text:
\begin{document}

\title{Classification of torus bundles that bound rational homology circles}
\author{Jonathan Simone}

\begin{abstract}
In this article, we completely classify torus bundles over the circle that bound 4-manifolds with the rational homology of the circle. Along the way, we classify certain integral surgeries along chain links that bound rational homology balls and explore a connection to 3-braid closures whose double branched covers bound rational homology 4-balls.
\end{abstract}

\maketitle

\section{Introduction}\label{intro}
In \cite{simone3}, two infinite families of $T^2$-bundles over $S^1$ are shown to bound rational homology circles ($\QQ S^1\times B^3s$). As an application, the $\QQ S^1\times B^3s$ were used to construct infinite families of rational homology 3-spheres ($\QQ S^3s$) that bound rational homology 4-balls ($\QQ B^4s$). The main purpose of this article is to show that the two families of torus bundles used in \cite{simone3} are the only torus bundles that bound $\QQ S^1\times B^3s$.

After endowing $T^2\times[0,1]=\mathbb{R}^2/\mathbb{Z}^2\times[0,1]$ with the coordinates $(\textbf{x},t)=(x,y,t)$, any torus bundle over $S^1$ is of the form $T^2\times[0,1]/(\textbf{x},1)\sim(\pm A\textbf{x},0)$, where $A\in SL(2,\ZZ)$, which is well-defined up to conjugation. The matrix $A$ is called the \textit{monodromy} of the torus bundle. Throughout, we will express the monodromy in terms of the generators $T=\begin{bmatrix} 1&1\\0&1\end{bmatrix}$ and $S=\begin{bmatrix} 0&1\\-1&0\end{bmatrix}$. A torus bundle is called \textit{elliptic} if $|\text{tr}A|<2$, \textit{parabolic} if $|\text{tr}A|=2$, and \textit{hyperbolic} if $|\text{tr}A|>2$. Moreover, a torus bundle is called \textit{positive} if tr$A>0$ and \textit{negative} if tr$A<0$. Torus bundles naturally arise as the boundaries of plumbings of $D^2$-bundles over $S^2$. Using these plumbing descriptions, it is easy to draw surgery diagrams for torus bundles (c.f. \cite{neumann}). Figure \ref{torusbundles} gives a complete list of torus bundles over $S^1$, along with their monodromies (up to conjugation) and surgery diagrams. To simplify notation, let $\textbf{T}_{\pm A(\textbf{a})}$ denote the hyperbolic torus bundle with monodromy $\pm A(\textbf{a})=\pm T^{-a_1}S\cdots T^{-a_n}S$, where $\textbf{a}=(a_1,\ldots,a_n)$.

\begin{figure}[h!]
\centering
 {\tabulinesep=1.5mm
 \begin{tabu}{|c@{\hskip.5in}c|c@{\hskip.5in}c|}
 \multicolumn{4}{c}{\textbf{Elliptic Torus Bundles}}\\\hline
\underline{Monodromy} & \underline{Surgery Diagram} & \underline{Monodromy} & \underline{Surgery Diagram} \\
\shortstack[c]{$ S$\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}}  & \includegraphics[scale=.5]{posellS.png} & \shortstack[c]{$-S$\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}}  & \includegraphics[scale=.5]{negellS.png} \\\hline
\shortstack[c]{$ T^{-1}S$\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}} & \includegraphics[scale=.5]{posellTS.png} & \shortstack[c]{$- T^{-1}S$\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}} & \includegraphics[scale=.5]{negellTS.png}  \\\hline
 \shortstack[c]{$ (T^{-1}S)^2$\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}} & \includegraphics[scale=.5]{posellTS2.png} &  \shortstack[c]{$- (T^{-1}S)^2$\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}} & \includegraphics[scale=.5]{negellTS2.png}\\\hline
\end{tabu}}

\vspace{.2in}

{\tabulinesep=1.2mm
\begin{tabu}{|c@{\hskip.5in}c|c@{\hskip.5in}c|}
\multicolumn{4}{c}{\textbf{Parabolic Torus Bundles}}\\\hline
\underline{Monodromy} & \underline{Surgery Diagram} & \underline{Monodromy} & \underline{Surgery Diagram}\\
\shortstack[c]{$T^n$\\ $(n\in\ZZ)$\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}} &  \includegraphics[scale=.5]{pospar.png} & \shortstack[c]{$-T^n$\\ $(n\in\ZZ)$\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}} &  \includegraphics[scale=.5]{negpar.png}\\\hline
\end{tabu}}

\vspace{.2in}

{\tabulinesep=1.2mm
 \begin{tabu}{|c@{\hskip.5in}c|}
\multicolumn{2}{c}{\textbf{Hyperbolic Torus Bundles $\textbf{T}_{\pm A(a_1,\ldots,a_n)}$}}\\\hline
\underline{Monodromy} & {\underline{Surgery Diagram}}\\
\shortstack[c]{$T^{-a_1}S\cdots T^{-a_n}S$ \\ $(a_i\ge2, a_1\ge3)$\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in} } & \includegraphics[scale=.5]{poshyp.png}\\\hline
\shortstack[c]{$-T^{-a_1}S\cdots T^{-a_n}S$ \\ $(a_i\ge2, a_1\ge3)$\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in}\\\hspace{1in} } & \includegraphics[scale=.5]{neghyp.png}\\\hline
\end{tabu}}
\caption{Mondromy and surgery diagrams of $T^2$-bundles over $S^1$}\label{torusbundles}
\end{figure}

\begin{thm} A torus bundle over $S^1$ bounds a $\QQ S^1\times B^3$ if and only if
	\begin{itemize}
		\item it is negative parabolic or
		\item it is positive hyperbolic of the form $\textbf{T}_{A(\textbf{a})}$, where\\ $\textbf{a}=(3+x_1,2^{[x_2]},\ldots,3+x_{2k+1},2^{[x_1]}, 3+x_2,2^{[x_3]},\ldots,3+x_{2k},2^{[x_{2k+1}]})$, \\
		$k\ge0$, and $x_i\ge0$ for all $i$.
	\end{itemize}
\label{mainthm}\end{thm}

Elliptic torus bundles and parabolic torus bundles that bound $\QQ S^1\times B^3s$ are rather simple to classify. Classifying hyperbolic torus bundles, which make up the ``generic" class of torus bundles, is much more involved and includes the bulk of the technical work. In \cite{simone3}, it is shown that $\textbf{T}_{A(\textbf{a})}$ indeed bounds a $\QQ S^1\times B^3$ when $\textbf{a}=(3+x_1,2^{[x_2]},\ldots,3+x_{2k+1},2^{[x_1]}, 3+x_2,2^{[x_3]},\ldots,3+x_{2k},2^{[x_{2k+1}]})$. To obstruct all other hyperbolic torus bundles from bounding $\QQ S^1\times B^3s$, we first consider a related class of $\QQ S^3s$.

Let $L_n^t$ denote the $n$-component link shown in Figure \ref{chainlink}, where $t$ denotes the number of half-twists. We call $L_n^t$ the \textit{n-component, t-half-twisted chain link}. If $t=0$, we call the chain link \textit{untwisted}. Consider the surgery diagrams for $\textbf{T}_{\pm A(\textbf{a})}$ given in Figure \ref{torusbundles}. By performing $m$-surgery along the meridians of the 0-framed unknots, we obtain $\QQ S^3s$, which can be realized as integer Dehn-surgery along certain chain links. This is shown in Figure \ref{chainlinks}. In the left side of each of the four diagrams, slide the unknot with framing $-a_1$ (or $-a_1\pm2$) twice over the blue $m$-framed unknot so that it no longer passes through the $0$-framed unknot. Then cancel the $0$-framed and $m$-framed unknots. When $n\ge2$, the resulting $\QQ S^3s$ are obtained by $(-a_1,\ldots,-a_n)$-surgery along the chain link $L_n^t$, where $t=2m$ or $t=2m-1$. We denote these $\QQ S^3s$ by $Y_{\textbf{a}}^t=S^3_{(-a_1,\ldots,-a_n)}(L_n^t)$, where $\textbf{a}=(a_1,\ldots,a_n)$. Note that by cyclically reordering or reversing the surgery coefficients, we obtain the same 3-manifold. When $n=1$, the resulting $\QQ S^3s$ are obtained by $(-a_1\pm 2)$-surgery along $L_1^t$, where $t=2m$ or $2m-1$. We denote these $\QQ S^3s$ by  $Y_{(a_1)}^t$. Note that $Y_{(a_1)}^t=S^3_{-a_1+2}(L_1^t)$ when $t$ is even, and $Y_{(a_1)}^t=S^3_{-a_1-2}(L_1^t)$ when $t$ is odd.

 \begin{lem}[\cite{simone3}] Let $Y$ be a $\mathbb{Q}S^1\times S^2$ that bounds a $\mathbb{Q}S^1\times B^3$, and let $K$ be a knot in $Y$ such that $[K]$ has infinite order in $H_1(Y;\mathbb{Z})$. Then any integer surgery on $Y$ along $K$ yields a $\QQ S^3$ that bounds a $\QQ B^4$. \label{rationalballobstruction}\end{lem}

By Lemma \ref{rationalballobstruction}, if $\textbf{T}_{A(\textbf{a})}$ bounds a $\QQ S^1\times B^3$, then $Y_{\textbf{a}}^{t}$ bounds a $\QQ B^4$ for all even $t$, and if $\textbf{T}_{-A(\textbf{a})}$ bounds a $\QQ S^1\times B^3$, then $Y_{\textbf{a}}^{t}$ bounds a $\QQ B^4$ for all odd $t$. Thus if $Y_{\textbf{a}}^t$ does not bound a $\QQ B^4$ for some even (or odd) $t$, then $\textbf{T}_{A(\textbf{a})}$ (or $\textbf{T}_{-A(\textbf{a})}$) does not bound a $\QQ S^1\times B^3$. Using this fact, we will obstruct most hyperbolic torus bundles from bounding $\QQ S^1\times B^3s$ by identifying the strings $\textbf{a}$ for which $Y_{\textbf{a}}^0$ and $Y_{\textbf{a}}^{-1}$ do not bound $\QQ B^4s$. Before writing down the result, we first recall and introduce some useful terminology.

Let $(b_1,\ldots,b_k)$ be a string of integers such that $b_i\ge2$ for all $i$. If $b_j\ge 3$ for some $j$, then we can write this string in the form $(2^{[m_1]},3+n_1,\ldots,2^{[m_j]},2+n_j)$, where $m_i,n_i\ge 0$ for all $i$ and $(\dots,2^{[t]},\ldots)$ denotes $(\dots,\overbrace{2,\ldots,2}^t,\ldots)$. The string $(c_1,\ldots,c_l)=(2+m_1, 2^{[n_1]},2+m_2,\ldots,3+m_j,2^{[n_j]})$ is called the \textit{linear-dual string} of $(b_1,\ldots,b_k)$. If $b_i=2$ for all $1\le i\le k$, then we define its linear-dual string to be $(k+1)$. As a topological interpretation of this, by Neumann \cite{neumann}, if $P$ is a plumbing of $D^2$-bundles over $S^2$ whose associated graph is a linear tree with Euler numbers $(-b_1,\ldots,-b_k)$, then the reversed-orientation plumbing $\overline{P}$ is a linear plumbing with Euler numbers $(-c_1,\ldots,-c_l)$. Finally, we define the linear-dual string of $(1)$ to be the empty string.

\begin{figure}[t!]
	\centering
	\includegraphics[scale=.6]{chainlink.png}
	\caption{The $n$-component, $t$-half-twisted chain link, $L_n^t$. The box labeled ``$t$" denotes $t$ half-twists.}
	\label{chainlink}
\end{figure}

\begin{figure}
	\centering
	\includegraphics[scale=.55]{chainlinksurgery.png}
	\caption{Surgering the hyperbolic torus bundle $\textbf{T}_{\pm A(\textbf{a})}$, where $\textbf{a}=(a_1,\ldots,a_n)$ to obtain the rational homology sphere $Y_{\textbf{a}}^t$. The blue boxes labeled $2m$ and $2m-1$ indicate the number of half-twists.}\label{chainlinks}
\end{figure}

Suppose $(a_1,\ldots,a_n)$ is of the form $(2^{[m_1]},3+n_1,\ldots,2^{[m_j]},3+n_j)$, where $m_i,n_i\ge 0$ for all $i$, we define its \textit{cyclic-dual} to be the string $(d_1,\ldots,d_m)=(3+m_1,2^{[n_1]},\ldots,3+m_j,2^{[n_j]})$. Notice that this only slightly differs from the definition of the linear-dual string. Topologically, by \cite{neumann}, if $P_{\pm\textbf{a}}$ is a plumbing of $D^2$-bundles over $S^2$ whose associated graph is a cycle with Euler numbers $(-a_1,\ldots,-a_n)$, then $\overline{P_{\pm\textbf{a}}}=P_{\pm\textbf{d}}$ is a cyclic plumbing with Euler numbers $(-d_1,\ldots,-d_m)$.  Moreover, $\partial P_{\pm\textbf{a}}=\overline{\textbf{T}}_{\pm A(\textbf{a})}$; hence $\overline{\textbf{T}}_{\pm A(\textbf{a})}=\textbf{T}_{\pm A(\textbf{d})}$. Finally, $(a_n,\ldots,a_1)$ is called the $\textit{reverse}$ of $(a_1,\ldots,a_n)$.

\begin{definition} Two strings are considered to be equivalent if one is a cyclic reordering and/or reverse of the other. Each string in each of the following sets is defined up to this equivalence. Moreover, in the following sets, strings of the form $(b_1,\ldots,b_k)$ and $(c_1,\ldots,c_l)$ are assumed to be linear-dual.
	\begin{itemize}
		\item $ \mathcal{S}_{1a}=\{(b_1,\ldots,b_k,2,c_l,\ldots,c_1,2)\,|\, k+l\ge3\}$
	\item $ \mathcal{S}_{1b}=\{(b_1,\ldots,b_k,2,c_l,\ldots,c_1,5)\,|\, k+l\ge2\}$
	\item $ \mathcal{S}_{1c}=\{(b_1,\ldots,b_k,3,c_l,\ldots,c_1,3)\,|\, k+l\ge2\}$
	\item $ \mathcal{S}_{1d}=\{(2,b_1+1,b_2,\ldots,b_{k-1},b_k+1,2,2,c_l+1,c_{l-1},\ldots,c_2,c_1+1,2)\,|\, k+l\ge3\}$
	\item $ \mathcal{S}_{1e}=\{(2, 3+x, 2, 3, 3, 2^{[x-1]},3,3)\,|\, x\ge0 \textup{ and } (3,2^{[-1]},3):=(4)\}$
	\item $ \mathcal{S}_{2a}=\{(b_1+3,b_2,\ldots,b_k,2,c_l,\ldots,c_1)\}$
	\item $ \mathcal{S}_{2b}=\{(3+x,b_1,\ldots,b_{k-1},b_k+1,2^{[x]},c_l+1,c_{l-1},\ldots,c_1)\,|\, x\ge0\text{ and } k+l\ge2\}$
	\item $\begin{aligned}[t] \mathcal{S}_{2c}=\{&(3+x_1,2^{[x_2]},3+x_3,2^{[x_4]},\ldots,3+x_{2k+1},2^{[x_1]}, 3+x_2,2^{[x_3]},\ldots,3+x_{2k},2^{[x_{2k+1}]})\,|\\ &\, k\ge0 \text{ and } x_i\ge 0 \text{ for all } i\}\end{aligned}$
	\item $ \mathcal{S}_{2d}=\{(2,2+x,2,3,2^{[x-1]},3,4)\,|\, x\ge0 \textup{ and } (3,2^{[-1]},3):=(4)\}$
	\item $ \mathcal{S}_{2e}=\{(2,b_1+1,b_2,\ldots,b_k,2,c_l,\ldots,c_2,c_1+1,2),(2,2,2,3)\,|\, k+l\ge3\}$
		\end{itemize}
	Moreover, $\mathcal{S}_1=\mathcal{S}_{1a}\cup\mathcal{S}_{1b}\cup\mathcal{S}_{1c}\cup\mathcal{S}_{1d}\cup\mathcal{S}_{1e}$, $\mathcal{S}_2=\mathcal{S}_{2a}\cup\mathcal{S}_{2b}\cup\mathcal{S}_{2c}\cup\mathcal{S}_{2d}\cup\mathcal{S}_{2e}$, and $\mathcal{S}=\mathcal{S}_1\cup\mathcal{S}_2$.
	\label{definition}
\end{definition}

\begin{definition} Let $\textbf{a}=(a_1,\ldots,a_n)$, where $a_i\ge 2$ for all $i$. Define $I(\textbf{a})$ to be the integer $I(\textbf{a})=\sum_{i=1}^n(a_i-3)$.\end{definition}

\begin{remark} It it easy to check that if $\textbf{a}\in\mathcal{S}$, then $-4\le I(\textbf{a})\le0$. Moreover, $\textbf{a}\in\mathcal{S}$ and $I(\textbf{a})=0$ if and only if $a\in\mathcal{S}_{2a}\cup\mathcal{S}_{2b}\cup\mathcal{S}_{2c}$ (c.f. Theorem \ref{latticethm} in Section \ref{lattice}). Moreover, we will see that $\mathcal{S}_1\cap \mathcal{S}_2=\emptyset$ (Corollary \ref{orderbound} in Section \ref{obstructions}).\label{ibounds}\end{remark}

\begin{thm} Let $\textbf{a}=(a_1,\ldots,a_n)$, where $n\ge1$, $a_i\ge 2$ for all $i$, and $a_j\ge 3$ for some $j$, and let $\textbf{d}$ be the cyclic-dual of $\textbf{a}$.
\begin{enumerate}
	\item Suppose $\textbf{d}\notin\mathcal{S}_{1a}\cup\{(6,2,2,2,6,2,2,2)\}$. Then $Y_{\textbf{a}}^{-1}$ bounds a $\QQ B^4$ if and only if $\textbf{a}\in\mathcal{S}_1$ or $\textbf{d}\in\mathcal{S}_{1b}\cup\mathcal{S}_{1c}\cup\mathcal{S}_{1d}\cup\mathcal{S}_{1e}$.\label{(1)}
	\item Suppose $\textbf{a}\notin\mathcal{S}_{1a}\cup\{(6,2,2,2,6,2,2,2)\}$. Then $Y_{\textbf{a}}^{1}$ bounds a $\QQ B^4$ if and only if $\textbf{d}\in\mathcal{S}_1$ or $\textbf{a}\in\mathcal{S}_{1b}\cup\mathcal{S}_{1c}\cup\mathcal{S}_{1d}\cup\mathcal{S}_{1e}$.\label{(2)}
	\item $Y_{\textbf{a}}^0$ bounds a $\QQ B^4$ if and only if $\textbf{a}\in \mathcal{S}_2$ or $\textbf{d}\in \mathcal{S}_2$.\label{(3)}
\end{enumerate}
\label{thm1}\end{thm}

\begin{remark} The hypothesis ``$a_j\ge3$ for some $j$" in Theorem \ref{thm1} ensures that $\textbf{T}_{\pm A(\textbf{a})}$ is a hyperbolic torus bundle. If we remove this condition from the theorem, then we would have an additional case: $a_i=2$ for all $i$. In this case, $Y_{\textbf{a}}^{-1}$ bounds a $\QQ B^4$ and $Y_{\textbf{a}}^0$ does not bound a $\QQ B^4$. This follows from Lemma \ref{rationalballobstruction} and Theorem \ref{mainthm}, and the fact that the corresponding torus bundles are the parabolic torus bundles with respective monodromies $-T^n$ and $T^n$ (c.f. \cite{simone3}).\label{parremark} \end{remark}

\begin{remark} We will see in Section \ref{spheres} (Lemma \ref{noboundlem}) that for certain strings $\textbf{d}$ that are the cyclic-duals of $(b_1,\ldots,b_k,2,c_l,\ldots,c_1,2)$, $Y_{\textbf{d}}^{-1}$ does not bound a $\QQ B^4$ (c.f. Theorem \ref{thm1}(\ref{(1)})). However, we are unable to prove this fact for all such strings. Moreover, we are unable to obstruct $Y^{-1}_{(6,2,2,2,6,2,2,2)}$ from bounding a $\QQ B^4$ or show that it indeed bounds a $\QQ B^4$. \end{remark}

Combined with Lemma \ref{rationalballobstruction}, Theorem \ref{thm1} obstructs most hyperbolic torus bundles from bounding $\QQ S^1\times B^3s$. In Section \ref{tbundles}, we will obstruct the rest by considering certain cyclic covers of $\QQ S^1\times B^3s$. The proof of Theorem \ref{thm1} relies on Donaldson's Diagonalization Theorem \cite{donaldson} and lattice analysis. From this analysis, it follows that if $\textbf{a}\notin\mathcal{S}_1\cup\{(6,2,2,2,6,2,2,2)\}$, then $Y^t_{\textbf{a}}$ does not bound a $\QQ B^4$ for all odd $t$ and if $\textbf{a}\notin\mathcal{S}_2$, then $Y^t_{\textbf{a}}$ does not bound a $\QQ B^4$ for all even $t$. Moreover, by Lemma \ref{rationalballobstruction} and Theorem \ref{mainthm}, if $\textbf{a}\in\mathcal{S}_{2c}$, then $Y^t_{\textbf{a}}$ bounds a $\QQ B^4$ for all even $t$. This leads to the following question.

\begin{question} For what values of $t$ and for which strings $\textbf{a}\in\mathcal{S}\setminus\mathcal{S}_{2c}$ does $Y^t_{\textbf{a}}$ bound a $\QQ B^4$?\label{questionballs}\end{question}

\subsection{Connection to 3-braids}\label{3braids} Let $\textbf{a}=(a_1,\ldots,a_n)$. If $t\in\{-1,0\}$, then it is easy to see that $Y_{\textbf{a}}^{t}$ is the double cover of $S^3$ branched over the closure of the 3-braid represented by $(\sigma_1\sigma_2)^{3t}\sigma_1\sigma_2^{-(a_1-2)}\cdots\sigma_1\sigma_2^{-(a_n-2)}$. In particular, using the techniques in \cite{akbulutkirby}, it is clear that $Y^{-1}_{\textbf{a}}$ and $Y^0_{\textbf{a}}$ are the double covers of $S^3$ branched over the links shown in the middle of Figure \ref{dbcbraid}. Notice that the $\ZZ_2$-action on $Y^t_{\textbf{a}}$ inducing these covers are the $180^{\circ}$ rotations shown at the top of Figure \ref{dbcbraid}. By isotoping the links, we obtain the desired 3-braid closures, shown at the bottom of Figure \ref{dbcbraid}. Note that, in the figure, the blue box labeled t indicates the number of full-twists, while all other boxes in the figure indicate the number of half-twists.

By working through some Kirby calculus, we can further see that for any $t$, $Y_{\textbf{a}}^{t}$ is the double cover of $S^3$ branched over the 3-braid closure represented by $(\sigma_1\sigma_2)^{3t}\sigma_1\sigma_2^{-(a_1-2)}\cdots\sigma_1\sigma_2^{-(a_n-2)}$. Notice that if $t=2m-1\ge -1$ is odd, then $Y_{\textbf{a}}^{t}$ can be realized as $(-1^{[m]})$-surgery along a link in $Y_{\textbf{a}}^{-1}$, as shown in the top of Figure \ref{oddt}, and if $t=2m\ge 0$ is even, then $Y_{\textbf{a}}^{t}$ can be realized as $(-1^{[m]})$-surgery along a link in $Y_{\textbf{a}}^{0}$, as shown in the top of Figure \ref{event}. Under the $\ZZ_2$-action, each of these surgery curves double covers a curve isotopic to the braid axis of the 3-braid. Thus each $-1$-surgery curve maps to a $-1/2$-surgery curve isotopic to the braid axis, as shown in the middle of Figures \ref{oddt} and \ref{event}. By blowing down these curves, we obtain the desired 3-braid closures at the bottom of the figures. Note that the same argument can be used when $t<-1$; the only difference is that the surgery curves would all have positive coefficients.

\begin{figure}
\centering
\includegraphics[scale=.525]{dbcbraid.png}
\caption{$Y^{-1}_{\textbf{a}}$ and $Y^{0}_{\textbf{a}}$ are the double covers of $S^3$ branched over the 3-braid closures represented by $(\sigma_1\sigma_2)^{3t}\sigma_1\sigma_2^{-(a_1-2)}\cdots\sigma_1\sigma_2^{-(a_n-2)}$, where $t=-1$ and $t=0$, respectively. The blue box labeled t indicates the number of full-twists, while all other boxes in all other diagrams indicated the number of half-twists.}\label{dbcbraid}
\end{figure}

\begin{figure}
\centering
\begin{subfigure}{.48\textwidth}
	\centering
	\includegraphics[scale=.51]{oddt.png}
	\caption{}\label{oddt}
\end{subfigure}
\begin{subfigure}{.48\textwidth}
	\centering
	\includegraphics[scale=.51]{event.png}
	\caption{}\label{event}
\end{subfigure}
\caption{This figure shows that when $t\ge -1$, $Y^{t}_{\textbf{a}}$ is the double cover of $S^3$ branched over the 3-braid closure represented by $(\sigma_1\sigma_2)^{3t}\sigma_1\sigma_2^{-(a_1-2)}\cdots\sigma_1\sigma_2^{-(a_n-2)}$. The same is true when $t<-1$.}\label{evenoddt}
\end{figure}

As a result of this characterization, we can use Theorem \ref{thm1}, Theorem \ref{mainthm}, and Lemma \ref{rationalballobstruction} to classify certain families of 3-braid closures admitting double branched covers that bound $\QQ B^4s$.

\begin{cor} Let $\textbf{a}=(a_1,\ldots,a_n)$, where $n\ge1$, $a_i\ge 2$ for all $i$ and $a_j\ge 3$ for some $j$, and let $\textbf{d}$ be the cyclic-dual of $\textbf{a}$.
	\begin{itemize}
		\item Suppose $\textbf{d}\notin\mathcal{S}_{1a}\cup\{(6,2,2,2,6,2,2,2)\}$. Then the double cover of $S^3$ branched over the 3-braid given by the representation $(\sigma_1\sigma_2)^{-3}\sigma_1\sigma_2^{-(a_1-2)}\cdots\sigma_1\sigma_2^{-(a_n-2)}$ bounds a $\QQ B^4$ if and only if $\textbf{a}\in\mathcal{S}_1$ or $\textbf{d}\in\mathcal{S}_{1b}\cup\mathcal{S}_{1c}\cup\mathcal{S}_{1d}\cup\mathcal{S}_{1e}$.
		\item  Suppose $\textbf{a}\notin\mathcal{S}_{1a}\cup\{(6,2,2,2,6,2,2,2)\}$. Then the double cover of $S^3$ branched over the 3-braid given by the representation $(\sigma_1\sigma_2)^{3}\sigma_1\sigma_2^{-(a_1-2)}\cdots\sigma_1\sigma_2^{-(a_n-2)}$ bounds a $\QQ B^4$ if and only if $\textbf{d}\in\mathcal{S}_1$ or $\textbf{a}\in\mathcal{S}_{1b}\cup\mathcal{S}_{1c}\cup\mathcal{S}_{1d}\cup\mathcal{S}_{1e}$.
		\item The double cover of $S^3$ branched over the 3-braid given by the representation\\ $\sigma_1\sigma_2^{-(a_1-2)}\cdots\sigma_1\sigma_2^{-(a_n-2)}$ bounds a $\QQ B^4$ if and only if $\textbf{a}\in\mathcal{S}_2$.
		\item If $\textbf{a}\in\mathcal{S}_{2c}$, then the double cover of $S^3$ branched over the 3-braid given by the representation $(\sigma_1\sigma_2)^{3t}\sigma_1\sigma_2^{-(a_1-2)}\cdots\sigma_1\sigma_2^{-(a_n-2)}$ bounds a $\QQ B^4$ for all even $t$.
		\end{itemize}
		\label{cor1}\end{cor}

The 3-braid knots corresponding to strings in $\mathcal{S}_{1a}\cup\mathcal{S}_{2a}\cup\mathcal{S}_{2b}\cup\mathcal{S}_{2c}$ (and their mirrors) were shown in \cite{lisca3braid} to be the only 3-braid knots of finite concordance order. Moreover, some of them were shown be slice knots and so the corresponding double branched covers are already known to bound $\QQ B^4s$. Many of the the remaining strings in $\mathcal{S}$ correspond to infinite-concordance order 3-braid knots. Thus, these give examples of infinite concordance order knots whose double branched covers bound $\QQ B^4s$. Rewording Question \ref{questionballs} in terms of 3-braids, a natural question is the following.

\begin{question} What other 3-braid closures admit double branched covers bounding $\QQ B^4s$?\end{question}

\subsection{Organization} In Section \ref{obstructions}, we will highlight some simple obstructions to $\QQ S^1\times S^2s$ bounding $\QQ S^1\times B^3s$, recall Heegaard Floer homology calculations of 3-braid closures due to Baldwin, and use these calculations to explore the orientation reversal of the 3-manifold $Y^t_{\textbf{a}}$. These obstructions and calculations will be used in Sections \ref{tbundles} and \ref{spheres}. In particular, in Section \ref{tbundles}, we will use the obstructions and other techniques to prove Theorem \ref{mainthm} and in Section \ref{spheres}, we will show that the $\QQ S^3s$ of Theorem \ref{thm1} do indeed bound $\QQ B^4s$ by explicitly constructing them. In Sections \ref{lattice}$-$\ref{p1=0}, we will use lattice analysis to prove that the $\QQ S^3s$ of Theorem \ref{thm1} are the only such $\QQ S^3s$ that bound $\QQ B^4s$. Finally, Section \ref{appendix} (the Appendix) provides some continued fraction calculations that are used in Sections \ref{obstructions} and \ref{spheres}.

\section{Obstructions}\label{obstructions}

In this section, we highlight some simple ways to obstruct a $\QQ S^1\times S^2$ from bounding a  $\QQ S^1\times B^3$, recall Baldwin's calculations of the Heegaard Floer homology of double covers of $S^3$ branched over certain 3-braid closures (\cite{baldwinfloertorusbundles}), and consider the effect that reversing orientation has on the rational homology spheres $Y^t_{\textbf{a}}$. We first make note of some simple obstructions, the first of which is a simple homology calculation.

\begin{lem} If $Y$ bounds a $\QQ S^1\times B^3$, then the torsion part of $H_1(Y)$ has square order.\label{firsthomologyobstruction}\end{lem}
\begin{proof}
It is well-known that if a $\QQ S^3$ bounds a $\QQ B^4$, then its first homology group has square order (see, for example, Lemma 3 in \cite{cassongordon86}). The proof involves exploring the long exact sequence of the pair. An analogous argument shows that the same is true of $\QQ S^1\times S^2$s that bound $\QQ S^1\times B^3s$.
\end{proof}

\begin{cor} For a fixed string $\textbf{a}$, $\textbf{T}_{A(\textbf{a})}$ and $\textbf{T}_{-A(\textbf{a})}$ do not both bound $\QQ S^1\times B^3s$. Consequently, $Y^0_{\textbf{a}}$ and $Y^{-1}_{\textbf{a}}$ do not both bound $\QQ B^4s$ and thus $\mathcal{S}_1\cap\mathcal{S}_2=\emptyset$. \label{orderbound}\end{cor}

\begin{proof}
	By Lemma \ref{order} in the Appendix (Section \ref{appendix}), $|\text{Tor}(H_1(\textbf{T}_{A(\textbf{a})}))|=|\text{Tor}(H_1(\textbf{T}_{- A(\textbf{a})}))|-4$. Thus $|\text{Tor}(H_1(\textbf{T}_{A(\textbf{a})}))|$ and  $|\text{Tor}(H_1(\textbf{T}_{-A(\textbf{a})}))|$ cannot simultaneously be squares and so by Lemma \ref{firsthomologyobstruction}, $\textbf{T}_{A(\textbf{a})}$ and $\textbf{T}_{-A(\textbf{a})}$ cannot simultaneously bound $\QQ S^1\times B^3s$. By construction, $|H_1(Y^0_{\textbf{a}})|=|\text{Tor}(H_1(\textbf{T}_{A(\textbf{a})}))|$ and $|H_1(Y^{-1}_{\textbf{a}})|=|\text{Tor}(H_1(\textbf{T}_{-A(\textbf{a})}))|$; the second statement of the corollary follows.
\end{proof}

\noindent The next obstruction is a consequence of Proposition 1.5 and Corollary 1.6 in \cite{cochranetal}.

\begin{lem}[\cite{cochranetal}] If $K\subset S^3$ is an alternating knot and $S^3_0(K)$ bounds a $\QQ S^1\times B^3$, then $\sigma(K)=0$.
\label{d1/2invtcor}\end{lem}

\subsection{Heegaard Floer homology calculations}\label{baldwinsection}
Let $\textbf{a}=(a_1,\ldots,a_n)$, where $a_i\ge2$ for all $1\le i\le n$ and $a_j\ge3$ for some $j$. As mentioned in Section \ref{3braids}, the rational sphere $Y^t_{\textbf{a}}$ is the double cover of $S^3$ branched over the closure of the 3-braid represented by the word $(\sigma_1\sigma_2)^{3t}\sigma_1\sigma_2^{-(a_1-2)}\cdots\sigma_1\sigma_2^{-(a_n-2)}$. In \cite{baldwinfloertorusbundles}, Baldwin calculated the Heegaard Floer homology of these 3-manifolds equipped with the unique self-conjugate spin$^c$ structure $\mathfrak{s}_0$. In particular, he showed that:

\[ HF^+(Y_{\textbf{a}}^{2m},\mathfrak{s}_0)=\begin{cases}
       \Big(\mathcal{T}^+_0\oplus\ZZ_{0}^{m}\Big)\Big\{(3n-\sum a_i)/4\Big\} & \text{if }m\ge0 \\
      \Big(\mathcal{T}^+_0\oplus\ZZ_{-1}^{-m}\Big)\Big\{(3n-\sum a_i)/4\Big\} & \text{if }m<0 \\
   \end{cases}
\] \[ HF^+(Y_{\textbf{a}}^{2m+1},\mathfrak{s}_0)=\begin{cases}
     \Big( \mathcal{T}^+_0\oplus\ZZ_{-1}^{m}\Big)\Big\{(3n+4-\sum a_i)/4\Big\} & \text{if }m\ge0 \\
     \Big( \mathcal{T}^+_{-2}\oplus\ZZ_{-2}^{-(m+1)}\Big)\Big\{(3n+4-\sum a_i)/4\Big\} &\text{if } m<0 \\
   \end{cases}
\]
$$\text{and}$$
$$ \{d(Y_{\textbf{a}}^{t},\mathfrak{s})\text{ }|\text{ }\mathfrak{s}\neq\mathfrak{s}_0\}=\{d(Y_{\textbf{a}}^{s},\mathfrak{s})\text{ }|\text{ }\mathfrak{s}\neq\mathfrak{s}_0\} \text{ for all } s,t\in\ZZ.$$

\subsection{Reversing Orientation}\label{reverseorientation} Again, let $\textbf{a}=(a_1,\ldots,a_n)$, where $a_i\ge2$ for all $1\le i\le n$ and $a_j\ge3$ for some $j$. As discussed in the introduction, reversing the orientation of the hyperbolic torus bundle $\textbf{T}_{\pm A(\textbf{a})}$ yields the hyperbolic torus bundle $\overline{\textbf{T}}_{\pm A(\textbf{a})}=\textbf{T}_{\pm A(\textbf{d})}$, where $\textbf{d}=(d_1,\ldots,d_m)$ is the cyclic-dual of $\textbf{a}$ (\cite{neumann}). Likewise, by reversing the orientation on $Y_{\textbf{a}}^{t},$ we obtain $\overline{Y_{\textbf{a}}^{t}}=Y_{\textbf{d}}^{s}$ for some integer $s$.

\begin{lem} Let $\textbf{a}=(a_1,\ldots,a_n)$ and $\textbf{d}=(d_1,\ldots,d_m)$ be cyclic-dual. Then $\overline{Y_{\textbf{a}}^{t}}=Y_{\textbf{d}}^{-t}$.
\label{revolemma}\end{lem}

\begin{proof} One way to see this is by performing blowups and blowdowns to the surgery diagram of $Y_{\textbf{a}}^{t}$ until all of the surgery coefficients are positive (they would in fact be $(d_1,\ldots,d_m)$) and then change the orientation of the surgery diagram by reflecting it through the page (c.f. Neumann \cite{neumann}). A more robust way to prove this fact is by using the Heegaard Floer homology computations outlined in Section \ref{baldwinsection}.

As mentioned before the statement of the lemma, $\overline{Y_{\textbf{a}}^{t}}=Y_{\textbf{d}}^{s}$ for some integer $s$. Let $I(\textbf{a})=\sum_{i=1}^n(a_i-3)$ and $I(\textbf{d})=\sum_{i=1}^m(d_i-3)$. Recall that if $(a_1,\cdots,a_n)$ is of the form $(2^{[m_1]},3+n_1,\ldots,2^{[m_k]},3+n_k)$, then $(d_1,\ldots,d_m)$ is of the form $(3+m_1,2^{[n_1]},\ldots,3+m_k,2^{[n_k]})$ (\cite{neumann}). Thus $I(\textbf{a})+I(\textbf{d})=0$.

By the discussion in Section \ref{baldwinsection}, for any string of integers $\textbf{x}=(x_1,\ldots,x_n)$, where $x_i\ge 2$ for all $i$ and $x_j\ge 3$ for some $j$, we have:

\[ HF^+_{red}(Y_{\textbf{x}}^{t},\mathfrak{s}_0)=\begin{cases}
	 \ZZ^{\frac{t}{2}} & t\ge 0 \text{ and even}\\
      \ZZ^{-\frac{t}{2}} & t\le-2 \text{ and even}\\
      \ZZ^{\frac{t-1}{2}} & t\ge 1 \text{ and odd}\\
      \ZZ^{-\frac{t+1}{2}} & t\le -1 \text{ and odd}\\
   \end{cases}.
\]

\noindent Thus the rank of $HF^{+}_{red}$ depends only on the twisting of the chain link. Since the rank of $HF^{+}_{red}$ is a 3-manifold invariant, it is clear that $\overline{Y_{\textbf{a}}^{t}}=Y_{\textbf{d}}^{t}$ or $\overline{Y_{\textbf{a}}^{t}}=Y_{\textbf{d}}^{-t}$. If $t$ is odd, we can appeal to the correction terms, which are given by

\[ d(Y_{\textbf{x}}^{t},\mathfrak{s}_0)=\begin{cases}
1-\frac{I(\textbf{x})}{4} & t\ge 1 \text{ and odd}\\
-1-\frac{I(\textbf{x})}{4} & t\le -1 \text{ and odd}\\
\end{cases}.
\]

\noindent Since $d(\overline{Y_{\textbf{x}}^{t}},\mathfrak{s}_0)=-d(Y_{\textbf{x}}^{t},\mathfrak{s}_0)$ and $I(\textbf{a})=-I(\textbf{d})$ it is clear that $d(\overline{Y_{\textbf{a}}^{t}},\mathfrak{s}_0)\neq d(Y_{\textbf{d}}^{t},\mathfrak{s}_0)$. Thus $\overline{Y_{\textbf{a}}^{t}}=Y_{\textbf{d}}^{-t}$ for all odd $t$.

When $t$ is even, the correction term formula is not helpful. When $t=0$, it is clear that $\overline{Y_{\textbf{a}}^{0}}=Y_{\textbf{d}}^{0}$. If $t=2m>0$ is even, then $Y^t_{\textbf{a}}$ has surgery diagram shown at the top of Figure \ref{event}, which can be viewed as $(-1^{[m]})$-surgery along a link in $Y^0_{\textbf{a}}$. By reversing the orientation, we obtain $(1^{[m]})$-surgery on $Y^0_{\textbf{d}}$, which is simply $Y^{-t}_{\textbf{d}}$. Thus if $t$ is even, $\overline{Y_{\textbf{a}}^{t}}=Y_{\textbf{d}}^{-t}$.

\end{proof}

\section{Torus bundles over $S^1$ that bound rational homology circles}\label{tbundles}

In this section, we will prove Theorem \ref{thm1}. By considering the obvious handlebody diagrams of the plumbings shown in Figure \ref{torusbundles}, it is rather straight forward to classify elliptic and parabolic torus bundles over $S^1$ that bound $\QQ S^1\times B^3s$. In fact, by using Kirby calculus, we will explicitly construct $\QQ S^1\times B^3s$ bounded negative parabolic torus bundles and use the obstructions in Section \ref{obstructions} to obstruct the rest from bounding $\QQ S^1\times B^3s$.

\begin{prop} No elliptic torus bundle bounds a $\QQ S^1\times B^3$.\label{ellipticprop}\end{prop}

\begin{proof} According to Figure \ref{torusbundles}, there are only six elliptic torus bundles; they have monodromies $\pm S$, $\pm T^{-1}S$, and $\pm (T^{-1}S)^2$. By Proposition \ref{firsthomologyobstruction}, if one of these torus bundles bounds a $\QQ B^4$, then the torsion part of its first homology group must be a square. By considering the surgery diagrams in Figure \ref{torusbundles}, it is easy to see that the only elliptic torus bundles  that have the correct first homology are those with monodromy $T^{-1}S$ and $-(T^{-1}S)^2$. Moreover, note that by reversing the orientation on the torus bundle with monodromy $T^{-1}S$, we obtain the torus bundle with monodromy $-(T^{-1}S)^2$. Thus we need only show that one of these torus bundles does not bound a $\QQ S^1\times B^3$. Consider the leftmost surgery diagram of the elliptic torus bundle with monodromy $T^{-1}S$ in Figure \ref{ellobstruction}. By blowing down the $1$-framed unknot, we obtain $0$-surgery on the right-handed trefoil. Since the signature of the right-handed trefoil is 2, by Corollary \ref{d1/2invtcor}, the elliptic torus bundle does not bound a $\QQ S^1\times B^3$.\end{proof}

\begin{prop} Every negative parabolic torus bundle bounds a $\QQ S^1\times B^3$. No positive parabolic torus bundle bounds a $\QQ S^1\times B^3.$\label{parabolicprop}\end{prop}

\begin{proof} By considering the surgery diagrams of the parabolic torus bundles in Figure \ref{torusbundles}, it is easy to see that positive parabolic torus bundles, which have monodromy $T^n$, have $b_1=2$. Thus, by the homology long exact sequence of the pair, it is easy to see that no such torus bundle can bound a $\QQ S^1\times B^3$. On the other hand, the negative parabolic torus bundles with monodromy  $-T^n$ bound obvious $\QQ S^1\times B^3s$, as shown in Figure \ref{parabolicrationalcircles}.\end{proof}

\begin{figure}[t!]
	\centering
	\includegraphics[scale=.6 ]{ellobstruction.png}
	\caption{The elliptic torus bundle with monodromy $T^{-1}S$ does not bound a rational homology circle.}\label{ellobstruction}
\end{figure}
\begin{figure}[t!]
\centering
\includegraphics[scale=.6]{parabolic.png}
\caption{$\QQ S^1\times B^3$ bounded by the negative parabolic torus bundle with monodromy $-T^n$.}\label{parabolicrationalcircles}
\end{figure}

Classifying hyperbolic torus bundles that bound $\QQ S^1\times B^3s$ is not as simple as the elliptic and parabolic cases. The hyperbolic torus bundles listed in Theorem \ref{mainthm} were shown to bound $\QQ S^1\times B^3s$ in \cite{simone3}.

\begin{prop}[\cite{simone3}] Let $\textbf{a}=(3+x_1,2^{[x_2]},\ldots,3+x_{2k+1},2^{[x_1]}, 3+x_2,2^{[x_3]},\ldots,3+x_{2k},2^{[x_{2k+1}]})$, where $k\ge0$ and $x_i\ge0$ for all $i$. Then $\textbf{T}_{A(\textbf{a})}$ bounds a $\QQ S^1\times B^3$.\label{hypprop}\end{prop}

It remains to obstruct all other hyperbolic torus bundles from bounding $\QQ S^1\times B^3s$. To do this, we first need to prove a few technical lemmas. The first considers the effect on the string $\textbf{a}$ (viewed as surgery coefficients) when we perform certain blowups and blowdowns to $\textbf{T}_{\pm A(\textbf{a})}$.

\begin{lem} Let $\textbf{a}=(a_1,\ldots,a_n)$ and consider the obvious surgery diagram of $\textbf{T}_{A(\textbf{a})}$ in Figure \ref{hypsurgery}. Let $(d_1,\ldots,d_j)$ be the linear-dual of $(a_{i+1},\ldots,a_{n})$, where $1\le i\le n-1$. Then $\textbf{T}_{ A(\textbf{a})}$ can be obtained by surgery along the link shown in Figure \ref{hypsurgery2}, which has chain link surgery coefficients  $(-(a_1-1),-a_2,\ldots,-a_{i-1},-(a_i-1),d_1,\ldots,d_j)$.\label{technicallem1}\end{lem}

\begin{proof} This is a simple exercise in Kirby calculus. The trick is to perform $+1$-blowups followed by successive $-1$-blowdowns. Starting with the diagram in Figure \ref{hypsurgery}, blow up the linking of the $-a_1$- and $-a_n$-framed unknots with a $+1$-framed unknot. If $-(a_n-1)=1$, then blow down the $-1$-framed unknot and any subsequent $-1$-framed unknots. Otherwise, blow up the linking of the $+1$-framed unknot and the $-(a_n-1)$-framed unknot. Continuing in this way, we obtain the result.
\end{proof}

\begin{cor} Let $\textbf{a}=(a_1,\ldots,a_n)\in\mathcal{S}_{2c}$, where $a_1\ge3$. Then $\textbf{T}_{A(\textbf{a})}$ has surgery diagram with chain link surgery coefficients $(-d_1,\ldots,-d_i,d_1,\ldots,d_i)$, for some $2\le i\le n$. In particular, $d_1=a_1-1\ge2$, $d_i=a_i-1\ge2$ and $d_j=a_j$ for all $2\le j\le i-1$.
	\label{halfreverse}\end{cor}

\begin{proof}
	Without loss of generality, assume $$\textbf{a}=(3+x_1,2^{[x_2]},\ldots,2^{[x_{2k}]},3+x_{2k+1},\textcolor{blue}{2^{[x_1]}, 3+x_2,2^{[x_3]},\ldots,3+x_{2k},2^{[x_{2k+1}]}}).$$
	The blue substring has linear-dual $(2+x_1,2^{[x_2]},3+x_3,\ldots,2^{[x_{2k}]},2+x_{2k+1})$. By Lemma \ref{technicallem1}, $\textbf{T}_{A(\textbf{a})}$ has a surgery diagram with chain link surgery coefficients of the form $$(-(2+x_1),-2^{[x_2]},\ldots,-2^{[x_{2k}]},-(2+x_{2k+1}),2+x_1,2^{[x_2]},3+x_3,\ldots,2^{[x_{2k}]},2+x_{2k+1}),$$
which is of the desired form.
\end{proof}

\begin{figure}
	\centering
	\begin{subfigure}{.45\textwidth}
		\centering
		\includegraphics[scale=.55]{hypsurgery.png}
		\caption{Obvious surgery diagram of $\textbf{T}_{A(\textbf{a})}$}\label{hypsurgery}
		\vspace{.3cm}
	\end{subfigure}
	\begin{subfigure}{.45\textwidth}
		\centering
		\includegraphics[scale=.55]{hypsurgery2.png}
		\caption{An alternate surgery diagram of $\textbf{T}_{A(\textbf{a})}$}\label{hypsurgery2}
		\vspace{.3cm}
	\end{subfigure}
	\caption{Surgery diagrams for $\textbf{T}_{A(\textbf{a})}$}
\end{figure}

We now use Lemma \ref{technicallem1} and Corollary \ref{halfreverse} to show that $\mathcal{S}_{2a}\cap \mathcal{S}_{2c}$ and $\mathcal{S}_{2b}\cap\mathcal{S}_{2c}$ are nonempty. We call a string $(a_1,\ldots,a_n)$ a \textit{palindrome} if $a_i=a_{n-(i-1)}$ for all $1\le i\le n$.

\begin{lem}
Let $\textbf{a}=(b_1+3,b_2,\ldots,b_k,2,c_l,\ldots,c_1)\in\mathcal{S}_{2a}$ and $\textbf{b}=(3+x,b_1,\ldots,b_{k-1},b_k+1,2^{[x]},c_l+1,c_{l-1},\ldots,c_1)\in\mathcal{S}_{2b}.$ Then $\textbf{a}\in \mathcal{S}_{2c}$ if and only if $(b_1+1,b_2,\ldots,b_k)$ is a palindrome and $\textbf{b}\in\mathcal{S}_{2c}$ if and only if $(b_1\ldots,b_k)$ is a palindrome.
\label{palindrome}\end{lem}

\begin{proof} Since $(2,c_1,\ldots,c_l)$ has linear-dual $(b_k+1,b_{k-1},\ldots,b_1)$, by Lemma \ref{technicallem1}, $\textbf{T}_{A(\textbf{a})}$ has surgery diagram with chain link surgery coefficients $$(-(b_1+2),-b_2,\ldots,-b_{k-1},-(b_k-1),b_k+1,b_{k-1},\ldots,b_1).$$
If $\textbf{a}\in\mathcal{S}_{2c}$, then since $b_1+3>3,$ by Corollary \ref{halfreverse}, $b_1+1=b_k$ and $b_i=b_{k-(i-1)}$ for all $2\le i\le k-1$. Thus $(b_1+1,b_2,\ldots,b_k)$ is a palindrome. The converse is clearly true.

