$\lambda_{2}$ is

$\displaystyle\partial_{2}\,J^{(\lambda_{2})}=Q_{2}^{(\lambda_{2})}\cdot
J^{(\lambda_{2})}\quad,\quad
Q_{2}^{(\lambda_{2})}=\left[\begin{array}[]{cc}{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
\frac{\epsilon-1}{z_{2}^{2}}&\frac{-1}{z_{2}}\end{array}\right]\,.$ (257)

The general solution is

$\displaystyle J_{1}^{(\lambda_{2})}=c_{1}(\epsilon)\quad,\quad
J_{2}^{(\lambda_{2})}=\frac{c_{1}(\epsilon)(1-\epsilon)\log(z_{2})}{z_{2}}+\frac{c_{2}(\epsilon)}{z_{2}}\,.$
(258)

The constants $c_{1}(\epsilon),c_{2}(\epsilon)$ will be fixed by matching with
the full solution vector $I$.

####  5.2.5 Full result

Let us combine the solution vectors from the previous two sections. Given
$J^{(\lambda_{1})}$ and $J^{(\lambda_{2})}$, the vectors
$I^{(\lambda,\bullet)}$ from (189) are determined by

$\displaystyle I^{(\lambda_{1})}$
$\displaystyle=\bigl{(}M^{(\lambda_{1})}\bigr{)}^{-1}\cdot\left[\begin{array}[]{c}J^{(\lambda_{1})}\\\\[-5.0pt]
{\color[rgb]{0.5,0.5,0.5}\cdot}\\\\[-5.0pt]
{\color[rgb]{0.5,0.5,0.5}\cdot}\end{array}\right]$ (262) $\displaystyle
I^{(\lambda_{2},0)}$
$\displaystyle=\bigl{(}M^{(\lambda_{2})}\bigr{)}^{-1}\cdot\left[\begin{array}[]{c}J^{(\lambda_{2})}\\\\[-5.0pt]
{\color[rgb]{0.5,0.5,0.5}\cdot}\\\\[-5.0pt]
{\color[rgb]{0.5,0.5,0.5}\cdot}\\\\[-5.0pt]
{\color[rgb]{0.5,0.5,0.5}\cdot}\end{array}\right]$ (267) $\displaystyle
I^{(\lambda_{2},1)}$ $\displaystyle=(P_{1,-1}-\lambda_{2})\cdot
I^{(\lambda_{2},0)}\,.$ (268)

Reinstating prefactors and gauge transformations, the expression (189) becomes

$\displaystyle\left[\begin{array}[]{c}I_{0010}(m)\\\\[5.0pt]
I_{0101}(m)\\\\[5.0pt] I_{1010}(m)\\\\[5.0pt] I_{0111}(m)\\\\[5.0pt]
I_{1111}(m)\end{array}\right]\,\overset{m\to
0}{=}\,z_{1}^{\lambda_{1}}G_{0}\cdot
G_{1}\cdot\left[\begin{array}[]{c}0\\\\[5.0pt] I_{0101}(0)\\\\[5.0pt]
I_{1010}(0)\\\\[5.0pt] \frac{1-2\epsilon}{t\epsilon}I_{0101}(0)\\\\[5.0pt]
I_{1111}(0)\end{array}\right]\,+\,z_{1}^{\lambda_{2}}G_{0}\cdot
G_{1}\cdot\left[\begin{array}[]{c}\frac{c_{1}}{\epsilon-1}\\\\[5.0pt]
0\\\\[5.0pt] 0\\\\[5.0pt]
\frac{c_{1}\log(z_{2}/z_{1})+c_{2}}{z_{2}}\\\\[5.0pt]
\frac{-2}{\epsilon}\frac{c_{1}[\epsilon\log(z_{2}/z_{1})-1]+\epsilon
c_{2}}{z_{2}}\end{array}\right]\,,$ (284)

where $c_{1}\to\frac{c_{1}}{\epsilon-1}$ has been rescaled to simplify the
expression. $c_{1}$ can be found by matching against the exact result for the
tadpole:

$\displaystyle\text{LHS:}\quad
I_{0010}(m)=-\Gamma(\epsilon-1)(m^{2})^{1-\epsilon}$
$\displaystyle\text{RHS:}\quad(-s)^{1-\epsilon}\cdot
z_{1}^{1-\epsilon}\cdot\frac{c_{1}}{\epsilon-1}=(-s)^{1-\epsilon}\cdot\left(\frac{m^{2}}{-s}\right)^{1-\epsilon}\cdot\frac{c_{1}}{\epsilon-1}$
(285) $\displaystyle\implies c_{1}=(1-\epsilon)\Gamma(\epsilon-1)\,.$

The constant $c_{2}$ can be found order by order in $\epsilon$ by matching
with the LHS at a particular phase space point. Expanding171717 Note that
leading $\epsilon$-poles for $I_{1111}(m)$ and $I_{1111}(0)$ are
$\epsilon^{-2}$ and $\epsilon^{-1}$ respectively. We therefore start the
Laurent series for $c_{2}$ at $O(\epsilon^{-2})$ to cancel the $\epsilon^{-2}$
poles on the RHS.  $c_{2}=\sum_{i=-2}^{2}c_{2,i}\epsilon^{i}$ and matching
with $I_{1111}(m)$ at $(s,t,m^{2})=(-1,-9,10^{-3})$ using AMFlow Liu:2022chg ,
we obtain

$\displaystyle(c_{2,-2},c_{2,-1},c_{2,0},c_{2,1},c_{2,2})=(1,-0.588997,-7.12155,-4.69031,-20.4068).$
(286)

After inserting the solutions for $c_{1},c_{2}$ into the LHS of (285), we find
good agreement at other phase space points. For instance, at
$(s,t,m^{2})=(-1,-8,10^{-3})$ we find that, up to $O(\epsilon^{3})$, the LHS
agrees with $I_{0111}(m)$ within $0.01\%$ accuracy, and with $I_{1111}(m)$
within $0.001\%$ accuracy.

###  5.3 From massive Bhabha to the massless double-box

Here we apply the Pfaffian-level restriction protocol to a more complicated
two-loop diagram to illustrate that the method is computationally cheap.

####  5.3.1 Setup

We study the massless limit of a planar 2-loop integral family contributing to
Bhabha scattering in $d=4-2\epsilon$ dimensions:

$\displaystyle\parbox[c]{0.0pt}{\includegraphics{figures/bhabha2}}\quad\xrightarrow{m\to
0}\quad\parbox[c]{0.0pt}{\includegraphics{figures/dbox}}$

Let

$\displaystyle
I_{\nu_{1}\ldots\nu_{9}}(m)=\int\frac{\mathrm{d}^{d}k_{1}\mathrm{d}^{d}k_{2}}{D_{1}^{\nu_{1}}\cdots
D_{9}^{\nu_{9}}}$ (287) $\displaystyle
D_{1}=k_{1}^{2}-m^{2}\,,\,D_{2}=(k_{1}+p_{1}+p_{2})^{2}-m^{2}\,,\,D_{3}=k_{2}^{2}-m^{2}$
$\displaystyle
D_{4}=(k_{2}+p_{1}+p_{2})^{2}-m^{2}\,,\,D_{5}=(k_{1}+p_{1})^{2}\,,\,D_{6}=(k_{1}-k_{2})^{2}$
(288) $\displaystyle
D_{7}=(k_{2}-p_{3})^{2}\,,\,D_{8}=(k_{2}+p_{1})^{2}\,,\,D_{9}=(k_{1}-p_{3})^{2}\,,$

with $D_{8}$ and $D_{9}$ added because of irreducible scalar products. The
kinematics are as in (169). Solutions for this integral family can be found in
Henn:2013woa (see also Duhr:2021fhk ).

As in the 1-loop example, we write two dimensionless variables

$\displaystyle z_{1}=\frac{m^{2}}{-s}\quad,\quad z_{2}=\frac{s}{t}\,,$ (289)

and study the massless limit $m^{2}\to 0$, i.e. the restriction

$\displaystyle z_{1}\to 0\,.$ (290)

For simplicity, in this section we only illustrate the holomorphic
restriction, i.e. we do not compute logarithmic corrections.

Inspired by Henn:2013woa , we choose the following unrestricted basis:

$\displaystyle G_{0}\cdot I(m)=\bigl{[}$ $\displaystyle I_{\scriptscriptstyle
202000000}\,,\,I_{\scriptscriptstyle 000022100}\,,\,I_{\scriptscriptstyle
001022000}\,,\,I_{\scriptscriptstyle 122000000}\,,\,I_{\scriptscriptstyle
012002000}\,,\,I_{\scriptscriptstyle 022001000}\,,\,$ $\displaystyle
I_{\scriptscriptstyle 001112000}\,,\,I_{\scriptscriptstyle
012011000}\,,\,I_{\scriptscriptstyle 100011200}\,,\,I_{\scriptscriptstyle
121200000}\,,\,I_{\scriptscriptstyle 011011100}\,,\,I_{\scriptscriptstyle
011012100}\,,\,$ $\displaystyle I_{\scriptscriptstyle
021011100}\,,\,I_{\scriptscriptstyle 022011100}\,,\,I_{\scriptscriptstyle
101111000}\,,\,I_{\scriptscriptstyle 101211000}\,,\,I_{\scriptscriptstyle
102011100}\,,\,I_{\scriptscriptstyle 101011200}\,,\,$ $\displaystyle
I_{\scriptscriptstyle 110011200}\,,\,I_{\scriptscriptstyle
111011100}\,,\,I_{\scriptscriptstyle 1110111-10}\,,\,I_{\scriptscriptstyle
111111100}\,,\,I_{\scriptscriptstyle 1111111-10}\bigr{]}^{T}\,,$ (291)

where $G_{0}$ is a diagonal matrix containing factors of $s$ which render the
master integrals unitless. We also choose a basis $J$ for the restricted
system, which is associated massless double-box topology $I_{1\ldots 1}(m=0)$,
where it is understood that $m=0$ is taken at the level of the integrand. The
$8\times 23-$dimensional binary matrix $B$ relating the two bases as

$\displaystyle J=B\cdot I(m=0)$ (292)

is given by

$\displaystyle
B=\left[\tiny\begin{array}[]{ccccccccccccccccccccccc}{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
\end{array}\right]\,.$ (301)

We have the following Pfaffian data for the massive and massless systems:

$\displaystyle\begin{array}[]{cccl}\text{System}&\text{Matrices}&\text{Rank}&\text{Variables}\\\
\hline\cr\text{Massive}&\\{P_{1},P_{2}\\}&23&\\{z_{1},z_{2}\\}\\\
\text{Massless}&\\{Q_{2}\\}&8&\\{z_{2}\\}\end{array}$ (305)

Our goal is to derive the massless differential equation matrix $Q_{2}$ as a
restriction of the massive system $\\{P_{1},P_{2}\\}$.

####  5.3.2 Normal form

The massive Pfaffian system (305) for the basis (291) is not in normal form.
The first issue is that the Pfaffians $P_{i}$ have higher order singularities
in the restriction limit:

$\displaystyle P_{1}$
$\displaystyle=z_{1}^{-3}\>P_{1,-3}+z_{1}^{-2}\>P_{1,-2}+z_{1}^{-1}\>P_{1,-1}+\ldots$
(306) $\displaystyle P_{2}$
$\displaystyle=z_{1}^{-2}\>P_{2,-2}+z_{1}^{-1}\>P_{2,-1}+P_{1,0}+\ldots\ .$
(307)

To cure it, we perform Moser reduction via two gauge transformations by

$\displaystyle G_{1}=\texttt{Diag[}\,\underbrace{z_{1},\ldots,z_{1}}_{\text{8
times}},1,\ldots,1\,\texttt{]}\quad,\quad
G_{2}=\texttt{Diag[}\,\underbrace{z_{1},\ldots,z_{1}}_{\text{12
times}},1,\ldots,1\,\texttt{]}$ (308)

whereby $P_{1}$ becomes logarithmically singular at $z_{1}=0$,

$\displaystyle G_{21}[P_{1}]=z_{1}^{-1}\>G_{21}[P_{1}]_{-1}+\ldots\ .$ (309)

Here we introduced the short-hand notation

$\displaystyle G_{i\ldots j}:=G_{i}\cdot\ldots\cdot G_{j}\ .$ (310)

The transformations (309) still leave $G_{21}[P_{2}]$ with an unwanted
singularity at $z_{1}=0$:

$\displaystyle
G_{21}[P_{2}]=z_{1}^{-1}\>G_{21}[P_{2}]_{-1}+G_{21}[P_{2}]_{0}+\ldots\ .$
(311)

This issue arises because the residue matrix $G_{21}[P_{1}]_{-1}$ is resonant,
meaning that its spectrum contains eigenvalues with integral difference:

$\displaystyle\texttt{Spec\leavevmode\resizebox{}{9.95863pt}{[}}\,G_{21}[P_{2}]_{-1}\,\texttt{\leavevmode\resizebox{}{9.95863pt}{]}}$
$\displaystyle=\bigl{\\{}-2,\>-2,\>-2,\>-1,\>-1,\>0,\>0,\>0\bigr{\\}}$
$\displaystyle\cup\>\bigl{\\{}-2-\epsilon,\>-2-\epsilon,\>-1-\epsilon,\>-1-\epsilon,\>-\epsilon,\>-\epsilon,\>-\epsilon\bigr{\\}}$
$\displaystyle\cup\>\bigl{\\{}-3-2\epsilon,\>-2-2\epsilon,\>-2-2\epsilon,\>-1-2\epsilon,\>-1-2\epsilon,\>-1-2\epsilon,\>-2\epsilon\bigr{\\}}$
$\displaystyle\cup\>\bigl{\\{}-2-4\epsilon\bigr{\\}}\ .$ (312)

These resonances are cured via an additional gauge transformation by a matrix
$G_{3}$, resulting in the non-resonant spectrum that reads

$\displaystyle\texttt{Spec\leavevmode\resizebox{}{9.95863pt}{[}}\,G_{321}[P_{1}]_{-1}\,\texttt{\leavevmode\resizebox{}{9.95863pt}{]}}=\bigl{\\{}0,\ldots,-\epsilon,\ldots,-2\epsilon,\ldots,-2-4\epsilon,\ldots\bigr{\\}}\,,$
(313)

where we only wrote unique eigenvalues. We review the algorithmic procedure to
construct such a $G_{3}$ in Appendix B 181818 In our basic computer
implementation, we perform sequential Jordan decompositions of the residue
matrix, which can be computed in a matter of seconds using Mathematica on a
laptop. .

With this final composition of gauge transformations applied to our Pfaffian
system, we now have that

$\displaystyle G_{321}[P_{2}]=G_{321}[P_{2}]_{0}+\ldots\ ,$ (314)

which means that the system is in normal form.

####  5.3.3 Restriction

Given our choice of restricted basis, the matrix $R$ from (48) is simply

$\displaystyle
R=\texttt{RowReduce\leavevmode\resizebox{}{9.95863pt}{[}}\,G_{321}[P_{1}]_{-1}\,\texttt{\leavevmode\resizebox{}{9.95863pt}{]}}=\left[\begin{array}[]{c|c}\mathbb{0}_{\scriptscriptstyle
15\times 8}&\mathbb{1}_{\scriptscriptstyle 15\times 15}\end{array}\right]\ .$
(316)

Had we chosen a restricted basis which was not a subset of $I(m=0)$ (cf.
(291)), say a Laporta basis where all non-zero propagator powers equal to $\pm
1$, then $R$ would generally depend on $z_{2}$ and $\epsilon$. Given the
matrix $B$ from (301), we are now ready to build the restriction
transformation matrix (48):

$\displaystyle M=\left[\begin{array}[]{c}B\cdot G_{321}\\\ \hline\cr
R\end{array}\right]\,.$ (319)

The upper matrix block encodes information about the restricted basis, and the
lower block contains linear relations (IBPs) which hold in the massless limit.

We finally compute the gauge transformation (52)

$\displaystyle\Bigl{(}\partial_{2}M+M\cdot G_{321}[P_{2}]_{0}\Bigr{)}\cdot
M^{-1}=\left[\begin{array}[]{cc}Q_{2}&\star\\\ \mathbb{0}_{\scriptscriptstyle
15\times 8}&\star\end{array}\right]\ ,$ (322)

and obtain the $8$-dimensional restricted Pfaffian matrix $Q_{2}$ associated
to the massless double-box topology. In other words, $Q_{2}$ satisfies
$\partial_{2}J=Q_{2}\cdot J$, with $J$ given in (292). We have verified the
form of $Q_{2}$ by an independent calculation using the IBP software Kira
Klappert:2020nbg .

###  5.4 Unequal to equal-mass sunrise

This example studies symmetry relations of Feynman integrals from the point of
view of Pfaffian-level restriction.

