aainstitutetext: Zhejiang Institute of Modern Physics, Zhejiang University,
Hangzhou, 310027, P. R. China bbinstitutetext: Beijing Computational Science
Research Center, Beijing 100084, Chinaccinstitutetext: Center of Mathematical
Science, Zhejiang University, Hangzhou, 310027, P. R. Chinaddinstitutetext:
Peng Huanwu Center for Fundamental Theory, Hefei, Anhui 230026, China

# Universal Treatment of Reduction for One-Loop Integrals in Projective Space

Bo Feng a Jianyu Gong a Tingfei Li<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

Recently a nice work about the understanding of one-loop integrals has been
done in Arkani-Hamed:2017ahv  using the tricks of the projective space
language associated to their Feynman parametrization. We find this language is
also very suitable to deal with the reduction problem of one-loop integrals
with general tensor structures as well as propagators with arbitrary higher
powers. In this paper, we show that how to combine Feynman parametrization and
embedding formalism to give a universal treatment of reductions for general
one-loop integrals, even including the degenerated cases, such as the
vanishing Gram determinant. Results from this method can be written in a
compact and symmetric form.

###### Keywords:

Projective space, One-loop integrals, Reduction

##  1 Introduction

In recent years we have witnessed enormous progress in computing and
understanding analytic structures of scattering amplitudes. At one-loop level,
it is well-known that a general one-loop integrals in
$D=4-2\epsilon$-dimension can always be reduced to a linear combination of
one-loop scalar integrals with no more than 5 external legs with reduction
coefficients as rational functions of the external variables 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 . These master
integrals at one-loop level (i.e., tadpoles, bubbles, triangles, boxes, and
pentagons) are well known. Therefore, the main problem of one-loop integrals
is to calculate the reduction coefficients. There are a host of methods to
deal with the reduction at integrand level and integral level, such as
Integration-By-Parts (IBP) Chetyrkin:1981qh ; Tkachov:1981wb , Passarino-
Veltman reduction Passarino:1978jh , Ossola-Papadopoulos-Pittau reduction
Ossola:2006us ; Ossola:2007bb ; Ellis:2007br , and unitarity cut Bern:1994zx ;
Bern:1994cg ; Britto:2004nc ; Britto:2005ha ; Campbell:1996zw ; Bern:1997sc ;
Denner:2005nn ; Anastasiou:2006gt ; Britto:2010um .

Although in practice, we will not meet many situations where propagators have
higher power, a complete reduction method should be able to deal with it. From
this point of view, IBP method is a complete method, because it treats these
complicated cases with the same framework as the one without higher poles.
Recently, combining unitarity cut and derivation over mass, the reduction
coefficients for higher pole cases can be calculated Feng:2021spv , except for
the tadpole coefficients.

Recently, we have proposed an improved PV-reduction method for one-loop
integrals Feng:2021enk ; Hu:2021nia . The reduction coefficients can be
expressed with the cofactors of the Gram matrices and have some symmetry. Thus
it is useful to understand these symmetries appearing in our middle recursion
relations and the final results. Notably, the analytical structure of one-loop
integrals is studied by investigating Feynman parametrization in the
projective space for its compactness and the close relation to geometry
Arkani-Hamed:2017ahv . Inspired by the geometric angle, we find it could be
convenient to do reduction for one-loop integrals in projective space. By our
study in this paper, one can see that the symmetry and simplicity of reduction
coefficients are illustrated clearly with the denotations in Arkani-
Hamed:2017ahv .

Motivated by the work Arkani-Hamed:2017ahv , we will develop an alternative
method for determining the reduction coefficients of one-loop integrals in
$D=4-2\epsilon$ dimension. The general tensor integrals with higher poles are
related to integrals $E_{n,k}[T]$ in projective space

$\displaystyle
E_{n,k}[T]\equiv\int_{\Delta}{{\langle{X}\mathrm{d}^{n-1}{X}\rangle}T[X^{k}]\over(XQX)^{n+k\over
2}}\,,$ (1)

where $\Delta$ is a simplex in $n$-dimensional space $X_{I}$, which is defined
by $H_{I}X=X_{I}>0,\forall I=1,2,\ldots,n$. The $T$ is a general tensor, which
is contracted with $k$ $X$’s.

The homogeneous coordinates $X_{I}$ are denoted by a square bracket
$X=[x_{1}:x_{2}:\ldots:x_{n}]$, and two coordinates are equivalent to each
other up to a scaling, i.e.,
$[x_{1}:x_{2}:\ldots:x_{n}]\sim[kx_{1}:kx_{2}:\ldots:kx_{n}]$ for any $k\neq
0$.

The measure in the projective space is given by the differential forms

$\displaystyle{\langle{X}\mathrm{d}^{n-1}{X}\rangle}={\epsilon^{I_{1},I_{2},\ldots,I_{n}}\over(n-1)!}X_{I_{1}}\text{d}X_{I_{2}}\wedge\text{d}X_{I_{3}}\wedge\ldots\wedge\text{d}X_{I_{n}};~{}~{}XQX=Q^{IJ}X_{I}X_{J}\,.$
(2)

As pointed out in Arkani-Hamed:2017ahv , the integral $E_{n,k}$ satisfies a
nice recursion relation, which will be recalled in Appendix A. It is exactly
the property we will use to do the reduction in this paper.

This paper is structured as follows. In section 2, we discuss how to write a
general one-loop integral as the sum of integrals $E_{n,k}$ in projective
space. In section 3, we derive recursion relations for $E_{n,k}[V^{i}\otimes
L^{k-i}]$ and dimension recursion relations for non-degenerate $Q$, and then
apply them to the reduction of one-loop integrals. In section 4, we discuss
the reduction framework for degenerate $Q$. In section 5, we show how to
obtain the general expression of reduction coefficient from $n$-gon tensor
integrals to $n$-gon scalar integrals while general expressions of reduction
coefficients are given in Appendix C. More reduction results are listed in
Appendix B.

##  2 One-loop integrals in projective space

In this paper, we will discuss the reduction of the most general one-loop
integrals

$\displaystyle\int{d^{D}k\over
i\pi^{D/2}}{k^{\mu_{1}}k^{\mu_{2}}...k^{\mu_{r}}\over\prod_{j=1}^{n}D_{j}^{v_{j}}}=\int{d^{D}k\over
i\pi^{D/2}}{k^{\mu_{1}}k^{\mu_{2}}...k^{\mu_{r}}\over\prod_{j=1}^{n}[(k-q_{j})^{2}-m_{j}^{2}]^{v_{j}}}\,,~{}~{}~{}~{}$
(3)

where $q_{j}=\sum_{i=1}^{j-1}p_{i}$.As pointed out in Feng:2021enk , we can
recover the tensor structure by multiplying each index with an auxiliary
vector $R_{i,\mu_{i}}$. Furthermore, we can combine these $R_{i}$ to
$R=\sum_{i=1}^{r}x_{i}R_{i}$ to simplify the expression (3) in a Lorentzian
invariant form to

$\displaystyle I^{(r)}_{\mathbf{v}_{n};D}\equiv\int{d^{D}k\over
i\pi^{D/2}}{(2R\cdot
k)^{r}\over\prod_{j=1}^{n}((k-q_{j})^{2}-m_{j}^{2})^{v_{j}}}\,.~{}~{}~{}~{}$
(4)

To recover the result of (3), one can expand $R$ and extract the coefficient
of $\prod_{i=1}^{r}x_{i}$ from the auxiliary formulas (4). Thus, we will
transform the general forms (4) into projective space suggested in Arkani-
Hamed:2017ahv . First, to make our formulas elegant, we denote $y^{\mu}\equiv
k^{\mu},y_{i}\equiv q_{i}$. Then (4) becomes

$\displaystyle I^{(r)}_{\mathbf{v}_{n};D}=\int{d^{D}y\over
i\pi^{D/2}}{(2R\cdot
y)^{r}\over\prod_{j=1}^{n}((y-y_{i})^{2}-m_{j}^{2})^{v_{j}}}\,.~{}~{}~{}~{}$
(5)

Then we put the whole formula into the embedding space with two higher
dimensions with

$\displaystyle y^{\mu}\longmapsto
Y^{M}=(Y^{+},Y^{-},Y^{\mu})=(1,y^{2},y^{\mu})\,,~{}~{}~{}~{}$ (6)

where we use the light-cone coordinates, i.e. the metric
$\eta_{+-}=\eta_{-+}=-\frac{1}{2}$, $\eta_{\mu\nu}=diag(+,----)$ while all
other entries vanish. For clarity, we will use the capital letters $M,N$ to
denote the components of vectors in the embedding space, Greek letters
$\mu,\nu$ to denote the components of Lorentzian vectors, and lower-case
letters $i,j$ for the external legs. We will also use capital letters $Y,X$
simultaneously to denote vectors in the embedding space and projective space
without ambiguity. Therefore, we can simplify the denominator of (5) into the
inner product of two vectors in the embedding space, and the quadratic
expression has been somehow linearized, for example,

$\displaystyle(y-y_{i})^{2}=-2Y\cdot Y_{i}\,.~{}~{}~{}~{}$ (7)

After defining

$\displaystyle\mathcal{Y}_{i}^{M}=(1,y_{i}^{2}-m_{i}^{2},y_{i}^{\mu})\,,~{}~{}~{}Y_{\infty}^{M}=(0,1,0,\ldots,0)\,,~{}~{}~{}\mathcal{R}^{M}=(0,0,R^{\mu})\,,~{}~{}~{}~{}$
(8)

it is easy to check that (5) becomes

$\displaystyle
I^{(r)}_{\mathbf{v}_{n};D}=\int[\mathrm{d}^{D}Y]\frac{(-2\,Y\cdot
Y_{\infty})^{v-D-r}(2Y\cdot\mathcal{R})^{r}}{\prod_{j=1}^{n}(-2\,Y\cdot\mathcal{Y}_{j}){}^{v_{j}}}\,,~{}~{}~{}~{}$
(9)

where $v=\sum_{j=1}^{n}v_{j}$. The projective space invariant measure is given
by
$\int[\mathrm{d}^{D}Y]\equiv\int\frac{\mathrm{d}^{D+2}Y\,\delta(Y^{2})}{i\pi^{D/2}\text{vol.}\mathrm{GL}(1)}$
($\mathrm{GL}(1)$ acts as an overall scaling of the $Y$ coordinates) and the
factor $(Y\cdot Y_{\infty})^{v-D-r}$ is necessary for the last expression to
be genuinely an integral over the projective light cone. Using the most
general Feynman parametrization

$\displaystyle{1\over
A_{1}^{m_{1}}A_{2}^{m_{2}}...A_{n}^{m_{n}}}={\Gamma(\sum_{i}m_{i})\over\prod\Gamma(m_{i})}\int_{0}^{1}dx_{1}...dx_{n}\delta\bigg{(}\sum_{i}x_{i}-1\bigg{)}{\prod
x_{i}^{m_{i}-1}\over(\sum_{i}x_{i}A_{i})^{\sum_{i}m_{i}}}\,,$

and putting the Feynman parameters into the projective space, (9) becomes

$\displaystyle
I^{(r)}_{\mathbf{v}_{n};D}={\Gamma(v)\over\prod\Gamma(v_{i})}\int_{\Delta}{\langle{X}\mathrm{d}^{n-1}{X}\rangle}X^{\mathbf{v}_{n}-1}\int[\mathrm{d}^{D}Y]\frac{(-2\,Y\cdot
Y_{\infty})^{v-D-r}(2Y\cdot\mathcal{R})^{r}}{(-2Y\cdot W)^{v}}\,,~{}~{}~{}~{}$
(10)

where $X=[x_{1}:x_{2}:\ldots:x_{n}]$, which is a vector in a different
projective (Feynman parametrization) space, $W=\sum_{j}x_{j}\mathcal{Y}_{j}$
and there is an additional factor
$X^{\mathbf{v}_{n}-1}\equiv\prod_{i=1}^{n}x_{i}^{v_{i}-1}=\prod_{i=1}^{n}(H_{i}X)^{v_{i}-1}$.
Then the Feynman parametrization integral has been written into the compact
form $\int_{\Delta}{\langle{X}\mathrm{d}^{n-1}{X}\rangle}$ (see Arkani-
Hamed:2017ahv  for more details). We can further simplify (10) using the
common trick as

$\displaystyle I^{(r)}_{\mathbf{v}_{n};D}$ $\displaystyle=$
$\displaystyle{\Gamma(v)\over\prod\Gamma(v_{i})}\int_{\Delta}{\langle{X}\mathrm{d}^{n-1}{X}\rangle}X^{\mathbf{v}_{n}-1}{(-)^{v+D+r}\Gamma(D)\over\Gamma(v)}\left(\mathcal{R}^{M}\frac{\partial}{\partial
W^{M}}\right)^{r}$ (11)
$\displaystyle~{}~{}~{}~{}\times\left(Y_{\infty}^{M}\frac{\partial}{\partial
W^{M}}\right)^{v-D-r}\int[\mathrm{d}^{D}Y]\frac{1}{(-2Y\cdot
W)^{D}}\,.~{}~{}~{}~{}$

Up to now, the last integral in (11) can be done easily. One way to solve it
is to translate it back to the form (5), which is111We have used the fact
$W_{-}=1$ by the Feynman parametrization. The integration result can be found
in the formula (A.44) in the book of Peskin and Schroeder Peskin:1995ev .

$\displaystyle\int{d^{D}y\over i\pi^{D/2}}{1\over(W_{-}(y-{w\over
W_{-}})^{2}+W_{+}-{w^{2}\over
W_{-}})^{D}}={\Gamma({D/2})\over(-)^{D}\Gamma(D)}(W\cdot W)^{-D/2}\,,$ (12)

and we have

$\displaystyle I^{(r)}_{\mathbf{v}_{n};D}$ $\displaystyle=$
$\displaystyle{(-)^{v+r}\Gamma(D/2)\over\prod\Gamma(v_{i})}\int_{\Delta}{\langle{X}\mathrm{d}^{n-1}{X}\rangle}X^{\mathbf{v}_{n}-1}\left(\mathcal{R}^{M}\frac{\partial}{\partial
W^{M}}\right)^{r}\left(Y_{\infty}^{M}\frac{\partial}{\partial
W^{M}}\right)^{v-D-r}(W\cdot W)^{-D/2}\,.~{}~{}~{}~{}~{}~{}~{}~{}$ (13)

The action of $\left(Y_{\infty}^{M}\frac{\partial}{\partial W^{M}}\right)$ can
be done easily after using the fact $Y_{\infty}\cdot Y_{\infty}=0$ and we get

$\displaystyle I^{(r)}_{\mathbf{v}_{n};D}$ $\displaystyle=$
$\displaystyle{(-)^{v+r}\Gamma(D/2)\over\prod\Gamma(v_{i})}\int_{\Delta}{\langle{X}\mathrm{d}^{n-1}{X}\rangle}X^{\mathbf{v}_{n}-1}{\Gamma(v-{D\over
2}-r)\over\Gamma(D/2)}(LX)^{v-D-r}$ (14)
$\displaystyle~{}~{}~{}~{}\left(\mathcal{R}^{M}\frac{\partial}{\partial
W^{M}}\right)^{r}(W\cdot W)^{-(v-D/2-r)}\,,~{}~{}~{}~{}$

where we have written $(-2Y_{\infty}\cdot W)=\sum_{i}x_{i}\equiv L\cdot X$
with $L=[1:1:...:1]$222The reason that we can ignore the action of
$\left(\mathcal{R}^{M}\frac{\partial}{\partial W^{M}}\right)$ on
$(-2Y_{\infty}\cdot W)$ is because $Y_{\infty}\cdot\mathcal{R}=0$.. The action
of $\left(\mathcal{R}^{M}\frac{\partial}{\partial W^{M}}\right)$ is more
complicated. By power counting, we have the general expansion

