# A Calabi–Yau threefold coming from two black holes

Dino Festi, Bert van Geemen  Dipartimenti di matematica “Federigo Enriques”,
Università degli studi di Milano, 20133, Milano, Italy<EMAIL_ADDRESS>Dipartimenti di matematica “Federigo Enriques”, Università degli studi di
Milano, 20133, Milano, Italy<EMAIL_ADDRESS>
###### Abstract.

In this paper, we show that a set of six square roots of homogeneous
polynomials in four variables, related to a binary system of black holes
studied by Stefan Weinzierl, is not rationalizable. We prove it by showing
that the variety $X$ associated to the product of four of the six square roots
is not unirational. In particular, we show that the smooth model of $X$ is a
Calabi–Yau threefold.

##  1\. Introduction

Experiments in high energy physics often require predictions based on the
computation of Feynman integrals, see for example [16]. Unfortunately, an
exact evaluation of these integrals is often unfeasible and so practitioners
resort to numerical integration. In order to achieve results of high enough
precision in a reasonable time, it is useful to “solve” the square root of
polynomials appearing in the arguments of these integrals. Motivated by this
problem, Marco Besier and the first author introduced the concept of
_rationalizability_ of a square root, cf. [2].

######  Definition 1.1.

Let $f\in\mathbb{C}[x_{1},\ldots,x_{n}]$ be a squarefree polynomial. We say
that $\sqrt{f}$ is rationalizable if there is a morphism of fields
$\phi\colon\mathbb{C}(x_{1},\ldots,x_{n})\to\mathbb{C}(x_{1},\ldots,x_{n})$
such that $\phi(z)=z$ for every $z\in\mathbb{C}$ and $\phi(f)=h^{2}$, for some
$h\in\mathbb{C}(x_{1},\ldots,x_{n})$.

In the same article, they show that the rationalizability of a square root is
equivalent to the unirationality of a certain algebraic variety associated to
it, [2, Theorem 1]. This equivalence is then used to give some criteria to
decide rationalizability of given square roots. The notion of
rationalizability can be extended to sets of square roots of polynomials, also
called _alphabets_.

######  Definition 1.2.

Let $\mathcal{A}:=\\{f_{1},\ldots,f_{r}\\}$ be a set of squarefree polynomials
in $\mathbb{C}[x_{1},\dots,x_{n}]$. We say that $\mathcal{A}$ is
rationalizable if there is a morphism of fields
$\phi\colon\mathbb{C}(x_{1},\ldots,x_{n})\to\mathbb{C}(x_{1},\ldots,x_{n})$
such that $\phi(z)=z$ for every $z\in\mathbb{C}$ and for $i=1,\ldots,r$ we
have $\phi(f_{i})=h_{i}^{2}$, for some
$h_{i}\in\mathbb{C}(x_{1},\ldots,x_{n})$.

The notion of rationalizability is further generalised to algebraic field
extensions in [6] by Andreas Hochenegger and the first author, allowing for
more results to decide rationalizability of given (sets of) polynomials.

Discussing these new developments, Stefan Weinzierl asked us about the
rationalizability of an alphabet of six square roots of homogeneous
polynomials in four variables related to the production of gravitational waves
from a binary system of two black holes [10]. After some straightforward
changes of variables and reordering of the roots, we write Weinzierl’s
alphabet as the set $\mathcal{A}$ of the square roots of the following six
homogeneous polynomials in $R:=\mathbb{C}[x,y,z,t]$:

$\displaystyle f_{1}$ $\displaystyle:=4x-z,$ $\displaystyle f_{2}$
$\displaystyle:=4y-z,$ $\displaystyle f_{3}$
$\displaystyle:=4x^{2}y-z(y-t)^{2},$ $\displaystyle f_{4}$
$\displaystyle:=4xy^{2}-z(x-t)^{2},$ $\displaystyle f_{5}$
$\displaystyle:=(x-y)^{2}-2(x+y)t+t^{2},$ $\displaystyle f_{6}$
$\displaystyle:=(x-y)^{2}-2(x+y)(z+t)+(z+t)^{2}.$

Unfortunately, this set seems too complicated for the methods presented in the
previous papers, and requires an ad-hoc study that we present in this paper.
As by-product, we also developed a strategy to _guess_ (but not
prove/disprove) the rationalizability of square roots in certain cases.

The rationalizability of $\mathcal{A}$ would imply the rationalizability of
the square root of every product of the polynomials $f_{i}$. Conversely, if
one can prove that the square root of one of the products is not
rationalizable, then $\mathcal{A}$ is not rationalizable. Weinzierl has
studied several products of two and three polynomials (see 2.1), proving the
rationalizability of all of them but leaving the rationalizability of the
whole alphabet open. In these notes we are going to show that the set is not
rationalizable.

######  Theorem 1.3.

The square root $\sqrt{f_{1}f_{2}f_{3}f_{4}}$ is not rationalizable and hence
neither is the set $\mathcal{A}$.

The proof of Theorem 1.3 is presented in § 3. In § 4 we propose a method to
guess the rationalizability of square roots in certain cases. This method
provided a decisive hint towards the non-rationalizability of
$\sqrt{f_{1}f_{2}f_{3}f_{4}}$, see § 4.6.

## Acknowledgements

We thank Stefan Weinzierl for posing the original question. We also thank
Sławomir Cynk and Bartosz Naskrȩcki for the insight into the modularity of the
threefold $X$. The first author is also grateful to Andreas Hochenegger and
Matthias Schütt for useful conversations about this work.

##  2\. The double octic

As stated above, Stefan Weinzierl studied the rationalizability of several
products of two and three polynomials, obtaining the result below.

######  Proposition 2.1 (Weinzierl, private communication).

The square roots of the following products are rationalizable:

$\begin{array}[]{ccc}f_{1}f_{5},&f_{2}f_{5},&f_{1}f_{2}f_{3},\\\
f_{1}f_{2}f_{4},&f_{1}f_{2}f_{5},&f_{1}f_{4}f_{3},\\\
f_{1}f_{3}f_{5},&f_{2}f_{3}f_{5},&f_{2}f_{4}f_{5}.\end{array}$

The proof of the proposition above is made easier by the relations between the
polynomials, stated below.

######  Lemma 2.2.

The involution of $R$ defined by switching $x$ and $y$, exchanges $f_{1}$ with
$f_{2}$ and $f_{3}$ with $f_{4}$; it fixes $f_{5}$ and $f_{6}$.

The automorphism of $R$ defined by $t\mapsto z+t$ sends $f_{5}$ to $f_{6}$ and
fixes $f_{1}$ and $f_{2}$.

###### Proof.

Immediate by direct computations. ∎

The next natural step in the study of $\mathcal{A}$ is then to consider the
product of four polynomials.

###  2.1. A product of four polynomials

We define the polynomial $f$ of degree $8$ as the product

(1) $f:=f_{1}f_{2}f_{3}f_{4}.$

Let $\mathbb{P}$ denote the weighted projective space $\mathbb{P}(1,1,1,1,4)$
with coordinates $x,y,z,t,w$ of weights $1,1,1,1,4$, respectively. We define
the variety $\mathcal{X}\subset\mathbb{P}$ as

(2) $\mathcal{X}\colon\;w^{2}=f\;.$

Then the rationalizabilty of $\sqrt{f}$ is equivalent to the unirationality of
$\mathcal{X}$ (cf. [2, Theorem 1.1]).

Notice that $\mathcal{X}$ is a threefold given as a double cover of
$\mathbb{P}^{3}$ branched along the octic surface
$\mathcal{V}\subset\mathbb{P}^{3}$ defined by

(3) $\mathcal{V}\colon\;f=0\;.$

If $\mathcal{V}$ were smooth, then $\mathcal{X}$ would also be smooth.
Moreover, a smooth double cover of $\mathbb{P}^{3}$ branched above an octic
surface (a _double octic threefold_ or _octic double solid_) is a Calabi–Yau
threefold (cf. 4.1), hence it is not unirational. This would already answer
the question about the rationalizabilty of $\mathcal{A}$. Unfortunately,
$\mathcal{V}$ is (very) singular, making hard to see whether there is a
desingularization of $\mathcal{X}$ that is a Calabi–Yau or not. This is why we
look for a more convenient birationally equivalent model.

###  2.2. A more convenient model

Consider the Cremona-like transformation of $\mathbb{P}^{3}$ given by

$\sigma\colon\;\mathbb{P}^{3}\dashrightarrow\mathbb{P}^{3}\,,\quad(x:y:z:y)\mapsto(xz:yz:4xy:tz).$

Let $X\subset\mathbb{P}$ be the threefold defined by

(4) $X\colon\;w^{2}=xy(x-z)(y-z)(yz-(x-t)^{2})(xz-(y-t)^{2})\;.$

Notice that $X$ is again a double cover of $\mathbb{P}^{3}$ ramified above an
octic surface, namely the surface

$B\colon\;xy(x-z)(y-z)(yz-(x-t)^{2})(xz-(y-t)^{2})=0\;\subset\mathbb{P}^{3}\,.$

######  Lemma 2.3.

The following statements hold.

  1. (1)

$\sigma$ is a rational involution; in particular it is a birational map.

  2. (2)

$\sigma$ induces a birational morphism between $X$ and $\mathcal{X}$.

###### Proof.

It easy to see that $\sigma\circ\sigma$ is the identity, proving the first
statement.

