\begin{proposition}\label{prop:two-def-same}
Let $v$ be a discrete valuation on $k$, and $O$ its valuation ring.
Let $\sX \to \Spec O$ be a smooth proper morphism with geometrically connected fibers,
and denote by $X$ the generic fiber.
Then we have
\[ \Im(\langle \cdot, \cdot \rangle :
\CH^1(X)_\Tor \times A_0(X) \to k^\times \otimes \Q/\Z)
\subset
\Im(O^\times \otimes \Q/\Z \to k^\times \otimes \Q/\Z).
\]
\end{proposition}

\begin{proposition}\label{prop:TateCv}
\begin{enumerate}
\item
Let $a, b \in k^\times$ and $m>0$.
If $[a],  [b] \in E_q(k)[m]$, we have
\[ \langle [a], [b] \rangle
= a^{-s_m([b])} \otimes \frac{1}{m}
= b^{-s_m([a])} \otimes \frac{1}{m}
= q^{-s_m([a])s_m([b])} \otimes \frac{1}{m^2}.
\]
\item
We have $E_q(k)_\Tor^\is = \{ [\zeta] \mid \zeta \in \mu(k) \}$.
\end{enumerate}
\end{proposition}

\begin{theorem} \label{theorem: intrinsic torsion}
  Let $E$ be an elliptic curve over $\Q$.
Then there is a pair $(A, B)$ of an abelian group $A$
and its subgroup $B$ in the following table,
together with
an isomorphism $\alpha : E(\Q)_\Tor \cong A$,
such that $\alpha(E(\Q)_\Tor^\is)=B$.
  $$ \extrarowheight3pt
  \begin{array}{ll} \hline\hline
    A & \text{order of}\ B \\ \hline
    0 & 1 \\
    \Z/2\Z & 1,2\\
    \Z/3\Z & 1,3\\
    \Z/4\Z & 1,2,4 \\
    \Z/5\Z & 1,5 \\
    \Z/6\Z & 1,2,3 \\
    \Z/7\Z & 1 \\
    \Z/8\Z & 1,2 \\
    \Z/9\Z & 1 \\
    \Z/10\Z & 1\\
    \Z/12\Z & 1 \\
    \Z/2\Z\times\Z/2\Z & 1,2\\
    \Z/4\Z\times\Z/2\Z & 1,2 \ \\
    \Z/6\Z\times\Z/2\Z & 1 \\
    \Z/8\Z\times\Z/2\Z & 1 \\ \hline\hline
  \end{array}
  $$
(For $A=\Z/4\Z \times \Z/2\Z$,
there are two $\Aut(A)$-conjugacy classes of order two subgroups,
both of which can be taken as $B$.)
  Moreover, for each possible pair $(A, B)$ as above,
there are infinitely many (mutually non-isomorphic) elliptic curves $E$ over $\Q$
for which there is an isomorphism $\alpha : E(\Q)_\Tor \cong A$
such that $\alpha(E(\Q)_\Tor^\is)=B$.
\end{theorem}

\begin{theorem}\label{proposition: pairing 1-mod}
  Assume that $N\in\{4,5,6,7,8,9,10,12\}$ and
  let $M$ be a positive divisor of $N$.
  Denote by $X_1(MN,N)$ the modular curve over $\Q$
  associated to the congruence subgroup
    $$
    \Gamma_1(MN,N):= \Gamma_0(MN) \cap \Gamma_1(N)
    =\left\{\M abcd\in\SL(2,\Z) : \
      a,d\equiv1\bmod N, \ MN|c \right\},
    $$
  on which the cusp $0$ is $\Q$-rational.
  Then $X_1(N)_M^+$ is isomorphic to $X_1(MN,N)$ over $\Q$.
\end{theorem}

Let $X$ be a geometrically irreducible smooth projective curve
over a field $k$. In \S \ref{sect:pairing},
we shall construct a biadditive symmetric pairing
\begin{equation}\label{eq:pairing-intro}
\langle \cdot , \cdot \rangle :
\operatorname{Pic}(X)_\Tor \times \operatorname{Pic}(X)_\Tor \to k^\times \otimes \mathbb{Q}/\mathbb{Z},
\end{equation}
where $\operatorname{Pic}(X)_\Tor$ is the
torsion part of the Picard group $\operatorname{Pic}(X)$ of $X$.
We then define the \emph{intrinsic subgroup}
of $\operatorname{Pic}(X)_\Tor$ by
\[
\operatorname{Pic}(X)_\Tor^\is
:= \{ a \in \operatorname{Pic}(X)_\Tor \mid \langle a, b \rangle = 0
\text{ for all } b \in \operatorname{Pic}(X)_\Tor \}.
\]

We remark that \eqref{eq:pairing-intro} is not entirely new.
Indeed, its finite coefficient analogue
(with values in $k^\times/(k^\times)^m$) is considered by Frey and R\"uck in \cite{FreyRuck}
(see Remark \ref{rem:FreyRuck}).
The first and second authors encountered with
a disguised version of \eqref{eq:pairing-intro} in \cite{YY}
(see Remark \ref{sect:genJac}).
Obviously, \eqref{eq:pairing-intro} is trivial if $k^\times \otimes \mathbb{Q}/\mathbb{Z}=0$,
e.g. when $k$ is finite, algebraically closed, or $k=\mathbb{R}$.
On the other hand,
$\operatorname{Pic}(X)_\Tor^\is$ appears to be a non-trivial new invariant
if $k$ is a global field or a $p$-adic field.

We will mainly consider the case where $X=E$ is an elliptic curve
so that we may identify $E(k)_\Tor= \operatorname{Pic}(E)_\Tor$
by $P \mapsto [P-O]$, where $O \in E(k)$ is the identity element.
Thus we write
$\langle P,Q\rangle:=\langle [P-O], [Q-O]\rangle$ for $P, Q \in E(k)_\Tor$
and
$E(k)_\Tor^\is:=\operatorname{Pic}(E)_\Tor^\is$.
Our main result is a classification of possible structures of
$(E(\mathbb{Q})_\Tor,E(\mathbb{Q})_\Tor^\is)$ for an elliptic curve $E$ over $\mathbb{Q}$.
A celebrated theorem of Mazur \cite[Theorem (8)]{Mazur} states
that $E(\mathbb{Q})_\Tor$ is either cyclic of order
$1, \ldots,10$, or $12$, or is isomorphic to $(\mathbb{Z}/2N\mathbb{Z}) \times (\mathbb{Z}/2\mathbb{Z})$,
$N=1, 2,3,4$,
and all cases are realized by infinitely many (mutually non-isomorphic) elliptic curves.
The classification of
$E(\mathbb{Q})_\Tor^\is$ is given in the following theorem,
whose proof will occupy \S \ref{sect:ref-mazur}--\ref{sect:ref-mazur2}.

\begin{equation}\label{eq:pairing-intro}
\langle \cdot , \cdot \rangle :
\Pic(X)_\Tor \times \Pic(X)_\Tor \to k^\times \otimes \Q/\Z,
\end{equation}

\begin{remark}\label{rem:FreyRuck}
As is seen from the proof,
there is an analogous biadditive pairing
\[
\langle \cdot, \cdot \rangle_n :
\Pic(X)[n] \times \Pic^0(X)/n \to k^\times/(k^\times)^n
\]
for each $n>0$,
which satisfies
$\langle d, e \rangle_n
=
\WP_n(d, e)
\langle e, d \rangle_n
$for any $d, e \in \Pic(X)[n]$.
This pairing has been constructed by Frey and R\"uck in \cite{FreyRuck}
when $k$ is a finite field
and shown to be perfect if $|k| \equiv 1 \bmod n$.
(This is called the
Lichtenbaum-Tate pairing in some literature,
see e.g. \cite[XI.9]{sil2}.)
\end{remark}

We will mainly consider the case where $X=E$ is an elliptic curve
so that we may identify $E(k)_\Tor= \Pic(E)_\Tor$
by $P \mapsto [P-O]$, where $O \in E(k)$ is the identity element.
Thus we write
$\gen{P,Q}:=\gen{[P-O], [Q-O]}$ for $P, Q \in E(k)_\Tor$
and
$E(k)_\Tor^\is:=\Pic(E)_\Tor^\is$.
Our main result is a classification of possible structures of
$(E(\Q)_\Tor,E(\Q)_\Tor^\is)$ for an elliptic curve $E$ over $\Q$.
A celebrated theorem of Mazur \cite[Theorem (8)]{Mazur} states
that $E(\Q)_\Tor$ is either cyclic of order
$1, \ldots,10$, or $12$, or is isomorphic to $(\Z/2N\Z) \times (\Z/2\Z)$,
$N=1, 2,3,4$,
and all cases are realized by infinitely many (mutually non-isomorphic) elliptic curves.
The classification of
$E(\Q)_\Tor^\is$ is given in the following theorem,
whose proof will occupy \S \ref{sect:ref-mazur}--\ref{sect:ref-mazur2}.

We briefly explain the outline of the proof.
Our approach is completely explicit.
As usual, we let $X_1(N)$ denote the modular curve associated to the
congruence subgroup $\Gamma_1(N)$. It possesses a model over $\Q$ on
which the cusp $0$ is $\Q$-rational. For a subfield $k$ of $\C$,
the non-cuspidal points in $X_1(N)(k)$ parameterize isomorphism
classes of pairs $(E, P)$ of an elliptic curve $E$ over $k$
and a $k$-rational point $P$ of order $N$.
More precisely, if the coordinates of a point
$\tau\in\H$ on this model of $X_1(N)$ is $k$-rational,
then the isomorphism class of $(\C/(\Z\tau+\Z),1/N)$ contains a pair $(E,P)$
of an elliptic curve $E$ over $k$ and a $k$-rational $N$-torsion point $P$.
A famous theorem of Mazur \cite[Theorem (7)]{Mazur} says that
$X_1(N)(\Q)$ has a non-cuspidal point precisely
when $N=1,\ldots,10$, or $N=12$.

Similarly, for a positive even
integer $N$, we let $X_1^0(N,2)$ denote the modular curve associated
to the congruence subgroup
\begin{equation}\label{eq:G10-N2}
\Gamma_1^0(N,2):=\Gamma_1(N)\cap\Gamma^0(2)
=\left\{\M abcd\in\SL(2,\Z):a,d\equiv 1\bmod N,~~2|b,~N|c\right\}.
\end{equation}
The modular curve $X_1^0(N,2)$ parameterizes isomorphism classes of
triples $(E, P, Q)$ of an elliptic curve $E$
and $k$-rational points $P, Q$
such that
$P$ and $Q$ are of order $N$ and $2$ respectively,
and $Q\notin\gen P$.
For $N \in \{4, 6, 8 \}$
and $\ul{M}=(M_1, M_2, M_3)$ with $M_1|N, M_2, M_3 \in \{ 1, 2\}$,
we will construct a smooth projective curve
$X_1^0(N, 2)_{\ul{M}}^\pm$ over $\Q$
equipped with a finite morphism to $X_1^0(N, 2)$,
by which one can interpret the conditions
\[ \langle P, \frac{N}{M_1}P \rangle
= \langle Q, \frac{2}{M_2}Q \rangle
= \langle P, \frac{2}{M_3}Q \rangle = 0
\]
for a triple $(E, P, Q)$ as above corresponding to a point of $X_1^0(N, 2)(\Q)$.
Therefore Theorem \ref{theorem: intrinsic torsion} is reduced to a study of the $\Q$-rational points of
$X_1(N)_M^\pm$ and $X_1^0(N, 2)_{\ul{M}}^\pm$.
As our construction of these curves are explicit,
this can be done by a (more or less) direct computation.

\begin{lemma} \label{lemma: ENt}
 \begin{enumerate}
 \item Assume that the characteristic of $k$ is not $2$.
 If $E$ is an elliptic curve over $k$ with a $k$-rational point
 $P$ of order two, then $(E,P)$ is isomorphic to $(E_{2,t,a},(0,0))$
 for some $a,t\in k^\times$, where
 \begin{equation} \label{eq: E2ta}
 E_{2,t,a}:
 \begin{cases}
 \displaystyle y^2=x\left(x^2+ax+\frac{a^2t}{4(t+1)}\right),
 &\text{if }t\neq-1, \\
 y^2=x(x^2+a), &\text{if }t=-1.
 \end{cases}
 \end{equation}
 Moreover, two elliptic curves $(E_{2,t,a},(0,0))$ and
 $(E_{2,t',a'},(0,0))$ are isomorphic over $k$ if and only
 if $t=t'\neq-1$ and $a/a'$ is a square in $k^\times$
 or $t=t'=-1$ and $a/a'\in(k^\times)^4$ .
 \item Assume that the characteristic of $k$ is not $3$.
 If $E$ is an elliptic curve over $k$ with a $k$-rational
 point $P$ of order $3$, then $(E,P)$ is isomorphic to $(E_{3,t},(0,0))$
 for some $t\neq-1$ in $k^\times$, where
 \begin{equation} \label{eq: E3t}
 E_{3,t}:y^2+xy+\frac t{27(t+1)}y=x^3
 \end{equation}
 or isomorphic to $(E_{3,-1,a},(0,0))$ for some $a\in k^\times$,
 where
 \begin{equation} \label{eq: E3a}
 E_{3,-1,a}:y^2+ay=x^3.
 \end{equation}
 Two elliptic curves $(E_{3,t},(0,0))$ and
 $(E_{3,t'},(0,0))$ with $t,t'\neq-1$ are isomorphic if and only
 if $t=t'$. Also, $(E_{3,-1,a},(0,0))\simeq(E_{3,-1,a'},(0,0))$
 if and only if $a/a'\in(k^\times)^3$.
 \end{enumerate}
\end{lemma}
\begin{proof}
 Consider first the case $N=2$.
We start with the equation \eqref{eq: E2}
and introduce a new parameter
 $t=4b/(a^2-4b)$ (so $b=a^2t/4(t+1)$). When $a\neq0$, we have
 $t\neq-1$ and the equation \eqref{eq: E2} in the new parameters is
 $y^2=x(x^2+ax+a^2t/4(t+1))$. When $a=0$, we change the notation $b$
 in \eqref{eq: E2} to $-a$ and get $y^2=x(x^2-a)$. Now it is easy to
 see that an isomorphism
 $\phi:(E_{2,t,a},(0,0))\to(E_{2,t',a'},(0,0))$ must be of the form
 $\phi(x,y)=(u^2x,u^3y)$ (c.f. \cite[Proposition 3.1]{sil}). From
 this, we immediately get the condition for $E_{2,t,a}$ and
 $E_{2,t',a'}$ to be isomorphic.

