aainstitutetext: Zhejiang Institute of Modern Physics, Zhejiang University,
Hangzhou, 310027, P. R. China bbinstitutetext:  Beijing Computational Science
Research Center, Beijing 100084, Chinaccinstitutetext: Hangzhou Institute of
Advanced Study, UCAS, Hangzhou, 310027, P.R. Chinaddinstitutetext: Center of
Mathematical Science, Zhejiang University, Hangzhou, 310027, P. R.
Chinaeeinstitutetext: Peng Huanwu Center for Fundamental Theory, Hefei, Anhui,
230026, Chinaffinstitutetext: University of Chinese Academy of Sciences,
100190 Beijing, China

# Reduction with Degenerate Gram matrix for One-loop Integrals

Bo Feng c,f Chang Hu a Tingfei Li a Yuekai Song<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

An improved PV-reduction (Passarino-Veltman) method for one-loop integrals
with auxiliary vector $R$ has been proposed in Feng:2021enk ; Hu:2021nia . It
has also been shown that the new method is a self-completed method in
Feng:2022uqp . Analytic reduction coefficients can be easily produced by
recursion relations in this method, where the Gram determinant appears in
denominators. The singularity caused by Gram determinant is a well-known fact
and it is important to address these divergences in a given frame. In this
paper, we propose a systematical algorithm to deal with this problem in our
method. The key idea is that now the master integral of the highest topology
will be decomposed into combinations of master integrals of lower topologies.
By demanding the cancellation of divergence for obtained general reduction
coefficients, we solve decomposition coefficients as a Taylor series of the
Gram determinant. Moreover, the same idea can be applied to other kinds of
divergences.

###### Keywords:

One-loop, PV-Reduction, Degenerate Gram Matrix

##  1 Introduction

The calculation of scattering amplitudes at loop level is an important issue
in High Energy Physics. A widely used method is to reduce a loop amplitude
into a linear combination of some scalar integrals under dimensional
regularization Brown:1952eu ; Melrose:1965kb ; Passarino:1978jh ;
tHooft:1978jhc ; vanNeerven:1983vr ; Stuart:1987tt ; vanOldenborgh:1989wn ;
Bern:1992em ; Bern:1993kr ; Fleischer:1999hq ; Binoth:1999sp ; Denner:2002ii ;
Duplancic:2003tv ; Denner:2005nn ; Ellis:2007qk ; Ossola_2007 ; Bern:1994cg

$\displaystyle
I^{1-loop}=\sum_{i_{d_{0}+1}}C_{i_{d_{0}+1}}I_{d_{0}+1}^{i_{d_{0}+1}}+\sum_{i_{d_{0}}}C_{i_{d_{0}}}I_{d_{0}}^{i_{d_{0}}}+\cdots+\sum_{i_{1}}C_{i_{1}}I_{1}^{i_{1}}\,,$
(1)

where $i_{s}$ is the set of propagators appearing in the basis. The
coefficient $C_{i_{s}}$ ($s=1,\cdots,d_{0}+1$) is simply a rational function
of some Lorentz invariant contractions of external momenta as well as the
masses and spacetime dimension, while the terms $I_{s}^{i_{s}}$ are the
$s$-gon master integrals. With the general expansion (1), the computation of
general one-loop amplitudes has been switched to determining those
coefficients $C_{i_{s}}$. Many tools have been invented to shovel the
brambles, such as integration-by-parts (IBP) Chetyrkin:1981qh ; Tkachov:1981wb
, PV (Passarino-Veltman) reduction Passarino:1978jh , OPP (Ossola-Papadopoulo-
Pittau) reduction Ossola:2006us ; Ossola:2007bb ; Ellis:2007br ; Ossola_2007 ,
unitarity cut method, etc Bern:1994zx ; Bern:1994cg ; Britto:2004nc ;
Britto:2005ha ; Campbell:1996zw ; Bern:1997sc ; Denner:2005nn ;
Anastasiou:2006gt ; Britto:2010um .

All these methods can be divided into two categories, i.e., the reduction at
the integrand level or the integral level. For reduction at the integrand
level, Ossola_2007  shows how to extract the coefficients of the 4-, 3-, 2-
and 1-point one-loop scalar integrals from the full one-loop integrand of
arbitrary scattering processes in an algebraical way. For the reduction at the
integral level, an efficient way is the unitarity cut method. The main idea is
to compare the imaginary part of two sides of (1). However, since the loop-
integral is well-defined only after using the dimensional regularization, the
unitarity cut method in pure 4D needs to be generalized to
$(4-2\epsilon)$-dimension, which has been done in Anastasiou:2006jv ;
Anastasiou:2006gt . Based on this generalization, the analytic expressions for
reduction coefficients (except the tadpole coefficients) have been derived in
a series of papers Britto:2006fc ; Britto:2007tt ; Britto:2008vq ;
Britto:2008sw ; Feng:2013sa .

Recently, we proposed a new framework for general one-loop tensor reduction by
employing an auxiliary vector $R$ and two kinds of differential operators
Feng:2021enk ; Hu:2021nia . Similar to other reduction methods, our method
also suffers from divergences for vanishing Gram determinant, which appears as
the inverse of Gram matrix in the recursion constructions of reduction
coefficients. It is well known that the vanishing Gram determinant indicates
external momenta are not completely independent of each other, thus the master
integral of the highest topology will not be in the basis anymore, and it can
be decomposed into combinations of master integrals of lower topologies.111In
this paper, the integrals of the highest topology refer to the integrals
containing the maximum number of propagators while lower topology means the
integrals containing fewer propagators, which is distinct from its definition
in mathematics. With this decomposition, the reduction coefficients will be
reorganized so that the divergences between different terms will cancel with
each other. To see this picture clearly, calculating analytic coefficients
will be important, which is exactly the merit of our new method. In this
paper, we will systematically study the degenerate case with vanishing Gram
determinant using the results in Feng:2021enk ; Hu:2021nia . It turns out that
the decomposition coefficients can easily be solved as a series in the Gram
determinant.

This paper is organized as follows. In section 2, we will review the improved
PV-reduction method with auxiliary vector $R$ in our previous work and briefly
discuss the general structure of reduction coefficients. In section 3, based
on the explicit computation for bubbles in the limit $K^{2}\to 0$, we
establish an algorithm to deal with the vanishing Gram determinant for general
topologies. We demonstrate the algorithm with the triangle in the main text
and the box and pentagon in Appendix A. In section 4, an important consistency
has been proved, i.e., our algorithm should be independent of the auxiliary
vector $R$ in reduction coefficients and the choice of tensor ranks. Finally,
we give a summary and some discussions in section 5.

##  2 Background

###  2.1 Review of Improved PV-Reduction

In this section, we briefly review how to reduce one-loop tensor integrals
using our improved PV-reduction method. After that, we will explain how to
deal with the degenerated Gram determinant in our framework.

For a general one-loop $m$-rank tensor integral with $(n+1)$ propagators

$\displaystyle I_{n+1}^{\mu_{1}\mu_{2}\cdots\mu_{m}}=\int{d^{D}\ell\over
i\pi^{D/2}}{\ell^{\mu_{1}}\ell^{\mu_{2}}...\ell^{\mu_{m}}\over
D_{0}\prod_{j=1}^{n}D_{j}}=\int{d^{D}\ell\over
i\pi^{D/2}}{\ell^{\mu_{1}}\ell^{\mu_{2}}...\ell^{\mu_{m}}\over(\ell^{2}-M_{0}^{2})\prod_{j=1}^{n}((\ell-
K_{j})^{2}-M_{j}^{2})}~{}~{}~{}~{}$ (2)

the algorithm of traditional PV (Passarino-Veltman) reduction method
Passarino:1978jh  is following. First, we use the external momenta to write
down the matched general tensor structure. For example, with rank $3$ tensor
integral of the form (2), we write down

$\displaystyle
I_{n+1}^{\mu_{1}\mu_{2}\mu_{3}}=a_{i}(g^{\mu_{1}\mu_{2}}K_{i}^{\mu_{3}}+g^{\mu_{1}\mu_{3}}K_{i}^{\mu_{2}}+g^{\mu_{2}\mu_{3}}K_{i}^{\mu_{1}})+b_{ijl}K_{i}^{\mu_{1}}K_{j}^{\mu_{2}}K_{l}^{\mu_{3}}~{}~{}~{}$
(3)

with unknown expansion coefficients $a_{i},b_{ijk}$. Secondly, we contract
both sides of (3) with various $g^{\mu\nu}$ and $K_{i}$ to establish enough
algebraic relations to solve these coefficients. It is easy to see that with
higher and higher tensor rank, there will be more and more different tensor
structures to be written down in (3) and more and more algebraic relations to
be established to fix them.

A nice observation made in Feng:2021enk ; Hu:2021nia  is that the complicated
tensor structure can be simply recovered from

$\displaystyle I^{(m)}_{n+1}\equiv\int{d^{D}\ell\over
i\pi^{D/2}}{(2R\cdot\ell)^{m}\over D_{0}\prod_{j=1}^{n}D_{j}}~{}~{}~{}~{}$ (4)

by writing $R=\sum_{i=1}^{m}x_{i}R_{i}$ and extracting terms with coefficients
$x_{1}...x_{m}$ (or taking the proper differentiation action). For example,

$\displaystyle(\ell\cdot
P)\ell^{2}\sim(\partial_{R}\cdot\partial_{R})(P\cdot\partial_{R})(2R\cdot\ell)^{3}\,.$
(5)

The vector $R$ in (4) is called auxiliary vector. Involving the auxiliary
vector $R$ not only greatly simplified the tensor structure but also provided
a very simple way to establish algebraic relations to solve the reduction
problem, as we will review shortly.

The reduction of (4) can be denoted as

$\displaystyle I_{n+1}^{(m)}=C^{(m)}_{n+1\to n+1}I_{n+1}+C^{(m)}_{n+1\to
n+1;\widehat{i}}I_{n+1;\widehat{i}}+C^{(m)}_{n+1\to
n+1;\widehat{ij}}I_{n+1;\widehat{ij}}+\ldots$ (6)

where $I_{n+1;\widehat{i_{1},\ldots,i_{a}}}$ represents the scalar integral
got by removing propagators $D_{i_{1}},D_{i_{2}},\ldots,D_{i_{a}}$ from the
integral (4). For example,

$I_{8;\widehat{0,2,5,6}}=\int{d^{D}\ell\over i\pi^{D/2}}{1\over D_{1}\cdot
D_{3}\cdot D_{4}\cdot D_{7}};\ \ \ I_{7;\widehat{1,3,5}}=\int{d^{D}\ell\over
i\pi^{D/2}}{1\over D_{0}\cdot D_{2}\cdot D_{4}\cdot D_{6}}.$ (7)

For later convenience, in this paper we define

$\displaystyle s_{ij}=K_{i}\cdot K_{j},~{}~{}~{}s_{0i}=R\cdot
K_{i},~{}~{}~{}f_{i}=M_{0}^{2}-M_{i}^{2}+K^{2}_{i}~{},~{}~{}~{}$ (8)

and the Gram matrix is given by $\boldsymbol{G}=[G_{ij}=s_{ij}]$ and its
determinant is denoted by $|G|$.

Now we review how to solve reduction coefficients in (6). Without loss of
generality, we only review how to calculate $C^{(m)}_{n+1\to
n+1;\widehat{r+1,r+2,...,n}}$. Other reduction coefficients can be got by a
label permutation and a proper momentum shift.222More details can be found in
Feng:2021enk ; Hu:2021nia . Having introduced the $R$, we can construct the
following two types of differential operators

$\displaystyle\mathcal{D}_{i}\equiv K_{i}\cdot{\partial\over\partial
R},~{}~{}i=1,...,n;~{}~{}~{}~{}\quad\mathcal{T}\equiv\eta^{\mu\nu}{\partial\over\partial
R^{\mu}}{\partial\over\partial R^{\nu}}.$ (9)

Acting with these two operators on both sides of (6) and comparing the
coefficients of each Master Integral.333Note that Master Integral contains no
$R$, the differential operators will directly act on reduction coefficients.
we get

$\displaystyle\mathcal{T}C_{n+1\to n+1;\widehat{r+1,...,n}}^{(m)}=$
$\displaystyle 4m(m-1)M_{0}^{2}C_{n+1\to n+1;\widehat{r+1,...,n}}^{(m-2)},$
(10)

and

$\displaystyle\mathcal{D}_{i}C_{n+1\to n+1;\widehat{r+1,...,n}}^{(m)}=$
$\displaystyle-mC_{n+1;\widehat{i}\to
n+1;\widehat{r+1,...,n}}^{(m-1)}+mf_{i}C_{n+1\to
n+1;\widehat{r+1,...,n}}^{(m-1)}.$ (11)

where $C_{n+1;\widehat{i}\to n+1;\widehat{r+1,...,n-1}}^{(m-1)}$ is the
reduction coefficient of integral $I^{(m-1)}_{n+1;\widehat{i}}$ to the Master
Integral $I_{n+1;\widehat{r+1,...,n}}$. After expanding reduction coefficients
according to its tensor structure

$\displaystyle C^{(m)}_{n+1\to n+1;\widehat{r+1,r+2,...,n}}=$
$\displaystyle\sum_{2a_{0}+\sum_{k=1}^{n}a_{k}=m}\Bigg{\\{}c^{(0,1,\cdots,r)}_{a_{1},\cdots,a_{n}}(m)(M_{0}^{2})^{a_{0}+r-n}\prod_{k=0}^{n}s_{0k}^{a_{k}}\Bigg{\\}},$
(12)

equations (10) and (11) will lead to algebraic recursion relations for
expansion coefficients $c^{(0,1,\cdots,r)}_{a_{1},\cdots,a_{n}}(m)$ as a
rational function of spacetime dimension $D$, rank $m$ and kinematics
$(K_{i}\cdot K_{j})$ and $M_{i}^{2}$. A very nice feature is that these
algebraic recursion relations can be solved to give the recursive construction
for expansion coefficients as following,

$\displaystyle\boldsymbol{c}^{(0,1,\cdots,r)}(a_{1},\cdots,a_{n};m)=\boldsymbol{T}^{-1}\widetilde{\boldsymbol{G}}^{-1}\boldsymbol{O}^{(0,1,\cdots,r)}(a_{1},\cdots,a_{n};m),$
(13) $\displaystyle c^{(0,1,\cdots,r)}_{\underbrace{0,\cdots,0}_{\text{n\
times}}}(2k)={2k-1\over
D+2k-n-2}\left[(4-\boldsymbol{{\alpha}}^{T}\widetilde{\boldsymbol{G}}^{-1}\boldsymbol{{\alpha}})c_{\underbrace{0,\cdots,0}_{\text{n\
times}}}^{(0,1,\cdots,r)}(2k-2)+\boldsymbol{{\alpha}}^{T}\widetilde{\boldsymbol{G}}^{-1}\boldsymbol{c}_{\underbrace{0,\cdots,0}_{\text{n-1\
times}}}^{(0,1,\cdots,r)}(2k-2)\right],$ (14)

where $\widetilde{\boldsymbol{G}}=[\beta_{ij}={s_{ij}\over M_{0}^{2}}]$ is the
$n\times n$ dimensionless Gram matrix and
$\boldsymbol{T}=\text{diag}(a_{1}+1,a_{2}+1,\cdots,a_{n}+1)$ is a diagonal
matrix. Other vectors are defined as

$\displaystyle\boldsymbol{{\alpha}}^{T}=\left(\frac{f_{1}}{M_{0}^{2}},\frac{f_{2}}{M_{0}^{2}},\cdots,\frac{f_{n}}{M^{2}_{0}}\right),$
(15)
$\displaystyle[\boldsymbol{c}^{(0,1,\cdots,r)}(a_{1},\cdots,a_{n};m)]_{i}=c^{(0,1,\cdots,r)}_{a_{1},a_{2},\cdots,a_{i}+1,\cdots,a_{n}}(m),$
(16)

and

$\displaystyle[\boldsymbol{O}^{(0,1,\cdots,r)}(a_{1},\cdots,a_{n};m)]_{i}=m\alpha_{i}c^{(0,1,\cdots,r)}_{a_{1},\cdots,a_{n}}(m-1)-m\delta_{0a_{i}}c^{(0,1,\cdots,r)}_{a_{1},\cdots,\widehat{a}_{i},\cdots,a_{n}}(m-1;\widehat{i})$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}-(m+1-\sum_{l=1}^{n}a_{l})c^{(0,1,\cdots,r)}_{a_{1},\cdots,a_{i}-1,\cdots,a_{n}}(m)$
$\displaystyle\boldsymbol{c}_{\underbrace{0,\cdots,0}_{\text{n-1\
times}}}^{(0,1,\cdots,r)}(m)=\left(0,0,\cdots,0,c_{\underbrace{0,\cdots,0}_{\text{n-1\
times}}}^{(0,1,\cdots,r)}(m;\widehat{r+1}),\cdots,c_{\underbrace{0,\cdots,0}_{\text{n-1\
times}}}^{(0,1,\cdots,r)}(m;\widehat{n-1})\right).$ (17)

With the known boundary conditions, one can get all reduction coefficients by
applying the recursions (13), (14) iteratively. Above formulas are easy to be
implemented into Mathematica and one can generate analytic expressions for
reduction coefficients for any tensor rank (Exactly because of these benefits
involving the auxiliary vector $R$, we call the method as "imporoved PV-
reduction method"). Knowing the analytic expressions for reduction
coefficients, we can discuss many topics. For example, we can find the
reduction for integrals with propagators having power higher than one as shown
in Feng:2022uqp . Another important application is for the case where some
reduction coefficients become divergent, as will be discussed in this paper.

Now we want to point out an important observation from (13), (14): with
${\boldsymbol{G}}^{-1}$ we will have $|G|$ appearing in denominators of all
reduction coefficients. When applying above PV-reduction method, we have made
a hidden assumption that $D\geq n$ and all $K_{i}$’s in propagators are
generic (i.e., linear independent). With this hidden assumption, $|G|\neq 0$
and above formulas are well-defined. However, in the practical applications,
we do meet the degeneration of the Gram determinant, i.e., $|G|=0$, which
means that these $K_{i}$’s are not linear independent anymore and the
kinematics lives in a smaller space. For this case, the formulas (13) and (14)
are ill-defined. Thus, for the completion of the improved PV-reduction method,
we must provide an algorithm to solve this issue. Briefly, $|G|=0$ means that
the scalar integrals of the highest topology are not in the basis anymore, and
they can be reduced to a combination of lower topologies.

Later we need to use the divergence behavior when $|G|\to 0$. Noticing that
$G^{-1}=G^{*}/|G|$, where $G^{*}$ is the adjugate matrix of Gram matrix, there
is a Gram determinant appearing in the denominator for each iteration, thus
one can conclude that the reduction coefficients have the highest divergent
part about the determinant of Gram matrix444In fact, only those reduction
coefficients with $m+r\geq n$ are nonzero. Because there are not enough
$\ell\cdot R$ in the numerator to cancel $(n-r)$ propagators for $m+r<n$.

$\displaystyle C^{(m)}_{n+1\to
n+1;\widehat{r+1,...,n}}\sim\left\\{\begin{matrix}&\frac{1}{|G|^{m}},\
r=n,n-1\\\ &\frac{1}{|G|^{m-n+r+1}},\ r<n-1.\end{matrix}\right.$ (18)
$\displaystyle
c^{(0,...,n)}_{a_{1},...,a_{n}}\sim|G|^{-\frac{m+\sum_{i}^{n}a_{i}}{2}}$

##  3 Reduction for degenerate Gram Matrix

In the previous section, we find that the reduction coefficients diverge when
$|G|\to 0$, while $I_{n+1}^{(m)}$ is still well-defined. The naive conflict in
(6) indicates that the expansion on the RHS is not proper. In other words, the
scalar integrals are not independent of each other anymore. As we will show in
this section, these divergences tell us a lot of information about the
degenerating behavior of the basis. We will use various examples, i.e.,
integrals of bubble, triangle, box, and pentagon, to demonstrate the strategy
to deal with the divergences in our framework.

###  3.1 The reduction of scalar bubble with degenerate Gram Matrix

Let us start with the simplest example, i.e., the reduction of the bubble with
tensor rank $1$. Our method gives

$\displaystyle I_{2}^{(1)}=$ $\displaystyle C^{(1)}_{2\to
2}~{}I_{2}+C^{(1)}_{2\to 2;\widehat{0}}~{}I_{2;\widehat{0}}+C^{(1)}_{2\to
2;\widehat{1}}~{}I_{2;\widehat{1}}\,,$ (19)

where

$\displaystyle C^{({1})}_{{2}\to{2}}$
$\displaystyle=\frac{\left(M_{0}^{2}-M_{1}^{2}+s_{11}\right)s_{01}}{s_{11}}\,,$
$\displaystyle C^{({1})}_{{2}\to{2};\widehat{{1}}}$
$\displaystyle=\frac{s_{01}}{s_{11}}\,,~{}~{}~{}C^{({1})}_{{2}\to{2};\widehat{{2}}}=-\frac{s_{01}}{s_{11}}\,.$
(20)

It is easy to see that with $|G|=k_{1}\cdot k_{1}=s_{11}\to 0$, all
coefficients in (20) become divergent. But $I_{2}^{(1)}$ is a definitely well-
defined integral even with $s_{11}=0$, so the divergence on the RHS in (19)
implies that the expansion is not proper. More explicitly, the three scalar
integrals $I_{2},I_{2;\widehat{0}},I_{2;\widehat{1}}$ are not linearly
independent anymore, i.e., we should have

$\displaystyle I_{2}=\sum_{i=0}^{1}\mathcal{B}_{2\to
2;\widehat{i}}I_{2;\widehat{i}}=\sum_{i=0}^{1}\sum_{a=0}^{\infty}\mathcal{B}^{(a)}_{2\to
2;\widehat{i}}s_{11}^{a}I_{2;\widehat{i}}\,,~{}~{}~{}$ (21)

where the decomposition coefficient $\mathcal{B}_{2\to 2;\widehat{i}}$ is
expanded as a Taylor series of $|G|=s_{11}$.

Now we show how to use the (21) to make the RHS of (19) well-defined under the
limit $|G|\to 0$. Putting (20) and (21) to (19), the RHS becomes

$\displaystyle\left(\frac{\left(M_{0}^{2}-M_{1}^{2}+s_{11}\right)s_{01}}{s_{11}}\sum_{a=0}^{\infty}\mathcal{B}^{(a)}_{2\to
2;\widehat{0}}s_{11}^{a}+\frac{s_{01}}{s_{11}}\right)I_{2;\widehat{0}}$
$\displaystyle+\left(\frac{\left(M_{0}^{2}-M_{1}^{2}+s_{11}\right)s_{01}}{s_{11}}\sum_{a=0}^{\infty}\mathcal{B}^{(a)}_{2\to
2;\widehat{1}}s_{11}^{a}-\frac{s_{01}}{s_{11}}\right)I_{2;\widehat{1}}.$ (22)

To cancel the divegence, we just need

$\displaystyle\mathcal{B}^{(0)}_{2\to
2;\widehat{0}}={-1\over(M_{0}^{2}-M_{1}^{2})}\,,~{}~{}~{}\mathcal{B}^{(0)}_{2\to
2;\widehat{1}}={1\over(M_{0}^{2}-M_{1}^{2})}\,.~{}~{}~{}$ (23)

Although we have only fixed the first coefficient in the above computation,
the smoothness of RHS in (19) does tell us the information of later
coefficients. To get them, it is natural to consider the reduction of higher
tensor rank. For rank $2$, we have

$\displaystyle C^{({2})}_{{2}\to{2}}$ $\displaystyle=$
$\displaystyle\frac{D\left(M_{0}^{2}-M_{1}^{2}\right){}^{2}s_{01}^{2}}{(D-1)s_{11}^{2}}+\frac{Ds_{01}^{2}+2M_{0}^{2}s_{00}+2M_{1}^{2}s_{00}}{D-1}$
$\displaystyle+\frac{2M_{0}^{2}\left((D-2)s_{01}^{2}+M_{1}^{2}s_{00}\right)-2DM_{1}^{2}s_{01}^{2}-M_{0}^{4}s_{00}-M_{1}^{4}s_{00}}{(D-1)s_{11}}-\frac{s_{11}s_{00}}{D-1}\,,$
$\displaystyle C^{({2})}_{{2}\to{2};\widehat{{1}}}$ $\displaystyle=$
$\displaystyle\frac{D\left(M_{0}^{2}-M_{1}^{2}\right)s_{01}^{2}}{(D-1)s_{11}^{2}}+\frac{(3D-4)s_{01}^{2}+M_{0}^{2}\left(-s_{00}\right)+M_{1}^{2}s_{00}}{(D-1)s_{11}}+\frac{s_{00}}{D-1}\,,$
$\displaystyle C^{({2})}_{{2}\to{2};\widehat{{2}}}$ $\displaystyle=$
$\displaystyle\frac{D\left(M_{1}^{2}-M_{0}^{2}\right)s_{01}^{2}}{(D-1)s_{11}^{2}}+\frac{-Ds_{01}^{2}+M_{0}^{2}s_{00}-M_{1}^{2}s_{00}}{(D-1)s_{11}}+\frac{s_{00}}{D-1}\,.~{}~{}~{}$
(24)

Now the denominators have $s_{11}^{2}$ in (24). When combining (21) and (24),
the smoothness under the limit requires the cancellation of poles ${1\over
s_{11}^{2}}$ and ${1\over s_{11}}$. One can easily check that the result (23)
will remove the pole ${1\over s_{11}^{2}}$ automatically, while the
cancellation of the pole ${1\over s_{11}}$ gives

$\displaystyle\mathcal{B}^{(1)}_{2\to 2;\widehat{0}}$ $\displaystyle=$
$\displaystyle-\frac{DM_{0}^{2}-(D-4)M_{1}^{2}}{D\left(M_{0}-M_{1}\right){}^{3}\left(M_{0}+M_{1}\right){}^{3}}\,,$
$\displaystyle\mathcal{B}^{(1)}_{2\to 2;\widehat{1}}$ $\displaystyle=$
$\displaystyle-\frac{(D-4)M_{0}^{2}-DM_{1}^{2}}{D\left(M_{0}-M_{1}\right){}^{3}\left(M_{0}+M_{1}\right){}^{3}}\,.~{}~{}~{}$
(25)

Now the idea is clear. To get the higher-order decomposition coefficients
$\mathcal{B}^{(a)}_{2\to 2;\widehat{i}}$, we just need to impose the
smoothness limit of tensor reduction with higher ranks. In fact, we can get
expansion coefficients $\mathcal{B}^{(a)}_{2\to 2;\widehat{i}}$ with $a\leq m$
for bubble through calculation with tensor rank $(m+1)$.

