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\title{Triangulations of the product of spheres with few vertices}
\author{Alexander Wang\addressmark{1}\thanks{\href{mailto:alxwang@umich.edu}{alxwang@umich.edu}} \and Hailun Zheng\addressmark{1}\thanks{\href{mailto:hailunz@umich.edu}{hailunz@umich.edu}}}
\address{Department of Mathematics, University of Michigan, Ann Arbor, MI, USA}
\received{2018-11-13}
\revised{}
\abstract{
    A small triangulation of sphere products can be found in lower dimensional cases by computer search and is known in few other cases: Klee and Novik constructed a balanced triangulation of $\mathbb{S}^1\times \mathbb{S}^{d-2}$ with $3d$ vertices and a centrally symmetric triangulation of $\mathbb{S}^i\times \mathbb{S}^{d-i-1}$ with $2d+2$ vertices for all $d\geq 3$ and $1\leq i\leq d-2$. In this paper, we provide an alternative centrally symmetric $(2d+2)$-vertex triangulation of $\mathbb{S}^{i}\times \mathbb{S}^{d-i-1}$. We also construct the first balanced triangulation of $\mathbb{S}^2\times \mathbb{S}^{d-3}$ with $4d$ vertices, using a sphere decomposition inspired by handle theory.}
\keywords{sphere products, minimal triangulations of manifolds, balanced complexes, centrally symmetric complexes}

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\begin{document}

\maketitle

\section{Introduction}\label{sec:intro}
%motivating questions

Minimal triangulations of manifolds are an important research object in combinatorial and computational topology. What is the minimal number of vertices required to triangulate a given manifold? How do we construct a vertex-minimal triangulation and is this triangulation unique?

%non-balanced case: overall
In this paper, we focus on the triangulation of sphere products. From a result of \cite{Brehm-Kuhnel}, it is known that a combinatorial triangulation of $\Sp^i\times\Sp^{d-i-1}$ has at least $2d-i+2$ vertices. In 1986, \cite{Kuhnel} constructed a triangulation of $\Sp^1 \times \Sp^{d-2}$ with $2d+1$ vertices for odd $d$. Later, two groups of researchers, \cite{Bagchi-Datta} as well as \cite{Chestnut-Sapir-Swartz}, found in 2008 that K\"uhnel's construction is indeed the unique minimal triangulation for odd $d$. For even $d$, they showed that the minimal triangulation requires $2d+2$ vertices and is not unique.

%non-balanced case: construction
The minimal triangulations of other sphere products are less well-understood. The best general result is from \cite{Klee-Novik-cs-manifolds}, where a centrally symmetric triangulation of $\Sp^i\times \Sp^{d-i-1}$ with $2d+2$ vertices is constructed as a subcomplex of the boundary of the $(d+1)$-cross-polytope. In general, a result of \cite{Brehm-Kuhnel} states that a triangulation of $\Sp^i \times \Sp^j$ requires at least $i + 2j + 4$ vertices for $i \geq j$.  In addition, the minimal triangulation of $\Sp^2\times \Sp^{d-3}$ for $d\leq 6$ as well as the minimal triangulation of $\Sp^3\times \Sp^3$ are found by the computer program BISTELLAR \cite{Lutz}, which has shown that this lower bound is not always tight, as a triangulation of $\Sp^2 \times \Sp^2$ requires at least $11$ vertices. In this paper, we give an alternative centrally symmetric $(2d+2)$-vertex triangulation of $\Sp^2\times \Sp^{d-3}$ for all $d\geq 5$. The construction is based on finding two shellable balls in the $d$-sphere whose intersection triangulates $\Sp^1\times \mathbb{D}^{d-2}$, where $\mathbb{D}^{d-2}$ is the $(d-2)$-dimensional disk. By an inductive argument, we also obtain the triangulation of other sphere products in higher dimensions, see \cref{ssec:triangulation_other}.

%balanced case: overall
In recent years, balanced triangulated manifolds have caught much attention. A $(d-1)$-dimensional simplicial complex is \emph{balanced} provided that its graph is $d$-colorable. Many important classes of complexes arise as balanced complexes, such as barycentric subdivisions of regular CW complexes and Coxeter complexes. As taking barycentric subdivisions of a complex would generate a lot of new vertices, one would ask if there is a more efficient way to construct the balanced triangulated manifold from a non-balanced one.

%balanced case: construction
In much of the same spirit as K\"uhnel's construction, \cite{Klee-Novik-balanced-LBT} provided a balanced triangulation of $\Sp^1\times \Sp^{d-2}$ with $3d$ vertices for odd $d$ and with $3d+2$ vertices otherwise. Furthermore, \cite{Zheng} showed that the number of vertices for the minimal triangulation is indeed $3d$ for odd $d$ and $3d+2$ otherwise. However, as of yet, no small balanced triangulations of $\Sp^i\times \Sp^{d-i-1}$ for $2\leq i\leq d-3$ exist in literature. In this paper, we construct the first balanced triangulation of $\Sp^2\times \Sp^{d-3}$ with $4d$ vertices. The construction uses a sphere decomposition inspired by handle theory. %Recently, \cite{Izmestiev-Klee-Novik} proved that any two balanced PL homeomorphic closed combinatorial manifolds can be connected using a sequence of cross-flips. In particular, given a balanced triangulated manifold, this allows us to computationally search for a minimal balanced triangulation. However, the computation complexity grows very fast as the dimension and the number of vertices increase, see \cite[Theorem 2.4]{Kubitzke-Venturello}. So far, we have been unable to find a smaller triangulation from our construction.


% The structure of the manuscript
The extended abstract is structured as follows. In \cref{sec:preliminaries}, we review the basics of simplicial complexes, balanced triangulations, and other relevant definitions. In \cref{sec:triangulations}, we present our centrally symmetric $(2d+2)$-vertex triangulation of $\Sp^2\times \Sp^{d-3}$ and construct other sphere products inductively. In \cref{sec:balanced}, the balanced triangulation of $\Sp^2\times \Sp^{d-3}$ with $4d$ vertices is constructed, followed by a discussion of its properties.

\section{Preliminaries}\label{sec:preliminaries}
A \textit{simplicial complex} $\Delta$ with vertex set $V$ is a collection of subsets
$\sigma\subseteq V$, called \textit{faces}, that is closed under inclusion, such that for every $v \in V$, $\{v\} \in \Delta$. For $\sigma\in \Delta$, let $\dim\sigma:=|\sigma|-1$ and define the \textit{dimension} of $\Delta$, $\dim \Delta$, as the maximum dimension of the faces of $\Delta$.  A face $\sigma \in \Delta$ is said to be a \textit{facet} provided that it is a face which is maximal with respect to inclusion.  We say that a simplicial complex $\Delta$ is \textit{pure} if all of its facets have the same dimension. If $\Delta$ is $(d-1)$-dimensional and $-1 \leq i \leq d-1$, then the \textit{$f$-number} $f_i = f_i(\Delta)$ denotes the number of $i$-dimensional faces of $\Delta$. The \textit{star} and \textit{link} of a face $\sigma$ in $\Delta$ is defined as follows:

\[
\st(\sigma, \Delta):= \{\tau \in\Delta: \sigma\cup\tau\in\Delta \}, \quad \lk(\sigma,\Delta):=\{\tau \in \st_\Delta \sigma: \tau \cap \sigma = \emptyset\}.
\]
When the context is clear, we may simply denote the star and link of $\sigma$ as $\st(\sigma)$ and $\lk(\sigma)$ respectively. We also define the \textit{restriction} of $\Delta$ to a vertex set $W$ as $\Delta[W]:=\{\sigma\in \Delta:\sigma\subseteq W\}$. A subcomplex $\Omega \subset \Delta$ is said to be \textit{induced} provided that for all faces $F \in \Delta$, if every vertex $v \in F$ is a vertex of $\Omega$, then $F$ is a face in $\Omega$. The \textit{$i$-skeleton} of a simplicial complex $\Delta$ is the subcomplex containing all faces of $\Delta$ which have dimension at most $i$.  In particular, the $1$-skeleton of $\Delta$ is the graph of $\Delta$.

Denote by $\sigma^d$ the $d$-simplex. A \emph{combinatorial $(d-1)$-sphere} (respectively, a \emph{combinatorial $(d-1)$-ball}) is a simplicial complex PL
homeomorphic to $\partial \sigma^d$ (respectively, $\sigma^{d-1}$). A closed \emph{combinatorial $(d-1)$-manifold} is a connected simplicial complex with the property that the link of each vertex is a combinatorial $(d-2)$-sphere. A simplicial complex $\Delta$ is a \textit{simplicial manifold}, if the geometric realization of $\Delta$ is homeomorphic to a manifold. The boundary complex of a simplicial $d$-ball is a simplicial $(d-1)$-sphere. In general, a simplicial manifold need not be combinatorial. 