Similarly, since $(2^{[x]},c_l+1,\ldots,c_1)$ has linear-dual $(2+x,b_k,\ldots,b_1)$, by Lemma \ref{technicallem1}, $\textbf{T}_{A(\textbf{b})}$ has surgery diagram with chain link surgery coefficients $$(-(x+2),-b_1,\ldots,-b_k,2+x,b_{k},\ldots,b_1).$$
If $\textbf{b}\in	\mathcal{S}_{2c}$, then since $x+3\ge3,$ by Corollary \ref{halfreverse}, $b_i=b_{k-(i-1)}$ for all $1\le i\le k$. Thus $(b_1,\ldots,b_k)$ is a palindrome. Again, the converse is clear.
\end{proof}

The last technical lemma shows that certain cyclic covers of $\QQ S^1\times B^3s$ are  $\QQ S^1\times B^3s$.

\begin{lem} Let $W$ be a $\QQ S^1\times B^3$ and let $\widetilde{W}$ be a $p$-fold cyclic cover of $W$, where $p$ is prime and not a divisor of $|\textup{Tor}(H_2(W;\ZZ))|$. If $\partial \widetilde{W}$ is a $\QQ S^1\times S^2$, then $\widetilde{W}$ is a $\QQ S^1\times B^3$.\label{coverlem}\end{lem}

\begin{proof} Let $Y=\partial W$ and $\widetilde{Y}=\partial \widetilde{W}$. Since $W$ is a $\QQ S^1\times B^3$ and $H_3(W;\ZZ)$ has no torsion, it follows that $H_3(W;\ZZ)=0$. By Poincar{\'e} duality and the Universal Coefficient Theorem, we have the following isomorphisms:
	$$H_1(W,Y;\ZZ_p)\cong H^3(W;\ZZ_p)\cong \text{Hom}(H_3(W;\ZZ),\ZZ_p)\oplus\text{Ext}(H_2(W;\ZZ),\ZZ_p).$$
	Since $p$ is relatively prime to $|\textup{Tor}(H_2(Y;\ZZ)|$, we have that $H_1(W,Y;\ZZ_p)=0$. By the proof of Theorem 1.2 of \cite{goldsmith}, since $p$ is prime, it follows that $H_1(\widetilde{W},\widetilde{Y};\ZZ_p)=0$. Once again, by Poincar{\'e} duality and the Universal Coefficient Theorem, we have the following isomorphisms:
	$$0=H_1(\widetilde{W},\widetilde{Y};\ZZ_p)\cong H^3(\widetilde{W};\ZZ_p)\cong \text{Hom}(H_3(\widetilde{W};\ZZ),\ZZ_p)\oplus\text{Ext}(H_2(\widetilde{W};\ZZ),\ZZ_p).$$
	Thus $H_3(\widetilde{W};\QQ)=0$. Moreover, if we change the coefficients in the above string of isomorphisms to $\QQ-$coefficients, we obtain $H_1(\widetilde{W},\widetilde{Y};\QQ)=0$. Thus the map $H_1(\widetilde{Y};\QQ)\to H_1(\widetilde{W};\QQ)$ induced by inclusion is surjective. Since $\widetilde{Y}$ is a $\QQ S^1\times S^2$, it follows that $\text{rank}(H_1(\widetilde{W};\QQ))\le 1$. Finally, since $\chi(\widetilde{W})=p\chi(W)=0$ and  $H_3(\widetilde{W};\QQ)=0$, we necessarily have that $H_1(\widetilde{W};\QQ)=\QQ$ and $H_2(\widetilde{W};\QQ)=0$. \end{proof}

We are now ready to finish the proof of Theorem \ref{mainthm}. The proof relies, in part, on Theorem \ref{thm1}. The proof of Theorem \ref{thm1} takes up the remainder of the paper.

\begin{prop} No negative hyperbolic torus bundle bounds a $\QQ S^1\times B^3$. If $\textbf{T}_{A(\textbf{a})}$ bounds a $\QQ S^1\times B^3$, then  $\textbf{a}\in\mathcal{S}_{2c}$. \label{hypobstruct1}\end{prop}

\begin{proof}
Let $W$ be a $\QQ S^1\times B^3$ bounded by some hyperbolic torus bundle $\textbf{T}_{\pm A(\textbf{a})}$, where $\textbf{a}=(a_1,\ldots,a_n)$. Let $p$ be an odd prime number that is not a factor of $|\text{Tor}(H_2(W;\ZZ))|$. Consider the obvious surgery diagram of $\textbf{T}_{\pm A(\textbf{a})}$ as in Figure \ref{precover}, where the box labeled $(-1\pm1)/2$ indicates the number of half-twists. Let $\mu_i$ denote the homology class of the meridian of the $-a_i$-framed surgery curve and let $\mu_0$ denote the homology class of the meridian of the 0-framed surgery curve. Then $H_1(\textbf{T}_{\pm A(\textbf{a})})$ is generated by $\mu_0,\ldots,\mu_n$. Consider the torus bundle $\textbf{T}_{(\pm A(\textbf{a}))^p}$, which has monodromy $\pm (T^{-a_1}S\cdots T^{-a_n}S)^p$; it has the surgery diagram shown in Figure \ref{covering} (for the case $p=3$). There is an obvious $\ZZ_p$-action on $\textbf{T}_{(\pm A(\textbf{a}))^p}$ obtained by rotating the chain link through the 0-framed unknot by an angle of $2\pi /p$, as indicated in Figure \ref{covering}. The quotient of $\textbf{T}_{(\pm A(\textbf{a}))^p}$ by this action is clearly $\textbf{T}_{\pm A(\textbf{a})}$ and the induced map $f:H_1(\textbf{T}_{\pm A(\textbf{a})})\to \ZZ_p$ satisfies $f(\mu_0)=1$ and $f(\mu_i)=0$ for all $1\le i\le n$. Let $i_*:H_1(\textbf{T}_{\pm A(\textbf{a})})\to H_1(W)$ be the map induced by inclusion and choose a basis $\{m_0,m_1,\ldots,m_k\}$ for $H_1(W)$ such that $m_0$ has infinite order and $m_i$ is a torsion element for all $1\le i\le k$. Since $H_1(W,\textbf{T}_{\pm A(\textbf{a})})$ is a torsion group,  $i_*(\mu_0)=\alpha m_0+\sum_{i=1}^k\beta_im_i$ for some $\alpha,\beta_i\in\ZZ$, where $\alpha\neq 0$. Since $p$ is relatively prime to $|\text{Tor}(H_2(W;\ZZ))|=|H_1(W,\textbf{T}_{\pm A(\textbf{a})})|$, there exists an integer $t$ such that $t\alpha=1\text{mod}p$. Define a map $g:H_1(W)\to\ZZ_p$ by $g(m_0)=t$ and $g(m_i)=0$ for all $1\le i\le k$. Then $g$ is a surjective homomorphism satisfying $f=g\circ i_*$. Let $\widetilde{W}$ be the $p$-fold cyclic cover of $W$ induced by $g$. Then $\partial \widetilde{W}=\textbf{T}_{(\pm A(\textbf{a}))^p}$ and by Lemma \ref{coverlem}, $\widetilde{W}$ is a $\QQ S^1\times B^3$.

Let $\textbf{a}^p$ denote the string obtained by concatenating $\textbf{a}$ with itself $p$ times. Since $\textbf{T}_{\pm A(\textbf{a})}$ and $\textbf{T}_{(\pm A(\textbf{a}))^p}$ bound $\QQ S^1\times B^3s$, namely $W$ and $\widetilde{W}$, by Lemma \ref{rationalballobstruction}, $Y_{\textbf{a}}^{\frac{-1\pm1}{2}}$ and $Y_{\textbf{a}^p}^{\frac{-1\pm1}{2}}$ bound $\QQ B^4s$. Hence by Theorem \ref{thm1}, $\textbf{a},\textbf{a}^p\in\mathcal{S}\cup\{(6,2,2,2,6,2,2,2)\}$. By Remark \ref{ibounds}, $-4\le I(\textbf{a}),I(\textbf{a}^p)\le 0$. Since $I(\textbf{a}^p)=pI(\textbf{a})$, if $I(\textbf{a})<0$, then by taking $p>3$, we have that $I(\textbf{a}^p)<-4$, which is a contradiction. Thus $I(\textbf{a}^p)=I(\textbf{a})=0$ and by Remark \ref{ibounds}, $\textbf{a},\textbf{a}^p\in\mathcal{S}_{2a}\cup\mathcal{S}_{2b}\cup\mathcal{S}_{2c}\cup\{(6,2,2,2,6,2,2,2)\}$. By Lemma \ref{rationalballobstruction} and Theorem \ref{thm1}, if $\textbf{a}=(6,2,2,2,6,2,2,2)$, then $\textbf{T}_{-A(\textbf{a})}$ might bound a $\QQ S^1\times B^3$, whereas $\textbf{T}_{A(\textbf{a})}$ does not. Consequently, since $p$ is odd, $\textbf{a}^p\in \mathcal{S}_1$, which is not possible by Remark \ref{ibounds}. If $\textbf{a}^p=(6,2,2,2,6,2,2,2)$, then $p=2$ and $\textbf{a}=(6,2,2,2)$, up to cyclic reordering, which contradicts the assumption that $p$ is odd. Therefore, $\textbf{a},\textbf{a}^p\in\mathcal{S}_{2a}\cup\mathcal{S}_{2b}\cup\mathcal{S}_{2c}$.

By Corollary \ref{orderbound}, $\mathcal{S}_1\cap\mathcal{S}_2=\emptyset$. Thus by Theorem \ref{thm1}, $\partial W$ cannot be a negative hyperbolic torus bundle. We may now assume that $\partial W=\textbf{T}_{A(\textbf{a})}$, where $\textbf{a},\textbf{a}^p\in\mathcal{S}_{2a}\cup\mathcal{S}_{2b}\cup\mathcal{S}_{2c}$. It remains to show that $\textbf{a}\in\mathcal{S}_{2c}$, or equivalently, $\textbf{a}\notin\mathcal{S}_{2a}\cup\mathcal{S}_{2b}\setminus\mathcal{S}_{2c}$. Assume the opposite, $\textbf{a}\in \mathcal{S}_{2a}\cup\mathcal{S}_{2b}\setminus\mathcal{S}_{2c}$. We will show that $\textbf{a}^p\notin\mathcal{S}_{2a}\cup\mathcal{S}_{2b}\cup\mathcal{S}_{2c}$, which is a contradiction.

\begin{figure}
	\centering
	\begin{subfigure}{.45\textwidth}
		\centering
		\includegraphics[scale=.55]{precover.png}
		\caption{A surgery diagram for $\textbf{T}_{\pm A(\textbf{a})}$}\label{precover}
	\end{subfigure}
	\begin{subfigure}{.45\textwidth}
	\centering
	\includegraphics[scale=.55]{covering.png}
	\caption{A surgery diagram for $\textbf{T}_{(\pm A(\textbf{a}))^3}$}\label{covering}
	\end{subfigure}
\caption{$\textbf{T}_{ (\pm A(\textbf{a}))^3}$ is a $3$-fold cyclic cover of $\textbf{T}_{\pm A(\textbf{a})}$. The boxes labeled $(-1\pm 1)/2$ indicate the number of half-twists. There is an obvious $\ZZ_3$-action on $\textbf{T}_{ (\pm A(\textbf{a}))^3}$ given by a rotation of $120^{\circ}$ radians through the 0-framed unknot. The quotient of $\textbf{T}_{ (\pm A(\textbf{a}))^3}$ by this action is $\textbf{T}_{\pm A(\textbf{a})}$.}
\end{figure}

We first show that $\textbf{a}^p\notin\mathcal{S}_{2b}$ for large $p$. Let $\textbf{b}=(3+x,b_1,\ldots,b_{k-1},b_k+1,2^{[x]},c_l+1,c_{l-1},\ldots,c_1)\in\mathcal{S}_{2b}$. We claim that for large $p$, $\textbf{b}\neq \overline{\textbf{b}}^p$, for some substring $\overline{b}$ of $\textbf{b}$. Assume otherwise and without loss of generality, let $\overline{\textbf{b}}=(3+x,b_1,\ldots,b_m)$ be a substring of $\textbf{b}$ such that $\overline{\textbf{b}}^p=\textbf{b}$, where $k<m$. Such a substring exists because $p$ is assumed to be large. Suppose $(b_1,\ldots,b_m)$ is of the form
$$(2+z_1,2^{[w_1]},3+z_2,2^{[w_2]},\ldots,3+z_t,2^{[w_t]}).$$
Then its linear-dual is
$$(2^{[z_1]},3+w_1,2^{[z_2]},\ldots,2^{[z_t]},2+w_t)=(c_1,\ldots,c'_{m'}),$$
where $c'_{m'}<c_{m'}$ for some $m'<l$. Since $\overline{\textbf{b}}^p=\textbf{b}$, we necessarily have that $c_1=b_m, c_2=b_{m-1},\ldots$. Hence $z_i=w_{t-(i-1)}$ for all $1\le i\le t$ and $m=m'$. Thus $c_{m+1}=3+x\ge 3$. However, the linear-dual of $(b_1,\ldots,b_m,b_{m+1})=(b_1,\ldots,b_m,3+x)$ is $(c_1,\ldots,c_m,2^{[x+1]} )$, implying that $c_{m+1}=2$, which is a contradiction. Consequently, for $p$ large, $\textbf{a}^p\notin\mathcal{S}_{2b}$. A very similar argument shows that $\textbf{a}^p\notin\mathcal{S}_{2a}$ for large $p$. Let $\textbf{b}=(b_1+3,b_2,\ldots,b_k,2,c_l,\ldots,c_1)\in\mathcal{S}_{2b}$ and, as above, let $\overline{\textbf{b}}=(2,c_l,\ldots,c_m)$ be a substring of $\textbf{b}$ such that $\overline{\textbf{b}}^p=\textbf{b}$. Continuing as in the previous case, we can arrive to a contradiction. Thus for large $p$, $\textbf{a}^p\notin\mathcal{S}_{2a}\cup\mathcal{S}_{2b}$.

Now assume $\textbf{a}^p\in\mathcal{S}_{2c}$. We will show that this implies $\textbf{a}\in\mathcal{S}_{2c}$, contradicting our assumption. First, let $\textbf{a}\in\mathcal{S}_{2b}$ so that $\textbf{a}^p$ is of the form
\begin{equation*}
\begin{split}
\textbf{a}^p=(&3+x,b_1,\ldots,b_{k-1},b_k+1,2^{[x]},c_l+1,c_{l-1},\ldots,c_1,\\
&\hspace{1.5in} \vdots \text{ }{\scriptstyle \frac{p-2}{2} \text{ times}}\\
&3+x,b_1,\ldots,b_{k-1},b_k+1,2^{[x]},c_l+1,c_{l-1},\ldots,c_1,\\
&3+x,b_1,\ldots,b_{k-1},b_k+1,\textcolor{blue}{2^{[x]},c_l+1,c_{l-1},\ldots,c_1,}\\
&\textcolor{blue}{3+x,b_1,\ldots,b_{k-1},b_k+1,2^{[x]},c_l+1,c_{l-1},\ldots,c_1,}\\
&\hspace{1.5in}\textcolor{blue}{ \vdots \text{ }{\scriptstyle \frac{p-2}{2} \text{ times}}}\\
&\textcolor{blue}{3+x,b_1,\ldots,b_{k-1},b_k+1,2^{[x]},c_l+1,c_{l-1},\ldots,c_1,}).
\end{split}
\end{equation*}

By applying Lemma \ref{technicallem1} using the linear-dual of the blue substring (c.f. proof of Lemma \ref{palindrome}), $\textbf{T}_{A(\textbf{a})^p}$ has chain link surgery coefficients
\begin{equation*}
\begin{split}
(&-(2+x),-b_1,\ldots,-b_{k-1},-(b_k+1),-2^{[x]},-(c_l+1),-c_{l-1},\ldots,-c_1,\\
&\hspace{2.2in} \vdots \text{ }{\scriptstyle \frac{p-2}{2} \text{ times}}\\
&-(3+x),-b_1,\ldots,-b_{k-1},-(b_k+1),-2^{[x]},-(c_l+1),-c_{l-1},\ldots,-c_1,\\
&\hspace{.6in}-(3+x),-b_1,\ldots,-b_{k-1},-b_k,\textcolor{blue}{2+x,b_k,\ldots ,b_{2},b_1+1,}\\
&\hspace{1.2in}\textcolor{blue}{2^{[x]},c_1+1,c_{2},\ldots,c_l,2^{[x]},b_k,\ldots,b_{2},b_1+1,}\\
&\hspace{2.2in}\textcolor{blue}{ \vdots \text{ }{\scriptstyle \frac{p-2}{2} \text{ times}}}\\
&\hspace{1.3in}\textcolor{blue}{2^{[x]},c_1+1,c_{2},\ldots,c_l,2^{[x]},b_k,\ldots,b_{2},b_1}).
\end{split}
\end{equation*}

By Corollary \ref{halfreverse}, this string is necessarily of the form $(-d_1,\ldots,-d_p,d_1,\ldots,d_p)$. It is clear that $(b_k,\ldots,b_2,b_1+1)$ and $(b_1,\ldots,b_{k-1},b_k+1)$ must be the same. Thus $(b_1,\ldots,b_k)$ is a palindrome. By Lemma \ref{palindrome}, $\textbf{a}\in\mathcal{S}_{2c}$.

Next let $\textbf{a}\in\mathcal{S}_{2a}$ so that $\textbf{a}^p$ is of the form
\begin{equation*}
\begin{split}
\textbf{a}^p=(&b_1+3,b_2,\ldots,b_k,2,c_l,\ldots,c_1,\\
&\hspace{.9in} \vdots  \text{ }{\scriptstyle \frac{p-2}{2} \text{ times}}\\
&b_1+3,b_2,\ldots,b_k,2,c_l,\ldots,c_1,\\
&b_1+3,b_2,\ldots,b_k,\textcolor{blue}{2,c_l,\ldots,c_1,}\\
&\textcolor{blue}{b_1+3,b_2,\ldots,b_k,2,c_l,\ldots,c_1,}\\
&\hspace{.9in}\textcolor{blue}{ \vdots  \text{ }{\scriptstyle \frac{p-2}{2} \text{ times}}}\\
&\textcolor{blue}{b_1+3,b_2,\ldots,b_k,2,c_l,\ldots,c_1}).
\end{split}
\end{equation*}

By applying Lemma \ref{technicallem1} using the linear-dual of the blue substring as above, $\textbf{T}_{A(\textbf{a})^p}$ has chain link surgery coefficients
\begin{equation*}
\begin{split}
(&-(b_1+2),-b_2,\ldots,-b_k,-2,-c_l,\ldots,-c_1,\\
&\hspace{1.2in} \vdots  \text{ }{\scriptstyle \frac{p-2}{2} \text{ times}}\\
&-(b_1+3),-b_2,\ldots,-b_k,-2,-c_l,\ldots,-c_1,\\
&\hspace{-.2in}-(b_1+3),-b_2,\ldots,-(b_k-1),\textcolor{blue}{b_k+1,\ldots,b_1+1,}\\
&\hspace{.05in}\textcolor{blue}{2,2,c_1,\ldots,c_{l-1},c_l+b_k-1,b_{k-1},\ldots,b_1+1,}\\
&\hspace{1.2in}\textcolor{blue}{ \vdots  \text{ }{\scriptstyle \frac{p-2}{2} \text{ times}}}\\
&\hspace{.05in}\textcolor{blue}{2,2,c_1,\ldots,c_{l-1},c_l+b_k-1,b_{k-1},\ldots,b_1+1}).
\end{split}
\end{equation*}

By Corollary \ref{halfreverse}, this string is necessarily of the form $(-d_1,\ldots,-d_p,d_1,\ldots,d_p)$. It is clear that $(b_1+2,b_2,\ldots,b_k)$  and $(b_k+1,b_{k-1},\ldots,b_2,b_1+1)$ must be the same. Thus $(b_1+1,b_2,\ldots,b_k)$ is a palindrome. By Lemma \ref{palindrome}, $\textbf{a}\in\mathcal{S}_{2c}$.
\end{proof}

\section{Surgeries on chain links bounding rational homology 4-balls}\label{spheres}

In this section, we will prove the necessary conditions of Theorem \ref{thm1}. Namely, we will show that the $\QQ S^3s$ of Theorem \ref{thm1} bound $\QQ B^4s$ by explicitly constructing such $\QQ B^4s$ via Kirby calculus. Notice that the necessary condition of Theorem \ref{thm1}(\ref{(2)}) follows from the the necessary condition of Theorem \ref{thm1}(\ref{(1)}) in light of Lemma \ref{revolemma}. Therefore, we need only show the following. Let $\textbf{a}$ be a string and let $\textbf{d}$ be its cyclic-dual, then:
\begin{itemize}
	\item if $\textbf{a}\in\mathcal{S}_{1a}$, then $Y_{\textbf{a}}^{-1}$ bounds a $\QQ B^4$;
	\item if $\textbf{a}\in \mathcal{S}_{1b}\cup\mathcal{S}_{1c}\cup\mathcal{S}_{1d}\cup\mathcal{S}_{1e}$, then $Y_{\textbf{a}}^{-1}$ and $Y_{\textbf{d}}^{-1}$ bound $\QQ B^4s$; and
	\item if $\textbf{a}\in\mathcal{S}_2$, then $Y_{\textbf{a}}^{0}$ and $Y_{\textbf{d}}^{0}$ bound $\QQ B^4s$.
\end{itemize}

Figures \ref{negativekirby} and \ref{positivekirby} exhibit the Kirby calculus needed to produce these $\QQ B^4s$. We will describe in detail the $\QQ B^4$ constructed in Figure \ref{negativekirby}a. The constructions in the other cases are similar. Notice that the leftmost figure of Figure \ref{negativekirby}a (without the $-1$-framed blue unknot) is a surgery diagram for  $Y_{\textbf{a}}^{-1}$, where $\textbf{a}=(b_1,\ldots,b_k,2,c_l,\ldots,c_1,2)\in\mathcal{S}_{1a}$. Thicken $Y_{\textbf{a}}^{-1}$ to the 4-manifold $Y_{\textbf{a}}^{-1}\times [0,1]$. By attaching a $-1$-framed 2-handle to $Y_{\textbf{a}}^{-1}\times\{1\}$ along the blue unknot in Figure \ref{negativekirby}a, we obtain a 2-handle cobordism from $Y_{\textbf{a}}^{-1}$ to a new 3-manifold, which we will show is $S^1\times S^2$. By performing a blowdown, we obtain the middle surgery diagram. Blowing down a second time, the surgery curves with framings $-b_1$ and $-c_1$ linking each other once and have framings $-(b_1-1)$ and $-(c_1-1)$, respectively. Since $(b_1,\ldots,b_k)$ and $(c_1,\ldots,c_l)$ are linear-dual, either $-(b_1-1)$ or $-(c_1-1)$ is equal to $-1$. We can thus blow down again. Continuing in this way, we can continue to blow down $-1$-framed unknots until we obtain $0$-surgery on the unknot, which is shown on the right side of the figure. Thus we have a 2-handle cobordism from $Y_{\textbf{a}}^{-1}$ to $S^1\times S^2$. By gluing this cobordism to $S^1\times B^3$, we obtain the desired $\QQ B^4$ bounded by $Y_{\textbf{a}}^{-1}$.

Suppose $\textbf{a}\in\mathcal{S}_{1b}\cup\mathcal{S}_{1c}\cup\mathcal{S}_{1d}\cup\mathcal{S}_{1e}$ and let $\textbf{d}$ be its cyclic-dual. Then by  Lemma \ref{revolemma}, $\overline{Y_{\textbf{d}}^{-1}}=Y_{\textbf{a}}^{1}$. To show that $Y_{\textbf{d}}^{-1}$ bounds a $\QQ B^4$, we will show that  $Y_{\textbf{a}}^{1}$ bounds a $\QQ B^4$. Figures \ref{1b}$-$\ref{1e} show that if $\textbf{a}\in \mathcal{S}_{1b}\cup\mathcal{S}_{1c}\cup\mathcal{S}_{1d}\cup\mathcal{S}_{2e}$, then  $Y_{\textbf{a}}^{-1}$ and $Y_{\textbf{a}}^{1}$ and bound $\QQ B^4s$.

Note that Figure \ref{1b} depicts a cobordism similar to the one constructed in Figure \ref{1a}, which was described in the previous paragraph. However, the cobordisms constructed in Figures \ref{1c}$-$\ref{1e} are slightly different. In Figure \ref{1d}, we have a 2-handle cobordism from $Y_{\textbf{a}}^{\pm1}$ to $S^1\times S^2\#L(-4,1)$, which bounds a $\QQ S^1\times B^3$, since $L(-4,1)$ bounds a $\QQ B^4$ (\cite{liscalensspace}). Gluing this $\QQ S^1\times B^3$ to the cobordism yields the desired $\QQ B^4$. The cobordisms depicted in Figures \ref{1c} and \ref{1e} are built out of two 2-handles. These cobordisms are from $Y_{\textbf{a}}^{\pm1}$ to $S^1\times S^2\# S^1\times S^2$. Gluing these cobordisms to $S^1\times B^3\natural S^1\times B^3$ yields the desired $\QQ B^4s$.

Lastly, suppose $\textbf{a}\in\mathcal{S}_2$. By Lemma \ref{revolemma}, $\overline{Y_{\textbf{a}}^{0}}=Y_{\textbf{d}}^{0}$. Thus once we show that $Y_{\textbf{a}}^{0}$ bounds a $\QQ B^4$, it will follow that $Y_{\textbf{d}}^{0}$ also bounds a $\QQ B^4$. Figures \ref{2a}$-$\ref{2e} show that if $\textbf{a}\in\mathcal{S}_2$, then $Y_{\textbf{a}}^{0}$ bounds a $\QQ B^4$. The $\QQ B^4s$ in almost all of the cases are constructed in very similar ways as in the negative cases. The last case, $Y^0_{(2,2,2,3)}$, is much simpler; Figure \ref{2f} shows that $Y^0_{(2,2,2,3)}=L(-4,1)$, which bounds a $\QQ B^4$.

