to express the sought after number of master integrals as the Euler
characteristic of a projective hypersurface complement. We found that this
algorithm performs well for small numbers of variables (edges): The examples
in table 2 require not more than a couple of minutes of runtime. For more
variables, however, the computations tend to rapidly become much more time
consuming and often impracticable. Apart from the results in table 2, we also
verified proposition 55 for the sunrise graphs $\mathrm{S}_{L}$ using Euler
for up to six loops.

Graph $G$ |  |  |  |  |
---|---|---|---|---|---
$\mathfrak{C}\left(G\right)$ massless | 4 | 11 | 3 | 4 | 20
$\mathfrak{C}\left(G\right)$ massive | 7 | 15 | 30 | 19 | 55
Table 2: Counts of master integrals according to (3.10) computed with
Macaulay2’s Euler for some graphs for massless and massive internal
propagators (as no symmetries are regarded, all masses can be assumed to be
different from each other). All external momenta are assumed to be non-
degenerate (non-zero and not on any internal mass shell) in both cases.

######  Example 57.

Consider the one-loop sunrise graph $\mathrm{S}_{1}$ with
$m_{1}^{2}=m_{2}^{2}=-p^{2}=1$, which is a non-degenerate kinematic
configuration. According to example 47, its Lee-Pomeransky polynomial is
$\mathcal{G}=(x_{1}+x_{2})(x_{1}+x_{2}+1)+x_{1}x_{2}$. The Macaulay2 script

⬇

load "CharacteristicClasses.m2"

R=QQ[x0,x1,x2]

I=ideal(x0*x1*x2*((x1+x2)*x0+(x1+x2)^2+x1*x2))

Euler(I)

computes the output $0$ for
$\chi(\mathbbm{V}(x_{0}x_{1}x_{2}\tilde{\mathcal{G}})\cap\mathbbm{P}^{2})$.
Using $\chi(\mathbbm{P}^{2})=3$ and (4.15), we conclude
$\mathfrak{C}\left(\mathrm{S}_{1}\right)=3-0=3$ in agreement with proposition
55.

Recall that the number of master integrals depends on the kinematical
configuration; in table 2 we give the results both for massless and for
massive internal propagators. In particular, note how the massless 2-loop
propagator $\mathrm{WS}_{3}^{\prime}$ from example 53 with only
$\mathfrak{C}\left(\mathrm{WS}_{3}^{\prime}\right)=3$ master integrals grows
to carry $\mathfrak{C}\left(\mathrm{WS}_{3}^{\prime}\right)=30$ master
integrals in the fully massive case.

Furthermore, Macaulay2 also provides an implementation (the command deRham) of
the algorithm [65] of Oaku and Takayama for the computation of the individual
de Rham cohomology groups. This uses $D$-modules and Gröbner bases and tends
to demand more resources than the method discussed above.

######  Example 58.

Consider again the massive one-loop sunrise from example 47. The program

⬇

load "Dmodules.m2"

R=QQ[x1,x2]

f=x1*x2*(x1+x2+(x1+x2)^2+x1*x2)

deRham f

computes the following cohomology groups of
$X=\mathbbm{G}_{m}^{2}\setminus\mathbbm{V}(\mathcal{G})$:
$H^{0}(X)\cong\mathbbm{Q}$, $H^{1}(X)\cong\mathbbm{Q}^{3}$ and
$H^{2}(X)\cong\mathbbm{Q}^{5}$. Hence,
$\mathfrak{C}\left(\mathrm{S}_{1}\right)=\chi(X)=5-3+1=3$ as in example 57.

The same functionality is provided by Singular’s deRham.lib library via the
command deRhamCohomology. The Singular analogue of example 58 is

⬇

LIB "deRham.lib";

ring R = 0,(x1,x2),dp;

list L = (x1*x2*(x1+x2+(x1+x2)^2+x1*x2));

deRhamCohomology(L);

###  4.4 Comparison to other approaches

We successfully reproduced all of our results above (the wheels
$\mathrm{WS}_{L}^{\prime}$ with $L\leq 6$, the sunrises $\mathrm{S}_{L}$ with
$L\leq 4$ loops and the graphs from figure 4.1 and table 2) with the program
Azurite [31], which provides an implementation of Laporta’s approach [49].
While it employs novel techniques to boost performance, in the end it solves
linear systems of equations between integrals obtained from annihilators of
the integrand of Baikov’s representation (2.48) in order to count the number
of master integrals.

The observed agreement with our results is to be expected, since the
identification of integral relations with parametric annihilators that we
elaborated on in section 2.3 works equally for the Baikov representation,
which can also be interpreted as a Mellin transform. Note, however, that we
must use the options Symmetry -> False and GlobalSymmetry -> False for Azurite
in order to switch off the identification of integrals that differ by a
permutation of the edges. The reason being that, in our approach, all edges
$e$ carry their own index $\nu_{e}$ and no relation between these indices for
different edges is assumed.

Unfortunately, due to the way Azurite treats subsectors, this can occasionally
lead to an apparent mismatch. However, this is rather a technical nuisance
than an actual disagreement.

$G=\\!\\!\\!\\!\vbox{\hbox{\includegraphics[scale={0.75}]{Graphs/lorenzo}}}$
$G/\left\\{1,2\right\\}=\vbox{\hbox{\includegraphics[scale={0.6}]{Graphs/lorenzosub}}}$
$\mapsto\begin{cases}\vbox{\hbox{\includegraphics[scale={0.4}]{Graphs/lorenzo3}}}&3\\\
\vbox{\hbox{\includegraphics[scale={0.4}]{Graphs/lorenzo4}}}&4\\\ \end{cases}$
$G^{\prime}=\vbox{\hbox{\includegraphics[scale={0.6}]{Graphs/lorenzovac}}}$

Figure 4.4: A graph $G$ with a massive loop (edges 3, 4 and 5), two massless
propagators $\left\\{1,2\right\\}$ and four external legs ($p_{1}^{2}\neq 0$
massive and $p_{2}^{2}=p_{3}^{2}=p_{4}^{2}=0$ massless). Since the only
momentum $p_{4}$ running through the graph after contracting $1$ and $2$ is
null ($p_{4}^{2}=0$), the associated integral is the same as for $G^{\prime}$.

######  Example 59.

For the graph $G$ in figure 4.4, the Euler characteristic gives
$\mathfrak{C}\left(G\right)=15$, whereas both Reduze [96] and Azurite produce
$16$ master integrals. The problem arises from the _subsector_ where the edges
$1$ and $2$ are contracted: As shown in figure 4.4, it does have a remaining
external momentum $p_{4}$, such that the momenta running through edges $3$ and
$4$ are different—however, since $p_{4}^{2}=0$, the graph polynomials (and
hence the Feynman integrals) are identical to those of the vacuum graph
$G^{\prime}$ in figure 4.4. Since edges $3$ and $4$ in $G^{\prime}$ have the
same mass, they can be combined and thus $G^{\prime}$ clearly has only a
single master integral: the product of two tadpoles.

But Azurite and Reduze instead consider the subsectors of
$G/\left\\{1,2\right\\}$ obtained by contracting a further edge (3 or 4), and
obtain the two tadpoles (see figure 4.4) consisting only of edges
$\left\\{4,5\right\\}$ and $\left\\{3,5\right\\}$, respectively, as master
integrals. Of course, these would be recognized as identical if symmetries
were allowed; but the point is that even without using symmetries, there is
only a single master integral for $G^{\prime}$ (as computed by the Euler
characteristic).

Our results are also consistent with the conclusions obtained within the
differential reduction approach [41]; indeed, we demonstrated in section 4.2
how the master integral counts of [42] for the sunrise graphs emerge directly
from the computation of the Euler characteristic. Let us point out again,
however, that some care is required for these comparisons, since those works
refer to _irreducible_ master integrals, which excludes integrals that can be
expressed with gamma functions. In particular, the fact that the two loop
sunrise $\mathrm{S}_{2}$ with one massless line has
$\mathfrak{C}\left(\mathrm{S}_{2}\right)=4$ master integrals (see proposition
56) is consistent with [40]. We are counting all master integrals and are not
concerned here with the much more subtle question addressed by the observation
that two of these integrals may be expressed with gamma functions.

Finally, let us note that also the work of Lee and Pomeransky [57] addresses a
different problem: Considering only integer indices $\nu\in\mathbbm{Z}^{N}$,
how many _top-level_ master integrals are there for a graph $G$? This means
that integrals obtained from subsectors (graphs $G/e$ with at least one edge
$e$ contracted) are discarded. Geometrically, the number of the remaining
master integrals is identified with the dimension of the cohomology group
$H^{N}(\mathbbm{C}^{N}\setminus\mathbbm{V}(\mathcal{G}))$.323232Actually, they
initially refer to a different, relative cohomology group; but in the
description of their implementation in Mint they seem to work with this total
cohomology group.  In most cases, the program Mint computes this number
correctly, which then agrees with the other mentioned methods.333333Occasional
mismatches are known, like for the graph $F_{9}$ from figure 4.1 that was
addressed in [13, section 4.1]. These discrepancies are due to an error in the
implementation of Mint that misses contributions from critical points at
infinity (we thank Yang Zhang for bringing this to our attention).  We refer
to [13, section 4] and [42, section 6] for detailed discussions of this
comparison. Note that the dimension (and a basis) of the top cohomology group
can also be computed with the command deRhamCohom from the Singular library
dmodapp.lib.

A direct comparison of Mint with our results is not possible, since the
concept of _top-level_ integrals does not make sense in our setting of
arbitrary, non-integer indices $\nu$. Here, there is no relation at all
between integrals of a quotient graph $G/e$ and integrals of $G$ (the former
do not depend on $\nu_{e}$ at all; the latter do).

######  Remark 60.

Using the inclusion-exclusion principle, one might be tempted to define

$\widehat{\mathfrak{C}}\left(G\right)\mathrel{\mathop{:}}=\sum_{\gamma\subseteq
G}(-1)^{\left\lvert\gamma\right\rvert}\mathfrak{C}\left(G/\gamma\right)=\mathfrak{C}\left(G\right)-\sum_{e}\mathfrak{C}\left(G/e\right)+\sum_{e<f}\mathfrak{C}\left(G/\left\\{e,f\right\\}\right)-\cdots$
(4.16)

as the number of _top-level_ master integrals, since it subtracts from all
master integrals $\mathfrak{C}\left(G\right)$ the integrals associated to
subsectors (and corrects for double counting). Note that if $\gamma$ contains
a loop, the corresponding term in the sum should be set to zero (we only
consider contractions with the same loop number as $G$). The reverse relation,

$\mathfrak{C}\left(G\right)=\sum_{\gamma\subseteq
G}\widehat{\mathfrak{C}}\left(G/\gamma\right),$ (4.17)

is consistent with the intuition that the total set of master integrals is
obtained as the union of all top-level masters. By
$\mathcal{G}_{G/\gamma}=\left.\mathcal{G}\right|_{x_{e}=0\forall e\in\gamma}$,
we find that

$\widehat{\mathfrak{C}}\left(G\right)=(-1)^{N}\chi\left(\mathbbm{A}^{N}\setminus\mathbbm{V}(\mathcal{G})\right)$
(4.18)

is the Euler characteristic of the hypersurface complement inside affine space
(as compared to the torus $\mathbbm{G}_{m}^{N}$ as ambient space). However,
this number can take negative values and may thus not be interpreted as a
dimension. For example,
$\widehat{\mathfrak{C}}\left(G/\left\\{1,2\right\\}\right)=-1$ for the graph
from figure 4.4, which is required for consistency of (4.17) to get

$1=\mathfrak{C}\left(G/\left\\{1,2\right\\}\right)=\widehat{\mathfrak{C}}\left(G/\left\\{1,2\right\\}\right)+\widehat{\mathfrak{C}}\left(G/\left\\{1,2,3\right\\}\right)+\widehat{\mathfrak{C}}\left(G/\left\\{1,2,4\right\\}\right)=-1+1+1.$

##  5 Outlook

We have studied linear relations between Feynman integrals that arise from
parametric annihilators of the integrand $\mathcal{G}^{s}$ in the Lee-
Pomeransky representation. Seen as a multivariate (twisted) Mellin transform,
the integration bijects these special partial differential operators with
relations of various shifts (in the indices) of a Feynman integral. In
particular, every classical IBP relation (derived in momentum space) is of
this type.

The question whether _all_ shift relations of Feynman integrals (equivalently,
all parametric annihilators of $\mathcal{G}^{s}$) follow from momentum space
relations remains open (see question 24). We showed that the well-known
lowering and raising operators with respect to the dimension are consequences
of the classical IBPs. A next step would be to clarify if the same applies to
the relations implied by the trivial annihilators (2.14). Similarly, question
23 asking whether the annihilator
$\operatorname{Ann}(\mathcal{G}^{s})=\operatorname{Ann}^{1}(\mathcal{G}^{s})$
is linearly generated, remains to be settled. A positive answer to either of
these would imply that the labor-intensive computation of the parametric
annihilators could be simplified considerably ($\operatorname{Mom}$ is known
explicitly, and $\operatorname{Ann}^{1}$ can be calculated through syzygies).

The main insight of this article is a statement on the number of master
integrals, which we define as the dimension of the vector space of the
corresponding family of Feynman integrals over the field of rational functions
in the dimension and the indices. Since we treat all indices $\nu_{a}$ as
independent variables, this definition does not account for symmetries
(automorphisms) of the underlying graph. A natural next step is to incorporate
these into our setup by studying the action of the corresponding permutation
group. The widely used partition of master integrals into _top-level_ and
_subsector_ integrals, however, seems to be more difficult to take into
account, as we touched on in remark 60.