$\displaystyle(\mathcal{R}\cdot\partial_{W})^{r}(W^{2})^{k}$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{r}\mathscr{C}^{k}_{r,i}(W^{2})^{k-{r+i\over
2}}(R^{2})^{r-i\over 2}(\mathcal{R}\cdot W)^{i}\,,~{}~{}$ (15)

where $k$ can be an arbitrary number and $i$ has the same parity as $r$ due to
the power of $R^{2}$ must be an integer. The expansion coefficients
$\mathscr{C}^{k}_{r,i}$ are determined by initial conditions
$\mathscr{C}^{k}_{0,i}=\delta_{i,0},\mathscr{C}^{k}_{1,i}=2k\delta_{i,1}$ and
the recursion relation

$\displaystyle\mathscr{C}^{k}_{r+1,i}=(i+1)\mathscr{C}^{k}_{r,i+1}+(2k-r-i+1)\mathscr{C}^{k}_{r,i-1}\,.~{}~{}~{}$
(16)

From the recursion relation, one can solve $\mathscr{C}^{k}_{r,i}$ for general
$r,i$ as

$\displaystyle\mathscr{C}^{k}_{r,i}=\frac{2^{r}r!\Gamma\big{(}{r-i+1\over
2}\big{)}}{\sqrt{\pi}i!(r-i)!}\prod_{j=1}^{{r+i\over
2}}(k+1-j)=\frac{2^{r}r!k!\Gamma\big{(}{r-i+1\over
2}\big{)}}{\sqrt{\pi}i!(r-i)!(k-{r+i\over 2})!}\,,~{}~{}~{}{r-i\over
2}\in\mathbb{N}\,.$ (17)

Plugging (15) into (14) we get

$\displaystyle I^{(r)}_{\mathbf{v}_{n};D}$ $\displaystyle=$
$\displaystyle{\Gamma(v-{D\over
2}-r)\over(-)^{v+r}\prod\Gamma(v_{i})}\sum_{i=0}^{r}\mathscr{C}^{D/2+r-v}_{r,i}(R^{2})^{r-i\over
2}\int_{\Delta}{{\langle{X}\mathrm{d}^{n-1}{X}\rangle}X^{\mathbf{v}_{n}-1}(\mathcal{R}\cdot
W)^{i}(LX)^{v-D-r}\over(W\cdot W)^{v-{D+r-i\over
2}}}\,.~{}~{}~{}~{}~{}~{}~{}~{}~{}$ (18)

Since the remaining integrals are over the projective space of Feynman
parameters, we should rewrite the formula by

$\displaystyle W\cdot W$ $\displaystyle=$
$\displaystyle(\sum_{a=1}^{n}x_{a}\mathcal{Y}_{a})\cdot(\sum_{b=1}^{n}x_{b}\mathcal{Y}_{b})=\sum_{a,b}x_{a}(\mathcal{Y}_{a}\cdot\mathcal{Y}_{b})x_{b}=XQX\,,$
$\displaystyle\mathcal{R}\cdot W$ $\displaystyle=$
$\displaystyle\sum_{b=1}^{n}x_{b}\mathcal{R}\cdot\mathcal{Y}_{b}=V\cdot
X\,,~{}~{}~{}V=[R\cdot q_{1}:R\cdot q_{2}:\ldots:R\cdot q_{n}]\,.~{}$ (19)

Then we get

$\displaystyle I^{(r)}_{\mathbf{v}_{n};D}$ $\displaystyle=$
$\displaystyle{\Gamma(v-{D\over
2}-r)\over(-)^{v+r}\prod\Gamma(v_{i})}\sum_{i=0}^{r}\mathscr{C}^{D/2+r-v}_{r,i}(R^{2})^{r-i\over
2}\int_{\Delta}{{\langle{X}\mathrm{d}^{n-1}{X}\rangle}X^{\mathbf{v}_{n}-1}(VX)^{i}(LX)^{v-D-r}\over(XQX)^{v-{D+r-i\over
2}}}\,.~{}~{}~{}~{}~{}~{}~{}~{}$ (20)

For general one loop integrals (3), one can calculate $Q$ as

$\displaystyle Q_{ij}={1\over
2}\left(m_{i}^{2}+m_{j}^{2}-q_{ij}^{2}\right)\,,~{}~{}~{}~{}q_{ij}=q_{i}-q_{j}.~{}~{}~{}$
(21)

Now the expression is written as the integration over the $X$-projective
space. Using the result of Arkani-Hamed:2017ahv

$\displaystyle
E_{n,k}[T]\equiv\int_{\Delta}{{\langle{X}\mathrm{d}^{n-1}{X}\rangle}T[X^{k}]\over(XQX)^{n+k\over
2}}\,,~{}~{}~{}~{}$ (22)

where $\Delta$ is a simplex in $n$-dimensional space defined by
$H_{i}X=X_{i}>0,\forall i=1,2,\ldots,n$ and $T$ is a $k$-th tensor contracted
with $k$ $X$’s, the general one loop integral in the projective space (20) can
be written as

$\displaystyle I^{(r)}_{\mathbf{v}_{n};D}$ $\displaystyle=$
$\displaystyle{\Gamma(v-D/2-r)\over(-)^{v+r}\prod\Gamma(v_{i})}\sum_{i=0}^{r}\mathscr{C}^{D/2+r-v}_{r,i}(R^{2})^{r-i\over
2}E_{n,2v-n-D-r+i}[\otimes_{j}H_{j}^{v_{j}-1}\otimes
V^{i}]\,,~{}~{}~{}~{}~{}~{}~{}~{}~{}$ (23)

where for simplicity, we write $E_{n,k}[V^{a}\otimes L^{k-a}]\equiv
E_{n,k}[V^{a}]$ by neglecting the power of $L$.

For the later use, we need to do symmetrization for the tensor
$\otimes_{j}H_{j}^{v_{j}-1}\otimes V^{i}$. To do so, we can use the same trick
as used in (3) and (4), i.e., $Z\equiv\sum_{i=1}^{v-n}z_{i}H_{i}$ and
$S=tV+Z$. Thus (23) can be obtained from

$\displaystyle{\Gamma(v-D/2-r)\over(-)^{v+r}\prod\Gamma(v_{i})}\sum_{i=0}^{r}\mathscr{C}^{D/2+r-v}_{r,i}(R^{2})^{r-i\over
2}E_{n,2v-n-D-r+i}[S^{v-n+i}]~{}~{}~{}~{}~{}~{}~{}~{}~{}$ (24)

after taking the coefficients of $t^{i}z^{\mathbf{v}_{n}-1}\equiv
t^{i}\prod_{i=1}^{n}z_{i}^{v_{i}-1}$. Taking care of numerical factors, the
final expression is

$\displaystyle I_{\mathbf{v}_{n};D}^{(r)}$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{r}{i!\Gamma(v-D/2-r)\over(-1)^{v+r}(v-n+i)!}\mathscr{C}^{D/2+r-v}_{r,i}(R^{2})^{r-i\over
2}E_{n,2v-n-
D-r+i}[S^{v-n+i}]\Big{|}_{t^{i}z^{\mathbf{v}_{n}-1}}\,.~{}~{}~{}~{}~{}~{}~{}$
(25)

A special case of (25) is that for $\mathbf{v}_{n}=\mathbf{1}_{n}$ and $r=0$,
we have

$\displaystyle I_{n;D}$ $\displaystyle=$
$\displaystyle(-1)^{n}{\Gamma(n-D/2)}E_{n,n-D}[L^{n-D}]\,,~{}~{}~{}$ (26)

where $I_{n;D}$ is the scalar integral of $n$ propagators in $D$ dimension.

Before ending this section, we want to point out in our discussion, for
example in (10), that the power $v-D-r$ could be positive or negative with the
arbitrary choice of $r$. Since we have kept the dimension $D$ arbitrary, we
can take $D$ properly (even a negative number) to make $v-D-r$ a positive
integer to make later discussion legitimate. At the end of reduction, we can
analytically continue $D$ to the proper dimension. We have checked with
several examples that such a continuation is allowed.

In this paper, we mainly discuss Feynman integrals in
$D=4-2\epsilon$-dimension space. At the one-loop level, the master integrals
are related to $E_{n,n-D}[L^{n-D}],n=1,2,\ldots,5$. So the main task of one-
loop integral reduction is to reduce general integral $E_{n,k}[S^{a}\otimes
L^{k-a}]$ to the basis $E_{n\leq 5,n-D}[L^{n-D}]$.

##  3 Reduction for non-degenerate $Q$

Having transformed our problem (3) to the form (25), in this section, we will
show how to use the tricks of integrals in projective space (see Arkani-
Hamed:2017ahv ) to generate recursion relations of $E_{n,k}[V^{a}\otimes
L^{k-a}]$ . By applying these recursion relations iteratively, one can reduce
a general one-loop integral to the basis with coefficients written by elegant
expressions. In other words, the PV-reduction can be done universally in the
new projective space form. As we will point out, the reduction coefficients
will have an interesting pattern other than the obvious permutation
symmetry333In the work Feng:2021enk ; Hu:2021nia , one can see that the
recursion relations, for example, the bubble tensors to bubble basis and the
triangle tensor to triangle basis, are similar, except the boundary
conditions. This similarity has given an explanation for the interesting
pattern we observed in this paper.. Moreover, the reduction process can be
carried out in Mathematica automatically.

###  3.1 Recursion relation

In this subsection, we derive the recursion relations of $E_{n,k}[V^{a}\otimes
L^{k-a}]$. We first consider the case $Q$ is non-degenerate. The key equation
is following (see Eq.(4.2) in Arkani-Hamed:2017ahv )

$\displaystyle{{\langle{X}\mathrm{d}^{n-1}{X}\rangle}T[X^{k}]\over(XQX)^{n+k\over
2}}$ $\displaystyle=$ $\displaystyle{1\over
n+k-2}\text{d}_{X}\left[{\langle(Q^{-1}T)[X^{k-1}]X\text{d}^{n-2}X\rangle\over(XQX)^{n+k-2\over
2}}\right]$ (27) $\displaystyle+{k-1\over
n+k-2}{{\langle{X}\mathrm{d}^{n-1}{X}\rangle}(tr_{Q}T)[X^{k-2}]\over(XQX)^{n+k-2\over
2}}\,,~{}~{}~{}$

where $\text{d}_{X}=\text{d}X^{I}{\partial\over\partial
X^{I}},tr_{Q}T=Q^{-1}_{I_{1}I_{2}}T^{I_{1}I_{2}\ldots I_{k}}$. The proof of
the formula can be found in the Appendix A. By integrating (27) we get

$\displaystyle
E_{n,k}[T]=\alpha_{n,k}E^{(b)}_{n-1,k-1}[(H_{b}Q^{-1}T)]+\beta_{n,k}E_{n,k-2}[tr_{Q}T]\,,~{}~{}$
(28)

where summing over $b$ is implicitly and to simplify our denotations, we have
defined

$\displaystyle\alpha_{n,k}$ $\displaystyle\equiv$ $\displaystyle{1\over
n+k-2}\,,~{}~{}~{}\beta_{n,k}\equiv{k-1\over n+k-2}\,.$ (29)

Let us give a little explanation for the first term on the right-hand side of
(28). When integrating a total derivative term, we should choose a patch. For
simplicity, we assume $X_{i}=1$. Then we get the contribution from the
boundary $X_{b}=0$ and $X_{b}=+\infty$ for $b\neq i$. By the dimensional
regularization, the term with $X_{b}=+\infty$ gives zero. For the term
$X_{b}=0$, in $\langle(Q^{-1}T)[X^{k-1}]X\text{d}^{n-2}X\rangle$, only when
the first index of $Q^{-1}$ takes the value $b$, the contribution is nonzero,
which is equivalent to write as $(H_{b}Q^{-1}T)$.

When we repeatedly do the recursion relation, there will be a set of $X_{b}$
setting to zero. Writing the index set as
$\mathbf{b}_{j}\equiv\\{b_{1},b_{2},\ldots,b_{j}\\}$ with $1\leq
b_{i}<b_{i+1}\leq n$, we define $X_{(\mathbf{b}_{j})}$ to be the new vector in
lower-dimensional projective space obtained by removing these components
belonging to the set $\mathbf{b}_{j}$ from the original vector $X$. With this
understanding, the meaning of

$\displaystyle E^{(\mathbf{b}_{j})}_{n-j,k}[T]$ $\displaystyle\equiv$
$\displaystyle\int_{\Delta_{\mathbf{b}_{j}}}{T[X^{k}_{(\mathbf{b}_{j})}]{\langle{X_{(\mathbf{b}_{j})}}\mathrm{d}^{n-1-j}{X_{(\mathbf{b}_{j})}}\rangle}\over(X_{(\mathbf{b}_{j})}Q_{(\mathbf{b}_{j})}X_{(\mathbf{b}_{j})})^{{n-j+k\over
2}}}$ (30)

is clear. The equation (30) represents the integral got by removing
propagators belonging to $\mathbf{b}_{j}$.

Now we consider the reduction of $E_{n,k}[T]$ with $T=V^{i}\otimes L^{k-i}$ as
given in (25). Since tensor $T$ is contracted with $k$ X’s, we can symmetrize
its last $k-1$ indices as below

$\displaystyle V^{i}\otimes L^{k-i}\to V\otimes{1\over(k-1)!}\sum_{\sigma\in
S_{k-1}}\sigma\left[V^{i-1}\otimes L^{k-i}\right]\,,$ (31)

where the $\sigma$ is the permutation acting on the tensor $V^{i-1}\otimes
L^{k-i}$. By applying (27), one has

$\displaystyle
E_{n,k}[V^{i+1}]=\alpha_{n,k}\left[(\overline{H_{b}V})E^{(b)}_{n-1,k-1}[V^{i}]+{i}(\overline{VV})E_{n,k-2}[V^{i-1}]+(k-i-1)(\overline{VL})E_{n,k-2}[V^{i}]\right]\,,$
(32)

where for simplicity the $L$ tensor part has been omitted. To make our formula
more compact, here we have defined $(\overline{AB})\equiv AQ^{-1}B$. Here we
want to remark a subtle point. In (32), the $Q$ is $n\times n$ matrix as
defined in $E_{n,k}$ and appears in front of $E^{(b)}_{n-1,k-1}$ . When we try
to iteratively use (32) for $E^{(b)}_{n-1,k-1}$, that $Q$ will become the
$(n-1)\times(n-1)$ matrix $Q_{(b)}$. Thus, for later convenience, we define
$(\overline{AB})_{(\mathbf{a}_{j})}=A_{(\mathbf{a}_{j})}(Q_{(\mathbf{a}_{j})})^{-1}B_{(\mathbf{a}_{j})}$
where matrix $(Q_{(\mathbf{a}_{j})})^{-1}$ is the inverse of the matrix
obtained by removing the rows and columns of the index set $\mathbf{a}_{j}$
from the original matrix $Q$.

As the main result of the whole paper, recursion relation (32) plays a crucial
role in the reduction of one-loop integrals. By comparing the power of $V$,
one sees that it has been reduced from the LHS to RHS. Furthermore, from (25)
one sees that the power is given by $v+i-n$, where $v$ contains the
contribution of higher power of propagators and $i$ contains the contribution
of the tensor numerator, thus (32) provides the universal reduction of both
cases. As shown in the Fig(1), after iteratively using (32), we get a linear
combination of $E^{(\mathbf{a}_{n-n^{\prime}})}_{n^{\prime}\leq
n,k^{\prime}<k}$, i.e., the scalar integral $I_{n^{\prime};D^{\prime}}$ in
dimension $D^{\prime}=n^{\prime}-k^{\prime}$. So starting with a general one-
loop integral, one can always reduce it to the scalar integrals in different
dimensions with coefficients being rational functions of external momenta. If
we prefer the scalar basis in a given $D$-dimensional space, we need to find
the formula to shift the dimension of the scalar basis to a fixed $D$.