Before ending this subsection, let us point out that the decomposition
coefficients given in (23) and (25) have a nice symmetry between
$M_{0}\leftrightarrow M_{1}$ and they become singular under the limit
$M_{1}\to M_{0}$. In that limit, $I_{2;\widehat{0}}$ and $I_{2;\widehat{1}}$
are not linearly independent of each other, and we should expand

$\displaystyle I_{2;\widehat{1}}$ $\displaystyle=$
$\displaystyle\sum_{a=0}^{\infty}\mathcal{B}^{(a)}_{(2;\widehat{1})\to(2;\widehat{0})}(M_{1}^{2}-M_{0}^{2})^{a}I_{2;\widehat{0}}\,.~{}~{}~{}~{}$
(26)

Putting (26) into (21) with known $\mathcal{B}^{(a)}_{2\to 2;\widehat{i}}$ and
requiring the smoothness under the limit $M_{1}\to M_{0}$, we can calculate
all the $\mathcal{B}^{(a)}_{(2;\widehat{1})\to(2;\widehat{0})}$. For example,
with results (23) and (25), we get

$\displaystyle\mathcal{B}^{(0)}_{(2;\widehat{1})\to(2;\widehat{0})}=1\,,~{}~{}~{}\mathcal{B}^{(1)}_{(2;\widehat{1})\to(2;\widehat{0})}=\frac{D-2}{2M_{0}^{2}}\,,~{}~{}~{}\mathcal{B}^{(2)}_{(2;\widehat{1})\to(2;\widehat{0})}=\frac{D^{2}-6D+8}{8M_{0}^{4}}\,.$
(27)

###  3.2 Algorithm for general case

Now we are ready to discuss the degeneration of scalar $(n+1)$-gon integral
when the Gram determinant $|G|\to 0$. The Gram matrices corresponding to lower
topologies don’t degenerate generally with $|G|\to 0$, thus the scalar
$(n+1)$-gon integral will become a linear combination of scalar $m$-gon
integrals with $m\leq n$, i.e.,

$\displaystyle
I_{n+1}=\sum_{|\mathbf{b}_{j}|=1}^{n}\mathcal{B}_{n+1;\widehat{\mathbf{b}_{j}}}I_{n+1;\widehat{\mathbf{b}_{j}}}=\sum_{|\mathbf{b}_{j}|=1}^{n}\sum_{a=0}^{\infty}\mathcal{B}^{(a)}_{n+1;\widehat{\mathbf{b}_{j}}}|G|^{a}I_{n+1;\widehat{\mathbf{b}_{j}}}\,,$
(28)

where at the second equation, the Taylor expansion of $|G|$ is given
explicitly. To determine these decomposition coefficients
$\mathcal{B}^{(a)}_{n+1;\widehat{\mathbf{b}_{j}}}$, we put (28) to (6) and get

$\displaystyle I_{n+1}^{(m)}$ $\displaystyle=$
$\displaystyle\sum_{|\mathbf{b}_{j}|=0}^{n}C^{(m)}_{n+1\to
n+1;\widehat{\mathbf{b}_{j}}}I_{n+1;\widehat{\mathbf{b}_{j}}}=C^{(m)}_{n+1\to
n+1}I_{n+1}+\sum_{|\mathbf{b}_{j}|=1}^{n}C^{(m)}_{n+1\to
n+1;\widehat{\mathbf{b}_{j}}}I_{n+1;\widehat{\mathbf{b}_{j}}}$ (29)
$\displaystyle=$
$\displaystyle\sum_{|\mathbf{b}_{j}|=1}^{n}\left(C^{(m)}_{n+1\to
n+1;\widehat{\mathbf{b}_{j}}}+C^{(m)}_{n+1\to
n+1}\mathcal{B}_{n+1;\widehat{\mathbf{b}_{j}}}\right)I_{n+1;\widehat{\mathbf{b}_{j}}}\,.~{}~{}~{}$

For later convenience, let us write

$\displaystyle\mathcal{F}^{(m)}_{n+1\to
n+1;\widehat{\mathbf{b}_{j}}}=\left(C^{(m)}_{n+1\to
n+1;\widehat{\mathbf{b}_{j}}}+C^{(m)}_{n+1\to
n+1}\mathcal{B}_{n+1;\widehat{\mathbf{b}_{j}}}\right)\,.~{}~{}~{}$ (30)

The smoothness of $I_{n+1}^{(m)}$ under the limit $|G|\to 0$ means that
$\mathcal{F}^{(m)}_{n+1\to n+1;\widehat{\mathbf{b}_{j}}}$ is a Taylor series
of $|G|$. Using this condition, we can get all the decomposition coefficients.

Now we show how to do it more directly instead of solving linear equations.
Since $C^{(m)}_{n+1\to n+1;\widehat{\mathbf{b}_{j}}}$ is a Laurent series
about the Gram determinant $|G|$, for example,

$\displaystyle C^{(m)}_{n+1\to n+1}=\sum_{i=-m}^{\infty}C^{(m;i)}_{n+1\to
n+1}|G|^{i}\,,~{}~{}~{}$ (31)

using (30), we can write

$\displaystyle\mathcal{B}_{n+1;\widehat{\mathbf{b}_{j}}}=$
$\displaystyle\frac{\mathcal{F}^{(m)}_{n+1\to
n+1;\widehat{\mathbf{b}_{j}}}-C^{(m)}_{n+1\to
n+1;\widehat{\mathbf{b}_{j}}}}{C^{(m)}_{n+1\to n+1}}=-\frac{C^{(m)}_{n+1\to
n+1;\widehat{\mathbf{b}_{j}}}}{C^{(m)}_{n+1\to
n+1}}+\frac{\mathcal{F}^{(m)}_{n+1\to
n+1;\widehat{\mathbf{b}_{j}}}}{C^{(m)}_{n+1\to n+1}}\,.$ (32)

In (32), $\mathcal{F}^{(m)}_{n+1\to n+1;\widehat{\mathbf{b}_{j}}}$ is regular
and $C^{(m)}_{n+1\to n+1}\sim|G|^{-m}$, so we just need to keep the first term
in the numerator of (32) to get the coefficients
$\mathcal{B}^{(k<m)}_{n+1;\widehat{\mathbf{b}_{j}}}$, i.e.,

$\displaystyle\mathcal{B}_{n+1;\widehat{\mathbf{b}_{j}}}=-{\frac{C^{(m)}_{n+1\to
n+1;\widehat{\mathbf{b}_{j}}}}{C^{(m)}_{n+1\to
n+1}}}+\mathcal{O}(|G|^{m})\,.~{}~{}~{}$ (33)

Using (33), we can get

$\displaystyle\mathcal{F}^{(m)}_{n+1\to n+1;\widehat{\mathbf{b}_{j}}}$
$\displaystyle=$ $\displaystyle\left(C^{(m)}_{n+1\to
n+1;\widehat{\mathbf{b}_{j}}}+\left(-{\frac{C^{(m^{\prime})}_{n+1\to
n+1;\widehat{\mathbf{b}_{j}}}}{C^{(m^{\prime})}_{n+1\to
n+1}}}+\mathcal{O}(|G|^{m^{\prime}})\right)C^{(m)}_{n+1\to n+1}\right)$ (34)
$\displaystyle=$ $\displaystyle{\frac{C^{(m)}_{n+1\to
n+1;\widehat{\mathbf{b}_{j}}}C^{(m^{\prime})}_{n+1\to
n+1}-C^{(m^{\prime})}_{n+1\to n+1;\widehat{\mathbf{b}_{j}}}C^{(m)}_{n+1\to
n+1}}{C^{(m^{\prime})}_{n+1\to
n+1}}}+\mathcal{O}(|G|^{m^{\prime}-m})\,.~{}~{}~{}$

Here we use the fact $C^{(m)}_{n+1\to n+1}\sim|G|^{-m}$. To get the accurate
expansion of $\mathcal{F}^{(m)}_{n+1\to n+1;\widehat{\mathbf{b}_{j}}}$ up to
$k$-th order of $|G|$, we set $m^{\prime}-m=k+1$, then we can write

$\displaystyle\mathcal{F}^{(m)}_{n+1\to
n+1;\widehat{\mathbf{b}_{j}}}=\frac{C^{(m)}_{n+1\to
n+1;\widehat{\mathbf{b}_{j}}}C^{(m+k+1)}_{n+1\to n+1}-C^{(m+k+1)}_{n+1\to
n+1;\widehat{\mathbf{b}_{j}}}C^{(m)}_{n+1\to n+1}}{C^{(m+k+1)}_{n+1\to
n+1}}+{\cal O}(|G|^{k+1})\,.~{}~{}~{}$ (35)

If we take $k=0$ in (35), it gives

$\displaystyle F^{(m)}_{n+1\to n+1;\widehat{\mathbf{b}_{j}}}=\lim_{|G|\to
0}\left(\frac{C^{(m)}_{n+1\to n+1;\widehat{\mathbf{b}_{j}}}C^{(m+1)}_{n+1\to
n+1}-C^{(m+1)}_{n+1\to n+1;\widehat{\mathbf{b}_{j}}}C^{(m)}_{n+1\to
n+1}}{C^{(m+1)}_{n+1\to n+1}}\right)\,,~{}~{}~{}$ (36)

which gives the reduction coefficients for $I_{n+1}^{(m)}$ when $|G|=0$. The
same result (36) has been obtained by methods in projective space in
Feng:2022rwj . For the expansion (28), if we care about only the zeroth order,
the result is also well known, which is given by Denner:1991kt ;
Melrose:1965kb ; Ossola_2007  (see Eq(5.8) in Ossola_2007 )

$\left|\begin{array}[]{ccccc}I_{n+1}&-I_{n+1,\widehat{0}}&-I_{n+1,\widehat{1}}&\ldots&-I_{n+1,\widehat{n}}\\\
1&Y_{0\,0}&Y_{0\,1}&\ldots&Y_{0\,n}\\\ 1&Y_{1\,0}&Y_{1\,1}&\ldots&Y_{1\,n}\\\
\vdots&\vdots&\vdots&\vdots&\vdots\\\
1&Y_{n\,0}&Y_{n\,1}&\ldots&Y_{n\,n}\end{array}\right|=0\,,~{}~{}~{}$ (37)

with

$\displaystyle
Y_{ij}=M_{i}^{2}+N_{j}^{2}-(K_{i}-K_{j})^{2},\,\,\,i,j=0,\ldots,n.\,\,\,$ (38)

A good feature of the result (33) is that we can get the higher-order
expansion, which may be helpful for a better numerical evaluation around the
point $|G|=0$.

Having established the general formula in (33) and (36), let us redo the
example of the bubble in the previous subsection:

  * •

Up to the order $s_{11}^{0}$: we have

$\displaystyle\mathcal{B}_{2;\widehat{1}}$ $\displaystyle=$
$\displaystyle-{\frac{C^{(1)}_{2\to 2;\widehat{1}}}{C^{(1)}_{2\to
2}}}=\frac{1}{M_{0}^{2}-M_{1}^{2}}+\mathcal{O}\left(|G|\right)\,,$
$\displaystyle\mathcal{B}_{2;\widehat{0}}$ $\displaystyle=$
$\displaystyle-{\frac{C^{(1)}_{2\to 2;\widehat{0}}}{C^{(1)}_{2\to
2}}}=\frac{1}{M_{1}^{2}-M_{0}^{2}}+\mathcal{O}\left(|G|\right)\,.$ (39)

after using the result (20).

  * •

Up to the order $s_{11}^{1}$: we have

$\displaystyle\mathcal{B}_{2;\widehat{1}}$ $\displaystyle=$
$\displaystyle-{\frac{C^{(2)}_{2\to 2;\widehat{1}}}{C^{(2)}_{2\to
2}}}=\frac{1}{M_{0}^{2}-M_{1}^{2}}+\frac{\Big{(}DM_{1}^{2}-(D-4)M_{0}^{2}\Big{)}|G|}{D\left(M_{0}^{2}-M_{1}^{2}\right){}^{3}}+\mathcal{O}\left(|G|^{2}\right)\,,$
$\displaystyle\mathcal{B}_{2;\widehat{0}}$ $\displaystyle=$
$\displaystyle-{\frac{C^{(2)}_{2\to 2;\widehat{0}}}{C^{(2)}_{2\to
2}}}=\frac{1}{M_{1}^{2}-M_{0}^{2}}+\frac{\Big{(}(D-4)M_{1}^{2}-DM_{0}^{2}\Big{)}|G|}{D\left(M_{0}^{2}-M_{1}^{2}\right){}^{3}}+\mathcal{O}\left(|G|^{2}\right)\,.$
(40)

after using the result (24).

  * •

Up to the order $s_{11}^{2}$: we have

$\displaystyle\mathcal{B}_{2;\widehat{1}}$ $\displaystyle=$
$\displaystyle-{\frac{C^{(3)}_{2\to 2;\widehat{1}}}{C^{(3)}_{2\to
2}}}=\frac{1}{M_{0}^{2}-M_{1}^{2}}+\frac{\Big{(}DM_{1}^{2}-(D-4)M_{0}^{2}\Big{)}|G|}{D\left(M_{0}^{2}-M_{1}^{2}\right){}^{3}}$
$\displaystyle+\frac{\Big{(}\left(D^{2}-10D+24\right)M_{0}^{4}-2\left(D^{2}-4D-12\right)M_{1}^{2}M_{0}^{2}+D(D+2)M_{1}^{4}\Big{)}|G|^{2}}{D(D+2)\left(M_{0}^{2}-M_{1}^{2}\right){}^{5}}$
$\displaystyle+\mathcal{O}\left(|G|^{3}\right)\,,$
$\displaystyle\mathcal{B}_{2;\widehat{0}}$ $\displaystyle=$
$\displaystyle-{\frac{C^{(3)}_{2\to 2;\widehat{0}}}{C^{(3)}_{2\to
2}}}=\frac{1}{M_{1}^{2}-M_{0}^{2}}+\frac{\Big{(}(D-4)M_{1}^{2}-DM_{0}^{2}\Big{)}|G|}{D\left(M_{0}^{2}-M_{1}^{2}\right){}^{3}}$
(41)
$\displaystyle-\frac{\Big{(}\left(D^{2}-10D+24\right)M_{1}^{4}-2\left(D^{2}-4D-12\right)M_{1}^{2}M_{0}^{2}+D(D+2)M_{0}^{4}\Big{)}|G|^{2}}{D(D+2)\left(M_{0}^{2}-M_{1}^{2}\right){}^{5}}$
$\displaystyle+\mathcal{O}\left(|G|^{3}\right)\,.$

after using the result in Hu:2021nia  (see Eq(4.11) to Eq(4.16)).

###  3.3 The degenerate triangle

In this subsection, we will present some results for the degenerate triangle.
First, we parameterize the Gram determinant as555One should be very careful
about the kinematics. For example, in the calculation of tensor triangle, we
cannot treat $s_{11},s_{12},s_{22}$ and $|G|$ as four independent Lorentz
invariant combinations. Instead, we should replace, for example, every
$s_{22}$ with $(|G|+s^{2}_{12})/s_{11}$.

$\displaystyle G_{\text{tri}}=s_{11}s_{22}-s_{12}^{2}=|G|\,,$ (42)

where $s_{ij}\equiv K_{i}\cdot K_{j}$ and the parameter $|G|$ controls the
speed of degeneration. We expand the Master Integral $I_{3}$ as

$\displaystyle I_{3}$ $\displaystyle=$
$\displaystyle\sum_{i_{1}=0}^{2}\mathcal{B}_{3;\widehat{i_{1}}}I_{3;\widehat{i_{1}}}+\sum_{0\leq
i_{1}<i_{2}}^{2}\mathcal{B}_{3;\widehat{i_{1}i_{2}}}I_{3;\widehat{i_{1}i_{2}}}\,.$
(43)

We will calculate several typical $\mathcal{B}_{3;\widehat{\mathbf{b}_{j}}}$
by (33), while others can be obtained by proper permutation.

  * •

$\mathcal{B}_{3;\widehat{2}}$: Using the tensor rank 1 we get

$\displaystyle\mathcal{B}_{3;\widehat{2}}=\frac{s_{11}}{M_{0}^{2}\left(s_{11}-s_{12}\right)-M_{2}^{2}s_{11}+s_{12}\left(M_{1}^{2}-s_{11}+s_{12}\right)}+\mathcal{O}\left(|G|\right)\,,~{}~{}~{}$
(44)

while using the tensor rank 2 we get

$\displaystyle\mathcal{B}_{3;\widehat{2}}$ $\displaystyle=$
$\displaystyle\frac{s_{11}}{M_{0}^{2}\left(s_{11}-s_{12}\right)-M_{2}^{2}s_{11}+s_{12}\left(M_{1}^{2}-s_{11}+s_{12}\right)}$
(45)
$\displaystyle-\frac{s_{11}|G|}{(D-1)\left(M_{0}^{2}\left(s_{11}-s_{12}\right)-M_{2}^{2}s_{11}+s_{12}\left(M_{1}^{2}-s_{11}+s_{12}\right)\right){}^{3}}\times$
$\displaystyle\Big{(}s_{11}\left(DM_{0}^{2}-DM_{2}^{2}-3M_{0}^{2}-2M_{1}^{2}+M_{2}^{2}\right)+(D-1)s_{12}^{2}+M_{1}^{4}+s_{11}^{2}+$
$\displaystyle
s_{12}\left(-DM_{0}^{2}+DM_{1}^{2}+M_{0}^{2}-M_{1}^{2}\right)+M_{0}^{4}-2M_{1}^{2}M_{0}^{2}+(1-D)s_{11}s_{12}\Big{)}$
$\displaystyle+\mathcal{O}\left(|G|^{2}\right)\,.~{}~{}~{}~{}~{}$

One can see that for zeroth order, results in (44) and (45) are same. Similar
results hold for other coefficients.

  * •

$\mathcal{B}_{3;\widehat{0}}$: Using the tensor rank 1 we get

$\displaystyle\mathcal{B}_{3;\widehat{0}}=\frac{s_{12}-s_{11}}{M_{0}^{2}\left(s_{11}-s_{12}\right)-M_{2}^{2}s_{11}+s_{12}\left(M_{1}^{2}-s_{11}+s_{12}\right)}+\mathcal{O}\left(|G|\right)\,,$
(46)

while using the tensor rank 2 we get

$\displaystyle\mathcal{B}_{3;\widehat{0}}=$
$\displaystyle\frac{s_{12}-s_{11}}{M_{0}^{2}\left(s_{11}-s_{12}\right)-M_{2}^{2}s_{11}+s_{12}\left(M_{1}^{2}-s_{11}+s_{12}\right)}$
(47)
$\displaystyle+\frac{|G|}{(D-1)\left(s_{11}-s_{12}\right)\left(M_{0}^{2}\left(s_{12}-s_{11}\right)+M_{2}^{2}s_{11}-s_{12}\left(M_{1}^{2}-s_{11}+s_{12}\right)\right){}^{3}}\times$
$\displaystyle\Big{(}s_{11}^{2}\big{(}DM_{0}^{2}M_{1}^{2}-DM_{2}^{2}M_{1}^{2}+DM_{2}^{4}-DM_{0}^{2}M_{2}^{2}-M_{1}^{4}-M_{0}^{2}M_{1}^{2}+3M_{2}^{2}M_{1}^{2}-$
$\displaystyle
2M_{2}^{4}+M_{0}^{2}M_{2}^{2}\big{)}+s_{12}s_{11}^{2}\left(-DM_{1}^{2}+DM_{2}^{2}-3M_{1}^{2}-5M_{2}^{2}\right)+s_{12}^{2}s_{11}\big{(}DM_{1}^{2}-$
$\displaystyle
DM_{2}^{2}+M_{1}^{2}+3M_{2}^{2}\big{)}+s_{12}s_{11}\big{(}DM_{1}^{4}-DM_{0}^{2}M_{1}^{2}-DM_{2}^{2}M_{1}^{2}+DM_{0}^{2}M_{2}^{2}-$
$\displaystyle
M_{1}^{4}+M_{0}^{2}M_{1}^{2}+M_{2}^{2}M_{1}^{2}-M_{0}^{2}M_{2}^{2}\big{)}+\left(2M_{1}^{2}+2M_{2}^{2}\right)s_{11}^{3}-s_{11}^{4}+4s_{12}s_{11}^{3}-$
$\displaystyle
6s_{12}^{2}s_{11}^{2}+4s_{12}^{3}s_{11}-s_{12}^{4}\Big{)}+\mathcal{O}\left(|G|^{2}\right)\,.$

  * •

$\mathcal{B}_{3;\widehat{12}}$: Using the tensor rank 1 we get

$\displaystyle\mathcal{B}_{3;\widehat{12}}=0+\mathcal{O}\left(|G|\right)\,,$
(48)

while using the tensor rank 2 we get

$\displaystyle\mathcal{B}_{3;\widehat{12}}=\frac{(D-2)s_{11}|G|}{(D-1)s_{12}\left(M_{0}^{2}\left(s_{11}-s_{12}\right)-M_{2}^{2}s_{11}+s_{12}\left(M_{1}^{2}-s_{11}+s_{12}\right)\right){}^{2}}+\mathcal{O}\left(|G|^{2}\right)\,.$
(49)

Finally using the tensor rank 3 we get

$\displaystyle\mathcal{B}_{3;\widehat{12}}$ $\displaystyle=$
$\displaystyle\frac{(D-2)s_{11}|G|}{(D-1)s_{12}\left(M_{0}^{2}\left(s_{11}-s_{12}\right)-M_{2}^{2}s_{11}+s_{12}\left(M_{1}^{2}-s_{11}+s_{12}\right)\right){}^{2}}$
(50)
$\displaystyle-\frac{(D-2)s_{11}|G|^{2}}{\left(D^{2}-1\right)s_{12}^{3}\left(M_{0}^{2}\left(s_{11}-s_{12}\right)-M_{2}^{2}s_{11}+s_{12}\left(M_{1}^{2}-s_{11}+s_{12}\right)\right){}^{4}}\times$
$\displaystyle\Big{(}s_{12}^{3}\left(-3DM_{0}^{2}+3DM_{1}^{2}-2M_{0}^{2}+2M_{1}^{2}\right)+s_{12}^{2}\big{(}DM_{0}^{4}-2DM_{1}^{2}M_{0}^{2}+$
$\displaystyle
DM_{1}^{4}+2M_{0}^{4}-4M_{1}^{2}M_{0}^{2}+2M_{1}^{4}\big{)}+s_{11}s_{12}^{2}\big{(}4DM_{0}^{2}-2DM_{1}^{2}-2DM_{2}^{2}-$
$\displaystyle
4M_{0}^{2}-4M_{1}^{2}-4M_{2}^{2}\big{)}+s_{11}^{2}s_{12}\left(DM_{2}^{2}-DM_{0}^{2}\right)+s_{11}s_{12}\big{(}-DM_{0}^{4}+$
$\displaystyle
DM_{1}^{2}M_{0}^{2}+DM_{2}^{2}M_{0}^{2}-DM_{1}^{2}M_{2}^{2}\big{)}+(2D+3)s_{12}^{4}+(-3D-2)s_{11}s_{12}^{3}$
$\displaystyle+(D+2)s_{11}^{2}s_{12}^{2}+\left(M_{0}^{4}-2M_{2}^{2}M_{0}^{2}+M_{2}^{4}\right)s_{11}^{2}\Big{)}+\mathcal{O}\left(|G|^{3}\right)\,.~{}~{}~{}~{}~{}~{}~{}$

