%~Mouliné par MaN_auto v.0.31.0 2024-02-05 12:49:30 \documentclass[CRMATH,Unicode]{cedram} \TopicFR{Géométrie algébrique, Systèmes dynamiques} \TopicEN{Algebraic geometry, Dynamical systems} %\usepackage{float} \usepackage{bbm} \usepackage{tikz-cd} \usepackage{mathtools} \usepackage[noadjust]{cite} \newcommand\C{{\mathbb C}} \newcommand\PP{{\mathbb P}} \newcommand\R{{\mathbb R}} \newcommand\A{{\mathbb A}} \newcommand\Z{{\mathbb Z}} \newcommand{\F}{\mathbb{F}} \newcommand*{\kbf}{\mathbf{k}} \newcommand*{\zebf}{\mathbf{0}} \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\Bir}{Bir} \DeclareMathOperator{\id}{id} \DeclareMathOperator{\Base}{Base} \DeclareMathOperator{\charpoly}{char} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\Pic}{Pic} \DeclareMathOperator{\Div}{div} \DeclareMathOperator{\mult}{mult} \hyphenation{ex-cep-tion-al com-mute non-iso-morph-ic} \graphicspath{{./figures/}} \newcommand*{\mk}{\mkern -1mu} \newcommand*{\Mk}{\mkern -2mu} \newcommand*{\mK}{\mkern 1mu} \newcommand*{\MK}{\mkern 2mu} \newcommand*\unn{\underline{n}} \hypersetup{urlcolor=purple, linkcolor=blue, citecolor=red} \newcommand*{\relabel}{\renewcommand{\labelenumi}{(\theenumi)}} \newcommand*{\romanenumi}{\renewcommand*{\theenumi}{\roman{enumi}}\relabel} \newcommand*{\Romanenumi}{\renewcommand*{\theenumi}{\Roman{enumi}}\relabel} \newcommand*{\alphenumi}{\renewcommand*{\theenumi}{\alph{enumi}}\relabel} \newcommand*{\Alphenumi}{\renewcommand*{\theenumi}{\Alph{enumi}}\relabel} \let\oldtilde\tilde \renewcommand*{\tilde}[1]{\mathchoice{\widetilde{#1}}{\widetilde{#1}}{\oldtilde{#1}}{\oldtilde{#1}}} \let\oldhat\hat \renewcommand*{\hat}[1]{\mathchoice{\widehat{#1}}{\widehat{#1}}{\oldhat{#1}}{\oldhat{#1}}} \let\oldexists\exists \renewcommand*{\exists}{\mathrel{\oldexists}} \makeatletter \def\@setafterauthor{% \vglue3mm% % \hspace*{0pt}% \begingroup\hsize=12cm\advance\hsize\abstractmarginL\raggedright \noindent %\hspace*{\abstractmarginL}\begin{minipage}[t]{10cm} \leftskip\abstractmarginL \normalfont\Small \@afterauthor\par \endgroup \vskip2pt plus 3pt minus 1pt } \makeatother \title[The ordinal of dynamical degrees of birational maps] {The ordinal of dynamical degrees \\ of birational maps of the projective plane} \alttitle{L'ordinal des degrés dynamiques des transformations birationelles du plan projectif} \author{\firstname{Anna} \lastname{Bot}} \address{Department of Mathematics and Computer Science, University of Basel, 4051 Basel, Switzerland} \email{annakatharina.bot@unibas.ch} \subjclass{37F10, 32H50, 14E07, 14E05} \keywords{Dynamical degree, rational projective surfaces, ordinals} \thanks{The author acknowledges support by the Swiss National Science Foundation Grant ``Geometrically ruled surfaces'' 200020-192217} \CDRGrant[Swiss National Science Foundation Grant]{200020-192217} \begin{abstract} We show that the ordinal of the dynamical degrees of all birational maps of the complex projective plane is $\omega^{\omega}$. \end{abstract} \begin{altabstract} Nous démontrons que l'ordinal des degrés dynamiques de toutes les transformations birationnelles du plan projectif complexe est $\omega^\omega$. \end{altabstract} \begin{document} \maketitle \section{Introduction} \label{section: introduction} Fix a smooth projective surface $X$ over a field $\kbf$. Any birational self-map $f$ of $X$ can be analysed by considering the dynamical system its iterates provide. A measure of how chaotic such a system becomes is the dynamical degree of $f\colon X\dashrightarrow X$, which is given by \begin{align*} \lambda(f)=\lim_{n \rightarrow \infty} (D\cdot(f^n)^\ast D)^{1/n}, \end{align*} where $D \subseteq X$ is any ample divisor and $D\cdot C$ denotes the intersection form. If $X$ is the projective plane, the definition agrees with $\lambda(f) = \lim_{n \rightarrow \infty} (\deg(f^n))^{1/n}$. Since the dynamical degree is invariant under conjugation, the set $\Lambda(\Bir_{\kbf}(X)) \subseteq \R$ of all possible dynamical degrees of all elements of $\Bir_{\kbf}(X)$ coincides with $\Lambda(\Bir_{\kbf}(\PP^2_{\kbf}))$ whenever $X$ is a rational projective surface. The dynamical degree has been considered in many different settings, often due to its close connection with entropy, see for example~\cite{MR889979, MR1704282, MR1867314, MR2026895, MR2354205, MR2496566, MR2648688, MR2997001}. {Since} any dynamical degree on a rational projective surface {is} an algebraic integer $\geq 1$ (see~\cite[Theorem~5.1]{MR1867314}), the set $\Lambda(\Bir_{\kbf}(\PP^2_{\kbf}))$ is at most countable. In fact, by~\cite[{Theorem~1}]{MR3818624} the set of all dynamical degrees of all rational self-maps of all projective varieties is at most countable. By contrast, the theory of ordinal numbers allows to make a qualitative statement about accumulation points of $\Lambda(\Bir_{\kbf}(\PP^2_{\kbf}))$. A totally ordered set is called \emph{well ordered} if every non-empty subset has a minimum. It is a nice exercise to prove that, assuming the axiom of choice, well orderedness is equivalent to any decreasing sequence becoming stationary; the statement can also be found in~\cite[{Chapter~III, \S6.5, Proposition~6}]{MR2102219}. Recall that a well ordered set corresponds to a unique ordinal, that is, its order type, and that we have a total order on ordinals. The ordinals of finite cardinality are in bijection with the natural numbers, and we denote by $\omega$ the smallest infinite ordinal, which is the order type of the natural numbers with the standard order. Blanc and Cantat showed in~\cite[Theorem~7.2]{MR3454379} that the set of all dynamical degrees of all projective surfaces is well ordered with respect to the standard order on $\R$. Yet, by~\cite[Theorem~B]{MR3454379}, the union of all $\Lambda(\Bir_{\kbf}(X))$ where $X$ is geometrically non-rational and where $\kbf$ is arbitrary is a discrete closed subset of $\R$ and thus {its} ordinal is at most $\omega$. To fully classify the ordinals of $\Lambda(\Bir_{\kbf}(X))$ for any smooth projective surface $X$, the following question remains: what is the ordinal of $\Lambda(\Bir_{\kbf}(\PP^2_{\kbf}))$? We show that it is $\omega^\omega$ if $\kbf=\C$. More generally, we have: \begin{theo}\label{thm: main theorem} For any field $\kbf$ and any projective geometrically rational surface $X$ defined over $\kbf$, the order type of $\Lambda(\Bir_{\kbf}(X)) \subseteq \R$ is bounded above by $\omega^\omega$. If, in addition, $\kbf$ contains the real algebraic numbers and $X$ is rational over $\kbf$, then $\Lambda(\Bir_{\kbf}(X))=\omega^\omega$. \end{theo} The second statement of Theorem~\ref{thm: main theorem} can be interpreted as follows: whenever $\kbf$ contains the real algebraic numbers and $X$ is rational over $\kbf$, there exists a sequence of birational transformations whose dynamical degrees are each limits of a sequence of dynamical degrees of other birational transformations. Now, infinitely many terms of these dynamical degrees are again each the limit of a sequence, most of whose members are again limits of sequences, and so on. One can prove the lower bound for $\Lambda(\Bir_{\kbf}(X))$ in two ways, either assuming $\kbf=\C$ and using the Weyl group as in Section~\ref{section: bounding from below, Weyl group} {(see Theorem~\ref{thm: main theorem, but with weyl group})}, or by explicitly constructing rational surface automorphisms with the desired dynamical degrees as in Section~\ref{section: bounding from below, realisations}. The first strategy was pointed out to the author by Curtis T McMullen. Section~\ref{section: bounding from below, Weyl group} results in the following additional statement: \begin{theo}\label{thm: main theorem for Weyl spectrum} The Weyl spectrum $\Lambda(W){=\bigcup_{n\geq 3} \Lambda(W(E_n))}$ has order type~$\omega^\omega$. \end{theo} For the rest of the paper, we omit the field $\kbf$ from the notation, and write $\Bir(X)$ instead of $\Bir_{\kbf}(X)$, and $\PP^2$ instead of $\PP^2_{\kbf}$. \section{Bounding from above} \label{section: bounding from above} Denote by $\Bir(\PP^2)$ the set of birational maps $f\colon\PP^2 \dashrightarrow \PP^2$, and by $\Bir_d(\PP^2)$ all birational maps of $\PP^2$ of degree~$d$, meaning that the map $f$ can be described by three homogeneous polynomials of degree~$d$ having no common divisor. Thus, $\Bir_d(\PP^2)$ is a subset of some projective space. One can see that it is locally closed, and therefore, $\Bir_d(\PP^2)$ can be seen as an algebraic variety and endowed with a Zariski topology, see also~\cite[Proposition~2.15]{MR3092478}. For any subset $S \subseteq \Bir(\PP^2)$, write $\Lambda(S)$ for the set of all dynamical degrees that elements of $S$ can attain. As the degree is greater {than} or equal to $1$, the set~$\Lambda(S)$ is contained in $\R^{\geq 1}$. Let $X$ be a topological space and $I \subseteq \R \cup \{-\infty, \infty\}$ a subset. A function $g \colon X \rightarrow I$ is called \emph{lower semicontinuous} if for any $r \in I$, the set $L(r) = \left\{ x \in X \mid g(x) \leq r \right\}$ is closed in $X$. The following result of Xie shows that the dynamical degree as a function $\lambda \colon \Bir_d(\PP^2) \rightarrow \R^{\geq 1}$ is lower semicontinuous. \begin{theo}[{\cite[{Theorem~1.5}]{MR3332894}}] \label{thm: junyi} Let $\kbf$ be an algebraically closed field and {$d \geq 1$} an integer. Then for any $\lambda < d$, the set $U_{\lambda}=\{f \in \Bir_d(\PP^2) \thinspace | \thinspace \lambda(f) > \lambda \}$ is a Zariski dense open subset of $\Bir_d(\PP^2)$. \end{theo} The proof of the next result uses the lower semicontinuity of the dynamical spectrum and the noetherianity of the Zariski topology. Such an argument was already used at the end of the proof of~\cite[{Theorem~7.2}]{MR3454379} to prove that the set $\Lambda(\Bir(\PP^2))$ is well ordered. Here we apply the same strategy and go one step further by describing the associated ordinals. \begin{theo}\label{thm: order type for general variety with morphism to Bir} Let $\kbf$ be an {algebraically closed field}, $d \geq 1$ an integer and $X \subseteq \Bir_d(\PP^2)$ a locally closed subset, with $n$ denoting the number of irreducible components of $X$. Then $\Lambda(X) \subseteq \R$ is well ordered and of order type smaller {than} or equal to $\omega^{\dim(X)}n+n$. \end{theo} \begin{proof} The set $\Lambda(\Bir(\PP^2))$ is well ordered by~\cite[Theorem~7.2]{MR3454379}, and it contains $\Lambda(X)$. Thus, $\Lambda(X)$ must be well ordered, too. We fix $d \geq 1$ and prove the second claim by induction on the dimension of $X$. If $\dim(X)=0$, then any irreducible component of $X$ is a point, and the claim follows. Suppose the claim is proved for any locally closed subvariety of $\Bir_d(\PP^2)$ of dimension $c-1 \geq 0$, and consider $X \subseteq \Bir_d(\PP^2)$ locally closed with $\dim(X)=c$. Decompose $X$ into its irreducible components $X_1\cup \ldots \cup X_n$. Pick $X_i$ with $\dim(X_i)=\dim(X)=c$, as any irreducible component of dimension smaller than $c$ has ordinal smaller {than} $\omega^{\dim(X)}+1$ by induction. Assume by contradiction that the order type of $\Lambda(X_i)$ is strictly greater than $\omega^{c}+1$. Then there exist $\nu_1 < \nu_2$ in $\Lambda(X_i)$ such that $\left\{ \nu \in \Lambda(X_i) \mid \nu <\nu_1 \right\}$ is of order type $\omega^c$. By lower semicontinuity (see Theorem~\ref{thm: junyi}), the set $L(\nu_1)=\{ x \in \Bir_d(\PP^2) \mid \lambda(x) \leq \nu_1 \}$ is closed in $\Bir_d(\PP^2)$. Thus $L(\nu_1)\cap X_i$ is closed in $X_i$. If the dimension of this set were strictly smaller than $c$, its image under the map $\lambda\colon \Bir_d(\PP^2)\rightarrow\R^{\geq 1}$ in $\Lambda(X_i)$ would by induction be smaller {than} or equal to $\omega^{c-1}m+m$ for some $m$ counting the irreducible components of $L(\nu_1)\cap X_i$. But $\omega^{c-1}m+m<\omega^c$, a contradiction. Thus, $L(\nu_1)\cap X_i=X_i$, which in turn implies that $\nu_2 \leq \nu_1$. This cannot be. Therefore, any irreducible component $X_i$ of $X$ satisfies $\Lambda(X_i)\leq \omega^c+1$ and hence $\Lambda(X)\leq \omega^c n +n$. This finishes the induction and proves the theorem. \end{proof} \begin{theo}\label{thm: bounding from above} Over any field $\kbf$, {the ordinal of $\Lambda(\Bir_d(\PP^2))$ is less than or equal to $\omega^{4d+6}$,} and the ordinal of $\Lambda(\Bir(\PP^2))$ is less than or equal to $\omega^{\omega}$. \end{theo} \begin{proof} {We may reduce to $\kbf$ being algebraically closed. If the result holds in that case, then we use that any birational map defined over some field is also defined over an algebraic closure, and deduce the claim.} By Theorem~\ref{thm: order type for general variety with morphism to Bir}, we see that $\Lambda(\Bir_d(\PP^2))$ is of order type less than or equal to $\omega^{\dim(\Bir_d(\PP^2))}=\omega^{4d+6}$, where $\dim(\Bir_d(\PP^2))=4d+6$ by~\cite[Theorem~1]{MR3319964}. {This proves the first claim.} As $\Bir(\PP^2)=\bigcup_{d\geq 1} \Bir_d(\PP^2)$, we find that $\Lambda(\Bir(\PP^2)) = \bigcup_{d\geq 1}\Lambda(\Bir_d(\PP^2))$. According to~\cite[{Theorem~1}]{MR5878}, the union of finitely many ordinals which are subsets of a common set is bounded above by the Hessenberg normal sum of the ordinals, here denoted by $\oplus$. Recall that the Hessenberg normal sum of two ordinals $\alpha=\sum_{k=0}^{t}\omega^k a_k$, $\beta=\sum_{k=0}^{t}\omega^k b_k$ expressed in Cantor normal form satisfies $\alpha \oplus \beta = \sum_{k=0}^{t}\omega^k (a_k+b_k)$. Consider $\bigcup_{1\leq d\leq e} \Bir_d(\PP^2)$ with $e \geq 1$. Then, by writing $\Lambda(\Bir_d(\PP^2))=\sum_{i=0}^{4d+6}\omega^i a_{d,i}$ in the Cantor normal form, we find \begin{align*} \bigcup_{1\leq d\leq e} \Lambda(\Bir_d(\PP^2)) \leq \bigcup_{1\leq d\leq e} \Lambda(\Bir_d(\PP^2))=\bigcup_{1\leq d\leq e} \sum_{i=0}^{4d+6}\omega^i a_{d,i} = \sum_{k=0}^{4e+6}\omega^k b_k, \end{align*} where $b_k$ are some sum of the $a_{d,i}$. But then $\bigcup_{1\leq d\leq e} \Lambda(\Bir_d(\PP^2)) \leq \omega^{4e+7}$. Thus, $\Lambda(\Bir(\PP^2))$ is of order type at most~$\omega^\omega$. \end{proof} \section{Arithmetic properties of auxiliary polynomials} \label{section: arithmetic properties of auxiliary polynomials} To bound $\Lambda(\Bir(\PP^2))$ from below, we first collect arithmetic facts about very specific, auxiliary polynomials and their largest real roots, all of which play a part in Sections~\ref{section: bounding from below, Weyl group} and~\ref{section: bounding from below, realisations}. This current section may thus be skipped if one would first like to understand why the polynomials below and their largest positive roots are of importance. \begin{defi}\label{def: auxiliary polynomials} Fix $d \geq 1$, $1 \leq m \leq 2d-1$ and $\unn=(n_2, \ldots, n_m)$. Abbreviate $I_m=\{2, \ldots, m\}$. Then, define \[ p_{d, \unn}(X) = (X^2-(d-1)X-1)\prod_{i=2}^{m}(X^{n_i}+1)+X\sum_{i=2}^{m}\prod_{j \in I_m\setminus \{i\}}(X^{n_j}+1). \] {Note that if $m=1$, we have $p_{d, \emptyset}(X)=X^2-(d-1)X-1$, with roots $\tfrac{1}{2}(d-1\pm \sqrt{d^2-2d+5})$. } \end{defi} The polynomial $p_{d,\unn}(X)$ appears as a factor in the characteristic polynomial of a certain matrix: fix $d\geq 1$, $1 \leq m \leq 2d-1$ and $\unn=(n_2, \ldots, n_{m})$, and consider the matrix \begin{align} J_d^{\unn}=\left( \begin{array}{c|c|c|c|c|c} \scalebox{0.75}{$d$}& \begin{smallmatrix} 0 &~& \phantom{-}d-1 \end{smallmatrix} & \begin{smallmatrix} 0 & \,\scalebox{1.005}{$\cdots$} & 0 & \phantom{-}1 \end{smallmatrix} & \begin{smallmatrix} 0 & \,\scalebox{1.005}{$\cdots$} & 0 & \phantom{-}1 \end{smallmatrix}& \cdots & \begin{smallmatrix} 0 &~& \,\scalebox{1.005}{$\cdots$} &~& 0 & \phantom{-}1 \end{smallmatrix} \\ \hline \begin{smallmatrix} ~\\~\\-(d-1) \\~\\ 0 \\~ \end{smallmatrix} & \begin{smallmatrix} ~&~\\~&~\\ 0 & -(d-2) \\~&~\\ 1 & \phantom{-}0 \\~&~ \end{smallmatrix}& \begin{smallmatrix} ~\\ 0 & \,\scalebox{1.005}{$\cdots$} & 0 &-1 \\~\\ 0 & \,\scalebox{1.005}{$\cdots$} & 0& \phantom{-}0 \end{smallmatrix} & \begin{smallmatrix} ~\\ 0 & \,\scalebox{1.005}{$\cdots$} & 0 &-1 \\~\\ 0 & \,\scalebox{1.005}{$\cdots$} & 0& \phantom{-}0 \end{smallmatrix} & \cdots & \begin{smallmatrix} ~\\ 0 &~& \,\scalebox{1.005}{$\cdots$} &~& 0 &-1 \\~\\ 0 &~&\,\scalebox{1.005}{$\cdots$} &~& 0& \phantom{-}0 \end{smallmatrix} \\ \hline \begin{smallmatrix} ~\\~\\-1 \\~\\ 0 \\ \vdots\\~\\ 0 \\~ \end{smallmatrix} & \begin{smallmatrix} ~\\ 0 &~& -1 &~\\~\\ 0 &~& \phantom{-}0&~\\ \vdots &~& \phantom{-}\vdots&~\\~\\ 0 &~& \phantom{-}0&~ \end{smallmatrix} & \begin{smallmatrix} 0 & \scalebox{1.005}{$\cdots$} & 0 & -1 \\~\\~&~&~& \phantom{-}0 \\~& \raisebox{0.15em}{$\mathbbm{1}_{n_2-1}$} &~& \phantom{-}\vdots \\~\\~&~&~& \phantom{-}0 \end{smallmatrix} & \zebf&\cdots & \zebf \\ \hline \begin{smallmatrix} ~\\~\\-1 \\~\\ 0 \\ \vdots\\~\\ 0 \\~ \end{smallmatrix} & \begin{smallmatrix} ~\\ 0 &~& -1 &~\\~\\ 0 &~& \phantom{-}0&~\\ \vdots &~& \phantom{-}\vdots&~\\~\\ 0 &~& \phantom{-}0&~ \end{smallmatrix} & \zebf & \begin{smallmatrix} 0 & \scalebox{1.005}{$\cdots$} & 0 & -1 \\~\\~&~&~& \phantom{-}0 \\~& \raisebox{0.15em}{$\mathbbm{1}_{n_3-1}$} &~& \phantom{-}\vdots \\~\\~&~&~& \phantom{-}0 \end{smallmatrix} & \cdots & \zebf \\ \hline \vdots& \vdots & \vdots & \vdots & \ddots & \vdots \\ \hline \begin{smallmatrix} ~\\~\\-1 \\~\\ 0 \\ \vdots\\~\\ 0 \\~ \end{smallmatrix} & \begin{smallmatrix} ~\\ 0 &~& -1 &~\\~\\ 0 &~& \phantom{-}0&~\\ \vdots &~& \phantom{-}\vdots&~\\~\\ 0 &~& \phantom{-}0&~ \end{smallmatrix} & \zebf & \zebf & \cdots & \begin{smallmatrix} 0 & \scalebox{1.005}{$\cdots$} & 0 & -1 \\~\\~&~&~& \phantom{-}0 \\~& \raisebox{0.15em}{$\mathbbm{1}_{n_{m}-1}$} &~& \phantom{-}\vdots \\~\\~&~&~& \phantom{-}0 \end{smallmatrix} \end{array}\right), \label{eq: block matrix} \tag{$\diamondsuit$} \end{align} with $\mathbbm{1}_{n_k-1}$ the identity matrix of dimension $n_k-1$. Here, $\zebf$ is short hand for the zero matrix, always of the respective suitable dimension. \begin{lemm}\label{lemma: characteristic poly of all Jonquieres} Consider $d\geq 1$, $1 \leq m \leq 2d-1$ and $\unn=(n_2, \ldots, n_{m})$. Then the characteristic polynomial $\charpoly_{d,\unn}(X)$ of $J_d^{\unn}$ satisfies $\charpoly_{d,\unn}(X)=(X-1)p_{d, \unn}(X)$. \end{lemm} \begin{proof} To calculate the characteristic polynomial of $J^{\unn}_d$ consider first the last $n_{m}+1$ rows of $X\mathbbm{1}-J^{\unn}_d$: \begin{align*} \left( \begin{array}{c|c|c|c|c|c} \begin{smallmatrix} 1 \\~\\ 0 \\~\\ \vdots \\~\\~\\ 0 \end{smallmatrix} & \begin{smallmatrix} 0 &~& 1 &~\\~\\ 0 &~& 0&~\\~\\ \vdots &~& \vdots&~\\~\\~\\ 0 &~& 0&~ \end{smallmatrix} & \zebf & \cdots & \zebf & \begin{smallmatrix} X & 0 & \cdots & 0 & 1 \\ ~\\ -1 & X &~&~& 0 \\ ~& -1& \ddots &~& \vdots \\ ~&~& \ddots & X & 0 \\~\\ ~&~&~& -1 & X \end{smallmatrix} \end{array}\right). \end{align*} With row manipulations this can be reduced to \begin{align*} \left( \begin{array}{c|c|c|c|c|c} \begin{smallmatrix} 1 \\~\\ 0 \\~\\ \vdots \\~\\~\\ 0 \end{smallmatrix} & \begin{smallmatrix} 0 &~& 1 &~\\~\\ 0 &~& 0&~\\~\\ \vdots &~& \vdots&~\\~\\~\\ 0 &~& 0&~ \end{smallmatrix} & \zebf & \cdots & \zebf & \begin{smallmatrix} \phantom{-}0 &~& \phantom{-}0 & \cdots & \phantom{-}0 & \phantom{-}X^{n_{m}}+1 \\ ~\\ -1 &~& \phantom{-}0 &~&~& \phantom{-}X^{n_{m}-1} \\ ~&~& -1 & \ddots &~& \vdots \\ ~&~&~& \ddots& \phantom{-}0 & \phantom{-}X^2 \\~\\ ~&~&~&~& -1 & X \end{smallmatrix} \end{array}\right). \end{align*} Repeat this for the other rows. Then, using Laplace expansion along the columns with the $-1$ entries, we find \begin{align} \charpoly_{d,\unn}(X)&=\det \left( \begin{array}{c|c} \begin{smallmatrix} X-d & \phantom{-}0 & -(d-1) \\ d-1 & \phantom{-}X & \phantom{-}d-2 \\ 0 & -1 & \phantom{-}X \\~ \end{smallmatrix} & \begin{smallmatrix} -1 &~&~&~& -1 &~&~& \scalebox{1.005}{$\cdots$}~& & -1 \\ \phantom{-}1 &~&~&~&\phantom{-}1 &~&~& \scalebox{1.005}{$\cdots$} &~& \phantom{-}1 \\ \phantom{-}0 &~&~&~& \phantom{-}0 &~&~& \scalebox{1.005}{$\cdots$} &~& \phantom{-}0 \\~ \end{smallmatrix} \\ \hline \begin{smallmatrix} ~\\~\\ 1 &~&~& 0 &~&~& \phantom{-}1 \\~\\1 &~&~& 0 &~&~& \phantom{-}1 \\ \vdots &~&~& \vdots &~&~& \phantom{-}\vdots\\~\\ 1 &~&~& 0 &~&~& \phantom{-}1 \end{smallmatrix} & \begin{smallmatrix} ~\\~\\ \phantom{--}X^{n_2}+1 &~&~&~\\~\\~& \phantom{-}X^{n_3}+1 &~&~\\~&~&\ddots &~\\~\\~&~&~& X^{n_{m}}+1 \end{smallmatrix} \end{array}\right) \nonumber \\ &=\det \left( \begin{array}{c|c} \begin{smallmatrix} X-d & \phantom{-}0 & -(d-1) \\ X-1 & \phantom{-}X & -1 \\ 0 & -1 & \phantom{-}X \\~ \end{smallmatrix} & \begin{smallmatrix} -1 &~&~&~& -1 &~&~& \scalebox{1.005}{$\cdots$}~& & -1 \\ \phantom{-}0 &~&~&~&\phantom{-}0 &~&~& \scalebox{1.005}{$\cdots$} &~& \phantom{-}0 \\ \phantom{-}0 &~&~&~& \phantom{-}0 &~&~& \scalebox{1.005}{$\cdots$} &~& \phantom{-}0 \\~ \end{smallmatrix} \\ \hline \begin{smallmatrix} ~\\~\\ 1 &~&~& 0 &~&~& \phantom{-}1 \\~\\1 &~&~& 0 &~&~& \phantom{-}1 \\ \vdots &~&~& \vdots &~&~& \phantom{-}\vdots\\~\\ 1 &~&~& 0 &~&~& \phantom{-}1 \end{smallmatrix} & \begin{smallmatrix} ~\\~\\ \phantom{--}X^{n_2}+1 &~&~&~\\~\\~& \phantom{-}X^{n_3}+1 &~&~\\~&~&\ddots &~\\~\\~&~&~& X^{n_{m}}+1 \end{smallmatrix} \end{array}\right) \label{eq: laplace} \end{align} If we apply Laplace expansion along the fourth column, we find that $\charpoly_{d,\unn}(X)$ is equal to \begin{multline*} \phantom{=}~\det\left( \begin{array}{c|c} \begin{smallmatrix} X-1 & \phantom{-}X & -1 \\ 0 & -1 & \phantom{-}X \\~ \end{smallmatrix} & \begin{smallmatrix} \phantom{-}0 &~&~&~& \scalebox{1.005}{$\cdots$} &~& \phantom{-}0 \\ \phantom{-}0 &~&~&~& \scalebox{1.005}{$\cdots$} &~& \phantom{-}0 \\~ \end{smallmatrix} \\ \hline \begin{smallmatrix} ~\\~\\ 1 &~&~& 0 &~& \phantom{-}1 \\~\\1 &~&~& 0 &~& \phantom{-}1 \\ \vdots &~&~& \vdots &~& \phantom{-}\vdots\\~\\ 1 &~&~& 0 &~& \phantom{-}1 \end{smallmatrix} & \begin{smallmatrix} ~\\~\\~& 0\phantom{-} & \scalebox{1.00}{$\cdots$} & \phantom{-}0\\~\\~& \phantom{-}X^{n_3}+1 &~&~\\~&~&\ddots &~\\~\\~&~&~& X^{n_{m}}+1 \end{smallmatrix} \end{array}\right)+(X^{n_2}+1)\det\left( \begin{array}{c|c} \begin{smallmatrix} X-d & \phantom{-}0 & -(d-1) \\ X-1 & \phantom{-}X & -1 \\ 0 & -1 & \phantom{-}X \\~ \end{smallmatrix} & \begin{smallmatrix} -1 &~&~& \scalebox{1.005}{$\cdots$}~& & -1 \\ \phantom{-}0 &~&~& \scalebox{1.005}{$\cdots$} &~& \phantom{-}0 \\ \phantom{-}0 &~&~& \scalebox{1.005}{$\cdots$} &~& \phantom{-}0 \\~ \end{smallmatrix} \\ \hline \begin{smallmatrix} ~\\~\\1 &~&~& 0 &~&~& \phantom{-}1 \\ \vdots &~&~& \vdots &~&~& \phantom{-}\vdots\\~\\ 1 &~&~& 0 &~&~& \phantom{-}1 \end{smallmatrix} & \begin{smallmatrix} ~\\~\\ \phantom{-}X^{n_3}+1 &~&~\\~&\ddots &~\\~\\~&~& X^{n_{m}}+1 \end{smallmatrix} \end{array}\right) \\ = (X-1)X\prod_{i \in I_m \setminus \{2\}}(X^{n_i}+1)+(X^{n_2}+1)\det(A_2). \end{multline*} Here, $A_2$ denotes the matrix obtained by deleting the fourth row and fourth column of the matrix in~\eqref{eq: laplace}, which corresponds to the cofactor of the entry $X^{n_2}+1$. Denote by $A_j$ the matrix obtained by also deleting the fourth row and fourth column of $A_{j-1}$. Then, with induction, we find \begin{align*} \det(A_j)&=(X-1)X\prod_{i >j}(X^{n_i}+1)+(X^{n_j}+1)\det(A_{j+1})\\ &=(X-1)X\sum_{k > j} \prod_{i \neq k}(X^{n_i}+1) + \prod_{i>j}(X^{n_i}+1)\det(A_{2d-1}). \end{align*} Note that \[ \det(A_{2d-1})=\det\left( \begin{smallmatrix} X-d & \phantom{-}0 & -(d-1) \\ X-1 & \phantom{-}X & -1 \\ 0 &-1 & \phantom{-}X \end{smallmatrix}\right) = X^3-dX^2+(d-2)X+1=(X-1)(X^2-(d-1)X-1). \] Thus, $\charpoly_{d,\unn}(X)=(X-1)\left((X^2-(d-1)X-1)\prod_{i=2}^{m}(X^{n_i}+1)+X\sum_{i=2}^{m} \prod_{j \in I_m \setminus \{i\}}(X^{n_j}+1)\right)=\linebreak (X-1)p_{d, \unn}(X)$, as claimed. \end{proof} Using this correspondence between the polynomials $p_{d,\unn}(X)$ and the characteristic polynomial of the matrices $J_{d}^{\unn}$ in~\eqref{eq: block matrix}, one can prove that there is at most one root of $p_{d, \unn}(X)$ whose absolute value is larger than $1$, and this root is real. To show this, we first prove the following lemma, for which we define $Q=\left( \begin{smallmatrix} 1 & &&\\& -1 &&\\[-0.5em] && \,\ddots\!\! &\\&&& -1 \end{smallmatrix}\right)$. Also, we write $A^T$ for the transpose of a matrix $A$. \begin{lemm}\label{lemma: adapted intersection form} For $d\geq 1$, $1\leq m\leq 2d-1$ and $\unn=(n_2, \ldots, n_m)$, the matrix $J_d^{\unn}$ satisfies $J_d^{\unn}Q(J_d^{\unn})^T=Q+H_{d,m}$, where \[ H_{d,m}=\left( \begin{array}{cc|ccc} \phantom{-}2d-1-m & -(2d-1-m) & 0 & \cdots & 0 \\ -(2d-1-m) & \phantom{-}2d-1-m & 0 & \cdots & 0 \\ \hline \phantom{-}0 & \phantom{-}0 &~&~&~\\ \phantom{-}\vdots & \phantom{-}\vdots &~& \zebf&~\\ \phantom{-}0 & \phantom{-}0 &~&~&~\\ \end{array}\right). \] The matrix $H_{d,m}$ is positive semidefinite. \end{lemm} \begin{proof} The first part of the claim follows by a direct calculation using $Q$ and $J_{d}^{\unn}$ given in~\eqref{eq: block matrix}. The characteristic polynomial of $H_{d,m}$ is equal to the product of $X-2(2d-1-m)$ and some power of $X$. Since $m\leq 2d-1$, all eigenvalues of $H_{d,m}$ are non-negative. Thus, $H_{d,m}$ is positive semidefinite. \end{proof} Equipped with this lemma, we can prove that there is at most one root of $p_{d,\unn}(X)$ whose absolute value is greater than $1$, and this must be a real root. The idea of the proof comes directly from the proof of~\cite[{Theorem~5.1, Assertion~(1)}]{MR1867314}. Yet for the purpose of this section, we are solely concerned with matrices and their characteristic polynomials, and thus do not need to apply~\cite[{Theorem~5.1}]{MR1867314} in its full generality. \begin{prop}\label{prop: root real and strictly larger than 1} Fix $d\geq 1$, $1\leq m\leq 2d-1$ and $\unn=(n_2,\ldots, n_m)$. Then $p_{d, \unn}(X)$ has at most one root whose modulus is strictly larger than $1$, and this root is real. \end{prop} \begin{proof} By Lemma~\ref{lemma: characteristic poly of all Jonquieres}, $(X-1)p_{d,\unn}(X)=\charpoly_{d,\unn}(X)$, where $\charpoly_{d,\unn}(X)$ is the characteristic polynomial of the matrix $J_d^{\unn}$ defined in~\eqref{eq: block matrix}. Thus, a root of $p_{d, \unn}(X)$ corresponds to an eigenvalue of $J_d^{\unn}$. Suppose by contradiction that we have two eigenvalues $\mu_1, \mu_2 \in \C$ of $J_d^{\unn}$ with $|\mu_1|, |\mu_2|>1$, and denote by $v_1, v_2$ some corresponding eigenvectors. We claim that on the subspace $\C v_1+ \C v_2$, the {hermitian} form corresponding to $Q$ is positive semidefinite. This is equivalent to proving that $v_i^TQ\overline{v_i}\geq 0$ for $i=1,2$ and $(v_1^TQ\overline{v_1})(v_2^TQ \overline{v_2})-|v_1^TQ\overline{v_2}|^2 \geq 0$. Abbreviate $H=H_{d,m}$ the matrix from Lemma~\ref{lemma: adapted intersection form}. Note that $\mu_i\overline{\mu_j}v_i^TQ\overline{v_j}=v_i^TJ_d^{\unn}Q(J_d^{\unn})^T\overline{v_j}\stackrel{\ref{lemma: adapted intersection form}}{=}v_i^TQ\overline{v_j}+v_i^TH\overline{v_j}$ for $i,j=1,2$, and thus $v_i^TQ\overline{v_i}=\tfrac{v_i^TH\overline{v_i}}{|\mu_i|^2-1}$ and $v_1^TQ\overline{v_2}=\tfrac{v_1^TH\overline{v_2}}{\mu_1\overline{\mu_2}-1}$. Since $H$ is positive semidefinite by Lemma~\ref{lemma: adapted intersection form}, we directly see that $v_i^TQ\overline{v_i}\geq 0$ holds. Furthermore, \[ (v_1^TQ\overline{v_1})(v_2^TQ \overline{v_2})-|v_1^TQ\overline{v_2}|^2=\tfrac{1}{(|\mu_1|^2-1)(|\mu_2|^2-1)}(v_1^TH\overline{v_1})(v_2^TH\overline{v_2})-\tfrac{1}{|\mu_1\overline{\mu_2}-1|^2}|v_1^TH\overline{v_2}|^2. \] Since $(|\mu_1|^2-1)(|\mu_2|^2-1)\leq|\mu_1\overline{\mu_2}-1|^2$ by expanding $|\mu_1-\mu_2|^2\geq 0$, we deduce \[ (v_1^TQ\overline{v_1})(v_2^TQ\overline{v_2})-|v_1^TQ\overline{v_2}|^2\geq\tfrac{1}{(|\mu_1|^2-1)(|\mu_2|^2-1)}((v_1^TH\overline{v_1})(v_2^TH\overline{v_2})-|v_1^TH\overline{v_2}|^2)\geq0, \] where the last inequality is due to $H$ being positive semidefinite by Lemma~\ref{lemma: adapted intersection form}. Thus, $Q$ is positive semidefinite on $\C v_1 + \C v_2$. But the signature of $Q$ is $(1, \dim(Q)-1)$, so there cannot be a subspace of dimension~$2$ on which $Q$ is positive semidefinite. Therefore, {$\C v_1=\C v_2$} and $\mu_1=\mu_2$. As $\overline{\mu_1}$ is also an eigenvalue of $J_d^{\unn}$, we additionally deduce $\mu_1=\overline{\mu_1}$. The root of largest absolute value must thus be real. This finishes the proof of the proposition. \end{proof} \begin{defi}\label{def: largest real roots of polynomials} For $d\geq 1$, $1\leq m\leq 2d-1$ and $\unn$ we denote by $\lambda_{d, \unn} \in \R$ the root of largest absolute value of $p_{d, \unn}(X)$. \end{defi} We are interested in describing $\lambda_{d,\unn}$; by the following proposition, it must be larger than $2$ under mild assumptions. \begin{prop}\label{prop: greater than 2} Fix $d\geq 4$, $1\leq m \leq 2d-1$ and $\unn=(n_2, \ldots, n_m)$ with $n_i \geq 2$ for all $2\leq i \leq m$. Then $\lambda_{d, \unn}$ is strictly larger than $2$. \end{prop} \begin{proof} The claim for $m=1$ follows by observing that the largest root $\lambda$ of $p_{d,\emptyset}(X)=X^2-(d-1)X-1$ satisfies $\lambda-\tfrac{1}{\lambda}=d-1$ and thus $22$. Since $p_{d,\unn}(\lambda_{d,\unn})=0$, we deduce $p_{d,\unn_k}(\lambda_{d, \unn})<0$. The polynomial $p_{d, \unn_k}(X)$ is monic, which implies that it must have a zero which is strictly larger than $\lambda_{d, \unn}$. Hence, $\lambda_{d,\unn_k}>\lambda_{d, \unn}$, which implies the claim. \end{proof} As a last property of the auxiliary polynomials, we prove that if we fix $n_2, \ldots, n_m \geq 2$ and consider all $\lambda_{d, \unn'}$ where $\unn'=(n_2, \ldots, n_m, n_{m+1})$ with $n_{m+1}\geq2$, then the sequence $(\lambda_{d, \unn'})_{n_{m+1}}$ has $\lambda_{d, \unn}$ as its limit. \begin{lemm}\label{lemma: adding one parameter is smaller} Fix $d\geq 4$, $2\leq m \leq 2d-1$ and $\unn=(n_2, \ldots, n_{m-1})$, and $\unn'=(n_2, \ldots, n_{m-1}, n_{m})$, with all $n_i\geq 2$. Then $\lambda_{d,\unn'}<\lambda_{d, \unn}$. \end{lemm} \begin{proof} The two polynomials $p_{d,\unn}(X)$ and $p_{d,\unn'}(X)$ are related as $p_{d, \unn'}(X)=(X^{n_{m}}+1)p_{d,\unn}(X)+X\prod_{j\in I_{m-1}}(X^{n_j}+1)$. By Proposition~\ref{prop: greater than 2}, we find $\lambda_{d, \unn'}>2$. Therefore, plugging in $\lambda_{d, \unn'}$, and using that it is a root of $p_{d,\unn'}(X)$, we find \[ p_{d,\unn}(\lambda_{d, \unn'})=-\lambda_{d,\unn'}\prod_{j\in I_{m-1}}(\lambda_{d, \unn'}^{n_j}+1)<0 \] As $p_{d, \unn}(X)$ is monic, this proves $\lambda_{d, \unn'}<\lambda_{d, \unn}$. \end{proof} \begin{prop}\label{prop: limit of polynomials} Fix $d\geq 4$, $2\leq m\leq 2d-1$ and $\unn=(n_2, \ldots, n_{m-1})$ with $n_i\geq 2$. Denote $\unn_{n_m}=(n_2, \ldots, n_{m-1}, n_m)$. Then the limit of the sequence $(\lambda_{d, \unn_{n_m}})_{n_m}$ exists and is equal to $\lambda_{d, \unn}$. \end{prop} \begin{proof} Fix some $\unn=(n_2, \ldots, n_{m-1})$ and $2 < \lambda < \lambda_{d, \unn}$. Note that the polynomials satisfy \[ p_{d, \unn_{n_m}}(X)=(X^{n_m}+1)p_{d,\unn}(X)+X\prod_{j \in I_{m-1}}(X^{n_j}+1). \] Note that since by Proposition~\ref{prop: root real and strictly larger than 1}, $\lambda_{d, \unn}$ is the only real root of $p_{d,\unn}(X)$ which is larger than $1$, on $(2,\lambda_{d, \unn})$, the polynomial $p_{d, \unn}(X)$ is negative. Thus, there exists $n_m(\lambda)\geq 2$ such that for all $n_m\geq n_m(\lambda)$, we have \[ p_{d,\unn_{n_m}}(\lambda)=(\lambda^{n_m}+1)p_{d,\unn}(\lambda)+\lambda\prod_{j \in I_{m-1}}(\lambda^{n_j}+1)<0. \] This implies that for $n_m\geq n_m(\lambda)$, we have $\lambda<\lambda_{d, \unn_{n_m}} < \lambda_{d, \unn}$, where the last inequality is due to Lemma~\ref{lemma: adding one parameter is smaller}. This proves the claim. \end{proof} To finish this section, we prove that we can construct a well ordered set out of the $\lambda_{d, \unn}$ such that we can precisely describe the ordinal of that set. \begin{prop}\label{prop: ordinal of helping sets} Consider for $d\geq 4$ and $m \leq 2d-1$ the inductively defined sets \begin{align*} \Lambda_{d,1} &= \{ \tfrac{1}{2}(d-1+\sqrt{d^2-2d+5}) \}, \\ \Lambda_{d,2}&=\{\lambda_{d, (n_2)} \thinspace | \thinspace 2\leq n_2 {\text{ and }} d-1 < \lambda_{d, (n_2)} \}, \\ \Lambda_{d,m} &= \left\{\lambda_{d, (n_2, \ldots, n_m)} \thinspace\middle | \thinspace \begin{aligned} 2\leq n_2 < \ldots < n_m {\text{ and }} \lambda_{d, (n_2, \ldots, n_{m-1}-1)} \in \Lambda_{d,m-1}{\text{ and } \lambda_{d, (n_2, \ldots, n_{m-1})} \in \Lambda_{d,m-1}} \\[-0.25em] {\text{ and } \lambda_{d, (n_2, \ldots, n_{m-1}-1)} < \lambda_{d, (n_2, \ldots, n_m)} < \lambda_{d, (n_2, \ldots, n_{m-1})}} \end{aligned} \right\}. \end{align*} The following holds: \begin{enumerate} \item The well ordering on the sets $\Lambda_{d,m}$ with respect to the lexicographic ordering on $(n_2, \ldots, n_m)$ agrees with the (well) ordering inherited from the one on the real line. \label{claim 1} \item The ordinal of $\Lambda_{d, m}$ with respect to this ordering is~$\omega^{m-1}$. \label{claim 2} \item The disjoint union $\sqcup_{d\geq 4} \Lambda_{d, 2d-1}$ has ordinal equal to $\omega^\omega$. \label{claim 3} \end{enumerate} \end{prop} \begin{proof} We prove Claims~\ref{claim 1} and~\ref{claim 2} simultaneously by induction. For $m=1$ it is clear. For $m=2$, the elements of $\Lambda_{d,2}$ indexed by $n_2$ form by Proposition~\ref{prop: increase one parameter, root is larger} an ascending chain of elements. Thus $\Lambda_{d, 2}$ is well ordered with ordinal~$\omega$. Now assume we know for some $m \in \{2,\ldots, 2d-2\}$ that the well ordering on $\Lambda_{d, m}$ given by the lexicographic ordering of the indexing set is the same as the one inherited from the real line and that $\Lambda_{d, m}$ has ordinal~$\omega^{m-1}$ with respect to this ordering. Define $J_m=\{(n_2, \ldots, n_m) \in \Z^{m-1} \thinspace | \thinspace 2 \leq n_2 < \ldots < n_m {\text{ and }}\lambda_{d, (n_2, \ldots, n_{m}-1)} \in \Lambda_{d, m} {\text{ and }} \lambda_{d, (n_2, \ldots, n_m)} \in \Lambda_{d, m}\}$ and for each $(n_2, \ldots, n_m) \in J_m$, define the open interval $I_{(n_2, \ldots, n_m)} = (\lambda_{d, (n_2, \ldots, n_{m}-1)},\linebreak \lambda_{d, (n_2, \ldots, n_m)})$. {These intervals serve the purpose of making the sequences in $\Lambda_{d,m}$ disjoint. The deduction on the ordinal will then be immediate.} {By Proposition~\ref{prop: limit of polynomials}, for each $(n_2, \ldots, n_m) \in J_m$, there exists $n_{m+1}(\lambda_{d, (n_2, \ldots, n_{m}-1)}) > n_m$ such that for $n_{m+1} \geq n_{m+1}(\lambda_{d, (n_2, \ldots, n_{m}-1)})$, the values $\lambda_{d, (n_2, \ldots, n_{m+1})}$ form a sequence with limit $\lambda_{d, (n_2, \ldots, n_m)}$. By Proposition~\ref{prop: increase one parameter, root is larger}, this sequence is strictly increasing. Furthermore, we can choose this sequence to lie entirely within the interval $I_{(n_2, \ldots, n_m)}$.} Thus, the sequence within $I_{(n_2, \ldots, n_m)}$ is well ordered. Moreover, we can write $\Lambda_{d, m+1}$ as a union of such sequences. {Since the intervals $I_{(n_2, \ldots, n_m)}$ are disjoint, so are the sequences making up $\Lambda_{d, m+1}$}. Therefore, $\Lambda_{d, m+1}$ is well ordered in regard to the lexicographic ordering on its indexing set, too, and this ordering agrees again with the ordering on $\R$. {This proves~\ref{claim 1}.} Each sequence has ordinal~$\omega$, and by induction, $\Lambda_{d, m}$ has ordinal~$\omega^{m-1}$; this implies that $\Lambda_{d, m+1}$ has ordinal~$\omega^m$, as claimed {in~\ref{claim 2}. } As for the last claim, note that for a fixed $d\geq 4$, the $\Lambda_{d, m}$ with $m\in \{1, \ldots, 2d-1\}$ are disjoint by construction and by Lemma~\ref{lemma: adding one parameter is smaller}. For $4\leq d< d'$, note that all elements of $\Lambda_{d', m}$ for all $m \in \{1,\ldots, 2d'-1\}$ are larger than $d'-1$ by construction. Furthermore, all elements of $\Lambda_{d, m}$ with $m\in \{2,\ldots, 2d-1\}$ are smaller than $\tfrac{1}{2}(d-1+\sqrt{d^2-2d+5})$, which is smaller than $d\leq d'-1$ for $d\geq 4$. {Therefore, in particular, the $\Lambda_{d, 2d-1}$ are disjoint, and their union has ordinal~$\omega^\omega$. This proves~\ref{claim 3}, and hence the proposition.} \end{proof} \begin{rema}\label{rem: Lambdas} Note that the sets $\Lambda_{d, m}$ do not contain any accumulation points in the open interval topology on $\R$. In fact, for any $m\in\{2,\ldots, 2d-1\}$, the accumulation points of the set $\Lambda_{d, m}$ lie in $\Lambda_{d, m-1}$ by Proposition~\ref{prop: limit of polynomials}, and the closure of $\Lambda_{d, m}$ is equal to $\bigsqcup_{1 \leq i \leq m}\Lambda_{d,i}$. \end{rema} \section{Bounding from below using the Weyl group} \label{section: bounding from below, Weyl group} We introduce the Weyl group in a brief way; for a more comprehensive introduction to Coxeter groups see~\cite{MR1066460} and~\cite{MR1890629}, and for an introduction to the Weyl group as it relates to this setting see~\cite{MR1007155} and~\cite{MR2354205}. The \emph{Weyl group} is defined as the Coxeter group $W_n=W(E_n)$ given by the Coxeter-Dynkin diagram $E_n$ on $n$ vertices as in Figure~\ref{figure: E_n}. \begin{figure}[h] \begin{center} \begin{tikzpicture} \path (0,1) node [style={draw,circle,minimum size=2mm,inner sep=0pt,outer sep=0pt,fill=black}, draw]{} node[above=1pt] {} (1,1) node [style={draw,circle,minimum size=2mm,inner sep=0pt,outer sep=0pt,fill=black}, draw]{} node[above=1pt] {} (2,1) node [style={draw,circle,minimum size=2mm,inner sep=0pt,outer sep=0pt,fill=black}, draw]{} node[above=1pt] {} (3,1) node [style={draw,circle,minimum size=2mm,inner sep=0pt,outer sep=0pt,fill=black}, draw]{} node[above=1pt] {} (4,1) node [style={draw,circle,minimum size=2mm,inner sep=0pt,outer sep=0pt,fill=black}, draw]{} node[above=1pt] {} (5,1) node [style={draw,circle,minimum size=2mm,inner sep=0pt,outer sep=0pt,fill=black}, draw] {} node[above=1pt] {} (2,0) node [style={draw,circle,minimum size=2mm,inner sep=0pt,outer sep=0pt,fill=black}, draw] {} node[below=1pt] {}; \draw (0,1) -- (1,1) -- (2,1) -- (3,1); \draw [dashed] (3,1) -- (4,1); \draw (4,1) -- (5,1); \draw (2,1) -- (2,0); \end{tikzpicture} \caption{The graph $E_n$ on $n$ vertices.} \label{figure: E_n} \end{center} \end{figure} Alternatively, fix $n\geq 3$. Write $\Z^{1,n}$ for the lattice $\Z^{n+1}$ with inner product $x \cdot x =x_0^2-x_1^2-\cdots -x_n^2$ and standard basis $(e_0,e_1,\ldots, e_n)$. Then the Weyl group $W_n$ is the subgroup of $O(\Z^{1,n})$ generated by the reflections~$x \mapsto x+(x\cdot \alpha_i) \alpha_i$, where $\alpha_0=e_0-e_1-e_2-e_3$ and $\alpha_i=e_i-e_{i+1}$ for $i=1, \ldots, n-1$. There exists an isometry \begin{align*} \Z^{1,n} &\longrightarrow \Pic(X), \quad e_0 \longmapsto L, \quad e_i \longmapsto \mathcal{E}_i, \quad 1 \leq i \leq n \end{align*} for a rational surface $X$ obtained as the blow-up of $\PP^2$ in $n$ points, where $L$ is the pullback of a general line and $\mathcal{E}_i$ are the exceptional curves. Via this isometry, $W_n$ is the subgroup of $\Aut(\Pic(X)) \cong \GL_{n+1}(\Z)$ generated by the permutation matrices with the upper left entry fixed as $1$, and by the matrix \begin{align*} A_0&=\left( \begin{array}{c|c} \begin{smallmatrix} 2& \phantom{-}1& \phantom{-}1& \phantom{-}1\\ -1& \phantom{-}0 & -1 & -1\\ -1 & -1 & \phantom{-}0 & -1 \\ -1 & -1 & -1 & \phantom{-}0 \end{smallmatrix} & \\[1em] \hline & \mathbbm{1} \end{array} \right). \end{align*} Even more, Nagata~\cite{MR126444} proved that the image of $\Aut(X)$ under the homomorphism $\Aut(X)\rightarrow \Aut(\Pic(X))$ lies in $W_n$. This homomorphism is injective whenever out of the $n$ points blown up we can pick four points where no three are collinear. Note that $W_n$ must not necessarily be equal to the subgroup consisting of elements preserving the intersection form and the anticanonical divisor: take for example the reflection determined by $3e_0 - \sum_{i=1}^{10}e_i + e_{11}$. The matrices with $m=2d-1$ introduced in~\eqref{eq: block matrix} are all elements of a Weyl group. \begin{lemm}\label{lemma: Jonquieres matrices are in W_n} For any $d\geq 2$ and $\unn=(n_2, \ldots, n_{2d-1})$ with $n_i \in \Z^{\geq 1}$ the matrix $J_d^{\unn}$ lies in $W_n$ with $n=\sum_{i =2}^{2d-1}n_i+2$. \end{lemm} \begin{proof} Fix some $\unn=(n_2, \ldots, n_{2d-1})$ and $J_{d}^{\unn}$ as in~\eqref{eq: block matrix}. Any permutation matrix with fixed upper left entry lies in $W_n$. Consider $P_1$ such a permutation matrix which by right multiplication permutes columns $3$, $n_2+3$ and $n_2+n_3+3$ of any matrix into second, third and fourth position, respectively. Similarly, there exists a permutation matrix $P'_1$ in $W_n$ which by multiplication from the left permutes rows $4$ and $n_2+4$ of a matrix into third and fourth position. Multiplying from the right and the left, we can bring $J_d^{\unn}$ into the following form: \begin{align*} P_1'J_d^{\unn}P_1=\left( \begin{array}{c|c|c} \begin{smallmatrix} \phantom{-}d & \phantom{-}d-1 & \phantom{-}1 & \phantom{-}1\\ -(d-1)&-(d-2) & -1 & -1 \\-1 & -1 & -1 & \phantom{-}0 \\ -1 & -1 & \phantom{-}0 & -1 \end{smallmatrix} & \ast & \zebf \\[1em] \hline \ast & \ast & \zebf \\ \hline \zebf & \zebf & \mathbbm{1} \end{array}\right). \end{align*} Then, multiplying by $A_0$ (defined above) from the right, we obtain \begin{align*} P_1'J_d^{\unn}P_1A_0=\left( \begin{array}{c|c|c} \begin{smallmatrix} \phantom{-}d & \phantom{-}d-1 & \phantom{-}1 & \phantom{-}1\\ -(d-1)&-(d-2) & -1 & -1 \\-1 & -1 & -1 & \phantom{-}0 \\ -1 & -1 & \phantom{-}0 & -1 \end{smallmatrix} & \ast & \zebf \\[1em] \hline \ast & \ast & \zebf \\ \hline \zebf & \zebf & \mathbbm{1} \end{array}\right)\left( \begin{array}{c|c} \begin{smallmatrix} \phantom{-}2& \phantom{-}1& \phantom{-}1& \phantom{-}1\\ -1& \phantom{-}0 & -1 & -1\\ -1 & -1 & \phantom{-}0 & -1 \\ -1 & -1 & -1 & \phantom{-}0 \end{smallmatrix} & \\[1em] \hline & \mathbbm{1} \end{array} \right)=\left( \begin{array}{c|c|c} \begin{smallmatrix} \phantom{-}d-1 & \phantom{-}d-2 & \phantom{-}0 & \phantom{-}0\\ -(d-2)&-(d-3) & \phantom{-}0 & \phantom{-}0 \\ \phantom{-}0 & \phantom{-}0 & \phantom{-}0 & \phantom{-}1 \\ \phantom{-}0 & \phantom{-}0 & \phantom{-}1 & \phantom{-}0 \end{smallmatrix} & \ast & \zebf \\[1em] \hline \ast & \ast& \zebf \\ \hline \zebf & \zebf & \mathbbm{1} \end{array}\right). \end{align*} The $2d-2$ columns and rows indicated by $\ast$ in $P_1'J_d^{\unn}P_1A_0$ above are of the form $e_0-e_i-e_j$ or $(e_0-e_i-e_j)^T$ for some $i \neq j$. Thus, there exist permutation matrices $P_2, P_2' \in W_n$, both with $1$ in the upper left entry, such that \begin{align*} P_2'P_1'J_d^{\unn}P_1A_0P_2=\left( \begin{array}{c|c|c} \begin{smallmatrix} \phantom{-}d-1 & \phantom{-}d-2 & \phantom{-}1 & \phantom{-}1\\ -(d-2)&-(d-3) & -1 & -1 \\ -1 & -1 & -1 & \phantom{-}0 \\ -1 & -1 & \phantom{-}0 & -1 \end{smallmatrix} & \ast & \zebf \\[1em] \hline \ast & \ast & \zebf \\ \hline \zebf & \zebf & \mathbbm{1} \end{array}\right). \end{align*} We can therefore inductively repeat the process of multiplying by $A_0$ from the right and permuting the rows and columns, and obtain that $P_{d-2}'\cdots P_1'J_d^{\unn}P_1A_0 \cdots P_{d-2}A_0$ is the identity matrix. Thus, $J_d^{\unn}$ is indeed an element of $W_n$. This proves the lemma. \end{proof} Define the \emph{Weyl spectrum} $\Lambda(W)$ as the set of all spectral radii of all elements of all $W_n$ with $n\geq 3$. By the following result of Uehara, rephrased to fit the framework of this article, the Weyl spectrum agrees with all possible dynamical degrees of rational surface automorphisms. \begin{theo}[{\cite[{Theorem~1.1}]{MR3477879}}] \label{thm: uehara} If the ground field is $\C$, then the Weyl spectrum $\Lambda(W)$ agrees with \[ \Lambda\bigl(\bigl\{f \in \Bir(\PP^2) \, \big| \exists \pi \colon X \rightarrow \PP^2\colon \pi^{-1}f\pi \in \Aut(X) \bigr\}\bigr). \] \end{theo} With Uehara's result, we can infer the lower bound~$\omega^\omega$ on the ordinal of $\Lambda(\Bir(\PP^2))$ by showing it for $\Lambda(W)$. \begin{prop}\label{prop: ordinal Weyl spectrum}\leavevmode \begin{enumerate} \item The Weyl spectrum $\Lambda(W)$ is a well ordered subset of $\R$ and its ordinal is greater {than} or equal to~$\omega^\omega$. \label{item 1} \item {The ordinal of $\Lambda(\Bir_d(\PP^2))$ is greater than or equal to $\omega^{2d-2}$.} \label{item 2} \end{enumerate} \end{prop} \begin{proof} The Weyl spectrum $\Lambda(W)$ is a subset of $\Lambda(\Bir(\PP^2)) \subseteq \R$, which is well ordered by~\cite[{Theorem~7.2}]{MR3454379}. Therefore, $\Lambda(W)$ is also well ordered. {By Lemma~\ref{lemma: Jonquieres matrices are in W_n}, all the matrices $J_d^{\unn}$ in~\eqref{eq: block matrix} with $m=2d-1$ belong to the respective Weyl group $W_n$ with $n=\sum_{i=2}^{2d-1}n_i+2$.} {Therefore, by Lemma~\ref{lemma: characteristic poly of all Jonquieres}, the disjoint sets $\Lambda_{d, 2d-1}$ defined in Proposition~\ref{prop: ordinal of helping sets} lie in $\Lambda(W)$ and thus $\bigsqcup_{d \geq 4} \Lambda_{d, 2d-1} \subseteq \Lambda(W)$. By Proposition~\ref{prop: ordinal of helping sets}~\ref{claim 3}, we deduce that the ordinal of $\Lambda(W)$ is at least~$\omega^\omega$, which proves~\eqref{item 1}. Claim~\eqref{item 2} follows from observing that $\Lambda_{d, 2d-1} \subseteq \Lambda(\Bir_d(\PP^2))$. This proves the proposition.} \end{proof} \begin{theo}\label{thm: main theorem, but with weyl group} If the base field is $\C$, then both the ordinal of $\Lambda(\Bir(\PP^2))$ and the ordinal of the Weyl spectrum $\Lambda(W)$ are equal to $\omega^\omega$. \end{theo} \begin{proof} Since by Theorem~\ref{thm: uehara}, we have the inclusion $\Lambda(W) \subseteq \Lambda(\Bir(\PP^2))$. Thus, by Proposition~\ref{prop: ordinal Weyl spectrum}~\ref{item 1}, the ordinal of $\Lambda(\Bir(\PP^2))$ is also greater {than} or equal to $\omega^\omega$. This {together with Theorem~\ref{thm: bounding from above}} proves the claims. \end{proof} \section{Bounding from below using explicit realisations} \label{section: bounding from below, realisations} {In this section, we} prove that the ordinal of $\Lambda(\Bir(\PP^2))$ can be bounded from below by $\omega^\omega$ without appealing to Theorem~\ref{thm: uehara}. {We construct suitable birational maps having the elements of $\Lambda_{d,2d-1}$ defined in Proposition~\ref{prop: ordinal of helping sets} as dynamical degrees. To this end, we show a general result (Proposition~\ref{prop: eigenvector implies existence on cuspidal cubic}): Suppose we are given a matrix, an eigenvalue and a corresponding eigenvector, satisfying some geometrically inspired conditions. Then, there exist points on a cuspidal cubic in $\PP^2$ such that their blow-up admits an automorphism having precisely this matrix encoding its action on the Picard group. In the terminology of~\cite{MR3477879}, we then prove the realisability of the orbit data explicitly. The proof we provide is more direct, and easier; it also shows that our realisations are Jonqui\`eres maps. We refer to~\cite{MR2825269,MR2354205, MR3498924, MR3477879} for similar constructions.} Note that in this section, for the constructions to work, the field of definition needs to contain the field of real algebraic integers; this implies further that $\kbf$ is of characteristic $0$. \subsection{Existence of maps with base points on a cuspidal cubic} \label{subsection: Existence of maps with base points on a cupsidal cubic} Consider a birational map $f\in \Bir(\PP^2)$. Assume that there exist points $p_1, \ldots, p_n \in \PP^2$ such that $f$ lifts to an automorphism of the blow-up $X$ of $\PP^2$ in the points $p_1, \ldots, p_n$. Call this automorphism $f$ as well. The automorphism $f$ induces an automorphism of the Picard group via pushforward, i.e. $f_\ast\colon [D] \mapsto [f(D)]$. This automorphism can be represented by an integer-valued matrix in the ordered basis $([L], [E_1], \ldots, [E_n])$, where $E_i$ denotes the exceptional curve above $p_i$. Denote this matrix by~$F$. We know that the spectral radius~$\rho(F)$ is equal to the dynamical degree~$\lambda(f)$; in addition, since $f$ preserves the canonical divisor, $(3,-1,\ldots, -1)^T$ is an eigenvector of $F$ with eigenvalue $1$. Setting \begin{align} Q=\left( \begin{smallmatrix} 1 & & & \\ & -1 & & \\[-0.5em] & & \,\ddots\! & \\ & & & -1 \end{smallmatrix} \right),\label{eq: Q} \tag{$\clubsuit$} \end{align} the fact that $f$ preserves the intersection form translates to $FQF^T=Q$, where $Q$ is the matrix given in~\eqref{eq: Q} and $F^T$ denotes the transpose of~$F$. If, conversely, we want to find a birational map of $\PP^2$ with corresponding matrix $F$, we can prove a partial converse by choosing appropriate points on a cuspidal cubic curve. Consider the singular cuspidal cubic $C = V(XY^2-Z^3)$ in $\mathbb{P}^2$ with unique singular point $[1:0:0]$. Set $C^\ast =C \setminus \{[1:0:0]\}$ the set of smooth points. The isomorphism $\A^1 \overset{\sim}{\rightarrow} C^\ast$, $t \mapsto [t^3:1:t]$ allows to parametrise the smooth points of $C$ with $\A^1$ and to endow $C^\ast$ with a group structure, for which $t_1+t_2+t_3=0$ if and only if the points $t_1, t_2, t_3$ on $C^\ast$ are collinear. \begin{lemm}\label{lemma: on curve of degree d, sum is zero} If for $3d \geq 3$ not necessarily distinct points $t_i \in C^\ast$, there exists a homogeneous polynomial $P$ of degree $d$ whose restriction to $C$ is not zero and such that for $p_i=[t_i^3:1:t_i]$ we have ${\Div_C(P)}=p_1+\cdots+p_{3d}$, then $\sum_{i=1}^{3d}t_i=0$. \end{lemm} \begin{proof} {This is a well known consequence of the group law on the Picard group of $C^\ast$. We provide a direct proof for the sake of completeness.} Suppose we have $3d$ points $p_i=[t_i^3 : 1:t_i]$ with $t_i\in C^\ast$ such that ${\Div_C(P)}=p_1+\cdots+p_{3d}$ for a homogeneous polynomial $P\in \C[X,Y,Z]$ of degree $d$ not a multiple of the defining equation of $C$. As $P(t^3,1,t)\in \C[t]$ is a polynomial of degree at most $3d$ and ${\Div_C(P)}=p_1+\cdots+p_{3d}$, we must have $P(t^3,1,t)=a\prod_{i=1}^{3d}(t-t_i)$ for some $a\in \C^\ast$. The coefficient of $t^{3d-1}$ has to be zero. Otherwise, we could find a term $X^{i}Y^{j}Z^k$ of $P$ with $3i+k=3d-1$ and $i+k\leq d$. This, however, would imply $2k \leq 1$, leading to the desired contradiction. Therefore, the coefficient of $t^{3d-1}$ is precisely equal to $-a\sum_{i=1}^{3d}t_i=0$, which finishes the proof. \end{proof} \begin{lemm}\label{lemma: kernel and eigenvector, arithmetic} Consider any matrix $F \in \GL_{n+1}(\Z)$ satisfying $FQF^T=Q$, where $Q$ is defined as in~\eqref{eq: Q}, and $F^T(3,1, \ldots, 1)^T=(3,1,\ldots, 1)^T$. Fix $a \in \C\setminus \{1\}$ and $b,v_1,\ldots, v_n \in \C$. If the vector $(b,v_1, \ldots, v_n)^T$ is an eigenvector of $F^T$ with respect to $a$, then we have the following system of equations: \begin{align} 3b&=F_{21}\frac{1}{a-1}(3v_1-b)+\cdots +F_{n+1,1}\frac{1}{a-1}(3v_n-b), \label{eq: kernel and eigenvector, 1} \tag{$\heartsuit$} \\ b&=F_{2,i+1}\frac{1}{a-1}(3v_1-b)+ \cdots + (F_{i+1,i+1}-a)\frac{1}{a-1}(3v_i-b)+\cdots +F_{n+1,i+1}\frac{1}{a-1}(3v_n-b). \label{eq: kernel and eigenvector, 2} \tag{$\spadesuit$} \end{align} \end{lemm} \begin{proof} Assume that the vector $(b,v_1, \ldots, v_n)^T$ is an eigenvector of $F^T$ with respect to the eigenvalue $a$. Then $(b,v_1, \ldots, v_n)^T$ satisfies \begin{align*} (F_{11}-a)b+F_{21}v_1+\cdots +F_{n+1,1}v_n&=0, \\ F_{1,i+1}b+F_{2,i+1}v_1+ \cdots + (F_{i+1,i+1}-a)v_i+\cdots +F_{n+1,i+1}v_n&=0, \quad 1 \leq i \leq n. \end{align*} By multiplying each equation by $3$, and subtracting and adding $\sum_{j=2}^{n+1}F_{j1}b$ from the first equation and $\sum_{j=2}^{n+1}F_{j,i+1}b$ from the remaining $n$ equations, the equations imply \begin{align*} 3(F_{11}-a)b+F_{21}(3v_1-b)+\cdots +F_{n+1,1}(3v_n-b)+\sum_{j=2}^{n+1}F_{j1}b&=0, \\ \Biggl(3F_{1,i+1}\mk+\!\sum_{j=2}^{n+1}F_{j,i+1}\mk-\mk a\Biggr)b\mk+\mk F_{2,i+1}(3v_1-b)\mk+\mk \cdots \mk+\mk (F_{i+1,i+1}-a) (3v_i-b)\mk+\mk \cdots \mk+\mk F_{n+1,i+1}(3v_n-b)&=0, \end{align*} where $1\leq i \leq n$. By assumption, since $F^T(3,1,\ldots, 1)^T=(3,1,\ldots, 1)^T$, we have $3F_{11}+\sum_{j=2}^{n+1}F_{j1}=3$ and $3F_{1,i+1}+\sum_{j =2}^{n+1}F_{j,i+1}=1$, and thus find \begin{align*} 3(1-a)b+F_{21}(3v_1-b)+\cdots +F_{n+1,1}(3v_n-b)&=0, \\ (1-a)b+F_{2,i+1}(3v_1-b)+ \cdots + (F_{i+1,i+1}-a) (3v_i-b)+\cdots +F_{n+1,i+1}(3v_n-b)&=0, \end{align*} which after dividing by $(a-1)$ implies~\eqref{eq: kernel and eigenvector, 1} and~\eqref{eq: kernel and eigenvector, 2}. \end{proof} \begin{prop}\label{prop: eigenvector implies existence on cuspidal cubic} Let $C \subseteq \PP^2$ be a singular cuspidal cubic given by $C=V(XY^2-Z^3)$, fix $n\geq 10$ and let $p_1, \ldots, p_n \in C^\ast$ be distinct points. Call $\pi \colon X \rightarrow \PP^2$ the blow-up of $\PP^2$ in the points $p_1, \ldots, p_n$. Denote by $E(p_i)=\pi^{-1}(p_i)$ the exceptional curves and by $L$ the pullback of a general line in $\PP^2$. Consider $F \in \Aut(\Pic(X)) \setminus \{\id\}$, viewing it as a matrix in the basis $([L], [E(p_1)], \ldots, [E(p_n)])$ with entries $F_{ij}$. Suppose that $F$ sends the anticanonical divisor $-K_X$ to itself and that it preserves the intersection form. Furthermore, assume that for each $i\in \{1,\ldots, n\}$ there exists an irreducible curve $C_i$ of $X$ in the class of $F([E(p_i)])$. Fix an eigenvalue $a \in \C\setminus \{1\}$ of $F$ and consider an eigenvector $(b,v_1, \ldots, v_n)^T$ of $F^T$. If $p_i=\frac{1}{a-1}(3v_i-b)$, then there exists an automorphism $f \in \Aut(X)$ such that the induced action on $\Pic(X)$ is equal to $F$, and the restriction of $f$ to the strict transform $\widetilde{C}$ of $C$ induces an automorphism of $C^\ast\cong \A^1$ given by $z \mapsto az +b$. \end{prop} \begin{proof} Note that $E(p_i)\cdot E(p_j)=-\delta_{ij}$, which implies $F([E(p_i)])\cdot F([E(p_j)])=-\delta_{ij}$, since $F$ preserves the intersection form. By assumption, there are irreducible curves $C_i$ lying in $F([E(p_i)])$, for which we therefore find that they are disjoint $(-1)$-curves. Hence there exists a contraction $\eta \colon X \rightarrow \PP^2$ of the curves $C_i$. As we blow up only points on $C^\ast$, the strict transform $\widetilde{C}$ of $C$ lies in $-K_X$. As $F$ preserves the intersection form and since $F(-K_X)=-K_X$ by assumption, we find $C_i\cdot\widetilde{C}=[C_i]\cdot(-K_X)= F([E(p_i)])\cdot F(-K_X)=[E(p_i)]\cdot (-K_X)=1$, and thus any $C_i$ meets $\widetilde{C}$ transversally in one point. From this, we deduce that the birational morphism $\eta$ maps $\widetilde{C}$ to an irreducible cubic curve $\eta(\widetilde{C})$. Additionally, the $\eta(C_i)$ cannot be singular points of $\eta(\widetilde{C})$, since the cusp has multiplicity~$2$ but $\widetilde{C}\cdot C_i=1$. Since none of the points blown up by $\eta$ equals the singular point and since an irreducible cubic curve can have at most one double point, $\eta(\widetilde{C})$ is again cuspidal. We can therefore choose an automorphism $\beta \in \Aut(\PP^2)$ which maps $\eta(\widetilde{C})$ to $C$, and by replacing $\eta$ with $\beta \circ \eta$ we may assume that $\eta(\widetilde{C})=C$. Consequently, the birational map $\pi \circ \eta^{-1}$ restricts to an automorphism of $C$, say given by $z \mapsto a'z+b'$. By assumption, there exists an eigenvalue $a\neq1$ of $F$. Any automorphism of $C$ given by $z \mapsto \hat{a}z$, where $\hat{a} \in \C^\ast$, extends to some automorphism $\gamma_{\hat{a}} \in \Aut(\PP^2)$. Hence, by replacing $\eta$ by $\gamma_{aa'^{-1}}^{-1}\circ \eta$, we may assume that $\pi \circ \eta^{-1}$ restricts to the automorphism $C^\ast \rightarrow C^\ast, z \mapsto az+ b'$. Denote $\varphi_F=\pi \circ \eta^{-1}$. We first prove $b'=b$. For this, consider the pullback $\eta^{-1}(L_{\PP^2})$ of a line $L_{\PP^2}$ not passing through any of the $\eta(C_i)$ via $\eta$, and not being tangent to $C$. Since $\eta^{-1}(L_{\PP^2})$ is disjoint from all the $C_i$, the Picard group of $X$ admits the basis $([\eta^{-1}(L_{\PP^2})], [C_1], \ldots, [C_n])$. Writing $F([L])$ in that basis, we see that $F([L])=d[\eta^{-1}(L_{\PP^2})]$ for some $d\in \Z^{\geq 1}$, since $F([L])\cdot [C_i]=F([L])\cdot F([E(p_i)])=[L]\cdot [E(p_i)]=0$. Then, since $F([L])$ has self-intersection~$1$, we deduce $d=1$ and thus $\eta^{-1}(L_{\PP^2}) \in F([L])$. Call $r_1, r_2, r_3$ the three points of intersection of $L_{\PP^2}$ and $C$. As $L_{\PP^2}$ and $C$ intersect in three points, the curve $\eta^{-1}(L_{\PP^2})$ has three points of intersection with $\widetilde{C}$. The images of $r_1, r_2, r_3$ under $\varphi_F|_{C^\ast}$ equal the images of these three points $\eta^{-1}(r_1), \eta^{-1}(r_2), \eta^{-1}(r_3)$ under $\pi$. Since $F([L]) = F_{11}[L]+F_{21}[E(p_1)]+\cdots + F_{n+1,1}[E(p_n)]$, the curve $\pi(\eta^{-1}(L_{\PP^2}))$ is a curve of degree~$F_{11}$ passing through $p_i$ with multiplicity~$-F_{i+1,1}$ and through $\varphi_F|_{C^\ast}(r_1), \varphi_F|_{C^\ast}(r_2), \varphi_F|_{C^\ast}(r_3)$ with multiplicity~$1$ each. Thus, by the above, on $C^\ast \cong \A^1$ we have the following equality: \[ 3b'\stackrel{\text{Lem.~\ref{lemma: on curve of degree d, sum is zero}}}{=}a(r_1+r_2+r_3)+3b'=\varphi_F|_{C^\ast}(r_1)+\varphi_F|_{C^\ast}(r_2)+\varphi_F|_{C^\ast}(r_3)\stackrel{\text{Lem.~\ref{lemma: on curve of degree d, sum is zero}}}{=}F_{21}p_1+\cdots + F_{n+1,1}p_n. \] By Lemma~\ref{lemma: kernel and eigenvector, arithmetic}~\eqref{eq: kernel and eigenvector, 1}, we know that $F_{21}p_1+\cdots + F_{n+1,1}p_n=3b$, and thus find $b'=b$. We analyse how $\varphi_F|_{C^\ast}$ maps the points $\eta(C_i)$. The curve $C_i$ is an element of $F([E(p_i)])=F_{1,i+1}[L]+\sum_{j=1}^{n}F_{j+1,i+1}[E(p_j)]$; this corresponds to the $(i+1)$-th column of $F$. By the argument laid out in the first paragraph of the proof, there is exactly one point lying in the intersection of $C_i$ and $\widetilde{C}$, say $\widetilde{p_i}$, and $\pi(\widetilde{p_i})=\varphi_F|_{C^\ast}(\eta(C_i))$. The point $\pi(\widetilde{p_i})$ lies on $\pi(C_i)$ and the curve $\pi(C_i)$ is a curve of degree~$F_{1,i+1}$ with multiplicity~$-F_{j+1,i+1}$ at $p_j$ and multiplicity~$1$ at $\pi(\widetilde{p_i})$. Considering the divisor on $C$ given by $\pi(C_i)$, we thus find \[ a\eta(C_i)+b=\varphi_F|_{C^\ast}(p_i)=\pi(\widetilde{p_i})\stackrel{\text{Lem.~\ref{lemma: on curve of degree d, sum is zero}}}{=}F_{2,i+1}p_1 + \cdots + F_{n+1,i+1}p_n. \] By Lemma~\ref{lemma: kernel and eigenvector, arithmetic}~\eqref{eq: kernel and eigenvector, 2}, this must be equal to $ap_i+b$, which implies $\eta(C_i)=p_i$. Thus, $\varphi_F$ lifts to an automorphism on $X$, whose induced action on $\Pic(X)$ is equal to $F$ and whose restriction to $\widetilde{C}$ is the same automorphism as $\varphi_F|_{C^\ast}\colon z\mapsto az +b$. \end{proof} \begin{rema} A similar result holds for an arbitrary irreducible cubic, but one has to take a bit more care, see for instance Diller~\cite{MR2825269} for the degree~$2$ case. However, only very few dynamical degrees outside the unit circle arise on a non-cuspidal cubic, see~\cite[Theorem~2]{MR2825269}. \end{rema} \subsection{Existence of Jonqui\`eres maps}\label{subsection: existence of jonquieres maps} {We take a similar set-up as in the previous section, describing the points blown up more precisely.} Consider a birational map $f \in \Bir(\PP^2)$. Write $\Base(f)=\{p_1, \ldots, p_m\}$ for the base points of $f$ and $\Base(f^{-1})=\{q_1, \ldots, q_m\}$ for the base points of its inverse. {Label} them so that the multiplicities satisfy $\mult_{p_1}(f^{-1}(L)) \geq \ldots \geq \mult_{p_m}(f^{-1}(L))$ and $\mult_{q_1}(f(L)) \geq \ldots \geq \mult_{q_m}(f(L))$. Fix this ordering of the points. Assume that the points of $\Base(f^{-1})$, and $\Base(f)$, respectively, are proper (often called not infinitely near), which by definition means that none of the base points lie on an exceptional curve above some other base point. Suppose that for $1 \leq i \leq m$, we have $n_i \in \Z^{\geq 0}$ such that $f^{n_i-1}(q_i)=p_i$. {Take the minimal $n_i$'s satisfying these conditions. Then} the points $f^s(q_t)$ for $1 \leq t \leq m$ and $0 \leq s \leq n_t-1$ are pairwise distinct. Call the tuple $(n_1, \ldots, n_m)$ the \emph{orbit data of $f$}, the set $O(q_t)=\{f^s(q_t) \thinspace | \thinspace 0 \leq s \leq n_t-1 \}$ the \emph{orbit of $q_t$} and write $O(f)=\bigcup_{1 \leq t \leq m}O(q_t)$. Moreover, denote $q_i^{j}=f^j(q_i)$, where $q_i^0=q_i$ and $q_i^{n_i-1}=p_i$; fix once and for all the ordering of the points $q_i^j$ of $O(f)$ induced by the lexicographic ordering on $(i,j)$. Note that if we blow up $O(f)$, then $f$ lifts to an automorphism on the blow-up $X$, and the ordering on $O(f)$ results in an ordered basis $([L], ([E(q_i^j)]))$ of the Picard group $\Pic(X)$, where $L$ is the pullback of a general line in $\PP^2$ and $E(q_i^j)$ denotes the exceptional curve above $q_i^j$. {Recall that a \emph{Jonqui\`eres map} of degree~$d$ is a map $j_d \in \Bir(\PP^2)$ with base point $p_1$ of multiplicity~$d-1$ and base points $p_2, \ldots, p_{2d-1}$ of multiplicity~$1$. Call the base points of the inverse $q_1, \ldots, q_{2d-1}$, where $q_1$ is the unique one of multiplicity~$d-1$ and all others have multiplicity~$1$. Write $j_d^{\unn}$ for a Jonqui\`eres map with orbit data $\unn=(n_1, \ldots, n_{2d-1})$ and $\lambda_{d,\unn}$ for its dynamical degree. Also, since it will make computations easier, we set $n_1=2$ and abbreviate $\unn=(n_2, \ldots, n_{2d-1})$. In the basis of $\Pic(X)$ described above, the matrix corresponding to $j_d^{\unn}$ takes precisely the form of $J_d^{\unn}$ defined in~\eqref{eq: block matrix} for $m=2d-1$. We call such a matrix $J_d^{\unn}$ a \emph{Jonqui\`eres matrix.} A priori, we do not know if these Jonqui\`eres maps $j_d^{\unn}$ can be realised. Using Section~\ref{subsection: Existence of maps with base points on a cupsidal cubic}, we have an ansatz for which points on the smooth locus $C^\ast$ of the cuspidal cubic we are looking for. Thus, we calculate an eigenvector of $(J^{\unn}_d)^T$ with respect to the eigenvalue $\lambda_{d, \unn}$ realising the spectral radius in the following lemma:} \begin{lemm}\label{lemma: eigenvector of transpose} Fix $\unn$ with $n_1=2$ and $n_i \geq2$ for all $i \in \{2, \ldots, 2d-1\}$. Abbreviate $\lambda=\lambda_{d, \unn}$. Then the vector $(\lambda+1, 1, \lambda, \frac{\lambda}{\lambda^{n_2}+1},\frac{\lambda^2}{\lambda^{n_2}+1}, \ldots, \frac{\lambda^{n_2}}{\lambda^{n_2}+1}, \ldots, \frac{\lambda^{n_{2d-1}}}{\lambda^{n_{2d-1}}+1})^T$ is an eigenvector of $(J^{\unn}_d)^T$ with respect to the eigenvalue $\lambda$. \end{lemm} \begin{proof} Since $\lambda$ is strictly larger than $2$ by Proposition~\ref{prop: greater than 2}, the vector is well-defined. We check directly that $(\lambda+1, 1, \lambda, \frac{\lambda}{\lambda^{n_2}+1},\frac{\lambda^2}{\lambda^{n_2}+1}, \ldots, \frac{\lambda^{n_2}}{\lambda^{n_2}+1}, \ldots, \frac{\lambda^{n_{2d-1}}}{\lambda^{n_{2d-1}}+1})^T$ is an eigenvector of $(J^{\unn}_d)^T$ with respect to the eigenvalue $\lambda$. Indeed, for all the entries where $(J_d^{\unn})^T$ only permutes the entries of the eigenvector, this follows by construction of the vector. The only equations we need to check are \begin{align*} d(\lambda+1)-(d-1)-\sum_{i=2}^{2d-1}\frac{\lambda}{\lambda^{n_i}+1} &= \lambda(\lambda+1), \\ (d-1)(\lambda+1)-(d-2)-\sum_{i=2}^{2d-1}\frac{\lambda}{\lambda^{n_i}+1} &= \lambda^2, \\ \lambda+1 -1-\frac{\lambda}{\lambda^{n_i}+1}&=\lambda\frac{\lambda^{n_i}}{\lambda^{n_i}+1}, \quad i\in \{2,\ldots, 2d-1\}. \end{align*} Multiplying the two upper equations by $\prod_{i =2}^{2d-1}(\lambda^{n_i}+1)$, we find the expression of the characteristic polynomial, which evaluates to $0$ at $\lambda$, and the last equation also holds, which can be seen after multiplying both sides by $\lambda^{n_i}+1$. \end{proof} {The following technical lemma will be used several times in the proposition thereafter. } \begin{lemm}\label{lemma: no three on a line} {Consider $d\geq 2$, $(n_1, n_2, \ldots, n_{2d-1})$ with $n_1=2$, $n_i\in \Z^{\geq 2}$ all pairwise distinct, and $\lambda >2$. Define $q_1^0=3-(\lambda+1)$, $q_1^1=3\lambda-(\lambda+1)$ and $q_i^j=3\tfrac{\lambda^{j+1}}{\lambda^{n_i}+1}-(\lambda+1)$ for $i\geq 2$. Then for any $i_1, i_2, i_3 \in \{1, \ldots, 2d-1\}$ and any $j_k \in \{0, \ldots, n_{i_k}-1\}$ with $k=1,2,3$, we have $q_{i_1}^{j_1}+q_{i_2}^{j_2}+q_{i_3}^{j_3}\neq 0$.} \end{lemm} \begin{proof} {We prove the claim by contradiction. We first prove it for when $i_1, i_2, i_3\neq 1$. Suppose that there exist $i_1, i_2, i_3 \in \{2, \ldots, 2d-1\}$ and any $j_k \in \{0, \ldots, n_{i_k}-1\}$ with $k=1,2,3$ such that $q_{i_1}^{j_1}+q_{i_2}^{j_2}+q_{i_3}^{j_3} = 0$. This implies that \[ \lambda+1=\tfrac{\lambda^{j_1+1}}{\lambda^{n_{i_1}}+1}+\tfrac{\lambda^{j_2+1}}{\lambda^{n_{i_2}}+1}+\tfrac{\lambda^{j_3+1}}{\lambda^{n_{i_3}}+1}. \] Rewrite it as $(\lambda+1)(\lambda^{n_{i_1}}+1)(\lambda^{n_{i_2}}+1)(\lambda^{n_{i_3}}+1)=\lambda^{j_1+1}(\lambda^{n_{i_2}}+1)(\lambda^{n_{i_3}}+1)+\lambda^{j_2+1}(\lambda^{n_{i_1}}+1)(\lambda^{n_{i_3}}+1)+\linebreak\lambda^{j_3+1}(\lambda^{n_{i_1}}+1)(\lambda^{n_{i_2}}+1)$, where after observing $\lambda^{j_k+1}<\lambda^{n_{i_k}}+1$, we deduce \[ (\lambda+1)(\lambda^{n_{i_1}}+1)(\lambda^{n_{i_2}}+1)(\lambda^{n_{i_3}}+1)< 3(\lambda^{n_{i_1}}+1)(\lambda^{n_{i_2}}+1)(\lambda^{n_{i_3}}+1). \] But this would imply $\lambda<2$, a contradiction.} {The cases where $i_k=1$ for all $k=1,2,3$ are done similarly: If $3q_1^0=0$, we would find $\lambda=2$, and if $3q_1^1=0$, then $2\lambda=1$, both times in contradiction to $\lambda>2$. Moreover, $2q_1^0+q_1^1=3\neq 0$ and $q_1^0+2q_1^1=3\lambda\neq 0$. } {We consider now the cases where --- up to labelling the indices --- $i_1=i_2=1$. Denote $i\coloneqq i_3$ and $j\coloneqq j_3$. If $2q_1^0+q_i^j=0$, then $3(\lambda+1)=6+3\tfrac{\lambda^{j+1}}{\lambda^{n_i}+1}$, which is equivalent to $(\lambda-1)(\lambda^{n_i}+1)=\lambda^{j+1}$. Using once more that $\lambda^{j+1}<\lambda^{n_i}+1$, we obtain $\lambda<2$, in contradiction to $\lambda>2$. If $2q_1^1+q_i^j=0$, a similar calculation leads to $1-\lambda=\tfrac{\lambda^{j+1}}{\lambda^{n_i}+1}$, which cannot be, as the left side is negative and the right side is positive. If $q_1^0+q_1^1+q_i^j=0$, a direct calculation results in $\tfrac{\lambda^{j+1}}{\lambda^{n_i}+1}=0$, again a contradiction. } {The cases where only one of the $i_k$ equals $1$ remain, say $i_3=1$. If $q_1^0+q_{i_1}^{j_1}+q_{i_2}^{j_2}=0$, then $3(\lambda+1)=3+3\tfrac{\lambda^{j_1+1}}{\lambda^{n_{i_1}}+1}+3\tfrac{\lambda^{j_2+1}}{\lambda^{n_{i_2}}+1}$. But this results, using again $\lambda^{j_k+1}<\lambda^{n_{i_k}}+1$, in $\lambda<2$, a contradiction. If $q_1^1+q_{i_1}^{j_1}+q_{i_2}^{j_2}=0$, we obtain $(\lambda^{n_{i_1}}+1)(\lambda^{n_{i_2}}+1)=\lambda^{j_1+1}(\lambda^{n_{i_2}}+1)+\lambda^{j_2+1}(\lambda^{n_{i_1}}+1)$. If both $j_1j_1+n_{i_2}$ and $n_{i_1}< j_1+n_{i_2}+2$. This implies $n_{i_1}=j_1+n_{i_2}+1$. We thus find $\lambda^{j_1+1}=1$, a contradiction. This proves the lemma.} \end{proof} With the choice on the points of Lemma~\ref{lemma: eigenvector of transpose} and a weak assumption on the orbits, we can ensure that {the Jonqui\`eres map} $j_d^{\unn}$ exists. Parts of the proof follow the same arguments as in the proof of~\cite[{Proposition~5.12}]{MR3454379}. \begin{prop}\label{prop: jonquieres maps exist} Let $d \geq 4$ and denote by $C\subseteq \PP^2$ the cuspidal cubic $C=V(XY^2-Z^3)$. Fix the orbit data $\unn=(n_2,\ldots, n_{2d-1})$ with $n_i\geq 2$ for each $i\in \{2, \ldots, 2d-1\}$ and all the $n_i$ pairwise distinct. Denote by $\lambda \in \R$ the eigenvalue $\lambda_{d, \unn}$ realising the spectral radius of the Jonqui\`eres matrix $J_d^{\unn}$ and~set \begin{align*} q_1^0=q_1&=\frac{1}{\lambda-1}(2-\lambda), \\ q_1^1=p_1&=\frac{1}{\lambda-1}(2\lambda-1), \\ q_i^0=q_i&=\frac{1}{\lambda-1}(3\frac{\lambda}{\lambda^{n_i}+1}-(\lambda+1)), \\ q_i^j&=\frac{1}{\lambda-1}(3\frac{\lambda^{j+1}}{\lambda^{n_i}+1} -(\lambda+1)), \quad 2 \leq i \leq 2d-1, 1 \leq j \leq n_i-1, \end{align*} where the points are identified with their corresponding point on $C^\ast \cong \A^1$. Then there exists a Jonqui\`eres map $j^{\unn}_d \in \Bir_d(\PP^2)$ with base points $p_1, p_2=q_2^{n_2-1}, \ldots, p_{2d-1}=q_{2d-1}^{n_{2d-1}-1}$, orbits $O(q_i)=\{q_i^j \thinspace | \thinspace 0 \leq j \leq n_i-1\}$ and dynamical degree~$\lambda$. \end{prop} \begin{proof} To prove the proposition, we show that there exists an automorphism realising the Jonqui\`eres matrix $J_d^{\unn}$ of~\eqref{eq: block matrix} on the blow-up $\pi\colon X \rightarrow \PP^2$ in the points $q_i^j$ by applying Proposition~\ref{prop: eigenvector implies existence on cuspidal cubic}. {By Lemma~\ref{lemma: Jonquieres matrices are in W_n}, any} $J_d^{\unn}$ sends $-K_X$ to itself and preserves the intersection form. Thus, to apply Proposition~\ref{prop: eigenvector implies existence on cuspidal cubic}, we need to prove the following two claims: that the points $q_i^j$ are all distinct, and that in each class of $J_d^{\unn}([E(q_i^j)])$ there exists an irreducible curve, where $E(q_i^j)=\pi^{-1}(q_i^j)$ denotes the exceptional curve above $q_i^j$. We start with proving that the points are distinct. Assume $q_1=p_1$. This is equivalent to $\lambda = 1$, a contradiction to $\lambda>2$ by Proposition~\ref{prop: greater than 2}. Now, take any $i\neq 1$ and assume $q_1=q_i^j$. This is equivalent to $\lambda^{n_i}+1=\lambda^{j+1}$, a contradiction to $j\leq n_i-1$ and $\lambda >2$. Similarly, we find that $p_1=q_i^j$ is equivalent to $\lambda^{n_i}+1=\lambda^{j}$, again a contradiction. Then, for a fixed $i \neq 1$, we can also immediately see that $q_i^{k} \neq q_i^{\ell}$ whenever $k \neq \ell$. Assume therefore that $i,j \neq 1$ and that we have $0\leq k \leq n_i-1$ and $0 \leq \ell \leq n_j-1$ such that $q_i^k=q_j^\ell$. Assume without loss of generality that $n_i>n_j$. This is equivalent to $\lambda^k(\lambda^{n_j}+1)=\lambda^\ell(\lambda^{n_i}+1)$. Since $n_i>n_j$ and $\lambda>2$, the equation yields $k>\ell$ and is equivalent to $\lambda^m(\lambda^{n_j}+1)=\lambda^{n_i}+1$, where $m=k-\ell$. As $\lambda>2$ it must follow that $\lambda^{m+n_j}<\lambda^m(\lambda^{n_j}+1)-1= \lambda^{n_i}$, and thus $m+n_j1$ contains the strict transform of a line through $q_1$ and $q_i$, we are left with securing the existence of an irreducible degree~$d-1$ curve through $q_1$ with multiplicity~$d-2$ and through $q_2, \ldots, q_{2d-1}$ with multiplicity~$1$ each. Start with the blow-up $\tau\colon X_1\rightarrow\PP^2$ of $\PP^2$ in $q_1$ to the first Hirzebruch surface $X_1=\F_1$. Blow up the point on $X_1$ corresponding to $q_2$ and blow down the strict transform of the line through $q_1$ and $q_2$, resulting in a birational map $\tau_2\colon X_1\dashrightarrow X_2$, where $X_2$ is either $\F_0$ or $\F_2$. By proving that for all $2\leq i0$, which are blown up by $\pi_2$, lie on one of these lines through $q_1$ and $q_k$, or on the curve of degree $d-1$ passing through $q_1$ with multiplicity~$d-2$ and through $q_2, \ldots, q_{2d-1}$ with multiplicity~$1$. {By combining Lemma~\ref{lemma: on curve of degree d, sum is zero} and Lemma~\ref{lemma: no three on a line}, no three can lie on a line}. Thus, assume by contradiction that some $q_i^j$ lies on a degree~$d-1$ curve through $q_1, \ldots, q_{2d-1}$ with all points having multiplicity~$1$ except $q_1$ with multiplicity~$d-2$. By Lemma~\ref{lemma: on curve of degree d, sum is zero}, this implies $q_i^j=-(d-2)q_1-\sum_{k =2}^{2d-1}q_k$. By replacing the $q_i$ with their explicit values as defined in the statement and using $\lambda\sum_{k =2}^{2d-1}\prod_{\ell \neq k}(\lambda^{n_\ell}+1)=-(\lambda^2-(d-1)\lambda-1)\prod_{k =2}^{2d-1}(\lambda^{n_k}+1)$, we find \begin{align*} (\lambda -1)((d-2)q_1+\sum_{k =2}^{2d-1}q_k)&=(d-2)(2-\lambda)+\sum_{k =2}^{2d-1}\Biggl(3\frac{\lambda}{\lambda^{n_k}+1}-(\lambda+1)\Biggr) \\ &=(d-2)(2-\lambda)+\frac{3}{\prod_{k =2}^{2d-1}(\lambda^{n_k}+1)}\lambda\sum_{k =2}^{2d-1}\prod_{\ell \neq k}(\lambda^{n_\ell}+1)-(2d-2)(\lambda+1) \\ &=(d-2)(2-\lambda)-3(\lambda^2-(d-1)\lambda-1)-(2d-2)(\lambda+1) \\ &= -3\lambda^2+(\lambda+1). \end{align*} Therefore, $q_i^j=-(d-2)q_1-\sum_{k =2}^{2d-1}q_k$ would imply $\lambda^2+1=0$ if $q_i^j=q_1$, $\lambda^2+\lambda=0$ if $q_i^j=p_1$ and $\lambda^2(\lambda^{n_i}+1)+\lambda^j=0$ otherwise. But for any of these equations, the left side is positive, since $\lambda>2$, leading to the desired contradiction. By the above arguments, there exist irreducible curves lying in $J_d^{\unn}([E(q_i^j)])$ on the blow-up $\pi \colon X\rightarrow \PP^2$ in the points $q_i^j$. Thus, applying Proposition~\ref{prop: eigenvector implies existence on cuspidal cubic}, there exists a Jonqui\`eres map $j_d^{\unn} \in \Bir_d(\PP^2)$ with base points $p_1,\ldots, p_{2d-1} \in C^\ast$ and orbits $O(q_i)=\{q_i^j\thinspace |\thinspace 0\leq j\leq n_i-1\}$ which lifts to an automorphism on $X$ such that the induced action on $\Pic(X)$ is equal to the Jonqui\`eres matrix $J_d^{\unn}$. As the spectral radius of $J_d^{\unn}$ is equal to $\lambda$ by assumption, $j_d^{\unn}$ has dynamical degree~$\lambda$. This finishes the proof. \end{proof} By Proposition~\ref{prop: eigenvector implies existence on cuspidal cubic}, the sets $\Lambda_{d, 2d-1}$ of Proposition~\ref{prop: ordinal of helping sets} lie in $\Lambda(\Bir_d(\PP^2))$. This implies Theorem~\ref{thm: main theorem} whenever the ground field $\kbf$ contains the real algebraic numbers. \begin{rema}\label{rem: last remark} As noted in Remark~\ref{rem: Lambdas}, the accumulation points of $\Lambda_{d, 2d-1}$ lie in $\Lambda_{d, 2d-2}$, which are not Salem numbers any more, but Pisot numbers. \emph{Salem numbers }are algebraic integers strictly larger than $1$ whose other Galois conjugates lie in the closure of the unit disk, with at least one of the conjugates on the boundary, and a \emph{Pisot number} is an algebraic integer strictly larger than $1$ whose other Galois conjugates lie in the open unit disk (see for example~\cite[1.1.2]{MR3454379}). One can see that the elements of $\Lambda_{d, 2d-1}$ are Salem numbers and the elements of $\Lambda_{d, m}$ with $m<2d-1$ are Pisot numbers by noting that $m=2d-1$ if and only if $H_{d,m}$ from Lemma~\ref{lemma: adapted intersection form} is the zero matrix. The main difference between Salem and Pisot numbers is that any birational map which can be realised as an automorphism on some blow-up must be Salem, and any birational map which cannot is Pisot, except for quadratic reciprocal integers, which can occur in both cases. Yet by~\cite[Theorem~D]{MR3454379}, the set $\Lambda(\Bir(\PP^2))$ is closed as soon as $\kbf$ is uncountable and algebraically closed (for example $\kbf=\C$), and thus the accumulation points $\Lambda_{d, 2d-2}$ and in fact all elements of the closure $\overline{\Lambda_{d, 2d-1}}=\bigsqcup_{1 \leq m \leq 2d-1}\Lambda_{d, m}$ are also realised as dynamical degrees of birational maps. \end{rema} \section*{Acknowledgements} I would like to thank Serge Cantat, St\'ephane Lamy and Konstantin Loginov for clarifying discussions, mostly, but not exclusively, in relation to ordinals. Furthermore, I am grateful to Curtis T McMullen for drawing my attention to how Uehara's article can shorten the proof of Theorem~\ref{thm: main theorem}, giving rise to Section~\ref{section: bounding from below, Weyl group}. Also, thank you to Gabriel Dill and again Stéphane Lamy for suggesting improvements on the first draft. As always, I thank my PhD advisor J\'er\'emy Blanc for suggesting the question and for the many helpful discussions. \bibliographystyle{crplain} \bibliography{CRMATH_Bot_20220859} \end{document}