###  3.2 Dimension recursion

As we have seen in (26), the integral $E_{n,k}$ corresponds to the scalar
$n$-gon diagram in $(n-k)$-dimensional space. To find dimension recursion
relations, we set $V=L$ in (32) and get

$\displaystyle
E_{n,k}=\alpha_{n,k}(\overline{H_{b}L})E^{(b)}_{n-1,k-1}+\beta_{n,k}(\overline{LL})E_{n,k-2}\,.~{}~{}~{}$
(33)

To reduce $E_{n,k}$, where $n-k=D+2s,s\in\mathbb{Z},s\not=0$, we can
iteratively use (33), which is established for the scalar integrals already.
Noticing that in the RHS of (33), the first term has the same dimension as the
LHS with one propagator removed, while the second term has two higher
dimensions with the same number of propagators. Depending on the sign of $s$,
we can take different manipulations.

  * •

$s>0$: For this case, we need to reduce an integral in a higher dimension to
$D$-dimension, so we solve the second term in the RHS of (33) and get

$\displaystyle
E_{n,k}={E_{n,k+2}-\alpha_{n,k+2}(\overline{H_{j}L})E^{(j)}_{n-1,k+1}\over{\beta}_{n,k+2}(\overline{LL})}\,.$
(34)

It is obviously that such a rewriting (34) is legitimate when and only when
$(\overline{LL})\not=0$. For $(\overline{LL})=0$, we have

$\displaystyle E_{n,k}=\alpha_{n,k}(\overline{H_{j}L})E_{n-1,k-1}^{(j)}\,.$
(35)

Both sides have the same dimension, but the RHS of (35) has one less
propagator. One well-known example of $(\overline{LL})=0$ is that the bubble
with null external momentum is not a basis anymore, and it is reduced to two
tadpoles. Having established (34) and (35), we can reduce $(D+2s)$-dimensional
integrals to $D$-dimensional iteratively using either (34) or (35) depending
on if $(\overline{LL})$ is zero or not at that step444Please remember that as
emphasized under (32), at each step $Q$ is different. .

  * •

$s<0$: For this case, we can use (33) directly.

$\displaystyle
E_{n,k}=\alpha_{n,k}(\overline{H_{j}L})E_{n-1,k-1}^{(j)}+\beta_{n,k}(\overline{LL})E_{n,k-2}\,.$
(36)

As pointed out already, the first term on the RHS corresponds to the scalar
integrals in the same dimension but with the $j$-th propagator removed and the
second term corresponds to scalar integrals in $D^{\prime}=n-k+2$ dimension.
For the boundary situation, i.e., $n=1$, the first term vanishes and the
second term gives a higher dimensional scalar basis. Repeating it, we can
reduce $E_{n,k}$ to scalar integrals in $D$-dimensional space.

00111222333444555$n$$i$ Figure 1: The reduction process of
$E_{n=5,k}[V^{i=5}]$, where the black points represent zero terms. The red,
orange and cyan arrows represent the first, second and third terms
respectively in (32).

###  3.3 Examples

To illustrate our method and avoid complicated computation in general cases,
we first consider the reduction of tensor bubbles

$\displaystyle I_{2}^{(r)}=\int{d^{D}\ell\over
i\pi^{D/2}}{(2R\cdot\ell)^{r}\over D_{1}^{v_{1}}D_{2}^{v_{2}}}\,,$ (37)

where

$\displaystyle D_{1}=(\ell-q_{1})^{2}-m_{1}^{2}\,,~{}~{}~{}~{}D_{2}=(\ell-
q_{2})^{2}-m_{2}^{2}\,.$ (38)

We set $q_{1}=0$ for simplicity, then

$\displaystyle L=\left[\begin{array}[]{c}1\\\ 1\\\
\end{array}\right],H_{1}=\left[\begin{array}[]{c}1\\\ 0\\\
\end{array}\right],H_{2}=\left[\begin{array}[]{c}0\\\ 1\\\
\end{array}\right],Q=\left[\begin{array}[]{cc}m_{1}^{2}&\frac{1}{2}\left(m_{1}^{2}+m_{2}^{2}-q_{2}^{2}\right)\\\
\frac{1}{2}\left(m_{1}^{2}+m_{2}^{2}-q_{2}^{2}\right)&m_{2}^{2}\\\
\end{array}\right]\,.$ (47)

Here we give some results for different choices of the rank and the powers to
illustrate our idea.

  * •

Tensor bubble with primary propagator

We first consider reducing a rank-1 bubble $I_{2}^{(1)}\equiv
I_{\\{1,1\\}}^{(1)}$. Since there are no higher poles, we just choose
$S=V,r=1,v=n=2$ in (25) and we have

$\displaystyle
I_{2}^{(1)}=-{\Gamma(1-D/2)}\mathscr{C}^{D/2-1}_{1,1}E_{2,2-D}[V]=-{\Gamma(1-D/2)}(D-2)E_{2,2-D}[V]\,.$
(48)

First, we use (32) to reduce $E_{2,2-D}[V]$ and get

$\displaystyle
E_{2,2-D}[V]={\alpha}_{2,2-D}(\overline{H_{i}V})E^{(i)}_{1,1-D}+{\beta}_{2,2-D}(\overline{VL})E_{2,-D}\,,$
(49)

where the first term $E^{(i)}_{1,1-D}$ corresponds to $D$-dimensional scalar
tadpoles generated by removing the $i$-th propagator of bubble $I_{2}$ while
the second term $E_{2,-D}$ corresponds to a $(D+2)$-dimensional bubble. We
need to lower the dimension of the second term further. Here we assume
$(\overline{LL})\not=0$, by using (34), we get

$\displaystyle
E_{2,-D}={E_{2,2-D}-{\alpha}_{2,2-D}(\overline{H_{i}L})E^{(i)}_{1,1-D}\over{\beta}_{2,2-D}(\overline{LL})}\,,$
(50)

where the two terms in the numerator correspond to the $D$-dimensional bubble
and two tadpoles. Plugging (50) and (49) to (48) and recognizing them as
master integral according to (26) we have

$\displaystyle
I_{2}^{(1)}=\sum_{i}-\frac{(\overline{LL})(\overline{H_{i}V})-(\overline{H_{i}L})(\overline{VL})}{(\overline{LL})}I_{2;\widehat{i}}+\frac{2(\overline{VL})}{(\overline{LL})}I_{2}\,.$
(51)

So the reduction coefficients are

$\displaystyle C^{(1)}_{2\to 2;\widehat{1}}$ $\displaystyle=$
$\displaystyle-\frac{(\overline{LL})(\overline{H_{1}V})-(\overline{H_{1}L})(\overline{VL})}{(\overline{LL})}=\frac{R\cdot
q_{2}}{q_{2}^{2}}\,,$ $\displaystyle C^{(1)}_{2\to 2;\widehat{2}}$
$\displaystyle=$
$\displaystyle-\frac{(\overline{LL})(\overline{H_{2}V})-(\overline{H_{2}L})(\overline{VL})}{(\overline{LL})}=-\frac{R\cdot
q_{2}}{q_{2}^{2}}\,,$ $\displaystyle C^{(1)}_{2\to 2}$ $\displaystyle=$
$\displaystyle\frac{2(\overline{VL})}{(\overline{LL})}=\frac{\left(m_{1}^{2}-m_{2}^{2}+q_{2}^{2}\right)R\cdot
q_{2}}{q_{2}^{2}}\,.$ (52)

  * •

Tensor bubble with massless legs

One can notice there is a pole of $q_{2}^{2}$ in the reduction coefficients of
$I_{2}^{(1)}$, which comes from $(\overline{LL})$

$\displaystyle(\overline{LL})=-\frac{4q_{2}^{2}}{-2\left(m_{1}^{2}+m_{2}^{2}\right)q_{2}^{2}+\left(m_{1}^{2}-m_{2}^{2}\right){}^{2}+q_{2}^{4}}\,.$
(53)

For $q_{2}^{2}=0$, we have $(\overline{LL})=0$, so we have

$\displaystyle
E_{2,2-D}[V]={\alpha}_{2,2-D}(\overline{H_{b}V})E^{(b)}_{1,1-D}+{\beta}_{2,2-D}(\overline{VL})E_{2,-D}\,.$
(54)

Here we need to reduce $E_{2,-D}$

$\displaystyle
E_{2,-D}=E^{(b)}_{1,-D-1}={{\alpha}_{2,-D}(\overline{H_{b}L})E^{(b)}_{1,1-D}\over{\beta}_{1,1-D}(\overline{LL})_{(b)}}\,,$
(55)

where we have used (35) and

$\displaystyle
E^{(b)}_{1,-D-1}={E^{(b)}_{1,1-D}\over{\beta}_{1,1-D}(\overline{LL})_{(b)}}\,.$
(56)

Using (26), we finally get

$\displaystyle I_{2}^{(1);q_{2}^{2}=0}=\left[{1-D\over
D}{(\overline{VL})(\overline{H_{b}L})\over(\overline{LL})_{(b)}}-(\overline{H_{b}V})\right]I_{2;\widehat{b}}\,.$
(57)

Explicitly, we have

$\displaystyle
I_{2}^{(1);q_{2}^{2}=0}=-\frac{2\left(Dm_{1}^{2}-(D-2)m_{2}^{2}\right)R\cdot
q_{2}}{D\left(m_{1}^{2}-m_{2}^{2}\right)^{2}}I_{2;\widehat{1}}+\frac{4m_{1}^{2}R\cdot
q_{2}}{D\left(m_{1}^{2}-m_{2}^{2}\right)^{2}}I_{2;\widehat{2}}\,.$ (58)

  * •

Scalar bubble with higher poles

Then we consider reducing scalar bubbles $I_{\\{v_{1},v_{2}\\}}$ with higher
poles $v=v_{1}+v_{2}=3$. Due to there is no tensor structure in the numerator,
we just set $S=Z=z_{1}H_{1}+z_{2}H_{2},r=0$ in (25)

$\displaystyle I_{\mathbf{v}_{2};v=3}$ $\displaystyle=$
$\displaystyle{-\Gamma(3-D/2)}E_{2,4-D}[Z]\Big{|}_{z^{\mathbf{v}_{2}-1}}\,,$
(59)

where $\mathbf{v}_{2}=\\{1,2\\},\\{2,1\\}$. First, we use (32) to reduce
$E_{2,4-D}[Z]$ and get

$\displaystyle
E_{2,4-D}[Z]={\alpha}_{2,4-D}(\overline{H_{i}Z})E^{(i)}_{1,3-D}+{\beta}_{2,4-D}(\overline{ZL})E_{2,2-D}\,,$
(60)

where the first term $E^{(i)}_{1,3-D}$ corresponds to $(D-2)$-dimensional
scalar tadpoles generated by removing the $i$-th propagator of bubble $I_{2}$
while the second term $E_{2,2-D}$ corresponds to wanted $D$-dimensional
bubble. We need to lift the dimension of the first term further. By using
(36), we get

$\displaystyle
E^{(i)}_{1,3-D}={\beta}_{1,3-D}(\overline{LL})E^{(i)}_{1,1-D}\,,$ (61)

where we have used $E^{(ij)}_{0,2-D}=0$, and the RHS corresponds to
$D$-dimensional scalar tadpoles. Plugging (61) and (60) to (59) and
recognizing them as master integrals according to (26) we have

$\displaystyle
I_{\mathbf{v}_{2};v=3}=-\frac{1}{4}(D-2)(\overline{LL})_{(i)}(\overline{H_{i}Z})\Big{|}_{z^{\mathbf{v}_{2}-1}}I_{2;\widehat{i}}+\frac{1}{2}(D-3)(\overline{ZL})\Big{|}_{z^{\mathbf{v}_{2}-1}}I_{2}\,.$
(62)

There are two configurations

$\displaystyle I_{\\{2,1\\}}$
$\displaystyle=-\frac{1}{4}(D-2)(\overline{LL})_{(i)}(\overline{H_{i}Z})\Big{|}_{z_{1}}I_{2;\widehat{i}}+\frac{1}{2}(D-3)(\overline{ZL})\Big{|}_{z_{1}}I_{2}$
$\displaystyle=-\frac{1}{4}(D-2)(\overline{LL})_{(i)}(\overline{H_{i}H_{1}})I_{2;\widehat{i}}+\frac{1}{2}(D-3)(\overline{H_{1}L})I_{2}\,,$
$\displaystyle I_{\\{1,2\\}}$
$\displaystyle=-\frac{1}{4}(D-2)(\overline{LL})_{(i)}(\overline{H_{i}Z})\Big{|}_{z_{2}}I_{2;\widehat{i}}+\frac{1}{2}(D-3)(\overline{ZL})\Big{|}_{z_{2}}I_{2}$
$\displaystyle=-\frac{1}{4}(D-2)(\overline{LL})_{(i)}(\overline{H_{i}H_{2}})I_{2;\widehat{i}}+\frac{1}{2}(D-3)(\overline{H_{2}L})I_{2}\,.$
(63)

Explicitly, using (47), we find the reduction coefficients are

$\displaystyle C_{\\{2,1\\}\to 2;\widehat{1}}$
$\displaystyle=\frac{D-2}{\left(\left(m_{1}-m_{2}\right){}^{2}-q_{2}^{2}\right)\left(\left(m_{1}+m_{2}\right){}^{2}-q_{2}^{2}\right)}\,,$
$\displaystyle C_{\\{2,1\\}\to 2;\widehat{2}}$
$\displaystyle=-\frac{(D-2)\left(m_{1}^{2}+m_{2}^{2}-q_{2}^{2}\right)}{2m_{1}^{2}\left(\left(m_{1}-m_{2}\right){}^{2}-q_{2}^{2}\right)\left(\left(m_{1}+m_{2}\right){}^{2}-q_{2}^{2}\right)}\,,$
$\displaystyle C_{\\{2,1\\}\to 2}$
$\displaystyle=\frac{(D-3)\left(m_{1}^{2}-m_{2}^{2}-q_{2}^{2}\right)}{\left(\left(m_{1}-m_{2}\right){}^{2}-q_{2}^{2}\right)\left(\left(m_{1}+m_{2}\right){}^{2}-q_{2}^{2}\right)}\,,$
(64) $\displaystyle C_{\\{1,2\\}\to 2;\widehat{1}}$
$\displaystyle=-\frac{(D-2)\left(m_{1}^{2}+m_{2}^{2}-q_{2}^{2}\right)}{2m_{2}^{2}\left(\left(m_{1}-m_{2}\right){}^{2}-q_{2}^{2}\right)\left(\left(m_{1}+m_{2}\right){}^{2}-q_{2}^{2}\right)}\,,$
$\displaystyle C_{\\{1,2\\}\to 2;\widehat{2}}$
$\displaystyle=\frac{D-2}{\left(\left(m_{1}-m_{2}\right){}^{2}-q_{2}^{2}\right)\left(\left(m_{1}+m_{2}\right){}^{2}-q_{2}^{2}\right)},$
$\displaystyle C_{\\{1,2\\}\to 2}$
$\displaystyle=-\frac{(D-3)\left(m_{1}^{2}-m_{2}^{2}+q_{2}^{2}\right)}{\left(\left(m_{1}-m_{2}\right){}^{2}-q_{2}^{2}\right)\left(\left(m_{1}+m_{2}\right){}^{2}-q_{2}^{2}\right)}\,.~{}~{}~{}$
(65)

  * •

Tensor bubble with higher poles

At last, we consider a combined case, $I^{(1)}_{\mathbf{v}_{2};v=3}$. Here we
need to set $S=tV+Z,Z=z_{1}H_{1}+z_{2}H_{2}$. Setting $v=3,r=1,n=2$ in (25),
we have

$\displaystyle
I^{(1)}_{\mathbf{v}_{2};v=3}=-\Gamma\left(3-D/2\right)E_{2,4-D}[S^{2}]\Big{|}_{tz^{\mathbf{v}_{2}-1}}\,.$
(66)