####  5.4.1 Setup

We examine the equal-mass limit of the 2-loop 3-mass sunrise integral family:

$\displaystyle\parbox[c]{0.0pt}{\includegraphics{figures/sunrise3m}}\quad\xrightarrow{m_{i}\to
m}\quad\parbox[c]{0.0pt}{\includegraphics{figures/sunrise}}$

The family of momentum space Feynman integrals is

$\displaystyle
I_{\nu_{1}\nu_{2}\nu_{3}}(m_{1},m_{2},m_{3})=\int\frac{\mathrm{d}^{d}k_{1}\mathrm{d}^{d}k_{2}}{\bigl{[}k_{1}^{2}-m_{1}^{2}\bigr{]}^{\nu_{1}}\bigl{[}k_{2}^{2}-m_{2}^{2}\bigr{]}^{\nu_{2}}\bigl{[}(k_{1}+k_{2}+p)^{2}-m_{3}^{2}\bigr{]}^{\nu_{3}}}\,.$
(323)

We choose the following $3$ dimensionless kinematic variables:

$\displaystyle z_{1}=\frac{m_{1}^{2}}{s},\quad z_{2}=\frac{m_{2}^{2}}{s},\quad
z_{3}=\frac{m_{3}^{2}}{s},\quad\text{where}\quad s=p^{2}\,.$ (324)

We are interested in the equal-mass limit, that is

$\displaystyle m_{i}\to m,\quad z_{i}\to z:=\frac{m^{2}}{s},\quad\text{for
$i=1,2,3$}\,.$ (325)

The details of the two Pfaffian systems before and after restriction are as
follows:

$\displaystyle\begin{array}[]{cccl}\text{System}&\text{Matrices}&\text{Rank}&\text{Variables}\\\
\hline\cr\text{Unequal
mass}&\\{P_{1},P_{2},P_{3}\\}&7&\\{z_{1},z_{2},z_{3}\\}\\\ \text{Equal
mass}&\\{Q_{z}\\}&3&\\{z\\}\end{array}$ (329)

For the unequal-mass integral family we choose the following basis of master
integrals:

$\displaystyle G_{0}\cdot
I(m_{1},m_{2},m_{3})=\bigl{[}I_{011},\,I_{101},\,I_{110},\,I_{111},\,I_{211},\,I_{121},\,I_{112}\bigr{]}^{T}\,,$
(330)

where, as before, $G_{0}$ is a diagonal matrix containing factors of $s$ which
render the integrals unitless. In the equal-mass case, our target basis reads

$\displaystyle
J(m)=\bigl{[}I_{011}(m,m,m),\,I_{111}(m,m,m),\,I_{211}(m,m,m)\bigr{]}^{T}\,,$
(331)

so we introduce the following binary matrix

$\displaystyle B=\left[\begin{array}[]{ccc ccc
c}1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\end{array}\right]$
(335)

relating the bases by $J(m)=B\cdot I(m,m,m)$.

####  5.4.2 Restriction

The unequal-mass Pfaffian system turns out to have the following denominators:

$\displaystyle\texttt{Denominator[}\,P_{i}\,\texttt{]}=\bigl{\\{}z_{1},z_{2},z_{3},$
(336) $\displaystyle\quad
1-4z_{1}+6z_{1}^{2}-4z_{1}^{3}+z_{1}^{4}-4z_{2}+4z_{1}z_{2}+4z_{1}^{2}z_{2}-4z_{1}^{3}z_{2}+$
$\displaystyle\qquad
6z_{2}^{2}+4z_{1}z_{2}^{2}+6z_{1}^{2}z_{2}^{2}-4z_{2}^{3}-4z_{1}z_{2}^{3}+z_{2}^{4}-4z_{3}+4z_{1}z_{3}+4z_{1}^{2}z_{3}-$
$\displaystyle\qquad
4z_{1}^{3}z_{3}+4z_{2}z_{3}-40z_{1}z_{2}z_{3}+4z_{1}^{2}z_{2}z_{3}+4z_{2}^{2}z_{3}+4z_{1}z_{2}^{2}z_{3}-4z_{2}^{3}z_{3}+$
$\displaystyle\qquad
6z_{3}^{2}+4z_{1}z_{3}^{2}+6z_{1}^{2}z_{3}^{2}+4z_{2}z_{3}^{2}+4z_{1}z_{2}z_{3}^{2}+6z_{2}^{2}z_{3}^{2}-4z_{3}^{3}-4z_{1}z_{3}^{3}-4z_{2}z_{3}^{3}+z_{3}^{4}\bigr{\\}}\,.$

The last polynomial reduces to $(z-1)^{3}(9z-1)$ under the identification
(325), so none of the denominators vanish in the equal-mass limit! This
example is hence not a restriction to a singular locus. In particular, we have
no residue matrix to utilize, so we need to slightly modify the restriction
procedure to derive the equal-mass rank-3 Pfaffian system.

To this end, we begin by observing that some of the MIs (330) become identical
in the equal-mass limit, e.g. $I_{011}(m,m,m)=I_{101}(m,m,m)$191919In the
momentum space representation (323), this can be shown by shifts $k_{i}\to
k_{i}+p$ in the loop momenta. These shifts have no effect on the integrated
result, because the integration contour is over all of Minkowski space.. Let
us collect the four such _symmetry relations_ arising in the equal-mass limit
into the matrix

$\displaystyle R=\left[\begin{array}[]{ccc ccc
c}1&-1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
1&{\color[rgb]{0.5,0.5,0.5}\cdot}&-1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&-1&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&-1\end{array}\right]\,.$
(341)

In analogy with the residue matrix relations from (55), we have

$\displaystyle R\cdot I(m,m,m)=0\,.$ (342)

Now the holomorphic restriction protocol can proceed as before. Namely, taking
the equal-mass limit202020 In practice, we take
$P_{z,0}=\bigl{(}P_{1}+P_{2}+P_{3}\bigr{)}\big{|}_{z_{i}\to z}\>$, which is
obtained from the unequal mass system (329) by shifting the variables
$\\{z_{1},z_{2},z_{3}\\}\mapsto\\{z,z+x_{1},z+x_{2}\\}$ and then taking the
smooth limit of $x_{1},x_{2}\to 0$.  of the initial Pfaffian system, we
compute

$\displaystyle\bigl{(}\partial_{z}M+M\cdot P_{z,0}\bigr{)}\cdot
M^{-1}=\left[\begin{array}[]{cc}{Q}_{z}&\star\\\
\mathbb{0}_{\scriptscriptstyle 4\times
3}&\star\end{array}\right]\>,\quad\text{with}\quad
M=\left[\begin{array}[]{c}B\\\ \hline\cr R\end{array}\right]$ (347)

and $B$ given in (335). This gives

$\displaystyle
Q_{z}=\left[\begin{array}[]{ccc}\frac{-2\epsilon}{z}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&3\\\
\frac{2\epsilon^{2}}{(z-1)z^{2}(9z-1)}&-\frac{(1+2\epsilon)(1+3\epsilon)(3z-1)}{(z-1)z(9z-1)}&-\frac{1+\epsilon-20z-30\epsilon
z+27z^{2}+45\epsilon z^{2}}{(z-1)z(9z-1)}\end{array}\right]\,,$ (351)

in agreement with an independent LiteRed calculation.

####  5.4.3 Deriving symmetry relations

In this previous calculation, we merely _guessed_ the symmetry relation matrix
(341). In the following we show how it can be derived.

Let us recall the secondary-like equation (29), which in the present context
reads

$\displaystyle\partial_{z}\Phi+P_{z,0}^{T}\cdot\Phi-\Phi\cdot
P_{z,0}^{T}=0\>,$ (352)

for some unknown matrix $\Phi$. The IntegrableConnections integrableconnection
library in Maple can find rational solutions to such PDEs via the function
RationalSolutions (see also the function EigenRing). We find five non-trivial
solutions

$\displaystyle\\{\Phi_{1},\ldots,\Phi_{5}\\}=$ (353)
$\displaystyle\left\\{\left[\tiny\begin{array}[]{ccccccc}{\color[rgb]{0.5,0.5,0.5}\cdot}&-1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&1&-1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&-1&1\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&-1\\\
\end{array}\right],\left[\tiny\begin{array}[]{ccccccc}{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1\\\
\end{array}\right],\left[\tiny\begin{array}[]{ccccccc}{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
1&-1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&-1&1\\\
\end{array}\right],\left[\tiny\begin{array}[]{ccccccc}{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
\end{array}\right],\left[\tiny\begin{array}[]{ccccccc}1&-1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
-1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&-1&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&-1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
\end{array}\right]\right\\}\,.$ (389)

We are interested in eigenspaces of these matrices, so let us first record
their eigenvalues:

$\displaystyle\texttt{Spec[}\,\Phi_{1}\,\texttt{]}$
$\displaystyle=\\{-1,-1,0,0,0,0,0\\}\>,$
$\displaystyle\texttt{Spec[}\,\Phi_{2}\,\texttt{]}$
$\displaystyle=\\{-1,-1,1,1,1,1,1\\}\>,$
$\displaystyle\texttt{Spec[}\,\Phi_{3}\,\texttt{]}$
$\displaystyle=\\{0,0,1,1,1,1,1\\}\>,$ (390)
$\displaystyle\texttt{Spec[}\,\Phi_{4}\,\texttt{]}$
$\displaystyle=\\{e^{2/3\pi\sqrt{-1}},e^{2/3\pi\sqrt{-1}},e^{4/3\pi\sqrt{-1}},e^{4/3\pi\sqrt{-1}},1,1,1\\}\>,$
$\displaystyle\texttt{Spec[}\,\Phi_{5}\,\texttt{]}$
$\displaystyle=\\{0,0,0,0,0,1,1\\}\>.$

Next we collect the eigenvectors of $\Phi_{i}$ corresponding to an eigenvalue
$\lambda$ into the rows of a matrix $V_{i}^{(\lambda)}\>$, meaning that it
satisfies

$\displaystyle\left[\Phi_{i}-\lambda\mathbb{1}\right]\cdot\left(V_{i}^{(\lambda)}\right)^{T}=0\quad,\quad
i=1,\ldots,5\,.$ (391)

Now comes our main observation: the rows of the symmetry relations matrix
(341) span the same linear space as the rows of (some of the)
$V_{i}^{(\lambda)}$ matrices! We have

$\displaystyle\texttt{RowReduce[}\,R\,\texttt{]}=\texttt{RowReduce[}\,\left[\begin{array}[]{c}V_{1}^{(-1)}\\\
V_{2}^{(-1)}\\\ V_{3}^{(0)}\\\ V_{5}^{(1)}\end{array}\right]\,\texttt{]}\>,$
(396)

where, as before, the RowReduce operation includes deletion of zero-rows. The
correspondence (396) is the reason why our choice of the symmetry relations
matrix (341) actually gave the correct restricted Pfaffian (351): the rows of
this $R$ span an invariant subspace $\mathcal{N}$ of the
$\mathcal{R}=\mathbb{C}(z)\langle\partial_{z}\rangle$-module $\mathcal{M}$
associated to the Pfaffian matrix $P_{z,0}$ (see the discussion around Theorem
2.6). Equation (351) is thus the Pfaffian matrix for the quotient
$\mathcal{R}$-module $\mathcal{M}/\mathcal{N}$ with respect to a basis $B$,
and can be determined by the restriction method as shown above.

###  5.5 Restriction of GKZ systems by the Macaulay matrix method

Next we present examples of Feynman integrals arising as restrictions of GKZ
systems. We employ the Macaulay matrix based algorithms from Section 4. For
simplicity, in all examples we omit $\Gamma$-prefactors coming from the Lee-
Pomeransky representation. To obtain Pfaffian matrices for the proper Feynman
integrals, one would need to include these prefactors and then send $\delta\to
0$.

Given $n$ integration variables in the Euler integral (2), for every example
below we fix the GKZ $\beta$-parameters to be the $(n+1)$-dimensional vector
$\beta=[\epsilon,-\epsilon\delta,\ldots,-\epsilon\delta]^{T}$. This follows
the convention of Chestnov:2022alh .

####  5.5.1 One-loop triangle

We first consider the one-loop triangle diagram with one internal mass $m$.
After rescaling the integration variables by $m^{2}$, the Lee-Pomeransky
polynomial associated to this diagram takes the form

$\displaystyle g(x;z)$
$\displaystyle=z_{1}x_{1}+z_{2}x_{2}+z_{3}x_{3}+z_{4}x_{1}^{2}+z_{5}x_{1}x_{2}+z_{6}x_{1}x_{3}$
(397) $\displaystyle z_{1}$ $\displaystyle=\ldots=z_{5}=1\quad,\quad
z_{6}=\frac{(p_{1}+p_{2})^{2}}{m^{2}}$ (398)

where $p_{1},p_{2}$ denote the incoming momenta of two massless particles.

We are interested in obtaining the Pfaffian system associated to the proper
Feynman integral as a restriction of the GKZ system:

$A=\left[\begin{array}[]{cccccc}1&1&1&1&1&1\\\
1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&2&1&1\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1\\\
\end{array}\right]\quad\xrightarrow[i\,\neq\,6]{z_{i}=1}\quad\parbox[c]{0.0pt}{\includegraphics{figures/tri1m}}$
(399)

With all variables $z_{1},\ldots,z_{6}$ being generic, the GKZ system has rank
$3$. However, an Euler characteristic computation (see Theorem 2.1) gives
$\chi=2$ for the case $z_{1}=\ldots=z_{5}=1$ and generic $z_{6}$. We thus seek
to drop the rank from $3$ to $2$ via a $\mathcal{D}_{Y}$-module restriction.

Following (Chestnov:2022alh, , Appendix A), we are free to immediately rescale
$z_{1}=\ldots z_{4}=1$ without changing the rank of GKZ system. The remaining
variables are thus $z_{5}$ and $z_{6}$, and we seek the restriction $z_{5}=1$.
We first find a standard basis when $z_{5}=1$ by applying Algorithm 4.7 with
the randomly chosen value $z_{6}=1/17$ and the algorithm parameters
$\gamma=2$, $k=3$. The output is a basis consisting of $2$ elements:

${\tt{RStd}}=[\partial_{6},1]^{T}\,,$ (400)

which agrees with the rank being $2$. We can obtain a Pfaffian matrix in the
variable $z_{6}$ for this restricted basis by Algorithm 4.11. The Macaulay
matrix succeeds at degree $D=1$, and we get

$P_{6}=\left[\begin{array}[]{cc}\frac{((4{\epsilon}{\delta}+{\epsilon}+1){z}_{6}-4{\epsilon}{\delta}-2{\epsilon}-1)}{{z}_{6}(1-{z}_{6})}&\frac{{\epsilon}^{2}{\delta}(3{\delta}+1)}{{z}_{6}(1-{z}_{6})}\\\
1&0\\\ \end{array}\right].$

####  5.5.2 Two-loop N-box

The two-loop N-box with one internal mass has the following Lee-Pomeransky
polynomial:

$\displaystyle g(z;x)=$ $\displaystyle
z_{1}x_{1}x_{2}+z_{2}x_{1}x_{4}+z_{3}x_{1}x_{5}+z_{4}x_{2}x_{3}+z_{5}x_{2}x_{4}+z_{6}x_{2}x_{5}+$
$\displaystyle
z_{7}x_{3}x_{4}+z_{8}x_{3}x_{5}+z_{9}x_{1}x_{2}x_{4}+z_{10}x_{1}x_{4}^{2}+z_{11}x_{1}x_{4}x_{5}+z_{12}x_{2}x_{3}x_{4}+$
(401) $\displaystyle
z_{13}x_{2}x_{3}x_{5}+z_{14}x_{2}x_{4}^{2}+z_{15}x_{2}x_{4}x_{5}+z_{16}x_{3}x_{4}^{2}+z_{17}x_{3}x_{4}x_{5}\,.$

We study it as a restriction of a GKZ system:

$A=\left[\begin{array}[]{ccccccccccccccccc}1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1\\\
1&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&2&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&2&1&2&1\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&1\\\
\end{array}\right]\quad\xrightarrow[i\,\neq\,9,\,13]{z_{i}\to
1}\quad\parbox[c]{0.0pt}{\includegraphics{figures/nbox1m}}$ (402)

The 17-variable GKZ system has rank $33$ for generic $\epsilon,\delta$. To
match with the proper Feynman integral, we seek the restriction

$\displaystyle z_{i}=1\,,\,i\neq 9,13$ (403)

with $z_{9}=1+\frac{s}{m^{2}}$ and $z_{13}=\frac{t}{m^{2}}$ being the
remaining kinematic variables. By computing the Euler characteristic with
restricted variables, we expect a rank drop from $33$ to $7$.

To begin with, we choose the simplex $\sigma=\left[1,2,3,4,5,10\right]$. The
remaining variables to restrict are thus $z_{i}=1$ for
$i\in\\{6,7,8,11,12,14,15,16,17\\}$, for which we apply Algorithm 4.7. Taking
parameters $\gamma=3,k=4$ and randomly chosen values $z_{9}=1/17,z_{13}=1/19$,
we are able to guess the basis

${\tt{RStd}}=[\partial_{8},\partial_{15},\partial_{16},\partial_{17},\partial_{9},\partial_{13},1]^{T}$
(404)

by running the algorithm212121It takes 7.32s on an m-PC(Intel(R) Core(TM)
i7-10700K CPU @ 3.80GHz, 8G memory) over the finite field of characteristic
$p=100000007$. Note that the basis is $7$-dimensional, as expected.

Next we use Algorithm 4.11 to obtain a Macaulay matrix, which succeeds222222It
takes 3.28s on the m-PC. at degree $D=1$. Since the Macaulay matrix is valid,
it means that the probabilistic basis from Algorithm 4.7 is indeed a basis.
Using FiniteFlow Peraro:2019svx , in a matter of seconds we obtain a solution
to equation (107) for $i=9,13$ via rational reconstruction over finite fields.
The result is a set of $7\times 7$ Pfaffian matrices $P_{9}$ and $P_{13}$. We
have checked that they satisfy integrability.