Our result shows that the number of master integrals is not only finite, but
identical to the Euler characteristic of the complement of the hyperspace
$\left\\{\mathcal{G}=0\right\\}$ determined by the Lee-Pomeransky polynomial
$\mathcal{G}$. This statement follows from a theorem of Loeser and Sabbah. We
exemplified several methods to compute this number and found agreement with
other established methods. We expect that, combining the available tools for
the computation of the Euler characteristic, it should be possible to compile
a program for the efficient calculation of the number of master integrals for
a wide range of Feynman graphs.

Let us conclude by emphasizing, once again, that the main objects of the
approach elaborated here—the $s$-parametric annihilators generating the
integral relations, and the Euler characteristic giving the number of master
integrals—are well studied objects in the theory of $D$-modules and
furthermore algorithms for their automated computation are available in
principle.

In particular, we hope that this parametric, $D$-module theoretic and
geometric approach can also shed light on the problems most relevant for
perturbative calculations in QFT: the construction of a basis of master
integrals, and the actual reduction of arbitrary integrals to such a basis.
For this perspective, we would like to point out that our approach of treating
the indices $\nu_{a}$ as free variables, in particular not tied to take
integer values, is desirable in order to deal with dimensionally regulated
integrals in position space, and for the ability to integrate out one-scale
subgraphs (both situations introduce non-integer indices). For a recent step
into this direction, see [89].

##  Appendix A Integral representations

In this appendix we add technical details on the material of section 2. We
summarize the various well-known parametric representations, including their
proofs, and the explicit relation to momentum space via propositions 2 and 20.
Furthermore, we give an alternative, algebraic proof for corollary 63.

###  A.1 Momentum space and Schwinger parameters

Figure A.1: A two-loop Feynman graph with a choice of loop momenta and the
resulting momentum flow.

As in section 3.2 we consider a connected Feynman graph $G$ with $N$ internal
edges, $E+1$ external legs and loop-number $L$, which is related by Euler’s
formula $L=N-V+1$ to the number $V$ of vertices of $G$. Let us consider the
case where each edge $e$ of $G$ is associated with a Feynman
propagator,343434We use the signature $(1,-1,\ldots,-1)$ for the Minkowski
metric.

$\frac{1}{\mathsf{D}_{e}}=\frac{1}{-k_{e}^{2}+m_{e}^{2}-\mathrm{{i}}\epsilon}\quad(1\leq
e\leq N),$

which depends on the mass $m_{e}$ of the particle $e$ and the $d$-dimensional
momentum $k_{e}\in\mathbbm{R}^{d}$ flowing through this edge. Enforcing
momentum conservation at each vertex fixes all $k_{e}$ in terms of $E$
independant external momenta $p_{1},\ldots,p_{E}$ and $L$ free loop momenta
$\ell_{1},\ldots,\ell_{L}$. Note that the actual number of external legs of
$G$ is $E+1$, since overall momentum conservation $\sum_{i=1}^{E+1}p_{i}=0$
imposes one relation among the external momenta. Taking only the inverse
Feynman propagators $\mathsf{D}_{e}$ as denominators, equation (2.1) defines
the Feynman integral associated to $G$.

######  Example 61.

The graph in figure A.1 has $V=5$ vertices, $N=6$ internal edges and $L=2$
loops. It depends on two independent external momenta $p_{1}$ and $p_{2}$. A
choice of loop momenta and the resulting momentum flow is depicted in figure
A.1. With all masses zero, this yields

$\mathcal{I}(\nu_{1},\ldots,\nu_{6})=\int_{\mathbbm{R}^{d}}\frac{\mathrm{d}^{d}\ell_{1}}{\mathrm{{i}}\pi^{d/2}}\int_{\mathbbm{R}^{d}}\frac{\mathrm{d}^{d}\ell_{2}}{\mathrm{{i}}\pi^{d/2}}\frac{1}{[-\ell_{1}^{2}-\mathrm{{i}}\epsilon]^{\nu_{1}}[-\ell_{2}^{2}-\mathrm{{i}}\epsilon]^{\nu_{2}}[-(\ell_{1}-\ell_{2})^{2}-\mathrm{{i}}\epsilon]^{\nu_{3}}}\\\
\times\frac{1}{[-(\ell_{2}+p_{1})^{2}-\mathrm{{i}}\epsilon]^{\nu_{4}}[-(\ell_{2}-p_{2})^{2}-\mathrm{{i}}\epsilon]^{\nu_{5}}[-(\ell_{1}-p_{2})^{2}-\mathrm{{i}}\epsilon]^{\nu_{6}}}.$

Typically, the number $\left\lvert\Theta\right\rvert=L(L+1)/2+LE$ of
independent scalar products $s_{\left\\{i,j\right\\}}$ in (2.28) is larger
than the number of edges in a graph $G$. We can then extend the initial set of
denominators (given as the inverse propagators of the graph) by a suitable
choice of additional quadratic (or linear) forms in the loop momenta, such
that we reach a set of $\left\lvert\Theta\right\rvert$ denominators with the
property that the matrix $\mathcal{A}$ defined by

$\mathsf{D}_{a}=\sum_{\left\\{i,j\right\\}\in\Theta}\mathcal{A}_{a}^{\left\\{i,j\right\\}}s_{\left\\{i,j\right\\}}+\lambda_{a}$

becomes invertible. This means that all loop-momentum dependent scalar
products can be written as linear combinations of the denominators, see
(2.29). The additional denominators introduced in this way are called
_irreducible scalar products_.

######  Example 62.

We again consider the graph from figure A.1 with $6$ internal edges labelled
as in example 61. In the massless case, the inverse propagators are just
$\mathsf{D}_{e}=-k_{e}^{2}$ and their explicit decomposition into the
$\left\lvert\Theta\right\rvert=7$ scalar products takes the form

$\begin{pmatrix}\mathsf{D}_{1}\\\ \mathsf{D}_{2}\\\ \mathsf{D}_{3}\\\
\mathsf{D}_{4}\\\ \mathsf{D}_{5}\\\ \mathsf{D}_{6}\\\
\end{pmatrix}=\begin{pmatrix}-\ell_{1}^{2}\\\ -\ell_{2}^{2}\\\
-(\ell_{1}-\ell_{2})^{2}\\\ -(\ell_{2}+p_{1})^{2}\\\ -(\ell_{2}-p_{2})^{2}\\\
-(\ell_{1}-p_{2})^{2}\\\
\end{pmatrix}=\underbrace{\begin{pmatrix}-1&0&0&0&0&0&0\\\ 0&0&-1&0&0&0&0\\\
-1&2&-1&0&0&0&0\\\ 0&0&-1&0&0&-2&0\\\ 0&0&-1&0&0&0&2\\\ -1&0&0&0&2&0&0\\\
\end{pmatrix}}_{\mathcal{A}}\begin{pmatrix}\ell_{1}^{2}\\\ \ell_{1}\ell_{2}\\\
\ell_{2}^{2}\\\ \ell_{1}p_{1}\\\ \ell_{1}p_{2}\\\ \ell_{2}p_{1}\\\
\ell_{2}p_{2}\\\ \end{pmatrix}+\underbrace{\begin{pmatrix}0\\\ 0\\\ 0\\\
-p_{1}^{2}\\\ -p_{2}^{2}\\\ -p_{2}^{2}\\\ \end{pmatrix}}_{\lambda}.$

The matrix $\mathcal{A}$ has rank $6$ and annihilates
$(0,0,0,1,0,0,0)^{\intercal}$. Thus we can choose
$\mathsf{D}_{7}=\ell_{1}p_{1}$ as an irreducible scalar product to complete
the basis of quadratic forms in the loop momenta. The matrix $\mathcal{A}$
then acquires an additional row $(0,0,0,1,0,0,0)$ and becomes invertible.

Note that we assume that the inverse propagators of $G$ (the initial set of
denominators) are linearly independent (that is, the
$N\times\left\lvert\Theta\right\rvert$ matrix $\mathcal{A}$ of the inverse
propagators has full rank $N$) in order to be able to extend them to a basis
of quadratic forms by choosing $\left\lvert\Theta\right\rvert-N$ irreducible
scalar products.353535If there are linear dependencies between the inverse
propagators, these relations imply that the Feynman integral can be expressed
in terms of contracted graphs with linearly independent inverse propagators.
For example, if $\alpha\mathsf{D}_{1}+\beta\mathsf{D}_{2}=1$, then iterated
use of
$1/(\mathsf{D}_{1}\mathsf{D}_{2})=\alpha/\mathsf{D}_{2}+\beta/\mathsf{D}_{1}$
allows one to ultimately eliminate one of $\mathsf{D}_{1}$ or
$\mathsf{D}_{2}$. Therefore, requesting linear independence is no restriction.

For each denominator $\mathsf{D}_{a}$ we introduce a scalar $x_{a}$, which is
known as Schwinger-, Feynman- or $\alpha$-parameter. In definition 1 we have
introduced the decomposition

$\sum_{a=1}^{N}x_{a}\mathsf{D}_{a}=-\sum_{i,j=1}^{L}\Lambda_{ij}\ell_{i}\ell_{j}+\sum_{i=1}^{L}2Q_{i}\ell_{i}+J,$
(A.1)

which determines a symmetric $L\times L$ matrix $\Lambda$, a vector $Q$ and a
scalar $J$. With their help we defined the polynomials $\mathcal{U}$,
$\mathcal{F}$ and $\mathcal{G}=\mathcal{U}+\mathcal{F}$ in (2.3). Explicitly,
from definition 1 and (2.26) we can read off that

$\displaystyle\Lambda_{ij}$
$\displaystyle=-\frac{1+\delta_{ij}}{2}\sum_{a=1}^{N}x_{a}\mathcal{A}_{a}^{\left\\{i,j\right\\}}\quad\text{for
$1\leq i,j\leq L$,}$ (A.2) $\displaystyle Q_{i}$
$\displaystyle=\frac{1}{2}\sum_{a=1}^{N}\sum_{j=L+1}^{M}x_{a}\mathcal{A}_{a}^{\left\\{i,j\right\\}}q_{j}\quad\text{for
$1\leq i\leq L$ and}$ (A.3) $\displaystyle J$
$\displaystyle=\sum_{a=1}^{N}x_{a}\lambda_{a}.$ (A.4)

Since $\Lambda_{ij}$ is an $L\times L$ matrix with entries that are linear in
the Schwinger parameters, the polynomial $\mathcal{U}$ is homogeneous of
degree $L$. By Cramer’s rule, $(\det\Lambda)\Lambda^{-1}_{ij}$ is homogeneous
of degree $L-1$ and the linearity of $Q$ and $J$ in the $x_{a}$ implies

######  Corollary 63.

$\mathcal{U}$ and $\mathcal{F}$ are homogeneous polynomials in the variables
$x_{1},\ldots,x_{N}$ with the degrees $\deg(\mathcal{U})=L$ and
$\deg(\mathcal{F})=L+1$. Hence, for $\mathcal{G}=\mathcal{U}+\mathcal{F}$, we
have

$\left(\sum_{a=1}^{N}x_{a}\partial_{a}\right)\mathcal{G}=L\mathcal{U}+(L+1)\mathcal{F}=(L+1)\mathcal{G}-\mathcal{U}=L\mathcal{G}+\mathcal{F}.$
(A.5)

Let us now come to the proof of proposition 2 following [63] and [57].

###### Proof of proposition 2.

We consider the Feynman integral defined in (2.1),

$\mathcal{I}(\nu_{1},\ldots,\nu_{N})=\left(\prod_{j=1}^{L}\int\frac{\mathrm{d}^{d}\ell_{j}}{\mathrm{{i}}\pi^{d/2}}\right)\prod_{i=1}^{N}\mathsf{D}_{i}^{-\nu_{i}}.$
(A.6)

Using the Schwinger trick to exponentiate each denominator,363636The integral
(A.7) converges only for $\operatorname{Re}(\nu_{a})>0$ and therefore
restricts the domain of convergence for the parametric integral. However, this
has no consequences for algebraic relations, see remark 4.

$\frac{1}{\mathsf{D}_{a}^{\nu_{a}}}=\frac{1}{\Gamma(\nu_{a})}\int_{0}^{\infty}x_{a}^{\nu_{a}-1}e^{-x_{a}\mathsf{D}_{a}}\mathrm{d}x_{a},$
(A.7)

the integral in (A.6) turns into

$\mathcal{I}(\nu_{1},\ldots,\nu_{N})=\left(\prod_{i=1}^{N}\int_{0}^{\infty}\frac{x_{i}^{\nu_{i}-1}\mathrm{d}x_{i}}{\Gamma\left(\nu_{i}\right)}\right)\left(\prod_{j=1}^{L}\int\frac{\mathrm{d}^{d}\ell_{j}}{\mathrm{{i}}\pi^{d/2}}\right)e^{-\sum_{a=1}^{N}x_{a}\mathsf{D}_{a}}.$

According to (A.1) and (2.3), we can complete the square in the exponent

$-\sum_{a=1}^{N}x_{a}\mathsf{D}_{a}=(\ell-\Lambda^{-1}Q)^{\intercal}\Lambda(\ell-\Lambda^{-1}Q)-\mathcal{F}/\mathcal{U}$

to perform the Gaußian integrals over the shifted loop momenta
$\ell^{\prime}\mathrel{\mathop{:}}=\ell-\Lambda^{-1}Q$ as373737Recall that our
metric has signature $(1,-1,\ldots,-1)$, so the integrations over the $d-1$
spacelike components are Euclidean and give $\sqrt{\pi^{L}\mathcal{U}}$ each.
The timelike integrations are understood as contour integrals and yield the
same factor after rotating the integration contour to the imaginary axis,
according to the Feynman $\mathrm{{i}}\epsilon$-prescription.

$\left(\prod_{j=1}^{L}\int\frac{\mathrm{d}^{d}\ell^{\prime}_{j}}{\mathrm{{i}}\pi^{d/2}}\right)e^{(\ell^{\prime})^{\intercal}\Lambda\ell^{\prime}}=(\det\Lambda)^{-d/2}=\mathcal{U}^{-d/2}.$

In summary, we therefore arrive at the integral representation (2.5):

$\mathcal{I}(\nu_{1},\ldots,\nu_{N})=\left(\prod_{i=1}^{N}\int_{0}^{\infty}\frac{x_{i}^{\nu_{i}-1}\mathrm{d}x_{i}}{\Gamma\left(\nu_{i}\right)}\right)\frac{e^{-\mathcal{F}/\mathcal{U}}}{\mathcal{U}^{d/2}}.$

We now multiply with
$1=\int_{0}^{\infty}\delta(\rho-\sum_{j=1}^{N}x_{j})\mathrm{d}\rho$ and
substitute $x_{a}\rightarrow\rho x_{a}$.383838Much more generally, we could
replace $\sum_{j=1}^{N}x_{j}$ in the $\delta$-constraint with any other
function as long as it is homogeneous of degree 1 and positive on
$\mathbbm{R}_{+}^{N}$.  The Jacobian $\rho^{N}$, the monomials
$x_{i}^{\nu_{i}-1}$ and
$\delta(\rho-\sum_{j}x_{j})\rightarrow\delta(1-\sum_{j}x_{j})/\rho$ contribute
the power $\rho^{\left\lvert\nu\right\rvert-1}$, whereas the homogeneity of
$\mathcal{F}$ and $\mathcal{U}$ from corollary 63 implies that
$\mathcal{U}\rightarrow\rho^{L}\mathcal{U}$ and
$\mathcal{F}/\mathcal{U}\rightarrow\rho\mathcal{F}/\mathcal{U}$. Overall, by
realising that the integral over $\rho$ is

$\int_{0}^{\infty}\rho^{\omega-1}e^{-\rho\mathcal{F}/\mathcal{U}}\mathrm{d}\rho=\Gamma(\omega)\left(\frac{\mathcal{U}}{\mathcal{F}}\right)^{\omega},$

we arrive at the first parametric formula (2.6). Similarly, we multiply the
integrand of (2.7) with
$1=\int_{0}^{\infty}\delta(\rho-\sum_{i}x_{i})\mathrm{d}\rho$ and substitute
$x_{i}\rightarrow\rho x_{i}$. Using
$\mathcal{U}\rightarrow\rho^{L}\mathcal{U}$ and
$\mathcal{F}\rightarrow\rho^{L+1}\mathcal{F}$ from corollary 63, the integral
over $\rho$ becomes

$\int_{0}^{\infty}\rho^{\omega-1}(\mathcal{U}+\rho\mathcal{F})^{-d/2}=\mathcal{U}^{-d/2}\left(\frac{\mathcal{U}}{\mathcal{F}}\right)^{\omega}\frac{\Gamma(\omega)\Gamma(\frac{d}{2}-\omega)}{\Gamma(\frac{d}{2})}$

and combines with the prefactors in (2.7) to reproduce (2.7). ∎

We conclude the section with the proof of proposition 20 following Grozin
[34]:

###### Proof of proposition 20.

The action of $\mathbf{o}_{j}^{i}$ on the integrand from (2.30) is

$\mathbf{o}_{j}^{i}f=d\delta_{ij}f+f\sum_{a=1}^{N}\frac{-\nu_{a}}{\mathsf{D}_{a}}q_{j}\frac{\partial\mathsf{D}_{a}}{\partial
q_{i}}.$

According to (2.26), the chain rule gives

$q_{j}\frac{\partial}{\partial
q_{i}}\mathsf{D}_{a}=q_{j}\frac{\partial}{\partial
q_{i}}\sum_{\left\\{k,m\right\\}\in\Theta}\mathcal{A}_{a}^{\left\\{k,m\right\\}}q_{k}q_{m}=\sum_{m=1}^{M}\mathcal{A}_{a}^{\left\\{i,m\right\\}}(1+\delta_{i,m})q_{j}q_{m}$

and we can express the scalar products $q_{j}q_{m}$ with
$\left\\{j,m\right\\}\in\Theta$ in terms of denominators using (2.29). The
remaining terms with $j,m>L$ are products of external momenta, so

$\displaystyle\mathbf{o}_{j}^{i}f$
$\displaystyle=d\delta_{ij}f-f\sum_{a,b=1}^{N}C_{aj}^{bi}\frac{\nu_{a}}{\mathsf{D}_{a}}\left(\mathsf{D}_{b}-\lambda_{b}\right)\quad\text{for}\quad
1\leq j\leq L\quad\text{and}$ $\displaystyle\mathbf{o}_{j}^{i}f$
$\displaystyle=-f\sum_{a,b=1}^{N}C_{aj}^{bi}\frac{\nu_{a}}{\mathsf{D}_{a}}\left(\mathsf{D}_{b}-\lambda_{b}\right)-f\sum_{a=1}^{N}\sum_{m=L+1}^{M}\mathcal{A}_{a}^{\left\\{i,m\right\\}}q_{j}q_{m}\frac{\nu_{a}}{\mathsf{D}_{a}}\quad\text{if}\quad
L<j\leq M.$

We conclude by noticing that multiplying the integrand $f$ with
$\nu_{a}/\mathsf{D}_{a}$ is equivalent to the action of the operator
$\hat{\mathbf{a}}^{+}$ defined in (2.17), whereas multiplication with
$\mathsf{D}_{b}$ lowers the index $\nu_{b}$ and corresponds to
$\mathbf{b}^{-}$. ∎

###  A.2 Algebraic proof for corollary 21

With the proof of corollary 21 we have shown that, for every momentum space
IBP relation, there is a corresponding annihilator in
$\operatorname{Ann}\left(\mathcal{G}^{-d/2}\right)$. The proof rests on the
inverse Mellin transform, which may be seen as a convenient but rather
abstract argument. As a more direct alternative, we prove the statement in a
purely algebraical way by use of properties of the graph polynomials.

######  Lemma 64.

The operators $\widetilde{\mathbf{O}}_{j}^{i}$ from (2.36) and (2.37)
corresponding to the momentum space IBP relations annhilate the parametric
integrand $\mathcal{G}^{-d/2}$:

$\widetilde{\mathbf{O}}_{j}^{i}\in\operatorname{Ann}\left(\mathcal{G}^{-d/2}\right)\quad\text{for
all $1\leq i\leq L$ and $1\leq j\leq M$.}$

###### Proof.

Let us first consider the case $j\leq L$. After acting with
$\widetilde{\mathbf{O}}_{j}^{i}$ from (2.36) on $\mathcal{G}^{-d/2}$ and
dividing by $(d/2)\mathcal{G}^{-d/2-1}$, we are left to prove the vanishing of

$2\mathcal{G}\delta_{i,j}-\sum_{a,b}C^{bi}_{aj}x_{a}\left(\partial_{b}\mathcal{G}-\lambda_{b}\Big{[}L+1-\sum_{c}x_{c}\partial_{c}\Big{]}\mathcal{G}\right)=2\mathcal{G}\delta_{i,j}-\sum_{a,b}C^{bi}_{aj}x_{a}\left(\partial_{b}\mathcal{G}-\lambda_{b}\mathcal{U}\right)$
(A.8)

where we exploited the homogeneity from (A.5). Using (2.3), we note that

$\partial_{b}\mathcal{G}-\lambda_{b}\mathcal{U}=\partial_{b}\left[\mathcal{U}\left(1+J+Q^{\intercal}\Lambda^{-1}Q\right)\right]-\lambda_{b}\mathcal{U}=\mathcal{G}\frac{\partial_{b}\mathcal{U}}{\mathcal{U}}+\mathcal{U}\partial_{b}\left(Q^{\intercal}\Lambda^{-1}Q\right)$
(A.9)

because $\partial_{b}J=\lambda_{b}$ according to (A.4). In order to evaluate
$\partial_{b}\mathcal{U}$ with Jacobi’s formula
$(\partial_{b}\mathcal{U})/\mathcal{U}=\sum_{r,s=1}^{L}\Lambda^{-1}_{r,s}\partial_{b}\Lambda_{s,r}$,
we use (A.2) to compute

$\sum_{b}\mathcal{A}^{b}_{\left\\{m,j\right\\}}\partial_{b}\Lambda_{s,r}=-\frac{1+\delta_{s,r}}{2}\sum_{b}\mathcal{A}^{b}_{\left\\{m,j\right\\}}\mathcal{A}_{b}^{\left\\{s,r\right\\}}=-\frac{1+\delta_{s,r}}{2}\delta_{\left\\{m,j\right\\},\left\\{r,s\right\\}}=-\frac{\delta_{m,r}\delta_{j,s}+\delta_{m,s}\delta_{j,r}}{2}$

which restricts $m$ to either $r$ or $s$. So in particular, $m\leq L$ and we
can use (A.2) in

$\sum_{a,b}C^{bi}_{aj}x_{a}\partial_{b}\Lambda_{r,s}=-\sum_{a,m}x_{a}\mathcal{A}_{a}^{\left\\{i,m\right\\}}\frac{1+\delta_{i,m}}{2}\left(\delta_{m,r}\delta_{j,s}+\delta_{m,s}\delta_{j,r}\right)=\Lambda_{i,r}\delta_{j,s}+\Lambda_{i,s}\delta_{j,r}$
(A.10)

which proves that for arbitrary $j$ (independent of whether $j\leq L$ or
$j>L$)

$\sum_{a,b}C^{bi}_{aj}x_{a}\frac{\partial_{b}\mathcal{U}}{\mathcal{U}}=\sum_{r,s=1}^{L}\Lambda^{-1}_{r,s}(\Lambda_{i,r}\delta_{j,s}+\Lambda_{i,s}\delta_{j,r})=2\delta_{i,j}.$
(A.11)

Via (A.9), this identity reduces the proof of (A.8) to showing that

$\sum_{a,b}C^{bi}_{aj}x_{a}\partial_{b}\left(Q^{\intercal}\Lambda^{-1}Q\right)=\sum_{a,b}C^{bi}_{aj}x_{a}\Big{[}2(\partial_{b}Q)^{\intercal}\Lambda^{-1}Q-Q^{\intercal}\Lambda^{-1}(\partial_{b}\Lambda)\Lambda^{-1}Q\Big{]}$
(A.12)

vanishes. The last term is easily evaluated with (A.10) and gives

$\sum_{a,b}C^{bi}_{aj}x_{a}Q^{\intercal}\Lambda^{-1}(\partial_{b}\Lambda)\Lambda^{-1}Q=2\sum_{r,s=1}^{L}(\Lambda^{-1}Q)_{r}(\Lambda^{-1}Q)_{s}\Lambda_{i,r}\delta_{j,s}=2Q_{i}(\Lambda^{-1}Q)_{j},$
(A.13)

whereas the derivative $\partial_{b}Q$ can be read off from (A.3) and the sum
over $b$ yields

$\sum_{b}\mathcal{A}_{\left\\{m,j\right\\}}^{b}(2\partial_{b}Q_{s})=\sum_{b}\mathcal{A}_{\left\\{m,j\right\\}}^{b}\sum_{r>L}\mathcal{A}_{b}^{\left\\{s,r\right\\}}q_{r}=\sum_{r>L}q_{r}\delta_{\left\\{m,j\right\\},\left\\{s,r\right\\}}=\sum_{r>L}q_{r}\delta_{m,r}\delta_{s,j}$
(A.14)

because $j\leq L<r$ excludes the possibility that $m=s$ and $r=j$. Thus with
(A.3),

$\sum_{a,b}C^{bi}_{aj}x_{a}(2\partial_{b}Q)^{\intercal}\Lambda^{-1}Q=(\Lambda^{-1}Q)_{j}\sum_{a}\sum_{m>L}x_{a}\mathcal{A}_{a}^{\left\\{i,m\right\\}}q_{m}=2Q_{i}(\Lambda^{-1}Q)_{j}$
(A.15)

cancels the contribution from (A.13) in (A.12) and finishes the proof in the
case $j\leq L$. If instead we have $j>L$, then we must replace
$\delta_{\left\\{m,j\right\\},\left\\{s,r\right\\}}=\delta_{j,r}\delta_{m,s}$
in (A.14) such that

$\sum_{a,b}C^{bi}_{aj}x_{a}(2\partial_{b}Q)^{\intercal}\Lambda^{-1}Q=\sum_{s=1}^{L}\sum_{a}\mathcal{A}_{a}^{\left\\{i,s\right\\}}(1+\delta_{i,s})q_{j}(\Lambda^{-1}Q)_{s}=-2\sum_{s=1}^{L}\Lambda_{i,s}(\Lambda^{-1}Q)_{s}=-2Q_{i}q_{j}$

where we used (A.2) once more. Now recall that (A.11) remains true and becomes
zero for $j>L$ because $\delta_{i,j}=0$ since $i\leq L$. For the same reason,
$\delta_{j,s}=0$ in (A.13) and therefore, using (A.9),

$-\sum_{a,b}C^{bi}_{aj}x_{a}(\partial_{b}\mathcal{G}-\lambda_{b}\mathcal{U})=-\mathcal{U}\sum_{a,b}C^{bi}_{aj}x_{a}(2\partial_{b}Q)^{\intercal}\Lambda^{-1}Q=2\mathcal{U}Q_{i}q_{j}.$

This is precisely cancelled by the additional contribution to (A.8) coming
from $\widetilde{\mathbf{O}}_{j}^{i}$ in (2.37) in the case $j>L$: The
additional term acts on $\mathcal{G}^{-d/2}$ as

$-\sum_{a}\sum_{m>L}\mathcal{A}_{a}^{\left\\{i,m\right\\}}q_{j}q_{m}x_{a}\Big{[}L+1-\sum_{c}x_{c}\partial_{c}\Big{]}\mathcal{G}=-\mathcal{U}q_{j}\sum_{a}\sum_{m>L}\mathcal{A}_{a}^{\left\\{i,m\right\\}}x_{a}q_{m}=-2\mathcal{U}q_{j}Q_{i}$

after dividing by $(d/2)\mathcal{G}^{-d/2-1}$. Note that here we used (A.5)
and (A.3). ∎

###  A.3 The Baikov representation

In this section we discuss the representation of Feynman integrals suggested
by Baikov in [7], whose complete form (2.48) was given by Lee in [52, 53]. We
will give some details on the derivation of this formula (see also [34,
section 9]), which was presented in [52] and applied in our discussion of the
lowering dimension shift in section 2.5.

Assume that $q_{1}$, …, $q_{M}$ are vectors in a Euclidean vector space and
write

$V_{n}\mathrel{\mathop{:}}=\begin{pmatrix}q_{n}\cdot q_{n}&\cdots&q_{n}\cdot
q_{M}\\\ \vdots&\ddots&\vdots\\\ q_{M}\cdot q_{n}&\cdots&q_{M}\cdot q_{M}\\\
\end{pmatrix}=\left(q_{i}\cdot q_{j}\right)_{n\leq i,j\leq
M}\quad\text{and}\quad G_{n}\mathrel{\mathop{:}}=\det V_{n}$ (A.16)

for their Gram matrices and determinants. Note that

$V_{n}=\begin{pmatrix}q_{n}^{2}&q_{\bullet}\cdot q_{n}\\\ q_{n}\cdot
q_{\bullet}&V_{n+1}\\\ \end{pmatrix}\quad\text{where}\quad q_{\bullet}\cdot
q_{n}\mathrel{\mathop{:}}=\begin{pmatrix}q_{n+1}\cdot q_{n}\\\ \vdots\\\
q_{M}\cdot q_{n}\\\ \end{pmatrix},\quad q_{n}\cdot
q_{\bullet}\mathrel{\mathop{:}}=\left(q_{\bullet}\cdot
q_{n}\right)^{\intercal}$

and thus, by adding $-(p_{n}\cdot p_{\bullet})V_{n+1}^{-1}$ times the lower
$M-n$ rows to the first row,

$\frac{G_{n}}{G_{n+1}}=q_{n}^{2}-\left\|\mathrm{pr}_{\operatorname{lin}\left\\{q_{n+1},\ldots,q_{M}\right\\}}^{\perp}(q_{n})\right\|^{2}=\left\|\mathrm{pr}_{\operatorname{lin}\left\\{q_{n+1},\ldots,q_{M}\right\\}^{\perp}}^{\perp}(q_{n})\right\|^{2}.$
(A.17)

Indeed, the formula
$\mathrm{pr}_{\operatorname{lin}\left\\{q_{n+1},\ldots,q_{M}\right\\}}^{\perp}(v)=\sum_{i,j=n+1}^{M}q_{i}\left(V^{-1}_{n+1}\right)_{i,j}(q_{j}\cdot
v)$ for the orthogonal projection of $v$ onto the space spanned by
$q_{n+1},\ldots,q_{M}$ shows that

$\displaystyle\left\|\mathrm{pr}_{\operatorname{lin}\left\\{q_{n+1},\ldots,q_{M}\right\\}}^{\perp}(q_{n})\right\|^{2}$
$\displaystyle=\sum_{i,j,k,l=n+1}^{M}\underbrace{(q_{i}\cdot
q_{k})}_{\left(V_{n+1}\right)_{i,k}}\left(V^{-1}_{n+1}\right)_{i,j}(q_{j}\cdot
q_{n})\left(V^{-1}_{n+1}\right)_{k,l}(q_{k}\cdot q_{n})$
$\displaystyle=\sum_{j,k,l=n+1}^{M}\delta_{k,j}(q_{j}\cdot
q_{n})\left(V^{-1}_{n+1}\right)_{k,l}(q_{l}\cdot
q_{n})=\sum_{k,l=n+1}^{M}(q_{k}\cdot
q_{n})\left(V^{-1}_{n+1}\right)_{k,l}(q_{l}\cdot q_{n})$
$\displaystyle=(q_{n}\cdot q_{\bullet})V^{-1}_{n+1}(q_{\bullet}\cdot q_{n}).$

Now assume our integrand $f$ only depends on the scalar products
$s_{i,j}=q_{i}\cdot q_{j}$, and we want to integrate out the first loop
momentum $q_{1}$. Let us decompose $q_{1}=q_{\perp}+q_{\parallel}$ into the
component
$q_{\parallel}\in\operatorname{lin}\left\\{q_{2},\ldots,q_{M}\right\\}$ that
lies in the space spanned by the other momenta, and the component $q_{\perp}$
in its orthogonal complement. According to (A.17), $G_{n}^{1/2}$ is the volume
of the parallelotope spanned by $q_{n},\ldots,q_{M}$. Hence, changing
coordinates from $q_{\parallel}$ to $(s_{1,2},\ldots,s_{1,M})$ yields

$\int_{\mathbbm{R}}\mathrm{d}s_{1,2}\
\cdots\int_{\mathbbm{R}}\mathrm{d}s_{1,M}=\sqrt{G_{2}}\int_{\mathbbm{R}^{M-1}}\mathrm{d}^{M-1}q_{\parallel}.$

The integral over the orthogonal component is, due to
$s_{1,1}=q_{1}^{2}=q_{\perp}^{2}+q_{\parallel}^{2}$, given by

$\displaystyle\int_{\mathbbm{R}^{d-M+1}}\mathrm{d}^{d-M+1}q_{\perp}$
$\displaystyle=\frac{\pi^{(d-M+1)/2}}{\Gamma\left(\frac{d-M+1}{2}\right)}\int_{0}^{\infty}\\!\\!\mathrm{d}q_{\perp}^{2}\left(q_{\perp}^{2}\right)^{(d-M-1)/2}=\frac{\pi^{(d-M+1)/2}}{\Gamma\left(\frac{d-M+1}{2}\right)}\int_{q_{\parallel}^{2}}^{\infty}\\!\\!\mathrm{d}s_{1,1}\left(\frac{G_{1}}{G_{2}}\right)^{(d-M-1)/2}.$

Note that the lower boundary $s_{1,1}=q_{\parallel}^{2}$ corresponds to
$0=q_{\perp}^{2}=G_{1}/G_{2}$. Altogether,

$\int_{\mathbbm{R}^{d}}\frac{\mathrm{d}^{d}q_{1}}{\pi^{d/2}}\
f(s)=\frac{\pi^{(1-M)/2}}{\Gamma\left(\frac{d-M+1}{2}\right)}\int_{G_{1}/G_{2}>0}\\!\\!\mathrm{d}^{M}s_{1,\bullet}\
\ f(s)\frac{G_{1}^{(d-M-1)/2}}{G_{2}^{(d-M)/2}}.$ (A.18)

Transforming the remaining loop integrations analogously, all but two of the
Gram determinants cancel, and we conclude that

$\prod_{i=1}^{L}\int\frac{\mathrm{d}^{d}\ell_{i}}{\pi^{d/2}}f(s)=\frac{\pi^{-LE/2-L(L-1)/4}}{\Gamma\left(\frac{d-L-E+1}{2}\right)\cdots\Gamma\left(\frac{d-E}{2}\right)}\int\mathrm{d}^{N}s_{\bullet,\bullet}\
f(s)\cdot\frac{G_{1}^{(d-M-1)/2}}{G_{L+1}^{(d-E-1)/2}}$ (A.19)

where $M=L+E$ is the sum of the number $L$ of loops and the number $E$ of
linearly independent external momenta. Note that $G_{L+1}=\det(p_{i}\cdot
p_{j})_{1\leq i,j\leq E}$ is the Gram determinant of the external momenta and
independent of the integration variables.

###### Proof of Theorem 29.

Since every denominator $\mathsf{D}_{e}$ is a linear combination of $s_{i,j}$
(and some loop momentum independent constant $\lambda_{e}$) according to
(2.26), an affine change of variables allows us to integrate over the values
of $\mathsf{D}_{e}$’s instead of $s_{i,j}$’s. This transformation only
introduces a constant Jacobian
$c^{\prime}=\det(\mathcal{A}^{a}_{\left\\{i,j\right\\}})$. We set
$f(s)=\prod_{e}\mathsf{D}_{e}^{-\nu_{e}}$ in the formula (A.19) above. The
(Euclidean) integration domain is determined, according to (A.17), by
$0<\|\mathrm{pr}_{\operatorname{lin}\left\\{q_{n+1},\ldots,q_{M}\right\\}^{\perp}}^{\perp}(q_{n})\|^{2}=G_{n}/G_{n+1}$
for $1\leq n\leq L$. Therefore, a point on the boundary of the integration
domain is determined by $G_{n}=0$ for some $1\leq n\leq L$, which is
equivalent to a linear dependence
$q_{n}\in\operatorname{lin}\left\\{q_{n+1},\ldots,q_{M}\right\\}$ and hence
implies $G_{1}=0$.

Note that we have to analytically continue (A.19) from Euclidean to Minkowski
space in order to obtain the Feynman integral (2.1). As Wick rotation turns
$\int\mathrm{d}^{d}\ell_{k}/(\mathrm{{i}}\pi^{d/2})$ exactly into the measure
$\int\mathrm{d}^{d}\ell_{k}/\pi^{d/2}$ on the left-hand side of (A.19), we
only have to remember that, due to our mostly-minus signature
$(1,-1,\cdots,-1)$ of the Minkowsi metric, the Euclidean scalar products on
the right-hand side of (A.19) receive a factor $(-1)$. For example, the
$M\times M$ determinant $G_{1}$ turns into $(-1)^{M}\mathsf{Gr}_{1}$;
similarly, $G_{L+1}$ becomes $(-1)^{E}\mathsf{Gr}$. Overall, analytic
continuation gives an additional factor of

$(-1)^{N}\cdot\frac{(-1)^{M(d-M-1)/2}}{(-1)^{E(d-E-1)/2}}=(-1)^{Ld/2}\cdot(-1)^{N+M(M-1)/2-E(E-1)/2}.$

We absorb the last factor, together with the Jacobian $c^{\prime}$, into the
constant prefactor $c$, and have thus finally arrived at (2.48). ∎

##  Appendix B The theory of Loeser-Sabbah

This section is devoted to the theorem 35, which was first stated in [58].
Beware that the original argument is flawed; a correct (but terse) proof was
given in [60]. Our aim here is to provide a simplified and more detailed
derivation.

Throughout we will consider modules $\mathscr{M}$ over the algebra
$\mathcal{D}^{N}_{k}={A}^{N}_{k}[x^{-1}]$ of differential operators (3.4) on
the torus in some number $N$ of variables $x_{i}$, over some field $k$ of
characteristic zero. To lighten the notation, let us abbreviate
$\theta_{i}\mathrel{\mathop{:}}=x_{i}\partial_{i}$ and set

$\mathscr{M}(\theta_{i},\ldots,\theta_{N})\mathrel{\mathop{:}}=k(\theta_{i},\ldots,\theta_{N})\otimes_{k[\theta_{i},\ldots,\theta_{N}]}\mathscr{M}$
(B.1)

for the _algebraic Mellin transform_ [59, section 1.2]. We begin with the
finite-dimensionality, which was proven in [59, Lemme 1.2.2]:

######  Lemma 65.

Let $\mathscr{M}$ denote a holonomic $\mathcal{D}^{N}_{k}$-module. Then, for
any $1\leq i\leq N$, its algebraic Mellin transform
$\mathscr{M}(\theta_{i},\ldots,\theta_{N})$ is a holonomic
$\mathcal{D}^{i-1}_{k(\theta_{i},\ldots,\theta_{N})}$-module.

######  Corollary 66.

The full Mellin transform $\mathscr{M}(\theta_{1},\ldots,\theta_{N})$ is a
finite-dimensional vector space over the field
$k(\theta_{1},\ldots,\theta_{N})$.

###### Proof of lemma 65.

Since
$\mathscr{M}(\theta_{i},\ldots,\theta_{N})=[\mathscr{M}(\theta_{i+1},\ldots,\theta_{N})](\theta_{i})$,
it suffices (by induction over $i$) to consider the case $i=N$. Introducing a
new indeterminate $\nu$, we extend the scalars from $k$ to $k[\nu]$ to obtain
a $\mathcal{D}^{N}_{k[\nu]}$-module
$\mathscr{M}[\nu]\mathrel{\mathop{:}}=\mathscr{M}\otimes_{k}k[\nu]$. It sits
in an exact sequence

$0\longrightarrow\mathscr{M}[\nu]\xrightarrow{\partial_{N}+\nu/x_{N}}\mathscr{M}[\nu]\xrightarrow{\sum_{j}\nu^{j}m_{j}\mapsto\sum_{j}(-x_{N}\partial_{N})^{j}m_{j}}\mathfrak{M}\longrightarrow
0$

of $\mathcal{D}^{N-1}_{k[\nu]}$-modules, where $\mathfrak{M}=\mathscr{M}$
denotes the initial module $\mathscr{M}$ with the action of $\nu$ defined as
$\nu m\mathrel{\mathop{:}}=-x_{N}\partial_{N}m$. Since $k(\nu)$ is flat, this
sequence remains exact after tensoring with $k(\nu)$ over $k[\nu]$. Through
identification of $\nu$ with $-\theta_{N}$, we conclude that

$\frac{\mathscr{M}\otimes_{k}k(\nu)}{(\partial_{N}+\nu/x_{N})(\mathscr{M}\otimes_{k}k(\nu))}\cong\mathfrak{M}\otimes_{k[\nu]}k(\nu)\cong\mathscr{M}\otimes_{k[\theta_{N}]}k(\theta_{N})=\mathscr{M}(\theta_{N})$

are isomorphic as $\mathcal{D}^{N-1}_{k(\nu)}$-modules. The left hand side is
the quotient $\mathscr{M}x_{N}^{\nu}/\partial_{N}\mathscr{M}x_{N}^{\nu}$ of
the $\mathcal{D}^{N}_{k(\nu)}$-module
$\mathscr{M}x_{N}^{\nu}\mathrel{\mathop{:}}=\mathscr{M}\otimes_{k}k(\nu)$
defined by the original action of $\mathcal{D}^{N-1}_{k}$ and $x_{N}^{\pm 1}$
on $\mathscr{M}$, but twisting the operator $\partial_{N}$ to act like
$\partial_{N}+\nu/x_{N}$.393939This just encodes the natural action
$\partial_{N}x_{N}^{\nu}m=x_{N}^{\nu}(\partial_{N}+\nu/x_{N})m$ on products of
elements $m$ of $\mathscr{M}$ with the function $x_{N}^{\nu}$—hence the
suggestive notation $\mathscr{M}x_{N}^{\nu}$.  The holonomicity of
$\mathscr{M}$ implies that $\mathscr{M}x_{N}^{\nu}$ is also
holonomic,404040Given a good filtration $\Gamma_{\bullet}$ of $\mathscr{M}$,
$\Gamma_{j}^{\prime}(\mathscr{M}x_{N}^{\nu})\mathrel{\mathop{:}}=(x_{N}^{-j}\Gamma_{2j}\mathscr{M})\otimes_{k}k(\nu)$
defines a filtration of $\mathscr{M}x_{N}^{\nu}$ with
$\dim_{k(\nu)}\Gamma_{j}^{\prime}(\mathscr{M}x_{N}^{\nu})\leq\dim_{k}\Gamma_{2j}\mathscr{M}\leq
c\cdot(2j)^{N}$ for some $c<\infty$.  and hence its push-forward
$\pi_{\ast}(\mathscr{M}x_{N}^{\nu})=\mathscr{M}x_{N}^{\nu}/\partial_{N}(\mathscr{M}x_{N}^{\nu})\cong\mathscr{M}(\theta_{N})$,
with respect the projection
$\pi\colon\mathbbm{G}_{m,k(\nu)}^{N}\relbar\joinrel\twoheadrightarrow\mathbbm{G}_{m,k(\nu)}^{N-1}$
that forgets the last coordinate, is also holonomic. ∎

Now we want to relate this dimension to the de Rham complex
$\operatorname{DR}(\mathscr{M})=K(\mathscr{M};\partial)$, which is a special
case of the Koszul complex:

######  Definition 67.

For commuting $k$-linear endomorphisms $s=(s_{1},\ldots,s_{N})$ of a
$k$-vector space $\mathscr{M}$, let

$K(\mathscr{M};s_{1},\ldots,s_{N})\mathrel{\mathop{:}}=\big{(}\Lambda^{\bullet}k\otimes_{k}\mathscr{M}[N],\mathrm{d}\big{)}$
(B.2)

denote the Koszul complex with $r$-forms sitting in degree $r-N$. The $r-N$
cochains

$K^{r-N}(\mathscr{M};s)=\Lambda^{r}\mathscr{M}=\bigoplus_{\left\lvert
I\right\rvert=r}\mathsf{e}_{I}\otimes\mathscr{M}$

have natural coordinates with respect to the basis
$\mathsf{e}_{I}\mathrel{\mathop{:}}=\mathsf{e}_{i_{1}}\wedge\cdots\wedge\mathsf{e}_{i_{r}}\in\Lambda^{r}k$
indexed by $r$-sets $I=\left\\{i_{1}<\cdots<i_{r}\right\\}$. The cochain map
reads $\mathrm{d}(\mathsf{e}_{I}\otimes m)=\sum_{i\notin
I}(\mathsf{e}_{i}\wedge\mathsf{e}_{I})\otimes s_{i}m$.

######  Remark 68.

Since $\theta_{j}\left(\prod_{i\in I}x_{i}\right)m=\left(\prod_{i\in
I\cup\left\\{j\right\\}}x_{i}\right)\partial_{j}m$ (for $j\notin I$), the rule

$K(\mathscr{M};\partial)\longrightarrow
K(\mathscr{M};\theta),\quad\mathsf{e}_{I}\otimes m\mapsto\mathsf{e}_{I}\otimes
x^{I}m$ (B.3)

defines a cochain map. It has an inverse, defined by $\mathsf{e}_{I}\otimes
m\mapsto\mathsf{e}_{I}\otimes x^{-I}m$. We thus conclude that
$\operatorname{DR}(\mathscr{M})=K(\mathscr{M};\partial)$ and
$K(\mathscr{M};\theta)$ are quasi-isomorphic and therefore share the same
Euler characteristic.

We prove theorem 35 by an induction over the number of variables. The base
case is

######  Theorem 69 ([58, Théorème 1]).

If $\mathscr{M}$ denotes a holonomic $\mathcal{D}^{1}_{k}$-module, then

$\dim_{k(\theta_{1})}\mathscr{M}(\theta_{1})=\chi(\mathscr{M})\mathrel{\mathop{:}}=\dim_{k}\frac{\mathscr{M}}{\partial_{1}\mathscr{M}}-\dim_{k}\ker(\partial_{1}).$
(B.4)

###### Proof.

We can pick a generator of $\mathscr{M}$ (by holonomicity, $\mathscr{M}$ is
cyclic as a $\mathcal{D}^{1}_{k}$-module) and extend it to a (finite) basis of
$\mathscr{M}(\theta_{1})$ as a vector space over $k(\theta_{1})$, due to
corollary 66. Let $\mathscr{N}\subset\mathscr{M}$ denote the
$k[\theta_{1}]$-module generated by such a basis, hence
$\mathscr{N}(\theta_{1})=\mathscr{M}(\theta_{1})$. By construction,
$\mathscr{M}={A}^{1}_{k}\mathscr{N}=\sum_{j\in\mathbbm{Z}}x_{1}^{j}\mathscr{N}$
is exhaustively filtered by the finitely generated $k[\theta_{1}]$-modules
$\mathscr{N}_{j}\mathrel{\mathop{:}}=\sum_{i=-j}^{j}x_{1}^{i}\mathscr{N}$.414141Due
to $\theta_{1}x_{1}^{i}\mathscr{N}=x_{1}^{i}(\theta_{1}+i)\mathscr{N}\subseteq
x_{1}^{i}\mathscr{N}\subseteq\mathscr{N}_{j}$, indeed $\mathscr{N}_{j}$ is a
$k[\theta_{1}]$-module.

Since
$\mathscr{M}(\theta_{1})=\mathscr{N}(\theta_{1})=\mathscr{N}_{1}(\theta_{1})$
is finitely generated, there is a non-zero polynomial $b(\theta_{1})\in
k[\theta_{1}]$ such that $b(\theta_{1})\mathscr{N}_{1}\subseteq\mathscr{N}$.
Therefore, using $(\theta_{1}-1)x_{1}=x_{1}\theta_{1}$,

$b(\theta_{1}\mp j)x_{1}^{\pm(j+1)}\mathscr{N}=x_{1}^{\pm
j}b(\theta_{1})x_{1}^{\pm 1}\mathscr{N}\subseteq x_{1}^{\pm
j}b(\theta_{1})\mathscr{N}_{1}\subseteq x_{1}^{\pm
j}\mathscr{N}\subseteq\mathscr{N}_{j}$

shows that the polynomials
$b_{j+1}(\theta_{1})\mathrel{\mathop{:}}=b(\theta_{1}+j)b(\theta_{1}-j)\in
k[\theta_{1}]$ have the property
$b_{j+1}(\theta_{1})\mathscr{N}_{j+1}\subseteq\mathscr{N}_{j}$. Let
$Z=b^{-1}(0)$ denote the zeroes of $b_{1}=b^{2}$, then note that the zeroes of
$b_{j+1}$ are $(Z+j)\cup(Z-j)$ and get pushed away from zero for increasing
$j$. In particular, there exists some $j_{0}\in\mathbbm{N}$ such that
$b_{j}(0)\neq 0$ for all $j>j_{0}$. For each such value of $j$, we can find
$u_{j},v_{j}\in k[\theta_{1}]$ such that
$1=u_{j}(\theta_{1})b_{j}(\theta_{1})+v_{j}(\theta_{1})\theta_{1}$; then

$m=1\cdot
m=u_{j}(\theta_{1})b_{j}(\theta_{1})m+v_{j}(\theta_{1})\theta_{1}m\in\mathscr{N}_{j-1}+v_{j}(\theta_{1})\theta_{1}m$

holds for every $m\in\mathscr{N}_{j}$. This proves
$\ker(\theta_{1})\cap\mathscr{N}_{j}\subseteq\mathscr{N}_{j-1}$ for all
$j>j_{0}$, and therefore $\ker(\theta_{1})\subseteq\mathscr{N}_{j_{0}}$.
Similarly, we conclude
$\mathscr{M}/(\theta_{1}\mathscr{M})\cong\mathscr{N}_{j_{0}}/(\mathscr{N}_{j_{0}}\cap\theta_{1}\mathscr{M}$).
But given some $m=\theta_{1}x\in\mathscr{N}_{j_{0}}$ with
$x\in\mathscr{N}_{j}$,
$x=u_{j}(\theta_{1})b_{j}(\theta_{1})x+v_{j}(\theta_{1})m\in\mathscr{N}_{j-1}+\mathscr{N}_{j_{0}}$
proves that
$\mathscr{N}_{j_{0}}\cap\theta_{1}(\mathscr{N}_{j})=\mathscr{N}_{j_{0}}\cap\theta_{1}(\mathscr{N}_{j-1})$
for all $j>j_{0}$. In consequence, we have proven that

$\ker(\partial_{1})=\ker(\theta_{1})=\ker\left(\left.\theta_{1}\right|_{\mathscr{N}_{j_{0}}}\right)\quad\text{and}\quad\frac{\mathscr{M}}{\partial_{1}(\mathscr{M})}\cong\frac{\mathscr{M}}{\theta_{1}(\mathscr{M})}\cong\frac{\mathscr{N}_{j_{0}}}{\theta_{1}(\mathscr{N}_{j_{0}})};$

in other words, the Koszul complexes
$\operatorname{DR}(\mathscr{M})=K(\mathscr{M};\partial_{1})$ and
$K(\mathscr{N}_{j_{0}};\theta_{1})$ are quasi-isomorphic (see remark 68). The
statement of the theorem thus reduces to the identity

$\dim_{k(\theta_{1})}\mathscr{N}_{j_{0}}(\theta_{1})=\chi\left(K(\mathscr{N}_{j_{0}};\theta_{1})\right)$

for a finitely generated $k[\theta_{1}]$-module $\mathscr{N}_{j_{0}}$. Since
both sides are additive under short exact sequences, this claim reduces (via a
finite free resolution) to the case of a free rank one $k[\theta_{1}]$-module,
i.e. $k[\theta_{1}]$, which is clear:
$k[\theta_{1}](\theta_{1})=k(\theta_{1})$ is of dimension one over
$k(\theta_{1})$, while $\ker(\theta_{1})=\left\\{0\right\\}$ is trivial and
$k[\theta_{1}]/(\theta_{1}k[\theta_{1}])=k$ is one-dimensional. ∎

With this starting point, we can now prove theorem 35 by induction. In fact,
the higher dimensional case can be seen as a straightforward corollary of the
univariate case above. In contrast to [60], our demonstration avoids any
reference to higher-dimensional lattices.

###### Proof of theorem 35.

Let $\mathscr{M}$ denote a holonomic $\mathcal{D}^{N}_{k}$-module, and suppose
we have proven theorem 35 for all holonomic modules in less than $N$
variables. In particular, we may invoke the claim for the
$\mathcal{D}^{N-1}_{k}$-modules $\ker\partial_{N}$ and
$\mathscr{M}/\partial_{N}\mathscr{M}$, as these are holonomic because they are
the cohomologies of the complex

$0\longrightarrow\mathscr{M}\xrightarrow{\partial_{N}}\mathscr{M}\longrightarrow
0$

which computes the push-forward of $\mathscr{M}$ along the projection
$\pi\colon\mathbbm{G}_{m,k}^{N}\longrightarrow\mathbbm{G}_{m,k}^{N-1}$ that
forgets the last coordinate. So we already know that

$\chi\left(\frac{\mathscr{M}}{\partial_{N}\mathscr{M}}\right)=\dim_{k^{\prime}}\left(\frac{\mathscr{M}}{\partial_{N}\mathscr{M}}(\theta^{\prime})\right)=\dim_{k^{\prime}}\frac{\mathscr{M}^{\prime}}{\partial_{N}\mathscr{M}^{\prime}}$

where $k^{\prime}\mathrel{\mathop{:}}=k(\theta^{\prime})$ and
$\mathscr{M}^{\prime}\mathrel{\mathop{:}}=\mathscr{M}(\theta^{\prime})$ with
$\theta^{\prime}\mathrel{\mathop{:}}=(\theta_{1},\ldots,\theta_{N-1})$.
Analogously,
$\chi(\ker\partial_{N})=\dim_{k^{\prime}}\ker(\partial_{N}^{\prime})$, where
$\partial_{N}^{\prime}$ denotes the action of $\partial_{N}$ on
$\mathscr{M}^{\prime}$. In conclusion, we know that

$\displaystyle\chi\left(\mathscr{M}/(\partial_{N}\mathscr{M})\right)-\chi\left(\ker\partial_{N}\right)$
$\displaystyle=\dim_{k^{\prime}}\mathscr{M}^{\prime}/(\partial_{N}\mathscr{M}^{\prime})-\dim_{k^{\prime}}\ker(\partial_{N}^{\prime})=\chi(\mathscr{M}^{\prime})$
$\displaystyle=\dim_{k^{\prime}(\theta_{N})}\mathscr{M}^{\prime}(\theta_{N})=\dim_{k(\theta)}\mathscr{M}(\theta),$

where we recognized the first line as the Euler characteristic of the de Rham
complex of the $\mathcal{D}^{1}_{k^{\prime}}$-module $\mathscr{M}^{\prime}$
and applied theorem 69 to get to the last line
($\mathscr{M}^{\prime}=\mathscr{M}(\theta^{\prime})$ is holonomic by lemma
65). So we only need to show that the left hand side is equal to
$\chi(\mathscr{M})$.

This is well-known and follows from the Grothendieck spectral
sequence.424242Let
$\pi^{N}\colon\mathbbm{G}_{m,k}^{N}\longrightarrow\left\\{\text{pt}\right\\}$
denote the projection to a point, such that $\pi^{N}=\pi^{N-1}\circ\pi$. The
identity $\pi^{N}_{+}=\pi^{N-1}_{+}\circ\pi_{+}$ of the corresponding push-
forwards in the derived category of $\mathcal{D}^{N}_{k}$-modules implies that
$\chi(\operatorname{DR}(\mathscr{M}))=\chi(\pi^{N}_{+}(\mathscr{M}))=\sum_{i}(-1)^{i}\chi(H^{i}(\pi_{+}\mathscr{M}))$,
where $H^{-1}(\pi_{+}\mathscr{M})=\ker\partial_{N}$ and
$H^{0}(\pi_{+}\mathscr{M})=\mathscr{M}/(\partial_{N}\mathscr{M})$.
Alternatively, an elementary way to obtain the identity
$\chi(\mathscr{M}/(\partial_{N}\mathscr{M}))-\chi(\ker\partial_{N})=\chi(\mathscr{M})$
is given by the long exact sequence

$\cdots\rightarrow H^{i+1}(\operatorname{DR}(\ker\partial_{N}))\rightarrow
H^{i}(\operatorname{DR}(\mathscr{M}))\rightarrow
H^{i}(\operatorname{DR}(\mathscr{M}/\partial_{N}\mathscr{M}))\rightarrow
H^{i+2}(\operatorname{DR}(\ker\partial_{N}))\rightarrow\cdots$ (B.5)

in de Rham cohomology [12, Ch. 2, Proposition 4.13]. ∎

For completeness, let us demonstrate (B.5), by following the standard
construction in the proof of Kashiwara’s theorem: First note that

$\mathscr{N}\mathrel{\mathop{:}}=\left\\{m\in\mathscr{M}\colon\partial_{N}^{k}m=0\quad\text{for
some}\quad k>0\right\\}\subseteq\mathscr{M}$

defines a $\mathcal{D}^{N}_{k}$-submodule of $\mathscr{M}$.434343One only
needs to check that $x_{N}^{\pm 1}\mathscr{N}\subseteq\mathscr{N}$, which
follows from $\partial_{N}^{k+1}x_{N}^{\pm
1}m=\big{[}(k+1)(\partial_{N}x_{N}^{\pm 1})+x_{N}^{\pm
1}\partial_{N})\partial_{N}^{k}m=0$ whenever $\partial_{N}^{k}m=0$.  The exact
sequence
$0\rightarrow\mathscr{N}\rightarrow\mathscr{M}\rightarrow\\!\mathscr{M}/\mathscr{N}\rightarrow
0$ of holonomic $\mathcal{D}^{N}_{k}$-modules induces a sequence of de Rham
complexes,
$0\rightarrow\operatorname{DR}(\mathscr{N})\rightarrow\operatorname{DR}(\mathscr{M})\rightarrow\operatorname{DR}(\mathscr{M}/\mathscr{N})\rightarrow
0$, which is also exact
($\mathscr{M}\mapsto\operatorname{DR}(\mathscr{M})=\Lambda^{\bullet}k\otimes_{k}\mathscr{M}$
is exact by flatness of $\Lambda^{\bullet}k$). Hence we get a long exact
sequence in cohomology:

$\cdots\rightarrow H^{i}(\operatorname{DR}(\mathscr{N}))\rightarrow
H^{i}(\operatorname{DR}(\mathscr{M}))\rightarrow
H^{i}(\operatorname{DR}(\mathscr{M}/\mathscr{N}))\rightarrow
H^{i+1}(\operatorname{DR}(\mathscr{N}))\rightarrow\cdots$ (B.6)

The key observation now is that $\partial_{N}$ is injective on
$\mathscr{M}/\mathscr{N}$ and surjective on $\mathscr{N}$.444444The first
statement is clear since $\ker\partial_{N}\subseteq\mathscr{N}$. The second
claim follows from the identity $x^{k}\partial^{k}=(\partial
x-k)x^{k-1}\partial^{k-1}=\cdots=\prod_{i=1}^{k}(\partial x-i)$, which implies
$0=x_{N}^{k}\partial_{N}^{k}m=(-1)^{k}(k!)m\mod\partial_{N}\mathscr{N}$
whenever $\partial_{N}^{k}m=0$.  We thus get short exact sequences

$\displaystyle
0\longrightarrow\ker\left(\partial_{N}\right)\lhook\joinrel\longrightarrow\mathscr{N}\xrightarrow{\partial_{N}}\mathscr{N}\longrightarrow
0\quad\text{and}$ $\displaystyle
0\longrightarrow\mathscr{M}/\mathscr{N}\xrightarrow{\partial_{N}}\mathscr{M}/\mathscr{N}\relbar\joinrel\twoheadrightarrow\mathscr{M}/\partial_{N}\mathscr{M}\longrightarrow
0$

of $\mathcal{D}^{N-1}_{k}$-modules. The induced short exact sequences of de
Rham complexes provide quasi-isomorphisms
$\operatorname{DR}(\ker\partial_{N})\simeq\operatorname{DR}(\mathscr{N})[1]$
and
$\operatorname{DR}(\mathscr{M}/\partial_{N}\mathscr{M})\simeq\operatorname{DR}(\mathscr{M}/\mathscr{N})$,
because the de Rham complex of a $\mathcal{D}^{N}_{k}$-module $\mathscr{N}$ is
the mapping cone of the map $\partial_{N}\colon
K(\mathscr{N};\partial^{\prime})\longrightarrow
K(\mathscr{N};\partial^{\prime})$. Hence we obtain (B.5) from (B.6) due to

$H^{i}(\operatorname{DR}(\mathscr{N}))\cong
H^{i+1}(\operatorname{DR}(\ker\partial_{N}))\quad\text{and}\quad
H^{i}(\operatorname{DR}(\mathscr{M}/\mathscr{N}))\cong
H^{i}(\operatorname{DR}(\mathscr{M}/(\partial_{N}\mathscr{M}))).$

To clarify this final step, first note that separating $\partial_{N}$ from
$\partial^{\prime}\mathrel{\mathop{:}}=(\partial_{1},\ldots,\partial_{N-1})$
yields an isomorphism of $k$-vector spaces

$\displaystyle\Phi\colon K^{\bullet-N}(\mathscr{N};\partial^{\prime})\oplus
K^{\bullet-1-N}(\mathscr{N};\partial^{\prime})$
$\displaystyle\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}K^{\bullet-N}(\mathscr{N};\partial)=\bigoplus_{N\notin
I}\mathsf{e}_{I}\otimes\mathscr{N}\oplus\bigoplus_{N\in
I}\mathsf{e}_{I}\otimes\mathscr{N}$ $\displaystyle x\oplus
y\phantom{\Lambda^{r-1}K(\mathscr{N};\partial^{\prime})}$ $\displaystyle\
\mapsto\ x\oplus(\mathsf{e}_{N}\wedge y).$

In this representation, the differential is given by

$\Phi^{-1}\left(\mathrm{d}\Phi(x\oplus
y)\right)={\mathrm{d}}^{\prime}(x)\oplus(\partial_{N}x-{\mathrm{d}}^{\prime}(y)),$

which is known as the mapping cone of $\partial_{N}\colon
K(\mathscr{N};\partial^{\prime})\longrightarrow
K(\mathscr{N};\partial^{\prime})$. Here we denote by ${\mathrm{d}}^{\prime}$
the differential of $K(\mathscr{N};\partial^{\prime})$.

If $\partial_{N}$ is surjective, we can find $x$ with $\partial_{N}x=y$ and
hence $\mathrm{d}\Phi(x\oplus 0)=\Phi({\mathrm{d}}^{\prime}x\oplus y)$ for
every $y$. Therefore, every element of $K(\mathscr{N};\partial)$ has a
representative of the form $\Phi(x\oplus 0)$, modulo exact forms. But such a
form is closed, $\mathrm{d}\Phi(x\oplus 0)=0$, if and only if
$x\in\ker\partial_{N}\cap\ker{\mathrm{d}}^{\prime}$.

######  Corollary 70.

If $\partial_{N}$ is surjective, then $K(\mathscr{N};\partial)$ and
$K(\ker\partial_{N};\partial^{\prime})[1]$ are quasi-isomorphic. If
$\partial_{N}$ is injective, then $K(\mathscr{N};\partial)$ and
$K(\mathscr{N}/\partial_{N}\mathscr{N};\partial^{\prime})$ are quasi-
isomorphic.

The proof of the second statement is very similar to the surjective case and
left as a straightforward exercise.

##  Appendix C A two-loop example

We demonstrate some main points of this article by a pedagogical example.
Consider the massless two-loop two-point graph with five propagators, graph
$\mathrm{WS}_{3}^{\prime}$ in figure 4.2. To this graph we associate the
family of integrals

$\mathcal{I}(\nu_{1},\ldots,\nu_{5})=\int\frac{\mathrm{d}^{d}l_{1}}{\mathrm{{i}}\pi^{d/2}}\int\frac{\mathrm{d}^{d}l_{2}}{\mathrm{{i}}\pi^{d/2}}\frac{1}{[-l_{1}^{2}]^{\nu_{1}}[-l_{2}^{2}]^{\nu_{2}}[-(l_{2}-p)^{2}]^{\nu_{3}}[-(l_{1}-p)^{2}]^{\nu_{4}}[-(l_{1}-l_{2})^{2}]^{\nu_{5}}}$
(C.1)

with two loop-momenta $q_{1}=l_{1},\,q_{2}=l_{2}$ and one external momentum
$q_{3}=p$. We normalize to $-p^{2}=1$. The graph polynomial
$\mathcal{G}=\mathcal{U}+\mathcal{F}$ is given by the Symanzik polynomials

$\displaystyle\mathcal{U}$
$\displaystyle=(x_{1}+x_{4})(x_{2}+x_{3})+x_{5}(x_{1}+x_{2}+x_{3}+x_{4})\quad\text{and}$
$\displaystyle\mathcal{F}$
$\displaystyle=x_{1}x_{2}(x_{3}+x_{4})+x_{3}x_{4}(x_{1}+x_{2})+x_{5}(x_{1}+x_{2})(x_{3}+x_{4}).$

By definition 6 the modified Feynman integral

$\widetilde{\mathcal{I}}(\nu_{1},\ldots,\nu_{5})=\mathcal{M}\\!\left\\{\mathcal{G}^{s}\right\\}$
(C.2)

is related to the Feynman integral by

$\mathcal{I}(\nu_{1},\ldots,\nu_{5})=\frac{\Gamma(-s)}{\Gamma(-s-\omega)}\widetilde{\mathcal{I}}(\nu_{1},\ldots,\nu_{5})$
(C.3)

with $\omega=\nu_{1}+\nu_{2}+\nu_{3}+\nu_{4}+\nu_{5}+2s$ and $s=-d/2$.