This leaves us with the cases
$(N, M) \in \{ (6,6), (8, 4), (9,3), (10, 2), (12, 2) \}$ with $\epsilon=-$.
(For $(N, M)=(9,3)$, the value of $\epsilon$ does not matter
because of Remark \ref{rem:plus-minus} (1).)
In all cases, there exists
an isomorphism $\alpha : E \to X_1(N)_M^\epsilon$,
where
$E$ is an elliptic curve over $\Q$
with finitely many $\Q$-rational points,
as exhibited in the following table,
where we set
$h_1(x)=(1-x)(1+x)^4(1+x-x^2)^2$ and
$h_2(x)=(1+x)(1+x+x^2)(1+x^2)^2(1-x)^3$.
 $$ \extrarowheight3pt
 \begin{array}{c|c|c|c} \hline\hline
 (N, M) & E & E(\Q) \setminus \{ O \} &
(s, t)=\alpha(x,y) \\ \hline
(6,6) & y^2=x^3-1 & (1,0) &
\left(-\frac{xy}{1-x^3},\frac1{1-x^3}\right)
\\
(8,4) & y^2=x^3+x & (0,0) &
\left(\frac y{x^2}(1-\frac{1}{x^2}),-\frac1{x^2}\right)
\\
(9,3) & y^2+y=x^3 & (0,0), (0,-1) &
\left(xy(y^2+y+1),y+1\right)
\\
(10,2) & y^2=x^3-x^2-x & (0,0) &
\left(yh_1(x),x\right)
\\
(12,2) & y^2=x^3+x^2+x & (0,0) &
\left(yh_2(x),x\right)
\\
 \hline\hline
 \end{array}
 $$
Since all points of $\alpha(E(\Q))$ are cusps
by Lemma \ref{lemma: uniformizers},
we get $Y_1(\Q)^\epsilon_M=\emptyset$ in these cases as well.
We are done.
\end{proof}

\begin{corollary}\label{cor:kodairatype}
Let $E$ be an elliptic curve over a number field $k$,
and $P \in E(k)_\Tor^\is$ an element of the intrinsic subgroup of order $m>0$.
Let $v$ be a finite place of $k$
such that the completion $k_v$ of $k$ at $v$ satisfies $\mu_m(k_v)= \{ 1 \}$.
If $E$ has split multiplicative reduction of Kodaira type $I_n$ at $v$,
then $n$ is divisible by $m^2$.
\end{corollary}
\begin{proof}
The assumption implies that
the base change of $E$ to $k_v$ is
isomorphic to the Tate curve $E_q=\G_m/q^\Z$
for some $q \in k_v$ such that $v(q)=n$
(\cite[Chapter V, Theorem 5.3]{sil})
Let us write the image of $P$ in $E_q(k_v)$ as $[a]$ with $a \in k_v$.
Since $\mu_m(k_v)= \{ 1 \}$,
we find from \eqref{eq:ex-TateCv} that
$E_q(k_v)[m]$ is a cyclic group of order $m$ generated by $[a]$.
In particular, $s:= s_m([a])$ is invertible in $\Z/m\Z$.
We have $\langle [a], [a] \rangle = 0$ since $P \in E(k)_\Tor^\is$.
On the other hand,
Proposition \ref{prop:TateCv} (1) shows
$\langle [a], [a] \rangle = q^{-s^2} \otimes (1/m^2)$.
By looking at its image under the map
$v \otimes \id_{\Q/\Z} : k_v^\times \otimes \Q/\Z \to \Q/\Z$,
we conclude that $-s^2 n/m^2 = 0$ holds in $\Q/\Z$,
that is, $m^2$ divides $n$.
\end{proof}

We briefly explain the outline of the proof.
Our approach is completely explicit.
As usual, we let $X_1(N)$ denote the modular curve associated to the
congruence subgroup $\Gamma_1(N)$. It possesses a model over $\mathbb{Q}$ on
which the cusp $0$ is $\mathbb{Q}$-rational. For a subfield $k$ of $\mathbb{C}$,
the non-cuspidal points in $X_1(N)(k)$ parameterize isomorphism
classes of pairs $(E, P)$ of an elliptic curve $E$ over $k$
and a $k$-rational point $P$ of order $N$.
More precisely, if the coordinates of a point
$\tau\in\mathbb H$ on this model of $X_1(N)$ is $k$-rational,
then the isomorphism class of $(\mathbb{C}/(\mathbb{Z}\tau+\mathbb{Z}),1/N)$ contains a pair $(E,P)$
of an elliptic curve $E$ over $k$ and a $k$-rational $N$-torsion point $P$.
A famous theorem of Mazur \cite[Theorem (7)]{Mazur} says that
$X_1(N)(\mathbb{Q})$ has a non-cuspidal point precisely
when $N=1,\ldots,10$, or $N=12$.

For a pair $(N, M)$ of $N \in \{4, \dots, 10, 12\}$ and a positive divisor $M$ of $N$,
we will construct a smooth projective curve $X_1 (N)^\pm_M$ over $\mathbb{Q}$ equipped with
a finite morphism $X_1(N)_M^\pm \to X_1(N)$
that enjoys the following property:
when $t \in X_1(N)(\mathbb{Q})$ corresponds to a pair $(E, P)$ as above (with $k=\mathbb{Q}$),
one has
\[
\langle P, (N/M)P \rangle = 0 \Leftrightarrow
t \in \operatorname{Im}(X_1(N)^\pm_M(\mathbb{Q}) \to X_1(N)(\mathbb{Q})).
\]
It is worth  mentioning that
$X_1(N)^+_M$ is isomorphic to the modular curve
$X_1(MN, N)$ associated to $\Gamma_1(MN, N):=\Gamma_0(MN) \cap \Gamma_1(N)$
(Theorem \ref{proposition: pairing 1-mod})
and $X_1(N)^-_M$ is its twist.

Similarly, for a positive even
integer $N$, we let $X_1^0(N,2)$ denote the modular curve associated
to the congruence subgroup
\begin{equation}\label{eq:G10-N2}
\Gamma_1^0(N,2):=\Gamma_1(N)\cap\Gamma^0(2)
=\left\{\M abcd\in\SL(2,\mathbb{Z}):a,d\equiv 1\bmod N,~~2|b,~N|c\right\}.
\end{equation}
The modular curve $X_1^0(N,2)$ parameterizes isomorphism classes of
triples $(E, P, Q)$ of an elliptic curve $E$
and $k$-rational points $P, Q$
such that
$P$ and $Q$ are of order $N$ and $2$ respectively,
and $Q\notin\gen P$.
For $N \in \{4, 6, 8 \}$
and $\underline{M}=(M_1, M_2, M_3)$ with $M_1|N, M_2, M_3 \in \{ 1, 2\}$,
we will construct a smooth projective curve
$X_1^0(N, 2)_{\underline{M}}^\pm$ over $\mathbb{Q}$
equipped with a finite morphism to $X_1^0(N, 2)$,
by which one can interpret the conditions
\[ \langle P, \frac{N}{M_1}P \rangle
= \langle Q, \frac{2}{M_2}Q \rangle
= \langle P, \frac{2}{M_3}Q \rangle = 0
\]
for a triple $(E, P, Q)$ as above corresponding to a point of $X_1^0(N, 2)(\mathbb{Q})$.
Therefore Theorem \ref{theorem: intrinsic torsion} is reduced to a study of the $\mathbb{Q}$-rational points of
$X_1(N)_M^\pm$ and $X_1^0(N, 2)_{\underline{M}}^\pm$.
As our construction of these curves are explicit,
this can be done by a (more or less) direct computation.

In the last section \S \ref{sect:high-dim},
we generalize the pairing \eqref{eq:pairing-intro}
and the intrinsic subgroup
to higher dimensional varieties.
This new construction enables us to prove that,
if $X$ has good reduction with respect to a discrete valuation of $k$,
there is a strong restriction on the values of the pairing
(see Proposition \ref{prop:two-def-same}).
Finally, as a sample for the case of bad reduction,
we compute the intrinsic subgroup of
Tate elliptic curves (see Proposition \ref{prop:TateCv}).
As an application,
we show that the intrinsic subgroup of an elliptic curve over a number field
imposes some constraint on its reduction type
(see Corollary \ref{cor:kodairatype}).

\subsection*{Acknowledgement}
It is Kenneth Ribet who formulated \eqref{eq:pairing-intro} as a pairing
and asked if it is symmetric.
The authors express deep gratitude to him for asking this question,
which was a starting point of our work.

The first author (T. Y.) is supported by JSPS KAKENHI Grant (25K06961).
The second author (Y. Y.) is supported by
Grant 113-2115-M-002-003-MY3 of the National Science and Technology
  Council of the Republic of China (Taiwan).
The third author (H. Y.) is supported by National Research Foundation of Korea(NRF) grant
funded by the Korea government(MSIT) (No. RS-2023-00239918  and No. 2020R1A5A1016126).
The fourth author (M. Y.) is supported by the National Research Foundation of Korea (NRF) grant
funded by the
Korea government (MSIT) (RS-2025-23525445).

\subsection*{Notation and convention}
For an abelian group $A$,
we write $A[n]$ and $A/n$ for the kernel and cokernel
of $A \to A, \ a \mapsto na$ for each $n \in \mathbb{Z}$,
and put $A_\Tor := \cup_{n > 0} A[n]$.
For a field $k$,
we set $\mu_n(k):=(k^\times)[n]$ and $\mu(k):=(k^\times)_\Tor$.

The identity element of an elliptic curve is denoted by $O$.

The authors say: “We briefly explain the outline of the proof. Our approach is completely explicit.” They use modular curves $X_1(N)$: non-cuspidal points in $X_1(N)(k)$ “parameterize isomorphism classes of pairs $(E,P)$ of an elliptic curve $E$ over $k$ and a $k$-rational point $P$ of order $N$.” Using Mazur’s theorem that $X_1(N)(\mathbb{Q})$ has a non-cuspidal point precisely for $N=1,\ldots,10$ or $N=12$, they restrict to these torsion orders.

For each $(N,M)$ with $N\in\{4,\dots,10,12\}$ and $M\mid N$, they “construct a smooth projective curve $X_1(N)^\pm_M$ over $\mathbb{Q}$ equipped with a finite morphism $X_1(N)_M^\pm\to X_1(N)$” such that for $t\in X_1(N)(\mathbb{Q})$ corresponding to $(E,P)$ one has
\[
\langle P,(N/M)P\rangle=0\Longleftrightarrow t\in\operatorname{Im}\bigl(X_1(N)^\pm_M(\mathbb{Q})\to X_1(N)(\mathbb{Q})\bigr).
\]
They note $X_1(N)^+_M$ is isomorphic to a modular curve $X_1(MN,N)$ and $X_1(N)^-_M$ is its twist.

For even $N$, they similarly use the modular curve $X_1^0(N,2)$, which “parameterizes isomorphism classes of triples $(E,P,Q)$” with $P$ of order $N$, $Q$ of order $2$, and $Q\notin\langle P\rangle$. For $N\in\{4,6,8\}$ and $\underline{M}=(M_1,M_2,M_3)$ with $M_1\mid N$ and $M_2,M_3\in\{1,2\}$, they construct curves $X_1^0(N,2)^{\pm}_{\underline{M}}$ with finite morphisms to $X_1^0(N,2)$ “by which one can interpret the conditions”
\[
\langle P,\tfrac{N}{M_1}P\rangle=\langle Q,\tfrac{2}{M_2}Q\rangle=\langle P,\tfrac{2}{M_3}Q\rangle=0.
\]
They conclude that “Therefore Theorem~\ref{theorem: intrinsic torsion} is reduced to a study of the $\mathbb{Q}$-rational points of $X_1(N)_M^\pm$ and $X_1^0(N,2)_{\underline{M}}^\pm$,” and since “our construction of these curves are explicit, this can be done by a (more or less) direct computation.”

