aainstitutetext: Center for Gravitation and Cosmology, College of Physical
Science and Technology, Yangzhou University,
No.180, Siwangting Road, Yangzhou, 225009, P.R. China.bbinstitutetext: Center
for Gravitation and Cosmology, College of Physical Science and Technology,
Yangzhou University,
No.180, Siwangting Road, Yangzhou, 225009, P.R. China.

# Note on tree NLSM amplitudes and soft theorems

Kang Zhou b Fang-Stars Wei<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

We use the universality of single soft behavior, together with the double copy
structure, to completely determine the tree amplitudes of non-linear sigma
model (NLSM). We first figure out the Adler’s zero for $4$-point NLSM
amplitudes, by considering kinematics. Then, we assume the universality of the
Adler’s zero, and use this requirement to construct general tree NLSM
amplitudes in the expanded formula, i.e., the formula of expanding NLSM
amplitudes to bi-adjoint scalar amplitudes. We also derive double soft factors
for tree NLSM amplitudes, based on the resulting expanded formula.

##  1 Introduction

Soft theorems describe the universal behavior of scattering amplitudes when
one or more external massless momenta are taken to near zero. This limit can
be achieved by re-scaling the massless momenta via a soft parameter as
$k^{\mu}\to\tau k^{\mu}$, and taking the limit $\tau\to 0$. Soft theorems then
state the factorization of amplitudes. For instance, when one of external
gravitons is taken to be soft, the $(n+1)$-point GR amplitude factorizes as
Cachazo:2014fwa ; Schwab:2014xua ; Afkhami-Jeddi:2014fia

$\displaystyle{\cal
A}_{n+1}\,\to\,\Big{(}\tau^{-1}\,S^{(0)}_{h}+\tau^{0}\,S^{(1)}_{h}+\tau\,S^{(2)}_{h}+\cdots\Big{)}\,{\cal
A}_{n}\,,~{}~{}~{}~{}$ (1)

where ${\cal A}_{n}$ is the sub-amplitude of ${\cal A}_{n+1}$, which is
generated from ${\cal A}_{n+1}$ by removing the soft external graviton. The
universal operators $S^{(0)}_{h}$, $S^{(1)}_{h}$, $S^{(2)}_{h}$ are called
soft factors, or soft operators, at leading, sub-leading, and sub-sub-leading
orders.

Soft theorems were exploited in the construction of tree amplitudes, such as
the related on-shell recursion relations, and the inverse soft theorem
program, and so on Cheung:2014dqa ; Luo:2015tat ; Cheung:2018oki ;
Elvang:2018dco ; Cachazo:2016njl ; Rodina:2018pcb ; Boucher-Veronneau:2011rwd
; Nguyen:2009jk . Impressively, in Rodina:2018pcb , it was shown that soft
theorems uniquely fix tree amplitudes. However, the previous constructions
mentioned above require the explicit forms of soft factors. To derive soft
factors, one need a certain expression for amplitudes, such as summations of
contributions from Feynman diagrams Low ; Weinberg , Britto-Cachzo-Feng-Witten
(BCFW) on-shell recursion relations Cachazo:2014fwa ; Casali:2014xpa ;
Britto:2004ap ; Britto:2005fq , Cachazo-He-Yuan (CHY) contour integrals
Schwab:2014xua ; Afkhami-Jeddi:2014fia ; Cachazo:2013gna ; Cachazo:2013hca ;
Cachazo:2013iea ; Cachazo:2014nsa ; Cachazo:2014xea , and so on. Then a
logical confusion is caused, the tree amplitude, and the soft factor, which
one determines another one?

On the other hand, the factorization in (1) has a natural physical picture.
Roughly speaking, in the soft limit the soft particle can be thought as
vanishing, leaving a lower-point amplitude with the soft external leg removed,
and the universal soft factors carried by the soft particle. Thus, to avoid
the logical confusion mentioned previously, it is natural to ask, can we take
the soft behavior in (1) as the principle, and use it to construct tree
amplitudes, without knowing the explicit forms of soft factors? In the recent
work of one of two authors for the present note, it was shown that such
construction can be realized at least for tree amplitudes of Yang-Mills-
scalar, pure Yang-Mills, Einstein-Yang-Mills and pure gravitational theories
Zhou:2022orv . The factorization behaviors, together with assuming the
universality of soft factors, and the double copy structure Kawai:1985xq ;
Bern:2008qj ; Chiodaroli:2014xia ; Johansson:2015oia ; Johansson:2019dnu ,
completely determine tree amplitudes of above theories.

This note is the continuation of the previous work in Zhou:2022orv , this time
applying the similar idea to consider the tree amplitudes of nonlinear sigma
model (NLSM). The tree NLSM amplitudes vanishing in the single soft limit (one
of external leg being soft), known as Adler’s zero Adler . For such case, the
factorization behavior in (1) does not occur. However, one can still talk
about the universality of the single soft behavior. As will be seen in
subsection.3.1 in section.3, it is easy to figure out the Adler’s zero for the
$4$-point tree NLSM amplitude, by considering kinematics. Then, the
universality of soft behavior indicates that all tree NLSM amplitudes vanish
in the single soft limit. The assumption of universality, together with other
physical conditions such as double copy structure and manifest permutation
invariance among external legs, uniquely fix the general tree NLSM amplitudes,
in the expanded formula, i.e., the formula of expanding NLSM amplitudes to bi-
adjoint scalar (BAS) amplitudes. We emphasize that the Adler’s zero was not
assumed at the beginning, it was deduced. Using the resulting expanded
formula, we also derive double soft factors for the tree NLSM amplitudes
Cachazo:2015ksa ; Du:2015esa , which describe the behavior when two external
scalars are taken to be soft simultaneously.

The remainder of this note is organized as follows. In section.2, we give a
quick introduction to the necessary background, including BAS tree amplitudes,
and expansions of other amplitudes to them. In section.3, we use the the
universality of single soft behavior, to construct the tree NLSM amplitudes in
the expanded formula. In section.4, we derive the double soft factor for tree
NLSM amplitudes, based on the expanded formula obtained in section.3. Finally,
we close with a brief summery in section.5.

##  2 Background

In this section we rapidly review the necessary background. In subsection.2.1,
we introduce the tree level amplitudes of bi-adjoint scalar (BAS) theory. Some
notations and technics which will be used in subsequent sections are also
included. In subsection.2.2, we rapidly review the expansions of tree
amplitudes to BAS amplitudes, including the choice of basis, as well as the
double copy structure for coefficients.

###  2.1 Tree level BAS amplitudes

The BAS theory describes the bi-adjoint scalar field $\phi_{a\bar{a}}$ with
the Lagrangian

$\displaystyle{\cal L}_{\rm BAS}={1\over
2}\,\partial_{\mu}\phi^{a\bar{a}}\,\partial^{\mu}\phi^{a\bar{a}}-{\lambda\over
3!}\,f^{abc}f^{\bar{a}\bar{b}\bar{c}}\,\phi^{a\bar{a}}\phi^{b\bar{b}}\phi^{c\bar{c}}\,,~{}~{}$
(2)

where the structure constant $f^{abc}$ and generator $T^{a}$ satisfy

$\displaystyle[T^{a},T^{b}]=if^{abc}T^{c}\,,$ (3)

and the dual color algebra encoded by $f^{\bar{a}\bar{b}\bar{c}}$ and
$T^{\bar{a}}$ is analogous. The tree level amplitudes of this theory contain
only propagators, and can be decomposed into double color ordered partial
amplitudes via the standard technic. Each double color ordered partial
amplitude is simultaneously planar with respect to two color orderings, arise
from expanding the full $n$-point amplitude to $\mathop{\rm
Tr}(T^{a_{\sigma_{1}}}\cdots T^{a_{\sigma_{n}}})$ and $\mathop{\rm
Tr}(T^{\bar{a}_{\bar{\sigma}_{1}}}\cdots T^{\bar{a}_{\bar{\sigma}_{n}}})$
respectively, where $\sigma_{i}$ and $\bar{\sigma}_{i}$ denote permutations
among all external scalars.

To calculate double color ordered partial amplitudes, it is convenient to
employ the diagrammatical method proposed by Cachazo, He and Yuan in
Cachazo:2013iea . For a BAS amplitude whose double color-orderings are given,
this method provides the corresponding Feynman diagrams as well as the overall
sign directly. Let us consider the $5$-point example ${\cal A}_{\rm
S}(1,2,3,4,5|1,4,2,3,5)$. In Figure.1, the first diagram satisfies both two
color orderings $(1,2,3,4,5)$ and $(1,4,2,3,5)$, while the second one
satisfies the ordering $(1,2,3,4,5)$ but not $(1,4,2,3,5)$. Thus, the first
diagram is allowed by the double color orderings $(1,2,3,4,5|1,4,2,3,5)$,
while the second one is not. It is easy to see other diagrams are also
forbidden by the ordering $(1,4,2,3,5)$, thus the first diagram in Figure.1 is
the only diagram contributes to the amplitude ${\cal A}_{\rm
S}(1,2,3,4,5|1,4,2,3,5)$.

Figure 1: Two $5$-point diagrams

The Feynman diagrams contribute to a given BAS amplitude can be obtained via
systematic diagrammatical rules. For the above example, one can draw a disk
diagram as follows.

  * •

Draw points on the boundary of the disk according to the first ordering
$(1,2,3,4,5)$.

  * •

Draw a loop of line segments which connecting the points according to the
second ordering $(1,4,2,3,5)$.

The obtained disk diagram is shown in the first diagram in Figure.2. From the
diagram, one can see that two orderings share the boundaries $\\{1,5\\}$ and
$\\{2,3\\}$. These co-boundaries indicate channels ${1/s_{15}}$ and
${1/s_{23}}$, therefore the first Feynman diagram in Figure.1. Then the BAS
amplitude ${\cal A}_{\rm S}(1,2,3,4,5|1,4,2,3,5)$ can be computed as

$\displaystyle{\cal A}_{\rm S}(1,2,3,4,5|1,4,2,3,5)={1\over s_{23}}{1\over
s_{51}}\,,$ (4)

up to an overall sign. The Mandelstam variable $s_{i\cdots j}$ is defined as

$\displaystyle s_{i\cdots j}\equiv K_{i\cdots j}^{2}\,,~{}~{}~{}~{}K_{i\cdots
j}\equiv\sum_{a=i}^{j}\,k_{a}\,,~{}~{}~{}~{}$ (5)

where $k_{a}$ is the momentum carried by the external leg $a$.

Figure 2: Diagram for ${\cal A}_{S}(1,2,3,4,5|1,4,2,3,5)$ and ${\cal
A}_{S}(1,2,3,4,5|1,2,4,3,5)$.