####  5.5.3 Two-loop diagonal-box

Here we consider the massless two-loop diagonal-box diagram, whose Lee-
Pomeransky polynomial takes the form

$\displaystyle g(z;x)=$ $\displaystyle
z_{1}x_{1}x_{4}+z_{2}x_{1}x_{5}+z_{3}x_{1}x_{6}+z_{4}x_{2}x_{4}+z_{5}x_{2}x_{5}+z_{6}x_{2}x_{6}+$
$\displaystyle
z_{7}x_{3}x_{4}+z_{8}x_{3}x_{5}+z_{9}x_{3}x_{6}+z_{10}x_{4}x_{5}+z_{11}x_{4}x_{6}+z_{12}x_{4}x_{7}+$
$\displaystyle
z_{13}x_{5}x_{7}+z_{14}x_{6}x_{7}+z_{15}x_{1}x_{3}x_{4}+z_{16}x_{1}x_{3}x_{5}+z_{17}x_{1}x_{3}x_{6}+$
(405) $\displaystyle
z_{18}x_{1}x_{4}x_{5}+z_{19}x_{2}x_{4}x_{6}+z_{20}x_{2}x_{4}x_{7}+z_{21}x_{2}x_{5}x_{7}+z_{22}x_{2}x_{6}x_{7}\,.$

We consider the following restriction of the GKZ system:

$A=\left[\begin{array}[]{cccccccccccccccccccccc}1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1\\\
1&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&1&1\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&1\\\
\end{array}\right]\quad\xrightarrow[i\,\leq\,18,\>19\,\leq\,j\,\leq\,22]{z_{i}=1,\>z_{j}=t/s}\quad\parbox[c]{0.0pt}{\includegraphics{figures/diabox}}$
(406)

The GKZ system has rank $115$ for generic $\epsilon,\delta$. Upon restricting
to the proper Feynman integral case,

$\displaystyle z_{1}=\ldots=z_{18}=1\quad,\quad
z_{19}=\ldots=z_{22}=z:=\frac{t}{s}$ (407)

we except a rank drop to $7$, according to the Euler characteristic.

We follow an analogous strategy as in the previous section. We guess RStd by
making the change of coordinates $z_{i}=y_{i}+1$ for $i=1,\ldots,18$,
$z_{i}=y_{i}+z$ for $i=19,20,21$, $z_{22}=z$ and taking the restriction to
$y_{i}=0$ ($i=1,\ldots,21$). For
$\sigma=[1,2,5,7,9,11,13,18],\,\gamma=3,\,k=4,p=100000007$ and the randomly
chosen value $z=1/7$, we find a $7$-dimensional basis232323It takes 40.56s on
an m-PC(Intel(R) Core(TM) i7-10700K CPU @ 3.80GHz, 8G memory)

$\displaystyle{\tt{RStd}}=[\partial_{21}\partial_{z},\partial_{z}^{2},\partial_{6},\partial_{17},\partial_{21},\partial_{z},1]^{T}\,,$
(408)

and a valid corresponding Macaulay matrix at degree $D=1$242424It takes 65.76s
on the m-PC.. Note that $\partial_{i}$ means $\partial/\partial y_{i}$. With
this Macaulay matrix, the $7\times 7$ Pfaffian $P_{z}$ is solved for in less
than $3$ minutes on a laptop via FiniteFlow.

####  5.5.4 Two-loop double-box

Our last Feynman integral example is the massless two-loop double-box. We
start from the $26$-term Lee-Pomeransky polynomial

$\displaystyle g(z;x)$
$\displaystyle=z_{1}x_{1}x_{2}+z_{2}x_{1}x_{3}+z_{3}x_{1}x_{6}+z_{4}x_{1}x_{7}+z_{5}x_{2}x_{3}+z_{6}x_{2}x_{4}+z_{7}x_{2}x_{5}+$
$\displaystyle
z_{8}x_{3}x_{4}+z_{9}x_{3}x_{5}+z_{10}x_{3}x_{6}+z_{11}x_{3}x_{7}+z_{12}x_{4}x_{6}+z_{13}x_{4}x_{7}+z_{14}x_{5}x_{6}+$
(409) $\displaystyle
z_{15}x_{5}x_{7}+z_{16}x_{1}x_{2}x_{5}+z_{17}x_{1}x_{2}x_{6}+z_{18}x_{1}x_{3}x_{5}+z_{19}x_{1}x_{3}x_{6}+z_{20}x_{1}x_{5}x_{6}+$
$\displaystyle
z_{21}x_{1}x_{5}x_{7}+z_{22}x_{2}x_{3}x_{5}+z_{23}x_{2}x_{3}x_{6}+z_{24}x_{2}x_{4}x_{6}+z_{25}x_{2}x_{5}x_{6}+z_{26}x_{3}x_{4}x_{7}\,,$

and compute the restriction

$A=\left[\begin{array}[]{cccccccccccccccccccccccccc}1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1\\\
1&1&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&1&1&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&1\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&1&1&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1\\\
\end{array}\right]\quad\xrightarrow[i\,\leq\,25]{z_{i}=1}\quad\parbox[c]{0.0pt}{\includegraphics{figures/dbox}}$
(410)

For generic $\epsilon$ and $\delta$, the rank of the GKZ system is a whopping
$238$. While the GKZ system has generic $z_{i}$, the proper massless double-
box Feynman integral has monomial coefficients

$\displaystyle z_{i}=1\text{ for all }i=1,\ldots,25\quad,\quad
z_{26}=-1-s/t=:z\,.$ (411)

The Euler characteristic suggests a rank $12$ system for the restricted
variables above. However, this Feynman integral is well-known to satisfy a
rank $8$ Pfaffian system - we comment on this below.

We fix the simplex $\sigma=[1,2,3,4,5,6,7,25]$ and then run Algorithm 4.7 with
parameters $\gamma=3,\,k=5,\,p=100000007$ and the randomly chosen value
$z=1/7$. The result is a $12$-dimensional basis252525It takes 23,815s CPU time
on a t-PC(AMD EPYC 7552 four 48-Core Processors @ 1.5GHz, 1T memory)

${\tt{RStd}}=[\partial_{{z}}\partial_{{15}},\partial_{{23}}\partial_{{z}},\partial_{{24}}\partial_{{z}},\partial_{{z}}^{2},\partial_{{13}},\partial_{{15}},\partial_{{21}},\partial_{{22}},\partial_{{23}},\partial_{{24}},\partial_{{z}},1]^{T}\,,$
(412)

whose size agrees with the Euler characteristic.

In step 4 of Algorithm 4.7, we need to compute the echelon form of a matrix of
size $132,145\times 33,649$. Algorithm 4.11 succeeds at finding a Macaulay
matrix262626It takes 501s CPU time on the t-PC. at degree $D=1$. The size of
$M_{{\tt{Ext}}}$ is $2926\times 10775$. Running FiniteFlow272727It took
89,021s CPU-hours on the t-PC. However, FiniteFlow ran in parallel over 96
threads, so the real time was about 20 min. to solve the matrix equation (107)
associated to $M_{{\tt{Ext}}}$, we finally obtain a $12\times 12$ Pfaffian
matrix $P_{z}$, which can be downloaded from dataAndProgramsOfThisPaper .

We suspect that symmetry relations (namely, reparametrizations of the Euler
_integrand_) would bridge the gap between $\chi=12$ and the expected rank $8$
for the Pfaffian system associated to this Feynman integral. By using the
secondary-like equation (29), it should be possible to block-diagonalize our
$12$-dimensional ODE dictated by $P_{z}$ to reach the minimal size of $8$.
Indeed, similar to Section 5.4, with the help of IntegrableConnections
integrableconnection library we found a $4$-dimensional invariant subspace of
symmetry relations that reduce the rank from $12$ down to $8$, as expected for
this family of Feynman integrals282828 It took about 70 CPU-hours and 4Gb of
RAM on the t-PC for IntegrableConnections to find these rational solutions. .
It is a future problem for us to give a more efficient implementation
combining the secondary-like equation with the methods of Barkatou-1993 ;
singer-1996 .

###  5.6 Restriction of Appell’s $F_{4}$ and Horn’s $H_{i}$ functions to
hypersurfaces

The Appell $F_{4}$ function has the following series representation:

$F_{4}(a,b,c_{1},c_{2};x,y)=\sum_{i,j=0}^{\infty}\frac{(a)_{i+j}(b)_{i+j}}{(1)_{i}(1)_{j}(c_{1})_{i}(c_{2})_{j}}x^{i}y^{j}.$
(413)

It appears in several contexts related to Feynman integrals Anastasiou:1999ui
; FCCZ-2020 . $F_{4}$ is annihilated by the following operators:

$\displaystyle f_{1}$ $\displaystyle=$
$\displaystyle\theta_{x}(\theta_{x}+c_{1}-1)-x(\theta_{x}+\theta_{y}+a)(\theta_{x}+\theta_{y}+b)$
(414) $\displaystyle f_{2}$ $\displaystyle=$
$\displaystyle\theta_{y}(\theta_{y}+c_{2}-1)-y(\theta_{x}+\theta_{y}+a)(\theta_{x}+\theta_{y}+b)\,,$
(415)

where the Euler operators are $\theta_{x}=x\partial_{x}$ and
$\theta_{y}=y\partial_{y}$. Let us assume that the parameters
$a,b,c_{1},c_{2}$ take on generic values. The operators $f_{1}$ and $f_{2}$
generate a holonomic ideal $\mathcal{I}$ in the $\mathcal{D}$-module ${\cal
D}_{2}$ whose coordinates are $z_{1}=x$, $z_{2}=y$. The left ${\cal
D}_{2}$-module ${\cal D}_{2}/\mathcal{I}$ is regular holonomic with holonomic
rank $4$, and the singular locus is

$xy((x-y)^{2}-2(x+y)+1)=0\,.$ (416)

Put $L=(x-y)^{2}-2(x+y)+1$. The point $(x,y)=(1/4,1/4)$ is on the singular
locus, and we will construct a Pfaffian system restricted to a codimension-$1$
stratum $W=V(L)\setminus\\{(1,0),(0,1)\\}$ containing that point. Note that
$V(L)$ and $xy=0$ intersect at $(1,0)$ and $(0,1)$.

Computing the ODE on $x=y$ by restricting the Pfaffian system, we see that the
set of exponents at the point $V(x-y)\cap W$ is

$\\{a,\,b,\,c_{1},\,c_{2}\\}\ \to\ \\{0,\,0,\,0,\,2(a+b-c_{1}-c_{2})+5\\}\,.$
(417)

Moreover, the residue matrix at this point is diagonal, meaning that there is
no logarithmic solution. By an analytic coordinate transformation, we can
locally transform $W$ and $V(x-y)$ into hyperplanes that intersect
transversally. Moreover, it can be expressed as a logarithmic connection with
the residue matrix along $W$ being the diagonal matrix with eigenvalues (417).
It follows from Proposition 3.3 that the holonomic rank $r$ on $W$ is $3$ for
generic parameter values. Applying the restriction algorithm from Theorem 4.2
with the shift $(x,y)\to(x+1/4,y+1/4)$, we obtain the standard basis
$S={\tt{RStd}}=[1,\partial_{x},\partial_{y}]^{T}$ (see Remark 4.12).

Let us write a modifed version of (105) modulo $L$:

$\displaystyle
M_{D}:=\left[\begin{array}[]{c|c}M_{{\tt{Ext}}}&M_{{\tt{RStd}}}\end{array}\right]\text{
mod $L$}\>.$ (419)

The left block matrix above has columns labeled by
${\tt{Ext}}=[\partial_{x}^{3},\partial_{x}^{2}\partial_{y},\partial_{x}\partial_{y}^{2},\partial_{y}^{3},\partial_{x}^{2},\partial_{x}\partial_{y},\partial_{y}^{2}]$.
We proceed to solve (116) for the unknown matrix $C$, which will produce the
Pfaffian matrices by (130). Equation (116) requires the following two matrices
as input:

$\displaystyle M_{{\tt{Ext}}}$
$\displaystyle={\tiny\left[\begin{array}[]{ccccccc}{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&-{x}^{2}+{x}&-2{y}{x}&-{y}^{2}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&-{x}^{2}&-2{y}{x}&-{y}^{2}+{y}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&-{x}^{2}+{x}&-2{y}{x}&-{y}^{2}&{\color[rgb]{0.5,0.5,0.5}\cdot}&(-{a}-{b}-3){x}+{c}_{1}&(-{a}-{b}-3){y}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&-{x}^{2}&-2{y}{x}&-{y}^{2}+{y}&{\color[rgb]{0.5,0.5,0.5}\cdot}&(-{a}-{b}-3){x}&(-{a}-{b}-3){y}+{c}_{2}+1\\\
-{x}^{2}+{x}&-2{y}{x}&-{y}^{2}&{\color[rgb]{0.5,0.5,0.5}\cdot}&(-{a}-{b}-3){x}+{c}_{1}+1&(-{a}-{b}-3){y}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
-{x}^{2}&-2{y}{x}&-{y}^{2}+{y}&{\color[rgb]{0.5,0.5,0.5}\cdot}&(-{a}-{b}-3){x}&(-{a}-{b}-3){y}+{c}_{2}&{\color[rgb]{0.5,0.5,0.5}\cdot}.\end{array}\right]}$
(426) $\displaystyle C_{{\tt{Ext}}}$
$\displaystyle=\left[\begin{array}[]{ccccccc}{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1\\\
\end{array}\right].$ (431)

Although we explained a general method to find $C$ by a syzygy computation in
Section 4.2, we employ a more efficient method for this case. As (116) is
written modulo $L=\underline{x^{2}}+\ldots$, we can eliminate all instances of
$x^{p}$ for $p\geq 2$ via division by $L$. Consequently, it is sufficient to
write $C$ as a rational ansatz being linear in $x$:

$\displaystyle
C=\frac{p_{ij}(a,b,c_{1},c_{2})x+q_{ij}(a,b,c_{1},c_{2},y)}{r(a,b,c_{1},c_{2})x+s(a,b,c_{1},c_{2},y)}\,.$
(432)

After this preprocessing, we solve the system of linear equations

$\displaystyle\texttt{Numerator[}\,C_{{\tt{Ext}}}-C\cdot
M_{{\tt{Ext}}}\,\texttt{]}=0$ (433)

for $p_{ij},q_{ij},r$ and $s$, which, by (130), yields the Pfaffian matrices
$P_{x}$ and $P_{y}$ in the fraction field $\mathbb{C}[x,y]$ (the matrices can
be downloaded at dataAndProgramsOfThisPaper ). As per oaku-1996 , $P_{x}$ and
$P_{y}$ induce a ${\cal D}$-module on an affine variety.

Let $[g_{1},\,g_{2},\,g_{3}]$ denote the linearly independent holomorphic
solutions at $(x,y)=(1/4,\,1/4)\in W$. The standard monomials on the singular
locus $W$ are given by
$[s_{1},\,s_{2},\,s_{3}]=[1,\,\partial_{x},\,\partial_{y}]$. We then have a
$3\times 3$ matrix $F_{ij}=(s_{i}\bullet g_{j})$. On $W$, it holds that
$(\partial_{x}\bullet F)\cdot F^{-1}=P_{x}$ and $(\partial_{y}\bullet F)\cdot
F^{-1}=P_{y}$. The matrices $P_{x}$ and $P_{y}$ are Pfaffian matrices on $W$.
Note that these equalities hold on $W$ but not outside of $V(L)$, because the
PDE system (414) is irreducible hattori-takayama .

Let us end our discussion on $F_{4}$ with a numerical analysis of the
restricted Pfaffian system. To this end, we parameterize $L=0$ by

$\displaystyle x=q_{x}(t),\,y=q_{y}(t)\quad\text{with}\quad
q_{x}(t)=t,\,q_{y}(t)=t+1\pm 2\sqrt{t}\,.$ (434)

Fixing the exponent parameters
$(a,\,b,\,c_{1},\,c_{2})=(-2/3,\,1/3,\,1/3,\,1/3)$, we thus obtain an ODE of
rank $3$ on $V(L)$:

$\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}g(t)=\left[P_{x}(q_{x}(t),q_{y}(t))\frac{\mathrm{d}q_{x}}{\mathrm{d}t}+P_{y}(q_{x}(t),q_{y}(t))\frac{\mathrm{d}q_{y}}{\mathrm{d}t}\right]\cdot
g(t)\,.$ (435)

A solution for the $3$-dimensional vector $g(t)$ with initial condition
$g=[1,\,0,\,0]^{T}$ is given in a python notebook at
dataAndProgramsOfThisPaper .

We note that the restriction of $F_{4}$ to its singular locus was also studied
in mitsuo1995connection from a different point of view, utilizing a double
cover map. However, it is not known that this method can be applied to Horn’s
functions $H_{1},\ldots,H_{7}$ (see erdelyi for definitions). With our general
algorithm, we are able to obtain rational restrictions of every system for
$H_{i}$ onto irreducible components of their singular loci. The relevant
hypersurfaces are given in the following table:

| Hypersurface
---|---
$H_{1}$ | $y^{2}-4xy+2y+1=0$
$H_{2}$ | $xy-y-1=0$
$H_{3}$ | $y^{2}-y+x=0$
$H_{4}$ | $y^{2}-2y-4x+1=0$
$H_{4}$ | $27y^{2}x-36xy-y+16x^{2}+8x+1=0$
$H_{6}$ | $y^{2}x-y-1=0$
$H_{7}$ | $4xy^{2}-y^{2}-2y-1=0$

The holonomic $\mathcal{D}$-modules for each $H_{i}$ are of rank $4$, and in
all cases the restricted standard basis is
${{\tt{RStd}}}=[1,\partial_{x},\partial_{y}]^{T}$. In other words, the
rational restrictions of each system is of rank $3$. See
dataAndProgramsOfThisPaper for programs for these restrictions.