###  C.1 From annihilators to integral relations

A set of generators of the annihilator ideal
$\operatorname{Ann}_{A_{5}[s]}\left(\mathcal{G}^{s}\right)$ can be derived in
Singular [24] with algorithms introduced in [4]. Using the command SannfsBM we
obtain a set of 13 generators. To give an impression, the first five of them
read

$\displaystyle P_{1}$
$\displaystyle=\partial_{1}-\partial_{2}+(x_{3}+x_{4}+1)(\partial_{3}-\partial_{4})+(x_{3}-x_{4})\partial_{5},$
$\displaystyle P_{2}$
$\displaystyle=(x_{2}-1)\partial_{1}+(-2x_{1}-3x_{2}-1)\partial_{2}+(-x_{4}+1)\partial_{3}+(2x_{3}+3x_{4}+1)\partial_{4}+(2x_{1}-x_{2}$
$\displaystyle-2x_{3}+x_{4})\partial_{5},$ $\displaystyle P_{3}$
$\displaystyle=(x_{1}+2)\partial_{1}+(x_{1}+2x_{2})\partial_{2}+x_{4}\partial_{3}+(-2x_{3}-3x_{4}-2)\partial_{4}+(-x_{1}+2x_{3}-x_{4})\partial_{5},$
$\displaystyle P_{4}$
$\displaystyle=\partial_{1}x_{1}+(x_{2}+1)\partial_{2}+(-x_{3}-x_{4}-1)\partial_{3}+(x_{4}+x_{5})\partial_{5}-s,$
$\displaystyle P_{5}$
$\displaystyle=(x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{1}+x_{2}+x_{3}+x_{4})\partial_{4}+(-x_{1}x_{2}-x_{1}x_{3}-x_{1}x_{5}$
$\displaystyle-x_{2}x_{3}-x_{2}x_{5}-x_{2}-x_{3}-x_{5})\partial_{5}.$

We notice that this set includes one generator which is quadratic in the
differential operators, reading

$\displaystyle P_{13}$
$\displaystyle=(x_{4}+x_{5}-2)\partial_{1}^{2}+(2x_{2}+x_{5})\partial_{2}^{2}+(-2x_{3}^{2}+2x_{3}x_{4}-x_{3}x_{5}+x_{4}x_{5})\partial_{5}^{2}-x_{3}x_{4}\partial_{3}\partial_{5}$
$\displaystyle+(2x_{3}^{2}+x_{3}x_{4}-2x_{4}^{2}+2x_{3}-2x_{4})\partial_{4}\partial_{5}+(x_{3}-x_{4}+(-2x_{2}-x_{4}-2x_{5}+2)\partial_{2}$
$\displaystyle+(x_{4}^{2}-x_{4})\partial_{3}+(-x_{4}^{2}+2x_{3}+3x_{4}+2)\partial_{4}+(x_{3}x_{4}+x_{3}x_{5}-x_{4}^{2}-x_{4}x_{5}-4x_{3}+2x_{4}$
$\displaystyle-
x_{5})\partial_{5})\partial_{1}+(x_{3}x_{4}\partial_{3}-x_{3}+x_{4}+(-x_{3}x_{4}-2x_{3}-2x_{4}-2)\partial_{4}+(-2x_{2}x_{3}+2x_{2}x_{4}$
$\displaystyle-
x_{3}x_{5}+x_{4}x_{5}+2x_{3}+x_{5})\partial_{5}+2)\partial_{2}+(-x_{3}+x_{4})\partial_{5}.$

Every operator in
${\operatorname{Ann}_{A_{5}[s]}\left(\mathcal{G}^{s}\right)}$ gives rise to an
integral relation. According to lemma 7, we just need to replace each $x_{i}$
by $\hat{\mathbf{i}}^{+}$ and each $\partial_{i}$ by $-\mathbf{i}^{-}$ to
obtain a shift relation between modified Feynman integrals. For example, for
the generator $P_{1}$, we obtain the shift operator

$\mathcal{M}\\!\left\\{P_{1}\right\\}=-\mathbf{1}^{-}+\mathbf{2}^{-}-(\hat{\mathbf{3}}^{+}+\hat{\mathbf{4}}^{+}+1)(\mathbf{3}^{-}-\mathbf{4}^{-})-(\hat{\mathbf{3}}^{+}-\hat{\mathbf{4}}^{+})\mathbf{5}^{-}$

satisfying

$\mathcal{M}\\!\left\\{P_{1}\right\\}\widetilde{\mathcal{I}}(\nu_{1},\ldots,\nu_{5})=0.$

For the (unmodified) Feynman integral $\mathcal{I}(\nu_{1},\ldots,\nu_{5})$ we
obtain the corresponding shift relation via (2.22). From $P_{1}$ we obtain the
operator

$\boldsymbol{O_{1}}=\left(\sum_{i=1}^{5}\mathbf{n}_{i}+3s\right)(\mathbf{1}^{-}-\mathbf{2}^{-}+\mathbf{3}^{-}-\mathbf{4}^{-})-(\hat{\mathbf{3}}^{+}+\hat{\mathbf{4}}^{+})(\mathbf{3}^{-}-\mathbf{4}^{-})-(\hat{\mathbf{3}}^{+}-\hat{\mathbf{4}}^{+})\mathbf{5}^{-}$
(C.4)

which satisfies

$\boldsymbol{O_{1}}\mathcal{I}(\nu_{1},\ldots,\nu_{5})=0.$ (C.5)

###  C.2 Linear annihilators

It is useful to consider the linear annihilators
$\operatorname{Ann}^{1}_{{A}^{N}[s]}(\mathcal{G}^{s})\subseteq\operatorname{Ann}_{{A}^{N}[s]}(\mathcal{G}^{s}).$
Recall that they are the annihilators which are linear differential operators,
being of the form $P=q+\sum_{i=1}^{N}p_{i}\partial_{i}$ with
$p,q_{1},\ldots,q_{N}\in\mathbbm{Q}[s,x_{1},\ldots,x_{N}]$. We compute the
generators of $\operatorname{Ann}^{1}(\mathcal{G}^{s})$ as generators of the
Syzygy-module of
$(\mathcal{G},\partial_{1}\mathcal{G},\ldots,\partial_{N}\mathcal{G})$, using
the Singular command syz. For our example we obtain the following 8
generators:

$\displaystyle L_{1}$
$\displaystyle=\partial_{1}-\partial_{2}+(x_{3}+x_{4}+1)\partial_{3}+(-x_{3}-x_{4}-1)\partial_{4}+(x_{3}-x_{4})\partial_{5},$
$\displaystyle L_{2}$
$\displaystyle=(-x_{2}-1)\partial_{1}+(x_{2}+1)\partial_{2}+(-x_{4}-1)\partial_{3}+(x_{4}+1)\partial_{4}+(x_{2}+x_{4}+2x_{5})\partial_{5}-2s,$
$\displaystyle L_{3}$
$\displaystyle=(2x_{1}+x_{2}+1)\partial_{1}+(x_{2}+1)\partial_{2}+(-2x_{3}-x_{4}-1)\partial_{3}+(-x_{4}-1)\partial_{4}+(-x_{2}+x_{4})\partial_{5},$
$\displaystyle L_{4}$
$\displaystyle=(x_{1}+x_{2}+1)\partial_{1}+(-x_{1}-x_{2}-1)\partial_{2}+\partial_{3}-\partial_{4}+(x_{1}-x_{2})\partial_{5},$
$\displaystyle L_{5}$
$\displaystyle=-2sx_{4}+(x_{4}+x_{5})\partial_{1}+(-2x_{2}-x_{4}-x_{5}-2)\partial_{2}+(2x_{3}x_{4}+x_{4}^{2}+2x_{3}+3x_{4}+2)\partial_{3}$
$\displaystyle+(x_{4}^{2}+x_{4})\partial_{4}+(-x_{4}^{2}-2x_{4}-x_{5})\partial_{5},$
$\displaystyle L_{6}$
$\displaystyle=(-x_{2}x_{3}+x_{2}x_{4}-x_{2}-x_{3}+2x_{4}+2x_{5})\partial_{1}+(-2x_{1}x_{3}+2x_{1}x_{4}-x_{2}x_{3}+x_{2}x_{4}+2x_{1}$
$\displaystyle-9x_{2}-x_{3}-2x_{5}-6)\partial_{2}+(-2x_{1}x_{3}-2x_{1}x_{4}-x_{2}x_{3}-x_{2}x_{4}-2x_{3}^{2}-4x_{3}x_{5}+2x_{4}^{2}$
$\displaystyle-4x_{4}x_{5}-2x_{1}-x_{2}+3x_{3}+6x_{4}-4x_{5}+6)\partial_{3}+(2x_{1}x_{3}+2x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}$
$\displaystyle+4x_{3}x_{4}+4x_{3}x_{5}+4x_{4}^{2}+4x_{4}x_{5}+2x_{1}+x_{2}+x_{3}+6x_{4}+4x_{5})\partial_{4}+(-2x_{4}^{2}-6x_{4}$
$\displaystyle-4x_{5})\partial_{5}+s(-4x_{4}+2),$ $\displaystyle L_{7}$
$\displaystyle=(2x_{1}x_{2}+2x_{2}^{2}+2x_{1}+4x_{2}+2)\partial_{1}+(x_{2}x_{4}+2x_{2}+x_{5})\partial_{3}+(-x_{2}x_{4}-2x_{2}-2x_{4}$
$\displaystyle-
x_{5}-2)\partial_{4}+(-2x_{2}^{2}-x_{2}x_{4}-2x_{2}x_{5}-2x_{2}-x_{5})\partial_{5},$
$\displaystyle L_{8}$
$\displaystyle=(x_{4}+x_{5})\partial_{1}+(2x_{1}x_{2}+2x_{1}x_{4}+2x_{1}x_{5}+2x_{2}^{2}+2x_{2}x_{4}+2x_{2}x_{5}+2x_{1}+2x_{2}+x_{4}$
$\displaystyle+x_{5})\partial_{2}+(-2x_{1}x_{3}-2x_{1}x_{4}-2x_{3}x_{5}+x_{4}^{2}-2x_{1}+2x_{3}-x_{4})\partial_{3}+(-x_{4}^{2}-2x_{4}x_{5}-x_{4}$
$\displaystyle-2x_{5})\partial_{4}+(-x_{4}^{2}-2x_{4}x_{5}-2x_{5}^{2}+x_{5})\partial_{5}+s(-2x_{2}+2x_{5}-2),$

Using Singular we find that for our two-loop example every generator ${P_{i}}$
can be expressed as a linear combination of the linear generators ${L_{i}}$
over $\operatorname{Ann}_{{A}^{5}[s]}(\mathcal{G}^{s})$. For instance, the
first five annihilators satisfy

$\displaystyle P_{1}$ $\displaystyle=L_{1},\quad
P_{2}=-2L_{1}-L_{3}+2L_{4},\quad P_{3}=2L_{1}+L_{3}-L_{4},\quad
2P_{4}=L_{2}+L_{3},$ $\displaystyle 2P_{5}$
$\displaystyle=-2(x_{1}+x_{2}+1)L_{1}+(1+x_{2}-x_{4}-x_{5})(L_{2}-L_{3})-2(x_{2}+x_{4}+x_{5})L_{4}+L_{5}+2L_{7}-L_{8}.$

We emphasize that such a relation exists for all the $P_{i}$. In particular
such a relation also exists for the quadratic $P_{13}$, which however is too
long to be shown here. As a consequence we can view
$\operatorname{Ann}_{{A}^{N}[s]}(\mathcal{G}^{s})$ as generated by the linear
$L_{i}$, which will simplify the discussion in section C.4.

###  C.3 From IBP relations to annihilators

Going in the other direction, we can derive annihilators from momentum-space
IBP relations. In the usual way, inserting the differential operators

$\mathbf{o}_{j}^{i}=\frac{\partial}{\partial q_{i}}q_{j}\quad\text{for}\quad
i\in\left\\{1,2\right\\}\quad\text{and}\quad j\in\left\\{1,2,3\right\\}$ (C.6)

we obtain six IBP relations $\mathbf{O}_{j}^{i}\mathcal{I}=0$ with the shift
operators

$\displaystyle\mathbf{O}_{1}^{1}$
$\displaystyle=-\hat{\mathbf{4}}^{+}\mathbf{1}^{-}-\hat{\mathbf{5}}^{+}\mathbf{1}^{-}+\hat{\mathbf{5}}^{+}\mathbf{2}^{-}+\hat{\mathbf{4}}^{+}-2\mathbf{n}_{1}-\mathbf{n}_{4}-\mathbf{n}_{5},$
$\displaystyle\mathbf{O}_{2}^{1}$
$\displaystyle=-\hat{\mathbf{1}}^{+}\mathbf{2}^{-}+\hat{\mathbf{1}}^{+}\mathbf{5}^{-}-\hat{\mathbf{4}}^{+}\mathbf{1}^{-}-\hat{\mathbf{4}}^{+}\mathbf{3}^{-}+\hat{\mathbf{4}}^{+}\mathbf{5}^{-}-\hat{\mathbf{5}}^{+}\mathbf{1}^{-}+\hat{\mathbf{5}}^{+}\mathbf{2}^{-}+\hat{\mathbf{4}}^{+}-\mathbf{n}_{1}+\mathbf{n}_{5},$
$\displaystyle\mathbf{O}_{3}^{1}$
$\displaystyle=-s\hat{\mathbf{1}}^{+}+s\hat{\mathbf{4}}^{+}-\hat{\mathbf{4}}^{+}\mathbf{1}^{-}-\hat{\mathbf{5}}^{+}\mathbf{1}^{-}+\hat{\mathbf{5}}^{+}\mathbf{2}^{-}-\hat{\mathbf{5}}^{+}\mathbf{3}^{-}+\hat{\mathbf{1}}^{+}\mathbf{4}^{-}+\hat{\mathbf{5}}^{+}\mathbf{4}^{-}-\mathbf{n}_{1}+\mathbf{n}_{4},$
$\displaystyle\mathbf{O}_{1}^{2}$
$\displaystyle=-\hat{\mathbf{2}}^{+}\mathbf{1}^{-}+\hat{\mathbf{2}}^{+}\mathbf{5}^{-}-\hat{\mathbf{3}}^{+}\mathbf{2}^{-}-\hat{\mathbf{3}}^{+}\mathbf{4}^{-}+\hat{\mathbf{3}}^{+}\mathbf{5}^{-}+\hat{\mathbf{5}}^{+}\mathbf{1}^{-}-\hat{\mathbf{5}}^{+}\mathbf{2}^{-}+\hat{\mathbf{3}}^{+}-\mathbf{n}_{2}+\mathbf{n}_{5},$
$\displaystyle\mathbf{O}_{2}^{2}$
$\displaystyle=-\hat{\mathbf{3}}^{+}\mathbf{2}^{-}+\hat{\mathbf{5}}^{+}\mathbf{1}^{-}-\hat{\mathbf{5}}^{+}\mathbf{2}^{-}+\hat{\mathbf{3}}^{+}-2s-2\mathbf{n}_{2}-\mathbf{n}_{3}-\mathbf{n}_{5}$
$\displaystyle\mathbf{O}_{3}^{2}$
$\displaystyle=\hat{\mathbf{2}}^{+}\mathbf{3}^{-}-\hat{\mathbf{3}}^{+}\mathbf{2}^{-}+\hat{\mathbf{5}}^{+}\mathbf{1}^{-}-\hat{\mathbf{5}}^{+}\mathbf{2}^{-}+\hat{\mathbf{5}}^{+}\mathbf{3}^{-}-\hat{\mathbf{5}}^{+}\mathbf{4}^{-}-\hat{\mathbf{2}}^{+}-\hat{\mathbf{3}}^{+}-\mathbf{n}_{2}+\mathbf{n}_{3}.$

Following the steps in the proof of corollary 21 we derive for each shift
opeartor $\mathbf{O}_{j}^{i}$ a parametric annihilator
$\widetilde{\mathbf{O}}_{j}^{i}$. We obtain

$\displaystyle\widetilde{\mathbf{O}}_{1}^{1}$
$\displaystyle=(x_{1}x_{4}+2x_{1}+x_{4}+x_{5})\partial_{1}+(x_{2}x_{4}-x_{5})\partial_{2}+x_{3}x_{4}\partial_{3}-3sx_{4}+(x_{4}^{2}+x_{4})\partial_{4}+(x_{4}x_{5}$
$\displaystyle+x_{5})\partial_{5}-2s,$
$\displaystyle\widetilde{\mathbf{O}}_{2}^{1}$
$\displaystyle=-3sx_{4}+(x_{1}x_{4}+x_{1}+x_{4}+x_{5})\partial_{1}+(x_{2}x_{4}+x_{1}-x_{5})\partial_{2}+(x_{3}x_{4}+x_{4})\partial_{3}+x_{4}^{2}\partial_{4}$
$\displaystyle+(x_{4}x_{5}-x_{1}-x_{4}-x_{5})\partial_{5},$
$\displaystyle\widetilde{\mathbf{O}}_{3}^{1}$
$\displaystyle=3sx_{1}-3sx_{4}+(-x_{1}^{2}+x_{1}x_{4}+x_{1}+x_{4}+x_{5})\partial_{1}+(-x_{1}x_{2}+x_{2}x_{4}-x_{5})\partial_{2}+(-x_{1}x_{3}$
$\displaystyle+x_{3}x_{4}+x_{5})\partial_{3}+(-x_{1}x_{4}+x_{4}^{2}-x_{1}-x_{4}-x_{5})\partial_{4}+(-x_{1}x_{5}+x_{4}x_{5})\partial_{5},$
$\displaystyle\widetilde{\mathbf{O}}_{1}^{2}$
$\displaystyle=-3sx_{3}+(x_{1}x_{3}+x_{2}-x_{5})\partial_{1}+(x_{2}x_{3}+x_{2}+x_{3}+x_{5})\partial_{2}+x_{3}^{2}\partial_{3}+(x_{3}x_{4}+x_{3})\partial_{4}$
$\displaystyle+(x_{3}x_{5}-x_{2}-x_{3}-x_{5})\partial_{5},$
$\displaystyle\widetilde{\mathbf{O}}_{2}^{2}$
$\displaystyle=-3sx_{3}+(x_{1}x_{3}-x_{5})\partial_{1}+(x_{2}x_{3}+2x_{2}+x_{3}+x_{5})\partial_{2}+(x_{3}^{2}+x_{3})\partial_{3}+x_{3}x_{4}\partial_{4}+(x_{3}x_{5}$
$\displaystyle+x_{5})\partial_{5}-2s,$
$\displaystyle\widetilde{\mathbf{O}}_{3}^{2}$
$\displaystyle=3sx_{2}-3sx_{3}+(-x_{1}x_{2}+x_{1}x_{3}-x_{5})\partial_{1}+(-x_{2}^{2}+x_{2}x_{3}+x_{2}+x_{3}+x_{5})\partial_{2}+(-x_{2}x_{3}$
$\displaystyle+x_{3}^{2}-x_{2}-x_{3}-x_{5})\partial_{3}+(-x_{2}x_{4}+x_{3}x_{4}+x_{5})\partial_{4}+(-x_{2}x_{5}+x_{3}x_{5})\partial_{5}.$

These operators are useful to compare both approaches as discussed next.

###  C.4 Comparing annihilators and IBP operators

According to corollary 21, every momentum-space IBP relation corresponds to a
parametric annihilator. For our two-loop example, this is given by the fact
that

$\widetilde{\mathbf{O}}_{j}^{i}\in\operatorname{Ann}_{{A}^{5}[s]}\left(\mathcal{G}^{s}\right)\quad\text{for}\quad
i\in\left\\{1,2\right\\}\quad\text{and}\quad j\in\left\\{1,2,3\right\\}.$

We may furthermore ask if the reverse is true: Can every annihilator of
$\mathcal{G}$ be derived from IBP relations? If the answer would be no, the
approach via parametric annihilators would provide new integral identities.
While this question remains open for the general case, we can test it for
simple Feynman graphs such as the present two-loop example.

In a first attempt, we could consider the shift relations obtained from the
generators ${P_{1},\ldots,P_{13}}$ and try to confirm that they are combined
IBP relations. If we use one of the well-known implementations of Laporta’s
algorithm to reproduce e.g. equation (C.5), we have to fix the values of
$\nu_{1},\ldots,\nu_{5}$ and do not answer the question for arbitrary values
of the $\nu_{i}$. We therefore approach the problem on the level of parametric
differential operators instead.

We find that actually not all parametric annihilators are contained in
$\operatorname{Mom}$; however, they turn out to still be consequences of the
momentum space IBP relations in the following sense: While we checked that
$P_{1}\notin\operatorname{Mom}$, we can find a polynomial
$q_{1}\in\mathbbm{Q}[s,x_{1}\partial_{1},\ldots,x_{5}\partial_{5}]$ such that
$q_{1}P_{1}\in\operatorname{Mom}$. Recall that, under the Mellin transform,
such a $q_{1}$ corresponds to a polynomial in the dimension and in the
$\nu_{e}$. The interesting question then is if we can find a polynomial
$q\in\mathbbm{Q}[s,x_{1}\partial_{1},\ldots,x_{N}\partial_{N}]$ for every
$P\in\operatorname{Ann}_{{A}^{N}[s]}(\mathcal{G}^{s})$ such that
$qP\in\operatorname{Mom}$. If we can find such a $q_{i}$ for every generator
$P_{i}$, we can express every annihilator in terms of the
$\widetilde{\mathbf{O}}_{j}^{i}$. The $q_{i}$ are the denominators of the
coefficients in such a linear combination.

In section C.2 we have seen for our example that
$\operatorname{Ann}_{{A}^{N}[s]}(\mathcal{G}^{s})$ is generated by the linear
annihilators $L_{i}$. As a consequence, it is sufficient to show that for each
$L_{i}$ there is a $\tilde{q}_{i}$ such that
$\tilde{q}_{i}L_{i}\in\operatorname{Mom}$ for $i=1,\ldots,8$. Indeed we can
construct such ${q_{1},\ldots,q_{8}}$ by an explicit Ansatz. For example we
obtain the identity

$\left(2\sum_{i=1}^{5}x_{i}\partial_{i}-6s\right)L_{1}=c_{1}\widetilde{\mathbf{O}}_{1}^{1}+c_{2}\widetilde{\mathbf{O}}_{1}^{2}+c_{3}\widetilde{\mathbf{O}}_{2}^{1}+c_{4}\widetilde{\mathbf{O}}_{2}^{2}+c_{5}\widetilde{\mathbf{O}}_{3}^{1}+c_{6}\widetilde{\mathbf{O}}_{3}^{2}$

with

$\displaystyle c_{1}$
$\displaystyle=-(-2\partial_{1}+\partial_{2}-\partial_{3}+2\partial_{4}+x_{5}(\partial_{1}(s+1)+\partial_{2}(-s-1)+\partial_{5}s)+x_{4}\partial_{4}+x_{3}(\partial_{2}(-s-1)$
$\displaystyle-\partial_{4}s+\partial_{5}s+3s^{2}+3s+x_{5}\partial_{5}(-s-1)+x_{4}\partial_{4}(-s-1))+x_{3}^{2}\partial_{3}(-s-1)+x_{2}(-\partial_{1}s$
$\displaystyle+\partial_{2}(-s-1)+\partial_{5}s+x_{3}\partial_{2}(-s-1))+x_{1}(\partial_{1}+x_{3}\partial_{1}(-s-1))),$
$\displaystyle c_{2}$
$\displaystyle=-(-\partial_{2}+\partial_{3}-2s^{2}-2s+x_{5}\partial_{5}(2s+1)+x_{4}(-\partial_{3}s+\partial_{4}(s+1)+\partial_{5}s)+x_{3}\partial_{3}+x_{2}\partial_{2}$
$\displaystyle+x_{1}(\partial_{1}(s+1)-\partial_{2}s+\partial_{5}s)),$
$\displaystyle c_{3}$
$\displaystyle=-(\partial_{1}+1-\partial_{4}+s+x_{5}(\partial_{1}(-s-1)+\partial_{2}(s+1)-\partial_{5}s)-x_{4}\partial_{4}+x_{3}(\partial_{2}(s+1)+\partial_{4}s$
$\displaystyle-\partial_{5}s-3s^{2}-3s+x_{5}\partial_{5}(s+1)+x_{4}\partial_{4}(s+1))+x_{3}^{2}\partial_{3}(s+1)+x_{2}(\partial_{1}s+\partial_{2}(s+1)$
$\displaystyle-\partial_{5}s+x_{3}\partial_{2}(s+1))+x_{1}(-\partial_{1}+x_{3}\partial_{1}(s+1)))$
$\displaystyle c_{4}$
$\displaystyle=-(-\partial_{1}+2\partial_{2}-2\partial_{3}+\partial_{4}+x_{5}\partial_{5}+x_{4}(-\partial_{3}+\partial_{5})-x_{3}\partial_{3}-x_{2}\partial_{2}+x_{1}(-\partial_{2}+\partial_{5}))$
$\displaystyle c_{5}$
$\displaystyle=-(\partial_{1}-\partial_{2}+\partial_{5})$ $\displaystyle
c_{6}$ $\displaystyle=-(\partial_{1}-\partial_{2}-\partial_{5}).$

Using these results and the expressions for the generators of
$\operatorname{Ann}_{{A}^{5}[s]}(\mathcal{G}^{s})$ in terms of the $L_{i}$, we
can derive every annihilator from the IBP operators. We have done this
computation with the same conclusion for several further graphs of low loop-
order. These computations support our conjectures phrased in questions 23 and
24.

###  C.5 The number of master integrals

The result of the previous subsection implies that the parametric approach and
momentum-space IBP lead to the same number of master integrals for this
example. Indeed, as mentioned in section 4.1, we compute the Euler
characteristic

$\mathfrak{C}\left(\mathcal{G}\right)=3$

and obtain the same number of master integrals with Azurite. The three master
integrals suggested by Azurite are

$I_{1}=\mathcal{I}(1,1,1,1,0),\quad
I_{2}=\mathcal{I}(0,1,0,1,1)\quad\text{and}\quad
I_{3}=\mathcal{I}(1,0,1,0,1).$

Notice that symmetries of the graph were not taken into account here, which in
Azurite is assured by setting Symmetry -> False and GlobalSymmetry -> False.
For an integral reduction in practice, one would of course make use of the
symmetry

$\mathcal{I}(\nu_{1},\nu_{2},\nu_{3},\nu_{4},\nu_{5})=\mathcal{I}(\nu_{2},\nu_{1},\nu_{4},\nu_{3},\nu_{5})$

and compute with one of the sets $\left\\{I_{1},I_{2}\right\\}$,
$\left\\{I_{1},I_{3}\right\\}$.

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