  * •

$\mathcal{B}_{3;\widehat{02}}$: Using the tensor rank 1 we get

$\displaystyle\mathcal{B}_{3;\widehat{02}}=0+\mathcal{O}\left(|G|\right)\,,$
(51)

while using the tensor rank 2 we get

$\displaystyle\mathcal{B}_{3;\widehat{02}}=$
$\displaystyle\frac{(D-2)s_{11}|G|}{(D-1)\left(s_{11}-s_{12}\right)\left(M_{0}^{2}\left(s_{12}-s_{11}\right)+M_{2}^{2}s_{11}-s_{12}\left(M_{1}^{2}-s_{11}+s_{12}\right)\right){}^{2}}$
(52) $\displaystyle+\mathcal{O}\left(|G|^{2}\right)\,.$

Finally using the tensor rank 3 we get

$\displaystyle\mathcal{B}_{3;\widehat{02}}=$
$\displaystyle\frac{(D-2)s_{11}|G|}{(D-1)\left(s_{11}-s_{12}\right)\left(M_{0}^{2}\left(s_{12}-s_{11}\right)+M_{2}^{2}s_{11}-s_{12}\left(M_{1}^{2}-s_{11}+s_{12}\right)\right){}^{2}}$
$\displaystyle-\frac{(D-2)s_{11}|G|^{2}}{\left(D^{2}-1\right)\left(s_{11}-s_{12}\right){}^{3}\left(M_{0}^{2}\left(s_{12}-s_{11}\right)+M_{2}^{2}s_{11}-s_{12}\left(M_{1}^{2}-s_{11}+s_{12}\right)\right){}^{4}}\times$
$\displaystyle\Big{(}s_{12}s_{11}\big{(}-2DM_{0}^{4}+3DM_{1}^{2}M_{0}^{2}+DM_{2}^{2}M_{0}^{2}-DM_{1}^{4}-DM_{1}^{2}M_{2}^{2}-4M_{0}^{4}+$
$\displaystyle
8M_{1}^{2}M_{0}^{2}-4M_{1}^{4}\big{)}+s_{11}^{3}\left(DM_{0}^{2}-DM_{2}^{2}-2M_{0}^{2}-6M_{1}^{2}-4M_{2}^{2}\right)+s_{11}^{2}\big{(}DM_{0}^{4}-$
$\displaystyle
DM_{1}^{2}M_{0}^{2}-DM_{2}^{2}M_{0}^{2}+DM_{1}^{2}M_{2}^{2}+2M_{0}^{4}-4M_{1}^{2}M_{0}^{2}+3M_{1}^{4}+M_{2}^{4}-2M_{1}^{2}M_{2}^{2}\big{)}$
$\displaystyle+s_{12}s_{11}^{2}\left(-5DM_{0}^{2}+2DM_{1}^{2}+3DM_{2}^{2}+2M_{0}^{2}+14M_{1}^{2}+8M_{2}^{2}\right)+s_{12}^{3}\big{(}2M_{1}^{2}-$
$\displaystyle
3DM_{0}^{2}+3DM_{1}^{2}-2M_{0}^{2}\big{)}+s_{12}^{2}s_{11}\big{(}7DM_{0}^{2}-5DM_{1}^{2}-2DM_{2}^{2}+2M_{0}^{2}-4M_{2}^{2}-$
$\displaystyle
10M_{1}^{2}\big{)}-(D+10)s_{12}s_{11}^{3}+(4D+14)s_{12}^{2}s_{11}^{2}-(5D+10)s_{12}^{3}s_{11}+3s_{11}^{4}+$
$\displaystyle(2D+3)s_{12}^{4}+s_{12}^{2}\left(DM_{0}^{4}-2DM_{1}^{2}M_{0}^{2}+DM_{1}^{4}+2M_{0}^{4}-4M_{1}^{2}M_{0}^{2}+2M_{1}^{4}\right)\Big{)}$
$\displaystyle+\mathcal{O}\left(|G|^{3}\right)\,.$ (53)

Note that the expansion of $\mathcal{B}_{n+1;\widehat{\mathbf{b}_{j}}}$ is
independent of $R$ although the numerator and the denominator depend on $R$ on
the RHS of (33). And it’s also independent of the tensor rank $m$ up to the
order $|G|^{k}$ as long as $m>k$. The above results confirm these nontrivial
points, and we will give a proof in the later section.

The reduction coefficients $F^{(m)}_{3\to 3;\widehat{\mathbf{b}_{j}}}$ for
$|G|=0$ (see (36)) are given in following:

  * •

$m=0$:

$\displaystyle F^{(0)}_{3\to 3;\widehat{0}}$ $\displaystyle=$
$\displaystyle\mathcal{B}^{(0)}_{3\to
3;\widehat{0}}=-\left(\frac{C^{(1)}_{3\to 3;\widehat{0}}}{C^{(1)}_{3\to
3}}\right)\Bigg{|}_{\left|G\right|=0}$ (54a) $\displaystyle=$
$\displaystyle\frac{s_{12}-s_{11}}{M_{1}^{2}\left(s_{11}-s_{12}\right)-M_{3}^{2}s_{11}+s_{12}\left(M_{2}^{2}-s_{11}+s_{12}\right)}\,,$
$\displaystyle F^{(0)}_{3\to
3;\widehat{1}}=-\frac{s_{12}}{M_{1}^{2}\left(s_{11}-s_{12}\right)-M_{3}^{2}s_{11}+s_{12}\left(M_{2}^{2}-s_{11}+s_{12}\right)}\,,$
(54b) $\displaystyle F^{(0)}_{3\to
3;\widehat{2}}=\frac{s_{11}}{M_{1}^{2}\left(s_{11}-s_{12}\right)-M_{3}^{2}s_{11}+s_{12}\left(M_{2}^{2}-s_{11}+s_{12}\right)}\,,$
(54c) $\displaystyle F^{(0)}_{3\to 3;\widehat{12}}=F^{(0)}_{3\to
3;\widehat{01}}=F^{(0)}_{3\to 3;\widehat{02}}=0\,.$ (54d)

  * •

$m=1$:

$\displaystyle F^{(1)}_{3\to 3;\widehat{0}}=$
$\displaystyle\frac{1}{(D-1)s_{11}\left(s_{11}-s_{12}\right)\left(\left(s_{12}-s_{11}\right)M_{1}^{2}+M_{3}^{2}s_{11}-s_{12}\left(M_{2}^{2}-s_{11}+s_{12}\right)\right){}^{2}}\times$
$\displaystyle\Big{(}s_{02}s_{11}^{5}+\left(-DM_{1}^{2}+M_{1}^{2}+DM_{3}^{2}-M_{3}^{2}\right)s_{01}s_{11}^{4}+(D-2)s_{12}s_{01}s_{11}^{4}+$
$\displaystyle\left(-DM_{1}^{2}+M_{1}^{2}-2M_{2}^{2}+DM_{3}^{2}-3M_{3}^{2}\right)s_{02}s_{11}^{4}+(D-5)s_{12}s_{02}s_{11}^{4}+$
$\displaystyle(7-3D)s_{12}^{2}s_{01}s_{11}^{3}+\big{(}DM_{3}^{4}-M_{3}^{4}-DM_{1}^{2}M_{3}^{2}+M_{1}^{2}M_{3}^{2}-DM_{2}^{2}M_{3}^{2}+$
$\displaystyle
M_{2}^{2}M_{3}^{2}+DM_{1}^{2}M_{2}^{2}-M_{1}^{2}M_{2}^{2}\big{)}s_{01}s_{11}^{3}+\big{(}3DM_{1}^{2}-3M_{1}^{2}-2DM_{2}^{2}+$
$\displaystyle
4M_{2}^{2}-DM_{3}^{2}+3M_{3}^{2}\big{)}s_{12}s_{01}s_{11}^{3}+(9-3D)s_{12}^{2}s_{02}s_{11}^{3}+(5-D)s_{12}^{4}s_{01}s_{11}$
$\displaystyle+\left(3DM_{1}^{2}-3M_{1}^{2}+4M_{2}^{2}-3DM_{3}^{2}+7M_{3}^{2}\right)s_{12}s_{02}s_{11}^{3}+\big{(}-3DM_{1}^{2}+$
$\displaystyle
3M_{1}^{2}+3DM_{2}^{2}-7M_{2}^{2}-4M_{3}^{2}\big{)}s_{12}^{2}s_{01}s_{11}^{2}+\big{(}DM_{2}^{4}-2M_{2}^{4}-DM_{1}^{2}M_{2}^{2}+$
$\displaystyle
M_{1}^{2}M_{2}^{2}-DM_{3}^{2}M_{2}^{2}+3M_{3}^{2}M_{2}^{2}-M_{3}^{4}+DM_{1}^{2}M_{3}^{2}-M_{1}^{2}M_{3}^{2}\big{)}s_{12}s_{01}s_{11}^{2}$
$\displaystyle+(3D-7)s_{12}^{3}s_{02}s_{11}^{2}+\big{(}-3DM_{1}^{2}+3M_{1}^{2}+DM_{2}^{2}-3M_{2}^{2}+2DM_{3}^{2}$
$\displaystyle-4M_{3}^{2}\big{)}s_{12}^{2}s_{02}s_{11}^{2}+\big{(}-DM_{2}^{4}+M_{2}^{4}+DM_{1}^{2}M_{2}^{2}-M_{1}^{2}M_{2}^{2}+DM_{3}^{2}M_{2}^{2}$
$\displaystyle-
M_{3}^{2}M_{2}^{2}-DM_{1}^{2}M_{3}^{2}+M_{1}^{2}M_{3}^{2}\big{)}s_{12}s_{02}s_{11}^{2}+\big{(}M_{2}^{4}-DM_{1}^{2}M_{2}^{2}+M_{1}^{2}M_{2}^{2}+$
$\displaystyle
DM_{3}^{2}M_{2}^{2}-3M_{3}^{2}M_{2}^{2}-DM_{3}^{4}+2M_{3}^{4}+DM_{1}^{2}M_{3}^{2}-M_{1}^{2}M_{3}^{2}\big{)}s_{02}s_{11}^{3}$
$\displaystyle+\big{(}DM_{1}^{2}-M_{1}^{2}-DM_{2}^{2}+3M_{2}^{2}+2M_{3}^{2}\big{)}s_{12}^{3}s_{01}s_{11}+(2-D)s_{12}^{4}s_{02}s_{11}$
$\displaystyle+\big{(}DM_{1}^{2}-M_{1}^{2}-DM_{2}^{2}+M_{2}^{2}\big{)}s_{12}^{3}s_{02}s_{11}-s_{12}^{5}s_{01}+(3D-9)s_{12}^{3}s_{01}s_{11}^{2}\Big{)}\,,$
$\displaystyle F^{(1)}_{3\to 3;\widehat{1}}$ $\displaystyle=$
$\displaystyle\frac{1}{(D-1)s_{11}s_{12}\left(M_{1}^{2}\left(s_{12}-s_{11}\right)+M_{3}^{2}s_{11}-s_{12}\left(M_{2}^{2}-s_{11}+s_{12}\right)\right){}^{2}}\times$
(55b)
$\displaystyle\Big{(}s_{11}s_{12}^{3}\left(DM_{1}^{2}-DM_{2}^{2}-M_{1}^{2}+M_{2}^{2}\right)s_{02}+s_{11}^{2}s_{12}^{2}\big{(}-2DM_{1}^{2}+2DM_{3}^{2}-$
$\displaystyle
4M_{3}^{2}\big{)}s_{02}+s_{11}^{3}s_{12}\left(DM_{1}^{2}-DM_{3}^{2}-M_{1}^{2}+M_{3}^{2}\right)s_{02}+(2-D)s_{11}s_{12}^{4}s_{02}$
$\displaystyle+s_{11}^{2}s_{12}\big{(}DM_{1}^{4}-DM_{2}^{2}M_{1}^{2}-DM_{3}^{2}M_{1}^{2}+DM_{2}^{2}M_{3}^{2}-M_{1}^{4}+M_{2}^{2}M_{1}^{2}+$
$\displaystyle
M_{3}^{2}M_{1}^{2}-M_{2}^{2}M_{3}^{2}\big{)}s_{02}+(D-1)s_{11}^{2}s_{12}^{3}s_{02}+\left(2M_{1}^{2}+2M_{3}^{2}\right)s_{11}s_{12}^{3}s_{01}$
$\displaystyle+s_{11}^{3}\left(-DM_{1}^{4}+2DM_{3}^{2}M_{1}^{2}-DM_{3}^{4}+2M_{1}^{4}-4M_{3}^{2}M_{1}^{2}+2M_{3}^{4}\right)s_{02}$
$\displaystyle+\left(-M_{1}^{4}+2M_{3}^{2}M_{1}^{2}-M_{3}^{4}\right)s_{11}^{2}s_{12}s_{01}-s_{12}^{5}s_{01}\Big{)}\,,$
$\displaystyle F^{(1)}_{3\to 3;\widehat{2}}$ $\displaystyle=$
$\displaystyle\frac{1}{(D-1)\left(M_{1}^{2}\left(s_{11}-s_{12}\right)-M_{3}^{2}s_{11}+s_{12}\left(M_{2}^{2}-s_{11}+s_{12}\right)\right){}^{2}}\times$
(55c)
$\displaystyle\Big{(}s_{11}^{2}\left(DM_{1}^{2}-DM_{3}^{2}-M_{1}^{2}+M_{3}^{2}\right)s_{01}+s_{11}\big{(}DM_{1}^{4}-DM_{2}^{2}M_{1}^{2}-$
$\displaystyle
DM_{3}^{2}M_{1}^{2}+DM_{2}^{2}M_{3}^{2}-M_{1}^{4}+M_{2}^{2}M_{1}^{2}+M_{3}^{2}M_{1}^{2}-M_{2}^{2}M_{3}^{2}\big{)}s_{01}$
$\displaystyle+s_{12}s_{11}\left(-2DM_{1}^{2}+2DM_{2}^{2}-4M_{2}^{2}\right)s_{01}+s_{12}^{2}\big{(}DM_{1}^{2}-DM_{2}^{2}-$
$\displaystyle
M_{1}^{2}+M_{2}^{2}\big{)}s_{01}+s_{12}\big{(}-DM_{1}^{4}+2DM_{2}^{2}M_{1}^{2}-DM_{2}^{4}+2M_{1}^{4}$
$\displaystyle-4M_{2}^{2}M_{1}^{2}+2M_{2}^{4}\big{)}s_{01}+(2-D)s_{12}s_{11}^{2}s_{01}+(D-1)s_{12}^{2}s_{11}s_{01}$
$\displaystyle+\left(2M_{1}^{2}+2M_{2}^{2}\right)s_{11}^{2}s_{02}+\left(-M_{1}^{4}+2M_{2}^{2}M_{1}^{2}-M_{2}^{4}\right)s_{11}s_{02}$
$\displaystyle-s_{11}^{3}s_{02}\Big{)}\,,$ $\displaystyle F^{(1)}_{3\to
3;\widehat{12}}$ $\displaystyle=$
$\displaystyle-\frac{(D-2)\left(s_{12}s_{01}-s_{11}s_{02}\right)}{(D-1)s_{12}\left(M_{1}^{2}\left(s_{11}-s_{12}\right)-M_{3}^{2}s_{11}+s_{12}\left(M_{2}^{2}-s_{11}+s_{12}\right)\right)}\,,$
$\displaystyle F^{(1)}_{3\to 3;\widehat{01}}$ $\displaystyle=$
$\displaystyle\frac{(D-2)s_{11}\left(s_{11}s_{02}-s_{12}s_{01}\right)}{(D-1)\left(s_{11}-s_{12}\right)s_{12}\left(M_{1}^{2}\left(s_{12}-s_{11}\right)+M_{3}^{2}s_{11}-s_{12}\left(M_{2}^{2}-s_{11}+s_{12}\right)\right)}\,,$
$\displaystyle F^{(1)}_{3\to 3;\widehat{02}}$ $\displaystyle=$
$\displaystyle\frac{(D-2)\left(s_{12}s_{01}-s_{11}s_{02}\right)}{(D-1)\left(s_{11}-s_{12}\right)\left(M_{1}^{2}\left(s_{12}-s_{11}\right)+M_{3}^{2}s_{11}-s_{12}\left(M_{2}^{2}-s_{11}+s_{12}\right)\right)}\,.$

One interesting point is that expressions of the first three coefficients are
complicated while the last three are relatively simple. The reason is that
$I_{n+1}$ is reduced to combinations of $I_{n}$ only (see (37)) at zeroth
order, thus coefficients of $I_{m\leq n-1}$ are not affected.

In this section, we present examples up to the triangle. A complete result of
$\cal{B}$ is listed in the attachment Mathematica file. For box and pentagon,
we will give some numerical results in Appendix A.

###  3.4 Two-loop Example: Sunset

Our idea to deal with degenerate Gram determinant is simple and it should be
applicable to higher loops. To demonstrate this point, in this subsection we
consider the simplest two-loop integrals, i.e., the integrals with sunset
topology. Involving two auxiliary vectors $R_{1},R_{2}$ we consider the tensor
reduction of following integrals

$\displaystyle
I_{a_{1},a_{2},a_{3}}^{(r_{1},r_{2})}\equiv\int{d^{D}\ell_{1}\over
i\pi^{D/2}}{d^{D}\ell_{2}\over i\pi^{D/2}}{(2\ell_{1}\cdot
R_{1})^{r_{1}}(2\ell_{2}\cdot R_{2})^{r_{2}}\over
D_{1}^{a_{1}}D_{2}^{a_{2}}D_{3}^{a_{3}}}\,,$ (56)

where the propagators are

$\displaystyle
D_{1}\equiv\ell_{1}^{2}-M_{1}^{2}\,,~{}~{}~{}D_{2}\equiv\ell_{2}^{2}-M_{2}^{2}\,,~{}~{}~{}D_{3}\equiv(\ell_{1}+\ell_{2}-K)^{2}-M_{3}^{2}\,.~{}~{}~{}$
(57)

It is found that the integrals (56) can be reduced to seven master
integrals666For our purpose, we consider only the case $a_{i}=1$., i.e.,

$\displaystyle I_{1,1,1}^{(r_{1},r_{2})}={\bf C}^{(r_{1},r_{2})}{\bf
J}=\sum_{i_{1},i_{2},j}s_{01}^{i_{1}}s_{0^{\prime}1}^{i_{2}}s_{00^{\prime}}^{j}s_{00}^{r_{1}-i_{1}-j\over
2}s_{0^{\prime}0^{\prime}}^{r_{2}-i_{2}-j\over
2}{\vec{{\alpha}}}^{(r_{1},r_{2})}_{i_{1},i_{2},j}~{}{\bf J}\,.$ (58)

where the seven master integrals are (we have written ${\bf C}$, ${\bf J}$ to
emphasize they are vector)

$\displaystyle\mathbf{J}$ $\displaystyle=\Bigg{\\{}\int{1\over
D_{1}D_{2}D_{3}},\int{(2\ell_{1}\cdot K)\over
D_{1}D_{2}D_{3}},\int{(2\ell_{2}\cdot K)\over
D_{1}D_{2}D_{3}},\int{(2\ell_{1}\cdot K)(2\ell_{2}\cdot K)\over
D_{1}D_{2}D_{3}}$ $\displaystyle\hskip 40.0pt\int{1\over
D_{2}D_{3}},\int{1\over D_{1}D_{3}},\int{1\over
D_{1}D_{2}}\Bigg{\\}}\,,~{}~{}~{}$ (59)

and kinematic variables are

$\displaystyle s_{01}=R_{1}\cdot K\,,~{}s_{0^{\prime}1}=R_{2}\cdot
K\,,~{}s_{11}=K^{2}\,,~{}s_{00}=R_{1}^{2}\,,~{}s_{0^{\prime}0^{\prime}}=R_{2}^{2}\,,~{}s_{00^{\prime}}=R_{1}\cdot
R_{2}\,.$ (60)

All the reduction coefficients could be found in Feng:2022iuc , where we have
generalized the improved PV-reduction method to the two-loop sunset topology.
For example:

$\displaystyle\mathbf{C}^{(1,0)}$ $\displaystyle=$
$\displaystyle\left\\{0,\frac{s_{01}}{s_{11}},0,0,0,0,0\right\\}\,,$
$\displaystyle\mathbf{C}^{(0,1)}$ $\displaystyle=$
$\displaystyle\left\\{0,0,\frac{s_{0^{\prime}1}}{s_{11}},0,0,0,0\right\\}\,.$
(61)

To study the reduction for degenerate Gram determinant ($s_{11}\to 0$), we
follow the same logic, i.e., the basis with the highest topology will be
decomposed to the linear combinations of other integrals in the basis.
However, there is something different between two-loop and one-loop integrals.
For one-loop integrals, there is only one master integral with the highest
topology in the basis. For two-loop integrals, in general, there are more than
one highest-topology master integrals. For example, for the sunset topology,
there are four master integrals $J_{1},J_{2},J_{3},J_{4}$ as given in (59).
Now we face the problem: whether just one or several master integrals of them
should be degenerate in the limit? In fact, from the tensor reduction results
given in (61), we can see $J_{2}$ and $J_{3}$ should be degenerate. There is
also a brute-force algorithm to determine the number of degenerate master
integrals. Starting with just one degenerate master integral, we solve the
decomposition coefficients using the idea given in the last subsection for
various choices of tensor rank. If it is not the right degenerate pattern, we
will find that it is impossible to cancel all divergences coming from
$s_{11}\to 0$. Then we take two master integrals to be degenerate, repeat the
same computations for several tensor structures and check the consistency.
Interacting the procedure until reaching the point that all divergences have
been canceled consistently, we get the correct degenerate pattern.

For our sunset topology, because there are five independent Lorentz invariant
contractions $R_{i}\cdot R_{j},R_{i}\cdot K$ comparing to two $R^{2},R\cdot K$
for one-loop bubble, there are three master integrals degenerated under the
limit, which can be taken as $J_{2},J_{3},J_{4}$, i.e., we have the expansion

$\displaystyle J_{i}=\sum_{k=0}^{\infty}{\cal B}_{i\to
j}^{(k)}s_{11}^{k}J_{j}\,,~{}~{}~{}i\in\text{deI}\,,~{}j\in\text{reI}\,,$ (62)

where $\text{deI}=\\{2,3,4\\}$ denotes the degenerate master integrals and
$\text{reI}=\\{1,5,6,7\\}$ denotes the remaining master integrals. Expansion
coefficients ${\cal B}_{i\to j}^{(k)}$ are determined by requiring

$\displaystyle
F^{(r_{1},r_{2})}_{j}\equiv\sum_{i\in\text{deI}}\sum_{k}C^{(r_{1},r_{2})}_{i}{\cal
B}_{i\to j}^{(k)}s_{11}^{k}+C^{(r_{1},r_{2})}_{j},~{}~{}~{}~{}\forall
j\in\text{reI}\,.$ (63)

to be finite. The $F^{(r_{1},r_{2})}_{j}$ is nothing but the reduction
coefficients for the degenerate case. Above relation can be written more
compactly as

$\displaystyle C^{(r_{1},r_{2})}_{i}\Big{|}_{s_{11}^{-k}}{\cal B}_{i\to
j}^{(k)}=-C^{(r_{1},r_{2})}_{j}\Big{|}_{s_{11}^{-k}}$ (64)

where we have treated $C_{j=1,5,6,7}$ and $C_{i=2,3,4}$ as row vector and
${\cal B}$ as a matrix. The symbol $~{}\Big{|}_{s_{11}^{-k}}$ means to take
the coefficients of the $k$-th pole $s_{11}^{-k}$.

Now we determine decomposition coefficients using the results from rank one to
three given in Feng:2022iuc .

  * •

For the tensor rank level one, i.e., $r_{1}+r_{2}=1$, there are two cases
given in (61). With the only divergent pole $s_{11}^{-1}$, by (64) we get
following equations:

$\displaystyle(1,0):~{}~{}\left(\begin{array}[]{ccc}s_{01}&0&0\end{array}\right)\left(\begin{array}[]{cccc}{\cal
B}_{2\to 1}^{(0)}&{\cal B}_{2\to 5}^{(0)}&{\cal B}_{2\to 6}^{(0)}&{\cal
B}_{2\to 7}^{(0)}\\\ {\cal B}_{3\to 1}^{(0)}&{\cal B}_{3\to 5}^{(0)}&{\cal
B}_{3\to 6}^{(0)}&{\cal B}_{3\to 7}^{(0)}\\\ {\cal B}_{4\to 1}^{(0)}&{\cal
B}_{4\to 5}^{(0)}&{\cal B}_{4\to 6}^{(0)}&{\cal B}_{4\to 7}^{(0)}\\\
\end{array}\right)=\left(\begin{array}[]{cccc}0&0&0&0\end{array}\right)$ (70)

and

$\displaystyle(0,1):~{}~{}\left(\begin{array}[]{ccc}0&s_{0^{\prime}1}&0\end{array}\right)\left(\begin{array}[]{cccc}{\cal
B}_{2\to 1}^{(0)}&{\cal B}_{2\to 5}^{(0)}&{\cal B}_{2\to 6}^{(0)}&{\cal
B}_{2\to 7}^{(0)}\\\ {\cal B}_{3\to 1}^{(0)}&{\cal B}_{3\to 5}^{(0)}&{\cal
B}_{3\to 6}^{(0)}&{\cal B}_{3\to 7}^{(0)}\\\ {\cal B}_{4\to 1}^{(0)}&{\cal
B}_{4\to 5}^{(0)}&{\cal B}_{4\to 6}^{(0)}&{\cal B}_{4\to 7}^{(0)}\\\
\end{array}\right)=\left(\begin{array}[]{cccc}0&0&0&0\end{array}\right)$ (76)

thus we can solve

$\displaystyle{\cal B}^{(0)}_{i\to
j}=0\,,~{}~{}~{}i=2,3;~{}~{}~{}j=1,5,6,7\,.$ (77)

  * •

For the tensor rank level two, i.e., $r_{1}+r_{2}=2$, there are three cases
$(r_{1},r_{2})=(2,0),(1,1),(0,2)$. There are also two divergent poles ${1\over
s_{11}^{2}}$ and ${1\over s_{11}}$. Explicitly, we have

$\displaystyle\left(\begin{array}[]{ccc}C^{(2,0)}_{2}&C^{(1,1)}_{2}&C^{(0,2)}_{2}\\\
C^{(2,0)}_{3}&C^{(1,1)}_{3}&C^{(0,2)}_{3}\\\
C^{(2,0)}_{4}&C^{(1,1)}_{4}&C^{(0,2)}_{4}\end{array}\right)\Bigg{|}_{s_{11}^{-2}}={D\over
D-1}\left(\begin{array}[]{ccc}M_{12,3}s_{01}^{2}&0&2M_{2}^{2}s_{0^{\prime}1}^{2}\\\
2M_{1}^{2}s_{01}^{2}&0&M_{12,3}s_{0^{\prime}1}^{2}\\\
-2s_{01}^{2}&s_{01}s_{0^{\prime}1}&-2s_{0^{\prime}1}^{2}\\\
\end{array}\right)\,,$ (84)
$\displaystyle\left(\begin{array}[]{ccc}C^{(2,0)}_{2}&C^{(1,1)}_{2}&C^{(0,2)}_{2}\\\
C^{(2,0)}_{3}&C^{(1,1)}_{3}&C^{(0,2)}_{3}\\\
C^{(2,0)}_{4}&C^{(1,1)}_{4}&C^{(0,2)}_{4}\end{array}\right)\Bigg{|}_{s_{11}^{-1}}=\left(\begin{array}[]{ccc}\frac{3Ds_{01}^{2}-M_{12,3}s_{00}}{D-1}&-\frac{2s_{01}s_{0^{\prime}1}}{D-1}&\frac{2\left(Ds_{0^{\prime}1}^{2}-M_{2}^{2}s_{0^{\prime}0^{\prime}}\right)}{D-1}\\\
\frac{2\left(Ds_{01}^{2}-M_{1}^{2}s_{00}\right)}{D-1}&-\frac{2s_{01}s_{0^{\prime}1}}{D-1}&\frac{3Ds_{0^{\prime}1}^{2}-M_{12,3}s_{0^{\prime}0^{\prime}}}{D-1}\\\
\frac{2s_{00}}{D-1}&-\frac{s_{0^{\prime}0}}{D-1}&\frac{2s_{0^{\prime}0^{\prime}}}{D-1}\\\
\end{array}\right)$ (91)

where we have defined

$\displaystyle M_{12,3}\equiv M_{1}^{2}+M_{2}^{2}-M_{3}^{2}$ (92)

for the left hand side of (64). The corresponding data for the right hand side
of (64) is following

$\displaystyle\left(\begin{array}[]{ccc}C^{(2,0)}_{1}&C^{(1,1)}_{1}&C^{(0,2)}_{1}\\\
C^{(2,0)}_{5}&C^{(1,1)}_{5}&C^{(0,2)}_{5}\\\
C^{(2,0)}_{6}&C^{(1,1)}_{6}&C^{(0,2)}_{6}\\\
C^{(2,0)}_{7}&C^{(1,1)}_{7}&C^{(0,2)}_{7}\end{array}\right)\Bigg{|}_{s_{11}^{-2}},\left(\begin{array}[]{ccc}C^{(2,0)}_{1}&C^{(1,1)}_{1}&C^{(0,2)}_{1}\\\
C^{(2,0)}_{5}&C^{(1,1)}_{5}&C^{(0,2)}_{5}\\\
C^{(2,0)}_{6}&C^{(1,1)}_{6}&C^{(0,2)}_{6}\\\
C^{(2,0)}_{7}&C^{(1,1)}_{7}&C^{(0,2)}_{7}\end{array}\right)\Bigg{|}_{s_{11}^{-1}}=\left(\begin{array}[]{ccc}0&0&0\\\
0&0&0\\\ 0&0&0\\\ 0&0&0\\\ \end{array}\right),$ (105)
$\displaystyle\left(\begin{array}[]{ccc}-\frac{2\left((D+2)M_{1}^{2}+D\left(M_{2}^{2}-M_{3}^{2}\right)\right)s_{01}^{2}}{D-1}&\frac{2\left(M_{1}^{2}+M_{2}^{2}-M_{3}^{2}\right)s_{01}s_{0^{\prime}1}}{D-1}&-\frac{2\left(DM_{1}^{2}+(D+2)M_{2}^{2}-DM_{3}^{2}\right)s_{0^{\prime}1}^{2}}{D-1}\\\
-\frac{4s_{01}^{2}}{D-1}&\frac{2s_{01}s_{0^{\prime}1}}{D-1}&-\frac{2Ds_{0^{\prime}1}^{2}}{D-1}\\\
-\frac{2Ds_{01}^{2}}{D-1}&\frac{2s_{01}s_{0^{\prime}1}}{D-1}&-\frac{4s_{0^{\prime}1}^{2}}{D-1}\\\
\frac{2Ds_{01}^{2}}{D-1}&-\frac{2s_{01}s_{0^{\prime}1}}{D-1}&\frac{2Ds_{0^{\prime}1}^{2}}{D-1}\\\
\end{array}\right)\,.$ (110)

Now we can use the (64) to solve wanted decomposition coefficients.

    * –

For the pole $s_{11}^{-2}$ using the (64) we can solve

$\displaystyle{\cal B}^{(0)}_{i\to j}=0\,,~{}~{}~{}i=2,3,4\,.$ (111)

It is worth to notice that for $i=2,3$ it is consistent with the result (77),
but for $i=4$ it is the new solution.

    * –

For the pole $s_{11}^{-1}$, using the data (91) and (110), by (64) we can
solve

$\displaystyle{\cal B}_{2\to 1}^{(1)}$ $\displaystyle=$
$\displaystyle\frac{2\left(2DM_{1}^{2}M_{12,3}-DM_{12,3}^{2}-6M_{1}^{2}M_{12,3}+2M_{12,3}^{2}+4M_{2}^{2}M_{1}^{2}\right)}{D\left(4M_{1}^{2}M_{2}^{2}-M_{12,3}^{2}\right)}\,,$
$\displaystyle{\cal B}_{3\to 1}^{(1)}$ $\displaystyle=$
$\displaystyle\frac{2\left(2DM_{2}^{2}M_{12,3}-DM_{12,3}^{2}-6M_{2}^{2}M_{12,3}+2M_{12,3}^{2}+4M_{1}^{2}M_{2}^{2}\right)}{D\left(4M_{1}^{2}M_{2}^{2}-M_{12,3}^{2}\right)}\,,$
$\displaystyle{\cal B}_{2\to 5}^{(1)}$ $\displaystyle=$
$\displaystyle\frac{4(D-2)M_{1}^{2}}{D\left(4M_{1}^{2}M_{2}^{2}-M_{12,3}^{2}\right)}\,,~{}{\cal
B}_{3\to
5}^{(1)}=\frac{2(D-2)M_{12,3}}{D\left(M_{12,3}^{2}-4M_{1}^{2}M_{2}^{2}\right)}\,,$
$\displaystyle{\cal B}_{2\to 6}^{(1)}$ $\displaystyle=$
$\displaystyle\frac{2(D-2)M_{12,3}}{D\left(M_{12,3}^{2}-4M_{1}^{2}M_{2}^{2}\right)}\,,~{}{\cal
B}_{3\to
6}^{(1)}=\frac{4(D-2)M_{2}^{2}}{D\left(4M_{1}^{2}M_{2}^{2}-M_{12,3}^{2}\right)}\,,$
$\displaystyle{\cal B}_{2\to 7}^{(1)}$ $\displaystyle=$
$\displaystyle\frac{2(D-2)\left(2M_{1}^{2}-M_{12,3}\right)}{D\left(M_{12,3}^{2}-4M_{1}^{2}M_{2}^{2}\right)}\,,{\cal
B}_{3\to
7}^{(1)}=\frac{2(D-2)\left(2M_{2}^{2}-M_{12,3}\right)}{D\left(M_{12,3}^{2}-4M_{1}^{2}M_{2}^{2}\right)}\,,$
$\displaystyle{\cal B}_{4\to 1}^{(1)}$ $\displaystyle=$
$\displaystyle-\frac{2M_{12,3}}{D}\,,~{}~{}{\cal B}_{4\to
5}^{(1)}=-\frac{2}{D}\,,~{}~{}{\cal B}_{4\to
6}^{(1)}=-\frac{2}{D}\,,~{}~{}{\cal B}_{4\to 7}^{(1)}=\frac{2}{D}\,.$ (112)

There is one thing we need to emphasize. The expressions of $C$’s have
different tensor structures of $R$ while the coefficients ${\cal B}$ are $R$
independent. Thus the same solution (112) solving (64) for various tensor
structures is very nontrivial thing. Thus for practical purpose, we can focus
on one tensor structure with the simplest expressions to solve ${\cal B}$.

  * •

For the tensor rank level 3: $r_{1}+r_{2}=3$, there are four cases
$(r_{1},r_{2})=(3,0),(2,1)$, $(1,2),(0,3)$. Now the expressions for
$C_{i},C_{j}$ of (64) becomes complicated because of various tensor
structures. As we mentioned before, we will give only the part of a particular
tensor structure. For example, for $C^{(3,0)}_{i}$ we give the part of the
tensor structure $s_{01}^{3}$ as following

$\displaystyle\left(\begin{array}[]{c}C^{(3,0)}_{2}\\\ C^{(3,0)}_{3}\\\
C^{(3,0)}_{4}\end{array}\right)\Bigg{|}_{s_{11}^{-3}}=\left(\begin{array}[]{c}\frac{(D+2)\left(4(1-2D)M_{2}^{2}M_{12,3}+(3D-2)M_{12,3}^{2}+8(D-1)M_{1}^{2}M_{2}^{2}\right)s_{01}^{3}}{(D-1)(3D-2)}\\\
\frac{2(D+2)M_{1}^{2}\left((5D-4)M_{12,3}+4(1-2D)M_{2}^{2}\right)s_{01}^{3}}{(D-1)(3D-2)}\\\
-\frac{2(D+2)\left((4D-2)M_{12,3}+(D-2)M_{1}^{2}+4(1-2D)M_{2}^{2}\right)s_{01}^{3}}{(D-1)(3D-2)}\\\
\end{array}\right)$ (119)

while for $C^{(1,2)}_{i}$ we give the part of the tensor structure
$s_{01}^{2}s_{0^{\prime}1}$ as following

$\displaystyle\left(\begin{array}[]{c}C^{(1,2)}_{2}\\\ C^{(1,2)}_{3}\\\
C^{(1,2)}_{4}\end{array}\right)\Bigg{|}_{s_{11}^{-3}}=\left(\begin{array}[]{c}\frac{2(D+2)M_{2}^{2}\left((2D-1)M_{12,3}-2(D-1)M_{1}^{2}\right)s_{01}^{2}s_{0^{\prime}1}}{(D-1)(3D-2)}\\\
-\frac{2(D+2)M_{1}^{2}\left((D-1)M_{12,3}+2(1-2D)M_{2}^{2}\right)s_{01}^{2}s_{0^{\prime}1}}{(D-1)(3D-2)}\\\
\frac{(D+2)\left(DM_{12,3}+4(D-1)M_{1}^{2}+4(1-2D)M_{2}^{2}\right)s_{01}^{2}s_{0^{\prime}1}}{(D-1)(3D-2)}\\\
\end{array}\right)$ (126)

Expressions of $C_{i},C_{j}$ for other divergent poles and the chosen tensor
structures are listed in Appendix B.

    * –

For the pole $s_{11}^{-3}$, it produce the same results as given by (111).

    * –

For the pole $s_{11}^{-2}$, using (64) we find the same results as given in
(112).

    * –

For the pole $s_{11}^{-1}$, using (64) we find expansion coefficients ${\cal
B}_{i\to j}^{(2)}$. Since their expressions are very long, we have collected
them into a attached Mathematica file. To give a flavor, we present ${\cal
B}_{4\to 1}^{(2)}$ as following

$\displaystyle{\cal B}_{4\to
1}^{(2)}=\frac{1}{D^{2}(D+2)\left(M_{12,3}^{2}-4M_{1}^{2}M_{2}^{2}\right)^{2}}\Big{[}\left(16D^{3}-208D^{2}+528D-288\right)M_{2}^{2}M_{1}^{2}M_{12,3}^{2}$
$\displaystyle+\big{(}4D^{3}-14D^{2}+28D-48\big{)}M_{12,3}^{4}-\big{(}8D^{3}-40D^{2}+80D-96\big{)}(M_{1}^{2}+M_{2}^{2})M_{12,3}^{3}$
$\displaystyle+\left(128D^{2}-416D+96\right)(M_{2}^{2}M_{1}^{4}+M_{1}^{2}M_{2}^{4})M_{12,3}+\left(384D-96D^{2}\right)M_{2}^{4}M_{1}^{4}\Big{]}\,.$
(127)

Now we can calculate the final reduction coefficients by (63). For example,
the reduction coefficients for rank $(0,1)$ is given by

$\displaystyle F^{(0,1)}_{1}$ $\displaystyle=$
$\displaystyle\frac{2s_{0^{\prime}1}\left(-2DM_{2}^{2}M_{12,3}+DM_{12,3}^{2}+6M_{2}^{2}M_{12,3}-2M_{12,3}^{2}-4M_{1}^{2}M_{2}^{2}\right)}{D\left(M_{12,3}^{2}-4M_{1}^{2}M_{2}^{2}\right)}\,,$
$\displaystyle F^{(1,0)}_{1}$ $\displaystyle=$
$\displaystyle\frac{2s_{01}\left(2DM_{1}^{2}M_{12,3}-DM_{12,3}^{2}-6M_{1}^{2}M_{12,3}+2M_{12,3}^{2}+4M_{2}^{2}M_{1}^{2}\right)}{D\left(4M_{1}^{2}M_{2}^{2}-M_{12,3}^{2}\right)}\,,$
$\displaystyle F^{(0,1)}_{5}$ $\displaystyle=$
$\displaystyle\frac{2(D-2)M_{12,3}s_{0^{\prime}1}}{D\left(M_{12,3}^{2}-4M_{1}^{2}M_{2}^{2}\right)}\,,\hskip
38.0ptF^{(1,0)}_{5}=\frac{4(D-2)M_{1}^{2}s_{01}}{D\left(4M_{1}^{2}M_{2}^{2}-M_{12,3}^{2}\right)}\,,$
$\displaystyle F^{(0,1)}_{6}$ $\displaystyle=$
$\displaystyle\frac{4(D-2)M_{2}^{2}s_{0^{\prime}1}}{D\left(4M_{1}^{2}M_{2}^{2}-M_{12,3}^{2}\right)}\,,\hskip
37.0ptF^{(1,0)}_{6}=\frac{2(D-2)M_{12,3}s_{01}}{D\left(M_{12,3}^{2}-4M_{1}^{2}M_{2}^{2}\right)}\,,$
$\displaystyle F^{(0,1)}_{7}$ $\displaystyle=$
$\displaystyle\frac{2(D-2)s_{0^{\prime}1}\left(2M_{2}^{2}-M_{12,3}\right)}{D\left(M_{12,3}^{2}-4M_{1}^{2}M_{2}^{2}\right)}\,,~{}~{}F^{(1,0)}_{7}=\frac{2(D-2)s_{01}\left(2M_{1}^{2}-M_{12,3}\right)}{D\left(M_{12,3}^{2}-4M_{1}^{2}M_{2}^{2}\right)}\,.$
(128)

In the attached Mathematica file we have present reduction coefficients for
other tensor ranks up to rank level two.

##  4 Self-consistence

From the previous section we see that although the decomposition coefficients
$\mathcal{B}_{n+1;\widehat{\mathbf{b}_{j}}}$ are computed using the tensor
reduction results in (33), by definition in (28), it should be independent of
the auxiliary $R$ and the tensor rank used in the formula. Thus to show the
correctness of the formula (33), we should prove this point.

More explicitly, there are two facts we want to prove. The first one is that
to get $\mathcal{B}_{n+1;\widehat{\mathbf{b}_{j}}}$ up to $|G|^{k}$ order, we
require $m$ in (33) to satisfy condition $m>k$. One can choose different
$m_{i}$ . As long as $m_{i}>k$, result given by (33) should be the same up to
$|G|^{k}$. The second one is that although at the RHS of (33),
$C^{(m)}_{n+1\to n+1;\widehat{\mathbf{b}_{j}}},\ C^{(m)}_{n+1\to n+1}$ are
functions of $R$, as long as $m>k$, the series of $|G|$ should be independent
of $R$ up to $|G|^{k}$.

In this section, we will prove these crucial consistencies using two different
methods.

###  4.1 Proof with recursion relation

The crucial input of our proof is the rather non-trivial relation for one-loop
integrals777This relation is observed in Feng:2022iuc . Here we rephrase it in
the language of projective space Feng:2022rwj .

$\displaystyle I_{n}^{(m)}$ $\displaystyle=$
$\displaystyle{1\over\left|G\right|}\left[A_{m}I_{n}^{(m-1)}+B_{m}I_{n}^{(m-2)}+\text{Lower
Terms}\right]$ (129) $\displaystyle=$
$\displaystyle{1\over\left|G\right|}\left[A_{m}I_{n}^{(m-1)}+B_{m}I_{n}^{(m-2)}+\mathcal{O}(1)\right]\,.$

where "Lower Terms" means the contributions from lower topologies, we also
denote it as $O(1)$ for it contains no poles of $\left|G\right|$, and

$\displaystyle A_{m}$ $\displaystyle=$ $\displaystyle
a_{m}(VQ^{*}L)={2(D+2m-n-2)\over D+m-n-1}(VQ^{*}L)\,,$ $\displaystyle B_{m}$
$\displaystyle=$ $\displaystyle b_{m}\left[(\mathop{\rm
det}Q)R^{2}-VQ^{*}V)\right]={4(m-1)((\mathop{\rm det}Q)R^{2}-VQ^{*}V)\over
D+m-n-1}\,.$ (130)

Here $Q^{*}$ is the adjoint matrix of $Q$ with component
$Q_{ij}\equiv{M_{i}^{2}+M_{j}^{2}-(K_{i}-K_{j})^{2}\over
2},i,j=0,\ldots,n-1,K_{0}=0$. The two vectors in the expression of
$A_{m},B_{m}$ are $L=\\{1,1,\ldots,1\\}$,
$V=\\{0,s_{01},s_{02},\ldots,s_{0,n-1}\\}$. One can prove the relation (129)
by showing that it satisfies the $\cal D$-type and $\cal T$-type differential
relations888Since we have shown the reduction coefficients can be completely
determined by $\cal D$-type and $\cal T$-type differential relations in
Feng:2021enk ; Hu:2021nia , checking relation (129) satisfies these two
recursions will be sufficient for our proof. . First, acting with $\cal
D$-type operator, we have

$\displaystyle{\cal
D}_{i}I_{n}^{(m)}=m(f_{i}I_{n}^{(m-1)}+I_{n;\widehat{0}}^{(m)}-I_{n;\widehat{i}}^{(m)})=mf_{i}I_{n}^{(m-1)}+\mathcal{O}(1)$
(131)

at one side, while applying ${\cal D}_{i}$ on the RHS of (129), we find

$\displaystyle{1\over\left|G\right|}{\cal
D}_{i}\left[A_{m}I_{n}^{(m-1)}+B_{m}I_{n}^{(m-2)}+\mathcal{O}(1)\right]$ (132)
$\displaystyle=$ $\displaystyle{1\over\left|G\right|}\left[({\cal
D}_{i}A_{m})I_{n}^{(m-1)}+((m-1)f_{i}+{\cal
D}_{i}B_{m})I_{n}^{(m-2)}+B_{m}(m-2)f_{i}I_{n}^{(m-3)}+\mathcal{O}(1)\right]~{}~{}~{}$

at another side. Identifying (132) and (131), we have

$\displaystyle I_{n}^{(m-1)}={1\over mf_{i}\left|G\right|-{\cal
D}_{i}A_{m}}\left[((m-1)f_{i}+{\cal
D}_{i}B_{m})I_{n}^{(m-2)}+B_{m}(m-2)f_{i}I_{n}^{(m-3)}+\mathcal{O}(1)\right]\,.~{}~{}~{}$
(133)

By our recursive assumption of the form (129), we just need to prove

$\displaystyle(mf_{i}\left|G\right|-{\cal
D}_{i}A_{m})A_{m-1}=\left|G\right|((m-1)f_{i}+{\cal D}_{i}B_{m}),$
$\displaystyle(mf_{i}\left|G\right|-{\cal
D}_{i}A_{m})B_{m-1}=\left|G\right|B_{m}(m-2)f_{i}\,,$ (134)

which can be checked using the identities

$\displaystyle{\cal D}_{i}A_{m}={2(D+2m-n-2)\over
D+m-n-1}\sum_{ab}s_{ia}Q^{*}_{ab}={2(D+2m-n-2)\over
D+m-n-1}f_{i}\left|G\right|,$ $\displaystyle{\cal
D}_{i}B_{m}={4(m-1)(2(\mathop{\rm
det}Q)s_{0i}-2\sum_{jk}s_{ik}Q^{*}_{kj}s_{0j})\over
D+m-n-1}={4(m-1)f_{i}(VQ^{*}L)\over D+m-n-1}\,.$ (135)

Next we consider the $\cal T$ operator. First, we have

$\displaystyle{\cal
T}I_{n}^{(m)}=4m(m-1)M_{0}^{2}I_{n}^{(m-2)}+4m(m-2)I_{n;\widehat{0}}^{(m-2)}=4m(m-1)M_{0}^{2}I_{n}^{(m-2)}+\mathcal{O}(1)\,.$
(136)

Acting with $\cal T$ on the RHS of (129), we get

$\displaystyle{{\cal
T}\over|G|}\left[A_{m}I_{n}^{(m-1)}+B_{m}I_{n}^{(m-2)}+\mathcal{O}(1)\right]$
(137) $\displaystyle=$
$\displaystyle{1\over|G|}\left[2a_{m}(m-1)\sum_{i,j}Q^{*}_{ij}f_{i}+2(D+2m-4)b_{m}\mathop{\rm
det}(Q)-2b_{m}\sum_{i,j=1}^{n-1}s_{ij}Q^{*}_{ij}\right]I_{n}^{(m-2)}$
$\displaystyle+{1\over|G|}\left[4(m-1)(m-2)M_{0}^{2}A_{m}-4b_{m}(m-2)\sum_{i,j=1}^{n-1}f_{i}Q^{*}_{ij}s_{0j}\right]I_{n}^{(m-3)}$
$\displaystyle+{1\over|G|}\left[4b_{m}(m-2)(m-3)M_{0}^{2}(\mathop{\rm
det}(Q)R^{2}-(VQ^{*}V))\right]I_{n}^{(m-4)}$

where we have used following algebraic results

$\displaystyle{\cal
T}\left[(VQ^{*}L)I_{n}^{(m)}\right]=2\sum_{i,j}Q^{*}_{ij}mf_{i}I_{n}^{(m-1)}+4m(m-1)M_{0}^{2}(VQ^{*}L)I_{n}^{(m-2)}+\mathcal{O}(1)\,,$
$\displaystyle{\cal
T}\left[R^{2}I_{n}^{(m)}\right]=2(D+2m)I_{n}^{(m)}+4m(m-1)M_{0}^{2}R^{2}I_{n}^{(m-2)}+\mathcal{O}(1)\,,$
$\displaystyle{\cal
T}\left[(VQ^{*}V)I_{n}^{(m)}\right]=2\sum_{i,j=1}^{n-1}s_{ij}Q^{*}_{ij}I_{n}^{(m)}+4m\sum_{i,j=1}^{n-1}f_{i}Q^{*}_{ij}s_{0j}I_{n}^{(m-1)}$
$\displaystyle\hskip
100.0pt+4m(m-1)M_{0}^{2}(VQ^{*}V)I_{n}^{(m-2)}+\mathcal{O}(1)\,.$ (138)

Comparing both sides, we just need to prove

$\displaystyle
4m(m-1)M_{0}^{2}\left|G\right|-\left[2a_{m}(m-1)\sum_{i,j}Q^{*}_{ij}f_{i}+2(D+2m-4)b_{m}\mathop{\rm
det}(Q)-2b_{m}\sum_{i,j=1}^{n-1}s_{ij}Q^{*}_{ij}\right]A_{m-2}$
$\displaystyle={-4b_{m}(m-2)\over|G|}\sum_{i,j=1}^{n-1}f_{i}Q^{*}_{ij}s_{0j}$
(139)

and

$\displaystyle
4m(m-1)M_{0}^{2}\left|G\right|-\left[2a_{m}(m-1)\sum_{i,j}Q^{*}_{ij}f_{i}+2(D+2m-4)b_{m}\mathop{\rm
det}(Q)-2b_{m}\sum_{i,j=1}^{n-1}s_{ij}Q^{*}_{ij}\right]B_{m-2}$ (140)
$\displaystyle=$
$\displaystyle\left[4a_{m}(m-1)(m-2)M_{0}^{2}(VQ^{*}L)+4b_{m}(m-2)(m-3)M_{0}^{2}(\mathop{\rm
det}(Q)R^{2}-(VQ^{*}V))\right]\,.$

One can check the two equations after some algebra.