First, we use (32) iteratively to pull out all $S$’s in the numerator

$\displaystyle E_{2,4-D}[S^{2}]$
$\displaystyle={\alpha}_{2,4-D}\left[(\overline{H_{i}S})E^{(i)}_{1,3-D}[S]+(\overline{SS})E_{2,2-D}+(2-D)(\overline{SL})E_{2,2-D}[S]\right]$
$\displaystyle={\alpha}_{2,4-D}\bigg{[}{\beta}_{1,3-D}(\overline{H_{i}S})(\overline{SL})_{(i)}E^{(i)}_{1,1-D}+(\overline{SS})E_{2,2-D}$
$\displaystyle{\hskip
15.0pt}+(2-D)(\overline{SL}){\alpha}_{2,2-D}\big{[}(\overline{H_{i}S})E^{(i)}_{1,1-D}+(1-D)(\overline{SL})E_{2,-D}\big{]}\bigg{]}\,.$
(67)

Among the four terms above, we only need to deal with the last term $E_{2,-D}$
since it corresponds to a $(D+2)$-dimensional bubble, which has been discussed
in (55). Finally, we have

$\displaystyle I^{(1)}_{\mathbf{v}_{2};v=3}$
$\displaystyle=-\frac{(D-2)\left((\overline{LL})(\overline{H_{i}Z})(\overline{ZL})+(\overline{LL})(\overline{ZL})_{(i)}(\overline{H_{i}Z})-(\overline{H_{i}L})(\overline{ZL})^{2}\right)}{4(\overline{LL})}\Big{|}_{tz^{\mathbf{v}_{2}-1}}I_{2;\widehat{i}}$
$\displaystyle{\hskip
15.0pt}+\frac{1}{2}\left(\frac{(D-2)(\overline{ZL})^{2}}{(\overline{LL})}-(\overline{ZZ})\right)\Big{|}_{tz^{\mathbf{v}_{2}-1}}I_{2}\,.$
(68)

One can get the reduction results for
$I_{\\{2,1\\}}^{(1)},I_{\\{1,2\\}}^{(1)}$. For example,

$\displaystyle I^{(1)}_{\\{2,1\\}}$
$\displaystyle=-\frac{(D-2)}{4(\overline{LL})}\bigg{[}(\overline{LL})(\overline{H_{i}V})(\overline{H_{1}L})+(\overline{LL})(\overline{H_{i}H_{1}})(\overline{VL})+(\overline{LL})(\overline{VL})_{(i)}(\overline{H_{i}H_{1}})$
$\displaystyle{\hskip
15.0pt}+(\overline{LL})(\overline{H_{1}L})_{(i)}(\overline{H_{i}V})-2(\overline{H_{i}L})(\overline{VL})(\overline{H_{1}L})\bigg{]}I_{2;\widehat{i}}$
$\displaystyle{\hskip
15.0pt}+\left[\frac{(D-2)(\overline{H_{1}L})(\overline{VL})}{(\overline{LL})}-(\overline{H_{1}V})\right]I_{2}\,,$
(69)

where for simplicity, we will not present explicit expressions for these
coefficients.

Note that the reduction coefficients in these examples are rational functions.
For some special masses and momenta configurations, denominators can become
zero, which leads to several kinds of divergences. Since only
$(\overline{LL})\equiv LQ^{-1}L$ appears in the denominators, all divergences
come from $Q$-matrix and its all sub-matrices, which have $\mathop{\rm
det}Q=0$ or $(\overline{LL})=0$. For example, the pole of $q_{2}^{2}$ in (52)
comes from $LQ^{-1}L$ (see (53)). The divergence of $C_{\\{2,1\\}\to
2;\widehat{2}}$ is given by
$m_{1}^{2}\left(\left(m_{1}-m_{2}\right){}^{2}-q_{2}^{2}\right)\left(\left(m_{1}+m_{2}\right){}^{2}-q_{2}^{2}\right)=0$,
which is corresponds to $\mathop{\rm det}Q_{(2)}=0$ or $\mathop{\rm det}Q=0$.
One can find the pole $(\overline{LL})=0$ comes from the dimension shifting
process (34), which can be addressed by employing (33) to reduce $E_{n,k}$ to
lower topology. To deal with the divergences coming from $\mathop{\rm
det}Q=0$, we need to consider the reduction method for degenerate $Q$
elaborated in the next section.

##  4 Reduction for degenerate $Q$

In this section, we generalize our reduction method to degenerate $Q$. The
basic idea is to generalize the recursion relation (27) to the formula (129).
When $Q$ is degenerate, the characteristic equation $Q\xi=0$ always has
solutions, and we denote the $\mathfrak{N}_{Q}$ as the null space spanned by
linearly independent $\xi$’s.

###  4.1 $\widetilde{Q}L\neq 0$

When $Q$ is degenerate, the recursion relation (27) in the last sections
breaks down for $\mathop{\rm det}{Q}=0$. Our idea is to consider the tensor
structure with one $L$ in the first place and make other $(k-1)$ indices
completely symmetric by summing over all permutations between $i$ $V$’s and
$(k-1-i)$ $L$’s. Using (129) we have

$\displaystyle E_{n,k}[(Q\widetilde{Q}L)\otimes V^{i}\otimes L^{k-1-i}]$
$\displaystyle=$ $\displaystyle
E_{n,k}\bigg{[}(Q\widetilde{Q}L)\otimes{1\over(k-1)!}\sum_{\sigma\in
S_{k-1}}\sigma\left[V^{i}\otimes L^{k-1-i}\right]\bigg{]}$ $\displaystyle=$
$\displaystyle{\alpha}_{n,k}(H_{b}\widetilde{Q}L)E_{n-1,k-1}^{(b)}[V^{i}]+{\beta}_{n,k}\left[{i\over
k-1}(V\widetilde{Q}L)E_{n,k-2}[V^{i-1}]+{k-1-i\over
k-1}(L\widetilde{Q}L)E_{n,k-2}[V^{i}]\right]\,.$

For a degenerate $Q$, we can always find a matrix
$\widetilde{Q}=[\xi_{1},\xi_{2},\ldots,\xi_{n}]$,$\xi_{i}\in\mathfrak{N}_{Q}$
so that $Q\widetilde{Q}=0$. Then the LHS of (4.1) vanishes. If $Q^{*}L\neq 0$,
where $Q^{*}$ is the adjugate matrix of $Q$, we can take
$\widetilde{Q}=Q^{*}$. With the denotation

$\displaystyle\dot{(\overline{AB})}=(A\widetilde{Q}B)\,,$ (71)

(4.1) becomes

$\displaystyle{\alpha}_{n,k}\dot{(\overline{H_{b}L})}E_{n-1,k+1}^{(b)}[V^{i}]+i{\beta}_{n,k}\dot{(\overline{VL})}E_{n,k}[V^{i-1}]+(k+1-i){\beta}_{n,k}\dot{(\overline{LL})}E_{n,k}[V^{i}]=0\,.$
(72)

Depending on the value of $\dot{(\overline{LL})}$, we have following two
cases:

  * •

(1) When $\dot{(\overline{LL})}\not=0$, (72) can be rewritten as

$\displaystyle
E_{n,k}^{\left|Q\right|=0}[V^{i}]={1\over(i-k-1)\dot{(\overline{LL})}}\left[\dot{(\overline{H_{b}L})}E_{n-1,k+1}^{(b)}[V^{i}]+i\dot{(\overline{VL})}E_{n,k}[V^{i-1}]\right]\,,$
(73)

where the first term in RHS corresponds to the lower topologies and the second
term has tensor rank reduced by one.

  * •

(2) When $\dot{(\overline{LL})}=0$, (72) can be rewritten as

$\displaystyle
E_{n,k}^{\left|Q\right|=0}[V^{i}]={-\dot{(\overline{H_{b}L})}\over(i+1)\dot{(\overline{VL})}}E^{(b)}_{n-1,k+1}[V^{i+1}]\,,~{}~{}~{}$
(74)

where although the tensor rank increased by one on the RHS, it belongs to the
lower topologies. One particular thing of (74) is that since $V$ depends on
the auxiliary $R$, we will always have $\dot{(\overline{VL})}\neq 0$. Another
tricky point is that although $R$ appears in the denominator in (74) in the
middle steps, it will be canceled in the final reduction coefficients.

###  4.2 $\widetilde{Q}L=0$

Since $\widetilde{Q}L=0$, every term in (72) vanishes. Now we can put $V$ in
the first place of tensor structure and using (129) to reach

$\displaystyle E_{n,k}[(Q\widetilde{Q}V)V^{i}]$ (75) $\displaystyle=$
$\displaystyle
A_{n,k}\left[(H_{j}\widetilde{Q}V)E^{(j)}_{n-1,k-1}[V^{i}]+{i}(V\widetilde{Q}V)E_{n,k-2}[V^{i-1}]+(k-i-1)(V\widetilde{Q}L)E_{n,k-2}[V^{i}]\right]\,.~{}~{}~{}$

Using $Q\widetilde{Q}=\widetilde{Q}L=0$, (75) becomes

$\displaystyle
0=A_{n,k}\left[(H_{j}\widetilde{Q}V)E^{(j)}_{n-1,k-1}[V^{i}]+{i}(V\widetilde{Q}V)E_{n,k-2}[V^{i-1}]\right]\,,$
(76)

and we have

$\displaystyle
E_{n,k}[V^{i}]={-\dot{(\overline{H_{b}V})}\over(i+1)\dot{(\overline{VV})}}E_{n-1,k+1}^{(b)}[V^{i+1}]\,.$
(77)

Again, although $R$ appears in denominator through $\dot{(\overline{VV})}$ in
middle steps, it will be canceled in the final reduction coefficients.

###  4.3 Dimension recursion

Having reduced to scalar integrals with different dimensions, we want to shift
the dimension to a given $D$. Depending on various situations, we have:

  * •

For $\widetilde{Q}L\neq 0$, we can always choose
$Q\widetilde{Q}=0,\dot{(\overline{LL})}\not=0$. Then using (73) for the case
$i=0$, the second term vanishes, and we have

$\displaystyle
E_{n,k}={-\dot{(\overline{H_{b}L})}\over(1+k)\dot{(\overline{LL})}}E_{n-1,k+1}^{(b)}\,.$
(78)

  * •

For $\widetilde{Q}L=0$ we use (77) with $i=1$. Then we shift $V\to L+\epsilon
K$ with a reference $K$ such that $K\not\perp\mathfrak{N}_{Q}$ and get

$\displaystyle E_{n,k}+\epsilon E_{n,k}[K]={-\dot{(\overline{H_{b}K})}\over
2\epsilon\dot{(\overline{KK})}}E_{n-1,k+1}^{(b)}[(L+\epsilon K)^{2}]\,.$ (79)

Comparing both sides, especially the $\epsilon$ term, we have

$\displaystyle
E_{n,k}={-\dot{(\overline{H_{b}K})}\over\dot{(\overline{KK})}}E_{n-1,k+1}^{(b)}[K]\,.~{}~{}~{}$
(80)

We lower the topology in the RHS, so we can reduce it further by using the
equation recursively. The dependence of the choice of $K$ in the middle steps
will vanish in the final reduction coefficients as shown in later examples.

###  4.4 Examples

In this section, we illustrate our method for degenerate $Q$. To avoid
unnecessary and complicated calculations and compare with the reduction
procedures for non-degenerate Q discussed in section (3.3), we focus on
bubbles with some special masses and momenta configurations.

  * •

$\widetilde{Q}L=0$: Massless scalar bubble with equal internal masses

To check the validity of our method, we first consider a scalar bubble with
$m_{1}=m_{2}=m$ and $q_{2}^{2}=0$, defined as

$\displaystyle I_{2}^{m_{1}=m_{2},q_{2}^{2}=0}=\int{d^{D}\ell\over
i\pi^{D/2}}{1\over(\ell^{2}-m^{2})[(\ell-q_{2})^{2}-m^{2}]}\,,$ (81)

which can be reduced to two tadpoles555One can check this by using FIRE or
direct calculation. .

We find $Q$ defined in (47) degenerates to a corank-1 matrix

$\displaystyle Q=\left[\begin{array}[]{cc}m^{2}&m^{2}\\\ m^{2}&m^{2}\\\
\end{array}\right]\,,~{}~{}~{}\widetilde{Q}=Q^{*}=\left[\begin{array}[]{cc}m^{2}&-m^{2}\\\
-m^{2}&m^{2}\\\ \end{array}\right].$ (86)

Since $Q\widetilde{Q}=0$ and $Q^{*}L=0$, using (26) we have

$\displaystyle I_{2}=\Gamma(2-D/2)E_{2,2-D}\,.$ (87)

One can see the RHS is not irreducible anymore by employing (80)

$\displaystyle
E_{2,2-D}={-\dot{(\overline{H_{b}K})}\over\dot{(\overline{KK})}}E_{1,3-D}^{(b)}[K]\,,$
(88)

where the reference vector $K$ satisfies $Q^{*}K\neq 0$ and the RHS
corresponds to integrals of tadpole topology acquired by removing one
propagator from $I_{2}$. We then use (32) to pull out the $K$

$\displaystyle
E_{1,3-D}^{(b)}[K]=\beta_{1,3-D}(\overline{KL})_{(b)}E_{1,1-D}^{(b)}\,.$ (89)

Note that the $Q$-matrix in (89) is just a number $Q_{(b)}=m^{2}$, so its
inverse in $(\overline{KL})_{(b)}$ is just $Q_{(b)}^{-1}={1\over m^{2}}$.
Combining (88),(89) and (87), we finally find

$\displaystyle I_{2}^{m_{1}=m_{2},q_{2}^{2}=0}={2-D\over
2}{\dot{(\overline{H_{b}K})}(\overline{KL})_{(b)}\over\dot{(\overline{KK})}}I_{2;\widehat{b}}={D-2\over
2m^{2}}I_{2;\widehat{1}}\,.$ (90)

One can check that the result above does not depend on the choice of $K=(a,b)$
as long as $a\neq b$.

  * •

$\widetilde{Q}L=0$: Massless tensor bubble with equal internal masses

Here we consider the reduction of the tensor bubble
$I_{2}^{(1);m_{1}=m_{2},q_{2}^{2}=0}$. Recalling the first equation of (48)

$\displaystyle I_{2}^{(1)}=-{\Gamma(1-D/2)}(D-2)E_{2,2-D}[V]\,,~{}~{}~{}$ (91)

we just need to reduce $E_{2,2-D}[V]$. Due to $\widetilde{Q}L=0$ we use (77)
to get

$\displaystyle E_{2,2-D}[V]={-1\over
2\dot{(\overline{VV})}}\dot{(\overline{H_{b}V})}E_{1,3-D}^{(b)}[V^{2}]\,,$
(92)

where the RHS corresponds to integrals of tadpoles with non-degenerate
$Q$-matrix $Q_{(b)}=m^{2}$. We can use (32) reduce it iteratively

$\displaystyle E_{1,3-D}^{(b)}[V^{2}]$
$\displaystyle={\alpha}_{1,3-D}\left[(\overline{VV})_{(b)}E^{(b)}_{1,1-D}+(1-D)(\overline{VL})_{(b)}E^{(b)}_{1,1-D}[V]\right]$
$\displaystyle={\alpha}_{1,3-D}\left[(\overline{VV})_{(b)}E^{(b)}_{1,1-D}+(1-D)(\overline{VL})_{(b)}\beta_{1,1-D}(\overline{VL})_{(b)}E_{1,-1-D}^{(b)}\right]\,.$
(93)

There are two terms in the last line, we need to reduce the second one
$E_{1,-1-D}^{(b)}$ since it corresponds to $(D+2)$-dimensional tadpole. Using
(34) to lower the dimension to $D$, we get

$\displaystyle
E^{(b)}_{1,-1-D}={E^{(b)}_{1,1-D}\over{\beta}_{1,1-D}(\overline{LL})_{(b)}}\,.$
(94)

Plugging (93),(94), (92) into (91), we have

$\displaystyle I_{2}^{(1);m_{1}=m_{2},q_{2}^{2}=0}$
$\displaystyle={\dot{(\overline{H_{b}V})}\left((1-D)(\overline{VL})_{(b)}^{2}+(\overline{LL})_{(b)}(\overline{VV})_{(b)}\right)\over
2\dot{(\overline{VV})}(\overline{LL})_{(b)}}I_{2;\widehat{b}}\,.$ (95)

Using

$\displaystyle\dot{(\overline{VV})}$ $\displaystyle=m^{2}(R\cdot
q_{2})^{2}\,,~{}\hskip 22.0pt\dot{(\overline{H_{b}V})}=\left\\{-m^{2}R\cdot
q_{2},m^{2}R\cdot q_{2}\right\\}\,,$ $\displaystyle(\overline{VV})_{(b)}$
$\displaystyle=\left\\{{(R\cdot q_{2})^{2}\over
m^{2}},0\right\\}\,,~{}~{}~{}(\overline{VL})_{(b)}=\left\\{{R\cdot q_{2}\over
m^{2}},0\right\\}\,,~{}~{}~{}(\overline{LL})_{(b)}={1\over m^{2}}\,,$ (96)

we find the reduction relation

$\displaystyle I_{2}^{(1);m_{1}=m_{2},q_{2}^{2}=0}={(D-2)R\cdot q_{2}\over
2m^{2}}I_{2;\widehat{1}}\,,$ (97)

where $I_{2;\widehat{1}}$ is just the tadpole $I_{1}[m]$ with mass $m$.