A $(d-1)$-dimensional simplicial complex $\Delta$ is called \textit{balanced} if the graph of $\Delta$ is $d$-colorable; that is, there exists a coloring map $\kappa: V\to \{1,2,\cdots,d\}$ such that $\kappa(x) \neq \kappa(y)$ for all edges $\{x,y\}\in\Delta$. A simplicial complex is \emph{centrally symmetric} or \emph{cs} if it is endowed with a free involution $\alpha: V(\Delta)\to V(\Delta)$ that induces a free involution on the set of all non-empty faces. 

Let $\partial C^*_d$ be the boundary complex of the $d$-cross-polytope. It is cs, and furthermore, it is a balanced vertex-minimal triangulation of the $(d-1)$-sphere. Label the vertex set of $\partial C^*_d$ as $\{x_1,\dots, x_d,y_1,\dots, y_d\}$ such that $x_i, y_i$ form a pair of antipodal vertices for all $i$. Every facet of $\partial C^*_d$ can be written in the form $u_1u_2\dots u_d$, where each $u_i \in \{x_i,y_i\}$. We say a facet has a switch at position $i$ if $u_i$ and $u_{i+1}$ have different labels. Let $B(i,d)$ be the pure subcomplex of $\partial C^*_d$ that contains all facets with at most $i$ switches. For example, $B(0,d)$ consists of the two disjoint facets $\{x_1,\dots, x_d\}$ and $\{y_1,\dots, y_d\}$. If $\Gamma$ is a subcomplex of $\partial C_d^*$, we let the \emph{complement} of $\Gamma$ in $\partial C_d^*$ be the complex generated by those facets that are not in $\Gamma$. Denote by $\mathcal{D}_d$ the dihedral group of order $2d$. 

\begingroup\emergencystretch=5em
The following lemma is essentially Theorem 1.2 in \cite{Klee-Novik-cs-manifolds}.
\begin{lemma}\label{lemma: B(i-d) properties}
	For $0\leq i<d-1$, the complex $B(i,d)$ satisfies the following properties:
	\begin{enumerate}
		\item $B(i,d)$ contains the entire $i$-skeleton of $\partial C_d^*$ as a subcomplex.
		\item The boundary of $B(i,d)$ is homeomorphic to $\Sp^i\times \Sp^{d-i-2}$.
		\item $B(i,d)$ is a balanced centrally symmetric combinatorial manifold whose integral (co)homology groups coincide with those of $\Sp^i$. Also, $B(0,d)\cong \mathbb{D}^{d-1}\times \Sp^0$ and $B(1,d)\cong \mathbb{D}^{d-2}\times \Sp^1$.
		\item The complement of $B(i,d)$ in $\partial C^*_d$ is simplicially isomorphic to $B(d-i-2,d)$.
		\item B(i,d) admits a vertex-transitive action of $\mathbb{Z}_2\times \mathcal{D}_d$ if $i$ is even and of $\mathcal{D}_{2d}$ if $i$ is odd.
	\end{enumerate}
\end{lemma}
\endgroup
Finally, we define shellability.
\begin{definition}
	Let $\Delta$ be a pure $d$-dimensional simplicial complex. A \emph{shelling} of $\Delta$ is a linear ordering of the facets $F_1,F_2,\dots, F_s$ such that $F_k\cap (\cup_{i=1}^{k-1} F_i)$ is a pure $(d-1)$-dimensional complex for all $2\leq k\leq s$, and $\Delta$ is called \emph{shellable} if it has a shelling.
\end{definition}
%Equivalently, $\Delta$ is called shellable if for each $j\geq 0$ there exists a face $r(F_j )\subseteq F_j$ such that $F_j\backslash \cup_{i<j}F_i=[r(F_j), F_j]$, where $[r(F_j), F_j]=\{G: r(F_i)\subseteq G\subseteq F_j\}$. The faces $r(F_j)$ are called the \emph{restriction} of $F_j$. Note that not every simplicial ball and simplicial sphere are shellable.

\section{The cs triangulations of the sphere products}\label{sec:triangulations}
It is known that for $i\leq j$, the minimal triangulation of $\Sp^i\times \Sp^j$ requires at least $i+2j+4$ vertices, see \cite{Brehm-Kuhnel}. Such triangulations are constructed by \cite{Lutz} in lower dimensional cases but not known in general. 
We aim at finding an alternative triangulation of $\Sp^2\times \Sp^{d-3}$ with $2d+2$ vertices for $d\geq 5$. The following theorem is Theorem 7 in \cite{Kreck}.

\begin{theorem}\label{thm: homology of product of spheres}
	Let $M$ be a simply connected codimension-1 submanifold of $\Sp^{d}$, where
	$d \geq 5$. If $M$ has the homology of $\Sp^i \times \Sp^{d-i-1}$ and $1 < i \leq \frac{d-1}{2}$, then $M$ is homeomorphic
	to $\Sp^i \times \Sp^{d-i-1}$.
\end{theorem} 
%Throughout, we write $\tilde{H}_*(\Delta)$ to denote the reduced homology of $\Delta$ with coefficients in $\Z$, and $\beta_i(\Delta)$ denotes the rank of $\tilde{H}_i(\Delta)$. 
\begin{proposition}\label{prop: criteria}
	Fix $d$ and $i\leq \frac{d-1}{2}$. Let $D_1$ and $D_2$ be two combinatorial $d$-balls such that
	\begin{enumerate}
	    \item $\partial (D_1\cup D_2)$ is $(d-1)$-dimensional submanifold of a combinatorial $d$-sphere.
	    \item $D_1\cap D_2=\partial D_1\cap \partial D_2$ is a path-connected combinatorial $(d-1)$-manifold (with boundary) that has the same homology as $\Sp^{i-1}$.
	    \item $\partial (D_1\cap D_2)$ has the same homology as $
\Sp^{i-1}\times \Sp^{d-i-1}$.
	\end{enumerate}
	 Then $\partial(D_1\cup D_2)$ triangulates $\Sp^i\times \Sp^{d-i-1}$ for $d\geq 5$.
\end{proposition}   
\begin{proof}
	First note that $D_1\cup D_2$ is the union of two combinatorial $d$-balls that intersect along the combinatorial $(d-1)$-manifold $D_1\cap D_2$. Hence $D_1\cup D_2$ is a combinatorial $d$-manifold, and $\partial (D_1\cup D_2)$ is a combinatorial $(d-1)$-manifold.
	
    Since $D_1\cap D_2=\partial D_1 \cap \partial D_2$, we have that the intersection of $\overline{\partial D_1\backslash \partial D_2}$ and $D_1\cap D_2$ is exactly $\partial (D_1\cap D_2)$. We apply the Mayer-Vietoris sequence on $(\overline{\partial D_1\backslash\partial D_2}, D_1\cap D_2,\partial D_1)$, and by condition (2) we obtain that $\overline{\partial D_1\backslash\partial D_2}$ has the same homology as $\Sp^{d-i-1}$. 
    
    Note that the intersection of \[\big(\overline{\partial D_1\backslash \partial D_2}\big)\cap\big(\overline{\partial D_2\backslash \partial D_1}\big)=\partial (\partial D_1\cap \partial D_2)=\partial (D_1\cap D_2).\] 
    We then apply the Mayer-Vietoris sequence to $(\overline{\partial D_1\backslash \partial D_2}, \overline{\partial D_2\backslash \partial D_1}, \partial(D_1\cup D_2))$. By condition (3), we find that $\partial (D_1\cup D_2)$ has the same homology as $\Sp^i\times \Sp^{d-i-1}$.
    
    Finally, the complex $D_1\cup D_2$ is simply connected, since the union of two simply connected open subsets $\mathrm{int} D_1, \mathrm{int}D_2$ with path-connected intersection $D_1\cap D_2$ is simply connected. We conclude from condition (1) and \cref{thm: homology of product of spheres} that $\partial (D_1\cup D_2)$ triangulates $\Sp^i\times \Sp^{d-i-1}$.
\end{proof}

The above proposition provides us with a general method of constructing a triangulation of $\Sp^i\times \Sp^{d-i-1}$. 

