{}_{b}{}^{cf}{}_{d}{}^{e}.$

In order to reach the action at the order of six gauge fields and four
derivatives, one should consider the expansion of the field $G^{{}^{\prime}}$
up to the level of three gauge field $F^{3}$. So the expansion of $\partial
G^{{}^{\prime}}$ up to this level, could be found as follows:

$\displaystyle\partial_{a}{G^{{}^{\prime}}}_{bc}$ $\displaystyle=$
$\displaystyle
e^{-\phi_{0}}\partial_{a}F_{bc}+e^{-2\phi_{0}}\bigg{(}\frac{1}{4}F_{de}F^{de}\
\partial_{a}F_{bc}+F_{c}{}^{d}F_{d}{}^{e}\
\partial_{a}F_{be}-F_{b}{}^{d}F_{d}{}^{e}\ \partial_{a}F_{ce}$ $\displaystyle-
F_{b}{}^{d}F_{c}{}^{e}\ \partial_{a}F_{de}+\frac{1}{2}F_{bc}F^{de}\
\partial_{a}F_{de}\bigg{)}+\cdots$

where dots refer to the terms at higher order of gauge field.

Inserting the above expansion into the nonzero invariant structurs $WW$ in
(37), one can find the action containing six gauge fields with four
derivatives that is compatible with S-duality. This action could be appear in
the simple form when one consider the total derivative terms. Actualy, at the
level of six gauge fields and four derivatives, there are 584 contractions
with structure $FFDFDFDFDF$. To find the total derivative terms, we note that
there are 724 total derivative terms with structure $D[FFFDFDFDF]$. Using
their coefficients to eliminate the terms with structure $FFFDDFDFDF$, one
finds 483 constraint equations. Applying these constraints, the total
derivative terms with structure $FFDFDFDFDF$ could be found. Considering these
terms and using the the on-shell condition $\partial_{a}F^{ab}=0$ , one can
find the action corresponding to six gauge fields with four derivatives as
following simple form.

