\section{Introduction}
\label{sec:intro}

Recall that a link is \emph{alternating} if it admits a diagram such that along any of its components, overcrossings and undercrossings alternate. Ozsv\'ath and Szab\'o defined \emph{quasi-alternating (QA) links}, a class of links that properly includes non-split alternating links and shares many of their properties: for instance, if $L$ is a QA link, then $\Sigma_2(L)$, the double cover of $S^3$ branched over $L$, is a Heegaard Floer $L$-space that bounds a negative-definite 4-manifold~\cite{greene-thin,osquasi}. The definition of QA links is recursive: the unknot is QA, and any link $L$ that admits a diagram with a \emph{QA crossing} is QA; by a QA crossing we mean a crossing $c$ such that 0- and $\infty$-resolutions of $L$ at $c$, denoted $L_0$ and $L_\infty$, are both QA links satisfying the identity $\det L = \det L_0 + \det L_\infty $. Note that determinants of QA links are strictly positive since the determinant of the unknot is~$1$.


Say that $S$ is a \emph{slice surface} for a link $L \subset S^3$ if $S$ is properly smoothly embedded in the 4-ball $B^4$, has no closed components, and $\pp S = L $; we do not require that $S$ be connected or orientable. If such $S$ can be smoothly isotoped rel boundary so that the radial distance function $B^4 \ra [0,1]$ induces a handle decomposition of $S$ with only 0- and 1-handles, then $S$ is a \emph{ribbon surface} for $L$. Following~\cite{DonaldOwens2012}, we say that $L$ is \emph{$\chi$-slice} (resp., \emph{$\chi$-ribbon}) if there exists a slice (resp., ribbon) surface $S$ for $L$ with the Euler characteristic $\chi(S)=1$; these definitions coincide with the usual definitions of slice and ribbon knots when $L$ has one component.

The long-standing question of Fox~\cite{fox} asking whether the sets of slice and ribbon knots coincide readily generalises to $\chi$-slice and $\chi$-ribbon links; we refer to this generalisation as the \emph{$\chi$-slice--ribbon conjecture}. For $L$ a $\chi$-slice link with a slice surface $S$, write $\Sigma_2(S)$ for the double cover of $B^4$ branched over $S$; then $\Sigma_2(L) = \pp \Sigma_2(S)$. By Proposition~2.6 in~\cite{DonaldOwens2012}, if $\det L \neq 0$, then $\Sigma_2(L)$ is a rational homology sphere (denoted $\mathbb{Q} S^3$) and $\Sigma_2(S)$ is a rational homology 4-ball (denoted $\mathbb{Q} B^4$). If $ \Sigma_2(L) $ also bounds a 4-manifold $X$ with negative-definite intersection form $Q_X$, an application of the Mayer--Vietoris exact sequence and Donaldson's theorem \cite{donaldson1987} shows that there exists an embedding of the lattice $\Lambda_X := (H_2(X;\ZZ) / \mathrm{Tors}, Q_X)$ into the standard negative-definite integral lattice of the same rank, denoted $\varphi_X : \Lambda_X \hookrightarrow (\ZZ^{\rk \, \Lambda_X}, -I)$. 
Thus, one can show that the $\chi$-slice--ribbon conjecture holds for a given set $\mathcal{L}$ of links with non-zero determinant by defining a subset $\mathcal{L}' \subset \mathcal{L}$ whose elements are obstructed from being $\chi$-slice via lattice-theoretic methods, and then constructing ribbon surfaces for all links in $\mathcal{L} \setminus \mathcal{L}'$. This strategy was ingeniously used by Lisca in the case of 2-bridge links~\cite{lisca-rational}
, inspiring much following work (e.g., \cite{greene-pretzel,lecuona-montesinos,lecuona-pretzel,simone2020classification}).

In this paper we describe our progress towards resolving the $\chi$-slice--ribbon conjecture for QA links that can be obtained as closures of 3-braids. In~\cite{simone2020classification}, the second author has classified a large number of QA 3-braid closures whose double branched covers bound $\mathbb{Q} B^4$s, including all alternating links among them, while in~\cite{brejevs2020ribbon}, the first author has constructed ribbon surfaces for all families of alternating links in the classification that do not contain non-slice knots. Building on this work, we are able to describe all QA 3-braid closures whose double branched covers bound $\mathbb{Q} B^4$s, with the exception of two links that are not $\chi$-slice, and to construct Euler characteristic one ribbon surfaces for all $\chi$-slice non-alternating QA 3-braid closures.

The classification of QA 3-braid closures due to Baldwin~\cite{baldwinquasi} consists of one infinite and two finite families. 
It is straightforward to establish that the $\chi$-slice--ribbon conjecture holds for the finite families, while the infinite family is of greater interest: it consists of closures of 3-braids of the form
\[
    \tag{$\star$}
    \label{eq:3-braid}
    (\sigma_1 \sigma_2)^{3t} \sigma_1 \sigma_2^{-(a_1 - 2)} \dots \sigma_1 \sigma_2^{-(a_n - 2)},
\]
where $\sigma_1$ and $\sigma_2$ are the standard generators of the braid group on three strands $B_3$, $n\ge1$, $t\in\{-1,0,1\}$, $a_i\ge2$ for all $i=1,\dots,n$, and some $a_j\ge3$. 
Following~\cite{simone2020classification}, links in this family whose double branched covers are unobstructed from bounding $\Q B^4$s by certain lattice-theoretic arguments, can be further assigned into subfamilies
\[
\mathcal{S}_1^t = \bigcup_{x \in \{a, b, c, d, e\}} \mathcal{S}_{1x}^t, \qquad \mathcal{S}_2^t = \bigcup_{x \in \{a, b, c, d, e\}} \mathcal{S}_{2x}^t, \quad  \textrm{and} \quad \mathcal{O}^{t}
\]
for $t \in \{-1, 0, 1\}$. We defer precise definitions to Section~\ref{sec:qa3bc}, writing $\mathcal{S}_i = \bigcup_{t\in  \{-1,0,1\}} \mathcal{S}_i^t$ for $i \in \{1,2\}$, $\mathcal{S}^t  = \mathcal{S}_{1}^t \cup \mathcal{S}_{2}^t$ and $\mathcal{O} = \bigcup_{t\in \{-1,0,1\}} \mathcal{O}^t$. Also, let $L_e^{-1} \in \mathcal{O}^{-1}$ and $L_e^{+1} \in \mathcal{O}^{1}$ denote the closure of the braid $(\sigma_1 \sigma_2)^{3t} (\sigma_1 \sigma_2^{-1})^6$ for $t = -1$ and $t = 1$, respectively. With this notation, we prove the following:

\begin{thm-intro}
\label{thm:1}
    Suppose that $L$ is a closure of a 3-braid of the form~\eqref{eq:3-braid} and $L \neq L_e^{\pm1}$. Then $\Sigma_2(L)$ bounds a $\Q B^4$ if and only if $L \in \mathcal{S}_{1}^{-1} \cup (\mathcal{S}_{1}^1 \setminus \mathcal{S}_{1a}^1) \cup \mathcal{S}_{2}^0$.
\end{thm-intro}

In fact, it was shown in~\cite{simone2020classification} that if $L \in \mathcal{S}_2 \cup \mathcal{S}^0 \cup \mathcal{O}^0$, then $\Sigma_2(L)$ bounds a $\Q B^4$ if and only if $L \in \mathcal{S}_2^0$. The second author has also exhibited $\Q B^4$s bounded by $\Sigma_2(L)$ for $L \in \mathcal{S}_1^{-1} \cup (\mathcal{S}_1^{1} \setminus \mathcal{S}_{1a}^{1})$, but was unable to obstruct the existence of $\Q B^4$s bounded by $\Sigma_2(L)$ for all $L \in \mathcal{S}_{1a}^1 \cup \mathcal{O}^{-1} \cup \mathcal{O}^{1}$; 
in this paper we reach the obstruction for all links in $\mathcal{S}_{1a}^1 \cup \mathcal{O}^{-1} \cup \mathcal{O}^1$ except $L_e^{-1}$ and $L_e^{+1}$. Our proof uses lattice-theoretic arguments in the spirit of Lisca~\cite{lisca-rational} coupled with the following condition due to Greene and Jabuka~\cite{greene-pretzel} and developed by Greene and Owens~\cite{greene-owens}, called \emph{cubiquity}: if a negative-definite manifold $X$ shares the boundary with a $\Q B^4$ and satisfies a certain condition on its Heegaard Floer homology, called \emph{sharpness}, then the image of the embedding $\varphi_X: \Lambda_X \hookrightarrow (\Z^{\rk\,\Lambda_X}, -I)$ intersects every unit cube in $\Z^{\rk\,\Lambda_X}$.

To resolve the $\chi$-slice--ribbon conjecture for all QA 3-braid closures, one would hope to construct Euler characteristic one ribbon surfaces for all links in Theorem~\ref{thm:1} and to determine whether $L_e^{-1}$ and $L_e^{+1}$ are $\chi$-slice. In~\cite{brejevs2020ribbon}, the first author proved that if $L \in \mathcal{S}_2^0 \setminus \mathcal{S}_{2c}^0$, then $L$ is $\chi$-ribbon. However, as shown in~\cite{ammmps2020branched,brejevs2020ribbon,sartori2010}, the $\mathcal{S}_{2c}^0$ subfamily also contains non-slice knots. In this paper we extend the construction of ribbon surfaces to $\mathcal{S}_1^{-1} \cup (\mathcal{S}_1^1 \setminus \mathcal{S}_{1a}^1)$ by finding a sequence of \emph{band moves} of length $n$ that yield the $(n+1)$-component unlink for each of the links in this set, as well as prove that $L_e^{-1}$ and $L_e^{+1}$ are not $\chi$-slice.

\begin{thm-intro}
\label{thm:2}
    Suppose that $L \in \mathcal{S}_{1}^{-1} \cup (\mathcal{S}_{1}^1 \setminus \mathcal{S}_{1a}^1) \cup (\mathcal{S}_{2}^0 \setminus \mathcal{S}_{2c}^0) $ and $L\neq L_e^{\pm1}$. Then $L$ is $\chi$-ribbon. 
    Moreover, if $L = L^{\pm1}_e$, then $L$ is not $\chi$-slice.
\end{thm-intro}

It follows from Theorem~1 and Theorem~2 that the $\chi$-slice--ribbon conjecture holds for all QA 3-braid closures, with the possible exception of the  $\mathcal{S}_{2c}^0$ subfamily.

\begin{cor-intro}
\label{cor:3}
  Let $L$ be the closure of a 3-braid of the form~\eqref{eq:3-braid} such that $L\not\in\mathcal{S}^0_{2c}$. Then $L$ is $\chi$-slice if and only if $L\in \mathcal{S}_{1}^{-1} \cup (\mathcal{S}_{1}^1 \setminus \mathcal{S}_{1a}^1) \cup (\mathcal{S}_{2}^0 \setminus \mathcal{S}_{2c}^0)$. In particular, the $\chi$-slice--ribbon conjecture holds for all QA 3-braid closures $L \notin \mathcal{S}_{2c}^0$.
\end{cor-intro}

\subsection{Proof Overview} In order to prove Theorem~\ref{thm:1}, we derive three results of wider interest that we consequently apply to links in $ \mathcal{S}_{1a}^1 \cup \mathcal{O}^{-1} \cup \mathcal{O}^1$. First, in Theorem~\ref{thm:sharp} we use the Heegaard Floer surgery triangle to show that certain families of cyclic plumbings are sharp, and hence are subject to the cubiquity obstruction. Such plumbings arise naturally as negative-definite manifolds bounded by $\Sigma_2(L)$ for $L \in \bigcup_{i \in \{-1,0\}} \mathcal{S}^i \cup \mathcal{O}^i$; from a lattice-theoretic point of view these plumbings may be studied via associated circular subsets in the sense of Lisca~\cite{lisca-3braids}. In Theorem~\ref{thm:cubandcontr} we establish conditions under which the operation of circular subset expansion preserves the property of not being cubiquitous. Finally, Theorem~\ref{thm:nocontraction} shows that all negative cyclic subsets $S$ with $I(S) \leq 4$ are expanded from a finite set of base examples. By combining Theorems~\ref{thm:sharp},~\ref{thm:cubandcontr} and~\ref{thm:nocontraction} with the verification of non-cubiquity for the base examples in $\mathcal{S}_{1a}^1 \cup \mathcal{O}^{-1} \cup \mathcal{O}^1$, Theorem~\ref{thm:1} follows.

The main part of the proof of Theorem~\ref{thm:2} consists of explicit constructions of ribbon surfaces using the technique of~\cite{brejevs2020ribbon}. We also prove that $L = L_e^{\pm1}$ is not $\chi$-slice by employing an extension of the Fox--Milnor condition due to Florens~\cite{florens} and general topological considerations to obstruct the existence of, respectively, orientable and unorientable Euler characteristic one slice surfaces for $L$.

\subsection{Organisation of the Paper} In Section~\ref{sec:qa3bc} we outline Baldwin's classification of QA 3-braid closures and summarise results from~\cite{simone2020classification} about $\Q B^4$s bounded by their double branched covers, along with results from~\cite{brejevs2020ribbon} about their $\chi$-slice and $\chi$-ribbon status. In Section~\ref{sec:cubiquitydef} we introduce the cubiquity obstruction from~\cite{greene-pretzel} and~\cite{greene-owens}, as well as a practical method of computing it. The goal of Section~\ref{sec:sharpdbc} is to show that double branched covers of links in $\bigcup_{i \in \{-1,0\}} \mathcal{S}^i \cup \mathcal{O}^i$ bound sharp manifolds; in fact, we prove a more general result concerning sharpness of cyclically plumbed 4-manifolds. Section~\ref{sec:latticebasic} contains a summary of fundamental results about linear and circular subsets of $\ZZ^n$, while Section~\ref{sec:cubiquitycontractions} establishes the relation between cubiquity and Lisca's notion of subset contraction. The bulk of technical lattice work is carried out in Section~\ref{sec:nocontractions}, where we develop the theory of negative cyclic subsets; the proof of Theorem~\ref{thm:1} follows in Section~\ref{sec:thm1}. Section~\ref{sec:ribbons} contains constructions of ribbon surfaces that yield the proof of Theorem~\ref{thm:2}. Finally, in Appendix~\ref{app:determinants} we prove auxiliary results about determinants of intersection forms of three classes of plumbed 4-manifolds that are required in Section~\ref{sec:nocontractions}.

\subsection{Acknowledgements} We would like to thank Brendan Owens for many helpful discussions, feedback on an earlier draft of this paper, and explanations of the algorithm for verifying cubiquity in Section~\ref{sec:cubiquitydef}. We also thank Frank Swenton for developing and maintaining \textsc{KLO} software.


\section{Quasi-alternating 3-Braid Closures}
\label{sec:qa3bc}
In~\cite{baldwinquasi}, Baldwin fully described all QA 3-braid closures:

\begin{thm}[Theorem~8.7 in~\cite{baldwinquasi}]
\label{thm:baldwinclassification}
If $L$ is a QA link with braid index at most 3, then $L$ is isotopic to the closure of one of the following:
    \begin{enumerate}
        \item the braid $(\sigma_1 \sigma_2)^{3t} \sigma_1 \sigma_2^{-(a_1-2)} \cdots \sigma_1 \sigma_2^{-(a_n-2)}$, where $a_i \ge 2$ for $i=1,\dots,n$, some $a_j \ge 3$, and $t\in\{-1,0,1\}$;
        \item the braid $(\sigma_1 \sigma_2)^{3t} \sigma_2^{m}$, where $t=1$ and $m\in\{-1,-2,-3\}$, or $t=-1$ and $m\in\{1,2,3\}$;
        \item the braid $(\sigma_1 \sigma_2)^{3t} \sigma_1^m \sigma_2^{-1}$, where $t\in\{0,1\}$ and $m\in\{-1,-2,-3\}$.
    \end{enumerate}
\end{thm}

Figure~\ref{fig:fam2} in Section~\ref{sec:ribbons} shows that all closures of braids in the family (2) are $\chi$-ribbon, while closures of braids in the family (3) are precisely the unknot, which is trivially $\chi$-ribbon, and the trefoil knot along with the Hopf link, whose double branched covers do not bound $\mathbb{Q} B^4$s since the determinants of these links are non-square. 
Hence, our main focus for the remainder of this paper is on the family (1). Observe that any 3-braid in the family (1) can be equivalently described by specifying $t \in \{ -1,0,1 \}$ and the \emph{associated string} $\mathbf{a} = (a_1, \dots, a_n)$. Since closures of such 3-braids with fixed $t$ whose associated strings are related by cyclic reorderings and reversals 
are isotopic, we only need to consider associated strings up to those two operations.

Any string of integers $(b_1, \dots, b_k)$ with $b_i \ge 2$ for all $i$ and some $b_j \ge 3$ can be written in the form $$(2^{[x_1]}, 3 + y_1, 2^{[x_2]}, 3+y_2, \dots, 2^{[x_m]}, 2 + y_m),$$ where $m\ge 1$, $x_i, y_i \ge0$ for all $i$, and $2^{[x_i]}$ denotes a substring consisting of the integer 2 repeated $x_i$ times. Given such string, we define its \emph{linear dual} to be the string $$(2+x_1, 2^{[y_1]}, 3+x_2, 2^{[y_2]}, 3+x_3, \dots, 3+x_m, 2^{[y_m]}).$$ The linear duals of the strings $(2^{[k]})$ for $k \ge 1$ and $(1)$ are defined to be $(k+1)$ and the empty string, respectively. The \emph{cyclic dual} of a string $$(2^{[x_1]}, 3+y_1, 2^{[x_2]}, 3+y_2, \dots, 2^{[x_m]}, 3+y_m)$$ with $m \ge 1$ and $x_i, y_i \ge 0$ for all $i$ is given by $$(3+x_1, 2^{[y_1]}, 3+x_2, 2^{[y_2]}, \dots, 3+x_m, 2^{[y_m]}).$$

Let us now define the following sets of strings, where in each case $(b_1, \dots, b_k)$ and $(c_1, \dots, c_l)$ are linear duals of each other:
\begin{itemize}
    \item $\mathcal{S}_{1a} = \{ (b_1, \dots, b_k, 2, c_l, \dots, c_1, 2) \mid k + l \ge 3 \}$;
    \item $\mathcal{S}_{1b} = \{ (b_1, \dots, b_k, 2, c_l, \dots, c_1, 5) \mid k + l \ge 2 \}$;
    \item $\mathcal{S}_{1c} = \{ (b_1, \dots, b_k, 3, c_l, \dots, c_1, 3) \mid k + l \ge 2 \}$;
    \item $\mathcal{S}_{1d} = \{ (2, b_1 + 1, b_2, \dots, b_{k-1}, b_k + 1, 2, 2, c_l + 1, c_{l-1}, \dots, c_2, c_1 + 1, 2) \mid k + l \ge 3 \}$;
    \item $\mathcal{S}_{1e} = \{ (2, 3 + x, 2, 3, 3, 2^{[x-1]}, 3, 3) \mid x \ge 0 \textrm{ and } k + l \ge 2 \}$;
    \item $\mathcal{S}_{2a} = \{ (b_1 + 3, b_2, \dots, b_k, 2, c_l, \dots, c_1) \}$;
    \item $\mathcal{S}_{2b} = \{ (3 + x, b_1, \dots, b_{k-1}, b_k + 1, 2^{[x]}, c_l + 1, c_{l-1}, \dots, c_1) \mid x \ge 0 \textrm{ and } k + l \ge 2 \}$;

    \item $\begin{aligned}[t] \mathcal{S}_{2c}=\{&(3+x_1,2^{[x_2]},3+x_3,2^{[x_4]},\dots,3+x_{2m+1},\\&2^{[x_1]}, 3+x_2,2^{[x_3]},3+x_4,\dots,2^{[x_{2m+1}]}) \mid m\ge0 \text{ and } x_i\ge 0 \text{ for all } i\};\end{aligned}$
    \item $\mathcal{S}_{2d} = \{ (2, 2+x, 2, 3, 2^{[x-1]}, 3, 4) \mid x \ge 1 \} \cup \{ (2, 2, 2, 4, 4) \}$;
    \item $\mathcal{S}_{2e} = \{ (2, b_1 + 1, b_2, \dots, b_k, 2, c_l, \dots, c_2, c_1 + 1, 2) \mid k + l \ge 3 \} \cup \{ (2,2,2,3) \} $;
    \item $\mathcal{O} = \{(6,2,2,2,6,2,2,2), (4,2,4,2,4,2,4,2), (3,3,3,3,3,3)\}$.
\end{itemize}

For $t \in \{ -1, 0, 1 \}$ and $x \in \{ a, b, c, d, e \}$, define $\mathcal{S}_{1x}^t$, $\mathcal{S}_{2x}^t$ and $\mathcal{O}^t$ to be the sets of closures of 3-braids in the family (1) from Theorem~\ref{thm:baldwinclassification} that contain the $(\sigma_1 \sigma_2)^{3t}$ factor and whose associated strings lie in, respectively, $\mathcal{S}_{1x}$, $\mathcal{S}_{2x}$ and $\mathcal{O}$. Set
\[\mathcal{S}_1^t = \bigcup_{x \in \{a, b, c, d, e\}} \mathcal{S}_{1x}^t \quad \textrm{and} \quad \mathcal{S}_2^t = \bigcup_{x \in \{a, b, c, d, e\}} \mathcal{S}_{2x}^t,\] and let $\mathcal{S}_i = \bigcup_{t\in  \{-1,0,1\}} \mathcal{S}_i^t$ for $i \in \{1,2\}$ and $\mathcal{S}^t = \mathcal{S}_{1}^t \cup \mathcal{S}_{2}^t$; also, by abuse of notation, let $\mathcal{O} = \bigcup_{t\in\{-1,0,1\}} \mathcal{O}^t$. Observe that alternating links in $\mathcal{S}_1 \cup \mathcal{S}_2 \cup \mathcal{O}$ are precisely contained in $\mathcal{S}^0 \cup \mathcal{O}^0$.

In~\cite{simone2020classification}, the second author has exhibited an obstruction to $\Sigma_2(L)$ bounding a $\mathbb{Q} B^4$ if $L$ is a QA 3-braid closure that does not lie in $ \mathcal{S}_1 \cup \mathcal{S}_2 \cup \mathcal{O}$. Moreover, he has shown that if $L \in \mathcal{S}_2 \cup \mathcal{S}^0 \cup \mathcal{O}^0$, then $\Sigma_2(L)$ bounds a $\mathbb{Q} B^4$ if and only if $L \in \mathcal{S}_{2}^0$, and explicitly constructed $\mathbb{Q} B^4$s bounded by $\Sigma_2(L)$ for $L \in \mathcal{S}_1^{-1} \cup (\mathcal{S}_1^1 \setminus \mathcal{S}_{1a}^1)$. While it is shown in Lemma~4.2 of~\cite{simone2020classification} that if $L\in\mathcal{S}_{1a}^1$ has odd determinant, then $\Sigma_2(L)$ does not bound a $\mathbb{Q} B^4$, the analysis does not yield the obstruction for the rest of links in $\mathcal{S}_{1a}^1 \cup \mathcal{O}^{-1} \cup \mathcal{O}^1$.

Subsequently, in~\cite{brejevs2020ribbon} the first author has constructed ribbon surfaces for all links in $\mathcal{S}_2^0 \setminus \mathcal{S}_{2c}^0$ and shown that $\mathcal{S}_{2c}^0$ contains infinitely many $\chi$-slice knots and links. However, $\mathcal{S}_{2c}^0$ also contains closures of braids $(\sigma_1 \sigma_2^{-1})^i$ for $i\ge3$ odd that are non-slice for $i \in \{7, 11, 17, 23\}$ by~\cite{ammmps2020branched,sartori2010}, as well as three more knots shown to be non-slice in~\cite{brejevs2020ribbon}; sliceness of these seven examples is rather challenging to obstruct, requiring an involved verification of the Herald--Kirk--Livingston condition~\cite{heraldkirklivingston} on their twisted Alexander polynomials, and it is not known if there are in fact infinitely many non-$\chi$-slice links in $\mathcal{S}_{2c}^0$. For this reason, in the present paper we focus on non-alternating QA 3-braid closures, contained in $\mathcal{S}^{-1} \cup \mathcal{S}^1 \cup \mathcal{O}^{-1} \cup \mathcal{O}^1$.

\section{The Cubiquity Obstruction}
\label{sec:cubiquitydef}

In this section we recall a refinement of the Donaldson's theorem obstruction to the existence of a $\mathbb{Q} B^4$ bounded by a given $\mathbb{Q} S^3$. The classical form of the obstruction states that if a $\mathbb{Q} S^3$ bounds both a $\mathbb{Q} B^4$ and a 4-manifold $X$ with negative-definite intersection form $Q_X$, then the lattice $\Lambda_X = (H_2(X;\ZZ) / \mathrm{Tors}, Q_X)$ admits an embedding $\varphi_X: \Lambda_X \hookrightarrow (\ZZ^{\rk\,\Lambda_X}, -I)$ into the negative-definite integral lattice of equal rank. In~\cite{greene-pretzel}, Greene and Jabuka have derived a more restrictive condition on such embeddings, dubbed \emph{cubiquity} in~\cite{greene-owens} and applicable when $X$ is \emph{sharp}, which is a property related to the Heegaard Floer homology of its boundary. This condition will prove fruitful in the following to obstruct the existence of $\mathbb{Q} B^4$s bounded by double branched covers of all but two links in $\mathcal{S}_{1a}^1 \cup \mathcal{O}^{-1} \cup \mathcal{O}^1$.

\subsection{Cubiquitous Subsets}

We begin with the lattice aspect of the refined obstruction. Hereafter, we denote the negative-definite integral lattice $(\ZZ^n, -I)$ simply by $\ZZ^n$. The next definition and the proposition following are due to Greene and Owens~\cite{greene-owens}.

\begin{defn}
    A subset $S \subset \ZZ^n$ is \emph{cubiquitous} if it has non-zero intersection with every unit cube in $\ZZ^n$, i.e., 
        \[
            S \cap (x + \{ 0, 1\}^n) \neq \varnothing~\textrm{for all}~x\in\ZZ^n.
        \]
    A lattice $\Lambda$ is cubiquitous if it admits an embedding into $\ZZ^{\rk\,\Lambda}$ whose image is cubiquitous; such embeddings are also called cubiquitous.
    \label{def:cubiquitous}
\end{defn}

\begin{prop}[Proposition~2.1 in~\cite{greene-owens}]
    Let $ \Lambda $ be a sublattice of $\ZZ^n$. The following conditions are equivalent:
    \begin{enumerate}
        \item $\Lambda$ is cubiquitous;
        \item every coset of $\Lambda$ is cubiquitous;
        \item every coset of $\Lambda$ contains a point of the unit cube $\{0,1\}^n$.
    \end{enumerate}
\end{prop}

Condition (3) is particularly useful as it enables us to check whether a lattice embedding is cubiquitous in the following way. Let $\Lambda$ be a lattice endowed with a fixed basis and suppose $\rk\,\Lambda = n$. Choose an orthonormal basis $\{e_1, \dots, e_n\}$ of $\ZZ^n$ and let $\varphi: \Lambda\hookrightarrow\ZZ^n$ be a lattice embedding represented with respect to the chosen bases by an integral matrix $B$. Let $D$ be the Smith normal form of $B$, i.e., the diagonal matrix $D=\mathrm{diag}(a_1,\dots,a_n) \in \mathrm{Mat}_n(\ZZ)$ such that $a_1 \geqslant 1$ and $a_i\,|\,a_{i+1}$ for $i=1,\dots, n-1$, satisfying the condition that $D=UBV$ for two matrices $U, V \in \mathrm{Mat}_n(\ZZ)$ which are invertible over $\ZZ$. Consider the commutative diagram
\begin{equation*}
    \xymatrix{
        \ZZ^n \ar[r]^{B} & \ZZ^n \ar[r] \ar[d]^{U} &\ZZ^n/B\ZZ^n \ar[d]^{\psi} \\
        \ZZ^n \ar[u]^{V} \ar[r]^{D} & \ZZ^n  \ar[r] & \ZZ^n/D\ZZ^n \ar[r]^<<<<<{\sim} & \displaystyle{\bigoplus_{i=1}^n \ZZ/a_i\ZZ},
    }
\end{equation*}
where the unlabelled arrows are canonical quotient maps and $\psi([x]) = [Ux]$ for all $[x]\in \ZZ^n/B\ZZ^n$. Every class in $\ZZ^n/D\ZZ^n$ is represented by a vector $y=(y_1,\dots,y_n)$ with $0\le y_i \le a_i$ for $i=1,\dots, n$, hence every class in $\ZZ^n/B\ZZ^n$ is represented by $ U^{-1}y $ for some such $y$. Clearly, for $z\in\{0,1\}^n$ we have that $ [U^{-1}x] = [z] $ if and only if $U^{-1}x - z \in \im\,B$. To verify that $\varphi$ is cubiquitous, it suffices to check that for every $y$ as above, there exists $z\in \{0,1\}^n$ such that $B^{-1}(U^{-1}y - z) \in \ZZ^n $, where $B^{-1}$ is the inverse of $B$ over $\mathbb{Q}$. This procedure is implemented in the accompanying \textsc{SageMath} notebook, which will be used in Section~\ref{sec:thm1} to prove Theorem~\ref{thm:1}.

\subsection{Sharp Manifolds}

Suppose $Y$ is a $\mathbb{Q} S^3$ equipped with a $\spinc$ structure $ \mathfrak{t} $. In~\cite{ozsz2003}, Ozsv\'ath and Szab\'o employ Heegaard Floer homology to associate to every such pair a rational number $ d(Y,\mathfrak{t})$, called the \emph{correction term}, or the \emph{$d$-invariant}. If $X$ is a negative-definite 4-manifold bounded by $Y$ and equipped with a $\spinc$ structure $\mathfrak{s}$, we have that 
    \begin{equation*}
        \tag{$\dagger$}
        \label{eq:ineq-sharp}
        c_1(\mathfrak{s})^2 + b_2(X) \le 4d(Y, \mathfrak{s}|_Y),
    \end{equation*}
where $c_1(\mathfrak{s})$ is the first Chern class of $\mathfrak{s}$, $b_2(X)$ is the second Betti number of $X$, and $\mathfrak{s}|_Y$ is the restriction of $\mathfrak{s}$ to $Y$~\cite{os-absolutelygraded}.

\begin{defn}
    A negative-definite 4-manifold $X$ with $\mathbb{Q} S^3$ boundary $ Y $ is \emph{sharp} if for every $\mathfrak{t} \in \Spinc(Y)$ there exists $\mathfrak{s} \in \Spinc(X)$ with $\mathfrak{t} = \mathfrak{s}|_Y$ such that equality is attained in~\eqref{eq:ineq-sharp}.
\end{defn}

We can now state the cubiquity obstruction precisely.

\begin{thm}[Theorem~6.1 in~\cite{greene-owens}]
    \label{thm:cbobstruction}
    Let $X$ be a sharp 4-manifold with the intersection lattice $\Lambda_X$. If $\pp X$ is a $\mathbb{Q} S^3$ that also bounds a $\mathbb{Q} B^4$, then $\Lambda_X$ admits a cubiquitous embedding into $\ZZ^{\rk\,\Lambda_X}$.
\end{thm}

In view of the above discussion, one can show that a $\mathbb{Q} S^3$ does not bound a $\mathbb{Q} B^4$ by constructing a sharp 4-manifold $X$ bounded by the $\mathbb{Q} S^3$, finding all embeddings of $\Lambda_X$ into the standard integral lattice of the same rank, and verifying that none of them are cubiquitous.

\section{Sharp Manifolds and Quasi-alternating 3-Braid Closures}
\label{sec:sharpdbc}


Let $X_{(a_1, \dots, a_n)}^t$ denote the 4-manifold with handlebody diagram given in Figure \ref{fig:P}, where $a_i\ge2$ for all $i$. Write $\textbf{a}=(a_1,\ldots,a_n)$, and let $Y^t_{\textbf{a}}=\partial X_{\textbf{a}}^t$. 
Note that if $\textbf{a}'$ is any cyclic reordering of $\textbf{a}$, then $X_{\textbf{a}}^t$ and $X_{\textbf{a}'}^t$ are diffeomorphic.
As discussed in Section~1.1 of~\cite{simone2020classification}, $Y_{\textbf{a}}^t$ is the double cover of $S^3$ branched over the closure of the 3-braid $(\sigma_1\sigma_2)^{3t}\sigma_1\sigma_2^{-(a_1-2)}\dots\sigma_1\sigma_2^{-(a_n-2)}$. Note that it follows from Theorem~4.1 in~\cite{baldwinquasi} that when $t\in\{-1,0,1\}$, $Y^t_{\textbf{a}}$ is an $L$-space. The goal of this section is to prove the following:

\begin{thm} Let $\textbf{a}=(a_1,\ldots,a_n)$ such that $a_i\ge2$ for all $i$ and $a_j\ge3$ for some $j$. Then $X_{\textbf{a}}^t$ is sharp if and only if $t\le0$.
\label{thm:sharp}
\end{thm}

\begin{figure}[h]
\centering
\begin{overpic}
[scale=.5]{2handlebody.png}
\put(-8,27){$-a_1$}
\put(-3,38){$-a_2$}
\put(-3,5){$-a_n$}
\put(6,20.6){$t$}
\put(73.5,20.6){$t$}
\put(53,35){$-a_1-2$}
\put(18,-7){$n\ge2$}
\put(80,-7){$n=1$}
\end{overpic}\vspace{.4cm}
\caption{The 4-manifold $X^t_{(a_1, \dots, a_n)}$ whose boundary is $Y^t_{(a_1, \dots, a_n)}$, the double cover of $S^3$ branched over the closure of the 3-braid $(\sigma_1\sigma_2)^{3t}\sigma_1\sigma_2^{-(a_1-2)}\ldots\sigma_1\sigma_2^{-(a_n-2)}$.}\label{fig:P}
\end{figure}

Although we will only need the sharpness of $X^{-1}_{\textbf{a}}$, we will prove the much more general result as it might be of independent interest. To prove Theorem~\ref{thm:sharp}, we will use induction. To this end, we start with the base cases.

\begin{lem} Let $n\ge2$ and $a_i=2$ for all $i$. Then $X^{-1}_{\textbf{a}}$ is sharp.
\label{lem:basesharp}
\end{lem}

\begin{proof}
Set $X=X_{\textbf{a}}^{-1}$ and $Y=\partial X$. 
Let $Q$ denote the intersection form of $X$.
It is easy to see that if $n\ge2$, then $|\det Q|=4$; hence $|H_1(Y)|=|\Spinc(Y)|=4$.
Moreover, note that $Y$ is the double cover of $S^3$ branched over the closure of the 3-braid $(\sigma_1\sigma_2)^{-3}\sigma_1^n$.

We claim that the $d$-invariants of $Y$ are $\{\frac{n}{4}-1,\frac{n}{4},0,0\}$. 
If $n=2$, then $Y=L(2,1)\#L(2,1)$; by Theorem~4.3 and Proposition~4.9 in~\cite{os-absolutelygraded}, the $d$-invariants are indeed $\{-\frac{1}{2},\frac{1}{2},0,0\}$. Now assume $n\ge3$.
By Theorem~6.2(2) in~\cite{baldwinquasi}, there is a $\spinc$ structure $\mathfrak{s}_0$ satisfying $d(Y,\mathfrak{s}_0)=\frac{n}{4}-1$.

To show that there is a $\spinc$ structure $\mathfrak{t}$ such that $d(Y,\mathfrak{t})=\frac{n}{4}$, we will construct a negative-definite plumbing $Z$ with $\partial Z=Y$ and $b_2(Z)=n$, and use the method of \cite{ozsvathszaboplumbed}. Namely, we will find a characteristic element $K\in H^2(Z)$ such that $\frac{K^2+n}{4}=\frac{n}{4}$, or $K^2=0$, and that satisfies the following: 
if $K=c_1(\mathfrak{s})$ and $\mathfrak{s}|_{Y}=\mathfrak{s'}|_{Y}$ for some $\spinc$ structure $\mathfrak{s}'$ on $Z$, then $K^2\ge c_1^2(\mathfrak{s'})$.
Consider Figure \ref{fig:X-1sharpbase}. The first handlebody diagram is that of $X$. Blow up the diagram with a $+1$-framed unknot as in the second diagram. We can then blow down $n-3$ successive $-1$-framed unknots to obtain the third diagram. After handle sliding as indicated by the green arrow, we obtain the fourth diagram. Finally, blow up the linking between the $(n-2)$-framed and $-1$-framed 2-handles with a $+1$-framed unknot and perform successive blow downs until we obtain the last diagram; call the resulting 4-manifold $Z$. Note that $\partial Z=Y$. By \cite{ozsvathszaboplumbed}, $Z$ is sharp. 
Since the framing of each 2-handle of $Z$ is even, the class $K=0$ is characteristic in $H^2(Z)$. Hence $K^2=0$. Since $Z$ is negative-definite, if $K=c_1(\mathfrak{s})$ and $\mathfrak{s}|_{Y}=\mathfrak{s'}|_{Y}$ for some $\spinc$ structure $\mathfrak{s}'$ on $Z$, then $K^2\ge c_1(\mathfrak{s}')^2$. Hence $d(Y,\mathfrak{s}|_Y)=\frac{n}{4}$. 

It is easy to see that $Y$ bounds a $\mathbb{Q} B^4$: blow down the third diagram in Figure \ref{fig:X-1sharpbase} two times and then change the resulting $0$-framed unknot into a dotted circle, as shown in Figure \ref{fig:ball}, to see a $\mathbb{Q} B^4$ bounded by $Y$. 
Hence there is a metaboliser of $\spinc$ structures for which the $d$-invariant vanishes (c.f.~Section~2.3 in~\cite{greene-pretzel}). 
Thus the remaining two $\spinc$ structures must have vanishing $d$-invariants.

\begin{figure}
    \centering
    \hspace{.5cm}
    \begin{overpic}
    [scale=.5]{X-1sharpbase}
    \put(2,35){\huge$X=$}
    \put(21,25){$-2$}
    \put(17,38){$-2$}
    \put(21,44){$-2$}
    \put(22,34.25){$-1$}
    \put(45,37){blow down}
    \put(65,25){$-2$}
    \put(61.5,38){$-2$}
    \put(65,44){$-1$}
    \put(75,47){\color{darkblue}$1$}
    \put(78,48.5){$-1$}
    \put(66,34.2){$-1$}
    \put(86,20){\shortstack[c]{blow down\\$n-3$ times}}
    \put(88.5,16.5){\color{darkblue}$n-2$}
    \put(79,10){$-1$}
    \put(100,10){$-1$}
    \put(89,-3){$-2$}
    \put(71,10){slide}
    \put(49,13.5){\color{darkblue}$n-2$}
    \put(58,13.5){$-1$}
    \put(63,13.5){$-2$}
    \put(61,0){$-2$}
    \put(37,2.5){\shortstack[c]{blow up\\ and\\\\\\blow\\down}}
    \put(28,14){$-2$}
    \put(22,14){$-2$}
    \put(17,14){$-2$}
    \put(1,14){$-2$}
    \put(26,0){$-2$}
    \put(8,0){$n-3$}
    \put(-10,8.2){\huge$Z=$}
    \end{overpic}
    \vspace{.2cm}
    \caption{$\partial X=\partial Z$.}
    \label{fig:X-1sharpbase}
\end{figure}

\begin{figure}
    \centering
    \begin{overpic}
    [scale=.5]{ball.png}
    \put(8,23.5){\color{darkblue}$n-2$}
    \put(-7,10){$-1$}
    \put(25,10){$-1$}
    \put(10,-4){$-2$}
    \put(35,2){\shortstack[c]{blow down\\ twice and change\\\\\\$0$-framed unknot\\to dotted circle}}
    \put(86.5,22){\color{darkblue}$n$}
    \end{overpic}
    \vspace{.2cm}
    \caption{$Y$ bounds a rational homology ball.}
    \label{fig:ball}
\end{figure}

It remains to show that for each $\spinc$ structure $\mathfrak{t}$ on $Y$, there exists a $\spinc$ structure $\mathfrak{s}$ on $X$ such that $\mathfrak{s}|_Y=\mathfrak{t}$ and $c_1(\mathfrak{s})^2 + b_2(X)= 4d(Y, \mathfrak{t})$, or $c_1(\mathfrak{s})^2= 4d(Y, \mathfrak{t})-n\in\{-n,-n,-4,0\}$. Thus we need to find characteristic elements $K_1,K_2,K_3,K_4\in H^2(X)$ whose respective squares are $0,-4,-n,$ and $-n$, and whose corresponding $\spinc$ structures $\mathfrak{s}_i$ for $1\le i\le 4$ satisfy $\mathfrak{s}_i|_{Y}\neq\mathfrak{s}_j|_{Y}$ for $i\neq j$. Set
\begin{align*}
    K_1=(0,\ldots,0)^T, &\quad K_2=(2,0,\ldots,0,2)^T,\\ \quad K_3=(2,0,\ldots,0)^T, &\quad K_4=(0,2,0,\ldots,0)^T.
\end{align*}
A quick calculation shows that $K_1^2=0$, $K_2^2=-4$, and $K_3^2=K_4^2=-n$. 
Let $\mathfrak{s}_1,\mathfrak{s}_2,\mathfrak{s}_3,$and $\mathfrak{s}_4$ be the unique $\spinc$ structures on $X$ satisfying $c_1(\mathfrak{s}_i)=K_i$ for $1\le i\le 4$. Recall that $\spinc$ structures on $Y$ are in a one-to-one correspondence with $2H^2(X,Y)$-orbits in the set of characteristic elements in $H^2(X)$; hence if $\mathfrak{s}_i=\mathfrak{s}_j$, then $K_i-K_j\in 2\,\im(Q)$, where $\im(Q)$ is the image of $Q$, viewed as a map $H^2(X,Y)\to H^2(X)$. It is easy to check that $\frac{1}{2}Q^{-1}(K_i-K_j)\not\in \ZZ^n$ for all $i\neq j$; consequently, $\mathfrak{s}_1|_Y,\mathfrak{s}_2|_Y,\mathfrak{s}_3|_Y,$ and $\mathfrak{s}_4|_Y$ are the four distinct $\spinc$ structures on $Y$. Hence $X$ is sharp.
\end{proof}

\begin{lem} Let $n\ge2$, $a_1=3$, and $a_i=2$ for all $i\neq1$. Then $X^{0}_{\textbf{a}}$ is sharp.
\label{lem:basesharp2}
\end{lem}

\begin{proof}
This follows in the same way as the proof of Lemma~\ref{lem:basesharp}. First notice that $Y^{0}_{\textbf{a}}$ is a lens space; indeed, by blowing up the obvious surgery diagram of $Y^{0}_{\textbf{a}}$ between the $-3$-framed unknot and an adjacent $-2$-framed unknot and then performing $n+1$ successive blowdowns, we obtain a surgery diagram consisting of a single unknot with framing $n$. Thus by using the formula in~\cite{os-absolutelygraded}, the $d$-invariants of $Y^{0}_{\textbf{a}}$ are $$\left\{\frac{-n+(2i-n)^2}{4n}\,\Big|\,0\le i< n\right\}.$$ 
As in the proof of Lemma~\ref{lem:basesharp}, we must find characteristic elements in $H^2(X_{\textbf{a}}^0)$ that square to the values in the set
$$D=\left\{\frac{4i^2}{n}-4i-1\,\Big|\,0\le i< n\right\}.$$
Consider the vectors $K_j=e_1+\sum_{i=2}^j(-1)^{i-1}2e_i$, where $1\le j\le n$ and $\{e_1,\ldots,e_n\}$ is the standard basis for $\ZZ^n$. Following as in the proof of Lemma~\ref{lem:basesharp}, it can be shown that $K_j^2\in D$ for all $j$ and that these vectors correspond to spin$^c$ structures that restrict to distinct spin$^c$ structures on $Y_{\textbf{a}}^0$. 
The result follows.
\end{proof}

\begin{definition}
Let $M$ be an oriented 3-manifold with torus boundary, and let $\gamma_0,\gamma_1,\gamma_2$ be simple closed curves in $\partial M$ 
such that
$$\#(\gamma_0\cap \gamma_1)=\#(\gamma_1\cap \gamma_2)=\#(\gamma_2\cap \gamma_0)=-1,$$ where $\#$ denotes algebraic intersection number and the orientation of $\pp M$ is induced by that of $M$.
Let $Y_i$ denote the 3-manifold obtained by gluing a solid torus to $M$ such that the meridian of the boundary of the solid torus is identified with $\gamma_i\subset \partial M$ for $i\in\{0,1,2\}$. Then $(Y_0,Y_1,Y_2)$ is called a \textit{surgery triad}.
\end{definition}

\begin{thm}[Theorem 2.2 in~\cite{ozsvathszabosharp}]
Let $(Y_0,Y_1,Y_2)$ be a surgery triad. Then there exists a long exact sequence 
$$\cdots\to HF^+(Y_0)\to HF^+(Y_1)\to HF^+(Y_2)\to\cdots$$
where the maps are induced from the obvious 2-handle cobordisms connecting $Y_i$ to $Y_{i+1}$, where $i\in\ZZ/3$.
\label{thm:triad}
\end{thm}

\begin{prop}[Proposition 2.6 in~\cite{ozsvathszabosharp}] 
Suppose $(Y_0,Y_1,Y_2)$ is a triple of $\mathbb{Q} S^3$s that form a surgery triad such that $Y_0$ and $Y_2$ are $L$-spaces. Let $W_i:Y_i\to Y_{i+1}$ denote the 2-handle cobordism connecting $Y_i$ to $Y_{i+1}$. If $-Y_2$ bounds a sharp 4-manifold $X_2$ and $X_0=X_2 \cup (-W_1) \cup (-W_0)$ is sharp, then $X_1=X_2\cup (-W_1)$ is also sharp.
\label{prop:os}
\end{prop}

\begin{remark}
Note that our orientation conventions differ from the conventions used in~\cite{ozsvathszabosharp}. As a result, we adapted the statement of Proposition 2.6 in~\cite{ozsvathszabosharp} to our conventions.
\end{remark}

Given a sequence of integers $(a_1,\ldots,a_n)$, their \emph{(Hirzebruch--Jung) continued fraction expansion} is given by  
\[
    [a_1,\ldots,a_n] = a_1 - \cfrac{1}{a_2 - \cfrac{1}{\dots - \cfrac{1}{a_n}}}.
\]
Given coprime integers $p>q\ge1$, there is a unique continued fraction expansion $[a_1,\ldots,a_n]=\frac{p}{q}$, where $a_i\ge2$ for all $i$.


\begin{proof}[Proof of Theorem~\ref{thm:sharp}]
We first assume that $t\in\{-1,0\}$.
If $n=1$, then $X^{t}_{\textbf{a}}$ is obtained by attaching a single 2-handle to $B^4$ along an unknot with framing $-a_1-1$ (see Figure \ref{fig:P}). Hence by \cite{ozsvathszaboplumbed}, $X^{t}_{\textbf{a}}$ is sharp.

We now assume that $n\ge2$. 
We will prove sharpness by using Theorem~\ref{thm:triad}, Proposition~\ref{prop:os}, and induction. First recall that $\partial X_{\textbf{a}}^{t}$ is an $L$-space for all $\textbf{a}$. If $a_i=2$ for all $i$, then $X^{-1}_{\textbf{a}}$ is sharp by Lemma~\ref{lem:basesharp}; if $a_j=3$ for some integer $j$ and $a_i=2$ for all $i\neq j$, then $X^{0}_{\textbf{a}}$ is sharp by Lemma~\ref{lem:basesharp2} (up to cyclic reordering).

Let $\textbf{a'}=(a_1,\ldots,a_{i-1},a_i-1,a_{i+1},\ldots,a_n)$ be arbitrary and inductively assume that $X^{t}_{\textbf{a'}}$ is sharp. Up to cyclic reordering, we may assume that $i=1$. Let $\textbf{a}=(a_1,\ldots,a_n)$ and let $\frac{p}{q}=[a_2,\ldots,a_n]$. We claim that $(Y^t_{\textbf{a'}},Y^t_{\textbf{a}},L(p,q))$ forms a surgery triad. 
Let $m$ be a meridian of the $-(a_1-1)$ surgery curve in the obvious surgery diagram of $Y^t_{\textbf{a'}}$ and let $T=\partial\nu(m)$. Then $M=Y^t_{\textbf{a'}}\setminus\mathring{\nu}(m)$ is a 3-manifold with torus boundary. Let $\gamma_2$ be the simple closed curve on $T$ that can be identified with the blackboard framing curve of $m$; let $\gamma_0$ be the simple closed curve on $T$ that bounds a disk in $\nu(m)$, oriented so that $\#(\gamma_2,\gamma_0)=-1$; and let $\gamma_1$ be the simple closed curve on $T$ satisfying $[\gamma_1]=-[\gamma_0]-[\gamma_2]\in H_2(T)$ (see Figure \ref{fig:triad}). Then $\gamma_0,\gamma_1$, and $\gamma_2$ satisfy the conditions of Theorem~\ref{thm:triad}. Moreover, using the notation of Theorem~\ref{thm:triad}, $Y_0$ is obtained by $\infty$-surgery on $m$, $Y_1$ is obtained by $1$-surgery on $m$, and $Y_2$ is obtained by $0$-surgery on $m$; hence $Y_0=Y^t_{\textbf{a}'}$, $Y_1=Y^t_{\textbf{a}}$, and $Y_2=L(p,q)$. We have thus shown that $(Y^t_{\textbf{a'}},Y^t_{\textbf{a}},L(p,q))$ forms a surgery triad.

\begin{figure}[h]
	\centering
 \begin{overpic}
 {triad.pdf}
 \put(38,2){\color{green}{$\gamma_0$}}
 \put(20,6){\color{blue}{$\gamma_2$}}
 \put(74,21){\color{red}{$\gamma_1$}}
 \end{overpic}
 \caption{Curves on $T$ defining a surgery triad.}
 \label{fig:triad}
\end{figure}

Figure \ref{fig:cobs} shows the $2$-handle cobordisms $W_i:Y_i\to Y_{i+1}$ for $i\in\{0,1\}$ inducing the long exact sequence maps in Theorem~\ref{thm:triad}. Following Section 5.5 in~\cite{gompfstipsicz}, the bottom boundary component $\partial_{-}W_i$ of $W_i$ (for $i=0,1$) has surgery diagram given by the black link and whose framings are in angular brackets. The blue framed knot denotes a 2-handle attached to $\partial_-W_i\times[0,1]$. The top boundary component $\partial_+W_i$ of $W_i$ has surgery given by the full diagram (i.e., the diagram obtained by ignoring the angular brackets). Hence it is clear, after performing blowdowns, that $\partial_-W_0=Y^t_{\textbf{a}'}$, $\partial_+W_0=Y^t_{\textbf{a}}$, $\partial_-W_1=Y^t_{\textbf{a}}$, and $\partial_+W_1=L(p,q)$, where $\frac{p}{q}=[a_2,\ldots,a_n]$. Note that $L(p,q)$ bounds a linear plumbing $X_2$ with weights $-a_2,\ldots,-a_n$, which is sharp by \cite{ozsvathszaboplumbed}. 
We claim that $X_{\textbf{a}}^{t}=(-W_1)\cup X_2$. If we flip the handlebody diagram of $W_1$ upside down and reverse its orientation, we obtain the first handlebody diagram in Figure \ref{fig:-cob1} (c.f. Section 5.5 in~\cite{gompfstipsicz}). Blowing down the first $\langle 1\rangle$-framed unknot yields the next diagram in Figure \ref{fig:-cob1}. Finally, after sliding the $-1$-framed blue $2$-handle over the $\langle -(a_1-1)\rangle$-framed unknot, we obtain the last diagram in Figure \ref{fig:-cob1}. With this description, it is clear that $X_{\textbf{a}}^t=(-W_1)\cup X_2$. 

\begin{figure}
    \centering
    \begin{overpic}
    [scale=.5]{w0w1}
    \put(18,12){\huge$W_0$}
    \put(-2,12){\color{darkblue}$1$}
    \put(-3,2){$\langle-a_n\rangle$}
    \put(-14.5,19){$\langle-(a_1-1)\rangle$}
    \put(-1.5,26){$\langle-a_2\rangle$}
    \put(85,12){\huge$W_1$}
    \put(63,9.5){\color{darkblue}$1$}
    \put(67.5,9.5){$\langle1\rangle$}
    \put(64,2){$\langle-a_n\rangle$}
    \put(53,19){$\langle-(a_1-1)\rangle$}
    \put(66.5,26){$\langle-a_2\rangle$}
    \end{overpic}
    \caption{The cobordisms $W_0$ and $W_1$.}
    \label{fig:cobs}
\end{figure}

Next, consider the handlebody diagram for $-W_0$ as shown in the left side of Figure~\ref{fig:-cob0}. Blowing down the $\langle 1\rangle$-framed unknot yields the right handlebody diagram for $-W_0$ shown in Figure \ref{fig:-cob0}. Notice that the bottom boundary of $-W_0$ is $\partial X_2$; indeed if we remove the $-1$-framed $2$-handle, we are left with the surgery diagram for $\partial X_{\textbf{a}}^{t}$.  Let $X_0:=(-W_0)\cup (-W_1)\cup X_2$; note that $X_0$ has the handlebody diagram given by the right diagram in Figure \ref{fig:-cob0}, except with the brackets removed from the framings. It is thus clear that $X_0=X_{\textbf{a'}}^{t}\#\overline{\mathbb{CP}^2}$. 
By the inductive hypothesis, $X_{\textbf{a'}}^{t}$ is sharp (see, for example, \cite{ozsvathszabosharp}); hence $X_0$ is also sharp. Thus by Proposition~\ref{prop:os}, $X_{\textbf{a}}^t$ is sharp.

Now let $t$ be arbitrary. Notice that for fixed $\textbf{a}$, the 4-manifolds $X^{2k+1}_{\textbf{a}}$ all have the same intersection form and, similarly, the the 4-manifolds $X^{2k}_{\textbf{a}}$ all have the same intersection form.
In~\cite{baldwinquasi}, Baldwin considers the $\spinc$ structure $\mathfrak{t}_0$ on $Y^t_{\textbf{a}}$ associated to a particular contact structure. In particular, he shows in Theorem 6.2 in~\cite{baldwinquasi} that 
\[
d(Y^t_{\textbf{a}},\mathfrak{t}_0)=
\begin{cases}
(3n-\sum_{i=1}^na_i)/4 \hspace{.2cm}\text{ if } \hspace{.2cm}t\ge0\\
-1+(3n-\sum_{i=1}^na_i)/4 \hspace{.2cm}\text{ if } \hspace{.2cm}t<0
\end{cases}.
\]
Since $X^{-1}_{\textbf{a}}$ is sharp, it follows that $$d(Y^{-1}_{\textbf{a}},\mathfrak{t}_0)=\max_{\substack{\mathfrak{s}\in\spinc(X^{-1}_{\textbf{a}})\\\mathfrak{s}|_Y=\mathfrak{t}_0}}\frac{c_1(\mathfrak{s})^2+n}{4}.$$
Now since the intersection form of $X^{2k+1}_{\textbf{a}}$ is the same as the intersection form of $X^{-1}_{\textbf{a}}$ for all $k$, it follows that if $X^{2k+1}_{\textbf{a}}$ is sharp, then $d(Y^{-1}_{\textbf{a}},\mathfrak{t}_0)=d(Y^{2k+1}_{\textbf{a}},\mathfrak{t}_0)$. But this only occurs if $k<0$; hence $X^{2k+1}_{\textbf{a}}$ is not sharp for all $k\ge0$. 
Now suppose $k<0$. By the remarks preceding Proposition 5.1 in~\cite{baldwinquasi}, it follows that  $$\{d(Y^t_{\textbf{a}},\mathfrak{t})\,|\,\mathfrak{t}\neq\mathfrak{t}_0\}=\{d(Y^{t+2i}_{\textbf{a}},\mathfrak{t})\,|\,\mathfrak{t}\neq\mathfrak{t}_0\}$$
for all $i\in\ZZ$. Hence it follows that $X^{2k+1}_{\textbf{a}}$ is sharp if $k<0$. A similar argument shows that $X^{2k}_{\textbf{a}}$ is sharp if and only if $k\le0$.
\end{proof}

\begin{figure}
    \centering
    \begin{overpic}
    [scale=.5]{negw1}
    \put(17,8){\huge$-W_1$}
    \put(0,6){\color{darkblue}$0$}
    \put(3,6){$\langle 1\rangle$}
    \put(7,6){$\langle 1\rangle$}
    \put(4,1){$\langle-a_n\rangle$}
    \put(-3,13){$\langle-(a_1-1)\rangle$}
    \put(6,18){$\langle-a_2\rangle$}
    \put(30.5,4.5){\shortstack[c]{blow\\down}}
    
    \put(54,8){\huge$-W_1$}
    \put(39,6){\color{darkblue}$-1$}
    \put(43,6){$\langle 0\rangle$}
    \put(41,1){$\langle-a_n\rangle$}
    \put(33,13){$\langle-(a_1-1)\rangle$}
    \put(43,18){$\langle-a_2\rangle$}
     \put(66,4.5){\shortstack[c]{slide\\and cancel}}
     
    \put(88,8){\huge$-W_1$}
    \put(75,1){$\langle-a_n\rangle$}
    \put(76,13){\color{darkblue}{$-a_1$}}
    \put(76,18){$\langle-a_2\rangle$}
    \end{overpic}
  \caption{The cobordism $-W_1$.}
    \label{fig:-cob1}
\end{figure}

\begin{figure}
\centering
    \begin{overpic}
    [scale=.5]{negw0}
    \put(22,14){\huge$-W_0$}
    \put(0,10.5){\color{darkblue}$0$}
    \put(4,10.5){$\langle 1\rangle$}
    \put(1.5,3){$\langle-a_n\rangle$}
    \put(-11,21){$\langle-(a_1-1)\rangle$}
    \put(3,27){$\langle-a_2\rangle$}
    \put(47,7){\shortstack[c]{blow\\down}}
    
    \put(82,14){\huge$-W_0$}
    \put(60,11){\color{darkblue}$-1$}
    \put(61,3){$\langle-a_n\rangle$}
    \put(59,21){$\langle-a_1\rangle$}
    \put(63,27){$\langle-a_2\rangle$}
    \end{overpic}
    \caption{The cobordism $-W_0$.}
    \label{fig:-cob0}
\end{figure}





\section{Good, Standard, and Cyclic Subsets}
\label{sec:latticebasic}

In this section we establish some fundamental definitions and results pertaining to several classes of finite subsets of $\ZZ^n$ that shall be used in the following sections. Consider the standard negative-definite intersection lattice $(\ZZ^n,-I)$ and let $\{e_1,\ldots,e_n\}$ be the standard orthonormal basis of $\ZZ^n$. Then, with respect to the product $\cdot$ given by $-I$, we have $e_i\cdot e_j=-\delta_{ij}$ for all $i,j$; unless indicated otherwise, we use this product in the remainder of the paper. We begin by recalling definitions and results from~\cite{lisca-rational} and~\cite{simone2020classification}.

Given a subset $S=\{v_1,\ldots,v_n\}\subset\ZZ^n$, the \emph{intersection graph} of $S$ is the weighted graph consisting of a vertex with weight $v_i\cdot v_i$ for each vector $v_i$, and an edge labelled $v_i\cdot v_j$ between each pair of vertices $v_i$ and $v_j$ with $v_i\cdot v_j\neq0$.
We consider two subsets $S_1, S_2\subset\ZZ^n$ to be the same if $S_2$ can be obtained by applying an element of $\Aut\ZZ^n$ to $S_1$.
Let $S=\{v_1,\ldots,v_n\}\subset\ZZ^n$ be a subset. 
We call each element $v_i\in S$ a \textit{vector} or \textit{vertex} depending on context and we call the string of integers $(a_1,\ldots,a_n)$ defined by $a_i=-v_i\cdot v_i$ the \textit{string associated} to $S$.
Two vectors $z,w\in S$ are called \textit{linked} if there exists $e\in \ZZ^n$ such that $e\cdot e=-1$ and $z\cdot e, w\cdot e\neq0$. A subset $S$ is called \textit{irreducible} if for every pair of vectors $v,w\in S$, there exists a finite sequence of vectors $v_1=v,v_2,\ldots, v_k=w\in S$ such that $v_i$ and $v_{i+1}$ are linked for all $1\le i\le k-1$; otherwise $S$ is called \emph{reducible}.


\begin{definition} A subset $S=\{v_1,\ldots,v_n\}\subset\mathbb{Z}^n$ is:
	\begin{itemize}
		\item \textit{good} if it is irreducible and $v_i\cdot v_j=
		\begin{cases}
		-a_i\le-2 & \text{if } i=j \\
		0 \text{ or } 1 & \text{if } |i-j|=1 \\
		0 & \text{otherwise;}
		\end{cases}
		$
		\vspace{.2cm}
		\item \textit{standard} if $v_i\cdot v_j= 
		\begin{cases}
		-a_i\le-2 & \text{if } i=j \\
		1 & \text{if } |i-j|=1 \\
		0 & \text{otherwise.}
		\end{cases}
		$
\end{itemize}
\end{definition}

Note that it follows from the definitions that standard subsets are good. If $S$ is a good subset, then a vector $v\in S$ is called: \textit{isolated} if $v\cdot w=0$ for all $w\in S\setminus\{v\}$; \textit{final} if there exists exactly one vector $w\in S\setminus\{v\}$ such that $v\cdot w=1$; and \textit{internal} otherwise. A \textit{component} of a good subset $G$ is a subset of $G$ corresponding to a connected component of the intersection graph of $G$.

	
\begin{definition} A subset $S=\{v_1,\ldots,v_n\}\subset\mathbb{Z}^n$ is:
\begin{itemize}
\item  \textit{negative cyclic} if either 
\begin{itemize}
      \item[(1)] $n=2$ and $v_i\cdot v_j=\begin{cases}
      -a_i\le-2 & \text{if } i=j \\
      0 & \text{if } i\neq j \\
      \end{cases}$ \hspace{.5cm} or \vspace{.2cm}
      \item[(2)] $n\ge3$ and there is a cyclic reordering of $S$ such that\\ $v_i\cdot v_j=\begin{cases}
      -a_i\le-2 & \text{if } i=j \\
      1 & \text{if } |i-j|=1 \\
      -1 & \text{if } i\neq j\in\{1,n\}\\
      0 & \text{otherwise;}
      \end{cases}$  \vspace{.2cm}
\end{itemize}
\item  \textit{positive cyclic} if $-a_k\le -3$ for some $k$ and either 
\begin{itemize}
      \item[(1)] $n=2$ and $v_i\cdot v_j=\begin{cases}
      -a_i\le-2 & \text{if } i=j \\
      2 & \text{if } i\neq j \\
      \end{cases}$ \hspace{.5cm} or \vspace{.2cm}
      \item[(2)] $n\ge3$ and there is a cyclic reordering of $S$ such that\\ $v_i\cdot v_j=\begin{cases}
      -a_i\le-2 & \text{if } i=j \\
      1 & \text{if } |i-j|=1 \\
      1 & \text{if } i\neq j\in\{1,n\}\\
      0 & \text{otherwise;}
      \end{cases}$   \vspace{.2cm}
\end{itemize}
\item \textit{cyclic} if $S$ is negative or positive cyclic. 
\end{itemize}
Finally, the indices of vertices are understood to be defined modulo $n$ (e.g., $v_{n+1}=v_1$).
\end{definition}

\begin{rem}
    \label{rem:definite}
    Note that if $S = \{ v_1, \dots, v_n \} \subset \ZZ^n $ is standard or cyclic, then the matrix $A = ((v_i \cdot v_j)_{ij})$ is weakly chained diagonally dominant, hence non-singular~\cite{WCDD}. If $S$ is good, then $A$ is block-diagonal with each block being weakly chained diagonally dominant, hence also non-singular. Thus, good, standard and cyclic subsets are comprised of linearly independent vectors.
\end{rem}

\begin{remark}
Consider an irreducible subset $S=\{v_1,\ldots,v_n\}\subset\ZZ^n$, where
\begin{center} $v_i\cdot v_j=\begin{cases}
      -a_i\le-2 & \text{if } i=j \\
      \pm 1 & \text{if } |i-j|=1 \\
      0,\pm 1 & \text{if } i\neq j\in\{1,n\}\\
      0 & \text{otherwise.}
      \end{cases}$
\end{center}
If $v_1\cdot v_n=\pm1$, then $S$ can be modified to be either positive or negative cyclic by changing the signs of select vertices. Similarly, if $v_1\cdot v_n=0$, then $S$ can be modified into a good subset by changing the signs of select vertices (c.f. Remark 5.4 in~\cite{simone2020classification}).
\label{rem:negate}
\end{remark}

If $S=\{v_1,\ldots,v_n\}$ is negative cyclic, then by negating $v_1,\ldots, v_k$ and cyclically reordering, we obtain the negative cyclic subset $S'=\{v_{k+1},\ldots,v_n,-v_1,\ldots,-v_k\}$ (c.f.~Remark 5.4 in~\cite{simone2020classification}). For simplicity, we say that $S'$ is a \textit{cyclic reordering} of $S$; the cyclic reordering of a positive cyclic subset is an honest cyclic reordering in which no vertices are negated. Two cyclic subsets $S$ and $S'$ are considered to be equivalent if $S'$ can be obtained from $S$ by cyclically reordering $S$, reversing $S$, and/or applying an element of $\Aut \ZZ^n$ to $S$.

Following \cite{lisca-3braids}, if $S=\{v_1,\ldots,v_n\}$ is cyclic, then we call $W=\sum_{i=1}^n v_i$ the \textit{Wu element} of $S$.

\begin{definition} Let $S=\{v_1,\ldots,v_n\}\subset\ZZ^n$ be a subset with $v_i\cdot v_i=-a_i$. We define the following notation: 
\begin{itemize}
    \item $I(S):=\displaystyle\sum_{i=1}^n(a_i -3)$;
    \item $E^S_i:=\{j \mid v_j\cdot e_i\neq0\}$;
    \item $ V^S_{i}:=\{j \mid v_i\cdot e_j\neq 0\}$;
    \item $p_i(S):=\#\{j \mid |E_j^S|=i\}$;
    \item if $G$ is good, then $c(G):=$ the number of connected components of $G$.
\end{itemize}
We will often drop the superscript $S$ from the above notation if the subset being considered is understood. \end{definition}


We now review some facts about standard and good subsets due to Lisca in~\cite{lisca-rational} and some facts about cyclic subsets due to the second author in~\cite{simone2020classification}. The first result follows from the proof of Lemma 2.5 in~\cite{lisca-rational}.

\begin{lem}
Let $S=\{v_1,\ldots,v_n\}\subset\ZZ^n$ and $K=\{(\alpha,\beta)\,|\, v_\alpha\cdot e_\beta\neq0\}$. Then 
$$2p_1(S)+p_2(S)+I(S)=\sum_{j=4}^n (j-3)p_j(S) +\sum_{(\alpha,\beta)\in K} ((v_\alpha\cdot e_\beta)^2-1).$$
\label{lem:count}
\end{lem}

\begin{proof}
First note that $I(S)=\sum_{i=1}^n(a_i-3)$ implies that $\sum_{i=1}^n a_i=3n+I(S)$.
Let $v_i=\sum_{j=1}^nm_{ij}e_j$ for $1\le i \le n$. Notice that $v_i\cdot e_j=-m_{ij}$ and $\sum_{i,j=1}^n m_{ij}^2=\sum_{i=1}^na_i$. Let $K=\{(i,j)\,|\, m_{ij}\neq0\}$. Following the proof of Lemma 2.5 in~\cite{lisca-rational}, we see that $|K|=\sum_{j=1}^n jp_j(S)$ and
\begin{align*}|K|&=\sum_{(i,j)\in K} m_{ij}^2-(m_{ij}^2-1)\\
&=\sum_{i,j} m_{ij}^2-\sum_{(i,j)\in K}(m_{ij}^2-1)\\
&=\sum_{i=1}^na_i-\sum_{(i,j)\in K}(m_{ij}^2-1)\\
&=3n+I(S)-\sum_{(i,j)\in K}(m_{ij}^2-1).
\end{align*}
Now since $n=\sum_{j=1}^np_j(S)$, the result follows.
\end{proof}


\begin{definition} Given a vector $v\in\ZZ^n$, define
$$\pi_{{i_1},\ldots,{i_k}}(v):=v+(v\cdot e_{i_1})e_{i_1}+\cdots+(v\cdot e_{i_k})e_{i_k}.$$
We call $v'=\pi_{{i_1},\ldots,{i_k}}(v)$ the \textit{$\{i_1,\ldots,i_k\}$-projection} of $v$, dropping the prefix when the set $\{ i_1, \dots, i_k \}$ is understood.
\end{definition}

\begin{lem}[Lemma 3.2 in \cite{lisca-rational}] Suppose $G$ is a good subset with $p_1(G)>0$ and let $i$ and $s$ be integers with $E_i=\{s\}$. Then $v_s$ is interna and $V_s=\{i,j\}$ for some integer $j$. Moreover, $E_j=\{s-1,s,s+1\}$ and $|v_{s-1}\cdot e_j|=|v_s\cdot e_j|=|v_{s+1}\cdot e_j|=1$.
\label{lem:lisca3.2}
\end{lem}

\begin{lem} Let $G\subset\ZZ^n$ be a good subset with $n$ components such that $|v_\alpha\cdot e_\beta|\le1$ for all integers $\alpha$ and $\beta$. Then $p_1(G)=0$. In addition, if $n\ge3$, then $p_2(G)=0$ and $I(G)\ge0$.
\label{lem:ncomps}
\end{lem}

\begin{proof}
By assumption, $v_s\cdot v_t=0$ for all $s\neq t$. 
By Lemma \ref{lem:lisca3.2}, if $E_i=\{s\}$, then $v_s$ is an internal vertex. Since each vertex of $G$ is isolated, we necessarily have $p_1(G)=0$.

Suppose $n\ge3$ and $p_2(G)\neq0$. Let $(i,s,t)$ be a triple of integers such that $E_i=\{s,t\}$. Since $v_s\cdot v_t=0$, there exists $j\in V_s\cap V_t$. If $a_s,a_t\ge3$, then
$G'=(S\setminus\{v_s,v_t\})\cup\{\pi_{i}(v_s),\pi_{i}(v_t)\}$ is a good subset;
hence $G'$ is made of $n$ linearly independent vectors in $\ZZ^{n-1}$, which is impossible. We may assume $a_s=2$ and so $V_s=\{i,j\}$. Since $v_s\cdot v_u=0$ for all $u$, we necessarily have that $E_j=\{s,t\}$. Moreover, since $G$ is irreducible and $n\ge3$, we have $|V_t|\ge3$. Now the subset
$G'=(S\setminus\{v_s,v_t\})\cup\{\pi_{i}(v_t)\}$
is good with $p_1(G')=1$, which once again contradicts Lemma \ref{lem:lisca3.2}.
\end{proof}

\subsection{Contractions and Bad Components}

The main operation we shall carry out on subsets is called \emph{contraction}. It amounts to passing from a good (resp., standard, cyclic) subset $S \subset \ZZ^n$ to a good (resp., standard, cyclic) subset $S' \subset \ZZ^{n-1}$, enabling one to use induction arguments.

\begin{definition}
Suppose $S=\{v_1,\ldots,v_n\} \subset \ZZ^n$, $n\ge 3$, is a good or cyclic subset and suppose there exists a triple of integers $(i,s,t)$ such that $E_i=\{s,s+1,t\}$, where $v_s\cdot v_{s+1}=1$, $v_s\cdot e_i=-v_{s+1}\cdot e_i$, $|v_u\cdot e_i|=1$ for $u\in\{s,s+1,t\}$, and $a_t\ge 3$. Then the subset $S'\subset \mathbb{Z}^{n-1}=\langle e_1,\ldots,e_{i-1},e_{i+1},\ldots, e_n\rangle$ defined by
$$S'=(S\setminus\{v_s,v_{s+1},v_t\})\cup\{v_s+v_{s+1},\pi_{i}(v_t)\}$$
is called a \textit{contraction} of $S$ along $e_i$ and we say that $S$ can be contracted. Furthermore, we call $S$ an \textit{expansion} of $S'$.
\end{definition}

\begin{remark}
If $S$ is negative cyclic and there exists an integer $i$ such that $E_i=\{1,t,n\}$, where $t \neq 1, n$, $v_1\cdot e_i=v_{n}\cdot e_i$, $|v_u\cdot e_i|=1$ for $u\in\{1,t,n\}$, and $a_t\ge 3$, then we define the contraction of $S$ along $e_i$ to be $S'=(S\setminus\{v_1,v_{t},v_n\})\cup\{v_1-v_{n},\pi_{i}(v_t)\}$. Alternatively, one could cyclically reorder $S$ so that $E_i=\{s,s+1,u\}$, where $s\neq n$, and apply the original definition of contraction.
\end{remark}

Notice that $S$ is good (resp., standard, cyclic) if and only if $S'$ is good (resp., standard, cyclic) and $I(S)=I(S')$. It is also often true that $p_j(S)=p_j(S')$ for all $j\neq 3$. The following lemma shows one such situation in which this is true. It will be needed for the proof of Theorem \ref{thm:1}.

\begin{lem} Let $S'=\{v'_1,\ldots v'_{n}\}\subset\ZZ^n$ be a good or cyclic subset. If $|v'_\alpha\cdot e_\beta|\le 1$ for all $1\le \alpha,\beta\le n$, then any expansion $S$ of $S'$ satisfies $p_i(S)=p_i(S')$ for all $i\neq3$.
\label{lem:noexpansion}
\end{lem}

\begin{proof}
Let $S=\{v_1,\ldots,v_{n+1}\} \subset \ZZ^{n+1}$ be an expansion of $S'$. Then there exists an integer $i$ such that $S$ can be contracted along $e_i$ to obtain $S'$. Thus, up to cyclic reordering, there exist integers $s$ and $t$ such that $E^{S}_i=\{s,s+1,t\}$, $v_s\cdot e_i=-v_{s+1}\cdot e_i=\pm1$, $|v_t\cdot e_i|=1$, and $a_t\ge3$. 
Up to the action of $\Aut\ZZ^{n+1}$, we may assume that $i=n+1$.
By definition, $S'=(S\setminus\{v_s,v_{s+1},v_t\})\cup\{v_s+v_{s+1},\pi_{n+1}(v_t)\}$.

Note that by construction $p_i(S)\neq p_i(S')$ for some $i\neq 3$ if and only if there exists an integer $j\in (V^S_s\cap V^S_{s+1})\setminus\{n+1\}$.
Thus we need only prove that $V^{S}_{s}\cap V^S_{s+1}=\{n+1\}$. Let $v_s\cdot e_j=x_j$ and $v_{s+1}\cdot e_j=y_j$ for all $1\le j\le n$. Then $|(v_s+v_{s+1})\cdot e_j|=|x_j+y_j|\le 1$ for all $j$; consequently, $y_j\in\{-x_j,-x_j-1,-x_j+1\}$ for all $j$. Let $R_1=\{j\,|\, y_j=-x_j\}$ and $R_2=\{j\,|\, y_j=-x_j\pm1\}$. Now, $1=v_s\cdot v_{s+1}=1+\sum_{j=1}^nx_jy_j=1+\sum_{j\in R_1}x_j^2+\sum_{j\in R_2}(x_j^2\pm x_j)$; hence $0=\sum_{i=1}^nx_j^2 + \sum_{j\in R_2}\pm x_j$. This equality holds only if $x_j=0$ for all $j\in R_1$ and for each $j\in R_2$, either $x_j=0$ or $y_j=0$. It follows that $V^{S}_{s}\cap V^S_{s+1}=\{n+1\}$. 
\end{proof}


If $G$ is a good subset, there is a notion of a \textit{bad component}, which was defined by Lisca in~\cite{lisca-rational}. First we must recall Lisca's notion of a \textit{$(-2)$-final contraction}, namely a contraction of a good subset in which $v_s$ is a final vertex with $a_s=2$ and there exists an integer $j$ satisfying $E_j=\{s,t\}$ (c.f.~Definition 3.4 in~\cite{lisca-rational}). The reverse operation is called a \textit{$(-2)$-final expansion}.

\begin{definition}
\label{def:badcomponent}
Let $G'=\{v_1,\ldots,v_n\}\subset\ZZ^n$, $n\ge3$ be a good subset, and suppose there exists $1<s<n$ such that $C'=\{v_{s-1},v_s,v_{s+1}\}\subset G'$ is a connected component of $G'$ with $v_{s\pm1}\cdot v_{s\pm1}=-2$, $v_s\cdot v_s<-2$ and $E^{G'}_i=\{s-1,s,s+1\}$ for some $i$. Let $G\subset \ZZ^m$, $m\ge n$, be a good subset obtained from $G'$ by a sequence of $(-2)$-final
expansions applied to $G'$. Then the connected component $C\subseteq G$ is called a \textit{bad component} of $G$. We denote the number of bad components of $G$ by $b(G)$.
\label{def:badcomp}
\end{definition}

\begin{remark}
The notions of a $(-2)$-final contraction and, consequently, a bad component are meaningful without requiring that $G$ be a good subset: it is sufficient that the intersection graph of $C$ be a path each of whose edges is labelled with $1$.
\end{remark}

The main takeaway from Definition~\ref{def:badcomponent} that we will often use is the following. If $C=\{v_1,\ldots,v_k\}\subseteq G=\{v_1,\ldots,v_k,v_{k+1},\ldots,v_n\}$ is a bad component, then: 
\begin{itemize}
    \item there is a unique vertex $v_s\in C$ such that for all $1\le t\neq s\le k$, $V_t\cap V_x=\varnothing$ for all $x\not\in\{1,\ldots,k\}$;
    \item $v_s$ is internal and $V_s=\{i,j_1,\ldots,j_m\}$, where $j_\alpha\not\in V_{t}$ for all $1\le t\neq s\le k$ and $E_i^G=E_i^C=\{s-1,s,s+1\}$;
    \item $|v_t\cdot e_l|\le1$ for all integers $l$ and $t\neq s$;
    \item if $l\in V_t$ for some $1\le t\neq s\le k$, then $|E_l^C|=|E_l^G|\in\{2,3\}$;
    \item there is a unique integer $m$ such that $|E_m^C|=2$, and $E_m^C=\{v_1,v_k\}$; and 
    \item $|V_{i}\cap V_{{i+1}}|=1$ for all $1\le i\le k-1$ and $|V_{i}\cap V_{j}|\in\{0,2\}$ whenever $i\neq j$ and $|i-j|\neq1$.
\end{itemize}
In light of these facts, we define the following:

\begin{definition}
Suppose $G$ is a good subset with a bad component $C$. Using the notation above, the \textit{core of $C$} is defined to be the subset $\{\pi_{i}(v_s)\}$.
\end{definition}

Notice that if $G'$ is the subset obtained from a good subset $G$ by replacing $C$ with its core, then $G'$ is good and $I(G')=I(G)+1$. Moreover, if we replace every bad component of a good subset $G$ with its core, we obtain a good subset $G'$ without bad components that contains at least $b(G)$ isolated vertices such that $I(G')=I(G)+b(G)$.


\begin{lem} Suppose $T$ is a good subset with $I(T)\le 4$ and $p_1(T)\ge1$. Let $i$ and $s$ be integers such that $E_i=\{s\}$. Then $T$ is standard and either: $I(T)=0$ and $|v_s\cdot e_i|=2$; or $I(T)=-3$ and $|v_s\cdot e_i|=1$.
\label{lem:p1contraction}
\end{lem}

\begin{proof}
By Lemma 2.4 and Corollary 3.5 in~\cite{lisca-rational}, if $I(T)<0$, then $T$ is standard with $I(T)=-3$ and $|v_s\cdot e_i|=1$. Suppose $I(T)\ge0$. Following as in the proof of Corollary 3.5 in~\cite{lisca-rational}, we can contract $T$ until it is a length 3 subset. It is easy to check that up to the action of $\Aut \ZZ^3$, the only length 3 standard subset $T'$ with $p_1(T')>0$ and $0\le I(T')\le 4$ is $\{e_1-e_2,-e_1+2e_3,e_1+e_2\}$, which is standard and has $I(T')=0$. It follows that $I(T)=0$ and $|v_s\cdot e_i|=2$.
\end{proof}

The final results of this section follow from the lattice analysis conducted in~\cite{lisca-rational} and~\cite{simone2020classification}. They will be used extensively in Section \ref{sec:nocontractions}.

\begin{definition}
Let $S$ be a good of cyclic subset. We call a pair of vertices $(v_x,v_{x+1})$ of $S$ an \textit{adjacency}. An adjacency is called \textit{contractible} if there exists an integer $k$ 
such that $k\in V_x\cap V_{x+1}$ and $S$ can be contracted along $e_k$; otherwise, an adjacency is called \emph{rigid}.
\end{definition}


\begin{lem} \hfill
\begin{enumerate}
    \item If $S$ is good with $p_1(S)>0$ and $n\ge 4$, then $I(S)\ge-3$ and $S$ has at least $n-3$ contractible adjacencies.
    \item If $S$ is cyclic with $p_1(S)>0$ and $n\ge 5$, then $I(S)\ge-4$ and $S$ has at least $n-4$ contractible adjacencies.
    \item If $S$ is good with no bad components or standard with $p_1(S)=0$, $I(S)<0$, and $n\ge 6$, then $I(S)\ge-2$ and $S$ has at least $n-5$ contractible adjacencies.
    \item If $S$ is cyclic with $p_1(S)=0$, $I(S)\le0$, and $n\ge9$, then $I(S)\ge-3$ and $S$ has at least $n-8$ contractible adjacencies.
\end{enumerate}
\label{lem:hascontractions}
\end{lem}

\begin{proof}
(1) follows from Corollary 3.5 in~\cite{lisca-rational}. (2) follows from the proofs of Lemmas~6.2 and~6.3 in~\cite{simone2020classification} in the case of cyclic subsets. (3) follows by examining the good subsets found in~\cite{lisca-rational} (see also Proposition 5.10 in~\cite{simone2020classification}). (4) follows by examining the cyclic subsets found in~\cite{simone2020classification}.
\end{proof}


Given a cyclic subset $S=\{v_1,\ldots,v_n\}\subset\ZZ^n$ with $I(S)=4$ and $n\ge22$, we will often remove some vertices and replace others with projections to form a new subset $S'\subset\ZZ^{n'}$, where $n'<n$, that is either good, standard, or cyclic, and has contractible adjacencies by Lemma~\ref{lem:hascontractions}. We will then argue that $S$ also has contractible adjacencies. The following example shows how this argument works; it will be referred to throughout Section \ref{sec:nocontractions}.

\begin{example} Let $S=\{v_1,\ldots,v_n\}\subset\ZZ^n$ be a negative cyclic subset with $I(S)=4$ and $n\ge22$. Suppose there exist integers $i,j,s,t$ such that $E_i=E_j=\{s,t\}$, $|v_s\cdot e_i|=|v_s\cdot e_j|=1$ and $a_s\ge3$. Let $$S'=(S\setminus\{v_s,v_t\})\cup\{\pi_{i}(v_s)\}\subset\ZZ^{n-1}.$$ 
Then $S'$ is a standard subset with $p_1(S')=1$. 
Relabel the vertices of $S'$ by $v_\alpha'=v_\alpha$ for all $\alpha\neq s,t$ and $v_s'=\pi_i(v_s)$.
By Lemma~\ref{lem:hascontractions}, since $n-1\ge 21$, $S'$ has at least $(n-1)-3\ge18$ contractible adjacencies. Let $(v'_x,v'_{x+1})$ be a contractible adjacency of $S'$ and assume up to cyclic reordering that $v_x\cdot v_{x+1}=1$; then there is an integer $k$ such that $E_k=\{x,x+1,y\}$, where $a_y\ge3$ and $v'_x\cdot e_k=-v'_{x+1}\cdot e_k$. Now $(v_x,v_{x+1})$ is not a contractible adjacency of $S'$ only if $v_t\cdot e_k\neq0$, i.e., $y=t$. By Lemma~\ref{lem:hascontractions}, $I(S')\ge-3$ and so $a_t\le 9$. It follows that at most 8 of the contactible adjacencies of $S'$ do not give rise to contactible adjacencies of $S'$; hence $S$ has at least $n-12\ge10$ contractible adjacencies. Thus $S$ has contractions.
\label{ex:hascontractions}
\end{example}








\section{Cubiquity and Contractions}
\label{sec:cubiquitycontractions}

In the following, we will say that a negative cyclic subset is cubiquitous to mean that it generates a cubiquitous lattice. Recall that a unit cube $C$ in $\ZZ^n$ is of the form $C=x+\{0,1\}^n$, where $x\in\ZZ^n$. We define the \textit{solid unit cube} $\overline{C}$ in $\RR^n$ to be $C=x+[0,1]^n$.

Given two vectors $x,y\in\RR^n$, we define $d(x,y)$ to be the Euclidean distance between $x$ and $y$. Moreover, $||x||$ denotes the length of $x$. The goal of this section is to prove the following. 

\begin{thm} Let $S=\{v_1,\ldots,v_n\} \subset \ZZ^n$ be a negative cyclic subset and let $S'$ be a subset obtained from $S$ via contraction. Let $W'$ be the Wu element of $S'$ and let $\Lambda'$ denote the lattice generated by $S'$. 
Suppose $S'$ is not cubiquitous and there exists a unit cube $C'\subset\ZZ^{n-1}$ such that:
\begin{itemize}
\item $C'\cap \Lambda'=\varnothing$;
\item $\frac{1}{2}W'\in \overline{C'}$; and 
\item $d(x,\frac{1}{2}W')\le\frac{1}{2}\sqrt{(n-1)+I(S')+11}$ for all $x\in C'$.
\end{itemize}
Then $S$ is also not cubiquitous and there exists a cube $C \subset \ZZ^n$ such that $C \cap \Lambda = \varnothing$, $\frac{1}{2}W \in \overline{C}$, and $d(x, \frac{1}{2} W) \leq \frac{1}{2}\sqrt{n+I(S)+11}$.
\label{thm:cubandcontr}
\end{thm}

Let $S=\{v_1,\ldots,v_n\}\subset \ZZ^n$ be a negative cyclic subset and let $\Lambda=\langle v_1,\ldots,v_n\rangle$ be the lattice generated by $S$. Assume $S$ admits a contraction. By Remark \ref{rem:negate} and the discussion following it, up to cyclic reordering and the action of $\Aut\ZZ^n$, we may assume that there exists an integer $t\not\in\{n-1,n\}$ such that $E^S_n=\{t,n-1,n\}$, where $v_{n-1}\cdot e_n=-v_{n}\cdot e_n$, $|v_u\cdot e_n|=1$ for all $u\in\{1,t,n\}$, and $a_t\ge 3$. 

Let $v_t':=\pi_{n}(v_t)$, $v'_{n-1}:=v_{n-1}+v_n$, $v_n'=\pi_{n}(v_n)$, and $v_i':=v_i$ for all $i\not\in\{t,n-1,n\}$.
Then $S'=\{v_1',\ldots,v_{n-1}'\}\subset\ZZ^{n-1}$ is the negative cyclic subset obtained by contracting $S$ along $e_n$. Recall that $I(S')=I(S)$. Let $\Lambda'=\langle v_1',\ldots,v_{n-1}'\rangle$ denote the lattice generated by $S'$.
Let $W=\sum_{i=1}^nv_i=\sum_{i=1}^n\lambda_i e_i$ and $W'=\sum_{i=1}^{n-1}v_i'=\sum_{i=1}^{n-1}\lambda_i' e_i$ be the Wu elements of $S$ and $S'$, respectively. Let $D=\text{span}_{[0,1]}\{v_1,\ldots,v_n\}\subset\RR^n$ and $D'=\text{span}_{[0,1]}\{v'_1,\ldots,v'_{n-1}\}\subset\RR^{n-1}$ denote the fundamental parallelotopes of $\Lambda$ and $\Lambda'$, respectively. Let $z=\frac{1}{2}W$ and $z'=\frac{1}{2}W'$ denote the centroids of $D$ and $D'$, respectively. 

Assume $\Lambda'$ is not cubiquitous and let $C'\subset \ZZ^{n-1}$ be a unit cube satisfying $C'\cap \Lambda'=\varnothing$ and $z'\in \overline{C'}$.
Up to the action of $\Aut\ZZ^n$, we may assume that $v_t\cdot e_n=1$. Moreover, up to cyclic reordering and reversal, we may assume that $v_{n-1}\cdot e_n=1$ and $v_n\cdot e_n=-1$. Define $C:=C'\times\{0,1\}\subset \ZZ^n$. Our goal is to show that, under the hypotheses of Theorem~\ref{thm:cubandcontr}, $C\cap\Lambda=\varnothing$, which implies that $\Lambda$ is also not cubiquitous.


We first define an intermediate subset. 
Let $\overline{S}=\{v_1,\ldots,v_{n-2},v_{n-1}+v_n\}$ and 
$\overline{\Lambda}$ be the lattice generated by $\overline{S}$.
Notice that 
$E_n^{\overline{S}}=\{t\}$ and
$S'$ is obtained from $\overline{S}$ by projecting $v_t$ onto the hyperplane $\langle e_1,\ldots,e_{n-1}\rangle=\ZZ^{n-1}$ (i.e., by replacing $v_t$ with $\pi_{n}(v_t)$).


\begin{lem}
$C\cap \overline{\Lambda}=\varnothing$.
\label{lem:Clambdabar}
\end{lem}

\begin{proof}
Let $x\in \overline{\Lambda}$. Then $$x=x_1v_1+\cdots+x_{n-2}v_{n-2}+x_{n-1}(v_{n-1}+v_n)=\sum_{i=1}^{n-1}x_iv_i'-x_te_n.$$
Notice that $\sum_{i=1}^{n-1}x_iv_i'\in\Lambda'$. If $x\in C$, then $x_t\in\{-1,0\}$ and $\sum_{i=1}^{n-1}x_iv_i'\in C'$, which contradicts the assumption that $C'\cap \Lambda'=\varnothing$. Hence $x\not\in C$.
\end{proof}

\begin{lem}
If $d(x,z)\le \frac{\sqrt{n+I(S)+11}}{2}$ for all $x\in C$, then $S$ is not cubiquitous.
\label{lem:DCempty}
\end{lem}

\begin{proof}
Recall that $v_t\cdot e_n=1$, $v_{n-1}\cdot e_n=1$ and $v_n\cdot e_n=-1$. 

First observe that since $z$ is the centroid of $D$, for all $w\in \Lambda$, there exists $y\in \Lambda\cap D$ such that $d(y,z)\le d(w,z)$. 
We will show that $d(y,z)> \frac{\sqrt{n+I(S)+11}}{2}$ for all $y\in \Lambda\cap D$, which will imply that $y\not\in C$. Consequently, $\Lambda\cap C=\varnothing$ and so $S$ is not cubiquitous.

Let $y=\sum_{i=1}^ny_iv_i\in \Lambda \cap D$ (i.e., $y_i\in\{0,1\}$ for all $i$). If $y_{n-1}=y_n$, then $y=\sum_{i=1}^{n-2}y_iv_i+y_{n-1}(v_{n-1}+v_n)\in\overline{\Lambda}$, implying that $x\not\in C$, by Lemma~\ref{lem:Clambdabar}. Now assume that $y_{n-1}\neq y_n$. If $y\cdot e_n = y_t+y_{n-1}-y_n\not\in\{0,1\}$, then it is clear that $y\not\in C=C'\times\{0,1\}$. Thus we also assume that $y_t+y_{n-1}-y_n\in\{0,1\}$. Since $y_t,y_{n-1},y_n\in\{0,1\}$ and $y_{n-1}\neq y_n$, there are two cases to consider:
\begin{enumerate}
    \item[1.] $y_t=1$, $y_{n-1}=0$, $y_n=1$ (i.e., $y\cdot e_n=0$); and
    \item[2.] $y_t=0$, $y_{n-1}=1$, $y_n=0$ (i.e., $y\cdot e_n=1$).
\end{enumerate}
\vspace{\baselineskip}
\underline{Case 1}.
Note that since $y\cdot e_n=0$, if $y\in C$, then $y\in C'\times\{0\}$.
Moreover, we have that $y=\sum_{i=1}^{n-1}y_iv_i'+v_n'\in \ZZ^{n-1}=\langle e_1, \dots, e_{n-1} \rangle$. Also note that $y_i-\frac{1}{2}=\pm\frac{1}{2}$ for all $i$ and $v_{n-1}'\cdot v_{n}'=(v_{n-1}+v_n)\cdot \pi_{n}(v_n)=v_n\cdot v_n+1=v_n'\cdot v_n'\le-1$. Define $\langle v,w\rangle$ to be the usual positive-definite dot product. First assume $t\neq 1$ and note that $v_t'\cdot v_n'=-1$. Then
\begin{align*}
    d(y,z')^2&=\Big|\Big|\sum_{i=1}^{n-1}\Big(y_i-\frac{1}{2}\Big)v'_i+v_n'\Big|\Big|^2\\
    &=\sum_{i=1}^{n-1}\Big(y_i-\frac{1}{2}\Big)^2\langle v'_i,v_i'\rangle+\sum_{i=1}^{n-2}2\Big(y_i-\frac{1}{2}\Big)\Big(y_{i+1}-\frac{1}{2}\Big)\langle v'_i, v'_{i+1}\rangle\\
    &\hspace{.5cm}+2\Big(y_{n-1}-\frac{1}{2}\Big)\Big(y_{1}-\frac{1}{2}\Big)\langle v'_{n-1}, v'_{1}\rangle+
    \langle v_n',v_n'\rangle\\
    &\hspace{.5cm}+2\Big(y_1-\frac{1}{2}\Big)\langle v'_1, v'_{n}\rangle+2\Big(y_t-\frac{1}{2}\Big)\langle v'_t,v'_{n}\rangle+2\Big(y_{n-1}-\frac{1}{2}\Big)\langle v'_{n-1}, v'_{n}\rangle\\
    &=\frac{3(n-1)+I(S')}{4}+\sum_{i=1}^{n-2}2\Big(y_i-\frac{1}{2}\Big)\Big(y_{i+1}-\frac{1}{2}\Big)(-1)\\
    &\hspace{.5cm}-\Big(y_{1}-\frac{1}{2}\Big)(1)+\langle v_n',v_n'\rangle\\
    &\hspace{.5cm}+2\Big(y_1-\frac{1}{2}\Big)(1)+1-\langle v_{n}',v_n'\rangle\\
    &=\frac{3(n-1)+I(S')}{4}+\sum_{i=1}^{n-2}2\Big(y_i-\frac{1}{2}\Big)\Big(y_{i+1}-\frac{1}{2}\Big)(-1)+\Big(y_{1}-\frac{1}{2}\Big)(1)+1\\
    &\ge \frac{n+I(S)+11}{4}.
\end{align*}
If $t=1$, then since $v_1\cdot e_n=1$, $v_n\cdot e_n=-1$, and $v_1\cdot v_n=-1$, we have that $v_1'\cdot v_n'=-2$. Now by a similar calculation as above, we see that $d(y,z')^2\ge\frac{n+I(S')+11}{4}$. 
Notice that $W=W'-(v_t\cdot e_n)e_n$; hence $d(z,z')=||\frac{1}{2}W-\frac{1}{2}W'||=\frac{1}{2}||e_n||=\frac{1}{2}$.
By construction, the vector $y-z'$ is orthogonal to the vector $z-z'$. Thus $d(y,z)^2=d(y,z')^2+d(z,z')^2\ge \frac{n+I(S)+11}{4}+\frac{1}{4}=\frac{n+I(S)+12}{4}$.
Now since $d(x,z)\le \frac{\sqrt{n+I(S)+11}}{2}$ for all $x\in C$, we see that $y\not\in C$.\\

\underline{Case 2}.
Note that since $y\cdot e_n=1$, if $y\in C$, then $y\in C'\times\{1\}$.
Moreover, we have that $y=\sum_{i=1}^{n-1}y_iv'_i-v_n'+e_n\in\langle e_1, \dots, e_{n-1}\rangle\times\{1\}$. 
By a calculation similar to the previous case, we see that $d(y,z)\ge\frac{\sqrt{n+I(S)+11}}{2}$, implying that $y\not\in C$.\end{proof}


\begin{lem} If $d(x',z')\le\frac{\sqrt{(n-1)+I(S')+11}}{2}$ for all $x'\in C'$, then $d(x,z)\le\frac{\sqrt{n+I(S)+11}}{2}$ for all $x\in C$.
\label{lem:centroiddistance}
\end{lem}

\begin{proof}
Recall from the proof of Lemma~\ref{lem:DCempty} that $d(z,z')=\frac{1}{2}$. Also notice that any vector in $C=C'\times\{0,1\}$ is of the form $(x',i)$, where $x'\in C'$ and $i\in \{0,1\}$. Let $x'\in C'$ and consider $x_0=(x',0)\in C'\times\{0\}\subset C$. Since the vector $x-z'$ is orthogonal to the vector $z-z'$, we have $d(x,z)^2=d(x,z')^2+d(z,z')^2\le \frac{(n-1)+I(S')+11}{4}+\frac{1}{4}=\frac{n+I(S)+11}{4}$; hence $d(x,z)\le\frac{\sqrt{n+I(S)+11}}{2}$.
Next let $x_1=(x',1)\in C'\times\{1\}\subset C$. It is easy to see that $d(x_1,z)=d(x_0,z)\le\frac{\sqrt{n+I(S)+11}}{2}$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:cubandcontr}]
Let $S$ and $S'$ be as in the statement of the theorem. By Lemma~\ref{lem:centroiddistance}, $d(x,z)\le\frac{\sqrt{n+I(S)+11}}{2}$ for all $x\in C$ and so by Lemma~\ref{lem:DCempty}, $S$ is not cubiquitous.
\end{proof}









\section{Subsets without Contractions}\label{sec:nocontractions}
The goal of this section is to prove the following. 

\begin{restatable}{thm}{hascontractions}
If $S$ is a negative cyclic subset with $I(S)=4$ that does not have contractions, then the length of $S$ is at most 21.
\label{thm:nocontraction}\end{restatable}

\begin{remark}
Theorem \ref{thm:nocontraction} also holds for cyclic subsets with $I(S)\le 4$; moreover, the smaller the value of $I(S)$, the smaller the possible length of $S$ becomes. In particular, all of the forthcoming results can be shown to hold for subsets with $I(S)\le 4$, but the proofs would need to be suitably adapted.
\end{remark}

Throughout the following, we will assume that $n\ge22$ and that $S$ has no contractions. We will also assume that $S$ is negative cyclic with $I(S)=4$, although many of the results do not require this assumption.

\subsection{Preliminary Lemmas}

\begin{lem} $p_1(S)=0$.
\label{lem:p1}
\end{lem}

\begin{proof} Suppose $p_1(S)>0$ and let $i,s$ be integers such that $E_i=\{s\}$. If $|V_s|\ge3$, then the subset $S'=(S\setminus\{v_s\})\cup\{\pi_i(v_s)\}\subset\ZZ^{n-1}$ is negative cyclic; consequently $S'$ consists of $n$ linearly independent vectors in $\ZZ^{n-1}$, which is a contradiction. Let $V_s=\{i,j\}$. Up to cyclic reordering, we may assume that $v_s\cdot v_{s\pm1}=1$; hence $j\in V_{s\pm1}$ and $v_{s-1}\cdot e_j=v_{s+1}\cdot e_j=-v_s\cdot e_j=\pm1$. Since $|E_i|=1$, we necessarily have that $E_j=\{s-1,s,s+1\}$. If $a_{s+1}=a_{s-1}=2$, then since $v_{s+1}\cdot v_{s-1}=0$, we have $V_{s-1}=V_{s+1}=\{j,k\}$ for some integer $k$. Since $|v_{s-2}\cdot v_{s-1}|=1$, we have that $k\in V_{s-2}$. But then $|v_{s-2}\cdot v_{s+1}|=1$, implying that $n=4$, which is a contradiction. Therefore $a_{s+1}\ge3$ or $a_{s-1}\ge3$. Consequently, $S$ can be contracted along $e_j$, which is a contradiction.
\end{proof}


\begin{lem} If $E_i=\{s,t\}$, then $|V_s|\ge 3$ or $|V_t|\ge 3$.
\label{lem:1}
\end{lem}

\begin{proof}
Assume $|V_s|=|V_t|=2$ and set $V_s=\{i,j\}$ and $V_t=\{i,k\}$.
First suppose $v_s\cdot v_t=0$. Then $j=k$ and $j\in V_{s\pm1},V_{t\pm1}$. Since $n>5$, up to cyclic reordering, we may assume $v_{t+1}\cdot v_s=0$. Then $i\in V_{t+1}$, implying that $|E_i|\ge 3$, which is a contradiction.
Next suppose $|v_s\cdot v_t|=1$; up to cyclic reordering, we may assume $t=s+1$. If $j=k$, then $j\in V_{s-1},V_{s+2}$ and we obtain a similar contradiction as above. If $j\neq k$, then $V_s\cap V_{s+1}=\{i\}$, $V_{s-1}\cap V_s=\{j\}$, and $V_{s+1}\cap V_{s+2}=\{k\}$, implying that $a_s=a_t=2$. The last two paragraphs in the proof of Proposition 7.7 in~\cite{simone2020classification} show that this is not possible.
\end{proof}

\begin{lem}
If $E_i=\{s,t\}$, then $|v_s\cdot e_i|,|v_t\cdot e_i|\le 2$. 
\label{lem:2}
\end{lem}

\begin{proof}
By Lemma~\ref{lem:1}, either $|V_s|\ge 3$ or $|V_t|\ge3$. First assume $|V_s|\ge 3$. Then the subset $T=(S\setminus\{v_s,v_t\})\cup\{\pi_{i}(v_s)\}$ is standard and so by Lemma~\ref{lem:hascontractions}, $I(T)\ge -3$. Thus $-3\le I(T)= I(S)-(a_t-3)-|v_s\cdot e_i|^2\le 4-(|v_t\cdot e_i|^2-2)-|v_s\cdot e_i|^2=6-|v_t\cdot e_i|^2-|v_s\cdot e_i|^2$, implying that $|v_s\cdot e_i|\le 2$ and $|v_t\cdot e_i|\le 2$. An analogous argument holds if $|V_t|\geq3$.
\end{proof}

\begin{lem} If $E_i=\{s,t\}$ and $|v_t\cdot e_i|=2$, then $|v_s\cdot e_i|=1$.
\label{lem:3}
\end{lem}

\begin{proof}
By Lemma~\ref{lem:2}, $|v_s\cdot e_i|\le2$. Assume $|v_s\cdot e_i|=2$. By Lemma~\ref{lem:1}, we may assume $|V_s|\ge 3$; hence $a_s\ge 6$ and $a_t\ge5$. Now the subset $T=(S\setminus\{v_s,v_t\})\cup\{\pi_{i}(v_s)\}\subset\ZZ^{n-1}$ is standard with $I(T)=I(S)-(a_t-3)-(v_s\cdot e_i)^2\le-2$. By Lemma~\ref{lem:hascontractions}, $I(T)\ge-3$ and so $a_{t}\le6$; hence $T$ has at least $n-6$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ has at least $n-8\ge14$ contractible adjacencies, which is a contradiction. 
\end{proof}

\begin{lem} If $E_i=\{s,s+1\},$ then $|V_s|,|V_{s+1}|\ge 3$. 
\label{lem:4}
\end{lem}

\begin{proof}
Up to cyclic reordering, we may assume that $v_{s-1}\cdot v_s=v_s\cdot v_{s+1}=1$. By Lemma~\ref{lem:1}, we may assume that $|V_{s+1}|\ge3$. Assume $|V_s|=2$ and let $V_s=\{i,j\}$. By Lemma~\ref{lem:2}, $|v_s\cdot e_i|\le 2$. Since $i\not\in V_{s-1}$, we have $j\in V_{s-1}$ and $v_s\cdot e_j=-v_{s-1}\cdot e_j=\pm1$. 

If $|E_j|=2$, then the subset 
$$T=(S\setminus\{v_{s},v_{s+1}\})\cup \{\pi_{i}(v_{s+1})\}$$
is standard with final vertex $v_{s-1}$ and $E^T_j=\{s-1\}$. But by Lemma \ref{lem:lisca3.2}, no such subset exists. Thus $|E_j|\ge3$. Since $E_i=\{s,s+1\}$ and $V_s=\{i,j\}$, it follows that $|E_j|=3$ and $E_j=\{s-1,s,s+1\}$. If $|v_{s+1}\cdot e_j|=1$, then we can contract, which is a contradiction. Thus we may assume that $|v_{s+1}\cdot e_j|=x\ge2$. 

Let $m\ge1$ be the smallest integer such that $a_{s-m}\ge3$. Suppose $m\ge2$. Then $V_{s-\alpha}=\{j_{\alpha-1},j_\alpha\}$ for $1\le \alpha\le m-1$, where $j_0:=j$. Since $|v_{s-\alpha}\cdot e_\alpha|=|v_{s-\alpha}\cdot e_{\alpha-1}|=1$, for $\alpha\geq1$, it follows that $|v_{s+1}\cdot e_{j_\alpha}|=x$ and  $E_{j_\alpha}=\{s-\alpha-1,s-\alpha,s+1\}$ for all $0\le \alpha\le m-2$. If $v_{s-m}=v_{s+1}$, then it follows that $|v_{s+1}\cdot e_{j_{m-1}}|=x\pm1$. In this case, $n=m+1$ and $I(S)=\sum_{i=s-m-1}^s(a_i-3)+(a_{s+1}-3)=-m+(m-1)x^2+(x\pm1)^2+1=(x^2-1)m+1\pm2x$. Since $n\ge22$, it follows that $I(S)>4$, which is a contradiction. Thus we may assume that $v_{s-m}\neq v_{s+1}$ and so $|v_{s+1}\cdot e_{j_\alpha}|=x$ for all $0\le \alpha\le m-2$ and $E_{j_{m-1}}=\{s-m-1,s-m,s+1\}$. Moreover note that $|V_{s-m}\cap V_{s+1}|\ge2$ and so $a_{s+1}\ge mx^2+2$.
Thus the subset $$T=(S\setminus\{v_{s-m},\ldots,v_{s+1}\})\cup\{\pi_{{j_{m-1}}}(v_{s-m})\}\subset\ZZ^{n-(m+1)}$$
is standard with $I(T)\le 4-(x^2-1)m<0$. By Lemma~\ref{lem:hascontractions}, $I(T)\ge-3$ and so $m\le 2$ and $a_{s+1}\le10$; hence $T$ has at least $n-8$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ has at least $n-13\ge9$ contractible adjacencies, which is a contradiction. 

We may now assume that $m=1$, so that $a_{s-1}\ge3$. 
First suppose $a_{s-1}=3$ and set $V_{s-1}=\{j,k,l\}$. Since $v_{s-1}\cdot v_{s+1}=0$, it follows that $k,l\in V_{s+1}$ and so $a_{s+1}\ge7$. 
Thus the subset $T$ defined above is still standard and $ I(T)\le 4-(a_{s+1}-3)\le0$. As above, if $I(T)<0$, then by Lemma~\ref{lem:hascontractions}, $I(T)\ge-3$ and so $T$ has at least $n-7$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ has at least $n-13\ge9$ contractible adjacencies, which is a contradiction. Thus we may assume that $I(T)=0$ and $a_{s+1}=7$. Hence $V_{s+1}=\{i,j,k,l\}$ and $|v_{s+1}\cdot e_k|=|v_{s+1}\cdot e_l|=1$. Since $|v_{s+1}\cdot v_{s+2}|=1$ we necessarily have that $|v_{s+2}\cdot e_k|=|v_{s+2}\cdot e_l|\pm1$. But then $v_{s+2}\cdot v_{s-1}\neq0$, which is a contradiction.

Next suppose $a_{s-1}\ge4$. If $a_{s+1}\ge7$, then the subset 
$$T=(S\setminus\{v_{s-1},v_s,v_{s+1}\})\cup\{\pi_{i,j}(v_{s+1})\}\subset\ZZ^{n-2}$$
is standard with $I(T)<0$. By Lemma~\ref{lem:hascontractions}, $T$ has at least $n-7$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ has contractible adjacencies, which is a contradiction. 
Thus $a_{s+1}=6$, $V_{s+1}=\{i,j,k\}$ for some integer $k$, and $|v_{s+1}\cdot e_j|=2$ and $|v_{s+1}\cdot e_k|=1$. Since $v_{s-1}\cdot v_{s+1}=0$, it follows that $|v_{s-1}\cdot e_k|=2$. Finally, we necessarily have that $k\in V_{s+2}$, $E_k=\{s-1,s+1,s+2\}$, and $|V_{s-1}|\ge3.$ Hence the subset 
$$C=(S\setminus\{v_{s-1},v_s,v_{s+2}\})\cup\{\pi_k(v_{s-1}+v_s),\pi_j(v_{s+1}+v_{s+1})\}\subset\ZZ^{n-2}$$
is cyclic with $I(C)=-2$. By Lemma~\ref{lem:hascontractions}, $C$ has at least $n-10$ contractible adjacencies. Following as in Example \ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction.
\end{proof}

\begin{lem} If $E_i=\{s,s+1\},$ then $a_{s},a_{s+1}\ge 3$, and we may assume $a_s=4$ and $a_{s+1}=3$.
\label{lem:5}
\end{lem}

\begin{proof}
Cyclically reorder $S$ so that $s=1$ and $s+1=2$. By Lemma~\ref{lem:4}, we may assume that $|V_1|,|V_2|\ge3$.
Let $R=(S\setminus\{v_1,v_2\})\cup\{\pi_{i}(v_1),\pi_{i}(v_2)\}$ and set $v_\alpha'=\pi_{i}(v_\alpha)$ for $\alpha=1,2$. Then $v_1'\cdot v_{2}'\in\{0,2\}$ if $|v_1\cdot e_i|=|v_2\cdot e_i|=1$ and $v_1'\cdot v_{2}'\in\{-1,3\}$ if $|v_1\cdot e_i|=2$ or $|v_2\cdot e_i|=2$. If $v_1'\cdot v_{2}'=0$ (resp., $v_1'\cdot v_{2}'=-1$), then $R$ is a standard (resp., cyclic) subset and thus contains $n$ linearly independent vectors in $\ZZ^{n-1}$, which is not possible. If $v_1'\cdot v_{2}'=3$, then, up to cyclic reordering and reversal, we may assume $v_1\cdot e_i=2$ and $v_2\cdot e_i=1$. It follows that either $a_1\ge 7$ or $a_2\ge 4$. In the former case, the subset 
$T_1=(S\setminus\{v_1,v_2\})\cup\{\pi_{i}(v_2)\}$ is standard with $I(T_1)\le -1$ and in the latter case, the subset $T_2=(S\setminus\{v_1,v_2\})\cup\{\pi_{i}(v_1)\}$ is standard with $I(T_2)\le -1$. In either case, by Lemma~\ref{lem:hascontractions}, $T$ at least $n-6$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction. Thus we may assume that $v_1'\cdot v_2'=2$, which implies that $V_1\cap V_2\neq\{i\}$. If $a_1=a_2=3$, then $V_1=V_2=\{i,j,k\}$, $v_1\cdot e_i=v_2\cdot e_i$, $v_1\cdot e_j=-v_2\cdot e_j$, and $v_1\cdot e_k=-v_2\cdot e_k$. In this case, it is easy to check that either $v_3\cdot v_1\neq0$ or $|v_3\cdot v_2|\neq 1$, neither of which are possible. Thus we may assume, up to cyclic reordering and reversal, that $a_1\ge 4$. Let $B$ be the intersection matrix for $R$. Since $R$ contains $n$ vectors in $\ZZ^{n-1}$, we necessarily have $\det B=0$. Thus by Proposition~\ref{prop:negdef2}(\ref{negdef2a}), we must have $a_2=3$ and by Proposition~\ref{prop:negdef2}(\ref{negdef2b}), we must have $a_1\le 5$. Let $V_2=\{i,j,k\}$. Since $v_1'\cdot v_2'=2$, it follows that $V_1\supset\{i,j,k\}$.

Suppose $a_1=5$ and let $V_1=\{i,j,k,l,m\}$. By Proposition~\ref{prop:negdef2}(\ref{negdef2c}), we must have that $a_3=a_4=2$, since overwise $\det B\neq0$. Since $v_2\cdot v_3=1$, up to the action of $\Aut\ZZ^n$, we may assume $j\in V_3$ and $k\not\in V_3$. Since $v_1\cdot v_3=0$, up to the action of $\Aut\ZZ^n$, we may assume that $V_3=\{j,m\}$. Since $|v_3\cdot v_4|=1$ and $a_3=2$, there are two cases: either $j\in V_4$ and $m\not\in V_4$; or $m\in V_4$ and $j\not\in V_4$. If $j\in V_4$, then since $v_2\cdot v_4=0$, $k\in V_4.$ Since $a_4=2$, either $j\in V_5$ or $k\in V_5$ (but not both). But then $v_2\cdot v_5\neq0$, which is a contradiction. A similar argument provides a contradiction if $m\in V_5$. Thus $a_1\neq 5$ and so $a_1=4$.
\end{proof}

\begin{lem} If $E_i=\{s,t\}$, then $v_s\cdot v_t=0$.
\label{lem:s,s+1}
\end{lem}

\begin{proof} 
Assume $|v_s\cdot v_t|=1$.
Cyclically reorder $S$ so that $s=1$ and $t=2$. By Lemma~\ref{lem:5}, $a_1=4$ and $a_2=3$. Moreover, by the proof of Lemma~\ref{lem:5}, $V_1=\{i,j,k,l\}$, $V_2=\{i,j,k\}$,
and $(v_2\cdot e_i,v_2\cdot e_j,v_2\cdot e_k)=(v_1\cdot e_i,-v_1\cdot e_j,-v_1\cdot e_k)$. Now either $j\in V_3$ or $k\in V_3$; up to the action of $\Aut\ZZ^n$, we may assume $j\in V_3$. Since $v_1\cdot v_3=0$, $l\in V_3$. Moreover, since $v_1\cdot v_n=-1$ and $v_3\cdot v_n=0$, we necessarily have that $l\in V_n$ and $v_n\cdot e_l=v_1\cdot e_l$. We claim that $E_l=\{n,1,3\}$. Suppose there exists an integer $u\not\in\{n,1,2,3\}$ such that $l\in V_u$. Then since $v_u\cdot v_1=0$, either $j\in V_u$ or $k\in V_u$. Assume $j\in V_u$; then since $v_u\cdot v_2=0$, we necessarily have that $k\in V_u$ and $(v_u\cdot e_j,v_u\cdot e_k)=\pm(v_2\cdot e_j,-v_2\cdot e_k)=\mp(v_1\cdot e_j,-v_1\cdot e_k)$, implying that $v_1\cdot v_u\neq0$, which is a contradiction. If $k\in V_u$, then we obtain a similar contradiction. Thus $E_l=\{n,1,3\}$. Since there do not exist contractions, $a_3=2$ and $V_3=\{j,l\}$. Since $v_3\cdot v_n=0$, we have that $j\in V_n$; since $v_2\cdot v_n=0$, we have $k\in V_n$. Moreover, since $v_3\cdot v_4=1$ and $v_2\cdot v_4=0$, we have $j\in V_4$ and $k\in V_4$. Since $E_l=\{n,1,3\}$ and $V_3=\{j,l\}$, it follows that $E_j=\{n,1,2,3,4\}$ which implies $E_k=\{n,1,2,4\}$. In fact, since $(v_4\cdot e_j,v_4\cdot e_k)=\pm (v_n\cdot e_j, v_n\cdot e_k)$ and $v_4\cdot v_n=0$, we have that $|V_4\cap V_n|\ge3$, and so $a_4\ge 3$ and $a_n\ge4$.


The subset
\[
    C=(S\setminus\{v_1,v_2,v_3,v_4,v_n\})\cup\{\pi_{j}(v_4),\pi_{j,l}(v_n)\}
\]
is negative cyclic with $I(C)=1$. Notice that $\pi_{j}(v_4)$ and $\pi_{j,l}(v_n)$ are adjacent vertices of $C$, and $E^C_k=\{n,4\}$. It is easy to see that Lemmas \ref{lem:4} and \ref{lem:5} apply to negative cyclic subsets with $I(C)\le 4$. By Lemma~\ref{lem:4}, $|V_n^C|,|V_4^C|\ge3$ and so by Lemma~\ref{lem:5}, $(\pi_{j,l}(v_n))^2=-3$ and $(\pi_{j}(v_4))^2=-4$ or vice versa. Hence $(a_4,a_n)\in\{(5,5),(4,6)\}$. If $a_4=a_n=5$, then $T=(S\setminus\{v_1,v_2,v_3,v_4,v_n\})\cup\{\pi_{j,k,l}(v_n)\}$ is a standard subset with $I(T)=-1$; by Lemma~\ref{lem:hascontractions}, $T$ at least $n-9$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction. If $a_4=4, a_n=6$, then $T=(S\setminus\{v_1,v_2,v_3,v_4,v_n\})\cup\{\pi_{j,k}(v_4)\}$ is a standard subset with $I(T)=-1$. By $T$ at least $n-6$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction. 
\end{proof}

\begin{lem} If $E_i=\{s,t\}$ and $|V_s|,|V_t|\ge3$, then $|v_s\cdot e_i|=|v_t\cdot e_i|=1$.
\label{lem:6}
\end{lem}

\begin{proof} 
By Lemma~\ref{lem:s,s+1}, $v_s\cdot v_t=0$.
Suppose either $|v_s\cdot e_i|\ge 2$ or $|v_t\cdot e_i|\ge 2$.
By Lemmas~\ref{lem:2} and~\ref{lem:3}, we may assume $|v_s\cdot e_i|=2$ and $|v_t\cdot e_i|=1$. 
The subset $T=(S\setminus\{v_s,v_t\})\cup\{\pi_{i}(v_s)\}$ is standard with $I(T)=4-(a_t-3)-4=3-a_t$. If $a_t\ge 4$, then $I(T)<0$. By Lemma~\ref{lem:hascontractions}, $T$ has at least $n-6$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction. Since $|V_t|\ge3$, we have that $a_t=3$. By an analogous argument, it is easy to see that $a_s\le 6$ and so $|V_s|=3$. 

Let $V_t=\{i,j,k\}$. Up to the action of $\Aut\ZZ^n$, we may assume $j\in V_s$ since $v_s\cdot v_t=0$. Notice that if $k\not\in V_s$, then $|v_s\cdot e_j|=2$, implying that $a_s\ge 8$, which is a contradiction. Thus $k\in V_s$ and so $a_s=6$ and $|v_s\cdot e_j|=|v_s\cdot e_k|=1$. Up to cyclic reordering we may assume that $v_{t+1}\cdot v_s=0$. Since $V_t=\{i,j,k\}$, either $j\in V_{t+1}$ or $k\in V_{t+1}$. Up to the action of $\Aut\ZZ^n$, we may assume the former. Then since $v_{t+1}\cdot v_s=0$, we have $k\in V_{s+1}$ and $(v_{t+1}\cdot e_j,v_{t+1}\cdot e_k)=\pm(v_{s}\cdot e_j,v_{s}\cdot e_k)$, implying that $|v_t\cdot v_{t+1}|\neq1$, which is a contradiction.
\end{proof}

\begin{lem} If $E_i=\{s,t\}$ and $|V_s|,|V_t|\ge 3$, then we may assume that $a_{s}=3$ and $|V_s\cap V_t|=2$. 
\label{lem:7}
\end{lem}

\begin{proof}
Cyclically reorder $S$ so that $s=1$. By Lemma~\ref{lem:s,s+1}, $v_1\cdot v_t=0$ and so $|V_1\cap V_t|\ge2$.
Let $R=(S\setminus\{v_1,v_t\})\cup\{\pi_{i}(v_1),\pi_{i}(v_t)\}$. By Lemma~\ref{lem:6}, $|v_1'\cdot v_{t}'|=1$, $-(v_1')^2=a_1-1$, and $-(v_t')^2=a_t-1$. By Proposition~\ref{prop:negdef1}(\ref{negdef1a}), $a_1=3$ or $a_t=3$. Up to cyclically reordering, we may assume that $a_1=3$ and let $V_1=\{i,j,k\}$. Assume $V_1\cap V_t=V_1$. Since $v_1\cdot v_t=0$, it follows that either $|v_t\cdot e_i|\ge2$ or $|v_t\cdot e_k|\ge2$. Moreover, by the proof of Lemma~\ref{lem:6}, $|V_t|\ge 4$, and so $a_t\ge 7$. Let $T$ be the standard subset defined in the proof of Lemma~\ref{lem:6}. Then $I(T)<0$ and so by Lemma~\ref{lem:hascontractions}, $T$ at least $n-6$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction. Hence $|V_1\cap V_t|=2$.
\end{proof}


In light of Lemma~\ref{lem:7}, we define the following sets:

\begin{definition}\hfill
\label{def:I,J}
\begin{itemize}
    \item $\mathcal{I}_1=\{i\,|\, E_i=\{s,t\} \text{ for some integers $s$ and $t$, } |V_s|=2, a_t=3\}$;
    \item $\mathcal{I}_2=\{i\,|\, E_i=\{s,t\} \text{ for some integers $s$ and $t$, }|V_s|=2, a_t\ge4\}$;
    \item $\mathcal{J}_1=\{i\,|\, E_i=\{s,t\} \text{ for some integers $s$ and $t$, } a_s=a_t=3\}$;
    \item $\mathcal{J}_2=\{i\,|\, E_i=\{s,t\} \text{ for some integers $s$ and $t$}, a_s=3, a_t\ge4\}$;
    \item $\mathcal{I}=\mathcal{I}_1\cup \mathcal{I}_2$ and  $\mathcal{J}=\mathcal{J}_1\cup \mathcal{J}_2$.
\end{itemize}
\end{definition}

Note that $|\mathcal{I}|+|\mathcal{J}|=p_2(S)$.
For each $i\in\mathcal{I}\cup\mathcal{J}$, we set $E_i=\{s(i),t(i)\}$.
By Lemma~\ref{lem:s,s+1}, for each $i\in \mathcal{I}\cup\mathcal{J}$, we may assume that $v_{s(i)}\cdot v_{t(i)}=0$. Hence there exists an integer $j(i)\in V_{s(i)}\cap V_{t(i)}$. If $i\in\mathcal{I}$, then since $|V_{s(i)}|=2$, we have $V_{s(i)}=V_{s(i)}\cap V_{t(i)}=\{i,j(i)\}$ and $E_{j(i)}=\{s(i)-1,s(i),s(i)+1,t(i)\}$. If $i\in\mathcal{I}_1$, then we set $V_{t(i)}=\{i,j(i),k(i)\}$. If $i\in\mathcal{J}$, then by Lemma~\ref{lem:7}, $V_{s(i)}\cap V_{t(i)}=\{i,j(i)\}$. Finally, if $i\in\mathcal{J}$, we set $V_{s(i)}=\{i,j(i),k(i)\}$.

\begin{lem} Let $i\in\mathcal{J}$. If $l\in V_{s(i)}\cup V_{t(i)}$ and $l\neq i$, then $|E_l|\ge 3$.
\label{lem:p1,I}
\end{lem}

\begin{proof}
By Lemma~\ref{lem:1} we have $p_1(S)=0$, hence $|E_l|\ge 2$. Suppose $l\in V_{t(i)}$ and assume $|E_l|=2$. Then  $T=(S\setminus\{v_{s(i)},v_{t(i)}\})\cup\{\pi_{i}(v_{s(i)})\}\subset\ZZ^{n-1}$ is a standard subset with $p_1(T)=1$. By Lemma~\ref{lem:hascontractions}, $T$ has at least $n-4$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ has at least $n-12\ge 10$ contractible adjacencies, which is a contradiction.
Thus $|E_l|\ge3$. An identical argument shows that if $l\in V_{s(i)}$, then $|E_l|\ge3$.
\end{proof}

 

\subsection{Examining $\mathcal{I}$ and $\mathcal{J}$}

\begin{lem}
If $i_1\neq i_2\in\mathcal{I}$, then $j(i_1)\neq j(i_2)$.
\label{lem:10}
\end{lem}

\begin{proof}
Assume $j(i_1)=j(i_2)$ and set $j:=j(i_1)=j(i_2)$. We will show that there exists a contraction. First notice that $E_{j}=\{s(i_\alpha)-1, s(i_\alpha),s(i_\alpha)+1,t(i_\alpha)\}$ for $\alpha=1,2$. Since $i_1\neq i_2$ and $v_{s(i_1)}\cdot v_{s(i_2)}\neq0$, we have that $s(i_1)\neq s(i_2)$ and so $s(i_2)=s(i_1)+1$ or $s(i_2)=s(i_1)-1$. Up to reversing $S$, we may assume $s(i_2)=s(i_1)+1$. Then $s(i_2)+1=s(i_1)+2=t(i_1)$ and $t(i_2)=s(i_1)-1$. Set $s:=s(i_1)$. Then $s-1=t(i_2)$, $s+1=s(i_2)$, and $s+2=t(i_1)$; hence $a_{s-1},a_{s+2}\ge3$. Since we may assume (again, up to cyclically reordering $S$) that $v_s\cdot v_{s+1}=v_{s+1}\cdot v_{s+2}=1$, we have that $v_s\cdot e_j=-v_{s+1}\cdot e_j$ and $v_{s+1}\cdot e_j=-v_{s+2}\cdot e_j$. Consequently, $|v_\alpha\cdot e_{i_1}|,|v_\alpha\cdot e_{i_2}|,|v_\alpha\cdot e_{j}|\le1$ for $\alpha=1,2$. Now since $i,j\in V_{s-1}\cap V_{s+2}$ and $a_{s-1},a_{s+2}\ge3$, we have that $|V_{s-1}|,|V_{s+2}|\ge3$.

Suppose $a_{s-1},a_{s+2}\ge4$. Then
\[
    T=(S\setminus\{v_{s-1},\ldots,v_{s+2}\})\cup\{\pi_{{i_2},j}(v_{s-1}),\pi_{{i_1},j}(v_{s+2})\}
\]
is standard. Hence $T$ has $n-2$ linearly independent vectors in $\ZZ^{n-3}$, which is not possible. Up to reversing $S$, we may assume as before that $a_{s+2}=3$. Let $V_{s+2}=\{i_1,j,k\}$. Since $v_{s-1}\cdot v_{s+2}=0$, $k\in V_{s-1}$ and $|v_{s-1}\cdot e_k|=|v_{s+2}\cdot e_k|=1$. Since $v_{s+2}\cdot v_{s+3}=1$, we have that $k\in V_{s+3}$ and $v_{s+2}\cdot e_k=-v_{s+3}\cdot e_k$. Finally, since $|E_{i_1}|=|E_{i_2}|=2$ and $|E_j|=4$, we must have $|E_k|=3$, and thus we can contract $S$.
\end{proof}

\begin{lem} Let $i_1 \in \mathcal{I}$ and $i_2 \in \mathcal{J}$. Then $j(i_1) \neq j(i_2)$.
\label{lem:pre8}
\end{lem}

\begin{proof} Let $i_1\in\mathcal{I}$ and $i_2 \in \mathcal{J}$.
Assume $j(i_1)=j(i_2)$ and set $j:=j(i_1)=j(i_2)$. Then $E_j=\{s(i_1)-1,s(i_1),s(i_1)+1,t(i_1)\}$. By definition, $s(i_2)\neq s(i_1)$, $t(i_2)\neq s(i_1)$, and by Lemma~\ref{lem:p1,I}, $s(i_2)\neq t(i_1)$ and $t(i_2)\neq t(i_1)$; thus $s(i_2),t(i_2)\in\{s(i_1)-1,s(i_1)+1\}$. Up to reversing $S$, we may assume that $s(i_2)=s(i_1)-1$ and $t(i_2)=s(i_1)+1$. Hence
\[
    C=(S\setminus\{v_{s(i_1)},v_{s(i_1)\pm1},v_{t(i_1)}\})\cup\{\pi_{{i_2}}(v_{s(i_1)\pm1}),\pi_{i_1}(v_{t(i_1)})\}
\]
is a cyclic subset and thus contains $n-1$ linearly independent vectors in $\ZZ^{n-2},$ which is impossible. Thus $j(i_1)\neq j(i_2)$.
\end{proof}

\begin{lem}
Let $i_1 \neq i_2\in\mathcal{J}$. Then $j(i_1)\neq j(i_2)$.
\label{lem:9}
\end{lem}

\begin{proof}
Assume $j(i_1)=j(i_2)$ and set $j:=j(i_1)=j(i_2)$.
Let $s_1:=s(i_1)$, $s_2:=s(i_2)$, $t_1:=t(i_1)$, $t_2:=t(i_2)$, $V_{s_1}=\{i_1,j,k_1\}$, and $V_{s_2}=\{i_2,j,k_2\}$. Then $E_j\supset\{s_1,s_2,t_1,t_2\}$. 

Suppose $v_{s_1}\cdot v_{s_2}=0$. Then $k_1=k_2$; set $k:=k_1=k_2$. Moreover, $(v_{s_1}\cdot e_j,v_{s_1}\cdot e_k)=\pm(v_{s_2}\cdot e_j,-v_{s_2}\cdot e_k)$. Note that this implies if $v_u\cdot v_{s_1}=0$ or $v_u\cdot v_{s_2}=0$ for some $u\neq t_1,t_2$, then $j,k\not\in V_u$. Since $k\not\in V_{t_1}\cup V_{t_2}$ (by Lemma~\ref{lem:7}), we necessarily have that $|v_{t_1}\cdot v_{s_2}|=|v_{t_2}\cdot v_{s_1}|=1$; hence $t_1\in\{s_2-1,s_2+1\}$ and $t_2\in\{s_1-1,s_1+1\}$. Suppose $t_1=s_2-1$ and $t_2=s_1+1$; the other cases are identical. Since $|v_{s_1-1}\cdot v_{s_1}|=1$, either $k\in V_{s_1-1}$ or $j\in V_{s_1-1}$. If $j,k\in V_{s_1-1}$, then $(v_{s_1-1}\cdot e_j,v_{s_1-1}\cdot e_k)=\pm(x(v_{s_1}\cdot e_j),-(x\pm1)(v_{s_1}\cdot e_k))$, where $x,x\pm1>0$; but then $|v_{s_1-1}\cdot v_{s_2}|=2x\pm1\ge3$, which is not possible. Thus $k\in V_{s_1-1}$ and $|v_{s_1-1}\cdot e_k|=1$ or $j\in V_{s_1-1}$ and $|v_{s_1-1}\cdot e_k|=1$ (but not both); consequently, $|v_{s_1-1}\cdot v_{s_2}|=1$ and so $s_2=s_1-2$ and $t_1=s_1-3$. Thus $E_j\subset\{s_1-3,s_1-2,s_1-1,s_1,s_1+1\}$ and $E_k\subset\{s_1-2,s_1-1,s_1\}$. If $k\not\in V_{s_1-1}$, then $E_k=\{s_1-2,s_1\}$, contradicting Lemma~\ref{lem:p1,I}. If $k\in V_{s_1-1}$, then it is clear that $j\not\in V_{s_1-1}$ and we can contract, which is another contradiction. 

Suppose $|v_{s_1}\cdot v_{s_2}|=1$. Up cyclic reordering of $S$, we may assume $v_{s_1}\cdot v_{s_2}=1$ and $s_2=s_1+1$. Then $k_1\neq k_2$. Suppose $t_2=s_1-1$. By Lemma~\ref{lem:s,s+1}, $t_1\neq s_1\pm1$. Hence the subset
\[C=(S\setminus\{v_{s_1-1},v_{s_1},v_{s_1+1},v_{t_1}\})\cup\{\pi_{{i_2}}(v_{s_1-1}),\pi_{{i_2}}(v_{s_1+1}),\pi_{{i_1}}(v_{t_1})\}\subset\ZZ^{n-2}\]
is cyclic (after possibly negating select vertices (c.f.~Remark \ref{rem:negate})) and thus has $n-1$ linearly independent vectors in $\ZZ^{n-2}$, which is impossible. Thus $t_2\neq s_1-1$; similarly, $t_1\neq s_1+2$. Hence, $k_1\in V_{t_2}$ and $k_2\in V_{t_1}$. Either $v_{t_2-1}\cdot v_{s_1}=0$ or $v_{t_2+1}\cdot v_{s_1}=0$; up to reversing $S$, we may assume that $v_{t_2+1}\cdot v_{s_1}=0$. If $a_{t_2}=3$, then it is easy to see that either $v_{t_2+1}\cdot e_j = \pm 1 $ or $ v_{t_2+1}\cdot e_{k_1}=\pm1$ (but not both). But then $v_{t_2+1}\cdot v_{s_1}\neq 0$, which is a contradiction. Thus $a_{t_2}\ge4$.
Now by Proposition~\ref{prop:negdef3}, the subset $$R=(S\setminus\{v_{s_1},v_{s_1+1},v_{t_1},v_{t_2}\})\cup\{\pi_{{i_1}}(v_{t_1}),\pi_{{i_2}}(v_{s_1+1}),\pi_{{i_2}}(v_{t_2})\}\subset\ZZ^{n-2}$$ contains $n-1$ linearly independent vectors in $\ZZ^{n-2}$, which is not possible.\qedhere
\end{proof}

\begin{lem} Let $i\in \mathcal{J}$ and set $s:=s(i)$, $t:=t(i)$, $j:=j(i)$, and $k:=k(i)$. Then $|E_j|,|E_k|\ge3$. 
If $|E_j|=3$, then $|E_k|=4$ and: 
\begin{itemize}
\item $k\neq j(i')$ for all $i'\in \mathcal{I}\cup\mathcal{J}$; and
\item $k\neq k(i')$ for all $i'\in\mathcal{J}$.
\end{itemize}
\label{lem:8}
\end{lem}

\begin{proof}
By Lemma~\ref{lem:p1,I}, $|E_j|,|E_k|\ge3$. Up to cyclic reordering, we may assume that $v_s\cdot v_{s\pm1}=1$.
Let $E_j=\{s,t,u\}$. First suppose that $|v_s\cdot v_u|=1$. Up to reversing $S$, we may assume that $u=s-1$. Since there do not exist contractions, we necessarily have that $v_{s-1}\cdot e_j\neq -v_{s}\cdot e_j$; hence $k\in V_{s-1}$ and so $E_k=\{s-1,s,s+1\}$.
Up to cyclic reordering, we may assume that $v_{s-1}\cdot v_s=1$, and up to the action of $\Aut\ZZ^n$, we may assume that $v_s\cdot e_j=v_s\cdot e_k=1$ and $v_{s-1}\cdot e_j\neq -1$. If $|v_{s-1}\cdot e_k|=1$, then we can contract along $e_k$. Hence we may further assume that either: $v_{s-1}\cdot e_k=m$ and $v_{s-1}\cdot e_j=-(m+1)$ for some $m\ge2$; or $v_{s-1}\cdot e_k=-m$ and $v_{s-1}\cdot e_j=m-1$ for some $m\ge2$. Since $v_{s-1}\cdot v_{s+1}=0$, there exists an integer $l\in V_{s-1}\cap V_{s+1}$; hence $|V_{s-1}|\ge 3$ and so the subset
$$ T=(S\setminus\{v_{s-1},v_s,v_{s+1},v_t\})\cup\{\pi_{k}(v_{s-1}),\pi_{i}(v_t)\}\subset\ZZ^{n-2}$$
is standard. If $I(T)<0$, then by Lemma~\ref{lem:hascontractions}, $T$ also has contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ has at least $n-13>0$ contractible adjacencies, which is a contradiction. Hence $I(T)\ge0$, which implies that $|v_{s-1}\cdot e_k|=2$ and $a_{s+1}=2$. Thus $V^S_{s+1}=\{k,l\}$, $E^S_l=\{s-1,s+1,s+2\}$ and $|v_{s-1}\cdot e_l|=2$. Since $l\in v_{s+2}$ and $v_{s+2}\cdot v_{s-1}=0$, we necessarily have that $a_{s-1}\ge10$. 
Consider the subset
$$
T=
\begin{cases}
(S\setminus\{v_{s-1},\ldots,v_{s+2},v_t\})\cup\{\pi_{k,l}(v_{s-1}),\pi_{i}(v_t)\} & \text{ if } t\neq s+2\\
(S\setminus\{v_{s-1},\ldots,v_{s+2}\})\cup\{\pi_{k,l}(v_{s-1})\} & \text{ if } t=s+2
\end{cases}
.$$
If $t=s+2$, then $T\subset\ZZ^{n-3}$ is standard with final vertex $\pi_{k,l}(v_{s-1})$ satisfying $E_j^T=\{s-1\}$; but no such subset exists by Lemma \ref{lem:lisca3.2}. 
If $t\neq s+2$, then $T$ is standard with $I(T)<0$ and so by Lemma~\ref{lem:hascontractions}, $T$ also has contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ has at least $n-8>0$ contractible adjacencies, which is a contradiction.

Thus we have shown that $u\neq s-1$. Similarly, $u\neq s+1.$ Hence $E_k\supseteq\{s-1,s,s+1,u\}$. Moreover, if $w\in E_k\setminus\{s-1,s,s+1,u\}$, then since $v_s\cdot v_w=0$, we have $j\in V_w$, which implies that $w=t$; but this contradicts Lemma~\ref{lem:7}. Thus $E_k=\{s-1,s,s+1,u\}$.

Suppose $k= j(i')$ for some $i'\in\mathcal{I}$. Then $E_{k}=\{s(i')-1,s(i'),s(i')+1,t(i')\}$. Since $s\neq s(i')$, we have that $s(i')\in\{s-1,s+1,u\}$. If $s(i')=u$, then $j=i'$, which contradicts the assumption that $|E_j|=3$. Thus, up to reversing $S$, we may assume that $s(i')=s+1$. Consequently, $u=s+2$ and $t(i')=s-1$. Since $k\in V_{s-1}\cap V_{s+2}$ and $v_{s+2}\cdot v_{s-1}=0$, there exists $m\in V_{s-1}\cap V_{s+2}$; hence $|V_{s-1}|,|V_{s+2}|\ge3$ and so
$$C=(S\setminus\{v_{s-1},\ldots,v_{s+2},v_t\})\cup\{\pi_{{i'}}(v_{s-1}),\pi_{k}(v_{s+2}),\pi_{{i}}(v_t)\}$$ is a negative cyclic subset with $p_1(C)=1$. 
By Lemma~\ref{lem:hascontractions}, $C$ has at least $n-6$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction.

Now suppose $k=j(i')$ for some $i'\in\mathcal{J}$. Up to reversal of $S$, we may assume that $s-1,u\in\{s(i'),t(i')\}$. First assume $u=s(i')$ and $s-1=t(i')$. Then $j=k(i')$ and $V_{u}=\{i',k,j\}$. Since $|E_{i'}|=2$, $|E_j|=3$, and $|E_k|=4$, and either $j\in V_{u\pm1}$ or $k\in V_{u\pm1}$, we must have that
$u=s+2$ and $t=s+3$. Up to cyclic reordering, we may assume that $v_{s+2}\cdot v_{s+3}=1$ and so $v_{s+2}\cdot e_j=-v_{s+3}\cdot e_j$; thus we can contract along $e_j$, which is not possible. Now assume $s-1=s(i')$ and $u=t(i')$. Set $V_{s-1}=\{i',k,k_1\}$. Then since $k\in V_s\cap V_u$ and $k_1\not\in V_u$ by Lemma~\ref{lem:7}, we necessarily have that $u=s+1$, which is a contradiction. Thus $k\neq j(i')$ for all $i'\in\mathcal{J}$.

Finally suppose $k=k(i')$ for some $i'\in\mathcal{J}$. By Lemma~\ref{lem:9}, $j\neq j(i')$ and so $|v_s\cdot v_{s(i')}|=1$. Up to reversal, we may assume $s(i')=s+1$. Since $k\in V_{s-1}\cap V_{s+1}$ and $v_{s-1}\cdot v_{s+1}=0$, we have that $j(i')\in V_{s-1}$. Now either $k\in V_{s+2}$ or $j(i')\in V_{s+2}$. In either case, since $v_{s-1}\cdot v_{s+2}=0$, there exists an integer $m\in V_{s-1}\cap V_{s+2}$ and so $|V_{s-1}|\ge3$.
The subset
$$G=(S\setminus\{v_s,v_{s+1},v_u,v_t,v_{t(i')}\})\cup\{\pi_{i}(v_t),\pi_{{i'}}(v_{t(i')})\}\subset\ZZ^{n-3}$$
is a good subset with $p_1(G)=1$. Note that $v_{s-1}$ is a final vertex of $G$ and $E_k^G=\{s-1\}$; but by Lemma \ref{lem:lisca3.2}, no such subset exists.
\end{proof}
 

\begin{prop} $p_4(S)\le p_2(S)+4$, and if there exists a pair $(\alpha,\beta)$ such that $|v_\alpha\cdot e_\beta|\ge2$, then $(\alpha,\beta)$ is the only such pair and $|v_\alpha\cdot e_\beta|=2$. 
\label{prop:bounds}
\end{prop}

\begin{proof}
By the discussion following Definition~\ref{def:I,J}, we know that for all $i\in\mathcal{I}$, $|E_{j(i)}|=4$. 
By Lemmas~\ref{lem:10}, \ref{lem:pre8}, and~\ref{lem:9}, for every $i\in\mathcal{I}\cup\mathcal{J}$, $j(i)\neq j(i')$ for all $i'\in\mathcal{I}\cup\mathcal{J}$. 
Hence $p_4(S)\ge |\mathcal{I}|$. By Lemma~\ref{lem:8}, for every $i'\in\mathcal{J}$, $|E_{j(i')}|,|E_{k(i')}|\ge3$ and if $|E_{j(i')}|=3$, then $|E_{k(i')}|=4$ and $k(i')\not\in\{j(i''),k(i'')\,|\,i''\in\mathcal{I}\cup\mathcal{J},i''\neq i'\}$. 
It follows that $\sum_{j=4}^n(j-3)p_j(S)\ge p_2(S)$. By Lemma~\ref{lem:count}, $$p_2(S)+4=\sum_{j=4}^n(j-3)p_j(S)+\sum_{(\alpha,\beta)\in K} ((v_\alpha\cdot e_\beta)^2-1),$$
where $K = \{ (\alpha,\beta)\mid v_\alpha\cdot e_\beta\neq0 \}$. It is clear that $p_2(S)+4\ge p_4(S)$. Since $\sum_{j=4}^n(j-3)p_j(S)\ge p_2(S)$, we also have that
$p_2(S)+4\ge p_2(S) +\sum_{(\alpha,\beta)\in K} ((v_\alpha\cdot e_\beta)^2-1)$, or $4\ge\sum_{(\alpha,\beta)\in K} ((v_\alpha\cdot e_\beta)^2-1)$. Thus there is at most one pair $(\alpha,\beta)$ such that $|v_\alpha\cdot e_\beta|\ge2$, and if there exists such a pair, then $|v_\alpha\cdot e_\beta|=2$.
\end{proof}

\begin{cor} If $i\in\mathcal{I}$, then $|v_{s(i)}\cdot e_i|=|v_{t(i)}\cdot e_i|=|v_{s(i)}\cdot e_j|=|v_{t(i)}\cdot e_j|=1$.
\label{cor:v_se_i,v_te_i=1}
\end{cor}

\begin{proof}
Set $s:=s(i)$, $t:=t(i)$, and $j:=j(i)$.
By Proposition~\ref{prop:bounds}, $|v_s\cdot e_i|,|v_t\cdot e_i|\le 2$. If $|v_s\cdot e_i|=2$, then by Proposition~\ref{prop:bounds}, $|v_\alpha\cdot e_\beta|\le1$ for all $(\alpha,\beta)\neq(s,i)$. Since $V_s=\{i,j\}$, it follows that $v_s\cdot v_t\neq0$, which contradicts Lemma~\ref{lem:s,s+1}. A similar argument shows that $|v_t\cdot e_i|=|v_s\cdot e_j|=|v_t\cdot e_j|=1$.
\end{proof}


\begin{lem} If $i\in \mathcal{J}$ and $|E_{k(i)}|=3$, then $|E_{j(i)}|=6$.
\label{cor:ej=6}
\end{lem}

\begin{proof}
Set $s:=s(i)$, $t:=t(i)$, $j:=j(i)$, and $k:=k(i)$. 
Let $E_k=\{s,u_1,u_2\}$.\\

\noindent\underline{Case 1: $|v_{u_1}\cdot v_s|=|v_{u_2}\cdot v_s|=0$}.
In this case, $u_1,u_2\not\in\{s-1,s+1\}$. Hence $E_j=\{s-1,s,s+1,t,u_1,u_2\}$ and by Lemma~\ref{lem:7}, $t\neq u_1,u_2$; hence $|E_j|=6$. \\

\noindent\underline{Case 2: $|v_{u_1}\cdot v_s|\neq|v_{u_2}\cdot v_s|$}.
Up to the action of $\Aut\ZZ^n$, we may assume that $|v_{u_1}\cdot v_s|=1$ and $|v_{u_2}\cdot v_s|=0$. Moreover, up to cyclic reordering and reversal of $S$, we may assume that $u_1=s-1$, $u_2\neq s+1$, and $v_{s-1}\cdot v_s=1$.  
First assume $j\not\in V_{s-1}$ so that $v_{s-1}\cdot e_k=-v_{s}\cdot e_k$. If $|V_{u_2}|\ge3$, then we can contract, which is a contradiction. If $V_{u_2}=\{j,k\}$, then since $j\not\in V_{s-1}$, we necessarily have that $u_2=s-2$ and $v_{s-2}\cdot e_k=-v_{s-1}\cdot e_k$; consequently we can contract, which is a contradiction. Hence we may assume that $j\in V_{s-1}$ and so $E_j=\{s-1,s,s+1,u_2,t\}$. By Lemma~\ref{lem:7}, it is clear that $u_2\neq t$ and so $|E_j|=5$. Moreover, since $v_{s-1}\cdot v_{s}=1$, by Proposition~\ref{prop:bounds}, either $|v_{s-1}\cdot e_j|=2$ and $|v_{s-1}\cdot e_k|=1$, or $|v_{s-1}\cdot e_j|=1$ and $|v_{s-1}\cdot e_k|=2$. 
By Proposition~\ref{prop:bounds}, $|v_{u_2}\cdot e_j|=|v_{u_2}\cdot e_k|=1$. Moreover, it is easy to see that either $(v_{s-1}\cdot e_j,v_{s-1}\cdot e_k)=\pm (2v_{u_2}\cdot e_j, v_{u_2}\cdot e_k)$ or $(v_{s-1}\cdot e_j,v_{s-1}\cdot e_k)=\pm (v_{u_2}\cdot e_j, 2v_{u_2}\cdot e_k)$. Since $|v_{u_2}\cdot v_{s-1}|\le1$, it follows from Proposition~\ref{prop:bounds} that $|V_{u_2}\cap V_{s-1}|\ge 4$; hence $|V_{u_2}|,|V_{s-1}|\ge 4$. Consider the subset 
$$R=(S\setminus\{v_{s-1},v_{s},v_t,v_{u_2}\})\cup\{\pi_{k}(v_{s-1}),\pi_{i}(v_t),\pi_{k}(v_{u_2})\}\subset\ZZ^{n-2}.$$
By Proposition~\ref{prop:negdef2}(\ref{negdef2c}), $R$ contains $n-1$ linearly independent vectors in $\ZZ^{n-2}$, which is not possible. \\


\noindent\underline{Case 3: $|v_{u_1}\cdot v_s|=|v_{u_2}\cdot v_s|=1$}.
Up to the action of $\Aut\ZZ^n$, we may assume $u_1=s-1$ and $u_2=s+1$. 
Since $|E_j|\ge3$, by Lemma~\ref{lem:8}, we necessarily have that $j\in V_{s-1}$ or $j\in V_{s+1}$. Up to reversing $S$, we may assume that $j\in V_{s+1}$. Hence by Proposition~\ref{prop:bounds}, either $|v_{s+1}\cdot e_j|=2$ or $|v_{s+1}\cdot e_k|=2$.
Moreover, up to cyclic reordering, we may assume that $v_{s-1}\cdot v_s=v_{s}\cdot v_{s+1}=1$.
If $j\in V_{s-1}$, then either $|v_{s-1}\cdot e_j|=2$ or $|v_{s-1}\cdot e_k|=2$, contradicting Proposition~\ref{prop:bounds}. Thus we may assume that $j\not\in V_{s-1}$ and so $E_j=\{s,s+1,t\}$. Since $v_{s-1}\cdot v_{s+1}=0$, it follows that $|V_{s-1}\cap V_{s+1}|\ge2$ and so $|V_{s+1}|\ge3$. Moreover, since $j\not\in V_{s-1}$, we have that $v_{s-1}\cdot e_k=-v_s\cdot e_k$; consequently, since there do not exist contractions, $|v_{s+1}\cdot e_k|=2$ and $|v_{s+1}\cdot e_j|=1$. 
If $a_{s-1}=2$ and $V_{s-1}=\{k,l\}$, then since $v_{s-1}\cdot v_{s+1}=0$, we have that $|v_{s+1}\cdot e_l|=2$, which contradicts Proposition~\ref{prop:bounds}. Hence $a_{s-1}\ge3$.
Thus the subset 
$$T=(S\setminus\{v_{s-1},v_s,v_{s+1},v_t\})\cup\{\pi_{k}(v_{s-1}),\pi_i(v_t)\}\subset\ZZ^{n-2}$$
is standard with $p_1(T)=1$; by Lemma~\ref{lem:hascontractions}, $T$ has at least $n-5$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction.
\end{proof}

\subsection{Further Examination of $\mathcal{J}$}

\begin{lem}
If $i\in\mathcal{J}_2$, then $|E_{j(i)}|\ge4$.
\label{lem:Ej>3}
\end{lem}

\begin{proof}
Let $i\in\mathcal{J}_2$ and set $j:=j(i)$, $k:=k(i)$, $s:=s(i)$, and $t:=t(i)$. Assume $E_j=\{s,t,u\}$. Up to reversal of $S$, we may assume $u\neq s-1$. First assume $u=s+1$. Since there do not exist contractions, up to cyclic reordering, we may assume that $v_s\cdot v_{s+1}=1$, $v_{s+1}\cdot e_j\neq -v_s\cdot e_j$. Hence $k\in V_{s+1}$ and $E_k=\{s-1,s,s+1\}$. Now by Proposition~\ref{prop:bounds}, either $|v_{s+1}\cdot e_k|=2$ and $|v_{s+1}\cdot e_j|=1$ or $|v_{s+1}\cdot e_k|=1$ and $|v_{s+1}\cdot e_j|=2$. In the latter case, we would be able to contract along $e_k$. Hence $|v_{s+1}\cdot e_k|=2$ and $|v_{s+1}\cdot e_j|=1$. 
Thus $T=(S\setminus\{v_s,v_{s+1},v_t\})\cup \{\pi_{i,j}(v_t)\}\subset \ZZ^{n-2}$ is a standard subset with final vertices $v_{s-1}$ and $v_{s+2}$ and $E_k=\{s-1\}$; but by Lemma \ref{lem:lisca3.2}, no such subset exists. Hence $u\neq s\pm1$ and $E_k=\{s-1,s,s+1,u\}$. Note that since either $v_{s-1}\cdot v_u=0$ or $v_u\cdot v_{s+1}=0$ and $k\in V_u\cap V_{s\pm1}$, it follows that $a_u\ge3$.

Let $x\ge 1$ be the smallest integer such that $a_{s+x}\ge 3$. Since $a_u\ge3$, such an integer $x$ exists and $u\not\in\{s+1,\ldots,s+x-1\}$. Set $V_{s+\alpha}=\{k_{\alpha-1},k_\alpha\}$ for all $1\le\alpha\le x-1$, where $k_0:=k$. Then $E_{k_\alpha}=\{s-1,s+\alpha,s+\alpha+1,u\}$ for all $0\le\alpha\le x-2$. If $u=s+x$, then $E_{k_{x-1}}\subseteq\{s-1,s+x-1,s+x\}$ and the subset
$$T=(S\setminus\{v_{s-1},\ldots,v_{s+x},v_t\})\cup\{\pi_{{k_0},\ldots,{k_{x-2}}}(v_{s-1})),\pi_{i,j}(v_t)\}\subset \ZZ^{n-(x+1)}$$ 
is standard with final vertex $\pi_{{k_0},\ldots,{k_{x-2}}}(v_{s-1})$ and $E^T_{k_{x-1}}=\{s-1\}$; once again by Lemma \ref{lem:lisca3.2}, no such subset exists. If $u\neq s+x$, then $E_{k_{x-1}}=\{s-1,s+x-1,s+x,u\}$ and so the subset $$T=(S\setminus\{v_{s-1},\ldots,v_{s+x},v_t,v_u\})\cup\{\pi_{{k_0},\ldots,{k_{x-2}}}(v_{s-1}), \pi_{{k_{x-1}}}(v_{s+x}),\pi_{i,j}(v_t)\}\subset \ZZ^{n-(x+1)}$$ is standard with final vertices $v_{u\pm 1}$, $I(T)\le 3-x$, and $E_{k_{x-1}}=\{s-1\}$. By Lemma~\ref{lem:p1contraction}, $I(T)=-3$ and so $x\leq6$. By Lemma~\ref{lem:hascontractions}, $T$ has at least $n-10$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction. Thus we have proved that $|E_j|\ge4$.
\end{proof}


\begin{lem}
If $i\in\mathcal{J}_1$, then $|E_{j(i)}|\ge4$. 
\label{lem:at=3}
\end{lem}

\begin{proof}
Let $V_{t(i)}=\{i,j(i),m(i)\}$. 
Set $s:=s(i)$, $t:=t(i)$, $j:=j(i)$, $k:=k(i)$, and $m:=m(i)$.
Assume that $|E_j|=3$ and let $E_j=\{s,t,u\}$.
By Lemma~\ref{lem:8}, $|E_k|=|E_m|=4$. Moreover, by the proof of Lemma~\ref{lem:8}, $u\not\in\{s\pm1,t\pm1\}$, $E_k=\{s-1,s,s+1,u\}$ and $E_m=\{t-1,t,t+1,u\}$.
By Proposition~\ref{prop:negdef1}(\ref{negdef1c}), we may assume that $a_{s+1}=2$ (otherwise, after cyclic reordering, the subset $(S\setminus\{v_s,v_t\})\cup\{\pi_{i}(v_s),\pi_{i}(v_t)\}$ would contain $n$ linearly independent vectors in $\ZZ^{n-1}$). 
Let $x\ge2$ be the smallest integer such that $a_{s+x}\ge3$ and set $V_{s+\alpha}=\{k_{\alpha-1},k_\alpha\}$ for $1\le\alpha\le x-1$, where $k_0:=k$. Then $E_{k_\alpha}=\{s-1,s+\alpha-1,s+\alpha,u\}$ for $1\le\alpha\le x-2$. Also note that $k_{x-1}\in V_{s-1}\cap V_{s+x-1}$ and $(v_{s-1}\cdot e_{k_0},\ldots,v_{s-1}\cdot e_{k_{x-2}})=\pm (v_{u}\cdot e_{k_0},\ldots,v_{u}\cdot e_{k_{x-2}})$. \\

\noindent\underline{Case 1: $u=s+x$}.
Let
$$T=(S\setminus\{v_{s-1},\ldots,v_{s+x},v_t\})\cup\{\pi_{{k_0},\ldots,{k_{x-1}}}(v_{s-1}),\pi_{i}(v_t)\}\subset\ZZ^{n-(x+1)}.$$
Note that $E^S_{k_{x-1}}=\{s-1,s+x-1,s+x\}$ or $E^S_{k_{x-1}}=\{s-1,s+x-1\}$. In the former case, by Proposition~\ref{prop:bounds}, we would have $|v_{s+x}\cdot e_{k_{x-2}}|=1$ and $|v_{s+x}\cdot e_{k_{x-1}}|=2$ and $(v_{s+x}\cdot e_{k_0},\ldots,v_{s+x}\cdot e_{k_{x-1}})=\pm(v_{s-1}\cdot e_{k_0},\ldots,v_{s-1}\cdot e_{k_{x-2}},2v_{s-1}\cdot e_{k_{x-1}})$. Since $v_{s-1}\cdot v_{s+x}=0$, we have $|V^S_{s-1}\cap V^S_{s+x}|\ge 2x+1$ and so $a_{s-1}\ge 2x+1$ and $a_{s+x}\ge 2x+5$. Hence $(\pi_{{k_0},\ldots,{k_{x-1}}}(v_{s-1}))^2\le-(x+1)\le -3$ and so $T$ is standard with $I(T)\le -2x\le -4$. But by \cite{lisca-rational}, no such subset exists.
Next suppose we are in the latter case: $E^S_{k_{x-1}}=\{s-1,s+x-1\}$; then $k_{x-1}\in \mathcal{I}$. Note that by Lemma~\ref{lem:8}, we must have that $x\ge3$ (since otherwise $k=j(k_1)$, where $k_1\in \mathcal{I}$). If $(\pi_{{k_0},\ldots,{k_{x-1}}}(v_{s-1}))^2\le -2$, then $T$ is standard with $E^T_j=\{t\}$ and $I(T)\le 4-2x$. By Lemma~\ref{lem:p1contraction}, $I(T)\ge -3$, implying that $x\le 3$. Hence $T$ has length at least $n-4$. By Lemma~\ref{lem:hascontractions}, $T$ has at least $n-8$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction. Now assume $(\pi_{{k_0},\ldots,{k_{x-1}}}(v_{s-1}))^2\ge -1$, so that $a_{s-1}\le x+1$. Then $a_{s-1}=x+1$ and there exists and integer $l\in V_{s-1}\cap V_{s+x}$ such that $|v_{s-1}\cdot e_l|=1$ and $|v_{s+x}\cdot e_l|=x-1$. By Proposition~\ref{prop:bounds}, $x\le 3$; hence $x=3$, $a_{s-1}=4$, and $V_{s-1}=\{k,k_1,k_2,l\}$. 
By Proposition~\ref{prop:bounds}, $|v_\alpha\cdot v_\beta|\le 1$ for all $(\alpha,\beta)\neq (s+3,l)$. Moreover, it is clear that $E_l=\{s-2,s-1,s+3\}$; since $v_{s-2}\cdot v_{s+3}=0$, there exist integers $l_1,l_2\in (V_{s-2}\cap V_{s+3})\setminus\{l\}$. Hence  $a_{s-2}\ge 3$ and $a_{s+3}\ge 10$ (since $V_{s+3}\supseteq\{j,k,k_1,l,l_1,l_2,m\}$). Thus
\[
    T'=(S\setminus\{v_{s-2},\ldots,v_{s+3},v_t\})\cup\{\pi_{l}(v_{s-2}),\pi_{i}(v_t)\}\subset\ZZ^{n-5}
\]
is a standard subset with $I(T')\le -4$; but no such subset exists by \cite{lisca-rational}.\\

\noindent\underline{Case 2: $u\neq s+x$}.
Since $|v_{s-x-1}\cdot v_{s+x}|=1,$ either $k_{x-2}\in V_{s+x}$ or $k_{x-1}\in V_{s+x}$. If $k_{x-2}\in V_{s+x}$, it follows that $k_\alpha\in V_{s+x}$ for all $0\le \alpha\le x-2$; consequently, $k\in V_{s+x}$. Since $u\neq s+x$, this implies that $t=s+x$, which contradicts Lemma~\ref{lem:7}. Thus we may assume that $k_{x-2}\not\in V_{s+x}$ and so $k_{x-1}\in V_{s+x}$ and $E_{k_{x-1}}=\{s-1,s+x-1,s+x,u\}$. Hence $V_{s-1}\cap V_u\supset\{k_0,\ldots,k_{x-1}\}$ and $V_u\supset\{j,k_0,\ldots,k_{x-1},m\}$. By Proposition~\ref{prop:bounds}, it follows that $|V_u\cap V_{s-1}|\ge 2x-1$ and so $a_{u}\ge 2x$. Consider the subset
$$T=(S\setminus\{v_{s-1},\ldots,v_{s+x},v_t,v_u\})\cup\{\pi_{{k_0},\ldots,{k_{x-1}}}(v_{s-1}),\pi_{{k_{x-1}}}(v_{s+x}),\pi_{i}(v_t)\}\subset\ZZ^{n-(x+1)}.$$
If $(\pi_{{k_0},\ldots,{k_{x-1}}}(v_{s-1}))^2\le-2$, then $T$ is standard with $E^T_j=\{t\}$ and $I(T)\le 1-2x$. By Lemma~\ref{lem:p1contraction}, $I(T)\ge -3$; hence $x=2$ and by Lemma~\ref{lem:hascontractions}, $T$ as at least $n-6$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction. Assume $(\pi_{{k_0},\ldots,{k_{x-1}}}(v_{s-1}))^2\ge-1$. Then $a_{s-1}=x+1$ and $V^S_{s-1}=\{k_0,\ldots,k_{x-1},l\}$. Consequently, $l\in V^S_u$ and either: $|v_u\cdot e_l|=x\pm1$ if $u=s-2$; or $|v_u\cdot e_l|=x$ if $u\neq s-2$. In the first case $E_l^S=\{s-2,s-1,s+x\}$ and, by Proposition~\ref{prop:bounds}, $x\le3$; hence the subset
$$T'=(S\setminus\{v_{s-2},\ldots,v_{s+2}, v_t\})\cup\{\pi_{k,{k_1},l}(v_{s-2}),\pi_{{k_1}}(v_{s+2}),\pi_{i}(v_t)\}\subset\ZZ^{n-(x+1)}$$
is standard with $E^{T'}_l=\{s+2\}$ and final vertex $\pi_{{k_1}}(v_{s+2})$, contradicting Lemma \ref{lem:lisca3.2}.
Suppose we are in the second case. Then $E^S_l=\{s-2,s-1,s+2,u\}$ and by Proposition~\ref{prop:bounds}, $x=2$.
Moreover, since $(v_{u}\cdot e_{k_1},v_{u}\cdot e_l)=\pm (v_{s+2}\cdot e_{k_1},2v_{s+2}\cdot e_l)$ and $|v_{s+2}\cdot v_u|\le1$, by Proposition~\ref{prop:bounds}, $|V_{s+2}\cap V_{u}|\ge4$. Hence $a_{s+2}\ge4$ and $a_u\ge9$. 
First assume $|V^S_{s-2}|\ge3$. Then
the subset
$$G=(S\setminus\{v_{s-2},\ldots,v_{s+2}, v_t,v_u\})\cup\{\pi_{k_1,l}(v_{s-2}),\pi_{{l}}(v_{s+2})\}\subset\ZZ^{n-5}$$
is good (since $m\in V^G_{t-1}\cap V^G_{t+1}$) with two components and $I(G)\le-4$; but by Lemma 6.6 in~\cite{acetogollalarsonlecuona}, no such subset exists.
Next assume $a_{s-2}=2$ and let $V^S_{s-2}=\{l,l_1\}$. Since $|v_{u}\cdot e_l|=2$, it follows that $u=s-3$ and $|v_{s-3}\cdot e_{l_1}|=1$. Moreover, $E_l=\{s-3,s-2, s-1,s+2\}$, $E_{l_1}=\{s-3,s-2,s+2\}$, and $(v_{s-3}\cdot e_{k_1},v_{s-3}\cdot e_l,v_{s-3}\cdot e_{l_1})=\pm (v_{s+2}\cdot e_{k_1},2v_{s+2}\cdot e_l,v_{s+2}\cdot e_{l_1})$. Hence, as above, $a_{s+2}\ge7$ and $a_{s-3}\ge13$. Thus the subset 
$$G'=(S\setminus\{v_{s-3},\ldots,v_{s+2}, v_t\})\cup\{\pi_{{k_1},l,{l_1}}(v_{s+2})\}\subset\ZZ^{n-6}$$
is good (since $m\in V^G_{t-1}\cap V^G_{t+1}$) with two components and $I(G)\le-7$. But, once again, by Lemma 6.6 in~\cite{acetogollalarsonlecuona}, no such subset exists. 

We have thus shown that $|E_j^S|\ge4.$\end{proof}


\begin{lem} Let $i\in\mathcal{J}$. Then at least one of the following holds:
\begin{itemize}
    \item $|E_{j(i)}|\ge5$;
    \item $|E_{k(i)}|\ge5$;
    \item or $|E_{j(i)}|=|E_{k(i)}|=4$, $|v_\alpha\cdot e_\beta|=2$, where $(\alpha,\beta)\in\{(s\pm1,e_j),(s\pm1,e_k)\}$, and $k(i)\not\in\{j(i')\,|\,i'\in\mathcal{I}\}\cup\{j(i'),k(i')\,|\,i'\in\mathcal{J}\}$.
\end{itemize}
\label{lem:j,k}
\end{lem}

\begin{proof}
Let $i\in\mathcal{J}$ and set $j:=j(i)$, $s:=s(i)$, $t:=t(i)$, and $k:=k(i)$. By Lemmas~\ref{lem:Ej>3} and~\ref{lem:at=3}, $|E_j|\ge4$. Moreover, by Lemma~\ref{cor:ej=6}, if $|E_k|=3$, then $|E_j|=6$ and we are done. 

Assume $|E_j|=|E_k|=4$ and set $E_{k}=\{s,u,u_1,u_2\}$. Up to relabelling the vertices, we may assume that $v_{u}\cdot v_{s}=0$ so that $j\in V_{u}$. If $|v_{u_\alpha}\cdot v_{s}|=0$ for $\alpha=1,2$, then $j\in V_{u_\alpha}$ for $\alpha=1,2$, and $j\in V_{s\pm1}$, which implies that $|E_j|=7$, which is a contradiction. Similarly, if $|v_{u_1}\cdot v_{s}|=1$ and $|v_{u_2}\cdot v_{s}|=0$ (or vice versa), then $|E_j|\ge5$, which is again a contradiction. Now assume $|v_{u_\alpha}\cdot v_s|=1$ for $\alpha=1,2$. Up to relabeling, we may assume that $u_1=s-1$ and $u_2=s+1$. Since $|E_{j}|=4$, up to reversal of $S$, we may assume that $E_j=\{s,s+1,t,u\}$ and, by Proposition~\ref{prop:bounds}, either: $|v_{s+1}\cdot e_j|=1$ and $|v_{s+1}\cdot e_k|=2$; or $|v_{s+1}\cdot e_j|=2$ and $|v_{s+1}\cdot e_k|=1$.

Since $E_k=\{s-1,s,s+1,u\}$, it is clear that $k\not\in\{j(i')\,|\,i'\in\mathcal{I}\}$. Suppose $k\in\{j(i'),k(i')\,|\,i'\in\mathcal{J}\}$. Note that $(v_{s+1}\cdot e_j,v_{s+1}\cdot e_k)=\pm(xv_{u}\cdot e_j,yv_{u}\cdot e_k)$, where $x,y\in\{1,2\}$. Hence, since $|v_u\cdot v_{s+1}|\le1$, by Proposition~\ref{prop:bounds}, we have $|V_{s+1}\cap V_u|\ge4$. Hence $a_{s+1}\ge 7$ and $a_u\ge 4$. Consequently, $s(i')=s-1$ and so $a_{s-1}=3$. Let $V_{s-1}=\{i',k,l\}$. If $|v_{s+1}\cdot e_k|=2$, then by Lemma~\ref{lem:7}, either $i'\in V_{s+1}$ and $|v_{s+1}\cdot e_{i'}|=2$ or $l\in V_{s+1}$ and $|v_{s+1}\cdot e_l|=2$; both cases contradict Proposition~\ref{prop:bounds}. Thus we may assume $|v_{s+1}\cdot e_k|=1$ and $|v_{s+1}\cdot e_j|=2$. Thus the subset
$$T=(S\setminus\{v_{s-1},v_s,v_{s+1},v_t,v_u\})\cup\{\pi_k(v_{s-1}),\pi_{j,k}(v_{s+1}),\pi_i(v_t)\}\subset\ZZ^{n-2}$$
is standard with $I(T)\le -4$; but by \cite{lisca-rational}, no such subset exists.
Thus $k\not\in\{j(i'),k(i')\,|\,i'\in\mathcal{J}\}$.
\end{proof}

In light of Lemmas~\ref{lem:Ej>3} and~\ref{lem:at=3}, we make the following definition.

\begin{definition} For $\alpha\ge4$, set
$$p_\alpha^1(S):=\#\{j(i)\,:\,|E_{j(i)}|=\alpha\,\, \text{for some}\,\,i\in\mathcal{I}\cup\mathcal{J}\}$$
$$p_\alpha^2(S):=\#\{l\,:\,|E_{l}|=\alpha,\,\, l\neq j(i)\,\,\text{for all}\,\,i\in\mathcal{I}\cup\mathcal{J}\}$$
\end{definition}

\begin{cor} $\sum_{j=4}^n p_j^1(S)\ge p_2(S)$. Moreover, if $|\mathcal{J}|\neq0$, then either:
\begin{itemize}
    \item $|v_\alpha\cdot e_\beta|=2$ for some integer pair $(\alpha,\beta)$, $p_4(S)=p_2(S)+1$, $p_j(S)=0$ for all $j\ge5$, and $|\mathcal{J}|=1$; or
    \item $|v_\alpha\cdot e_\beta|\leq1$ for all integer pairs $(\alpha,\beta)$, $p_j(S)\ge1$ for some $j\ge5$, $p_j(S)=0$ for all $j\ge7$, and $4\ge p_5^1(S)+2p_6^1(S)+p_4^2(S)+2p_5^2(S)+3p_6^2(S)$.
\end{itemize}
\label{cor:jcor}
\end{cor}

\begin{proof}
By Lemmas~\ref{lem:Ej>3} and~\ref{lem:at=3}, and the discussion preceding Lemma~\ref{lem:p1,I} $|E_{j(i)}|=4$ for all $i\in\mathcal{I}$ and $|E_{j(i)}|\ge4$ for all $i\in\mathcal{J}$. Hence by Lemmas~\ref{lem:10}, \ref{lem:pre8}, and~\ref{lem:9}, $\sum_{j=4}^np_j^1(S)\ge p_2(S)$.

Suppose there exists an integer $i\in\mathcal{J}$ such that $|E_{j(i)}|=|E_{k(i)}|=4$. Then by Lemma~\ref{lem:j,k}, $|v_\alpha\cdot e_\beta|=2$, where $(\alpha,\beta)\in\{(s\pm1,e_j),(s\pm1,e_k)\}$ and $k(i)\not\in\{j(i')\,|\,i'\in\mathcal{I}\}\cup\{j(i'),k(i')\,|\,i'\in\mathcal{J}\}$. It then follows from Lemmas~\ref{lem:count}, \ref{lem:Ej>3}, and~\ref{lem:at=3} that $|\mathcal{J}|=1$; consequently, by Lemma~\ref{lem:count}, $p_4(S)=p_2(S)+1$ and $p_j(S)=0$ for all $j\ge5$. 

Next assume that for all $i\in\mathcal{J}$, either $|E_{j(i)}|\ge5$ or $|E_{k(i)}|\ge5$. It follows that $p_2(S)+4\ge \sum_{j=4}^n(j-3)p_j(S)\ge p_2(S)+2$. 
Consequently, by Lemma~\ref{lem:count}, $p_j(S)=0$ for all $j\ge 7$ and $|v_\alpha\cdot e_\beta|\le1$ for all integers $\alpha$ and $\beta$. Moreover, $p_2(S)+4=\sum_{j=4}^n(j-3)p_j^1(S)+\sum_{j=4}^n(j-3)p_j^2(S)\ge p_2(S)+\sum_{j=4}^n(j-4)p_j^1(S)+\sum_{j=4}^n(j-3)p_j^2(S)$, which implies that $4\ge p_5^1(S)+2p_6^1(S)+p_4^2(S)+2p_5^2(S)+3p_6^2(S)$.\end{proof}

\begin{prop}
$|\mathcal{J}|\le2$.
\label{prop:J<2}
\end{prop}

\begin{proof}
Suppose $|\mathcal{J}|\ge3$ and let $i_1,i_2,i_3\in\mathcal{J}$ be distinct integers. Note that by Corollary~\ref{cor:jcor}, $|v_\alpha\cdot e_\beta|\le1$ for all integer pairs $(\alpha,\beta)$ and $\sum_{j=4}^n p_j(S)\ge p_2(S)$; hence by Lemma~\ref{lem:count}, $4\ge p_5^1(S)+2p_6^1(S)+2p_5^2(S)+3p_6^2(S)$.

We first claim that $|\mathcal{J}_1|\le 1$. Suppose otherwise that $i_1,i_2\in\mathcal{J}_1$. Then $V_{s(i_\alpha)}=\{i_\alpha, j(i_\alpha),k(i_\alpha)\}$ and there exist integers $m(i_\alpha)$ such that $V_{t(i_\alpha)}=\{i_\alpha, j(i_\alpha),m(i_\alpha)\}$ for $\alpha\in\{1,2\}$. Note that Lemma~\ref{lem:j,k} also now applies to $k(i_\alpha)$ as well as $m(i_\alpha)$, for each $\alpha$. Hence, either: $|E_{j(i_\alpha)}|\ge5$; or $|E_{k(i_\alpha)}|\ge5$ and $|E_{m(i_\alpha)}|\ge5$. Moreover, either $|E_{j(i_3)}|\ge5$ or $|E_{k(i_3)}|\ge5$. One can check case-by-case that the inequality $4\ge p_5^1(S)+2p_6^1(S)+2p_5^2(S)+3p_6^2(S)$ is necessarily violated. Thus we may assume $|\mathcal{J}_1|\le 1$.


Let 
$$G=(S\setminus\{v_{s(i_\alpha)},v_{t(i_\alpha)}\,|\,1\le\alpha\le3\})\cup\{\pi_{i_\alpha}(v_{s(i_\alpha)}) \,|\, 1\le\alpha\le3\}\subset\ZZ^{n-3}.$$
Then $I(G)\le -1$ and by following the proof of Proposition~5.3 in~\cite{lisca-rational}, it is easy to see that $G$ is good and has no bad components. By Lemma~\ref{lem:hascontractions}, $G$ has as at least $n-8$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction.
\end{proof}


\subsection{Further Examination of $\mathcal{I}$}

\begin{lem}[Lemma 7.14 in~\cite{simone2020classification}]
If $i\in\mathcal{I}$, then $a_{s(i)-1}\ge 3$ or $a_{s(i)+1}\ge 3$.
\label{lem:11}
\end{lem}

In light of Lemma~\ref{lem:11}, let $m(i)\ge1$ denote 
the smallest integer such that $a_{s(i)\pm m(i)}\ge3$ when $a_{s(i)\mp1}\ge3$. If $m(i)\ge2$, then there exist distinct integers $j(i)_1,\ldots,j(i)_{m(i)-1}\not\in\{i,j(i)\}$ such that $V_{s(i)\pm\alpha}=\{j(i)_{\alpha-1},j(i)_\alpha\}$ for $1\le\alpha\le m-1$, where $j(i)_0:=j(i)$. 
Given an integer $i\in\mathcal{I}$, up to reversal of $S$, we will assume throughout this section that $a_{s(i)-1}\ge 3$ and $a_{s(i)+m(i)}\ge3$.  




\begin{lem}
$s(i)+m(i)\neq t(i)$ for all $i\in\mathcal{I}$.
\label{lem:t=s+m}
\end{lem}

\begin{proof}
Set $s:=s(i)$, $t:=t(i)$, $m:=m(i)$, and $j_\alpha:=j(i)_\alpha$ for all $1\le\alpha\le m-1$. Assume $t=s+m$. Then $m\ge 2$ (since $v_s\cdot v_t=0$). Note that $j_0,j_1,\ldots,j_{m-1}\in V_{s-1}$, $i,j_0,j_1,\ldots,j_{m-2}\in V_{s+m}$, $E_{j_\alpha}=\{s+\alpha,s+\alpha+1,s-1,s+m\}$ for all $0\le \alpha\le m-2$, and either $E_{j_{m-1}}=\{s-1,s+m-1\}$ or $E_{j_{m-1}}=\{s-1,s+m-1,s+m\}$.

By Proposition~\ref{prop:bounds}, if $E_{j_{m-1}}=\{s-1,s+m-1,s+m\}$, then $|v_{s+m}\cdot e_{j_\alpha}|=1$ for all $0\le \alpha\le m-2$ and $|v_{s+m}\cdot e_{j_{m-1}}|=2$. It is easy to see that $(v_{s-1}\cdot e_{j_0},\ldots,v_{s-1}\cdot e_{j_{m-2}},v_{s-1}\cdot e_{j_{m-1}})=\pm(v_{s+m}\cdot e_{j_0},\ldots,v_{s+m}\cdot e_{j_{m-2}},2v_{s+m}\cdot e_{j_{m-1}})$. Since $|v_{s-1}\cdot v_{s+m}|\le1$, we have that $|V_{s-1}\cap V_{s+m}|\ge 2m$; hence $a_{s+m}\ge 2m+4$. But then the subset 
$$T=(S\setminus\{v_{s-1},\ldots,v_{s+m}\})\cup\{\pi_{{j_0},\ldots,{j_{m-1}}}(v_{s-1})\}\subset\ZZ^{n-(m+1)}$$
is standard with $I(T)\le 3-2m\le -1$. By \cite{lisca-rational}, $I(T)\ge-3$, implying that $m\le3$. By Lemma~\ref{lem:hascontractions}, $G$ has as at least $n-9$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction.
Thus we may now assume that $E_{j_{m-1}}=\{s-1,s+m-1\}$. Note that $(v_{s-1}\cdot e_{j_0},\ldots,v_{s-1}\cdot e_{j_{m-2}})=\pm(v_{s+m}\cdot e_{j_0},\ldots,v_{s+m}\cdot e_{j_{m-2}})$.
By Lemma~\ref{lem:10}, $m\ge 3$.
Moreover, we may assume that $v_{s+\alpha}\cdot v_{s+\alpha+1}=1$ for all $-1\le\alpha\le s-m-1$. 

If $|v_{s-1}\cdot v_{s+m}|=1$, then $n=m+2$ and $S$ has associated string $(a_{s+m},a_{s-1},2^{[m]})$. But then since $(v_{s-1}\cdot e_{j_0},\ldots,v_{s-1}\cdot e_{j_{m-2}})=\pm(v_{s+m}\cdot e_{j_0},\ldots,v_{s+m}\cdot e_{j_{m-2}})$ and $m\ge3$, there must exist an integer $k\in V_{s-1}\cap V_{s+m}$ such that $E_k=\{s-1,s+m\}$, 
contradicting Lemma~\ref{lem:s,s+1}. Thus $v_{s-1}\cdot v_{s+m}=0$.

We next claim that $m\ge4$. Suppose $m=3$ and let
$$S'=(S\setminus\{v_{s-1},\ldots,v_{s+3}\})\cup\{\pi_{{j},{j_2}}(v_{s-1}),\pi_{{i},{j}}(v_{s+3})\}\subset\ZZ^{n-3}.$$
Then $S'$ is cyclic with $I(S')=3$ and adjacent vertices $\pi_{{j},{j_2}}(v_{s-1})$ and  $\pi_{{i},{j}}(v_{s+3})$ satisfying $E_{j_1}^{S'}=\{s-1,s+3\}$. By Lemma~\ref{lem:5},
we may assume that $a_{s-1}=5$ and $a_{s+3}=6$. Hence $V_{s-1}=\{j,j_1,j_2,k,k_1\}$ and $V_{s+3}=\{i,j,j_1,k,k_1,m\}$ for some integers $k,k_1$ and $m$. But then it is clear that $E_m=\{s-2,s+3,s+4\}$, $v_{s+3}\cdot e_m=-v_{s+4}\cdot e_m$ (assuming $v_{s+3}\cdot v_{s+4}=1$), and $a_{s-2}\ge3$, and hence $S$ has a contraction. Thus we have that $m\ge4$.

Next notice that $p_4(S)\ge p_2(S)+m-2\ge p_2(S)+2$. Thus by Proposition~\ref{prop:bounds}, $|v_\alpha\cdot e_\beta|\le1$ for all integers $\alpha$ and $\beta$. 
Since $v_{s-1}\cdot v_{s+m}=0$, there exist integers $k_0,\ldots,k_{m-2}\in V_{s-1}\cap V_{s+m}$. Hence $V_{s-1}\supseteq\{j_0,\ldots,j_{m-1},k_0,\ldots,k_{m-2}\}$ and $V_{s+m}\supseteq\{i,j_0,\ldots,j_{m-2},k_0,\ldots,k_{m-2}\}$ and so $a_{s-1},a_{s+m}\ge2m-1$.
Assume $a_{s-1}=2m-1$. Since $|v_{s-2}\cdot v_{s-1}|=1$, we may assume, up to the action of $\Aut\ZZ^n$, that $k_1,\ldots,k_x\in V_{s-2}$, where $x\le m-2$ is odd and $|v_{s-2}\cdot e_{k_\alpha}|=1$ for all $1\le\alpha\le x$. If $a_{s+m}=2m-1$, then it follows that $v_{s+m}\cdot v_{s-2}\neq0$ and so $n=m+3$. But then $|\{i,j_0,\ldots,j_{m-1},k_1,\ldots,k_{m-2}\}|=2m\le m+3$, implying that $m\le3$, which is a contradiction. Thus $a_{s+m}\ge 2m$.
Let $$T=(S\setminus\{v_{s-1},\ldots,v_{s+m}\})\cup\{\pi_{{j_0},\ldots,{j_{m-1}}}(v_{s-1})\}\subset\ZZ^{n-(m+1)}.$$ Then $T$ is standard with $I(T)\le 4-(a_{s+m}-3)\le 7-2m$. By \cite{lisca-rational}, $I(T)\ge -3$ and so $m\le 5$. By Lemma~\ref{lem:hascontractions}, by Lemma~\ref{lem:hascontractions}, $S$ has as at least $n-11$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction.
We may now assume that $a_{s-1}\ge2m$ and similarly, $a_{s+m}\ge2m$.


Let
\begin{align*}
S'=(S&\setminus\{v_{s-1},\ldots,v_{s+m}\})\\&\cup\{\pi_{{j_0},\ldots,{j_{m-1}}}(v_{s-1}), \pi_{i,{j_0},\ldots,e_{{m-2}}}(v_{s+m})\}\subset\ZZ^{n-(m+1)}.
\end{align*}
Then $|v_{s-1}'\cdot v_{s+m}'|= m-1$, $-(\pi_{i,{j_0},\ldots,{j_{m-2}}}(v_{s+m}))^2=a_{s+m}-m\ge m-1$, and \linebreak[4]$-(\pi_{{j_0},\ldots,{j_{m-1}}}(v_{s-1}))^2=a_{s-1}-m\ge m-1$. By Proposition~\ref{prop:negdef2}(\ref{negdef2a}), $a_{s-1}=a_{s+m}=2m$ and $a_u=2$ for all $u\neq s-1,s+m$. Since $I(S)=4$, we have that $n=4m-8$ and $S$ has associated string of the form $(a_{s-1},\ldots,a_{s+m},\ldots,a_{s-2})=(2m,2^{[m]},2m,2^{[3m-10]})$. Since $m\le 5$, we necessarily have that $S$ has length at most 12, which is a contradiction.
\end{proof}


\begin{lem}
Let $i\in\mathcal{I}$. Then $a_{s(i)-1}\ge m(i)+2$.
\label{lem:12}
\end{lem}

\begin{proof} 
Set $s=s(i)$, $j=j(i)$, $j(i)_\alpha=j_\alpha$, and $m=m(i)$.
Since our standing assumption is that $a_{s-1}\ge3$, we may assume $m\ge2$. Further assume that $a_{s-1}<m+2$.
By Lemma~\ref{lem:t=s+m}, $t\neq s+m$ and so $V_t\cap V_{s-1}=\{j,j_1,\ldots,j_{m-1}\}$. Moreover, it is easy to see that $(v_{s-1}\cdot e_j,v_{s-1}\cdot e_{j_1},\ldots,v_{s-1}\cdot e_{j_{m-1}})=\pm(v_{t}\cdot e_j,v_{t}\cdot e_{j_1},\ldots,v_{t}\cdot e_{j_{m-1}})$. Since $m\ge2$ and $|v_{s-1}\cdot v_{t}|\le1$, we necessarily have that $|V_{s-1}\cap V_t|\ge3$; hence $a_{s-1}=m+1$ and $V_{s-1}=\{j,j_1,\ldots,j_{m-1},k\}$ for some integer $k$ with $k\in V_t$.

First suppose $v_{t}\cdot v_{s-1}=0$. By Proposition~\ref{prop:bounds}, since $m\ge2$, we necessarily have that $|v_t\cdot e_{k}|=2$ and $m=2$. By Proposition~\ref{prop:bounds}, $|v_\alpha\cdot e_\beta|\le1$ for all $(\alpha,\beta)\neq(t,k)$. It is clear that $E_k=\{s-2,s-1,s+2,t\}$ and $(v_{s+2}\cdot e_{j_1},2v_{s+2}\cdot e_k)=\pm (v_{t}\cdot e_{j_1},v_{t}\cdot e_k)$. Thus since $|v_t\cdot v_{s+2}|\le1$, $|V_t\cap V_{s+2}|\ge 4$; hence $|V_{s+2}|\ge 4$ and $a_t\ge 9$. Suppose $a_{s-2}\ge3$; then the subset 
$$T=(S\setminus\{v_{s-2},\ldots,v_{s+2},v_t\})\cup\{\pi_k(v_{s-2}),\pi_{j_1,k}(v_{s+2})\}\subset\ZZ^{n-4}$$
is standard with $I(T)\le -3$. By Lemma~\ref{lem:hascontractions}, it follows that $I(T)=-3$ and $T$ has at least $n-7$ contractible adjacencies; hence by following Example \ref{ex:hascontractions}, $S$ necessarily has contractible adjacencies, which is a contradiction.
Thus we may assume that $a_{s-2}=2$. It follows that since $j,j_1\not\in\mathcal{J}$ and $a_t,a_{s+2}\ge4$, we have $k\not\in\{j(i'),k(i')\,|\,i'\in\mathcal{J}\}$, so $p_4^2(S)\ge2$ and hence by Corollary~\ref{cor:jcor}, $\sum_{j=5}^np_j(S)\ge p_2(S)+2$.

Next suppose $|v_t\cdot v_{s-1}|=1$ so that $t=s-2$. 
By Proposition~\ref{prop:bounds}, we necessarily have that $|v_t\cdot e_{k}|\le 2$. If $|v_t\cdot e_{k}|=2$, then $m=3$ and, as above, we see that $p^2_4(S)\ge 2$. By Corollary~\ref{cor:jcor}, $\sum_{j=5}^np_j(S)\ge p_2(S)+2$, which contradicts Lemma~\ref{lem:count}. 
Thus we may assume $|v_t\cdot e_{k}|=1$, which implies that $m=2$. We have $V_{s-1}=\{j,j_1,k\}$ and $V_{s-2}\supset\{j,j_1,k\}$. Since $j_1\in V_{s+2}$ and $v_{s-1}\cdot v_{s+2}=0$, we have $E_k=\{s-2,s-1,s+2\}$ and $(v_{s-2}\cdot e_{j_1}, v_{s-2}\cdot e_{k})=\pm(v_{s+2}\cdot e_{j_1}, v_{s+2}\cdot e_{k})$. Thus there exists $l\in V_{s-2}\cap V_{s+2}$ and so $|V_{s-2}|\ge5$ and $|V_{s+2}|\ge 3$. Let $$C=(S\setminus\{v_{s-2},\ldots,v_{s+2}\}\cup\{\pi_{i,j,{j_1}}(v_{s-2}),\pi_{{j_1}}(v_{s+2})\}\subset\ZZ^{n-3}.$$ Then $C$ is cyclic with $I(C)=2$ and $\pi_{i,j,{j_1}}(v_{s-2})$ and $\pi_{{j_1}}(v_{s+2})$ are adjacent vertices. Moreover, $E_k^C=\{s-2,s+2\}$. Following the proof of Lemma~\ref{lem:s,s+1}, we can further form a standard subset $T\subset\ZZ^{n-7}$ with $I(T)=-3$. By Lemma~\ref{lem:hascontractions}, $T$ has at least $n-11$ contractible adjacencies; hence by following the argument in Example \ref{ex:hascontractions}, $S$ has contractible adjacencies, which is a contradiction.
\end{proof}


\begin{lem} If $i\in\mathcal{I}$, then $a_{t(i)}\le6$. Moreover, if $a_{t(i)}=3$ and $V_{t(i)}=\{i,j(i),k(i)\}$, then $t(i)\neq s(i)\pm2$, $E_{k(i)}=\{s-1,s+1,t-1,t,t+1\}$, and $k(i)\not\in\{j(i'),k(i')\,|\,i'\in\mathcal{J}\}.$
\label{lem:38}
\end{lem}

\begin{proof} Set $s=s(i)$, $t=t(i)$, $j=j(i)$, $j_\alpha=j(i)_\alpha$ for all $1\le \alpha\le m(i)-1$, and $m=m(i)$.
Suppose $a_{t}\ge7$. Let 
$$T=(S\setminus\{v_{s-1},\ldots,v_{s+m},v_t\})\cup\{\pi_{j,{j_1},\ldots,{j_{m-1}}}(v_{s-1}),\pi_{{j_{m-1}}}(v_{s+m})\}\subset\ZZ^{n-(m+1)}.$$
By Lemma~\ref{lem:12}, $T$ is standard and $I(T)=I(S)-\sum_{\alpha=0}^{m-1}(a_{s+\alpha}-3)-(a_t-3)-m-1\le I(S)-5<0$. By~\cite{lisca-rational}, $I(T)\ge-3$ and so $a_{t}\le 10$ and $m\le 4$. By Lemma~\ref{lem:hascontractions}, $S$ necessarily has contractions, which is a contradiction.

Next suppose $a_{t}=3$ and let $V_t=\{i,j,k\}$. First assume $t=s-2$. It is clear that $E_k=\{s-3,s-2,s+1\}$ or $E_k=\{s-3,s-2,s-1,s+1\}$. Since $v_{s-3}\cdot v_{s+1}=0$, it follows that $|V_{s-3}\cap V_{s+1}|\ge2$ and so $a_{s+1}\ge3$. Moreover since $v_{s-2}\cdot v_{s+1}=0$ and $|v_{s+1}\cdot e_j|=1$, we have that $|v_{s+1}\cdot e_k|=1$. Hence if $E_k=\{s-3,s-2,s+1\}$, we can contract. We now assume that $E_k=\{s-3,s-2,s-1,s+1\}$. Then $(v_{s-1}\cdot e_j,v_{s-1}\cdot e_k)=\pm(v_{s+1}\cdot e_j,2v_{s+1}\cdot e_k)$. By Proposition~\ref{prop:bounds}, $|V_{s-1}\cap V_{s+1}|\ge5$ and $|V_{s-3}\cap V_{s-1}|\ge3$; hence $|V_{s+1}|\ge5$, $|V_{s-3}|\ge3$, and $a_{s+1}\ge 8$. The subset
$$C=(S\setminus\{v_{s-3},\ldots,v_{s+1}\})\cup\{\pi_k(v_{s-3}),\pi_{j,k}(v_{s+1})\}\subset\ZZ^{n-3}$$
is cyclic with $I(C)\le -3$. By Lemma~\ref{lem:hascontractions}, $C$ has as at least $n-9$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction. Thus $t\neq s-2$. A similar argument shows that $t\neq s+2$. It follows that $E_k=\{s-1,s+1,t-1,t,t+1\}$. Moreover, note that $(v_{s-1}\cdot e_j,v_{s-1}\cdot e_k)=\pm(v_{s+1}\cdot e_j,v_{s+1}\cdot e_k)$.

Finally, assume $k\in\{j(i'),k(i')\,|\,i'\in\mathcal{J}\}.$ It follows that $s(i')\in\{s-1,s+1,t-1,t+1\}$. If $s(i')=s-1$, then since $a_{s-1}=3$ and $v_{s-1}\cdot v_{s+1}=0$, it follows that $s-1=t(i')$ and $|v_{s-1}\cdot e_{i'}|=2$, which contradicts Lemma~\ref{lem:6}. Thus $s(i')\neq s-1$ and, similarly, $s(i')\neq s+1$. If $s(i')=t-1$, then $t(i')=t+1$ and the subset 
$$C=(S\setminus\{v_{t-1},v_t,v_{t+1}\})\cup\{\pi_{i'}(v_{t-1}),\pi_{i'}(v_{t+1})\}\subset\ZZ^{n-1}$$
is cyclic with $p_1(C)=1$; by Lemma~\ref{lem:hascontractions}, $S$ necessarily has contractions, which is a contradiction. Thus $k\not\in\{j(i'),k(i')\,|\,i'\in\mathcal{J}\}.$
\end{proof}


\begin{lem}
Let $i\in \mathcal{I}$.  If $|v_{t(i)}\cdot v_{s(i_1)}|=|v_{t(i)}\cdot v_{s(i_2)}|=1$ for some $i_1\neq i_2\in\mathcal{I}$, then $a_{t(i)}\ge5$.
\label{lem:14'}
\end{lem}

\begin{proof} 
Set $j:=j(i)$, $s:=s(i)$, $t:=t(i)$, $j_\alpha:=j(i_\alpha)$, $s_\alpha:=s(i_\alpha)$, and $t_\alpha:=t(i_\alpha)$ for $\alpha\in\{1,2\}$. Up to reversing $S$, we may assume that $t-1=s_1$ and $t+1=s_2$.
Assume $a_t=4$. Then $V_t=\{i,j,j_1,j_2\}$.
Since $j_1\in V_{t-2}\cap V_t$ and $v_{t-2}\cdot v_t=0$, either $j\in V_{t-2}$ or $j_2\in V_{t-2}$. \\

\noindent\underline{Case 1: $j\in V_{t-2}$}. Since $j\in V_{t-2}$, we have that $t-2=s+1$ and so $s=t-3$. Hence $E_i=\{t-3,t\}$ and $E_j=\{t-4,t-3,t-2,t\}$. 
Since $j_2\in V_{t+2}\cap V_t$ and $v_{t+2}\cdot v_t=0$, it follows that $j_1\in V_{t+2}$ and so $t_1=t+2$. Similarly, 
since $j\in V_{t-4}\cap V_t$ and $v_{t-4}\cdot v_t=0$, we have $j_2\in V_{t-4}$ and so $t-4=t_2$. Since $i_1,j_1,j_2\in V_{t+2}$ and $i_1,j_1,j_2\not\in V_{t+3}$, it follows that $a_{t+2}\ge4$. 
Moreover, since $j\in V_{t-2}\cap V_{t-4}$ and $v_{t-2}\cdot v_{t-4}=0$, it follows that $|V_{t-2}\cap V_{t-4}|\ge2$ and so $a_{t-2}\ge3$. Hence the subset 
$$C=(S\setminus\{v_{t-4},v_{t-2},\ldots,v_{t+2}\})\cup\{\pi_{{i_2}}(v_{t-4}),\pi_{{j_1}}(v_{t-2}),\pi_{{i_1},{j_1}}(v_{t+2})\}\subset\ZZ^{n-3}$$
is cyclic with $p_1(C)=1$. By Lemma~\ref{lem:hascontractions}, $C$ has as at least $n-7$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction.\\

\noindent\underline{Case 2: $j_2\in V_{t-2}$}.
Since $j_2\in V_{t-2}$, we have $t-2=t_2$. Now, since $j_2\in V_{t+2}\cap V_t$ and $v_t\cdot v_{t+2}=0$, either $j\in V_{t+2}$ or $j_1\in V_{t+2}$. We obtain a similar contradiction as in Case 1 if $j\in V_{t+2}$. Thus we may assume that $j_1\in V_{t+2}$ so that $t+2=t_1$. As above, it is clear that $a_{t\pm2}\ge4$. Hence the subset 
$$C=(S\setminus\{v_{t-2},\ldots,v_{t+2}\})\cup\{\pi_{{i_2},e_{j_1}}(v_{t-2}),\pi_{{i_1},e_{j_1}}(v_{t+2})\}\subset\ZZ^{n-3}$$
is cyclic with $p_1(C)=1$. By Lemma~\ref{lem:hascontractions}, $C$ has as at least $n-7$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction.
\end{proof}

\begin{lem} Let $i\in\mathcal{I}$ and suppose $t(i)=s(i')\pm1$ for some $i'\in\mathcal{I}$. Then $a_{t(i)}\ge4$ and if $a_{t(i)}=4$, then $v_{t(i')}\cdot v_{s(i')-1}=v_{t(i')}\cdot v_{s(i')+1}=0$.
\label{lem:55}
\end{lem}

\begin{proof} Set $j=j(i)$ and $j'=j(i')$. By Lemma~\ref{lem:10}, $j\neq j'$. Up to reversal of $S$, we may assume that $t(i)=s(i')-1$. It follows from Lemma~\ref{lem:38} that $a_{t(i)}\ge3$; moreover, if $a_{t(i)}=3$ so that $V_{t(i)}=\{i,j,j'\}$, it follows from Lemma~\ref{lem:38} that $|E_{j'}|=5$, which is a contradiction. 
Now assume $a_{t(i)}=4$ and set $V_{t(i)}=\{i,j,j',k\}$ for some integer $k$.
First assume $v_{t(i')}\cdot v_{s(i')-1}\neq0$ so that $t(i')=s(i')-2$. By Lemma~\ref{lem:38}, $a_{s(i')-2}\ge4$. Since $j'\in V_{s(i')-2}\cap V_{s(i')+1}$, there exists $l\in V_{s(i')-2}\cap V_{s(i')+1}$; since also $j \in V_{s(i')+1}$ or $k \in V_{s(i')+1}$, it follows that $a_{s(i')+1}\ge3$. If $v_{s(i')-2}\cdot v_{s(i)}=0$, then it follows that $j,k\in V_{s(i)-1}\cap V_{s(i)+1}$ and $(v_{s(i)-1}\cdot e_j,v_{s(i)-1}\cdot e_k)=\pm (v_{s(i)+1}\cdot e_j,v_{s(i)+1}\cdot e_k)$; hence $|V_{s(i)-1}\cap V_{s(i)+1}|\ge3$. Thus the subset
\begin{equation*}
    \begin{split}
      C=(S&\setminus\{v_{s(i')-2},\ldots,v_{s(i')+1},v_{s(i)-1},v_{s(i)},v_{s(i)+1}\})\\
      &\cup\{ \pi_{{i',j'}}(v_{s(i')-2}),\pi_{{j'}}(v_{s(i')-1}),\pi_{{j'}}(v_{s(i')+1}),\pi_{j}(v_{s(i)-1}), \pi_{j}(v_{s(i)+1})\}\subset\ZZ^{n-4}
    \end{split}
\end{equation*}
is cyclic; hence it contains $n-3$ linearly independent vectors in $\ZZ^{n-4}$, which is not possible. If $v_{s(i')-2}\cdot v_{s(i)}\neq0$, on the other hand, then $s(i)=s(i')-3$ and $j,j',k\in V_{s(i')-2}$; hence $a_{s(i')-2}\ge5$. Moreover, since $k\in V_{s(i')-4}\cap V_{s(i')+1}$, it follows that $|V_{s(i')-4}\cap V_{s(i')+1}|\ge2$ and so $a_{s(i')-4}\ge3$. Thus the subset
\begin{equation*}
    \begin{split}
      C=(S&\setminus\{v_{s(i')-4},\ldots,v_{s(i')+1}\})\\
      &\cup\{\pi_j(v_{s(i')-4}),\pi_{i',j',j}(v_{s(i')-2}),\pi_{j'}(v_{s(i')+1})\}\subset \ZZ^{n-4}
    \end{split}
\end{equation*} 
is cyclic; hence it contains $n-3$ vectors in $\ZZ^{n-4}$, which is impossible. Thus $v_{t(i')}\cdot v_{s(i')-1}=0$.

Next assume $v_{t(i')}\cdot v_{s(i')+1}\neq0$ so that $t(i')=s(i')+2$. Since $j'\in V_{s(i')-1}\cap V_{s(i')+1}$, either $j\in V_{s(i')+1}$ or $k\in V_{s(i')+1}$. If $j\in V_{s(i')+1}$, then it follows that $s(i')+2=s(i)$, which is not possible. Thus $k\in V_{s(i')+1}$ and so $E_k\supset\{s(i')-2,s(i')-1,s(i')+1\}$. It follows that $|V_{s(i')-2}\cap V_{s(i')+1}|\ge2$ and so $a_{s(i')+1}\ge3$. It is also easy to see that either $a_{s(i')-2}\ge3$ or $a_{s(i')+1}\ge4$. Assume the former and let
\begin{equation*}
    \begin{split}
        G=(S&\setminus\{v_{s(i')-2},\ldots,v_{s(i')+2}\})\\
        &\cup\{\pi_{k}(v_{s(i')-2}),\pi_{{j'},k}(v_{s(i')-1}),\pi_{{j'}}(v_{s(i')+1})\}\subset\ZZ^{n-2}.
    \end{split}
\end{equation*}
Then $G$ is good and $E_k^G=\{s(i')+1\}$, where $\pi_{{j'}}(v_{s(i')+1})$ is an isolated vertex. But by Lemma \ref{lem:lisca3.2}, no such subset exists. 
\end{proof}



\begin{lem} Let $i\in\mathcal{I}$ and suppose $t(i)=s(i')\pm1$ for some $i'\in\mathcal{I}$ and $a_{t(i)}=4$. If $v_{t(i')}\cdot v_{s(i)}=0$, then $s(i)\neq s(i')\pm2$.
\label{lem:56}
\end{lem}

\begin{proof}
Set $j=j(i)$, $j'=j(i'),$ and $V_{t(i)}=\{i,j,j',k\}$. Up to reversal of $S$, we may assume that $t(i)=s(i')-1$.
Assume $s(i)=s(i')+2$. Then $E_{j}=\{s(i')-1,s(i')+1,s(i')+2,s(i')+3\}$ and by Lemma~\ref{lem:t=s+m} (applied to $i$), $a_{s(i')+1}\ge3$. Since $t(i')\neq s(i')\pm2$ (by Lemma~\ref{lem:55}) and $v_{t(i')}\cdot v_{s(i)}=0$, we have $k\in V_{t(i')}$.

First assume $a_{s(i')+1}=3$. Then there is an integer $m\neq k$ such that $V_{s(i')+1}=\{j,j',m\}$. Moreover, $E_k=\{s(i')-2,s(i')-1,s(i')+3,t(i')\}$, $E_m=\{s(i')+1,s(i')+3,t(i')\}$, and $V_{s(i')+3}\supset\{j,k,m\}$. It is easy to see that either $a_{s(i')-2}\ge3$ or $a_{s(i')+3}\ge5$. In the former case, the subset $$T=(S\setminus\{v_{s(i')-1},\ldots,v_{s(i')+3},v_{t(i')}\})\cup \{\pi_{k}(v_{s(i')-2}),\pi_{j,m}(v_{s(i')+3})\}\subset \ZZ^{n-4}$$
is standard with $E_k^T=\{s(i')+3\}$; by Lemma~\ref{lem:p1contraction}, $I(T)=-3$ and so $a_{t(i')}=8$, contradicting Lemma~\ref{lem:38}. In the latter case, the subset $$T=(S\setminus\{v_{s(i')-1},\ldots,v_{s(i')+3},v_{t(i')}\})\cup \{\pi_{j,k,m}(v_{s(i')+3})\}\subset\ZZ^{n-5}$$
is standard with $E_k^T=\{s(i')-2\}$; by Lemma~\ref{lem:p1contraction}, $I(T)=-3$ and so $a_{t(i')}=8$, contradicting Lemma~\ref{lem:38}.

Next assume $a_{s(i')+1}\ge4$. If $a_{t(i')}\ge5$, then let 
\begin{equation*}
    \begin{split}
        R=(S&\setminus\{v_{s(i')-1},\ldots,v_{s(i')+3}, v_{t(i')}\})\\
        &\cup\{\pi_{i,{j}}(v_{s(i')-1}),\pi_{j,{j'}}(v_{s(i')+1}),\pi_{j}(v_{s(i')+3}),\pi_{{i'},{j'}}(v_{t(i')})\}\subset\ZZ^{n-3}.
    \end{split}
\end{equation*}
Then by Proposition~\ref{prop:negdef3}, $R$ contains $n-2$ linearly independent vectors in $\ZZ^{n-3}$, which is not possible. 
We may now assume that $a_{t(i')}=4$; set $V_{t(i')}=\{i',j',k,m\}$. Further suppose $v_{t(i')}\cdot v_{s(i')-2}=v_{t(i')}\cdot v_{s(i')+3}=0$. Then $E_m\supseteq\{s(i')-2,s(i')+1,s(i')+3,t(i')-1,t(i'),t(i')+1\}$. 
By Lemma~\ref{lem:count} and Proposition~\ref{prop:bounds}, $|E_m|=6$ and $|v_\alpha\cdot v_\beta|\le1$ for all $\alpha,\beta$. Moreover, $(v_{s(i')+1}\cdot e_j,v_{s(i')+1}\cdot e_m)=\pm (v_{s(i')+3}\cdot e_j,v_{s(i')+3}\cdot e_m)$; hence $|V_{s(i')+1}\cap V_{s(i')+3}|\ge4$ and so $a_{s(i')+1}, a_{s(i')+3}\ge5$. Moreover, it is easy to see that $a_{s(i')-2}\ge 3$. Thus the subset
$$T=(S\setminus\{v_{s(i')-2},\ldots,v_{s(i')+3},v_{t(i')}\})\cup \{\pi_{k}(v_{s(i')-2}),\pi_{j,k}(v_{s(i')+3})\}\subset\ZZ^{n-5}$$ 
is standard with $I(T)\le -1$ and so $a_{s(i')+1}\le 7$. By Lemma~\ref{lem:hascontractions}, $T$ has as at least $n-10$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction

Thus either $|v_{t(i')}\cdot v_{s(i')-2}|=1$ or $|v_{t(i')}\cdot v_{s(i')+3}|=1$. First suppose $|v_{t(i')}\cdot v_{s(i')-2}|=1$ so that $t(i')=s(i')-3$, $E_k\subset\{s(i')-3,s(i')-2,s(i')-1,s(i')+1,s(i')+3\}$, and the subset 
\begin{align*}
C=(S&\setminus\{v_{s(i')-2},\ldots,v_{s(i')+3}\})\\
&\cup \{\pi_{{i'},{j'}}(v_{s(i')-3}),\pi_{j,{j'}}(v_{s(i')+1}),\pi_{j,k}(v_{s(i')+3})\}\subset\ZZ^{n-3}
\end{align*}
is cyclic with $I(C)=0$. If $E_k=\{s(i')-3,s(i')-2,s(i')-1,s(i')+1,s(i')+3\}$, then $|\pi_{j,{j'}}(v_{s(i')+1})\cdot e_k|=2$, but by Lemmas 7.1 and 7.10 in~\cite{simone2020classification}, 
no such subset exists. If $E_k\neq\{s(i')-3,s(i')-2,s(i')-1,s(i')+1,s(i')+3\}$, then $p_1(C)=1$, but again by \cite{simone2020classification}, no such subset exists. 
Next suppose $|v_{t(i')}\cdot v_{s(i')+3}|=1$ so that $t(i')=s(i')+4$. First suppose $a_{s(i')+3}=3$ and set $V_{s(i')+3}=\{j,k,l\}$. Then $E_l=\{s(i')-2,s(i')+1,s(i')+3\}$ and since $m\in V_{s(i')+1}\cap V_{s(i')+5}$, it is easy to see that $a_{s(i')+1}\ge5$. Now
$$T=(S\setminus\{v_{s(i')-2},\ldots,v_{s(i')+4}\})\cup \{\pi_{k}(v_{s(i')-2}),\pi_{j,{j'},l}(v_{s(i')+1})\}\subset\ZZ^{n-5}$$
is a standard subset with $E_l=\{s(i')-3\}$, $|\pi_{k}(v_{s(i')-2})\cdot e_l|=1$, and $I(T)=0$; but by Lemma~\ref{lem:p1contraction}, no such subset exists.
Thus we may assume $a_{s(i')+3}\ge4$. Note that $a_{s(i')+1}\ge4$. Then 
\begin{align*}
T=(S&\setminus\{v_{s(i')-2},\ldots,v_{s(i')+4}\})\\
&\cup \{\pi_{k}(v_{s(i')-2}),\pi_{j,{j'}}(v_{s(i')+1}),\pi_{j,{k}}(v_{s(i')+1})\}\subset\ZZ^{n-5}
\end{align*}
is a standard subset and thus contains $n-4$ linearly independent vectors in $\ZZ^{n-5}$, which is not possible.
\end{proof}



\begin{lem} 
Let $i\in\mathcal{I}$ and suppose $t(i)=s(i')\pm1$ for some $i'\in\mathcal{I}$ and $a_{t(i)}=4$. Then $|v_{t(i')}\cdot v_{s(i)}|=1$.
\label{lem:57}
\end{lem}

\begin{proof}
Set $j=j(i)$, $j'=j(i'),$ and $V_{t(i)}=\{i,j,j(i'),k\}$. Up to reversal of $S$, we may assume that $t(i)=s(i')-1$.
Assume $t(i')\neq s(i)\pm 1$.
By Lemma~\ref{lem:55}, $t(i')\neq s(i')\pm2$ and by Lemma~\ref{lem:56}, $s(i)\neq s(i')+2$.
Note that $k\in V_{t(i')}\cap V_{s(i')+1}$ and $(v_{t(i')}\cdot e_{j'},v_{t(i')}\cdot e_k)=\pm(v_{s(i')+1}\cdot e_{j'},v_{s(i')+1}\cdot e_k)$.
Hence $|V_{t(i')}\cap V_{s(i')+1}|\ge3$.

 

First suppose $s(i)\neq s(i')-3$. Then $E_k\supset\{s(i')-2,s(i')-1,s(i')+1,t(i'),s(i)-1,s(i)+1\}$; by Proposition~\ref{prop:bounds}, we have that $|E_k|=6$ and $|v_\alpha\cdot v_\beta|\le1$ for all $\alpha,\beta$.
Thus $|V_{t(i')}\cap V_{s(i')+1}|\ge4$ and so $a_{t(i')}\ge 5$ and $a_{s(i')+1}\ge 4$. Similarly, $(v_{s(i)-1}\cdot e_j,v_{s(i)-1}\cdot e_k)=(v_{s(i)+1}\cdot e_j,-v_{s(i)+1}\cdot e_k)$, implying that $a_{s(i)\pm1}\ge3$. Let 
\begin{equation*}
    \begin{split}
        R=(&S\setminus\{v_{s(i')-1},v_{s(i')},v_{s(i')+1},v_{s(i)-1},v_{s(i)},v_{s(i)-1}, v_{t(i')}\})\\
        &\cup\{\pi_{{j'}}(v_{s(i')+1}),\pi_{j}(v_{s(i)-1}),\pi_{j}(v_{s(i)+1}),\pi_{{i'},{j'}}(v_{t(i')})\}\subset\ZZ^{n-4}.
    \end{split}
\end{equation*}
Then by Proposition~\ref{prop:negdef3}, $R$ contains $n-3$ linearly independent vectors in $\ZZ^{n-4}$, which is not possible.

Next suppose $s(i)=s(i')-3$. Then by Lemma~\ref{lem:t=s+m} (applied to $i$), $a_{s(i')-2}=a_{s(i)+1}\ge3$ and since $j,k\in V_{s(i')-4}$ and $v_{s(i')-4}\cdot v_{s(i')+1}=0$, it follows that $a_{s(i')-4}\ge3$. Recall that $|V_{t(i')}\cap V_{s(i')+1}|\ge3$.
If $|V_{t(i')}\cap V_{s(i')+1}|\ge4$, then $a_{t(i')}\ge5$ and $a_{s(i')}\ge 4$. In this case, the subset $R\subset\ZZ^{n-4}$ defined above yields a contradiction. We may now assume $|V_{t(i')}\cap V_{s(i')+1}|=3$. Then there exists $m\in V_{t(i')}\cap V_{s(i')+1}$ such that either $|v_{t(i')}\cdot e_m|=2$ or $|v_{s(i')+1}\cdot e_m|=2$. 
In the former case, $a_{t(i')}\ge7$, contradicting Lemma~\ref{lem:38}.
In the latter case, the subset 
\begin{equation*}
    \begin{split}
       T=(S&\setminus\{v_{s(i')-4},\ldots,v_{s(i')+1},v_{t(i')}\})\\
       &\cup \{\pi_{j}(v_{s(i')-4}),\pi_{j}(v_{s(i')-2}),\pi_{{i'},{j'}}(v_{t(i')})\}\subset\ZZ^{n-4} \end{split}
\end{equation*}
is standard with $I(T)\le -2$. By Lemma~\ref{lem:hascontractions}, $T$ has as at least $n-9$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction
\end{proof}


\begin{lem} Let $i\in\mathcal{I}$ and suppose $t(i)=s(i')\pm1$ for some $i'\in\mathcal{I}$ and $a_{t(i)}=4$. Then $a_{t(i')}\ge5$.
\label{lem:58}
\end{lem}

\begin{proof} Up to reversal of $S$, we may assume that $t(i)=s(i')-1$.
Using the same notation as in the proof of Lemma~\ref{lem:57}, we have that $V_{t(i)}=V_{s(i')-1}=\{i,j,j',k\}$.
By Lemma~\ref{lem:57}, $v_{t(i')}\cdot v_{s(i)}=1$ and so $t(i')=s(i)\pm1$ (and $t(i')\neq s(i)\mp1$).

Assume that $a_{t(i')}=4$ and set $V_{t(i')}=\{i',j,j',k'\}$.
By Lemma~\ref{lem:55}, $v_{t(i')}\cdot v_{s(i')-1}=v_{t(i')}\cdot v_{s(i')+1}=0$ and $v_{s(i')-1}\cdot v_{s(i)-1}=v_{s(i')-1}\cdot v_{s(i)+1}=0$. Thus $E_{k}=\{s(i')-2, s(i')-1, s(i')+1, s(i)\mp1\}$ and $E_{k'}=\{s(i)-2, s(i)-1,s(i)+1,s(i')+1\}$. Moreover, $(v_{s(i)\mp1}\cdot e_k,v_{s(i)\mp1}\cdot e_{k'})=\pm(v_{s(i')+1}\cdot e_k,-v_{s(i')+1}\cdot e_{k'})$. 

We next claim that $s(i')+2\neq s(i)\mp1$. Otherwise, $V_{s(i')+1}\cap V_{s(i)\mp1}=\{k,k',l\}$ for some integer $l$, where $(v_{s(i)\mp1}\cdot e_k,v_{s(i)\mp1}\cdot e_{l})=\pm(v_{s(i')+1}\cdot e_k,-v_{s(i')+1}\cdot e_{l})$. Since $v_{s(i)-2}\cdot v_{s(i)+1}=0$, we have that $l\in V_{s(i)-2}$ and $(v_{s(i)+1}\cdot e_{k'},v_{s(i)+1}\cdot e_{l})=\pm(v_{s(i)-2}\cdot e_{k'},-v_{s(i)-2}\cdot e_{l})$. Consequently, $|v_{s(i')+1}\cdot v_{s(i)-1}|=2$, which is a contradiction. Since $s(i')+2\neq s(i)\mp1$, we have that $j',k,k'\not\in V_{s(i')+2}$ and so $|V_{s(i')+1}|\ge4$. A similar argument shows that $|V_{s(i)\mp1}|\ge4$.


We next claim that either $a_{s(i')+1}\ge5$ or $a_{s(i)\mp1}\ge5$. Assume $a_{s(i)\mp1}=a_{s(i')+1}=4$ and let $V_{s(i')+1}=\{j',k,k',l\}$ and $V_{s(i')\mp1}=\{j,k,k',l'\}$ for some integers $l$ and $l'$. Since $s(i')+2\neq s(i)\mp1$, it follows that $l\neq l'$. Up to cyclic reordering, we may assume that $v_{s(i')+1}\cdot v_{s(i')+2}=1$. Since $j',k,k'\not\in V_{s(i')+2}$, we have that $l\in V_{s(i')+2}$ and $v_{s(i')+1}\cdot e_l=-v_{s(i')+2}\cdot e_l$. Moreover, $E_l=\{s(i')-2,s(i')+1,s(i')+2\}$. Similarly, $E_{l'}=\{s(i)\pm2,s(i)\mp1,s(i)\mp2\}$. 
Hence $V_{s(i')-2}\supseteq\{k,l\}$ and $V_{s(i)\mp2}\supseteq\{k',l'\}$; it follows that $|v_{s(i')-2}\cdot v_{s(i)\mp1}|=|v_{s(i)\pm2}\cdot v_{s(i')+1}|=1$, implying that $n=8$, which is a contradiction.
Thus, either $a_{s(i)\mp1}\ge5$ or $a_{s(i')+1}\ge5$. 

We next claim that $a_{s(i)\pm2}=a_{s(i')-2}=2$. Assume $a_{s(i)\pm2}\ge3$. If $a_{s(i')+1}\ge5$, then
\begin{equation*}
\begin{split}
      T=(S&\setminus\{v_{s(i)-2},\ldots,v_{s(i)+1},v_{s(i')-1},v_{s(i')},v_{s(i')+1}\})\\
      &\cup\{\pi_{{k'}}(v_{s(i)-2}),\pi_{j,{k'}}(v_{s(i)+1})\}\subset\ZZ^{n-5}
\end{split}
\end{equation*}
is standard with $-3\le I(T)\le-1$. By Lemma~\ref{lem:hascontractions}, $T$ has as at least $n-11$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction. Thus $a_{s(i')+1}=4$ and $a_{s(i)\mp1}\ge5$. By a similar argument, since $a_{s(i)\mp1}\ge5$, we must have that $a_{s(i)\pm2}=2$. Let $V_{s(i')\mp1}=\{j',k,k',l\}$. Then $(v_{s(i)\mp1}\cdot e_{k'},v_{s(i)\mp1}\cdot e_l)=\pm (v_{s(i')-2}\cdot e_{k'},v_{s(i')-2}\cdot e_l)$, implying that $a_{s(i')-2}\ge 3$, which is a contradiction.
Thus $a_{s(i')-2}=a_{s(i)\pm2}=2$. 


Set $V_{s(i')-2}=\{k,l\}$ and $V_{s(i)\pm2}=\{k',l'\}$. 
Either $k\in V_{s(i')-3}$ or $l\in V_{s(i')-3}$. First suppose $k\in V_{s(i')-3}$. Then $l\in V_{s(i')-3}$ since otherwise, $l\in\mathcal{I}$ and $k=j(l)$, contradicting Lemma~\ref{lem:14'}. Moreover, $s(i')-3=s(i)\mp1$. Thus $|v_{s(i')-3}\cdot e_k|=1$ and $|v_{s(i')-3}\cdot e_l|=2$. It is also easy to see that $(v_{s(i')-3}\cdot e_k,v_{s(i')-3}\cdot e_l,-v_{s(i')-3}\cdot e_{k'})=\pm(v_{s(i')+1}\cdot e_k,v_{s(i')+1}\cdot e_l,-v_{s(i')+1}\cdot e_{k'})$. Thus by Proposition~\ref{prop:bounds}, $|V_{s(i')-3}\cap V_{s(i')+1}|\ge 4$ and so $a_{s(i')+1}\ge 6$. Thus the subset 
$$T=(S\setminus\{v_{s(i')-5},\ldots,v_{s(i')+1}\})\cup\{\pi_{{j'},{k},{k'},l}(v_{s(i')+1})\}\subset \ZZ^{n-6}$$
is standard with $I(T)\le -3$. By Lemma~\ref{lem:hascontractions}, $T$ has as at least $n-11$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction

Thus $k\not\in V_{s(i')-3}$ and $l\in V_{s(i')-3}$. By a similar argument, we have that $k'\not\in V_{s(i)\pm3}$ and $l'\in V_{s(i)\pm3}$. Hence $|E_k|=|E_l|=|E_{k'}|=|E_{l'}|=4$. Moreover, notice that $k,k',l,l'\in V_{s(i')+1}\cap V_{s(i)\pm1}$. It is clear that $k,l,k',l'\not\in \{j(\alpha)\,|\,\alpha\in\mathcal{I}\}$. Moreover, since $E_k=\{s(i')-2,s(i')-1,s(i')+1,s(i)\pm1\}$ and $a_{s(i')-2},a_{s(i')-1},a_{s(i')+1},a_{s(i)\pm1}\neq 3$, $k\not\in\{j(\alpha),k(\alpha)\,|\,\alpha\in\mathcal{J}\}$. Similarly, $k'\not\in\{j(\alpha),k(\alpha)\,|\,\alpha\in\mathcal{J}\}$. If $l\in\{j(\alpha),k(\alpha)\,|\,\alpha\in\mathcal{J}\}$, then $s(i')-3=s(\alpha)$ and either $s(i')+1=t(\alpha)$ or $s(i')\pm1=t(\alpha)$. Assume the former case. Then the subset 
\begin{equation*}
\begin{split}
      T=(S&\setminus\{v_{s(i')-3},v_{s(i')-1},v_{s(i')},v_{s(i')+1},v_{s(i)-1},v_{s(i)},v_{s(i)+1}\})\\
      &\cup\{\pi_{\alpha}(v_{s(i')-3}),\pi_{j}(v_{s(i)\mp1}),\pi_{{i'},j}(v_{s(i)\pm1})\}\subset\ZZ^{n-4}
\end{split}
\end{equation*}
is standard with $p_1(T)>0$. By Lemma~\ref{lem:hascontractions}, $T$ has as at least $n-8$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction. Hence $s(i')+1\neq t(\alpha)$. A similar argument shows that $s(i')\pm1\neq t(\alpha)$. Thus $l\not\in\{j(\alpha),k(\alpha)\,|\,\alpha\in\mathcal{J}\}$.

Now by Lemmas~\ref{lem:Ej>3} and~\ref{lem:at=3} we have that $p_4(S)\ge p_2(S)+4$. Moreover, by Lemma~\ref{lem:count} and Corollary~\ref{cor:jcor}, $p_4(S)= p_2(S)+4$ and $|\mathcal{J}|=0$. If $a_{s(i')-3}=2$ and $V_{s(i')-3}=\{l,m\}$, then by Lemma~\ref{lem:t=s+m}, $m\in V_{s(i')-4}$; it follows that $|E_{m}|=4$ and $p_4(S)\ge p_2(S)+5$, which is not possible. Thus $a_{s(i')-3}\ge3$. Moreover, since $l\in V_{s(i')-3}\cap V_{s(i')+1}$, it follows that $a_{s(i')+1}\ge6$. Hence the subset
\begin{equation*}
\begin{split}
      T=(S&\setminus\{v_{s(i')-3},\ldots,v_{s(i')+1},v_{s(i)\mp1},v_{s(i)},v_{s(i)\pm1},v_{s(i)\pm2}\})\\
      &\cup\{\pi_{l}(v_{s(i')-3},\pi_{{j'},k,{k'},l}(v_{s(i')+1})\}\subset\ZZ^{n-7}
\end{split}
\end{equation*}
is standard with $I(T)\le -1$. By Lemma~\ref{lem:hascontractions}, $T$ has as at least $n-12$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction.
\end{proof}


\begin{prop} Let $i\in\mathcal{I}$ and suppose $t(i)=s(i')\pm1$ for some $i'\in\mathcal{I}$. Then $a_{t(i)}\ge5$.
\label{prop:59}
\end{prop}

\begin{proof}
Assume $a_{t(i)}=4$ and $V_{t(i)}=\{i,j,j',k\}$ as in the proof of Lemma~\ref{lem:58}. Up to reversal of $S$, we may assume that $t(i)=s(i')-1$. By Lemmas~\ref{lem:57} and~\ref{lem:58}, $t(i')=s(i)\pm1$ and $a_{t(i')}\ge5$. Recall that $E_k=\{s(i')-2,s(i')-1,s(i')+1,s(i)\mp1\}$. Since $v_{s(i')-2}\cdot v_{s(i')+1}=0$, there exists and integer $l\in V_{s(i')-2}\cap V_{s(i')+1}$; hence $a_{s(i')+1}\ge3$.\\

\noindent\underline{Case 1: $a_{s(i')+1}=3$}.
In this case, $V_{s(i')+1}=\{j',k,l\}$.
Suppose $a_{s(i')-2}=2$; then $V_{s(i')-2}=\{k,l\}$. Since either $v_{s(i)\mp1}\cdot v_{s(i')-2}=0$ or $v_{s(i)\mp1}\cdot v_{s(i')+1}=0$, we have $l\in V_{s(i)\mp1}$. But then either $|v_{s(i)\mp1}\cdot v_{s(i')-2}|=2$ or $|v_{s(i)\mp1}\cdot v_{s(i')+1}|=2$, which is a contradiction. Hence $a_{s(i')-2}\ge3$. Moreover, since $k\in V_{s(i)\mp1}$, $v_{s(i)-1}\cdot v_{s(i)+1}=0$, and $j\in V_{s(i)-1}\cap V_{s(i)+1}$, it is clear that $a_{s(i)\mp1}\ge3$. 
First suppose that $|v_{s(i')+1}\cdot v_{s(i)\mp1}|=1$. Then the subset
\begin{equation*}
    \begin{split}
      C=(S&\setminus\{v_{s(i')-2},\ldots,v_{s(i')+1},v_{s(i)-1},\ldots,v_{s(i)+1}\})\\
      &\cup\{\pi_{k}(v_{s(i')-2}),\pi_{{i'},j,{j'}}(v_{s(i)\pm1}),\pi_{j}(v_{s(i)\mp1})\}\subset\ZZ^{n-4}
    \end{split}
\end{equation*}
is a cyclic subset with $p_1(C)=1$ (i.e., $E_k^C=\{s(i)\mp1\}$). By Lemma~\ref{lem:hascontractions}, $C$ has as at least $n-9$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction. Next suppose $v_{s(i')+1}\cdot v_{s(i)\mp1}=0$. Then $l\in V_{s(i)\mp1}$ and $(v_{s(i)\mp1}\cdot e_k,v_{s(i)\mp1}\cdot e_l)=\pm (v_{s(i')-2}\cdot e_k,v_{s(i')-2}\cdot e_l)$. Since $|v_{s(i')-1}\cdot v_{s(i)\mp1}|\le1$, it follows that $|V_{s(i')-1}\cap V_{s(i)\mp1}|\geq3$ and so $a_{s(i)\mp1}\ge4$. If $a_{s(i')-2}\ge4$, then the subset
\begin{equation*}
    \begin{split}
      T=(S&\setminus\{v_{s(i')-2},\ldots,v_{s(i')+1},v_{s(i)-1},\ldots,v_{s(i)+1}\})\\
      &\cup\{\pi_{{i'},j,{j'}}(v_{s(i)\pm1}),\pi_{j,k}(v_{s(i)\mp1})\}\subset\ZZ^{n-5}
    \end{split}
\end{equation*}
is standard with $I(T)\le-1$. By Lemma~\ref{lem:hascontractions}, $T$ has as at least $n-10$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction. If $a_{s(i')-2}=3$ and $V_{s(i')-2}=\{k,l,m\}$ for some integer $m$, then $V_{s(i')-2}\cap V_{s(i)\mp1}=\{k,l,m\}$ and so $|v_{s(i')-2}\cdot v_{s(i)\mp1}|=1$. Moreover, $(v_{s(i')-2}\cdot e_k, v_{s(i')-2}\cdot e_l,v_{s(i')-2}\cdot e_m)=\pm(v_{s(i)\mp1}\cdot e_k, v_{s(i)\mp1}\cdot e_l,-v_{s(i)\mp1}\cdot e_m)$. Thus the subset 
\begin{equation*}
    \begin{split}
      G=(S&\setminus\{v_{s(i')-2},\ldots,v_{s(i')+1},v_{s(i)-1},\ldots,v_{s(i)+1}\})\\
      &\cup\{\pi_{k}(v_{s(i')-2}),\pi_{{i'},j,{j'}}(v_{s(i)\pm1}),\pi_{j,k}(v_{s(i)\mp1})\}\subset\ZZ^{n-5}
    \end{split}
\end{equation*}
is good. But then $G$ has $n-4$ linearly independent vectors in $\ZZ^{n-5}$, which is not possible.\\
 
\noindent\underline{Case 2: $a_{s(i')+1}\ge5$}.
In this case, the subset 
\begin{equation*}
    \begin{split}
      T=(S&\setminus\{v_{s(i')-1},\ldots,v_{s(i')+1},v_{s(i)-1},\ldots,v_{s(i)+1}\})\\
      &\cup\{\pi_{{i'},j,{j'}}(v_{s(i)\pm1}),\pi_{j}(v_{s(i)\mp1})\}\subset\ZZ^{n-4}
    \end{split}
\end{equation*}
is standard with $-3\le I(T)\le -1$. By Lemma~\ref{lem:hascontractions}, $T$ has as at least $n-9$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction.\\

\noindent\underline{Case 3: $a_{s(i')+1}=4$}.
First suppose $a_{s(i')-2}=2$ so that $V_{s(i')-2}=\{k,l\}$. Further assume that $|v_{s(i)\mp1}\cdot v_{s(i')-2}|=1$ so that $s(i)\mp1=s(i')-3$; then $E_l=\{s(i')-2,s(i')+1\}$. Since $a_{s(i')-3}\ge3$, the subset
\begin{equation*}
    \begin{split}
        T=(S&\setminus\{v_{s(i')-5},\ldots,v_{s(i')+1}\})\\
        &\cup\{\pi_{j}(v_{s(i')-3}),v_{s(i')-2},\pi_{{j'},l}(v_{s(i')+1})\}\subset\ZZ^{n-4}
    \end{split}
\end{equation*}
is standard with $p_1(T)=1$. By Lemma~\ref{lem:hascontractions}, $T$ has as at least $n-8$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction.
Thus we may assume $v_{s(i)\mp1}\cdot v_{s(i')-2}=0$ and so  $l\in V_{s(i')+1}\cap V_{s(i)\mp1}$ and $(v_{s(i')+1}\cdot e_k,v_{s(i')+1}\cdot e_l)=\pm(v_{s(i)\mp1}\cdot e_k,v_{s(i)\mp1}\cdot e_l)$. If $v_{s(i')+1}\cdot v_{s(i)\mp1}=0$, then $|V_{s(i)\mp1}\cap V_{s(i')+1}|\ge4$ and so $a_{s(i')+1}\ge5$, which is a contradiction. Thus $|v_{s(i)\mp1}\cdot v_{s(i')+1}|=1$ (i.e., $s(i)\mp1=s(i)-1=s(i')+2$) and $|V_{s(i')+1}\cap V_{s(i')+2}|=3$; hence $a_{s(i')+2}\ge4$. 
Now
\begin{equation*}
    \begin{split}
      C=(S&\setminus\{v_{s(i')-1},\ldots,v_{s(i')+4}\})\\
      &\cup\{\pi_{j,k}(v_{s(i')+2}),\pi_{{i'},j,c{j'}}(v_{s(i')+4})\}\subset\ZZ^{n-4}
    \end{split}
\end{equation*}
is cyclic with $p_1(C)=1$. By Lemma~\ref{lem:hascontractions}, $C$ has as at least $n-9$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction.

We have shown that $a_{s(i')-2}\ge3$. By the first argument in Case 1, we must have that $a_{s(i)\mp1}=3$. Let $V_{s(i)\mp1}=\{j,k,z\}$, where $z\in V_{s(i)\mp1}\cap V_{s(i)\pm1}$. 
If $|v_{s(i)\mp1}\cdot v_{s(i')-2}|=1$ so that $s(i)\mp1=s(i)+1=s(i')-3$, then $E_z=\{s(i')-5,s(i')-3,s(i')+1\}$. In this case, the subset 
\begin{equation*}
    \begin{split}
      G=(S&\setminus\{v_{s(i')-5},\ldots,v_{s(i')+1}\})\\
      &\cup\{\pi_{{i'},j,{j'}}(v_{s(i')-5}),\pi_{k}(v_{s(i')-2})\}\subset\ZZ^{n-5}
    \end{split}
\end{equation*}
is good with $E^G_{z}=\{s(i')-5\}$, where $\pi_{{i'},j,{j'}}(v_{s(i')-5})$ is a final vertex. But by~\cite{lisca-rational}, no such subset exists. 
If $|v_{s(i)\mp1}\cdot v_{s(i')+1}|=1$ so that $s(i)\mp1=s(i)-1=s(i')+2$, then the subset
\begin{equation*}
    \begin{split}
      C=(S&\setminus\{v_{s(i')-2},\ldots,v_{s(i')+4}\})\\
      &\cup\{\pi_{k}(v_{s(i')-2}),\pi_{{j'},k}(v_{s(i')+1}),\pi_{{i'},j,{j'}}(v_{s(i')+4})\}\subset\ZZ^{n-5}
    \end{split}
\end{equation*}
is cyclic and contains $n-4$ linearly independent vectors in $\ZZ^{n-5}$, which is not possible. 
Thus we may assume $v_{s(i)\mp1}\cdot v_{s(i')+1}=v_{s(i)\mp1} \cdot v_{s(i')-2} =0$. But then $z\in V_{s(i')-2}\cap V_{s(i')+1}$ and $(v_{s(i')-2}\cdot e_k,v_{s(i')-2}\cdot e_z)=\pm (v_{s(i')+1}\cdot e_k,v_{s(i')+1}\cdot e_z)$. Since $v_{s(i')+1}\cdot v_{s(i')-2}=0$ and $a_{s(i')+1}=4$, we necessarily have that $|v_{s(i')-2}\cdot e_l|=2$, $E_l=\{s(i')-2,s(i')+1,s(i')+2,s(i)\mp2\}$ and $E_z=\{s(i')-2,s(i')+1,s(i)\mp1,s(i)\mp2\}$. By Corollary~\ref{cor:jcor}, $|\mathcal{J}|=0$; hence $p_4(S)\ge p_2(S)+3$ and $|v_{s(i')-2}\cdot e_l|=2$. But this contradicts Lemma~\ref{lem:count}.
\end{proof}


\subsection{Upper Bound on $|\mathcal{I}|$}


Recall that by Lemma~\ref{lem:11}, either $a_{s(i)-1}\ge3$ or $a_{s(i)+1}\ge3$ for all $i\in\mathcal{I}$. We defined $m(i)\ge1$ to be the smallest integer such that $a_{s(i)\pm m(i)}\ge3$ when $a_{s(i)\mp1}\ge3$.
Moreover, there exist distinct integers $j(i)_1,\ldots,j(i)_{m(i)-1}\not\in\{i,j(i)\}$ such that $V_{s(i)\pm\alpha}=\{j(i)_{\alpha-1},j(i)_\alpha\}$ for $1\le\alpha\le m(i)-1$, where $j(i)_0:=j(i)$. 


Let $\mathcal{I}'\subset\mathcal{I}$ and define \begin{equation*}
\begin{split}
      G_{\mathcal{I}'}=(S&\setminus\{v_{s(i)\mp1},v_{s(i)},v_{s(i)\pm1},\ldots,v_{s(i)\pm m(i)}, v_{t(i)}\,|\,i\in\mathcal{I}'\})\\
      &\cup\{\pi_{{j(i)_0},\ldots,{j(i)_{m(i)-1}}}(v_{s(i)\mp1}),\pi_{{j(i)_{m(i)-1}}}(v_{s(i)\pm m(i)})\,|\, i\in\mathcal{I}'\}
\end{split}
\end{equation*}

Moreover, if $v_\alpha\in S\setminus\{v_{t(i)}\,|\,i\in\mathcal{I}'\}$, then denote the corresponding vertex in $G_{\mathcal{I}'}$ by $v_\alpha'$. That is, set $v_{s(i)\mp1}':=\pi_{{j(i)_0},\ldots,{j(i)_{m(i)-1}}}(v_{s(i)\mp1})$, $v_{s(i)\pm m(i)}':=\pi_{{j(i)_{m(i)-1}}}(v_{s(i)\pm m(i)})$, and $v_u':=v_u$ for all $u\not\in\{s(i)\mp1,s(i),s(i)\pm1,\ldots,s(i)\pm m(i),t(i)\,|\,i\in\mathcal{I}'\}$.

\begin{lem}
Let $C=\{v_{u_1}',\ldots,v_{u_k}'\}$ be a connected component of $G_{\mathcal{I}'}$. 
If $v_{u_1}'=v_{u_1}$ or $v_{u_k}=v_{u_k}'$, then $C$ is not a bad component.
\label{lem:notbad}
\end{lem}

\begin{proof}
Assume $C$ is a bad component. By the discussion following Definition \ref{def:badcomp}, there exists a unique vertex $v_{u_y}'$ of $C$ such that $|V_{u_y}^{G_{\mathcal{I}'}}\cap V^{G_{\mathcal{I}'}}_x|\neq0$ for some $x\not\in\{u_1,\ldots,u_k\}$; moreover, $v_{u_y}'$ is an internal vertex of $C$.
Up to reversal of $S$ we may assume that $v_{u_1}=v'_{u_1}$. Then $u_1-1=t(i)$ for some $i\in\mathcal{I}$ and by construction $i,j(i)\not\in V^S_{u_1}$.
Hence there exists an integer $k_1\in V^C_{u_1}\cap V^S_{t(i)}$.
By construction, either: $u_\alpha\notin\{s(i)\mp1,s(i)\pm m(i)\}$ for all $1\le \alpha\le k$; or there exists an integer $z$ such that $u_z=s(i)\mp1$ and $u_{z\pm1}=s(i)\pm m(i)$. \\

\noindent\underline{Case 1}. Assume $u_\alpha\notin\{s(i)\mp1,s(i)\pm m(i)\}$ for all $1\le \alpha\le k$. By the discussion following Definition \ref{def:badcomp}, $|E_{k_1}^{G_{\mathcal{I}'}}|=|E_{k_1}^C|\in\{2,3\}$ and there exists a vertex $v_{u_{m_1}}\neq v_{u_y}$ of $C$ such that $k_1\in V^C_{u_{m_1}}$. 
Since $v_{t(i)}\cdot v_{u_{m_1}}=0$, there exists an integer $k_2\in V_{t(i)}\cap V^C_{u_{m_1}}$ such that $(v_{u_{m_1}}\cdot e_{k_1},v_{u_{m_1}}\cdot e_{k_2})=\pm (v_{t(i)}\cdot e_{k_1},-v_{t(i)}\cdot e_{k_2})$.
Since $|E^{G_{\mathcal{I}'}}_{k_2}|=|E^C_{k_2}|\in\{2,3\}$, there exists a vertex $v_{u_{m_2}}$, where $u_{m_2}\not\in\{u_{m_1},u_y\}$, such that $k_2\in V_{u_{m_2}}$. By the discussion following Definition \ref{def:badcomp}, if $k_1\in V_{u_{m_2}}^C$, then $v_{u_{m_1}}\cdot v_{u_{m_2}}=0$, and so $(v_{u_{m_1}}\cdot e_{k_1},v_{u_{m_1}}\cdot e_{k_2})=\pm (v_{u_{m_2}}\cdot e_{k_1},-v_{u_{m_2}}\cdot e_{k_2})$. Since $v_{t(i)}\cdot v_{u_{m_2}}=0$, there exists an integer $k_3\in V^S_{t(i)}\cap V^C_{u_{m_2}}$ such that $k_3\neq k_1$ and $(v_{u_{m_2}}\cdot e_{k_2},v_{u_{m_2}}\cdot e_{k_3})=\pm (v_{t(i)}\cdot e_{k_2},-v_{t(i)}\cdot e_{k_3})$. Continuing inductively, we eventually see that either $|v_{t(i)}\cdot v_{u_1}|\neq1$ or $v_{t(i)}\cdot v_{u_x}\neq0$ for some integer $x\neq1$, which is a contradiction.  \\

\noindent\underline{Case 2}. Assume that there exists an integer $z$ such that $u_z=s(i)\mp1$ and $u_{z\pm1}=s(i)\pm m(i)$. Since $v_{u_z}\cdot v_{t(i)}=v_{u_{z\pm1}}\cdot v_{t(i)}=0$, there exist integers $l_1\in V^S_{u_z}\cap V^S_{t(i)}$ and $l_2\in V_{u_{z+1}}\cap V_{t(i)}$. 
Note that, by the discussion following Definition \ref{def:badcomp}, either $l_1\in V_{u_z}^C$ or $l_2\in V_{u_{z\pm1}}^C$.
First suppose $l_1\in V_{u_z}^C$ and $l_2\in V_{u_{z\pm1}}^C$. If $l_1\neq l_2$, then arguing as in Case 1 will lead to a contradiction. Assume $l_1=l_2$ and set $l:=l_1=l_2.$ Then $|E^C_{l}|=3$; set $E_l^C=\{u_z,u_{z\mp1},u_t\}$. If $t\neq 1$, then arguing as in Case 1 will once again lead to a contradiction. Suppose $t=1$. Since $(v_{u_z}\cdot e_j,v_{u_z}\cdot e_l)=\pm(v_{u_{z\pm1}}\cdot e_j,-v_{u_{z\pm1}}\cdot e_l)$, it follows that there exist integers $l_3$ and $l_4$ in either $V_{t(i)}\cap V_{u_z}^C$ or $V_{t(i)}\cap V_{u_{z\pm1}}^C$; assume the former (the latter is analogous). We can argue towards a contradiction as in Case 1 unless $z=y$ and $l_3,l_4\not\in V^S_y\setminus V^C_y$. Then it follows from the discussion following Definition \ref{def:badcomp} that $E^C_l=\{z\mp1,z,z\pm1\}$; hence $z\mp1=t=1$. It follows that $C$ must be of the form 
$$\{e_1+\ldots+e_{k-1},-e_{k-1}+a,e_{k-1}-e_{k-2},\ldots,e_{2}-e_1\},$$
where $l=k$, and $a$ is some vector with $a\cdot e_{l_3},a\cdot e_{l_4}\neq 0$. By construction, there exists an integer $i'\neq i$ in $\mathcal{I}$ such that $|v_{t(i')}\cdot v_{u_k}|=1$; hence either $1\in V_{t(i'')}$ or $2\in V_{t(i'')}$. In either case, we may again argue as in Case 1 and arrive at a contradiction.

Next suppose that $l_1\not\in V_{u_z}^C$ (an analogous argument handles the case $l_2\not\in V_{u_{z\pm1}}^C$). Then $u_z=u_y$. If $|E_{l_2}|=3$, then there exists a vertex $v_{u_m}$ such that $l_2\in V_{u_m}^C$. Arguing as in Case 1 yields a contradiction. If $|E_{l_2}|=2$, it follows that $u_{z\pm1}=k$ and $E^C_{l_2}=\{1,k\}$. Now by construction, there exists an integer $i'\neq i$ in $\mathcal{I}$ such that $|v_{t(i')}\cdot v_{u_k}|=1$; hence there exists an integer $p\in V^S_{t(i)}\cap V^C_{u_k}$. Arguing as above, we arrive at a contradiction.
\end{proof}


\begin{lem} $G_{\mathcal{I}'}$ is good for all subsets $\mathcal{I}'\subseteq\mathcal{I}$.\label{lem:Ggood}
\end{lem}

\begin{proof} Set $G:=G_{\mathcal{I}'}$.
By Lemma~\ref{lem:12}, every element of $G$ has square at most $-2$. Thus we need only show that $G$ is irreducible. Suppose $G$ is reducible; then $G=G_1\cup\ldots\cup G_k$, where $k\ge2$ and $G_1,\ldots,G_k$ are good subsets such that for all $1\le\alpha\neq\beta\le k$, $G_\alpha\cup G_\beta$ is reducible. Note that each $G_\alpha$ has length at least two since no good subsets of length one exist. Let $G_2'=G_2\cup \cdots \cup G_k$ and let $i_1,i_2\in\mathcal{I}$, $t_1:=t(i_1)$ and $t_2:=t(i_2)$, such that $v_{t_1}$ and $v_{t_2}$ were the vectors removed from $S$ to form $G_1$, bounding $G_1$ on either side. 
By Lemma~\ref{lem:38}, $a_{t_1},a_{t_2}\le6$. Let $G_1=\{v'_{z_1},\ldots,v'_{z_m}\}$ and $G_2'=\{v'_{y_1},\ldots,v'_{y_t}\}$. We claim that $|V^S_{z_1}\cap V^S_{t_1}|\neq0$. If $t_1\not\in\{s(i)\mp1,s(i)\pm m(i)\,|\, i\in\mathcal{I}\}$, then $z_1=t_1+1$ and so it is clear that $|V^S_{z_1}\cap V^S_{t_1}|\neq0$. If $t_1=s(i)\mp1$ for some $i\in\mathcal{I}$, then $z_1=s(i)\pm m(i)$; hence by Lemma~\ref{lem:t=s+m}, $j_{m(i)-1}\in V^S_{t_1}\cap V^S_{z_1}$. Similarly, if $t_1=s(i)\pm m(i)$, then $z_1=s(i)\mp1$ and so $|V^S_{z_1}\cap V^S_{t_1}|\neq0$. By a similar argument, it follows that $|V^S_{y_t}\cap V^S_{t_1}|\neq0$, $|V^S_{z_m}\cap V^S_{t_2}|\neq0$, and $|V^S_{y_1}\cap V^S_{t_2}|\neq0$. Let $k\in V^S_{z_1}\cap V^S_{t_1}$ and $a\in V^S_{y_t}\cap V^S_{t_1}$. Then since $G$ is reducible, $v'_{y_\alpha}\cdot e_k=v'_{z_\alpha}\cdot e_a=0$ for all $v'_{y_\alpha}\in G'_2$ and $v'_{z_\alpha}\in G_1$. If $v'_\alpha\cdot e_k=0$ for all $v'_\alpha\in G\setminus\{v'_{z_1}\}$, then $E_k^{G_1}=\{z_1\}$. By Lemma \ref{lem:lisca3.2}, no such subset exists since $v'_{z_1}$ is a final vertex of $G_1$ and $G_1$ is good. Thus there exists an integer $p\neq1$ such that $v_{z_p}\cdot e_k\neq 0$ and consequently, there exists $k_1\in V_{z_p}\cap V_{t_1}$. Similarly, there exists an integer $a_1\in V_{t_1}\cap V_{y_q}$ for some integer $q$. Thus $a_{t_1}=6$.
Similarly, $a_{t_2}=6$.

Let $\mathcal{\widehat{I}}=\{i_1,i_2\}$. Then by Lemma~\ref{lem:t=s+m}, $\widehat{G}:=G_{\mathcal{\widehat{I}}}$
has exactly two connected components, and $I(\widehat{G})\le -2$. Moreover, by Lemma~\ref{lem:notbad}, $\widehat{G}$ has no bad components. Since $a_{t_1},a_{t_2}=6$, it is easy to see that $m(i_1),m(i_2)\le 2$ and so $\widehat{G}$ has length at least $n-6$. If $\widehat{G}$ is irreducible (and hence good), then by Lemma~\ref{lem:hascontractions}, $G$ has at least $n-11$ contractible adjacencies; following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction.
Assume $\widehat{G}$ is reducible; then $\widehat{G}=\widehat{G}_1\cup \widehat{G}_2$, where $\widehat{G}_1$ and $\widehat{G}_2$ are standard subsets. 
By \cite{lisca-rational}, $I(\widehat{G}_1),I(\widehat{G}_2)\ge-3$. Since $I(\widehat{G})=I(\widehat{G}_1)+I(\widehat{G}_2)\le -2$, we may assume $I(\widehat{G}_1)\le -1$.


If $|E^{\widehat{G}_1}_{k_1}|=1$, then by Lemma \ref{lem:lisca3.2}, we would have $E^{\widehat{G}_1}_k=\{z_{p-1},z_p,z_{p+1}\}$; but then $v_{t_1}\cdot v_{z_{p+1}}\neq 0$, which is a contradiction. Thus there exists an integer $r\not\in \{z_1,z_p\}$ such that $k_1\in V^S_{z_r}$. Since $v_{t_1}\cdot v_{z_r}=0$, $k\in V_{z_r}^S$ and $(v_{z_p}\cdot e_k,v_{z_p}\cdot e_{k_1})=\pm (v_{z_r}\cdot e_k, v_{z_r}\cdot e_{k_1})$. It follows that if $|v_{z_p}\cdot v_{z_r}|=1$, then $|V^{\widehat{G}_1}_{z_p}\cap V^{\widehat{G}_1}_{z_r}|\ge3$, and if $|v_{z_p}\cdot v_{z_r}|=0$, then $|V^{\widehat{G}_1}_{z_p}\cap V^{\widehat{G}_1}_{z_r}|\ge4$. But by the classification of standard subsets with $I<0$ (c.f. Proposition 5.10 in~\cite{simone2020classification}), no such subset exists.
\end{proof}



\begin{lem}
Let $i_1\in\mathcal{I}$. If there exists an integer $i_2\in\mathcal{I}$ such that $t(i_1)=s(i_2)\pm1$, then $|\mathcal{I}_2|\le2$.
\label{lem:103}
\end{lem}


\begin{proof}
Up to reversal of $S$, we may assume that $t(i_1)=s(i_2)-1$. By Proposition~\ref{prop:59}, $a_{t(i_1)}\ge5$. Assume $|\mathcal{I}_2|\ge3$ and let $i_3\in\mathcal{I}_2\setminus\{i_1,i_2\}$. Note that by Proposition~\ref{prop:59}, if $|v_{t(i_\alpha)}\cdot v_{s(i_\beta)}|=1$ for any $\alpha,\beta\in\{1,2,3\}$, then $a_{t(i_\alpha)}\ge5$. Let $\mathcal{I}'=\{i_1,i_2,i_3\}$. Then by Lemma~\ref{lem:Ggood}, $G_{\mathcal{I}'}$ is good with $I(G_{\mathcal{I}'})\le -2$ and at most 3 components.
Moreover, in light of Lemma~\ref{lem:notbad}, it is clear that $G_{\mathcal{I}'}$ has at most one bad component. If $G_{\mathcal{I}'}$ has no bad components, then by Corollary 5.4 in~\cite{lisca-rational}, $I(G_{\mathcal{I}'})\ge-3$, implying that $G_{\mathcal{I}'}$ has at least $n-7$ vertices.  By Lemma~\ref{lem:hascontractions}, $G_{\mathcal{I}'}$ has as at least $n-12$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction.  

We may now assume that $b(G_{\mathcal{I}'})=1$. Since $I(G_{\mathcal{I}'})+b(G_{\mathcal{I}'})\le -1$, by Proposition~4.10 in~\cite{lisca-sumsoflensspaces} and Lemma~5.1 in~\cite{lisca-rational}, $G_{\mathcal{I}'}$ has two components. Let $G'$ be the good subset obtained from $G_{\mathcal{I}'}$ be replacing the bad component with its core. Then $I(G')\le-1$ and has no bad components. Once again by Corollary~5.4 and Lemma~5.1 in~\cite{lisca-rational}, $I(G')\ge -3$ and so $G'$ has at least $n-7$ vertices.  By Lemma~\ref{lem:hascontractions}, $G'$ has as at least $n-12$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction.
\end{proof}


\begin{prop} $|\mathcal{I}_1|,|\mathcal{I}_2|\le 2$.
\label{prop:|I|}
\end{prop}

\begin{proof}
Suppose $|\mathcal{I}_1|\ge3$. It follows from Lemma~\ref{lem:38} that $p_5(S)\ge5$. In light of Corollary~\ref{cor:jcor}, this is not possible by Lemma~\ref{lem:count}.

By Lemma~\ref{lem:103}, we may assume that for all $i, i'\in\mathcal{I}_2$, we have $v_{t(i)}\cdot v_{s(i')}=0$. 
Assume $|\mathcal{I}_2|\ge3$ and let $\mathcal{I}'=\{i_1,i_2,i_3\}\subseteq\mathcal{I}_2$ be distinct. Set $G:=G_{\mathcal{I}'}$. By Lemma~\ref{lem:Ggood}, $G$ is good and by Lemma~\ref{lem:notbad}, $G$ has no bad components. By Proposition~\ref{prop:59}, $I(G)\le 4-2|\mathcal{I}_2|\le-2$. By Corollary 5.4 in~\cite{lisca-rational}, $I(G)\ge -3$, implying that $|\mathcal{I}_2|=3$ and $G$ has at least $n-6$ vectors.  By Lemma~\ref{lem:hascontractions}, $G$ has as at least $n-11$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction.
\end{proof}



\subsection{Rigid Adjacencies}

Recall that an adjacency is called \textit{contractible} if there exists an integer $k$ such that $E_k=\{x,x+1,y\}$, where $a_y\ge3$ and $v_x\cdot e_k=-v_{x+1}\cdot e_k$ if $v_x\cdot v_{x+1}=1$ and $v_x\cdot e_k=v_{x+1}\cdot e_k$ if $v_x\cdot v_{x+1}=-1$, and otherwise it is called \emph{rigid}. In our setting, since $S$ has no contractions, every adjacency of $S$ is rigid.

We say that an integer $p$ \emph{contributes a rigid adjacency} $(v_x,v_{x+1})$ if $x,x+1\in E_p$. Notice that each integer $p$ satisfying $|E_p|=m\ge4$ can contribute at most $m-1$ rigid adjacencies; this maximum occurs, for example, when $E_p=\{x,x+1,\ldots,x+m-1\}$ for some integer $x$ and $V_{x+\alpha}\cap V_{x+\alpha+1}=\{p\}$ for all $0\le\alpha\le m-2$. By Lemma~\ref{lem:s,s+1}, an integer $p$ satisfying $|E_p|=2$ cannot contribute a rigid adjacency. Finally, if an integer $p$ satisfying $|E_p|=3$ contributes a rigid adjacency, then there are a few possibilities to consider.

\begin{lem} Let $p$ be an integer satisfying $|E_p|=3$ that contributes a rigid adjacency $(v_x,v_{x+1})$. Then there exists an integer $k$ satisfying $|E_k|\ge 4$ contributing the same rigid adjacency $(v_x,v_{x+1})$.
\label{lem:nonCA}
\end{lem}

\begin{proof}
By definition, $E_p=\{x,x+1,u\}$ for some integer $u$. Note that either $u\neq x-1$ or $u\neq x+2$. By cyclic reordering, we may assume that $v_x\cdot v_{x+1}=v_{x+1}\cdot v_{x+2}=1$. Since $(v_{x},v_{x+1})$ is a rigid adjacency, by Proposition \ref{prop:bounds}, one of the following holds: 
\begin{itemize}
    \item[1.] $v_x\cdot e_p=-v_{x+1}\cdot e_p=\pm1$ and either $|v_u\cdot e_p|=2$ or $a_u=2$;
    \item[2.] $v_x\cdot e_p=v_{x+1}\cdot e_p=\pm1$; or
    \item[3.] $|v_x\cdot e_p|=2$ or $|v_{x+1}\cdot e_p|=2$.\\
\end{itemize}

\noindent\underline{Case 1}. 
Up to reversal of $S$, we may assume that $u\neq x-1$.
First suppose $|v_u\cdot e_p|=2$. Then by Proposition \ref{prop:bounds} $|V_u\cap V_{x}|\ge3$; hence $a_u\ge 6$ and $|V_x|\ge3$. If $a_{x+1}\ge3$, then the subset 
$$G=(S\setminus\{v_x,v_{x+1},v_u\})\cup\{\pi_p(v_x),\pi_p(v_{x+1})\}\subset\ZZ^{n-1}$$
is good with $I(G)\le -1$. 
By Lemma~\ref{lem:hascontractions}, $G$ has as at least $n-6$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction. Thus we may assume that $a_{x+1}=2$. Let $V_{x+1}=\{p,k\}$ for some integer $k$. If $u\neq x+2$, then $E_k=\{x+1,x+2,u\}$, $v_{x+1}\cdot e_k=-v_{x+2}\cdot e_k$, and $|v_u\cdot e_k|=1$; hence $S$ has a contraction, which is a contradiction. If $u=x+2$, then $E_k=\{x+1,x+2\}$, which contradicts Lemma \ref{lem:s,s+1}. 

Next suppose $a_u=2$ so that $V_u=\{p,k\}$ for some integer $k$. If $u=x+2$, then it is clear that $k\in V_{x}$ and $k\in V_{x+3}$. Moreover, since there are no contractions, we have that $v_{x+1}\cdot e_p=v_{x+2}\cdot e_p$. Consequently, $k\in V_{x+1}$ and so $E_k=\{x,x+1,x+2,x+3\}$, implying that $k$ contributes the rigid adjacency $(v_x,v_{x+1})$. If $u\neq x+2$, it follows that $E_k=\{x,x+1,u-1,u,u+1\}$. Thus $k$ contributes the rigid adjacency $(v_x,v_{x+1}).$\\

\noindent\underline{Case 2}.
Up to reversal of $S$ we may once again assume that $u\neq x-1$.
Since $v_x\cdot e_p=v_{x+1}\cdot e_p$ and $v_x\cdot v_{x+1}=1$, there is an integer $l\in V_x\cap V_{x+1}$ such that $v_x\cdot e_l$ and $v_{x+1}\cdot e_l$ have opposite sign.
If $|E_l|\ge4$, then we are done. Thus we now assume that $|E_l|=3$ and set $E_l=\{x,x+1,t\}$.
Notice that if $|V_{x+1}|=2$, then $V_{x+1}=\{p,l\}$ and since $u\neq x+2$, we have $E_l=\{x,x+1,x+2,u\}$, which is a contradiction. Thus $|V_{x+1}|\ge3$; similarly, $|V_x|\ge3$. By Proposition~\ref{prop:bounds}, $|v_x\cdot e_l|\in\{1,2\}$. 

Assume first that $|v_x\cdot e_l|=2$; then $|v_{x+1}\cdot e_l|=1$. Suppose $a_t=2$ and $V_t=\{l,k\}$ for some integer $k$. If $u=x+2$, then it follows that $t=x-1$ and $E_k=\{x-2,x-1,x,x+1\}$; hence $k$ contributes to the rigid adjacency $(v_x,v_{x+1})$. On the other hand, if $u \neq x+2$, then by Proposition~\ref{prop:bounds}, $t\neq x+2$ and so $p\in V_t$, which contradicts that $V_t = \{ l,k \}$.
Next suppose $a_t\ge 3$; in this case, the subset 
$$T=(S\setminus\{v_x, v_{x+1},v_t\})\cup\{\pi_{l}(v_x),\pi_l(v_{x+1})\}\subset\ZZ^{n-1}$$
is standard with $I(T)\le -1$. 
By Lemma~\ref{lem:hascontractions}, $T$ has as at least $n-6$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction.

Thus we may assume that $|v_x\cdot e_l|=1$ and similarly, $|v_{x+1}\cdot e_l|=1$. Since $v_x\cdot e_l$ and $v_{x+1}\cdot e_l$ have opposite sign and $(v_x,v_{x+1})$ is a rigid adjacency, we necessarily have that  $|v_t\cdot e_l|=2$ or $a_t=2$ and $t\not\in\{x-1,x+2\}$. Hence by Case~1, there exists an integer $k\in V_x\cap V_{x+1}$ such that $|E_k|=5$. \\


\noindent\underline{Case 3}. 
Up to reversal of $S$, we may assume that $|v_x\cdot e_p|=2$. By Proposition~\ref{prop:bounds}, $|v_{x+1}\cdot e_p|=1$.
Assume $v_x\cdot e_p$ and $v_{x+1}\cdot e_p$ have the same sign. Then by Proposition~\ref{prop:bounds}, there exists an integer $k\in V_x\cap V_{x+1}$ such that $v_x\cdot e_{k}=-v_{x+1}\cdot e_{k}=\pm1$. By Lemma~\ref{lem:s,s+1}, $|E_{k}|\ge3$. If $|E_k|\ge4$, then we are done. If $|E_k|=3$, then by Case~2, we are done.

Next assume that $v_x\cdot e_p$ and $v_{x+1}\cdot e_p$ have opposite sign. 
Then there exists an integer $k\in V_x\cap V_{x+1}$ such that $v_x\cdot e_k=v_{x+1}\cdot e_k$. By Lemma \ref{lem:s,s+1}, $|E_k|\ge3$. If $|E_k|\ge4$, then we are done; hence we may assume that $|E_k|=3$. If $|V_x|=2$, then  $u= x-1$, since otherwise $E_k=\{x-1,x,x+1,u\}$. It is clear that $(v_{x-1}\cdot e_p,v_{x-1}\cdot e_k)=(v_{x+1}\cdot e_p,v_{x+1}\cdot e_k)$; hence $|V_{x-1}\cap V_{x+1}|\ge4$. 
The subset 
$$T=(S\setminus\{v_{x-1},v_x,v_{x+1}\})\cup\{\pi_{k,p}(v_{x-1})\}\subset\ZZ^{n-2}$$
is standard with $I(T)\le -1$. By Lemma~\ref{lem:hascontractions}, $T$ has as at least $n-7$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction. We may thus assume that $|V_x|\ge3$.

If $|V_{x+1}|\ge3$ and $|V_u|\ge 3$, then the subset 
$$T=(S\setminus\{v_x,v_{x+1},v_u\})\cup\{\pi_p(v_x),\pi_p(v_{x+1})\}\subset\ZZ^{n-1}$$
is standard with $I(T)\le -1$.
By Lemma~\ref{lem:hascontractions}, $T$ has as at least $n-6$ contractible adjacencies. Following as in Example~\ref{ex:hascontractions}, we see that $S$ also has contractible adjacencies, which is a contradiction.
Thus either $|V_{x+1}|=2$ or $|V_u|=2$.
First assume that $|V_{x+1}|=2$ so that $V_{x+1}=\{k,p\}$. If $u=x+2$, then it follows from Proposition~\ref{prop:bounds} that $E_k=\{x,x+1\}$, which is a contradiction. If $u\neq x+2$, then $E_k=\{x,x+1,x+2,u\}$, which is again a contradiction.
Next assume $a_{u}=2$ and set $V_u=\{p,l\}$ for some integer $l$. It follows that $u=x-1$ and $E_l=\{x-2,x-1,x,x+1\}$; hence $l$ contributes to the rigid adjacency $(v_x,v_{x+1})$. 
\end{proof}

In light of Lemma~\ref{lem:nonCA}, for every rigid adjacency $(v_x,v_{x+1})$, there exists an integer $k$ such that $|E_k|\ge4$ and $x,x+1\in E_k$. This observation will be used in the next two results. 
By assumption, all of the adjacencies of the subset $S$ are rigid. 

\begin{lem} $|v_\alpha\cdot e_\beta|\le1$ for all $\alpha,\beta$.
\label{lem:coeffic1}
\end{lem}

\begin{proof}
Assume there exists a pair $(\alpha,\beta)$ such that $|v_{\alpha}\cdot e_\beta|>1$. By Proposition~\ref{prop:bounds}, $(\alpha,\beta)$ is the only such pair and $|v_{\alpha}\cdot e_\beta|=2$. By Corollary~\ref{cor:jcor}, $|\mathcal{J}|\le1$. Moreover, by Lemmas~\ref{lem:count} and~\ref{lem:38}, $|\mathcal{I}_2|=0$ and so by Proposition~\ref{prop:|I|}, $|\mathcal{I}|\le2$.

First assume $|\mathcal{J}|=0$. Then $p_2(S)=|\mathcal{I}|\le 2$ and $p_4(S)\ge p_2(S)$. Thus by Lemma~\ref{lem:count}, $p_j(S)=0$ for all $j\ge5$ and $p_4(S)=p_2(S)+1\le 3$. By the discussion preceding the statement of this lemma, we know that there are at most $3(3)=9$ rigid adjacencies, which is a contradiction.

Next assume $|\mathcal{J}|=1$ and set $\mathcal{J}=\{i\}$. Then by Corollary~\ref{cor:jcor}, $p_4(S)=p_2(S)+1\le4$, $p_j(S)=0$ for all $j\ge5$, and $|E_{j(i)}|=|E_{k(i)}|=4$. Since $j(i)$ and $k(i)$ contribute at most three rigid adjacencies each,
$S$ has at most $2(3)+3+3=12$ rigid adjacencies, which is a contradiction.
\end{proof}


\begin{lem} If $i\in\mathcal{I}_1$, then together $j(i)$ and $k(i)$ contribute at most 5 rigid adjacencies.
\label{lem:I1contribution}
\end{lem}

\begin{proof}
By Lemma~\ref{lem:38}, $t(i)\neq s(i)\pm2$ and $E_{k(i)}=\{s(i)-1,s(i)+1,t(i)-1,t(i),t(i)+1\}$. Since $E_{j(i)}=\{s(i)-1,s(i),s(i)+1,t(i)\}$, $j(i)$ contributes only the rigid adjacencies $(v_{s(i)-1},v_{s(i)})$ and $(v_{s(i)},v_{s(i)+1})$. Similarly, $k(i)$ contributes the rigid adjacencies $(v_{t(i)-1},v_{t(i)})$ and $(v_{t(i)},v_{t(i)+1})$. Up to reversal, we may assume that $v_{t(i)}\cdot v_{s(i)+2}=0$; hence $t(i)-1\neq s(i)+2$. If $|v_{t(i)}\cdot v_{s(i)-2}|=1$, then $t(i)+1=s(i)-2$ and $k(i)$ also contributes the rigid adjacency $(v_{t(i)+1},v_{s(i)-1})$. Hence $j(i)$ and $k(i)$ contribute at most 5 rigid adjacencies.
\end{proof}

\begin{lem} If $i\in\mathcal{J}$, then together $j(i)$ and $k(i)$ contribute at most $\frac{|E_{j(i)}|+|E_{k(i)}|+1}{2}$ rigid adjacencies. Moreover, if $i_1,i_2\in\mathcal{J}$ and $k:=k(i_1)=k(i_2)$, then together $j(i_1),j(i_2)$ and $k$ contribute at most $\frac{|E_{j(i_1)}|+|E_{j(i_2)}|+|E_k|+7}{3}$ rigid adjacencies.
\label{lem:Jcontribution}
\end{lem}

\begin{proof}
Set $j:=j(i)$, $k:=k(i)$, $s:=s(i)$, and $t:=t(i)$. By Lemmas~\ref{lem:Ej>3} and~\ref{lem:at=3}, $|E_j|\ge4$ and by Lemma~\ref{lem:8}, $|E_k|\ge3$. 
By Lemma~\ref{lem:coeffic1}, $|v_\alpha\cdot e_\beta|\le1$ for all integers $\alpha$ and $\beta$. Consequently, $k\in V_{s\pm1}$ if and only if $j\not\in V_{s\pm1}$, and $k\in V_u$ for some $u\not\in\{s-1,s+1,t\}$ if and only if $j\in V_u$. 
Hence there are exactly three integers $\alpha\in E_j\cup E_k$ such that $|V_\alpha\cap \{j,k\}|=1$, namely $s-1$, $s+1$, and $t$; all other integers $\alpha\in E_j\cup E_k$ satisfy $j,k\in V_\alpha\cap \{j,k\}$.
Consequently, $|E_j\cup E_k|$ contains exactly $\frac{|E_j|+|E_k|-3}{2}+3=\frac{|E_j|+|E_k|+3}{2}$ distinct integers. Since $|E_j|,|E_k|\le 7$ by Corollary~\ref{cor:jcor}, $n\ge22$ is strictly greater than the number of distinct integers in $|E_j\cup E_k|$. Consequently, $j$ and $k$ contribute at most $\frac{|E_j|+|E_k|+3}{2}-1=\frac{|E_j|+|E_k|+1}{2}\le 7$ rigid adjacencies.

Suppose $i_1,i_2\in\mathcal{J}$ and $k:=k(i_1)=k(i_2)$. By Lemma~\ref{lem:9}, $j(i_1)\neq j(i_2)$ and so $|v_{s(i_1)}\cdot v_{s(i_2)}|=1$. Up to reversal of $S$, we may assume $s(i_2)=s(i_1)+1$. 
There are exactly: four integers $\alpha\in E_{j(i_1)}\cup E_{j(i_2)}\cup E_k$ such that $|V_{\alpha}\cap\{j(i_1),j(i_2),k\}|=1$, namely $s(i_1)-1,s(i_2)+1,t(i_1),$ and $t(i_2)$; two integers $\alpha\in E_{j(i_1)}\cup E_{j(i_2)}\cup E_k$ such that $|V_{\alpha}\cap\{j(i_1),j(i_2),k\}|=2$, namely $s(i_1)$ and $s(i_2)$; and all other integers in $E_{j(i_1)}\cup E_{j(i_2)}\cup E_k$ satisfy $|V_{\alpha}\cap\{j(i_1),j(i_2),k\}|=3$. 
Hence $E_{j(i_1)}\cup E_{j(i_2)}\cup E_k$ contains exactly $$\frac{|E_{j(i_1)}|+ |E_{j(i_2)}|+|E_k|-8}{3}+\frac{4}{2}+\frac{4}{1}=\frac{|E_{j(i_1)}|+ |E_{j(i_2)}|+|E_k|+10}{3}$$ distinct integers. As a result, $j(i_1),$ $j(i_2)$, and $k$ contribute at most $\frac{|E_{j(i_1)}|+ |E_{j(i_2)}|+|E_k|+7}{3}$ rigid adjacencies.
\end{proof}


We are now ready to prove Theorem \ref{thm:nocontraction}, which we recall here for convenience.

\hascontractions*



\begin{proof}
Let $n\ge 22$. We will first show that $S$ necessarily has a contraction. Assume the opposite; then, by definition, $S$ has at least 22 rigid adjacencies. 
By Propositions \ref{prop:|I|} and~\ref{prop:J<2}, $|\mathcal{I}_1|,|\mathcal{I}_2|\le2$ and $|\mathcal{J}|\le2$. By Lemma~\ref{lem:coeffic1}, $|v_\alpha\cdot e_\beta|\le 1$ for all $\alpha$ and $\beta$. \\

\noindent\underline{Case 1: $|\mathcal{J}|=2$}.
Set $\mathcal{J}=\{i_1,i_2\}$. Moreover, set $j_1:=j(i_1),$ $j_2:=j(i_2)$, $k_1:=k(i_1),$ and $k_2:=k(i_2)$
By Lemma~\ref{lem:9}, $j_1\neq j_2$ and by Lemmas~\ref{lem:Ej>3} and~\ref{lem:at=3}, $|E_{j_1}|,|E_{j_2}|\ge4$. Moreover, by Lemma~\ref{lem:8}, $|E_{k_1}|,|E_{k_2}|\ge3$.

Suppose $k_1=k_2$ and set $k:=k_1=k_2$.
Then $|v_{s_1}\cdot v_{s_2}|=1$. Up to reversal of $S$, we may assume that $s_2=s_1+1$. It follows that $|E_k|\neq3$, since otherwise we would be able to contract along $e_k$. Hence $|E_k|\ge4$. Moreover, if $k\in V_{s_1-1}$, then $|E_{j_1}|=|E_k|-2$, and if $j_1\in V_{s_1-1}$, then $|E_{j_1}|=|E_k|+1$. Similarly, if $k\in V_{s_2+1}$, then $|E_{j_2}|=|E_k|-2$, and if $j_2\in V_{s_2+1}$, then $|E_{j_2}|=|E_k|+1$.
Thus if $|E_k|\ge7$, then $|E_{j_1}|,|E_{j_2}|\ge5$, which contradicts Lemma~\ref{lem:count}. 
Moreover, if $|E_k|=5$ then we necessarily have that $|E_{j_1}|=|E_{j_2}|=6$ (since $|E_{j_1}|,|E_{j_2}|\ge4$), which contradicts Lemma~\ref{lem:count}. Thus we may assume $|E_k|\in\{4,6\}$.

First assume $|E_k|=4$; then $|E_{j_1}|=|E_{j_2}|=5$. Hence by Lemma~\ref{lem:Jcontribution}, $j_1,j_2,$ and $k$ contribute at most 7 rigid adjacencies. It follows from Lemmas~\ref{lem:count} and~\ref{lem:38} that $|\mathcal{I}_1|=0$; thus $|\mathcal{I}|\le2$ and so by Lemma~\ref{lem:count}, $p_4(S)=p_2(S)\le4$, implying that $S$ has at most $7+3(3)=16$ rigid adjacencies. 
Next assume that $|E_k|=6$. Then by Lemma~\ref{lem:count}, we necessarily have that $|E_{j_1}|=|E_{j_2}|=4$. Once again by Lemma~\ref{lem:Jcontribution}, $j_1,j_2$, and $k$ contribute at most 7 rigid adjacencies, and by Lemmas~\ref{lem:count} and~\ref{lem:38}, $|\mathcal{I}_1|=0$; hence by Lemma~\ref{lem:count}, $p_4(S)=p_2(S)+1\le 5$, implying that $S$ has at most $7+3(3)=16$ rigid adjacencies, which is a contradiction.

Next suppose $k_1\neq k_2$. It follows from Lemmas~\ref{lem:count} and~\ref{lem:38} that $|\mathcal{I}_1|=0$. By Lemma~\ref{lem:Jcontribution}, $j_1$, $j_2$, $k_1$ and $k_2$ contribute at most $\frac{|E_{j_1}|+|E_{j_2}|+|E_{k_1}|+|E_{k_2}|+2}{2}$ rigid adjacencies. 
Since $|E_{j_1}|,|E_{j_2}|\ge4$, it follows from Lemmas~\ref{lem:count}, \ref{lem:8}, and~\ref{lem:j,k}, that the maximum value of this expression is 10 and that $p_4(S)=p_2(S)\le4$. 
Thus $S$ has at most $10+2(3)=16$ rigid adjacencies, which is again a contradiction.\\

\noindent\underline{Case 2: $|\mathcal{J}|=1$}. 
Set $\mathcal{J}=\{i\}$.
By Lemmas~\ref{lem:count} and~\ref{lem:38} and Corollary~\ref{cor:jcor}, $|\mathcal{I}_1|\le1$.
First assume $|\mathcal{I}_1|=1$ and set $\mathcal{I}_1=\{i_1\}$. By Lemma~\ref{lem:count}, Lemma~\ref{lem:38}, and Corollary~\ref{cor:jcor}, $p_5(S)=2$ and $p_4(S)=p_2(S)\le 4$. Moreover by Lemmas~\ref{lem:Ej>3} and~\ref{lem:at=3}, and Lemma~\ref{cor:ej=6}, $(|E_{j(i)}|,|E_{k(i)}|)\in\{(4,5),(5,4)\}$. Hence by Lemma~\ref{lem:Jcontribution}, $j(i)$ and $k(i)$ contribute at most 5 rigid adjacencies, and by Lemma~\ref{lem:I1contribution}, $j(i_1)$ and $k(i_1)$ contribute at most 5 rigid adjacencies. Consequently, there are at most $5+5+2(3)=16$ rigid adjacencies, which is a contradiction. 
Next assume $|\mathcal{I}_1|=0$. Now arguing as in Case 1, we see that $S$ has at most 17 rigid adjacencies, which is a contradiction.\\

\noindent\underline{Case 3: $|\mathcal{J}|=0$}. 
Note that by Lemma~\ref{lem:count}, $p_5(S)\le2$.
By Proposition~\ref{prop:|I|}, $|\mathcal{I}_1|,|\mathcal{I}_2|\le2$. 
First suppose $|\mathcal{I}_1|=2$; then by Lemmas~\ref{lem:count} and~\ref{lem:38}, $p_5(S)=2$ and $p_4(S)=p_2(S)\le 4$. By Lemma~\ref{lem:I1contribution}, for each $i\in\mathcal{I}_1$, $j(i)$ and $k(i)$ contribute at most 5 rigid adjacencies. Thus there are at most $2(5)+2(3)=16$ rigid adjacencies, which is a contradiction.
Next suppose $|\mathcal{I}_1|=1$; by Lemma~\ref{lem:38}, $p_5(S)\ge1$. If $p_5(S)=2$, then by Lemma~\ref{lem:count}, $p_4(S)=p_2(S)\le 3$; hence there are at most $5+4+2(3)=15$ rigid adjacencies, which is a contradiction. If $p_5(S)=1$, then $p_4(S)=p_2(S)+2\le5$; hence there are $5+4(3)=17$ rigid adjacencies, which is a contradiction.
Finally suppose $|\mathcal{I}_1|=0$; then $0\le p_5(S)\le2$. Arguing as above, it is routine to see that the number of rigid adjacencies is maximized when $p_j(S)=0$ for all $j\ge5$ and $p_4(S)=6$. In this case, we have $6(3)=18$ rigid adjacencies, which is a contradiction.\qedhere
\end{proof}









\section{Proof of Theorem~\ref{thm:1}}
\label{sec:thm1}

As discussed in Section~\ref{sec:intro}, to complete the proof of Theorem~\ref{thm:1} we need to obstruct the double branched covers of QA 3-braid closures in families $\mathcal{S}_{1a}^1$, $\mathcal{O}^{-1} \setminus \{ L_e^{-1} \}$ and $\mathcal{O}^{-1} \setminus \{ L_e^{+1} \}$ from bounding $\mathbb{Q} B^4$s. In fact, since the associated strings of links in $\mathcal{O}^{-1} \setminus \{ L_e^{-1} \}$ are precisely the cyclic duals of the associated strings of links in $\mathcal{O}^{1} \setminus \{L_e^{+1}\}$, the double branched covers of links in $\mathcal{O}^{-1} \setminus \{ L_e^{-1} \}$ are related to those of links in $\mathcal{O}^{1} \setminus \{ L_e^{+1} \}$ by changing the orientation (see, e.g., Theorem~7.3 in~\cite{Neumann1981ACF} and Section~2 in~\cite{simone2020classification}); thus, it is sufficient to deal with the case $\mathcal{S}_{1a}^1 \cup \mathcal{O}^{-1} \setminus \{L_e^{-1}\}$.

Now, $\Sigma_2(L)$ for $L \in \mathcal{O}^{-1} \setminus \{ L_e^{-1} \}$ bounds a sharp manifold by Theorem~\ref{thm:sharp}, and we carry out a computation in the accompanying \textsc{SageMath} notebook to directly verify that none of the embeddings of the associated lattices are cubiquitous. Thus, by Theorem~\ref{thm:cbobstruction} and the discussion above, if $L\in \mathcal{O}^{\pm1} \setminus \{ L_e^{\pm1} \}$, then $\Sigma_2(L)$ does not bound a $\mathbb{Q} B^4$.

In order to obstruct the existence of $\mathbb{Q} B^4$s bounded by $\Sigma_2(L)$ for $L \in \mathcal{S}_{1a}^1$, it is again sufficient to derive the same obstruction for closures of braids of the form~\eqref{eq:3-braid} with $t=-1$ and associated strings in $ \mathcal{S}_{1a}^* = \{ \mathbf{a} : \mathbf{a} \textrm{ is the cyclic dual of some string in } \mathcal{S}_{1a} \} $. Recall that $\mathcal{S}_{1a} = \{ (b_1, \dots, b_k, 2, c_l, \dots, c_1, 2) : k + l \ge3 \}$, where $(b_1, \dots, b_k)$ and $(c_1, \dots, c_l)$ are linear duals. It is straightforward to show that
\[
    \mathcal{S}_{1a}^* = \{ (c_1' + b_1', b_2', \dots, b_{k'-1}', b_{k'}'+ c_{l'}', c_{l'-1}', \dots, c_2') : k' + l' \ge2 \}
\]
for linear duals $(b_1', \dots, b_{k'}')$ and $(c_1',\dots, c_{l'}')$; moreover, $I(\mathbf{a})=4$ for all $\mathbf{a} \in \mathcal{S}_{1a}^*$. Thus, Theorem~\ref{thm:nocontraction} implies that if $S$ is a negative cyclic subset that does not admit contractions and the associated string of $S$ lies in $\mathcal{S}_{1a}^*$, then the length of $S$ is at most~21. By explicitly determining all negative cyclic subsets with associated strings in $\mathcal{S}_{1a}^*$ and length at most~21, we find that the smallest such subset without contractions has length~12; we postpone computational details to Subsection~\ref{subsec:computations}. After using Theorems~\ref{thm:cbobstruction} and~\ref{thm:cubandcontr} coupled with the discussion above to verify that none of the negative cyclic subsets with associated strings in $\mathcal{S}_{1a}^*$ and length at most 12 are cubiquitous or admit expansions to cubiquitous subsets of higher rank, we conclude that no $\Sigma_2(L)$ for $L \in \mathcal{S}_{1a}^1$ bounds a $\mathbb{Q} B^4$.\hfill$\square$

\begin{rem}
    Obstructing the existence of a $\mathbb{Q} B^4$ bounded by $\Sigma_2(L_e^{\pm1})$ appears particularly challenging as all negative cyclic subsets with the associated string $(3^{[6]})$ are cubiquitous. On the other hand, we have not been able to construct any such $\mathbb{Q} B^4$ explicitly.
\end{rem}

\subsection{Computational Details}
    \label{subsec:computations}
    The proof of Theorem~\ref{thm:1} involves a significant computational component that we carry out in~\textsc{SageMath} and~\textsc{GAP}; the scripts are available on the first author's website.\footnote{\texttt{https://sites.google.com/view/vbrej/}} There are three main computational steps:
        \begin{enumerate}
            \setlength{\itemindent}{1pc}
            \item[Step 1.] Enumerating equivalence classes of strings in $\mathcal{S}_{1a}^*$ of length at most 21 with respect to cyclic reordering and reversal.
            \item[Step 2.] For a representative $\mathbf{a}$ of each class in Step 1, finding all negative cyclic subsets whose associated strings equal $\mathbf{a}$.
            \item[Step 3.] Verifying that all subsets from Step 2 are not cubiquitous and do not admit expansions to cubiquitous subsets of higher rank.
        \end{enumerate}

For Step 1, recall that if $(b_1', \dots, b_{k'}') = (2+x_1, 2^{[y_1]}, 3+x_2, \dots, 3+x_m, 2^{[y_m]})$ for some $m\ge1$ and $x_i, y_i \ge0$ for all $i$, then its linear dual can be written as $$(c_1',\dots, c_{l'}') = (2^{[x_1]}, 3+y_1,2^{[x_2]}, \dots, 2^{[x_m]}, 3+y_m).$$ Notice that $k' = m + \sum_{i=1}^m y_i$ and $l' = m + \sum_{i=1}^m x_i$, so the length of any string in $\mathcal{S}_{1a}^*$ is given by $n = k' + l' - 2 = 2m  - 2 + \sum_{i=1}^m (x_i + y_i)$. Thus, one can produce all strings in $\mathcal{S}_{1a}^*$ with given length $n$ by generating all ordered partitions of $\sum_{i=1}^m (x_i + y_i) = n + 2 - 2m$ of length $2m$ into non-negative integers for $m = 1, \dots, \lfloor \frac{n}{2} \rfloor + 1$, then constructing $(b_1', \dots, b_{k'}')$ and $(c_1', \dots, c_{l'}')$ by viewing those partitions as tuples of the form $(x_1, \dots, x_m, y_1, \dots, y_m)$. Reducing each string to its lexicographically minimal form with respect to cyclic reordering and reversal (this can be done efficiently via, e.g., Booth's algorithm~\cite{booth}) and storing unique minimal forms, we obtain the representatives desired in Step~2. The number of these representatives of length $n=13,\dots,21$ is presented in Table~\ref{tab:s1aduals}, which suggests that this number is roughly exponential in $n$.

\begin{table}[h]
\begin{tabular}{c|ccccccccc}

$n$             & 13   & 14   & 15   & 16    & 17    & 18    & 19     & 20     & 21 \\
$\#$ of strings & 1188 & 4160 & 8255 & 16508 & 32895 & 65790 & 131321 & 262656 & 524799 
\end{tabular}
\caption{Number of strings in $\mathcal{S}_{1a}^*$ of length $n=13,\dots,21$.}
\label{tab:s1aduals}
\end{table}

To find all negative cyclic subsets in Step 2, we use the algorithm of Plesken~\cite{plesken95} for solving the matrix equation $XX^T = A$ over the integers for definite $A$, 
implemented in \textsc{GAP} as the \textsc{OrthogonalEmbeddings} routine. In our case, $A = ((v_i \cdot v_j)_{ij})$ for the sought negative cyclic subsets $\{ v_1, \dots, v_n \} \subset \ZZ^n$ (cf.~Remark~\ref{rem:definite}). The algorithm relies on first finding candidate columns $X_i$ of $X$ and is drastically more efficient in our setting upon restricting to columns that satisfy $X_i^T A^{-1}X_i = 1$, as outlined in Section~3(a) of~\cite{plesken95}; in fact, by examining the proofs in Section~\ref{sec:nocontractions} leading up to Proposition~\ref{prop:bounds}, we discover that we can restrict to $X_i$ with entries bounded in absolute value by $2$ when $n\geq15$. 
After modifying the original \textsc{GAP} code to make use of these restrictions, we are able to find all embeddings of interest for $n\leq21$ within several hours of runtime on the first author's personal computer.

For Step 3, the naive approach would be to directly check that every negative cyclic subset $S$ from Step 2 with rank $n \leq 21$ is not cubiquitous, and that all $S$ with $n = 21$ satisfy the conditions of Theorem~\ref{thm:cubandcontr}. However, this is very expensive computationally; instead, we seek to find the smallest possible $n_c\leq21$ such that all $S$ of rank $n_c$ satisfy the conditions of Theorem~\ref{thm:cubandcontr}, and such that for every $S$ with rank $n > n_c$ one of the following holds:
    \begin{enumerate}
        \item $S$ admits a contraction;
        \item $S$ is not cubiquitous and satisfies the conditions of Theorem~\ref{thm:cubandcontr};
        \item $S$ is not cubiquitous and does not admit expansions by Lemma~\ref{lem:noexpansion}.
    \end{enumerate}

As discussed earlier in this section, all subsets from Step 2 with $n = 13, \dots, 21$ admit contractions. When $n=12$, there exist three non-contractible subsets $S_1^{12}$, $S_2^{12}$ and $S_3^{12}$ with associated strings $(2,4,2,4,2,6,2,4,2,4,2,6)$, $(2^{[3]},4,2,8,2^{[3]},4,2,8)$ and $(2^{[5]},10,2^{[5]},10)$, for each of which the Wu element is given, up to the action of $\Aut\ZZ^{12}$, by $W = \sum_{i=1}^5 2e_i$. Since there does not exist $S$ with $n=13$ and $W = \sum_{i=1}^5 2e_i + e_{13}$, and since $S_1^{12}$, $S_2^{12}$ and $S_3^{12}$ satisfy the setting of Lemma~\ref{lem:noexpansion}, it follows that these three subsets do not admit expansions; we verify non-cubiquity of $S_1^{12}$, $S_2^{12}$ and $S_3^{12}$ using the procedure described in Section~\ref{sec:cubiquitydef}. We deal with four more non-contractible subsets with $n=10$ analogously. Next, there are six non-contractible subsets with $n=8$ and one with $n=6$ for which the argument above does not apply, yet we explicitly construct the desired small cubes around $\frac{1}{2}W$ from the statement of Theorem~\ref{thm:cubandcontr} that are not hit by the lattice generated by $S$. The next smallest non-contractible subset occurs when $n=4$, and indeed we can set $n_c = 4$ as we are able to check that all $S$ of rank $4$ satisfy the conditions of Theorem~\ref{thm:cubandcontr}.

\section{Proof of Theorem~2}
\label{sec:ribbons}


In~\cite{brejevs2020ribbon}, the first author showed that if $L\in \mathcal{S}_0^{2}\setminus\mathcal{S}_{2c}^0$, then $L$ is $\chi$-ribbon. Following the conventions established in Section~2 of~\cite{brejevs2020ribbon}, we exhibit $\chi$-ribbon surfaces for links in $\mathcal{S}_1^{-1}\cup( \mathcal{S}_1^1\setminus\mathcal{S}_{1a}^1)$ in Figures \ref{fig:S1ad-1} to \ref{fig:S1ed+1}.

It remains to show that if $L=L_e^{\pm1}$, then $L$ is not $\chi$-slice. 
Up to isotopy, there are two distinct orientations on $L$: in one case, all strands in the braid whose closure yields $L$ are oriented in the same direction, and in the other, one of the strands is reversed. Routine computations show that in the first case, the signature of $L$ is non-zero, so by Lemma~3.2 in~\cite{DonaldOwens2012}, $L$ does not bound an oriented Euler characteristic one slice surface. In the second case, the Alexander polynomial of $L$ is given by $\Delta_L(t) = (t - 1)^2 \cdot (1 - 6t + 19t^2 - 29t^3 + 19t^4 - 6t^5 + t^6)$; the degree six factor is irreducible in $\ZZ[t^{\pm 1}]$, hence $\Delta_L(t)$ does not factor (up to units) as $f(t) \cdot f(t^{-1})$ for some $f \in \ZZ[t^{\pm 1}]$. Thus, by Remark~4.10 in~\cite{florens}, $L$ also does not bound an oriented Euler characteristic one slice surface in this case.

Next notice that $L$ is a three-component link and each pair of components forms a Hopf link. 
If $L=L_1\cup L_2\cup L_3$ is $\chi$-slice, then any Euler characteristic one surface $F$ bounded by $L$ must be non-orientable; hence $F$ must be the disjoint union of a disk and two M\"obius bands. Without loss of generality, assume $L_1$ bounds the disk. But then $L_1\cup L_2$ bounds the disjoint union of a disk and M\"obius band; that is, the Hopf link is $\chi$-slice. This is clearly not possible, however, since the determinant of the Hopf link is not a square.\hfill$\square$


       
		\begin{figure}[h]
                \vspace{2cm}
			\centering
			\def0.9\textwidth{0.6\textwidth}
			\makebox[\textwidth][c]{\input{qa2_all.pdf_tex}}
			\caption{
				Band moves exhibiting $\chi$-ribbon surfaces for closures of braids in family (2) from Theorem~\ref{thm:baldwinclassification} for $m \in \{-1,-2,-3\}$. Here and further, bands are shown in red and annotated with the number of half-twists in the band with respect to the blackboard framing.
			} \label{fig:fam2}
		\end{figure}
  
	\afterpage{\clearpage}
		\begin{figure}[p]
			\centering
			\def0.9\textwidth{1.1\textwidth}
			\makebox[\textwidth][c]{\input{s1ad-1.pdf_tex}}
			\caption{
				Band moves for the family $ \mathcal{S}_{1a}^{-1} $. For the definition of blue and chartreuse rectangles in all following figures see Section~2 in~\cite{brejevs2020ribbon}.
			} \label{fig:S1ad-1}
		\end{figure}

	\afterpage{\clearpage}
		\begin{figure}[p]
			\centering
			\def0.9\textwidth{1.1\textwidth}
			\makebox[\textwidth][c]{\input{s1bd-1.pdf_tex}}
			\caption{
				Band moves for the family $ \mathcal{S}_{1b}^{-1} $.
			} \label{fig:S1bd-1}
		\end{figure}

	\afterpage{\clearpage}
		\begin{figure}[p]
			\centering
			\def0.9\textwidth{1.1\textwidth}
			\makebox[\textwidth][c]{\input{s1bd+1.pdf_tex}}
			\caption{
				Band moves for the family $ \mathcal{S}_{1b}^1 $.
			} \label{fig:S1bd+1}
		\end{figure}

	\afterpage{\clearpage}
		\begin{figure}[p]
			\centering
			\def0.9\textwidth{1.1\textwidth}
			\makebox[\textwidth][c]{\input{s1cd-1.pdf_tex}}
			\caption{
				Band moves for the family $ \mathcal{S}_{1c}^{-1} $.
			} \label{fig:S1cd-1}
		\end{figure}

	\afterpage{\clearpage}
		\begin{figure}[p]
			\centering
			\def0.9\textwidth{1.1\textwidth}
			\makebox[\textwidth][c]{\input{s1cd+1.pdf_tex}}
			\caption{
				Band moves for the family $ \mathcal{S}_{1c}^1 $.
			} \label{fig:S1cd+1}
		\end{figure}

	\afterpage{\clearpage}
		\begin{figure}[p]
			\centering
			\def0.9\textwidth{1.1\textwidth}
			\makebox[\textwidth][c]{\input{s1dd-1.pdf_tex}}
			\caption{
				Band moves for the family $ \mathcal{S}_{1d}^{-1} $.
			} \label{fig:S1dd-1}
		\end{figure}

	\afterpage{\clearpage}
		\begin{figure}[p]
			\centering
			\def0.9\textwidth{1.1\textwidth}
			\makebox[\textwidth][c]{\input{s1dd+1.pdf_tex}}
			\caption{
				Band moves for the family $ \mathcal{S}_{1d}^1 $.
			} \label{fig:S1dd+1}
		\end{figure}
		
        \begin{sidewaysfigure}
            \centering
            \def0.9\textwidth{0.95\textwidth}
           
            \makebox[\textwidth][c]{\input{s1ed-1.pdf_tex}}
            \captionsetup{width=.8\linewidth}
		  \vspace*{0.5em}
            \caption{
                Band moves for the family $ \mathcal{S}_{1e}^{-1} $. In step (2) we flype the tangle between the two purple rectangles $x$ times to cancel the crossings contained in the rectangles.
            } \label{fig:S1ed-1}
        \end{sidewaysfigure}

        \begin{sidewaysfigure}
            \centering
            \def0.9\textwidth{0.9\textwidth}
           
            \makebox[\textwidth][c]{\input{s1ed+1.pdf_tex}}
            \captionsetup{width=.8\linewidth}
		  \vspace*{0.5em}
            \caption{
                Band moves for the family $ \mathcal{S}_{1e}^{1} $. In step (2) we flype the tangle between the two purple rectangles $x$ times to cancel the crossings contained in the rectangles.
            } \label{fig:S1ed+1}
        \end{sidewaysfigure}

\newpage
