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.


\begin{thm}
\label{thm:main}
Let $\beta(n)\colonequals \max_{w\in S_n}|\supp(\mathfrak S_w)|$. Then
\[
\lim_{n\to\infty}\frac{\ln(\beta(n))}{n\ln(n)} = 1.
\]
More precisely,
\[
-\ln(4)-1 \leq \liminf_{n\to\infty}\frac{\ln(\beta(n)) - n\ln(n)}n \leq \limsup_{n\to\infty}\frac{\ln(\beta(n)) - n\ln(n)}n \leq -1.
\]
\end{thm}

Schubert polynomials $\mathfrak S_w(x_1, \dots, x_n)$, indexed by permutations $w\in S_n$, are lifts of Schubert cycles in the cohomology of the flag variety \cite{ls82}. The specialization $\mathfrak S_w(1,\dots,1)$ is equal to the number of reduced pipe dreams of $w$, and has a geometric interpretation as the degree of the matrix Schubert variety of $w$. Writing $u(n)\colonequals \max_{w\in S_n}\mathfrak S_w(1, \dots,1)$, Stanley \cite{stanley17} observed that
\[
\frac14 \leq \liminf_{n\to\infty} \frac{\log_2(u(n))}{n^2} \leq \limsup_{n\to\infty}\frac{\log_2(u(n))}{n^2} \leq \frac12
\]
and asked whether $\lim_{n\to\infty}\frac{\log_2(u(n))}{n^2}$ exists, and, if so, what the value of this limit is. His question remains open, but see \cite{mpp19,gao21,gl24,mppy25,zhang25} for recent progress on this problem and its variants.

We study the growth rate of the maximal sizes of \emph{supports} of Schubert polynomials. Write $\mathrm{supp}(\mathfrak S_w)\colonequals \{\alpha \in \ZZ_{\geq 0}^n\colon \mathbf x^\alpha\textup{ appears in } \mathfrak S_w\}$.

The key ingredient in our proof of Theorem~\ref{thm:main} is the observation that a certain \emph{layered} permutation has support of size asymptotically at least $n!/4^n$. Layered permutations have previously appeared in Merzon--Smirnov's conjecture \cite{ms16} that $\mathfrak S_w(1,\dots,1)$ is maximized at a layered permutation. Guo--Lin conjectured \cite[Prob 5.3]{gl24} that there exist permutations simultaneously maximizing $\mathfrak S_w(1, \dots,1)$ and $|\supp(\mathfrak S_w)|$.

We will use the following (crude) estimates for the factorial function, whose proofs we include for completeness.
\begin{lem}
\label{lem:factorial-estimate}
The inequalities
\[
e\left(\frac ne\right)^n \leq n! \leq e\left(\frac{n+1}e\right)^{n+1}
\]
hold for all positive integers $n$.
\end{lem}
\begin{proof}
Estimate $\ln(n!) = \sum_{k=1}^n\ln(k)$ by the integrals
\[
\int_1^n \ln(x)\,dx \leq \sum_{k=1}^n \ln(k) \leq \int_1^{n+1}\ln(x)\,dx.
\]
The definite integrals evaluate to $n\ln(n) - n + 1$ and $(n+1)\ln(n+1) - (n+1) + 1$. The claim follows.
\end{proof}
\begin{lem}
\label{lem:floor-fact-est}
The inequality
\[
\lfloor k\rfloor! \geq \frac1{2k}\left(\frac ke\right)^k
\]
holds for all real numbers $k>1$.
\end{lem}
\begin{proof}
Using Lemma~\ref{lem:factorial-estimate}, compute that
\[
\lfloor k\rfloor! \geq \frac1{k+1} \lceil k\rceil! \geq \frac1{k+1} \cdot \frac{\lceil k\rceil^{\lceil k\rceil}}{e^{\lceil k\rceil-1}} \geq \frac1{k+1}\cdot \frac{k^k}{e^k}.
\]
As $k > 1$, the claim follows.
\end{proof}
\section{A layered permutation with large Schubitope}
Our main technical result is as follows.
\begin{thm}
\label{thm:tech-main}
Let $w = w(b_1, \dots, b_m)$ be a layered permutation in $S_n$. If $b_m \leq n/2$, then
\[
|\supp(\mathfrak S_w)| \geq b_m!\cdot |\supp(\mathfrak S_{w'})|, \qquad \textup{ where } w'\colonequals w(b_1, \dots, b_{m-1})\in S_{n-b_m}.
\]
\end{thm}
Our proof of Theorem~\ref{thm:tech-main} will use the following lemma.
\begin{lem}
\label{lem:last-block}
Let $w\in S_n$ be a layered permutation of the form $w = w(1, \dots, 1, b_m)$ for $b_m \leq n/2$. For any subset $S\subseteq D(w)$, there exists a diagram $C$ satisfying $C\leq D(w)$ and $C\cap D(w) = S$.
\end{lem}
\begin{proof}
Each column $D_j$ of $D(w)$ is of the form $\{n-b_m + 1, \dots, n-c\}$ for some $c \geq 1$. As $|D_j| \leq b_m-1 < n-b_m$, it follows that for any subset $S_j$ of $D_j$ there exists a set $C_j\leq D_j$ with $C_j \cap D_j = S_j$. Applying this column by column to $D(w)$, the claim follows.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:tech-main}]
For each of the $b_m!$ many vectors $\alpha\in\ZZ_{\geq0}^{b_m}$ satisfying $(0,\dots,0) \leq \alpha \leq (b_m - 1, b_m - 2, \dots, 1, 0)$, we construct an embedding $i_\alpha\colon \mathcal S_{D(w')} \hookrightarrow \mathcal S_{D(w)}$ so that $\img(i_\alpha)\cap\img(i_{\alpha'}) = \emptyset$ for all $\alpha\neq\alpha'$.

For each $\alpha$ satisfying $(0,\dots,0)\leq\alpha\leq (b_m - 1, b_m - 2, \dots, 1)$, use Lemma~\ref{lem:last-block} to choose a diagram $C_\alpha \leq D(w_{\mathrm{LB}})$ whose weight satisfies $\wt(C_\alpha)_{n-b_i + j} = \alpha_j$. Then define the map
\begin{align*}
i_\alpha\colon \mathcal S_{D(w')}&\to \mathcal S_{D(w)}\\
\wt(C')&\mapsto\wt((C', C_\alpha));
\end{align*}
the map $i_\alpha$ is well-defined and injective because it is in fact the translation map $\gamma \mapsto \gamma +\wt(C_\alpha)$. Furthermore, any $\gamma' \in \mathcal S_{D(w')}$ satisfies $\gamma'_{n-b_m+j} = 0$ for all $j\geq 0$, so any vector $\gamma\in\img(i_\alpha)$ satisfies $\gamma_{n-b_m+j} = \alpha_j$. In particular, the $\img(i_\alpha)$ are disjoint subsets of $\mathcal S_{D(w)}$.
\end{proof}
\begin{cor}
\label{cor:upshot}
Fix an integer $n\geq 3$ and let $c\colonequals \max\{k \colon \lfloor n/2^k\rfloor \geq 1\}$ and $d \colonequals n - \sum_{k=1}^c \lfloor n/2^k\rfloor$. Let $w$ be the layered permutation with blocks of size
\[
\underbrace{1,\dots,1}_{d\textup{ many}}, \left\lfloor \frac n{2^c}\right\rfloor, \left\lfloor \frac n{2^{c-1}}\right\rfloor, \dots, \left\lfloor\frac n4\right\rfloor,\left\lfloor \frac n2\right\rfloor.
\]
Then
\[
|\supp(\mathfrak S_w)| \geq \prod_{k=1}^c\left\lfloor \frac n{2^k}\right\rfloor!.
\]
In particular,
\[
\ln(|\supp(\mathfrak S_w)|) \geq \left(n - \frac1{\ln(2)}\ln(n) - 2\right)\ln(n) - n\ln(4) - (n-2).
\]
\end{cor}
\begin{proof}
From the inequality
\[
\left\lfloor \frac n{2^k}\right\rfloor \leq \frac n{2^k} = \frac12\left(n - \sum_{i=1}^{k-1} \frac n{2^i}\right) \leq \frac12\left(n - \sum_{i=1}^{k-1} \left\lfloor\frac n{2^i}\right\rfloor\right),
\]
Theorem~\ref{thm:tech-main} may be repeatedly applied to the layered permutation $w$ to obtain
\[
|\supp(\mathfrak S_w)| \geq \prod_{k=1}^c\left\lfloor \frac n{2^k}\right\rfloor!.
\]
Lemma~\ref{lem:floor-fact-est} implies
\begin{align*}
\prod_{k=1}^c\left \lfloor \frac n{2^k}\right\rfloor! &\geq \prod_{k=1}^c\left(\frac{2^{k-1}}n \left(\frac n{2^k e}\right)^{\frac n{2^k}}\right) \\&\geq \frac 1{n^c} \cdot \underbrace{\prod_{k=1}^c \left(\frac1{2^k}\right)^{\frac n{2^k}}}_{\geq 2^{-2n}}\cdot \underbrace{\prod_{k=1}^c \left(\frac ne\right)^{\frac n{2^k}}}_{\geq \left(\frac ne\right)^{n-2}}\label{eqn:estimate}\tag{$*$}\\
&\geq \frac{n^{n-c-2}}{4^n\cdot e^{n-2}}\\&\geq \frac{n^{n - \log_2(n) - 2}}{4^n \cdot e^{n-2}}
\end{align*}
where the estimates in~\eqref{eqn:estimate} follow from
\[
\sum_{k=1}^c k\cdot \frac n{2^k} \leq \sum_{k=1}^\infty k \cdot \frac n{2^k} = 2n
\]
and
\[
\frac n2 + \dots + \frac n{2^c} = n - \frac n{2^c} \geq n-2.
\]
\end{proof}

\newtheorem*{thm:main}{Theorem~\ref{thm:main}}
\begin{thm:main}
Let $\beta(n)\colonequals \max_{w\in S_n}|\supp(\mathfrak S_w)|$. Then
\[
\lim_{n\to\infty}\frac{\ln(\beta(n))}{n\ln(n)} = 1.
\]
More precisely,
\[
-\ln(4)-1 \leq \liminf_{n\to\infty}\frac{\ln(\beta(n)) - n\ln(n)}n \leq \limsup_{n\to\infty}\frac{\ln(\beta(n)) - n\ln(n)}n \leq -1.
\]
\end{thm:main}
\begin{proof}[Proof of Theorem~\ref{thm:main}]
By Lemma~\ref{lem:upper-bound}, the inequality
\[
\ln(|\supp(\mathfrak S_w)|) \leq \ln(n!) \leq (n+1)\ln(n+1) - n
\]
holds for all $w\in S_n$. By Corollary~\ref{cor:upshot}, there exists $w\in S_n$ with
\[
\ln(|\supp(\mathfrak S_w)|) \geq \left(n-\frac1{\ln(2)}\ln(n) - 2\right)\ln(n) - n\ln(4) - (n-2).
\]
The inequalities
\[
-\ln(4)-1 \leq \liminf_{n\to\infty}\frac{\ln(\beta(n)) - n\ln(n)}n \leq \limsup_{n\to\infty}\frac{\ln(\beta(n)) - n\ln(n)}n \leq -1
\]
follow.
\end{proof}

\begin{prop}
\label{prop:groth-precise}
Fix an integer $n$, let $k$ be the unique integer so that $\binom{k+1}2\leq n<\binom{k+2}2$, and let $b \colonequals n - \sum_{i=1}^k i$. Let $w$ be the layered permutation with blocks of size $1, 2, \dots, k, b$. Then
\[
|\supp(\mathfrak G_w)| \geq \frac{n!}{n^{k+1}}.
\]
In particular,
\[
\ln(|\supp(\mathfrak G_w)|) \geq n\ln(n) - n - (\sqrt{2n}+1)\ln(n)
\]
\end{prop}
\begin{proof}
Let $\mathbf c = \mathrm{wt}(D(w))$ denote the Lehmer code of $w$, and let $\mathbf d = \mathrm{wt}(\overline{D(w)})$. Proposition~\ref{prop:fireworks-grothendieck} implies that $[\mathbf c, \mathbf d] \subseteq \supp(\mathfrak G_w)$, so
\begin{equation}
\label{eqn:crude-grothendieck}
|\supp(\mathfrak G_w)| \geq \prod_{i=1}^n (d_i - c_i + 1).
\end{equation}

\newtheorem*{thm:groth}{Theorem~\ref{thm:groth}}
\begin{thm:groth}
Let $\beta^{\mathfrak G}(n)\colonequals\max_{w\in S_n}|\supp(\mathfrak G_w)|$. Then
\[
\lim_{n\to\infty}\frac{\ln(\beta^{\mathfrak G}(n)) - n\ln(n)}n = -1.
\]
\end{thm:groth}
\begin{proof}[Proof of Theorem~\ref{thm:groth}]
By Lemma~\ref{lem:upper-bound}, the inequality
\[
\ln(|\supp(\mathfrak G_w)|)\leq \ln(n!) \leq (n+1)\ln(n+1)-n
\]
holds for all $w\in S_n$. By Proposition~\ref{prop:groth-precise}, there exists $w\in S_n$ with
\[
\ln(|\supp(\mathfrak G_w)|) \geq n\ln(n) - n - (\sqrt{2n}+1)\ln(n).
\]
The claim follows.
\end{proof}

Theorem~\ref{thm:main} answers a problem \cite[Prob 5.5]{gl24} posed by Guo--Lin. We do not know if the limit $\lim_{n\to\infty}\frac{\ln(\beta(n)) - n\ln(n)}n$ exists, nor do we have a conjecture for its value.

The key ingredient in our proof of Theorem~\ref{thm:main} is the observation that a certain \emph{layered} permutation has support of size asymptotically at least $n!/4^n$. Layered permutations have previously appeared in Merzon--Smirnov's conjecture \cite{ms16} that $\mathfrak S_w(1,\dots,1)$ is maximized at a layered permutation. Guo--Lin conjectured \cite[Prob 5.3]{gl24} that there exist permutations simultaneously maximizing $\mathfrak S_w(1, \dots,1)$ and $|\mathrm{supp}(\mathfrak S_w)|$.

Grothendieck polynomials $\mathfrak G_w$ are inhomogeneous deformations of Schubert polynomials; they are generating functions for (possibly nonreduced) pipe dreams of $w$. We are able to produce a layered permutation with support of size asymptotically at least $n!/e^{\sqrt{2n} \cdot \ln(n)}$, i.e., of size $n!$ up to a subexponential factor. In particular, the maximal sizes of supports of Grothendieck polynomials satisfy the following more precise asymptotics.
\begin{thm}
\label{thm:groth}
Let $\beta^{\mathfrak G}(n)\colonequals\max_{w\in S_n}|\mathrm{supp}(\mathfrak G_w)|$. Then
\[
\lim_{n\to\infty}\frac{\ln(\beta^{\mathfrak G}(n)) - n\ln(n)}n = -1.
\]
\end{thm}

The post-theorem introduction indicates that “the key ingredient in our proof of Theorem~\ref{thm:main} is the observation that a certain \emph{layered} permutation has support of size asymptotically at least $n!/4^n$.”

\label{thm:main}
Let $\beta(n)\colonequals \max_{w\in S_n}|\mathrm{supp}(\mathfrak S_w)|$. Then
\[
\lim_{n\to\infty}\frac{\ln(\beta(n))}{n\ln(n)} = 1.
\]
More precisely,
\[
-\ln(4)-1 \leq \liminf_{n\to\infty}\frac{\ln(\beta(n)) - n\ln(n)}n \leq \limsup_{n\to\infty}\frac{\ln(\beta(n)) - n\ln(n)}n \leq -1.
\],

Question:
For each positive integer \(n\), let \(S_n\) be the symmetric group on \(n\) letters, and for \(w\in S_n\) let \(\mathfrak S_w(x_1,\dots,x_n)\) denote the corresponding Schubert polynomial. Define its support by
\[
\mathrm{supp}(\mathfrak S_w)\colonequals \{\alpha\in \mathbb Z_{\ge 0}^n : \mathbf x^\alpha \text{ appears in } \mathfrak S_w\},
\]
and set
\[
\beta(n)\colonequals \max_{w\in S_n} |\mathrm{supp}(\mathfrak S_w)|.
\]
Using \(\ln\) for the natural logarithm, as \(n\to\infty\), which asymptotic statement holds for \(\beta(n)\)?

The quantity \(\beta(n)\) satisfies
\[
\lim_{n\to\infty}\frac{\ln(\beta(n))}{n\ln(n)}=1,
\]
and more precisely,
\[
-\ln(4)-1\le \liminf_{n\to\infty}\frac{\ln(\beta(n))-n\ln(n)}{n}
\le \limsup_{n\to\infty}\frac{\ln(\beta(n))-n\ln(n)}{n}
\le -1.
\]

Choices:
B.
The quantity \(\beta(n)\) satisfies
\[
\lim_{n\to\infty}\frac{\ln(\beta(n))}{n\ln(n)}=1,
\]
and more precisely,
\[
-\ln(4)-1\le \liminf_{n\to\infty}\frac{\ln(\beta(n))-n\ln(n)}{n}
\le \limsup_{n\to\infty}\frac{\ln(\beta(n))-n\ln(n)}{n}
\le 0.
\]
C.
The quantity \(\beta(n)\) satisfies
\[
\lim_{n\to\infty}\frac{\ln(\beta(n))}{n\ln(n)}=1.
\]
D.
The quantity \(\beta(n)\) satisfies
\[
\lim_{n\to\infty}\frac{\ln(\beta(n))}{n\ln(n)}=1,
\]
and more precisely,
\[
-\ln(4)-1\le \liminf_{n\to\infty}\frac{\ln(\beta(n))-n\ln(n)}{n}
\le \limsup_{n\to\infty}\frac{\ln(\beta(n))-n\ln(n)}{n}
\le -\ln(4)-1.
\]
E.
The quantity \(\beta(n)\) satisfies
\[
\limsup_{n\to\infty}\frac{\ln(\beta(n))}{n\ln(n)}=1,
\]
and more precisely,
\[
-\ln(4)-1\le \liminf_{n\to\infty}\frac{\ln(\beta(n))-n\ln(n)}{n}
\le \limsup_{n\to\infty}\frac{\ln(\beta(n))-n\ln(n)}{n}
\le -1.
\]


\label{l1}Let $\Omega$ be a bounded domain in $\mathbb{R}^n$, $f_{\pm} :
  \bar{\Omega} \rightarrow \mathbb{R}^n$ be $C^2$-maps such that $f_+
  |_{\partial \Omega}   = f_- |_{\partial \Omega}  $. then
  \begin{equation}
    \int_{\Omega} \det f_+' (x) \mathrm{d} x = \int_{\Omega} \det f_-' (x) \mathrm{d}
    x \text{.} \label{e1}
  \end{equation}

\begin{theorem}
  \label{l1}Let $\Omega$ be a bounded domain in $\mathbb{R}^n$, $f_{\pm} :
  \bar{\Omega} \rightarrow \mathbb{R}^n$ be $C^2$-maps such that $f_+
  |_{\partial \Omega}   = f_- |_{\partial \Omega}  $. then
  \begin{equation}
    \int_{\Omega} \det f_+' (x) \mathd x = \int_{\Omega} \det f_-' (x) \mathd
    x \text{.} \label{e1}
  \end{equation}
\end{theorem}

\begin{equation}
  \int_B \det R' (x) \mathd x = \int_B \mathd x = \mathrm{Vol} (B) > 0 \text{,}
  \label{e2}
\end{equation}

In a recent paper {\cite{MR4725395}}, to respond inquiry from some readers
about the solution of Exercise 1.2.1 in his book {\cite{MR2435520}}, Krylov
gives a proof of the following theorem, see \citet[Lemma 1]{MR4725395}, where $f_\pm$ only need to be $C^1$.

\begin{abstract}
 It is known that the integral of the Jacobian determinant of a smooth map $f
 : \bar{\Omega} \rightarrow \mathbb{R}^n$ depends only on $f |_{\partial
 \Omega} $ and this result leads to an analytic proof of the
 Brouwer fixed point theorem. In this note we provide two new proofs of this
 result, one by classical analysis and one by differential forms and Stokes
 formula.
\end{abstract}

In a recent paper {\cite{MR4725395}}, to respond inquiry from some readers
about the solution of Exercise 1.2.1 in his book {\cite{MR2435520}}, Krylov
gives a proof of the following theorem, see \citet[Lemma 1]{MR4725395}, where $f_\pm$ only need to be $C^1$.

Krylov's proof of Theorem \ref{l1} in {\cite{MR4725395}} is based on the
observation: for small $t$, $f_{\pm}^t = \mathrm{id} + t f_{\pm}$ are
diffeomorphisms and $f^t_+ (\Omega) = f_-^t (\Omega)$ because $f_+ |_{\partial
\Omega} = f_- |_{\partial \Omega} $, thus by the changing
variable formula
\[ \int_{\Omega} \det (I + t f_+' (x)) \mathd x = \mathrm{Vol} (f_+^t (\Omega))
 = \mathrm{Vol} (f_-^t (\Omega)) = \int_{\Omega} \det (I + t f_-' (x)) \mathd
 x \text{.} \]
Since the two sides are polynomials in $t$, comparing the coefficients of
$t^n$ gives (\ref{e1}). To make the argument rigorous some issues including
why $f_{\pm}^t (\partial \Omega) = \partial f_{\pm}^t (\Omega)$ and why
$\partial f_+^t (\Omega) = \partial f_-^t (\Omega)$ implies $f_+^t (\Omega) =
f_-^t (\Omega)$ need to be handled. These are clarified via a careful point
set analysis in a long remark following the proof. One of the advantages of
this argument is that the maps $f_{\pm}$ only need to be $C^1$.

Theorem \ref{l1} also appears in {\cite{MR2597943,MR1046451}}. In
{\cite{MR1046451}}, Kulpa defines
\[ I (h^1, \ldots, h^n) = \int_{\Omega} \det h' (x) \mathd x \]
for smooth $h : \bar{\Omega} \rightarrow \mathbb{R}^n$ and manages to show
\begin{equation}
 I (f^1_+, f^2_+, \ldots, f_+^n) = I (f^1_-, f^2_+, \ldots, f_+^n) \label{e3}
\end{equation}
using Fubini theorem and projections $\Pi_i : \mathbb{R}^n \rightarrow
\mathbb{R}^{n - 1}$ given by deleting the $i$-th component. Then (\ref{e1})
follows by applying (\ref{e3}) successively to replace the remaining columns
of $\det f_+' (x)$ by those of $\det f_-' (x)$. While in {\citet[pp.\
464]{MR2597943}}, (\ref{e1}) is proved using the facts that $L (P) = \det P$
is a null Lagrangian and the energy $\int_{\Omega} L (w' (x), w (x), x) \mathd
x$ of null Lagrangian $L : \mathbb{M}^{m \times n} \times \mathbb{R}^m \times
\bar{\Omega} \rightarrow \mathbb{R}$ depends only on boundary conditions.

Let $f^i_{\pm}$ be the $i$-th component of $f_{\pm}$ and $\Phi_{\pm} =
(f_{\pm}^2, \ldots, f^n_{\pm})$. Then $\Phi_+ = \Phi_-$ on $\partial \Omega$.
Let
\[ A_{\pm} = \left( \frac{\partial (f_{\pm}^2, \ldots, f_{\pm}^n)}{\partial
 (x^2, \ldots, x^n)}, - \frac{\partial (f_{\pm}^2, f_{\pm}^3, \ldots,
 f_{\pm}^n)}{\partial (x^1, x^3, \ldots, x^n)}, \ldots, (- 1)^{n + 1}
 \frac{\partial (f_{\pm}^2, \ldots, f_{\pm}^n)}{\partial (x^1, \ldots, x^{n
 - 1})} \right) \text{,} \]
then the $i $-th component of $A_{\pm}$ is just the cofactor of
$\partial_{x^i} f_{\pm}^1$ in $\det f_{\pm}' (x)$. By Jacobi identity (see
e.g.\ {\citet[Eq.\ (5)]{MR610487}} or {\citet[Eq.\ (0.3)]{MR4701455}}), we have
$\mathrm{div} A_{\pm} = 0$. Thus
\begin{equation}
 \mathrm{div} (f_{\pm}^1 A_{\pm}) = \nabla f_{\pm}^1 \cdummy A_{\pm} +
 f_{\pm}^1 \mathrm{div} A_{\pm} = \nabla f_{\pm}^1 \cdummy A_{\pm} = \det
 f_{\pm}' (x) \text{.} \label{d}
\end{equation}
For $a \in \partial \Omega$, let $\eta : (u^1, \ldots, u^{n - 1}) \mapsto x$
be a local parametrization of $\partial \Omega$ at $a = \eta (\alpha)$ such
that
\[ N(a) = \left( \frac{\partial (x^2, \ldots, x^n)}{\partial (u^1, \ldots, u^{n -
 1})}, - \frac{\partial (x^1, x^3, \ldots, x^n)}{\partial (u^1, \ldots, u^{n
 - 1})}, \ldots, (- 1)^{n + 1} \frac{\partial (x^1, \ldots, x^{n -
 1})}{\partial (u^1, \ldots, u^{n - 1})} \right)_{\alpha} \]
is an outward normal vector of $\partial \Omega$ at $a$. Applying the chain
rule to the composition $y_{\pm} = \Phi_{\pm} (\eta (u))$, we see that
\[ \left( \begin{array}{ccc}
 \dfrac{\partial y_{\pm}^2}{\partial u^1} & \cdots & \dfrac{\partial
 y_{\pm}^2}{\partial u^{n - 1}}\\
 & & \\
 \dfrac{\partial y_{\pm}^n}{\partial u^1} & \cdots & \dfrac{\partial
 y_{\pm}^n}{\partial u^{n - 1}}
 \end{array} \right)_{\alpha} = \left( \begin{array}{ccc}
 \dfrac{\partial y_{\pm}^2}{\partial x^1} & \cdots & \dfrac{\partial
 y_{\pm}^2}{\partial x^n}\\
 & & \\
 \dfrac{\partial y_{\pm}^n}{\partial x^1} & \cdots & \dfrac{\partial
 y_{\pm}^n}{\partial x^n}
 \end{array} \right)_a \left( \begin{array}{ccc}
 \dfrac{\partial x^1}{\partial u^1} & \cdots & \dfrac{\partial
 x^1}{\partial u^{n - 1}}\\
 & & \\
 \dfrac{\partial x^n}{\partial u^1} & \cdots & \dfrac{\partial
 x^n}{\partial u^{n - 1}}
 \end{array} \right)_{\alpha} \text{,} \]
where the matrix on the left is the Jacobian matrix $(\Phi_{\pm} \circ \eta)'
(\alpha)$.

Because $\Phi_+ = \Phi_-$ on $\partial \Omega$ and $\eta (u) \in \partial
\Omega$ for $u$ near $\alpha$, we have $\Phi_+ \circ \eta = \Phi_- \circ \eta$
near $\alpha$. Using the Cauchy-Binet formula we deduce
\begin{align*}
 (A_+ \cdummy N) (a) & = \sum_{i = 1}^n \left. \frac{\partial (f_+^2,
 \ldots, f_+^n)}{\partial (x^1, \ldots, \hat{x}^i, \ldots, x^n)} \right|_a
 \cdummy \left. \frac{\partial (x^1, \ldots, \hat{x}^i, \ldots,
 x^n)}{\partial (u^1, \ldots, u^{n - 1})} \right|_{\alpha}\\
 & = \det (\Phi_+ \circ \eta)' (\alpha) = \det (\Phi_- \circ \eta)'
 (\alpha) = (A_- \cdummy N) (a) \text{.}
\end{align*}
Therefore
\begin{equation}
 A_+ \cdummy \nu = \frac{A_+ \cdummy N}{| N |} = \frac{A_- \cdummy N}{| N |}
 = A_- \cdummy \nu \label{e5}
\end{equation}
on $\partial \Omega$, where $\nu = N / | N |$ is outward unit normal vector
field on $\partial \Omega$.

Since $f^1_+ = f^1_-$ on $\partial \Omega$, using the divergence theorem we
deduce from (\ref{d}) and (\ref{e5})
\begin{align*}
 \int_{\Omega} \det f_+' (x) \mathd x & = \int_{\Omega} \mathrm{div} (f_+^1
 A_+) \mathd x = \int_{\partial \Omega} f_+^1 A_+ \cdummy \nu \mathd \sigma\\
 & = \int_{\partial \Omega} f_-^1 A_- \cdummy \nu \mathd \sigma =
 \int_{\Omega} \det f_-' (x) \mathd x \text{.}
\end{align*}
\begin{example}
 Let $A$ be an $n \times n$ matrix, $\varphi \in C_0^{\infty} (\Omega,
 \mathbb{R}^n)$. Set $f_+ (x) = A x + \varphi (x)$, $f_- (x) = A x$, we get
 the formula stated in {\citet[pp. 21]{MR3887613}}:
 \[ \int_{\Omega} \det (A + \varphi' (x)) \mathd x = \det (A) \cdummy m
 (\Omega) \text{.} \]
\end{example}

To conclude this note, we present another proof using differential forms. Let
$f^i$ be the components of $f:\bar{\Omega}\to\mathbb{R}^n$, then it is well known that
\begin{align*}
 \det f' (x) \mathd x^1 \wedge \cdots \wedge \mathd x^n & = \mathd f^1
 \wedge \mathd f^2 \wedge \cdots \wedge \mathd f^n \text{.}\\
 & = \mathd (f^1 \mathd f^2 \wedge \cdots \wedge \mathd f^n) \text{,}
\end{align*}
where in the second equality we used $\mathd^2 = 0$. Recall the Stokes formula
\[ \int_{\Omega} \mathd \omega = \int_{\partial \Omega} i^{\ast} \omega
 \text{,} \]
where $\omega$ is an $(n - 1)$-form on $\Omega$ and $i^{\ast}$ is the pull
back induced by the embedding $i : \partial \Omega \rightarrow \Omega$.
Because $i^{\ast} \circ \mathd = \mathd \circ i^{\ast}$, we conclude
\begin{align*}
 \int_{\Omega} \det f' (x) \mathd x & = \int_{\Omega} \det f' (x) \mathd
 x^1 \wedge \cdots \wedge \mathd x^n\\
 & = \int_{\Omega} \mathd (f^1 \mathd f^2 \wedge \cdots \wedge \mathd f^n)
 = \int_{\partial \Omega} i^{\ast} (f^1 \mathd f^2 \wedge \cdots \wedge
 \mathd f^n)\\
 & = \int_{\partial \Omega} (f^1 \circ i) \mathd (f^2 \circ i) \wedge
 \cdots \wedge \mathd (f^n \circ i) \text{.}
\end{align*}
Clearly the right hand side depends only on $f \circ i = f |_{\partial \Omega}
 $.

Using this theorem, the \emph{no-retraction theorem} follows immediately, see \citet[Corollary
1]{MR4725395}. Let $B$ be the closed unit ball in
$\mathbb{R}^n$. If there was a smooth retraction $R : B \rightarrow \partial
B$, since $R |_{\partial B}   = \mathrm{id} |_{\partial B}  $,
Theorem \ref{l1} with $f_+ = R$, $f_- = \mathrm{id}$ yields a contradiction
\begin{equation}
  \int_B \det R' (x) \mathrm{d} x = \int_B \mathrm{d} x = \mathrm{Vol} (B) > 0 \text{,}
  \label{e2}
\end{equation}
because $\det R' (x) = 0$, a consequence of $R (B) \subset \partial B$. That the two
integrals in (\ref{e2}) are equal also follows from a version of the changing
variable formula given in {\citet[Theorem 3.1]{MR3711061}}: \emph{For smooth closed
bounded domain $D$ in $\mathbb{R}^n$ and smooth map $\varphi : B \rightarrow
D$, if $\varphi : \partial B \rightarrow \partial D$ is a diffeomorphism, then
for continuous $f : D \rightarrow \mathbb{R}$ there holds}
\[ \int_D f (y) \mathrm{d} y = \pm \int_B f (\varphi (x)) \det \varphi' (x) \mathrm{d}
   x \text{.} \]
The first equality in (\ref{e2}) follows by letting $f = 1$, $D = B$ and
$\varphi = R$ in this formula. It is well known that the no-retraction theorem
is equivalent to the famous Brouwer fixed point theorem.

Krylov's proof of Theorem \ref{l1} in {\cite{MR4725395}} is based on the
observation: for small $t$, $f_{\pm}^t = \mathrm{id} + t f_{\pm}$ are
diffeomorphisms and $f^t_+ (\Omega) = f_-^t (\Omega)$ because $f_+ |_{\partial
\Omega}   = f_- |_{\partial \Omega}  $, thus by the changing
variable formula
\[ \int_{\Omega} \det (I + t f_+' (x)) \mathrm{d} x = \mathrm{Vol} (f_+^t (\Omega))
   = \mathrm{Vol} (f_-^t (\Omega)) = \int_{\Omega} \det (I + t f_-' (x)) \mathrm{d}
   x \text{.} \]
Since the two sides are polynomials in $t$, comparing the coefficients of
$t^n$ gives (\ref{e1}). To make the argument rigorous some issues including
why $f_{\pm}^t (\partial \Omega) = \partial f_{\pm}^t (\Omega)$ and why
$\partial f_+^t (\Omega) = \partial f_-^t (\Omega)$ implies $f_+^t (\Omega) =
f_-^t (\Omega)$ need to be handled. These are clarified via a careful point
set analysis in a long remark following the proof. One of the advantages of
this argument is that the maps $f_{\pm}$ only need to be $C^1$.

Theorem \ref{l1} also appears in {\cite{MR2597943,MR1046451}}. In
{\cite{MR1046451}}, Kulpa defines
\[ I (h^1, \ldots, h^n) = \int_{\Omega} \det h' (x) \mathrm{d} x \]
for smooth $h : \bar{\Omega} \rightarrow \mathbb{R}^n$ and manages to show
\begin{equation}
  I (f^1_+, f^2_+, \ldots, f_+^n) = I (f^1_-, f^2_+, \ldots, f_+^n) \label{e3}
\end{equation}
using Fubini theorem and projections $\Pi_i : \mathbb{R}^n \rightarrow
\mathbb{R}^{n - 1}$ given by deleting the $i$-th component. Then (\ref{e1})
follows by applying (\ref{e3}) successively to replace the remaining columns
of $\det f_+' (x)$ by those of $\det f_-' (x)$. While in {\citet[pp.\
464]{MR2597943}}, (\ref{e1}) is proved using the facts that $L (P) = \det P$
is a null Lagrangian and the energy $\int_{\Omega} L (w' (x), w (x), x) \mathrm{d}
x$ of null Lagrangian $L : \mathbb{M}^{m \times n} \times \mathbb{R}^m \times
\bar{\Omega} \rightarrow \mathbb{R}$ depends only on boundary conditions.

The purpose of this note is to present another two proofs of Theorem \ref{l1}.
The first one is performed by playing classical analysis, which is written for
readers not familiar with differential forms and in our opinion is very
elementary and transparent. The second one is quite short, which depends on
differential form and Stokes formula on manifolds. As demonstrated in the
paragraph after Theorem \ref{l1}, our note is useful for understanding the
Brouwer fixed point theorem from the analytic point of view. In addition to
{\cite{MR2597943,MR4725395,MR1046451,MR3711061}} mentioned above, other
analytic proofs of the Brouwer fixed point theorem can be found in \cite[pp.\ 467--470]{MR117523} and
{\cite{MR610487,MR1699248,MR505523,MR600910}}.

Krylov's proof of Theorem~\ref{l1} is based on the observation that for small $t$, $f_{\pm}^t=\mathrm{id}+t f_{\pm}$ are diffeomorphisms and $f_+^t(\Omega)=f_-^t(\Omega)$ because $f_+|_{\partial\Omega}=f_-|_{\partial\Omega}$. Hence, by the change-of-variables formula,
\[\int_{\Omega}\det(I+t f_+'(x))\,dx=\mathrm{Vol}(f_+^t(\Omega))=\mathrm{Vol}(f_-^t(\Omega))=\int_{\Omega}\det(I+t f_-'(x))\,dx.\]
Since both sides are polynomials in $t$, comparing the coefficients of $t^n$ gives \eqref{e1}. To make this rigorous, one must handle issues such as why $f_{\pm}^t(\partial\Omega)=\partial f_{\pm}^t(\Omega)$ and why $\partial f_+^t(\Omega)=\partial f_-^t(\Omega)$ implies $f_+^t(\Omega)=f_-^t(\Omega)$; these are addressed by “a careful point set analysis”. The same paragraph notes an advantage: the argument only requires $f_{\pm}$ to be $C^1$.

Another proof approach (Kulpa) defines $I(h^1,\dots,h^n)=\int_{\Omega}\det h'(x)\,dx$ and shows
\[I(f_+^1,f_+^2,\ldots,f_+^n)=I(f_-^1,f_+^2,\ldots,f_+^n)\tag{\ref{e3}}\]
using Fubini’s theorem and coordinate projections $\Pi_i$. Then \eqref{e1} follows by applying \eqref{e3} successively to replace the remaining columns of $\det f_+'(x)$ by those of $\det f_-'(x)$. A further cited route proves \eqref{e1} by using that $L(P)=\det P$ is a null Lagrangian, so the energy $\int_{\Omega}L(w'(x),w(x),x)\,dx$ “depends only on boundary conditions”.

Krylov's proof of the main theorem is based on the observation that for small $t$, $f_{\pm}^t=\mathrm{id}+t f_{\pm}$ are diffeomorphisms and $f_+^t(\Omega)=f_-^t(\Omega)$ because $f_+|_{\partial\Omega}=f_-|_{\partial\Omega}$. Hence, by the change-of-variables formula,
\[\int_{\Omega}\det(I+t f_+'(x))\,dx=\mathrm{Vol}(f_+^t(\Omega))=\mathrm{Vol}(f_-^t(\Omega))=\int_{\Omega}\det(I+t f_-'(x))\,dx.\]
Since both sides are polynomials in $t$, comparing the coefficients of $t^n$ gives
\begin{equation}
    \int_{\Omega} \det f_+' (x) \mathd x = \int_{\Omega} \det f_-' (x) \mathd
    x \text{.} \label{e1}
  \end{equation}
To make this rigorous, one must handle issues such as why $f_{\pm}^t(\partial\Omega)=\partial f_{\pm}^t(\Omega)$ and why $\partial f_+^t(\Omega)=\partial f_-^t(\Omega)$ implies $f_+^t(\Omega)=f_-^t(\Omega)$; these are addressed by “a careful point set analysis”. The same paragraph notes an advantage: the argument only requires $f_{\pm}$ to be $C^1$.

Another proof approach (Kulpa) defines $I(h^1,\dots,h^n)=\int_{\Omega}\det h'(x)\,dx$ and shows
\begin{equation}
  I (f^1_+, f^2_+, \ldots, f_+^n) = I (f^1_-, f^2_+, \ldots, f_+^n) \label{e3}
\end{equation}
using Fubini’s theorem and coordinate projections $\Pi_i$. Then the equation above gives \eqref{e1} by applying the preceding identity successively to replace the remaining columns of $\det f_+'(x)$ by those of $\det f_-'(x)$. A further cited route proves the equation above by using that $L(P)=\det P$ is a null Lagrangian, so the energy $\int_{\Omega}L(w'(x),w(x),x)\,dx$ “depends only on boundary conditions”.

Question: Let \(\Omega\) be a bounded domain in \(\mathbb{R}^n\). Let \(f_{\pm}:\overline{\Omega}\to\mathbb{R}^n\) be \(C^2\) maps such that their boundary values agree, i.e. \(f_+|_{\partial\Omega}=f_-|_{\partial\Omega}\). Here \(f_\pm'(x)\) denotes the Jacobian matrix of \(f_\pm\) at \(x\), and \(\det f_\pm'(x)\) its Jacobian determinant. Which statement holds for every such pair of maps?

\[\int_{\Omega} \det f_+'(x)\,dx = \int_{\Omega} \det f_-'(x)\,dx.\]

Choices:
B. \[\det f_+'(x)=\det f_-'(x)\quad\text{for almost every }x\in\Omega.\]
C. \[\text{If }f_-\text{ is constant on }\overline\Omega,\text{ then }\int_{\Omega} \det f_+'(x)\,dx=0.\]
D. \[\int_{\Omega} \bigl|\det f_+'(x)\bigr|\,dx = \int_{\Omega} \bigl|\det f_-'(x)\bigr|\,dx.\]
E. \[\int_{\Omega} \det\!igl(I+f_+'(x)\bigr)\,dx = \int_{\Omega} \det\!igl(I+f_-'(x)\bigr)\,dx.\]


Harmonic functions have long been a central object of study in geometric analysis. Among the most celebrated and foundational results in the field is Yau’s Liouville theorem \cite{Yau75}, which asserts that on a complete Riemannian manifold with nonnegative Ricci curvature, any bounded harmonic function must be constant. This theorem not only captures a fundamental rigidity phenomenon, but also inaugurated a far-reaching program aimed at understanding how analytic growth conditions reflect the large-scale geometry of the underlying manifold.

\medskip
Since Yau’s work, the study of harmonic functions and spaces of harmonic functions has developed extensively. Notable contributions include the work of Li and Tam \cite{LiTam1989}, as well as Colding and Minicozzi \cite{ColdingMinicozzi97, ColdingMinicozziAnnals}, who investigated the structure and finite dimensional properties of spaces of harmonic functions of polynomial growth. Further important developments include results of Ni and Tam \cite{NiTam2003} on pluriharmonic functions, and more recently Liu’s work on holomorphic functions via three--circle theorems \cite{LiuThreeCircle}. Collectively, these results demonstrate that under nonnegative Ricci curvature, harmonic functions exhibit rigidity properties closely resembling those of the Euclidean setting.

\medskip
By contrast, Liouville-type phenomena for higher-order elliptic equations have received comparatively little attention. This gap is striking given the intrinsic analytic and geometric relevance of higher-order operators, such as the Paneitz operator in conformal geometry or the operators arising in the study of polyharmonic flows for curves. Aside from a few notable exceptions, there are very few results addressing Liouville-type properties in this broader context. Among these, we mention the works of Branding \cite{Branding18, Branding2021}, where Liouville-type theorems are obtained for biharmonic \emph{maps} under smallness assumptions on certain Sobolev norms.

\medskip
Recently, the authors proved the following result (see \cite{BravoCortissoz2025}, as well as the related work \cite{WangZhu2025}).

\medskip
Recently, the authors proved the following result (see \cite{BravoCortissoz2025}, as well as the related work \cite{WangZhu2025}).

The purpose of the present paper is to extend Theorem~\ref{thm:biharmonic} to the general polyharmonic setting. Recall that a $k$-polyharmonic function on a Riemannian manifold $(M,g)$ is a function $u\in C^{2k}(M)$ satisfying
\[
\Delta^k u = 0,
\]
where $\Delta$ denotes the Laplacian.

\begin{theorem}
\label{thm:main}
Let $(M,g)$ be a complete Riemannian manifold with
$\mbox{Ric}_g\geq 0$.
Then any $k$-polyharmonic function $u\in C^{2k}(M)$ satisfying
\[
u=o(r^{2(k-1)}) \quad \mbox{as}\quad r\rightarrow \infty,
\]
where $r=d(p,x)$, is $(k-1)$-polyharmonic.
In particular, any polyharmonic function of sublinear growth must be constant.
\end{theorem}

\begin{lemma}[Cutoff with explicit bounds]\label{lem:cutoff}
Let $(M^n,g)$ be a complete Riemannian manifold
with $\mbox{Ric}_g\geq 0$
Set $B_\rho:=B_\rho(p)$. Let $r$ be such that and for $0<\frac{1}{8}R\leq r\leq R$.
Then there exists $\chi\in C_c^\infty(B_R)$ such that
\[
0\le \chi\le 1,\qquad
\chi\equiv 1\ \text{on } B_r,\qquad
\mbox{supp}\left(\chi\right)\subset B_R,
\]
and
\[
\left|\nabla \chi\right|\le \frac{C(n)}{R-r},
\quad
|\Delta\chi|\le \frac{C(n)}{(R-r)^{2}}
\quad \text{on}\quad B_R.
\]
\end{lemma}
\begin{proof}
 Take $\chi=\phi$, where $\phi$ is as in Corollary 2.3 in
 \cite{BianchiSetti2018}. Notice that given $r$, we can write $R=\gamma r$ with $\gamma>1$, and in this case,
 we obtain the following estimates for the gradient and Laplacian of the cutoff function.
 \[
 \left|\nabla \chi\right|\leq \frac{C_1}{R}=\frac{\left(\gamma-1\right)C_1}{\left(\gamma-1\right) R}\leq
 \frac{7C_1}{R-r},
 \]
 where we used that $r\geq \frac{1}{8}R$.

\begin{lemma}\label{cor:L2-Estimate Laplacian Biharmonic}
Let $(M,g)$ be an $n$–dimensional complete Riemannian manifold with $\mbox{Ric}_g\ge 0$.
Let $a>0$ and let $u\in C^\infty(B_{a})$ be Biharmonic, i.e.\ $\Delta^2 u=0$ on $B_{a}$.
Then, there is a constant $C>0$ such that for every $s,r\in(0,a)$, with $0<\frac{1}{8}r<s < r<a$
\begin{equation}\label{eq:L2-HF-polh}
\int_{B_s}|\Delta u|^2 dV \leq \frac{C}{(r-s)^4}\int_{B_{r}} u^2 dV.
\end{equation}

\subsection{Proof of Main Theorem \ref{thm:main}}
\label{sect:proof_main_thm}
We are ready to prove our main result. Let $u$ be a $k$-polyharmonic function.
By Corollary \ref{cor:L2-Estimate Laplacian k-1} we can estimate
\begin{equation*}
\label{ineq:Cacciopoli_laplacian}
\int_{B_{r}}|\Delta^{k-1} u|^2\,dV
\leq \frac{C'}{r^{4(k-1)}}\int_{B_{2^{k-1}r}}u^2\,dV.
\end{equation*}
By the volume doubling property (which holds if $\mbox{Ric}\geq 0$), we have
\begin{eqnarray*}
\frac{1}{\mbox{Vol}\left(B_{r}\right)}\int_{B_{r}}|\Delta^{k-1} u|^2\,dV
&\leq& \frac{C'}{r^{4(k-1)}}\frac{1}{\mbox{Vol}\left(B_{2^{k-1}r}\right)}\int_{B_{2^{k-1}r}}u^2\,dV,
\end{eqnarray*}
and hence, since $\Delta^{k-1} u$ is harmonic, then $|\Delta^{k-1} u|$ is subharmonic, by the mean value inequality (Lemma \ref{lemma:mean_value_ineq}),
\begin{eqnarray*}
\sup_{x\in B_{r/2}}\left|\Delta^{k-1} u\right|^2 &\leq&
\frac{C''}{r^{4(k-1)}}\frac{1}{\mbox{Vol}\left(B_{2^{k-1}r}\right)}\int_{B_{2^{k-1}r}}u^2\,dV\\
&\leq& \frac{C''}{r^{4(k-1)}}\sup_{x\in B_r\left(p\right)}\left|u\left(x\right)\right|^2
=\frac{1}{r^{4(k-1)}}o\left(r^{4(k-1)}\right)
\end{eqnarray*}
and hence if $u=o(r^{2(k-1)})$, letting $r\rightarrow \infty$, it
follows that
\[
\Delta^{k-1} u \equiv 0,
\]
i.e., $u$ is $(k-1)$-polyharmonic. For the last statement, let $u$ be a polyharmonic function of
sublinear growth, i.e.
\[
\sup_{B_r}|u| = o(r) \quad \text{as } r\to\infty.
\]
For every $k\ge2$ we have
\[
\frac{\sup_{B_r}|u|}{r^{2(k-1)}}
= o\big(r^{1-2(k-1)}\big) \;\longrightarrow\;0.
\]
In particular $u = o\big(r^{2(k-1)}\big)$ for each $k\ge2$.
Assume that $u$ is $k$–polyharmonic for some $k\ge2$. By the first
part of the theorem, $u$ is then $(k-1)$–polyharmonic. Iterating
this argument for $k,k-1,\dots,2$, we conclude that $u$ is harmonic
and it is sublinear growth.

As stated in the introduction, inequality \eqref{eq:Cacciopoli_Laplacian}
does not appear in \cite{BravoCortissoz2025}. However,
the estimate can be inferred from the proofs of Lemmas 5 and 6
in \cite{BravoCortissoz2025} as follows. From the proof of the local
energy estimate for the Laplacian, Lemma 5, for a cutoff function
$\varphi\in C_c^\infty(B_{2r})$ satisfying for $0<\frac{1}{8}r<s<r$
\[
\varphi\equiv1 \text{ on }B_s,\quad 0\le\varphi\le1,\quad
|\nabla\varphi|\le \frac{C_0}{r-s}\varphi,\quad
|\Delta\varphi|\le \frac{C_0}{(r-s)^2},
\]
we have
\begin{lemma}
\label{lem:laplacian_estimate}
Let $(M,g)$ be an $n$–dimensional complete Riemannian manifold with $\mathrm{Ric}_g\ge 0$. Let $B_{\rho}\colon = B_{\rho}\left(p\right)$.
Fix $a>0$ and let $u\in C^\infty(B_{a})$ be biharmonic, i.e.\ $\Delta^2 u=0$ on $B_{2a}$. Then there exists a constant $C=C(n)>0$ such that, for every pair of radii $0<\dfrac{1}{8}r<s<r\le a$,
\[
\int_{B_s}|\Delta u|^2\,dV
\leq \frac{C}{(r-s)^2}\int_{B_r}|\nabla u|^2\varphi^2\,dV
\;+\; \frac12\int_{B_r} |\Delta u|^2\,dV.
\]
\end{lemma}

From the proof of Lemma 6, the Cacciopoli inequality
for a biharmonic function, for a cutoff
$\varphi\in C_c^\infty(B_{2r})$ such that $0\leq \varphi\leq 1$,
$\varphi\equiv 1$ on $B_s$, and for $\dfrac{1}{8}r<s<r$, it also
satisfies
\[
\left|\nabla \varphi\right|^2\leq \frac{C}{\left(r-s\right)^2}\varphi\quad
 \text{ and } \quad
 \frac{\left|\Delta\left(\varphi^4\right)\right|^2}{\varphi^4}\leq \frac{C}{\left(r-s\right)^4},
\]
we have
\begin{lemma}
\label{lem:Caccio-1-biharmonic-Br}
Let $(M,g)$ be a smooth Riemannian manifold
such that $\mbox{Ric}\geq 0$.
Let $0<\dfrac{1}{8}r<s<r<\infty$
and let $B_\rho:=B_\rho(p)$. If $u\in C^\infty(B_{2r})$ is \emph{biharmonic}, i.e.\ $\Delta^2 u=0$ in $B_{2r}$, then there exists a constant $C=C(n)>0$ such that
\begin{equation*}\label{eq:Caccio-int-Br}
\int_{B_r} |\nabla u|^2\varphi^2\,dV \;\le\; \frac{C}{(r-s)^2}\int_{B_r} u^2\,dV.
\end{equation*}
\end{lemma}

\begin{equation}
\label{eq:Cacciopoli_Laplacian}
\int_{B_s}|\Delta u|^2\,dV \leq \frac{C}{(r-s)^4}\int_{B_r}u^2\,dV,
\end{equation}

\begin{theorem}
\label{thm:biharmonic}
Let $(M,g)$ be a complete Riemannian manifold with a pole and such that
$\mbox{Ric}_g\geq 0$.
Then any biharmonic function $u\in C^{4}(M)$ of subquadratic growth is harmonic.
In particular, any biharmonic function of sublinear growth must be constant.
\end{theorem}

The purpose of the present paper is to extend Theorem~\ref{thm:biharmonic} to the general polyharmonic setting. Recall that a $k$-polyharmonic function on a Riemannian manifold $(M,g)$ is a function $u\in C^{2k}(M)$ satisfying
\[
\Delta^k u = 0,
\]
where $\Delta$ denotes the Laplacian.

\medskip
Our main result is the following.

\begin{theorem}
\label{thm:main}
Let $(M,g)$ be a complete Riemannian manifold with
$\mbox{Ric}_g\geq 0$.
Then any $k$-polyharmonic function $u\in C^{2k}(M)$ satisfying
\[
u=o(r^{2(k-1)}) \quad \mbox{as}\quad r\rightarrow \infty,
\]
where $r=d(p,x)$, is $(k-1)$-polyharmonic.
In particular, any polyharmonic function of sublinear growth must be constant.
\end{theorem}

\medskip
As is well known, Theorem~\ref{thm:main} has long been established in the Euclidean setting. In $\mathbb{R}^n$, the critical growth rate that separates genuinely $k$-polyharmonic behavior from lower-order behavior is precisely $r^{2(k-1)}$, as follows from Almansi’s decomposition. This shows that Theorem~\ref{thm:main} is sharp and constitutes a natural geometric extension of the classical Euclidean theory.

\medskip
We now briefly describe the main analytic difficulty underlying the proof of Theorem~\ref{thm:main}. We begin by recalling the strategy used in the proof of Theorem~\ref{thm:biharmonic}. In \cite{BravoCortissoz2025}, the argument relied on an energy estimate for the $L^2$-norm of $\Delta u$ together with a Caccioppoli-type inequality,
\begin{equation}
\label{eq:Cacciopoli}
\int_{B_s}|\nabla u|^2\,dV \leq \frac{C}{(r-s)^2}\int_{B_r}u^2\,dV,
\end{equation}
valid for both harmonic and biharmonic functions. This estimate was then used to derive
\begin{equation}
\label{eq:Cacciopoli_Laplacian}
\int_{B_s}|\Delta u|^2\,dV \leq \frac{C}{(r-s)^4}\int_{B_r}u^2\,dV,
\end{equation}
which, combined with a mean value inequality, yields the desired rigidity.

\medskip
At first glance, these arguments suggest the possibility of an inductive approach to Theorem~\ref{thm:main}. However, the proof of \eqref{eq:Cacciopoli} is fundamentally based on the Bochner formula, and this approach does not extend in any straightforward way to polyharmonic functions beyond the biharmonic case. To overcome this obstacle, we establish a direct analogue of \eqref{eq:Cacciopoli_Laplacian} for \emph{arbitrary polyharmonic functions}. This new estimate is the most delicate part of the argument and forms the analytic backbone of the inductive procedure. Although the biharmonic case serves as the base step, the proof requires a genuinely new estimate that may be of independent interest.

\medskip
Taken together with Yau’s theorem and our previous work on biharmonic functions, the results of this paper complete, on manifolds with nonnegative Ricci curvature, a natural hierarchy of Yau--Liouville type theorems for the Laplacian and its iterates. They also raise the question of whether similar rigidity phenomena persist for more general higher-order elliptic operators, as they do in the Euclidean setting.