\subsection{A triangulation of $\Sp^2\times \Sp^{d-3}$}
Let $\tau$ be a face of $\partial C_d^*$ and let $\kappa(\tau)$ count the number of $y$ labels in $\tau$. Define $\Gamma_j$ to be the union of facets $\tau$ in $\partial C_d^*$ that have at most 2 switches and with $\kappa(\tau)=j$. Hence for $1\leq j\leq d-1$, the complex $\Gamma_j$ consists of $d$ facets  $\tau_j^k=\{x_1,\dots,x_d\}\backslash\{x_{k}, \dots x_{k+j-1}\}\cup \{y_{k},\dots, y_{k+j-1}\}$ for $1\leq k\leq d$.

\begin{lemma}\label{lm: shellable ball}
	The complex $\cup_{k=0}^{i}\Gamma_k$ is a shellable $(d-1)$-ball for all $0\leq i\leq \left\lceil \frac{d+1}{2}\right\rceil$.
\end{lemma}
\begin{proof}
	The $(d-1)$-ball $\cup_{k=0}^{i}\Gamma_k$ has a shelling order $\{x_1.\dots. x_d\},\tau_1^1,\dots, \tau_1^d,\dots \tau_i^1,\dots,\tau_i^d$.
\end{proof}

We propose the candidates $D_1,D_2\subseteq \partial C_{d+1}^*$ that satisfy the conditions in \cref{prop: criteria}.
\begin{construction}\label{def: D_1-D_2}
	For $d\geq 3$, define two simplicial $d$-balls $D_1,D_2$ as a subcomplex of the octahedral $d$-sphere on vertex set $\{x_1, y_1,\dots, x_{d+1},y_{d+1}\}$ as follows: 
	\begin{enumerate}
		\item For $d$ is odd, let $m=\frac{d-1}{2}$. Define $D_1=(\cup_{k=0}^{m+1}\Gamma_k)*\{x_{d+1}\}$ and $D_2=(\cup_{k=m}^{d}\Gamma_k)*\{y_{d+1}\}$. In particular, $D_1\cap D_2=\Gamma_{m}\cup \Gamma_{m+1}$ is cs.
		\item For $d$ is even, let $m=\frac{d}{2}$ and $\gamma:=\cup_{i=1}^{m} \tau_{m-1}^i$ be a subcomplex of $\Gamma_{m-1}$. By the definition, $\tau_j^k$ and $\tau_{d-j}^{k+j}$ are antipodal facets for any $k,j$. So $-\gamma=\cup_{i=m}^{d-1}\tau_{m+1}^i\subseteq \Gamma_{m+1}$. In this case we let \[D_1=\big((\cup_{k=0}^{m}\Gamma_k) \cup (-\gamma)\big)*\{x_{d+1}\}, \quad D_2=\big((\cup_{k=m}^{d}\Gamma_k)\cup \gamma\big)*\{y_{d+1}\}.\] 
		In particular, $D_1\cap D_2=\Gamma_m\cup \gamma\cup (-\gamma)$ is centrally symmetric.
	\end{enumerate} 
\end{construction}

Next we show that $\partial(D_1\cap D_2)\cong \Sp^1\times \Sp^{d-3}$. Given two facets $F_1, F_2\in \partial C_d^*$, let $d(F_1,F_2)$ be the distance from $F_1$ to $F_2$ in the facet-ridge graph of $\partial C_d^*$. 
\begin{lemma}\label{lm: criteria of sphere disc product}
	Let $\Delta$ be a combinatorial $(d-1)$-manifold in $\partial C_d^*$ whose facet-ridge graph is a $2d$-cycle. Enumerate its facets as $\sigma_1,\sigma_2,\dots \sigma_{2d}$ such that $\sigma_i, \sigma_{i+1}$ are adjacent for $1\leq i\leq 2d$. If $\sigma_i=-\sigma_{d+i}$ for all $i$, then $\Delta$ triangulates $\Sp^1\times \mathbb{D}^{d-2}$.
\end{lemma}
\begin{proof}
	Let $\sigma_1=\{u_1,\dots, u_d\}$. By the assumption, $\sigma_{d+1}=\{-u_1,\dots, -u_d\}$. Since $d(\sigma_1, \sigma_{d+1})=d$ in $\partial C_d^*$, the sequence $\sigma_1,\sigma_2,\dots, \sigma_{d+1}$ gives the shortest path from $\sigma_1$ to $\sigma_{d+1}$. So it follows that there is an ordering of the vertices, say $(u_1,\dots, u_d)$, such that $\sigma_{i+1}=\sigma_i\backslash\{u_{i}\}\cup \{-u_i\}$. Together with $\sigma_i=-\sigma_{i+d}$ for all $i$, we see that $\Delta\cong B(1,d)$ as defined in \cite{Klee-Novik-cs-manifolds}. Hence as $B(1,d)$, $\Delta$ also triangulates $\Sp^1\times \mathbb{D}^{d-2}$.
\end{proof}
\begin{lemma}\label{lm: D_1-D_2}
	The complex $D_1\cap D_2$ constructed above triangulates $\Sp^1\times \mathbb{D}^{d-2}$.
\end{lemma}
\begin{proof}
For odd $d$ and $m=\frac{d-1}{2}$, we enumerate the facets of $D_1\cap D_2=\Gamma_m\cup\Gamma_{m+1}$ as $(\sigma_1\dots, \sigma_{2d}):=$
	\[(\tau_{m}^1,\tau_{m+1}^1, \tau_m^2, \tau_{m+1}^2,\dots, \tau_m^d, \tau_{m+1}^d).\] 
	Each $\sigma_i$ has exactly two adjacent facets $\sigma_{i-1}, \sigma_{i+1}$, and so the facet-ridge graph of  $D_1\cup D_2$ is a $2d$-cycle. Furthermore, $\tau_m^j=-\tau_{m+1}^{j+m}$ by the definition. So the claim follows from \cref{lm: criteria of sphere disc product}. The proof is similar for $d$ is even.
\end{proof}
\begin{theorem}\label{cor: intersection of D_1 D_2 is a manifold}
	The complex $\partial (D_1\cup D_2)$ triangulates $\Sp^2\times \Sp^{d-3}$ for $d\geq 5$.
\end{theorem}
\begin{proof}
	This follows from \cref{lm: shellable ball,lm: D_1-D_2,prop: criteria}.
\end{proof}

\begin{property} For the complex $\partial (D_1\cup D_2)$ in \cref{def: D_1-D_2}:
	\begin{enumerate}
	\item It has $2d+2$ vertices.
	\item It contains the 2-skeleton of $\partial C_{d+1}^*$.
	\item It admits vertex-transitive actions by the group $\mathbb{Z}_2\times \mathcal{D}_d$ if $d$ is odd, and by $\mathbb{Z}_2$ if $d$ is even.
\end{enumerate}
\end{property}
\begin{remark}
For $d$ is odd, three types of vertex-transitive action on $\partial(D_1\cup D_2)$ are given by
	\begin{itemize}
		\item $D$ maps $x_j$ to $y_j$, and $y_j$ to $x_j$, for $1\leq j\leq d+1$. 
		\item $R$ fixes $x_{d+1}, y_{d+1}$, and maps $x_j, y_j$ to $x_{d-j+1}, y_{d-j+1}$ respectively, for $1\leq j\leq d$. 
		\item $S$ fixes $x_{d+1}, y_{d+1}$, and maps $x_j, y_j$ to $x_{j+1}, y_{j+1}$ (modulo $d$) respectively.
	\end{itemize}
Comparing with the group actions on $B(2, d+1)$ in \cite{Klee-Novik-cs-manifolds}, we see that $\partial(D_1\cup D_2)$ and $\partial B(2,d+1)$ are combinatorially distinct.
\end{remark}


\begin{remark}\label{rk: variation of constructions}
	There are many other ways to construct $D_1', D_2'$ as the subcomplex of $\partial C_{d+1}^*$ that satisfies the conditions in \cref{prop: criteria}. For example, when $d=2m+1$, let $\tau=\{x_1,\dots x_d\}$, $-\tau=\{y_1,\dots, y_d\}$. It is possible to construct a simplicial $(d-1)$-ball $B$ in $\partial C_d^*$ and simplicial $d$-balls \[D_1'=(B\cup \Gamma_m\cup \Gamma_{m+1})*\{x_{d+1}\},\; D_2'=\Big((-B)\cup \Gamma_{m}\cup \Gamma_{m+1}\Big)*\{y_{d+1}\}\] such that \[B\cup (-B)=\partial C_d^*,\;\; B\supseteq\{\sigma: |\sigma\cap \tau|\geq m\} , \;\; -B=\{\sigma: |\sigma\cap (-\tau)|\geq m\}.\] Furthermore, $D_1'\cap D_2'=\Gamma_m\cup \Gamma_{m+1}$. When $d=2m$, $D_1', D_2'$ can also be defined in the same spirit as \cref{def: D_1-D_2}. 
\end{remark}