  * •

$\widetilde{Q}L\neq 0$: Scalar bubble

We then consider the scalar bubble $I_{2}^{(1)}$ with degenerate $Q$ but
$m_{1}\not=m_{2}$. Here we will show that it can be reduced to two tadpoles
using our method. The equation $\mathop{\rm det}Q=0$ gives two solutions

$\displaystyle q_{2}^{2}=(m_{1}\pm m_{2})^{2}\,,~{}~{}~{}$ (98)

which are just the poles in the reduction coefficients of bubbles (see
(65))666The condition is the threshold (or the Landau poles) of the integrals.
It appears is because with higher power of propagators, the singularity of
Landau poles becomes higher. Thus, it will appear in both master integrals and
the reduction coefficients.. Here we choose

$\displaystyle\widetilde{Q}=Q^{*}=\left(\begin{array}[]{cc}m_{2}^{2}&\pm
m_{1}m_{2}\\\ \pm m_{1}m_{2}&m_{1}^{2}\\\ \end{array}\right)\,.$ (101)

One can check that $\widetilde{Q}L\not=0$ for $m_{1}\not=m_{2}$. Due to
$\dot{(\overline{LL})}\not=0$, we can use (73) to reduce $E_{2,2-D}$ in the
expansion

$\displaystyle I_{2}=\Gamma(2-D/2)E_{2,2-D}\,,$ (102)

and we get

$\displaystyle
E_{2,2-D}={-\dot{(\overline{H_{b}L})}\over(3-D)\dot{(\overline{LL})}}E_{1,3-D}^{(b)}={\dot{(\overline{H_{b}L})}{\beta}_{1,3-D}\over(D-3)\dot{(\overline{LL})}}(\overline{LL})_{(b)}E_{1,1-D}^{(b)}\,.$
(103)

Plugging (103) into (102), we have

$\displaystyle
I_{2}^{m_{1}=m_{2},q_{2}^{2}=0}={(D-2)\dot{(\overline{H_{b}L})}(\overline{LL})_{(b)}\over
2(D-3)\dot{(\overline{LL})}}I_{2;\widehat{b}}\,.$ (104)

One can find the explicit expression easily. For
$q_{2}^{2}=(m_{1}+m_{2})^{2}$, the explicit expression is

$\displaystyle
I_{2}^{m_{1}=m_{2},q_{2}^{2}=0}=\frac{D-2}{2(D-3)m_{2}\left(m_{1}+m_{2}\right)}I_{2;\widehat{1}}+\frac{D-2}{2(D-3)m_{1}\left(m_{1}+m_{2}\right)}I_{2;\widehat{2}}\,.$
(105)

  * •

$\widetilde{Q}L\neq 0$: Tensor bubble

Here we discuss the reduction of the rank-1 bubble with $q_{2}^{2}=(m_{1}\pm
m_{2})^{2}$. First, we expand it using (25)

$\displaystyle I_{2}^{(1)}=-{\Gamma(1-D/2)}(D-2)E_{2,2-D}[V]\,.$ (106)

Choosing $\widetilde{Q}=Q^{*}$, due to $\dot{(\overline{LL})}=LQ^{*}L\not=0$,
we can use (73) to reduce the RHS to

$\displaystyle
E_{2,2-D}[V]={1\over(D-2)\dot{(\overline{LL})}}\left[\dot{(\overline{H_{b}L})}E_{1,3-D}^{(b)}[V]+\dot{(\overline{VL})}E_{2,2-D}\right]\,,$
(107)

where the second term has been discussed in the last example. We just need to
reduce the first term $E_{1,3-D}^{(b)}[V]$, which corresponds to a tadpole.
Using (32), we find

$\displaystyle
E_{1,3-D}^{(b)}[V]={\beta}_{1,3-D}(\overline{VL})_{(b)}E_{1,1-D}^{(b)}\,,$
(108)

where the RHS is a $D$-dimensional scalar tadpole. Plugging (108),(107) into
(106) we get

$\displaystyle I_{2}^{(1);q_{2}^{2}=(m_{1}\pm
m_{2})^{2}}={\dot{(\overline{H_{b}L})}(\overline{VL})_{(b)}\over\dot{(\overline{LL})}}I_{2;\widehat{b}}+{2\over
D-2}{\dot{(\overline{VL})}\over\dot{(\overline{LL})}}I_{2}^{q_{2}^{2}=(m_{1}\pm
m_{2})^{2}}\,.$ (109)

Then we refer to the result (104), and find

$\displaystyle I_{2}^{(1);q_{2}^{2}=(m_{1}\pm
m_{2})^{2}}={\dot{(\overline{H_{b}L})}(\overline{VL})_{(b)}\over\dot{(\overline{LL})}}I_{2;\widehat{b}}+{\dot{(\overline{H_{b}L})}(\overline{LL})_{(b)}\dot{(\overline{VL})}\over(D-3)\dot{(\overline{LL})}^{2}}I_{2;\widehat{b}}\,.$
(110)

For $q_{2}^{2}=(m_{1}+m_{2})^{2}$, the explicit expression is

$\displaystyle
I_{2}^{(1);q_{2}^{2}=(m_{1}+m_{2})^{2}}=\frac{\left((D-2)m_{1}+(D-3)m_{2}\right)R\cdot
q_{2}}{(D-3)m_{2}\left(m_{1}+m_{2}\right){}^{2}}I_{2;\widehat{1}}+\frac{R\cdot
q_{2}}{(D-3)\left(m_{1}+m_{2}\right){}^{2}}I_{2;\widehat{2}}\,.$ (111)

##  5 General expression of $C^{(r)}_{n\to n}$

It seems hard to solve (32), but if we only care the reduction coefficients to
the same topology, we can ignore the first term for it contributes only to
lower topologies. Then by iteratively using (32), keeping only the second and
the third term, one find that

$\displaystyle E_{n,k}[V^{i}]=\sum_{j}{i!\over
j!(i-j)!!}(\overline{VV})^{{i-j\over
2}}(\overline{VL})^{j}{\prod_{l=1}^{j}(k-i-l+1)\over\prod_{l=1}^{{i+j\over
2}}(k+n-2l)}E_{n,k-i-j}+\textrm{Lower topology}\,.~{}~{}~{}$ (112)

Although the first term has the same topology, with different choices of
$i,j$, the dimension is different, thus we need to reduce it further.

To simplify our denotation, we define

$\displaystyle\mathscr{E}_{n,k,i;j}={i!\over
j!(i-j)!!}{(k-i)!(k+n-i-j-2)!!\over(k-i-j)!(k+n-2)!!}\,.$ (113)

Thus (112) becomes

$\displaystyle
E_{n,k}[V^{i}]=\sum_{j}\mathscr{E}_{n,k,i;j}(\overline{VV})^{{i-j\over
2}}(\overline{VL})^{j}E_{n,k-i-j}+\textrm{Lower topology}\,.$ (114)

Here we give the computation for tensor reduction with all propagators power
one777The general results for arbitrary tensor structure, general powers as
well as the coefficients for lower topologies are given in Appendix C.. Thus,
we need to reduce all $E_{n,n-D-(r-i)}[V^{i}]$ to $E_{n,n-D}$. After using
(34) to repeatedly lift $k$ we have

$\displaystyle E_{n,k-2s}$ $\displaystyle=$
$\displaystyle{E_{n,k}\over(\overline{LL})^{s}\prod_{l=1}^{s}({\beta}_{n,k-2s+2l})}+\sum_{l=1}^{s}{-{\alpha}_{n,k-2s+2l}(\overline{H_{j}L})\over\prod_{p=1}^{l}{\beta}_{n,k-2s+2p}(\overline{LL})^{l}}E^{(j)}_{n-1,k-1-2s+2l}$
(115) $\displaystyle=$
$\displaystyle{\mathscr{K}^{+}_{n,k,s}\over(\overline{LL})^{s}}E_{n,k}+\sum_{j=1}^{s}{\mathscr{K}^{+}_{n,k,s;j}\over(\overline{LL})^{j}}(\overline{H_{b}L})E^{(b)}_{n-1,k-1-2(s-j)}\,,~{}~{}~{}~{}$

where we have defined

$\displaystyle\mathscr{K}^{+}_{n,k,s}={1\over\prod_{l=1}^{s}({\beta}_{n,k-2s+2l})}\,,~{}~{}~{}~{}~{}\mathscr{K}^{+}_{n,k,s;j}={-{\alpha}_{n,k-2s+2j}\over\prod_{p=1}^{j}{\beta}_{n,k-2s+2p}}\,.$
(116)

Since the second term in (115) belongs to lower topologies, which can be
ignored, finally we have

$\displaystyle
E_{n,n-D-(r-i)}[V^{i}]=\sum_{j}\mathscr{E}_{n,n-D-(r-i),i;j}(\overline{VV})^{i-j\over
2}(\overline{VL})^{j}E_{n,n-D-(r+j)}+Lower$ (117) $\displaystyle=$
$\displaystyle\sum_{j}\mathscr{E}_{n,n-D-(r-i),i;j}(\overline{VV})^{i-j\over
2}(\overline{VL})^{j}{\mathscr{K}^{+}_{n,n-D,{r+j\over
2}}\over(\overline{LL})^{r+j\over 2}}E_{n,n-D}+Lower$ $\displaystyle=$
$\displaystyle\sum_{j}\mathscr{E}_{n,n-D-(r-i),i;j}(\overline{VV})^{i-j\over
2}(\overline{VL})^{j}{\mathscr{K}^{+}_{n,n-D,{r+j\over
2}}\over(\overline{LL})^{r+j\over
2}}{(-)^{n}\over\Gamma(n-D/2)}I_{n;D}+Lower\,.$

So

$\displaystyle I_{n;D}^{(r)}$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{r}{\Gamma(n-D/2-r)\over(-1)^{n+r}}\mathscr{C}^{D/2+r-n}_{r,i}(R^{2})^{r-i\over
2}E_{n,n-D-(r-i)}[V^{i}]$ (118) $\displaystyle=$
$\displaystyle\sum_{i=0}^{r}{\Gamma(n-D/2-r)\over(-1)^{n+r}}\mathscr{C}^{D/2+r-n}_{r,i}(R^{2})^{r-i\over
2}\times$
$\displaystyle\sum_{j}\mathscr{E}_{n,n-D-(r-i),i;j}(\overline{VV})^{i-j\over
2}(\overline{VL})^{j}{\mathscr{K}^{+}_{n,n-D,{r+j\over
2}}\over(\overline{LL})^{r+j\over 2}}{(-)^{n}\over\Gamma(n-D/2)}I_{n;D}+Lower$
$\displaystyle=$
$\displaystyle{(-)^{r}\Gamma(n-D/2-r)\over\Gamma(n-D/2)}\sum_{i=0}^{r}\sum_{j=0}^{i}\mathscr{C}^{D/2+r-n}_{r,i}\mathscr{E}_{n,n-D-(r-i),i;j}\mathscr{K}^{+}_{n,n-D,{r+j\over
2}}\times$ $\displaystyle(R^{2})^{r-i\over 2}{(\overline{VV})^{i-j\over
2}(\overline{VL})^{j}\over(\overline{LL})^{r+j\over 2}}I_{n;D}+Lower\,.$

From it, we read out reduction coefficient

$\displaystyle C^{(r)}_{n\to n}$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{r}\sum_{j=0}^{i}c^{(r)}_{n\to
n;i,j}(R^{2})^{r-i\over 2}{(\overline{VV})^{i-j\over
2}(\overline{VL})^{j}\over(\overline{LL})^{r+j\over 2}}$ (119)

with

$\displaystyle c^{(n)}_{n\to
n;i,j}={(-)^{r}\Gamma(n-D/2-r)\over\Gamma(n-D/2)}\mathscr{C}^{D/2+r-n}_{r,i}\mathscr{E}_{n,n-D-(r-i),i;j}\mathscr{K}^{+}_{n,n-D,{r+j\over
2}}\,,$ (120)

where we require $r,i,j$ having the same parity.

##  6 Discussion

Although the main target of the paper Arkani-Hamed:2017ahv  is to understand
general one-loop integrals from a geometric point of view, it does contain
many important and valuable results. In this paper, we have elaborated on the
method to compute reduction coefficients for general one-loop integrals.

The essential idea of the method is to put the whole one-loop Feynman
integrals in the projective space, in which integrals have compact forms and
geometry properties. Using a vital recursion relation of $E_{n,k}[T]$, one
achieves the wanted reduction. The advantage and the most promising point of
this method is that we can solve any one-loop integrals with higher poles and
tensor structures at the same time, which is demonstrated by some examples in
section 3 and 4. The language of projective space simplifies the reduction
process a lot by keeping the elegant contractions like
$(\overline{HL}),(\overline{LL})$ in these recursion relations without
expansion, thus making the whole reduction process a symbolic calculation. For
some programs based on the traditional IBP method, like FIRE, LiteRed,Kira,
etc Smirnov:2008iw ; Smirnov:2014hma ; Lee:2013mka ; Smirnov:2019qkx ;
vonManteuffel:2012np ; Maierhofer:2017gsa ; Gerlach:2022qnc , they do
reduction by solving linear equations where the determinant of the Gram matrix
appears with the full-expanded form, which makes the final results too
complicated to read. The appearance of (reduced) Gram matrix has also been
observed in our recent work Feng:2021enk ; Hu:2021nia .

One obvious idea is to generalize the above method to the reduction of two-
loops and higher-loops. Recently, using the improved PV-reduction method with
auxiliary vectors, we have shown how to do the general tensor reduction for
two-loop sunset integrals Feng:2022iuc . From the results in Feng:2021enk ;
Hu:2021nia  and results in this paper, we see that these two methods have some
correspondences. In other words, they treat the same thing from different but
related angles. Since the improved PV-reduction method has some results for
two-loop reductions, the projective space method should be generalized to two-
loop too.