As shown above, if $\textbf{a}\in\mathcal{S}_{1b}\cup\mathcal{S}_{1c}\cup\mathcal{S}_{1d}\cup\mathcal{S}_{1e}$, then $Y_{\textbf{d}}^{-1}$ bounds a $\QQ B^4$. However, as the next results will show, if $\textbf{a}\in\mathcal{S}_{1a}$, then $Y_{\textbf{d}}^{-1}$ does not necessarily bound a $\QQ B^4$. The key is that $|H_1(Y_{\textbf{a}}^{-1})|$ can be either even or odd when $\textbf{a}\in\mathcal{S}_{1a}$, but, in all other cases, $H_1(Y_{\textbf{a}}^{-1})$ has even order. Recall that $[b_1,\ldots,b_k]$ represents the Hirzebruch-Jung continued fraction (see the Appendix (Section \ref{appendix}) for details).

\begin{prop} Let $\textbf{a}=(b_1,\ldots,b_k,2,c_l,\ldots,c_1,2)\in\mathcal{S}_{1a}$, where $[b_1,\ldots,b_k]=\frac{p}{q}$. Then $|H_1(Y^{-1}_{\textbf{a}})|=|\textup{Tor}(H_1(\textbf{T}_{-A(\textbf{a})}))|=p^2$.\label{1(a)order}
\end{prop}

\begin{proof}
	The proof can be found in the Appendix (Section \ref{appendix}).
\end{proof}

\begin{lem} Let $\textbf{a}=(2,b_1,\ldots,b_k,2,c_l,\ldots,c_1)\in\mathcal{S}_{1a}$, where $[b_1,\ldots,b_k]=\frac{p}{q}$, and let $\textbf{d}=(d_1,\ldots,d_m)$ be the cyclic-dual of $\textbf{a}$. If $p$ is odd, then $Y_{\textbf{d}}^{-1}$ does not bound a $\QQ B^4$.\label{noboundlem}\end{lem}

\begin{proof} To prove the lemma, we will show that $\overline{Y_{\textbf{d}}^{-1}}$ does not bound a $\QQ B^4$. By Lemma \ref{revolemma}, $\overline{Y_{\textbf{d}}^{-1}}=Y_{\textbf{a}}^{1}$, where $\textbf{a}=(b_1,\ldots,b_k,2,c_l,\ldots,c_k,2)$, where $(b_1,\ldots,b_k)$ and $(c_1,\ldots,c_l)$ are linear-dual strings. Notice that, $\displaystyle\frac{\sum (a_i-3)}{4}=-1$. By the calculations in Section \ref{baldwinsection}, $d(Y_{\textbf{a}}^{1},\mathfrak{s}_0)=1-\displaystyle\frac{\sum (a_i-3)}{4}=2$. Since $p$ is odd, by Proposition \ref{1(a)order}, the torsion part of $H_1(Y_{\textbf{a}}^{1})=H_1(Y_{\textbf{a}}^{-1})$ has odd order. Thus $\mathfrak{s}_0$ extends over any $\QQ B^4$ bounded by $Y_{\textbf{a}}^{1}$. Thus if $Y_{\textbf{a}}^{1}$ bounds a $\QQ B^4$, then $d(Y_{\textbf{a}}^{1},\mathfrak{s}_0)=0$, which is not possible.\end{proof}

\begin{remark} By Lemma \ref{rationalballobstruction} and Theorem \ref{mainthm}, we already know that if  $\textbf{a}\in\mathcal{S}_{2c}$, then $Y^0_{\textbf{a}}$  bounds a $\QQ B^4$. However, by \cite{simone3}, the $\QQ B^4s$ constructed via Theorem \ref{mainthm} necessarily admit handlebody decompositions with 3-handles. On the other hand, the $\QQ B^4s$ constructed in this section do not contain 3-handles. Thus $Y^0_{\textbf{a}}$ bounds a $\QQ B^4s$ without 3-handles, even though $\textbf{T}_{A(\textbf{a})}$ only bounds $\QQ S^1\times B^3s$ containing 3-handles.\end{remark}

\begin{figure}[h!]
\centering
\begin{subfigure}{\textwidth}
\centering
\vspace{.3in}
\includegraphics[scale=.55]{1a.png}
\caption{If $\textbf{a}\in\mathcal{S}_{1a}$, then  $Y^{-1}_{\textbf{a}}$ bounds a $\QQ B^4$}\label{1a}
\end{subfigure}\vspace{.8cm}
\begin{subfigure}{\textwidth}
\centering
\includegraphics[scale=.55]{1b.png}
\caption{If $\textbf{a}\in\mathcal{S}_{1b}$, then  $Y^{-1}_{\textbf{a}}$ and $Y^{1}_{\textbf{a}}$ bound  $\QQ B^4s$}\label{1b}
\end{subfigure}\vspace{.8cm}
\begin{subfigure}{\textwidth}
\centering
\includegraphics[scale=.55]{1c.png}
\caption{If $\textbf{a}\in\mathcal{S}_{1c}$, then  $Y^{-1}_{\textbf{a}}$ and $Y^{1}_{\textbf{a}}$ bound  $\QQ B^4s$}\label{1c}
\end{subfigure}
\caption{The 3-manifolds in Theorem \ref{thm1}(\ref{(1)}) bound rational balls}
\end{figure}

\begin{figure}\ContinuedFloat
\begin{subfigure}{\textwidth}
\centering
\vspace{.3in}
\includegraphics[scale=.55]{1d.png}
\caption{If $\textbf{a}\in\mathcal{S}_{1d}$, then  $Y^{-1}_{\textbf{a}}$ and $Y^{1}_{\textbf{a}}$ bound  $\QQ B^4s$}\label{1d}
\end{subfigure}\vspace{1cm}
\begin{subfigure}{\textwidth}
\centering
\includegraphics[scale=.5]{1e.png}
\caption{If $\textbf{a}\in\mathcal{S}_{1e}$, then  $Y^{-1}_{\textbf{a}}$ and $Y^{1}_{\textbf{a}}$ bound  $\QQ B^4s$}\label{1e}
\end{subfigure}
\caption{The 3-manifolds in Theorem \ref{thm1}(\ref{(1)}) bound rational balls (continued)}\label{negativekirby}
\end{figure}

\newpage

\begin{figure}[h!]
\begin{subfigure}{\textwidth}
\centering
\vspace{1.5cm}
\includegraphics[scale=.55]{2a.png}
\caption{If $\textbf{a}\in\mathcal{S}_{2a}$, then  $Y^{0}_{\textbf{a}}$ bounds a $\QQ B^4s$}\label{2a}
\end{subfigure}\vspace{1cm}
\begin{subfigure}{\textwidth}
\centering
\includegraphics[scale=.55]{2b.png}
\caption{If $\textbf{a}\in\mathcal{S}_{2b}$, then  $Y^{0}_{\textbf{a}}$ bounds a $\QQ B^4s$}\label{2b}
\end{subfigure}\vspace{1cm}
\begin{subfigure}{\textwidth}
\centering
\includegraphics[scale=.5]{2c.png}
\caption{If $\textbf{a}\in\mathcal{S}_{2c}$, then  $Y^{0}_{\textbf{a}}$ bounds a $\QQ B^4s$}\label{2c}
\end{subfigure}
\caption{The 3-manifolds in Theorem \ref{thm1}(\ref{(3)}) bound rational balls (continued)}\label{positivekirby}
\end{figure}

\begin{figure}[h!]\ContinuedFloat
	\begin{subfigure}{\textwidth}
		\centering
		\vspace{1in}
		\includegraphics[scale=.55]{2d.png}
		\caption{If $\textbf{a}\in\mathcal{S}_{2d}$, then  $Y^{0}_{\textbf{a}}$ bounds a $\QQ B^4s$}\label{2d}
	\end{subfigure}\vspace{1cm}
	\begin{subfigure}{\textwidth}
		\centering
		\includegraphics[scale=.55]{2e.png}
		\caption{If $\textbf{a}\neq(3,2,2,2)\in\mathcal{S}_{2e}$, then  $Y^{0}_{\textbf{a}}$ bounds a $\QQ B^4s$}\label{2e}
	\end{subfigure}\vspace{1cm}
\begin{subfigure}{\textwidth}
	\centering
	\includegraphics[scale=.55]{2f.png}
	\caption{If $\textbf{a}=(3,2,2,2)\in\mathcal{S}_{2e}$, then  $Y^{0}_{\textbf{a}}$ bounds a $\QQ B^4s$}\label{2f}
\end{subfigure}
\caption{The 3-manifolds in Theorem \ref{thm1}(\ref{(3)}) bound rational balls (continued)}\label{positivekirby}
\end{figure}

\clearpage

\section{Cyclic Subsets}\label{lattice}
The remainder of the sections are dedicated to proving the sufficient conditions of Theorem \ref{thm1}. In fact, we will prove something more general. We will show that if $t$ is odd and $Y_{\textbf{a}}^t$ bounds a $\QQ B^4$, then $\textbf{a}\in\mathcal{S}_1\cup\{(6,2,2,2,6,2,2,2)\}$ or $\textbf{d}\in\mathcal{S}_1\cup\{(6,2,2,2,6,2,2,2)\}$, and if $t$ is even and $Y_{\textbf{a}}^t$  bounds a $\QQ B^4$, then $\textbf{a}\in\mathcal{S}_2$ or $\textbf{d}\in\mathcal{S}_2$. For convenience, we recall the definition of these sets.\\

\noindent\textbf{Definition \ref{definition}.} Two strings are considered to be equivalent if one is a cyclic reordering and/or reverse of the other. Each string in each of the following sets is defined up to this equivalence. Moreover, in the following sets, strings of the form $(b_1,\ldots,b_k)$ and $(c_1,\ldots,c_l)$ are assumed to be linear-dual.
	\begin{itemize}
		\item $ \mathcal{S}_{1a}=\{(b_1,\ldots,b_k,2,c_l,\ldots,c_1,2)\,|\, k+l\ge3\}$
		\item $ \mathcal{S}_{1b}=\{(b_1,\ldots,b_k,2,c_l,\ldots,c_1,5)\,|\, k+l\ge2\}$
		\item $ \mathcal{S}_{1c}=\{(b_1,\ldots,b_k,3,c_l,\ldots,c_1,3)\,|\, k+l\ge2\}$
		\item $ \mathcal{S}_{1d}=\{(2,b_1+1,b_2,\ldots,b_{k-1},b_k+1,2,2,c_l+1,c_{l-1},\ldots,c_2,c_1+1,2)\,|\, k+l\ge3\}$
		\item $ \mathcal{S}_{1e}=\{(2, 3+x, 2, 3, 3, 2^{[x-1]},3,3)\,|\, x\ge0 \textup{ and } (3,2^{[-1]},3):=(4)\}$
		\item $ \mathcal{S}_{2a}=\{(b_1+3,b_2,\ldots,b_k,2,c_l,\ldots,c_1)\}$
		\item $ \mathcal{S}_{2b}=\{(3+x,b_1,\ldots,b_{k-1},b_k+1,2^{[x]},c_l+1,c_{l-1},\ldots,c_1)\,|\, x\ge0\text{ and } k+l\ge2\}$
		\item $\begin{aligned}[t] \mathcal{S}_{2c}=\{&(3+x_1,2^{[x_2]},3+x_3,2^{[x_4]},\ldots,3+x_{2k+1},2^{[x_1]}, 3+x_2,2^{[x_3]},\ldots,3+x_{2k},2^{[x_{2k+1}]})\,|\\ &\, k\ge0 \text{ and } x_i\ge 0 \text{ for all } i\}\end{aligned}$
		\item $ \mathcal{S}_{2d}=\{(2,2+x,2,3,2^{[x-1]},3,4)\,|\, x\ge0 \textup{ and } (3,2^{[-1]},3):=(4)\}$
		\item $ \mathcal{S}_{2e}=\{(2,b_1+1,b_2,\ldots,b_k,2,c_l,\ldots,c_2,c_1+1,2),(2,2,2,3)\,|\, k+l\ge3\}$
	\end{itemize}
Moreover, $\mathcal{S}_1=\mathcal{S}_{1a}\cup\mathcal{S}_{1b}\cup\mathcal{S}_{1c}\cup\mathcal{S}_{1d}\cup\mathcal{S}_{1e}$, $\mathcal{S}_2=\mathcal{S}_{2a}\cup\mathcal{S}_{2b}\cup\mathcal{S}_{2c}\cup\mathcal{S}_{2d}\cup\mathcal{S}_{2e}$, and $\mathcal{S}=\mathcal{S}_1\cup\mathcal{S}_2$.\\

First suppose $n=1$ and let $\textbf{a}=(a_1)$, where $a_1\ge 3$. Then $L_1^0$ and $L_1^{-1}$ are both the unknot and so $Y_{(a_1)}^{0}=L(a_1-2,1)$ and $Y_{(a_1)}^{-1}=L(a_1+2,1)$ (c.f. Figure \ref{chainlinks}). By Lisca's classification of lens spaces that bound $\QQ B^4s$ (\cite{liscalensspace}), the only such lens spaces that bound $\QQ B^4s$ are $L(1,1)=S^3$ and $L(4,1)$. Thus $Y^{-1}_{(a_1)}$ does not bound a $\QQ B^4$ for all $a_1\ge3$ and $Y_{(a_1)}^0$ bounds a $\QQ B^4s$ if and only if $a_1=3$ or $a_1=6$. In the former case, $\textbf{a}=(3)\in\mathcal{S}_{2c}$, and in the latter case, $\textbf{d}=(2,2,2,3)\in\mathcal{S}_{2e}$.

We now assume the length of $\textbf{a}$ is at least 2. Many of the arguments in the next three sections are similar to arguments given by Lisca in \cite{liscalensspace}. We begin by recalling relevant definitions and results and introducing new terminology for our purposes.

Throughout, we will consider the standard negative definite intersection lattice $(\ZZ^n,-I_n)$. Let $\{e_1,\ldots,e_n\}$ be the standard basis of $\ZZ^n$. Then, with respect to the product $\cdot$ given by $-I_n$, we have $e_i\cdot e_j=-\delta_{ij}$ for all $i,j$.

\begin{definition} A subset $S=\{v_1,\ldots,v_n\}\in\mathbb{Z}^n$ is:
\begin{itemize}

\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}
      $
      \vspace{.2cm}
\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 $-a_i\le -3$ for some $i$ 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}
Each vector $v_i$ is called a \textit{vertex} of $S$ and $v_i\cdot v_{i+1}=\pm 1$ is called a \textit{positive/negative intersection}. If $S$ is standard, then the vertices $v_1, v_n$ are called \textit{final} and all other vertices are called \textit{internal}. If $S$ is cyclic, then the indices of each vertex are understood to be defined mod$n$ (e.g. $v_{n+1}=v_1$). The string of integers $(a_1,\ldots,a_n)$ defined by $a_i=-v_i\cdot v_i$ is called the \textit{string associated} to $S$. Notice that if $S$ is linear, then its associated string $(a_1,\ldots,a_n)$ is well-defined up to reversal (i.e. $(a_1,\ldots,a_n)$ and $(a_n,\ldots,a_1)$ are both strings associated to $S$). Moreover, if $S$ is cyclic, then its associated string is well-defined up to reversal and cyclic-reordering. Finally, 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 $S_1$.  \label{standarddef}\end{definition}

\begin{remark} By standard linear algebra, it is easy to see that if $S$ is standard or cyclic, then $\{v_1,\ldots,v_n\}$ forms a linearly independent set in $\mathbb{Z}^n$ (c.f. Remark 2.1 in \cite{liscalensspace}).\label{LIremark}\end{remark}

\begin{remark} In \cite{liscalensspace}, Lisca classified all standard subsets of $\ZZ ^n$. The results in the next three sections rely in part on his classification of standard subsets. We will review his classification in Section \ref{liscaclassification}.\end{remark}

\begin{remark} Suppose $S=\{v_1,\ldots,v_n\}$ is a cyclic subset. Then by replacing $v_k$ with $v_k'=-v_k$, we obtain a new subset $\hat{S}=\{v_1,\ldots,v_{k-1},v_k',v_{k+1},\ldots,v_n\}$ such that $v_{k-1}\cdot v_{k}'=-v_{k-1}\cdot v_{k}$ and $v_k'\cdot v_{k+1}=-v_k\cdot v_{k+1}$. Notice that $S$ and $\hat{S}$ have the same associated strings. Thus we can change the number of positive and negative intersections of $S$ without changing the associated string. Conversely, any subset of the form $S=\{v_1,\ldots,v_n\}$ where $n\ge 3$, and
\begin{center} $v_i\cdot v_j=\begin{cases}
      -a_i\le-2 & \text{if } i=j \\
      \pm 1 & \text{if } |i-j|=1 \\
      \pm 1 & \text{if } i\neq j\in\{1,n\}\\
      0 & \text{otherwise}
      \end{cases}$\end{center}
can modified into a positive or negative cyclic subset by changing the signs of select vertices. In particular, for any negative cyclic subset, the negative intersection can be moved at will by negating select vertices. \label{remarkneg}\label{moveneg}\end{remark}

\begin{definition} Let $S=\{v_1,\ldots,v_n\}\subset\ZZ^n$ be a subset with $v_i\cdot v_i=-a_i$. Then we define $$I(S):=\displaystyle\sum_{i=1}^n(a_i -3).$$\end{definition}

\begin{example} The subset $S=\{e_1-e_2,e_2-e_3,\ldots,e_{n-1}-e_n,e_n+e_1\}\subset\ZZ^n$, $n\ge 2$, is a negative cyclic subset with associated string $(2^{[n]})$. Moreover, $I(S)=-n$, $p_2(S)=n$, and $p_j(S)=0$ for all $j\neq 2$. When $n=4$, there is an alternate subset with associated string $(2,2,2,2)$, namely $S'=\{e_1-e_2,e_2-e_3,-e_2-e_1,e_1+e_4\}$, which satisfies $p_1(S')=p_3(S')=2$. This latter subset will be used to construct the family strings in $\mathcal{S}_{1a}$. \label{example}\end{example}

Let $\textbf{a}=(a_1,\ldots,a_n)$. The rational sphere $Y^t_{\textbf{a}}$ is the boundary of the negative definite 2-handlebody $P^t$ whose handlebody diagram is given in Figure \ref{2handlebody}. Let $Q_{P^t}$ denote the intersection form of $P^t$. Note that $Q_{P^t}$ depends only on the parity of $t$. Further suppose $Y^t_{\textbf{a}}$ bounds a rational homology ball $B$. Then the closed 4-manifold $X^t=P^t\cup B$ is negative definite. By Donaldson's Diagonalization Theorem \cite{donaldson}, the intersection lattice $(H_2(X^t),Q_{X^t})$ is isomorphic to the standard negative definite lattice $(\ZZ^n,-I_n)$. Thus the intersection lattice $(H_2(P^t),Q_{P^t})$ must embed in $(\ZZ^n,-I_n)$. The existence of such an embedding implies the existence of a cyclic subset $S\subset \ZZ^n$ with associated string $(a_1,\ldots,a_n)$. Thus our goal is to classify all cyclic subsets of $\ZZ^n$, where $n\ge 2$.

\begin{figure}
	\centering
	\includegraphics[scale=.6]{2handlebody.png}
	\caption{A 4-manifold with boundary $Y^t_{\textbf{a}}$}\label{2handlebody}
\end{figure}

Recall that by reversing the orientation of $Y^t_{\textbf{a}}$, we obtain the $\overline{Y^t_{\textbf{a}}}=Y^{-t}_{\textbf{d}}$, where $\textbf{d}=(d_1,\ldots,d_m)$ is the cyclic-dual of $(a_1,\ldots,a_n)$ (Section \ref{reverseorientation}). In particular, $(a_1,\ldots,a_n)$ is of the form $(2^{[m_1]},3+n_1,\ldots,2^{[m_k]},3+n_k)$ if and only if $(d_1,\ldots,d_m)$ is of the form $(3+m_1,2^{[n_1]},\ldots,3+m_k,2^{[n_k]})$. If $S$ and $\overline{S}$ denote the cyclic subsets associated to $(a_1,\ldots,a_n)$ and $(d_1,\ldots,d_m)$, respectively, then $I(S)+I(\overline{S})=0$. Thus by possibly changing orientation, we may assume that $I(S)\le 0$.

The following theorem is the main result of our lattice analysis. The theorem, coupled with Example \ref{example}, classifies all cyclic subsets $S$ satisfying $I(S)\le 0$.

\begin{thm} Let $S$ be a cyclic subset containing at least one vertex with square at most $-3$ such that $I(S)\le 0$. Then $S$ is either negative with associated string in $\mathcal{S}_1\cup\{(6,2,2,2,6,2,2,2)\}$ or positive with associated string in $\mathcal{S}_2$. Moreover, $-4\le I(S)\le 0$.
\label{latticethm}\end{thm}

The sufficient conditions of Theorem \ref{thm1} follow from Theorem \ref{latticethm}. Coupled with the calculations in Section \ref{spheres}, Theorem \ref{latticethm} completes the proof of Theorem \ref{thm1}. The proof of Theorem \ref{latticethm} will span the next three sections. The proof will begin in earnest in Section \ref{p1>0}. In the remainder of this section, we introduce some more definitions and notation, and explore standard subsets.

\begin{definition} Let $S=\{v_1,\ldots,v_n\}\subset\mathbb{Z}^n$ be a subset. Then
	$$E^S_i:=\{j\text{ }|\text{ } v_j\cdot e_i\neq0\}$$
	 $$ V^S_{i}:=\{j\text{ }|\text{ } v_i\cdot e_j\neq 0\}$$
	  $$p_i(S):=\Big|\{j\text{ }|\text{ } |E_j^S|=i\}\Big|$$ \end{definition}

In some cases we will drop the superscript $S$ from the above notation if the subset being considered is understood. The following lemma will be important in future sections.

\begin{lem}[Lemma 2.5 in \cite{liscalensspace}] If $S=\{v_1,\ldots,v_n\}\subset\ZZ^n=\langle e_1,\ldots,e_n\rangle$ is any subset with $I(S)\le0$, then $$2p_1(S)+p_2(S)\ge\displaystyle\sum_{j=4}^n(j-3)p_j(S).$$ Moreover, by the proof of Lemma 2.5 in \cite{liscalensspace}, if $|v_i\cdot e_j|\le1$ for all $i,j$, then $$2p_1(S)+p_2(S)=\sum_{j=4}^n(j-3)p_j(S)-I(S).$$ \label{keylem1}\end{lem}

 To prove Theorem \ref{latticethm}, we will apply two strategies. The first will be to reduce certain  cyclic subsets to  standard subsets and appeal to Lisca's classification of standard subsets in \cite{liscalensspace}. The second will be to reduce (via \textit{contractions}) certain  cyclic subsets to a small list of base cases. In the upcoming subsection, we will recall Lisca's classification of  standard subsets and in the subsequent subsection, we will describe how to perform contractions and list the relevant base cases.

\subsection{Lisca's Classification}\label{liscaclassification} In Section \ref{p2>0}, we will construct standard subsets from cyclic subsets satisfying $p_1=0$ and $p_2>0$, thus reducing the problem of classifying such cyclic subsets to Lisca's classification of standard subsets in \cite{liscalensspace}. In this section, we collect some facts about  standard subsets proved by Lisca. These can be found in Sections 3-7 in \cite{liscalensspace}. In particular, the `moreover' statements in Proposition \ref{liscasummary2} are obtained by examining the proofs of Lemmas 7.1-7.3 in \cite{liscalensspace}.

\begin{prop} Let $T=\{v_1,\ldots,v_n\}$ be a  standard subset with $I(T)<0$. Then:
\begin{enumerate}
\item $I(T)\in\{-1,-2,-3\}$;
\item $|v_i\cdot e_j|\le1$ for all $i,j$;
\item $p_1(T)=1$ if and only if $I(T)=-3$ and if $p_1(T)=0$, then $p_2(T)>0$;
\item If $I(T)=-3$, then $p_1(T)=p_2(T)=1$ and $p_3(T)=n-2$;
\item If $I(T)=-2$, then $p_2(T)=3$, $p_4(T)=1$, and $p_3(T)=n-4$; and
\item If $I(T)=-1$, then $p_2(T)=2$, $p_4(T)=1$ and $p_3(T)=n-3$.
\end{enumerate}
\label{liscasummary1}\end{prop}

\begin{prop} Let $T$ be standard with $I(T)<0$. Let $x,y\ge0$.
\begin{enumerate}
\item If $I(T)=-3$, then: if $E_i=\{s\}$, then $v_s$ is internal (i.e. $1< s< n$) and $v_s\cdot v_s=-2$; if $|E_j|=2$, then $E_j=\{1,n\}$; either $v_1\cdot v_1=-2$ or $v_n\cdot v_n=-2$; and $v_1\cdot e_j=-v_n\cdot e_j$. Moreover, $T$ or its reverse has associated string of the form $(b_1,\ldots,b_k,2,c_l,\ldots,c_1)$, where $(b_1,\ldots,b_k)$ and $(c_1,\ldots,c_l)$ are linear-dual strings.
\item If $I(T)=-2$, then (up to reversal) $T$ has associated string of the form:
\begin{enumerate}[(a)]
\item $(2^{[x]}, 3, 2+y, 2+x, 3, 2^{[y]})$;
\item $(2^{[x]}, 3+y,2,2+x,3,2^{[y]})$; or
\item $(b_1,\ldots ,b_{k-1},b_k+1,2,2,c_l+1,c_{l-1},\ldots,c_1)$, where $(b_1,\ldots,b_k)$ and $(c_1,\ldots,c_l)$ are  linear-dual strings.
\end{enumerate}
Moreover, up to the action of Aut$\ZZ^n$, the corresponding embeddings are of the form:
\begin{enumerate}[(a)]
\item $\{e_{x+4}-e_{x+3},e_{x+3}-e_{x+2},\ldots,e_5-e_4,e_4-e_2-e_3,e_2+e_1+\sum_{i=x+5}^{x+y+4}e_i,\\-e_2-e_4-\sum_{i=5}^{x+4}e_i,e_2-e_1-e_3,e_1-e_{x+5},e_{x+5}-e_{x+6},\ldots,e_{x+y+3}-e_{x+y+4}\}$
\item $\{e_{x+4}-e_{x+3},e_{x+3}-e_{x+2},\ldots,e_5-e_4,e_4-e_2-e_3-\sum_{i=x+5}^{x+y+4}e_i,e_2+e_1,\\-e_2-e_4-\sum_{i=5}^{x+4}e_i,e_2-e_1-e_3,e_3-e_{x+5},e_{x+5}-e_{x+6},\ldots,e_{x+y+3}-e_{x+y+4}\}$
\item $\{u_1,\ldots,u_{k-1}, u_k +e_4-e_2-e_3, e_2+e_1,-e_2-e_4,e_2-e_1-e_3+w_1,w_2,\ldots,w_l\}$, where $k+l\ge 3$, $u_k=0$ or $w_1=0$, $|E_1|=|E_4|=2$, there exists an integer $i$ such that $|E_i|=3$ and $u_1\cdot e_i=-u_2\cdot e_i=-w_l\cdot e_i=1$, and there exists and integer $j\neq i$ such that $|E_j|=2$ and $u_1\cdot e_j, w_l\cdot e_j\neq 0$.
\end{enumerate}
\item If $I(T)=-1$, then (up to reversal) $T$ has associated string of the form:
\begin{enumerate}[(a)]
\item $(2+x,2+y,3,2^{[x]},4, 2^{[y]})$;
\item $(2+x,2,3+y,2^{[x]},4,2^{[y]})$; or
\item $(3+x,2,3+y,3,2^{[x]},3,2^{[y]})$.
\end{enumerate}
Moreover, up to the action of Aut$\ZZ^n$, the corresponding embeddings are of the form:
\begin{enumerate}[(a)]
\item $\{e_2+e_4+\sum_{i=5}^{x+4}e_i,e_1-e_2+\sum_{i=x+5}^{x+y+4}e_i,e_2-e_3-e_4,e_4-e_5,e_5-e_6,\ldots,e_{x+3}-e_{x+4},\\e_{x+4}-e_1-e_2-e_3,e_1-e_{x+5},e_{x+5}-e_{x+6},\ldots,e_{x+y+3}-e_{x+y+4}\}$
\item $\{e_2+e_4+\sum_{i=5}^{x+4}e_i,e_1-e_2,e_2-e_3-e_4-\sum_{i=x+5}^{x+y+4}e_i,e_4-e_5,\ldots,e_{x+3}-e_{x+4},\\e_{x+4}-e_1-e_2-e_3,e_3-e_{x+5},e_{x+5}-e_{x+6},\ldots,e_{x+y+3}-e_{x+y+4}\}$
\item $\{e_1-e_2-e_5-\sum_{i=6}^{x+5}e_i,e_2+e_3,-e_2-e_1-e_4+\sum_{i=x+6}^{x+y+5}e_i,-e_5+e_2-e_3,\\e_5-e_6,e_6-e_7,\ldots,e_{x+4}-e_{x+5},e_{x+5}+e_1-e_4,e_4-e_{x+6},e_{x+6}-e_{x+7},\ldots, \\e_{x+y+4}-e_{x+y+5}\}$
\end{enumerate}
\end{enumerate}
\label{liscasummary2}
\end{prop}

\subsection{Contractions, expansions, and base cases}\label{basecases}
In this section, we discuss how to reduce cyclic subsets with $p_1>0$ or $p_1=p_2=0$ to a short list of base cases.

\begin{definition} Suppose $S=\{v_1,\ldots,v_n\}$, $n\ge 3$, is a cyclic subset and suppose there exist integer $i,s$, and $t$ such that $E_i=\{s,\tilde{s},t\}$, where $\tilde{s}\in\{s\pm1\}$, $V_{\tilde{s}}\cap V_s=\{i\}$, $|v_u\cdot e_i|=1$ for all $u\in E_i$, and $a_t\ge 3$. After possibly cyclically reordering and reindexing $S$, we may assume that $s\notin\{ 1,n\}$. Let $S'\subset \mathbb{Z}^{n-1}=\langle e_1,\ldots,e_{i-1},e_{i+1},\ldots, e_n\rangle$ be the subset defined by
\begin{center}
$S'=S\setminus\{v_s,v_{\tilde{s}},v_t\}\cup\{v_s+v_{\tilde{s}},\pi_{e_i}(v_t)\},$
\end{center}

\noindent where $\pi_{e_i}(v_t)=v_t+(v_t\cdot e_i)e_i$. We say that $S'$ is obtained from $S$ by a \textit{contraction} and $S$ is obtained from $S'$ by an \textit{expansion}.\end{definition}

Since $s\notin\{1,n\}$ and $|v_{\tilde{s}}\cdot e_i|=|v_{s}\cdot e_i|=1$, we have that $v_{s-1}\cdot e_i=-v_{s}\cdot e_i$. Thus
\begin{center}
	$(v_s +v_{\tilde{s}})\cdot v_u=\begin{cases}
	1 & \text{if } \tilde{s}=s+1\text{ and } u\in\{s-1,s+2\} \\
	1 & \text{if } \tilde{s}=s-1\text{ and } u\in\{s-2,s+1\} \\
	0 & \text{otherwise}
	\end{cases}$
\end{center}

\noindent Moreover, $(\pi_{e_i}(v_t))^2=v_t^2+1\le -2$ and
\begin{center}
	$\pi_{e_i}(v_t)\cdot v_u=\begin{cases}
	1 & \text{if } u=t\pm1 \\
	0 & \text{otherwise}
	\end{cases}.$
\end{center}

\noindent Therefore, $S'$ is positive/negative cyclic subset if and only if $S$ is positive/negative cyclic. Moreover, $I(S')=I(S)$, $p_j(S')=p_j(S)$ for all $j\neq 3$, and $p_3(S')=p_3(S)-1$.

\begin{definition} Using the notation above, if $v_t\cdot v_s=v_t\cdot v_{\tilde{s}}=0$ and $a_{\tilde{s}}=2$, then we say
	\begin{itemize}
		\item $S'$ obtained by a \textit{contraction of $S$ rooted at $v_t$ relative to $e_i$}.
	\end{itemize}
If $v_t\cdot v_s=1$ (so that $t=s\pm1$ if $\tilde{s}=s\mp1$) and $a_{\tilde{s}}=2$, then we say
\begin{itemize}
	\item $v_s$ is the \textit{center of $S$ relative to $e_i$} and
	\item $S'$ is obtained by a \textit{contraction of $S$ centered at $v_s$}.
\end{itemize}
In both cases, we say $S$ is obtained by a \textit{$-2$-expansion of $S$}.
\end{definition}

Note that a subset obtained by a contraction of $S$ centered at $v_s$ is unique. Indeed, if $E_i=\{s-1,s,s+1\}$, $a_{s-1}=2$, $a_{s+1}\ge3$, then $V_{s-1}\cap V_s=\{i\}$ and the only contraction centered at $v_s$ is $S\setminus\{v_s,v_{s-1},v_{s+1}\}\cup\{v_{(s-1,s)},\pi_{e_i}(v_{s+1})\}$. Similarly, if $E_i=\{s-1,s,s+1\}$, $a_{s-1}=2$, $a_{s+1}\ge3$, then $V_{s-1}\cap V_s=\{i\}$ and the only contraction centered at $v_s$ is $S\setminus\{v_s,v_{s-1},v_{s+1}\}\cup\{v_{(s,s+1)},\pi_{e_i}(v_{s-1})\}$. Now let $S$ have associated string $(a_1,\ldots,a_n)$. Then, under the contraction centered at $v_s$, the associated string changes as follows:
\begin{center}
 $(a_1,\ldots,a_{s-2},2,\bm{a_{s}},a_{s+1},a_{s+2},\ldots,a_n)\to(a_1,\ldots,a_{s-2},\bm{a_{s}},a_{s+1}-1,a_{s+2},\ldots, a_n)$ or \\
 $(a_1,\ldots,a_{s-2},a_{s-1},\bm{a_{s}},2,a_{s+2},\ldots,a_n)\to(a_1,\ldots,a_{s-2},a_{s-1}-1,\bm{a_{s}},a_{s+1},\ldots, a_n)$.
\end{center}

Notice that two strings $(b_1,\ldots,b_k)$ and $(c_l,\ldots,c_1)$ are reverse linear dual if and only if $(b_1,\ldots,b_{k-1})$ and $(c_l-1,\ldots,c_1)$ or $(b_1,\ldots,b_{k}-1)$ and $(c_{l-1},\ldots,c_1)$ are reverse linear-dual. Thus the substrings on either side of $a_s$ in the associated string of $S$ are reverse linear-dual if and only if the substrings on either side of $a_s$ in the associated string of the contraction of $S$ centered at $v_s$ are reverse linear-dual.

More generally, let $S=\{v_1,\ldots,v_n\}$ and consider a sequence of contractions $S^0=S$, $S^1$, $S^2,\ldots, S^m$ such that $S^k$ is obtained from $S^{k-1}$ by performing a contraction centered at $v_{s}^{(k-1)}\in S^{k-1}$, where $v_s^{(0)}=v_s$. We call such a sequence of contractions \textit{the sequence of contractions centered at $v_s$} and call the reverse sequence of expansions \textit{a sequence of $-2$-expansions centered at $v^{(m)}_s$}. Notice that for all $1\le k\le m$, $v_s^{(k)}=v^{(k-1)}_s+v^{(k-1)}_{\tilde{s}}$, where $v^{(k-1)}_{\tilde{s}}$ is the unique vertex of $S^{k-1}$ adjacent to $v^{(k-1)}_s$ with square $-2$. We have proven the following.

\begin{lem} Let $S'$ be obtained from $S$ by a sequence of contractions centered at $v$ and let $v^2=-a$. Then $S$ has associated string of the form $(b_1,\ldots,b_k,a,c_l,\ldots,c_1)$, where $(b_1,\ldots,b_k)$ and $(c_l,\ldots,c_1)$ are reverse linear-dual, if and only if $S'$ has associated string of the form $(b'_1,\ldots,b'_{k'},a,c'_{l'},\ldots,c'_1)$, where $(b'_1,\ldots,b'_{k'})$ and $(c'_{l'},\ldots,c'_1)$ are reverse linear-dual.\label{centeredlem}\end{lem}

When $p_1(S)>0$ or $p_1(S)=p_2(S)=0$, we will be able to sequentially perform contractions until we arrive at certain base cases. We will now list all cyclic subsets of length 2 and 3. First, it can be concretely checked case-by-case that the only cyclic subsets of length 2 (up to the action of Aut$\ZZ^2$) are:
\begin{itemize}
\item $\{e_1-e_2,e_2+e_1\}$, which is negative cyclic with associated string $(2,2)$; and
\item $\{2e_1,-e_1+e_2\}$, which is positive cyclic with associated string $(4,2)$.
\end{itemize}

\noindent Similarly, it can be verified that the only cyclic subsets of length 3 are:
\begin{itemize}
\item $\{e_1-e_2,e_2-e_3,e_3+e_1\}$, which is negative cyclic with associated string $(2,2,2)$;
\item $\{2e_1-e_3,-e_1-e_3,e_3+e_2\}$ which is positive cyclic with associated string $(5,2,2)$; and
\item $\{e_1-e_2-e_3,e_3-e_1-e_2,e_2-e_3-e_1\}$ which is positive cyclic with associated string $(3,3,3)$.
\end{itemize}

Notice that the second and third vertices of the subset with associated string $(5,2,2)$ are both centers relative to $e_3$. If we perform a contraction centered at either vertex relative to $e_3$, we obtain the subset with associated string $(4,2)$. Note that when $n=3$, centers are not unique, but when $n\ge 4$, centers are necessarily unique.

The subset with associated string $(3,3,3)$ falls under a larger class of base cases, namely subsets satisfying $p_3(S)=n$. These subsets are given by:
\begin{itemize}
\item $\{e_1-e_2-e_3,e_3-e_4-e_5,\dots,e_{2k-1}-e_{2k}-e_{2k+1},e_{2k+1}-e_1-e_2,e_2-e_3-e_4,\\\ldots,e_{2k}-e_{2k+1}-e_1\}.$
\end{itemize}
Note that these are positive cyclic of odd length and have associated strings of the form $(3^{[2k+1]})$. In Section \ref{p3=n}, we will see that any  cyclic subset with $p_1(S)=p_2(S)=0$ can be contracted to one of these base cases.

\begin{remark} For notation convenience, we will usually denote cyclic subsets by $S$ and standard subsets by $T$. Moreover, $S'$ will be reserved for contractions of $S$.\end{remark}

\section{Lattice Analysis Case I: $p_1(S)>0$}\label{p1>0}

Throughout this section, we will assume that $S=\{v_1,\ldots,v_n\}$ is a cyclic subset with $I(S)\le0$ and $p_1(S)>0$. Thus there exist integers $i$ and $s$ such that $E_i=\{s\}$. Lemmas \ref{lem1.3}$-$\ref{lem1.2} will ensure that we can contract such subsets.

\begin{lem} Let $S$ be a cyclic subset of length 4 such that $I(S)\le0$ and $E_i=\{s\}$ for some integers $i$ and $s$. If $a_{s+1}\ge 3$ or $a_{s-1}\ge3$, then $S$ is positive and has associated string of the form $(6,2,2,2)$. If $a_{s\pm1}=2$, then $S$ is either: negative and has associated string of the form $(2,2,2,2)$ or  $(2,2,2,5)$; or positive and has associated string of the form $(2,2,2,3)$ or $(2,2,2,6)$.
\label{lem1.3}\end{lem}

\begin{proof} If $|V_s|=1$, then since $E_i=\{s\}$, we obtain $v_s\cdot v_{s+1}=0$, which is a contradiction. Thus $|V_s|\ge2$.

Suppose $a_{s-1}\ge 3$. If $|V_s|\ge 3$, then let $R\subset\ZZ^3$ be the subset obtained by replacing $v_s$ by $v_s+(v_s\cdot e_i)e_i$.  Then $R$ is a cyclic subset and by Remark \ref{LIremark}, $R$ is made of four linearly independent vectors in $\ZZ^3$, which is not possible. Thus $|V_s|=2$. Let $V_s=\{i,j\}$. Then $E_j=\{s-1,s,s+1\}$, since otherwise, we would necessarily have that $|E_i|>1$. Moreover, since $V_{s-1}\cap V_s=V_{s+1}\cap V_s=\{j\}$, we necessarily have that $|v_{s-1}\cdot e_j|=|v_{s}\cdot e_j|=|v_{s+1}\cdot e_j|=1$. If $S$ is positive cyclic, then it is clear that $v_{s-1}\cdot e_j=v_{s+1}\cdot e_j=-v_{s}\cdot e_j$. If $S$ is negative cyclic, then by possibly moving the negative intersection (c.f. Remark \ref{moveneg}), we may assume that $v_{s-1}\cdot e_j=v_{s+1}\cdot e_j=-v_{s}\cdot e_j$. Thus we may perform a contraction of $S$ centered at $v_s$ relative to $e_j$ to obtain a length 3 cyclic subset $S'$ with $I(S')=I(S)\le0$ and $p_1(S')>0$. By considering the base cases in Section \ref{basecases}, it is clear that $S'=\{2e_1-e_3,-e_1-e_3,e_3+e_2\}$ (up to the action of Aut$\ZZ^3$), which has associated string $(5,2,2)$. Thus $s=1$, $i=2$, $j=4$, and $S=\{2e_1-e_3-e_4,-e_1-e_3,e_3-e_4,e_4+e_2\}$. Therefore, $S$ is positive and has associated string $(6,2,2,2)$.

Now suppose $a_{s-1}=a_{s+1}=2$. Without loss of generality, assume $s=j=4$. Let $T=\{v_1,v_2,v_3\}\subset\ZZ^3=\langle e_1,e_2,e_3\rangle$ be the length 3 standard subset obtained by removing $v_s$ from $S$. Then $T$ has associated string of the form $(2,a_2,2)$. Since $I(S)\le0$, we must have $a_2\le 6$. It is easy to see that $a_2\neq6$, since, otherwise, $v_2=2e_1-e_2-e_3$ (up to the action of Aut$\ZZ^3$), implying that $v_1\cdot v_2\neq \pm1$, which is a contradiction. If $a_2=5$, then $T$ is of the form $\{e_1-e_2,e_2+2e_3,-e_2-e_1\}$ and therefore, $S$ must be of the form $\{e_1-e_2,e_2+2e_3,-e_2-e_1, e_1+e_4\}$  (up to the action of Aut$\ZZ^3$). Thus $S$ is negative with associated string $(2,5,2,2)$ (equivalently, $(2,2,2,5)$). If $a_2\le 4$, then $I(T)<0$. By Lemma \ref{liscasummary2}, the only such length 3 standard subset has associated string $(2,2,2)$. Moreover, $T$ is of the form $T=\{e_1-e_2,e_2-e_3,-e_2+e_1\}$ (c.f. Lemma 2.4 in \cite{liscalensspace}). Since $v_3\cdot v_4=\pm1$, either $1\in V^S_4$, $2\in V^S_4$, or both.  If $1,2\in V^S_4$, then since $v_2\cdot v_4=0$, we must have $3\in V^S_4$; thus $|V^S_4|=4$. Moreover, since $v_1\cdot v_4=\pm1$, we must have that $v_4\cdot e_1=v_4\cdot e_2\pm1$, implying that $a_4\ge 7$, which is not possible. Thus either $1\in V^S_4$ or $2\in V^S_4$, but not both. If $1\in V^S_4$, then $S$ is negative and of the form $\{e_1-e_2,e_2-e_3,-e_2-e_1,e_1+e_4\}$ or $\{e_1-e_2,e_2-e_3,-e_2-e_1,e_1+2e_4\}$, which have associated strings $(2,2,2,2)$ and $(2,2,2,5)$ (note that we found the latter subset above). If $2\in V^S_4$, then $3\in V^S_4$ and $S$ is positive and of the form $\{e_1-e_2,e_2-e_3,-e_2-e_1,e_2+e_3+e_4\}$ or $\{e_1-e_2,e_2-e_3,-e_2-e_1,e_2+e_3+2e_4\}$, which have associated strings $(2,2,2,3)$ and $(2,2,2,6)$.
\end{proof}

\begin{lem} Let $S$ be a cyclic subset of length at least 5 such that $E_i=\{s\}$ for some $i$ and $s$. Then $|V_s|=2$. Moreover, if $V_s=\{i,j\}$, then $E_j=\{s-1,s,s+1\}$ and $v_{s-1}\cdot e_j=v_{s+1}\cdot e_j=-v_{s}\cdot e_j=\pm1$.\label{lem1.1}\end{lem}

\begin{proof} First note that if $|V_s|=1$, then since $E_i=\{s\}$, we obtain $v_s\cdot v_{s+1}=0$, which is a contradiction. Now suppose $|V_s|\ge3$. Then by replacing $v_s$ with $v_s'=v_s+(v_s\cdot e_i)e_i$ and relabeling $v_u'=v_u$ for all $u\neq s$, we obtain a subset $R=\{v'_1,...,v'_{s-1},v_{s}',v'_{s+1},...,v'_n\}\subset \mathbb{Z}^{n-1}=\langle e_1,\ldots,e_{i-1},e_{i+1},\ldots,e_n\rangle$. Let $(a_1',\ldots,a_n')$ be the string associated to $R$, where $-a_s':=v_s'\cdot v_s'\le -2$, and $a_j'=a_j$ for all $j\neq i$. If $S$ is negative cyclic, then so is $R$ and thus by Remark \ref{LIremark}, $R$ is made of $n$ linearly independent vectors in $\mathbb{Z}^{n-1}$, which is not possible. If $S$ is  positive cyclic and if either $a_s'\ge 3$ or $a_i\ge 3$ for some $i\neq s$, then $R$ is also positive cyclic, and we obtain a similar contradiction. Now suppose $S$ is positive cyclic, $a_s'=2$ and $a_t'=a_t=2$ for all $t\neq s$. Let $T$ be the subset obtained by removing $v_s$ from $S$. Then $T$ has associated string $(2^{[n-1]})$ and so $I(T)=-(n-1)\le -4$. If $|E_k^{S}|\ge2$ for all $k\in V_s^{S}$, where $k\neq i$, then $T$ is a standard subset of $\ZZ^{n-1}$ with $I(T)\le -4$, which contradicts Proposition \ref{liscasummary1}. If $|E_k^{S}|=1$ for some $k\in V_s^{S}$ such that $k\neq i$, then by Remark \ref{LIremark}, $T$ consists of $n-1$ linearly independent vectors in $\ZZ^{m}$, where $m<n-1$, which is not possible. Thus $|V_s|=2$.  Let $V^S_s=\{i,j\}$. Then, as in the proof of Lemma \ref{lem1.3}, $E_j=\{s-1,s,s+1\}$ and $v_{s-1}\cdot e_j=v_{s+1}\cdot e_j=-v_{s}\cdot e_j=\pm1$.
\end{proof}

\begin{lem} Let $S$ be a cyclic subset of length at least 5 such that $I(S)\le0$ and $E_i=\{s\}$ for some $i$ and $s$. Then either $a_{s-1}\ge 3$ or $a_{s+1}\ge3$. Moreover, if $a_{s\pm1}\ge 3$, then $S$ is positive with associated string $(2,3,2,3,2)$ or $(2,3,5,3,2)$.\label{lem1.2}\end{lem}

\begin{proof} By Lemma \ref{lem1.1}, we have $V_s=\{i,j\}$ and $E_j=\{s-1,s,s+1\}$. Assume that $a_{s-1}=a_{s+1}=2$. Then $V_{s-1}=\{j,k\}$ for some $k$, $V_{s+1}=\{j,k'\}$ for some $k'$, and $|v_{s\pm1}\cdot e_j|=|v_{s-1}\cdot e_k|=|v_{s+1}\cdot e_{k'}|=1$. Since $v_{s-1}\cdot v_{s+1}=0$, we must have $k=k'$. Since $|v_{s-2}\cdot v_{s-1}|=1$ and $j\notin V_{s-2}$, we must have $k\in V_{s-2}$. But then $v_{s-2}\cdot v_{s+1}\neq0$, which is a contradiction.

Now suppose $a_{s-1},a_{s+1}\ge 3$ and let $R$ be the subset obtained by removing $v_s$ and replacing $v_{s\pm1}$ with $v_{s\pm 1}'=v_{s\pm1}+(v_{s\pm1}\cdot e_j)\cdot e_j$. Note that $v_{s-1}'\cdot v_{s+1}'=\pm1$. As in the proof of Lemma \ref{lem1.1}, $R$ is either cyclic or $S$ is positive cyclic and $R$ has associated string of the form $(2^{[n-1]})$. In the former case, by Remark \ref{LIremark}, $R\subset\ZZ^{n-2}$ contains $n-1$ linearly independent vectors, which is not possible. In the latter case, let $T\subset\ZZ^{n-1}$ be the standard subset obtained from $S$ by only removing $v_s$. Then $T$ has associated string $(3,2,\ldots,2,3)$. By Proposition \ref{liscasummary2}, the only such standard subset is $\{e_4+e_3-e_2,e_2+e_1,-e_2-e_4,e_2+e_3-e_1\}$ (up to the action of Aut$\ZZ^4$), which has associated string $(3,2,2,3)$. Thus $j=3$, $|v_s\cdot e_3|=1$. Since $I(S)\le0$, $S$ is of the form $\{-e_2-e_4,e_2+e_3-e_1,e_5-e_3,e_4+e_3-e_2,e_2+e_1 \}$ or $\{-e_2-e_4,e_2+e_3-e_1,2e_5-e_3,e_4+e_3-e_2,e_2+e_1 \}$, which are positive and have associated strings $(2,3,2,3,2)$ and $(2,3,5,3,2)$, respectively.

\end{proof}

Let $S=\{v_1,\ldots,v_n\}$ be a cyclic subset such that $n\ge6$, $I(S)\le0$ and $E^S_i=\{s\}$ for some integers $i$ and $s$. By Lemma \ref{lem1.1}, we may assume that $V^S_s=\{i,j\}$ and $E^S_j=\{s-1,s,s+1\}$ for some integer $j$. Thus $v_s$ is the center vertex of $S$ relative to $e_j$. By Lemma \ref{lem1.2}, we may further assume that $a_{s+1}\ge3$ and $a_{s-1}=2$ and so $V^S_{s-1}=\{j,j_1\}$ for some integer $j_1$. Let $S'=\{v_1',\ldots,v_{s-2}',v'_s,v_{s+1}',\ldots,v'_n\}$ be the contraction of $S$ centered at $v_s$, where $v_x'=v_x$ for all $x\notin\{ s-1,s,s+1\}$, $v_{s}'=v_{s-1}+v_s$, and $v_{s+1}'=\pi_{e_j}(v_t)$. Since $V_{s}^{S'}=\{i,j_1\}$ and $E_{j_1}^{S'}=\{s-2,s,s+1\}$, $v_{s}'$ is the center vertex of $S'$ relative to $e_{j_1}$ and by Lemma \ref{lem1.2}, either $(v_{s-2}')^2\le-3$ or $(v_{s+1}')^2\le-3$. If $(v_{s-2}')^2\le-3$ and $(v_{s+1}')^2\le-3$, then by Lemma \ref{lem1.2}, $S'$ is positive and has associated string of the form $(2,3,2,3,2)$ or $(2,3,5,3,2)$. If $(v_{s-2}')^2=-2$ or $(v_{s+1}')^2=-2$, then we can perform the contraction centered at $v_s'$ relative to $e_{j_1}$, as above. Continuing in this way, we have a sequence of contractions centered at $v_s$, which ends in a subset $\hat{S}$ either of length $4$ or of length 5 with associated string $(2,3,2,3,2)$ or $(2,3,5,3,2)$. Let $\hat{v}_s$ denote the resulting center vertex of $\hat{S}$. Then $V_s^{\hat{S}}=\{i,k\}$ for some integer $k$ and $|E_k^{\hat{S}}|=3$.

Suppose that $\hat{S}$ has length 4. By considering the length 4 cyclic subsets in the proof of Lemma \ref{lem1.3}, it is clear that $\hat{S}$ is either negative and of the form:
\begin{itemize}
	\item $\{e_1-e_2,e_2-e_3,-e_2-e_1,e_1+e_4\}$ with associated string $(2,\textbf{2},2,\textbf{2})$; or
	\item $\{e_1-e_2,e_2-e_3,-e_2-e_1,e_1+2e_4\}$ with associated string $(2,\textbf{2},2,\textbf{5})$;
\end{itemize}
or positive and of the form:
\begin{itemize}
	\item $\{2e_1-e_3-e_4,-e_1-e_3,e_3-e_4,e_4+e_2\}$ with associated string $(6,2,2,\textbf{2})$.
\end{itemize}

\noindent Each bold number in the above strings corresponds to a vertex $\hat{v}_m$ satisfying $E^{\hat{S}}_{\alpha}=\{m\}$ for some integers $\alpha, m$. In particular, one of the bold numbers in each of the above strings corresponds to $\hat{v}_s$. In the first two cases, notice that the substrings in between the bold numbers (i.e. $(2)$ and $(2)$) are reverse linear-dual. Thus, by Lemma \ref{centeredlem}, $S$ has  associated string of the form $(b_1,\ldots,b_k,2,c_l,\ldots,c_1,2)$ or $(b_1,\ldots,b_k,2,c_l,\ldots,c_1,5)$, where $(b_1,\ldots,b_k)$ and $(c_l,\ldots,c_1)$ are reverse linear-dual. Similarly, the third string is of the form $(b_1+3,b_2,\ldots,b_k,2,c_l,\ldots,c_1)$, where $(b_1,\ldots,b_k)$ and $(c_l,\ldots,c_1)$ are reverse linear-dual and so $S$ has associated string of the same form. Note that the strings $(5,2,2)$ and $(4,2)$ also fall under this family (recall that the linear-dual of $(1)$ is the empty string).

Now suppose $\hat{S}$ has length 5. Then by the proof of Lemma \ref{lem1.2}, $\hat{S}$ is positive and of the form
\begin{itemize}
	\item $\{-e_2-e_4,e_2+e_3-e_1,e_5-e_3,e_4+e_3-e_2,e_2+e_1 \}$ with associated string $(2,3,\textbf{2},3,2)$ or
	\item $\{-e_2-e_4,e_2+e_3-e_1,2e_5-e_3,e_4+e_3-e_2,e_2+e_1 \}$ with associated string $(2,3,\textbf{5},3,2)$.
\end{itemize}

\noindent As above, the bold numbers in these two strings correspond to the vertex $\hat{v}_s$. Notice that after performing a $-2$-expansion centered at $\hat{v}_s$, the first and last entries in each string remain unchanged. Moreover, the substrings adjacent to the bold numbers are $(3)$ and $(3)$; notice $(3-1)=(2)$ and $(3-1)=(2)$ are reverse linear-dual strings. Thus, as above, $S$ has associated string of the form $(2,b_1+1,b_2,\ldots,b_k,2,c_l,\ldots,c_2,c_1+1,2)$ or $(2,b_1+1,b_2,\ldots,b_k,5,c_l, \ldots,c_2,c_1+1,2)$, where $(b_1,\ldots,b_k)$ and $(c_l,\ldots,c_1)$ are reverse linear-dual strings.

\begin{remark} Consider the length 5 subsets above. We can perform contractions to obtain the cyclic subsets of Lemma \ref{lem1.3} with associated strings $(2,2,2,3)$ and $(2,2,2,6)$. However, these do not fall under the general formulas listed above. Moreover, the string $(2,2,2,6)$ is also the associated string of a different subset, as seen in Lemma \ref{lem1.3}. This string already appeared in first set of cases we considered and so we will not count this string again. \end{remark}

\noindent Combining all of these cases, we have proven the following.

\begin{prop} Let $S$ be a cyclic subset with $I(S)\le0$ and $p_1(S)>0$. Then $S$ is either negative with associated string in $\mathcal{S}_{1a}\cup\mathcal{S}_{1b}$ or positive with associated string in $\mathcal{S}_{2a}\cup\mathcal{S}_{2b}\cup\mathcal{S}_{2e}$.

\label{prop1}
\end{prop}

\section{Lattice Analysis Case II: $p_1(S)=0$}\label{p1=0}

In this section, we will assume that $S=\{v_1,\ldots,v_n\}$ is  cyclic with $I(S)\le0$ and  $p_1(S)=0$. By Lemma \ref{keylem1}, $p_2(S)\ge\sum_{j=4}^n(j-3)p_j(S)$. If $p_2(S)=0$, then the inequality is necessarily an equality and so $p_j(S)=0$ for all $4\le j\le n$. Thus, in this case, $I(S)=0$ and $p_3(S)=n$. Therefore, we have two cases to consider: $p_2(S)=0$ and $p_2(S)>0$.

\subsection{Case IIa}\label{p3=n}
Let $S$ be cyclic and $p_1(S)=p_2(S)=0$. Then as shown above, $I(S)=0$ and $p_3(S)=n$. The next two lemmas provide some general properties of $S$.

\begin{lem} If $S$ is cyclic and $p_1(S)=p_2(S)=0$, then $|v_i\cdot e_j|\le 1$ for all $i,j$. \label{lem2a.1} \end{lem}
\begin{proof}
Let $v_i=\displaystyle\sum_{j=1}^nm_{ij}e_j$ for each $i$, where $m_{ij}=v_i\cdot e_j$. Then since $I(S)=0$, we have that $3n=-\displaystyle\sum_{i=1}^nv_i^2=\sum_{i,j} m_{ij}^2\ge \sum_{i,j}|m_{ij}|\ge 3n$. Thus $m^2_{ij}=|m_{ij}|$ for all $i,j$ and so $|v_i\cdot e_j|=|m_{ij}|\le1$ for all $i,j$.
\end{proof}

\begin{lem} If $S$ is cyclic and $p_1(S)=p_2(S)=0$, then $S$ is positive cyclic.\label{lem2a.2}\end{lem}

\begin{proof} Again, let $v_i=\displaystyle\sum_{j=1}^nm_{ij}e_j$. By Lemma \ref{lem2a.1}, $|m_{ij}|\le 1$ for all $i,j$. Let $\displaystyle\sum_{i=1}^nv_i=\sum_{i=1}^n\lambda_ie_i$. Then since $p_3(S)=n$,  $\lambda_i\in\{\pm1, \pm 3\}$ for all $i$. Now, if $S$ is negative, then $-3n=\displaystyle \sum_{i=1}^nv_i^2=\Big(\sum_{i=1}^nv_i\Big)^2-2\sum_{i<j}v_i\cdot v_j = (-\sum_{i=1}^n \lambda_i^2)-2(n-2)$ or $\displaystyle\sum_{i=1}^n\lambda_i^2=n+4$. Thus there must exist $j$ such that $\lambda_j=\pm 3$. But then $n-1\le\displaystyle\sum_{i\neq j}\lambda_i^2=n-5$, which is impossible. Thus $S$ must be positive.\end{proof}

As in Section \ref{p1>0}, we would like to be able to contract any cyclic subset $S$ with $p_3(S)=n$ to a base case. The base cases in question, which were mentioned in Section \ref{basecases}, have odd length and associated strings of the form $(3^{[2k+1]})$, namely subsets of the form $$(\star)\{e_1-e_2-e_3,e_3-e_4-e_5,\dots,e_{2k-1}-e_{2k}-e_{2k+1},e_{2k+1}-e_1-e_2,e_2-e_3-e_4,\ldots,e_{2k}-e_{2k+1}-e_1\},$$
where $k\ge1$. The following result of Lisca states that these are the only such subsets.

\begin{lem}[Proposition 3.21 in \cite{lisca3braid}] If $S$ satisfies $a_i=3$ for all $i$, then $S$ has odd length and is of the form $(\star)$, up to the action of Aut$\ZZ^{2k+1}$.\label{lem2a.3}\end{lem}

Now suppose $S$ does not have associated string of the form $(3^{[2k+1]})$. Then $n\ge4$ and since $I(S)=0$, there is an integer $s$ such that $a_s=2$.

\begin{lem} Suppose $S$ is cyclic and $p_1(S)=p_2(S)=0$. Let $s$ be an integer such that $a_s=2$ and let $V_s=\{i,j\}$. Then $E_i=\{s-1,s,t\}$ and $E_j=\{s,s+1,t\}$, where $t\neq s\pm1$. Moreover, $a_t\ge 3$ and if $t=s\pm2$, then $a_{s\pm1}\ge 3$. \label{lem2a.4}\end{lem}

\begin{proof} Without loss of generality, assume $i\in V_{s-1}$. Then $j\notin V_{s-1}$ because otherwise by Lemma \ref{lem2a.1}, $v_{s-1}\cdot v_s\neq 1$. Suppose that $i\in V_{s+1}$. Then $j\notin V_{s+1}$, since once again by Lemma \ref{lem2a.1}, $v_{s+1}\cdot v_s\neq 1$. Thus $E_j=\{s,t,u\}$, where $t\neq u\notin\{s-1, s, s+1\}$. But then $i\in V_u\cap V_t$, which implies $|E_i|\ge5$, which is a contradiction. Thus $i\notin V_{s+1}$, $j\in V_{s+1}$, and $E_i=\{s-1, s, t\}$, where $t\neq s+1$. Since $v_t\cdot v_s=0$ and $a_s=2$, we necessarily have that $j\in V_t$. Thus $E_j=\{s, s+1,  t\}$.

 If $n\ge 5$, then we may assume that $t\neq s+2$ and so $v_{t}\cdot v{s+1}=0$. Since $i\notin V_{s+1}$, there must be an integer $l\in V_t\cap V_{s+1}\setminus\{j\}$. Thus $|V_t|\ge 3$ and so $a_t\ge 3$. If $n=4$ and $V_t=\{i,j\}$, then since $I(S)=0$, we would have that $a_{s\pm1}=4$ and so $i,j\in V_{s\pm1}$, contradicting the first part of the lemma. Thus $a_t\ge3$.
Now suppose $t=s-2$ and $a_{s-1}=2$. Let $V_{s-1}=\{i,k\}$. Then $k\notin V_{s-2}\cup V_s$. Since $|E_k|=3$, there exists $u\notin\{s-2,s-1,s\}$ such that $k\in V_u$. Thus $i\in V_u$, which implies that $|E_i|\ge 4$, which is not possible. Thus $a_{s-1}\ge3$. Similarly, if $t=s+2$, then $a_{s+1}\ge 3$.
\end{proof}

By Lemma \ref{lem2a.4}, for every triple $i,j,s$ such that $V_s=\{i,j\}$, we have that $E_i=\{s-1,s,t\}$ and $E_j=\{s,s+1,t\}$. We will now consider two separate cases involving these triples, which will yield two kinds of families of cyclic subsets.

\begin{lem} Let $S$ be cyclic with $p_1(S)=p_2(S)=0$. Suppose there exists a triple $i,j,s$ such that $V_s=\{i,j\}$, $E_i=\{s-1,s,t\}$, $E_j=\{s,s+1,t\}$, and $t=s-2$ or $t=s+2$. Then $S$ has associated string of the form $(3+x,b_1,\ldots,b_{k-1},b_k+1,2^{[x]},c_l+1,c_{l-1},\ldots,c_1)$, where $(b_1,\ldots,b_k)$ and $(c_1,\ldots,c_l)$ are linear-dual strings. \label{lem2a.6}\end{lem}

\begin{proof}
	Let $i,j,s$ be as in the statement of the lemma. Without loss of generality, assume that $t=s-2$. Then $E^S_i=\{s-2,s-1,s\}$, $E_j^S=\{s-2,s,s+1\}$ and, by Lemma \ref{lem2a.4}, $a_{s-1}\ge3$ and $a_t\ge3$. Thus $v_s$ is the center of $S$ relative to $e_i$. Perform the contraction of $S$ centered $v_{s-1}$ to obtain a positive cyclic subset $S'=\{v_1',\ldots,v_{s-2}',v_{s-1}',v_{s+1}',\ldots,v_n'\}$, where $v_x'=v_x$ for all $x\notin\{ s-2,s-1,s\}$, $v_{s-1}'=v_{s-1}+v_s$, and $v_{s-2}'=\pi_{e_i}(v_{s-2})$. Moreover, $I(S')=0$ and $p_3(S')=n-1$. Now the vertices $v_{s-2}'$, $v_{s-1}'$, and $v_{s+1}'$ are adjacent in $S'$, $E_j^{S'}=\{s-2,s-1,s+1\}$, and $(v_{s-1}')^2=v_{s-1}^2\le-3$. Thus $v_{s-1}'$ is the center vertex of $S'$ relative to $e_j$. If $(v_{s-2}')^2=-2$ or $(v_{s+1}')^2=-2$, then we again contract $S'$ centered at $v_s'$. Continuing  in this way, we have a sequence of contractions centered at $v_{s-1}$ which terminates in a positive subset $\tilde{S}$ such that the resulting center vertex $\tilde{v}_{s-1}$ has adjacent vertices whose squares are both at most $-3$. Reindex $\tilde{S}$ chronologically and let $u=s-1$ under the new indexing. Then $\tilde{v}_u^2=v_{s-1}^2\le -3$, $\tilde{v}_{u\pm1}^2\le-3$ and there is an integer $l$ such that $E^{\tilde{S}}_l=\{u-1,u,u+1\}$.

	Let $R$ be the subset obtained by removing $\tilde{v}_u$ and replacing $\tilde{v}_{u\pm1}$ with $\tilde{v}'_{u\pm1}=\tilde{v}_{u\pm1}+(\tilde{v}_{u\pm1}\cdot e_l)e_l$. Then $I(R)\le -2$, $p_1(R)=0$, and $p_2(R)>0$. Since $\tilde{v}_u\cdot \tilde{v}_{u\pm1}=1$, we have that $\tilde{v}_{u-1}\cdot e_l=\tilde{v}_{u+1}\cdot e_l=\pm1$ and so $\tilde{v}'_{u-1}\cdot\tilde{v}'_{u+1}=1$. Thus if there exists a vertex of $R$ with square at most $-3$, then $R$ is a positive cyclic subset. However, by Propositions \ref{lem2b.6} and \ref{prop-1,-2} in Section \ref{p2>0}, positive cyclic subsets with $p_1=0$ and $p_2>0$ must satisfy $I=-1$. Thus every vertex of $R$ must have square equal to $-2$ and so $\tilde{S}$ has associated string of the form $(3+x,3,2^{[x]},3)$, where $-(\tilde{v}_u)^2=3+x$. Notice that the vertices adjacent to $3+x$ are both $(3)$ and $(3-1)=(2)$ and $(3-1)=(2)$ are reverse linear-dual strings.  Thus by Lemma \ref{centeredlem}, $S$ is of the form $(3+x,b_1,\ldots, b_{k-1},b_k+1,2^{[x]},c_l+1,c_{l-1},\ldots,c_1)$, where $(b_1,\ldots,b_k)$ and $(c_1,\ldots,c_l)$ are linear-dual strings.
\end{proof}

\begin{lem} Let $S=\{v_1,\ldots,v_n\}$ be cyclic with $p_1(S)=p_2(S)=0$. If $n\ge 6$, further assume that for every triple $i,j,s$ such that $V_s=\{i,j\}$, $E_i=\{s-1,s,t\}$, and $E_j=\{s,s+1,t\}$, we have that $t\neq s\pm2$. Then $S$ has associated string of the form
\begin{center} $(3+x_1,2^{[x_2]},3+x_3,2^{[x_4]},\ldots,3+x_{2k+1},2^{[x_1]}, 3+x_2,2^{[x_3]},\ldots,3+x_{2k},2^{[x_{2k+1}]}),$ \end{center}
where $x_i\ge0$ for all $i$ and $k\ge 1$.
\label{lem2a.5}\end{lem}

\begin{proof}
If $n=4$, then $t=s+2=s-2$ and if $n=5$, then $t\in\{s-2,s+2\}$. By Lemma \ref{lem2a.6}, $S$ must have associated string of the form $(4,3,2,3)$, $(5,3,2,2,3)$, or $(4,2,3,2,4)$ (up to cyclic reordering). Each of these strings fall under the general formula in the statement of Lemma \ref{lem2a.5}. Since $|E_y|=3$ for all $y$, the associated subsets must be of the form
\begin{itemize}
	\item $\{e_1-e_2-e_3-e_4,e_3-e_1-e_2,e_2-e_4,e_4-e_3-e_1\}$,
	\item $\{e_1-e_2-e_3-e_4-e_5,e_3-e_1-e_2,e_2-e_4,e_4-e_5,e_5-e_3-e_1\}$, and
	\item $\{e_1-e_2-e_3-e_4,e_3-e_5,e_5-e_1-e_2,e_2-e_4,e_4-e_3-e_1-e_5\}$, respectively.
\end{itemize}

\noindent We leave the proof to the reader. Note that all three of these subsets can be contracted to the subset ($\star$), where $k=1$, which has associated string $(3,3,3)$.

Now assume $n\ge 6$ and let $V^S_s=\{i,j\}$ and $a_s=2$. By Lemma \ref{lem2a.4}, $E^S_i=\{s-1,s,t\}$ and $E^S_j=\{s,s+1,t\}$, where $t\neq s\pm1$ and by the hypothesis, $t\neq s\pm2$. Since $v_t\cdot v_{s+1}=0$, $i\notin V_{s+1}^S$, we must have that $a_t\ge3$. Let $l\ge 1$ be the smallest integer such that $a_{s+l}\ge3$. It is easy to see that $V_{s+\alpha}=\{j_{\alpha-1},j_{\alpha}\}$, for all $1\le \alpha\le l-1$, where $j_0:=j$, $j_{\alpha}\notin\{i, j\}$ for all $1\le \alpha\le l-1$, and $t\neq s+\alpha$ for all $0\le \alpha\le l-1$. Similarly, let $m\ge 1$ be the smallest integer such that $a_{s-m}\ge 3$. Then $V_{s-\beta}=\{i_{\beta-1},i_{\beta}\}$ for all $1\le \beta\le m-1$, where $i_0:=i$, the set $\{i_{\beta},  j_{\alpha}\}$ has $m+l$ distinct elements, and $t\neq s-\beta$ for all $0\le \beta\le m-1$. Now since $v_{s+l-1}\cdot v_{s+l}=1$, we must have that $V_{s+l-1}\cap V_{s+l}=\{j_{l-1}\}\text{ or }\{j_{l-2}\}$. If $V_{s+l-1}\cap V_{s+l}=\{j_{l-2}\}$, then $j_{\alpha}\in V_{s+l}$ for all $0\le j\le l-1$ and, in particular, $j\in V_{s+l}$. Since $E^S_j=\{s,s+1,t\}$, we necessarily have $s+l=t$; consequently, since $|E^S_{j_{l-1}}|=3$, there exists an integer $u\notin \{s-m,\ldots,s+l\}$ such that $j_{l-1}\in V_u$. Finally, since $v_{s+l-1}\cdot v_u=0$, we necessarily have $j_{l-2}\in V_u$, implying that $|E_{j_{l-2}}|\ge 4$, which is not possible. Thus $V_{s+l-1}\cap V_{s+l}=\{j_{l-1}\}$ and $t\neq s+l$. Similarly, $V_{s-m+1}\cap V_{s-m}=\{i_{m-1}\}$ and $t\neq s-m$. Moreover, $E_{j_{\alpha}}=\{s+\alpha, s+\alpha +1, t\}$ and $E_{i_{\beta}}=\{s-\beta, s-\beta -1, t\}$ for all $0\le\alpha\le l-1$ and $0\le\beta\le m-1$.

We next claim that $t\notin \{s+l+1, s-(m+1)\}$. Assume that $t=s+l+1$. Then $j_{l-2},j_{l-1}, s+l-1$ is a triple satisfying $V^S_{s+l-1}=\{j_{l-2},j_{l-1}\}$, $E^S_{j_{l-2}}=\{s+l-2, s+l-1, s+l+1\}$, and $E^S_{j_{l-1}}=\{s+l-1, s+l,s+l+1\}$, which contradicts an assumption in the statement of the lemma. Thus $t\neq s+l+1$ and, similarly, $t\neq s-(m+1)$.

Now, since $j_{l-1}\in V_{t}\cap V_{s+l}$, $j_{m-1}\in V_{t}\cap V_{s-m}$, and $v_t\cdot v_{t\pm1}=1$, there exist integers $k_1\in V_{t}\cap V_{s+l}$, $k_2\in V_{t}\cap V_{s-m}$, $k_3\in V_t\cap V_{t-1}$, and $k_4\in V_t\cap V_{t-1}$ such that $k_1,k_2,k_3,k_4\notin \{i_0,\ldots,i_{m-1}, j_0,\ldots, j_{l-1}\}$. Moreover, these integers cannot all be the same since $|E^S_y|=3$ for all $y$. Therefore, $|V_t|\ge s+l+2$.

Since $V_{s+l-1} \cap V_{s+l}=\{j_{l-1}\}$ and $v_{s+l-1}\cdot v_{s+l}=1$, we have that $v_{s+l-1}\cdot e_{j_{l-1}}=-v_{s+l-1}\cdot e_{j_{l-1}}$. Thus we may perform a contraction rooted at $v_t$ relative to $e_{j_{l-1}}$ to obtain a cyclic subset  $S'=\{v_1',\ldots,v_{s+l-2}',v_{s+l}',\ldots,v_n'\}$, where $v_x'=v_x$ for all $x\notin\{s+l-1,s+l,t\}$, $v_{s+l}'=v_{s+l-1}+v_{s+l}$, and $v_{t}'=\pi_{e_{j_{l-1}}}(v_{t})$. Since $E_{j_{l-2}}^{S'}=\{s+l-2, s+1, t\}$ and $|V_{t}^{S'}|\ge s+l+1$, we may perform a contraction rooted at $v_t'$ relative to $e_{j_{l-2}}$. Continuing in this way, we can perform  rooted contractions relative to $e_{i_{\beta}}$ and $e_{j_{\alpha}}$ for every $0\le \alpha\le l-1$ and $0\le \beta \le m-2$ until we obtain a length $n-(l+m-1)$ positive cyclic subset of the form $\tilde{S}=\{\tilde{v}_1,\ldots,\tilde{v}_{s-m},\tilde{v}_{s+l},\ldots,\tilde{v}_n\}$, where $\tilde{v}_x=v_x$ for all $x\notin\{s+l, t\}$, $(\tilde{v}_t)^2\le -3$, and $\tilde{v}_{s+l}^2=v_{s+l}^2$. Notice that at each step of this rooted contraction process, the square of the vertex originally labeled $v_t$ decreases by 1. We call this sequence of contractions a \textit{rooted sequence of contractions}. The opposite sequence of $-2$-expansions is called a \textit{rooted sequence of expansions}.

Now if there exists a vertex in $\tilde{S}$ with square equal to $-2$ and the length of $\tilde{S}$ is 4 or 5, then as we saw above, we can continue contracting until we obtain $(3,3,3)$. If the length of $\tilde{S}$ is at least 6, then the hypotheses of the statement of lemma is still satisfied and we can perform another rooted sequence of contractions. Continue in this way, we will obtain a cyclic subset $\hat{S}$ that has no vertex with square $-2$. Since $I(\hat{S})=0$, it must have associated string is of the form $(3,\ldots,3)$. By Lemmas \ref{lem2a.2} and \ref{lem2a.3}, $\hat{S}$ has odd length and is a positive cyclic subset of the form ($\star$). To recover $S$, we must perform sequences of rooted expansions starting with ($\star$).

For each vertex in ($\star$), there is a sequence of expansions rooted at that vertex. More precisely, for every integer $i$, there is a unique sequence of expansions rooted at $v_{\frac{i}{2}}$ if $i$ is even and at $v_{\frac{2k+i+1}{2}}$ if $i$ is odd (up to the action of Aut$\ZZ^{2k+2}$). For example, for $e_2$, we replace $v_1$ and $v_{k+2}$ with $v_1^*=e_1-e_2-e_3-e_{2k+2}$ and $v_{k+2}^*=e_{2k+2}-e_3-e_4$ and add the vertex $v_x=e_2-e_{2k+2}$ between $v_{k+1}$ and $v_{k+2}^*$. The corresponding associated string is $(4,3^{[k]},2,3^{[k]})$. Continuing with this rooted sequence of expansions, we obtain a subset with associated string $(3+x,3^{[k]},2^{[x]},3^{[k]})$. We can continue performing rooted $-2$-expansions at each vertex in ($\star$) to obtain subsets with associated strings of the form $(3+x_1,2^{[x_2]},3+x_3,2^{[x_4]},\ldots,3+x_{2k+1},2^{[x_1]}, 3+x_2,2^{[x_3]},\ldots,3+x_{2k},2^{[x_{2k+1}]}).$
\end{proof}

\begin{remark} The positive cyclic subsets of Lemma \ref{lem2a.6} are obtained by performing $-2$-expansions starting with ($\star$), where $k=1$. These kinds of expansions cannot be applied when $k\ge 2$ because there are no triples $i,j,s$ such that $V_s=\{i,j\}$ and $t=s-2$ or $t=s+2$, where $E_i=\{s-1,s,t\}$ and $E_j=\{s,s+1,t\}$.\end{remark}

\noindent To summarize, we have proven the following.

\begin{prop} Let $S$ be a cyclic subset with $p_1(S)=p_2(S)=0$. Then $S$ is positive with associated string in $\mathcal{S}_{2b}\cup\mathcal{S}_{2c}$.

\label{prop2}
\end{prop}

\subsection{Case IIb: $p_2(S)>0$}\label{p2>0}\hfill

Throughout this section, we will consider cyclic subsets satisfying $p_1(S)=0$ and $p_2(S)>0$. By the discussion in Section \ref{basecases}, there are unique length 2 and 3 cyclic subsets satisfying these assumptions, namely the negative cyclic subsets with associated strings $(2,2)$ and $(2,2,2)$, respectively. Thus we will focus on subsets of length at least 4.

\begin{lem} Let $S$ be cyclic, $p_1(S)=0$, $p_2(S)>0$, and $n\ge4$. Let $i,s,t$ be integers such that $E_i=\{s,t\}$. Then either $a_s\ge 3$ or $a_t\ge 3$.\label{lem2b.1}\end{lem}

\begin{proof}
Assume $a_s=a_t=2$.

Suppose $v_s\cdot v_t=0$. Then $V_s=V_t=\{i,j\}$, for some $j$, and $E_j\supseteq\{s-1,s,s+1,t-1,t,t+1\}$. If $n\ge 5$, then either $v_{s-1}\cdot v_t=0$ or $v_{s+1}\cdot v_t=0$, and so $i\in V_{s-1}$ or $i\in V_{s+1}$, which is a contradiction. If $n=4$, then $t\pm1=s\mp1$. Since $v_{t-1}\cdot v_{t+1}=0$, there exists an integer $k$ such that $k\in V_{t\pm1}$. Moreover, there exists a fourth integer $m$ such that $m\in V_{t+1}$ or $V_{t-1}$, but not both, since $v_{t-1}\cdot v_{t+1}=0$. Thus $p_1(S)>0$, contradicting the hypothesis.

Now suppose $|v_s\cdot v_t|=1$ and, without loss of generality, let $t=s+1$. Since $a_s=a_{s+1}=2$, we have that $V_s=\{i,j\}$ and $V_{s+1}=\{i,i_1\}$, where $i_1\neq j$. Let $l\ge 2$ be the smallest integer such that $a_{s+l}\ge3$. Then it is easy to see that $V_{s+\alpha}=\{i_{\alpha-1},i_{\alpha}\}$, for all $1\le \alpha\le l-1$, where $i_0:=i$, $i_{\alpha}\notin\{ i, j\}$ for all $1\le \alpha\le l-1$ and the $i_{\alpha}'s$ are all distinct. Similarly, let $m\ge 1$ be the smallest integer such that $a_{s-m}\ge 3$. Then, as above, $V_{s-\beta}=\{j_{\beta-1},j_{\beta}\}$ for all $1\le \beta\le m-1$, where $j_0:=j$ and the set $\{j_{\beta}, i, i_{\alpha}\}$ has $m+l$ distinct elements. Now since $|v_{s+l-1}\cdot v_{s+l}|=1$, we must have that $V_{s+l-1}\cap V_{s+l}=\{i_{l-1}\}$ and $|v_{s+l}\cdot e_{i_{l-1}}|=1$. Similarly, $V_{s-m+1}\cap V_{s-m}=\{j_{m-1}\}$ and $|v_{s-m}\cdot e_{j_{m-1}}|=1$. Moreover, $E_{i_{\alpha}}=\{s+\alpha, s+\alpha +1\}$ and $E_{j_{\beta}}=\{s-\beta, s-\beta -1\}$ for all $\alpha$ and $\beta$.

If $v_{s-m}=v_{s+l}=v_u$, then $\{i_{l-1}, j_{m-1}\}\subset V_u$. Since $|v_u\cdot e_{i_{l-1}}|=|v_u\cdot e_{j_{m-1}}|=1$ and $a_u\ge 3$, we must have that $|V_u|\ge3$. Thus there is an integer $p$ such that $E_p=\{u\}$, which contradicts $p_1(S)=0$. Now suppose that $v_{s-m}\neq v_{s+l}$. Let $T\subset\ZZ^{n-(m+l)}=\langle e_1,\ldots,e_n\rangle/\langle e_{i_0},\ldots,e_{i_{l-1}},e_{j_0},\ldots,e_{j_{m-1}}\rangle$ be the subset obtained from $S$ by removing the vertices $\{v_{s-(m-1)},\ldots,v_{s+l-1}\}$, replacing $v_{s-m}$ with $v_{s-m}'=v_{s-m}+(v_{s-m}\cdot e_{j_{m-1}})e_{j_{m-1}}$, and replacing $v_{s+l}$ with $v_{s+l}'=v_{s+l}+(v_{s+l}\cdot e_{i_{l-1}})e_{i_{l-1}}$. Since $|v_{s+l}\cdot e_{i_{l-1}}|=|v_{s-m}\cdot e_{j_{m-1}}|=1$ and $a_{s-m},a_{s+l}\ge 3$, we have that $(v_{s-m}')^2, (v_{s+l}')^2\le -2$. Thus $T\subset\mathbb{Z}^{n-(l+m)}$ is a standard subset made of $n-(l+m-1)$ vectors. However, by Remark \ref{LIremark}, these vectors are linearly independent in $\ZZ^{n-(l+m)}$, which is not possible.
\end{proof}

\begin{lem}  Let $S$ be cyclic, $p_1(S)=0$, $p_2(S)>0$, and $n\ge4$. Let $i,s,t$ be integers such that $E_i=\{s,t\}$. If $a_s=2$ or $a_t=2$, then $v_s\cdot v_t=0$.\label{lem2b.2}\end{lem}

\begin{proof} Let $a_s=2$ and $V_s=\{i,j\}$. Then by Lemma \ref{lem2b.1}, $a_t\ge 3$. Assume $|v_s\cdot v_t|=1$ and without loss of generality, assume $t=s+1$. Then $\{s-1,s\}\subseteq E_j$. If there exists an integer $u\notin\{s-1,s,s+1\}$ such that $u\in E_j$, then we necessarily have that $i\in V_u$, implying that $|E_i|\ge 3$, which is not possible. Thus either $E_j=\{s-1,s\}$ or $E_j=\{s-1,s,s+1\}$.

If $E_j=\{s-1,s\}$, then by Lemma \ref{lem2b.1}, $a_{s-1}\ge3$. Moreover, since $|v_s\cdot e_i|=|v_s\cdot e_j|=1$, $V_s\cap V_{s-1}=\{j\}$, and $V_s\cap V_{s+1}=\{i\}$, we have $|v_{s+1}\cdot e_i|=|v_{s-1}\cdot e_j|=1$. Let $T\subset\ZZ^{n-2}=\langle e_1,\ldots,e_n\rangle/\langle e_i,e_j\rangle$ be the subset obtained by removing $v_s$, replacing $v_{s+1}$ with $v_{s+1}'=v_{s+1}+(v_{s+1}\cdot e_i)e_i$, and replacing $v_{s-1}$ with $v_{s-1}'=v_{s-1}+(v_{s-1}\cdot e_j)e_j$. Then $(v_{s\pm1}')^2\le -2$ and $v_{s-1}'\cdot v_{s+1}'=0$. Thus $T$ is standard with final vertices $v_{s-1}'$ and $v_{s+1}'$. By Remark \ref{LIremark}, $T\subset\ZZ^{n-2}$ contains $n-1$ linearly independent vectors, which is impossible.

If $E_j=\{s-1,s,s+1\}$, then $v_{s-1}\cdot v_{s+1}=0$, there exists an integer $k\notin\{i,j\}$ such that $k\in V_{s-1}\cap V_{s+1}$. Moreover, $|v_{s-1}\cdot e_j|=1$ and since $V_{s+1}\cap V_s=\{i,j\}$ and $|v_{s+1}\cdot v_s|=1$, we have that $|v_{s+1}\cdot e_i|=x$ and $|v_{s+1}\cdot e_j|=x\pm1$, where $x,x\pm1\neq0$. Thus $a_{s+1}\ge x^2+(x\pm1)^2+1\ge6$. If $|v_{s+1}\cdot e_i|=x\ge2$, let $T\subset\ZZ^{n-1}=\langle e_1,\ldots e_n\rangle/\langle e_i\rangle$ be the subset obtained by removing $v_s$ and replacing $v_{s+1}$ with $v_{s+1}+(v_{s+1}\cdot e_i)e_i$. Then $T$ is standard and $0\ge I(S)=I(T)+x^2+(a_s-3)=I(T)+x^2-1.$ Thus $I(T)\le 1-x^2<0$ and so by Proposition \ref{liscasummary1}, we necessarily have that $I(T)=-3$ and $p_1(T)=1$. But then $p_1(S)=p_1(T)=1$, which contradicts our assumption that $p_1(S)=0$. If $|v_{s+1}\cdot e_i|=1$, then $|v_{s+1}\cdot e_j|=2$. Since  $|v_{s-1}\cdot e_j|=1$, and $v_{s-1}\cdot v_s$, we necessarily have that $a_{s-1}\ge 3$. Let $T'\subset\ZZ^{n-2}=\langle e_1,\ldots,e_n\rangle/\langle e_i,e_j\rangle$ be the subset obtained by removing $v_s$ and $v_{s+1}$, and replacing $v_{s-1}$ with $v_{s-1}+(v_{s-1}\cdot e_j)e_j$. Then $T'$ is a standard subset such that $0\ge I(S)=I(T')+1+(a_s-3)+(a_{s+1}-3)\ge I(T')+3$. By Proposition \ref{liscasummary1}, we necessarily have that $I(T')=-3$ and $p_1(T')=1$. Thus $a_{s+1}=6$ and $V^S_{s+1}=\{i,j,k\}$. This implies that $|E_k^{T'}|=1$. But $k\in V_{s-1}^{T'}$ and $v_{s-1}$ is a final vertex of $T'$. By Proposition \ref{liscasummary2}, no such standard subset exists.

\end{proof}

\begin{lem}  Let $S$ be cyclic, $p_1(S)=0$, $p_2(S)>0$, and $n\ge 4$. Then $|v_j\cdot e_k|\le 1$ for all integers $j$ and $k$.\label{lem2b.5}\end{lem}
\begin{proof}
First assume that there is a triple $\{i,s,t\}$ such that $E_i=\{s,t\}$, $a_s\ge 3$ and $a_t\ge 3$.  Further assume that $v_s\cdot v_t=0$. Then $|V_s\cap V_t|\ge 2$. Moreover, since $|v_t\cdot v_{t\pm1}|=1$ and either $v_{t+1}\cdot v_s=0$ or $v_{t-1}\cdot v_s=0$, we must have $|V_s|\ge 3$. Similarly, $|V_t|\ge 3$. Let $T\subset\ZZ^{n-1}=\langle e_1,\ldots,e_n\rangle/\langle e_i\rangle$ be the subset given by $T=\{v_1',...,v_{t-1}',v_{t+1}',...,v_{n}'\}$, where $v_s'=v_s+(v_s\cdot e_i)e_i$ and $v_u'=v_u$ for all $u\neq s,t$. Then $(v_s')^2\le -2$ and $v'_{t-1}\cdot v'_{t+1}=0$, and so $T$ is standard. Let $|v_s\cdot e_i|=x\ge 1$. Then $$ 0\ge I(S)=I(T)+x^2+(a_t-3)\ge I(T)+x^2\ge I(T)+1.$$ Thus $I(T)\le -1$ and so by Proposition \ref{liscasummary1}, $I(T)\in\{-1,-2,-3\}$. Thus $0\ge I(T)+x^2\ge x^2-3$ and so $v_s\cdot e_i=x=\pm1$. Moreover, if $|v_t\cdot e_k|\ge 2$ for some $k$, then since $|V_t|\ge 3$, we would have $a_t\ge 6$ and so the above inequality would become $0\ge I(T)+4$ or $I(T)\le -4$, which is impossible. Finally, by Proposition \ref{liscasummary1}, since $T$ is standard and $I(T)<0$, $|v'_j\cdot e_k|\le 1$ for all $j,k$. Thus  $|v_j\cdot e_k|\le 1$ for all $j,k$. A similar argument shows the same is true if $|v_s\cdot v_t|=1$.

Now suppose there does not exist a triple $\{i,s,t\}$ such that $E_i=\{s,t\}$, $a_s\ge 3$ and $a_t\ge 3$. Then by combining Lemma \ref{lem2b.2} and the proof Lemma 4.4 in \cite{liscalensspace}, we must have that $I(S)=0$, $p_2(S)=p_4(S)$ and $p_j(S)=0$ for all $j=5,...,n$. Thus $p_3(S)=n-2p_2(S)$. Let $m_{ij}:=v_i\cdot e_j$. Then $$3n=\sum a_i= \sum_{i,j}m_{ij}^2\ge\sum_{i,j} |m_{ij}|\ge\sum ip_i(S)=2p_2(S)+4p_2(S)+3(n-2p_2(S))=3n.$$ Thus $|v_i\cdot e_j|=|m_{ij}|\le1$ for all $i,j$.
\end{proof}

Lemma \ref{lem2b.5} serves to simplify the proofs of the remaining results. In particular, the lemma implies that $a_k=|V_k|$ for all $k$. These facts will now be standing assumptions throughout the remainder of the section.

\begin{prop}  Let $S$ be cyclic, $p_1(S)=0$, and $p_2(S)>0$. Let $n\ge 4$ and suppose for every triple $\{i,s,t\}$ such that $E_i=\{s,t\}$, we have that $a_s\ge3$, and $a_t\ge 3$. Then $S$ is negative with associated string in $\mathcal{S}_{1c}$.
	\label{lem2b.6}\end{prop}

\begin{proof} Let $i,s,t$ be such that $E_i=\{s,t\}$. Then, by the hypothesis, $a_s,a_t\ge3$.
Let $T=\{v_1',\ldots,v_{s-1}',v_{s+1}',\ldots,v_n'\}\subset \ZZ^{n-1}=\langle e_1,\ldots,e_n\rangle/\langle e_i\rangle$, where $v_u'=v_u$ for all $u\neq s, t$ and $v_t'=v_t+(v_t\cdot e_i)e_i$. Then $S$ is a standard subset and $0\ge I(T)=I(S)+(a_s-3)+1$. Thus $I(T)<0$ and so by Proposition \ref{liscasummary1}, $I(T)\in\{-3,-2,-1\}$. We claim that $I(T)=-3$. Suppose otherwise. We can work case-by-case and show that $T$ cannot be any of the  standard subsets with $I=-1,-2$ listed in Proposition \ref{liscasummary2}. We will go through the first case carefully. The remaining five cases follow similarly. Suppose $I(T)=-2$ and $T$ is of the form

\begin{center}
$\{e_{x+4}-e_{x+3},e_{x+3}-e_{x+2},\ldots,e_5-e_4,e_4-e_2-e_3,e_2+e_1+\sum_{i=x+5}^{x+y+4}e_i,$\\$-e_2-e_4-\sum_{i=5}^{x+4}e_i,e_2-e_1-e_3,e_1-e_{x+5},e_{x+5}-e_{x+6},\ldots,e_{x+y+3}-e_{x+y+4}\}$
\end{center}

Since $|E^T_{x+4}|=|E^T_{x+y+4}|=2$ and $e_{x+4}$ and $e_{x+y+4}$ are both part of vertices of square $-2$ (namely, the first and last vertices), by the main hypothesis in the statement of the proposition, we must have $ x+4, x+y+4\in V_s^S$. Let $v_{u_1}'=e_2+e_1+\sum_{j=x+5}^{x+y+4}e_j$ and $v_{u_2}'=-e_2-e_4-\sum_{j=5}^{x+4}e_j$ be the two ``large" vertices of $T$. Since $v_s\cdot v_{u_1}=v_s\cdot v_{u_2}=0$, $x+y+4\in V_{u_1}^S$, and $x+4\in V_{u_2}^S$, one of the integers in $\{2,4,5,\ldots,x+y+4\}$ must be in $V^S_s$, which in turn, necessarily implies at least one more integer must be in $V_s^S$. Thus $a_s=|V_s^S|\ge5$ and so $I(S)=I(T)+(a_s-3)+1\ge1$, which is a contradiction.

Since $I(T)=-3$ and $T$ has final vertices $v'_{s-1}$ and $v'_{s+1}$, there is an integer $k$ such that $E_k^{T}=\{s-1,s+1\}$ and either $(v_{s-1}')^2=-2$ or $(v_{s+1}')^2=-2$, by Proposition \ref{liscasummary2}. If $k\notin V_s$, then $E_k^S=\{s-1,s+1\}$ and either $a_{s-1}=2$ or $a_{s+1}=2$, which contradicts the standing hypothesis. Thus, $k\in V_s^S$. Moreover, by Proposition \ref{liscasummary2}, there exist integers $l$ and $u$ such that $E_l^{T}=\{u\}$, $u\neq s\pm1$, and $(v'_u)^2=-2$. Since $p_1(S)=0$, we have that $l\in V^S_s$ and so $V^S_u\cap V^S_s=\{l,k\}$. Since $v_u^2\le 3$, by the standing hypothesis, we have that $t=u$, $l=i$ and, consequently, $a_t=3$. By similarly considering the standard subset $T'$ obtained by removing $v_t$ and replacing $v_s$ by $v_s+(v_s\cdot e_i)e_i$, we also obtain $a_s=3$.

Now, by Proposition \ref{liscasummary2}, $T$ has associated string of the form $(b_1,\ldots,b_k,2,c_l,\ldots,c_1)$, where $(b_1,\ldots,b_k)$ and $(c_1,\ldots,c_l)$ are linear-dual strings and the entry in between these two strings corresponds to $v'_t$. Thus $S$ has associated string $(b_1,\ldots,b_k,3,c_l,\ldots,c_1,3)\in\mathcal{S}_{1c}$ with $I(S)=-2$. It remains to show that $S$ is a negative cyclic subset. Once again by Proposition \ref{liscasummary2}, we have that $v'_{x}\cdot v'_{x+1}=1$ for all $x\neq s-1, s$ and $v'_{s-1}\cdot e_k=-v'_{s+1}\cdot e_k$. Now since $V_s\cap V_{s\pm1}=\{k\}$, we necessarily have that $v_{s-1}\cdot v_{s}=-v_{s+1}\cdot v_{s}$. Thus $S$ is a negative cyclic subset.
\end{proof}

\begin{prop} Let $S$ be cyclic, $p_1(S)=0$, $p_2(S)>0$, $I(S)<0$, and $n\ge4$. Suppose there exists $i,s,t$ such that $E_i=\{s,t\}$, $a_s=2$, and  $a_t\ge 3$. Then $S$ is either negative with associated string in $\mathcal{S}_{1d}\cup\mathcal{S}_{1e}$ or positive with associated string in $\mathcal{S}_{2d}$

\label{prop-1,-2}
\end{prop}

\begin{proof} Let $V_s=\{i,j\}$. By Lemma \ref{lem2b.2}, that $v_s\cdot v_t=0$. Let $T\subset\ZZ^{n-1}=\langle e_1,\ldots,e_n\rangle/\langle e_i\rangle$ denote the subset obtained by removing $v_s$, replacing $v_t$ with $v_t'=v_t+(v_t\cdot e_i)e_i$, and setting $v_u':=v_u$ for all $u\neq s,t$. Then $T$ is  standard with final vertices $v'_{s-1}$ and $v'_{s+1}$, and $I(T)=I(S)<0$. Thus by Proposition \ref{liscasummary1}, $I(T)\in\{-1,-2,-3\}$. Moreover, note that $E_j^{T}=\{s-1,s+1,t\}$, since otherwise, we would necessarily have $|E^S_i|\ge3$.

If $I(T)=-3$, then by Proposition \ref{liscasummary2}, there exists $u,k$ such that $E^{T}_k=\{u\}$, $u\neq s-1,s+1$, and $(v_u')^2=-2$. Now since $p_1(S)=0$, $k\in V^S_s=\{i,j\}$ and so $k=j$. But, $|E_j^{T}|=3$, which is a contradiction. Thus $I(S)=I(T)\in\{-2,-1\}$.

Suppose $I(S)=I(T)=-1$. Then $T$ is one of the subsets listed in Proposition \ref{liscasummary2}(3). As noted above, the final vertices of $T$ satisfy $V^{T}_{s-1}\cap V^{T}_{s+1}\neq \emptyset$. It is easy to see that the only subsets in Proposition \ref{liscasummary2}(3) satisfying this condition are those with $y=0$, which have associated strings $(2+x,2,3,2^{[x]},4)$ and $(3+x, 2, 3, 3, 2^{[x]},3)$. Consider the first case. Since $j\in V^S_{s-1}\cap V^S_{s+1}\cap V^S_t$, we have that $j=x+4$ and $v'_t$ is the second to last vertex in $T$. Thus $S$ has associated string $(2,2+x,2,3,2^{[x-1]},3,4)\in\mathcal{S}_{2d}$ if $x>0$ and $(2,2,2,4,4)\in\mathcal{S}_{2d}$ if $x=0$. Moreover, since $v'_{s\pm1}\cdot e_{x+4}=-1$, $S$ is positive cyclic. In the second case, either $j=x+5$ and $v'_t$ is the second to last vertex in $T$ or $j=1$ and $v'_t$ is the third vertex in $T$. Thus $S$ has associated string $(2, 3+x, 2, 3, 3, 2^{[x-1]},3,3)\in\mathcal{S}_{1e}$ (where $(3,2^{[-1]},3)$ is defined to be $(4)$) or $(2, 3+x, 2, 4, 3, 2^{[x]},3)\in\mathcal{S}_{2d}$. Moreover, in the former case, since $v'_{s-1}\cdot e_{x+5}=-v'_{s+1}\cdot e_{x+5}$, $S$ is necessarily negative cyclic, and in the latter case, since $v'_{s\pm1}\cdot e_1=-1$, $S$ is necessarily positive cyclic. Finally notice that the strings $(2,2+x,2,3,2^{[x-1]},3,4)$ and $(2, 3+x, 2, 4, 3, 2^{[x]},3)$ are reverse cyclic reorderings of each other.

Suppose $I(S)=I(T)=-2$. Then, by similarly examining the subsets found in Proposition \ref{liscasummary2}, there is only one candidate for $T$, namely the subset with associated string of the form $(d_1,\ldots,d_{p-1},d_p+1,2,2,e_q+1,e_{q-1},\ldots,e_1)$. Proceeding as in the previous cases, it is easy to see that $S$ is negative cyclic with associated string of the form $(2,d_1,d_2+1,d_3,\ldots,d_{p-1},d_p+1,2,2,e_q+1,e_{q-1},\ldots,e_1)$, where $(d_1,\dots,d_p)$ and $(e_1,\ldots,e_q)$ are linear-dual strings with $l+k\ge 3$. Without loss of generality, assume $d_1=2$. Then we can rewrite this string as $(2,2,b_1+1,b_2,\ldots,b_{k-1},b_k+1,2,2,c_l+1,c_{l-1},\ldots,c_1+1)$, where $(b_1,\ldots,b_k)=(d_2,\ldots,d_p)$ and $(c_1,\ldots,c_l)=(e_1-1,\ldots,c_l)$ are linear-dual. Cyclically reordering it, we obtain $(2,b_1+1,b_2,\ldots,b_{k-1},b_k+1,2,2,c_l+1,c_{l-1},\ldots,c_1+1,2)\in\mathcal{S}_{1d}.$
\end{proof}

\begin{remark} As mentioned in the above proof, if $I(S)=I(T)=-3$, we would have $p_1(S)\neq0$. If we consider this case and perform the same analysis as in the other cases, we would obtain the positive cyclic families that found in Proposition \ref{prop1}.
\end{remark}

In Proposition \ref{prop-1,-2}, we assumed there existed $i,s,t$ such that $E_i=\{s,t\}$, $a_s=2$, and $a_t\ge 3$, and we explored the cases in which $I(S)<0$. The final case to consider is $I(S)=0$. In this case, using the notation in the proof, $I(T)=0$ and so we can no longer rely on Lisca's classification. The proof of this requires deeper analysis than the previous results, but it is the last of the cases we must consider. Before we dive into that result, we prove a quick lemma.

\begin{lem}  Let $S$ be cyclic, $p_1(S)=0$, $p_2(S)>0$, and $n\ge4$. Suppose there exists $i,s,t$ such that $E_i=\{s,t\}$, $a_s=2$, and $a_t\ge 3$. Then either $a_{s-1}\ge3$ or $a_{s+1}\ge3$.\label{lem2b.3}\end{lem}
\begin{proof}
Let $V_s=\{i,j\}$ and suppose $a_{s-1}=a_{s+1}=2$. Then since $v_{s-1}\cdot v_{s+1}=0$, $V_{s-1}=V_{s+1}=\{j,k\}$ for some $k$. If $n=4$, then it is easy to see that the only cyclic subset satisfying $a_{s-1}=a_s=a_{s+1}=2$ necessarily has $p_1=1$. If $n\ge5$, then either $s-2\neq t$ or $s+2\neq t$. Assume $t\neq s-2$. If $j\in V_{s-2}$, then since $v_{s-2}\cdot v_s=0$, $i\in V_{s-2}$, which is a contradiction. If $k\in V_{s-2}$, then since $v_{s-2}\cdot v_{s+1}=0$, we must have $j\in V_{s-2}$, which is, again, not possible. Thus either $a_{s-1}\ge3$ or $a_{s+1}\ge 3$.
\end{proof}

\begin{prop}  Let $S$ be cyclic, $p_1(S)=0$, $p_2(S)>0$, and $n\ge 4$. Suppose $I(S)=0$ and that there exists $i,s,t$ such that $E_i=\{s,t\}, a_s=2,$ and $a_t\ge3$. Then $S$ is either positive with associated string in $\mathcal{S}_{2c}$ or negative with associated string $(6,2,2,2,6,2,2,2)$.
 \label{prop2b.6}\end{prop}

\begin{proof} By Lemma \ref{lem2b.3}, we may assume that $a_{s-1}\ge 3$. Let $l\ge 1$ be the smallest integer such that $a_{s+l}\ge3$. By Lemma \ref{lem2b.2}, $t\neq s\pm1$ and so $E_j=\{s-1,s,s+1,t\}$. Moreover, as in the proof of Lemma \ref{lem2b.1}, $V_{s+\alpha}=\{j_{\alpha-1},j_{\alpha}\}$ for all $1\le\alpha\le l-1$, where $j_0=j$ and the $j_{\alpha}$ are all distinct. There are two main cases to consider.\\

\noindent \textbf{Case 1}: Assume $t\neq s+l$.

Then $j_{l-1}\in V_{s+l}$, $i,j, j_1,...,j_{l-1}\in V_t$, and $j,j_1,...,j_{l-1}\in V_{s-1}$. Moreover, $E_{j_{\alpha}}=\{s+\alpha, s+\alpha +1, s-1, t\}$ for all $1\le\alpha\le l-1$. Further note that either $v_t\cdot e_{j_{\alpha}}=v_{s-1}\cdot e_{j_{\alpha}}$ for all $\alpha$ or $v_t\cdot e_{j_{\alpha}}=-v_{s-1}\cdot e_{j_{\alpha}}$ for all $\alpha$. Thus there exists $k_0,...,k_{l-1}\in V_t\cap V_{s-1}\setminus\{j,j_1,...,j_{l-1}\}$ if $v_{s-1}\cdot v_t=0$ and there exists $k_0,...,k_{l-2}\in V_t\cap V_{s-1}\setminus\{j,j_1,...,j_{l-1}\}$ if $|v_{s-1}\cdot v_t|=1$. In either case, $|V_{s-1}|\ge 2l-1$ and $|V_t|\ge 2l$. Let $T\in\mathbb{Z}^{n-l-1}=\langle e_1,\ldots,e_n\rangle/\langle e_i,e_{j_0},\ldots,e_{j_{l-1}}\rangle$ be the subset obtained by removing $v_{s+\alpha}$ for all $0\le \alpha\le l-1$, removing $v_t$, replacing $v_{s-1}$ with $v_{s-1}'=v_{s-1}+\sum_{\alpha=0}^{l-1} (v_{s-1}\cdot e_{j_{\alpha}})e_{j_{\alpha}}$, replacing $v_{s+l}$ with $v_{s+l}'=v_{s+l}+(v_{s+l}\cdot e_{j_{l-1}})e_{j_{l-1}}$, and setting $v_u':=v_u$ for all $u\neq s+\alpha, t, s-1$, $0\le \alpha\le l$. Then $(v_{s+l}')^2\le -2$ and $|v_{s-1}'\cdot v_{s+l}'|=1$. Moreover, $0= I(S)= I(T)+\sum_{\alpha=0}^{l-1}(a_{s+\alpha}-3)+(a_t-3)+l+1= I(T) +a_t-2\ge I(T)+1$ and so $I(T)<0$. Finally note that if $(v_{s-1}')^2\le -2$, then $T$ is a standard subset with final vertices $v'_{t\pm1}$. By examining the standard subsets in Proposition \ref{liscasummary2}, it is clear that adjacent vertices in $T$ must satisfy $|V_u^{T}\cap V_{u+1}^{T}|=1$. Thus, since $v_{s-1}'$ and $v_{s+l}'$ are adjacent in $T$, $|V_{s-1}^S\cap V_{s+l}^S|=2$.

Suppose $l\ge 3$. Then $(v_{s-1}')^2\le -(l-1)\le -2$ and so $T$ is standard and $0= I(S)= I(T)+a_t-2\ge I(T) +2l-2\ge I(T)+4$. In particular, $I(T)\le -4$, which contradicts Proposition \ref{liscasummary1}. Thus $l\in\{1,2\}$.\\

\noindent\textbf{Case 1a:} Let $l=2$.

\underline{Suppose $v_t\cdot v_{s-1}=0$}. Then $V^S_t\supseteq\{i, j, j_1, k, k_1\}$, $V^S_{s-1}\supseteq\{j, j_1, k, k_1\}$, and $(v_{s-1}')^2\le-2$. Thus $T$ is standard and $0=I(T)+a_t-2\ge I(T)+3$. By Proposition \ref{liscasummary1}, $I(T)=-3$ and, moreover, $a_t=5$ so that $V_t=\{i, j, j_1, k, k_1\}$. By Proposition \ref{liscasummary2}, there exists a $p,x$ such that $E_p^{T}=\{x\}$, where $v_x'$ is internal (i.e. $x\neq t\pm1$) and $(v_x')^2=-2$. Since $p_1(S)=0$, we necessarily have that $p= k$ or $p= k_1$ and so $x=s-1$ and $V^S_s=\{j,j_1,k,k_1\}$. Since $v_{s-1}\cdot v_{s+2}=0$ and $j_1\in V_{s-1}\cap V_{s+2}$, we may assume that $k\in V_{s-1}\cap V_{s+2}$. Note that either: $v_t\cdot e_{j_1}=v_{s+2}\cdot e_{j_1}$ and $v_t\cdot e_{k}=v_{s+2}\cdot e_{k}$; or $v_t\cdot e_{j_1}=-v_{s+2}\cdot e_{j_1}$ and $v_t\cdot e_{k}=-v_{s+2}\cdot e_{k}$. Thus, since $t\neq s+2$, there exists $m\in V_t\cap V_{s+2}$ such that $m\neq i,j,j_1,k,k_1$, implying that $|V_t|\ge 6$, which is a contradiction.

\underline{Suppose $|v_t\cdot v_{s-1}|=1$}. Thus $V^S_t\supseteq\{i, j, j_1, k, k_1\}$ and $V^S_{s-1}\supseteq\{j, j_1, k\}$. If $|v_t\cdot v_{s+2}|=1$, then $n=5$. If $|V^S_{s-1}|\ge 4$, then $T$ is a subset with two vertices, each with weight at most $-2$. Since there is no such standard subset by Proposition \ref{liscasummary2}, we must have $|V_{s-1}|=3$. Now it is easy to see that $S$ is of the form $$\{e_j+e_{j_1}+e_k,e_i-e_j,e_{j}-e_{j_1},e_{j_1}-e_k-e_m,-e_i-e_j-e_{j_1}+e_k-e_m\},$$ which is positive cyclic and has associated string $(3,2,2,3,5)\in\mathcal{S}_{2c}$.

Now suppose $v_t\cdot v_{s+2}=0$. We claim that $a_t=5$. Assume otherwise, so that $a_t=4$ and $V_t=\{i,j,j_1,k\}$. Then, since $v_t\cdot v_{s+2}=0$, $k\in V_{s+2}$ and moreover, either $v_{s-1}\cdot e_k=v_{s+2}\cdot e_k$ and $v_{s-1}\cdot e_{j_1}=e_{s+2}\cdot e_{j_1}$ or $v_{s-1}\cdot e_k=-v_{s+2}\cdot e_k$ and $v_{s-1}\cdot e_{j_1}=-e_{s+2}\cdot e_{j_1}$. Thus, since $v_{s-1}\cdot v_{s+2}=0$, there exists $m_1,m_2\in V_{s-1}\cap V_{s+2}$ such that $m_1,m_2\neq k$. Thus $|V_{s-1}|\ge 5$ and $T$ is standard with $I(T)=-2$ and $|V^{T}_{s-1}\cap V^{T}_{s+2}|\ge 3$, which is not possible since $v_{s-1}'$ and $v_{s+2}'$ are adjacent in $T$. Thus $a_t=5$.

Let  $V_t=\{i,j,j_1,k,p\}$. Then since $v_t\cdot v_{s+2}=0$, $k\in V_{s+2}$ or $p\in V_{s+2}$ and since $v_{s-1}\cdot v_{s+2}=0$, there exists $m\in V_{s-1}\cap V_{s+2}$ such that $m\notin V_t$. Thus $|V_{s-1}|\ge4$ and $T$ is standard and satisfies $I(T)=-3$ and, by Proposition \ref{liscasummary2}, there exists $u\neq t\pm1$ such that $E_p^{T}=\{u\}$ and $(v'_u)^2=-2$. If $u\neq s+2$, then $(v_u)^2=-2$ and so $k\in V^S_u$, since $v_u\cdot v_t=0$. But then $|v_u\cdot v_{s-1}|=1$, which is not possible. Thus $u=s+2$ and so $V_{s+2}\supseteq\{j_1,k,p\}$. This, along with Proposition \ref{liscasummary2}, implies that $T$ has associated string of the form $(a_{s-1}-2,2,d_1,\ldots,d_p)$, where $(d_1,\ldots,d_p)$ and $(a_{s-1}-2)$ are linear-dual. Thus the string is $(a_{s-1}-2,2,2^{[a_{s-1}-3]})$ and so the string associated to $S$ must be of the form $(5,a_{s-1},2,2,3,2^{[a_{s-1}-3]})$ or $(5,3+x,2,2,3,2^{[x]})$, where $x\ge 1$. By writing down the actual subset, it is clear that $S$ is positive cyclic. Note, if $x=0$, we obtain the string $(5,3,2,2,3)$, whose reverse we found above in the case $|v_t\cdot v_{s+2}|=1$. Thus we have uncovered the family $(5,3+x,2,2,3,2^{[x]})\in\mathcal{S}_{2c}$ for all $x\ge 0$.\\

\noindent\textbf{Case 1b:} Let $l=1$.

Since $a_{s-1}\ge 3$, we have that $(v'_{s-1})^2\le-2$. Thus $T$ is standard and $a_t\le5$, as above. Moreover note that $E_j^S=\{s-1,s,s+1,t\}$.

\underline{Suppose that $a_t=5$}. Then $I(T)=-3$ and so Proposition \ref{liscasummary2}, $p_1(T)=p_2(T)=1$ and $p_l(T)=0$ for all $l\ge 4$. Thus $p_l(S)=0$ for all $l\ge 5$ and if $|E_x^S|=4$, then $x\in V_t$. Since $I(S)=0$, we necessarily have that $p_2(S)=p_4(S)\in\{2,3\}$. Now there exists integers $m$ and $u$ such that $E_m^{T}=\{u\}$. Since $p_1(S)=0$, $m\in V^S_t\setminus\{i,j\}$. Let $R\subset\ZZ^{n-2}$ be the subset obtained by removing $v_s$ and $v_t$, and replacing $v_u$ by $v_u+(v_t\cdot e_m)e_m$. Then $I(R)=-2$, $p_1(R)=0$, and $p_2(R)\le 1$. However, no such subset exists by Lemma \ref{keylem1}.

\underline{Suppose that $a_t=4$}. Let $V_t=\{i,j,k,m\}$. First note that since $a_t=4$, $T$ is standard with $I(T)=-2$. If $|v_t\cdot v_{s\pm1}|=1$, then $T$ contains exactly two vertices, which is not possible by Proposition \ref{liscasummary2}. Assume $v_t\cdot v_{s-1}=0$ and $v_t\cdot v_{s+1}=1$ (or vice versa) and let $k\in V_{s-1}$. Then $V_t^S\cap V^S_{s+1}=\{j\}$ or $\{j,k,m\}$. In the latter case, since $v_{s-1}\cdot v_t=v_{s-1}\cdot v_{s+1}=0$ and $|V_{s-1}^S\cap V_{s+1}^S|=2$, up to the action of Aut$\ZZ^n$, we have that $v_{s-1}=e_j-e_k+f$, $v_s=e_i-e_j$, $v_{s+1}=e_j+e_k-e_m+g$, and $v_t=-e_i-e_j-e_k-e_m$, where $f$ and $g$ are some vectors in $\ZZ^n$. Let $T'\subset\ZZ^{n-1}=\langle e_1,\ldots, e_n\rangle/\langle e_i\rangle$ be the subset obtained from $S$ by removing $v_s$, replacing $v_{s+1}$ with $v_{s+1}''=v_{s+1}+(v_{s+1}\cdot e_j)$, replacing $v_t$ with $v_t''=v_t+(v_t\cdot e_i)e_i$, and setting $v_u'':=v_u$ for all other $u$. Then regardless of whether $V_t^S\cap V^S_{s+1}=\{j\}$ or $\{j,k,m\}$, $T'$ is standard with terminal vertices $v_t''$ and $v''_{s+1}$ and $I(T')=-1$. Moreover, $E^{T'}_j=\{t,s-1\}$, where $v_t''$ and $v_{s-1}''$ are (up to a reflection) the first and second-to-last vertices of $T'$ and $(v''_t)^2=-3$. Now, by analyzing the subsets in Proposition \ref{liscasummary2} with $I=-1$, it is clear that no such subset exists. We have now shown that $v_t\cdot v_{s\pm1}=0$.

We next claim that $k,m\notin V_{t-1}\cap V_{t+1}$. Assume $k\in V_{t-1}\cap V_{t+1}$. Then $|E_k^{T}|\ge3$. Now, since $|E_m^{T}|\ge 2$ (since $p_1(T)=0$), there exists $u\neq s+1$ such that $m\in V_u$. Moreover, $u\neq t\pm1$ since otherwise, $|V_t\cap V_{t\pm1}|=2$, which is impossible. Thus we have that $n\ge7$. Now since $v_t\cdot v_u=0$, we must either have $j\in V_u$ or $k\in V_u$. If $j\in V_u$, then $u=s-1$, which implies $j,k,m\in V_{s-1}\cap V_{t}$, which is not possible. If $k\in V_u$, then $|E_k^{T}|=4$. But, by inspecting the embeddings in Proposition \ref{liscasummary2} with $I=-2$, it is clear that the the vertices $v_{t+1}', v'_{s-1},v'_u,v'_{t-1}$ must have three adjacencies among themselves, which is not possible since $T$ has length at least 5 and $v_{t\pm1}'$ are the terminal vertices. Thus either $k\in V_{t-1}$ and $m\in V_{t+1}$ or vice versa and so $T=\{w_1,\ldots w_{n-2}\}$ is a standard subset such that there exists $k,m$ with $E_k^{T}\supseteq\{1,a\}$ and $E_m^{T}\supseteq\{a+1,n-2\}$ for some $a\neq 1,n-2$. Now by analyzing the subsets in Proposition \ref{liscasummary2}(2), it is clear that the only such subsets have associated strings of the form $(2^{[x]}, 3, 2+y, 2+x,3,2^{[y]})$ and, using the notation in Proposition \ref{liscasummary2}(2)(a), $k=x+4$ and $m=1$. Now it is easy to write down the explicit subset $S$, which is positive and has associated string of the form $(2^{[x]},3,3+y,2,3+x,3,2^{[y]},4)\in\mathcal{S}_{2c}$.

\underline{Suppose $a_t=3$}. Let $V_t=\{i,j,k\}$. Then $I(T)=-1$. If $v_{s\pm 1}\cdot v_t=0$, then $k\in V_t, V_{s-1}, V_{s+1}$ and so we would necessarily have that $|V^{S}_{s-1}\cap V^{S}_{s+1}|\ge4$, which is not possible. Without loss of generality, assume  $|v_{s+1}\cdot v_t|=1$. If $|v_{s-1}\cdot v_t|=1$, then $n=4$ and we would necessarily have $E^S_k=\{t\}$, contradicting the standing assumption that $p_1(S)=0$. Assume $n\ge5$ so that $v_{s-1}\cdot v_t=0$. Then $E^S_k=\{ s-1,t,t+1\}$. Thus by reindexing $T$, we have that $E^{T}_k=\{2,m\}$. By analyzing the subsets in Proposition \ref{liscasummary2}, there is one such family of subsets, which has associated string $(2+x,2+y,3,2^{[x]},4,2^{[y]})$ and, using the notation of Proposition \ref{liscasummary2}, $k=1$. Thus $S$ is positive and has associated string of the form $(3+x,2,3+y,3,2^{[x]},4,2^{[y]},3)\in\mathcal{S}_{2c}$. Notice that this family is the same one we found when $a_t=4$.\\

\noindent \textbf{Case 2}: Assume $t=s+l$.

By Lemma \ref{lem2b.2}, $l\neq1$. Moreover, $V_t\supseteq\{i,j,j_1,...,j_{l-2}\}$, $V_{s-1}\supseteq\{j,j_1,...,j_{l-1}\}$, and $j_{l-1}\notin V_t$. Thus $E_{j_{l-1}}=\{s-1,s+l-1\}$. Once again, either $v_t\cdot e_{j_{\alpha}}=v_{s-1}\cdot e_{j_{\alpha}}$ or $v_t\cdot e_{j_{\alpha}}=-v_{s-1}\cdot e_{j_{\alpha}}$ for all $0\le\alpha\le l-2$. Thus, there exists $k_0,k_1...,k_{l-2}\in V_t\cap V_{s-1}$ if $v_{s-1}\cdot v_t=0$ and $k_0,...,k_{l-3}\in V_{s-1}\cap V_t$ if $|v_{s-1}\cdot v_t|=1$. However, in the latter case, $n=l+2$ and so $a_{s-1}+a_t=l+6$, since $I(S)=0$. On the other hand, since $|V_t|,|V_{s-1}|\ge 2l-2$, we have that $l+6=a_{s-1}+a_t\ge 4l-4$. Thus $l=2,3$. If $l=3$, then $a_{s-1}+a_t=9$ and so $|V_t|=4$ and $|V_{s-1}|=5$ or vice versa, which implies that $p_1(S)=1$, a contradiction. If $l=2$, then $n=4=|\{i,j,j_1,k\}|$ and so we have a subset with associated string $(2,2,2,2)$, which does not satisfy $I=0$. Thus $v_{s-1}\cdot v_t=0$ and so $|V_{s-1}|, |V_{t}|\ge2l-1$.

\underline{Suppose $l=2$}. Then $V_{s-1}\supseteq\{j,j_1,k\}$ and $V_t\supseteq\{i,j,k\}$. Since $E^S_j=\{s-1,s,s+1,s+2\}$ and $V_{s+1}^S=\{j,j_1\}$, we necessarily have that $|E_{j_1}^S|=2$. Let $R\subset\ZZ^{n-2}$ be the subset obtained by removing $v_s$ and $v_{s+1}$, replacing $v_{s-1}$ by $v_{s-1}+(v_{s-1}\cdot e_j)e_j+(v_{s-1}\cdot e_{j_1})e_{j_1}$, and replacing $v_{s+2}$ by $v_{s+2}+(v_{s+2}\cdot e_i)e_i+(v_{s+2}\cdot e_{j})e_{j}$. Then $I(R)=-2$, $p_2(R)=p_2(S)-2$, $p_4(R)=p_4(S)-1$, and $p_l(R)=p_l(S)$ for all $l\neq 2,4$, which contradicts the equality in Lemma \ref{keylem1}.

\underline{Suppose $l\ge3$} and define the subset $T\subset \mathbb{Z}^{n-l-1}=\langle e_1,\ldots,e_n\rangle/\langle e_i,e_{j_0},\ldots,e_{j_{l-1}}\rangle$ by removing $v_{s-1},...,v_{s-l-1}$, replacing $v_t$ by $v_t'=v_t+(v_t\cdot e_i)e_i+\sum_{\alpha=0}^{l-2}(v_t\cdot e_{j_{\alpha}})e_{j_{\alpha}}$, and setting $v'_u:=v_u$ for all other values of $u$. Then $T$ is  standard with terminal vertices $v_{s-2}'$ and $v_t'$, and $0= I(S)=I(T)+\sum_{\alpha=0}^{l-1}(a_{s+\alpha}-3) +(a_{s-1}-3)+ l=I(T)+(a_{s-1}-3)\ge I(T)+2l-4\ge2l-7$, since $I(T)\ge -3$. Thus $l=3$ and so $V_{s-1}\supseteq\{j,j_1,j_2,k,k_1\}$ and $V_t\subseteq\{i,j,j_1,k,k_1\}$. Notice, $|V_{s-1}|\le 6$, since otherwise, we would contradict the above inequalities. Moreover, an analogous argument shows $|V_t|\le 6$.

If $|V_{s-1}|=|V_t|=5$, then since $|v_{s-2}\cdot v_{s-1}|=1$, exactly one of $k$ or $k_1$ is in $V_{s-2}$ and so $|v_{s-2}\cdot v_t|=1$. Thus $n=6$ and so since $|\{i,j,j_1,j_2,k,k_1\}|=6$ we must have $a_{s-2}=-1$, by Lemma \ref{lem2b.5}, which is a contradiction.

Let $|V_{s-1}|=6$ and $V_{s-1}=\{j,j_1,j_2,k,k_1,f\}$. Then we necessarily have $I(T)=-3$ and so, by Proposition \ref{liscasummary2}, $|V_{s-2}\cap V_t|=2$ and $a_{s-2}=2$ if and only if  $(v_t')^2\le3$. Moreover, there exists $x,u$ such that $E_x^{T}=\{u\}$ with $u\neq s-2, t$. It is clear that $x=f$ and so $f\notin V_{s-2}$. Thus either $k$ or $k_1\in V_{s-2}$ and since $v_{s-2}\cdot v_t=0$, there exists $g\in V_{s-2}\cap V_t$, where $g\neq i,j,j_1,k,k_1$. Thus $|V_t|=6$ and $a_{s-2}=2$. Without loss of generality, assume $k\in V_{s-2}$ so that $V_{s-2}=\{k,g\}$. Now, since $V_t=\{i,j,j_1,k,k_1,g\}$ and $E^S_{j_2}=\{s-1,s+2\}$, by a similar argument, $V_{t+1}=V_{s+4}=\{k_1,f\}$ and, moreover, $E^{T}_f=\{s+4\}$ and $E^{T}_g=\{s-2\}$. Thus $V_{s-3}=V_{s+4}=\{k,k_1\}$ and $S$ is of the form $$\{e_j+e_{j_1}+e_{j_2}+e_k+e_{k_1}+e_f,e_i-e_j,e_j-e_{j_1},e_{j_1}-e_{j_2},-e_i-e_j-e_{j_1}+e_k+e_{k_1}+e_g,e_f-e_{k_1},e_{k_1}-e_k, e_k-e_g\},$$ which is a negative cyclic subset with associated string $(6,2,2,2,6,2,2,2)$.
\end{proof}

\section{Appendix}\label{appendix}

Given a sequence of integers $(a_1,\ldots,a_n)$ the (Hirzebruch-Jung) continued fraction expansion is given by
\begin{center}
	$[a_1,\ldots,a_n]=\displaystyle a_1-\frac{1}{a_2-\displaystyle\frac{1}{\cdot\cdot\cdot-\displaystyle\frac{1}{a_n}}}$
\end{center}\bigskip
If $a_i\ge2$ for all $1\le i\le n$, then this fraction is well-defined and the numerator is greater than the denominator. In fact, for coprime $p>q>0\in\mathbb{Z}$, there exists a unique continued fraction expansion $[a_1,\ldots,a_n]=\frac{p}{q}$, where $a_i\ge2$ for all $1\le i\le n$. Moreover, by reversing the order of the continued fraction, we have $[a_n,\ldots,a_1]=\frac{p}{q'}$, where $q'$ is the unique integer such that $1\le q'<p$ and $qq'\equiv 1\text{mod}p$.

\begin{lem} Let $\frac{p}{q}=[a_1,\ldots,a_n]$, $\frac{s}{r}=[a_1,\ldots,a_{n-1}]$, and $\textbf{a}=(a_1,\ldots,a_n)$.\\ Then $|\textup{Tor}(H_1(\textbf{T}_{\pm A(\textbf{a})}))|=p-(r\pm2)$.\label{order}
\end{lem}

\begin{proof}
	Let $\textbf{a}=(a_1,\ldots,a_n)$. By Theorem 6.1 of \cite{neumann}, hyperbolic torus bundles are of the form $\textbf{T}_{\pm A(\textbf{a})}=T^2\times[0,1]/(\textbf{x},1)\sim(\pm A\textbf{x},0)$, where $$A=A(\textbf{a})=\begin{pmatrix}
	p&q\\
	-s&-r\\
	\end{pmatrix}, \frac{p}{q}=[a_1,...,a_n], \textrm{ and } \frac{s}{r}=[a_1,...,a_{n-1}].$$
	Note that, since $A\in SL_2(\ZZ)$, we have that $qs-pr=1$. Moreover, since $\textbf{T}_{\pm A(\textbf{a})}$ is hyperbolic, $\text{tr} A(\textbf{a})=p-r>2$. Now, by Lemma 10 in \cite{sakuma}, $|\textup{Tor}(H_1(\textbf{T}_{\pm A(\textbf{a})}))|=|\text{tr}(\pm A(\textbf{a}))-2|=|\pm(p-r)-2|=|\pm (p-(r\pm2))| =p-(r\pm2)$.

\end{proof}

\begin{lem} Let $(b_1,\ldots,b_k)$ and $(c_1,\ldots,c_l)$ be linear-dual strings, where $l+k\ge 2$, let $x\ge 1$ be an integer, and let $[b_1,\ldots,b_k]=\frac{p}{q}$. Then $\displaystyle[b_1,\ldots,b_k,x+1,c_l,\ldots,c_1]=\frac{xp^2}{xpq+1}$ and $\displaystyle[c_1,\ldots,c_l,x+1,b_k,\ldots,b_1]=\frac{xp^2}{xp^2-xpq+1}$.\label{contfraccalc}
\end{lem}

\begin{proof} First assume $[b_1,\ldots,b_k]=\frac{p}{q}$ and $[b_1,\ldots,b_k,x+1,c_l,\ldots,c_1]=\frac{xp^2}{xpq+1}$. Then since $(xpq+1)(xp^2-xpq+1)=xp^2(xpq-q^2+1)+1$, we have that $[c_1,\ldots,c_l,x+1,b_k,\ldots,b_1]=\frac{xp^2}{xp^2-xpq+1}$. We will now prove that $[b_1,\ldots,b_k,x+1,c_l,\ldots,c_1]=\frac{xp^2}{xpq+1}$.

	Let $n=k+l+1$ be the length of $(b_1,\ldots,b_k,x+1,c_l,\ldots,c_1)$. We proceed by induction on $n$. If $n=3$, then $k=1$, $l=1$, $(b_1)=\frac{2}{1}$, and $[2,x+1,2]=\frac{4x}{2x+1}=\frac{x2^2}{x(2)(1)+1}$. Suppose the lemma is true for all length $n-1$ continued fractions and consider $[b_1,\ldots,b_k,x+1,c_l,\ldots,c_1]$. By definition of linear-dual strings, either $b_1=2$ and $c_1\ge 3$ or vice versa.

	Assume that $b_1=2$. Then the strings $(b_2,\ldots,b_k)$ and $(c_1-1,\ldots,c_l)$ are linear-dual and by the inductive hypothesis, $$[b_2,\ldots,b_k,x+1,c_l,\ldots,c_1-1]=\frac{xm^2}{xmn+1}\text{ and}$$ $$[c_1-1,c_2,\ldots,c_l,x+1,b_k,\ldots,b_2]=\frac{xm^2}{xm^2-xmn+1},$$
	where $[b_2,\ldots,b_k]=\frac{m}{n}.$ Thus $$[c_1,c_2,\ldots,c_l,x+1,b_k,\ldots,b_2]=1+\frac{xm^2}{xm^2-xmn+1}=\frac{2xm^2-xmn+1}{xm^2-xmn+1}.$$
	Since $(2xmn-xn^2+2)(xm^2-xmn+1)=(2xm^2-xmn+1)(xmn-xn^2+1)+1$, we have that $$[b_2,\ldots,b_k,x+1,c_l,\ldots,c_1]=\frac{2xm^2-xmn+1}{2xmn-xn^2+2}.$$
	Thus $$[b_1,\ldots,b_k,x+1,c_l,\ldots,c_1]=2-\frac{2xmn-xn^2+2}{2xm^2-xmn+1}=\frac{x(2m-n)^2}{x(2m-n)m+1}$$
	$$\text{and }[b_1,\ldots,b_k]=2-\frac{n}{m}=\frac{2m-n}{m}.$$
	Setting $p=2m-n$ and $q=m$ yields the result.

	Now suppose $c_1=2$. Then $(b_1-1,\ldots,b_k)$ and $(c_2,\ldots,c_l)$ are linear-dual and: $$[b_1-1,\ldots,b_k,x+1,c_l,\ldots,c_2]=\frac{xm^2}{xmn+1}\text{ and}$$ $$[c_2,\ldots,c_l,x+1,b_k,\ldots,b_1-1]=\frac{xm^2}{xm^2-xmn+1},$$ where $[b_1-1,\ldots,b_k]=\frac{m}{n}$. Thus $$[c_1,\ldots,c_l,x+1,b_k,\ldots,b_1-1]=2-\frac{xm^2-xmn+1}{xm^2}=\frac{xm^2+xmn-1}{xm^2}.$$
	Since $(xmn+xn^2+1)xm^2=(xm^2+xmn-1)(xmn+1)+1$, we have that $$[b_1-1,\ldots,b_k,x+1,c_l,\ldots,c_2,c_1]=\frac{xm^2+xmn-1}{xmn+xn^2+1}.$$
	Thus $$[b_1,\ldots,b_k,x+1,c_l,\ldots,c_2,c_1]=1+\frac{xm^2+xmn-1}{xmn+xn^2+1}=\frac{x(m+n)^2}{x(m+n)n+1}$$
	$$\text{and }[b_1,\ldots,b_k]=1+\frac{m}{n}=\frac{m+n}{n}.$$
	 Setting $p=m+n$ and $q=n$ yields the result.
\end{proof}

\begin{prop} Let $[b_1,\ldots,b_k]=\frac{p}{q}$ and let $\textbf{a}=(a_1,\ldots,a_n)\in\mathcal{S}_{1a}$. Then $|\textup{Tor}(H_1(\textbf{T}_{-A(\textbf{a})}))|=p^2$.\label{1(a)orderb}
\end{prop}

\begin{proof}
	Let $\textbf{a}=(2,b_1,\ldots,b_k,2,c_l,\ldots,c_1)$, where $(b_1,\ldots,b_k)$ and $(c_1,\ldots,c_l)$ are linear-dual (up to cyclic reordering). By Lemma \ref{contfraccalc}, $[b_1,\ldots,b_k,2,c_l,\ldots,c_1]=\frac{p^2}{pq+1}$ and so $$\displaystyle[2,b_1,\ldots,b_k,2,c_l,\ldots,c_1]=2-\frac{pq+1}{p^2}=\frac{2p^2-pq-1}{p^2}.$$ By Proposition \ref{order}, $|\textup{Tor}(H_1(\textbf{T}_{-A(\textbf{a})}))|=|2p^2-pq-1-(\alpha-2)|$, where $\alpha$ is the denominator of $[2,b_1,\ldots,b_k,2,c_l,\ldots,c_2]$. By Lemma \ref{contfraccalc},
	$$[c_1,\ldots,c_l,2,b_k,\ldots,b_1]=\frac{p^2}{p^2-pq+1}$$  and so
	$$[c_2,\ldots,c_l,2,b_k,\ldots,b_1]=\frac{p^2-pq+1}{(c_1-1)p^2-c_1pq+c_1}.$$
	Thus
	$$[b_1,\ldots,b_k,2,c_l,\ldots,c_2]=\frac{p^2-pq+1}{s}\text{ for some $s$}.$$
	Now, it is clear that $\alpha=p^2-pq+1$ and so $|\textup{Tor}(H_1(\textbf{T}_{-A(\textbf{a})}))|=|2p^2-pq-1-(p^2-pq+1-2)|=p^2$.
\end{proof}

\bibliographystyle{plain}
\bibliography{Bibliography}

\end{document}