\subsection{The triangulation of other sphere products}\label{ssec:triangulation_other}

The goal of this section is to construct a triangulation of $\Sp^i\times \Sp^{d-i-1}$ as a subcomplex of $\partial C_{d+2}^*$ from a given triangulation  of $\Sp^{i-1}\times \Sp^{d-i-1}$ in $\partial C_{d+1}^*$, for $i\leq \frac{d-1}{2}$.
\begin{proposition}\label{prop: construct sphere products inductively}
	Let $D_1$ and $D_2$ be cones over two combinatorial $(d-1)$-balls in $\partial C_{d}^*$ whose coning points are $x_{d+1}, y_{d+1}$ respectively. Furthermore, 
	\begin{enumerate}
	    \item The union of $D_1\backslash \{x_{d+1}\}$ and $D_2\backslash \{y_{d+1}\}$ covers $\partial C_d^*$.
	    \item $D_1\cap D_2$ is a path-connected combinatorial $(d-1)$-manifold that has the same homology as $\Sp^{i-1}$ for some $2\leq i\leq d-2$.
	    \item $\partial (D_1\cap D_2)$ has the same homology as $\Sp^{i-1}\times \Sp^{d-i-1}$.
	\end{enumerate}
	Let \[E_1=(\st(y_{d+1}, \partial C_{d+1}^*)\cup D_1)*\{x_{d+2}\}, \quad E_2=(\st(x_{d+1}, \partial C_{d+1}^*)\cup D_2)*\{y_{d+2}\}.\]
	Then the union of $E_1\backslash\{x_{d+2}\}$ and $E_2\backslash \{y_{d+2}\}$ covers $\partial C_{d+1}^*$, $E_1\cap E_2$ is a combinatorial $d$-manifold that has the same homology as $\Sp^i$ and $\partial (E_1\cup E_2)$ triangulates $\Sp^{i+1}\times \Sp^{d-i-1}$.
\end{proposition}
\begin{proof}By condition (1), $E_1\cap E_2=D_1\cup D_2$. Then we use the Mayer-Vietoris sequence on the triple $(D_1, D_2, D_1\cup D_2)$ and conclude from \cref{thm: homology of product of spheres}. The proof is similar to that of \cref{prop: criteria}.
%	Since the subcomplexes $\st(y_{d+1}, \partial C_{d+1}^*)$, $D_1$ and the intersection $\st(y_{d+1}, \partial C_{d+1}^*)\cap D_1=D_1\backslash \{x_{d+1}\}$ are all simplicial balls, it follows that their union is a simplicial $d$-ball. Hence $E_1$ is a simplicial $(d+1)$-ball. Similarly $E_2$ is also a simplicial $(d+1)$-ball. Furthermore, since $D_1\cup D_2\supseteq \partial C_d^*$, 
%	\[E_1\cap E_2=(\st(y_{d+1}, \partial C_{d+1}^*)\cup D_1)\cap (\st(x_{d+1}, \partial C_{d+1}^*)\cup D_2)=D_1\cup D_2,\]which is a path-connected combinatorial $d$-manifold. Hence $E_1\cup E_2$ is a combinatorial $(d+1)$-manifold, and \[(E_1\backslash\{x_{d+2}\})\cup (E_2\backslash \{y_{d+2})\supseteq \st(y_{d+1}, \partial C_{d+1}^*)\cup \st(x_{d+1}, \partial C_{d+1}^*)=\partial C_{d+1}^*.\]
%	
%	The conditions together with Proposition \ref{prop: criteria} imply that $\partial (D_1\cup D_2)$ triangulates $\Sp^i\times \Sp^{d-i-1}$. Also using the Mayer-Vietoris sequence on the triple $(D_1, D_2, D_1\cup D_2)$, it follows that $E_1\cap E_2=D_1\cup D_2$ has the same homology as $\Sp^i$. Since $\partial(E_1\cap E_2)=\partial(D_1\cup D_2)$ triangulates $\Sp^i\times \Sp^{d-i-1}$, we conclude from Proposition \ref{prop: criteria} that $\partial (E_1\cup E_2)$ triangulates $\Sp^{i+1}\times \Sp^{d-i-1}$.
\end{proof}
\begin{construction}
    We take our base construction $\partial (D_1'\cup D_2')$ as given in \cref{rk: variation of constructions} (Note that the complexes $D_1, D_2$ in \cref{def: D_1-D_2} does not satisfy the condition that the union of $D_1\backslash\{x_{d+1}\}$ and $D_2\backslash \{y_{d+1}\}$ covers $\partial C_d^*$.) and apply the above proposition inductively. This gives us a family of cs triangulations of sphere products $\Sp^2\times \Sp^{d-3}$, $\Sp^3\times \Sp^{d-3}$, $\dots, \Sp^{d-3}\times \Sp^{d-3}$. Each triangulation of $\Sp^i\times \Sp^{d-3}$ has $2d+2i-2$ vertices.
\end{construction}


%By the work of \cite{Lutz}, the minimal triangulation of $\Sp^i\times \Sp^j$ has $i+2j+4$ vertices for $(i,j)=(2,3)$ or $(3,3)$; the minimal triangulation is also not unique. Unfortunately this is not true for all $(i,j)$: it requires at least 11 vertices to triangulate $\Sp^2\times \Sp^2$, see \cite{Kuhnel-Lasmann}. Our construction is obtained by setting $D_1, D_2$ (as required in Proposition \ref{prop: criteria}) as subcomplexes in the boundary of the cross-polytope. Can we choose other $D_1, D_2$ to reduce the number of vertices in the triangulation?
%\begin{problem}
%Find a triangulation of $\Sp^i\times \Sp^j$ with less than $2i+2j+4$ vertices.
%\end{problem}

\section{A balanced triangulation of $\Sp^2\times \Sp^{d-3}$}\label{sec:balanced}

In this section, we present our main construction for a balanced triangulation of $\Sp^2 \times \Sp^{d-3}$.  The geometric intuition of our construction comes from handle theory. The sphere $\Sp^d$ admits the following decomposition, see \cite{Rourke}:
\[
\Sp^{d-1} = (\Sp^1 \times \mathbb{D}^{d-2}) \cup (\mathbb{D}^2 \times \Sp^{d-3}).\]


Let $S$ be a triangulated $(d-1)$-sphere that has the decomposition $S=B_1\cup_{\partial B_1=\partial B_2} B_2$, where $B_1\cong \Sp^1\times \mathbb{D}^{d-2}$, $B_2\cong \mathbb{D}^2\times \Sp^{d-3}$, and $\partial B_1\cong \partial B_2\cong \Sp^1 \times \Sp^{d-3}$. Note that $\Sp^2 \times \Sp^{d-3}$ admits the decomposition into $(\mathbb{D}^2 \times \Sp^{d-3}) \cup (\mathbb{D}^2 \times \Sp^{d-3}) \cong B_2 \cup B_2$. Then, from $S$ we can form a triangulation of $\Sp^2\times \Sp^{d-2}$ in the following way: take two copies of $B_2$ and denote them as $B_2$ and $B_2'$. If $\partial B_2$ is an induced subcomplex in $B_2$, then we glue $B_2$ and $B_2'$ along their boundaries. The resulting complex is homeomorphic to $\Sp^2\times \Sp^{d-3}$. However, if $\partial B_2$ is not an induced subcomplex of $B_2$, then usually we cannot glue $B_2$ and $B_2'$ by identifying their boundaries directly and still obtain a triangulated manifold. An alternative method is to find a complex $N\cong \partial B_2\times \mathbb{D}^1$ with $\partial N=\partial B_2\cup\partial B_2'$ so that $N$ serves as a tubular neighborhood of both $\partial B_2$ and $\partial B_2'$. Finally the complex $B_2\cup N\cup B_2'$ is a triangulation of $\Sp^2\times \Sp^{d-3}$.