##  6 Conclusion

In this work we studied the process of taking limits on Feynman integrals and,
more generally, Euler integrals. Our philosophy has been to study such
integrals from the perspective of the differential equations that they obey;
in particular their Pfaffian PDE systems. The structure of Pfaffian systems,
in turn, is richly encoded in the algebraic language of holonomic
$\mathcal{D}$-modules. It is thereby possible to formalize the notion of a
limit as a _restriction_ of a $\mathcal{D}$-module. In our work, the
motivating example of a $\mathcal{D}$-module has been the GKZ hypergeometric
system, but our results and methods apply directly to Feynman integrals as
well.

A restriction is generally accompanied by a drop in rank. Regarding the
Pfaffian system for some Euler integral, this means that we expect a smaller
and simpler Pfaffian system to survive in the limit. The main result of this
publication is the development of efficient algorithms for computing these
smaller Pfaffian systems. We presented two such restriction algorithms, each
one having unique benefits. The first algorithm, working at the level of
Pfaffian matrices, only requires sequential applications of certain gauge
transformations. This algorithm is able to extract logarithmic corrections in
the limiting variable, a feature that we expect to be especially useful for
particle physics phenomenology. The second algorithm, working at the level of
$\mathcal{D}$-modules, computes the restricted Pfaffian system by passing
through the _Macaulay matrix_. This algorithm benefits from the use of linear
algebra over finite fields, making it quite efficient, and it does not require
the unrestricted Pfaffian system, which can be hard to obtain, as input. While
traditional algorithms have been limited to restriction onto hyperplanes, we
have shown that the Macaulay matrix based method is able to restrict onto
hypersurfaces.

In future work, it would be interesting to test our methods on state-of-the-
art examples relevant for current phenomenological applications. While this
paper only attempted to compute logarithmic corrections to the one-loop case
of Bhabha scattering, there is in principle no computational barrier
preventing the push to higher loops and more points. For the sake of increased
accuracy, it is then alluring to also include the mixed power-and-log
corrections of the form $z_{1}^{a}\log^{b}(z_{1})$, $a,b\in\mathbb{Z}_{>0}\>$,
as discussed at the end of Section 3.4. On the mathematical side, we believe
that it would be interesting to investigate restriction _ideals_ using the
Macaulay matrix method. This would also aid in the computation of symbol
alphabets at two loops, as they can be obtained from restrictions of GKZ
systems Dlapa:2023cvx .

## Acknowledgements

We deeply appreciate discussions had with our mentor Prof. Pierpaolo Mastrolia
and colleagues Manoj K. Mandal and Federico Gaparotto. This work gives answers
to questions raised in the previous paper Chestnov:2022alh , written in
collaboration with them.

S.M. and N.T. are supported in part by the JST CREST Grant Number JP19209317.
S.M. is supported by JSPS KAKENHI Grant Number 22K13930. N.T. is supported in
part by JSPS KAKENHI Grant Number JP21K03270. V.C. is supported by the
Diagrammalgebra Stars Wild-Card Grant UNIPD, and in part by the European
Research Council (ERC) under the European Union’s research and innovation
programme grant agreement 101040760 (ERC Starting Grant FFHiggsTop).

##  Appendix A Modules and tensors in a nutshell

This exposition is independent of the main text and contains a pedagogical
review of tensor products in the context of $\mathcal{D}$-modules, along with
several instructive examples of tensor products also appearing in the text.

Let $R$ be a ring with $1$ and ${\cal M}$ be an abelian group with the group
operation denoted additively by “+”. Consider a map from $R\times{\cal M}$ to
${\cal M}$, whose image $a\,m$ for given $a\in R$ and $m\in{\cal M}$ is
denoted multiplicatively292929 Explicitly, we consider a map
$\psi\,:\,R\times{\cal M}\to{\cal M}$ and write its image as multiplication:
$\psi(a,m)\equiv a\,m\>$. We use the same notation for multiplication inside
of the ring $R$.  and is referred to as the result of the action of $a$ on
$m$. Then the abelian group ${\cal M}$ forms a left $R$-module when it
satisfies the following conditions:

  1. 1.

For any $a\in R$ and $m_{i}\in{\cal M}$, we have
$a\,(m_{1}+m_{2})=a\,m_{1}+a\,m_{2}$.

  2. 2.

For any $a_{i}\in R$ and $m\in{\cal M}$, we have
$(a_{1}\,a_{2})\,m=a_{1}\,(a_{2}\,m)\>$, and
$(a_{1}+a_{2})\,m=a_{1}\,m+a_{2}\,m\>$.303030 In other words,
$\psi(a_{1}\,a_{2},m)=\psi\bigl{(}a_{1},\psi(a_{2},m)\bigr{)}$ and
$\psi(a_{1}+a_{2},m)=\psi(a_{1},m)+\psi(a_{2},m)\>$.

  3. 3.

For any $m\in{\cal M}$, we have $1\,m=m$.

When $R$ is a (commutative) field, a left $R$-module is nothing but an
$R$-vector space.

A right $R$-module is defined analogously by the following conditions:
$(m_{1}+m_{2})\,a=m_{1}\,a+m_{2}\,a\>$,
$m\,(a_{1}\,a_{2})=(m\,a_{1})\,a_{2}\>$,
$m\,(a_{1}+a_{2})=m\,a_{1}+m\,a_{2}\>$, and $m=m\,1$.

######  Example A.1.

Consider the ring of univariate differential operators with polynomial
coefficients ${\cal D}=\mathbb{C}\langle z,\partial_{z}\rangle$. The quotient
set ${\cal M}={\cal D}/(z\,{\cal D})$ forms a right ${\cal D}$-module by
defining the right action $[m\,a]$ of some $a\in{\cal D}$ on $[m]\in{\cal M}$
as the product (i.e. composition) of differential operators in ${\cal D}$ and
a subsequent reduction to the equivalence class313131 Here and in the
following the square brackets
$[\mathchoice{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\displaystyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\textstyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\scriptstyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\scriptscriptstyle\bullet$}}}}}]$
denote the equivalence class of some element
$\mathchoice{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\displaystyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\textstyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\scriptstyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\scriptscriptstyle\bullet$}}}}}$
under some relation. Sometimes we will omit these brackets for the sake of
brevity if it does not lead to confusion.  by $z\,{\cal D}\>$.

$\blacksquare$

######  Example A.2.

Let ${\cal D}_{n}$ be the Weyl algebra:

${\cal D}_{n}=\mathbb{C}\langle
z_{1},\ldots,z_{n},\partial_{1},\ldots,\partial_{n}\rangle\>,$

that is, the ring of differential operators in $n$ variables
$z_{1},\ldots,z_{n}\>$. The set of length $k$-vectors ${\cal D}_{n}^{k}$ is an
example of a left ${\cal D}_{n}$-module: The addition is defined by
$(f_{1},\ldots,f_{k})+(g_{1},\ldots,g_{k})=(f_{1}+g_{1},\ldots,f_{k}+g_{k})$
and the action of $h\in{\cal D}_{n}$ on $(f_{1},\ldots,f_{k})$ is given by
$(hf_{1},\ldots,hf_{k})\>$. ${\cal D}_{n}^{k}$ is called the free ${\cal
D}_{n}$-module of rank $k$. $\blacksquare$

Let us now briefly review the construction of a Gröbner basis in the free
module ${\cal D}_{n}^{k}$ that we have just introduced above. We fix a term
order $\prec$ in ${\cal D}_{n}$. For a $k$-vector
$f=(0,\ldots,0,f_{i},\ldots,f_{k})$ with $f_{i}\not=0\>$, the index $i$ is
called the top index of $f$. Consider two elements $f$ and $g$ in ${\cal
D}_{n}^{k}\>$. We say that $f$ is larger than $g$ in the position over term
(POT) order of ${\cal D}_{n}^{k}\>$, when the top index of $f$ is smaller than
the top index of $g\>$, or when the top indices of $f$ and $g$ are both $i$
and the corresponding components $f_{i}\succ g_{i}\>$. The reduction (or the
division) algorithm follows from this definition of the total order. Namely,
the $S$-pair $S(f,g)$ of $f$ and $g$ is a $(0,\ldots,0)$ vector when the top
indices of $f$ and $g$ are not the same. When the top indices are the same and
equal to $i$, then the $S$-pair of $f$ and $g$ is defined by
$h_{1}f-h_{2}\,g\>$, where $h_{j}\in{\cal D}_{n}$ are chosen so that
$h_{1}f_{i}-h_{2}\,g_{i}$ is the $S$-pair of $f_{i}$ and $g_{i}$ in ${\cal
D}_{n}$.

######  Example A.3.

Consider the free module ${\cal D}_{2}^{2}$ in two variables $z_{1}$ and
$z_{2}\>$. When $f=(\partial_{1},z_{1})$ and $g=(0,z_{1})\>$, then
$S(f,g)=(0,0)$ as the top indices of $f$ and $g$ are not the same. If
$f=(z_{1}\,\partial_{1},z_{1})$ and $g=(z_{2}\,\partial_{1},z_{1})$, then the
$S$-pair becomes $S(f,g)=z_{2}\,f-z_{1}\,g=(0,z_{1}z_{2}-z_{1}^{2})\>$. For
further examples and details we refer the interested reader to adams and
(dojo, , Sections 6.6-7).

$\blacksquare$

######  Definition A.4.

For a left ${\cal D}_{n}$-module ${\cal M}$ and a right ${\cal D}_{n}$-module
$P$, let us define the tensor product $P\otimes_{{\cal D}_{n}}{\cal M}$ as
follows. We denote by $p\otimes m$ the element of the direct product $(p,m)$,
where $p\in P$ and $m\in{\cal M}$. Then $P\otimes_{{\cal D}_{n}}{\cal M}$ is
the set of a finite formal sum of elements of this form modulo the equivalence
relation $\sim$ given by:

  1. 1.

$p_{1}\otimes m+p_{2}\otimes m\sim(p_{1}+p_{2})\otimes m$, $m\in{\cal M}$,
$p_{i}\in P$ (linearity in the left slot).

  2. 2.

$p\otimes m_{1}+p\otimes m_{2}\sim p\otimes(m_{1}+m_{2})$, $p\in P$,
$m_{i}\in{\cal M}$ (linearity in the right slot).

  3. 3.

$pf\otimes m\sim p\otimes fm$ for $f\in{\cal D}_{n}$, $m\in{\cal M}$, and
$p\in P\>$.323232 Hence the subscript ${\cal D}_{n}$ in $\otimes_{{\cal
D}_{n}}$ means that any element $f\in{\cal D}_{n}$ can be carried across the
$\otimes$ sign.

In other words, if we denote by $F(P\otimes{\cal M})$ the set of formal finite
sums of $p\otimes m$, then the tensor product is just the factor
$P\otimes_{{\cal D}_{n}}{\cal M}:=F(P\otimes{\cal M})/\sim\>$.

######  Remark A.5.

When the left entry $P$ is also a left ${\cal D}_{n-1}$-module, the tensor
product $P\otimes_{{\cal D}_{n}}{\cal M}$ can be turned into a left ${\cal
D}_{n-1}$-module by specifying the action of $g\in{\cal D}_{n-1}$ on $p\otimes
m$ as $g\,(p\otimes m)=(g\,p)\otimes m\>$.

######  Example A.6.

Let ${\cal M}$ be a left ${\cal D}_{n}$-module. In order to illustrate the
various concepts introduced so far, here we show a proof of the following
well-known isomorphism between two left ${\cal D}_{n-1}$-modules written as
quotients:

${\cal D}_{n}\big{/}(z_{n}\,{\cal D}_{n})\otimes_{{\cal D}_{n}}{\cal
M}\simeq{\cal M}\big{/}(z_{n}\,{\cal M})$ (436)

We start with showing that the right ${\cal D}_{n}$-module ${\cal
D}_{n}\big{/}(z_{n}\,{\cal D}_{n})$ (see Example A.1), while being also a left
${\cal D}_{n-1}$-module, is in fact not a left ${\cal D}_{n}$-module. Indeed,
suppose that it is a left ${\cal D}_{n}$-module. We denote by
$[\mathchoice{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\displaystyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\textstyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\scriptstyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\scriptscriptstyle\bullet$}}}}}]$
the equivalence class of some element
$\mathchoice{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\displaystyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\textstyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\scriptstyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\scriptscriptstyle\bullet$}}}}}$
in ${\cal D}_{n}\big{/}(z_{n}\,{\cal D}_{n})$. On one hand, $[z_{n}]=[0]$ in
${\cal D}_{n}\big{/}(z_{n}\,{\cal D}_{n})\>$, so that the action of the
derivative $\partial_{n}$ should also be333333 Note that the condition
$a([0]+[0])=a([0])=a[0]+a[0]$ implies $a[0]=[0]$ for any $a\in{\cal D}_{n}$.
$\partial_{n}[z_{n}]=[0]$ in ${\cal D}_{n}\big{/}(z_{n}\,{\cal D}_{n})$. On
the other hand, $\partial_{n}[z_{n}]=[z_{n}\partial_{n}+1]$ and
$[z_{n}\partial_{n}]=[0]$ in ${\cal D}_{n}\big{/}(z_{n}\,{\cal D}_{n})\>$,
leading to $\partial_{n}[z_{n}]=[1]\neq[0]$ in ${\cal
D}_{n}\big{/}(z_{n}\,{\cal D}_{n})\>$, which is a contradiction. The following
technical observation will become useful shortly: any non-zero element of
${\cal D}_{n}\big{/}(z_{n}\,{\cal D}_{n})\otimes_{{\cal D}_{n}}{\cal M}$ can
be expressed as $[1]\otimes m$ for some $m\not\in z_{n}\,{\cal M}\>$. Indeed,
in a single product any factor $f\in{\cal D}_{n}$ can be moved from the left
to the right slot: $[f]\otimes m=[1]\otimes fm\>$. After performing this
operation termwise, any finite sum can be united into a single product:
$[1]\otimes m_{1}+[1]\otimes m_{2}=[1]\otimes(m_{1}+m_{2})\>$ according to
Definition A.4. For two non-zero elements $[1]\otimes m_{1}$ and $[1]\otimes
m_{2}\>$, where $m_{i}\not\in z_{n}\,{\cal M}\>$, it can be shown that the
condition $[1]\otimes m_{1}\not=[1]\otimes m_{2}$ is equivalent to
$m_{1}\not=m_{2}$ in ${\cal M}$. We also note that $[0]\otimes m$ and
$[f]\otimes 0$ are $0$ by definition.

We define a ${\cal D}_{n-1}$-morphism $\varphi$ from ${\cal
D}_{n}\big{/}(z_{n}\,{\cal D}_{n})\otimes_{{\cal D}_{n}}{\cal M}$ to ${\cal
M}/(z_{n}{\cal M})$ by $\varphi\bigl{(}[f]\otimes m\bigr{)}:=[fm]$ and extend
this definition on finite sums $\varphi\bigl{(}[f_{1}]\otimes
m_{1}+[f_{2}]\otimes m_{2}\bigr{)}=[f_{1}m_{1}]+[f_{2}m_{2}]\>$ by linearity.
Let us now prove that $\varphi$ is indeed an isomorphism.

#####  $\varphi$ is well-defined:

We prove that if $t_{1}\sim t_{2}$, $t_{i}\in{\cal D}_{n}\big{/}(z_{n}\,{\cal
D}_{n})\otimes_{{\cal D}_{n}}{\cal M}$, then $\varphi(t_{1})=\varphi[t_{2}]$.
From the definition of $\varphi$ it follows that if $t\sim 0\otimes 0$, then
$\varphi(t)=[0]$. Any $0$ in ${\cal D}_{n}\big{/}(z_{n}\,{\cal
D}_{n})\otimes_{{\cal D}_{n}}{\cal M}$ can be expressed as $[z_{n}]\otimes m$,
$m\in{\cal M}$ after a simplification by “+”. Note that $\varphi$ satisfies
$\varphi(t_{1}+t_{2})=\varphi(t_{1})+\varphi(t_{2})$ for any $t_{i}\in{\cal
D}_{n}\big{/}(z_{n}\,D_{n})\otimes_{{\cal D}_{n}}{\cal M}$. Since
$\varphi\bigl{(}[z_{n}]\otimes m\bigr{)}=[z_{n}m]$, $z_{n}m\in z_{n}{\cal M}$,
then $[0]\otimes m$ is sent to $[0]$ by $\varphi$.

#####  $\varphi$ is ${\cal D}_{n-1}$-morphism:

Let $c_{i}\in{\cal D}_{n-1}$, $f_{i}\in{\cal D}_{n}$, and $m_{i}\in{\cal M}$.

$\displaystyle\varphi\bigl{(}c_{1}\,([f_{1}]\otimes
m_{1})+c_{2}\,([f_{2}]\otimes m_{2})\bigr{)}$
$\displaystyle=\varphi\bigl{(}[c_{1}\,f_{1}]\otimes
m_{1}+[c_{2}\,f_{2}]\otimes m_{2}\bigr{)}$ (Remark A.5)
$\displaystyle=[c_{1}\,f_{1}\,m_{1}]+[c_{2}\,f_{2}\,m_{2}]$ (definition of
$\varphi$) $\displaystyle=c_{1}\,[f_{1}\,m_{1}]+c_{2}\,[f_{2}\,m_{2}]$ (action
of $c_{i}$ on ${\cal M}\big{/}(z_{n}\,{\cal M})$)
$\displaystyle=c_{1}\,\varphi\bigl{(}[f_{1}]\otimes
m_{1}\bigr{)}+c_{2}\,\varphi\bigl{(}[f_{2}]\otimes m_{2}\bigr{)}\qquad$
(definition of $\varphi$)

#####  $\varphi$ is surjective:

Clearly the image $\varphi\bigl{(}[1]\otimes{\cal M}\bigr{)}$ covers the whole
${\cal M}\big{/}z_{n}\,{\cal M}$.