## Acknowledgments

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

##  Appendix A Recursion relations of $E_{n,k}[T]$

In this section, we recall the proof of the relation (27) given in Arkani-
Hamed:2017ahv . Let us calculate directly

$\displaystyle\text{d}_{X}\left[{\langle
Q^{-1}T[X^{k-1}]X\text{d}^{n-2}X\rangle\over(XQX)^{{n+k-2\over
2}}}\right]={1\over(XQX)^{{n+k-2\over 2}}}\times$ (121)
$\displaystyle\Big{[}-(n+k-2){(XQ\text{d}X)\over(XQX)}\langle
Q^{-1}T[X^{k-1}]X\text{d}^{n-2}X\rangle+(n-1)\langle
Q^{-1}T[X^{k-1}]\text{d}^{n-1}X\rangle$ $\displaystyle+$
$\displaystyle(k-1)X_{I_{1}}{1\over(n-2)!}Q^{-1}_{I_{1}i_{1}}T^{i_{1}i_{2}i_{3}\ldots
i_{k}}X_{i_{2}}X_{i_{3}}\cdots\text{d}X_{i_{k}}X_{I_{2}}\wedge\text{d}^{n-2}X\epsilon^{I_{1}I_{2}\ldots
I_{k}}\Big{]}\,,~{}~{}~{}$

where we require $T$ is completely symmetric of the last $(k-1)$ indices
$i_{2}i_{3}\ldots i_{k}$. For simplicity, we denote

$\displaystyle A_{I}=Q^{-1}_{Is}T^{si_{2}i_{3}\ldots
i_{k}}X_{i_{2}}X_{i_{3}}\cdots
X_{i_{k}}\,,~{}~{}~{}~{}B_{Ir}=(Q^{-1}T[X^{k-2}])_{Ir}\,.~{}~{}~{}$ (122)

Thus (121) becomes

$\displaystyle\text{d}_{X}\left[{\langle
Q^{-1}T[X^{k-1}]X\text{d}^{n-2}X\rangle\over(XQX)^{{n+k-2\over
2}}}\right]={1\over(XQX)^{{n+k-2\over 2}}}\times$
$\displaystyle\Big{[}-(n+k-2){(XQ\text{d}X)\over(XQX)}\langle
AX\text{d}^{n-2}X\rangle+(n-1)\langle
A\text{d}^{n-1}X\rangle+(k-1)\text{d}X_{I}\langle B^{\
I}X\text{d}^{n-2}X\rangle\Big{]}\,.~{}~{}~{}~{}~{}~{}$ (123)

Using the fact

$\displaystyle XQA=XQQ^{-1}T[X^{k-1}]=T[X^{k}]\,,~{}~{}~{}~{}B_{I}^{\
I}=(tr_{Q}T)[X^{k-2}]\,,$ (124)

and the Schouten identity

$\displaystyle\text{d}X^{I}\langle AX\text{d}X^{n-2}\rangle+A^{I}\langle
X\text{d}^{n-1}X\rangle-X^{I}\langle
A\text{d}^{n-1}X\rangle=0\,,~{}~{}~{}~{}\forall A^{I}$ (125)

to simplify the first term in (123) as

$\displaystyle XQ\text{d}X\langle
AX\text{d}^{n-2}X\rangle=QX\text{d}X\left(-A\langle
X\text{d}^{n-1}X\rangle+X\langle A\text{d}^{n-1}X\rangle\right)$
$\displaystyle=-(XQA)\langle X\text{d}^{n-1}X\rangle+(QXX)\langle
A\text{d}^{n-1}X\rangle\,,$ (126)

and the third term in (123) as

$\displaystyle\text{d}X_{I}\langle B^{\ I}X\text{d}^{n-2}X\rangle=X^{I}\langle
B_{\ I}\text{d}^{n-1}X\rangle-B_{I}^{\ I}\langle X\text{d}^{n-1}X\rangle\,,$
(127)

the (123) becomes

$\displaystyle\text{d}_{X}\left[{\langle
Q^{-1}T[X^{k-1}]X\text{d}^{n-2}X\rangle\over(XQX)^{{n+k-2\over
2}}}\right]=(n+k-2){T[X^{k}]\langle
X\text{d}^{n-1}X\rangle\over(XQX)^{{n+k\over
2}}}-(k-1)tr_{Q}T[X^{k-2}]{\langle{X}\mathrm{d}^{n-1}{X}\rangle}\,,~{}~{}~{}~{}~{}~{}$
(128)

which is nothing but the wanted (27).

The derivation above can be generalized to the case ${\rm det}Q=0$, where the
$Q^{-1}$ does not exist. In this case, we can replace $Q^{-1}$ by arbitrary
matrix $\widetilde{Q}$ in (121) and repeat the same derivations to reach the
similar expression likes (128), except $Q^{-1}$ replaced by $\widetilde{Q}$.
Rearranging (128) we get

$\displaystyle
E_{n,k}[(Q\widetilde{Q}T)]=\alpha_{n,k}E^{(b)}_{n-1,k-1}[(H_{b}\widetilde{Q}T)]+\beta_{n,k}E_{n,k-2}[tr_{\widetilde{Q}}T]\,,~{}~{}~{}~{}$
(129)

where $(Q\widetilde{Q}T)$ has the same rank as $T$

$\displaystyle(Q\widetilde{Q}T)^{I_{1}I_{2},\ldots,I_{k}}=Q^{I_{1}J_{1}}\widetilde{Q}_{J_{1}J_{2}}T^{J_{2},I_{2},I_{3},\ldots,I_{k}}\,,~{}~{}~{}~{}(tr_{\widetilde{Q}}T)^{I_{2},I_{3},\ldots,I_{k}}=\widetilde{Q}_{I_{1}I_{2}}T^{I_{1}I_{2}\ldots
I_{k}}\,.$ (130)

In (129), the LHS is the first term in RHS of (128), while the first term on
the RHS is the boundary contribution of LHS of (128).

##  Appendix B More results

###  B.1 Bubbles

Here list the results for rank from $r=1$ to $r=4$, where we set $q_{1}=0$.

  * •

$r=1$

$\displaystyle
I_{2}^{(1)}=\sum_{i}-\frac{(\overline{LL})(\overline{H_{i}V})-(\overline{H_{i}L})(\overline{VL})}{(\overline{LL})}I_{{2;\widehat{i}}}+\frac{2(\overline{VL})}{(\overline{LL})}I_{2}\,.$
(131)

So we have

$\displaystyle C^{(1)}_{2\to 2;\widehat{2}}$ $\displaystyle=$
$\displaystyle-\frac{(\overline{LL})(\overline{H_{2}V})-(\overline{H_{2}L})(\overline{VL})}{(\overline{LL})}=-\frac{R\cdot
q_{2}}{q_{2}^{2}}\,,$ $\displaystyle C^{(1)}_{2\to 2;\widehat{1}}$
$\displaystyle=$
$\displaystyle-\frac{(\overline{LL})(\overline{H_{1}V})-(\overline{H_{1}L})(\overline{VL})}{(\overline{LL})}=\frac{R\cdot
q_{2}}{q_{2}^{2}}\,,$ $\displaystyle C^{(1)}_{2\to 2}$ $\displaystyle=$
$\displaystyle\frac{2\left(VQ^{-1}L\right)}{\left(LQ^{-1}L\right)}=\frac{\left(m_{1}^{2}-m_{2}^{2}+q_{2}^{2}\right)R\cdot
q_{2}}{q_{2}^{2}}\,.~{}~{}~{}~{}$ (132)

  * •

$r=2$

$\displaystyle C^{(2)}_{2\to 2;{\widehat{i}}}$ $\displaystyle=$
$\displaystyle\frac{2R^{2}(\overline{H_{i}L})}{(D-1)(\overline{LL})}+\frac{2D(\overline{H_{i}L})(\overline{VL})^{2}}{(D-1)(\overline{LL})^{2}}-\frac{2(\overline{H_{i}V})\left((\overline{VL})+(\overline{VL})_{(i)}\right)}{(\overline{LL})_{(i)}}$
$\displaystyle-\frac{2(\overline{H_{i}L})\big{(}-D(\overline{VL})^{2}+(\overline{VL})^{2}+(\overline{LL})_{(i)}(\overline{VV})\big{)}}{(D-1)(\overline{LL})_{(i)}(\overline{LL})}\,,$
$\displaystyle C^{(2)}_{2\to 2}$ $\displaystyle=$
$\displaystyle\frac{4\left(R^{2}-(\overline{VV})\right)}{(D-1)(\overline{LL})}+\frac{4D(\overline{VL})^{2}}{(D-1)(\overline{LL})^{2}}\,.~{}~{}~{}~{}$
(133)

the exact form is

$\displaystyle C^{(2)}_{2\to 2;\widehat{2}}$ $\displaystyle=$
$\displaystyle\frac{\left(m_{1}^{2}-m_{2}^{2}+q_{2}^{2}\right)R^{2}}{(D-1)q_{2}^{2}}-\frac{D\left(m_{1}^{2}-m_{2}^{2}+q_{2}^{2}\right)\left(R\cdot
q_{2}\right){}^{2}}{(D-1)q_{2}^{4}}\,,$ $\displaystyle C^{(2)}_{2\to
2;\widehat{1}}$ $\displaystyle=$
$\displaystyle-\frac{R^{2}\left(m_{1}^{2}-m_{2}^{2}-q_{2}^{2}\right)}{(D-1)q_{2}^{2}}-\frac{\left(-Dm_{1}^{2}+Dm_{2}^{2}-3Dq_{2}^{2}+4q_{2}^{2}\right)\left(R\cdot
q_{2}\right){}^{2}}{(D-1)q_{2}^{4}}\,,$ $\displaystyle C^{(2)}_{2\to 2}$
$\displaystyle=$
$\displaystyle-\frac{R^{2}\left(-2m_{1}^{2}\left(m_{2}^{2}+q_{2}^{2}\right)+\left(m_{2}^{2}-q_{2}^{2}\right){}^{2}+m_{1}^{4}\right)}{(D-1)q_{2}^{2}}$
$\displaystyle+\frac{\left(-2m_{1}^{2}\left(Dm_{2}^{2}-(D-2)q_{2}^{2}\right)+D\left(m_{2}^{2}-q_{2}^{2}\right){}^{2}+Dm_{1}^{4}\right)\left(R\cdot
q_{2}\right){}^{2}}{(D-1)q_{2}^{4}}\,.$

  * •

$r=3$

$\displaystyle C^{(3)}_{2\to 2;{\widehat{i}}}$
$\displaystyle=-\frac{4(D+1)(\overline{H_{i}V})\left((\overline{VL})_{(i)}(\overline{VL})+(\overline{VL})^{2}+(\overline{VL})_{(i)}^{2}\right)}{D(\overline{LL})_{(i)}^{2}}$
$\displaystyle{\hskip
15.0pt}-\frac{4(\overline{H_{i}V})\left(-2(\overline{VV})-(\overline{VV})_{(i)}+3R^{2}\right)}{D(\overline{LL})_{(i)}}+\frac{4(D+2)(\overline{H_{i}L})(\overline{VL})^{3}}{(D-1)(\overline{LL})^{3}}$
$\displaystyle{\hskip
15.0pt}+\frac{4(\overline{H_{i}L})(\overline{VL})\left(D(\overline{VL})^{2}+(\overline{VL})^{2}-3(\overline{LL})_{(i)}(\overline{VV})+3R^{2}(\overline{LL})_{(i)}\right)}{D(\overline{LL})_{(i)}^{2}(\overline{LL})}$
$\displaystyle{\hskip
15.0pt}+\frac{4(\overline{H_{i}L})(\overline{VL})}{(D-1)D(\overline{LL})_{(i)}(\overline{LL})^{2}}\Big{(}D^{2}(\overline{VL})^{2}+D(\overline{VL})^{2}-3D(\overline{LL})_{(i)}(\overline{VV})$
$\displaystyle{\hskip 15.0pt}{\hskip
15.0pt}-2(\overline{VL})^{2}+3DR^{2}(\overline{LL})_{(i)}\Big{)}\,,$
$\displaystyle C^{(3)}_{2\to 2}$
$\displaystyle=\frac{24R^{2}(\overline{VL})}{(D-1)(\overline{LL})^{2}}+\frac{8(D+2)(\overline{VL})^{3}}{(D-1)(\overline{LL})^{3}}-\frac{24(\overline{VV})(\overline{VL})}{(D-1)(\overline{LL})^{2}}\,.$
(134)

The exact form is

$\displaystyle C^{(3)}_{2\to 2;\widehat{2}}$
$\displaystyle=-\frac{(D+2)\left(m_{1}^{2}-m_{2}^{2}\right){}^{2}\left(R\cdot
q_{2}\right){}^{3}}{(D-1)q_{2}^{6}}+\frac{3R^{2}R\cdot q_{2}}{D-1}$
$\displaystyle{\hskip
15.0pt}+\frac{\left(2D(D+2)m_{2}^{2}-2\left(D^{2}-2D+4\right)m_{1}^{2}\right)\left(R\cdot
q_{2}\right){}^{3}+3D\left(m_{1}^{2}-m_{2}^{2}\right){}^{2}R^{2}R\cdot
q_{2}}{(D-1)Dq_{2}^{4}}$ $\displaystyle{\hskip 15.0pt}-\frac{R\cdot
q_{2}\left(6R^{2}\left((D-2)m_{1}^{2}+Dm_{2}^{2}\right)+D(D+2)\left(R\cdot
q_{2}\right){}^{2}\right)}{(D-1)Dq_{2}^{2}}\,,$ $\displaystyle C^{(3)}_{2\to
2;\widehat{1}}$
$\displaystyle=\frac{3R^{2}\left(4(D-1)m_{2}^{2}q_{2}^{2}-Dm_{1}^{4}+2Dm_{2}^{2}m_{1}^{2}-Dm_{2}^{4}+Dq_{2}^{4}\right)R\cdot
q_{2}}{(D-1)Dq_{2}^{4}}$ $\displaystyle{\hskip 15.0pt}\frac{\left(R\cdot
q_{2}\right){}^{3}}{(D-1)Dq_{2}^{6}}\Big{[}-4\left(D^{2}+D-2\right)m_{2}^{2}q_{2}^{2}-2Dm_{1}^{2}\left((D+2)m_{2}^{2}-2(D-1)q_{2}^{2}\right)$
$\displaystyle{\hskip
15.0pt}+D(D+2)m_{1}^{4}+D(D+2)m_{2}^{4}+D(7D-10)q_{2}^{4}\Big{]}\,,$
$\displaystyle C^{(3)}_{2\to 2}$
$\displaystyle=\frac{\left(m_{1}^{2}-m_{2}^{2}+q_{2}^{2}\right)\left(R\cdot
q_{2}\right){}^{3}}{(D-1)q_{2}^{6}}\Big{[}-2m_{1}^{2}\left((D+2)m_{2}^{2}-(D-4)q_{2}^{2}\right)$
$\displaystyle{\hskip
15.0pt}+(D+2)\left(m_{2}^{2}-q_{2}^{2}\right){}^{2}+(D+2)m_{1}^{4}\Big{]}$
$\displaystyle{\hskip
15.0pt}-\frac{3R^{2}\left(m_{1}^{2}-m_{2}^{2}+q_{2}^{2}\right)\left(-2m_{1}^{2}\left(m_{2}^{2}+q_{2}^{2}\right)+\left(m_{2}^{2}-q_{2}^{2}\right){}^{2}+m_{1}^{4}\right)R\cdot
q_{2}}{(D-1)q_{2}^{4}}\,.$ (135)

  * •

$r=4$

$\displaystyle C^{(4)}_{2\to 2;\widehat{i}}$
$\displaystyle=\frac{8(D+2)(\overline{H_{i}L})(\overline{VL})^{2}\left(\left(D^{2}+3D-4\right)(\overline{VL})^{2}+6D(\overline{LL})_{(i)}\left(R^{2}-(\overline{VV})\right)\right)}{(D-1)D(D+1)(\overline{LL})_{(i)}(\overline{LL})^{3}}$
$\displaystyle{\hskip
15.0pt}+\frac{8(\overline{H_{i}L})}{D(D+1)(\overline{LL})_{(i)}^{3}(\overline{LL})}\Big{(}\left(D^{2}+4D+3\right)(\overline{VL})^{4}+3(\overline{LL})_{(i)}^{2}\left(R^{2}-(\overline{VV})\right){}^{2}$
$\displaystyle{\hskip
15.0pt}+6(D+1)(\overline{LL})_{(i)}(\overline{VL})^{2}\left(R^{2}-(\overline{VV})\right)\Big{)}+\frac{8(\overline{H_{i}L})}{D\left(D^{2}-1\right)(\overline{LL})_{(i)}^{2}(\overline{LL})^{2}}\times$
$\displaystyle{\hskip
15.0pt}\Big{(}\left(D^{3}+4D^{2}-D-4\right)(\overline{VL})^{4}+6\left(D^{2}+D-2\right)(\overline{LL})_{(i)}(\overline{VL})^{2}\left(R^{2}-(\overline{VV})\right)$
$\displaystyle{\hskip
15.0pt}+3D(\overline{LL})_{(i)}^{2}\left(R^{2}-(\overline{VV})\right){}^{2}\Big{)}+\frac{8(D+2)(D+4)(\overline{H_{i}L})(\overline{VL})^{4}}{\left(D^{2}-1\right)(\overline{LL})^{4}}$
$\displaystyle{\hskip
15.0pt}-\frac{8(\overline{H_{i}V})}{D(\overline{LL})_{(i)}^{2}}\Big{(}(\overline{VL})\left(-5(\overline{VV})-(\overline{VV})_{(i)}+6R^{2}\right)+(\overline{VL})_{(i)}\left(-3(\overline{VV})-3(\overline{VV})_{(i)}+6R^{2}\right)\Big{)}$
$\displaystyle{\hskip
15.0pt}-\frac{8(D+3)\left((\overline{VL})+(\overline{VL})_{(i)}\right)\left((\overline{VL})^{2}+(\overline{VL})_{(i)}^{2}\right)(\overline{H_{i}V})}{D(\overline{LL})_{(i)}^{3}}\,,$
$\displaystyle C^{(4)}_{2\to 2}$
$\displaystyle=\frac{96(D+2)(\overline{VL})^{2}\left(R^{2}-(\overline{VV})\right)}{\left(D^{2}-1\right)(\overline{LL})^{3}}+\frac{48\left(R^{2}-(\overline{VV})\right){}^{2}}{\left(D^{2}-1\right)(\overline{LL})^{2}}+\frac{16\left(D^{2}+6D+8\right)(\overline{VL})^{4}}{\left(D^{2}-1\right)(\overline{LL})^{4}}\,.$
(136)

###  B.2 Triangles

For triangle topology, we have presented results for scalar triangles with
higher poles. Here we present some examples, including the tensor triangles
without higher poles and with higher poles.