Our approach of constructing a balanced triangulation of $\Sp^2\times \Sp^{d-3}$ is by finding suitable balanced candidates of $B_2$ and $N$ as described above. 
%{\color{blue}Help! I am thinking of the terminology. The deleted connected sum seems to be a bad notation. What about $\Diamond$-connected sum? Is it better or worse?}
\begin{definition}\label{Definition: connected sum}
	Consider $(\Gamma_1, \sigma_1)$ and $(\Gamma_2, \sigma_2)$, where $\Gamma_i$ is the boundary complex of the $d$-cross-polytope, and $\sigma_i$ is a fixed facet of $\Gamma_i$. Let $\kappa$ be the coloring map on $\Gamma_1\cup \Gamma_2$. If $e_i$ is an edge in $\Gamma_i$ but not in $\pm \sigma_i$ and $\kappa(e_1)=\kappa(e_2)$, then the $\Diamond$-connected sum $(\Gamma_1\#\Gamma_2,\sigma_1\#\sigma_2)$ is obtained by deleting $e_i$ from $\Gamma_i$, and gluing $\Gamma_1-e_1$ with $\Gamma_2-e_2$ by identifying $\st(e_1)[V(\sigma_1)]$ with $\st(e_2)[V(\sigma_2)]$, and $\st(e_1)[V(-\sigma_1)]$ with $\st(e_2)[V(-\sigma_2)]$.
\end{definition}
\begin{figure}[h]
	\centering
	\subfloat[$\Gamma_1$ and $\Gamma_2$; here $\sigma_1=\{y_1,y_2,y_3\}$ and $\sigma_2=\{x_1,y_2,y_3\}$]{\includegraphics{3-crosspolytope}}
	\hspace{0.8cm}
	\subfloat[$(\Gamma_1\#\Gamma_2,\sigma_1\#\sigma_2)$]{\includegraphics[scale=1.2]{connectedsum}}
	\caption{The $\Diamond$-connected sum $(\Gamma_1\#\Gamma_2,\sigma_1\#\sigma_2)$: delete the edge $\{y_3x_1'\}$ in both $\Gamma_1$ and $\Gamma_2$, then glue $\Gamma_1$ and $\Gamma_2$ along the 4-cycle $(y_3,x_2',x_1',y_2)$.}
\end{figure}
The following properties of the $\Diamond$-connected sum justify the notation $(\Gamma_1\#\Gamma_2,\sigma_1\#\sigma_2)$ in the definition. 
\begin{property} Let $\Gamma_1$ and $\Gamma_2$ be two $d$-crosspolytopes. Furthermore, $\Gamma_1$ has antipodal facets $\sigma_1=\{x_1,\dots,x_d\}, \;-\sigma_1=\{y_1,\dots,y_d\}$, and $\Gamma_2$ has antipodal facets $\sigma_2=\{x_{d+1},\dots,x_{2d}\},\;-\sigma_2=\{y_{d+1},\dots,y_{2d}\}$. Then $(\Gamma_1\#\Gamma_2,\sigma_1\#\sigma_2)$ satisfies the following properties:
	\begin{enumerate}
		\item The complex is a balanced triangulation of $\Sp^{d-1}$.
		\item The restriction of $(\Gamma_1\#\Gamma_2,\sigma_1\#\sigma_2)$ to $V(\sigma_1)\cup V(\sigma_2)$ is the usual connected sum of simplices $\sigma_1\# \sigma_2$.
		\item The link of every edge $e=\{x_i,y_j\}$ in $(\Gamma_1\#\Gamma_2,\sigma_1\#\sigma_2)$ is the boundary complex of a $(d-2)$-crosspolytope.
	\end{enumerate}	
\end{property}
%\begin{proof}
	%Part 1 is clear from the construction. For part 2, if $e_1=\{x_i,y_j\}$ in $\Gamma_1$, then the link $\lk(e_1,\Gamma_1)$ is the boundary of a $(d-2)$-cross-polytope containing the antipodal facets $\sigma_1\backslash\{x_i,x_j\}$ and $(-\sigma_1)\backslash\{y_i,y_j\}$. Similarly, for $e_2=\{x_k,y_l\}$ in $\Gamma_2$, the link $\lk(e_2,\Gamma_2)$ has antipodal facets $\sigma_2\backslash\{x_k,x_l\}$ and $(-\sigma_2)\backslash\{y_k,y_l\}$. Hence the restriction of $(\Gamma_1\#\Gamma_2,\sigma_1\#\sigma_2)$ to $V(\sigma_1)\cup V(\sigma_2)$ is obtained by taking the union of $\sigma_1$ and $\sigma_2$ and identifying $\sigma_1\backslash\{x_j\}$ with $\sigma_2\backslash\{x_l\}$. In this manner, we get the connected sum $\sigma_1\#\sigma_2$.
	%
	%For part 3, let $\Delta$ denote the boundary complex on which $\Gamma_1$ and $\Gamma_2$ are glued together. If $e\notin \Delta$, there is nothing to prove. Otherwise, assume without loss of generality that $e=\{x_1,y_2\}$ and the edge $e'=\{x_1, y_3\}$ is deleted from $\Gamma_1$ to form $(\Gamma_1\#\Gamma_2,\sigma_1\#\sigma_2)$. Then, $\lk(e, \st(e',\Gamma_1))=\{y_3\}*\Sigma$, where $\Sigma$ is the boundary of the crosspolytope on vertices $\{x_4,y_4,\dots,x_d,y_d\}$. Hence, by construction, the link of $e$ in $(\Gamma_1\#\Gamma_2,\sigma_1\#\sigma_2)$ must be the suspension of $\Sigma$; here the suspension vertices are the antipodal vertices of $y_3$ in $\Gamma_1$ and $\Gamma_2$ respectively.
%\end{proof}

The above properties ensure that it is possible to take the  $\Diamond$-connected sum inductively. %To form $(\Gamma_1\#\dots \#\Gamma_k, \sigma_1\#\dots \#\sigma_k)$ from $(\Gamma_1\#\dots \#\Gamma_{k-1}, \sigma_1\#\dots \#\sigma_{k-1})$ and $(\Gamma_k,\sigma_k)$, we take an edge $e_1\in (\Gamma_1\#\dots \#\Gamma_{k-1}, \sigma_1\#\dots \#\sigma_{k-1})$ but not in $\sigma_1\#\dots \sigma_{k-1}$, then take an edge $e_2\in \Gamma_k\backslash \sigma_k$ so that $\kappa(e_1)=\kappa(e_2)$, and then construct as in Definition \ref{Definition: connected sum}. 
Also recall that if $\Gamma$ is a pure simplicial complex, then as long as there exist two facets $F$ and $F'$ on $\Gamma$ and a map $\phi: F\to F'$ so that $v$ and $\phi(v)$ do not have a common neighbor for every $v \in F$, then we can remove $F, F'$ and identify $\partial F$ with $\partial F'$ to get $\Gamma^\phi$. This is called a \emph{handle addition}. Similarly, assume that there are two edges $e_1$ and $e_2$ of the same color in $(\Gamma_1\#\dots \#\Gamma_k, \sigma_1\#\dots \#\sigma_k)$ but not in $ A:=\sigma_1\#\sigma_2\dots\#\sigma_k$ or $-A$. Note that $\st(e_i)$ is a cross-polytope with antipodal facets $\st(e_i)[V(A)]$ and $\st(e_i)[V(-A)]$ for $i=1,2$. If the identification maps 
\[\phi: \st(e_1)[V(A)] \to \st(e_2)[V(A)] \quad \mathrm{and}\quad \phi': \st(e_1)[V(-A)] \to \st(e_2)[V(-A)]\] 
are well-defined, then the maps $\phi$ and $\phi'$ naturally extend to a map \[\bar{\phi}: \st(e_1)\to \st(e_2)\] if for every $v\in \st(e_1)$, $v$ and $\phi(v)$(or $\phi'(v)$) do not have a common neighbor. In this way we obtain a balanced simplicial complex $((\Gamma_1\#\Gamma_2\dots \#\Gamma_k)^{\bar{\phi}}, (\sigma_1\#\sigma_2\dots\#\sigma_k)^\phi)$ by removing $e_1,e_2$ and identifying $\lk(e_1)$ with $\bar{\phi}(\lk(e_1))=\lk(e_2)$. We call this the $\Diamond$-handle addition. 
%Note that as long as the handle addition is well-defined, \[f_0((\Gamma_1\#\Gamma_2\dots \#\Gamma_k)^{\bar{\phi}})=2f_0((\sigma_1\#\sigma_2\dots\#\sigma_k)^\phi)=2k\]
%regardless of the dimension of $\Gamma_i$

We are now ready to construct a balanced triangulation of $\Sp^2\times \Sp^{d-3}$ with $4d$ vertices. We will write $\Gamma_1\#\Gamma_2$ to denote the $\Diamond$-connected sum if $\sigma_1$ and $\sigma_2$ are clear from the context. Also, to simplify notation, we will sometimes write $x_1\dots x_m$ to denote the face $\{x_1,\dots,x_m\}$.