As another example, let us consider the BAS amplitude ${\cal A}_{\rm
S}(1,2,3,4,5|1,2,4,3,5)$. The corresponding disk diagram is shown in the
second configuration in Figure.2, and one can see two orderings have co-
boundaries $\\{3,4\\}$ and $\\{5,1,2\\}$. The co-boundary $\\{3,4\\}$
indicates the channel ${1/s_{34}}$. The co-boundary $\\{5,1,2\\}$ indicates
the channel ${1/s_{512}}$ which is equivalent to $1/s_{34}$, as well as sub-
channels ${1/s_{12}}$ and ${1/s_{51}}$. Using the above decomposition, one can
calculate ${\cal A}_{\rm S}(1,2,3,4,5|1,2,4,3,5)$ as

$\displaystyle{\cal A}_{\rm S}(1,2,3,4,5|1,2,4,3,5)={1\over
s_{34}}\Big{(}{1\over s_{12}}+{1\over s_{51}}\Big{)}\,,$ (6)

up to an overall sign.

The overall sign, determined by the color algebra, can be fixed by following
rules.

  * •

Each polygon with odd number of vertices contributes a plus sign if its
orientation is the same as that of the disk and a minus sign if opposite.

  * •

Each polygon with even number of vertices always contributes a minus sign.

  * •

Each intersection point contributes a minus sign.

We now apply these rules to previous examples. In the first diagram in
Figure.2, the polygons are three triangles, namely $51A$, $A4B$ and $B23$,
which contribute $+$, $-$, $+$ respectively, while two intersection points $A$
and $B$ contribute two $-$. In the second one in Figure.2, the polygons are
$512A$ and $A43$, which contribute two $-$, while the intersection point $A$
contributes $-$. Then we arrive at the full results

$\displaystyle{\cal A}_{\rm S}(1,2,3,4,5|1,4,2,3,5)$ $\displaystyle=$
$\displaystyle-{1\over s_{23}}{1\over s_{51}}\,,$ $\displaystyle{\cal A}_{\rm
S}(1,2,3,4,5|1,2,4,3,5)$ $\displaystyle=$ $\displaystyle-{1\over
s_{34}}\Big{(}{1\over s_{12}}+{1\over s_{51}}\Big{)}\,.$ (7)

In the reminder of this note, we adopt another convention for the overall
sign. If the line segments form a convex polygon, and the orientation of the
convex polygon is the same as that of the disk, then the overall sign is $+$.
For instance, the disk diagram in Figure.3 indicates the overall sign $+$
under the new convention, while the old convention gives $-$ according to the
square formed by four line segments. Notice that the diagrammatical rules
described previously still give the related sign between different disk
diagrams. For example, two disk diagrams in Figure.2 shows that the relative
sign between ${\cal A}_{\rm S}(1,2,3,4,5|1,4,2,3,5)$ and ${\cal A}_{\rm
S}(1,2,3,4,5|1,2,4,3,5)$ is $+$. The advantage of the new convention is that
when removing a soft external scalar, the resulting sub-amplitude carries the
same sign as the original one.

Figure 3: The overall sign $+$ under the new convention.

When considering the soft limit, the $2$-point channels play the central role.
Since the partial BAS amplitude carries two color orderings, if the $2$-point
channel contributes $1/s_{ab}$ to the amplitude, external legs $a$ and $b$
must be adjacent to each other in both two orderings. Suppose the first color
ordering is $(\cdots,a,b,\cdots)$, then $1/s_{ab}$ is allowed by this
ordering. To denote if it is allowed by another one, we introduce the symbol
$\delta_{ab}$ whose ordering of two subscripts $a$ and $b$ is determined by
the first color ordering111The Kronecker symbol will not appear in this paper,
thus we hope the notation $\delta_{ab}$ will not confuse the readers.. The
value of $\delta_{ab}$ is $\delta_{ab}=1$ if another color ordering is
$(\cdots,a,b,\cdots)$, $\delta_{ab}=-1$ if another color ordering is
$(\cdots,b,a,\cdots)$, due to the ani-symmetry of the structure constant,
i.e., $f^{abc}=-f^{bac}$, and $\delta_{ab}=0$ otherwise. From the definition,
it is straightforward to see $\delta_{ab}=-\delta_{ba}$, and a simple but
useful identity

$\displaystyle\sum_{b\neq a}\,\delta_{ab}=0\,.~{}~{}$ (8)

Before ending this subsection, we discuss the single soft behavior of BAS
amplitudes at the leading order. Consider the double color ordered BAS
amplitude ${\cal A}_{\rm S}(1,\cdots,n|\sigma_{n})$. We re-scale $k_{i}$ as
$k_{i}\to\tau k_{i}$, and expand the amplitude in $\tau$. The leading order
contribution manifestly aries from $2$-point channels $1/s_{1(i+1)}$ and
$1/s_{(i-1)i}$ which provide the $1/\tau$ order contributions, namely,

$\displaystyle{\cal A}^{(0)}_{\rm S}(1,\cdots,n|\sigma_{n})$ $\displaystyle=$
$\displaystyle{1\over\tau}\Big{(}{\delta_{i(i+1)}\over
s_{i(i+1)}}+{\delta_{(i-1)i}\over s_{(i-1)i}}\Big{)}\,{\cal A}_{\rm
S}(1,\cdots,i-1,\not{i},i+1,\cdots,n|\sigma_{n}\setminus i)$ (9)
$\displaystyle=$ $\displaystyle S^{(0)}_{s}(i)\,{\cal A}_{\rm
S}(1,\cdots,i-1,\not{i},i+1,\cdots,n|\sigma_{n}\setminus i)\,,~{}~{}~{}$

where $\not{i}$ stands for removing the leg $i$, $\sigma_{n}\setminus 1$ means
the color ordering generated from $\sigma_{n}$ by eliminating $i$. The leading
soft operator $S^{(0)}_{s}(i)$ for the scalar $i$ is extracted as

$\displaystyle S^{(0)}_{s}(i)={1\over\tau}\,\Big{(}{\delta_{i(i+1)}\over
s_{i(i+1)}}+{\delta_{(i-1)i}\over s_{(i-1)i}}\Big{)}\,,~{}~{}~{}~{}$ (10)

which acts on external scalars which are adjacent to $i$ in two color
orderings.

###  2.2 Expanding tree level amplitudes to BAS basis

Tree level amplitudes which contain only massless particles can be expanded to
double color ordered BAS amplitudes, due to the observation that each Feynman
diagram for pure propagators can be mapped to at least one disk diagram whose
polygons are all triangles. An illustrative example is given in Figure.4.
Since each tree amplitude can be expanded by tree Feynman diagrams, and each
Feynman diagram contributes propagators together with a numerator without any
pole, one can conclude that each tree amplitude can be expanded to double
color ordered BAS amplitudes, with coefficients which are polynomials depend
on Lorentz invariants created by external kinematical variables. To realize
the expansion, one need to find the basis consists of BAS amplitudes. Such
basis can be determined by the well known Kleiss-Kuijf (KK) relation
Kleiss:1988ne

$\displaystyle{\cal A}_{\rm
S}(1,\vec{\boldsymbol{{\alpha}}},n,\vec{\boldsymbol{{\beta}}}|\sigma_{n})=(-)^{|\boldsymbol{{\alpha}}|}\,{\cal
A}_{\rm
S}(1,\vec{\boldsymbol{{\alpha}}}\shuffle\vec{\boldsymbol{{\beta}}}^{T},n|\sigma_{n})\,.~{}~{}~{}$
(11)

Here $\vec{\boldsymbol{{\alpha}}}$ and $\vec{\boldsymbol{{\beta}}}$ are two
ordered subsets of external scalars, and $\vec{\boldsymbol{{\beta}}}^{T}$
stands for the ordered set generated from $\vec{\boldsymbol{{\beta}}}$ by
reversing the original ordering. The $n$-point BAS amplitude ${\cal A}_{\rm
S}(1,\vec{\boldsymbol{{\alpha}}},n,\vec{\boldsymbol{{\beta}}}|\sigma_{n})$ at
the l.h.s of (11) carries two color orderings, one is
$(1,\vec{\boldsymbol{{\alpha}}},n,\vec{\boldsymbol{{\beta}}})$, another one is
denoted by $\sigma_{n}$. The symbol $\shuffle$ means summing over all possible
shuffles of two ordered sets $\vec{\boldsymbol{{\beta}}}_{1}$ and
$\vec{\boldsymbol{{\beta}}}_{2}$, i.e., all permutations in the set
$\vec{\boldsymbol{{\beta}}}_{1}\cup\vec{\boldsymbol{{\beta}}}_{2}$ while
preserving the orderings of $\vec{\boldsymbol{{\beta}}}_{1}$ and
$\vec{\boldsymbol{{\beta}}}_{2}$. For instance, suppose
$\vec{\boldsymbol{{\beta}}}_{1}=\\{1,2\\}$ and
$\vec{\boldsymbol{{\beta}}}_{2}=\\{3,4\\}$, then

$\displaystyle\vec{\boldsymbol{{\beta}}}_{1}\shuffle\vec{\boldsymbol{{\beta}}}_{2}=\\{1,2,3,4\\}+\\{1,3,2,4\\}+\\{1,3,4,2\\}+\\{3,1,2,4\\}+\\{3,1,4,2\\}+\\{3,4,1,2\\}\,.~{}~{}~{}~{}$
(12)

The analogous KK relation holds for another color ordering $\sigma_{n}$. The
KK relation implies that different double color ordered BAS amplitudes are not
independent, thus the basis can be chosen as BAS amplitudes ${\cal A}_{\rm
S}(1,\sigma_{1},n|1,\sigma_{2},n)$, with $1$ and $n$ are fixed at two ends in
each color ordering. We call such basis the KK BAS basis. Based on the
discussion above, the KK BAS basis can provide any structure of massless
propagators, thus any amplitude which includes only massless particles can be
expanded to this basis222The well known Bern-Carrasco-Johansson (BCJ) relation
Bern:2008qj ; Chiodaroli:2014xia ; Johansson:2015oia ; Johansson:2019dnu
links BAS amplitudes in the KK basis together, and the independent BAS
amplitudes can be obtained by fixing three legs at three particular positions
in the color orderings. However, in the BCJ relation, coefficients of BAS
amplitudes depend on Mandelstam variables, this character leads to poles in
coefficients when expanding to BCJ basis. On the other hand, when expanding to
KK basis, one can find the expanded formula in which the coefficients contain
no pole. In this paper, we choose the KK basis since we hope all poles of tree
amplitudes are included in basis, and coefficients only serve as numerators..
In other words, the basis provides propagators, and the coefficients in
expansions provide numerators. From this point of view, one can regard the BAS
KK basis as the complete set of different structures of propagators, and
forget the corresponding Lagrangian in (2).

Figure 4: Map between Feynman diagram and disk diagram.

In this note, we will consider the expansion of NLSM amplitudes. The color
ordered NLSM amplitude ${\cal A}_{\rm N}(1,\sigma_{n-2},n)$ can be expanded to
KK BAS basis as

$\displaystyle{\cal A}_{\rm
N}(1,\sigma_{n-2},n)=\sum_{\sigma^{\prime}_{n-2}}\,{\cal
C}(\sigma^{\prime}_{n-2},k_{i})\,{\cal A}_{\rm
S}(1,\sigma^{\prime}_{n-2},n|1,\sigma_{n-2},n)\,,~{}~{}~{}$ (13)

where $\sigma_{n-2}$ and $\sigma^{\prime}_{n-2}$ are permutations among
$(n-2)$ external legs in $\\{2,\cdots,n-1\\}$. The double copy structure
Kawai:1985xq ; Bern:2008qj ; Chiodaroli:2014xia ; Johansson:2015oia ;
Johansson:2019dnu  indicates that the coefficient ${\cal
C}(\sigma^{\prime}_{n-2},k_{i})$ depends on momenta $k_{i}$ carried by
external scalars, permutations $\sigma^{\prime}_{n-2}$, but is independent of
the permutation $\sigma_{n-2}$333Originally, the double copy means the GR
amplitude can be factorized as ${\cal A}_{G}={\cal A}_{Y}\times{\cal
S}\times{\cal A}_{Y}$, where the kernel ${\cal S}$ is obtained by inverting
propagators. Our assumption that the coefficients depend on only one color
ordering is equivalent to the original version, see in Zhou:2022orv .. Thus,
suppose we replace $(1,\sigma_{n-2},n)$ by the more general ordering
$\sigma_{n}$ among all external legs, without fixing $1$ and $n$ at any
position, the expansion in (13) still holds. The coefficients ${\cal
C}(\sigma^{\prime}_{n-2},k_{i})$ will be constructed in the next section.

##  3 Expanded NLSM amplitudes

Tree amplitudes can be expressed in various manners, the expansions of tree
amplitudes to KK BAS basis discussed in subsection.2.2 provide one of them. To
determine the tree NLSM amplitudes, it is sufficient to fix coefficients
${\cal C}(\sigma^{\prime}_{n-2},k_{i})$ in (13). This is the goal of the
present section.

The standard NLSM Lagrangian in the Cayley parametrization is given as

$\displaystyle{\cal L}_{\rm N}={1\over 8\lambda^{2}}\,{\rm
Tr}(\partial_{\mu}{\rm U}^{\dagger}\partial^{\mu}{\rm U})\,,~{}~{}$ (14)

with

$\displaystyle{\rm
U}=(\mathbb{I}+\lambda\Phi)\,(\mathbb{I}-\lambda\Phi)^{-1}\,,$ (15)

where $\mathbb{I}$ is the identity matrix, and $\Phi=\phi_{I}T^{I}$, with
$T^{I}$ the generators of $U(N)$. Fields $\phi_{I}$ describe massless scalars,
and the accompanied generators $T^{I}$ indicates the color ordering for the
corresponding partial tree amplitudes. From the Lagrangian in (14), we see the
mass dimension of coupling constant $\lambda$ is $(2-d)/2$, in $d$ dimensional
space-time. The mass dimension of the $n$-point amplitude is $d-{d-2\over
2}n$, the coupling constants contribute $(2-d)(n-2)/2$, thus the kinematic
part must have mass dimension $2$.

Thus, our purpose is to seek the $n$-point amplitude (kinematic part) ${\cal
A}_{\rm N}(\sigma_{n})$ for pure massless scalars which carry the color
ordering $\sigma_{n}$ among $n$ external scalars, and has mass dimension $2$.
As will be seen, the requirements mentioned above, together with the
universality of soft behavior, and the permutation symmetry among external
scalars, are sufficient to determine the tree NLSM amplitudes completely,
without using any other information. In this sense, in the reminder of this
section one can forget the traditional Lagrangian and Feynman rules, only
concentrate on the color ordering and mass dimension. Since the propagators
contribute the mass dimension $-2(n-3)$, the mass dimension of numerators can
be fixed as $2(n-2)$. This is the mass dimension of coefficients ${\cal
C}(\sigma^{\prime}_{n-2},k_{i})$ in the expansion (13), which plays the
important role in the subsequent subsections.

We will argue that the $3$-point NLSM tree amplitude does not exist, while the
$4$-point one has the vanishing single soft behavior at the $\tau^{0}$ order,
via the general consideration of Mandelstam variables. Such general discussion
also yields the statement that the un-vanishing tree NLSM amplitudes only have
the even number of external legs. Then, by imposing the universality of single
soft behavior observed from the $4$-point case, we will construct the general
${\cal C}(\sigma^{\prime}_{n-2},k_{i})$ for $n\geqq 6$. The whole process only
uses the mass dimension, the universality of soft behavior, as well as the
permutation invariance among external scalars.

###  3.1 $4$-point NLSM amplitude

For the $3$-point case, the numerator has mass dimension $2$. However, one can
never use three on-shell massless momenta satisfying momentum conservation to
construct any un-vanishing Lorentz invariant with mass dimension $2$. Thus,
the $3$-point NLSM amplitude does not exist.

The simplest NLSM amplitude is the $4$-point one ${\cal A}_{\rm
N}(\sigma_{4})$. It can be expanded to KK BAS basis, namely,

$\displaystyle{\cal A}_{\rm N}(\sigma_{4})=C_{1}\,{\cal A}_{\rm
S}(1,2,3,4|\sigma_{4})+C_{2}\,{\cal A}_{\rm
S}(1,3,2,4|\sigma_{4})\,.~{}~{}~{}~{}$ (16)

The coefficients $C_{1}$ and $C_{2}$ have mass dimension $4$. The $4$-point
case is special, since one can choose a particular external leg $i$, then use
on-shell and momentum conservation conditions to reduce any $k_{a}\cdot k_{b}$
to $k_{i}\cdot k_{j}$, with $a,b,j\in\\{1,2,3,4\\}$. For instance, suppose we
choose $i=3$, then we have

$\displaystyle k_{1}\cdot k_{2}=k_{3}\cdot k_{4}\,,~{}~{}~{}~{}k_{1}\cdot
k_{4}=k_{3}\cdot k_{2}\,,~{}~{}~{}~{}k_{2}\cdot k_{4}=k_{3}\cdot k_{1}\,.$
(17)

Therefore, when considering the soft behavior for $k_{i}\to\tau k_{i}$, one
can always express $C_{1}$ as

$\displaystyle C_{1}=\sum_{a,b\in\\{1,2,3,4\\}\setminus
i}\,{\alpha}_{ab}\,(k_{i}\cdot{\cal K}_{a})\,(k_{i}\cdot{\cal
K}_{b})\,,~{}~{}~{}$ (18)

where ${\cal K}_{a}$ and ${\cal K}_{b}$ are combinations of external momenta,
${\alpha}_{ab}$ has mass dimension $0$ and depend on subscripts $a$ and $b$.
Thus the coefficient $C_{1}$ is obviously at the $\tau^{2}$ order. On the
other hand, the leading soft behavior of ${\cal A}_{\rm
S}(1,2,3,4|\sigma_{4})$ is at the $\tau^{-1}$ order. Combining $C_{1}$ and
${\cal A}_{\rm S}(1,2,3,4|\sigma_{4})$ together, we see the soft behavior of
the first term at the r.h.s of (16) starts at the $\tau^{1}$ order. The
consideration for the second term at the r.h.s of (16) is analogous. Thus, we
conclude that the single soft behavior of $4$-point NLSM amplitude does not
exist at $\tau^{-1}$ and $\tau^{0}$ orders. In other words, in the limit
$\tau\to 0$, the $4$-point NLSM amplitude vanishes. Such behavior is known as
Adler’s zero.

The vanishing of the $4$-point amplitude in the single soft limit is another
evidence for the vanishing of $3$-point one. Suppose the $n$-point and
$(n+1)$-point NLSM amplitudes exist, by taking an external leg to be soft, one
can always factorize the later one as the product of the former one and an un-
vanishing leading soft factor. Thus, the vanishing of $(n+1)$-point amplitude
in the single soft limit indicates the vanishing of $n$-point amplitude.

In this note, one of basic assumptions is the universality of soft behaviors.
As the consequence of universality, the vanishing of single soft behavior at
$\tau^{-1}$ and $\tau^{0}$ orders holds for any NLSM amplitude with arbitrary
number of external legs. If we assume the existence of the $4$-point
amplitude, then the $5$-point one must vanish, otherwise the vanishing of
$5$-point amplitude in the single soft limit will imply the vanishing of
$4$-point one. One can generalize the above argument further and conclude that
the number of external legs for each un-vanishing NLSM amplitude should be
even.

For now, we can not give the explicit formula of ${\cal
C}(\sigma^{\prime}_{2},k_{i})$ for the $4$-point case. In subsection.3.2, we
will use the universality of soft behaviors to determine the $n$-point NLSM
amplitudes in the expanded formula, with $n\geq 6$. In the next section, we
will use such expanded formula to derive the double soft factors for NLSM
amplitudes at $\tau^{0}$ and $\tau^{1}$ orders. Imposing the universality of
double soft factor, the $4$-point amplitude can be generated by removing two
soft legs from $6$-point one, then ${\cal C}(\sigma^{\prime}_{2},k_{i})$ for
the $4$-point case can be fixed through such manipulation, as will be shown in
subsection.4.1 in section.4.

###  3.2 $n$-point case

As pointed out at the end of subsection.3.1, the universality of soft
behaviors indicates that each NLSM amplitude vanishes at $\tau^{-1}$ and
$\tau^{0}$ orders when one of external legs being soft. This subsection aims
to find ${\cal C}(\sigma^{\prime}_{n-2},k_{i})$ for the $n$-point NLSM
amplitude with $n\geq 6$, by imposing the above requirement, as well as the
permutation symmetry among external legs in $\\{2,\cdots,n-1\\}$ which is
evident in (13).