For the tensor triangles without higher poles, we have:

  * •

$r=1$:

$\displaystyle C^{(1)}_{{3}\to{3;\widehat{i}}}$ $\displaystyle=$
$\displaystyle\frac{(\overline{H_{i}L})(\overline{VL})}{(\overline{LL})}-(\overline{H_{i}V})\,,$
$\displaystyle C^{(1)}_{{3}\to{3}}$ $\displaystyle=$
$\displaystyle\frac{2(\overline{VL})}{(\overline{LL})}\,.~{}~{}~{}~{}$ (137)

The expression is the same as (132) for bubble topology. This phenomenon is
not an accident, and will be persistent to other topologies.

  * •

$r=2$:

$\displaystyle C^{(2)}_{{3}\to{3;\widehat{ij}}}$ $\displaystyle=$
$\displaystyle\frac{(\overline{H_{i}L})(\overline{H_{j}L})_{(i)}(\overline{VL})^{2}}{(\overline{LL})_{(i)}(\overline{LL})}-\frac{(\overline{H_{j}L})_{(i)}(\overline{H_{i}V})\left((\overline{VL})_{(i)}+(\overline{VL})\right)}{(\overline{LL})_{(i)}}$
$\displaystyle+(\overline{H_{i}V})(\overline{H_{j}V})_{(i)}+(i\leftrightarrow
j)\,,$ $\displaystyle C^{(2)}_{{3}\to{3;\widehat{i}}}$ $\displaystyle=$
$\displaystyle\frac{2(\overline{H_{i}L})\left((D-2)(\overline{VL})^{2}+(\overline{LL})_{(i)}\left(R^{2}-(\overline{VV})\right)\right)}{(D-2)(\overline{LL})_{(i)}(\overline{LL})}$
$\displaystyle+\frac{2(D-1)(\overline{H_{i}L})(\overline{VL})^{2}}{(D-2)(\overline{LL})^{2}}\\-\frac{2(\overline{H_{i}V})\left((\overline{VL})_{(i)}+(\overline{VL})\right)}{(\overline{LL})_{(i)}}\,,$
$\displaystyle C^{(2)}_{{3}\to{3}}$ $\displaystyle=$
$\displaystyle\frac{4\left((D-1)(\overline{VL})^{2}+(\overline{LL})\left(R^{2}-(\overline{VV})\right)\right)}{(D-2)(\overline{LL})^{2}}\,.~{}~{}~{}~{}$
(138)

Again, the last two coefficients are very similar to these given in (133).

For the tensor with higher poles, we have:

  * •

$v=4,r=1$:

$\displaystyle C^{(1)}_{\mathbf{v}_{3}\to 3;\widehat{ij}}$ $\displaystyle=$
$\displaystyle\frac{1}{8}(D-2)(\overline{LL})_{(ij)}\left((\overline{H_{i}S})(\overline{H_{j}S})_{(i)}+(i\leftrightarrow
j)\right)\,,$ $\displaystyle C^{(1)}_{\mathbf{v}_{3}\to 3;\widehat{i}}$
$\displaystyle=$ $\displaystyle{D-3\over
4}\left(\frac{(\overline{H_{i}L})(\overline{SL})^{2}}{(\overline{LL})}-(\overline{H_{i}S})(\overline{SL})_{(i)}-(\overline{H_{i}S})(\overline{SL})\right)\,,$
$\displaystyle C^{(1)}_{\mathbf{v}_{3}\to 3}$ $\displaystyle=$
$\displaystyle\frac{(D-3)(\overline{SL})^{2}}{2(\overline{LL})}-\frac{1}{2}(\overline{SS})\,.~{}~{}~{}~{}$
(139)

Here we have suppressed the process to take the coefficient of
$tz^{\mathbf{v}_{3}-1}$,i.e, $|_{tz^{\mathbf{v}_{3}-1}}$. The similarity of
the above expression with the one given in (59) is obvious.

  * •

$v=4,r=2$:

$\displaystyle C^{(2)}_{\mathbf{v}_{3}\to 3;\widehat{ij}}$ $\displaystyle=$
$\displaystyle\frac{(D-2)(\overline{H_{i}L})(\overline{H_{j}L})_{(i)}(\overline{SL})^{3}}{6(\overline{LL})_{(i)}(\overline{LL})}-\frac{(D-2)(\overline{H_{j}L})_{(i)}(\overline{H_{i}S})\left((\overline{SL})_{(i)}^{2}+(\overline{SL})(\overline{SL})_{(i)}+(\overline{SL})^{2}\right)}{6(\overline{LL})_{(i)}}$
$\displaystyle+\frac{1}{6}(D-2)(\overline{H_{i}S})(\overline{H_{j}S})_{(i)}\left((\overline{SL})_{(i)}+(\overline{SL})+(\overline{SL})_{(ij)}\right)+(i\leftrightarrow
j)\,,$ $\displaystyle C^{(2)}_{\mathbf{v}_{3}\to 3;\widehat{i}}$
$\displaystyle=$
$\displaystyle\frac{(\overline{H_{i}L})(\overline{SL})\left((D-2)(\overline{SL})^{2}+3(\overline{LL})_{(i)}\left(R^{2}-(\overline{SS})\right)\right)}{3(\overline{LL})_{(i)}(\overline{LL})}+\frac{(D-1)(\overline{H_{i}L})(\overline{SL})^{3}}{3(\overline{LL})^{2}}$
$\displaystyle+\frac{(\overline{H_{i}S})\left((\overline{LL})_{(i)}\left((\overline{SS})_{(i)}+2(\overline{SS})-3R^{2}\right)-(D-2)\left((\overline{SL})_{(i)}^{2}+(\overline{SL})(\overline{SL})_{(i)}+(\overline{SL})^{2}\right)\right)}{3(\overline{LL})_{(i)}}\,,$
$\displaystyle C^{(2)}_{\mathbf{v}_{3}\to 3}$ $\displaystyle=$
$\displaystyle\frac{2(D-1)(\overline{SL})^{3}}{3(\overline{LL})^{2}}+\frac{2(\overline{SL})\left(R^{2}-(\overline{SS})\right)}{(\overline{LL})}\,.$
(140)

Here, when calculating the reduction coefficients, we take the coefficient of
$z^{\mathbf{v}_{3}-1}$ for terms contains one $R^{2}$ and one $S$ while we
take the coefficient of $t^{2}z^{\mathbf{v}_{3}-1}$ for terms contains three
$S$’s.

###  B.3 Boxes

For the box topology, we present three cases : (1) $r=1,v-n=0$, (2)
$r=0,v-n=1$, (3) $r=1,v-n=1$. As pointed out in the triangle topology, the
reduction coefficients have some similarity between different topology. In
fact, the similarity is classified by the pair $(r,v-n)$ as one can check by
using the results listed in the Appendix and the main body of the paper.

  * •

$r=1,v-n=0$:

$\displaystyle C^{(1)}_{{4}\to{4;\widehat{i}}}$ $\displaystyle=$
$\displaystyle\frac{(\overline{H_{i}L})(\overline{VL})}{(\overline{LL})}-(\overline{H_{i}V}),$
$\displaystyle C^{(1)}_{{4}\to{4}}$ $\displaystyle=$
$\displaystyle\frac{2(\overline{VL})}{(\overline{LL})}\,.$ (141)

  * •

$r=0,v-n=1$:

$\displaystyle C_{\mathbf{v}_{4}\to 4;\widehat{ijk}}$ $\displaystyle=$
$\displaystyle-\frac{1}{16}(D-2)(\overline{LL})_{(ijk)}\left((\overline{H_{j}L})_{(i)}(\overline{H_{k}L})_{(ij)}(\overline{H_{i}Z})+\text{permutations}\
\text{of}\ (ijk)\right)\,,$ $\displaystyle C_{\mathbf{v}_{4}\to
4;\widehat{ij}}$ $\displaystyle=$
$\displaystyle\frac{1}{8}(D-3)(\overline{LL})_{(ij)}\left((\overline{H_{j}L})_{(i)}(\overline{H_{i}Z})+(i\leftrightarrow
j)\right)\,,$ $\displaystyle C_{\mathbf{v}_{4}\to 4;\widehat{i}}$
$\displaystyle=$
$\displaystyle-\frac{1}{4}(D-4)(\overline{LL})_{(i)}(\overline{H_{i}Z})\,,$
$\displaystyle C_{\mathbf{v}_{4}\to 4}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(D-5)(\overline{ZL})\,.$ (142)

  * •

$r=1,v-n=1$:

$\displaystyle C^{(1)}_{\mathbf{v}_{4}\to 4;\widehat{ijk}}$ $\displaystyle=$
$\displaystyle-\frac{1}{16}(D-2)(\overline{LL})_{(ijk)}\left((\overline{H_{k}L})_{(ij)}(\overline{H_{i}S})(\overline{H_{j}S})_{(i)}+\text{permutations}\
\text{of}\ (ijk)\right)\,,$ $\displaystyle C^{(1)}_{\mathbf{v}_{4}\to
4;\widehat{ij}}$ $\displaystyle=$
$\displaystyle\frac{1}{8}(D-3)(\overline{LL})_{(ij)}\left((\overline{H_{i}S})(\overline{H_{j}S})_{(i)}+(i\leftrightarrow
j)\right)\,,$ $\displaystyle C^{(1)}_{\mathbf{v}_{4}\to 4;\widehat{i}}$
$\displaystyle=$
$\displaystyle-\frac{(D-4)\left((\overline{LL})(\overline{H_{i}S})\left((\overline{SL})_{(i)}+(\overline{SL})\right)-(\overline{H_{i}L})(\overline{SL})^{2}\right)}{4(\overline{LL})}\,,$
$\displaystyle C^{(1)}_{\mathbf{v}_{4}\to 4}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left(\frac{(D-4)(\overline{SL})^{2}}{(\overline{LL})}-(\overline{SS})\right)\,.$
(143)

###  B.4 Pentagons

For pentagon topology, the reduction coefficients are similar, so we just
present one example, i.e., scalar pentagon with higher poles.

  * •

$v=6$

$\displaystyle C_{\mathbf{v}_{5}\to 5;\widehat{ijkl}}$ $\displaystyle=$
$\displaystyle\frac{1}{32}(D-2)(\overline{LL})_{(ijkl)}\Big{(}(\overline{H_{i}L})_{(jk)}(\overline{H_{j}Z})(\overline{H_{k}L})_{(j)}(\overline{H_{l}L})_{(ijk)}+\text{permutations}\
\text{of}\ (ijkl)\Big{)}\,,$ $\displaystyle C_{\mathbf{v}_{5}\to
5;\widehat{ijk}}$ $\displaystyle=$
$\displaystyle-\frac{1}{16}(D-3)(\overline{LL})_{(ijk)}\Big{(}(\overline{H_{i}L})_{(j)}(\overline{H_{k}L})_{(ij)}(\overline{H_{j}Z})+\text{permutations}\
\text{of}\ (ijk)\Big{)}\,,$ $\displaystyle C^{(1)}_{5\to 5;\widehat{ij}}$
$\displaystyle=$
$\displaystyle\frac{1}{8}(D-4)(\overline{LL})_{(ij)}\left((\overline{H_{j}L})_{(i)}(\overline{H_{i}Z})+(i\leftrightarrow
j)\right)\,,$ $\displaystyle C_{\mathbf{v}_{5}\to 5;\widehat{i}}$
$\displaystyle=$
$\displaystyle-\frac{1}{4}(D-5)(\overline{LL})_{(i)}(\overline{H_{i}Z})\,,$
$\displaystyle C_{\mathbf{v}_{5}\to 5}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(D-6)(\overline{ZL})\,.$ (144)

One thing we want to point out is that these coefficients have a manifest
permutation symmetry. Using these observations, the expression can be very
compact, as shown above.

##  Appendix C General expression of reduction coefficients

One can solve the recursion relations (32) in section 3 iteratively888For the
degenerate case discussed in the section 4, one can do similar computation,
although it will be more complicated. and get general expressions for the
final reduction coefficients. Here we present only the final results for non-
degenerate $Q$ without derivation details. We define

$\displaystyle\mathscr{F}^{n,k,i}_{i-n_{v},s}=\mathscr{P}^{k,i}_{i-n_{v}+1,s}\prod_{l=0}^{s}{\alpha}_{n,k-2l}$
(145)

with

$\displaystyle\mathscr{P}^{k,i}_{i-d,s}=(i+1-d)\mathscr{P}^{k,i}_{i-d+2,s-1}+(k-i+1+d-2s)\mathscr{P}^{k,i}_{i-d+1,s-1}\,,~{}~{}~{}$
(146)

where we set $\mathscr{P}^{k,i}_{i-d,s}=0$ for either $d<0$ or $s<0$. With the
initial condition $\mathscr{P}^{k,i}_{0,0}=1$, (146) can be solved. For
simplicity, we define