Now we employ (129) to prove self-consistence, i.e., to show that

$\displaystyle\mathcal{B}_{n;\widehat{\mathbf{b}_{j}}}=-{\frac{C^{(m)}_{n\to
n;\widehat{\mathbf{b}_{j}}}}{C^{(m)}_{n\to
n}}}+\mathcal{O}(\left|G\right|^{m})$ (141)

is independent of the rank choice $m$ and $R$. First let us show the
independence of rank, which is equivalent to the relation

$\displaystyle{\frac{C^{(m+1)}_{n\to
n;\widehat{\mathbf{b}_{j}}}}{C^{(m+1)}_{n\to n}}}-{\frac{C^{(m)}_{n\to
n;\widehat{\mathbf{b}_{j}}}}{C^{(m)}_{n\to
n}}}=\mathcal{O}(\left|G\right|^{m})\,.~{}~{}~{}$ (142)

To see it, first we rewrite (142) to

$\displaystyle{C^{(m+1)}_{n\to n;\widehat{\mathbf{b}_{j}}}C^{(m)}_{n\to
n}-C^{(m)}_{n\to n;\widehat{\mathbf{b}_{j}}}C^{(m+1)}_{n\to n}\over
C^{(m)}_{n\to n}C^{(m+1)}_{n\to n}}=\mathcal{O}(\left|G\right|^{m})\,.$ (143)

Using the divergence behavior of denominator

$\displaystyle C^{(m)}_{n\to n}C^{(m+1)}_{n\to
n}=\mathcal{O}(\left|G\right|^{-2m-1})\,,$ (144)

we just need to show the leading divergent behavior of numerator is

$\displaystyle{C^{(m+1)}_{n\to n;\widehat{\mathbf{b}_{j}}}C^{(m)}_{n\to
n}-C^{(m)}_{n\to n;\widehat{\mathbf{b}_{j}}}C^{(m+1)}_{n\to
n}}=\mathcal{O}(\left|G\right|^{-m-1})\,.~{}~{}~{}~{}$ (145)

To prove it, we assume the equation holds for all $m<m^{\prime}$, while
$m^{\prime}=1,2$ can be checked explicitly, then we prove it holds for
$m=m^{\prime}$. Using (129), the LHS of (145) becomes

$\displaystyle{C^{(m^{\prime}+1)}_{n\to
n;\widehat{\mathbf{b}_{j}}}C^{(m^{\prime})}_{n\to n}-C^{(m^{\prime})}_{n\to
n;\widehat{\mathbf{b}_{j}}}C^{(m^{\prime}+1)}_{n\to n}}$ (146)
$\displaystyle=$
$\displaystyle{1\over\left|G\right|}\left[A_{m^{\prime}+1}C^{(m^{\prime})}_{n\to
n;\widehat{\mathbf{b}_{j}}}+B_{m^{\prime}+1}C^{(m^{\prime}-1)}_{n\to
n;\widehat{\mathbf{b}_{j}}}+\mathcal{O}(1)\right]C^{(m^{\prime})}_{n\to n}$
$\displaystyle-C^{(m^{\prime})}_{n\to
n;\widehat{\mathbf{b}_{j}}}{1\over\left|G\right|}\left[A_{m^{\prime}+1}C^{(m^{\prime})}_{n\to
n}+B_{m^{\prime}+1}C^{(m^{\prime}-1)}_{n\to n}+\mathcal{O}(1)\right]$
$\displaystyle=$
$\displaystyle{B_{m^{\prime}+1}\over\left|G\right|}\left[{C^{(m^{\prime})}_{n\to
n;\widehat{\mathbf{b}_{j}}}C^{(m^{\prime}-1)}_{n\to
n}-C^{(m^{\prime}-1)}_{n\to n;\widehat{\mathbf{b}_{j}}}C^{(m^{\prime})}_{n\to
n}}\right]+\mathcal{O}(\left|G\right|^{-1})$ $\displaystyle=$
$\displaystyle\mathcal{O}(\left|G\right|^{-m^{\prime}-1})+\mathcal{O}(\left|G\right|^{-1})=\mathcal{O}(\left|G\right|^{-m^{\prime}-1})\,.$

Having shown the independence of the rank choice, now we prove the first $m$
terms in (141) don’t rely on $R$, i.e., the derivation over $R$ is zero up to
order $\mathcal{O}(\left|G\right|^{m})$:

$\displaystyle{\partial\over\partial R^{\mu}}\left[{\frac{C^{(m)}_{n\to
n;\widehat{\mathbf{b}_{j}}}}{C^{(m)}_{n\to n}}}\right]={{\partial\over\partial
R^{\mu}}C^{(m)}_{n\to n;\widehat{\mathbf{b}_{j}}}C^{(m)}_{n\to
n}-C^{(m)}_{n\to n;\widehat{\mathbf{b}_{j}}}{\partial\over\partial
R^{\mu}}C^{(m)}_{n\to n}\over(C^{(m)}_{n\to
n})^{2}}\leq\mathcal{O}(\left|G\right|^{m})\,.$ (147)

Similarly, since the leading divergent behavior is $\left|G\right|^{-2m}$, we
just need to show that

$\displaystyle{{\partial\over\partial R^{\mu}}C^{(m)}_{n\to
n;\widehat{\mathbf{b}_{j}}}C^{(m)}_{n\to n}-C^{(m)}_{n\to
n;\widehat{\mathbf{b}_{j}}}{\partial\over\partial R^{\mu}}C^{(m)}_{n\to
n}}\leq\mathcal{O}(\left|G\right|^{-m})\,.$ (148)

Due to reduction coefficients are functions of $s_{0i},s_{00}$, after acting
${\partial\over\partial R^{\mu}}$, the LHS of (148) can be wrote as the form

$\displaystyle
L^{\mu}\equiv\sum_{i=1}^{n-1}\widetilde{{\alpha}}_{i}K_{i}^{\mu}+\widetilde{{\alpha}}_{n}R^{\mu}\,.~{}~{}~{}$
(149)

Noticing the limit $|G|\to 0$ means these $K_{i}$ are not linearly
independent, thus the limit can be parameterized as

$\displaystyle K_{1}=\sum_{i=2}^{n-1}a_{i}K_{i}+tK_{0}~{}~{}~{}~{}$ (150)

where $t$ is the same order of $|G|$ and $K_{0}$ is another linear independent
momentum. Putting it back to (149) we get

$\displaystyle
L^{\mu}=\sum_{i=2}^{n-1}(\widetilde{{\alpha}}_{i}+\widetilde{{\alpha}}_{1}a_{i})K_{i}^{\mu}+\widetilde{{\alpha}}_{1}tK_{0}^{\mu}+\widetilde{{\alpha}}_{n}R^{\mu}\equiv\sum_{i=2}^{n-1}{\alpha}_{i}K_{i}^{\mu}+{\alpha}_{1}tK_{0}^{\mu}+{\alpha}_{n}R^{\mu}\,.~{}~{}~{}$
(151)

Let us contract $L^{\mu}$ with $K_{0}$, $K_{i},i=2,...,n-1$ and $R$, we will
get $n$ equations in the matrix form as

$\displaystyle
G(K_{0},K_{2},...,K_{n-1},R)\left(\begin{array}[]{c}t{\alpha}_{1}\\\
{\alpha}_{2}\\\ \vdots\\\
{\alpha}_{n}\end{array}\right)=\left(\begin{array}[]{c}b_{1}\\\ b_{2}\\\
\vdots\\\ b_{n}\end{array}\right)$ (160)

where $G(K_{0},K_{2},...,K_{n-1},R)$ is the non-degenerate Gram matrix and

$\displaystyle b_{1}$ $\displaystyle=$ $\displaystyle
K_{0}^{\mu}\left[{{\partial\over\partial R^{\mu}}C^{(m)}_{n\to
n;\widehat{\mathbf{b}_{j}}}C^{(m)}_{n\to n}-C^{(m)}_{n\to
n;\widehat{\mathbf{b}_{j}}}{\partial\over\partial R^{\mu}}C^{(m)}_{n\to
n}}\right]$ $\displaystyle b_{i}$ $\displaystyle=$ $\displaystyle
K_{i}^{\mu}\left[{{\partial\over\partial R^{\mu}}C^{(m)}_{n\to
n;\widehat{\mathbf{b}_{j}}}C^{(m)}_{n\to n}-C^{(m)}_{n\to
n;\widehat{\mathbf{b}_{j}}}{\partial\over\partial R^{\mu}}C^{(m)}_{n\to
n}}\right]$ $\displaystyle=$ $\displaystyle{\cal D}_{i}C^{(m)}_{n\to
n;\widehat{\mathbf{b}_{j}}}C^{(m)}_{n\to n}-C^{(m)}_{n\to
n;\widehat{\mathbf{b}_{j}}}{\cal D}_{i}C^{(m)}_{n\to n},~{}~{}~{}i=2,...,n-1$
$\displaystyle b_{n}$ $\displaystyle=$ $\displaystyle
R^{\mu}\left[{{\partial\over\partial R^{\mu}}C^{(m)}_{n\to
n;\widehat{\mathbf{b}_{j}}}C^{(m)}_{n\to n}-C^{(m)}_{n\to
n;\widehat{\mathbf{b}_{j}}}{\partial\over\partial R^{\mu}}C^{(m)}_{n\to
n}}\right]=0$ (161)

where we have used the homogeneity of $R$ for $b_{n}$. Thus to show
$t{\alpha}_{1}$ and all other ${\alpha}_{i}\leq O(\left|G\right|^{-m})$, we
just need to show $b_{i}\leq\mathcal{O}(\left|G\right|^{-m})$.

We will prove it recursively by assuming the equation holds for all
$m<m^{\prime}$, where the $m^{\prime}=1,2$ can be checked explicitly. For the
case $m=m^{\prime}$, using

$\displaystyle{\cal D}_{i}C^{(m)}_{n\to
n;\widehat{\mathbf{b}_{j}}}=mf_{i}C^{(m-1)}_{n\to
n;\widehat{\mathbf{b}_{j}}}+\mathcal{O}(1)\,,$ (162)

we have

$\displaystyle b_{i}$ $\displaystyle=$ $\displaystyle{\cal
D}_{i}C^{(m^{\prime})}_{n\to n;\widehat{\mathbf{b}_{j}}}C^{(m^{\prime})}_{n\to
n}-C^{(m^{\prime})}_{n\to n;\widehat{\mathbf{b}_{j}}}{\cal
D}_{i}C^{(m^{\prime})}_{n\to n}$ (163) $\displaystyle=$ $\displaystyle
m^{\prime}f_{i}\left[C^{(m^{\prime}-1)}_{n\to
n;\widehat{\mathbf{b}_{j}}}C^{(m^{\prime})}_{n\to n}-C^{(m^{\prime})}_{n\to
n;\widehat{\mathbf{b}_{j}}}C^{(m^{\prime}-1)}_{n\to
n}\right]+\mathcal{O}(\left|G\right|^{-m^{\prime}})$ $\displaystyle=$
$\displaystyle\mathcal{O}(\left|G\right|^{-m^{\prime}})~{}~{}~{}$

where in the second line we have used the result (145). Above argument has
shown that $b_{i}\leq O(\left|G\right|^{-m})$ $i=2,..,n-1$ for $m=m^{\prime}$.
For $b_{1}$, we need to be more careful. Using (150), we have

$\displaystyle b_{1}$ $\displaystyle=$ $\displaystyle{1\over
t}\left(K_{1}-\sum_{i=2}^{n-1}a_{i}K_{i})\right)\left[{{\partial\over\partial
R^{\mu}}C^{(m)}_{n\to n;\widehat{\mathbf{b}_{j}}}C^{(m)}_{n\to
n}-C^{(m)}_{n\to n;\widehat{\mathbf{b}_{j}}}{\partial\over\partial
R^{\mu}}C^{(m)}_{n\to n}}\right]\,.$ (164)

Naively we have ${1\over t}$ in the front, which will make the divergence
higher. However, when $t\to 0$, the combination
$\left(K_{1}-\sum_{i=2}^{n-1}a_{i}K_{i})\right)\sim t\to 0$, thus the
divergent order estimation given in (163) is also true for $b_{1}$. Thus we
finish the proof for $R$ independence.

###  4.2 Another proof

Now we will give another proof from different point of view. For
simplification, we will constrict to the decomposition coefficient
$\mathcal{B}_{n+1;\widehat{\mathbf{b}_{j}}}$ with
$\mathbf{b}_{j}=\\{r+1,\cdots,n\\}$. For given tensor rank $m$, the reduction
coefficients have the expansion

$\displaystyle C^{(m)}_{n+1\to n+1;\widehat{r+1,r+2,...,n}}=$
$\displaystyle\sum_{2a_{0}+\sum_{k=1}^{n}a_{k}=m}\Bigg{\\{}c^{(0,1,\cdots,r)}_{a_{1},\cdots,a_{n}}(m)(M_{0}^{2})^{a_{0}+r-n}\prod_{k=0}^{n}s_{0k}^{a_{k}}\Bigg{\\}}$
(165) $\displaystyle C^{(m)}_{n+1\to n+1}=$
$\displaystyle\sum_{2a_{0}+\sum_{k=1}^{n}a_{k}=m}\Bigg{\\{}c^{(0,1,\cdots,n)}_{a_{1},\cdots,a_{n}}(m)(M_{0}^{2})^{a_{0}+r-n}\prod_{k=0}^{n}s_{0k}^{a_{k}}\Bigg{\\}}.$

Thus we have the expansion of ${\cal F}$ in (30)

$\begin{aligned} \mathcal{F}^{(m)}_{n+1\to
n+1;\widehat{r+1,...,n}}=&\sum_{2a_{0}+\sum
a_{i}=m}\Bigg{\\{}c^{(0,...,r)}_{a_{1},...,a_{n}}(m)+c^{(0,...,n)}_{a_{1},...,a_{n}}(m)\mathcal{B}_{n+1;\widehat{r+1,...,n}}\Bigg{\\}}(R\cdot
R)^{a_{0}}\prod(R\cdot K_{i})^{a_{i}}\\\ \equiv&\sum_{2a_{0}+\sum
a_{i}=m}\mathcal{F}^{(m)}_{n+1\to
n+1;\widehat{r+1,...,n}}(a_{1},...,a_{n})\times(R\cdot R)^{a_{0}}\prod(R\cdot
K_{i})^{a_{i}}.\end{aligned}~{}~{}~{}$ (166)

Since $c^{(0,...,n)}_{a_{1},...,a_{n}}(m)\sim|G|^{-(\frac{m+\sum a_{i}}{2})}$,
smoothness under the limit gives

$\mathcal{B}_{n+1;\widehat{r+1,...,n}}=-\frac{c^{(0,...,r)}_{a_{1},...,a_{n}}(m)}{c^{(0,...,n)}_{a_{1},...,a_{n}}(m)}+\mathcal{O}(|G|^{\frac{m+\sum
a_{i}}{2}}).~{}~{}~{}$ (167)

The difference between (33) and (167) is that we have used expansion
coefficients in (167) instead of the whole reduction coefficients in (33).
Thus $\mathcal{B}_{n+1;\widehat{r+1,...,n}}$ from (167) will be independent of
$R$ manifestly, but we need to show the same result will be obtained taking
different $a_{i}^{\prime}$ and $m$. Overall, the irrelevancy of the tensor
rank $m$ and the expansion indices lead to following condition

$\displaystyle\frac{c^{(0,...,r)}_{a_{1},...,a_{n}}(m)}{c^{(0,...,n)}_{a_{1},...,a_{n}}(m)}-\frac{c^{(0,...,r)}_{a^{\prime}_{1},...,a^{\prime}_{n}}(m^{\prime})}{c^{(0,...,n)}_{a^{\prime}_{1},...,a^{\prime}_{n}}(m^{\prime})}\leq|G|^{(Min[\frac{m+\sum
a_{i}}{2},\frac{m^{\prime}+\sum a^{\prime}_{i}}{2}])}$ (168)
$\displaystyle\Rightarrow$ $\displaystyle
c^{(0,...,r)}_{a_{1},...,a_{n}}(m)c^{(0,...,n)}_{a^{\prime}_{1},...,a^{\prime}_{n}}(m^{\prime})-c^{(0,...,n)}_{a_{1},...,a_{n}}(m)c^{(0,...,r)}_{a^{\prime}_{1},...,a^{\prime}_{n}}(m^{\prime})\leq|G|^{-Max[\frac{m+\sum
a_{i}}{2},\frac{m^{\prime}+\sum a^{\prime}_{i}}{2}]}$

holds for any nonzero
$c^{(0,...,n)}_{a_{1},...,a_{n}}(m),c^{(0,...,n)}_{a^{\prime}_{1},...,a^{\prime}_{n}}(m^{\prime})$.

Before to prove (168), let us list several useful relations. Using the ${\cal
T}$ operator Feng:2021enk ; Hu:2021nia  we have relation

$\displaystyle
4m(m+1)c^{(0,1,\cdots,r)}_{a_{1},\cdots,a_{n}}(m-1)=(m+1-\sum_{k=1}^{n}a_{k})(D+m+\sum_{k=1}^{n}a_{k}-1)c^{(0,1,\cdots,r)}_{a_{1},\cdots,a_{n}}(m+1)$
$\displaystyle+\sum_{0<i<j}2(a_{i}+1)(a_{j}+1){\beta}_{ij}c^{(0,1,\cdots,r)}_{a_{1},\cdots,a_{i}+1,\cdots,a_{j}+1,\cdots,a_{n}}(m+1)$
$\displaystyle+\sum_{a=1}^{n}(a_{i}+1)(a_{i}+2){\beta}_{ii}c^{(0,1,\cdots,r)}_{a_{1},\cdots,a_{j}+2,\cdots,a_{n}}(m+2).$
(169)

Using the $\mathcal{D}$-relation (13) twice, we have

$c^{(0,...,r)}_{a_{1},...,a_{n}}(m+1)=\frac{(m+1)\Bigg{\\{}4m\cdot
c^{(0,...,r)}_{a_{1},...,a_{n}}(m-1)-\boldsymbol{{\alpha}}^{T}\widetilde{\boldsymbol{G}}^{-1}{\boldsymbol{O}}^{(0,...,r)}(a_{1},...,a_{n};m)\Bigg{\\}}}{(m+1-\sum_{k=1}^{n}a_{k})(D+m-n-1)}.$
(170)

where vector $\boldsymbol{O}^{(0,...,r)}(a_{1},...,a_{n};m)$ is given in (17).
Combining with the component form of (13) we have

$c^{(0,...,r)}_{a_{1},...,a_{i}+1,...,a_{n}}(m)=\frac{1}{a_{i}+1}\sum_{k}\frac{\widetilde{\boldsymbol{G}}^{*}_{ik}}{|\widetilde{\boldsymbol{G}}|}[\boldsymbol{O}^{(0,...,r)}(a_{1},...,a_{n};m)]_{k}.~{}~{}~{}$
(171)

From (170) and (171) we can read

$\displaystyle
c^{(0,...,n)}_{a_{1},...,a_{i}+1,...,a_{n}}(m)=\frac{N_{i}(a_{1},...,a_{n};m)}{|G|}\left\\{A_{i}\cdot
c^{(0,...,n)}_{a_{1},...,a_{n}}(m-1)+\sum_{k}B_{ik}\cdot
c^{(0,...,n)}_{a_{1},...,a_{k}-1,...,a_{n}}(m)\right\\},$ (172) $\displaystyle
c^{(0,...,r)}_{a_{1},...,a_{i}+1,...,a_{n}}(m)=\frac{N_{i}(a_{1},...,a_{n};m)}{|G|}\left\\{A_{i}\cdot
c^{(0,...,r)}_{a_{1},...,a_{n}}(m-1)+\sum_{k}B_{ik}\cdot
c^{(0,...,r)}_{a_{1},...,a_{k}-1,...,a_{n}}(m)\right\\}+\frac{X_{i}^{lower}}{|G|},$
$\displaystyle
c^{(0,...,n)}_{a_{1},...,a_{n}}(m+1)=\frac{M(a_{1},...,a_{n};m)}{|G|}\left\\{D\cdot
c^{(0,...,n)}_{a_{1},...,a_{n}}(m-1)+\sum_{k}E_{k}\cdot
c^{(0,...n)}_{a_{1},...,a_{k}-1,...,a_{n}}(m)\right\\},$ $\displaystyle
c^{(0,...,r)}_{a_{1},...,a_{n}}(m+1)=\frac{M(a_{1},...,a_{n};m)}{|G|}\left\\{D\cdot
c^{(0,...,r)}_{a_{1},...,a_{n}}(m-1)+\sum_{k}E_{k}\cdot
c^{(0,...r)}_{a_{1},...,a_{k}-1,...,a_{n}}(m)\right\\}+\frac{Y^{lower}}{|G|}.$

where $A,B_{k},D,E_{k}$ are finite under
$|G|\sim|\widetilde{\boldsymbol{G}}|\to 0$. $X_{i}^{lower},Y^{lower}$ are
those contributions from the reduction of lower topologies, which are also
finite under $|G|\to 0$. $M_{i},N$ are constants decided by
$a_{1},...,a_{n},m$. We also need to mention that under $|G|\to 0$:

$\displaystyle
A_{i}:A_{j}=(\sum_{\mu}\widetilde{\boldsymbol{G}}^{*}_{i\mu}\cdot\boldsymbol{\alpha}_{\mu}):(\sum_{\mu}\widetilde{\boldsymbol{G}}^{*}_{j\mu}\cdot\boldsymbol{\alpha}_{\mu}),$
(173) $\displaystyle
B_{i{k_{i}}}:B_{j{k_{2}}}=\widetilde{\boldsymbol{G}}^{*}_{ik_{1}}:\widetilde{\boldsymbol{G}}^{*}_{ik_{2}},$
$\displaystyle
D:E_{k_{1}}:E_{k_{2}}=(\sum_{\mu\nu}\boldsymbol{\alpha}_{\mu}\cdot\widetilde{\boldsymbol{G}}^{*}_{\mu\nu}\cdot\boldsymbol{\alpha}_{\nu}):(\sum_{\mu}\widetilde{\boldsymbol{G}}^{*}_{k_{1}\mu}\cdot\boldsymbol{\alpha}_{\mu}):(\sum_{\mu}\widetilde{\boldsymbol{G}}^{*}_{k_{2}\mu}\cdot\boldsymbol{\alpha}_{\mu}).$

There are also some useful properties of $\widetilde{\boldsymbol{G}}$. Since
$|\widetilde{\boldsymbol{G}}|\to 0$ means the corank of
$\widetilde{\boldsymbol{G}}$ is one, we have

$\displaystyle\sum_{\mu}\widetilde{\boldsymbol{G}}^{*}_{i\mu}\cdot\boldsymbol{v_{\mu}}/\sum_{\mu}\widetilde{\boldsymbol{G}}^{*}_{j\mu}\cdot\boldsymbol{v_{\mu}}=\widetilde{\boldsymbol{G}}^{*}_{ik}/\widetilde{\boldsymbol{G}}^{*}_{jk},$
(174)
$\displaystyle\widetilde{\boldsymbol{G}}^{*}_{ik_{1}}\widetilde{\boldsymbol{G}}^{*}_{jk_{2}}-\widetilde{\boldsymbol{G}}^{*}_{ik_{2}}\widetilde{\boldsymbol{G}}^{*}_{jk_{1}}=0,$
$\displaystyle(\sum_{\mu}\widetilde{\boldsymbol{G}}^{*}_{i\mu}\cdot\boldsymbol{v_{\mu}})\cdot(\sum_{\mu}\widetilde{\boldsymbol{G}}^{*}_{j\mu}\cdot\boldsymbol{v_{\mu}})=\widetilde{\boldsymbol{G}}_{ij}\cdot\boldsymbol{{\alpha}}^{T}\widetilde{\boldsymbol{G}}^{-1}\boldsymbol{\alpha}.$

Having above preparations, we will prove (168) by induction with index
$\kappa=Max[\frac{m+\sum a_{i}}{2},\frac{m^{\prime}+\sum a^{\prime}_{i}}{2}]$.
Suppose it is true for $\kappa<\kappa^{\prime}$, all we need to prove are
following four relations:

  * •

(A) With the same $m$ but different $a_{i},a_{j},a_{0}$:
$\frac{c^{(0,...,r)}_{a_{1},...,a_{i}+1,..,a_{n}}(m)}{c^{(0,...,n)}_{a_{1},...,a_{i}+1,...,a_{n}}(m)}-\frac{c^{(0,...,r)}_{a_{1},..,a_{j}-1,..,a_{n}}(m)}{c^{(0,...,n)}_{a_{1},...,a_{j}-1,...,a_{n}}(m)}\leq|\widetilde{\boldsymbol{G}}|^{(\frac{m+\sum
a_{i}-1}{2})}\\\ $
Proof of (A):
Above relation is equal to

$c^{(0,...,r)}_{a_{1},...,a_{i}+1,..,a_{n}}(m)c^{(0,...,n)}_{a_{1},...,a_{j}-1,...,a_{n}}(m)-c^{(0,...,r)}_{a_{1},...,a_{j}-1,..,a_{n}}(m)c^{(0,...,n)}_{a_{1},...,a_{i}+1,...,a_{n}}(m)\leq|\widetilde{\boldsymbol{G}}|^{\frac{m+\sum
a_{i}+1}{2}}.$ (175)

Using (172), we have

$\displaystyle
c^{(0,...,r)}_{a_{1},...,a_{i}+1,..,a_{n}}(m)c^{(0,...,n)}_{a_{1},...,a_{j}-1,...,a_{n}}(m)-c^{(0,...,r)}_{a_{1},...,a_{j}-1,..,a_{n}}(m)c^{(0,...,n)}_{a_{1},...,a_{i}+1,...,a_{n}}(m)$
(176) $\displaystyle=$
$\displaystyle\frac{N_{i}(a_{1},...,a_{n};m)}{|\widetilde{\boldsymbol{G}}|}\times\Bigg{\\{}A_{i}\cdot[c^{(0,...,r)}_{a_{1},...,a_{n}}(m-1)c^{(0,...,n)}_{a_{1},...,a_{j}-1,...,a_{n}}(m)-c^{(0,...,n)}_{a_{1},...,a_{n}}(m-1)c^{(0,...,r)}_{a_{1},...,a_{j}-1,...,a_{n}}(m)]$
$\displaystyle+\sum_{k}B_{ik}\cdot[c^{(0,...,r)}_{a_{1},...,a_{k}-1,...,a_{n}}(m)c^{(0,...,n)}_{a_{1},...,a_{j}-1,...,a_{n}}(m)-c^{(0,...,n)}_{a_{1},...,a_{k}-1,...,a_{n}}(m)c^{(0,...,r)}_{a_{1},...,a_{j}-1,...,a_{n}}]\Bigg{\\}}$
$\displaystyle+\frac{X^{lower}\cdot
c^{(0,...,n)}_{a_{1},...,a_{j}-1,...,a_{n}}(m)}{|\widetilde{\boldsymbol{G}}|}\leq\frac{1}{|\widetilde{\boldsymbol{G}}|}|\widetilde{\boldsymbol{G}}|^{\frac{m+\sum
a_{i}-1}{2}}\sim|\widetilde{\boldsymbol{G}}|^{\frac{m+\sum a_{i}+1}{2}}.$

Thus we have done the proof.