#####  $\varphi$ is injective:

Due to the linearity of $\varphi$, it is enough to prove that if
$\varphi\bigl{(}t\bigr{)}=0$ for some $t\in{\cal D}_{n}\big{/}(z_{n}\,{\cal
D}_{n})\otimes_{{\cal D}_{n}}{\cal M}$ , then necessarily $t=0$. Suppose that
$t\not=0$. Then there exists $m\in{\cal M}$ such that $t\sim[1]\otimes m$.
Since $\varphi\bigl{(}[1]\otimes m\bigr{)}=[m]=0$, we have $m\in z_{n}\,{\cal
M}$. Writing $m=z_{n}\,m^{\prime}$, $m^{\prime}\in{\cal M}$, we have
$[1]\otimes(z_{n}\,m^{\prime})\sim[z_{n}]\otimes m^{\prime}\sim 0\otimes
m^{\prime}$, which means that $t=0$.

Thus we constructed an explicit isomorphism $\varphi$ between ${\cal
D}_{n}\big{/}(z_{n}\,{\cal D}_{n})\otimes_{{\cal D}_{n}}{\cal M}$ and ${\cal
M}\big{/}(z_{n}\,{\cal M})$ and proved the identity (436). $\blacksquare$

######  Example A.7.

Another example of tensor product is:

${\cal N}=\mathbb{C}[z_{1},\ldots,z_{n}]\big{/}\langle
z_{n}\rangle\otimes_{\mathbb{C}[z_{1},\ldots,z_{n}]}{\cal M}\>.$

For $1\leq i\leq n-1$ we define the action of $\partial_{i}$ on the product
$[f]\otimes m\>$ as $\partial_{i}([f]\otimes m):=[\partial_{i}f]\otimes
m+[f]\otimes(\partial_{i}m)\>$, where
$[f]\in\mathbb{C}[z_{1},\ldots,z_{n}]\big{/}\langle z_{n}\rangle$ and
$m\in{\cal M}\>$. Multiplication by $h\in\mathbb{C}[z_{1},\ldots,z_{n-1}]$ is
defined by $h\,([f]\otimes m):=[h\,f]\otimes m\>$ (similar to Remark A.5).
Then we can prove that ${\cal N}$ is a left ${\cal D}_{n-1}$-module

${\cal N}\simeq{\cal M}\big{/}(z_{n}\,{\cal M})\>.$ (437)

Indeed, the isomorphism $\varphi$ from ${\cal N}$ to ${\cal
M}\big{/}(z_{n}\,{\cal M})$ is given by $\varphi\bigl{(}[f]\otimes
m\bigr{)}:=[f\,m]\>$. Note that for $[f]=[z_{n}\,h]\sim[0]$ we have
$\varphi\bigl{(}[z_{n}\,h]\otimes m\bigr{)}=[z_{n}\,h\,m]\sim[0]\>$, which
makes the map $\varphi$ is well-defined. It is left to the reader to prove
that $\varphi$ is an isomorphism. $\blacksquare$

######  Example A.8.

Our final example of tensor product is
$\mathbb{C}[z][1/f]\otimes_{\mathbb{C}[z]}{\cal D}_{n}$ , where
$\mathbb{C}[z]=\mathbb{C}[z_{1},\ldots,z_{n}]$ and $f$ is a polynomial in
$z\>$. Since the left entry
$\mathbb{C}[z][1/f]=\mathbb{C}[z_{1},\ldots,z_{n},1/f]$ can be viewed as a
left ${\cal D}_{n}$-module, we can regard the tensor product
$\mathbb{C}[z][1/f]\otimes_{\mathbb{C}[z]}{\cal D}_{n}$ as a left ${\cal
D}_{n}$-module by
$\partial_{i}(g\otimes\ell)=(\partial_{i}g)\otimes\ell+g\otimes(\partial_{i}\ell)\>$,
for $g\in\mathbb{C}[z][1/f]$ and $\ell\in{\cal D}_{n}\>$. It is an exercise
for the reader to describe the action of a general element of ${\cal D}_{n}$
explicitly. Furthermore, we can introduce the ring structure on
$\mathbb{C}[z][1/f]\otimes_{\mathbb{C}[z]}{\cal D}_{n}$ by defining the
product “$\cdot$” between two elements as
$(g_{1}\otimes\ell_{1})\cdot(g_{2}\otimes\ell_{2})=g_{1}\otimes\ell_{1}\,(g_{2}\otimes\ell_{2})\>$,
where $g_{i}\in\mathbb{C}[z][1/f]$ and $\ell_{i}\in{\cal D}_{n}\>$. Roughly
speaking, $\mathbb{C}[z][1/f]\otimes_{\mathbb{C}[z]}{\cal D}_{n}$ is
$\mathbb{C}\langle
z_{1},\ldots,z_{n},1/f,\partial_{1},\ldots,\partial_{n}\rangle\>$.
$\blacksquare$

In Proposition 2.3, we construct a zero-dimensional $\mathcal{R}$-module from
a holonomic $\mathcal{D}_{Y}$-module by a tensor product. The zero-dimensional
one can be expressed as a quotient by a left ideal as the right hand side of
(438) below. Let us sketch a proof of the isomorphism. Take a left ideal
$\mathcal{I}$ in the Weyl algebra ${\cal D}_{n}\>$, and let ${\cal M}:={\cal
D}_{n}\big{/}\mathcal{I}\>$ be a left ${\cal D}_{n}$-module. Let ${\cal
R}_{n}=\mathbb{C}(z)\langle\partial_{1},\ldots,\partial_{n}\rangle$ be the
ring of differential operators with rational function coefficients, where
$\mathbb{C}(z):=\mathbb{C}(z_{1},\ldots,z_{n})$ is the field of rational
functions. The left ${\cal R}_{n}$-modules ${\cal R}_{n}\otimes_{{\cal
D}_{n}}{\cal M}$ and ${\cal R}_{n}\big{/}({\cal R}_{n}\,\mathcal{I})$ are
isomorphic as left ${\cal R}_{n}$-modules. Here the left action of $h\ni{\cal
R}_{n}$ on $f\otimes g\in{\cal R}_{n}\otimes_{{\cal D}_{n}}{\cal M}$ is
defined by $h\,f\otimes g\>$. The isomorphism is given by

${\cal R}_{n}\otimes_{{\cal D}_{n}}{\cal M}\ni f\otimes g\mapsto fg\in{\cal
R}_{n}\big{/}({\cal R}_{n}\,\mathcal{I})\>,$ (438)

and we can easily check that it is an isomorphism of left ${\cal
R}_{n}$-modules by noting that ${\cal R}_{n}$ is an associative algebra. Note
that $f_{1}f_{2}\otimes g=f_{1}\otimes f_{2}\,g$ in ${\cal
R}_{n}\otimes_{{\cal D}_{n}}{\cal M}$ when $f_{2}\in{\cal D}_{n}\>$. We can
show analogously that ${\cal R}_{n}\otimes_{{\cal D}_{n}}{\cal
D}_{n}^{m}\big{/}\mathcal{J}$ is isomorphic to ${\cal R}_{n}^{m}\big{/}({\cal
R}_{n}\,\mathcal{J})$ as left ${\cal R}_{n}$-modules, where $\mathcal{J}$ is
some left submodule of ${\cal D}_{n}\>$.

######  Remark A.9.

We referred to the algorithm of constructing a _cyclic vector_ in Section 2.
It is a way to construct an ideal given a Pfaffian system.

In order to apply the cyclic vector to ${\cal R}_{n}^{m}\big{/}({\cal
R}_{n}\,\mathcal{J})$, we first compute the Weyl closure of $\mathcal{J}$, and
then apply the algorithm of leykin-cyclic to the closure. This procedure is
exact, but it needs huge computational resources. We present an easy but
inexact method. For simplicity, we illustrate the case of $m=2$, $n=1$
($\partial=\partial_{1}$, $z=z_{1}$), and ${\cal R}_{1}\mathcal{J}$ is
generated by

$(\partial,0)-(a,b),\quad(0,\partial)-(c,d),\quad a,b,c,d\in\mathbb{C}(z).$

These generators induce a system of differential equations for two unknown
functions $u^{1}$ and $u^{2}$:

$\partial\bullet u^{1}-au^{1}-bu^{2}=0,\quad\partial\bullet
u^{2}-cu^{1}-du^{2}=0.$ (439)

Differentiating both equations w.r.t. $z$, we obtain two more identities:

$u^{1}_{zz}-a_{z}u^{1}-au^{1}_{z}-b_{z}u^{2}-bu^{2}_{z}=0,u^{2}_{zz}-c_{z}u^{1}-cu^{1}_{z}-d_{z}u^{2}-du^{2}_{z}=0$
(440)

where ${u}_{z}$ and ${u}_{zz}$ denote the first and second derivatives w.r.t.
$z$. We regard $u^{1},u^{2},u^{1}_{z},u^{2}_{z},u^{1}_{zz}$ and $u^{2}_{zz}$
as independent variables in the polynomial ring
$\mathbb{C}(z)[u^{1},u^{2},u^{1}_{z},u^{2}_{z},u^{1}_{zz},u^{2}_{zz}]$ and
eliminate $u^{2},u^{2}_{z},u^{2}_{zz}$ from (439) and (440). Then we obtain an
ODE $u^{1}_{zz}+c_{1}(z)u^{1}_{z}+c_{2}(z)u^{1}=0$ for the function $u^{1}$.
In some cases, the left ideal
$\mathcal{I}=\langle\partial^{2}+c_{1}\partial+c_{2}\rangle$ in ${\cal R}_{1}$
gives an isomorphism of ${\cal R}_{1}^{2}\big{/}({\cal R}_{n}\,\mathcal{J})$
and ${\cal R}_{1}/\mathcal{I}$.

######  Remark A.10.

As a historical concluding remark, let us rephrase the notion of a _solution_
to a system of differential equations in the language of homological algebra,
which is a starting point of the theory of ${\cal D}$-modules. Let us denote
by ${\cal O}^{an}(U)$ the set of holomorphic functions343434 The superscript
“an” stands for “analytic”. On the other hand, ${\cal O}^{alg}(U)$ is the set
of rational functions that do not have poles on $U$. They are called regular
functions on $U$. We omit the superscript “alg”, that stands for “algebraic”,
in this paper.  on an open set $U\>$, and by

$\mathrm{Hom}_{{\cal D}_{n}}({\cal M},{\cal O}^{an}(U))\>$ (441)

the set of left ${\cal D}_{n}$-morphisms from the left ${\cal D}_{n}$-module
${\cal M}$ to ${\cal O}^{an}(U)\>$. The correspondence between a homomorphism
$\varphi\in\mathrm{Hom}_{{\cal D}_{n}}({\cal M},{\cal O}^{an}(U))$ and a
solution $f$ is given by $f=\varphi\bigl{(}[1]\bigr{)}\>$. In fact, take
${\cal M}:={\cal D}_{n}\big{/}\mathcal{I}$, and let $\ell$ be an element of
the left ideal $\mathcal{I}\>$, so that $[\ell]\sim[0]$ in ${\cal M}\>$. Since
$\varphi$ is a left ${\cal D}_{n}$-morphism, we have:

$0=\varphi\bigl{(}[0]\bigr{)}=\varphi\bigl{(}[\ell]\bigr{)}=\varphi\bigl{(}\ell\,[1]\bigr{)}=\ell\mathchoice{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\displaystyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\textstyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\scriptstyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\scriptscriptstyle\bullet$}}}}}\varphi\bigl{(}[1]\bigr{)}\>,$
(442)

which means that $\varphi\bigl{(}[1]\bigr{)}\in{\cal O}^{an}(U)$ is a solution
to the differential equation
$\ell\mathchoice{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\displaystyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\textstyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\scriptstyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\scriptscriptstyle\bullet$}}}}}f=0\>$.
Conversely, let $f$ be a solution of $\mathcal{I}\>$, meaning that for any
$\ell\in\mathcal{I}$ it satisfies
$\ell\mathchoice{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\displaystyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\textstyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\scriptstyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\scriptscriptstyle\bullet$}}}}}f=0\>$.
Then for and other element $m\in{\cal D}_{n}\>$, the mapping $m\mapsto
m\bullet f\in{\cal O}^{an}(U)$ defines a left ${\cal D}_{n}$-morphism, i.e. an
element of $\mathrm{Hom}_{{\cal D}_{n}}({\cal M},{\cal O}^{an}(U))\>$. The
derived functor of $\mathrm{Hom}$ gives cohomological solution spaces in the
theory of ${\cal D}$-modules Hotta-Tanisaki-Takeuchi-2008 .

##  Appendix B Normal form by Moser reduction

In Section 3.2, it was emphasised that a Pfaffian system must be in _normal
form_ before the restriction protocol can proceed. A given set of Pfaffian
matrices $P_{1},\ldots,P_{m}$ is in normal form w.r.t. $z_{1}$ when $P_{1}$
satisfies two conditions:

  1. 1.

_Logarithmic_ : $P_{1}$ has a simple pole at $z_{1}=0$.

  2. 2.

_Non-resonant_ : The eigenvalues of the residue matrix $P_{1,-1}(z^{\prime})$
have no integral difference, and $0$ is the unique integer eigenvalue.

As explained later in this section, the non-resonance condition additionally
implies that $P_{2},\ldots,P_{m}$ are finite at $z_{1}=0$. This is important,
because in the restriction protocol of Section 3.2 we use the matrices
$P_{2}(0,z^{\prime}),\ldots,P_{m}(0,z^{\prime})$, where
$z^{\prime}=(z_{2},\ldots,z_{m})$.

Normal form can be obtained by following the Moser reduction scheme Moser ,
which was later made algorithmic by Barkatou Barkatou1995rational (see also
Gerard-Ramis-1983 ). For CAS implementations, see e.g. ISOLDE
barkatou2013isolde and Fuchsia Gituliar:2017vzm . For the reader’s
convenience, in this section we review Moser reduction, largely following
Barkatou-Jaroschek-Maddah-2017 .

###  B.1 Logarithmic $P_{1}$

Let $\mathbb{1}_{r}$ denote the $r$-dimensional identity matrix. The following
theorem is a result of Moser ; Barkatou1995rational .

######  Theorem B.1.

Let $P(z)$ be an $r\times r$ matrix, depending on a single variable $z$, with
entries in the field of rational functions $\mathbb{C}(z)$. Assume that a
system of ODEs

$\frac{dI}{dz}=P(z)\cdot I$ (443)

is regular singular at $z=0$. After a finite sequence of gauge transforms
$P\to G[P]=G^{-1}(P\cdot G-\partial_{z}G)$ with $G$ of the form

$\displaystyle G=\text{a constant matrix,}\quad\text{or}\quad
G=\left[\begin{array}[]{cc}z\mathbb{1}_{r^{\prime}}&\mathbb{0}\\\
\mathbb{0}&\mathbb{1}_{r-r^{\prime}}\end{array}\right]\,,\ \text{for}\
r^{\prime}\in\mathbb{N}\,,$ (446)

the ODEs (443) can be transformed into a system with only simple poles at
$z=0$. Such a sequence of gauge transforms can be found algorithmically.

######  Remark B.2.

The gauge transformation matrices can be constructed without algebraic
extensions. In other words, if the entries of $P(z)$ belong to $K(z)$ for some
subfield $K\subset\mathbb{C}$, then $G$ can be taken to have coefficients in
$K$. This is useful when the system (443) contains additional parameters
independent of $z$.

The theorem above pertains to ODEs, but Pfaffian systems are PDEs in general.
Fortunately, we can still apply the theorem to the PDE case by treating one
variable at a time. Consider a Pfaffian system in $m$ variables:
$\partial_{i}I=P_{i}(z)\cdot I\>$ , for $i=1,\ldots,m\>$. Assume that the
$P_{i}$ have higher-order poles along $z_{1}=0\>$. Treat

$\displaystyle\frac{\mathrm{d}I}{\mathrm{d}z_{1}}=P_{1}(z_{1},z^{\prime})\cdot
I$ (447)

as an ODE in $z_{1}$ with parameters $z^{\prime}=(z_{2},\ldots,z_{m})\>$. By
Theorem B.1 and Remark B.2, we can find a gauge transformation matrix $G\>$
such that $G[P_{i}]$ only has a simple pole along $z_{1}=0\>$. We may thus
assume that the Pfaffian matrices take the form:

$\displaystyle P_{1}(z)=\sum_{n=-1}^{\infty}P_{1,n}(z^{\prime})\,z_{1}^{n}\,.$
(448)

This solves the problem of finding a logarithmic set of Pfaffian matrices.