\medskip
Beyond its analytic content, Theorem~\ref{thm:main} may be interpreted as a rigidity principle for complete manifolds with nonnegative Ricci curvature. It shows that, at the level of critical growth, such manifolds are analytically indistinguishable from Euclidean space with respect to iterates of the Laplacian. In particular, the behavior of polyharmonic function in
nonnegative Ricci curvature seems to mirror Almansi’s decomposition in $\mathbb{R}^n$.
\subsection{}

The interplay between the behavior of polyharmonic functions and curvature remains largely unexplored and may reveal further rigidity phenomena. As shown in this paper, when $\mbox{Ric}_g\geq 0$, the qualitative behavior closely parallels that of Euclidean space. In contrast, in negatively curved settings the situation is markedly different. In particular, the authors showed in \cite{BravoCortissoz} that there exist curvature regimes in which every bounded biharmonic function must be harmonic, while nonconstant bounded harmonic functions still exist. This naturally leads to the question of how $k$-polyharmonic functions behave in such settings.

\subsection{}

In a previous version of this manuscript, we imposed one of the following additional assumptions:
\begin{itemize}
\item[(H1)] the geodesic spheres centered at $p$ are mean-convex with respect to the radial vector field, or
\item[(H2)] the square of the distance function from $p$, denoted by $d$, is convex.
\end{itemize}
Either assumption yields a lower bound for the Laplacian of the distance function, which is needed to construct suitable cutoff functions. However, we later observed that the condition $\mbox{Ric}_g\geq 0$, together with the existence of a pole, already implies (H1), rendering it redundant (see \cite{BravoCortissoz2025}). Subsequently, inspired by the recent work \cite{WangZhu2025} and with the generous assistance of Professor Wang, we adopted the cutoff technology developed in \cite{BianchiSetti2018}, which requires only the assumption $\mbox{Ric}_g\geq 0$. We are grateful to Professors Wang and Zhu for their help.

\subsection{Organization of the paper}
The material of this paper is organized as follows. In Section
\ref{section:technical}, we collect some facts about cutoff functions
and the hole-filling technique needed in the proof of the main
$L^2$-estimates needed in the proof of our main result. These
$L^2$ estimates are presented and proved in Section \ref{sect:L2estimates}.
Finally, Theorem \ref{thm:main} is proved in Section \ref{sect:proof_main_thm}.

To prove Theorem~\ref{thm:main}, the introduction proposes an inductive strategy inspired by the biharmonic case (Theorem~\ref{thm:biharmonic}). In \cite{BravoCortissoz2025}, the biharmonic argument uses an energy estimate for the $L^2$-norm of $\Delta u$ together with a Caccioppoli-type inequality
\[
\int_{B_s}|\nabla u|^2\,dV \leq \frac{C}{(r-s)^2}\int_{B_r}u^2\,dV,
\]
which then yields
\[
\int_{B_s}|\Delta u|^2\,dV \leq \frac{C}{(r-s)^4}\int_{B_r}u^2\,dV.
\]
Combined with a mean value inequality, this gives the rigidity conclusion in the biharmonic case.

For Theorem~\ref{thm:main}, the text notes that while this suggests induction, the proof of the basic Caccioppoli inequality is “fundamentally based on the Bochner formula,” and “does not extend in any straightforward way to polyharmonic functions beyond the biharmonic case.” The stated remedy is to “establish a direct analogue of \eqref{eq:Cacciopoli_Laplacian} for \emph{arbitrary polyharmonic functions}.” This new $L^2$ estimate is described as “the most delicate part of the argument” and as “the analytic backbone of the inductive procedure,” with the biharmonic case as the base step but requiring “a genuinely new estimate.”

The introduction also indicates the technical tools used to implement this: cutoff-function technology under $\mathrm{Ric}_g\ge0$ (adopting the approach of \cite{BianchiSetti2018}) and the “hole-filling technique,” and it outlines where the proof components appear: cutoff/hole-filling facts in Section~\ref{section:technical}, the main $L^2$ estimates in Section~\ref{sect:L2estimates}, and the proof of Theorem~\ref{thm:main} in Section~\ref{sect:proof_main_thm}.

To prove Theorem~\ref{thm:main}, the introduction proposes an inductive strategy inspired by the biharmonic case. We first prove the following theorem.
\begin{theorem}
\label{thm:biharmonic}
Let $(M,g)$ be a complete Riemannian manifold with a pole and such that
$\mbox{Ric}_g\geq 0$.
Then any biharmonic function $u\in C^{4}(M)$ of subquadratic growth is harmonic.
In particular, any biharmonic function of sublinear growth must be constant.
\end{theorem}
In Bravo, Cortissoz, \emph{TITLE UNKNOWN} (2025), the biharmonic argument uses an energy estimate for the $L^2$-norm of $\Delta u$ together with a Caccioppoli-type inequality
\[
\int_{B_s}|\nabla u|^2\,dV \leq \frac{C}{(r-s)^2}\int_{B_r}u^2\,dV,
\]
which then yields the estimate
\begin{equation}
\label{eq:Cacciopoli_Laplacian}
\int_{B_s}|\Delta u|^2\,dV \leq \frac{C}{(r-s)^4}\int_{B_r}u^2\,dV,
\end{equation}
Combined with a mean value inequality, this gives the rigidity conclusion in establishing the main theorem.

For Theorem~\ref{thm:main}, the text notes that while this suggests induction, the proof of the basic Caccioppoli inequality is “fundamentally based on the Bochner formula,” and “does not extend in any straightforward way to polyharmonic functions beyond the biharmonic case.” The stated remedy is to “establish a direct analogue of the equation above for \emph{arbitrary polyharmonic functions}.” This new $L^2$ estimate is described as “the most delicate part of the argument” and as “the analytic backbone of the inductive procedure,” with the base step provided by the theorem above but requiring “a genuinely new estimate.”

The introduction also indicates the technical tools used to implement this: cutoff-function technology under $\mathrm{Ric}_g\ge0$ (adopting the approach of \cite{BianchiSetti2018}) and the “hole-filling technique,” and it outlines where the proof components appear: cutoff/hole-filling facts appear later, the main $L^2$ estimates are proved later, and the proof of Theorem~\ref{thm:main} appears later.

Question: Let $(M,g)$ be a complete Riemannian manifold with nonnegative Ricci curvature, $\mathrm{Ric}_g\ge 0$, and assume that $M$ has a pole, meaning that there exists a point $p\in M$ such that the exponential map $\exp_p:T_pM\to M$ is a global diffeomorphism. Let $r(x)=d(p,x)$. Suppose $u\in C^4(M)$ is biharmonic, i.e. $\Delta^2 u=0$ on $M$, and has subquadratic growth in the sense that $u=o(r^2)$ as $r\to\infty$. Which statement about $u$ is valid under these hypotheses?

The function $u$ is harmonic, i.e. $\Delta u=0$ on $M$. In particular, if a biharmonic function on $M$ has sublinear growth, meaning $u=o(r)$ as $r\to\infty$, then it must be constant.

Choices:
B. The function $u$ is harmonic, i.e. $\Delta u=0$ on $M$, provided one strengthens the growth assumption to $u=O(r^2)$ as $r\to\infty$. In particular, every biharmonic function on $M$ with at most quadratic growth is harmonic, and hence every biharmonic function with sublinear growth is constant.
C. The function $u$ satisfies that $\Delta u$ is constant on $M$. In particular, if a biharmonic function on $M$ has sublinear growth, meaning $u=o(r)$ as $r\to\infty$, then it must be constant.
D. There exists a constant $C=C(u)$ such that $\Delta u\equiv C$ on $M$. In particular, if a biharmonic function on $M$ has sublinear growth, meaning $u=o(r)$ as $r\to\infty$, then necessarily $C=0$, so $u$ is constant.
E. The function $u$ is harmonic, i.e. $\Delta u=0$ on $M$, even without assuming that $M$ has a pole; the condition $\mathrm{Ric}_g\ge 0$ together with the subquadratic growth $u=o(r^2)$ already suffices. In particular, any biharmonic function of sublinear growth must be constant.

little-o subquadratic threshold replaced by big-O quadratic growth

dropped the conclusion $\Delta u=0$ to the weaker conclusion that $\Delta u$ is constant

replaced vanishing of $\Delta u$ by dependence-on-$u$ constant Laplacian

removed the pole hypothesis used in the stated biharmonic theorem