The authors say: “We briefly explain the outline of the proof. Our approach is completely explicit.” They use modular curves $X_1(N)$: non-cuspidal points in $X_1(N)(k)$ “parameterize isomorphism classes of pairs $(E,P)$ of an elliptic curve $E$ over $k$ and a $k$-rational point $P$ of order $N$.” Using Mazur’s theorem that $X_1(N)(\mathbb{Q})$ has a non-cuspidal point precisely for $N=1,\ldots,10$ or $N=12$, they restrict to these torsion orders.

For each $(N,M)$ with $N\in\{4,\dots,10,12\}$ and $M\mid N$, they “construct a smooth projective curve $X_1(N)^\pm_M$ over $\mathbb{Q}$ equipped with a finite morphism $X_1(N)_M^\pm\to X_1(N)$” such that for $t\in X_1(N)(\mathbb{Q})$ corresponding to $(E,P)$ one has
\[
\langle P,(N/M)P\rangle=0\Longleftrightarrow t\in\operatorname{Im}\bigl(X_1(N)^\pm_M(\mathbb{Q})\to X_1(N)(\mathbb{Q})\bigr).
\]
They note $X_1(N)^+_M$ is isomorphic to a modular curve $X_1(MN,N)$ and $X_1(N)^-_M$ is its twist.

For even $N$, they similarly use the modular curve $X_1^0(N,2)$, which “parameterizes isomorphism classes of triples $(E,P,Q)$” with $P$ of order $N$, $Q$ of order $2$, and $Q\notin\langle P\rangle$. For $N\in\{4,6,8\}$ and $\underline{M}=(M_1,M_2,M_3)$ with $M_1\mid N$ and $M_2,M_3\in\{1,2\}$, they construct curves $X_1^0(N,2)^{\pm}_{\underline{M}}$ with finite morphisms to $X_1^0(N,2)$ “by which one can interpret the conditions”
\[
\langle P,\tfrac{N}{M_1}P\rangle=\langle Q,\tfrac{2}{M_2}Q\rangle=\langle P,\tfrac{2}{M_3}Q\rangle=0.
\]
They conclude that, in establishing the main theorem, everything is reduced to a study of the $\mathbb{Q}$-rational points of $X_1(N)_M^\pm$ and $X_1^0(N,2)_{\underline{M}}^\pm$, and since “our construction of these curves are explicit, this can be done by a (more or less) direct computation.”

\label{theorem: intrinsic torsion}
  Let $E$ be an elliptic curve over $\mathbb{Q}$.
Then there is a pair $(A, B)$ of an abelian group $A$
and its subgroup $B$ in the following table,
together with
an isomorphism $\alpha : E(\mathbb{Q})_\Tor \cong A$,
such that $\alpha(E(\mathbb{Q})_\Tor^\is)=B$.
  $$ \extrarowheight3pt
  \begin{array}{ll} \hline\hline
    A & \text{order of}\ B \\ \hline
    0 & 1 \\
    \mathbb{Z}/2\mathbb{Z} & 1,2\\
    \mathbb{Z}/3\mathbb{Z} & 1,3\\
    \mathbb{Z}/4\mathbb{Z} & 1,2,4 \\
    \mathbb{Z}/5\mathbb{Z} & 1,5 \\
    \mathbb{Z}/6\mathbb{Z} & 1,2,3 \\
    \mathbb{Z}/7\mathbb{Z} & 1 \\
    \mathbb{Z}/8\mathbb{Z} & 1,2 \\
    \mathbb{Z}/9\mathbb{Z} & 1 \\
    \mathbb{Z}/10\mathbb{Z} & 1\\
    \mathbb{Z}/12\mathbb{Z} & 1 \\
    \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z} & 1,2\\
    \mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z} & 1,2 \ \\
    \mathbb{Z}/6\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z} & 1 \\
    \mathbb{Z}/8\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z} & 1 \\ \hline\hline
  \end{array}
  $$
(For $A=\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$,
there are two $\operatorname{Aut}(A)$-conjugacy classes of order two subgroups,
both of which can be taken as $B$.)
  Moreover, for each possible pair $(A, B)$ as above,
there are infinitely many (mutually non-isomorphic) elliptic curves $E$ over $\mathbb{Q}$
for which there is an isomorphism $\alpha : E(\mathbb{Q})_\Tor \cong A$
such that $\alpha(E(\mathbb{Q})_\Tor^\is)=B$.,

Question:
Let $E$ be an elliptic curve over $\mathbb{Q}$ with identity element $O$. Via the identification $P \mapsto [P-O]$, identify $E(\mathbb{Q})_{\mathrm{Tor}}$ with $\operatorname{Pic}(E)_{\mathrm{Tor}}$. Let
\[
\langle\cdot,\cdot\rangle: E(\mathbb{Q})_{\mathrm{Tor}}\times E(\mathbb{Q})_{\mathrm{Tor}}\to \mathbb{Q}^\times\otimes \mathbb{Q}/\mathbb{Z}
\]
be the induced biadditive symmetric pairing, and define the intrinsic subgroup by
\[
E(\mathbb{Q})_{\mathrm{Tor}}^{\mathrm{is}}:=\{P\in E(\mathbb{Q})_{\mathrm{Tor}}\mid \langle P,Q\rangle=0\text{ for all }Q\in E(\mathbb{Q})_{\mathrm{Tor}}\}.
\]
Which existence statement holds for the pair consisting of the torsion subgroup $E(\mathbb{Q})_{\mathrm{Tor}}$ and its intrinsic subgroup $E(\mathbb{Q})_{\mathrm{Tor}}^{\mathrm{is}}$?

For every elliptic curve $E/\mathbb{Q}$, there exist an abelian group $A$ and a subgroup $B\le A$, together with an isomorphism $\alpha:E(\mathbb{Q})_{\mathrm{Tor}}\xrightarrow{\sim} A$, such that $\alpha(E(\mathbb{Q})_{\mathrm{Tor}}^{\mathrm{is}})=B$, where $A$ and the possible orders of $B$ are exactly as follows: $A=0$ with $|B|=1$; $A=\mathbb{Z}/2\mathbb{Z}$ with $|B|=1$ or $2$; $A=\mathbb{Z}/3\mathbb{Z}$ with $|B|=1$ or $3$; $A=\mathbb{Z}/4\mathbb{Z}$ with $|B|=1,2,$ or $4$; $A=\mathbb{Z}/5\mathbb{Z}$ with $|B|=1$ or $5$; $A=\mathbb{Z}/6\mathbb{Z}$ with $|B|=1,2,$ or $3$; $A=\mathbb{Z}/7\mathbb{Z}$ with $|B|=1$; $A=\mathbb{Z}/8\mathbb{Z}$ with $|B|=1$ or $2$; $A=\mathbb{Z}/9\mathbb{Z}$ with $|B|=1$; $A=\mathbb{Z}/10\mathbb{Z}$ with $|B|=1$; $A=\mathbb{Z}/12\mathbb{Z}$ with $|B|=1$; $A=\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ with $|B|=1$ or $2$; $A=\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ with $|B|=1$ or $2$; $A=\mathbb{Z}/6\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ with $|B|=1$; and $A=\mathbb{Z}/8\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ with $|B|=1$. For $A=\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$, there are two $\operatorname{Aut}(A)$-conjugacy classes of subgroups of order $2$, and either class can occur as $B$. Moreover, for every such possible pair $(A,B)$, there exist infinitely many mutually non-isomorphic elliptic curves $E/\mathbb{Q}$ for which some isomorphism $\alpha:E(\mathbb{Q})_{\mathrm{Tor}}\xrightarrow{\sim}A$ satisfies $\alpha(E(\mathbb{Q})_{\mathrm{Tor}}^{\mathrm{is}})=B$.

Choices:
B. For every elliptic curve $E/\mathbb{Q}$, there exist an abelian group $A$ and a subgroup $B\le A$, together with an isomorphism $\alpha:E(\mathbb{Q})_{\mathrm{Tor}}\xrightarrow{\sim} A$, such that $\alpha(E(\mathbb{Q})_{\mathrm{Tor}}^{\mathrm{is}})=B$, where $A$ and the possible orders of $B$ are exactly as follows: $A=0$ with $|B|=1$; $A=\mathbb{Z}/2\mathbb{Z}$ with $|B|=1$ or $2$; $A=\mathbb{Z}/3\mathbb{Z}$ with $|B|=1$ or $3$; $A=\mathbb{Z}/4\mathbb{Z}$ with $|B|=1,2,$ or $4$; $A=\mathbb{Z}/5\mathbb{Z}$ with $|B|=1$ or $5$; $A=\mathbb{Z}/6\mathbb{Z}$ with $|B|=1,2,$ or $3$; $A=\mathbb{Z}/7\mathbb{Z}$ with $|B|=1$; $A=\mathbb{Z}/8\mathbb{Z}$ with $|B|=1$ or $2$; $A=\mathbb{Z}/9\mathbb{Z}$ with $|B|=1$ or $3$; $A=\mathbb{Z}/10\mathbb{Z}$ with $|B|=1$; $A=\mathbb{Z}/12\mathbb{Z}$ with $|B|=1$; $A=\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ with $|B|=1$ or $2$; $A=\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ with $|B|=1$ or $2$; $A=\mathbb{Z}/6\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ with $|B|=1$; and $A=\mathbb{Z}/8\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ with $|B|=1$. For $A=\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$, there are two $\operatorname{Aut}(A)$-conjugacy classes of subgroups of order $2$, and either class can occur as $B$. Moreover, for every such possible pair $(A,B)$, there exists at least one elliptic curve $E/\mathbb{Q}$ for which some isomorphism $\alpha:E(\mathbb{Q})_{\mathrm{Tor}}\xrightarrow{\sim}A$ satisfies $\alpha(E(\mathbb{Q})_{\mathrm{Tor}}^{\mathrm{is}})=B$.
C. For every elliptic curve $E/\mathbb{Q}$, there exist an abelian group $A$ and a subgroup $B\le A$, together with an isomorphism $\alpha:E(\mathbb{Q})_{\mathrm{Tor}}\xrightarrow{\sim} A$, such that $\alpha(E(\mathbb{Q})_{\mathrm{Tor}}^{\mathrm{is}})=B$, where $A$ and the possible orders of $B$ are among the following: $A=0$ with $|B|=1$; $A=\mathbb{Z}/2\mathbb{Z}$ with $|B|=1$ or $2$; $A=\mathbb{Z}/3\mathbb{Z}$ with $|B|=1$ or $3$; $A=\mathbb{Z}/4\mathbb{Z}$ with $|B|=1,2,$ or $4$; $A=\mathbb{Z}/5\mathbb{Z}$ with $|B|=1$ or $5$; $A=\mathbb{Z}/6\mathbb{Z}$ with $|B|=1,2,$ or $3$; $A=\mathbb{Z}/7\mathbb{Z}$ with $|B|=1$; $A=\mathbb{Z}/8\mathbb{Z}$ with $|B|=1$ or $2$; $A=\mathbb{Z}/9\mathbb{Z}$ with $|B|=1$; $A=\mathbb{Z}/10\mathbb{Z}$ with $|B|=1$; $A=\mathbb{Z}/12\mathbb{Z}$ with $|B|=1$; $A=\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ with $|B|=1$ or $2$; $A=\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ with $|B|=1$ or $2$; $A=\mathbb{Z}/6\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ with $|B|=1$; and $A=\mathbb{Z}/8\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ with $|B|=1$.
D. For every elliptic curve $E/\mathbb{Q}$, there exist an abelian group $A$ and a subgroup $B\le A$, together with an isomorphism $\alpha:E(\mathbb{Q})_{\mathrm{Tor}}\xrightarrow{\sim} A$, such that $\alpha(E(\mathbb{Q})_{\mathrm{Tor}}^{\mathrm{is}})=B$, where $A$ and the possible orders of $B$ are exactly as follows: $A=0$ with $|B|=1$; $A=\mathbb{Z}/2\mathbb{Z}$ with $|B|=1$ or $2$; $A=\mathbb{Z}/3\mathbb{Z}$ with $|B|=1$ or $3$; $A=\mathbb{Z}/4\mathbb{Z}$ with $|B|=1,2,$ or $4$; $A=\mathbb{Z}/5\mathbb{Z}$ with $|B|=1$ or $5$; $A=\mathbb{Z}/6\mathbb{Z}$ with $|B|=1,2,$ or $3$; $A=\mathbb{Z}/7\mathbb{Z}$ with $|B|=1$; $A=\mathbb{Z}/8\mathbb{Z}$ with $|B|=1$ or $2$; $A=\mathbb{Z}/9\mathbb{Z}$ with $|B|=1$; $A=\mathbb{Z}/10\mathbb{Z}$ with $|B|=1$; $A=\mathbb{Z}/12\mathbb{Z}$ with $|B|=1$; $A=\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ with $|B|=1$ or $2$; $A=\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ with $|B|=1,2,$ or $4$; $A=\mathbb{Z}/6\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ with $|B|=1$; and $A=\mathbb{Z}/8\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ with $|B|=1$. For $A=\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$, there are two $\operatorname{Aut}(A)$-conjugacy classes of subgroups of order $2$, and either class can occur as $B$. Moreover, for every such possible pair $(A,B)$, there exist infinitely many mutually non-isomorphic elliptic curves $E/\mathbb{Q}$ for which some isomorphism $\alpha:E(\mathbb{Q})_{\mathrm{Tor}}\xrightarrow{\sim}A$ satisfies $\alpha(E(\mathbb{Q})_{\mathrm{Tor}}^{\mathrm{is}})=B$.
E. For every elliptic curve $E/\mathbb{Q}$, there exist an abelian group $A$ and a subgroup $B\le A$, together with an isomorphism $\alpha:E(\mathbb{Q})_{\mathrm{Tor}}\xrightarrow{\sim} A$, such that $\alpha(E(\mathbb{Q})_{\mathrm{Tor}}^{\mathrm{is}})=B$, where $A$ and the possible orders of $B$ are exactly as follows: $A=0$ with $|B|=1$; $A=\mathbb{Z}/2\mathbb{Z}$ with $|B|=1$ or $2$; $A=\mathbb{Z}/3\mathbb{Z}$ with $|B|=1$ or $3$; $A=\mathbb{Z}/4\mathbb{Z}$ with $|B|=1,2,$ or $4$; $A=\mathbb{Z}/5\mathbb{Z}$ with $|B|=1$ or $5$; $A=\mathbb{Z}/6\mathbb{Z}$ with $|B|=1,2,$ or $3$; $A=\mathbb{Z}/7\mathbb{Z}$ with $|B|=1$; $A=\mathbb{Z}/8\mathbb{Z}$ with $|B|=1$ or $2$; $A=\mathbb{Z}/9\mathbb{Z}$ with $|B|=1$; $A=\mathbb{Z}/10\mathbb{Z}$ with $|B|=1$; $A=\mathbb{Z}/12\mathbb{Z}$ with $|B|=1$; $A=\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ with $|B|=1$ or $2$; $A=\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ with $|B|=1$ or $2$; $A=\mathbb{Z}/6\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ with $|B|=1$; and $A=\mathbb{Z}/8\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ with $|B|=1$. For $A=\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$, there are two $\operatorname{Aut}(A)$-conjugacy classes of subgroups of order $2$, and both classes occur simultaneously for every elliptic curve $E/\mathbb{Q}$ with $E(\mathbb{Q})_{\mathrm{Tor}}\cong A$. Moreover, for every such possible pair $(A,B)$, there exist infinitely many mutually non-isomorphic elliptic curves $E/\mathbb{Q}$ for which some isomorphism $\alpha:E(\mathbb{Q})_{\mathrm{Tor}}\xrightarrow{\sim}A$ satisfies $\alpha(E(\mathbb{Q})_{\mathrm{Tor}}^{\mathrm{is}})=B$.