###  B.2 Non-resonant $P_{1,-1}$

Next we show how to cure the resonant spectrum of $P_{1,-1}(z^{\prime})$ . The
first thing to note is that the eigenvalues of $P_{1,-1}(z^{\prime})$ are
independent of $z^{\prime}$ (haraoka2020linear, , Theorem 12.1). Therefore, if
$P_{1,-1}(z^{\prime})$ is non-resonant at a point, then it is non-resonant
everywhere. Fix a generic value $z^{\prime}=a^{\prime}$, and set
$P:=P_{1,-1}(a^{\prime})$. Suppose that $P$ is in Jordan normal form, namely
in block-diagonal form with Jordan cells $J_{1},\ldots,J_{k}$. Let $r_{i}$
denote the size of each $J_{i}$. Subdivide the matrix $P_{1}(z)$ into cells as

$P_{1}=\left[\begin{array}[]{c|c|c|c}(P_{1})_{11}&(P_{1})_{12}&\dots&(P_{1})_{1k}\\\
\hline\cr(P_{1})_{21}&(P_{1})_{22}&&\vdots\\\ \hline\cr\vdots&&\ddots&\\\
\hline\cr(P_{1})_{k1}&\dots&&(P_{1})_{kk}\end{array}\right]$ (449)

and construct the gauge transformation matrix

$\displaystyle
G=\left[\begin{array}[]{c|c|c|c}z_{1}\mathbb{1}_{r_{1}}&\mathbb{0}&\dots&\mathbb{0}\\\
\hline\cr\mathbb{0}&\mathbb{1}_{r_{2}}&&\vdots\\\ \hline\cr\vdots&&\ddots&\\\
\hline\cr\mathbb{0}&\dots&&\mathbb{1}_{r_{k}}\end{array}\right]\,.$ (454)

By assumption, $P_{1}$ has a simple pole along $z_{1}=0$, and the same will
hold true for $G[P_{1}]$. A short calculation shows that the residue matrix of
$G[P_{1}]$ evaluated at $z^{\prime}=a^{\prime}$ is of the form

$G[P]=\left[\begin{array}[]{c|c|c|c|c}J_{1}-\mathbb{1}_{r_{1}}&(P_{1,0})_{12}&(P_{1,0})_{13}&\cdots&(P_{1,0})_{1k}\\\
\hline\cr\mathbb{0}&J_{2}&\mathbb{0}&\dots&\vdots\\\
\hline\cr\vdots&&\ddots&&\vdots\\\
\hline\cr\mathbb{0}&\dots&&&J_{k}\end{array}\right]\,.$ (455)

We have thereby shifted the eigenvalue corresponding to the Jordan block
$J_{1}$ by $1$. Repeating this procedure sufficiently many times for each
$J_{i}$, we will render $P$ non-resonant at $z^{\prime}=a^{\prime}$, and
therefore non-resonant everywhere.

Let us now show that non-resonance implies finiteness of $P_{2},\ldots,P_{m}$.
The following lemma is well-known.

######  Lemma B.3.

Suppose that $A,B\in M_{r}(\mathbb{C})$ are $r\times r$ matrices with their
entries in $\mathbb{C}$. Let $\lambda(A),\lambda(B)$ denote the sets of
eigenvalues for $A,B$. Consider the linear map

$T_{A,B}:M_{r}(\mathbb{C})\ni X\mapsto A\cdot X-X\cdot B\in
M_{r}(\mathbb{C})\,.$ (456)

Then the set of eigenvalues of $T_{A,B}$ equals the set of eigenvalue
differences $\lambda(A)-\lambda(B)$.

It follows from the integrability condition (18) and (448) that

$\displaystyle\left[P_{1,-1}(z^{\prime}),\,P_{i,n}(z^{\prime})\right]=n\,P_{i,n}(z^{\prime})\quad,\quad
i=2,\ldots m\quad,\quad n<0\,.$ (457)

We can rewrite this in terms of the map $T_{A,B}$ as

$\displaystyle T_{P_{1,-1},\,P_{1,-1}}(P_{i,n})=n\,P_{i,n}(z^{\prime})\,,$
(458)

meaning that $P_{i,n}$ is an eigenvector of $T_{P_{1,-1},\,P_{1,-1}}$ with
eigenvalue $n$. On the other hand, Lemma B.3 shows that any eigenvalue of
$T_{P_{1,-1},\,P_{1,-1}}$ must be of the form $\lambda_{a}-\lambda_{b}$, where
$\lambda_{a,b}$ are eigenvalues of $P_{1,-1}$. If $P_{1,-1}$ is non-resonant,
we cannot have that $\lambda_{a}-\lambda_{b}=n$. Consequently,
$P_{2,n}(z^{\prime})=0$ for $n<0$.

Summarizing the previous two sections, we have a

######  Theorem B.4.

The algorithm for transforming a Pfaffian system into normal form w.r.t.
$z_{1}$ is as follows:

  1. 1.

Apply the Moser reduction scheme from Theorem B.1 over the field
$\mathbb{C}(z^{\prime})$ to ensure that $P_{1}$ has a simple pole at
$z_{1}=0$.

  2. 2.

Apply gauge transformations of the form (455) until $P_{1,-1}$ becomes non-
resonant.

This theorem can be sequentially applied to several variables, in case one
seeks to restrict more than one variable.

###  B.3 Jordan normal form without field extensions

We have already noted that Moser reduction can be performed using only linear
algebra without field extensions. Step $1$ of Theorem B.4 is therefore
efficient to implement. Step $2$, on the other hand, apparently requires one
to solve polynomials equations, as this is the standard way of computing
Jordan normal forms. Here we argue that algebraic extensions can _also_ be
avoided in step $2$, wherefore Theorem B.4 does not require them at all. Fix a
subfield $K\subset\mathbb{C}$. The purpose of this subsection is to prove the
following lemma.

######  Lemma B.5.

There exists an invertible matrix $Q\in K^{r\times r}$ such that $Q^{-1}PQ$ is
block diagonalized as

$\displaystyle
Q^{-1}PQ=\left[\begin{array}[]{c|c|c|c}B_{1}&\mathbb{0}&\dots&\mathbb{0}\\\
\hline\cr\mathbb{0}&B_{2}&&\vdots\\\ \hline\cr\vdots&&\ddots&\\\
\hline\cr\mathbb{0}&\dots&&B_{k}\end{array}\right]$ (463)

where each $B_{i}$ is non-resonant. We write
_$\texttt{Spec[}\,B_{i}\,\texttt{]}$_ for the set of eigenvalues of the block
$B_{i}$. Then, for any pair of indices $i,j$, either
_$(\texttt{Spec[}\,B_{i}\,\texttt{]}+\mathbb{Z})\cap\texttt{Spec[}\,B_{j}\,\texttt{]}=\varnothing$_
or there exists a unique integer $m$ such that
_$\texttt{Spec[}\,B_{i}\,\texttt{]}+m=\texttt{Spec[}\,B_{j}\,\texttt{]}$_.

Once this lemma is proved, we can conclude that the gauge transformations
involving Jordan normal forms from step $2$ of Theorem B.4 can be
algorithmically chosen from $GL_{r}(K[z_{1}^{\pm}])$, which is free of field
extensions. Indeed, each Jordan block $J_{i}$ of $P$ is replaced by the block
$B_{i}$. Let us begin with a simple lemma.

######  Lemma B.6.

Let $f(t)\in K[t]$ be an irreducible polynomial. If $\alpha$ is a root of
$f(t)=0$, no integral shift of $\alpha$ is a root.

###### Proof.

Set $f(t)=a_{0}+\cdots+a_{N-1}t^{N-1}+t^{N}$ and fix an integer $m\neq 0$ such
that $f(\alpha+m)=0$. We prove that the existence of $m$ leads to a
contradiction. A swift computation shows that

$\displaystyle 0=f(\alpha+m)=f(\alpha)+(\text{terms of order }\leq N-1\text{
in }\alpha):=f(\alpha)+g(\alpha)=g(\alpha)\,,$ (464)

where in the last step we used that $\alpha$ is a root of $f$. The coefficient
of $\alpha^{N-1}$ in $g(\alpha)$ is $Nm$, which is clearly non-zero. This
means that $g(\alpha)=0$, i.e. $\alpha$ is also a root of $g(t)$. But this is
a contradiction, because $f(t)$ is the minimal polynomial353535 The minimal
polynomial of $\alpha$ over $K$ is the polynomial with the smallest degree,
whose roots contain $\alpha$. . $\Box$

Suppose we are given a pair of irreducible polynomials $f_{1},f_{2}\in K[t]$.
If they share a root $\alpha$ modulo $\mathbb{Z}$, that means that
$f_{1}(t+s)=f_{2}(t)$ has an integral solution as a system of equations in
$s$. This is seen by expanding both sides and collecting powers of $t$,
leading to a polynomial system for $s$ of the form
$g_{0}(s)=\ldots=g_{N-1}(s)=0$. The GCD $g(s)$ is used whether we have an
integral solution or not.

Let us recall a well-known fact from linear algebra (see e.g. (Dummit-Foote-
PID-Jordan, , Chapter 12)). Given an $r\times r$ matrix $P$ with entries in
$K$, we can equip $K^{r}$ with the structure of a $K[t]$-module by letting $t$
act on $K^{r}$ via left multiplication by $P$. Consider an exact sequence

$K[t]^{r}\quad\overset{t\mathbb{1}_{r}-P}{\longrightarrow}\quad
K[t]^{r}\quad\longrightarrow\quad K^{r}\quad\longrightarrow\quad 0\,.$ (465)

The last morphism sends $f(t)\boldsymbol{e}_{i}$ to
$f(P)\cdot\boldsymbol{e}_{i}$, where $f(t)\in K[t]$ and $\boldsymbol{e}_{i}$
is the standard unit vector. The fundamental theorem of principal ideal
domains shows that $t\mathbb{1}_{r}-P$ is transformed into a canonical form

$\displaystyle
L(t)(t\mathbb{1}_{r}-P)R(t)=\texttt{Diag[}\,1,\,\ldots,\,1,\,d_{1}(t),\,\ldots,\,d_{p}(t)\,\texttt{]}$
(466)

for some invertible polynomial matrices $L(t),R(t)\in GL(r,K[t])$ with
$d_{1}(t)|\cdots|d_{p}(t)$. According to the Chinese remainder theorem, we
thus have an isomorphism (which can be made algorithmically explicit)

$\displaystyle K^{r}\overset{\varphi}{\leftarrow}\
\bigoplus_{i=1}^{p}K[t]/d_{i}(t)\ \simeq\
\bigoplus_{i=1}^{p}\bigoplus_{j=1}^{q_{i}}K[t]/d_{ij}(t)^{m_{ij}}$ (467)

where $d_{i}(t)=\prod_{j=1}^{q_{i}}d_{ij}(t)^{m_{ij}}$ is the irreducible
decomposition. The morphism $\varphi$ sends any element
$(f_{i}(t))_{i=1}^{p}\in\bigoplus_{i=1}^{p}K[t]/d_{i}(t)$ to
$\sum_{i=1}^{p}f(t)v_{i}(t)|_{t\mapsto P}$ where $v_{i}(t)$ is the
$(r-p+i)$-th column vector of $L(t)^{-1}$ and the subscript $t\mapsto P$
stands for the operation of replacing an expression $t^{a}v$ $(v\in K^{r})$ by
$P^{a}v$. Each subspace $K[t]/d_{ij}(t)^{m_{ij}}$ is invariant under
multiplication by $t$, i.e. left multiplication by $P$. This means that we can
find a basis of $K^{r}$ such that the left multiplication by $P$ is block-
diagonalized into a direct sum of multiplication by $t$ on
$K[t]/d_{ij}(t)^{m_{ij}}$. It is not difficult to check that the $d_{ij}(t)$
share a common root modulo $\mathbb{Z}$, as explained in the previous
paragraph. Finally, any basis of each block $K[t]/d_{ij}(t)^{m_{ij}}$ is sent
to a set of column vectors $B_{ij}$. The matrix $Q$ is obtained by aligning
these $B_{ij}$ to form a square matrix.

######  Remark B.7.

In the examples of Pfaffian systems for Feynman integrals discussed in this
paper, any eigenvalue of the residue matrix is an integral linear combination
of $\varepsilon$ and $\varepsilon\delta$. Since the Euler integral (2) admits
a meromorphic continuation as in (berkesch2014euler, , Theorem 2.5), we expect
this to be true in general.

##  Appendix C Improvements on restriction algorithms

Here we remark on subtle improvements to steps in the restriction algorithms
presented in the main body of this text.

###  C.1 $\mathcal{D}$-module-level restriction

The $b$-function played a central role in the $\mathcal{D}$-module restriction
protocol of Section 4.1. However, it is well-known that the $b$-function is
computationally expensive to obtain. When we can find a logarithmic connection
for ${\cal D}^{\prime}/\mathcal{I}$ via another method (recall the definition
of $\mathcal{D}^{\prime}$ below Remark 4.3), such as the Macaulay matrix
method, we can utilize the following proposition to compute the $b$-function
more easily:

######  Proposition C.1.

Let $S=[s_{1}=1,s_{2},\ldots,s_{r}]^{T}$ be a set of standard monomials for
$\mathcal{I}$ such that $\mathrm{ord}\,s_{i}=0$. We denote by
${\overline{\mathcal{I}}}={\cal R}^{\prime}\mathcal{I}\cap{\cal D}^{\prime}$
the Weyl closure of $\mathcal{I}$. Assume that the Pfaffian system with
respect to $S$ has the form

$\displaystyle\partial_{x}S$ $\displaystyle=$
$\displaystyle\left(\frac{P_{-1}}{x}+P_{0}+xP_{1}+\ldots\right)\cdot
S\quad\mathrm{mod}\ {\cal R}^{\prime}\mathcal{I}$ $\displaystyle\partial_{y}S$
$\displaystyle=$ $\displaystyle\left(Q_{0}+xQ_{1}+\ldots\right)\cdot
S\quad\mathrm{mod}\ {\cal R}^{\prime}\mathcal{I}$

Then the eigenvalues of $P_{-1}$ contain the roots of the $b$-function of
${\overline{\mathcal{I}}}$ for the weight $w=(1,0)$.

Proof. We note that $x\partial_{x}S=P_{-1}\cdot S$ modulo
$F_{-1}+{\overline{\mathcal{I}}}$. Here $F_{-1}$ is the filtration defined in
(94). Let $\varphi(s)$ be the characteristic polynomial of $P_{-1}$. By the
Cayley-Hamilton theorem, we have $\varphi(P_{-1})=0$. Hence, we have
$\varphi(x\partial_{x})S=0$ modulo $F_{-1}+{\overline{\mathcal{I}}}$. In
particular, we have $\varphi(x\partial_{x})=0$ modulo
$F_{-1}+{\overline{\mathcal{I}}}$ because $s_{1}=1$. Therefore, the
$b$-function is a factor of $\varphi(s)$. $\Box$

If we have Pfaffian matrices for a logarithmic connection with the singular
locus $x=0$, we can accelerate Algorithm 4.4 with help from the residue matrix
$P_{-1}$ as follows. Let $S$ be a column vector consisting of standard
monomials of a Gröbner basis of $\mathcal{I}$ in ${\cal
R}^{\prime}=\mathbb{C}(x,y)\langle\partial_{x},\partial_{y}\rangle$. Suppose
that

$x\partial_{x}S-(P_{-1}+P_{0}x+\ldots)\cdot S\equiv 0\quad\mathrm{mod}\
{\overline{\mathcal{I}}}$ (468)

where $xP=P_{-1}+P_{0}x+\ldots$ is the Pfaffian matrix in the $x$ direction
multiplied by $x$. Taking the limit $x=0$ in (468), we have $P_{-1}\cdot
S\equiv 0\quad\mathrm{mod}\ {\overline{\mathcal{I}}}+x{\cal D}^{\prime}$,
which gives a relation among standard monomials modulo
${\overline{\mathcal{I}}}+x{\cal D}^{\prime}$.

We can obtain higher-order relations by applying $\partial_{x}^{j}$ to (468)
and taking the limit $x=0$. For example, applying $\partial_{x}$ in
conjunction with the relation $\partial_{x}S\equiv P\cdot S$,

$\partial_{x}S+x\partial_{x}^{2}S\equiv\big{(}P_{-1}+O(x)\big{)}\cdot
S+\big{(}P_{-1}+P_{0}x+O(x^{2})\big{)}\cdot
x^{-1}\big{(}P_{-1}+P_{0}x+O(x^{2})\big{)}\cdot S\,.$ (469)

When $P_{-1}\cdot S\equiv 0$, the limit $x=0$ yields

$\partial_{x}S\equiv P_{-1}\cdot S+(P_{-1}\cdot P_{0}+P_{0}\cdot P_{-1})\cdot
S\equiv 0\quad\mathrm{mod}\ {\overline{\mathcal{I}}}+x{\cal D}^{\prime}\,.$
(470)

Let ${\cal P}$ be the set of the relations coming from $P_{-1}\cdot S\equiv 0$
and higher-order obtained via the Pfaffian equation (468).

######  Algorithm C.2 (Rational restriction to $x=0$ using the residue matrix
$P_{-1}$).

Input: Generators $\\{f_{1},\ldots,f_{\mu}\\}$ of a holonomic ideal
$\mathcal{I}$ in ${\cal D}^{\prime}$. The relations ${\cal P}$. Holonomic rank
$r$ of the restriction to $x=0$ of $\mathcal{I}$. A non-negative integer
$\gamma$ such that $\gamma\geq\max(s_{0}+1,s_{1}+1)$.
Output: Generators of a left submodule $\mathcal{J}\cap{\cal D}^{\gamma}$ of
${\cal D}^{\gamma}$ such that ${\cal R}^{\gamma}/{\cal R}(\mathcal{J}\cap{\cal
D}^{\gamma})={\cal R}\otimes_{\cal D}{\cal D}^{\prime}/(\mathcal{I}+x{\cal
D}^{\prime})$.