Coefficients ${\cal C}(\sigma^{\prime}_{n-2},k_{i})$ have mass dimension
$2(n-2)$ and depend on color orderings $(1,\sigma^{\prime}_{n-2},n)$ carried
by ${\cal A}_{\rm S}(1,\sigma^{\prime}_{n-2},n|\sigma_{n})$. Consider
$k_{2}\to\tau k_{2}$, and expand ${\cal A}_{\rm N}(\sigma_{n})$ in $\tau$. In
order to construct ${\cal A}_{\rm N}(\sigma_{n})$ which vanishes at
$\tau^{-1}$ and $\tau^{0}$ orders, the simplest idea is to require each term
in ${\cal C}(\sigma^{\prime}_{n-2},k_{i})$ to contain the factor
$(k_{2}\cdot{\cal K}_{a_{2}})\,(k_{2}\cdot{\cal K}_{b_{2}})$, similar as in
(18) for the $4$-point case. Then, situations $k_{i}\to\tau k_{i}$ for other
$i\in\\{2,\cdots,n-1\\}$ yield the conclusion that each term in ${\cal
C}(\sigma^{\prime}_{n-2},k_{i})$ also contains $(k_{i}\cdot{\cal
K}_{a_{i}})\,(k_{i}\cdot{\cal K}_{b_{i}})$. The total number of $k_{i}$
required by the above construction is $2(n-2)$, which satisfies the correct
mass dimension of ${\cal C}(\sigma^{\prime}_{n-2},k_{i})$. Then, each ${\cal
C}(\sigma^{\prime}_{n-2},k_{i})$ can be decomposed as the linear combination
of building blocks $\prod_{j=1}^{n-2}\,k_{a_{j}}\cdot k_{b_{j}}$, where the
set $\\{a_{j},b_{j}\\}$ contains two $i$ for each $i\in\\{2,\cdots,n-1\\}$,
since when expanding to KK basis the coefficients do not contain any pole.
When taking $k_{1}\to\tau k_{1}$ or $k_{n}\to\tau k_{n}$ and expanding in
$\tau$, the coefficients ${\cal C}(\sigma^{\prime}_{n-2},k_{i})$ constructed
in the above way are at the $\tau^{0}$ order, which makes the single soft
behavior of ${\cal A}_{\rm N}(\sigma_{n})$ to be the $\tau^{-1}$ order,
therefore violates the universality of the soft behavior. Consequently, the
above naive construction is not correct. Notice that for $n\geq 6$ cases one
can not use on-shell and momentum conservation conditions to reduce the
general $k_{a}\cdot k_{b}$ to $k_{j}\cdot\widetilde{\cal K}$, as what we did
for the $n=4$ case in subsection.3.1, where $a,b,j\in\\{1,\cdots,n\\}$ and
$\widetilde{\cal K}$ is a combination of external momenta.

To find the correct answer, we consider $k_{2}\to\tau k_{2}$ and express the
leading order contribution of ${\cal A}_{\rm N}(\sigma_{n})$ as

$\displaystyle{\cal A}_{\rm
N}^{(0)}(\sigma_{n})=\sum_{\sigma^{\prime}_{n-2}}\,{\cal
C}^{(0)}(\sigma^{\prime}_{n-2},k_{i})\,S^{(0)}_{\rm S}(2)\,{\cal A}_{\rm
S}(1,\sigma^{\prime}_{n-2}\setminus 2,n|\sigma_{n}\setminus 2)\,,$ (19)

where the soft theorem (9) for external BAS scalar has been used. The leading
order contribution ${\cal C}^{(0)}(\sigma^{\prime}_{n-2},k_{i})$ is obtained
from ${\cal C}(\sigma^{\prime}_{n-2},k_{i})$ as follows. For each ${\cal
K}_{a}\cdot{\cal K}_{b}$ contained in ${\cal C}(\sigma^{\prime}_{n-2},k_{i})$,
where ${\cal K}_{a}$ and ${\cal K}_{b}$ are again two combinations of external
momenta, we turn it to ${\cal K}_{a}\cdot{\cal K}_{b}\to{\cal
K}^{(0)}_{a}\cdot{\cal K}^{(0)}_{b}$. The combinatory momentum ${\cal
K}^{(0)}_{a}$ is defined as

$\displaystyle{\cal K}^{(0)}_{a}=\begin{cases}\displaystyle~{}\tau
k_{2}~{}~{}~{}~{}&{\rm if}~{}{\cal K}_{a}=k_{2}\,,\\\ \displaystyle~{}{\cal
K}_{a}-k_{2}~{}~{}~{}~{}~{}&{\rm otherwise}\,.\end{cases}~{}~{}~{}$ (20)

The definition of ${\cal K}^{(0)}_{b}$ is analogous. The universality of soft
behavior requires ${\cal A}_{\rm N}^{(0)}(1,\cdots,n)$ to be the $\tau^{1}$
order. The simplest and most natural solution to the above condition can be
obtained by assuming the independence of BAS amplitudes in KK BAS basis. This
assumption indicates that at $\tau^{-1}$ and $\tau^{0}$ orders, coefficients
for different ${\cal A}_{\rm S}(1,\sigma^{\prime}_{n-2}\setminus
2,n|\sigma_{n}\setminus 2)$ vanish individually, namely,

$\displaystyle 0$ $\displaystyle=$ $\displaystyle{\cal
C}^{(0)}(2\shuffle\\{\sigma_{3},\cdots,\sigma_{n-1}\\},k_{i})\,S^{(0)}_{\rm
S}(2)\,{\cal A}_{\rm S}(1,{\not
2}\shuffle\\{\sigma_{3},\cdots,\sigma_{n-1}\\},n|\sigma_{n}\setminus 2)$ (21)
$\displaystyle=$ $\displaystyle C(1,{\not
2})\,{1\over\tau}\,\Big{(}{\delta_{12}\over s_{12}}+{\delta_{2\sigma_{3}}\over
s_{2\sigma_{3}}}\Big{)}\,{\cal A}_{\rm S}(1,{\not
2},\sigma_{3},\cdots,\sigma_{n-1},n|\sigma_{n}\setminus 2)$
$\displaystyle+\sum_{j=3}^{n-2}\,C(\sigma_{j},{\not
2})\,{1\over\tau}\,\Big{(}{\delta_{\sigma_{j}2}\over
s_{\sigma_{j}2}}+{\delta_{2\sigma_{j+1}}\over s_{2\sigma_{j+1}}}\Big{)}\,{\cal
A}_{\rm S}(1,\sigma_{3},\cdots,\sigma_{j},{\not
2},\sigma_{j+1},\cdots,\sigma_{n-1},n|\sigma_{n}\setminus 2)$
$\displaystyle+C(\sigma_{n-1},{\not
2})\,{1\over\tau}\,\Big{(}{\delta_{\sigma_{n-1}2}\over
s_{\sigma_{n-1}2}}+{\delta_{2n}\over s_{2n}}\Big{)}\,{\cal A}_{\rm
S}(1,\sigma_{3},\cdots,\sigma_{n-1},{\not 2},n|\sigma_{n}\setminus
2)\,,~{}~{}~{}$

therefore

$\displaystyle 0=C(1,{\not 2})\,\Big{(}{\delta_{12}\over
s_{12}}+{\delta_{2\sigma_{3}}\over
s_{2\sigma_{3}}}\Big{)}+C(\sigma_{n-1},{\not
2})\,\Big{(}{\delta_{\sigma_{n-1}2}\over s_{\sigma_{n-1}2}}+{\delta_{2n}\over
s_{2n}}\Big{)}+\sum_{j=3}^{n-2}\,C(\sigma_{j},{\not
2})\,\Big{(}{\delta_{\sigma_{j}2}\over
s_{\sigma_{j}2}}+{\delta_{2\sigma_{j+1}}\over
s_{2\sigma_{j+1}}}\Big{)}\,.~{}~{}~{}~{}$ (22)

Here we abbreviated ${\cal
C}^{(0)}(2\shuffle\\{\sigma_{3},\cdots,\sigma_{n-1}\\},k_{i})$ as $C(a,{\not
2})$ where $(a,{\not 2})$ emphasizes the position of external leg $2$ in the
color ordering $(\cdots,a,2,\cdots)$. The symbol $\shuffle$ in the first line
of (21) means the summation over particular permutations which preserves the
ordering $(\sigma_{3},\cdots,\sigma_{n-1})$, as explained below (11). The
notation ${\not 2}$ means delating the leg $2$, thus all ${\cal A}_{\rm
S}(1,{\not
2}\shuffle\\{\sigma_{3},\cdots,\sigma_{n-1}\\},n|\sigma_{n}\setminus 2)$ are
the same BAS amplitude. Notice that since our first attempt is failed, we must
assume $C(a,{\not 2})$ are at the $\tau^{0}$ or $\tau^{1}$ order.

Since the ordering $\sigma_{n}$ is arbitrary, which means the values of
$\delta_{ab}$ have not been fixed, to obtain the un-zero solution of
$C(a,{\not 2})$ to the equation (22), the only way is to employ the identity

$\displaystyle\sum_{i\neq 2}\,\delta_{i2}=0\,,$ (23)

which is the special case of identity (8). In other words, the equation (22)
should be reduced to

$\displaystyle 0=\widetilde{C}\,\sum_{i\neq 2}\,\delta_{i2}\,,~{}~{}~{}~{}$
(24)

where $\widetilde{C}$ is a Lorentz invariant constructed from external
momenta. The formula (24) requires that all $\delta_{i2}$ have a common
coefficient $\widetilde{C}$. Applying this condition to $\delta_{12}$, we
immediately find $C(1,{\not 2})=2(k_{2}\cdot k_{1})\,\widetilde{C}$. Then,
applying the same condition to $\delta_{\sigma_{3}2}$, we have

$\displaystyle\widetilde{C}\,={C(\sigma_{3},{\not 2})-C(1,{\not 2})\over
s_{2\sigma_{3}}}={C(\sigma_{3},{\not 2})-2\,(k_{2}\cdot
k_{1})\,\widetilde{C}\over s_{2\sigma_{3}}}\,,$ (25)

and the only solution is $C(\sigma_{3},{\not 2})=2(k_{2}\cdot
K_{1\sigma_{3}})\,\widetilde{C}$, where $K_{a_{1}\cdots
a_{m}}\equiv\sum_{i=1}^{m}k_{a_{i}}$. Repeating the above process recursively,
we get

$\displaystyle C(a,{\not 2})=2(k_{2}\cdot
K_{1\sigma_{3}\cdots\sigma_{a}})\,\widetilde{C}\,,~{}~{}~{}~{}$ (26)

for arbitrary $a\in\\{3,\cdots,n-1\\}$.

Some remarks are in order. First, when expanding the NLSM amplitude to KK
basis, coefficients do not contain any pole. Thus, $\widetilde{C}$ is a
Lorentz invariant without any pole, otherwise one can not ensure that non of
$C(a,{\not 2})$ contains any pole. Secondly, we have assumed that $C(a,{\not
2})$ are at the $\tau^{0}$ or $\tau^{1}$ order previously. Since $C(a,{\not
2})$ includes the factor $k_{2}\cdot K_{1\sigma_{3}\cdots\sigma_{a}}$
accompanied with $\tau$ under the re-scaling $k_{2}\to\tau k_{2}$, we now
exclude the $\tau^{0}$ case and conclude that $C(a,{\not 2})$ and
$\widetilde{C}$ are at the $\tau^{1}$ order. Finally, for a given color
ordering $(1,\sigma_{3},\cdots,\sigma_{n-1},n)$ carried by ${\cal A}_{\rm
S}(1,\sigma_{3},\cdots,\sigma_{n-1},n|\sigma_{n}\setminus 2)$, $\widetilde{C}$
is independent of the position of leg $2$ in original orderings
$(1,2\shuffle\\{\sigma_{3},\cdots,\sigma_{n-1}\\},n)$.

The solution (26) indicates that coefficients ${\cal
C}(\sigma^{\prime}_{n-2},k_{i})$ contain the component $k_{2}\cdot X_{2}$,
where the combinatory momentum $X_{i}$ is defined as the summation of momenta
carried by external legs at the l.h.s of $i$ in the color ordering. This
result can be generalized as ${\cal C}(\sigma^{\prime}_{n-2},k_{i})\propto
k_{i}\cdot X_{i}$ for each $i\in\\{2,\cdots,n-1\\}$, because of the
permutation symmetry among external legs $i\in\\{2,\cdots,n-1\\}$ in the
expansion (13). Since coefficients ${\cal C}(\sigma^{\prime}_{n-2},k_{i})$
have mass dimension $2(n-2)$, the above information fixes them completely,
namely,

$\displaystyle{\cal
C}(\sigma^{\prime}_{n-2},k_{i})={\alpha}\,\prod_{i=2}^{n-1}\,k_{i}\cdot
X_{i}\,,~{}~{}~{}$ (27)

where ${\alpha}$ is an constant with mass dimension $0$. Since ${\cal
C}(\sigma^{\prime}_{n-2},k_{i})$ does not contain any pole, ${\alpha}$ is
independent of any kinematical variable. Furthermore, ${\alpha}$ is also
independent of the color ordering $(1,\sigma^{\prime}_{n-2},n)$, due to the
permutation invariance. Substituting the solution (27) into (13), the expanded
formula of $n$-point NLSM amplitude is found to be

$\displaystyle{\cal A}_{\rm
N}(\sigma_{n})=\sum_{\sigma^{\prime}_{n-2}}\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot
X_{i}\Big{)}\,{\cal A}_{\rm
S}(1,\sigma^{\prime}_{n-2},n|\sigma_{n})\,.~{}~{}~{}~{}$ (28)

Here we have fixed the constant ${\alpha}$ as ${\alpha}=1$, via an overall re-
scaling of the amplitude. The expression in (28) is the desired expanded
formula of NLSM amplitude with $n\geq 6$. In subsection.4.1, we will show that
(28) is also correct for the $4$-point case.

It is not hard to verify that ${\cal A}_{\rm N}(\sigma_{n})$ given in (28)
vanishes at $\tau^{-1}$ and $\tau^{0}$ orders under the re-scaling
$k_{1}\to\tau k_{1}$ or $k_{n}\to\tau k_{n}$, thus the expanded formula (28)
satisfies the correct single soft behavior for any external leg being soft.
Taking $k_{2}\to\tau k_{2}$, $\widetilde{C}$ in the solution (26) can be
calculated from(28) as

$\displaystyle\widetilde{C}={\tau\over 2}\,\prod_{i=3}^{n-1}\,k_{i}\cdot
K_{1\sigma_{3}\cdots\sigma_{i-1}}\,,$ (29)

which satisfies our expectations: $\widetilde{C}$ is a Lorentz invariant
without any pole and is independent of the position of inserting the leg $2$.
The situations for other $k_{i}\to\tau k_{i}$ are analogous.

The solution (27) is obtained by assuming the independence of BAS amplitudes
in KK BAS basis. Indeed, such independence is violated by the well known Bern-
Carrasco-Johansson (BCJ) relations among BAS amplitudes. BCJ relations allow
more solutions to equation (22). For instance, one can modify the expanded
formula in (28) by adding terms which vanish automatically, such as

$\displaystyle{\cal A}_{\rm N}(\sigma_{n})$ $\displaystyle=$
$\displaystyle\sum_{\sigma^{\prime}_{n-2}}\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot
X_{i}\Big{)}\,{\cal A}_{\rm S}(1,\sigma^{\prime}_{n-2},n|\sigma_{n})$ (30)
$\displaystyle+(k_{2}\cdot X_{2})\,{\cal T}\,{\cal A}_{\rm
S}(1,\sigma^{\prime}_{n-2},n|\sigma_{n})\,,~{}~{}~{}$

where ${\cal T}$ is a Lorentz invariant with mass dimension $2(n-3)$. The
above modification is guaranteed by the fundamental BCJ relation

$\displaystyle 0=(k_{2}\cdot X_{2})\,{\cal A}_{\rm
S}(1,2\shuffle\\{\sigma_{3},\cdots,\sigma_{n-1}\\},n|\sigma_{n})\,.$ (31)

Thus, when considering BCJ relations, the expanded formula (28) should be
understood as an equivalent class rather than an unique expression.

The disadvantage of the modified formula (30) is the breaking of manifest
permutation symmetry among legs in $\\{2,\cdots,n-1\\}$. Let us only focus on
the expanded formulas those manifest such permutation invariance, and ask if
the BCJ relations lead to different solutions to equation (22) which satisfy
this symmetry, and the coefficients ${\cal C}(\sigma^{\prime}_{n-2},k_{i})$ do
not contain any pole. Quite surprisingly, for the $6$-point and $8$-point
cases, such new solution which is un-equivalent to (28) can not be found.
Thus, we conjecture that this situation is general, and claim that the
expansion in (28) is the only solution which manifests the desired permutation
invariance, although the general proof is absent.

##  4 Double soft factors

As discussed in subsection.3.1 in the previous section, using the single soft
theorem, one can generate the $n$-point tree amplitude from the $(n+1)$-point
one, by removing the soft external leg. The tree NLSM amplitude do not have
such luxuries, since the number of external legs must be even. However, it is
natural to think the $n$-point NLSM amplitude as the resulting object of
removing two soft legs from the $(n+2)$-point one. This picture leads to the
consideration of the double soft behavior of NLSM tree amplitudes.

The double soft theorems for tree NLSM amplitudes have been obtained via
different methods Cachazo:2015ksa ; Du:2015esa . The leading soft factor is at
the $\tau^{0}$ order, while the sub-leading one is the $\tau^{1}$ order. Such
leading order behavior is coincide with the picture of generating the
$n$-point amplitude from the $(n+2)$-point one, since such manipulation
forbids the vanishing of the $(n+2)$-point amplitude in the limit $\tau\to 0$.
In this section, we propose an efficient new approach to derive the leading
and sub-leading double soft factors, based on the expansion in (28).

Without losing of generality, we consider $k_{a}\to\tau k_{a},\,k_{b}\to\tau
k_{b}$ and expand ${\cal A}_{\rm N}(1,\cdots,n,a,b)$ by $\tau$. We chose the
basis as BAS amplitudes whose external legs $1$ and $n$ are fixed at two ends
in the first color ordering, thus the $n+2$-point amplitude ${\cal A}_{\rm
N}(1,\cdots,n,a,b)$ can be expanded as

$\displaystyle{\cal A}_{\rm N}(1,\cdots,n,a,b)=(k_{a}\cdot X_{a})\,(k_{b}\cdot
X_{b})\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot X_{i}\Big{)}\,{\cal A}_{\rm
S}(1,a\shuffle b\shuffle 2\shuffle\cdots\shuffle
n-1,n|1,\cdots,n,a,b)\,.~{}~{}~{}$ (32)

According to the meaning of $\shuffle$ mentioned below (11), one can rewrite
the expansion in (28) as

$\displaystyle{\cal A}_{\rm
N}(\sigma_{n})=\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot X_{i}\Big{)}\,{\cal
A}_{\rm S}(1,2\shuffle\cdots\shuffle n-1,n|\sigma_{n})\,.$ (33)

We used this notation in (32), to emphasize the positions of special legs $a$
and $b$ in the first color ordering carried by ${\cal A}_{\rm
S}(1,\sigma_{1},\cdots,\sigma_{n},n|1,\cdots,n,a,b)$. For later convenience,
in the reminder of this section we denote the BAS amplitude ${\cal A}_{\rm
S}(1,\sigma_{1},\cdots,\sigma_{n},n|1,\cdots,n,a,b)$ as ${\rm
A}(\sigma_{1},\cdots,\sigma_{n})$, where
$\\{\sigma_{1},\cdots,\sigma_{n}\\}=\\{a,b,2,\cdots,n-1\\}$.

###  4.1 Leading order

Figure 5: First type of Feynman diagrams.

To derive the double soft factor at leading order, let us consider Feynman
diagrams for BAS amplitudes ${\rm A}(a\shuffle b\shuffle
2\shuffle\cdots\shuffle n-1)$ those external legs $a$ and $b$ are coupled to a
common vertex. Such diagrams allowed by the second color ordering are shown in
Figure.5. Among terms at the r.h.s of the expansion (32), contributions from
the first configuration in Figure.5 are contained in

$\displaystyle(k_{a}\cdot k_{1})\,(k_{b}\cdot
K_{1a})\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot X_{i}\Big{)}\,{\rm
A}(a,b,2\shuffle\cdots\shuffle n-1)$ (34) $\displaystyle+$
$\displaystyle(k_{b}\cdot k_{1})\,(k_{a}\cdot
K_{1b})\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot X_{i}\Big{)}\,{\rm
A}(b,a,2\shuffle\cdots\shuffle n-1)\,,~{}~{}$

while contributions from the second configuration are included in

$\displaystyle(k_{b}\cdot k_{n})\,(k_{a}\cdot
K_{bn})\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot X_{i}\Big{)}\,{\rm
A}(2\shuffle\cdots\shuffle n-1,a,b)$ (35) $\displaystyle+$
$\displaystyle(k_{a}\cdot k_{n})\,(k_{b}\cdot
K_{an})\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot X_{i}\Big{)}\,{\rm
A}(2\shuffle\cdots\shuffle n-1,b,a)\,.~{}~{}$

After taking $k_{a}\to\tau k_{a},\,k_{b}\to\tau k_{b}$ and expanding in
$\tau$, the leading order contribution from (34) can be calculated as

$\displaystyle L_{1}$ $\displaystyle=$ $\displaystyle\tau^{2}\,(k_{a}\cdot
k_{1})\,\big{(}k_{b}\cdot(k_{1}+\tau
k_{a})\big{)}\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot
X^{(0)}_{i}\Big{)}\,{1\over s^{(0)}_{ab}}\,{1\over s^{(0)}_{ab1}}\,{\rm
A}(2\shuffle\cdots\shuffle n-1)$ (36) $\displaystyle-\tau^{2}\,(k_{b}\cdot
k_{1})\,\big{(}k_{a}\cdot(k_{1}+\tau
k_{b})\big{)}\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot
X^{(0)}_{i}\Big{)}\,{1\over s^{(0)}_{ab}}\,{1\over s^{(0)}_{ab1}}\,{\rm
A}(2\shuffle\cdots\shuffle n-1)$ $\displaystyle=$
$\displaystyle{(k_{a}-k_{b})\cdot k_{1}\over 4(k_{a}+k_{b})\cdot
k_{1}}\,\Big{[}\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot X^{(0)}_{i}\Big{)}\,{\rm
A}(2\shuffle\cdots\shuffle n-1)\Big{]}$ $\displaystyle=$
$\displaystyle{(k_{a}-k_{b})\cdot k_{1}\over 4(k_{a}+k_{b})\cdot k_{1}}\,{\cal
A}_{\rm N}(1,\cdots,n)\,,~{}~{}$

where $X^{(0)}_{i}$ arise from $X_{i}$ by deleting $k_{a}$ and $k_{b}$,
$s^{(0)}_{ab}$ and $s^{(0)}_{ab1}$ stand for leading order contributions of
$s_{ab}$ and $s_{ab1}$ respectively therefore
$s^{(0)}_{ab}=2\tau^{2}\,k_{a}\cdot k_{b}$,
$s^{(0)}_{ab1}=2\tau\,(k_{a}+k_{b})\cdot k_{1}$. Here the relative $-$ sign in
the first equality can be figured out via the diagrammatic rules, and a more
direct way to see it is employing the anti-symmetry of structure constant
$f^{abc}$ of Lie group which indicates a $-$ sign when swamping legs $a$ and
$b$. Because of this $-$ sign, terms at the $\tau^{-1}$ order cancel each
other, leaving the un-vanishing $L_{1}$ at the $\tau^{0}$ order. The last
equality uses the observation

$\displaystyle{\cal A}_{\rm
N}(1,\cdots,n)=\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot X^{(0)}_{i}\Big{)}\,{\rm
A}(2\shuffle\cdots\shuffle n-1)\,,~{}~{}$ (37)

based on the expansion (28) and the definition of $X^{(0)}_{i}$.

The consideration for (35) is analogous and gives

$\displaystyle L_{2}$ $\displaystyle=$ $\displaystyle{(k_{b}-k_{a})\cdot
k_{n}\over 4(k_{a}+k_{b})\cdot k_{n}}\,{\cal A}_{\rm N}(1,\cdots,n)\,,~{}~{}$
(38)

which is also at the $\tau^{0}$ order. We claim that (36) and (38) provide the
full double soft behavior of ${\cal A}_{\rm N}(1,\cdots,n,a,b)$ at leading
order. The reason can be explained as follows. For diagrams those legs $a$ and
$b$ are coupled to different vertices, the double soft limit can be achieved
by taking single soft limits for $a$ and $b$ successively. Since the single
soft behavior for $a$ or $b$ vanishes at the $\tau^{0}$ order, one can
conclude that taking two single soft limits consecutively contributes nothing
at the $\tau^{0}$ order. Thus, the leading double soft factor can be found by
combining (36) and (38) together, which gives

$\displaystyle{\cal A}^{(0)}_{\rm N}(1,\cdots,n,a,b)=S^{(0)}_{\rm
N}(a,b)\,{\cal A}_{\rm N}(1,\cdots,n)\,,~{}~{}$ (39)

where

$\displaystyle S^{(0)}_{\rm N}(a,b)={(k_{a}-k_{b})\cdot k_{1}\over
4(k_{a}+k_{b})\cdot k_{1}}+{(k_{b}-k_{a})\cdot k_{n}\over 4(k_{a}+k_{b})\cdot
k_{n}}\,.~{}~{}$ (40)

This result is the same as that obtained in Cachazo:2015ksa ; Du:2015esa  via
different approaches.

The universality of leading soft factor (40) indicates that the expanded
formula (28) is valid for the $4$-point NLSM amplitude. The reason is as
follows. The universality imposes

$\displaystyle{\cal A}^{(0)}_{\rm N}(1,2,3,4,a,b)=S^{(0)}_{\rm N}(a,b)\,{\cal
A}_{\rm N}(1,2,3,4)\,,~{}~{}$ (41)

which is a special case of (39). Since the relation (39) is based on the
expansion (37), ${\cal A}_{\rm N}(1,2,3,4)$ in (41) must satisfy the expansion
(37) which is equivalent to (28).

###  4.2 Sub-leading order

The sub-leading double soft factor is more complicate, since Feynman diagrams
those $a$ and $b$ are coupled to different vertices also have un-vanishing
contributions at the $\tau^{1}$ order. We will consider the corresponding
diagrams one by one.

Let us start with the first configuration in Figure.5, and express
corresponding terms contained in (34) as

$\displaystyle\Big{[}(k_{a}\cdot k_{1})\,(k_{b}\cdot
K_{1a})\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot X_{i}\Big{)}-(k_{b}\cdot
k_{1})\,(k_{a}\cdot K_{1b})\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot
X_{i}\Big{)}\Big{]}\,{1\over s_{ab}}\,{1\over s_{ab1}}\,{\cal M}$ (42)
$\displaystyle=$ $\displaystyle{(k_{a}-k_{b})\cdot k_{1}\over
2s_{ab1}}\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot X_{i}\Big{)}\,{\cal
M}\,.~{}~{}$

The relative $-$ sign was interpreted in subsection.4.1 around (36). When
$k_{a}\to\tau k_{a},\,k_{b}\to\tau k_{b}$, (42) behaves as

$\displaystyle{(k_{a}-k_{b})\cdot k_{1}\over 4(k_{a}+k_{b})\cdot
k_{1}+4\tau\,k_{a}\cdot k_{b}}\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot
X_{i}(\tau)\Big{)}\,{\cal M}(\tau)~{}~{}$ (43)

Our purpose is to extract the $\tau^{1}$ order terms from (43). First, one can
expand ${\cal M}(\tau)$ as

$\displaystyle{\cal M}(\tau)={\cal
M}(0)+\tau\,{\partial\over\partial\tau}\,{\cal
M}(\tau)\big{|}_{\tau=0}+\cdots\,$ (44)

and pick up the second term. One can observe that the parameter $\tau$ enters
${\cal M}(\tau)$ only through the combinatory momentum
$k_{1}+\tau(k_{a}+k_{b})$, thus

$\displaystyle{\partial\over\partial\tau}={1\over\tau}\,(k_{a}+k_{b})\cdot{\partial\over\partial(k_{a}+k_{b})}=(k_{a}+k_{b})\cdot{\partial\over\partial
k_{1}}\,,$ (45)

which leads to

$\displaystyle{\partial\over\partial\tau}\,{\cal
M}(\tau)\big{|}_{\tau=0}=(k_{a}+k_{b})\cdot{\partial\over\partial k_{1}}\,{\rm
A}(2\shuffle\cdots\shuffle n-1)\,,$ (46)

where the observation ${\cal M}(0)={\rm A}(2\shuffle\cdots\shuffle n-1)$ was
used. Consequently, the first piece is found to be

$\displaystyle{\cal P}_{1;1}=\tau\,{(k_{a}-k_{b})\cdot k_{1}\over
4(k_{a}+k_{b})\cdot k_{1}}\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot
X^{(0)}_{i}\Big{)}\,\Big{[}(k_{a}+k_{b})\cdot\partial_{k_{1}}\,{\rm
A}(2\shuffle\cdots\shuffle n-1)\Big{]}\,,~{}~{}$ (47)

where

$\displaystyle\partial_{k_{1}}\equiv{\partial\over\partial k_{1}}\,.$ (48)

Secondly, one can expand the denominator in (43) and pick up the second terms
at the r.h.s of

$\displaystyle{1\over 4(k_{a}+k_{b})\cdot k_{1}+4\tau\,k_{a}\cdot
k_{b}}={1\over 4(k_{a}+k_{b})\cdot k_{1}}-{\tau\,k_{a}\cdot k_{b}\over
4\big{(}(k_{a}+k_{b})\cdot k_{1}\big{)}^{2}}+\cdots\,,$ (49)

which gives the second piece

$\displaystyle{\cal P}_{2;1}=-\tau\,{\big{(}(k_{a}-k_{b})\cdot
k_{1}\big{)}\,(k_{a}\cdot k_{b})\over 4\big{(}(k_{a}+k_{b})\cdot
k_{1}\big{)}^{2}}\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot
X^{(0)}_{i}\Big{)}\,{\rm A}(2\shuffle\cdots\shuffle n-1)\,.~{}~{}$ (50)

Finally, each $X_{i}(\tau)$ contains $\tau\,(k_{a}+k_{b})$, thus the third
piece is found to be

$\displaystyle{\cal P}_{3;1}$ $\displaystyle=$
$\displaystyle\tau\,{(k_{a}-k_{b})\cdot k_{1}\over 4(k_{a}+k_{b})\cdot
k_{1}}\,\Big{[}\sum_{j=2}^{n-1}\,\Big{(}k_{j}\cdot(k_{a}+k_{b})\,{\prod_{i=2}^{n-1}\,k_{i}\cdot
X^{(0)}_{i}\over k_{j}\cdot X^{(0)}_{j}}\Big{)}\Big{]}\,{\rm
A}(2\shuffle\cdots\shuffle n-1)$ (51) $\displaystyle=$
$\displaystyle\tau\,{(k_{a}-k_{b})\cdot k_{1}\over 4(k_{a}+k_{b})\cdot
k_{1}}\,\Big{[}(k_{a}+k_{b})\cdot\partial_{k_{1}}\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot
X^{(0)}_{i}\Big{)}\Big{]}\,{\rm A}(2\shuffle\cdots\shuffle n-1)\,.~{}~{}$

Similarly, considering the second configuration in Figure.5 yields

$\displaystyle{\cal P}_{1;2}=\tau\,{(k_{b}-k_{a})\cdot k_{n}\over
4(k_{a}+k_{b})\cdot k_{n}}\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot
X^{(0)}_{i}\Big{)}\,\Big{[}(k_{a}+k_{b})\cdot\partial_{k_{n}}\,{\rm
A}(2\shuffle\cdots\shuffle n-1)\Big{]}\,,~{}~{}$ (52) $\displaystyle{\cal
P}_{2;2}=-\tau\,{\big{(}(k_{b}-k_{a})\cdot k_{n}\big{)}\,(k_{a}\cdot
k_{b})\over 4\big{(}(k_{a}+k_{b})\cdot
k_{n}\big{)}^{2}}\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot
X^{(0)}_{i}\Big{)}\,{\rm A}(2\shuffle\cdots\shuffle n-1)\,,~{}~{}$ (53)

as well as

$\displaystyle{\cal P}_{3;2}$ $\displaystyle=$
$\displaystyle\tau\,{(k_{b}-k_{a})\cdot k_{n}\over 4(k_{a}+k_{b})\cdot
k_{n}}\,\Big{[}\sum_{j=2}^{n-1}\,\Big{(}k_{j}\cdot(k_{a}+k_{b})\,{\prod_{i=2}^{n-1}\,k_{i}\cdot
X^{(0)}_{i}\over k_{j}\cdot X^{(0)}_{j}}\Big{)}\Big{]}\,{\rm
A}(2\shuffle\cdots\shuffle n-1)$ (54) $\displaystyle=$
$\displaystyle\tau\,{(k_{b}-k_{a})\cdot k_{n}\over 4(k_{a}+k_{b})\cdot
k_{n}}\,\Big{[}(k_{a}+k_{b})\cdot\partial_{k_{n}}\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot
X^{(0)}_{i}\Big{)}\Big{]}\,{\rm A}(2\shuffle\cdots\shuffle n-1)\,.~{}~{}$

The process is paralleled to those for obtaining (47), (50) and (51). To
derive (54), one should use momentum conservation to replace the explicit form
of $X_{i}$ by the equivalent one ${\cal
X}_{i}=X_{i}-\big{(}k_{a}+k_{b}+\sum_{i=1}^{n}\,k_{i}\big{)}$. The special
expressions for $X_{i}$ imply that the individual results in (51) and (54) are
inconsistent with momentum conservation. In (54), suppose one use momentum
conservation to remove $k_{n}$ in $X^{(0)}_{i}$, then $k_{i}\cdot X^{(0)}_{i}$
will be annihilated by the operator $\partial_{k_{n}}$. Similar phenomenon
happens for (51). However, after combining (51) and (54) together, the
resulting object is consistent with momentum conservation. One can use
momentum conservation to modify the explicit expression of any $X_{i}$ freely,
the combination ${\cal P}_{4;1}+{\cal P}_{4;2}$ always gives the correct
result. Similarly, the individual pieces (47) and (52) are inconsistent with
momentum conservation, while the combination ${\cal P}_{1;1}+{\cal P}_{1;2}$
does not have this problem.

Figure 6: Second type of Feynman diagrams.

Now we turn to a new type of Feynman diagrams which can be seen in Figure.6.
We focus on the first configuration in Figure.6, and the second one can be
treated similarly. At the r.h.s of (32), contributions from such diagrams are
included in

$\displaystyle(k_{b}\cdot k_{1})\,(k_{a}\cdot
K_{1b})\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot X_{i}\Big{)}\,{\rm
A}(b,a,2\shuffle\cdots\shuffle n-1)\,,$ (55)

thus can be expressed as

$\displaystyle(k_{b}\cdot k_{1})\,(k_{a}\cdot
K_{1b})\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot X_{i}\Big{)}\,{1\over
s_{b1}}\,{1\over s_{ab1}}\,{\cal N}$ $\displaystyle=$
$\displaystyle{k_{a}\cdot K_{1b}\over
2s_{ab1}}\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot X_{i}\Big{)}\,{\cal
N}\,.~{}~{}$ (56)

Taking $k_{a}\to\tau k_{a},\,k_{b}\to\tau k_{b}$ turns (56) to

$\displaystyle{k_{a}\cdot k_{1}+\tau\,k_{a}\cdot k_{b}\over
4(k_{a}+k_{b})\cdot k_{1}+4\tau\,k_{a}\cdot
k_{b}}\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot X_{i}(\tau)\Big{)}\,{\cal
N}(\tau)\,,~{}~{}$ (57)

and one can extract four pieces at the $\tau^{1}$ order. The first one,
expanding ${\cal N}(\tau)$ by $\tau$ gives

$\displaystyle{\cal P}_{4;1}=-\tau\,{k_{a}\cdot k_{1}\over 4(k_{a}+k_{b})\cdot
k_{1}}\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot
X^{(0)}_{i}\Big{)}\,\Big{[}(k_{a}+k_{b})\cdot\partial_{k_{1}}\,{\rm
A}(2\shuffle\cdots\shuffle n-1)\Big{]}\,,~{}~{}$ (58)

where the observation ${\cal N}(0)=-{\rm A}(2\shuffle\cdots\shuffle n-1)$ is
used, and the $-$ sign can be verified via diagrammatical rules. The second
piece arises from expanding the denominator of (57), which is given by

$\displaystyle{\cal P}_{5;1}=-\tau\,{k_{a}\cdot k_{b}\over 4(k_{a}+k_{b})\cdot
k_{1}}\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot X^{(0)}_{i}\Big{)}\,{\rm
A}(2\shuffle\cdots\shuffle n-1)\,.~{}~{}$ (59)

The third one is obtained by expanding the denominator of (57), which is found
to be

$\displaystyle{\cal P}_{6;1}=\tau\,{(k_{a}\cdot k_{1})\,(k_{a}\cdot
k_{b})\over 4\big{(}(k_{a}+k_{b})\cdot
k_{1}\big{)}^{2}}\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot X^{(0)}_{i}\Big{)}\,{\rm
A}(2\shuffle\cdots\shuffle n-1)\,.~{}~{}$ (60)

The final one comes from $\tau\,(k_{a}+k_{b})$ contained in $X_{i}(\tau)$ thus
is given as

$\displaystyle{\cal P}_{7;1}=-\tau\,{k_{a}\cdot k_{1}\over 4(k_{a}+k_{b})\cdot
k_{1}}\,\Big{[}(k_{a}+k_{b})\cdot\partial_{k_{1}}\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot
X^{(0)}_{i}\Big{)}\Big{]}\,{\rm A}(2\shuffle\cdots\shuffle n-1)\,.~{}~{}$ (61)

Considering the second configuration in Figure.6 gives analogous results

$\displaystyle{\cal P}_{4;2}=-\tau\,{k_{b}\cdot k_{n}\over 4(k_{a}+k_{b})\cdot
k_{n}}\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot
X^{(0)}_{i}\Big{)}\,\Big{[}(k_{a}+k_{b})\cdot\partial_{k_{n}}\,{\rm
A}(2\shuffle\cdots\shuffle n-1)\Big{]}\,,~{}~{}$ (62) $\displaystyle{\cal
P}_{5;2}=-\tau\,{k_{a}\cdot k_{b}\over 4(k_{a}+k_{b})\cdot
k_{n}}\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot X^{(0)}_{i}\Big{)}\,{\rm
A}(2\shuffle\cdots\shuffle n-1)\,,~{}~{}$ (63) $\displaystyle{\cal
P}_{6;2}=\tau\,{(k_{b}\cdot k_{n})\,(k_{a}\cdot k_{b})\over
4\big{(}(k_{a}+k_{b})\cdot
k_{n}\big{)}^{2}}\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot X^{(0)}_{i}\Big{)}\,{\rm
A}(2\shuffle\cdots\shuffle n-1)\,,~{}~{}$ (64)

and

$\displaystyle{\cal P}_{7;2}=-\tau\,{k_{b}\cdot k_{n}\over 4(k_{a}+k_{b})\cdot
k_{n}}\,\Big{[}(k_{a}+k_{b})\cdot\partial_{k_{n}}\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot
X^{(0)}_{i}\Big{)}\Big{]}\,{\rm A}(2\shuffle\cdots\shuffle n-1)\,.~{}~{}$ (65)

Combinations ${\cal P}_{4;1}+{\cal P}_{4;2}$ and ${\cal P}_{7;1}+{\cal
P}_{7;2}$ allow the expressions of $X^{(0)}_{i}$ to be rewritten via momentum
conservation, while individual pieces do not.

Figure 7: Third type of Feynman diagrams.

Then we consider the configuration in Figure.7, which corresponds to

$\displaystyle-(k_{b}\cdot k_{1})\,(k_{a}\cdot
k_{n})\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot X_{i}\Big{)}\,{\rm
A}(b,2\shuffle\cdots\shuffle n-1,a)\,,$ (66)

at the r.h.s of (32). Here we have used on-shell and momentum conservation
conditions to rewrite $X_{a}$ as $-k_{n}$. Under the re-scaling $k_{a}\to\tau
k_{a},\,k_{b}\to\tau k_{b}$, the leading order contributions from these terms
are the $\tau^{1}$ order, provides

$\displaystyle{\cal P}_{8}={\tau\over
4}\,\Big{[}\Big{(}k_{b}\cdot\partial_{k_{1}}+k_{a}\cdot\partial_{k_{n}}\Big{)}\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot
X^{(0)}_{i}\Big{)}\Big{]}\,{\rm A}(2\shuffle\cdots\shuffle n-1)\,.~{}~{}$ (67)

Only one of operators $\partial_{k_{1}}$ and $\partial_{k_{n}}$ is effective,
since the definition of $X^{(0)}_{i}$ implies that $k_{1}$ and $k_{n}$ can not
be included in $X^{(0)}_{i}$ simultaneously. The formula (67) gives correct
result for any expression of $X^{(0)}_{i}$, thus is consistent with momentum
conservation.

Figure 8: Fourth type of Feynman diagrams.

Next, we deal with the configuration in Figure.8 in which legs $a$ and $b$
coupled to a common vertex then coupled to an internal line. At the r.h.s of
(32), this type of Feynman diagrams correspond to

$\displaystyle(k_{a}\cdot K_{1\cdots i})\,(k_{b}\cdot K_{1\cdots
ia})\,\Big{(}\prod_{j=2}^{n-1}\,k_{j}\cdot X_{j}\Big{)}\,{\rm
A}(2\shuffle\cdots\shuffle i,a,b,i+1\shuffle\cdots\shuffle n-1)$ (68)
$\displaystyle+$ $\displaystyle(k_{b}\cdot K_{1\cdots i})\,(k_{a}\cdot
K_{1\cdots ib})\,\Big{(}\prod_{j=2}^{n-1}\,k_{j}\cdot X_{j}\Big{)}\,{\rm
A}(2\shuffle\cdots\shuffle i,b,a,i+1\shuffle\cdots\shuffle n-1)\,.$

When $k_{a}\to\tau k_{a},\,k_{b}\to\tau k_{b}$, the leading order
contributions correspond to Figure.8 are at the $\tau^{1}$ order, and can be
calculated as

$\displaystyle\tau\,{(k_{a}-k_{b})\cdot K_{1\cdots i}\over 2s_{1\cdots
i}^{2}}\,\Big{(}\prod_{j=2}^{n-1}\,k_{j}\cdot X^{(0)}_{j}\Big{)}\,{\cal
M}_{L}\,{\cal M}_{R}$ (69) $\displaystyle=$ $\displaystyle-{\tau\over
4}\,\Big{(}(k_{a}-k_{b})\cdot\partial_{k_{1}}\,{1\over s_{1\cdots
i}}\Big{)}\Big{(}\prod_{j=2}^{n-1}\,k_{j}\cdot X^{(0)}_{j}\Big{)}\,{\cal
M}_{L}\,{\cal M}_{R}$ $\displaystyle=$ $\displaystyle-{\tau\over
4}\,\Big{(}(k_{b}-k_{a})\cdot\partial_{k_{n}}\,{1\over s_{i+1\cdots
n}}\Big{)}\Big{(}\prod_{j=2}^{n-1}\,k_{j}\cdot X^{(0)}_{j}\Big{)}\,{\cal
M}_{L}\,{\cal M}_{R}\,.$

The last equality is obtained by replacing $K_{1\cdots i}$ with $-K_{i+1\cdots
n}$ via momentum conservation. Notice that since $s_{1\cdots i}$ is defined as
$s_{1\cdots i}=K_{1\cdots i}^{2}$ which contains $k_{1}^{2}$, we have
$\partial_{k_{1,\mu}}\,s_{1\cdots i}=2K^{\mu}_{1\cdots i}$, although
$k_{1}^{2}=0$ and we usually omit it when expressing $s_{1\cdots i}$
explicitly. The analogous situation holds for
$\partial_{k_{n,\mu}}\,s_{i+1\cdots n}$. Summing over all possible $i$ leads
to the piece

$\displaystyle{\cal P}_{9}=-{\tau\over
4}\,\Big{(}\prod_{j=2}^{n-1}\,k_{j}\cdot
X^{(0)}_{j}\Big{)}\,\Big{[}\Big{(}(k_{a}-k_{b})\cdot\partial_{k_{1}}+(k_{b}-k_{a})\cdot\partial_{k_{n}}\Big{)}\,{\rm
A}(2\shuffle\cdots\shuffle n-1)\Big{]}\,.~{}~{}$ (70)

The operator
$(k_{a}-k_{b})\cdot\partial_{k_{1}}+(k_{b}-k_{a})\cdot\partial_{k_{n}}$ is
introduced for two reasons. First, this operator is consistent with momentum
conservation. Secondly, this operator annihilates terms from Feynman diagrams
in Figure.9 which also contribute to ${\rm A}(2\shuffle\cdots\shuffle n-1)$,
leaves only desired contributions correspond to the configuration in Figure.8.

Figure 9: Feynman diagrams which should be excluded.

Figure 10: Fifth type of Feynman diagrams.

Finally, we should consider diagrams provided in Figure.10. Let us focus on
the first configuration in Figure.10. At the r.h.s of (32), the corresponding
terms are

$\displaystyle(k_{b}\cdot k_{1})\,(k_{a}\cdot K_{b1\cdots
i})\,\Big{(}\prod_{j=2}^{n-1}\,k_{j}\cdot X_{j}\Big{)}\,{\rm
A}(1,b,2\shuffle\cdots\shuffle i,a,i+1\shuffle\cdots\shuffle n-1)\,.$ (71)

When $k_{a}\to\tau k_{a},\,k_{b}\to\tau k_{b}$, leading contributions are at
the $\tau^{1}$ order and are given by

$\displaystyle-\tau\,{k_{a}\cdot K_{1\cdots i}\over 2s_{1\cdots
i}^{2}}\,\Big{(}\prod_{j=2}^{n-1}\,k_{j}\cdot X^{(0)}_{j}\Big{)}\,{\cal
N}_{L}\,{\cal N}_{R}$ (72) $\displaystyle=$ $\displaystyle{\tau\over
4}\,\Big{(}k_{a}\cdot\partial_{k_{1}}\,{1\over s_{1\cdots
i}}\Big{)}\,\Big{(}\prod_{j=2}^{n-1}\,k_{j}\cdot X^{(0)}_{j}\Big{)}\,{\cal
N}_{L}\,{\cal N}_{R}\,.$

Summing over all possible $i$, we get

$\displaystyle{\cal P}_{10;1}={\tau\over
4}\,\Big{(}\prod_{j=2}^{n-1}\,k_{j}\cdot
X^{(0)}_{j}\Big{)}\,\Big{(}k_{a}\cdot\partial_{k_{1}}\,{\rm
A}(2\shuffle\cdots\shuffle n-1)\Big{)}\,.~{}~{}$ (73)

Applying the same manipulation to the second configuration in Figure.10 gives

$\displaystyle{\cal P}_{10;2}={\tau\over
4}\,\Big{(}\prod_{j=2}^{n-1}\,k_{j}\cdot
X^{(0)}_{j}\Big{)}\,\Big{(}k_{b}\cdot\partial_{k_{n}}\,{\rm
A}(2\shuffle\cdots\shuffle n-1)\Big{)}\,.~{}~{}$ (74)

Again, the combination ${\cal P}_{10;1}+{\cal P}_{10;2}$ is consistent with
momentum conservation while individual pieces are not.

Now we are ready to figure out the sub-leading double soft operator. To
realize this, we need to regroup the sub-leading contributions at the
$\tau^{1}$ order as

$\displaystyle{\cal A}^{(1)}_{\rm N}(1,\cdots,n,a,b)=S^{(1)}_{\rm
N}(a,b)\,{\cal A}_{\rm N}(1,\cdots,n)\,,$ (75)

where $S^{(1)}_{\rm N}(a,b)$ is an operator at the $\tau^{1}$ order. We first
add ${\cal P}_{2;1}$, ${\cal P}_{5;1}$, ${\cal P}_{6;1}$, ${\cal P}_{2;2}$,
${\cal P}_{5;2}$, ${\cal P}_{6;2}$ in (50), (59), (60), (53), (63) and (64)
together to get

$\displaystyle{\cal R}_{1}=-\tau\,(k_{a}\cdot k_{b})\,\Big{(}{k_{a}\cdot
k_{1}\over 4\big{(}(k_{a}+k_{b})\cdot k_{1}\big{)}^{2}}+{k_{b}\cdot k_{n}\over
4\big{(}(k_{a}+k_{b})\cdot k_{n}\big{)}^{2}}\Big{)}\,{\cal A}_{\rm
N}(1,\cdots,n)\,.~{}~{}$ (76)

Then, by combining ${\cal P}_{3;1}$, ${\cal P}_{7;1}$, ${\cal P}_{3;2}$,
${\cal P}_{7;2}$, ${\cal P}_{8}$ in (51), (61), (54), (65) and (67) together,
we obtain

$\displaystyle{\cal R}_{21}$ $\displaystyle=$
$\displaystyle\tau\,\Big{[}\,{\cal J}\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot
X^{(0)}_{i}\Big{)}\Big{]}\,{\rm A}(2\shuffle\cdots\shuffle n-1)\,,~{}~{}$ (77)

where

$\displaystyle{\cal J}$ $\displaystyle=$ $\displaystyle{(k_{a}\cdot
k_{1})\,(k_{b}\cdot\partial_{k_{1}})-(k_{b}\cdot
k_{1})\,(k_{a}\cdot\partial_{k_{1}})\over 4(k_{a}+k_{b})\cdot
k_{1}}+{(k_{b}\cdot k_{n})\,(k_{a}\cdot\partial_{k_{n}})-(k_{a}\cdot
k_{n})\,(k_{b}\cdot\partial_{k_{n}})\over 4(k_{a}+k_{b})\cdot k_{n}}$ (78)
$\displaystyle=$ $\displaystyle{k_{a}\cdot J_{1}\cdot k_{b}\over
4(k_{a}+k_{b})\cdot k_{1}}+{k_{b}\cdot J_{n}\cdot k_{a}\over
4(k_{a}+k_{b})\cdot k_{n}}\,.$

In the second line at the r.h.s, $J^{\mu\nu}_{a}$ is the angular momentum
operator of the scalar particle $a$ which is defined by

$\displaystyle J_{a}^{\mu\nu}\equiv k_{a}^{\mu}{\partial\over\partial
k_{a,\nu}}-k_{a}^{\nu}{\partial\over\partial k_{a,\mu}}\,.$ (79)

Putting pieces ${\cal P}_{1;1}$, ${\cal P}_{4;1}$, ${\cal P}_{10;1}$, ${\cal
P}_{1;2}$, ${\cal P}_{4;2}$, ${\cal P}_{10;2}$, ${\cal P}_{9}$ in (47), (58),
(73), (52), (62), (74), (70) together, we find

$\displaystyle{\cal R}_{22}=\tau\,\Big{(}\prod_{i=2}^{n-1}\,k_{i}\cdot
X^{(0)}_{i}\Big{)}\,\Big{(}{\cal J}\,{\rm A}(2\shuffle\cdots\shuffle
n-1)\Big{)}\,.~{}~{}$ (80)

One can combine ${\cal R}_{21}$ and ${\cal R}_{22}$ to get

$\displaystyle{\cal R}_{2}={\cal R}_{21}+{\cal R}_{22}=\tau\,{\cal J}\,{\cal
A}_{\rm N}(1,\cdots,n)\,.$ (81)

Consequently, we can express ${\cal A}^{(1)}_{\rm N}(1,\cdots,n,a,b)$ as

$\displaystyle{\cal A}^{(1)}_{\rm N}(1,\cdots,n,a,b)={\cal R}_{1}+{\cal
R}_{2}=S^{(1)}_{\rm N}(a,b)\,{\cal A}^{(1)}_{\rm N}(1,\cdots,n)\,,$ (82)

where the soft operator $S^{(1)}_{\rm N}(a,b)$ is given as

$\displaystyle S^{(1)}_{\rm N}(a,b)$ $\displaystyle=$
$\displaystyle-\tau\,\Big{[}(k_{a}\cdot k_{b})\,\Big{(}{k_{a}\cdot k_{1}\over
4\big{(}(k_{a}+k_{b})\cdot k_{1}\big{)}^{2}}+{k_{b}\cdot k_{n}\over
4\big{(}(k_{a}+k_{b})\cdot k_{n}\big{)}^{2}}\Big{)}$ (83)
$\displaystyle+{k_{b}\cdot J_{1}\cdot k_{a}\over 4(k_{a}+k_{b})\cdot
k_{1}}+{k_{a}\cdot J_{n}\cdot k_{b}\over 4(k_{a}+k_{b})\cdot k_{n}}\Big{]}\,,$

which is again the same as the result found in Cachazo:2015ksa ; Du:2015esa .

##  5 Summery

In this note, we determined tree NLSM amplitudes by assuming the universality
of single soft behavior. The assumption of double copy structure was also
used, which tells us coefficients ${\cal C}(\sigma^{\prime}_{n-2},k_{i})$ in
(13) are independent of the ordering $\sigma_{n-2}$. We first observed the
Adler’s zero for $4$-point NLSM amplitudes, by considering kinematics. Then,
we figured out the expanded formula of general tree NLSM amplitudes, which
manifests the permutation invariance among external legs, by using the
universality of the Adler’s zero. Notice again the Adler’s zero was not
assumed at the beginning, it serves as a deduced property. We also re-derived
double soft factors for tree NLSM amplitudes at leading and sub-leading
orders, via the resulting expanded formula. The obtained double soft factors
are coincide with those in literatures.

Our soft bootstrap method, based on universality of soft behaviors and double
copy structure, are proven to be useful for constructing tree amplitudes of
Yang-Mills-scalar, Yang-Mills, Einstein-Yang-Mills, gravity Zhou:2022orv , as
well as non-linear sigma model. It is interesting to apply this method to a
wider range of theories. Another potential future direction is to generalize
this method from the tree level to the loop level.

## Acknowledgments

It is our great pleasure to thank Prof.Yijian Du for useful discussions and
comments.

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