%{\color{blue} I rewrite the main construction to make it more readable. Check if there are any holes in the proof!}
\begin{construction}\label{main construction}
	Let $d\geq 3$. Take two $d$-crosspolytopes $P$ and $P'$. The vertex sets of $P$ and $P'$ are $\{x_1,\dots,x_d,y_1,\dots, y_d\}$ and $\{x'_1,\dots,x'_d,y'_1,\dots, y'_d\}$ respectively. We let $\sigma_i = x_1\dots x_iy_{i+1}\dots y_d$ for $1 \leq i \leq d$ and let $\sigma_i = y_1\dots y_ix_{i+1}\dots x_d$ for $d+1 \leq i \leq 2d$. Then the complex $\Delta_1:=\cup_{i=1}^{2d}\sigma_i$ is exactly $B(1,d)$. We further partition the boundary of $P$ as $\partial P=\Delta_1 \cup_{\partial\Delta_1} \Delta_2$. By Lemma 1.1, $\Delta_2\cong B(d-3,d)$ and $\Delta_1\cap \Delta_2$ is homeomorphic to $\Sp^1\times \Sp^{d-3}$.
	
	Next, define a simplicial map $f: \partial P\to \partial P'$ induced by the following bijection on the vertex sets:
	\[x_i \mapsto x'_{i+1}, \;y_i \mapsto y'_{i+1} \;\; \mathrm{for}\; 1\leq i\leq d-1;\; x_d \mapsto y'_1, \;y_d \mapsto x'_1. \]
	
	By \cref{lemma: B(i-d) properties}, the complex $\Delta_1$ admits a vertex-transitive action by the dihedral group $\mathcal{D}_{2d}$ of order $4d$, where a generator is given by the map we have chosen (Theorem 1.2 of \cite{Klee-Novik-cs-manifolds}). Hence $f$ is a simplicial isomorphism and $f(\Delta_1) \cong B(1,d)$. For each $i$, there is a unique $d$-cross-polytope $\Gamma_i$ containing $\sigma_i$ and $f(\sigma_i)$ as antipodal facets. Next, we check that we can take the $\Diamond$-connected sum of $\Gamma_i$ and $\Gamma_{i+1}$ inductively. Without loss of generality, assume that $1 \leq i \leq d$; otherwise, we can relabel by switching $x$ and $y$. Note that for $i\leq d-2$, %\[\sigma_i = x_1x_2\dots x_iy_{i+1}y_{i+2}\dots y_d, \;\;\sigma_{i+1} = x_1x_2\dots x_{i+1}y_{i+2}y_{i+3}\dots y_d.\] %\[\mathrm{and}\;\; f(\sigma_i) = x'_2x'_3\dots x'_{i+1}y'_{i+2}y'_{i+3}\dots y'_dy'_1, \;\; f(\sigma_{i+1}) = x'_2x'_3\dots x'_{i+2}y'_{i+3}y'_{i+4}\dots y'_dy'_1.\] 
	\[\sigma_i \cap \sigma_{i+1} = x_1x_2\dots x_iy_{i+2}\dots y_d,\; \mathrm{and} \; f(\sigma_i) \cap f(\sigma_{i+1}) = x'_2x'_3\dots x'_{i+1}y'_{i+3}\dots y'_dy'_1.\]  The missing indices are $i+1$ and $i+2$ respectively, so we let $e_i=x'_{i+1}y_{i+2}$. It follows that $\Gamma_i \cap \Gamma_{i+1} = \st(e_i,\Gamma_i) =\st(e_i,\Gamma_{i+1})$ and hence the $\Diamond
	$-connected sum is well defined.  Similarly, $\Gamma_{d-1}\cap \Gamma_{d}=\st(\{x'_d,x_1\}, \Gamma_d )$ and $\Gamma_d\cap \Gamma_{d+1}=\st(\{y_1',x_2\}, \Gamma_d)$. Inductively, we form a complex $\Gamma=((\Gamma_1\#\Gamma_2\dots\#\Gamma_{2d})^{\bar{\phi}}, \Delta_1)$ which contains $\Delta_1$ and $f(\Delta_1)$ as subcomplexes.
	
	We partition $\Gamma$ as $\Gamma=\Delta_1\cup f(\Delta_1)\cup N$, so that $N \cap \Delta_1=\partial \Delta_1$ and $N \cap f(\Delta_1)=\partial f(\Delta_1)$. $N$ is then the tubular neighborhood that we would like to construct. Finally, let $\Sigma=\Delta_2\cup_{\partial \Delta_1} N\cup _{\partial f(\Delta_1)} f(\Delta_2)$. (This is well defined as by \cref{lemma: B(i-d) properties}, $\partial \Delta_1 \cong \partial \Delta_2$.) As shown in \cref{S0S2}, when $d=3$, $\sigma$ gives the minimal balanced triangulation of $\Sp^0\times \Sp^2$.
\end{construction}

\begin{figure}[h]
	\centering
	\subfloat[$\Delta_1$]{\includegraphics{P_B(1,d)}}
	\subfloat[$f(\Delta_1)$]{\includegraphics{fb(1,3)}}
	\hspace{0.3cm}
	%\subfloat[$\Gamma$]{\includegraphics[scale=1.5]{twocrosspolytopes}}
	\subfloat[$\Gamma$]{\includegraphics[scale=1.4]{twocrosspolytopes}}
	%\caption{The complexes $\Delta_1$ and $f(\Delta_1)$ when $d=3$: in the case we cannot define $\Gamma_2\#\Gamma_3$, since the pairs of edges $\sigma_2\cap\sigma_3=\{y_3x_1\}$ and $f(\sigma_1)\cap f(\sigma_2)=\{x_1'y_3'\}$ (as colored in green in the picture) share the same colors. Similarly, $\Gamma_5\#\Gamma_6$ is not well-defined. However, in this case the triangulation $\Gamma$ still exist: we need switch the positions of two pairs of vertices $(x_3, y_2)$ and $(x_2,y_3)$, see (c) above. }
	\caption{The complexes $\Delta_1$ and $f(\Delta_1)$ when $d = 3$, and the resulting $\Gamma$ constructed using the previously described sequence of connected sums.}\label{S0S2}
\end{figure}