$\displaystyle L_{F^{2}(\partial F{})^{4}}^{S}$ $\displaystyle\sim$
$\displaystyle\frac{3}{2}F^{ab}F^{cd}\partial_{c}F_{e}{}^{f}\partial_{d}F^{hi}\partial^{e}F_{ab}\partial_{f}F_{hi}+\frac{1}{4}F^{ab}F^{cd}\partial_{d}F_{c}{}^{f}\partial_{e}F^{hi}\partial^{e}F_{ab}\partial_{f}F_{hi}$
(38)
$\displaystyle-3F^{ab}F^{cd}\partial_{b}F_{e}{}^{f}\partial_{d}F^{hi}\partial^{e}F_{ac}\partial_{f}F_{hi}-\frac{1}{2}F^{ab}F^{cd}\partial_{d}F_{b}{}^{f}\partial_{e}F^{hi}\partial^{e}F_{ac}\partial_{f}F_{hi}$
$\displaystyle+3F_{a}{}^{c}F^{ab}\partial_{c}F^{hi}\partial_{d}F_{e}{}^{f}\partial^{e}F_{b}{}^{d}\partial_{f}F_{hi}-3F_{a}{}^{c}F^{ab}\partial_{c}F_{e}{}^{f}\partial_{d}F^{hi}\partial^{e}F_{b}{}^{d}\partial_{f}F_{hi}$
$\displaystyle+\frac{1}{2}F_{a}{}^{c}F^{ab}\partial_{c}F_{d}{}^{f}\partial_{e}F^{hi}\partial^{e}F_{b}{}^{d}\partial_{f}F_{hi}-\frac{1}{2}F^{ab}F^{cd}\partial_{d}F^{hi}\partial^{e}F_{ac}\partial_{f}F_{hi}\partial^{f}F_{be}$
$\displaystyle+\frac{1}{4}F^{ab}F^{cd}\partial_{d}F^{hi}\partial^{e}F_{ab}\partial_{f}F_{hi}\partial^{f}F_{ce}-F_{a}{}^{c}F^{ab}\partial_{d}F^{hi}\partial^{e}F_{b}{}^{d}\partial_{f}F_{hi}\partial^{f}F_{ce}$
$\displaystyle+\frac{1}{2}F_{a}{}^{c}F^{ab}\partial_{c}F^{hi}\partial^{e}F_{b}{}^{d}\partial_{f}F_{hi}\partial^{f}F_{de}+F^{ab}F^{cd}\partial_{b}F_{ac}\partial_{e}F^{hi}\partial_{f}F_{hi}\partial^{f}F_{d}{}^{e}$
$\displaystyle-\frac{1}{2}F^{ab}F^{cd}\partial_{c}F_{ab}\partial_{e}F^{hi}\partial_{f}F_{hi}\partial^{f}F_{d}{}^{e}+F_{a}{}^{c}F^{ab}\partial_{c}F_{b}{}^{d}\partial_{e}F^{hi}\partial_{f}F_{hi}\partial^{f}F_{d}{}^{e}$
$\displaystyle+3F^{ab}F^{cd}\partial_{d}F_{hi}\partial_{e}F_{f}{}^{i}\partial^{e}F_{ac}\partial^{h}F_{b}{}^{f}+3F^{ab}F^{cd}\partial_{d}F_{fi}\partial^{e}F_{ac}\partial_{h}F_{e}{}^{i}\partial^{h}F_{b}{}^{f}$
$\displaystyle-2F^{ab}F^{cd}\partial_{d}F_{e}{}^{i}\partial^{e}F_{ac}\partial_{h}F_{fi}\partial^{h}F_{b}{}^{f}-\frac{3}{2}F^{ab}F^{cd}\partial_{d}F_{hi}\partial_{e}F_{f}{}^{i}\partial^{e}F_{ab}\partial^{h}F_{c}{}^{f}$
$\displaystyle+3F_{a}{}^{c}F^{ab}\partial_{e}F_{hi}\partial^{e}F_{b}{}^{d}\partial_{f}F_{d}{}^{i}\partial^{h}F_{c}{}^{f}-2F_{a}{}^{c}F^{ab}\partial_{e}F_{d}{}^{i}\partial^{e}F_{b}{}^{d}\partial_{f}F_{hi}\partial^{h}F_{c}{}^{f}$
$\displaystyle+3F_{a}{}^{c}F^{ab}\partial^{e}F_{b}{}^{d}\partial_{f}F_{ei}\partial_{h}F_{d}{}^{i}\partial^{h}F_{c}{}^{f}-\frac{3}{2}F^{ab}F^{cd}\partial_{d}F_{fi}\partial^{e}F_{ab}\partial_{h}F_{e}{}^{i}\partial^{h}F_{c}{}^{f}$
$\displaystyle+F^{ab}F^{cd}\partial_{d}F_{e}{}^{i}\partial^{e}F_{ab}\partial_{h}F_{fi}\partial^{h}F_{c}{}^{f}-4F_{a}{}^{c}F^{ab}\partial_{e}F_{d}{}^{i}\partial^{e}F_{b}{}^{d}\partial_{h}F_{fi}\partial^{h}F_{c}{}^{f}$
$\displaystyle-3F_{a}{}^{c}F^{ab}\partial_{c}F_{hi}\partial_{e}F_{f}{}^{i}\partial^{e}F_{b}{}^{d}\partial^{h}F_{d}{}^{f}-3F_{a}{}^{c}F^{ab}\partial_{c}F_{fi}\partial^{e}F_{b}{}^{d}\partial_{h}F_{e}{}^{i}\partial^{h}F_{d}{}^{f}$
$\displaystyle+2F_{a}{}^{c}F^{ab}\partial_{c}F_{e}{}^{i}\partial^{e}F_{b}{}^{d}\partial_{h}F_{fi}\partial^{h}F_{d}{}^{f}+2F^{ab}F^{cd}\partial_{c}F_{f}{}^{i}\partial_{d}F_{hi}\partial^{e}F_{ab}\partial^{h}F_{e}{}^{f}$
$\displaystyle-4F^{ab}F^{cd}\partial_{b}F_{f}{}^{i}\partial_{d}F_{hi}\partial^{e}F_{ac}\partial^{h}F_{e}{}^{f}+4F_{a}{}^{c}F^{ab}\partial_{c}F_{hi}\partial_{d}F_{f}{}^{i}\partial^{e}F_{b}{}^{d}\partial^{h}F_{e}{}^{f}$
$\displaystyle-4F_{a}{}^{c}F^{ab}\partial_{c}F_{f}{}^{i}\partial_{d}F_{hi}\partial^{e}F_{b}{}^{d}\partial^{h}F_{e}{}^{f}-\frac{3}{2}F_{ab}F^{ab}\partial_{e}F_{fi}\partial^{e}F^{cd}\partial^{h}F_{c}{}^{f}\partial^{i}F_{dh}$