$\displaystyle
E_{n-l,k-l}^{(\mathbf{b}_{l})}=E_{n-l,k-l}^{(b_{1},b_{2},b_{3},\ldots,b_{l})},~{}~{}~{}~{}~{}(\overline{WV})_{(\mathbf{b}_{l})}=(\overline{WV})_{(b_{1},b_{2},b_{3},\ldots,b_{l})}$
$\displaystyle(\overline{HL})_{\widehat{\mathbf{b}_{l}}}=(\overline{H_{b_{1}}L})(\overline{H_{b_{2}}L})_{(b_{1})}(\overline{H_{b_{3}}L})_{(b_{1}b_{2})}\cdots(\overline{H_{b_{l}}L})_{(\mathbf{b}_{l-1})}$
$\displaystyle(\overline{LL})^{(s;\mathbf{j}_{l})}_{({\mathbf{b}_{l}})}=(\overline{LL})^{j_{1}}(\overline{LL})_{(b_{1})}^{j_{2}}(\overline{LL})_{(b_{1},b_{2})}^{j_{3}}\cdots(\overline{LL})_{(\mathbf{b}_{l-1})}^{j_{l}}(\overline{LL})_{(\mathbf{b}_{l})}^{s-j_{1}-j_{2}-j_{3}-\cdots-
j_{l}}$ (147)

and

$\displaystyle\sum_{\\{\mathbf{l,j,s}\\}_{a}}^{i}$ $\displaystyle=$
$\displaystyle\sum_{l_{1}=0}^{i-1}\sum_{j_{1}=0}^{l_{1}}\sum^{i-l_{1}-1}_{s_{1}=\lceil{i-l_{1}-1\over
2}\rceil}\sum_{l_{2}=0}^{l_{1}-1}\sum_{j_{2}=0}^{l_{2}}\sum^{l_{1}-l_{2}-1}_{s_{2}=\lceil{l_{1}-l_{2}-1\over
2}\rceil}\cdots\sum_{l_{a}=0}^{l_{a-1}-1}\sum_{j_{a}=0}^{l_{a}}\sum^{l_{a-1}-l_{a}-1}_{s_{a}=\lceil{l_{a-1}-l_{a}-1\over
2}\rceil}$ $\displaystyle\mathcal{F}^{n,k,i}_{\\{\mathbf{l,s}\\}_{a}}$
$\displaystyle=$
$\displaystyle\mathscr{F}^{n,k,i}_{l_{1},s_{1}}\mathscr{F}^{n-1,k-1-2s_{1},l_{1}}_{l_{2},s_{2}}\mathscr{F}^{n-2,k-2-2s_{1}-2s_{2},l_{2}}_{l_{3},s_{3}}\cdots\mathscr{F}^{n-a,k-a-2s_{1}-2s_{2}-\ldots-2s_{a-1},l_{a-1}}_{l_{a},s_{a}}$
$\displaystyle\mathcal{S}_{l_{0},\\{\mathbf{l,j,s}\\}_{a};\mathbf{b}_{a}}$
$\displaystyle=$
$\displaystyle\prod_{p=1}^{a}(\overline{SS})_{(\mathbf{b}_{p-1})}^{-1+l_{p-1}-l_{p}-s_{p}}(\overline{SL})_{(\mathbf{b}_{p-1})}^{1-l_{p-1}+l_{p}+2s_{p}}(\overline{SS})_{(\mathbf{b}_{a})}^{l_{a}-j_{a}\over
2}(\overline{SL})_{(\mathbf{b}_{a})}^{j_{a}}\,,$ (148)

where specially

$\displaystyle\sum_{\\{\mathbf{l,j,s}\\}_{0}}^{i}=\sum_{j_{0}=0}^{i},~{}~{}~{}~{}~{}\mathcal{F}^{n,k,i}_{\\{\mathbf{l,s}\\}_{0}}=1,~{}~{}~{}~{}~{}(\overline{HS})_{\widehat{\mathbf{b}_{0}}}=1\,,$
$\displaystyle\mathcal{S}_{i,\\{\mathbf{l,j,s}\\}_{0};\mathbf{b}_{0}}=(\overline{SS})^{{i-j_{0}\over
2}}(\overline{SL})^{j_{0}},~{}~{}~{}~{}~{}\mathscr{E}_{n,k-2\left|\mathbf{s}_{0}\right|,l_{0};j_{0}}=\mathscr{E}_{n,k,i;j_{0}}\,.$
(149)

Then we denote some notations used in dimension shifting as follows

$\displaystyle\mathscr{K}^{-}_{n,k,s}$ $\displaystyle=$
$\displaystyle{1\over\prod_{l=1}^{s}({\beta}_{n,k-2s+2l})},~{}~{}~{}~{}~{}\mathscr{K}^{-}_{n,k,s;j}={-{\alpha}_{n,k-2s+2j}\over\prod_{p=1}^{j}{\beta}_{n,k-2s+2p}}\,,$
$\displaystyle\mathscr{K}^{+}_{n,k,s}$ $\displaystyle=$
$\displaystyle\prod_{p=1}^{s}{\beta}_{n,k+2s+2-2p},~{}~{}~{}~{}~{}~{}\mathscr{K}^{+}_{n,k,s;j}={\alpha}_{n,k+2s-2j}\prod_{p=1}^{j}{\beta}_{n,k+2s+2-2p}\,,$
(150)

and

$\displaystyle\mathcal{K}^{-;\mathbf{b}_{l}}_{n,k,2s;l}=\sum_{j_{1}=1}^{s}\sum_{j_{2}=1}^{s-j_{1}}\sum_{j_{3}=1}^{s-j_{1}-j_{2}}\cdots\sum_{j_{l}=1}^{s-j_{1}-j_{2}-\ldots-
j_{l-1}}{1\over(\overline{LL})^{(s;\mathbf{j}_{l})}_{(\mathbf{b}_{l})}}\times$
$\displaystyle{\mathscr{K}^{-}_{n,k,s;j_{1}}\mathscr{K}^{-}_{n-1,k-1,s-j_{1};j_{2}}\mathscr{K}^{-}_{n-2,k-2,s-j_{1}-j_{2};j_{3}}\cdots\mathscr{K}^{-}_{n-l,k-l,s-j_{1}-j_{2}-j_{3}-\cdots-
j_{l}}}$
$\displaystyle\mathcal{K}^{+;\mathbf{b}_{l}}_{n,k,2s;l}=\sum_{j_{1}=0}^{s}\sum_{j_{2}=0}^{s-j_{1}}\sum_{j_{3}=0}^{s-j_{1}-j_{2}}\cdots\sum_{j_{l}=0}^{s-j_{1}-j_{2}-\ldots-
j_{l-1}}\times$
$\displaystyle{\mathscr{K}^{+}_{n,k,s;j_{1}}\mathscr{K}^{+}_{n-1,k-1,s-j_{1};j_{2}}\mathscr{K}^{+}_{n-2,k-2,s-j_{1}-j_{2};j_{3}}\cdots\mathscr{K}^{+}_{n-l,k-l,s-j_{1}-j_{2}-j_{3}-\cdots-
j_{l}}}$
$\displaystyle\mathcal{K}^{\mathbf{b}_{l}}_{n,k,2s;l}=\mathcal{K}^{Sign(s);\mathbf{b}_{l}}_{n,k,\left|2s\right|;l}\,.$
(151)

The final expression for general reduction coefficients is

$\displaystyle C^{(r)}_{\mathbf{v}_{n}\to n;\widehat{\mathbf{b}_{a}}}$
$\displaystyle=$ $\displaystyle\sum_{a_{1}+a_{2}=a,\atop\sigma\in
S[\mathbf{b}_{a}]}\sigma\Bigg{[}\sum_{i=0}^{r}\sum_{\\{\mathbf{l,j,s}\\}_{a_{1}}}^{\xi(n,v,i)}c^{n,v,r}_{i,a_{1},a_{2},\\{\mathbf{l,j,s}\\}_{a_{1}}}(R^{2})^{{r-i\over
2}}\times$ (152)
$\displaystyle\mathcal{S}_{\xi,\\{\mathbf{l,j,s}\\}_{a_{1}};\mathbf{b}_{a_{1}}}(\overline{HS})_{\widehat{\mathbf{b}_{a_{1}}}}(\overline{HL})_{(\mathbf{b}_{a_{1}})\widehat{\mathbf{b}_{a_{2}}}}\mathcal{K}^{(\mathbf{b}_{a_{1}})\mathbf{b}_{a_{2}}}_{n-a_{1},{-\mu-2\left|\mathbf{s}_{a_{1}}\right|-l_{a_{1}}-j_{a_{1}}};a_{2}}\Bigg{]}\Bigg{|}_{t^{i}z^{\mathbf{v}_{n}-1}}\,,$

where we have defined $\mu(n,v,r,i)=r-i+2n-2v,\xi(n,v,i)=v-n+i$, and the
dimensionless factor is

$\displaystyle c^{n,v,r}_{i,a_{1},a_{2},\\{\mathbf{l,j,s}\\}_{a_{1}}}$
$\displaystyle=$
$\displaystyle{(-)^{r+v+n+a_{1}+a_{2}}\Gamma(v-D/2-r)i!\over\Gamma(n-a_{1}-a_{2}-D/2)(v-n+i)!}\times$
(153)
$\displaystyle\mathcal{F}^{n,n-D-\mu(n,v,r,i),\xi(n,v,i)}_{\\{\mathbf{l,s}\\}_{a_{1}}}\mathscr{E}_{n-a_{1},n-a_{1}-D-\mu(n,v,r,i)-2\left|\mathbf{s}_{a_{1}}\right|,l_{a_{1}};j_{a_{1}}}\,.$

## References

  * (1) N. Arkani-Hamed and E. Y. Yuan, One-Loop Integrals from Spherical Projections of Planes and Quadrics, arXiv:1712.09991.
  * (2) L. M. Brown and R. P. Feynman, Radiative corrections to Compton scattering, Phys. Rev. 85 (1952) 231–244.
  * (3) D. B. Melrose, Reduction of Feynman diagrams, Nuovo Cim. 40 (1965) 181–213.
  * (4) G. Passarino and M. J. G. Veltman, One Loop Corrections for e+ e- Annihilation Into mu+ mu- in the Weinberg Model, Nucl. Phys. B 160 (1979) 151–207.
  * (5) G. ’t Hooft and M. J. G. Veltman, Scalar One Loop Integrals, Nucl. Phys. B 153 (1979) 365–401.
  * (6) W. L. van Neerven and J. A. M. Vermaseren, LARGE LOOP INTEGRALS, Phys. Lett. B 137 (1984) 241–244.
  * (7) R. G. Stuart, Algebraic Reduction of One Loop Feynman Diagrams to Scalar Integrals, Comput. Phys. Commun. 48 (1988) 367–389.
  * (8) G. J. van Oldenborgh and J. A. M. Vermaseren, New Algorithms for One Loop Integrals, Z. Phys. C 46 (1990) 425–438.
  * (9) Z. Bern, L. J. Dixon, and D. A. Kosower, Dimensionally regulated one loop integrals, Phys. Lett. B 302 (1993) 299–308, [hep-ph/9212308]. [Erratum: Phys.Lett.B 318, 649 (1993)].
  * (10) Z. Bern, L. J. Dixon, and D. A. Kosower, Dimensionally regulated pentagon integrals, Nucl. Phys. B 412 (1994) 751–816, [hep-ph/9306240].
  * (11) J. Fleischer, F. Jegerlehner, and O. V. Tarasov, Algebraic reduction of one loop Feynman graph amplitudes, Nucl. Phys. B 566 (2000) 423–440, [hep-ph/9907327].
  * (12) T. Binoth, J. P. Guillet, and G. Heinrich, Reduction formalism for dimensionally regulated one loop N point integrals, Nucl. Phys. B 572 (2000) 361–386, [hep-ph/9911342].
  * (13) A. Denner and S. Dittmaier, Reduction of one loop tensor five point integrals, Nucl. Phys. B 658 (2003) 175–202, [hep-ph/0212259].
  * (14) G. Duplancic and B. Nizic, Reduction method for dimensionally regulated one loop N point Feynman integrals, Eur. Phys. J. C 35 (2004) 105–118, [hep-ph/0303184].
  * (15) A. Denner and S. Dittmaier, Reduction schemes for one-loop tensor integrals, Nucl. Phys. B 734 (2006) 62–115, [hep-ph/0509141].
  * (16) R. K. Ellis and G. Zanderighi, Scalar one-loop integrals for QCD, JHEP 02 (2008) 002, [arXiv:0712.1851].
  * (17) G. Ossola, C. G. Papadopoulos, and R. Pittau, Reducing full one-loop amplitudes to scalar integrals at the integrand level, Nuclear Physics B 763 (Feb, 2007) 147–169.
  * (18) Z. Bern, L. J. Dixon, D. C. Dunbar, and D. A. Kosower, Fusing gauge theory tree amplitudes into loop amplitudes, Nucl. Phys. B 435 (1995) 59–101, [hep-ph/9409265].
  * (19) K. G. Chetyrkin and F. V. Tkachov, Integration by Parts: The Algorithm to Calculate beta Functions in 4 Loops, Nucl. Phys. B 192 (1981) 159–204.
  * (20) F. V. Tkachov, A Theorem on Analytical Calculability of Four Loop Renormalization Group Functions, Phys. Lett. B 100 (1981) 65–68.
  * (21) G. Ossola, C. G. Papadopoulos, and R. Pittau, Reducing full one-loop amplitudes to scalar integrals at the integrand level, Nucl. Phys. B 763 (2007) 147–169, [hep-ph/0609007].
  * (22) G. Ossola, C. G. Papadopoulos, and R. Pittau, Numerical evaluation of six-photon amplitudes, JHEP 07 (2007) 085, [arXiv:0704.1271].
  * (23) R. K. Ellis, W. T. Giele, and Z. Kunszt, A Numerical Unitarity Formalism for Evaluating One-Loop Amplitudes, JHEP 03 (2008) 003, [arXiv:0708.2398].
  * (24) Z. Bern, L. J. Dixon, D. C. Dunbar, and D. A. Kosower, One loop n point gauge theory amplitudes, unitarity and collinear limits, Nucl. Phys. B 425 (1994) 217–260, [hep-ph/9403226].
  * (25) R. Britto, F. Cachazo, and B. Feng, Generalized unitarity and one-loop amplitudes in N=4 super-Yang-Mills, Nucl. Phys. B 725 (2005) 275–305, [hep-th/0412103].
  * (26) R. Britto, E. Buchbinder, F. Cachazo, and B. Feng, One-loop amplitudes of gluons in SQCD, Phys. Rev. D 72 (2005) 065012, [hep-ph/0503132].
  * (27) J. M. Campbell, E. W. N. Glover, and D. J. Miller, One loop tensor integrals in dimensional regularization, Nucl. Phys. B 498 (1997) 397–442, [hep-ph/9612413].
  * (28) Z. Bern, L. J. Dixon, and D. A. Kosower, One loop amplitudes for e+ e- to four partons, Nucl. Phys. B 513 (1998) 3–86, [hep-ph/9708239].
  * (29) C. Anastasiou, R. Britto, B. Feng, Z. Kunszt, and P. Mastrolia, Unitarity cuts and Reduction to master integrals in d dimensions for one-loop amplitudes, JHEP 03 (2007) 111, [hep-ph/0612277].
  * (30) R. Britto and E. Mirabella, Single Cut Integration, JHEP 01 (2011) 135, [arXiv:1011.2344].
  * (31) B. Feng and H. Wang, Reduction of one-loop integrals with higher poles by unitarity cut method, arXiv:2104.00922.
  * (32) B. Feng, T. Li, and X. Li, Analytic tadpole coefficients of one-loop integrals, JHEP 09 (2021) 081, [arXiv:2107.03744].
  * (33) C. Hu, T. Li, and X. Li, One-loop Feynman integral reduction by differential operators, Phys. Rev. D 104 (2021), no. 11 116014, [arXiv:2108.00772].
  * (34) M. E. Peskin and D. V. Schroeder, An Introduction to quantum field theory.  Addison-Wesley, Reading, USA, 1995.
  * (35) A. V. Smirnov, Algorithm FIRE – Feynman Integral REduction, JHEP 10 (2008) 107, [arXiv:0807.3243].
  * (36) A. V. Smirnov, FIRE5: a C++ implementation of Feynman Integral REduction, Comput. Phys. Commun. 189 (2015) 182–191, [arXiv:1408.2372].
  * (37) R. N. Lee, LiteRed 1.4: a powerful tool for reduction of multiloop integrals, J. Phys. Conf. Ser. 523 (2014) 012059, [arXiv:1310.1145].
  * (38) A. V. Smirnov and F. S. Chuharev, FIRE6: Feynman Integral REduction with Modular Arithmetic, Comput. Phys. Commun. 247 (2020) 106877, [arXiv:1901.07808].
  * (39) A. von Manteuffel and C. Studerus, Reduze 2 - Distributed Feynman Integral Reduction, arXiv:1201.4330.
  * (40) P. Maierhöfer, J. Usovitsch, and P. Uwer, Kira—A Feynman integral reduction program, Comput. Phys. Commun. 230 (2018) 99–112, [arXiv:1705.05610].
  * (41) M. Gerlach, F. Herren, and M. Lang, tapir: A tool for topologies, amplitudes, partial fraction decomposition and input for reductions, arXiv:2201.05618.
  * (42) B. Feng and T. Li, PV-Reduction of Sunset Topology with Auxiliary Vector, arXiv:2203.16881.