Next to prove $\Sigma$ indeed triangulates $\Sp^2\times \Sp^{d-3}$, we check that $\Sigma$ satisfies all the conditions as described in \cref{thm: homology of product of spheres}.
\begin{lemma}\label{lm: simply connected}
	The complex $\Sigma$ in \cref{main construction} is simply connected codimension-1 submanifold of $\Sp^d$ for $d\geq 5$.
\end{lemma}
%\begin{proof}
%	Since $B(d-3,d)$ contains the 2-skeleton of $\partial C_d^*$, it follows that both $\Delta_2$ and $f(\Delta_2)$ are simply connected. Then $A:=\Delta_2\cup \Gamma'$ and $B:=\Gamma'\cup f(\Delta_2)$ are also simply connected. Write $\Sigma=\mathrm{int}(A)\cup \mathrm{int}(B)$. It is easy to see that both $\Sigma$ and $A\cap B=\mathrm{int}(\Gamma')$ are path connected. The result follows since the union of two simply connected open subsets with path-connected intersection is simply connected.
%\end{proof}
\begin{proposition}
	The complex $\Sigma$ in \cref{main construction} is a balanced triangulation of $\Sp^2\times \Sp^{d-3}$ for $d\geq 3$.
\end{proposition}                                                               
\begin{proof}
Applying the Mayer-Vietoris sequence to the triple $(\Delta_2\cup f(\Delta_2), N, \Sigma)$, we find that $\Sigma$ has the same homology as $\Sp^2 \times \Sp^{d-3}$, and so the result follows by \cref{lm: simply connected} and \cref{thm: homology of product of spheres}.
%	By the theorem and the lemma above, it suffices to check that $\Sigma$ has the same homology as $\Sp^2\times \Sp^{d-3}$ for $d\geq 5$. Applying the Mayer-Vietoris sequence on the triple $(\Delta_2\cup f(\Delta_2), \Gamma', \Sigma)$, for all $i$,
%	\[\cdots \to H_{i+1}(\Sigma)\to H_i(\partial \Delta_2\cup\partial f(\Delta_2))\to H_i(\Gamma')\oplus H_i(\Delta_2\cup f(\Delta_2))\to H_i(\Sigma)\to \cdots.\]
%	Since $\Gamma'\cong \partial \Delta_2 \times [0,1]$ , $\frac{1}{2}\beta_i(\partial \Delta_2\cup\partial f(\Delta_2))=\beta_i(\Gamma')=1$ for $i=1, d-3, d-2$, and zero otherwise. Also by the lemma, $\beta_i(\Delta_2\cup f(\Delta_2))=2\beta_i(\Delta_2)=2$ for $i=d-3$, and zero otherwise. Hence, we obtain that
%	\[0\to H_{d-1}(\Sigma)\to H_{d-2}(\partial \Delta_2\cup\partial f(\Delta_2))\to H_{d-2}(\Gamma')\oplus H_{d-2}(\Delta_2\cup f(\Delta_2))=0,\]
%	which implies that $\beta_{d-1}(\Sigma)=1$. Also
%	\[0\to H_{d-2}(\Sigma)\to H_{d-3}(\partial \Delta_1\cup\partial f(\Delta_1))\stackrel{i}{\rightarrow} H_{d-3}(\Gamma')\oplus H_{d-3}(\Delta_2\cup f(\Delta_2))\to H_{d-3}(\Sigma)\to 0.\]
%	Since the map $i$ must be injective, it follows that $\beta_{d-2}(\Sigma)=0$ and $\beta_{d-3}(\Sigma)=1$.
%	Finally, 
%	\[0\to H_2(\Sigma)\to H_1(\partial \Delta_2\cup\partial f(\Delta_2))\stackrel{i'}{\rightarrow}H_1(\Gamma')\to H_1(\Sigma)\to 0.\]
%	Again, the map $i'$ is surjective, which forces $\beta_2(\Sigma)=1$ and $\beta_1(\Sigma)=0$. It is clear to see that the other Betti numbers of $\Sigma$ are zero. Finally, the balancedness of $\Sigma$ follows from the construction.
\end{proof}
\begin{property}
For $d \geq 5$, the complex $\Sigma$ in \cref{main construction} satisfies:
	\begin{enumerate}
	\item $f_0(\Sigma)=4d$;
	\item $f_1(\Sigma)=4d(2d-3)$;
	\item $f_{d-1}(\Sigma)=(d+2)2^d-8d$;
	\item $\Aut(\Sigma$) admits a vertex-transitive action of $\Z_2\times\mathcal{D}_{2d}$.
	\end{enumerate}
\end{property}
%\begin{proof}
%   The proof is clear from face enumeration.
	%The complex $\Sigma$ has $4d$ vertices since $f_0(\Sigma)=f_0(\Delta_2)+f_0(f(\Delta_2))$. By the construction, there are $2d$ edges $e_1=\{x_2', y_3\},\dots, e_{2d}=\{x_1', y_2\}$ deleted from the cross-polytopes $\Gamma_1, \Gamma_2, \dots, \Gamma_{2d}$ to form $\Gamma$. Each $\Gamma_i$ and $\lk(e_i,\Gamma_i)$ are $(d-1)$-dimensional and $(d-3)$-dimensional octahedral spheres respectively, so we have that $f_1(\Gamma_i)=2d(d-1)$ and $f_1(\st(e_i,\Gamma_i))= 2(d-1)(d-2)+1$. Thus,
	%\begin{align*}
	%f_1(\Gamma_1\#\Gamma_2 \dots \Gamma_{2d})^\phi) &= \sum_{i=1}^{2d}\Big(f_1(\Gamma_i)- f_1(\st(e_i,\Gamma_i))-1\Big) \\
	%&=4d^2(d-1)-4d(d-1)(d-2)-4d \\
	%&=4d(2d-3)
	%\intertext{It follows from $f_1(\Delta_1)=f_1(\Delta_2)$ that $f_1(\Sigma)=f_1(\Gamma)=4d(2d-3)$. Similarly, since the facets in each $\st(e_i,\Gamma_i)$ are disjoint,}
	%f_{d-1}(\Gamma)&=\sum_{i=1}^{2d} \Big(f_{d-1} (\Gamma_i) -2f_{d-1}(\st(e_i,\Gamma_i))\Big)\\&=2d(2^d-2^{d-1})\\&=d2^d
	%\end{align*}
	%It follows that $f_{d-1}(\Sigma)=f_{d-1}(\Gamma)-2f_{d-1}(\Delta_1)+2f_{d-1}(\Delta_2)=d2^d-4d+(2^{d+1}-4d)=(d+2)2^d-8d$.
%\end{proof}

%Note that the construction for $d = 3$ satisfies the same face number relations, with the exception of $f_1$.  In this case, there are $24$ edges rather than $36$, since edges are lost when we replace $\Delta_1$ with $\Delta_2$.  This does not happen when $d \geq 4$ as by Theorem 1.2 in \cite{Klee-Novik-cs manifolds}, we have that for $d \geq 4$, $B(d-3,d)$ contains all edges in the cross-polytope.

%{\color{blue}I moved the proof inside the proof for the proposition above, since the proposition makes two claims.  Perhaps it would be better to group the proof with the below proposition - let me know which makes more sense.}

%{\color{blue} I slightly modified your proof. Remember that we use "D" to represent i-dimensional ball before. We need to change that notation later. I think it is OK that we use $\mathcal{D}_d$ for the dihedral group.}


%\begin{proposition}
%	For $d \geq 4$, $\Aut(\Sigma$) admits a vertex-transitive action of $\Z_2\times \mathcal{D}_{2d}$.
	%For $d > 3$, the automorphism group of $\Sigma$ contains $\Z_2 \times \mathcal{D}_{2d}$, where $\mathcal{D}_n$ is the dihedral group of order $2n$.
%\end{proposition}
%\begin{proof}
	%A simplicial map $g$ on the simplicial complex $\Sigma$ is an isomorphism if it gives a bijection on the facets of $\Sigma$.  A necessary condition for $g$ to be an automorphism is that it sends the missing edges in $\Sigma$ to missing edges in $\Delta$.  Define the following three permutations, modified from the proof of Theorem 1.2(b) in \cite{Klee-Novik-cs manifolds}:
%	The proof is similar to that of Theorem 1.2(b) in \cite{Klee-Novik-cs manifolds}, where we consider instead the following three permutations.
%	\begin{itemize}
%		\item $D$ maps $x_j$ to $y_j$, $y_j$ to $x_j$, $x_j'$ to $y_j'$ and $y_j'$ to $x_j'$.
%		\item $E'$ maps $x_j$ to $x_{d-j+1}'$, $y_j$ to $y_{d-j+1}'$, $x_j'$ to $x_{d-j+1}$ and $y_j'$ to $y_{d-j+1}$.
%		\item $R'$ maps $x_d$ to $y_1$, $y_d$ to $x_1$, $x_j$ to $x_{j+1}$, $y_j$ to $y_{j+1}$, and similarly for $x_j'$ and $y_j'$.
%	\end{itemize}
	%The maps $D$ and $E'$ have order $2$, whereas $R'$ has order $2d$.  Also note that $E'$ is the permutation $E$ from \cite{Klee-Novik-cs manifolds} composed with a switching between the prime and nonprime vertices.  We know that of the edges in $\Sigma$, the only missing edges are edges between antipodal vertices in $\Gamma_i$ and the edges deleted when we join $\Gamma_i$ and $\Gamma_{i+1}$; they are $\{x_i'y_{i+1}\}, \{y_i'x_{i+1}\}$ for $1\leq i\leq d-1$ together with $\{x_d' x_1\}, \{y_d'y_1\}$.  It is straightforward to check that $D$, $E'$, and $R'$ are bijections on the vertices of $\Sigma$, and additionally fix setwise the set of missing edges. Since $E'R' = R'^{-1}E'^{-1}$, $E'$ and $R'$ generate $\mathcal{D}_{2d}$, and since $D$ commutes with both $E'$ and $R'$, we have that the three together generate $\Z_2 \times \mathcal{D}_{2d}$.
	