1:$w=(1,0)$

2:$k=\gamma-1$

3:repeat

4: $\mathcal{J}={\cal
D}\cdot\Bigl{\\{}v_{k}\,\bigl{(}\,:\partial_{x}^{j}f_{i}:{\big{|}_{x=0}}\bigr{)}\Bigm{|}\mathrm{ord}\,_{(-w,w)}(\partial_{x}^{j}f_{i})\leq
m,1\leq i\leq N,j\in\mathbb{N}_{0}\Bigr{\\}}$

5: $\cup\ \bigl{\\{}v_{k}(p)\bigm{|}p\in{\cal
P},\mathrm{ord}\,_{(-w,w)}(p)\leq k\bigr{\\}}$

6: $k:=k+1$

7:until $\mathrm{rank}\bigl{(}{\cal D}^{\gamma}/\mathcal{J}\cap{\cal
D}^{\gamma}\bigr{)}=r$

8:return $\mathcal{J}$

######  Theorem C.3.

Algorithm C.2 stops and gives the correct answer.

Proof. Rows of the Pfaffian system belong to the Weyl closure
${\overline{\mathcal{I}}}=({\cal R}^{\prime}\mathcal{I})\cap{\cal
D}^{\prime}$. Then the elements of ${\cal P}$ belong to
${\overline{\mathcal{I}}}+x{\cal D}^{\prime}$, and this gives an expression
for ${\cal R}\otimes_{\cal D}({\cal D}^{\prime}/(\mathcal{I}+x{\cal
D}^{\prime})$ by Proposition 4.1. Combining this fact with the proof of
Theorem 4.5, we reach the desired conclusion. $\Box$

######  Example C.4.

To illustrate the difference between Algorithm 4.4 and Algorithm C.2, we
consider the Feynman integral for the one-loop triangle with one internal mass
(for comparison, see the example from Section 5.5.1). The Lee-Pomeransky
polynomial takes the form

$\displaystyle g(z;x)$
$\displaystyle=z_{1}x_{1}+z_{2}x_{2}+z_{3}x_{3}+z_{4}x_{1}^{2}+z_{5}x_{1}x_{2}+z_{6}x_{1}x_{3}$
(471) $\displaystyle z_{1}$
$\displaystyle=z_{2}=z_{3}=z_{4}=z_{5}=1\quad\text{and}\quad z_{6}\text{
generic}.$ (472)

The associated GKZ system has the data

$\displaystyle A=\left[\begin{array}[]{cccccc}1&1&1&1&1&1\\\
1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&2&1&1\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1\\\
\end{array}\right]\quad,\quad\beta=[b_{1},b_{2},b_{3},b_{4}]\,.$ (477)

The GKZ case with generic $z_{1},\ldots,z_{6}$ is of holonomic rank $3$. The
Feynman integral, subject to (472), has rank $2$. We proceed to compute this
drop in rank via a $\mathcal{D}$-module restriction.

To begin, choose the simplex $\sigma=[1,2,3,4]$. The remaining variables are
$z_{5}$ and $z_{6}$, and we seek the restriction $z_{5}=1$. Shifting $z_{5}\to
z_{5}+1$, we can alternatively restrict $z_{5}=0$. The generators of the GKZ
system then read (see (Chestnov:2022alh, , Appendix A) for how to rewrite
generators given a simplex $\sigma$)

$\displaystyle f_{1}$ $\displaystyle=$
$\displaystyle({z}_{5}-{z}_{6}+1)\underline{{\partial}_{5}\partial_{6}}+{b}_{4}{\partial}_{5}-{b}_{3}\partial_{6}$
$\displaystyle f_{2}$ $\displaystyle=$
$\displaystyle\big{[}(-{z}_{6}+1){z}_{5}-{z}_{6}+1\big{]}\underline{{\partial}_{5}\partial_{6}}+{z}_{6}(1-{z}_{6})\partial_{6}^{2}+({b}_{4}{z}_{5}+{b}_{4}){\partial}_{5}$
$\displaystyle+\big{[}(-{b}_{1}+{b}_{2}+{b}_{3}+2{b}_{4}-1){z}_{6}+2{b}_{1}-{b}_{2}-2{b}_{3}-2{b}_{4}+1\big{]}\partial_{6}$
$\displaystyle-{b}_{4}({b}_{3}-{b}_{1}+{b}_{2}+{b}_{4})$ $\displaystyle f_{3}$
$\displaystyle=$ $\displaystyle-
z_{5}({z}_{5}+1)\underline{{\partial}_{5}^{2}}-{z}_{6}{z}_{5}{\partial}_{5}\partial_{6}+\big{[}(-{b}_{1}+{b}_{2}+2{b}_{3}+{b}_{4}-1){z}_{5}+{b}_{1}-{b}_{4}\big{]}{\partial}_{5}+{b}_{3}{z}_{6}\partial_{6}$
$\displaystyle-{b}_{3}({b}_{3}-{b}_{1}+{b}_{2}+{b}_{4})$

where we underlined the leading terms according to the weight $w=(1,0)$. By
the Macaulay matrix method, we find Pfaffian matrices $P_{5}$ and $P_{6}$
written in the basis $S=[\partial_{5},\partial_{6},1]^{T}$. These matrices
have denominators ${z}_{5}({z}_{5}+1)({z}_{6}-{z}_{5}-1)$ and
${z}_{6}({z}_{6}-1)({z}_{6}-{z}_{5}-1)$ respectively. Since the Pfaffian
system is logarithmic along $z_{5}=0$, we may apply Algorithm C.2 with
$x=z_{5}$ and $y=z_{6}$. The residue matrix at $x=0$ is

$\displaystyle
P_{-1}=\left[\begin{array}[]{ccc}{b}_{1}-{b}_{4}&{b}_{3}{y}&{b}_{3}({b}_{1}-{b}_{2}-{b}_{3}-{b}_{4})\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
\end{array}\right].$ (481)

In $(x,y)$ variables, the basis reads $S=[\partial_{y},\partial_{x},1]^{T}$.
The only non-zero element of $P_{-1}\cdot S$ is thus

$({b}_{1}-{b}_{4}){\partial}_{x}+{b}_{3}{y}{\partial}_{y}+{b}_{3}({b}_{1}-{b}_{2}-{b}_{3}-{b}_{4}).$
(482)

Next we execute Algorithm C.2 with $k=1,\,\gamma=2$ and the target holonomic
rank $r=2$. The operator (482) is included in $\mathcal{J}$. The POT Gröbner
basis for $\mathcal{J}$ in ${\cal D}^{2}$ is the set

$\displaystyle\\{$ $\displaystyle[0,\
y(1-{y})\partial_{{y}}^{2}+((-{b}_{1}+{b}_{2}+{b}_{3}+2{b}_{4}-1){y}+2{b}_{1}-{b}_{2}-{b}_{3}-2{b}_{4}+1)\partial_{{y}}$
$\displaystyle\ \quad+{b}_{4}({b}_{1}-{b}_{2}-{b}_{3}-{b}_{4})]^{T},$
$\displaystyle[{b}_{1}-{b}_{4},\
{b}_{3}{y}\partial_{{y}}+{b}_{3}({b}_{1}-{b}_{2}-{b}_{3}-{b}_{4})]^{T}\ \\}.$

This is a rank $2$ system when $b_{1}-b_{4}\not=0$, wherefore the terminating
condition for the algorithm is met.

Incidentally, the $b$-function (the indicial polynomial) along $x=0$ is
${s}({s}-{b}_{1}+{b}_{4}-1)$. When $s=1$, it is equal to $-(b_{1}-b_{4})$,
which appears in the second element of the Gröbner basis. $\blacksquare$

###  C.2 Pfaffian-level restriction

Here we collect technical remarks on the computational side of the restriction
method based on equations (48, 52) of Section 3.2.

####  C.2.1 Details on the $M$-matrix

We start with a matrix-based argument that motivates the bottom-left
$\mathbb{0}$-block in the gauge-transformed Pfaffian matrix (52). Following
the row-decomposition of (48), let us focus on the first $r^{\prime}$ columns
of $M^{-1}$:

$\displaystyle
M^{-1}=\left[\begin{array}[]{c|c}\>{\color[rgb]{0.57421875,0.09375,0.09375}V}&\>{\color[rgb]{0.32421875,0.015625,0.4765625}\star}\end{array}\right]=\vbox{\hbox{
\parbox[c]{0.0pt}{\includegraphics{figures/full-mat-brace2}} }}\,.$ (484)

The column block $V$ satisfies the two identities

$\displaystyle\mathbb{1}_{\scriptscriptstyle r\times r}=M\cdot
M^{-1}=\left[\begin{array}[]{c}{\color[rgb]{0.57421875,0.09375,0.09375}B}\\\
\hline\cr{\color[rgb]{0.32421875,0.015625,0.4765625}R}\end{array}\right]\cdot\left[\begin{array}[]{c|c}\>{\color[rgb]{0.57421875,0.09375,0.09375}V}&\>{\color[rgb]{0.32421875,0.015625,0.4765625}\star}\end{array}\right]=\left[\begin{array}[]{c|c}\>{\color[rgb]{0.57421875,0.09375,0.09375}B}\cdot{\color[rgb]{0.57421875,0.09375,0.09375}V}&\>\star\\\
\hline\cr\>{\color[rgb]{0.32421875,0.015625,0.4765625}R}\cdot{\color[rgb]{0.57421875,0.09375,0.09375}V}&\>\star\end{array}\right]\implies\begin{cases}B\cdot
V=\mathbb{1}_{\scriptscriptstyle r^{\prime}\times r^{\prime}}\>,\\\ R\cdot
V=\mathbb{0}_{\scriptscriptstyle(r-r^{\prime})\times
r^{\prime}}\>.\end{cases}$ (490)

Next, note that since the matrix $R$ forms a basis for the row-span of
$P_{1}(0,z^{\prime})$, we can represent it as a product $R=T\cdot
P_{1}(0,z^{\prime})\>$, where $T$ is some $(r-r^{\prime})\times r$ matrix (of
row operations). If all non-zero rows of $P_{1}(0,z^{\prime})$ are already
independent, we may choose $T$ as a binary matrix, for example

$\displaystyle R=T\cdot
P_{1}(0,z^{\prime})=\left[\begin{array}[]{ccccc}1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}\\\
{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&{\color[rgb]{0.5,0.5,0.5}\cdot}&1&{\color[rgb]{0.5,0.5,0.5}\cdot}\end{array}\right]\cdot\vbox{\hbox{
\parbox[c]{0.0pt}{\includegraphics{figures/square-mat}} }}=\vbox{\hbox{
\parbox[c]{0.0pt}{\includegraphics{figures/R-mat-brace}} }}\>,$ (494)

but in general $T$ might be more complicated and depend on $z^{\prime}$. We
plug this decomposition into the LHS of (52) and focus on its bottom block

$\displaystyle\bigl{(}\partial_{i}R+R\cdot P_{i}(0,z^{\prime})\bigr{)}\cdot
M^{-1}$ $\displaystyle\quad=\bigl{(}(\partial_{i}T)\cdot
P_{1}(0,z^{\prime})+T\cdot[P_{i}(0,z^{\prime}),\,P_{1}(0,z^{\prime})]+T\cdot
P_{1}(0,z^{\prime})\cdot P_{i}(0,z^{\prime})\bigr{)}\cdot M^{-1}$
$\displaystyle\quad=\bigl{(}(\partial_{i}T)+T\cdot
P_{i}(0,z^{\prime})\bigr{)}\cdot\bigl{(}P_{1}(0,z^{\prime})\cdot
M^{-1}\bigr{)}\>.$ (495)

At this stage we recall that the second condition of (490) implies

$\displaystyle R\cdot V=\mathbb{0}\Longleftrightarrow T\cdot
P_{1}(0,z^{\prime})\cdot V=\mathbb{0}\implies P_{1}(0,z^{\prime})\cdot
V=\mathbb{0}\>,$ (496)

so that the second factor in (495) indeed has a $\mathbb{0}$-block

$\displaystyle P_{1}(0,z^{\prime})\cdot
M^{-1}=P_{1}(0,z^{\prime})\cdot\left[\begin{array}[]{c|c}\>V&\>\star\end{array}\right]=\left[\begin{array}[]{c|c}\>\mathbb{0}&\>\star\end{array}\right]\>,$
(499)

which leads to the $\mathbb{0}$-block in the bottom-left corner of (52).

####  C.2.2 An algorithmic choice of the $B$-matrix

Using the notation of Section 3.2, it is easy to see that one possible choice
of the $r^{\prime}\times r$ basis matrix $B$ appearing in (48) is simply the
nullspace of the $r\times r$ residue matrix $P_{1}(0,z^{\prime})$

$\displaystyle
B:=\texttt{NullSpace[}\,P_{1}(0,z^{\prime})\,\texttt{]}=\texttt{NullSpace[}\,R\,\texttt{]}\>,$
(500)

where $R:=\texttt{RowReduce[}\,P_{1}(0,z^{\prime})\,\texttt{]}$ is the matrix
of independent rows of $P_{1}(0,z^{\prime})$. Now, recall that the restricted
Pfaffian system is actually an integrable connection on a space given as a
quotient by the rows of $R$ (see Proposition 3.3). Hence, we are free to add
any linear combination of rows from the $R$-matrix to simplify our choice of
the basis matrix $B$ without changing the restricted Pfaffian system. In
practice, we can do such a simplification by row reducing the auxiliary block
matrix

$\displaystyle\texttt{RowReduce\leavevmode\resizebox{}{9.95863pt}{[}}\,\left[\begin{array}[]{c|c}\mathbb{1}_{r^{\prime}}&B\\\
\hline\cr\mathbb{0}_{(r-r^{\prime})\times
r^{\prime}}&R\end{array}\right]\,\texttt{\leavevmode\resizebox{}{9.95863pt}{]}}$
(503)

and then extracting the top right $r^{\prime}\times r$ block.

####  C.2.3 Inversion of the $M$-matrix

To find the restricted Pfaffian $Q_{i}$ in (52) it is enough to only partially
invert the $M$-matrix (48). Indeed, using the column decomposition (505) to
find $Q_{i}$ we only need to determine the $V$-block from the system (490) in
two steps. First find (any basis of) the right nullspace of $R$

$\displaystyle\widetilde{V}:=\texttt{NullSpace[}\,R\,\texttt{]}\,,$ (504)

and then rotate it from the right to fix the normalization

$\displaystyle V:=\widetilde{V}\cdot(B\cdot\widetilde{V})^{-1}\ .$ (505)

This second step will require inversion of a smaller $r^{\prime}\times
r^{\prime}$ matrix $(B\cdot\widetilde{V})$, which is easier than inverting the
full $r\times r$ matrix $M$.

With the column block (505), the Pfaffian $Q_{i}$ can be extracted from the
full restriction gauge transformation as follows:

$\displaystyle\bigl{(}\partial_{i}B+B\cdot P_{i}(0,z^{\prime})\bigr{)}\cdot
V=Q_{i}\ .$ (506)

##  Appendix D Proofs

###  D.1 Theorem 2.1

Setting $\mathcal{O}(X):=\mathbb{C}[z_{1},\dots,z_{m},x_{1}^{\pm
1},\dots,x_{n}^{\pm 1},\frac{1}{g}]$, we define an action of $\mathcal{D}_{Y}$
on $f=f(z,x)\in\mathcal{O}(X)$ by

$\displaystyle\frac{\partial}{\partial z_{i}}\bullet f$
$\displaystyle=\frac{\partial f}{\partial z_{i}}+\beta_{0}\left({1\over
g(z;x)}{\partial g(z;x)\over\partial z_{i}}\right)f\ ,$ (507)
$\displaystyle\frac{\partial}{\partial x_{i}}\bullet f$
$\displaystyle=\frac{\partial f}{\partial x_{i}}+\beta_{0}\left({1\over
g(z;x)}{\partial g(z;x)\over\partial x_{i}}\right)f-\beta_{i}\frac{f}{x_{i}}\
.$ (508)

The symbol $\mathcal{O}(X)\,g^{\beta_{0}}x_{1}^{-\beta_{1}}\dots
x_{n}^{-\beta_{n}}$ denotes the left $\mathcal{D}_{Y}$-module $\mathcal{O}(X)$
endowed with this action. By definition, $M_{A}(\beta;Y)$ is the $0$-th
cohomology group of the direct image $\mathcal{D}_{Y}$-module
$\int_{\pi}\mathcal{O}(X)\,g^{\beta_{0}}x_{1}^{-\beta_{1}}\dots
x_{n}^{-\beta_{n}}$. The first claim is an immediate consequence of the
preservation of regular holonomicity (Hotta-Tanisaki-Takeuchi-2008, , Theorem
6.1.5).

Let us prove the second claim. Since $M_{A}(\beta;Y)$ is holonomic, it is an
integrable connection on a generic open subset of $Y$. We fix a generic $z\in
Y$. Then the restriction of $M_{A}(\beta;Y)$ to $z$ is the $n$-th twisted
cohomology group on $\pi^{-1}(z)$ with its twist given by the integrand of
(2). In view of (Agostini:2022cgv, , Theorem 2.6), its dimension is given by
$|\chi(\pi^{-1}(z))|$.