Approximation theorems (Runge \cite{Run}, Mergelyan \cite{Mer}, Arakelian \cite{Ar}) play a central role in the field of complex approximation, since they give necessary and sufficient conditions for functions to be uniformly approximated on a given set by model functions, such as polynomials, rational functions or more generally entire/meromorphic functions. Our goal in the present paper is to give a new characterization of closed sets $F\subset \mathbb{C}$ in the complex plane, for which uniform approximation by entire functions is possible.

To this end, define the classes
\begin{align*}
H(F)=&\,\lbrace f:U\to\mathbb{C} \text{ holomorphic for some open set $U\supset F$}\rbrace,\\
 A(F)=&\,\lbrace f:F\to\mathbb{C} \text{ holomorphic in }F^\circ\text{ and continuous in F}\rbrace,
\end{align*}
where $F^\circ$ is the interior of $F$.
\begin{dfn}\label{def:hole}
Any bounded connected component $B$ of $\mathbb{C}\setminus F$ is called a hole of $F$.
\end{dfn}
Arakelian's theorem \cite{Ar} gives necessary and sufficient conditions such that every function $f\in A(F)$ can be uniformly approximated on $F$ by entire functions..
\begin{dfn}[Arakelian set]\label{def:Arak}
Let $F\subset \mathbb{C}$ be a closed set. We say that $F$ is an Arakelian set, if and only if the following points hold:\\[5pt]
1. $F$ is without holes;\\[5pt]
2. The union of all holes of $F\cup K$ is bounded, for any compact subset $K\subset \mathbb{C}$.
\end{dfn}
\begin{rmk}\label{rem:K}
Point 2. in Definition \ref{def:Arak} is equivalent to the one where $K$ is replaced by $\overline{D}_0(n)=\{z\in\mathbb{C}: |z|\leq n\}$, for any $n\in\mathbb{N}$.
\end{rmk}
\begin{rmk}
In fact, Definition \ref{def:Arak} appears in a later work \cite{RR}, where a new proof of Arakelian's theorem was given (see also \cite{Got}). The original conditions in \cite{Ar} are:\\[5pt]
1. $\mathbb{C}\setminus F$ is connected;\\[5pt]
2. $\mathbb{C}\setminus F$ is locally connected at infinity.\\[5pt]
Nevertheless, they are easily seen to be equivalent.
\end{rmk}

\begin{dfn}[Uniform approximation set]
Let $F\subset \mathbb{C}$ be a closed set. We say that $F$ is a set of uniform approximation, if for every $\epsilon>0$ and $f\in A(F)$, there exists an entire function $g$ such that $|f(z)-g(z)|<\epsilon$, for all $z\in F$.
\end{dfn}

\begin{abstract}
Arakelian's classical approximation theorem \cite{Ar} gives necessary and sufficient conditions such that functions can be uniformly approximated in (unbounded) closed sets $F\subset \mathbb{C}$ by entire functions. The conditions are purely topological and concern the connectedness of the complement of $F$. We give a new characterization of Arakelian sets in terms of logarithmic branches of functions $f\in A(F)$, which are continuous in $F$ and holomorphic in its interior $F^\circ$. Our proof is based on a contradiction argument and the counterexample function that we use is furnished by the Weierstrass factorization theorem.
\end{abstract}

Approximation theorems (Runge \cite{Run}, Mergelyan \cite{Mer}, Arakelian \cite{Ar}) play a central role in the field of complex approximation, since they give necessary and sufficient conditions for functions to be uniformly approximated on a given set by model functions, such as polynomials, rational functions or more generally entire/meromorphic functions. Our goal in the present paper is to give a new characterization of closed sets $F\subset \mathbb{C}$ in the complex plane, for which uniform approximation by entire functions is possible.

To this end, define the classes
\begin{align*}
H(F)=&\,\lbrace f:U\to\mathbb{C} \text{ holomorphic for some open set $U\supset F$}\rbrace,\\
 A(F)=&\,\lbrace f:F\to\mathbb{C} \text{ holomorphic in }F^\circ\text{ and continuous in F}\rbrace,
\end{align*}
where $F^\circ$ is the interior of $F$.
\begin{dfn}\label{def:hole}
Any bounded connected component $B$ of $\mathbb{C}\setminus F$ is called a hole of $F$.
\end{dfn}
Arakelian's theorem \cite{Ar} gives necessary and sufficient conditions such that every function $f\in A(F)$ can be uniformly approximated on $F$ by entire functions..
\begin{dfn}[Arakelian set]\label{def:Arak}
Let $F\subset \mathbb{C}$ be a closed set. We say that $F$ is an Arakelian set, if and only if the following points hold:\\[5pt]
1. $F$ is without holes;\\[5pt]
2. The union of all holes of $F\cup K$ is bounded, for any compact subset $K\subset \mathbb{C}$.
\end{dfn}
\begin{rmk}\label{rem:K}
Point 2. in Definition \ref{def:Arak} is equivalent to the one where $K$ is replaced by $\overline{D}_0(n)=\{z\in\mathbb{C}: |z|\leq n\}$, for any $n\in\mathbb{N}$.
\end{rmk}
\begin{rmk}
In fact, Definition \ref{def:Arak} appears in a later work \cite{RR}, where a new proof of Arakelian's theorem was given (see also \cite{Got}). The original conditions in \cite{Ar} are:\\[5pt]
1. $\mathbb{C}\setminus F$ is connected;\\[5pt]
2. $\mathbb{C}\setminus F$ is locally connected at infinity.\\[5pt]
Nevertheless, they are easily seen to be equivalent.
\end{rmk}

\begin{dfn}[Uniform approximation set]
Let $F\subset \mathbb{C}$ be a closed set. We say that $F$ is a set of uniform approximation, if for every $\epsilon>0$ and $f\in A(F)$, there exists an entire function $g$ such that $|f(z)-g(z)|<\epsilon$, for all $z\in F$.
\end{dfn}

\subsection{Main result}

We can now state the main result:
\begin{thm}\label{thm:main}
A closed set $F\subset \mathbb{C}$ is an Arakelian set, if and only if
for every $f\in A(F)$ with no zeros in $F$, there exists $g\in A(F)$ such that $e^g=f$ in $F$.
\end{thm}
At first glance, the existence of logarithmic branches of functions $f\in A(F)$ might seem to have little to do with the conditions in Definition \ref{def:Arak}. Intuitively, it is easier to make the connection by first stating yet another characterization of Arakelian sets, proved in \cite{F}, in terms of simply connected neighborhoods of $F$, that is to say, open sets $V\supset F$ whose connected components are simply connected.
\begin{thm}\label{thm:F}
A closed set $F\subset \mathbb{C}$ is an Arakelian set if and only if it possesses a neighborhood basis of simply connected open sets.
\end{thm}
At least one of the directions in Theorem \ref{thm:main} is clearer now. If a function $f\in A(F)$ could be extended to a neighborhood of $F$, then the existence of an intermediate simply connected neighborhood would allow for a logarithmic branch of $f$ to be well-defined in the latter.

Theorem \ref{thm:Ar} has been successfully extended to relatively closed subsets of planar domains $G\subset \mathbb{C}$ \cite{Ar2} and non-planar Riemann surfaces of finite genus \cite{GH,Sch}. One of the outstanding open problems in complex approximation is to characterize sets of uniform approximation in Riemann surfaces of infinite genus, where Arakelian sets (Definition \ref{def:Arak}) are not always sets of uniform approximation, see \cite{BG} for a counterexample. Moreover, it was shown by Scheinberg \cite{Sch} that such a characterization, if it exists, cannot be purely topological, but it has to take into account the complex analytic structure of the given Riemann surface.

The one direction of the characterization in Theorem \ref{thm:main} is contained in \cite{GP}. We include the proof here for the sake of completeness.
\begin{prp}[Gauthier-Pouryayevali \cite{GP}]\label{prop:GP}
Suppose that the closed set $F\subset \mathbb{C}$ is an Arakelian set. For every function $f\in A(F)$ with no zeros in $F$, there exists a function $g\in A(F)$ such that $e^g=f$ in $F$.
\end{prp}
\begin{proof}
By Tietze's extension theorem, there exists a continuous extension of $f$ in $\mathbb{C}$, which we
denote by $\widetilde{f}:\mathbb{C}\to\C$. Our assumption implies that the open set $U=\mathbb{C}\setminus \widetilde{f}^{-1}(\{0\})$ contains $F$.
Thus, by Theorem \ref{thm:F} there exists a simply connected open set $V$ with $F\subset V\subset U$.
If we consider the covering map $\exp: \mathbb{C}\to\mathbb{C}\setminus\{0\}$, then the latter implies that
$\widetilde{f}_{\big|V}:V\to\mathbb{C}\setminus\{0\}$ can be lifted
to a continuous function $\widetilde{g}:V\to\mathbb{C}$, such that $\widetilde{f}_{\big|V}=e^{\widetilde{g}}$.
The function $g=\widetilde{g}_{\big|F}$ is obviously continuous and $e^g=f$ in $F$.
Since $f_{\big|F^0}$ is holomorphic, by Lemma \ref{lem:Car}, $g$ is also holomorphic in $F^0$.
Thus, $g\in A(F)$ and the proof of the proposition is complete.
\end{proof}
The reverse direction is contained in the following proposition that, combined with Proposition \ref{prop:GP}, completes the proof of Theorem \ref{thm:main}.
\begin{prp}\label{prop:FNP}
Let $F\subset\mathbb{C}$ be a closed set, such that for every $f\in A(F)$ with no zeros in $F$, there exists a $g\in A(F)$ satisfying $e^g=f$ in $F$. Then $F$ is an Arakelian set.
\end{prp}
\begin{proof}
{\it Step 1. $F$ has no holes.} Arguing by contradiction, suppose that $F$ has a hole $B$. Let $\zeta\in B$ and let $f(z)=z-\zeta$. Since $f$ has no zeros in $F$, there exists $g\in A(F)$, such that $e^g=f$ in $F$.

\begin{thm}[Arakelian \cite{Ar}]\label{thm:Ar}
A closed set $F\subset \mathbb{C}$ is a set of uniform approximation if and only if $F$ is an Arakelian set.
\end{thm}

\begin{thm}\label{thm:main}
A closed set $F\subset \mathbb{C}$ is an Arakelian set, if and only if
for  every $f\in A(F)$ with no zeros in $F$, there exists $g\in A(F)$ such that $e^g=f$ in $F$.
\end{thm}

\subsection{Main result}

We can now state the main result:
\begin{thm}\label{thm:main}
A closed set $F\subset \mathbb{C}$ is an Arakelian set, if and only if
for  every $f\in A(F)$ with no zeros in $F$, there exists $g\in A(F)$ such that $e^g=f$ in $F$.
\end{thm}
At first glance, the existence of logarithmic branches of functions $f\in A(F)$ might seem to have little to do with the conditions in Definition \ref{def:Arak}. Intuitively, it is easier to make the connection by first stating yet another characterization of Arakelian sets, proved in \cite{F}, in terms of simply connected neighborhoods of $F$, that is to say, open sets $V\supset F$ whose connected components are simply connected.
\begin{thm}\label{thm:F}
A closed set $F\subset \mathbb{C}$ is an Arakelian set if and only if it possesses a neighborhood basis of simply connected open sets.
\end{thm}
At least one of the directions in Theorem \ref{thm:main} is clearer now. If a function $f\in A(F)$ could be extended to a neighborhood of $F$, then the existence of an intermediate simply connected neighborhood would allow for a logarithmic branch of $f$ to be well-defined in the latter.

We should give credit here to the influential work of Gauthier-Pouryayevali \cite{GP}, which inspired us to pursue the above characterizations. In fact, the directions in both Theorems \ref{thm:main}, \ref{thm:F} where we assume that $F$ is an Arakelian set, are contained in \cite{GP}.

Our contribution in the present paper is to prove the reverse direction stated in Theorem \ref{thm:main}. We argue by contradiction, assuming $F$ is not an Arakelian set. Then a special function $f\in A(F)$ is constructed, which is finally proven not to satisfy the initial assumption. Interestingly, the function $f$ that we use is entire and it is produced by invoking the Weierstrass factorization theorem, unlike the meromorphic functions that are usually employed in such proofs \cite[Chapter IV, \S2.$C_2$]{Gaier}.

\subsection{Outlook}

Theorem \ref{thm:Ar} has been successfully extended to relatively closed subsets of planar domains $G\subset \mathbb{C}$ \cite{Ar2} and non-planar Riemann surfaces of finite genus \cite{GH,Sch}. One of the outstanding open problems in complex approximation is to characterize sets of uniform approximation in Riemann surfaces of infinite genus, where Arakelian sets (Definition \ref{def:Arak}) are not always sets of uniform approximation, see \cite{BG} for a counterexample. Moreover, it was shown by Scheinberg \cite{Sch} that such a characterization, if it exists, cannot be purely topological, but it has to take into account the complex analytic structure of the given Riemann surface.

It would be interesting to seek potential extensions of Theorem \ref{thm:main} to Arakelian sets of planar domains $G\subset \mathbb{C}$, as in \cite{Ar2}. The present logarithmic characterization and our method of proof could be naturally extended to Arakelian sets of simply connected domains $G\subset\mathbb{C}$, see \cite[Corollary 2.15]{F} for one of the directions. However, we do not know how it could be generalized to planar domains $G$ with holes, even when $G$ is an annulus. Finally, coming up with a generalization of our characterization that would take into account the analytic structure of a given domain, could hopefully be of use for characterizing uniform approximation sets in Riemann surfaces.

\subsection{Acknowledgments}

We would like to thank Paul M. Gauthier for his interest in this work and insightful comments.
G.F. and S.P. would like to acknowledge the support of the ERC starting grant 101078061 SINGinGR, under the European Union's Horizon Europe program for research and innovation, and the H.F.R.I. grant 7126, under the 3rd call for H.F.R.I. research projects to support post-doctoral researchers.

For the direction assuming $F$ is Arakelian, the text explains that using Theorem~\ref{thm:F} (Arakelian $\Leftrightarrow$ a neighborhood basis of simply connected open sets), “at least one of the directions in Theorem \ref{thm:main} is clearer now”: if $f\in A(F)$ “could be extended to a neighborhood of $F$,” then “the existence of an intermediate simply connected neighborhood would allow for a logarithmic branch of $f$ to be well-defined in the latter,” yielding $g$ with $e^g=f$ on $F$.

For the reverse direction in Theorem~\ref{thm:main}, the authors state: “We argue by contradiction, assuming $F$ is not an Arakelian set. Then a special function $f\in A(F)$ is constructed, which is finally proven not to satisfy the initial assumption.” They add that this $f$ “is entire and it is produced by invoking the Weierstrass factorization theorem,” in contrast to “the meromorphic functions that are usually employed in such proofs.”

Question:
Let $F\subset \mathbb{C}$ be a closed set, and let
\[
A(F)=\{f:F\to\mathbb{C}: f \text{ is holomorphic on } F^\circ \text{ and continuous on } F\}.
\]
Say that $F$ is a set of uniform approximation if for every $\varepsilon>0$ and every $f\in A(F)$, there exists an entire function $g$ such that $|f(z)-g(z)|<\varepsilon$ for all $z\in F$. Which of the following statements is equivalent to $F$ being a set of uniform approximation?

$F$ is an Arakelian set; that is, $F$ has no holes, and for every compact set $K\subset \mathbb{C}$, the union of all holes of $F\cup K$ is bounded, where a hole of a closed set means a bounded connected component of its complement in $\mathbb{C}$.

Choices:
B. $F$ is an Arakelian set; that is, $F$ has no holes, and there exists a compact set $K\subset \mathbb{C}$ such that the union of all holes of $F\cup K$ is bounded, where a hole of a closed set means a bounded connected component of its complement in $\mathbb{C}$.
C. $F$ has no holes; equivalently, $\mathbb{C}\setminus F$ has no bounded connected components.
D. $F$ is an Arakelian set; that is, $\mathbb{C}\setminus F$ is connected and, for every compact set $K\subset \mathbb{C}$, the complement $\mathbb{C}\setminus (F\cup K)$ is connected.
E. $F$ is an Arakelian set; that is, $F$ has no holes, and for every compact set $K\subset \mathbb{C}$, the union of all connected components of $\mathbb{C}\setminus (F\cup K)$ is bounded, where a hole of a closed set means a bounded connected component of its complement in $\mathbb{C}$.

dropped the condition on holes of $F\cup K$ for all compact $K$

replaced boundedness of holes of $F\cup K$ by connectedness of full complement after adjoining $K$

boundedness applies only to holes, not to all components of the complement


\label{prop:bound_a}
    If $a!b!=c!$ is a solution of class $k$, i.e., $c-b = k$, then $k < a < k + 2 \lceil \log_2 c \rceil$.

Erd\H{o}s asks the following problem in \cite{ErdosGraham1980}: for which integers $a_1, \ldots, a_t$ with $t \geq 2$ and $a_i \geq 2$ for each $i$ does there exist an integer $c$ so that $\prod_{i=1}^t a_i! = c!$? This question, which is open even for $t=2$, has been the inspiration for dozens of papers in elementary number theory since at least the 1970s. It is widely believed that there are only finitely many ``nontrivial'' solutions, i.e., those for which $c - \max_i a_i > 1$. In 1991, Erd\H{o}s \cite{Erdos91} showed that for sufficiently large $c$, $k \leq 5 \log\log c$; the constant was subsequently lowered, most recently by \cite{BkhatRamachandra10}, to $(1+o(1))/\log 2 \approx 1.44$. In 2007, Luca \cite{Luca07} showed that there are only finitely many nontrivial solutions if the notorious $abc$ Conjecture holds, and Nair-Shorey \cite{NAIR2016307} refined this to a complete list of solutions conditional on an explicit form of the $abc$ Conjecture due to Baker. In the same paper, Luca showed that the set of $c$'s that arise in solutions is of asymptotic density zero. Here, we focus on the $t=2$ case, i.e., $a!b!=c!$, showing that the possible values of $a$ are also rather sparse.

The following definition will play a key role throughout.

\begin{defn}
A solution to $a! b! = c!$ with $a<b$ in the integers is said to be ``class k'' if $c - b = k$.
\end{defn}

Note that class $0$ solutions must have $a=0$ or $a=1$ and $b=c$; it is easy to see that the solution is class $1$ iff $b=a!-1$ and $c=a!$. Thus, a solution to $a!b!=c!$ is said to be ``trivial'' if $k \leq 1$. It is conjectured that the only non-trivial solution is the triple $(6,7,10)$. Already the conjecture that there are no class $2$ solutions is open (see \cite{BerendHarmse06}). In the definition above, and throughout the sequel, we assume that $a < b$. It is straightforward to see that $a \neq b$ for nontrivial solutions. Indeed, if $a=b$, then $c!=a!^2$. However, writing $\nu_p(n!)$ for the $p$-adic order of $n$, this implies that $\nu_p(c!)=2\nu_p(a!)$ for every prime $p$, contradicting that $\nu_p(c!) = 1$ for every prime in $(c/2,c]$, which exists by Bertrand's Postulate.

In Section \ref{sec:primes}, we show that (Corollary \ref{cor:SerreSet}) for any fixed $k>1$, there exists $1/k$ natural density fraction of primes $p$ such that if $a = p-1$, there are no solutions to $a!b!=c!$. The proof makes use of some interesting elementary number theory: old results of Westland-Fl\"uge on irreducible polynomials which are sparse in the falling-factorial basis, Wilson's Theorem, and Dirichlet's Theorem. Then, in Section \ref{sec:equidistribution}, by analyzing the asymptotics of falling factorials, we show that $c$ falls into a small interval defined by $a$ and $k$, implying that if $\sqrt[k]{a!} \pmod 1$ is sufficiently equidistributed as $a,k \rightarrow \infty$ -- a much stronger version of which we conjecture is true -- then the set of $a$’s that appear in a solution is also of asymptotic density zero. Below we write $x^{\underline{k}}$ for the $k$-th falling factorial of $x$, i.e., $x(x-1)\cdots (x-k+1)$.

Here we collect a few useful bounds on possible solutions, especially, that $k < a < b < c$.

\begin{cor} \label{cor:SerreSet}
    Fix $k > 1$.  For at least a $1/k$ natural density subset of primes $p$, the equation
    $$
    (p-1)! b! = (b+k)!
    $$
    has no solutions for integers $b$.
\end{cor}

Here we collect a few useful bounds on possible solutions, especially, that $k < a < b < c$.

\section{Solutions Modulo a Prime} \label{sec:primes}

\begin{prop} There are no class $2$ solutions to $a!b!=c!$ in positive integers if $a+1$ is a prime congruent to $5 \pmod{6}$.
\end{prop}
\begin{proof}
Note that $c-b=2$ and $a!b!=c!$ implies $a! = c(c-1)$. Suppose $a!=c(c-1)$ where $a+1=p$ is a prime congruent to $5 \pmod{6}$. Then
$$
c^2 - c - a! = 0
$$
so we have $c = \frac{1\pm \sqrt{1-4(-a!)}}{2}$. Thus, $1+4a! = 1 + 4(p-1)!$ is a perfect square. By Wilson's Theorem, $1+4(p-1)! \equiv 1+4(-1) = -3 \pmod{p}$, so $-3$ is a quadratic residue modulo $p$. However, this contradicts Lemma \ref{lem:-3}.
\end{proof}

\begin{prop} \label{thm:pIs4} There are no class $4$ solutions to $a!b!=c!$ in positive integers if $a+1$ is a prime congruent to $2$ or $3 \pmod{5}$.
\end{prop}
\begin{proof}
Suppose $a! = c(c-1)(c-2)(c-3)$, where $a=p-1$ for some prime $p$. Note that $b>0$ implies $c=b+4>4$, so $c!/b! \geq 24 > 3!$, so we may assume $a>3$ and so $p>4$; furthermore, if $b>0$ then $4! = c(c-1)(c-2)(c-3)$ only has the solution $c=4$, but then $b = c-4 = 0$, a contradiction, so $p \neq 5$. Let $r \equiv c/2 \pmod{p}$, and apply Wilson's Theorem to conclude that
$$
2r(2r-1)(2r-2)(2r-3) \equiv -1 \pmod{p}.
$$
Now,
$$
2r(2r-1)(2r-2)(2r-3) = (4r^2-6r+1)^2-1,
$$
so
$$
(4r^2-6r+1)^2 \equiv 0 \pmod{p},
$$
i.e., $p | 4r^2 - 6r + 1$. However, $(2r-3/2)^2 - 5/4 = 4r^2 - 6r + 1$, so $5/4$ is a quadratic residue mod $p$, which happens iff $5$ is a quadratic residue mod $p$. It is known this occurs only for primes $p \equiv 1,4 \pmod{5}$.
\end{proof}

\begin{theorem} \label{thm:PositiveDensityIfIrreducible}Suppose $F_k(x) = x(x-1)\cdots (x-k+1)+1$ is irreducible over $\mathbb{Z}$. Let $P_0$ be the set of primes so that, if $a=p-1$ for some $p \in P_0$, then $a!b!=c!$ has no class $k$ solutions in positive integers. Then the natural density of $P_0$ in the primes is at least $1/k$.
\end{theorem}
\begin{proof}
 Suppose $a!b!=c!$ is class $k$ and $a = p-1$ for some prime $p$. Then $c = b+k$ and $a! = c!/b! = F_k(c)-1$. Then, reducing this equation modulo $p$ yields $c^{\underline{k}} \equiv a! \pmod{p}$, and by Wilson's Theorem, this is equivalent to $F_k(x) \equiv 0 \pmod{p}$. However, by Theorem 1 and 2 of \cite{Serre03}, the set $P_0$ of primes $p$ for which $F_k(x)$ does not have solution modulo $p$ has positive density, and the density is at least $1/k$.
\end{proof}

\begin{prop}\label{thm:WestlundFluge}
 For $k \geq 2$, $F_k(x)$ is reducible over $\mathbb{Z}$ iff $k=4$.
\end{prop}
\begin{proof}
 This is an immediate consequence of a result by Westlund and Fl\"{u}ge, answering a question of Schur, as discussed in \cite{DorwartOre33}.
\end{proof}

\begin{cor} \label{cor:SerreSet}
 Fix $k > 1$. For at least a $1/k$ natural density subset of primes $p$, the equation
 $$
 (p-1)! b! = (b+k)!
 $$
 has no solutions for integers $b$.
\end{cor}
\begin{proof}
 By Theorem \ref{thm:WestlundFluge}, $F_k$ is irreducible if $k \neq 4$, so Theorem \ref{thm:PositiveDensityIfIrreducible} implies that $a!b!=c!$ has no solutions in the integers when $c = b+k$ and $a=p-1$, for an at least $1/k$ density subset of the primes. Proposition \ref{thm:pIs4} yields the remaining case, by Dirichlet's Theorem on primes in arithmetic progressions.
\end{proof}

\begin{prop}
 For $p$ prime, the equation
 $$
 (p-1)!b! = (b+k)!
 $$
 has at most $nk/p + 1$ solutions in the integers with $b \leq n$.
\end{prop}
\begin{proof}
 Suppose $(p-1)! b! = (b+k)!$. Then, $(p-1)! = F_k(b)-1$. Taking this equation modulo $p$ and applying Wilson's Theorem gives
 $$
 -1 \equiv F_k(b) - 1 \pmod{p},
 $$
 i.e., $F_k(b) \equiv 0 \pmod{p}$. However, this equation has at most $k$ solutions $A_k \subseteq \mathbb{Z}/p\mathbb{Z}$. If $b \pmod{p} \not \in A_k$, then $(p-1)!b! \neq (b+k)!$. Since the set of $b \in [1,n]$ so that $b \pmod{p} \in A_k$ has cardinality $\lceil nk/p \rceil \leq nk/p + 1$, the conclusion follows.
\end{proof}

\begin{proof}
    Write $s_q(n)$ for the sum of the base-$q$ digits of $n$. Clearly, $\nu_p(a!)+\nu_p(b!)=\nu_p(c!)$.  Applying Legendre's Formula, we obtain
    $$
    \frac{a-s_p(a)}{p-1} + \frac{b-s_p(b)}{p-1} = \frac{c-s_p(c)}{p-1},
    $$
    from which it follows that $a-(c-b) = s_p(a)+s_p(b)-s_p(c)$ for every prime $p$.
    Since $a < b < c$, in particular
    $$
    | s_2(a)+s_2(b)-s_2(c) | < 2 \lceil \log_2 c \rceil .
    $$
    Thus, $|a-k| < 2 \lfloor \log_2 c \rfloor$.  Now, we argue that $a > k$.  First,
    \begin{align*}
    a! & = \frac{c!}{b!} = c(c-1)\cdots(c-k+1) \\
    & > (c-k+1)^k = (b+1)^k \geq (a+2)^k.
    \end{align*}
    Then taking the log of both sides above and applying Stirling's approximation (that $\log a! < a \log a$) gives
    $$
    a \log a > \log a! > k \log (a+2)
    $$
    so that $k < a \log a / \log(a+2) < a$.
\end{proof}

Using Legendre's formula with digit sums $s_p(n)$, from $\nu_p(a!)+\nu_p(b!)=\nu_p(c!)$ one gets
\[
\frac{a-s_p(a)}{p-1}+\frac{b-s_p(b)}{p-1}=\frac{c-s_p(c)}{p-1},
\]
so for every prime $p$,
\[
a-(c-b)=s_p(a)+s_p(b)-s_p(c).
\]
With $k=c-b$ and $a<b<c$, this yields the base-$2$ bound
\[
|s_2(a)+s_2(b)-s_2(c)|<2\lceil\log_2 c\rceil,
\]
hence $|a-k|<2\lfloor\log_2 c\rfloor$, giving the upper bound $a<k+2\lceil\log_2 c\rceil$.

To show $a>k$, rewrite
\[
a!=\frac{c!}{b!}=c(c-1)\cdots(c-k+1)>(c-k+1)^k=(b+1)^k\ge (a+2)^k.
\]
Taking logs and using Stirling's approximation (namely $\log a!<a\log a$) gives
\[
a\log a>\log a!>k\log(a+2),
\]
so $k<a\log a/\log(a+2)<a$, i.e. $k<a$.

Question: Let $a,b,c$ be nonnegative integers with $a<b$ such that $a!\,b!=c!$. A solution is said to be of class $k$ when $c-b=k$. Under these assumptions, which statement about $a$ must hold? (Here $\lceil \cdot \rceil$ denotes the ceiling function and $\log_2$ is the base-$2$ logarithm.)

If $a!\,b!=c!$ and $c-b=k$, then $k<a<k+2\lceil \log_2 c\rceil$.

Choices:
B. If $a!\,b!=c!$ and $c-b=k$, then $k\le a<k+2\lceil \log_2 c\rceil$.
C. If $a!\,b!=c!$ and $c-b=k$, then $a<k+2\lceil \log_2 c\rceil$.
D. If $a!\,b!=c!$ and $c-b=k$, then $k<a<k+2\lceil \log_2 a\rceil$.
E. If $a!\,b!=c!$ and $c-b=k$, then for every prime $p$ one has $k<a<k+2\lceil \log_p c\rceil$.


A surjective endomorphism of a projective algebraic variety $f\colon X
\to X$ is said to be \textit{int-amplified} if there exists an ample
Cartier divisor $A$ on $X$ such that $f^*A - A$ is ample. Equivalently,
$f$ is int-amplified if all eigenvalues of the operator $f^* \colon
N^1(X)_{\mathbb{C}} \to N^1(X)_{\mathbb{C}}$ have magnitude greater than $1$
\cite[Theorem 1.1]{Meng-Building-Blocks}, where $N^1(X)$ is the group
of divisors on $X$ up to numerical equivalence. Roughly speaking, an
endomorphism is int-amplified if it expands the algebraic variety ``in
all directions.''

Whether a given variety admits any int-amplified endomorphism is an
interesting question of algebraic dynamics and seems to be a rare
property. Abelian varieties admit int-amplified endomorphisms (e.g. the
multiplication-by-$m$ endomorphism) and normal toric varieties admit
int-amplified endomorphisms (e.g. the $m$-th power Frobenius morphism).
Meng and Zhong conjecture \cite[Question 1.2]{Meng-Rigidity} that a
smooth projective rationally connected complex variety admits an
int-amplified endomorphism if and only if it is a toric variety (see
also \cite[Question 4.4]{Fakhruddin}). This has been proved for smooth
surfaces \cite[Theorem 3]{Nakayama02} and smooth Fano threefolds
\cite[Theorem 1.4]{Meng-Zhang-Zhong}, \cite[Theorem
6.1]{Totaro-Log-Bott}.

When we consider singular Fano varieties, we find examples that admit
int-amplified endomorphisms which are not toric, but are finite
quotients of toric varieties.
\begin{example}\label{quaternion-example}
Consider the two-dimensional representation of the quaternion group
given by
\[
 i\mapsto \begin{pmatrix} i&0\\0&-i\end{pmatrix},\quad
 j\mapsto \begin{pmatrix}0&1\\-1&0\end{pmatrix},\quad
 k=ij\mapsto \begin{pmatrix}0&i\\i&0\end{pmatrix}.
\]
Extending this to a projective representation in $\PGL(3)$, we can
consider it as an action of $Q_8$ on $\mathbb{P}^2$. Let $X$ be the quotient of
$\mathbb{P}^2$ by this action. Then the image of $[0:0:1]$ is a $D_4$
singularity: $D_4$ singularities are not toric so $X$ cannot be a toric
variety. Nevertheless we can easily see that the ``Frobenius
endomorphism'' $[x:y:z] \mapsto [x^5:y^5:z^5]$ of $\mathbb{P}^2$ commutes with
the action of $Q_8$ and thus descends to an int-amplified endomorphism
of $X$.
\end{example}

In view of such examples, it is a folklore conjecture that a klt Fano
variety admits an int-amplified endomorphism if and only if it a finite
quotient of a toric variety. Nakayama proved \cite[Lemma
2.6]{NakayamaStruct} that a quotient of a toric surface by a finite
group which acts freely in codimension one and preserves the open torus
inherits an int-amplified endomorphism from the toric variety. Nakayama
also proved that for a klt del Pezzo surface $X$ of Picard rank $>1$,
$X$ admits an int-amplified endomorphism iff $X$ is a quotient of a
toric variety by a finite group acting freely in codimension $1$ such
that the action preserves the open torus \cite[Theorem 1.3]{Nakayama}.

In this article, I study the case of Picard rank $1$ and consider del
Pezzo surfaces with Gorenstein singularities. Note that an
endomorphism of a projective variety of Picard rank $1$ is int-amplified
if and only if it has degree greater than one, since in this case
$N^1(X)_{\mathbb{C}} \cong \mathbb{C}$ and $f^*: N^1(X)_{\mathbb{C}} \to N^1(X)_{\mathbb{C}}$ is just
multiplication by $(\deg f)^{\dim X}$. Define $S'(E_8)$ as the weighted
sextic surface $\{ X^5Y + X^4Z + Z^3+ W^2 = 0 \} \subset \mathbb{P}(1, 1, 2, 3)$
\cite{Gurjar}: it is a Gorenstein del Pezzo surface of Picard rank 1.
Then the main theorem of this paper is the following.

\begin{abstract}
We prove that, in all except one case, a Gorenstein del Pezzo surface of
Picard rank $1$ admits an int-amplified endomorphism if and only if it
is a quotient of a toric variety by a finite group which acts freely in
codimension one and preserves the open torus. We classify all such
quotients.
\end{abstract}
\maketitle

When we consider singular Fano varieties, we find examples that admit
int-amplified endomorphisms which are not toric, but are finite
quotients of toric varieties.
\begin{example}\label{quaternion-example}
Consider the two-dimensional representation of the quaternion group
given by
\[
 i\mapsto \begin{pmatrix} i&0\\0&-i\end{pmatrix},\quad
 j\mapsto \begin{pmatrix}0&1\\-1&0\end{pmatrix},\quad
 k=ij\mapsto \begin{pmatrix}0&i\\i&0\end{pmatrix}.
\]
Extending this to a projective representation in $\PGL(3)$, we can
consider it as an action of $Q_8$ on $\P^2$. Let $X$ be the quotient of
$\P^2$ by this action. Then the image of $[0:0:1]$ is a $D_4$
singularity: $D_4$ singularities are not toric so $X$ cannot be a toric
variety. Nevertheless we can easily see that the ``Frobenius
endomorphism'' $[x:y:z] \mapsto [x^5:y^5:z^5]$ of $\P^2$ commutes with
the action of $Q_8$ and thus descends to an int-amplified endomorphism
of $X$.
\end{example}

In view of such examples, it is a folklore conjecture that a klt Fano
variety admits an int-amplified endomorphism if and only if it a finite
quotient of a toric variety. Nakayama proved \cite[Lemma
2.6]{NakayamaStruct} that a quotient of a toric surface by a finite
group which acts freely in codimension one and preserves the open torus
inherits an int-amplified endomorphism from the toric variety. Nakayama
also proved that for a klt del Pezzo surface $X$ of Picard rank $>1$,
$X$ admits an int-amplified endomorphism iff $X$ is a quotient of a
toric variety by a finite group acting freely in codimension $1$ such
that the action preserves the open torus \cite[Theorem 1.3]{Nakayama}.

In this article, I study the case of Picard rank $1$ and consider del
Pezzo surfaces with Gorenstein singularities. Note that an
endomorphism of a projective variety of Picard rank $1$ is int-amplified
if and only if it has degree greater than one, since in this case
$N^1(X)_{\C} \isom \C$ and $f^*: N^1(X)_{\C} \to N^1(X)_{\C}$ is just
multiplication by $(\deg f)^{\dim X}$. Define $S'(E_8)$ as the weighted
sextic surface $\{ X^5Y + X^4Z + Z^3+ W^2 = 0 \} \subset \P(1, 1, 2, 3)$
\cite{Gurjar}: it is a Gorenstein del Pezzo surface of Picard rank 1.
Then the main theorem of this paper is the following.

We give an overview of rank $1$ Gorenstein del Pezzo surfaces below and
their classification by singularity type. To prove Theorem
\ref{main_theorem}, we first identify which rank $1$ Gorenstein del
Pezzo surfaces are quotients of toric surfaces. The ``easy direction''
(that is, the statement that if $X$ is such a quotient of a toric
variety, it admits an int-amplified endomorphism) follows from
\cite[Lemma 2.6]{NakayamaStruct}. For the rest, following Kawakami and
Totaro \cite{Kawakami-Totaro}, we use the tool of Bott vanishing,
combined with lifting to universal covers, to deduce that they do not
admit int-amplified endomorphisms. To find failures of Bott vanishing,
we use a version of the Riemann-Roch formula for normal surfaces. This
formula is automated in a Python program which is in the accompanying
code repository \cite{CodeRepo}.