$\displaystyle+2F_{ab}F^{ab}\partial^{e}F^{cd}\partial_{f}F_{ei}\partial^{h}F_{c}{}^{f}\partial^{i}F_{dh}+2F_{a}{}^{c}F^{ab}\partial_{e}F_{c}{}^{f}\partial^{e}F_{b}{}^{d}\partial_{f}F_{hi}\partial^{i}F_{d}{}^{h}$
$\displaystyle+\frac{1}{2}F_{ab}F^{ab}\partial_{e}F_{c}{}^{f}\partial^{e}F^{cd}\partial_{h}F_{fi}\partial^{i}F_{d}{}^{h}+2F^{ab}F^{cd}\partial_{e}F_{c}{}^{f}\partial^{e}F_{ab}\partial_{i}F_{fh}\partial^{i}F_{d}{}^{h}$
$\displaystyle-4F^{ab}F^{cd}\partial_{e}F_{b}{}^{f}\partial^{e}F_{ac}\partial_{i}F_{fh}\partial^{i}F_{d}{}^{h}-4F_{a}{}^{c}F^{ab}\partial_{e}F_{c}{}^{f}\partial^{e}F_{b}{}^{d}\partial_{i}F_{fh}\partial^{i}F_{d}{}^{h}$
$\displaystyle+F_{ab}F^{ab}\partial_{e}F_{c}{}^{f}\partial^{e}F^{cd}\partial_{i}F_{fh}\partial^{i}F_{d}{}^{h}-2F_{a}{}^{c}F^{ab}\partial^{e}F_{b}{}^{d}\partial^{f}F_{ce}\partial_{i}F_{fh}\partial^{i}F_{d}{}^{h}$
$\displaystyle+F_{ab}F^{ab}\partial^{e}F^{cd}\partial^{f}F_{ce}\partial_{i}F_{fh}\partial^{i}F_{d}{}^{h}+3F^{ab}F^{cd}\partial_{b}F_{ac}\partial_{d}F^{ef}\partial_{f}F_{hi}\partial^{i}F_{e}{}^{h}$
$\displaystyle-\frac{3}{2}F^{ab}F^{cd}\partial_{c}F_{ab}\partial_{d}F^{ef}\partial_{f}F_{hi}\partial^{i}F_{e}{}^{h}+3F_{a}{}^{c}F^{ab}\partial_{c}F_{b}{}^{d}\partial_{d}F^{ef}\partial_{f}F_{hi}\partial^{i}F_{e}{}^{h}$
$\displaystyle-\frac{1}{4}F_{ab}F^{ab}\partial^{e}F^{cd}\partial_{f}F_{hi}\partial^{f}F_{cd}\partial^{i}F_{e}{}^{h}+3F^{ab}F^{cd}\partial_{b}F_{ac}\partial_{f}F_{hi}\partial^{f}F_{d}{}^{e}\partial^{i}F_{e}{}^{h}$
$\displaystyle-\frac{3}{2}F^{ab}F^{cd}\partial_{c}F_{ab}\partial_{f}F_{hi}\partial^{f}F_{d}{}^{e}\partial^{i}F_{e}{}^{h}+3F_{a}{}^{c}F^{ab}\partial_{c}F_{b}{}^{d}\partial_{f}F_{hi}\partial^{f}F_{d}{}^{e}\partial^{i}F_{e}{}^{h}$
$\displaystyle+\frac{3}{4}F_{ab}F^{ab}\partial^{e}F^{cd}\partial^{f}F_{cd}\partial_{h}F_{fi}\partial^{i}F_{e}{}^{h}+F^{ab}F^{cd}\partial_{d}F_{c}{}^{f}\partial^{e}F_{ab}\partial_{i}F_{fh}\partial^{i}F_{e}{}^{h}$
$\displaystyle-2F^{ab}F^{cd}\partial_{d}F_{b}{}^{f}\partial^{e}F_{ac}\partial_{i}F_{fh}\partial^{i}F_{e}{}^{h}+2F_{a}{}^{c}F^{ab}\partial_{c}F_{d}{}^{f}\partial^{e}F_{b}{}^{d}\partial_{i}F_{fh}\partial^{i}F_{e}{}^{h}$
$\displaystyle+F^{ab}F^{cd}\partial_{b}F_{ac}\partial^{f}F_{d}{}^{e}\partial_{i}F_{fh}\partial^{i}F_{e}{}^{h}-\frac{1}{2}F^{ab}F^{cd}\partial_{c}F_{ab}\partial^{f}F_{d}{}^{e}\partial_{i}F_{fh}\partial^{i}F_{e}{}^{h}$
$\displaystyle+F_{a}{}^{c}F^{ab}\partial_{c}F_{b}{}^{d}\partial^{f}F_{d}{}^{e}\partial_{i}F_{fh}\partial^{i}F_{e}{}^{h}-F^{ab}F^{cd}\partial_{d}F_{hi}\partial_{e}F_{c}{}^{f}\partial^{e}F_{ab}\partial^{i}F_{f}{}^{h}$
$\displaystyle+2F^{ab}F^{cd}\partial_{d}F_{hi}\partial_{e}F_{b}{}^{f}\partial^{e}F_{ac}\partial^{i}F_{f}{}^{h}-2F_{a}{}^{c}F^{ab}\partial_{c}F_{hi}\partial_{e}F_{d}{}^{f}\partial^{e}F_{b}{}^{d}\partial^{i}F_{f}{}^{h}$
$\displaystyle-\frac{1}{4}F_{ab}F^{ab}\partial_{e}F_{cd}\partial^{e}F^{cd}\partial_{h}F_{fi}\partial^{i}F^{fh}-\frac{1}{2}F^{ab}F^{cd}\partial_{d}F_{ce}\partial^{e}F_{ab}\partial_{i}F_{fh}\partial^{i}F^{fh}$
$\displaystyle-\frac{1}{2}F^{ab}F^{cd}\partial_{e}F_{cd}\partial^{e}F_{ab}\partial_{i}F_{fh}\partial^{i}F^{fh}+F^{ab}F^{cd}\partial_{d}F_{be}\partial^{e}F_{ac}\partial_{i}F_{fh}\partial^{i}F^{fh}$
$\displaystyle+F^{ab}F^{cd}\partial_{e}F_{bd}\partial^{e}F_{ac}\partial_{i}F_{fh}\partial^{i}F^{fh}-F_{a}{}^{c}F^{ab}\partial_{c}F_{de}\partial^{e}F_{b}{}^{d}\partial_{i}F_{fh}\partial^{i}F^{fh}$
$\displaystyle+F_{a}{}^{c}F^{ab}\partial_{d}F_{ce}\partial^{e}F_{b}{}^{d}\partial_{i}F_{fh}\partial^{i}F^{fh}+2F_{a}{}^{c}F^{ab}\partial_{e}F_{cd}\partial^{e}F_{b}{}^{d}\partial_{i}F_{fh}\partial^{i}F^{fh}$
$\displaystyle-\frac{1}{4}F_{ab}F^{ab}\partial_{e}F_{cd}\partial^{e}F^{cd}\partial_{i}F_{fh}\partial^{i}F^{fh}.$

Because of the presence of the nonlinear $SL(2,R)$ invariant structures in the
final S-dual form of the action (37), one can derive new couplings at the
order of N gauge fields with four derivatives by expanding the invariant
structures. We see that the consistency of the action (4) with the S-duality
predicts the couplings (37) and the couplings at the level of more gauge
fields in the Einstein frame. It would be interesting to confirm these
couplings by direct calculations555The S-matrix elements may be constrained by
both manifest Lorentz and duality invariance. Considering the duality
invariance, as we did in this paper, may be interpreted as an intermediate
step towards an analysis of the scattering amplitude [15]..

Acknowledgments:

The authors would like to kindly thank M.R.Garousi and A.A.Tseytlin for useful
comments and discussions on related topics.

Apendix

In this appendix we are going to find all identities that should be taken in
to account when one wants to write the action (4) in terms of $SL(2,R)$
invariant structures $WW$. In fact, the terms containing four dual gauge
fields and four derivatives should be written in terms of four gauge fields
and four derivatives.

Let us begin with the simple case, where we have four dual gauge fields with
zero derivative. There are two different contracted forms for these fields as
$\tilde{F}_{a}{}^{c}\tilde{F}^{ab}\tilde{F}_{b}{}^{d}\tilde{F}_{cd}$ and
$\tilde{F}_{ab}\tilde{F}^{ab}\tilde{F}_{cd}\tilde{F}^{cd}$. Using the identity
(25) one can write any of these terms in terms of four gauge fields
contraction
$(\tilde{F_{1}}\tilde{F_{2}}\tilde{F_{3}}\tilde{F_{4}}\mathop{\longrightarrow}\epsilon_{1}\epsilon_{2}\epsilon_{3}\epsilon_{4}F_{1}F_{2}F_{3}F_{4})$.
The result in this case is not dependent on the Levi-Civita paring choices
${\cal E},{\cal E}^{\prime}$ and ${\cal E}^{\prime\prime}$.

$\displaystyle{\tilde{F}}_{a}{}^{c}{\tilde{F}}^{ab}{\tilde{F}}_{b}{}^{d}{\tilde{F}}_{cd}\mathop{\longrightarrow}{F}_{a}{}^{c}{F}^{ab}{F}_{b}{}^{d}{F}_{cd}\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,\,\,\,\,\,{\tilde{F}}_{ab}{\tilde{F}}^{ab}{\tilde{F}}_{cd}{\tilde{F}}^{cd}\mathop{\longrightarrow}F_{ab}F^{ab}F_{cd}F^{cd}.$

In the next case where we have four dual gauge fields and two derivatives, the
result is dependent on the Levi-Civita paring choices. Consider all
contraction of this case with arbitrary coefficients $y_{i}$ as following:

$\displaystyle
y_{1}\tilde{F}^{ab}\tilde{F}^{cd}\partial_{b}\tilde{F}_{de}\partial_{c}\tilde{F}_{a}{}^{e}+y_{2}\tilde{F}_{a}{}^{c}\tilde{F}^{ab}\partial_{b}\tilde{F}^{de}\partial_{c}\tilde{F}_{de}+y_{3}\tilde{F}^{ab}\tilde{F}^{cd}\partial_{c}\tilde{F}_{a}{}^{e}\partial_{d}\tilde{F}_{be}$
$\displaystyle+y_{4}\tilde{F}^{ab}\tilde{F}^{cd}\partial_{b}\tilde{F}_{a}{}^{e}\partial_{d}\tilde{F}_{ce}+y_{5}\tilde{F}_{a}{}^{c}\tilde{F}^{ab}\partial_{d}\tilde{F}_{b}{}^{d}\partial_{e}\tilde{F}_{c}{}^{e}+y_{6}\tilde{F}^{ab}\tilde{F}^{cd}\partial_{b}\tilde{F}_{ac}\partial_{e}\tilde{F}_{d}{}^{e}$
$\displaystyle+y_{7}\tilde{F}^{ab}\tilde{F}^{cd}\partial_{c}\tilde{F}_{ab}\partial_{e}\tilde{F}_{d}{}^{e}+y_{8}\tilde{F}_{a}{}^{c}\tilde{F}^{ab}\partial_{c}\tilde{F}_{b}{}^{d}\partial_{e}\tilde{F}_{d}{}^{e}+y_{9}\tilde{F}_{ab}\tilde{F}^{ab}\partial_{c}\tilde{F}^{cd}\partial_{e}\tilde{F}_{d}{}^{e}$
$\displaystyle+y_{10}\tilde{F}^{ab}\tilde{F}^{cd}\partial_{d}\tilde{F}_{ce}\partial^{e}\tilde{F}_{ab}+y_{11}\tilde{F}^{ab}\tilde{F}^{cd}\partial_{e}\tilde{F}_{cd}\partial^{e}\tilde{F}_{ab}+y_{12}\tilde{F}^{ab}\tilde{F}^{cd}\partial_{d}\tilde{F}_{be}\partial^{e}\tilde{F}_{ac}$
$\displaystyle+y_{13}\tilde{F}^{ab}\tilde{F}^{cd}\partial_{e}\tilde{F}_{bd}\partial^{e}\tilde{F}_{ac}+y_{14}\tilde{F}_{a}{}^{c}\tilde{F}^{ab}\partial_{c}\tilde{F}_{de}\partial^{e}\tilde{F}_{b}{}^{d}+y_{15}\tilde{F}_{a}{}^{c}\tilde{F}^{ab}\partial_{d}\tilde{F}_{ce}\partial^{e}\tilde{F}_{b}{}^{d}$
$\displaystyle+y_{16}\tilde{F}_{a}{}^{c}\tilde{F}^{ab}\partial_{e}\tilde{F}_{cd}\partial^{e}\tilde{F}_{b}{}^{d}+y_{17}\tilde{F}_{ab}\tilde{F}^{ab}\partial_{d}\tilde{F}_{ce}\partial^{e}\tilde{F}^{cd}+y_{18}\tilde{F}_{ab}\tilde{F}^{ab}\partial_{e}\tilde{F}_{cd}\partial^{e}\tilde{F}^{cd}.$
(39)

By selecting the Levi-Civita paring choices ${\cal E},{\cal E}^{\prime}$ and
${\cal E}^{\prime\prime}$, one can find the first term of above dual gauge
content in terms of the following different gauge fields combination,
respectively:

$\displaystyle-\frac{1}{4}F^{bc}F^{de}\partial_{a}F_{de}\partial^{a}F_{bc}+F^{bc}F^{de}\partial^{a}F_{bc}\partial_{e}F_{da}-F_{b}{}^{d}F^{bc}\partial^{a}F_{c}{}^{e}\partial_{e}F_{da}=Y_{1},$
$\displaystyle-2F^{ab}F^{cd}\partial_{b}F_{ac}\partial_{e}F_{d}{}^{e}+F^{ab}F^{cd}\partial_{c}F_{ab}\partial_{e}F_{d}{}^{e}+\frac{1}{4}F^{ab}F^{cd}\partial_{e}F_{cd}\partial^{e}F_{ab}$
$\displaystyle+2F^{ab}F^{cd}\partial_{d}F_{be}\partial^{e}F_{ac}-F^{ab}F^{cd}\partial_{e}F_{bd}\partial^{e}F_{ac}-F_{a}{}^{c}F^{ab}\partial_{d}F_{ce}\partial^{e}F_{b}{}^{d}=Y_{1}^{{}^{\prime}},$
$\displaystyle-\frac{1}{2}F_{a}{}^{c}F^{ab}\partial_{b}F^{ed}\partial_{c}F_{ed}-2F_{a}{}^{c}F^{ab}\partial_{c}F_{ed}\partial^{d}F_{b}{}^{e}-F_{a}{}^{c}F^{ab}\partial_{d}F_{ce}\partial^{d}F_{b}{}^{e}$
$\displaystyle+\frac{1}{4}F_{ab}F^{ab}\partial_{d}F_{ce}\partial^{d}F^{ce}+F^{ab}F^{ce}\partial_{b}F_{a}{}^{d}\partial_{e}F_{cd}-\frac{1}{2}F_{ab}F^{ab}\partial^{d}F^{ce}\partial_{e}F_{cd}=Y_{1}^{{}^{\prime\prime}}.$
(40)

As $Y_{1},Y^{\prime}_{1}$ and $Y^{{}^{\prime\prime}}_{1}$ are the expressions
for a term containing dual gauge fields in terms of the gauge fields, they
should be equal to each other. From
$Y_{1}=Y^{\prime}_{1}=Y^{{}^{\prime\prime}}_{1}$, one can find the following
independent identities

$\displaystyle 0$ $\displaystyle=$ $\displaystyle
4F^{ab}F^{cd}\partial_{b}F_{ac}\partial_{e}F_{d}{}^{e}-2F^{ab}F^{cd}\partial_{c}F_{ab}\partial_{e}F_{d}{}^{e}+2F^{ab}F^{cd}\partial_{d}F_{ce}\partial^{e}F_{ab}$
$\displaystyle-F^{ab}F^{cd}\partial_{e}F_{cd}\partial^{e}F_{ab}-4F^{ab}F^{cd}\partial_{d}F_{be}\partial^{e}F_{ac}+2F^{ab}F^{cd}\partial_{e}F_{bd}\partial^{e}F_{ac}\,\,,$
$\displaystyle 0$ $\displaystyle=$ $\displaystyle
2F_{a}{}^{c}F^{ab}\partial_{b}F^{de}\partial_{c}F_{de}-4F^{ab}F^{cd}\partial_{b}F_{a}{}^{e}\partial_{d}F_{ce}+4F^{ab}F^{cd}\partial_{d}F_{ce}\partial^{e}F_{ab}$
(41)
$\displaystyle-F^{ab}F^{cd}\partial_{e}F_{cd}\partial^{e}F_{ab}+8F_{a}{}^{c}F^{ab}\partial_{c}F_{de}\partial^{e}F_{b}{}^{d}-4F_{a}{}^{c}F^{ab}\partial_{d}F_{ce}\partial^{e}F_{b}{}^{d}$
$\displaystyle+4F_{a}{}^{c}F^{ab}\partial_{e}F_{cd}\partial^{e}F_{b}{}^{d}+2F_{ab}F^{ab}\partial_{d}F_{ce}\partial^{e}F^{cd}-F_{ab}F^{ab}\partial_{e}F_{cd}\partial^{e}F^{cd}.$

To write the second term in (39) in terms of the gauge fields, one finds three
expressions $Y_{2},Y^{\prime}_{2}$ and $Y^{{}^{\prime\prime}}_{2}$ according
to the Levi-Civita paring choices ${\cal E},{\cal E}^{\prime}$ and ${\cal
E}^{\prime\prime}$. Considering the relation
$Y_{2}=Y^{\prime}_{2}=Y^{{}^{\prime\prime}}_{2}$ and the identities (41), the
another independent identity appears as following:

$\displaystyle 0$ $\displaystyle=$ $\displaystyle
4F^{ab}F^{cd}\partial_{b}F_{de}\partial_{c}F_{a}{}^{e}+2F_{a}{}^{c}F^{ab}\partial_{b}F^{de}\partial_{c}F_{de}-4F_{a}{}^{c}F^{ab}\partial_{d}F_{b}{}^{d}\partial_{e}F_{c}{}^{e}$
(42)
$\displaystyle-8F^{ab}F^{cd}\partial_{b}F_{ac}\partial_{e}F_{d}{}^{e}+8F_{a}{}^{c}F^{ab}\partial_{c}F_{b}{}^{d}\partial_{e}F_{d}{}^{e}-2F_{ab}F^{ab}\partial_{c}F^{cd}\partial_{e}F_{d}{}^{e}$
$\displaystyle-4F^{ab}F^{cd}\partial_{d}F_{ce}\partial^{e}F_{ab}+F^{ab}F^{cd}\partial_{e}F_{cd}\partial^{e}F_{ab}+8F^{ab}F^{cd}\partial_{d}F_{be}\partial^{e}F_{ac}$
$\displaystyle-4F^{ab}F^{cd}\partial_{e}F_{bd}\partial^{e}F_{ac}+4F_{a}{}^{c}F^{ab}\partial_{e}F_{cd}\partial^{e}F_{b}{}^{d}-F_{ab}F^{ab}\partial_{e}F_{cd}\partial^{e}F^{cd}.$

Doing the same procedure for the third term in (39) led to the following
identity that is independent of previous identities.

$\displaystyle 0$ $\displaystyle=$ $\displaystyle
2F^{ac}F^{de}\partial_{d}F_{a}{}^{b}\partial_{e}F_{cb}-2F_{a}{}^{d}F^{ac}\partial_{e}F_{c}{}^{e}\partial_{b}F_{d}{}^{b}-2F^{ac}F^{de}\partial_{c}F_{ad}\partial_{b}F_{e}{}^{b}$
(43)
$\displaystyle-F^{ac}F^{de}\partial_{d}F_{ac}\partial_{b}F_{e}{}^{b}+4F_{a}{}^{d}F^{ac}\partial_{d}F_{c}{}^{e}\partial_{b}F_{e}{}^{b}-F_{ac}F^{ac}\partial_{d}F^{de}\partial_{b}F_{e}{}^{b}$
$\displaystyle-F^{ac}F^{de}\partial_{e}F_{db}\partial^{b}F_{ac}-2F^{ac}F^{de}\partial_{e}F_{cb}\partial^{b}F_{ad}-4F_{a}{}^{d}F^{ac}\partial_{d}F_{eb}\partial^{b}F_{c}{}^{e}$
$\displaystyle+2F_{a}{}^{d}F^{ac}\partial_{e}F_{db}\partial^{b}F_{c}{}^{e}-F_{ac}F^{ac}\partial_{e}F_{db}\partial^{b}F^{de}.$

After finding the other terms in (39) in terms of the gauge fields
($Y_{i},Y^{{}^{\prime}}_{i}$ and $Y^{{}^{\prime\prime}}_{i}$ where
$i=3,\cdots,18$), one finds that the relations
$Y_{i}=Y^{{}^{\prime}}_{i}=Y^{{}^{\prime\prime}}_{i}$ do not produce any new
independent indentity.

Let us now consider the third case where we have four gauge fields and four
derivatives. One can find all contracted form of four dual gauge fields with
four derivatives by using ”xAct” as following:

$\displaystyle
z_{1}\partial_{c}\tilde{F}^{ef}\partial^{c}\tilde{F}^{ab}\partial_{d}\tilde{F}_{ef}\partial^{d}\tilde{F}_{ab}+z_{2}\partial_{a}\tilde{F}^{ab}\partial_{c}\tilde{F}^{ef}\partial_{d}\tilde{F}_{ef}\partial^{d}\tilde{F}_{b}{}^{c}+z_{3}\partial_{b}\tilde{F}_{ef}\partial^{c}\tilde{F}^{ab}\partial_{d}\tilde{F}_{c}{}^{f}\partial^{e}\tilde{F}_{a}{}^{d}$
$\displaystyle+z_{4}\partial_{b}\tilde{F}_{c}{}^{f}\partial^{c}\tilde{F}^{ab}\partial_{d}\tilde{F}_{ef}\partial^{e}\tilde{F}_{a}{}^{d}+z_{5}\partial_{c}\tilde{F}_{df}\partial^{c}\tilde{F}^{ab}\partial_{e}\tilde{F}_{b}{}^{f}\partial^{e}\tilde{F}_{a}{}^{d}+z_{6}\partial_{a}\tilde{F}^{ab}\partial_{c}\tilde{F}_{b}{}^{c}\partial_{d}\tilde{F}^{de}\partial_{f}\tilde{F}_{e}{}^{f}$
$\displaystyle+z_{7}\partial_{a}\tilde{F}^{ab}\partial_{c}\tilde{F}_{d}{}^{e}\partial^{d}\tilde{F}_{b}{}^{c}\partial_{f}\tilde{F}_{e}{}^{f}+z_{8}\partial_{a}\tilde{F}^{ab}\partial_{d}\tilde{F}_{c}{}^{e}\partial^{d}\tilde{F}_{b}{}^{c}\partial_{f}\tilde{F}_{e}{}^{f}+z_{9}\partial_{a}\tilde{F}^{ab}\partial_{b}\tilde{F}^{cd}\partial^{e}\tilde{F}_{cd}\partial_{f}\tilde{F}_{e}{}^{f}$
$\displaystyle+z_{10}\partial_{a}\tilde{F}^{ab}\partial^{d}\tilde{F}_{b}{}^{c}\partial^{e}\tilde{F}_{cd}\partial_{f}\tilde{F}_{e}{}^{f}+z_{11}\partial^{c}\tilde{F}^{ab}\partial^{e}\tilde{F}_{a}{}^{d}\partial_{f}\tilde{F}_{ce}\partial^{f}\tilde{F}_{bd}+z_{12}\partial_{c}\tilde{F}_{df}\partial^{c}\tilde{F}^{ab}\partial^{e}\tilde{F}_{a}{}^{d}\partial^{f}\tilde{F}_{be}$
$\displaystyle+z_{13}\partial^{c}\tilde{F}^{ab}\partial_{d}\tilde{F}_{cf}\partial^{e}\tilde{F}_{a}{}^{d}\partial^{f}\tilde{F}_{be}+z_{14}\partial^{c}\tilde{F}^{ab}\partial^{e}\tilde{F}_{a}{}^{d}\partial_{f}\tilde{F}_{cd}\partial^{f}\tilde{F}_{be}+z_{15}\partial^{c}\tilde{F}^{ab}\partial_{d}\tilde{F}_{ef}\partial^{d}\tilde{F}_{ac}\partial^{f}\tilde{F}_{b}{}^{e}$
$\displaystyle+z_{16}\partial_{c}\tilde{F}_{a}{}^{d}\partial^{c}\tilde{F}^{ab}\partial_{e}\tilde{F}_{df}\partial^{f}\tilde{F}_{b}{}^{e}+z_{17}\partial^{c}\tilde{F}^{ab}\partial^{d}\tilde{F}_{ac}\partial_{e}\tilde{F}_{df}\partial^{f}\tilde{F}_{b}{}^{e}+z_{18}\partial_{c}\tilde{F}_{a}{}^{d}\partial^{c}\tilde{F}^{ab}\partial_{f}\tilde{F}_{de}\partial^{f}\tilde{F}_{b}{}^{e}$
$\displaystyle+z_{19}\partial^{c}\tilde{F}^{ab}\partial^{d}\tilde{F}_{ac}\partial_{f}\tilde{F}_{de}\partial^{f}\tilde{F}_{b}{}^{e}+z_{20}\partial_{a}\tilde{F}^{ab}\partial_{b}\tilde{F}^{cd}\partial_{d}\tilde{F}_{ef}\partial^{f}\tilde{F}_{c}{}^{e}+z_{21}\partial^{c}\tilde{F}^{ab}\partial_{d}\tilde{F}_{ef}\partial^{d}\tilde{F}_{ab}\partial^{f}\tilde{F}_{c}{}^{e}$
$\displaystyle+z_{22}\partial_{a}\tilde{F}^{ab}\partial_{d}\tilde{F}_{ef}\partial^{d}\tilde{F}_{b}{}^{c}\partial^{f}\tilde{F}_{c}{}^{e}+z_{23}\partial^{c}\tilde{F}^{ab}\partial^{d}\tilde{F}_{ab}\partial_{e}\tilde{F}_{df}\partial^{f}\tilde{F}_{c}{}^{e}+z_{24}\partial_{a}\tilde{F}^{ab}\partial^{d}\tilde{F}_{b}{}^{c}\partial_{e}\tilde{F}_{df}\partial^{f}\tilde{F}_{c}{}^{e}$
$\displaystyle+z_{25}\partial^{c}\tilde{F}^{ab}\partial^{d}\tilde{F}_{ab}\partial_{f}\tilde{F}_{de}\partial^{f}\tilde{F}_{c}{}^{e}+z_{26}\partial_{a}\tilde{F}^{ab}\partial^{d}\tilde{F}_{b}{}^{c}\partial_{f}\tilde{F}_{de}\partial^{f}\tilde{F}_{c}{}^{e}+z_{27}\partial_{b}\tilde{F}_{ef}\partial^{c}\tilde{F}^{ab}\partial^{d}\tilde{F}_{ac}\partial^{f}\tilde{F}_{d}{}^{e}$
$\displaystyle+z_{28}\partial_{a}\tilde{F}^{ab}\partial_{c}\tilde{F}_{ef}\partial^{d}\tilde{F}_{b}{}^{c}\partial^{f}\tilde{F}_{d}{}^{e}+z_{29}\partial_{a}\tilde{F}^{ab}\partial_{c}\tilde{F}_{b}{}^{c}\partial_{e}\tilde{F}_{df}\partial^{f}\tilde{F}^{de}+z_{30}\partial_{b}\tilde{F}_{ac}\partial^{c}\tilde{F}^{ab}\partial_{e}\tilde{F}_{df}\partial^{f}\tilde{F}^{de}$
$\displaystyle+z_{31}\partial_{c}\tilde{F}_{ab}\partial^{c}\tilde{F}^{ab}\partial_{e}\tilde{F}_{df}\partial^{f}\tilde{F}^{de}+z_{32}\partial_{a}\tilde{F}^{ab}\partial_{c}\tilde{F}_{b}{}^{c}\partial_{f}\tilde{F}_{de}\partial^{f}\tilde{F}^{de}+z_{33}\partial_{c}\tilde{F}_{ab}\partial^{c}\tilde{F}^{ab}\partial_{f}\tilde{F}_{de}\partial^{f}\tilde{F}^{de},$

where $z_{i}$ are some unknown coefficients.