	%By Theorem 1.2(b) of \cite{Klee-Novik-cs manifolds}, we have that facets in $\Delta_1$ and $f(\Delta_1)$, as well as those in $\Delta_2$ and $f(\Delta_2)$, are mapped bijectively by $g=D,E'$ or $R'$. Therefore, it suffices to show that the facets in the tubular neighborhood $N$ are mapped bijectively. Note that any facet $F$ in $N$ must also be contained in some $\Gamma_i$.  Therefore, the only way in which $g(F)$ could not be a facet of $\Sigma$ is if $g(F)$ is in the star of an edge which is deleted. However, as we observed above, $g$ gives bijection on the missing edges of $\Sigma$, i.e., $g(F)\in \st(g(e))$ for some missing edge $e$ if and only if $g\in \st(e)$. Hence $g$ is a bijection on the facets of $\Sigma$, and so $g\in \Aut(\Sigma)$.
%	\end{proof}
	
%	For the case of $d = 3$, we can directly compute the automorphism group.  The construction consists of two disjoint octahedra, with an additional possibility of interchanging the two octahedra.  Then, the automorphism group is given by $\Z_2 \times O_h \times O_h$, where $O_h$ denotes the octahedral group of order $48$.  This group has order $2 \cdot 48^2 = 4608$.
	
\begin{remark}
	Working with Lorenzo Venturello, we created a CROSSFLIP program for balanced complexes to attempt to reduce the number of vertices of a given triangulation. However, the complexity of finding shellable subcomplexes in the $d$-cross-polytope grows exponentially with $d$, and so the program is highly inefficient for $d > 4$. \cite{Klee-Novik-cs-manifolds} showed that a balanced triangulation of a non-sphere $(d-1)$-manifold requires at least $3d$ vertices. It is not known that apart from the sphere bundle over the circle, if there are other manifolds that admit balanced triangulations with $3d$ vertices.  
\end{remark}
%\begin{problem}
%	Find a small balanced triangulation of $\Sp^i\times \Sp^j$.
%\end{problem}
%\begin{problem}
%	Determine the sharp lower bound on the number of vertices required for a balanced triangulation of $\Sp^i\times \Sp^j$.
%\end{problem}



	%Thus, it suffices to show that $F \in$ st$(e) \iff g(F) \in $ st$(g(e))$.  Since $g$ is a bijection on the vertices, it is invertible and without loss of generality we may prove only one direction.  Note that a facet $F$ must be of the form $u_1u_2\dots u_d$, so in order to be in the star of an edge, the vertices of the edge must be a subset of $F$.  Then, suppose $g(F) \in $ st$(g(e))$ but $F \notin $ st$(e)$.  Then, some edge in $F$ must be missing, but all edges exist in $g(F)$, and so a missing edge must have been sent to a non-missing edge under $g$, which is a contradiction, and therefore $g$ is an automorphism.

%\begin{property}
%  $\Sigma$ has the following $f$-numbers:
%   \begin{enumerate}
%		\item $f_0 = 4d$
%		\item $f_1 = 4d(2d-3)$
%		\item $f_{d-1} = d2^d$
%	\end{enumerate}
%\end{property}
%\begin{proof}
%The complex $\Sigma$ is constructed by adding edges to two copies of $B(1,d)$, but no additional vertices, and since $B(1,d)$ has $2d$ vertices we must have that $\Sigma$ has $4d$ vertices.
%
%We know that $\Sigma$ can be constructed from the (circularly) connected sum of $\Gamma_i$ for $1 \leq i \leq 2d$.  We know that each $\Gamma_i$ is the boundary of a $d$-cross-polytope, and so has the same number of edges.  We would like to construct $\Gamma'_i$ by deleting edges from $\Gamma_i$ such that each $\Gamma'_i$ has the same number of edges and the union forms $\Sigma$ exactly (i.e., there are no missing nor repeated edges).  Specifically, any time we glue $\Gamma_i$ and $\Gamma_{i+1}$ along a boundary, the boundary is double-counted, so we must delete the boundary from one of the original structures, which we can choose without loss of generality to be $\Gamma_i$.  The boundary we glue along is given by the star of an edge $e$ shared by $\Gamma_i$ and $\Gamma_{i+1}$, so we must delete this edge from both $\Gamma_i$ and $\Gamma_{i+1}$.  As a result, we have that $\Gamma'_i$ is given by $\Gamma_i$ with an edge and the star of a different edge removed.
%Since we have $2d$ such $\Gamma'_i$, the total number of edges is given by $2d(f_1(\Gamma'_i)) = 2d(f_1(\Gamma_i) - f_1($st$(e)) - 1)$.  Since $\Gamma_i$ is a $d$-cross-polytope, we know that the number of edges in $\Gamma_i$ is given by $2d(d-1)$.  It remains to find the number of edges in the star of an edge in a $d$-cross-polytope.
%
%We can count the edges in the star as follows: label the vertices in the cross-polytope as $x_1\dots x_d$ and $y_1 \dots y_d$, where an edge exists for all pairs of vertices not of the form $\{x_j,y_j\}$.  Without loss of generality, pick the edge $e = \{x_1,x_2\}$.  Then, an edge $e'$ is in the star of $e$ provided that the union of the vertices in both $e$ and $e'$ do not have vertices which share the same index but have different letters (e.g., $\{x_1,x_2\}$ and $\{x_1,y_3\}$ is allowed, but $\{x_1,x_2\}$ and $\{y_1,x_3\}$ is not since both $x_1$ and $y_1$ are in the vertex set).  Then, we can enumerate the edges in the star as follows: fix $x_1$ as a vertex.  Then, ignoring the edge $e$ itself for the moment, we cannot choose $x_1$, $x_2$, $y_1$, or $y_2$ as the other vertex.  Therefore, there are $2d-4$ vertices we can choose as the other vertex, with a similar argument for when we fix $x_2$ as the first vertex, for a total of $4d-8$ edges.  Then, there are $\binom{2d-4}{2}$ ways to choose two vertices where neither vertex is already used, since $x_1$, $x_2$, $y_1$, and $y_2$ are disallowed.  Subtracting out $d-2$ ways to choose two vertices of the form $\{x_j,y_j\}$, which do not exist as edges, the remaining choices of two vertices define an edge in the star.  Adding back the original edge, which must be in the star, we have that the total number of edges in the star is given by $4d-8 + \binom{2d-4}{2} - (d-2) = 3d-6 + (d-2)(2d-5) = (d-2)(2d-2) = 2(d-1)(d-2) + 1$.  Thus, the total number of edges in $\Sigma$ is given by $2d(f_1(\Gamma_i) - f_1($st$(e)) - 1) = 2d(2d(d-1) - 2(d-1)(d-2) - 1 - 1) = 4d(2(d-1)-1) = 4d(2d-3)$.
%
%Similarly, we can create $\Gamma'_i$ from $\Gamma_i$ analogously such that each $\Gamma'_i$ has the same number of facets, and when glued together create $\Sigma$ exactly (i.e., there are no missing nor repeated edges).  When $\Gamma_i$ is glued together with $\Gamma_{i+1}$, the facets in the star of the edge to be removed are deleted from both $\Gamma_i$ and $\Gamma_{i+1}$, so each $\Gamma'_i$ consists of $\Gamma_i$ with the stars of two edges deleted.  We find that for a facet to be in the star of an edge, the union of the vertices must be in the complex.  However, a facet must contain exactly one vertex of each index (e.g. a facet for $d = 3$ may look like $x_1x_2y_3$, but not $x_1x_3$ or $x_1y_1x_2x_3$).  Therefore, in order for a facet to be in the star, the vertex set of the edge must be a subset of the vertex set of the facet, otherwise, the union would add new vertices, and by the pigeonhole principle, at least one index will be repeated.  Note that the two stars that we intend to delete are completely disjoint, as the edges to be removed are of the form $\{x_i,y'_{i+1}\}$ and $\{x_{i+1},y'_{i+2}\}$, so any facet must contain both $y'_{i+1}$ and $x_{i+1}$, which is disallowed.
% 
%  Without loss of generality, consider our edge to be $e = \{x_1,x_2\}$.  Then, any facet must contain $\{x_1,x_2,\dots\}$, and for each of the remaining $d-2$ indices $i$, we may either choose $x_i$ or $y_i$, for a total of $2^{d-2}$ choices.  We must do this for both sides of $\Gamma'_i$ since it will be joined with both $\Gamma'_{i-1}$ and $\Gamma'_{i+1}$, so we have $2\cdot2^{d-2} = 2^{d-1}$ facets in both stars.  Since the total number of facets for $\Gamma_i = \partial C_d$ is $2^d$, the total number of vertices is given by $2d(f_{n-1}(\Gamma_i) - 2f_{n-1}($st$(e))) = 2d(2^d - 2^{d-1}) = d2^d$.
%   
%\end{proof}

\acknowledgements{The authors are partially supported by the Mathematics REU program at the University of Michigan during Summer 2018.  We would also like to thank Lorenzo Venturello for his contributions to the computational aspect of this project.}

%\appendix
%\section{Appendix: Computational Results}
%{\color{blue} When you submit this as your report, you may want to move Section 4 to the end of the paper and add Appendix: computational results. Perhaps also give your code there. (I don't know whether they will be interested in seeing your code, but I saw sometimes people do.) Also do you have a name for your program, say "CROSSFLIP"? }
%This appendix will briefly discuss our computation results.  

%\nocite{*}
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%\bibliography{biblio}\label{sec:biblio}
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