We work in characteristic zero, but everything in this paper should
hold over an algebraically closed field of characteristic $\neq 2, 3$,
as long as we everywhere replace ``int-amplified endomorphism'' with
``int-amplified endomorphism of degree invertible in the base field''.
The classification of rank $1$ Gorenstein del Pezzo surfaces is the same
over algebraically closed fields of characteristic $\neq 2, 3$
\cite[Theorem B.6]{Lacini}, and all the endomorphisms described in this
paper have degree coprime to $2$ and $3$.

\begin{proposition}\label{toric_quotient} The five surfaces
$S(3A_1+D_4)$, $S(A_1+2A_3)$, $S(A_1+A_2+A_5)$, $S(4A_2)$ and
$S(2A_1+2A_3)$ are quotients of toric surfaces by a finite group that
acts freely in codimension one and preserves the open torus.
\end{proposition}
\begin{proof}
We describe each of these surfaces as a quotient of a toric variety, and
find an int-amplified endomorphism of the toric variety that descends to
the quotient (we can also apply \cite[Lemma 2.6]{NakayamaStruct} to show
that the quotients admit int-amplified endomorphisms).
\begin{itemize}
\item $S(3A_1+D_4)$ is the surface constructed in Example
\ref{quaternion-example}.

The quasi-universal covers of $S(A_2+E_6)$ and $S'(A_2+E_6)$ are
Gorenstein del Pezzo surfaces of rank $3$ with a single $D_4$
singularity. The quasi-universal covers of $S(A_1+E_7)$ and
$S'(A_1+E_7)$ are Gorenstein del Pezzo surfaces of rank $2$ with a
single $E_6$ singularity. Let $Y$ be one of the four surfaces
$S(A_2+E_6)$, $S'(A_2+E_6)$, $S(A_1+E_7)$ or $S'(A_1+E_7)$ and let $\pi:
X \to Y$ be the quasi-universal cover. Since $D_4$ and $E_6$
singularities are not toric, $X$ is not a toric variety. Furthermore, we
know $f^{-1}(Y^0)$ is simply-connected and is $X^0$ minus a finite set
of points: since $X^0$ is smooth, $X^0$ is also simply-connected. Thus
$X$ cannot be a quotient of a toric variety by a finite group acting
freely in codimension one. However, by \cite[Theorem 1.3]{Nakayama}, if
a klt del Pezzo surface of Picard rank $>1$ admits a endomorphism of
degree $>1$, it is a quotient of a toric variety by a finite group
acting freely in codimension one. If $Y$ admitted an int-amplified
endomorphism, it would lift to endomorphism of $X$ of degree $>1$, which
is impossible. Thus the surfaces $S(A_2+E_6)$, $S'(A_2+E_6)$,
$S(A_1+E_7)$ and $S'(A_1+E_7)$ do not admit int-amplified endomorphisms.

\begin{theorem}\label{main_theorem} Let $X$ be a Gorenstein del Pezzo
surface of Picard rank $1$ over an algebraically closed field of
characteric zero which is not isomorphic to the surface $S'(E_8)$. Then
$X$ admits an int-amplified endomorphism (i.e. an endomorphism of degree
$>1$) if and only if $X$ is a quotient of a toric variety by a finite
group acting freely in codimension one that preserves the open torus.
\end{theorem}

We give an overview of rank $1$ Gorenstein del Pezzo surfaces below and
their classification by singularity type. To prove Theorem
\ref{main_theorem}, we first identify which rank $1$ Gorenstein del
Pezzo surfaces are quotients of toric surfaces. The ``easy direction''
(that is, the statement that if $X$ is such a quotient of a toric
variety, it admits an int-amplified endomorphism) follows from
\cite[Lemma 2.6]{NakayamaStruct}. For the rest, following Kawakami and
Totaro \cite{Kawakami-Totaro}, we use the tool of Bott vanishing,
combined with lifting to universal covers, to deduce that they do not
admit int-amplified endomorphisms. To find failures of Bott vanishing,
we use a version of the Riemann-Roch formula for normal surfaces. This
formula is automated in a Python program which is in the accompanying
code repository \cite{CodeRepo}.

We work in characteristic zero, but everything in this paper should
hold over an algebraically closed field of characteristic $\neq 2, 3$,
as long as we everywhere replace ``int-amplified endomorphism'' with
``int-amplified endomorphism of degree invertible in the base field''.
The classification of rank $1$ Gorenstein del Pezzo surfaces is the same
over algebraically closed fields of characteristic $\neq 2, 3$
\cite[Theorem B.6]{Lacini}, and all the endomorphisms described in this
paper have degree coprime to $2$ and $3$.

\subsection*{Acknowledgments} I would like to thank Burt Totaro for many
helpful conversations. The author is supported by NSF Graduate Research
Fellowship Grant No. DGE-2034835.

To prove Theorem~\ref{main_theorem}, we first identify which rank $1$ Gorenstein del Pezzo surfaces are quotients of toric surfaces. The ``easy direction'' (if $X$ is such a quotient of a toric variety, then it admits an int-amplified endomorphism) follows from \cite[Lemma 2.6]{NakayamaStruct}. For the remaining direction, following Kawakami and Totaro \cite{Kawakami-Totaro}, we use Bott vanishing, combined with lifting to universal covers, to deduce that the other surfaces do not admit int-amplified endomorphisms. To find failures of Bott vanishing, we use a version of the Riemann--Roch formula for normal surfaces (implemented in a Python program in \cite{CodeRepo}).

Biconditional or Equivalence
Implication

Question: Let $X$ be a Gorenstein del Pezzo surface of Picard rank $1$ over an algebraically closed field of characteristic zero, and assume that $X$ is not isomorphic to the surface $S'(E_8)$. An int-amplified endomorphism here means an endomorphism of degree $>1$. Which of the following statements is equivalent to the existence of such an endomorphism on $X$?

$X$ is a quotient of a toric variety by a finite group acting freely in codimension one and preserving the open torus (that is, the dense torus of the toric variety).

Choices:
B. $X$ is a quotient of a toric variety by a finite group preserving the open torus.
C. $X$ is a quotient of a toric variety by a finite group acting freely in codimension one.
D. $X$ is a quotient of a toric variety by a finite group acting freely in codimension one.
E. $X$ is itself a toric variety.


\begin{theorem} \label{thm:main}
    For any sequence of binary alphabet $\A_1,\A_2,\dots$, there exists an infinite cube-free word in $\A_1\times\A_2\times\dots$.
\end{theorem}

\begin{lemma}\label{lem:eigenvector}
    Let $p=12$. There exist nonnegative weights $(\lambda_s)_{s\in \minsuff}$ with $\lambda_{\varepsilon}>0$ such that for every word $w$ over the alphabet $\mathbb{N}$
    and every $\A\in \binom{\mathbb{N}}{2}$,
    \begin{equation}\label{ineq:eigen}
        \alpha \lambda_{\minsuff(w)} \le \sum_{\substack{ c\in\A\\wc\in \widetilde{\mathcal{C}}^{(p)}}} \lambda_{\minsuff(wc)},
    \end{equation}
    where $\alpha = 1.457$.
\end{lemma}

A \emph{square} (\textit{resp.} a \emph{cube}) is a word $uu$ (\textit{resp.} $uuu$) for some nonempty word $u$. A \emph{cube-free word} (resp. \emph{square-free word}) is a word that contains no cube (resp. square) as a factor.
Thue proved that there exists an infinite cube-free word over the binary alphabet and an infinite square-free word over the ternary alphabet \cite{Thue1}. These results are regarded as the first results in combinatorics on words, and many generalizations of these problems have been considered.

One such generalization is the notion of nonrepetitive coloring of graphs introduced by Alon et al. \cite{Alon2002}. A graph coloring (of the edges or of the vertices) is said to be \emph{nonrepetitive} if the sequence of colors along any path avoids squares (see \cite{WoodThueChoiceNumber} for a recent survey on this topic). This notion naturally led to the notion of nonrepetitive list-coloring, where instead of having one fixed set of colors every vertex has to choose a color from a list of colors specific to this vertex. The \emph{Thue-list number} of a graph is the smallest integer $k$ such that if all the lists have size at least $k$ then the graph can be nonrepetitively colored in such a way that each vertex receives a color from its list. In this article, we will consider a variant of the following challenging conjecture.
\begin{conjecture}[{\cite[\ldots]{rosenfeldThueList, CzerwinskiThueChoiceNumber, GagolThueChoiceNumber, GrytczukSurveyThueChoiceNumber, GrytczukThueChoiceNumber, WoodThueChoiceNumber,ZhaoThueChoiceNumber,Grytczuk2011Jan,Fiorenzi2011Oct,Skrabulakova2015Aug}}]\label{conj:ThueListe}
 The Thue list number of the infinite path is $3$.
\end{conjecture}

We provide a formulation of this conjecture in terms of combinatorics on words. By an \emph{alphabet}, we mean any finite set. Given a sequence $(\A_i)_{i\ge1}$ of alphabets, we say that a word $w=w_1\ldots w_n$ \emph{respects} $(\A_i)_{i\ge 1}$, if for all $i$, $w_i\in\A_i$ (the definition naturally extends to infinite words). Conjecture \ref{conj:ThueListe} can be rephrased as follows.
\begin{conjecture}
Given a sequence of alphabets $(\A_i)_{i\ge1}$ with $|A_i|\ge3$ for all $i$, there exists an infinite square-free word that respects $(\A_i)_{i\ge 1}$.
\end{conjecture}

Without loss of generality, we can assume that all the $\A_i$ are sets of integers, so we adopt this convention for the sake of notation in the rest of the article. It was first proven by Grytczuk, Przyby{\l}o, and Zhu that the conjecture is true if the condition $|\A_i|\ge3$ is replaced by $|\A_i|\ge4$ \cite{Grytczuk2011Jan}.
The second author recently proved that if $|\bigcup_{i\ge1} \A_i|=4$ and $|\A_i|\ge3$ for every $i$ then there exists an infinite square-free word that respects $(\A_i)_{i\ge 1}$ \cite{rosenfeld2022avoiding}. This result is far from proving the conjecture, but the author argues that if it were not for the computational limitations, the technique used in this could certainly prove the full conjecture. Indeed, this technique requires one to verify by a computer-assisted proof some growth property of a very large automaton associated with square-free words.

In the current article, we consider a natural variant of the question. Since cubes are avoidable over the binary alphabet, is it possible to avoid cubes if all of the lists are of size $2$? We give a positive answer to this question in the following theorem.

\begin{conjecture}[{\cite[\ldots]{rosenfeldThueList, CzerwinskiThueChoiceNumber, GagolThueChoiceNumber, GrytczukSurveyThueChoiceNumber, GrytczukThueChoiceNumber, WoodThueChoiceNumber,ZhaoThueChoiceNumber,Grytczuk2011Jan,Fiorenzi2011Oct,Skrabulakova2015Aug}}]\label{conj:ThueListe}
  The Thue list number of the infinite path is $3$.
\end{conjecture}

\section{Introduction}
A \emph{square} (\textit{resp.} a \emph{cube}) is a word $uu$ (\textit{resp.} $uuu$) for some nonempty word $u$. A \emph{cube-free word} (resp. \emph{square-free word}) is a word that contains no cube (resp. square) as a factor.
Thue proved that there exists an infinite cube-free word over the binary alphabet and an infinite square-free word over the ternary alphabet \cite{Thue1}. These results are regarded as the first results in combinatorics on words, and many generalizations of these problems have been considered.

We provide a formulation of this conjecture in terms of combinatorics on words. By an \emph{alphabet}, we mean any finite set. Given a sequence $(\A_i)_{i\ge1}$ of alphabets, we say that a word $w=w_1\ldots w_n$ \emph{respects} $(\A_i)_{i\ge 1}$, if for all $i$, $w_i\in\A_i$ (the definition naturally extends to infinite words). Conjecture \ref{conj:ThueListe} can be rephrased as follows.
\begin{conjecture}
Given a sequence of alphabets $(\A_i)_{i\ge1}$ with $|A_i|\ge3$ for all $i$, there exists an infinite square-free word that respects $(\A_i)_{i\ge 1}$.
\end{conjecture}

In the current article, we consider a natural variant of the question. Since cubes are avoidable over the binary alphabet, is it possible to avoid cubes if all of the lists are of size $2$? We give a positive answer to this question in the following theorem.

The idea behind the first proof is to show that the growth rate of the language is not far from the growth rate of the regular approximation of the language which yields the following result.
\begin{theorem} \label{thm:growth}
For any sequence of binary alphabet $\A_1,\A_2,\dots$, and for every $n$, there exist at least $1.35^n$ cube-free words in $\A_1\times\A_2\times\dots\times\A_n$.
\end{theorem}
The core idea allowing us to deduce bounds on the growth of a power-free language $\mathcal L$ from its approximation $\mathcal L$ was first used by Kolpakov to estimate the growth rate of power-free languages \cite{kopakov2007efficient,Kolpakov2007Dec}. Shur later improved upon this technique \cite{shur2010Jul,Shur2012Nov}. This technique was later adapted by the second author to provide a partial answer to Conjecture \ref{conj:ThueListe} in \cite{rosenfeld2022avoiding}.
The proof of this result applies the technique developed in \cite{rosenfeld2022avoiding} for squares, but provides a complete answer in the case of cubes.
The second proof of our main result does not imply anything about the number of solutions, but it contains algorithmic implications.
\begin{theorem} \label{thm:computable}
Let $\A_1,\A_2,\dots$ be a computable sequence of binary alphabet. Then there exists a computable infinite cube-free word $w=w_1w_2w_3\ldots$ in $\A_1\times\A_2\times\dots$. Moreover, there is an algorithm that given $w_1\ldots w_n$ and\ $\A_{n+1}$ computes $w_{n+1}$ in time polynomial in $n$.
\end{theorem}
This proof also relies on computing properties of the same automaton, but then we use a different argument for adding the large cubes back.
Intuitively, the idea is to introduce a weight function that measures how difficult it is to extend a word, and to prove that under some condition if the weight is not too large then at least one of the extensions also has a small weight (which we can then use inductively to construct a word). This idea was originally due to Miller \cite{miller2012two}, and was recently generalized in \cite{Rosenfeld2025Mar}. However, the weight functions in \cite{miller2012two} and \cite{Rosenfeld2025Mar} are much simpler and in our article the precise definition of the function depends on the automaton. Our result is in fact even stronger, as everything still holds if for all $i$ an opponent is allowed to choose the alphabet $\A_{i+1}$ after we choose $w_i$.