As the previous cases, we do the same procedure to find the above dual gauge
fields in terms of relevant gauge fields. Different choices of Levi-Civita
paring led to different results that shoud be equal to each other
$Z_{i}=Z^{{}^{\prime}}_{i}=Z^{{}^{\prime\prime}}_{i}$ and $i=1,\cdots,33$.
These equalities led to ten independent identities which we present three of
them in the following. To investigate the S-duality behavior of amplitude and
actions at the presence of a $D$-brane at the level of four gauge fields and
four(/N) derivatives, one has to use these identities as we have done to write
the action (4) in terms of the S-dual invariant structures (37).

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\partial_{c}F^{ef}\partial^{c}F^{ab}\partial_{d}F_{ef}\partial^{d}F_{ab}-2\partial_{c}F_{df}\partial^{c}F^{ab}\partial_{e}F_{b}{}^{f}\partial^{e}F_{a}{}^{d}-4\partial^{c}F^{ab}\partial^{e}F_{a}{}^{d}\partial_{f}F_{ce}\partial^{f}F_{bd}$
$\displaystyle+8\partial_{c}F_{df}\partial^{c}F^{ab}\partial^{e}F_{a}{}^{d}\partial^{f}F_{be}+4\partial^{c}F^{ab}\partial^{e}F_{a}{}^{d}\partial_{f}F_{cd}\partial^{f}F_{be}-4\partial^{c}F^{ab}\partial_{d}F_{ef}\partial^{d}F_{ac}\partial^{f}F_{b}{}^{e}$
$\displaystyle+4\partial^{c}F^{ab}\partial_{d}F_{ef}\partial^{d}F_{ab}\partial^{f}F_{c}{}^{e}\,\,\,,$
$\displaystyle 0$ $\displaystyle=$
$\displaystyle\partial_{c}F^{ef}\partial^{c}F^{ab}\partial_{d}F_{ef}\partial^{d}F_{ab}-4\partial^{c}F^{ab}\partial_{d}F_{ef}\partial^{d}F_{ac}\partial^{f}F_{b}{}^{e}+4\partial_{c}F_{a}{}^{d}\partial^{c}F^{ab}\partial_{e}F_{df}\partial^{f}F_{b}{}^{e}$
$\displaystyle-4\partial_{c}F_{a}{}^{d}\partial^{c}F^{ab}\partial_{f}F_{de}\partial^{f}F_{b}{}^{e}+8\partial^{c}F^{ab}\partial^{d}F_{ac}\partial_{f}F_{de}\partial^{f}F_{b}{}^{e}+4\partial^{c}F^{ab}\partial_{d}F_{ef}\partial^{d}F_{ab}\partial^{f}F_{c}{}^{e}$
$\displaystyle-2\partial^{c}F^{ab}\partial^{d}F_{ab}\partial_{f}F_{de}\partial^{f}F_{c}{}^{e}-2\partial_{c}F_{ab}\partial^{c}F^{ab}\partial_{e}F_{df}\partial^{f}F^{de}+\partial_{c}F_{ab}\partial^{c}F^{ab}\partial_{f}F_{de}\partial^{f}F^{de}\,\,\,,$
$\displaystyle 0$ $\displaystyle=$
$\displaystyle\partial_{c}F^{ef}\partial^{c}F^{ab}\partial_{d}F_{ef}\partial^{d}F_{ab}+4\partial_{a}F^{ab}\partial_{c}F^{ef}\partial_{d}F_{ef}\partial^{d}F_{b}{}^{c}-4\partial_{c}F_{df}\partial^{c}F^{ab}\partial_{e}F_{b}{}^{f}\partial^{e}F_{a}{}^{d}$
$\displaystyle+8\partial_{c}F_{df}\partial^{c}F^{ab}\partial^{e}F_{a}{}^{d}\partial^{f}F_{be}+4\partial^{c}F^{ab}\partial_{d}F_{ef}\partial^{d}F_{ac}\partial^{f}F_{b}{}^{e}-4\partial_{c}F_{a}{}^{d}\partial^{c}F^{ab}\partial_{e}F_{df}\partial^{f}F_{b}{}^{e}$
$\displaystyle+4\partial_{c}F_{a}{}^{d}\partial^{c}F^{ab}\partial_{f}F_{de}\partial^{f}F_{b}{}^{e}-8\partial^{c}F^{ab}\partial^{d}F_{ac}\partial_{f}F_{de}\partial^{f}F_{b}{}^{e}+8\partial_{a}F^{ab}\partial_{d}F_{ef}\partial^{d}F_{b}{}^{c}\partial^{f}F_{c}{}^{e}\
$
$\displaystyle+2\partial^{c}F^{ab}\partial^{d}F_{ab}\partial_{f}F_{de}\partial^{f}F_{c}{}^{e}+2\partial_{c}F_{ab}\partial^{c}F^{ab}\partial_{e}F_{df}\partial^{f}F^{de}-\partial_{c}F_{ab}\partial^{c}F^{ab}\partial_{f}F_{de}\partial^{f}F^{de}\,\,\,.$

These identities and the other ones that appear in this section could be
checked explicitly by replacing $F_{ab}=\partial_{a}A_{b}-\partial_{b}A_{a}$
(or by substituting the component of $F_{ab}$).

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