The map $\sigma$ sends the surface $\mathcal{V}$ to the surface
$B^{\prime}\subset\mathbb{P}^{3}$ defined by

$B^{\prime}\colon\;x^{3}y^{3}z^{4}(x-z)(y-z)(yz-(x-t)^{2})(xz-(y-t)^{2})=0.$

Let $X^{\prime}\subset\mathbb{P}(1,1,1,1,8)$ be the double cover of
$\mathbb{P}^{3}$ branched above $B^{\prime}$ and let $w^{\prime}$ be the
coordinate of weight $8$. As $B^{\prime}$ is birationally equivalent to
$\mathcal{V}$ (via $\sigma$), we have that $X^{\prime}$ is birationally
equivalent to $\mathcal{X}$; on the other hand, the change of variable
$w^{\prime}=xyz^{2}w$ shows that $X$ and $X^{\prime}$ are also birationally
equivalent, concluding the proof of the second statement. ∎

######  Corollary 2.4.

$\mathcal{X}$ is unirational if and only if $X$ is.

###### Proof.

By 2.3, $X$ and $\mathcal{X}$ are birationally equivalent. As unirationality
is preserved by birational morphisms, the statement follows. ∎

As $X$ is a double cover of $\mathbb{P}^{3}$, its singularities can only come
from the singularities of the branch locus $B$.

######  Lemma 2.5.

The surface $B$ has six irreducible components, named $B_{1},\dots,B_{6}$,
defined as follows:

$\displaystyle B_{1}$ $\displaystyle:x=0\,,$ $\displaystyle B_{2}$
$\displaystyle:y=0\,,$ $\displaystyle B_{3}$ $\displaystyle:x-z=0\,,$
$\displaystyle B_{4}$ $\displaystyle:y-z=0\,,$ $\displaystyle B_{5}$
$\displaystyle:yz-(x-t)^{2}=0\,,$ $\displaystyle B_{6}$
$\displaystyle:xz-(y-t)^{2}=0\,.$

The components $B_{1},B_{2},B_{3},B_{4}$ are planes. The components
$B_{5},B_{6}$ are singular quadrics, with a single node in $(1:0:0:1)$ and
$(0:1:0:1)$, respectively.

###### Proof.

By direct computations. ∎

###  2.3. The singular locus

In this section we study the singular locus of the branch locus $B$. From 2.5
we know that $B$ has six irreducible components: four planes and two quadrics;
each quadric has a single node. The singular locus of $B$ is given by the
intersection of the irreducible components (the two nodes lying in these
intersections).

######  Definition 2.6.

For $i,j\in\\{1,\dots,6\\},\,i<j$, we define $B_{i,j}:=B_{i}\cap B_{j}$.

######  Lemma 2.7.

The singular locus of $B$ is the union of 18 rational curves. It is non-
reduced. More specifically, the following statements hold:

  * •

the intersections $B_{1,2},B_{1,3},B_{1,4},B_{2,3},B_{2,4},B_{3,4}$ are lines;

  * •

the intersections $B_{1,6}$, resp. $B_{2,5}$, is a double line, that is,
$B_{1}$ and $B_{6}$, resp. $B_{2}$ and $B_{5}$, are tangent in their
intersection;

  * •

$B_{1,5},B_{2,6},B_{3,5},B_{4,6}$ are smooth conics;

  * •

$B_{3,6}$ and $B_{4,5}$ are the union of two incident lines, we write

    * –

$B_{3,6}=B_{3,6}^{1}\cup B_{3,6}^{2}$, with the two lines meeting in
$(0:1:0:1)$,

    * –

$B_{4,5}=B_{4,5}^{1}\cup B_{4,5}^{2}$, with the two lines meeting in
$(1:0:0:1)$;

  * •

$B_{5,6}$ is the union of two smooth conics meeting in $(1:1:0:1),(1:1:4:3)$,
we write $B_{5,6}=B_{5,6}^{1}\cup B_{5,6}^{2}$.

###### Proof.

By direct computations. ∎

######  Remark 2.8.

Notice that the equations defining the irreducible components of
$B_{3,6},B_{4,5},$ and $B_{5,6}$ can be retrieved from Table 1.

######  Definition 2.9.

Let $\mathcal{B}$ denote the set of the irreducible components of the singular
locus of $B$, that is, $\mathcal{B}$ is the set of the 18 rational curves
defined in 2.7.

######  Lemma 2.10.

The curves in $\mathcal{B}$ intersect in 16 points.

In Table 1, for each intersection point we indicate its multiplicity in $B$,
the components on which it lies, and the singular curves on which it lies.

###### Proof.

By direct computations. ∎

$(1:0:0:1)$ | $4$ | $B_{2},B_{4},B_{5}$ | $B_{2,4},B_{2,5},B_{4,5}^{1},B_{4,5}^{2}$
---|---|---|---
$(0:1:0:1)$ | $4$ | $B_{1},B_{3},B_{6}$ | $B_{1,3},B_{1,6},B_{3,6}^{1},B_{3,6}^{2}$
$(0:-1:-1:1)$ | $3$ | $B_{1},B_{4},B_{5}$ | $B_{1,4},B_{1,5},B_{4,5}^{2}$
$(-1:0:-1:1)$ | $3$ | $B_{2},B_{3},B_{6}$ | $B_{2,3},B_{2,6},B_{3,6}^{2}$
$(1:0:0:0)$ | $3$ | $B_{2},B_{4},B_{6}$ | $B_{2,4},B_{2,6},B_{4,6}$
$(0:1:0:0)$ | $3$ | $B_{1},B_{3},B_{5}$ | $B_{1,3},B_{1,5},B_{3,5}$
$(0:0:1:0)$ | $4$ | $B_{1},B_{2},B_{5},B_{6}$ | $B_{1,2},B_{1,5},B_{1,6},B_{2,5},B_{2,6},B_{5,6}^{2}$
$(0:0:0:1)$ | $4$ | $B_{1},B_{2},B_{3},B_{4}$ | $B_{1,2},B_{1,3},B_{1,4},B_{2,3},B_{2,4},B_{3,4}$
$(0:1:1:1)$ | $4$ | $B_{1},B_{4},B_{5},B_{6}$ | $B_{1,4},B_{1,5},B_{1,6},B_{4,5}^{1},B_{4,6},B_{5,6}^{1}$
$(1:0:1:1)$ | $4$ | $B_{2},B_{3},B_{5},B_{6}$ | $B_{2,3},B_{2,5},B_{2,6},B_{3,5},B_{3,6}^{1},B_{5,6}^{1}$
$(1:1:0:1)$ | $2$ | $B_{5},B_{6}$ | $B_{5,6}^{1},B_{5,6}^{2}$
$(1:1:1:0)$ | $4$ | $B_{3},B_{4},B_{5},B_{6}$ | $B_{3,4},B_{3,5},B_{3,6}^{2},B_{4,5}^{2},B_{4,6},B_{5,6}^{2}$
$(1:1:1:2)$ | $4$ | $B_{3},B_{4},B_{5},B_{6}$ | $B_{3,4},B_{3,5},B_{3,6}^{1},B_{4,5}^{1},B_{4,6},B_{5,6}^{2}$
$(1:1:4:3)$ | $2$ | $B_{5},B_{6}$ | $B_{5,6}^{1},B_{5,6}^{2}$
$(1:4:1:3)$ | $3$ | $B_{3},B_{5},B_{6}$ | $B_{3,5},B_{3,6}^{2},B_{5,6}^{1}$
$(4:1:1:3)$ | $3$ | $B_{4},B_{5},B_{6}$ | $B_{4,5}^{2},B_{4,6},B_{5,6}^{1}$
Table 1. For each point, we indicate its multiplicity on $B$, the irreducible
components of $B$ on which it lies, and the singular curves on which it lies.

##  3\. Resolution of the singularities

In this section we show that by suitably resolving the singularities of $X$ we
obtain a Calabi–Yau threefold, hence $X$ is not unirational. We follow the
procedure to desingularize $X$ described in [4, §2] and, later, in [12, §4.1].
Some concrete applications can be found in [3, Example 3] and [5, Example
5.2].

######  Definition 3.1.

Let $U$ be a smooth threefold and $S\subset U$ a surface. We call $S$ _an
arrangement_ if locally it looks like an union of planes, that is, $S$ is the
union of smooth irreducible surfaces $S_{1},...,S_{r}$ such that:

  1. (1)

for any $i\neq j$, the surfaces $S_{i},S_{j}$ either intersect transversally
along a smooth curve $C_{i,j}$ or are disjoint;

  2. (2)

the curves $C_{i,j},\,C_{k,l}$ either intersect transversally or coincide.

If $U=\mathbb{P}^{3}$ and the surfaces $S_{1},\dots,S_{r}$ are of degree
$d_{1},\ldots,d_{r}$ such that $d_{1}+\cdots+d_{r}=8$, then $S$ is called an
_octic_ arrangement.

An irreducible curve $C\subset S$ is said to be $q$-_fold_ if it lies on
exactly $q$ of the surfaces $S_{1},\dots,S_{r}$; a point $P\in S$ is said to
be $p$-_fold_ if it lies on exactly $p$ of the surfaces $S_{1},\dots,S_{r}$.

######  Remark 3.2.

One sees immediately that $B$ is not an octic arrangement: the two quadric
components $B_{5}$ and $B_{6}$ are not smooth (cf. 2.5); moreover, they are
tangent to $B_{2}$ and $B_{1}$, respectively, and the intersections
$B_{3,6},B_{4,5}$ and $B_{5,6}$ are reducible.

As noted in 3.2, the branch locus $B$ of $X\subset\mathbb{P}^{3}$ is not an
octic arrangement. We want to show that by blowing up $\mathbb{P}^{3}$ in some
of the curves in $\mathcal{B}$ we obtain threefold $U^{(5)}$ such that the
strict transform $B^{(5)}\subset U^{(5)}$ of $B^{(4)}$ is an arrangement. Then
we proceed as in [4] and [12, Chapter 4].