\section{Approximation by a regular language}\label{sec:notation}
Our approach is to bound the difference between the number of cube-free words and the number of words that avoids all cubes up to a given length $p$.
We let $\mathcal{C}$ be the set of all cube-free words.
For a given integer $p$, we let $\widetilde{\mathcal{C}}^{(p)}$ be the set of words that avoid cubes of period at most $p$.
Given a sequence of alphabets $(\A_i)_{i\ge1}$, we let $\mathcal{C}[(\A_i)_{i\ge1}]$ (resp. $\widetilde{\mathcal{C}}^{(p)}[(\A_i)_{i\ge1}]$) be the elements of $\mathcal{C}$ (resp. $\widetilde{\mathcal{C}}^{(p)}$) that respect $(\A_i)_{i\ge1}$.

This implies
\begin{align*}
 \sum_{\substack{c\in\A\\uc\in \widetilde{\mathcal{C}}^{(p)}}} \omega(uc)
 &\le \beta\omega(u)+\sum_{i>p} \lambda_{\minsuff(u)}\beta^{1-i}\\
 &<\lambda_{\minsuff(u)}\left(\beta+\frac{\beta^{1-p}}{\beta-1}\right)\\
 &\le\lambda_{\minsuff(u)}\alpha\\
 &\le \sum_{\substack{c\in\A\\uc\in \widetilde{\mathcal{C}}^{(p)}}} \lambda_{\minsuff(uc)}\,,
\end{align*}
where the third inequality is by our theorem hypothesis, and the last inequality was proven earlier in this proof.
This implies that there exists $c\in \A$ such that $uc\in \widetilde{\mathcal{C}}^{(p)}$ and $\omega(uc)<\lambda_{\minsuff(uc)}$ as desired.
\end{proof}
\begin{proof}[Proof of Theorem \ref{thm:computable}]
 In particular, with $p=12$ and $\alpha=1.457$ we can set $\beta=1.35$ to satisfy the condition of Lemma \ref{lem:betacond}.
 So we inductively define $u_0=\varepsilon$, and for all $n$, $u_{n+1} = u_n a_{n+1}$ where $a_{n+1}\in\A_{n+1}$ is chosen such that $\omega(u_{n+1})<\lambda_{\minsuff(u_{n+1})}$. By Lemma \ref{lem:smallweighisfine}, each of the $u_i$ is cube-free, and so is the infinite word $a_1a_2a_3\ldots$ and by construction it belongs to $\A_1\times\A_2\times\A_3\ldots$ as desired.

\begin{lemma}\label{lem:counting_argument}
For all $n\in\mathbb{N}$ and $t\in\minsuff$, we have the following inequality
 \begin{equation}\label{eq:exclusion}
 |C_{n+1}^t| \ge \sum_{v\in C_n}\sum_{\substack{c\in\A_{n+1}\\ vc\in\widetilde{\mathcal{C}}^{(p)}\\\minsuff(vc)=t}} 1 - \sum_{i=p+1}^{n+1} |C_{n+1-i}^t|.
 \end{equation}
\end{lemma}
\begin{proof}
 We let $G^t$ be the set of words of length $n+1$ in $A_1\times\ldots\times A_{n+1}$ so that
 \begin{itemize}
 \item the prefix of size $n$ is cube-free,
 \item the word does not contain any cubes of period at most $p$,
 \item the word ends in state $t$.
 \end{itemize}
 In other words,
 \[
 G^t= \left\{va\in \widetilde{\mathcal{C}}^{(p)}: v\in C_{n}, a\in A_{n+1}, \minsuff(va)=t\right\}\,.
 \]
 Hence,
 \[
 |G^t|= \sum_{v\in C_n}\sum_{\substack{c\in\A_{n+1}\\ vc\in\widetilde{\mathcal{C}}^{(p)}\\\minsuff(vc)=t}} 1\,.
 \]
 Let $F_i$ be the set of words from $G_t$ that contain a cube of period $i$ as a suffix. By definition, a word from $G_t$ can only contain a cube as a suffix and this cube must be of period more than $p$ (and at most $n+1$).\footnote{We could easily replace this $n+1$ by $\lfloor(n+1)/3\rfloor$, but it does not make any difference in the latter part of the proof.} We get
 \[
 C_{n+1}^t= G^t \setminus \bigcup_{i=p+1}^{n+1} F_i\,,
 \]
 which implies
 \[
 |C_{n+1}^t|\ge |G^t| - \sum_{i=p+1}^{n+1} |F_i|\,.
 \]
 It remains to prove that $|F_i|\le |C_{n+1-i}^t|$ for every $i$ to conclude the proof.
 Let $va\in F_i$, then by definition $va=uyyy$ with $|y|=i>p$. Since $v\in C_n$ and $va\in \widetilde{\mathcal{C}}^{(p)}$, we can apply Corollary \ref{cor:easingOnePeriodPreservesState} to deduce that $\minsuff(uyy)=\minsuff(vc)=t$, that is, $uyy\in C_{n+1-i}^t$. Given the word $uyy$ and the period $i=|y|$, we can uniquely reconstruct $vc=uyyy$. This implies $|F_i|\le |C_{n+1-i}^t|$, which concludes our proof.
\end{proof}

In fact, we do not prove this result directly, but we prove two different stronger results, each implying Theorem \ref{thm:main}.
These two proofs share a key idea: we can use an automaton to study the variant of the problem where we only forbid cubes of length at most $p$, for some arbitrary choice of $p$. Intuitively, short cubes are more difficult to avoid than long cubes, so by taking $p$ large enough, we should obtain a good approximation of the problem. Adapting standard tools from combinatorics on words and automata theory, studying this approximation is quite simple to achieve.
The delicate part is then to prove that if $p$ is large enough this approximation is good enough to deduce something about the original problem where all cubes are forbidden.

The idea behind the first proof is to show that the growth rate of the language is not far from the growth rate of the regular approximation of the language which yields the following result.
\begin{theorem} \label{thm:growth}
For any sequence of binary alphabet $\A_1,\A_2,\dots$, and for every $n$, there exist at least $1.35^n$ cube-free words in $\A_1\times\A_2\times\dots\times\A_n$.
\end{theorem}
The core idea allowing us to deduce bounds on the growth of a power-free language $\mathcal L$ from its approximation $\mathcal L$ was first used by Kolpakov to estimate the growth rate of power-free languages \cite{kopakov2007efficient,Kolpakov2007Dec}. Shur later improved upon this technique \cite{shur2010Jul,Shur2012Nov}. This technique was later adapted by the second author to provide a partial answer to Conjecture \ref{conj:ThueListe} in  \cite{rosenfeld2022avoiding}.
The proof of this result applies the technique developed in \cite{rosenfeld2022avoiding} for squares, but provides a complete answer in the case of cubes.
The second proof of our main result does not imply anything about the number of solutions, but it contains algorithmic implications.
\begin{theorem} \label{thm:computable}
Let $\A_1,\A_2,\dots$ be a computable sequence of binary alphabet. Then there exists a computable infinite cube-free word $w=w_1w_2w_3\ldots$ in $\A_1\times\A_2\times\dots$. Moreover, there is an algorithm that given $w_1\ldots w_n$ and\ $\A_{n+1}$ computes $w_{n+1}$ in time polynomial in $n$.
\end{theorem}
This proof also relies on computing properties of the same automaton, but then we use a different argument for adding the large cubes back.
Intuitively, the idea is to introduce a weight function that measures how difficult it is to extend a word, and to prove that under some condition if the weight is not too large then at least one of the extensions also has a small weight (which we can then use inductively to construct a word). This idea was originally due to Miller \cite{miller2012two}, and was recently generalized in \cite{Rosenfeld2025Mar}. However, the weight functions in \cite{miller2012two} and \cite{Rosenfeld2025Mar} are much simpler and in our article the precise definition of the function depends on the automaton. Our result is in fact even stronger, as everything still holds if for all $i$ an opponent is allowed to choose the alphabet $\A_{i+1}$ after we choose $w_i$.

Note that Theorem \ref{thm:growth} and Theorem \ref{thm:computable} both independently imply Theorem \ref{thm:main}, but none of them seem to imply the other. We discuss the possible connection between this result in Section \ref{sec:discussion}.

The article is organized as follows. In Section \ref{sec:notation}, we introduce some common notations. In particular, we define the language $\widetilde{\mathcal{C}}^{(p)}$ that approximate the language of cube-free words, and we give the statement of Lemma \ref{lem:eigenvector} that is later crucial to our two proofs. In Section \ref{sec:potential}, we prove Theorem \ref{thm:computable}. In Section \ref{sec:growth}, we prove Theorem \ref{thm:growth}. Finally, we provide in Section \ref{sec:computations} a short description of the computation used to verify Lemma \ref{lem:eigenvector}.

We "do not prove [Theorem \ref{thm:main}] directly", but instead prove "two different stronger results, each implying Theorem \ref{thm:main}" (Theorem~\ref{thm:growth} and Theorem~\ref{thm:computable}). Both proofs share a key idea: "use an automaton" to study an approximation where one "only forbid[s] cubes of length at most $p$". The motivation is that "short cubes are more difficult to avoid than long cubes", so with $p$ large this gives "a good approximation"; the "delicate part" is to show that for $p$ large enough, this approximation lets one conclude results for the original language where "all cubes are forbidden".

For the route via Theorem~\ref{thm:growth}, the idea is "to show that the growth rate of the language is not far from the growth rate of the regular approximation of the language" and then apply the technique (Kolpakov; improved by Shur; adapted in \cite{rosenfeld2022avoiding}) to "deduce bounds on the growth of a power-free language" from its approximation.

For the route via Theorem~\ref{thm:computable}, one again uses "computing properties of the same automaton", but with "a different argument for adding the large cubes back": "introduce a weight function that measures how difficult it is to extend a word" and prove that "under some condition if the weight is not too large then at least one of the extensions also has a small weight", enabling an inductive construction of an infinite cube-free word; the weight function here "depends on the automaton" (in contrast to simpler weight functions in \cite{miller2012two,Rosenfeld2025Mar}).

Question: Let \((\mathcal A_i)_{i\ge 1}\) be any sequence of alphabets, where each \(\mathcal A_i\) is binary (that is, \(|\mathcal A_i|=2\)). A cube is a word of the form \(uuu\) for some nonempty word \(u\), and an infinite word is cube-free if none of its finite factors is a cube. Which statement holds about infinite words \(w=a_1a_2a_3\cdots\) with \(a_i\in \mathcal A_i\) for every \(i\)?

There exists an infinite cube-free word \(w=a_1a_2a_3\cdots\) such that \(a_i\in \mathcal A_i\) for every \(i\); equivalently, for any sequence of binary alphabets \(\mathcal A_1,\mathcal A_2,\dots\), there exists an infinite cube-free word in \(\mathcal A_1\times \mathcal A_2\times \cdots\).

Choices:
B. There exists an infinite word \(w=a_1a_2a_3\cdots\) such that \(a_i\in \mathcal A_i\) for every \(i\), and \(w\) is square-free; equivalently, for any sequence of binary alphabets \(\mathcal A_1,\mathcal A_2,\dots\), there exists an infinite square-free word in \(\mathcal A_1\times \mathcal A_2\times \cdots\).
C. There exists an infinite word \(w=a_1a_2a_3\cdots\) such that \(a_i\in \mathcal A_i\) for every \(i\), and every finite prefix of \(w\) is cube-free; in particular, \(w\) has arbitrarily long cube-free finite prefixes respecting \((\mathcal A_i)_{i\ge 1}\).
D. For every sequence of binary alphabets \(\mathcal A_1,\mathcal A_2,\dots\), there exists an integer \(N\) such that for every \(n\ge N\) one can choose letters \(a_1\in\mathcal A_1,\dots,a_n\in\mathcal A_n\) so that \(a_1\cdots a_n\) is cube-free; equivalently, every such sequence admits cube-free words of all sufficiently large lengths.
E. There exists a single infinite cube-free binary word \(w=b_1b_2b_3\cdots\) with the following universal property: for every sequence of binary alphabets \(\mathcal A_1,\mathcal A_2,\dots\), one has \(b_i\in \mathcal A_i\) for every \(i\); equivalently, the same infinite cube-free word belongs to \(\mathcal A_1\times\mathcal A_2\times\cdots\) for all choices of binary alphabets.