  * •

(B) With same $a_{0}$, but different $m,a_{i}$:
$\frac{c^{(0,...,r)}_{a_{1},...,a_{i}+1,..,a_{n}}(m)}{c^{(0,...,n)}_{a_{1},...,a_{i}+1,...,a_{n}}(m)}-\frac{c^{(0,...,r)}_{a_{1},...,a_{n}}(m-1)}{c^{(0,...,n)}_{a_{1},...,a_{n}}(m-1)}\leq|\widetilde{\boldsymbol{G}}|^{(\frac{m+\sum
a_{i}}{2})}$

Proof of (B):
Change $c^{(0,...,r)/(0,...,n)}_{a_{1},...a_{j}-1,...,a_{m}}(m)$ in (176) into
$c^{(0,...,r)/(0,...,n)}_{a_{1},...,a_{n}}(m-1)$, we will reach the proof of
(B).

  * •

(C) With the same $m,a_{0}$ but different $a_{i},a_{j}$:
$\frac{c^{(0,...,r)}_{a_{1},...,a_{i}+1,..,a_{n}}(m)}{c^{(0,...,n)}_{a_{1},...,a_{i}+1,...,a_{n}}(m)}-\frac{c^{(0,...,r)}_{a_{1},.,,a_{j}+1,..,a_{n}}(m)}{c^{(0,...,n)}_{a_{1},...,a_{j}+1,...,a_{n}}(m)}\leq|\widetilde{\boldsymbol{G}}|^{(\frac{m+\sum
a_{i}+1}{2})}$.

Poof of (C):

By (172), we have

$\displaystyle
c^{(0,...,r)}_{a_{1},...,a_{i}+1,..,a_{n}}(m)c^{(0,...,n)}_{a_{1},...,a_{j}+1,...,a_{n}}(m)-c^{(0,...,r)}_{a_{1},...,a_{j}+1,..,a_{n}}(m)c^{(0,...,n)}_{a_{1},...,a_{i}+1,...,a_{n}}(m)$
(177) $\displaystyle=$
$\displaystyle\frac{N_{i}(a_{1},...,a_{n};m)N_{j}(a_{1},...,a_{n};m)}{|\widetilde{\boldsymbol{G}}|^{2}}\times\Bigg{\\{}\sum_{k}[A_{i}B_{jk}-A_{j}B_{ik}]$
$\displaystyle\times$
$\displaystyle[c^{(0,...,r)}_{a_{1},...,a_{n}}(m-1)c^{0,...,n)}_{a_{1},...,a_{k}-1,...,a_{n}}(m)-c^{0,...,r)}_{a_{1},...,a_{k}-1,...,a_{n}}(m)c^{(0,...,n)}_{a_{1},...,a_{n}}(m-1)]$
$\displaystyle+$
$\displaystyle\sum_{k_{1}<k_{2}}[B_{ik_{1}}B_{jk_{2}}-B_{jk_{1}}B_{ik_{2}}]$
$\displaystyle\times$
$\displaystyle[c^{(0,...,r)}_{a_{1},...,a_{k_{1}}-1,...,a_{n}}(m)c^{(0,...,n)}_{a_{1},...,a_{k_{2}}-1,...,a_{n}}(m)-c^{(0,...,r)}_{a_{1},...,a_{k_{2}-1},...,a_{n}}(m)c^{(0,...,n)}_{a_{1},...,a_{k_{1}}-1,...,a_{n}}(m)]\Bigg{\\}}$
$\displaystyle+$
$\displaystyle\frac{1}{|\widetilde{\boldsymbol{G}}|}\times\left\\{X_{i}^{lower}C^{(0,...,n)}_{a_{1},...,a_{j}+1,...,a_{n}}(m)-X_{j}^{lower}c^{(0,...,n)}_{a_{1},...,a_{i}+1,...,a_{n}}(m)\right\\}.$

From (173) and (174), we know

$\displaystyle\lim_{|\widetilde{\boldsymbol{G}}|\to
0}(A_{i}B_{jk}-A_{j}B_{ik})=\lim_{|\widetilde{\boldsymbol{G}}|\to
0}(B_{ik_{1}}B_{jk_{2}}-B_{jk_{1}}B_{ik_{2}})=0.$ (178)

For the last line of (177), using the following expression of $X^{lower}_{i}$
and $c^{(0,...,n)}_{a_{1},...,a_{i}+1,...,a_{n}}(m)$

$\displaystyle
c^{(0,...,n)}_{a_{1},...,a_{i}+1,...,a_{n}}(m)=\frac{1}{(a_{i}+1)|\widetilde{\boldsymbol{G}}|}\sum_{\mu}\widetilde{\boldsymbol{G}}^{*}_{i\mu}\cdot\boldsymbol{O}_{\mu}^{(0,...,n)}(a_{1},...,a_{n};m),$
(179) $\displaystyle
X^{lower}_{i}=-\frac{m}{a_{i}+1}\sum_{\mu}\widetilde{\boldsymbol{G}}^{*}_{i\mu}\cdot\delta_{0a_{\mu}}c^{(0,...,r)}_{a1,...,\widehat{a_{\mu}},...,a_{n}}(m-1;\widehat{\mu}),$

then by the first line of (174), we know

$X_{i}^{lower}C^{(0,...,n)}_{a_{1},...,a_{j}+1,...,a_{n}}(m)-X_{j}^{lower}c^{(0,...,n)}_{a_{1},...,a_{i}+1,...,a_{n}}(m)=0\cdot|\widetilde{\boldsymbol{G}}|^{-\frac{m+\sum
a_{k}+1}{2}}+\mathcal{O}(|\widetilde{\boldsymbol{G}}|^{-\frac{m+\sum
a_{k}-1}{2}}).$ (180)

Thus we have

$\displaystyle
c^{(0,...,r)}_{a_{1},...,a_{i}+1,..,a_{n}}(m)c^{(0,...,n)}_{a_{1},...,a_{j}+1,...,a_{n}}(m)-c^{(0,...,r)}_{a_{1},...,a_{j}+1,..,a_{n}}(m)c^{(0,...,n)}_{a_{1},...,a_{i}+1,...,a_{n}}(m)$
(181) $\displaystyle\leq$
$\displaystyle\frac{1}{|\widetilde{\boldsymbol{G}}|^{2}}\cdot\mathcal{O}(|\widetilde{\boldsymbol{G}}|)\cdot\mathcal{O}(|\widetilde{\boldsymbol{G}}|^{-\frac{m+\sum
a_{k}-1}{2}})+\frac{1}{|\widetilde{\boldsymbol{G}}|}\cdot\mathcal{O}(|\widetilde{\boldsymbol{G}}|^{-\frac{m+\sum
a_{k}-1}{2}})\sim|\widetilde{\boldsymbol{G}}|^{-\frac{m+\sum a_{k}+1}{2}}.$

  * •

(D) With the same $a_{0}$, but different $m,a_{i}$:
$\frac{c^{(0,...,r)}_{a_{1},...,a_{i}+1,..,a_{n}}(m)}{c^{(0,...,n)}_{a_{1},...,a_{i}+1,..,a_{n}}(m)}-\frac{c^{(0,...,r)}_{a_{1},...,a_{i},..,a_{n}}(m+1)}{c^{(0,...,n)}_{a_{1},...,a_{i},..,a_{n}}(m+1)}\leq|\widetilde{\boldsymbol{G}}|^{\frac{m+\sum
a_{i}+1}{2}}$

Proof of (D):

By (172), we have

$\displaystyle
c^{(0,...,r)}_{a_{1},...,a_{i}+1,..,a_{n}}(m)c^{(0,...,n)}_{a_{1},...,a_{i},..,a_{n}}(m+1)-c^{(0,...,r)}_{a_{1},...,a_{i},..,a_{n}}(m+1)c^{(0,...,n)}_{a_{1},...,a_{i}+1,..,a_{n}}(m)$
(182) $\displaystyle=$
$\displaystyle\frac{N_{i}(a_{1},...,a_{n};m)M(a_{1},...,a_{n};m)}{|\widetilde{\boldsymbol{G}}|^{2}}\times\Bigg{\\{}\sum_{k_{1}<k_{2}}[E_{k_{1}}B_{ik_{2}}-E_{k_{2}}B_{ik_{1}}]$
$\displaystyle\times$
$\displaystyle[c^{(0,...,r)}_{a_{1},...,a_{k_{1}}-1,...,a_{n}}(m)c^{(0,...,n)}_{a_{1},...,a_{k_{2}}-1,...,a_{n}}(m)-c^{(0,...,r)}_{a_{1},...,a_{k_{2}}-1,...,a_{n}}(m)c^{(0,...,n)}_{a_{1},...,a_{k_{1}}-1,...,a_{n}}(m)]$
$\displaystyle+$ $\displaystyle\sum_{k}[A_{i}E_{k}-D\cdot B_{ik}]$
$\displaystyle\times$
$\displaystyle[c^{(0,...,n)}_{a_{1},...,a_{n}}(m-1)c^{(0,...,r)}_{a_{1},...,a_{k}-1,...,a_{n}}(m)-c^{(0,...,r)}_{a_{1},...,a_{n}}(m-1)c^{(0,...,n)}_{a_{1},...,a_{k}-1,...,a_{n}}(m)]\Bigg{\\}}$
$\displaystyle+$
$\displaystyle\frac{1}{|\widetilde{\boldsymbol{G}}|}[Y^{lower}c^{(0,...,n)}_{a_{1},...,a_{i}+1,...,a_{n}}(m)-X^{lower}_{i}c^{(0,...,n)}_{a_{1},...,a_{n}}(m+1)].$

Similar to the proof of (C), we have

$\displaystyle\lim_{|\widetilde{\boldsymbol{G}}|\to
0}(E_{k_{1}}B_{ik_{2}}-E_{k_{2}}B_{ik_{1}})=\lim_{|\widetilde{\boldsymbol{G}}|\to
0}(A_{i}E_{k}-D\cdot B_{ik})=0.$ (183)

Then using the following expressions for $Y^{lower}$ and
$c^{(0,...,n)}_{a_{1},...,a_{n}}(m+1)$

$\displaystyle Y^{lower}=$ $\displaystyle\frac{m(m+1)}{(m+1-\sum
a_{\alpha})(D+m-n-1)}\cdot\sum_{\mu}\widetilde{\boldsymbol{G}}^{*}_{i\mu}\cdot\delta_{0a_{\mu}}c^{(0,...,r)}_{a1,...,\widehat{a_{\mu}},...,a_{n}}(m-1;\widehat{\mu})$
(184) $\displaystyle c^{(0,...,n)}_{a_{1},...,a_{n}}(m+1)=$
$\displaystyle\frac{m+1}{(m+1-\sum
a_{\alpha})(D+m-n-1)|\widetilde{\boldsymbol{G}}|}\times$
$\displaystyle\Bigg{\\{}4m|\widetilde{\boldsymbol{G}}|\cdot
c^{(0,...,n)}_{a_{1},...,a_{n}}(m-1)-\sum_{\mu\nu}\alpha_{\nu}\widetilde{\boldsymbol{G}}^{*}_{\mu\nu}O_{\nu}^{(0,...,n)}(a_{1},...,a_{n};m)\Bigg{\\}}$

and combining with (179) and the property (174), the leading order of last
line of (182) is

$\displaystyle
Y^{lower}c^{(0,...,n)}_{a_{1},...,a_{i}+1,...,a_{n}}(m)-X^{lower}_{i}c^{(0,...,n)}_{a_{1},...,a_{n}}(m+1)=0\cdot|\widetilde{\boldsymbol{G}}|^{-\frac{m+\sum
a_{k}+1}{2}}+\mathcal{O}(|\widetilde{\boldsymbol{G}}|^{-\frac{m+\sum
a_{k}-1}{2}}).$ (185)

thus we arrive

$\displaystyle
c^{(0,...,r)}_{a_{1},...,a_{i}+1,..,a_{n}}(m)c^{(0,...,n)}_{a_{1},...,a_{i},..,a_{n}}(m+1)-c^{(0,...,r)}_{a_{1},...,a_{i},..,a_{n}}(m+1)c^{(0,...,n)}_{a_{1},...,a_{i}+1,..,a_{n}}(m)$
(186) $\displaystyle\leq$
$\displaystyle\frac{1}{|\widetilde{\boldsymbol{G}}|^{2}}\cdot\mathcal{O}(|\widetilde{\boldsymbol{G}}|)\cdot\mathcal{O}(|\widetilde{\boldsymbol{G}}|^{-\frac{m+\sum
a_{k}-1}{2}})+\frac{1}{|\widetilde{\boldsymbol{G}}|}\cdot\mathcal{O}(|\widetilde{\boldsymbol{G}}|^{-\frac{m+\sum
a_{k}-1}{2}})\sim|\widetilde{\boldsymbol{G}}|^{-\frac{m+\sum a_{k}+1}{2}}.$

Having above four relations (A)(B)(C)(D), we can prove (168) iteratively.
Eventually, we have proved that $\mathcal{B}_{n+1;\mathbf{b}_{j}}$ is
independent of the rank $m$ and $R$ by the recursion relation of
$c^{(0,...,r)}_{a_{1},...,a_{n}}(m)$.

##  5 Discussion

In this paper, we show how to do the reduction for one-loop tensor integrals
when Gram determinant degenerates. In this case, the highest topological
master basis will be reduced to an expansion of the basis with lower
typologies. By demanding the cancellation of divergent parts, we can solve the
decomposition coefficients. We have proved that the results are independent of
the tensor rank $m$ and the auxiliary vector $R$, thus shown our method’s
self-consistency.

One advantage of our method is that we have the series expansion around the
degenerate point $|G|$, which will be helpful for some situations, for
example, the improvement of numerical accuracy. The same idea can also be used
to deal with other kinds of singularities, such as soft limits or massless
limits. Furthermore, as the improved PV-reduction method has the applied to
two-loop sunset topology recently in Feng:2022iuc , we show that the same
method works well for the sunset topology in the limit $K^{2}\to 0$ as given
in the section 3.4.

## Acknowledgments

This work is supported by Chinese NSF funding under Grant No.11935013,
No.11947301, No.12047502 (Peng Huanwu Center).

##  Appendix A Expansion for degenerate Gram determinant

In this appendix, we will give the decomposition for box and pentagon as the
series of $|G|$. Since the analytic expressions with higher and higher orders
of $|G|$ are complicated, we will present only numerical results to
demonstrate the independence of decomposition coefficients for the rank choice
and $R$.

In following, we list some numerical results for box and pentagon. Here we set
$K_{0}=0$, and choose the Gram Matrices as:

$\displaystyle\text{box}:\mathbf{G}$ $\displaystyle=$
$\displaystyle\frac{1}{71}\left(\begin{array}[]{ccc}2&5&11\\\ 5&17&23\\\
11&23&65+\frac{357911}{9}t\\\ \end{array}\right)\,,$ (190)
$\displaystyle\text{pentagon}:\mathbf{G}$ $\displaystyle=$
$\displaystyle\frac{1}{71}\left(\begin{array}[]{cccc}2&5&11&41\\\
5&17&23&47\\\ 11&23&31&59\\\ 41&47&59&\frac{1283}{17}+\frac{25411681}{306}t\\\
\end{array}\right)\,,$ (195)

where $\mathbf{G}_{ij}=s_{ij}=K_{i}\cdot K_{j}$. With this parameterization,
we have $\text{det}(\mathbf{G})=t$. The chosen squares of masses are

$\displaystyle\text{box}:\left(\begin{array}[]{cccc}M_{0}^{2},&M_{1}^{2},&M_{2}^{2},&M_{3}^{2}\end{array}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{71}\left(\begin{array}[]{cccc}3,&7,&13,&19\end{array}\right)\,,$
(198)
$\displaystyle\text{pentagon}:\left(\begin{array}[]{ccccc}M_{0}^{2},&M_{1}^{2},&M_{2}^{2},&M_{3}^{2},&M_{4}^{2}\end{array}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{71}\left(\begin{array}[]{ccccc}3,&7,&13,&19,&29\end{array}\right)\,.$
(201)

Firstly we present some reduction coefficients needed in (33) and (36). For
$I_{4}^{(2)}$ we have

$\displaystyle C^{({2})}_{{4}\to{4}}$ $\displaystyle=$
$\displaystyle\frac{s_{03}s_{01}}{384300851763(D-3)t^{2}}\times\big{(}-12041426688574Dt^{2}-32933538576Dt-$
(202a) $\displaystyle
20155392D+29975466437514t^{2}+73134102096t+40310784\big{)}$
$\displaystyle+\frac{s_{03}^{2}}{128100283921(D-3)t^{2}}\times\big{(}128100283921Dt^{2}+463852656Dt+$
$\displaystyle 419904D-256200567842t^{2}-856838934t-839808\big{)}$
$\displaystyle+\frac{s_{02}s_{03}}{384300851763(D-3)t^{2}}\times\big{(}2818206246262Dt^{2}+6493937184Dt+$
$\displaystyle 2519424D-7686017035260t^{2}-16273497348t-5038848\big{)}$
$\displaystyle+\frac{s_{01}^{2}}{10376122997601(D-3)t^{2}}\times\big{(}2546761744633401Dt^{2}+$
$\displaystyle 4709032163712Dt+2176782336D+779424512213633527t^{3}-$
$\displaystyle 3722722351028181t^{2}-8800212589632t-4353564672\big{)}$
$\displaystyle+\frac{s_{02}s_{01}}{10376122997601(D-3)t^{2}}\times\big{(}-1192101242168826Dt^{2}-$
$\displaystyle 1690742931120Dt-544195584D-458485007184490310t^{3}+$
$\displaystyle 1496467516765122t^{2}+2876350319856t+1088391168\big{)}$
$\displaystyle+\frac{s_{02}^{2}}{10376122997601(D-3)t^{2}}\times\big{(}139501209189969Dt^{2}+$
$\displaystyle 137764238832Dt+34012224D+91697001436898062t^{3}-$
$\displaystyle 25363856216358t^{2}-69403953654t-68024448\big{)}$
$\displaystyle+\frac{s_{00}\left(-128100283921t^{2}-534719034t-419904\right)}{228705129(D-3)t}\,,$

$\displaystyle C^{({2})}_{{4}\to{4};\widehat{{3}}}$ $\displaystyle=$
$\displaystyle\frac{s_{02}s_{01}}{16238064159(D-3)t^{2}}\times\big{(}2145318534Dt+839808D+$
(202b) $\displaystyle
1281002839210t^{2}-3652839666t-1679616\big{)}+\frac{(357911t+648)s_{00}}{3t57911(D-3)}$
$\displaystyle+\frac{s_{03}s_{01}}{1804229351(D-3)t^{2}}\times(100930902Dt+93312D-251253522t-$
$\displaystyle
186624)+\frac{s_{02}^{2}}{16238064159(D-3)t^{2}}\times\big{(}-183608343Dt-52488D-$
$\displaystyle
256200567842t^{2}+57981582t+104976\big{)}+\frac{s_{01}^{2}}{16238064159(D-3)t^{2}}\times$
$\displaystyle\big{(}-5411614320Dt-3359232D-2177704826657t^{2}+10436684760t+$
$\displaystyle
6718464\big{)}+\frac{s_{03}^{2}(-3221199Dt-5832D+6442398t+11664)}{1804229351(D-3)t^{2}}$
$\displaystyle+\frac{s_{02}s_{03}(-23622126Dt-11664D+64423980t+23328)}{1804229351(D-3)t^{2}}\,,$
$\displaystyle C^{({2})}_{{4}\to{4};\widehat{{23}}}$ $\displaystyle=$
$\displaystyle\frac{s_{01}^{2}(48317985t+10368)}{71t(715822t+81)}-\frac{32s_{02}s_{01}}{71t}+\frac{s_{03}s_{01}(-7158220t-2592)}{71t(715822t+81)}+\frac{2s_{02}^{2}}{71t}$
(202c)
$\displaystyle+\frac{162s_{03}^{2}}{71t(715822t+81)}+\frac{4s_{02}s_{03}}{71t}\,.$

For $I_{5}^{(2)}$, we have

$\displaystyle C^{({2})}_{{5}\to{5}}=$
$\displaystyle\frac{s_{04}s_{01}}{32933430093533811(D-4)t^{2}}\times\big{(}143357283936558942Dt^{2}+23554798645968Dt+$
$\displaystyle
801715968D-514019810871625756t^{2}-65771935818336t-2405147904\big{)}$
$\displaystyle+\frac{s_{04}^{2}}{645753531245761(D-4)t^{2}}\times\big{(}645753531245761Dt^{2}+150030564624Dt+$
$\displaystyle 8714304D-1937260593737283t^{2}-290912924088t-26142912\big{)}$
$\displaystyle+\frac{s_{00}}{552094181406(D-4)t}\times\left(645753531245761t^{2}+309209334408t+8714304\right)$
$\displaystyle+\frac{s_{02}s_{04}}{32933430093533811(D-4)t^{2}}\times\big{(}484315148434320750Dt^{2}+100670508862704Dt+$
$\displaystyle
5158867968D-1554974503239792488t^{2}-219630109481280t-15476603904\big{)}$
$\displaystyle-\frac{s_{03}s_{04}}{10977810031177937(D-4)t^{2}}\times\big{(}167895918123897860Dt^{2}+36157366074384Dt+$
$\displaystyle
1934575488D-528226388559032498t^{2}-75984992058960t-5803726464\big{)}$
$\displaystyle+\frac{s_{01}^{2}}{30232888825864038498(D-4)t^{2}}\times\big{(}143213926652622383058Dt^{2}+$
$\displaystyle
13789009133462592Dt+331910410752D-16409682740640811134241t^{3}-$
$\displaystyle
556117775080599407112t^{2}-43600316518303488t-995731232256\big{)}$
$\displaystyle+\frac{s_{02}s_{01}}{15116444412932019249(D-4)t^{2}}\times\big{(}483830833285886429250Dt^{2}+$
$\displaystyle
67656883239451296Dt+2135771338752D+804074454291399745577809t^{3}-$
$\displaystyle
1392037972235862072480t^{2}-183993225050551104t-6407314016256\big{)}$
$\displaystyle+\frac{s_{03}s_{01}}{1679604934770224361(D-4)t^{2}}\times\big{(}-55909340735257987380Dt^{2}-$
$\displaystyle 8237128089551472Dt-266971417344D-65638730962563244536964t^{3}+$
$\displaystyle
172423941884993136132t^{2}+22589794735742304t+800914252032\big{)}$
$\displaystyle+\frac{s_{02}^{2}}{60465777651728076996(D-4)t^{2}}\times\big{(}3269127251931665062500Dt^{2}+$
$\displaystyle
599522136237504000Dt+27486448533504D-968171281697807856920219t^{3}-$
$\displaystyle
11503282926679737573096t^{2}-1435853145074044608t-82459345600512\big{)}$
$\displaystyle+\frac{s_{03}^{2}}{6718419739080897444(D-4)t^{2}}\times\big{(}392876448409920992400Dt^{2}+$
$\displaystyle
77937877710875520Dt+3865281825024D+16409682740640811134241t^{3}-$
$\displaystyle
1277220411366240783636t^{2}-174378410197442208t-11595845475072\big{)}$
$\displaystyle+\frac{s_{02}s_{03}}{3359209869540448722(D-4)t^{2}}\times\big{(}-377765815778770185000Dt^{2}-$
$\displaystyle
72109190275233120Dt-3435806066688D+16409682740640811134241t^{3}+$
$\displaystyle
1264318255811950478856t^{2}+166745623518606528t+10307418200064\big{)}\,,$
(203a) $\displaystyle C^{({2})}_{{5}\to{5};\widehat{{4}}}$ $\displaystyle=$
$\displaystyle\frac{s_{01}^{2}}{1391553384233823(D-4)t^{2}}\times\big{(}3703193448768Dt+112435776D-$
(203b) $\displaystyle 645753531245761t^{2}-13919908971456t-337307328\big{)}$
$\displaystyle+\frac{s_{02}s_{01}}{1391553384233823(D-4)t^{2}}\times\big{(}33381800628840Dt+1446999552D+$
$\displaystyle 63283846062084578t^{2}-113719508702928t-4340998656\big{)}$
$\displaystyle+\frac{s_{03}s_{01}}{154617042692647(D-4)t^{2}}\times\big{(}-4023685569540Dt-180874944D-$
$\displaystyle 5166028249966088t^{2}+13937595501432t+542624832\big{)}$
$\displaystyle+\frac{s_{04}s_{01}(33848359092Dt+1629504D-121366188456t-4888512)}{9095120158391(D-4)t^{2}}$
$\displaystyle+\frac{s_{02}^{2}}{2783106768467646(D-4)t^{2}}\times\big{(}122937240222144Dt+9311127552D-$
$\displaystyle 38099458343499899t^{2}-416021948215128t-27933382656\big{)}$
$\displaystyle+\frac{s_{03}^{2}}{309234085385294(D-4)t^{2}}\times\big{(}15130216514124Dt+1309377312D+$
$\displaystyle 645753531245761t^{2}-49174347196548t-3928131936\big{)}$
$\displaystyle+\frac{s_{04}^{2}(7775974386Dt+903312D-23327923158t-2709936)}{9095120158391(D-4)t^{2}}$
$\displaystyle+\frac{s_{02}s_{03}}{154617042692647(D-4)t^{2}}\times\big{(}-14408118186828Dt-1163890944D+$
$\displaystyle 645753531245761t^{2}+47688373738392t+3491672832\big{)}$
$\displaystyle+\frac{s_{02}s_{04}(114352564500Dt+10485504D-367147967088t-31456512)}{9095120158391(D-4)t^{2}}$
$\displaystyle+\frac{s_{03}s_{04}(-118926667080Dt-11796192D+374161591044t+35388576)}{9095120158391(D-4)t^{2}}$
$\displaystyle+\frac{s_{00}(25411681t+2952)}{25411681(D-4)t}\,,$
$\displaystyle C^{({2})}_{{5}\to{5};\widehat{{34}}}$ $\displaystyle=$
$\displaystyle\frac{s_{01}^{2}(1568612244768t-8455536)}{85697t(25411681t+443556)}+\frac{s_{02}s_{01}(-760215848796t-108819072)}{85697t(25411681t+443556)}$
(203c)
$\displaystyle+\frac{276s_{03}s_{01}}{85697t}-\frac{999s_{03}^{2}}{85697t}+\frac{1776s_{02}s_{03}}{85697t}+\frac{s_{04}s_{01}(-3659282064t-1102896)}{5041t(25411681t+443556)}$
$\displaystyle+\frac{s_{02}^{2}(70517414775t-350113536)}{85697t(25411681t+443556)}-\frac{611388s_{04}^{2}}{5041t(25411681t+443556)}$
$\displaystyle+\frac{s_{02}s_{04}(457410258t-7096896)}{5041t(25411681t+443556)}+\frac{18s_{03}s_{04}}{5041t}\,.$

When $t=\text{det}(\mathbf{G})\to 0$, the basis with the highest typology is
decomposed to the combinations of lower ones. Let us consider the coefficients
$\mathcal{B}_{4;\widehat{3}}$ and $\mathcal{B}_{5;\widehat{4}}$ by (33) for
different $m$:

  * •

$\mathcal{B}_{4;\widehat{3}}$

    * –

$m=2$:

$\displaystyle\mathcal{B}_{4;\widehat{3}}$ $\displaystyle=$
$\displaystyle-{C^{(2)}_{4\to 4;\widehat{3}}\over C^{(2)}_{4\to
4}}+\mathcal{O}\left(t^{2}\right)=\frac{71}{72}-\frac{25411681(36D-61)t}{1679616(D-2)}+\mathcal{O}\left(t^{2}\right)\,.$
(204)

    * –

$m=3$:

$\displaystyle\mathcal{B}_{4;\widehat{3}}$ $\displaystyle=$
$\displaystyle-{C^{(3)}_{4\to 4;\widehat{3}}\over C^{(3)}_{4\to
4}}+\mathcal{O}\left(t^{3}\right)=\frac{71}{72}-\frac{25411681(36D-61)t}{1679616(D-2)}$
(205)
$\displaystyle+\frac{9095120158391\left(432D^{2}-468D+121\right)t^{2}}{13060694016((D-2)D)}+\mathcal{O}\left(t^{3}\right)\,.$

    * –

$m=4$:

$\displaystyle\mathcal{B}_{4;\widehat{3}}$ $\displaystyle=$
$\displaystyle-{C^{(4)}_{4\to 4;\widehat{3}}\over C^{(4)}_{4\to
4}}+\mathcal{O}\left(t^{4}\right)=\frac{71}{72}-\frac{25411681(36D-61)t}{1679616(D-2)}$
(206)
$\displaystyle+\frac{9095120158391\left(432D^{2}-468D+121\right)t^{2}}{13060694016((D-2)D)}$
$\displaystyle-\frac{3255243551009881201\left(15552D^{3}+28512D^{2}+16596D+50215\right)t^{3}}{304679870005248D\left(D^{2}-4\right)}$
$\displaystyle+\mathcal{O}\left(t^{4}\right)\,.$

  * •

$\mathcal{B}_{5;\widehat{4}}$

    * –

$m=2$:

$\displaystyle\mathcal{B}_{5;\widehat{4}}$ $\displaystyle=$
$\displaystyle-{C^{(2)}_{5\to 5;\widehat{4}}\over C^{(2)}_{5\to
5}}+\mathcal{O}\left(t^{2}\right)=\frac{1207}{164}-\frac{30671898967(41D-36)t}{19849248(D-3)}+\mathcal{O}\left(t^{2}\right)\,.$

    * –

$m=3$:

$\displaystyle\mathcal{B}_{5;\widehat{4}}$ $\displaystyle=$
$\displaystyle-{C^{(3)}_{5\to 5;\widehat{4}}\over C^{(3)}_{5\to
5}}+\mathcal{O}\left(t^{3}\right)=\frac{1207}{164}-\frac{30671898967(41D-36)t}{19849248(D-3)}$
$\displaystyle+\frac{779424512213633527\left(1681D^{2}+3977D+17049\right)t^{2}}{2402394183936(D-3)(D-1)}+\mathcal{O}\left(t^{3}\right)\,.$

    * –

$m=4$:

$\displaystyle\mathcal{B}_{5;\widehat{4}}$ $\displaystyle=$
$\displaystyle-{C^{(4)}_{5\to 5;\widehat{4}}\over C^{(4)}_{5\to
5}}+\mathcal{O}\left(t^{4}\right)=\frac{1207}{164}-\frac{30671898967(41D-36)t}{19849248(D-3)}$
(209)
$\displaystyle+\frac{779424512213633527\left(1681D^{2}+3977D+17049\right)t^{2}}{2402394183936(D-3)(D-1)}$
$\displaystyle-\frac{19806487067953459039028887t^{3}}{290766572870141952((D-3)(D-1)(D+1))}\times$
$\displaystyle\big{(}68921D^{3}+670719D^{2}+4586014D+13861761\big{)}+\mathcal{O}\left(t^{4}\right)\,.$

To make them look more intuitive, we choose $D=4$, and here we list:

Table 1: $\mathcal{B}_{4;\widehat{3}}$ rank order |  $t^{0}$ |  $t^{1}$ |  $t^{2}$ |  $t^{3}$
---|---|---|---|---
$m=2$ |  $-\frac{71}{72}$ |  $\frac{2109169523}{3359232}$ |  / |  /
$m=3$ |  $-\frac{71}{72}$ |  $\frac{2109169523}{3359232}$ |  $-\frac{46939915137455951}{104485552128}$ |  /
$m=4$ |  $-\frac{71}{72}$ |  $\frac{2109169523}{3359232}$ |  $-\frac{46939915137455951}{104485552128}$ |  $\frac{5104609261966063899030919}{14624633760251904}$
Table 2: $\mathcal{B}_{5;\widehat{4}}$ rank order |  $t^{0}$ |  $t^{1}$ |  $t^{2}$ |  $t^{3}$
---|---|---|---|---
$m=2$ |  $\frac{1207}{164}$ |  $-\frac{122687595868}{620289}$ |  / |  /
$m=3$ |  $\frac{1207}{164}$ |  $-\frac{122687595868}{620289}$ |  $\frac{15550298443174202497177}{2402394183936}$ |  /
$m=4$ |  $\frac{1207}{164}$ |  $-\frac{122687595868}{620289}$ |  $\frac{15550298443174202497177}{2402394183936}$ |  $-\frac{62520186560835559083105672288737}{290766572870141952}$

From these numerical results, one can see that as the series of $t$,
$\mathcal{B}_{4;\widehat{3}}$ and $\mathcal{B}_{5;\widehat{4}}$ are
independent of the choice of rank $m$. And they are also independent of the
auxiliary vector $R$, although $s_{0i}=R\cdot K_{i}$ and $s_{00}=R\cdot R$
appear in the expression of $C^{(3)}_{4\to 4;\widehat{3}}$, $C^{(3)}_{4\to
4}$, etc.

We can also use $\eqref{F-m-2}$ to derive
$F^{(m)}_{n+1;\widehat{\mathbf{b}_{j}}}$. Some results are:

$\displaystyle F^{(3)}_{4;\widehat{3}}$ $\displaystyle=$
$\displaystyle\frac{\left(-1239447D^{2}-1774432D+652916\right)s_{03}s_{01}^{2}}{23328D\left(D^{2}-4\right)}+\frac{(-197593D-206184)s_{03}^{2}s_{01}}{46656D\left(D^{2}-4\right)}$
(210a)
$\displaystyle+\frac{\left(862224D^{2}+1492491D-274912\right)s_{02}s_{03}s_{01}}{23328D\left(D^{2}-4\right)}-\frac{94501s_{03}^{3}}{559872D\left(D^{2}-4\right)}$
$\displaystyle+\frac{(274912D+455323)s_{02}s_{03}^{2}}{186624D\left(D^{2}-4\right)}+\frac{\left(-1199616D^{2}-2674144D-644325\right)s_{02}^{2}s_{03}}{186624D\left(D^{2}-4\right)}$
$\displaystyle+\frac{\left(-7774713D^{3}-9053352D^{2}+13507608D+2542936\right)s_{01}^{3}}{17496D\left(D^{2}-4\right)}$
$\displaystyle+\frac{\left(10816992D^{3}+12443673D^{2}-19600544D-3195852\right)s_{02}s_{01}^{2}}{23328D\left(D^{2}-4\right)}$
$\displaystyle+\frac{\left(-7524864D^{3}-10046784D^{2}+11389039D+2955304\right)s_{02}^{2}s_{01}}{46656D\left(D^{2}-4\right)}$
$\displaystyle+\frac{\left(10469376D^{3}+17994240D^{2}-7538496D-3393445\right)s_{02}^{3}}{559872D\left(D^{2}-4\right)}$
$\displaystyle+\frac{(759D-242)s_{00}s_{01}}{108(D-2)D}+\frac{(121-1056D)s_{00}s_{02}}{432(D-2)D}+\frac{121s_{00}s_{03}}{432(D-2)D}\,,$
$\displaystyle F^{(3)}_{4;\widehat{23}}$ $\displaystyle=$
$\displaystyle\frac{\left(-222655929D^{2}+225327659D+51781152\right)s_{02}s_{03}s_{01}}{2916D\left(D^{2}-4\right)}$
(210b)
$\displaystyle+\frac{\left(159295600D^{2}-46664537D-85762533\right)s_{02}s_{03}^{2}}{23328D\left(D^{2}-4\right)}+\frac{(28045D-22791)s_{00}s_{03}}{54(D-2)D}$
$\displaystyle+\frac{\left(887216D^{2}-259919001D+121362075\right)s_{02}^{2}s_{03}}{23328D\left(D^{2}-4\right)}+\frac{(781D-22791)s_{00}s_{02}}{54(D-2)D}$
$\displaystyle+\frac{\left(1053906747D^{3}-444555708D^{2}-186234704D-478975656\right)s_{01}^{3}}{2187D\left(D^{2}-4\right)}$
$\displaystyle+\frac{\left(-8000067D^{3}+607223737D^{2}-1031570076D+601955892\right)s_{02}s_{01}^{2}}{2916D\left(D^{2}-4\right)}$
$\displaystyle+\frac{\left(-752051667D^{3}+144661577D^{2}+657043940D-122980236\right)s_{03}s_{01}^{2}}{2916D\left(D^{2}-4\right)}$
$\displaystyle+\frac{\left(5565264D^{3}+9169579D^{2}+362503351D-556647384\right)s_{02}^{2}s_{01}}{5832D\left(D^{2}-4\right)}$
$\displaystyle+\frac{\left(269794320D^{3}+32842115D^{2}-349326177D+38835864\right)s_{03}^{2}s_{01}}{5832D\left(D^{2}-4\right)}$
$\displaystyle+\frac{\left(-7742976D^{3}-15727920D^{2}-3271609D+639173595\right)s_{02}^{3}}{69984D\left(D^{2}-4\right)}$
$\displaystyle+\frac{\left(-193574400D^{3}-97997040D^{2}+264828935D+17799771\right)s_{03}^{3}}{69984D\left(D^{2}-4\right)}$
$\displaystyle+\frac{(91164-79378D)s_{00}s_{01}}{27(D-2)D}\,,$ $\displaystyle
F^{(3)}_{4;\widehat{123}}$ $\displaystyle=$
$\displaystyle\frac{\left(117036897D^{2}+80887886D+52970828\right)s_{01}^{3}}{486D(D+2)}-\frac{5041s_{00}s_{01}}{3D}$
(210c)
$\displaystyle+\frac{\left(-8231953D^{2}+40086032D-66571446\right)s_{02}s_{01}^{2}}{648D(D+2)}+\frac{5041s_{00}s_{02}}{24D}$
$\displaystyle+\frac{\left(-79814153D^{2}-62992336D+13600618\right)s_{03}s_{01}^{2}}{648D(D+2)}+\frac{5041s_{00}s_{03}}{24D}$
$\displaystyle+\frac{\left(22906304D^{2}+79098331D+246242768\right)s_{02}^{2}s_{01}}{5184D(D+2)}$
$\displaystyle+\frac{\left(114531520D^{2}+130637515D-17179728\right)s_{03}^{2}s_{01}}{5184D(D+2)}$
$\displaystyle+\frac{\left(-1206875892D^{2}-5693290277D-9613489460\right)s_{02}^{3}}{2115072D(D+2)}$
$\displaystyle+\frac{\left(-10269182412D^{2}-15210859589D-1023625460\right)s_{03}^{3}}{8087040D(D+2)}$
$\displaystyle+\frac{\left(-10021508D^{2}+107731211D+75877132\right)s_{02}s_{03}^{2}}{41472D(D+2)}$
$\displaystyle+\frac{\left(10021508D^{2}+56192027D-107373300\right)s_{02}^{2}s_{03}}{41472D(D+2)}$
$\displaystyle+\frac{(-101288813D-22906304)s_{02}s_{03}s_{01}}{2592D(D+2)}\,,$
$\displaystyle F^{(3)}_{5;\widehat{4}}$ $\displaystyle=$
$\displaystyle\frac{s_{01}^{3}}{88197652332(D-3)\left(D^{2}-1\right)}\times\big{(}-1343847503375D^{3}+10417729271235D^{2}-$
(210d) $\displaystyle
17337620658673D-1794280981875\big{)}+\frac{s_{02}s_{01}^{2}}{29399217444(D-3)\left(D^{2}-1\right)}\times$
$\displaystyle\big{(}-1633291581025D^{3}+12077972337027D^{2}-68514754134455D+$
$\displaystyle
93475183460901\big{)}+\frac{s_{03}s_{01}^{2}}{9799739148(D-3)\left(D^{2}-1\right)}\times\big{(}392816962525D^{3}$
$\displaystyle-4603678621302D^{2}+27041270862119D-33850184121582\big{)}$
$\displaystyle+\frac{\left(14623481275D^{2}-89044148172D+94194083969\right)s_{04}s_{01}^{2}}{192151748(D-3)\left(D^{2}-1\right)}$
$\displaystyle+\frac{s_{02}^{2}s_{01}}{29399217444(D-3)\left(D^{2}-1\right)}\times\big{(}-1985077460015D^{3}+30582780834201D^{2}-$
$\displaystyle
243862525327933D+854168652203259\big{)}+\frac{(580029-231855D)s_{00}s_{01}}{28577(D-3)(D-1)}$
$\displaystyle+\frac{(5019844059-1432168335D)s_{04}^{2}s_{01}}{5651522(D-3)\left(D^{2}-1\right)}+\frac{s_{03}^{2}s_{01}}{3266579716(D-3)\left(D^{2}-1\right)}\times$
$\displaystyle\big{(}-114823419815D^{3}+2531886930939D^{2}-34017449607547D+$
$\displaystyle
122326329454647\big{)}+\frac{s_{02}s_{03}s_{01}}{4899869574(D-3)\left(D^{2}-1\right)}\times\big{(}477423692915D^{3}-$
$\displaystyle 8829762425970D^{2}+90569439300937D-324098013014706\big{)}$
$\displaystyle+\frac{\left(17773154165D^{2}-276974011536D+978449946715\right)s_{02}s_{04}s_{01}}{96075874(D-3)\left(D^{2}-1\right)}$
$\displaystyle+\frac{\left(-12823668195D^{2}+313094031390D-1106429051883\right)s_{03}s_{04}s_{01}}{96075874(D-3)\left(D^{2}-1\right)}$
$\displaystyle+\frac{s_{02}^{3}}{88197652332(D-3)\left(D^{2}-1\right)}\times\big{(}-2412632605249D^{3}+76689368763465D^{2}-$
$\displaystyle
1007290897198703D+6188766114569991\big{)}+\frac{s_{03}^{3}}{3266579716(D-3)\left(D^{2}-1\right)}\times$
$\displaystyle\big{(}33563768869D^{3}-1300394788050D^{2}+30312717181523D-352004287844976\big{)}$
$\displaystyle+\frac{2384439363s_{04}^{3}}{5651522(D-3)\left(D^{2}-1\right)}+\frac{s_{02}s_{03}^{2}}{3266579716(D-3)\left(D^{2}-1\right)}\times$
$\displaystyle\big{(}-139554617929D^{3}+5018030800863D^{2}-98440205350997D+$
$\displaystyle
919471445452659\big{)}+\frac{(39776661783-1740635361D)s_{02}s_{04}^{2}}{5651522(D-3)\left(D^{2}-1\right)}$
$\displaystyle+\frac{(1255901463D-45451057824)s_{03}s_{04}^{2}}{5651522(D-3)\left(D^{2}-1\right)}+\frac{(2522217-281793D)s_{00}s_{02}}{28577(D-3)(D-1)}$
$\displaystyle+\frac{s_{02}^{2}s_{03}}{9799739148(D-3)\left(D^{2}-1\right)}\times\big{(}580253411389D^{3}-19518756588588D^{2}+$
$\displaystyle
316580122515497D-2391505497411054\big{)}+\frac{(203319D-2723796)s_{00}s_{03}}{28577(D-3)(D-1)}$
$\displaystyle+\frac{\left(21601218139D^{2}-722503219422D+7448642198855\right)s_{02}^{2}s_{04}}{192151748(D-3)\left(D^{2}-1\right)}$
$\displaystyle+\frac{\left(11245370571D^{2}-552200043258D+9805499547885\right)s_{03}^{2}s_{04}}{192151748(D-3)\left(D^{2}-1\right)}$
$\displaystyle+\frac{\left(-15585689037D^{2}+639672478488D-8561581798443\right)s_{02}s_{03}s_{04}}{96075874(D-3)\left(D^{2}-1\right)}$
$\displaystyle+\frac{22707s_{00}s_{04}}{1681(D-3)(D-1)}\,,$ $\displaystyle
F^{(3)}_{5;\widehat{34}}=$
$\displaystyle\frac{s_{01}^{3}}{36248334956758812852(D-3)\left(D^{2}-1\right)}\times\big{(}-8299431779977556629537D^{3}+$
$\displaystyle 22726783007443171457382D^{2}+12750306686537356621621D-$
$\displaystyle
25457369478990513141690\big{)}+\frac{s_{02}s_{01}^{2}}{81640394046754083(D-3)\left(D^{2}-1\right)}\times$
$\displaystyle\big{(}28475597577033317899D^{3}-90607604692549882710D^{2}+$
$\displaystyle 107185135379627366777D-85969113572839070718\big{)}$
$\displaystyle+\frac{s_{03}s_{01}^{2}}{72569239152670296(D-3)\left(D^{2}-1\right)}\times\big{(}-5435043687063540225D^{3}+$
$\displaystyle 34743610215547035316D^{2}-133703649094040295147D+$
$\displaystyle
138355148225398767416\big{)}+\frac{s_{04}s_{01}^{2}}{473834443879200168(D-3)\left(D^{2}-1\right)}\times$
$\displaystyle\big{(}2027743034533980625D^{3}-47755513091825901372D^{2}+$
$\displaystyle 143945054906768879195D-118509140188920876912\big{)}$
$\displaystyle+\frac{s_{02}^{2}s_{01}}{4412994272797518(D-3)\left(D^{2}-1\right)}\times\big{(}95273008629079876D^{3}-$
$\displaystyle
5459007779794224849D^{2}+34251474265548951092D-68798575805871096567\big{)}$
$\displaystyle+\frac{s_{03}^{2}s_{01}}{217925643101112(D-3)\left(D^{2}-1\right)}\times\big{(}14312658358171485D^{3}-$
$\displaystyle
308183885271005346D^{2}+2120141865244853789D-4345132269866178904\big{)}$
$\displaystyle+\frac{s_{04}^{2}s_{01}}{9290871448611768(D-3)\left(D^{2}-1\right)}\times\big{(}-8402381828589725D^{3}-$
$\displaystyle
228340458232788810D^{2}+2175040781006931191D-4392791821329079248\big{)}$
$\displaystyle+\frac{(20997586789D-30664084565)s_{00}s_{01}}{1906477614(D-3)(D-1)}+\frac{s_{02}s_{03}s_{01}}{980665393955004(D-3)\left(D^{2}-1\right)}\times$
$\displaystyle\big{(}-178531580572981155D^{3}+2760744792435584546D^{2}-$
$\displaystyle 16899392445525068877D+34463496611388041614\big{)}$
$\displaystyle+\frac{s_{02}s_{04}s_{01}}{6403168160529732(D-3)\left(D^{2}-1\right)}\times\big{(}44888698974740075D^{3}-$
$\displaystyle
2281481676512594142D^{2}+17131774468846926949D-34844020585587415026\big{)}$
$\displaystyle+\frac{s_{03}s_{04}s_{01}}{355731564473874(D-3)\left(D^{2}-1\right)}\times\big{(}134329873732898840D^{2}-$
$\displaystyle 1075850083783373373D+2183705381853832549\big{)}$
$\displaystyle+\frac{s_{02}^{3}}{178905173221521(D-3)\left(D^{2}-1\right)}\times\big{(}10367549408524951D^{3}-$
$\displaystyle
244395249217448376D^{2}+2493217982690141465D-6695874076061568024\big{)}$
$\displaystyle-\frac{s_{03}^{3}}{654431360664(D-3)\left(D^{2}-1\right)}\times\big{(}12563664070467D^{3}-480258600687861D^{2}+$
$\displaystyle
11117705113357806D-37576216670725964\big{)}+\frac{(286248073-76106817D)s_{00}s_{03}}{5725158(D-3)(D-1)}$
$\displaystyle+\frac{s_{04}^{3}}{9290871448611768(D-3)\left(D^{2}-1\right)}\times\big{(}591889764258953D^{3}+$
$\displaystyle
18615866319928209D^{2}+438017795639564794D-2088530127407051460\big{)}$
$\displaystyle+\frac{s_{02}s_{03}^{2}}{5889882245976(D-3)\left(D^{2}-1\right)}\times\big{(}470145534426423D^{3}-$
$\displaystyle
16661705329346940D^{2}+279904962791498257D-883445043516938732\big{)}$
$\displaystyle+\frac{s_{02}s_{04}^{2}}{251104633746264(D-3)\left(D^{2}-1\right)}\times\big{(}-186005811461743D^{3}-$
$\displaystyle
122956210936788D^{2}+238775662666263895D-941769244106471412\big{)}$
$\displaystyle+\frac{\left(-378033081851837D^{2}-28380809735275779D+119557455493102352\right)s_{03}s_{04}^{2}}{27900514860696(D-3)\left(D^{2}-1\right)}$
$\displaystyle+\frac{s_{02}^{2}s_{03}}{53008940213784(D-3)\left(D^{2}-1\right)}\times\big{(}-5864446929424329D^{3}+$
$\displaystyle
183078746487862324D^{2}-2359327728622119111D+6895583935338277052\big{)}$
$\displaystyle+\frac{s_{02}^{2}s_{04}}{346117197866472(D-3)\left(D^{2}-1\right)}\times\big{(}993713336121961D^{3}-$
$\displaystyle
104743917618759156D^{2}+2191839897969601319D-7146277328453200380\big{)}$
$\displaystyle+\frac{\left(-467242505501283D^{2}+30910888602825155D-116182101038199856\right)s_{03}^{2}s_{04}}{4273051825512(D-3)\left(D^{2}-1\right)}$
$\displaystyle+\frac{\left(1961957761174202D^{2}-64920756785680566D+228237749746197820\right)s_{02}s_{03}s_{04}}{4807183303701(D-3)\left(D^{2}-1\right)}$
$\displaystyle+\frac{(64066921D-274208177)s_{00}s_{04}}{37381914(D-3)(D-1)}+\frac{(268004836D-1108569860)s_{00}s_{02}}{25763211(D-3)(D-1)}\,,$
(210e) $\displaystyle F^{(3)}_{5;\widehat{234}}=$
$\displaystyle\frac{s_{01}^{3}}{16806910847891893632\left(D^{2}-1\right)}\times\big{(}45979283969009274300559D^{2}+$
$\displaystyle 1047032357312884852901D-46656230678599317469812\big{)}$
$\displaystyle+\frac{s_{02}s_{01}^{2}}{33767531499168\left(D^{2}-1\right)}\times\big{(}-74104146458248832D^{2}+$
$\displaystyle 29506403105945531D+36745597196241525\big{)}$
$\displaystyle+\frac{s_{03}s_{01}^{2}}{1770919429734144\left(D^{2}-1\right)}\times\big{(}-900085462820883747D^{2}-$
$\displaystyle 1780257440638046161D+3064860211892055028\big{)}$
$\displaystyle+\frac{\left(16379947016073091D^{2}+598222244103749657D-688591797903427356\right)s_{04}s_{01}^{2}}{3723697983359232\left(D^{2}-1\right)}$
$\displaystyle+\frac{\left(-46832281406738D^{2}+487441428989732D-919312210645983\right)s_{02}^{2}s_{01}}{114079498308\left(D^{2}-1\right)}$
$\displaystyle+\frac{\left(-5962412169596571D^{2}+57241427343714135D-108319299169698748\right)s_{03}^{2}s_{01}}{10636152731136\left(D^{2}-1\right)}$
$\displaystyle+\frac{\left(-9882296771733875D^{2}+317362498527208511D-611771709694482252\right)s_{04}^{2}s_{01}}{2482465322239488\left(D^{2}-1\right)}$
$\displaystyle+\frac{\left(188034785817840D^{2}-970303447937245D+1748833608329821\right)s_{02}s_{03}s_{01}}{101403998496\left(D^{2}-1\right)}$
$\displaystyle+\frac{\left(-114853210406800D^{2}+975040680624469D-1817277951053637\right)s_{02}s_{04}s_{01}}{662108460768\left(D^{2}-1\right)}$
$\displaystyle+\frac{\left(66634569689337D^{2}-976812872646037D+1863724399749532\right)s_{03}s_{04}s_{01}}{588540854016\left(D^{2}-1\right)}$
$\displaystyle+\frac{\left(-2271838998767D^{2}+36591900808730D-87535035935325\right)s_{02}^{3}}{4624844526\left(D^{2}-1\right)}$
$\displaystyle+\frac{\left(11671435834413D^{2}-1899510805030945D+5546345527961140\right)s_{03}^{3}}{191642391552\left(D^{2}-1\right)}$
$\displaystyle+\frac{\left(16098192433184611D^{2}+144789759004827209D-565364111032193004\right)s_{04}^{3}}{4964930644478976\left(D^{2}-1\right)}$
$\displaystyle+\frac{\left(-990342600288D^{2}+33466366766187D-91942066376347\right)s_{02}s_{03}^{2}}{1218066048\left(D^{2}-1\right)}$
$\displaystyle+\frac{\left(16181230755488D^{2}+500766696393835D-1664193721989627\right)s_{02}s_{04}^{2}}{882811281024\left(D^{2}-1\right)}$
$\displaystyle+\frac{\left(-31681674345033D^{2}-481937156164675D+1698523705808332\right)s_{03}s_{04}^{2}}{784721138688\left(D^{2}-1\right)}$
$\displaystyle+\frac{\left(772076303514D^{2}-17465983786156D+44961327781741\right)s_{02}^{2}s_{03}}{685162152\left(D^{2}-1\right)}$
$\displaystyle+\frac{\left(-317817093958D^{2}+16432322501716D-46449248275737\right)s_{02}^{2}s_{04}}{4473705816\left(D^{2}-1\right)}$
$\displaystyle+\frac{\left(829966618209D^{2}+30291424183827D-96937318177204\right)s_{03}^{2}s_{04}}{7069559808\left(D^{2}-1\right)}$
$\displaystyle+\frac{(94917907364339-31480868378483D)s_{02}s_{03}s_{04}}{3976627392\left(D^{2}-1\right)}-\frac{115943s_{00}s_{01}}{49284(D-1)}$
$\displaystyle-\frac{5041s_{00}s_{02}}{333(D-1)}+\frac{5041s_{00}s_{03}}{296(D-1)}-\frac{85697s_{00}s_{04}}{32856(D-1)}\,,$
(210f) $\displaystyle
F^{(3)}_{5;\widehat{1234}}=\frac{25411681(D-2)\left(46s_{01}+296s_{02}-333s_{03}+51s_{04}\right){}^{3}}{13384745856(D+1)}\,.$
(210g)

These results have been checked with FIRE6 Smirnov:2019qkx .

##  Appendix B Divergence of reduction coefficients of two-loop sunset
topology

In this part, we give the related $C_{i},C_{j}$ coefficients for rank level 3
with the chosen tensor structure, which have been used to solve ${\cal B}$ by
(64).

  * •

The pole of $s_{11}^{-3}$

$\displaystyle\left(\begin{array}[]{c}C^{(3,0)}_{2}\\\ C^{(3,0)}_{3}\\\
C^{(3,0)}_{4}\end{array}\right)\Bigg{|}_{s_{11}^{-3}}$
$\displaystyle=\left(\begin{array}[]{c}\frac{(D+2)\left(4(1-2D)M_{2}^{2}M_{12,3}+(3D-2)M_{12,3}^{2}+8(D-1)M_{1}^{2}M_{2}^{2}\right)s_{01}^{3}}{(D-1)(3D-2)}\\\
\frac{2(D+2)M_{1}^{2}\left((5D-4)M_{12,3}+4(1-2D)M_{2}^{2}\right)s_{01}^{3}}{(D-1)(3D-2)}\\\
-\frac{2(D+2)\left((4D-2)M_{12,3}+(D-2)M_{1}^{2}+4(1-2D)M_{2}^{2}\right)s_{01}^{3}}{(D-1)(3D-2)}\\\
\end{array}\right)\,,$ (217)
$\displaystyle\left(\begin{array}[]{c}C^{(2,1)}_{2}\\\ C^{(2,1)}_{3}\\\
C^{(2,1)}_{4}\end{array}\right)\Bigg{|}_{s_{11}^{-3}}$
$\displaystyle=\left(\begin{array}[]{c}\frac{2(D+2)M_{2}^{2}\left((2D-1)M_{12,3}-2(D-1)M_{1}^{2}\right)s_{01}^{2}s_{0^{\prime}1}}{(D-1)(3D-2)}\\\
-\frac{2(D+2)M_{1}^{2}\left((D-1)M_{12,3}+2(1-2D)M_{2}^{2}\right)s_{01}^{2}s_{0^{\prime}1}}{(D-1)(3D-2)}\\\
\frac{(D+2)\left(DM_{12,3}+4(D-1)M_{1}^{2}+4(1-2D)M_{2}^{2}\right)s_{01}^{2}s_{0^{\prime}1}}{(D-1)(3D-2)}\\\
\end{array}\right)\,,$ (224)
$\displaystyle\left(\begin{array}[]{c}C^{(1,2)}_{2}\\\ C^{(1,2)}_{3}\\\
C^{(1,2)}_{4}\end{array}\right)\Bigg{|}_{s_{11}^{-3}}$
$\displaystyle=\left(\begin{array}[]{c}\frac{2(D+2)M_{2}^{2}\left((4D-2)M_{1}^{2}-(D-1)M_{12,3}\right)s_{01}s_{0^{\prime}1}^{2}}{(D-1)(3D-2)}\\\
-\frac{2(D+2)M_{1}^{2}\left((1-2D)M_{12,3}+2(D-1)M_{2}^{2}\right)s_{01}s_{0^{\prime}1}^{2}}{(D-1)(3D-2)}\\\
-\frac{(D+2)\left(-DM_{12,3}+(8D-4)M_{1}^{2}-4(D-1)M_{2}^{2}\right)s_{01}s_{0^{\prime}1}^{2}}{(D-1)(3D-2)}\\\
\end{array}\right)\,,$ (231)
$\displaystyle\left(\begin{array}[]{c}C^{(0,3)}_{2}\\\ C^{(0,3)}_{3}\\\
C^{(0,3)}_{4}\end{array}\right)\Bigg{|}_{s_{11}^{-3}}$
$\displaystyle=\left(\begin{array}[]{c}-\frac{2(D+2)M_{2}^{2}\left((4-5D)M_{12,3}+(8D-4)M_{1}^{2}\right)s_{0^{\prime}1}^{3}}{(D-1)(3D-2)}\\\
\frac{(D+2)\left(4M_{1}^{2}\left((1-2D)M_{12,3}+2(D-1)M_{2}^{2}\right)+(3D-2)M_{12,3}^{2}\right)s_{0^{\prime}1}^{3}}{(D-1)(3D-2)}\\\
\frac{2(D+2)\left(2(1-2D)M_{12,3}+(8D-4)M_{1}^{2}-(D-2)M_{2}^{2}\right)s_{0^{\prime}1}^{3}}{(D-1)(3D-2)}\\\
\end{array}\right)\,.$ (238)
$\displaystyle\left(\begin{array}[]{c}C^{(r_{1},r_{2})}_{1}\\\
C^{(r_{1},r_{2})}_{5}\\\ C^{(r_{1},r_{2})}_{6}\\\
C^{(r_{1},r_{2})}_{7}\end{array}\right)\Bigg{|}_{s_{11}^{-3}}$
$\displaystyle=\left(\begin{array}[]{c}0\\\ 0\\\ 0\\\ 0\\\
\end{array}\right),\forall r_{1}+r_{2}=3\,.$ (247)

  * •

The pole of $s_{11}^{-2}$

$\displaystyle\left(\begin{array}[]{c}C^{(3,0)}_{2}\\\ C^{(3,0)}_{3}\\\
C^{(3,0)}_{4}\end{array}\right)\Bigg{|}_{s_{01}^{3}s_{11}^{-2}}$
$\displaystyle=\left(\begin{array}[]{c}\frac{4\left(\left(-3D^{2}-5D+2\right)M_{3}^{2}+D(5D-2)M_{1}^{2}-D(D+2)M_{2}^{2}\right)}{(D-1)(3D-2)}\\\
\frac{2(D+2)\left(2(5D-4)M_{1}^{2}+(4-5D)M_{2}^{2}-3DM_{3}^{2}\right)}{(D-1)(3D-2)}\\\
-\frac{2\left(5D^{2}+6D-8\right)}{3D^{2}-5D+2}\\\ \end{array}\right)\,,$ (254)
$\displaystyle\left(\begin{array}[]{c}C^{(2,1)}_{2}\\\ C^{(2,1)}_{3}\\\
C^{(2,1)}_{4}\end{array}\right)\Bigg{|}_{s_{01}^{2}s_{0^{\prime}1}s_{11}^{-2}}$
$\displaystyle=\left(\begin{array}[]{c}-\frac{2(D+2)\left((2D-1)M_{1}^{2}+(2-4D)M_{2}^{2}-M_{3}^{2}\right)}{(D-1)(3D-2)}\\\
\frac{2\left(\left(D^{2}-16D+12\right)M_{1}^{2}+(D+2)\left((4D-3)M_{2}^{2}+M_{3}^{2}\right)\right)}{(D-1)(3D-2)}\\\
-\frac{D^{2}-10D+8}{3D^{2}-5D+2}\\\ \end{array}\right)\,,$ (261)
$\displaystyle\left(\begin{array}[]{c}C^{(1,2)}_{2}\\\ C^{(1,2)}_{3}\\\
C^{(1,2)}_{4}\end{array}\right)\Bigg{|}_{s_{01}s_{0^{\prime}1}^{2}s_{11}^{-2}}$
$\displaystyle=\left(\begin{array}[]{c}\frac{2\left(\left(4D^{2}+5D-6\right)M_{1}^{2}+\left(D^{2}-16D+12\right)M_{2}^{2}+(D+2)M_{3}^{2}\right)}{(D-1)(3D-2)}\\\
\frac{2(D+2)\left((4D-2)M_{1}^{2}+(1-2D)M_{2}^{2}+M_{3}^{2}\right)}{(D-1)(3D-2)}\\\
-\frac{D^{2}-10D+8}{3D^{2}-5D+2}\\\ \end{array}\right)\,,$ (268)
$\displaystyle\left(\begin{array}[]{c}C^{(0,3)}_{2}\\\ C^{(0,3)}_{3}\\\
C^{(0,3)}_{4}\end{array}\right)\Bigg{|}_{s_{0^{\prime}1}^{3}s_{11}^{-2}}$
$\displaystyle=\left(\begin{array}[]{c}-\frac{2(D+2)\left((5D-4)M_{1}^{2}+(8-10D)M_{2}^{2}+3DM_{3}^{2}\right)}{(D-1)(3D-2)}\\\
-\frac{4\left(\left(3D^{2}+5D-2\right)M_{3}^{2}+D(D+2)M_{1}^{2}+(2-5D)DM_{2}^{2}\right)}{(D-1)(3D-2)}\\\
-\frac{2\left(5D^{2}+6D-8\right)}{3D^{2}-5D+2}\\\ \end{array}\right)\,.$ (275)
$\displaystyle\left(\begin{array}[]{c}C^{(3,0)}_{1}\\\ C^{(3,0)}_{5}\\\
C^{(3,0)}_{6}\\\
C^{(3,0)}_{7}\end{array}\right)\Bigg{|}_{s_{01}^{3}s_{11}^{-2}}$
$\displaystyle=\left(\begin{array}[]{c}-\frac{2(D+2)\left((7D-4)M_{1}^{4}+2M_{1}^{2}\left((D-4)M_{2}^{2}+(2-5D)M_{3}^{2}\right)-\left(M_{2}^{2}-M_{3}^{2}\right)\left((5D-4)M_{2}^{2}+3DM_{3}^{2}\right)\right)}{(D-1)(3D-2)}\\\
-\frac{4(D+2)(2D-1)\left(DM_{1}^{2}-2\left(M_{2}^{2}+M_{3}^{2}\right)\right)}{(D-1)D(3D-2)}\\\
-\frac{2(D+2)\left(\left(3D^{2}+2D-4\right)M_{1}^{2}+D\left((4-5D)M_{2}^{2}-3DM_{3}^{2}\right)\right)}{(D-1)D(3D-2)}\\\
\frac{2(D+2)\left(\left(7D^{2}-10D+4\right)M_{1}^{2}+D\left((4-5D)M_{2}^{2}-3DM_{3}^{2}\right)\right)}{(D-1)D(3D-2)}\\\
\end{array}\right)\,,$ (284)
$\displaystyle\left(\begin{array}[]{c}C^{(2,1)}_{1}\\\ C^{(2,1)}_{5}\\\
C^{(2,1)}_{6}\\\
C^{(2,1)}_{7}\end{array}\right)\Bigg{|}_{s_{01}^{2}s_{0^{\prime}1}s_{11}^{-2}}$
$\displaystyle=\left(\begin{array}[]{c}\frac{2(D+2)\left((2D-1)M_{1}^{4}-2M_{1}^{2}\left((D+1)M_{2}^{2}+DM_{3}^{2}\right)-\left(M_{2}^{2}-M_{3}^{2}\right)\left((4D-3)M_{2}^{2}+M_{3}^{2}\right)\right)}{(D-1)(3D-2)}\\\
\frac{2(D+2)\left(D(2D-1)M_{1}^{2}+(4-7D)M_{2}^{2}-DM_{3}^{2}\right)}{(D-1)D(3D-2)}\\\
-\frac{2(D+2)\left((4-5D)M_{1}^{2}+D\left((4D-3)M_{2}^{2}+M_{3}^{2}\right)\right)}{(D-1)D(3D-2)}\\\
-\frac{2(D+2)\left((2D-1)M_{1}^{2}+(3-4D)M_{2}^{2}-M_{3}^{2}\right)}{(D-1)(3D-2)}\\\
\end{array}\right)\,,$ (293)
$\displaystyle\left(\begin{array}[]{c}C^{(1,2)}_{1}\\\ C^{(1,2)}_{5}\\\
C^{(1,2)}_{6}\\\
C^{(1,2)}_{7}\end{array}\right)\Bigg{|}_{s_{01}s_{0^{\prime}1}^{2}s_{11}^{-2}}$
$\displaystyle=\left(\begin{array}[]{c}-\frac{2(D+2)\left((4D-3)M_{1}^{4}+2M_{1}^{2}\left((D+1)M_{2}^{2}-2(D-1)M_{3}^{2}\right)+(1-2D)M_{2}^{4}+2DM_{2}^{2}M_{3}^{2}-M_{3}^{4}\right)}{(D-1)(3D-2)}\\\
-\frac{2(D+2)\left(D(4D-3)M_{1}^{2}+(4-5D)M_{2}^{2}+DM_{3}^{2}\right)}{(D-1)D(3D-2)}\\\
-\frac{2(D+2)\left((7D-4)M_{1}^{2}+D\left((1-2D)M_{2}^{2}+M_{3}^{2}\right)\right)}{(D-1)D(3D-2)}\\\
\frac{2(D+2)\left((4D-3)M_{1}^{2}+(1-2D)M_{2}^{2}+M_{3}^{2}\right)}{(D-1)(3D-2)}\\\
\end{array}\right)\,,$ (302)
$\displaystyle\left(\begin{array}[]{c}C^{(0,3)}_{1}\\\ C^{(0,3)}_{5}\\\
C^{(0,3)}_{6}\\\
C^{(0,3)}_{7}\end{array}\right)\Bigg{|}_{s_{0^{\prime}1}^{3}s_{11}^{-2}}$
$\displaystyle=\left(\begin{array}[]{c}\frac{2(D+2)\left((5D-4)M_{1}^{4}-2M_{1}^{2}\left((D-4)M_{2}^{2}+(D-2)M_{3}^{2}\right)-\left(M_{2}^{2}-M_{3}^{2}\right)\left((7D-4)M_{2}^{2}-3DM_{3}^{2}\right)\right)}{(D-1)(3D-2)}\\\
\frac{2(D+2)\left(\left(-3D^{2}-2D+4\right)M_{2}^{2}+3D^{2}M_{3}^{2}+D(5D-4)M_{1}^{2}\right)}{(D-1)D(3D-2)}\\\
-\frac{4(D+2)(2D-1)\left(DM_{2}^{2}-2M_{1}^{2}-2M_{3}^{2}\right)}{(D-1)D(3D-2)}\\\
-\frac{2(D+2)\left(\left(-7D^{2}+10D-4\right)M_{2}^{2}+3D^{2}M_{3}^{2}+D(5D-4)M_{1}^{2}\right)}{(D-1)D(3D-2)}\\\
\end{array}\right)\,.$ (311)

  * •

The pole of $s_{11}^{-1}$

$\displaystyle\left(\begin{array}[]{c}C^{(3,0)}_{2}\\\ C^{(3,0)}_{3}\\\
C^{(3,0)}_{4}\end{array}\right)\Bigg{|}_{s_{01}^{3}s_{11}^{-1}}$
$\displaystyle=\left(\begin{array}[]{c}\frac{13D^{2}+16D-20}{(D-1)(3D-2)}\\\
\frac{2\left(5D^{2}+6D-8\right)}{(D-1)(3D-2)}\\\ 0\\\ \end{array}\right)\,,$
(318) $\displaystyle\left(\begin{array}[]{c}C^{(2,1)}_{2}\\\ C^{(2,1)}_{3}\\\
C^{(2,1)}_{4}\end{array}\right)\Bigg{|}_{s_{01}^{2}s_{0^{\prime}1}s_{11}^{-1}}$
$\displaystyle=\left(\begin{array}[]{c}\frac{2\left(2D^{2}-9D+6\right)}{3D^{2}-5D+2}\\\
\frac{2\left(2D^{2}-9D+6\right)}{3D^{2}-5D+2}\\\ 0\\\ \end{array}\right)\,,$
(325) $\displaystyle\left(\begin{array}[]{c}C^{(1,2)}_{2}\\\ C^{(1,2)}_{3}\\\
C^{(1,2)}_{4}\end{array}\right)\Bigg{|}_{s_{01}s_{0^{\prime}1}^{2}s_{11}^{-1}}$
$\displaystyle=\left(\begin{array}[]{c}\frac{2\left(2D^{2}-9D+6\right)}{3D^{2}-5D+2}\\\
\frac{2\left(2D^{2}-9D+6\right)}{3D^{2}-5D+2}\\\ 0\\\ \end{array}\right)\,,$
(332) $\displaystyle\left(\begin{array}[]{c}C^{(0,3)}_{2}\\\ C^{(0,3)}_{3}\\\
C^{(0,3)}_{4}\end{array}\right)\Bigg{|}_{s_{0^{\prime}1}^{3}s_{11}^{-1}}$
$\displaystyle=\left(\begin{array}[]{c}\frac{2\left(5D^{2}+6D-8\right)}{(D-1)(3D-2)}\\\
\frac{13D^{2}+16D-20}{(D-1)(3D-2)}\\\ 0\\\ \end{array}\right)\,.$ (339)
$\displaystyle\left(\begin{array}[]{c}C^{(3,0)}_{1}\\\ C^{(3,0)}_{5}\\\
C^{(3,0)}_{6}\\\
C^{(3,0)}_{7}\end{array}\right)\Bigg{|}_{s_{01}^{3}s_{11}^{-1}}$
$\displaystyle=\left(\begin{array}[]{c}-\frac{8(D+2)\left((3D-2)M_{1}^{2}+(1-2D)M_{3}^{2}\right)}{(D-1)(3D-2)}\\\
\frac{8D^{2}-60D+40}{3D^{2}-5D+2}\\\
-\frac{2\left(5D^{2}+6D-8\right)}{3D^{2}-5D+2}\\\
\frac{2\left(5D^{2}+6D-8\right)}{(D-1)(3D-2)}\\\ \end{array}\right)\,,$ (348)
$\displaystyle\left(\begin{array}[]{c}C^{(2,1)}_{1}\\\ C^{(2,1)}_{5}\\\
C^{(2,1)}_{6}\\\
C^{(2,1)}_{7}\end{array}\right)\Bigg{|}_{s_{01}^{2}s_{0^{\prime}1}s_{11}^{-1}}$