###  3.1. The first blow up

In this subsection we illustrate the first step of the resolution process, the
blow up of $\mathbb{P}^{3}$ along the line $B_{1,6}$.

######  Definition 3.3.

We define

$\sigma^{(1)}\colon\;U^{(1)}:=\operatorname{Bl}_{1,6}\mathbb{P}^{3}:=\operatorname{Bl}_{B_{1,6}}\mathbb{P}^{3}\to\mathbb{P}^{3}=:U^{(0)}$

to be the blow up of $\mathbb{P}^{3}=:U^{(0)}$ along the line $B_{1,6}$.

We denote by $B^{(1)}\subset U^{(1)}$ the strict transform of
$B\subset\mathbb{P}^{3}$ and by $B_{i}^{(1)}\subset B^{(1)}$ the strict
transform of its components. Also, let $K:=K_{\mathbb{P}^{3}}$ and $K^{(1)}$
denote the canonical divisors of $\mathbb{P}^{3}$ and $U^{(1)}$, respectively.

######  Lemma 3.4.

The divisor $B^{(1)}\subset U^{(1)}$ is linearly equivalent to $-2K^{(1)}$ and
it inherits all the singularities of $B$, cf. 2.7, except:

  * •

$B^{(1)}_{6}$ is smooth,

  * •

$B_{1}^{(1)}$ and $B_{6}^{(1)}$ intersect transversally in a line,

  * •

$B_{3}^{(1)}$ and $B_{6}^{(1)}$ intersect in two disjoint lines.

###### Proof.

The branch surface $B_{6}$ is a singular quadric with a node in $P:=(0:1:0:1)$
which is tangent to the branch plane $B_{1}$ in the line $B_{1,6}\colon
x=y-t=0$. The node $P$ also lies on the branch surfaces $B_{1}$ (and hence on
the tangency line $B_{1,6}$) and $B_{3}$. For convenience, we introduce the
variable $u:=y-t$, so that we can write

$B_{1}\colon x=0,\;\;B_{3}\colon x-z=0,\;\;B_{6}\colon
xz=u^{2},\;\;B_{1,6}\colon x=u=0,\;\;P=(0:0:0:1)$

in $\mathbb{P}^{3}(x,u,z,t)$.

Identify the subset $\\{t\neq 0\\}\subset\mathbb{P}^{3}$ with
$\mathbb{A}^{3}(x,u,z)$ and consider the blow up
$\sigma\colon\tilde{\mathbb{A}}^{3}\to\mathbb{A}^{3}$ of $\mathbb{A}^{3}$ in
$\mathbb{A}^{3}\cap B_{1,6}$:

$\tilde{\mathbb{A}}^{3}:=\\{((x,u,z),(v_{0}:v_{1}))\;:\;v_{0}x-v_{1}u\\}\subset\mathbb{A}^{3}\times\mathbb{P}^{1}.$

The exceptional divisor is then

$E:=\\{((0:0:z),(v_{0}:v_{1}))\\}\cong\mathbb{A}^{1}\times\mathbb{P}^{1}.$

We denote the standard affine open subsets of $\tilde{\mathbb{A}}^{3}$ by

$\displaystyle\tilde{\mathbb{A}}^{3}_{0}:=\\{v_{1}\neq 0\\}$
$\displaystyle=\mathbb{A}^{3}(x,z,v_{0}),\;(u=v_{0}x);$
$\displaystyle\tilde{\mathbb{A}}^{3}_{1}:=\\{v_{0}\neq 0\\}$
$\displaystyle=\mathbb{A}^{3}(u,z,v_{1}),\;(x=v_{1}u).$

Notice that

$\displaystyle E_{0}$
$\displaystyle:=E\cap\tilde{\mathbb{A}}^{3}_{0}=\\{x=0\\},$ $\displaystyle
E_{1}$ $\displaystyle:=E\cap\tilde{\mathbb{A}}^{3}_{1}=\\{u=0\\}.$

The inverse image of $B_{1}\colon x=0$ in $\tilde{\mathbb{A}}^{3}$ lies in
$\\{v_{1}u=0\\}$; since $u$ is not identically zero on $B_{1}$, the strict
transform ${B}_{1}^{(1)}$ of $B_{1}$ lies in $v_{1}=0$. Hence,
${B}_{1}^{(1)}\cap\tilde{\mathbb{A}}^{3}_{0}=\emptyset$. Therefore, to study
${B}_{1}^{(1)}$ we only need to consider $\tilde{\mathbb{A}}^{3}_{1}$. The
preimages of $B_{3}$ and $B_{6}$ are given by

$\displaystyle\sigma^{-1}(B_{3})$ $\displaystyle\colon\;v_{1}u=z,$
$\displaystyle\sigma^{-1}(B_{6})$ $\displaystyle\colon\;u(v_{1}z-u)=0;$

denote by $B^{(1)}_{i}$ the strict transform of $B_{i}$ in
$\tilde{\mathbb{A}}^{3}_{1}$. As $E_{1}=\\{u=0\\}$ we have

(5)
$\sigma^{*}(B_{1}+B_{2}+B_{3})\cap\tilde{\mathbb{A}}^{3}_{1}=2E_{1}+B^{(1)}_{1}+B^{(1)}_{3}+B^{(1)}_{6}\,.$

By construction, $\tilde{\mathbb{A}}^{3}\subset U^{(1)}$ and the projection
$\sigma^{(1)}\colon U^{(1)}\to\mathbb{P}^{3}$ restricted to
$\tilde{\mathbb{A}}^{3}$ is $\sigma$. From (5) it follows that $E$ is not in
the branch locus of $X^{(1)}$.

Then, to compute the canonical divisor $K_{X^{(1)}}$ of $X^{(1)}$, notice that
the branch locus of $X$ on $\mathbb{P}^{3}$ is $B=B_{1}+\cdots+B_{6}$, a
divisor of degree 8 and hence

(6) $B=-2K_{\mathbb{P}^{3}}.$

Its strict transform is

(7) $B^{(1)}={\sigma^{(1)}}^{*}B-2E.$

By adjunction

(8) ${\sigma^{(1)}}^{*}K_{\mathbb{P}^{3}}=K^{(1)}-E.$

Equations (6), (7), and (8) yield

(9) $B^{(1)}=-2K^{(1)},$

proving the first statement.

The second statement follows immediately from the definition of blow up,
noticing that the node $P$ of $B_{1}$ lies on $B_{1,6}$, and this point is
also the point of intersection of the two components of $B_{3,6}$. ∎

###  3.2. The second blow up

We proceed by blowing the strict transform of $B_{2,5}$.

######  Definition 3.5.

We define

$\sigma^{(2)}\colon\;U^{(2)}\to U^{(1)}$

as the blow up of $U^{(1)}$ along $B^{(1)}_{2,5}$, the strict transform of
$B_{2,5}$ in $U^{(1)}$.

We denote by $B^{(2)}$ and $B_{i}^{(2)}$ the strict transforms in $U^{(2)}$ of
$B^{(1)}$ and $B_{i}^{(1)}$, respectively. We also denote by $K^{(2)}$ the
canonical divisor of $U^{(2)}$.

We obtain a result analogous to 3.4.

######  Lemma 3.6.

The divisor $B^{(2)}$ is linearly equivalent to $-2K^{(2)}$ and it inherits
all the singularities of $B^{(1)}$, cf. 3.4, except:

  * •

$B^{(2)}_{6}$ is smooth,

  * •

$B_{2}^{(2)}$ and $B_{5}^{(2)}$ intersect transversally in a line,

  * •

$B_{4}^{(2)}$ and $B_{5}^{(2)}$ intersect in two disjoint lines.

###### Proof.

Let $E$ be the exceptional divisor of $U^{(2)}$ introduced by the blow up
$\sigma^{(2)}$. Then, reasoning as in the proof of 3.4, one can show that
$B^{(2)}={\sigma^{(2)}}^{*}B^{(1)}-2E$ and
${\sigma^{(2)}}^{*}K^{(1)}=K^{(2)}-E$. As above, these two equalities and (9)
yield

$B^{(2)}=-2K^{(2)}.$

Again, the second statement follows immediately form the definition of blow
up, noticing that the node of $B_{5}^{(1)}$ lies on $B_{2,5}^{(1)}$, and this
node is also the intersection point of the two components of $B_{4,5}^{(1)}$.
∎

###  3.3. The third blow up

We proceed as above, blowing up the strict transform of one of the components
of $B_{5,6}$. Recall that $B_{5,6}$ has two irreducible components meeting in
two points. The strict transform $B_{5,6}^{(2)}\subset U^{(2)}$ stays
reducible, with the two components still meeting in two points. This feature
prevents $B^{(2)}$ from being an arrangement. In order to solve this
obstruction, we blow up one of the two components.

######  Definition 3.7.

We define

$\sigma^{(3)}\colon\;U^{(3)}\to U^{(2)}$

as the blow up of $U^{(2)}$ along $(B^{1}_{5,6})^{(2)}$, the strict transform
of $(B^{1}_{5,6})^{(1)}$ in $U^{(2)}$.

We denote by $B^{(3)}$ and $B_{i}^{(3)}$ the strict transforms in $U^{(3)}$ of
$B^{(2)}$ and $B_{i}^{(2)}$, respectively. We also denote by $K^{(3)}$ the
canonical divisor of $U^{(3)}$.

######  Lemma 3.8.

The divisor $B^{(3)}$ is linearly equivalent to $-2K^{(3)}$ and it inherits
all the singularities of $B^{(2)}$, cf. 3.6, except that $B_{5}^{(3)}$ and
$B_{6}^{(3)}$ intersect in a single smooth conic.

###### Proof.

As for 3.6. ∎

###  3.4. The fourth and fifth blow ups

We proceed our plan of resolving the singularities of $B$ in order to obtain
an arrangement. In order to do so, we have to make sure that the strict
transform of $B_{3}$ and $B_{6}$, respectively $B_{4}$ and $B_{5}$, meet in a
smooth irreducible curve. As $B_{3}\cap B_{6}$, resp. $B_{4}\cap B_{5}$, meet
in two incident lines, and so do their strict transforms in $U^{(3)}$, it is
enough to blow up one component of each intersection. Therefore, we define the
blow up

$\sigma^{(4)}\colon\;U^{(4)}\to U^{(3)}$

of $U^{(3)}$ along $(B^{1}_{3,6})^{(3)}$. Subsequently, we define

$\sigma^{(5)}\colon\;U^{(5)}\to U^{(4)}$

as the blow up of $U^{(4)}$ along $(B^{1}_{4,5})^{(4)}$. For $j=4,5$, we
define $B^{(j)}$ and $B_{i}^{(j)}$ as the strict transforms in $U^{(j)}$ of
$B^{(j-1)}$ and $B_{i}^{(j)}$, respectively. Moreover, for $j=4,5$ we denote
by $K^{(j)}$ the canonical divisor of $U^{(j)}$.

######  Lemma 3.9.

The divisor $B^{(5)}$ is linearly equivalent to $-2K^{(5)}$ and it is an
arrangement in the sense of 3.1; it only has 2-fold curves and at most 4-fold
points.

###### Proof.

The first statement is proven as in the previous cases.

To see that $B^{(5)}$ is an arrangement, note that $B^{(5)}$ is isomorphic to
$B=B^{(0)}$ outside the curves that have been blown up:
$B_{1,6},B_{2,5},B_{5,6}^{1},B_{3,6}^{1},B_{4,5}^{1}$. As noted in 3.2, these
are exactly the loci not meeting the conditions for $B$ to be an arrangement.
Indeed, $B^{(5)}$ is the union of six smooth surfaces:

  * •

$B^{(5)}_{1},\ldots,B^{(5)}_{4}$ are strict transforms of planes;

  * •

$B^{(5)}_{5}$ and $B^{(5)}_{6}$ are the strict transforms of quadrics with a
node blown up (also) along a line containing the node, hence they are smooth;

  * •

for every $i,j\in\\{1,\ldots,6\\},i\neq j$, the intersection $B^{(5)}_{i}\cap
B^{(5)}_{j}$ is a smooth curve: either a line or a conic.

As $B$ only has 2-fold curves and at most 4-fold points, so does $B^{(5)}$.
Notice that some of the 4-fold points have not been affected by the blow ups,
e.g., $(0:0:0:1)$, hence $B^{(5)}$ does have 4-fold points. ∎

###  3.5. Further blow ups

Following [4] and [12, §4.1], we proceed by blowing up the 4-fold points and
then the double curves of $B^{(5)}\subset U^{(5)}$, eventually obtaining a
smooth surface $\tilde{B}:=B^{(n)}\subset U^{(n)}=:U$. Let $\tilde{X}$ be the
double cover of $U$ ramified above $\tilde{B}$.

For the convenience of the reader, we also recall the definition of a
Calabi–Yau threefold: a smooth threefold $Y$ is a Calabi–Yau threefold if and
only if its canonical divisor $K_{Y}$ is trivial and
$H^{i}(Y,\mathcal{O}_{Y})=0$ for $i=1,2$; see also 4.1.

######  Proposition 3.10.

The threefold $\tilde{X}$ is smooth and birationally equivalent to $X$.
Moreover, it is a Calabi–Yau threefold.

###### Proof.

Let $\sigma=\sigma^{(n)}\circ\ldots\circ\sigma^{(1)}$ be the composition of
all the blow ups

$\sigma\colon\;U=U^{(n)}\to U^{(n-1)}\cdots U^{(1)}\to
U^{(0)}=\mathbb{P}^{3}\;.$

Then $\sigma$ is a birational map between $U$ and $\mathbb{P}^{3}$, inducing a
birational map between $\tilde{B}$ and $B$. Therefore $\tilde{X}$ and $X$ are
birationally equivalent.

It remains to prove that $\tilde{X}$ is a Calabi–Yau threefold, i.e. it is
smooth, has trivial canonical divisor, and
$H^{i}(\tilde{X},\mathcal{O}_{\tilde{X}})=0$ for $i=1,2$.

As $\tilde{X}$ is the double cover of $U$, which is smooth, branched above the
smooth (reducible) surface $\tilde{B}$, we have that $\tilde{X}$ is also
smooth.

After every blow up we have that $B^{(k)}=-2K^{(k)}$, where $K^{(k)}$ is the
canonical divisor of $U^{(k)}$. Hence

$\tilde{B}=B^{(n)}=-2K^{(n)}=-2K_{U},$

yielding $K_{\tilde{X}}=0$.

Let $\pi\colon\tilde{X}\to U$ denote the double cover projection. As $\pi$ is
finite, it is affine and hence $H^{i}(\tilde{X},\mathcal{O}_{\tilde{X}})\cong
H^{i}(U,\pi_{*}\mathcal{O}_{\tilde{X}})$ for each $i\geq 0$. As $\tilde{X}$ is
the double cover of $U$ branched above $\tilde{B}=-2K_{U}$, it follows that

$\pi_{*}\mathcal{O}_{\tilde{X}}=\mathcal{O}_{U}\oplus\mathcal{O}_{U}(K_{U}),$

and hence

$H^{i}(X,\mathcal{O}_{\tilde{X}})\cong H^{i}(U,\mathcal{O}_{\tilde{X}})\oplus
H^{i}(U,\mathcal{O}_{\tilde{X}}(K_{U})).$

By Serre duality we have that

$H^{i}(U,\mathcal{O}_{U}(K_{U}))\cong H^{3-i}(U,\mathcal{O}_{U}).$

So we are only left to compute $H^{i}(U,\mathcal{O}_{U})$ for $i=1,2$.

To see this, consider the sequence of blow ups

$U:=U^{(n)}\to U^{(n-1)}\to\cdots U^{(1)}\to U^{(0)}:=\mathbb{P}^{3},$

where $\sigma^{(\ell)}\colon U^{(\ell)}\to U^{(\ell-1)}$ is the blow up of
$U^{(\ell-1)}$ along a curve or a point, say $Z^{(\ell-1)}$ or $P^{(\ell-1)}$,
respectively.

If $\sigma^{(\ell)}$ is the blow up along the curve $Z^{(\ell-1)}$, then
$\operatorname{codim}_{U^{(\ell-1)}}Z^{(\ell-1)}=2$. From [15, Theorem 7.31],
it follows that

$H^{i}(U^{(\ell)},\mathbb{Z})\cong H^{i}(U^{(\ell-1)},\mathbb{Z})\oplus
H^{i-2}(Z^{(\ell-1)},\mathbb{Z}).$

Then, for $i=1$, we have

$H^{1}(U^{(\ell)},\mathbb{Z})\cong H^{1}(U^{(\ell-1)},\mathbb{Z})\,;$

for $i=2$,

$H^{2}(U^{(\ell)},\mathbb{Z})\cong H^{2}(U^{(\ell-1)},\mathbb{Z})\oplus
H^{0}(Z^{(\ell-1)},\mathbb{Z})\cong
H^{2}(U^{(\ell-1)},\mathbb{Z})\oplus\mathbb{Z},$

as $Z^{\ell-1}$ is a curve and hence
$H^{0}(U^{(\ell-1)},\mathbb{Z})\cong\mathbb{Z}$. Moreover,
$H^{0}(U^{(\ell-1)},\mathbb{Z})$ is of type $(0,0)$, and hence its
contribution to $H^{2}(U^{\ell},\mathbb{Z})$ is of type $(1,1)$.

The above reasoning applies also if $\sigma^{(\ell)}$ is the blow up along the
point $P^{(\ell)}$. In this case
$\operatorname{codim}_{U^{(\ell)}}P^{(\ell)}=3$. Then, from [15, Theorem
7.31], it follows that

$H^{i}(U^{(\ell)},\mathbb{Z})\cong H^{i}(U^{(\ell-1)},\mathbb{Z})\oplus
H^{i-2}(P^{(\ell-1)},\mathbb{Z})\oplus H^{i-4}(P^{(\ell-1)},\mathbb{Z}).$

From this it follows that

$\displaystyle H^{1}(U^{(\ell)},\mathbb{Z})$ $\displaystyle\cong
H^{1}(U^{(\ell-1)},\mathbb{Z})\,,$ $\displaystyle
H^{2}(U^{(\ell)},\mathbb{Z})$ $\displaystyle\cong
H^{2}(U^{(\ell-1)},\mathbb{Z})\oplus\mathbb{Z}\,,$

exactly as in the previous case. Also in this case, the contribution to
$H^{2}(U^{(\ell)},\mathbb{Z})$ comes from
$H^{0}(P^{(\ell-1)},\mathbb{Z})\cong\mathbb{Z}$ and it is of type $(1,1)$.

For $\ell=1$ we then have

$\displaystyle H^{1}(U^{(1)},\mathbb{Z})$ $\displaystyle\cong
H^{1}(\mathbb{P}^{3},\mathbb{Z})=0\,,$ $\displaystyle
H^{2}(U^{(1)},\mathbb{Z})$ $\displaystyle\cong
H^{2}(\mathbb{P}^{3},\mathbb{Z})\oplus
H^{0}(Z^{(1)},\mathbb{Z})\cong\mathbb{Z}^{2}\,.$

As $H^{2}(\mathbb{P}^{3},\mathbb{Z})$ is of type $(1,1)$, so is
$H^{2}(U^{(1)},\mathbb{Z})$. Then, proceeding by induction, we obtain

$\displaystyle H^{1}(U,\mathbb{Z})\cong H^{1}(\mathbb{P}^{3},\mathbb{Z})$
$\displaystyle=0\,,$ $\displaystyle
H^{2}(U,\mathbb{Z})=H^{2}(U^{(n)},\mathbb{Z})$
$\displaystyle\cong\mathbb{Z}^{n+1}\,,$

with $H^{2}(U,\mathbb{Z})$ purely of type $(1,1)$. From
$H^{1}(U,\mathbb{Z})=0$ it follows that $0=h^{1}(U)=h^{1,0}+h^{0,1}$ and hence

$0=h^{0,1}=\dim H^{1}(U,\mathcal{O}_{U}),$

concluding that $H^{1}(U,\mathcal{O}_{U})=0$; from
$H^{2}(U,\mathbb{Z})\cong\mathbb{Z}^{n+1}$, of type $(1,1)$, it follows that

$n+1=h^{2}(U)=h^{2,0}(U)+h^{1,1}(U)+h^{0,2}=h^{2,0}(U)+(n+1)+h^{0,2},$

yielding $h^{2,0}=h^{0,2}=0$. In particular, $h^{0,2}=0$ means that
$H^{2}(U,\mathcal{O}_{U})=0$, concluding the proof.

∎

We are now ready to prove the main theorem.

###### Proof of Theorem 1.3.

From [2, Proposition 18], the square root $\sqrt{f_{1}f_{2}f_{3}f_{4}}$ is
rationalizable if and only if $Y$ is unirational. As $Y$ is birationally
equivalent to $X$, then the condition above is equivalent to the
unirationality of $X$. From 3.10 we know that $X$ is birationally equivalent
to $\tilde{X}$ and that $\tilde{X}$ is a Calabi–Yau threefold, hence it is not
unirational. Therefore $\sqrt{f_{1}f_{2}f_{3}f_{4}}$ is non-rationalizable.

The non-rationalizability of $\mathcal{A}$ follows immediately from the first
statement and [2, Proposition 47]. ∎

######  Remark 3.11.

Notice that in order to prove the theorem, it would have been enough to show
that $\tilde{X}$ has a trivial, and in fact even just effective, canonical
divisor. Indeed, this would already imply that $h^{3,0}=1\gneq 0$ and hence
$\tilde{X}$ cannot be unirational.

##  4\. Guessing unirationality via point counting

In this section we show how to use modularity and point counting to _guess_
the unirationality of a given threefold $Y$ defined over $\mathbb{Q}$, cf.
4.21. The technique reported below is very quick to employ but subject to
assumptions that are often hard to check, making this method not viable to
give a proof. A consistency check can be made under the assumption that the
smooth model $\tilde{Y}$ of $Y$ is a K3 surface or a Calabi–Yau threefold, cf.
4.25 and 4.31, respectively.

###  4.1. Introduction to modularity of Calabi–Yau varieties

In what follows, $Y$ will denote a (not necessarily smooth) complex projective
variety of dimension $d$ that can be defined over $\mathbb{Q}$. In this
subsection we will mostly follow [12, Chapter 1].

######  Definition 4.1.

We say that $Y$ is a _Calabi–Yau_ variety if it is smooth,
$H^{i}(Y,\mathcal{O}_{Y})=0$ for every $0<i<d$, and the canonical divisor
$K_{Y}\sim 0$ is trivial.

From the second condition and by Serre duality we get that

$h^{0,0}(Y)=h^{d,0}(Y)=h^{0,d}(Y)=h^{d,d}(Y)=1,$

where $h^{p,q}(Y):=\dim_{\mathbb{C}}H^{q}(Y,\Omega^{p}_{Y})$ is the $p,q$-th
Hodge number of $Y$.

######  Example 4.2.

A Calabi–Yau variety of dimension $d=1$ (equipped with a rational point) is an
_elliptic curve_. It has the following Hodge diamond.

$\begin{array}[]{ccc}&1&\\\ 1&{}\hfil&1\\\ &1&\end{array}$

######  Example 4.3.

A Calabi–Yau variety of dimension $d=2$ is a _K3 surface_. It has the
following Hodge diamond.

$\begin{array}[]{ccccc}&&1&&\\\ &0&&0&\\\ 1&&20&&1\\\ &0&&0&\\\
&&1&&\end{array}$

######  Example 4.4.

A Calabi–Yau variety of dimension $d=3$ is an _Calabi–Yau threefold_ , from
now on abbreviated into _CY 3-fold_. It has the following Hodge diamond.

$\begin{array}[]{ccccccc}&&&1&&&\\\ &&0&&0&&\\\ &0&&h^{1,1}(Y)&&0&\\\
1&&h^{2,1}(Y)&&h^{1,2}(Y)&&1\\\ &0&&h^{2,2}(Y)&&0&\\\ &&0&&0&&\\\ &&&1&&&\\\
\end{array}$

######  Remark 4.5.

CY manifolds are Kähler, hence $h^{1,1}(Y)>0$. For CY 3-folds, all $2$-cycles
are algebraic, hence $H^{2}(Y,\mathbb{Z})\cong\operatorname{Pic}Y$, that is,
$h^{1,1}(Y)=h^{2}(Y)=\rho(Y)$, the _Picard number_ of $Y$.

######  Definition 4.6.

We say that a CY 3-fold $Y$ is _rigid_ if $h^{2,1}(Y)=0$, otherwise we call it
non-rigid.

As $Y$ is defined over $\mathbb{Q}$, it admits a model over $\mathbb{Z}$.

######  Definition 4.7.

Let $Y$ be a CY variety. We say that a prime $p$ is _of good reduction_ for
$Y$ if the reduction $\bar{Y}_{p}$ of $Y$ over $\overline{\mathbb{F}}_{p}$ is
again a CY variety. So, in particular, $\bar{Y}_{p}$ is smooth.

Let $p$ be a prime of good reduction for $Y$. Then the Frobenius map
$\operatorname{Frob}_{p}\colon\overline{\mathbb{F}}_{p}\to\overline{\mathbb{F}}_{p},\;x\mapsto
x^{p}$, acts on $\bar{Y}_{p}$; it induces an action on the cohomology groups,
in particular on $H_{\acute{e}t}^{d}(\bar{Y}_{p},\mathbb{Q}_{\ell})\cong
H_{\acute{e}t}^{d}(Y,\mathbb{Q}_{\ell})$, where $\ell\neq p$ is a prime. We
can then consider the trace of the pullback of $\operatorname{Frob}_{p}$ on
$H_{\acute{e}t}^{d}(Y,\mathbb{Q}_{\ell})$, denoted by

$\operatorname{tr}(\operatorname{Frob}_{p}^{*}\,|\,H_{\acute{e}t}^{d}(Y,\mathbb{Q}_{\ell})).$

######  Definition 4.8.

We recall that the _$L$ -function_ of $Y$ (or $H^{d}(Y))$ is a series

$L(Y,s):=\sum_{k=1}^{\infty}\frac{a_{k}(Y)}{k^{s}},$

and one has

  * •

$a_{1}(Y)=1$,

  * •

for every prime $p$ of good reduction,
$a_{p}=\operatorname{tr}(\operatorname{Frob}_{p}^{*}\,|\,H_{\acute{e}t}^{d}(Y,\mathbb{Q}_{\ell}))$.

######  Theorem 4.9 (Lefschetz trace formula).

Let $Y$ be a smooth variety over $\mathbb{F}_{p}$ and, for $r\geq 1$, let
$\\#Y_{p^{r}}$ denote the number of points of $Y$ over $\mathbb{F}_{p^{r}}$.
Then

$\\#Y_{p^{r}}=\sum_{i=0}^{2d}(-1)^{i}\operatorname{tr}((\operatorname{Frob}_{p}^{*})^{r}\,|\,H_{\acute{e}t}^{d}(\bar{Y},\mathbb{Q}_{\ell})).$

######  Theorem 4.10 (Weil–Deligne).

Let $Y$ be a smooth complex variety that can be defined over $\mathbb{Q}$ and
let $p$ be a prime of good reduction for $Y$. Then, for every $i=0,...,2d$ the
quantity
$\operatorname{tr}(\operatorname{Frob}_{p}^{*}\,|\,H_{\acute{e}t}^{i}(Y,\mathbb{Q}_{\ell}))$
is an integer and

$|\operatorname{tr}(\operatorname{Frob}_{p}^{*}\,|\,H_{\acute{e}t}^{i}(Y,\mathbb{Q}_{\ell}))|\leq
h^{i}(Y)p^{i/2}.$

So, in particular,

$|a_{p}(Y)|\leq h^{d}(Y)p^{d/2}.$

Using the knowledge of the Hodge diamond it is possible to give formulas to
compute $a_{p}(Y)$ when $Y$ is a CY variety that can be defined over
$\mathbb{Q}$, see below.

######  Example 4.11.

Let $E$ be an elliptic curve, $p$ a prime of good reduction and $\ell\neq p$
another prime, then

  * •

$a_{1}(E)=1$,

  * •

$a_{p}(E)=\operatorname{tr}(\operatorname{Frob}^{*}_{p}\,|\,H_{\acute{e}t}^{1}(E,\mathbb{Q}_{\ell}))$,

  * •

$\\#E_{p}=\sum_{i=0}^{2}(-1)^{i}\operatorname{tr}(\operatorname{Frob}^{*}_{p}\,|\,H_{\acute{e}t}^{i}(E,\mathbb{Q}_{\ell}))=1-a_{p}(E)+p$,

from which it follows that $a_{p}(E)=p+1-\\#E_{p}$.

######  Example 4.12.

Let $Y$ be a complex K3 surface that can be defined over $\mathbb{Q}$, $p$ a
prime of good reduction and $\ell\neq p$ another prime, then

  * •

$a_{1}(Y)=1$,

  * •

$a_{p}(Y)=\operatorname{tr}(\operatorname{Frob}^{*}_{p}\,|\,H_{\acute{e}t}^{2}(Y,\mathbb{Q}_{\ell}))$,

  * •

$\\#Y_{p}=\sum_{i=0}^{4}(-1)^{i}\operatorname{tr}(\operatorname{Frob}^{*}_{p}\,|\,H_{\acute{e}t}^{i}(Y,\mathbb{Q}_{\ell}))=1-0+a_{p}(Y)-0+p^{2}$,

from which it follows that $a_{p}(Y)=\\#Y_{p}-1-p^{2}$. Notice that the same
computations hold for any smooth surface with $h^{1}=h^{3}=0$.

######  Theorem 4.13 (Modularity of singular K3 surfaces [11]).

Let $S$ be a K3 surface defined over $\mathbb{Q}$ having maximal Picard number
over $\overline{Q}$, and let $q=e^{2\pi iz}$ with $\operatorname{Im}(z)>0$.
Then there exists a newform

$f(q)=\sum_{k=1}^{\infty}b_{k}q^{k}\in S^{\textrm{new}}_{3}(\Gamma(N))$

of weight $3$ with respect to a congruence subgroup
$\Gamma\leq\operatorname{PSL}_{2}(\mathbb{Z})$ and
$N=-m^{2}\cdot\det\operatorname{Pic}S$ for some $m\in\mathbb{N}$ (cf. [13,
Prop. 7.1]), such that

$a_{p}(Y)=b_{p}$

for all the primes $p$ of good reduction for $Y$.

######  Example 4.14.

Let $Y$ be a CY 3-fold that can be defined over $\mathbb{Q}$, $p$ a prime of
good reduction and $\ell\neq p$ another prime, then

  * •

$a_{1}(Y)=1$,

  * •

$a_{p}(Y)=\operatorname{tr}(\operatorname{Frob}^{*}_{p}\,|\,H_{\acute{e}t}^{3}(Y,\mathbb{Q}_{\ell}))$,

  * •

$\displaystyle\\#Y_{p}$
$\displaystyle=\sum_{i=0}^{6}(-1)^{i}\operatorname{tr}(\operatorname{Frob}^{*}_{p}\,|\,H_{\acute{e}t}^{i}(Y,\mathbb{Q}_{\ell}))=$
$\displaystyle=1-0+\operatorname{tr}(\operatorname{Frob}_{p}^{*}\,|\,H^{2}(Y,\mathbb{Q}_{\ell}))-a_{p}(Y)+\operatorname{tr}(\operatorname{Frob}_{p}^{*}\,|\,H^{4}(Y,\mathbb{Q}_{\ell}))-0+p^{3}.$

By definition of CY 3-fold, we have that $H^{2}(Y,\mathbb{C})=H^{1,1}(Y)$ and
hence $\operatorname{Pic}Y=H^{2}(Y,\mathbb{Z})$. Then

$\operatorname{tr}(\operatorname{Frob}_{p}^{*}\,|\,H^{2}(Y,\mathbb{Q}_{\ell}))=k_{p}(Y)p,$

where $k_{p}(Y)\in\mathbb{Z}$ and $|k_{p}(Y)|\leq h^{2}(Y)=h^{1,1}(Y)$. The
equality $k_{p}(Y)=h^{1,1}(Y)$ holds if and only if
$H_{\acute{e}t}^{2}(Y,\mathbb{Q}_{\ell})=\operatorname{Pic}(Y_{\overline{Q}})\otimes_{\mathbb{Z}}\mathbb{Q}_{\ell}$
is generated by divisors defined over $\mathbb{Q}$. By Poincaré duality it
follows that

$\operatorname{tr}(\operatorname{Frob}_{p}^{*}\,|\,H^{4}(Y,\mathbb{Q}_{\ell}))=k_{p}(Y)p^{2},$

yielding

$a_{p}(Y)=1+(p+p^{2})k_{p}(Y)+p^{3}-\\#Y_{p}.$

######  Theorem 4.15 (Modularity of rigid CY 3-folds [8]).

Let $Y$ be a rigid CY 3-fold, and $q=e^{2\pi iz}$, with
$\operatorname{Im}(z)>0$. Then, for some integer $N$, there exists a new form

$f(q)=\sum_{k=1}^{\infty}b_{k}q^{k}\in S_{4}^{\textrm{new}}(\Gamma_{0}(N))$

of weight $4$ with respect to $\Gamma_{0}(N)$, such that

$b_{p}=a_{p}(Y)$

for all the primes $p$ of good reduction for $Y$. The level $N$ is only
divisible by the primes of bad reduction of $Y$; the maximal exponent $e_{p}$
of a prime dividing $N$ is bounded:

  1. (1)

$e_{p}\leq 2$ for $p>3$;

  2. (2)

$e_{3}\leq 5$;

  3. (3)

$e_{2}\leq 8$.

######  Remark 4.16.

A similar result should hold for every smooth complex variety $Y$ of dimension
$2k-1$ with $h^{2k-1}(Y,\mathbb{C})=2$ and $h^{2k-1,0}(Y)=1$ if we replace
$S_{4}^{\textrm{new}}$ with $S_{2k}$, see [8, §2]. So in particular it should
hold for possibly singular threefolds $Z$ such that its desingularization
$\tilde{Z}$ has $h^{3}(\tilde{Z},\mathbb{C})=2$ and $h^{3,0}(\tilde{Z})=1$.

###  4.2. Non-unirationality via point counting

Let $Y$ be a complex variety, possibly non-smooth, and let $\tilde{Y}\to Y$ be
a smooth resolution of singularities.

######  Assumption 4.17.

The resolution $\tilde{Y}\to Y$ is defined over $\mathbb{Q}$ and $\tilde{Y}$
is projective.

######  Definition 4.18.

By assumption, $Y$ has a non-singular projective model $\tilde{Y}$ that can be
defined over $\mathbb{Q}$ and hence over $\mathbb{Z}$. This means that there
exists a $\mathbb{Z}_{p}$-scheme
$\mathcal{Y}\to\operatorname{Spec}\mathbb{Z}_{p}$ such that its generic fiber
$\mathcal{Y}_{\eta}$ is isomorphic to $\tilde{Y}$.

We say that a prime $p$ is of good reduction for $Y$ if
$\mathcal{Y}_{p}/\mathbb{F}_{q}$ is a smooth variety, for any $q=p^{m}$.

Notice that $\mathcal{Y}_{p}$ can be viewed as the reduction $\tilde{Y}_{p}$
of $\tilde{Y}$ modulo $p$.

######  Assumption 4.19.

For every prime of good reduction $p$,

$\\#\mathcal{Y}_{p}(\mathbb{F}_{p})\equiv\\#Y_{p}(\mathbb{F}_{p})\mod p\,.$

######  Lemma 4.20.

[1, Theorem 1.1] Let $\mathcal{Y}\to\operatorname{Spec}\mathbb{Z}_{p}$ be a
smooth $\mathbb{Z}_{p}$-scheme of dimension $d$. Let $\mathcal{Y}_{\eta}$
denote the fiber above $\eta=(0)\in\operatorname{Spec}\mathbb{Z}_{p}$, and
$\mathcal{Y}_{p}$ the closed fiber above $(p)$. If
$h^{k,0}(\mathcal{Y}_{\eta})=0$ for every $k=1,...,d$, then
$\\#\mathcal{Y}_{p}(\mathbb{F}_{p})\equiv 1\mod p$.

Using 4.20, one can make a first guess about the unirationality of $Y$ via
point counting under 4.19, as shown by the following result.

######  Corollary 4.21.

Let $Y$ be defined as before and let $p$ be a prime of good reduction for $Y$.
Under 4.19, if $\\#Y_{p}(\mathbb{F}_{p})\not\equiv 1\mod p$, then $Y$ is not
unirational.

###### Proof.

Assume that $Y$ is unirational, then it follows that $\tilde{Y}$ and hence
$\mathcal{Y}_{\eta}$ are unirational too. In particular,
$h^{k,0}(\mathcal{Y}_{\eta})=0$ for $k=1,2,3$, Then, as $p$ is a prime of good
reduction for $Y$, c.f. 4.18, we have that
$\\#\mathcal{Y}_{p}(\mathbb{F}_{p})\equiv 1\mod p$ by 4.20. Moreover, by 4.19,
it holds that
$\\#\mathcal{Y}_{p}(\mathbb{F}_{p})\equiv\\#Y_{p}(\mathbb{F}_{p})\mod p$. So,
in particular,

$1\equiv\\#\mathcal{Y}_{p}(\mathbb{F}_{p})\equiv\\#Y_{p}(\mathbb{F}_{p})\not\equiv
1\mod p\;,$

a contradiction. This proves that $Y$ is not unirational. ∎

We can apply the above tools to the more specific cases of K3 surfaces and
Calabi–Yau threefolds, for which, in some cases, we can use the modularity
theorems to perform a consistency check. In what follows, we will need the
following definition.

######  Definition 4.22.

Let $B\in\mathbb{N}$ a positive integer. Then we define the sets:

$\displaystyle\Sigma(B)$ $\displaystyle=\\{p\textrm{ prime }:\;p\leq B\textrm{
and }p\textrm{ is of good reduction for }Y\\}\,,\;and$
$\displaystyle\Sigma_{0}(B)$
$\displaystyle=\\{p\in\Sigma(B)\;:\;\\#Y_{p}(\mathbb{F}_{p})\not\equiv 1\mod
p\\}.$

###  4.3. Point counting for K3 surfaces

In this subsection we assume that $\dim Y=2$ and $Y$ satisfies assumptions
4.17 and 4.19. We denote by $\tilde{Y}$ a fixed minimal smooth model of $Y$.

######  Lemma 4.23.

If $\tilde{Y}$ is a K3 surface, then the following congruence holds:

$a_{p}(\tilde{Y})\equiv\\#Y_{p}(\mathbb{F}_{p})-1\mod p$

for every prime of good reduction $p$.

###### Proof.

As $\tilde{Y}$ is a K3 surface and $p$ is a prime of good reduction, we know
that

$a_{p}(\tilde{Y})=\\#\mathcal{Y}_{p}(\mathbb{F}_{p})-1-p^{2},$

see 4.12. By 4.19, this yields

$a_{p}(\tilde{Y})\equiv\\#Y_{p}(\mathbb{F}_{p})-1\mod p,$

concluding the proof. ∎

If we suspect that $Y$ is not unirational (for example by 4.21) and, moreover,
that its desingularization $\tilde{Y}$ is a singular K3 surface, i.e. a K3
surface with maximal Picard number over $\overline{\mathbb{Q}}$, we can use
4.23 to do a consistency check via point counting, instead of explicitly
computing the cohomology of $\tilde{Y}$. To do so, we consider the set of
primes of good reduction $\Sigma(B)$, cf. 4.22, and we compute the number of
points $\\#Y(\mathbb{F}_{p})$ for $p\in\Sigma(B)$. In the examples below, this
set does not need to be very large, we took $B=100$. So, upon computing the
number of points $\\#Y_{p}(\mathbb{F}_{p})$ for a finite set $\Sigma(B)$ of
primes of good reduction, we can look for levels $N$, congruence subgroups
$\Gamma\leq\operatorname{PSL}_{2}(\mathbb{Z})$ and modular forms

$f(q)=\sum_{k=1}^{\infty}b_{k}q^{k}\in S_{3}^{\textrm{new}}(\Gamma(N))$

such that $b_{p}\equiv\\#Y_{p}(\mathbb{F}_{p})-1\mod p$ for any
$p\in\Sigma(B)$.

######  Assumption 4.24.

For some $N\in\mathbb{N}$, subgroup
$\Gamma\leq\operatorname{PSL}_{2}(\mathbb{Z})$, and a positive integer $B$
such that $\\#\Sigma_{0}(B)\geq 10$, there exists a newform $f\in
S_{3}^{\textrm{new}}(\Gamma(N))$ such that

$b_{p}\equiv\\#Y_{p}(\mathbb{F}_{p})-1\mod p$

for every $p\in\Sigma(B)$.

######  Conjecture 4.25.

If $Y$ satisfies 4.24, then $Y$ is birationally equivalent to a smooth surface
$\tilde{Y}$ with $h^{2,0}(\tilde{Y})\geq 1$. In particular, $Y$ is not
unirational.

######  Remark 4.26.

It is important to take _all_ the primes up to a certain bound $B$: given a
surface $Y$ such that $\tilde{Y}$ is a singular K3 surface, it might still
happen that all for the primes $p$ satisfying a certain congruence one has
$\\#Y_{p}(\mathbb{F}_{p})\equiv 1\mod p$, giving the false impression that $Y$
is unirational. In fact this happens for the surface $S$ in § 4.5: in this
case we find that $\\#Y_{p}(\mathbb{F}_{p})\equiv 1\mod p$ for all the primes
$p\equiv 5,7\mod 8$. The same happens for the Fermat quartic and all the
primes $p\equiv 3\mod 4$.

######  Remark 4.27.

If $\tilde{Y}$ is a K3 surface with Picard number $20$ over
$\overline{\mathbb{Q}}$, then the level $N$ of the form $f$ determines the
discriminant of $\operatorname{Pic}\tilde{Y}$ modulo squares, cf. Theorem
4.13. In this case we expect that $b_{p}\equiv\\#Y_{p}(\mathbb{F}_{p})-1\mod
p$ for all primes $p$ of good reduction.

If $\tilde{Y}$ is a K3 surface with Picard number $20$ over ${\mathbb{Q}}$,
then we expect even more:
$\\#\mathcal{Y}_{p}(\mathbb{F}_{p})=1+20p+b_{p}+p^{2}$.

###  4.4. Point counting for Calabi–Yau threefolds

In this subsection we assume that $\dim Y=3$ and $Y$ satisfies assumptions
4.17 and 4.19. We denote by $\tilde{Y}$ a fixed smooth model of $Y$.

######  Remark 4.28.

If $Y$ has only ordinary double points, then 4.19 is satisfied. Indeed, an
ordinary double point gets resolved after one blow up, introducing a
$\mathbb{P}^{1}\times\mathbb{P}^{1}$ as exceptional divisor. Then the claim
follows from the computations in [12, §1.6.1].

######  Lemma 4.29.

If $\tilde{Y}$ is a Calabi–Yau threefold, then the following congruence holds:

$a_{p}(\tilde{Y})\equiv 1-\\#Y_{p}(\mathbb{F}_{p})\mod p$

for every prime of good reduction $p$.

###### Proof.

As $\tilde{Y}$ is a CY 3-fold, we know that

$a_{p}(\tilde{Y})=1+(p+p^{2})k_{p}(\tilde{Y})+p^{3}-\\#\mathcal{Y}_{p}(\mathbb{F}_{p}),$

cf. 4.14. By 4.19, this yields

$\displaystyle a_{p}(\tilde{Y})$ $\displaystyle\equiv
1+(k_{p}(\tilde{Y})-\alpha_{p})p+(k_{p}(\tilde{Y})-\beta_{p})p^{2}+p^{3}-\\#Y_{p}(\mathbb{F}_{p})\mod
p$ $\displaystyle\equiv 1-\\#Y_{p}(\mathbb{F}_{p})\mod p\,.$

∎

Analogously to the previous section, 4.29 tells us that if we suspect that $Y$
is not unirational (for example by 4.21) and that, moreover, its
desingularization $\tilde{Y}$ is a rigid CY 3-fold, then we can do a
consistency check via point counting instead of explicitly computing the
cohomology of $\tilde{Y}$. To do so, it is enough to consider the primes in
the set $\Sigma(B)$, cf. 4.22, for some bound $B$. Also in the example below,
we take $B=100$.

We can then look for levels $N$ and modular forms

$f(q)=\sum_{k=1}^{\infty}b_{k}q^{k}\in S_{4}^{\textrm{new}}(\Gamma_{0}(N))$

such that $b_{p}\equiv 1-\\#Y_{p}(\mathbb{F}_{p})\mod p$ for any
$p\in\Sigma(B)$. Note that in this case the congruence subgroup is fixed, and
Theorem 4.15 gives us a hint about the possible values for $N$, making the
search more practical.

######  Assumption 4.30.

For some $N\in\mathbb{N}$ and a positive integer $B$ such that
$\\#\Sigma_{0}(B)\geq 10$, there exists a newform $f\in
S_{4}^{\textrm{new}}(\Gamma_{0}(N))$ such that

$b_{p}\equiv 1-\\#Y_{p}(\mathbb{F}_{p})\mod p$

for every $p\in\Sigma(B)$.

######  Conjecture 4.31.

If $Y$ satisfies 4.30, then $Y$ is birationally equivalent to a smooth
threefold $\tilde{Y}$ with $h^{3,0}(\tilde{Y})\geq 1$. In particular, $Y$ is
not unirational.

###  4.5. Application to two surfaces

Let $Q\subset\mathbb{P}^{3}$ be the quartic surface defined by

$Q\colon\;-x_{3}^{2}(x_{4}^{2}-x_{1}x_{2})+(x_{1}+x_{2})(x_{1}x_{4}^{2}+x_{2}x_{4}^{2}-4x_{1}x_{2}x_{4}+x_{1}^{2}x_{2}+x_{1}x_{2}^{2})=0,$

and $S\subset\mathbb{P}(3,1,1,1)$ the double sextic defined by

$S\colon\;y_{0}^{2}=(y_{1}^{2}+y_{2}^{2}-2y_{3}^{2})(y_{1}^{2}y_{2}^{2}-y_{3}^{4}),$

both studied in [7]. The two surfaces have several singular points, so we
cannot immediately conclude that they are birationally equivalent to K3
surfaces. After counting points of the reductions modulo all the odd primes
smaller than $100$, one sees that
$\\#Q_{p}(\mathbb{F}_{p}),\\#S_{p}(\mathbb{F}_{p})\not\equiv 1\mod p$ for many
primes $p$. Moreover $Q$ and $S$ both satisfy 4.24 with

$\displaystyle f(q)$
$\displaystyle=q-6q^{5}+9q^{9}+10q^{13}-30q^{17}+11q^{25}+42q^{29}-70q^{37}+O(q^{41})\in
S_{3}^{\textrm{new}}(\Gamma_{1}(16)),$ $\displaystyle g(q)$
$\displaystyle=q-2q^{2}-2q^{3}+4q^{4}+4q^{6}-8q^{8}-5q^{9}+14q^{11}-8q^{12}+O(q^{14})\in
S_{3}^{\textrm{new}}(\Gamma_{1}(8))$

respectively. Therefore we expect $Q$ and $S$ not to be unirational.

The prediction for $S$ is confirmed by [7, Theorem], where it is shown that
its smooth model $\tilde{S}$ a K3 surface whose Picard lattice has rank $20$
and discriminant $-8$. In the same paper, the authors suggest that $Q$ and $S$
are birationally equivalent, which contradicts the heuristic associating them
to two distinct modular forms, moreover of levels differing by $2$, which is
not a square. Below we show that the suggestion is indeed wrong.

######  Proposition 4.32.

The surfaces $Q$ and $S$ are not birationally equivalent.

###### Proof.

Let $\tilde{Q}$ and $\tilde{S}$ denote the smooth models of $Q$ and $S$,
respectively. It is easy to check that $Q$ has only ADE singularities, and
hence $\tilde{Q}$ is a K3 surface. The same holds for $S$ and $\tilde{S}$.
Recall that if two K3 surfaces are birationally equivalent then they are
isomorphic. This means that if $Q$ and $S$ are birationally equivalent,
$\tilde{Q}$ and $\tilde{S}$ are isomorphic; in particular, they have isometric
Picard lattices.

From [7, Theorem] we know that $\operatorname{Pic}\tilde{S}$ has rank $20$ and
discriminant $-8$. By projecting from a node of $Q$, we obtain a model of $Q$
as double sextic. It is then easy to find some $-2$-curves on this model.
Using these curves and the exceptional divisors we generate a lattice
$\Lambda\subseteq\operatorname{Pic}\tilde{Q}$ of rank $20$, discriminant $-16$
and index $[\operatorname{Pic}\tilde{Q}:\Lambda]\leq 2$. It follows that
$\operatorname{Pic}\tilde{Q}$ has rank $20$ and discriminant either $-16$ or
$-4$. In any case,
$\det\operatorname{Pic}\tilde{Q}\not\cong\det\operatorname{Pic}\tilde{S}$,
showing that $Q$ and $S$ are not birationally equivalent. ∎

######  Remark 4.33.

Note that the level of $f$ is a square multiple of
$\det\operatorname{Pic}\tilde{Q}$, as predicted by Theorem 4.13. The same
holds for $g$ and $S$.

###  4.6. Application to the double octic

Let $X$ be the double octic defined in (4). Notice that it is reasonable to
assume that $X$ satisfies 4.17 and 4.19. We proceed then with the point
counting, obtaining Table 2.

$p$ | 3 | 5 | 7 | 11 | 13 | 17 | 19 | 23
---|---|---|---|---|---|---|---|---
$\\#X_{p}(\mathbb{F}_{p})$ | 46 | 180 | 500 | 1716 | 2732 | 6060 | 8132 | 13932
$1-\\#X_{p}(\mathbb{F}_{p})\mod p$ | 0 | 1 | 5 | 1 | 12 | 10 | 1 | 7
$p$ | 29 | 31 | 37 | 41 | 43 | 47 | 53 | 59
$\\#X_{p}(\mathbb{F}_{p})$ | 27492 | 33476 | 55580 | 75276 | 86612 | 112380 | 159492 | 219492
$1-\\#X_{p}(\mathbb{F}_{p})\mod p$ | 1 | 5 | 32 | 1 | 34 | 45 | 39 | 48
$p$ | 61 | 67 | 71 | 73 | 79 | 83 | 89 | 97
$\\#X_{p}(\mathbb{F}_{p})$ | 241916 | 317300 | 376716 | 409532 | 517892 | 599172 | 735132 | 948380
$1-\\#X_{p}(\mathbb{F}_{p})\mod p$ | 11 | 13 | 11 | 72 | 33 | 6 | 9 | 87
Table 2. Point counting on $X_{p}$ for every odd prime smaller than 100.

For many primes $p$ we see that $X_{p}(\mathbb{F}_{p})\not\equiv 1\mod p$,
hence by 4.21 we expect $X$ not to be unirational. In an attempt to gather
more evidence towards the non-unirationality of $X$, we look for a modular
form satisfying 4.30. We then find that the unique newform
$f(q)=\sum_{k=1}^{\infty}b_{k}q^{k}\in S_{4}^{\textrm{new}}(\Gamma_{0}(6))$,
with

(10)
$f(q)=q-2q^{2}-3q^{3}+4q^{4}+6q^{5}+6q^{6}-16q^{7}-8q^{8}+9q^{9}-12q^{10}+12q^{11}+O(q^{12}),$

satisfies 4.30, giving yet a stronger hint, cf. 4.31. The heuristics
suggesting that $X$ is not unirational are indeed confirmed by 3.10.

######  Remark 4.34.

In this particular case, the evidence is even stronger: indeed, one can check
that

(11) $b_{p}=1-8p+4p^{2}+p^{3}-\\#X_{p}(\mathbb{F}_{p})$

for every prime $3<p<100$, suggesting that $\tilde{X}$ is a _rigid_ CY 3-fold.
This is confirmed by B. Naskrȩcki in a private communication: he proves that
the $L$-function of $\tilde{X}$ is indeed (10) and deduces the rigidity of
$\tilde{X}$. The same result was suggested also by S. Cynk.

In [14], van den Bogaart and Edixhoven compute the cohomology of varieties
defined over $\mathbb{Q}$ and satisfying even stronger conditions on the
number of rational points of their reduction modulo a prime $p$. In
particular, they show that if $Y/\mathbb{Q}$ is a smooth variety such that
$\\#Y_{p}(\mathbb{F}_{p})$ behaves polynomially for all primes $p$, then all
the cohomology groups of odd degree vanish and they determine the dimension of
the cohomology groups of even degree.

######  Remark 4.35.

The fact that (11) does not hold for $p=3$ is consistent with the level of $f$
being $6$, suggesting that $p=3$ is a prime of bad reduction for $X$, cf.
Theorem 4.15. Indeed, if $X_{3}$ denotes the reduction of $X$ modulo $3$,
then, looking at Table 1, one can see that the four singular points

$(1:1:1:0),(1:1:4:3),(1:4:1:3),(4:1:1:3)\in X$

collapse to the singular point $(1:1:1:0)\in X_{3}$. One can then check that
$X_{3}$ has a non-ordinary singularity in $(1:1:1:0)$, whereas $X$ has an
ordinary one.

######  Remark 4.36.

The same newform $f$ in (10) appears also in [9] and [12, §5.7], in relation
to other CY 3-folds.

## References

  * [1] Pierre Berthelot, Hélène Esnault, and Kay Rülling, _Rational points over finite fields for regular models of algebraic varieties of Hodge type $\geq 1$_, Ann. of Math. (2) 176 (2012), no. 1, 413–508. MR 2925388
  * [2] Marco Besier and Dino Festi, _Rationalizability of square roots_ , J. Symbolic Comput. 106 (2021), 48–67. MR 4199812
  * [3] Sławomir Cynk, _Cohomologies of a double covering of a non-singular algebraic 3-fold_ , Math. Z. 240 (2002), no. 4, 731–743. MR 1922727
  * [4] Sławomir Cynk and Tomasz Szemberg, _Double covers and Calabi-Yau varieties_ , Singularities Symposium—Łojasiewicz 70 (Kraków, 1996; Warsaw, 1996), Banach Center Publ., vol. 44, Polish Acad. Sci. Inst. Math., Warsaw, 1998, pp. 93–101. MR 1677335
  * [5] Sławomir Cynk and Duco van Straten, _Infinitesimal deformations of double covers of smooth algebraic varieties_ , Math. Nachr. 279 (2006), no. 7, 716–726. MR 2226407
  * [6] Dino Festi and Andreas Hochenegger, _Rationalizability of field extensions with a view towards Feynman integrals_ , J. Geom. Phys. 178 (2022), Paper No. 104536. MR 4420531
  * [7] Dino Festi and Duco van Straten, _Bhabha scattering and a special pencil of K3 surfaces_ , Commun. Number Theory Phys. 13 (2019), no. 2, 463–485. MR 3951114
  * [8] Fernando Q. Gouvêa and Noriko Yui, _Rigid Calabi-Yau threefolds over $\mathbb{Q}$ are modular_, Expo. Math. 29 (2011), no. 1, 142–149. MR 2785550
  * [9] Klaus Hulek, Jeroen Spandaw, Bert van Geemen, and Duco van Straten, _The modularity of the Barth-Nieto quintic and its relatives_ , Adv. Geom. 1 (2001), no. 3, 263–289. MR 1874236
  * [10] Philipp Alexander Kreer, Robert Runkel, and Stefan Weinzierl, _Feynman integrals for binary systems of black holes_ , 15th International Symposium on Radiative Corrections: Applications of Quantum Field Theory to Phenomenology and LoopFest XIX: Workshop on Radiative Corrections for the LHC and Future Colliders, 10 2021.
  * [11] Ron Livné, _Motivic orthogonal two-dimensional representations of ${\rm Gal}(\overline{\mathbb{Q}}/{\mathbb{Q}})$_, Israel J. Math. 92 (1995), no. 1-3, 149–156. MR 1357749
  * [12] Christian Meyer, _Modular Calabi-Yau threefolds_ , Fields Institute Monographs, vol. 22, American Mathematical Society, Providence, RI, 2005. MR 2176545
  * [13] Matthias Schütt, _CM newforms with rational coefficients_ , arXiv:0511228v5, 2005.
  * [14] Theo van den Bogaart and Bas Edixhoven, _Algebraic stacks whose number of points over finite fields is a polynomial_ , Number fields and function fields—two parallel worlds, Progr. Math., vol. 239, Birkhäuser Boston, Boston, MA, 2005, pp. 39–49. MR 2176584
  * [15] Claire Voisin, _Hodge theory and complex algebraic geometry. I_ , english ed., Cambridge Studies in Advanced Mathematics, vol. 76, Cambridge University Press, Cambridge, 2007, Translated from the French by Leila Schneps. MR 2451566
  * [16] Stefan Weinzierl, _Feynman integrals: A comprehensive treatment for students and researchers_ , UNITEXT for Physics, Springer International Publishing, 2022.