###  D.2 Theorem 2.6

Let $\varphi\in{\rm End}_{\mathcal{R}}(\mathcal{M})$. Writing
$\overline{\mathcal{K}}$ for the algebraic closure of $\mathcal{K}$, the
action of $\mathcal{R}$ on $\mathcal{K}$ extends to that on
$\overline{\mathcal{K}}$. We can regard $\varphi$ as an element of ${\rm
End}_{\overline{\mathcal{K}}\otimes_{\mathcal{K}}\mathcal{R}}(\overline{\mathcal{M}})$
where we set
$\overline{\mathcal{M}}:=\overline{\mathcal{K}}\otimes_{\mathcal{K}}\mathcal{M}$.
We first prove that any eigenvalue of $\varphi$ is in $\mathbb{C}$. Let us
take an eigenvector $v\in\overline{\mathcal{M}}$ of $\varphi$. Since
$\overline{\mathcal{M}}$ is finite-dimensional over $\overline{\mathcal{K}}$,
there exists a minimal $k$ such that ${\rm
Span}_{\overline{\mathcal{K}}}\\{v,\partial_{1}v,\ldots,\partial_{1}^{k}v\\}={\rm
Span}_{\overline{\mathcal{K}}}\\{v,\partial_{1}v,\ldots\\}$. It follows that
there exist $a_{0},a_{1},\dots,a_{k}\in\overline{\mathcal{K}}$ such that
$\partial_{1}^{k}v=\sum_{j=0}^{k}a_{j}\partial_{1}^{j}v$. It is
straightforward to see that

$\varphi(\partial_{1}^{k}v)=\sum_{j=0}^{k}\binom{k}{j}(\partial_{1}^{k-j}\alpha)\partial_{1}^{j}v=\sum_{j=0}^{k-1}\binom{k}{j}(\partial_{1}^{k-j}\alpha)\partial_{1}^{j}v+\sum_{j=0}^{k-1}a_{j}\alpha(\partial_{1}^{j}v)$
(509)

holds true. On the other hand, the fact that $\varphi$ commutes with the
action of $\partial_{1}$ implies the following identity:

$\varphi(\partial_{1}^{k}v)=\sum_{j=0}^{k-1}a_{j}\varphi(\partial_{1}^{j}v)=\sum_{j=0}^{k-1}a_{j}\sum_{\ell=0}^{j}\binom{j}{\ell}(\partial_{1}^{j-\ell}\alpha)(\partial_{1}^{\ell}v).$
(510)

Comparing the coefficients of $\partial_{1}^{k-1}v$ in (509) and (510), we
obtain that $\partial_{1}\alpha=0$. Similarly, we can prove that
$\partial_{i}\alpha=0$ for all $i=1,\dots,n$. Since such a function $\alpha$
must be a constant function, it must belong to $\mathbb{C}$. Now, suppose that
$\dim_{\mathbb{C}}{\rm End}_{\mathcal{R}}(\mathcal{M})\geq 2$ and take a
morphism $\varphi\in{\rm End}_{\mathcal{R}}(\mathcal{M})$ which is linearly
independent from ${\rm id}_{\mathcal{M}}$ over $\mathbb{C}$. For any
eigenvalue $\alpha\in\mathbb{C}$ of $\varphi$, $\alpha\cdot{\rm
id}_{\mathcal{M}}-\varphi$ has a non-trivial kernel, which is a non-trivial
$\mathcal{R}$-submodule of $\mathcal{M}$. This is a contradiction.

###  D.3 Theorem 2.7

We set $\mathcal{M}:=\mathcal{R}\otimes_{\mathcal{D}_{Y}}M_{A}(\beta;Y)$. Let
$Y_{0}$ be a non-empty Zariski open subset of $Y$. Any
$\mathcal{D}_{Y_{0}}$-submodule $N$ of $M_{A}(\beta;Y)|_{Y_{0}}$ gives rise to
an $\mathcal{R}$-submodule $\mathcal{M}$. Conversely, any
$\mathcal{R}$-submodule $\mathcal{N}$ of $\mathcal{M}$ is of the form
$\mathcal{N}=\mathcal{R}\otimes_{\mathcal{D}_{Y_{0}}}N$ for some non-empty
Zariski open subset $Y_{0}$ of $Y$ and a $\mathcal{D}_{Y_{0}}$-submodule $N$
of $M_{A}(\beta;Y)|_{Y_{0}}$. Shrinking $Y_{0}$ appropriately if necessary, we
may assume that both $M_{A}(\beta;Y)|_{Y_{0}}$ and $N$ are free as
$\mathcal{O}_{Y_{0}}$-modules, i.e., they are integrable connections. Now, the
sheaf $\mathcal{L}$ of flat sections of $M_{A}(\beta;Y)|_{Y_{0}}$ admits an
invariant Hermitian form. This can be shown by adopting the argument of
(goto2020homology, , p. 559-560) since our exponent $\beta$ is real and
generic. Then, the sheaf $\mathcal{L}_{N}$ of flat sections of $N$ admits an
orthogonal complement $\mathcal{L}_{N^{\prime}}$ which is the sheaf of flat
sections of a $\mathcal{D}_{Y_{0}}$-submodule $N^{\prime}$ of
$M_{A}(\beta;Y)|_{Y_{0}}$ in view of (Deligne, , Théorème 5.9). The direct sum
decomposition $M_{A}(\beta;Y)|_{Y_{0}}=N\oplus N^{\prime}$ gives rise to the
corresponding decomposition
$\mathcal{M}=\mathcal{N}\oplus\mathcal{N}^{\prime}$, which proves that
$\mathcal{M}$ is a semi-simple $\mathcal{R}$-module.

###  D.4 Theorem 2.8

It is enough to show that $\dim_{\mathbb{C}}{\rm
End}_{\mathcal{R}}(\mathcal{M})>1$ if $\mathcal{M}$ is not simple. By the
assumption that $\mathcal{M}$ is semi-simple, there exist a pair of
$\mathcal{R}$-submodules $\mathcal{N},\mathcal{N}^{\prime}$ such that
$\dim_{\mathcal{K}}\mathcal{N},\dim_{\mathcal{K}}\mathcal{N}^{\prime}\geq 1$
and $\mathcal{M}=\mathcal{N}\oplus\mathcal{N}^{\prime}$. The identity morphism
${\rm id}_{\mathcal{N}}:\mathcal{N}\to\mathcal{N}$ extends uniquely to an
$\mathcal{R}$-endomorphism $\varphi:\mathcal{M}\to\mathcal{M}$ by the
requirement $\varphi(\mathcal{N}^{\prime})=0$. In the same way, the identity
morphism ${\rm
id}_{\mathcal{N}^{\prime}}:\mathcal{N}^{\prime}\to\mathcal{N}^{\prime}$
extends to an $\mathcal{R}$-endomorphism
$\varphi^{\prime}:\mathcal{M}\to\mathcal{M}$. It is obvious from the
construction that $\varphi$ and $\varphi^{\prime}$ are linearly independent,
which shows the desired inequality $\dim_{\mathbb{C}}{\rm
End}_{\mathcal{R}}(\mathcal{M})>1$.

###  D.5 Proposition 3.4

In order to simplify the discussion, we only prove the case when $m=1\>$, i.e.
when $Y^{\prime}$ is a single point and $Y\simeq\mathbb{A}^{1}\>$. The case of
$m\geq 2$ can be proved in the same way.

We identify $\mathcal{D}_{Y^{\prime}}$ with the field of complex numbers
$\mathbb{C}$. It is easy to see that the following morphism is well-defined:

$\displaystyle\varphi:\mathcal{M}^{\prime}=\mathbb{C}^{r}/\mathcal{N}^{\prime}\ni[v]\mapsto[v]\in\mathcal{D}_{Y}^{r}/(z_{1}\mathcal{D}_{Y}^{r}+\mathcal{N})=\iota^{*}\mathcal{M}\>,$
(511)

where the submodule $\mathcal{N}$ is defined in Proposition 3.3. Let us take a
constant column vector $v\in\mathbb{C}^{r}$ and consider an element of the
form $v\,\partial_{1}^{n}\in\mathcal{D}_{Y}^{r}\>$ for integer $n\geq 1$. A
straightforward computation shows the identity
$v\,\partial_{1}^{n}\equiv\bigl{(}\mathbb{1}_{r}\,n-P_{1}(0)^{T}\bigr{)}^{-1}\cdot\sum_{k=1}^{n}\binom{n}{k}\bigl{(}\partial_{1}^{k}P_{1}^{T}|_{z_{1}=0}\bigr{)}\cdot
v\,\partial_{1}^{n-k}$ in $\iota^{*}\mathcal{M}$, which proves that $\varphi$
is surjective. Finally, we prove that $\varphi$ is injective. Let us assume
that $\varphi([v])=[0]$. Since the constant vector $v$ does not contain
$\partial_{1}\>$, there exist an element $p\in\mathcal{D}_{Y}^{r}$ and a
column vector $w\in\mathbb{C}^{r}$ such that
$v=z_{1}p+(\mathbb{1}_{r}\,z_{1}\partial_{1}-P_{1}(z_{1})^{T})\cdot w\>$.
Substituting $z_{1}=0$, we obtain $v=-P_{1}(0)^{T}\cdot w\>$, which proves
that $[v]=[0]$.

###  D.6 Theorem 3.6

We prove the case $m=2$. The proof can easily be generalized to the case of
$m>2$.

Let $I(z)=\sum_{k=0}^{\infty}I_{k}(z_{2})\>z_{1}^{k}$ be a vector of
holomorphic functions and let $P_{1}\>$, $P_{2}$ be the $r\times r$ Pfaffian
matrices given by (40). We expand $P_{1}$ as shown in (54) and $P_{2}$ as

$P_{2}(z)=\sum_{k=0}^{\infty}P_{2,k}(z_{2})\>z_{1}^{k}\>.$ (512)

Then the integrability condition (18) is equivalent to

$\displaystyle\partial_{2}\,P_{1,-1}$
$\displaystyle=\bigl{[}P_{2,0},P_{1,-1}\bigr{]}\>,$
$\displaystyle\partial_{2}\,P_{1,k}-(k+1)P_{2,k+1}$
$\displaystyle=\sum_{\begin{subarray}{c}i+j=k\\\ i\geq
0,\>j\geq-1\end{subarray}}\bigl{[}P_{2,i},P_{1,j}\bigr{]}\>.$ (513)

The fact that $I(z_{1},z_{2})$ is a solution of (19) is equivalent to
$\\{I_{k}(z_{2})\\}_{k}$ being subject to the following relations:

$\displaystyle\bigl{(}(k+1)-P_{1,-1}\bigr{)}\cdot I_{k+1}$
$\displaystyle=\sum_{\begin{subarray}{c}i+j=k\\\ i,\>j\geq
0\end{subarray}}P_{1,i}\cdot I_{j}\>,\quad$ for $k\geq-1\>$, (514)
$\displaystyle\partial_{2}I_{k}$
$\displaystyle=\sum_{\begin{subarray}{c}i+j=k\\\ i,\>j\geq
0\end{subarray}}P_{2,i}\cdot I_{j}\>,$ for $k\geq 0\>$. (515)

If $I(z_{1},z_{2})$ is a solution of (19), it is clear that $I_{0}(z_{2})$ is
a solution to the restricted Pfaffian equation (55). Conversely, let
$I_{0}(z_{2})$ be a solution of the restricted Pfaffian equation (55). We can
uniquely extend $I_{0}(z_{2})$ to the solution
$\sum_{k=0}^{\infty}I_{k}(z_{2})\,z_{1}^{k}$ by requiring that the
coefficients $I_{k}(z_{2})$ solve the linear system (514), that is:

$\displaystyle\bigl{(}(k+1)-P_{1,-1}(z_{2})\bigr{)}\cdot
I_{k+1}(z_{2})=\sum_{\begin{subarray}{c}i+j=k\\\ i,\>j\geq
0\end{subarray}}P_{1,i}(z_{2})\cdot I_{j}(z_{2})\>,$ (516)

with $z_{2}$ kept as a parameter. By assumption, the residue matrix
$P_{1,-1}(z_{2})$ is non-resonant, so the only integer eigenvalue it has is
$0\>$, which means that the matrix on the LHS of (516) is invertible and there
is a unique $I_{k+1}(z_{2})$ that solves it for each $k\geq-1\>$. So by
construction of (516), the condition (514) is satisfied and it is left to show
that the other condition (515) is satisfied as well. To this end, consider the
identity

$\bigl{(}(k+1)-P_{1,-1}(z_{2})\bigr{)}\cdot\partial_{2}I_{k+1}(z_{2})=\bigl{(}(k+1)-P_{1,-1}(z_{2})\bigr{)}\
\cdot\sum_{\begin{subarray}{c}i+j=k+1\\\ i,\>j\geq
0\end{subarray}}P_{2,i}\cdot I_{j}(z_{2})\>,$ (517)

which holds by induction in $k$ from acting with $\partial_{2}$ on (516) and
the integrability condition (513). Due to invertibility of the common matrix
factor here, this equation implies (515).

Finally, we prove the convergence of
$I(z_{1},z_{2})=\sum_{k=0}^{\infty}I_{k}(z_{2})z_{1}^{n}$. Let
$f(t;z_{2})=\texttt{Det[}\,t-P_{1,-1}(z_{2})\,\texttt{]}$ be the
characteristic polynomial of $P_{1,-1}$. The leading coefficient of
$f(t;z_{2})$ in $t$ is $1$, and thus there exists a constant $C>0$ such that
$f(t;z_{2})\geq C\,t^{r}$ for any $t>0$ sufficiently large. To be more
precise, $z_{2}$ belongs to a small region in which we have a uniform bound on
the coefficients of $f(t;z_{2})$. Therefore, the matrix norm of the cofactor
matrix of $\bigl{(}t-P_{1,-1}(z_{2})\bigr{)}$ can be estimated by
$C^{\prime}\,t^{r-1}$ for some constant $C^{\prime}>0$ and for any $t>0$
sufficiently large. Taking into account Clamer’s formula for matrix inverse,
we obtain

$\left\lVert\bigl{(}t-P_{1,-1}(z_{2})\bigr{)}^{-1}\right\rVert\leq
C_{1}\,t^{-1}$ (518)

for any $t>0$ sufficiently large. Here
$\left\lVert\mathchoice{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\displaystyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\textstyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\scriptstyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{1.0}{$\scriptscriptstyle\bullet$}}}}}\right\rVert$
stands for a matrix norm. On the other hand, convergence of the expansion (54)
and Cauchy’s inequality implies that there are positive constants $C_{2}$ and
$s$ such that

$\left\lVert P_{1,k}(z_{2})\right\rVert\leq C_{2}\,s^{k}\>,$ (519)

for any $k\geq 0$. We claim that the inequality

$\left\lVert I_{k}(z_{2})\right\rVert\leq R^{n}\>\left\lVert
I_{0}(z_{2})\right\rVert$ (520)

is true with $R=\max(C_{1}C_{2},s)\>$. Indeed, if the estimate (520) is true
up to $k$, then

$\displaystyle\left\lVert I_{k+1}(z_{2})\right\rVert$
$\displaystyle\leq\left\lVert\bigl{(}(k+1)-P_{1,-1}(z_{2})\bigr{)}^{-1}\right\rVert\>\sum_{i+j=k}\left\lVert
P_{i}(z_{2})\right\rVert\>\left\lVert I_{j}(z_{2})\right\rVert$ (521)
$\displaystyle\leq
C_{1}\,C_{2}\,\frac{1}{k+1}\sum_{i+j=k}s^{i}\,R^{j}\>\left\lVert
I_{0}(z_{2})\right\rVert$ (522) $\displaystyle\leq R^{k+1}\>\left\lVert
I_{0}(z_{2})\right\rVert\>.$ (523)

This proves the convergence of $I(z_{1},z_{2})\>$. Uniqueness of $I$ is clear
from (514) and the assumption of the theorem.

###  D.7 Theorem 3.7

We use the same notation as in Appendix D.6. Suppose we construct a matrix
$G(z_{1},z_{2})\in GL(r;\mathbb{C}\\{\\!\\{z_{1},z_{2}\\}\\!\\})$ such that
the gauge transformation $I(z_{1},z_{2})=G(z_{1},z_{2})\cdot J(z_{1},z_{2})$
turns the system (19) into

$\displaystyle\partial_{1}J(z_{1},z_{2})$
$\displaystyle=\frac{\Gamma}{z_{1}}J(z_{1},z_{2})\>,$
$\displaystyle\partial_{2}J(z_{1},z_{2})$ $\displaystyle=0\>,$ (524)

for some constant matrix $\Gamma\>$. This system can be solved by the matrix
exponential: $J(z_{1},z_{2})=z_{1}^{\Gamma}\>$.

The transformation matrix $G$ is also subject to its own Pfaffian system:

$\displaystyle\partial_{1}G$ $\displaystyle=T_{P_{1},\>\Gamma/z_{1}}(G)\>,$
$\displaystyle\partial_{2}G$ $\displaystyle=P_{2}\cdot G\>.$ (525)

Here $T_{A,B}:M(r;\mathbb{C})\to M(r;\mathbb{C})$ is a linear map defined by
$T_{A,B}(G):=A\cdot G-G\cdot B\>$. Note that this is a Pfaffian system of the
rank $r^{2}\>$.

######  Lemma D.1.

The system (D.7) is integrable.

###### Proof.

For a matrix $A\in M(r;\mathbb{C})$, let us denote by
$T_{A}:M(r;\mathbb{C})\to M(r;\mathbb{C})$ the linear map given by the left
multiplication with $A\>$, explicitly $T_{A}(G):=A\cdot G\>$, so that
$T_{A}(G)=T_{A,\mathbb{0}}(G)\>$. We want to prove the following integrability
condition: