# Bistellar Cluster Algebras and Piecewise Linear Invariants

Alastair Darby  Department of Pure Mathematics
Xi’an Jiaotong–Liverpool University
Suzhou
215123
P. R. China<EMAIL_ADDRESS>,  Fang Li  Department of
Mathematics
Zhejiang University
Hangzhou
310027
P. R. China<EMAIL_ADDRESS>and  Zhi Lü  School of Mathematical Sciences
Fudan University
Shanghai
200433
P. R. China<EMAIL_ADDRESS>
###### Abstract.

Inspired by the ideas and techniques used in the study of cluster algebras we
construct a new class of algebras, called _bistellar cluster algebras_ , from
closed oriented triangulated even-dimensional manifolds by performing middle-
dimensional bistellar moves. This class of algebras exhibit the algebraic
behaviour of middle-dimensional bistellar moves but do not satisfy the
classical cluster algebra axiom: “every cluster variable in every cluster is
exchangeable”. Thus the construction of bistellar cluster algebras is quite
different from that of a classical cluster algebra.

Secondly, using bistellar cluster algebras and the techniques of combinatorial
topology, we construct a direct system associated with a set of PL
homeomorphic PL manifolds of dimension 2 or 4, and show that the limit of this
direct system is a PL invariant.

###### 2010 Mathematics Subject Classification:

57Q15, 13F60, 57Q25.

##  1\. Introduction

Finding invariants of manifolds is important in solving problems such as the
_smooth 4- dimensional Poincaré conjecture_ (see [10]) which asks if there are
different smooth structures on the 4-sphere. In this paper we describe a new
invariant of Piecewise Linear manifolds, which in the case of 4 dimensional
manifolds (and lower), are the same as smooth manifolds. This has the
potential of helping to answer the smooth Poincaré conjecture if different
candidates for the smooth/PL structures can be shown to have distinct
invariants. The invariants we have found for 2- and 4-dimensional smooth/PL
manifolds are inspired by the work in the field of cluster algebras along with
classical work in PL manifolds as described below.

###  1.1. Background and motivation

Cluster algebras were first introduced and studied systematically in a series
of papers [6, 7, 1, 9]. The original motivation and aim for cluster algebras
was to provide an algebraic framework for the study of total positivity
phenomena in semisimple Lie groups and of Lusztig’s and Kashiwara’s canonical
bases of quantum groups. In recent years, it has been shown that this kind of
algebraic structure for cluster algebras can arise in various mathematical
settings, such as total positivity, Lie theory, quiver representations,
Teichmüller theory, Poisson geometry, discrete dynamical systems, tropical
geometry, algebraic combinatorics, topology of surfaces and so on (see, for
example, [4, 3, 8]).

Roughly speaking, a cluster algebra $\mathcal{A}$ of rank $r$ is a subalgebra
of an ambient field $\mathcal{F}$ of rational functions in $r$ indeterminates,
which is defined via a distinguished family of generators. These generators
are called cluster variables and are grouped into $r$-subsets called clusters.
These clusters satisfy the exchange property: for any cluster ${\bf x}$ and
any element $x\in{\bf x}$, there is another cluster ${\bf x}^{\prime}$ and
$x^{\prime}\in{\bf x}^{\prime}$ with ${\bf x}^{\prime}=({\bf
x}\setminus\\{x\\})\cup\\{x^{\prime}\\}$ such that $x$ and $x^{\prime}$ are
related by an exchange relation

$xx^{\prime}\,=\,M^{+}+M^{-},$

where $M^{+}$ and $M^{-}$ are two monomials without common divisors in the
$r-1$ variables ${\bf x}\cap{\bf x}^{\prime}$, and the exponents in $M^{+}$
and $M^{-}$ are encoded in an $(r\times r)$ integral matrix $B=(b_{ij})$
(usually skew-symmetric) called the exchange matrix. The cluster algebra
$\mathcal{A}$ can also be obtained from a pair $({\bf x},B)$, called a seed,
by seed mutations, where ${\bf x}$ is an initial cluster and $B$ is an initial
exchange matrix, so that all cluster variables can be constructed inductively,
and each cluster variable $x$ is a rational function of the variables in the
initial cluster ${\bf x}$. With this point of view, the cluster algebra
$\mathcal{A}$ becomes a subring of the field formed by all rational functions
of the variables in the initial cluster ${\bf x}$.

In their work [5], Fomin, Shapiro and Thurston showed that any triangulation
$T$ of a connected oriented Riemann surface $S$ produces a cluster algebra
$\mathcal{A}_{T}$, which is briefly stated as follows: they regard the set of
all edges (or $1$-dimensional faces) of $T$ as an initial cluster ${\bf x}(T)$
and define a skew-symmetric matrix $B(T)$ by looking at the (signed)
adjacencies between the edges of $T$, which serves as an initial exchange
matrix of the cluster algebra $\mathcal{A}_{T}$. Then the required cluster
algebra $\mathcal{A}_{T}$ is obtained from the initial seed $({\bf
x}(T),B(T))$ by seed mutations. In particular, they showed that mutations of
matrices in algebra completely agree with 1-moves (or flips) in combinatorial
topology. It is well-known that any connected oriented Riemann surface $S$ has
a unique PL structure and any two triangulations with the same number of
vertices of $S$ are connected by 1-moves. Thus, for any two triangulations $T$
and $T^{\prime}$ with the same number of vertices of $S$, both
$\mathcal{A}_{T}$ and $\mathcal{A}_{T^{\prime}}$ are isomorphic. This means
that this class of cluster algebras are invariant up to isomorphism under the
actions of 1-moves.

The following question naturally arises:

######  Question 1.1.

What kinds of analogues of cluster algebras can be produced from high-
dimensional oriented triangulated manifolds?

Although the 2-dimensional case provides us with intuition on the study of the
higher dimensional cases, we note that the problem is more complicated than it
may seem at first glance. This heavily depends upon the combinatorial topology
of the various moves in the higher dimensional cases.

Actually, Question 1.1 is directly related to the problem of how to give an
algebraic description of various moves (especially bistellar moves) in the
higher dimensional cases. Pachner [11, 12] tells us that two PL manifolds are
PL homeomorphic if and only if one can be transformed into the other by a
finite sequence of bistellar moves. We therefore pay more attention to the
case of bistellar moves in general.

###  1.2. Our work and statements of main results

Let $K$ be a connected oriented triangulated closed manifold of dimension
$n\geq 2$. To extend the theory of Fomin, Shapiro and Thurston in the
2-dimensional case to the general case, we need to consider the following
questions:

  1. (1)

What will our cluster variables will be?

  2. (2)

How should we define the exchange matrices?

  3. (3)

How should the matrix mutations which correspond to the bistellar moves be
defined?

  4. (4)

How can we define the exchange relations among seeds?

These questions are closely related to each other. We first investigate how
the $f$-vector of $K$ changes under the action of bistellar moves. Together
with a theorem of Pachner [11], we show that the $f$-vector of $K$ does not
change when $n$ is even and we only perform middle-dimensional bistellar moves
(see Proposition 2.3). Based upon this, we shall pay more attention to the
case which $n=2h$ is even, and study the algebraic behaviour of performing
middle-dimensional bistellar moves on $K$.

The definition of the exchange matrix of $K$ depends upon the choice of
cluster variables. On the other hand, in order to make sure that the
definition of the exchange matrix of $K$ is well-defined, we need to fix the
orientations of all $n$-simplices (or lower-dimensional simplices) of $K$, as
shown in [5, Definition 4.1] for the 2-dimensional case. Indeed, we can use an
orientation class of $H_{n}(K)$ to fix the orientations of all $n$-simplices.
However, there is no canonical way to fix orientations of simplices of
dimension less than $n$. This leads us to choose all $(n-1)$-simplices of $K$
as cluster variables. We give a generalized boundary operator for oriented
simplices, and then use it to define the pair orderings among all
$(n-1)$-simplices of an oriented $n$-simplex. When $n=2$, Fomin, Shapiro and
Thurston use clockwise and counter-clockwise more visually to define such pair
orderings among three edges of an oriented triangle, as shown in [5,
Definition 4.1]. Furthermore, we can define the exchange matrix $B(K)$ of $K$.
Our definition is a direct generalization of [5, Definition 4.1]. It should be
pointed out that our definition can still be carried out even if $n$ is odd.

As we will see, if $n=2h$, when we perform an $h$-dimensional bistellar move
on $K$, then we will change $\binom{h+1}{2}$ cluster variables (that is,
$(n-1)$-simplices of $K$) into $\binom{h+1}{2}$ new cluster variables. When
$n=2$, $\binom{h+1}{2}=1$ and so we replace a cluster variable by a new one.
This is the case of a classical cluster algebra. However, if $n>2$, we need to
replace more cluster variables when we perform an $h$-dimensional bistellar
move on $K$. This results in an essential difference from the 2-dimensional
case.

To define the matrix mutation such that it agrees with the corresponding
middle-dimensional bistellar move, we analyze the combinatorial structure of
an $h$-dimensional bistellar move ${\bf bm}_{\alpha}$ on $K$. This associates
to a bistellar pair $(\alpha,\beta)$, and two sub-complexes $\Lambda_{\alpha}$
and $\Lambda_{\beta}$ of $K$ and ${\bf bm}_{\alpha}K$ respectively (see
Defintion 2.1 for the meaning of ${\bf bm}_{\alpha}K$). When we perform an
$h$-dimensional bistellar move ${\bf bm}_{\alpha}$ on $K$, actually the
resulting manifold ${\bf bm}_{\alpha}K$ is obtained from $K$ by replacing
$\Lambda_{\alpha}$ with $\Lambda_{\beta}$, and in particular, only
$\binom{h+1}{2}$ cluster variables (that is, $(n-1)$-simplices) in
$\Lambda_{\alpha}$ will be replaced by $\binom{h+1}{2}$ cluster variables in
$\Lambda_{\beta}$. Note that both $\Lambda_{\alpha}$ and $\Lambda_{\beta}$
have $\binom{h+1}{1}\binom{h+1}{1}+\binom{h+1}{2}$ simplices of dimension
$n-1$ (that is, cluster variables). Then we can choose a canonical
combinatorial equivalence $\sigma$ between $\Lambda_{\alpha}$ and
$\Lambda_{\beta}$, which can be regarded as a permutation of order 2, and
sends the $\binom{h+1}{2}$ cluster variables in $\Lambda_{\alpha}$ into the
$\binom{h+1}{2}$ cluster variables in $\Lambda_{\beta}$. Moreover, we define
the required matrix mutation

$\mu_{\alpha}\colon B(K)\longrightarrow B({\bf bm}_{\alpha}K)$

along _multi-directions_ (see Definition 4.5), so that the matrix mutation
$\mu_{\alpha}$ agrees with the bistellar move ${\bf bm}_{\alpha}$ for exchange
matrices, stated as follows:

Proposition 4.7. $\mu_{\alpha}(B(K))=B({\bf bm}_{\alpha}K)$.

When we perform a seed mutation we must know how to define the exchange
relations. The canonical combinatorial equivalence $\sigma$ helps us clearly
understand the correspondence between seeds $\Sigma_{K}$ and $\Sigma_{{\bf
bm}_{\alpha}K}$, so that with the matrix mutation together, we can define the
$\binom{h+1}{2}$ exchange relations (see (5.3)), giving further the definition
of a seed mutation $\Phi_{\alpha}\colon\Sigma_{K}\to\Sigma_{{\bf
bm}_{\alpha}K}$ (see also Definition 5.3). Since all middle-dimensional moves
that we perform are bistellar, we call $\Phi_{\alpha}$ the bistellar seed
mutation, and call those $\binom{h+1}{2}$ cluster variables in
$\Lambda_{\alpha}$ and $\Lambda_{\beta}$ the exchangeable bistellar cluster
variables. We further show that $\Phi_{\alpha}$ is invertible (see Proposition
5.9).

Using what we have shown above, we can define the required bistellar cluster
algebra $\mathcal{A}_{K}$ of $K$. Briefly speaking, $\mathcal{A}_{K}$ is
obtained from the initial seed $\Sigma_{K}$ via bistellar seed mutations by
performing all possible bistellar $h$-moves on $K$. To give the definition of
$\mathcal{A}_{K}$ more clearly, we introduce the bistellar exchange graph
determined by $K$, which gives the information on all possible middle-
dimensional bistellar moves on $K$. Moreover, in a similar way to classical
cluster algebra, we first define the bistellar cluster algebra
$\mathcal{A}_{[K]}$ of $[K]$ (see Definition 5.18), where $[K]$ is a set, each
of which can be obtained from $K$ by a finite sequence of middle-dimensional
bistellar moves. Then we give the definition of $\mathcal{A}_{K}$, which is
the reduction of $\mathcal{A}_{[K]}$ to the field $\mathbb{F}\mathcal{X}(K)$
with $\Sigma_{K}$ as initial seed by using the bistellar seed mutations, where
$\mathbb{F}\mathcal{X}(K)$ is the field of all rational functions in all
cluster variables given by $K$.

Now our first main result is stated as follows:

Theorem 5.21. Let $K$ and $K^{\prime}$ be two connected oriented triangulated
closed manifolds of dimension $2h$. If $K^{\prime}$ is obtained from $K$ by
doing a finite sequence of bistellar $h$-moves, then
$\mathcal{A}_{K}\cong\mathcal{A}_{K^{\prime}}$.

######  Remark 1.2.

Comparing with classical cluster algebras, our bistellar cluster algebras
$\mathcal{A}_{K}$ have the following properties:

  1. (A)

Each $\mathcal{A}_{K}$ is of finite type, that is, there are only finitely
many cluster variables. In particular, all exchangeable bistellar cluster
variables may not cover all the cluster variables.

  2. (B)

Generally $\mathcal{A}_{K}$ is some kind of quotient algebra with an ideal
determined by exchange relations. This is an essential difference from
classical cluster algebras. The main reason why there is an ideal is because
the moves used are bistellar and exchangeable bistellar cluster variables may
produce the same exchange relations with each other. So our algebra
$\mathcal{A}_{K}$, even for $\dim K=2$, does not satisfy the cluster algebra
axiom: every cluster variable in every cluster is exchangeable. Certainly our
algebra $\mathcal{A}_{K}$ is computable (see the example in Section 5.3).

Using the bistellar cluster algebras we have constructed, we further consider
the problem of PL invariants for closed PL manifolds. Let $K$ be a closed PL
manifold of dimension $n$. When $n=2,4$, we show that a set $[K]$ of all
closed PL manifolds PL homeomorphic to $K$ can form a direct system by
associating with the bistellar cluster algebras of closed PL manifolds, and
then show that the limit $\mathcal{A}_{K}^{PL}$ of this direct system is a PL
invariant. This is the second main result of this paper, which is stated as
follows:

Theorem 6.5. Let $K$ and $K^{\prime}$ be two closed PL manifolds of dimension
$n=2,4$, such that they are PL homeomorphic. Then
$\mathcal{A}_{K}^{PL}\cong\mathcal{A}_{K^{\prime}}^{PL}$.

##  2\. Bistellar Moves and Pachner’s Theorem

We begin by reviewing some standard background material regarding the
combinatorial topology of triangulated and PL manifolds (see [2] for more
information).

###  2.1. Triangulated Manifolds and Bistellar Moves

An _abstract simplicial complex_ $K$ on a finite set $S$ is a collection in
the power set $2^{S}$ such that for each $\alpha\in K$, any subset (including
the empty set) of $\alpha$ belongs to $K$. Each $\alpha$ in $K$ is called a
_simplex_ (or a _face_) of $K$ and has dimension $|\alpha|-1$, where
$|\alpha|$ is the cardinality of $\alpha$. The dimension of $K$ is defined as
$\max\\{\dim\alpha\mid\alpha\in K\\}$.

Let $K_{1}$ and $K_{2}$ be two abstract simplicial complexes on vertex sets
$S_{1}$ and $S_{2}$ respectively. A map $\phi\colon K_{1}\to K_{2}$ is said to
be _simplicial_ if $\phi$ is an extension of a map $f\colon S_{1}\to S_{2}$
such that $\phi(\alpha)=f(\alpha)\in K_{2}$, for any $\alpha\in K_{1}$. The
two simplicial complexes $K_{1}$ and $K_{2}$ are said to be _combinatorially
equivalent_ if there are simplicial maps $\phi\colon K_{1}\to K_{2}$ and
$\psi\colon K_{2}\to K_{1}$ such that
$\psi\circ\phi=\operatorname{id}_{K_{1}}$ and
$\phi\circ\psi=\operatorname{id}_{K_{2}}$.

It is well-known (see [2] for example) that there is a bijection, given by
geometric realization, between abstract simplicial complexes up to
combinatorial equivalence and geometric simplicial complexes up to
homeomorphism. Thus, we may identify abstract simplicial complexes with
geometric simplicial complexes.

The _join_ of two simplicial complexes $K_{1}$ and $K_{2}$ on vertex sets
$S_{1}$ and $S_{2}$ is given by

$K_{1}\ast K_{2}\,\mathbin{\raisebox{0.3pt}{:}\\!=}\,\\{\alpha\in S_{1}\sqcup
S_{2}\mid\alpha=\alpha_{1}\cup\alpha_{2},\ \alpha_{1}\in K_{1},\ \alpha_{2}\in
K_{2}\\}.$

Given a face $\alpha$ of a simplicial complex $K$, the _link_ of $\alpha$ in
$K$ is the subcomplex given by

$\operatorname{Link}_{K}\alpha\,\mathbin{\raisebox{0.3pt}{:}\\!=}\,\\{\alpha^{\prime}\in
K\mid\alpha\cup\alpha^{\prime}\in K,\ \alpha\cap\alpha^{\prime}=\emptyset\\}.$

A _triangulated manifold_ is a simplicial complex $K$ whose geometric
realization $|K|$ is a topological manifold.

######  Definition 2.1 ([11, 2]).

Let $K$ be a triangulated manifold of dimension $n$. Suppose that $\alpha\in
K$ is an $(n-h)$-simplex such that
$\operatorname{Link}_{K}\alpha=\partial\beta$, where $\beta\notin K$. Then the
operation

${\operatorname{{\bf
bm}}}_{\alpha}K\,\mathbin{\raisebox{0.3pt}{:}\\!=}\,(K\setminus(\alpha\ast\partial\beta))\cup(\partial\alpha\ast\beta)$

is called the bistellar $h$-move at $\alpha$. Note that $\dim\beta=h$.

The pair $(\alpha,\beta)$ plays an important role in performing the operation
${\operatorname{{\bf bm}}}_{\alpha}K$ and we call the ordered pair
$(\alpha,\beta)$ a _bistellar pair_ of type $h$ in $K$. Clearly, if
$(\alpha,\beta)$ is a bistellar pair of type $h$ in $K$, then $(\beta,\alpha)$
is a bistellar pair of type $n-h$ in ${\operatorname{{\bf bm}}}_{\alpha}K$ and
we can perform a bistellar $(n-h)$-move at $\beta$ on ${\operatorname{{\bf
bm}}}_{\alpha}K$; in particular,

${\operatorname{{\bf bm}}}_{\beta}{\operatorname{{\bf bm}}}_{\alpha}K\,=\,K.$

Thus, ${\operatorname{{\bf bm}}}_{\beta}$ is the inverse operation of
${\operatorname{{\bf bm}}}_{\alpha}$. It should be pointed out that there may
not be a bistellar pair of a certain type in a triangulated manifold, for
example, the boundary complex of an $(n+1)$-simplex has no bistellar pairs of
type $h$, for all $h>0$. However, we can always perform a bistellar $0$-move
on any triangulated manifold.

It is easy to see that a bistellar $0$-move adds a new vertex, a bistellar
$n$-move deletes a vertex, and all other bistellar moves do not change the
number of vertices of $K$.

We now study the relation between the numbers of faces of $K$ and
${\operatorname{{\bf bm}}}_{\alpha}K$. By $\mathcal{F}_{i}^{K}$ we denote the
set of all $i$-dimensional simplices of $K$ and also set
$\mathcal{F}(K)\mathbin{\raisebox{0.3pt}{:}\\!=}\mathcal{F}_{\dim K-1}^{K}$
for the codimension 1 faces. Each $n$-dimensional simplicial complex $K$
determines an integral vector $f(K)=(f_{0},f_{1},...,f_{n})$, called the
$f$-vector of $K$, where $f_{i}=|\mathcal{F}_{i}^{K}|$ is the number of
$i$-dimensional simplices in $K$. The $f$-vector of $K$ defines the $h$-vector
$h(K)=(h_{0},h_{1},...,h_{n+1})$ by

$h_{0}t^{n+1}+h_{1}t^{n}+\cdots+h_{n+1}\,=\,(t-1)^{n+1}+f_{0}(t-1)^{n}+\cdots+f_{n}.$

In addition, the $h$-vector of $K$ also defines the _$g$ -vector_
$g(K)=(g_{0},g_{1},\dots,g_{\left\lfloor{\frac{n+1}{2}}\right\rfloor})$ by

$g_{0}\,=\,1\text{ and }g_{i}\,=\,h_{i}-h_{i-1},\quad\text{for
$i=1,\dots,\left\lfloor{\frac{n+1}{2}}\right\rfloor$.}$

######  Theorem 2.2 (Pachner [11]).

Let $K$ be a triangulated manifold of dimension $n$. Assume that
$(\alpha,\beta)$ is a bistellar pair of type $h$ with $0\leq
h\leq\left\lfloor{\frac{n}{2}}\right\rfloor$ in $K$. If
$h\leq\left\lfloor{\frac{n-1}{2}}\right\rfloor$, then

$g_{i}({\operatorname{{\bf
bm}}}_{\alpha}K)\,=\,\begin{cases}g_{i}(K)+1,&\text{if $i=h+1$};\\\
g_{i}(K),&\text{if $i\not=h+1$}.\end{cases}$

If $n=2h$, then $g_{i}({\operatorname{{\bf bm}}}_{\alpha}K)=g_{i}(K)$ for all
$i$.

For our purpose we need to further analyze when $f({\bf bm}_{\alpha}K)=f(K)$.

######  Proposition 2.3.

Let $K$ be a triangulated manifold of dimension $n$. Assume that
$(\alpha,\beta)$ is a bistellar pair of type $h$ in $K$. Then
$f({\operatorname{{\bf bm}}}_{\alpha}K)=f(K)$ if and only if $n=2h$.

###### Proof.

First it is easy to see that if $g_{i}({\operatorname{{\bf
bm}}}_{\alpha}K)=g_{i}(K)$ for all $i$, then $f({\operatorname{{\bf
bm}}}_{\alpha}K)=f(K)$. Thus, if $n=2h$, by Theorem 2.2 we find that
$f({\operatorname{{\bf bm}}}_{\alpha}K)=f(K)$.

Assume that $f({\operatorname{{\bf bm}}}_{\alpha}K)=f(K)$. By the definition
of bistellar moves, the bistellar $h$-move at $\alpha$ on $K$ only changes the
local structure of $K$. Actually, since
$\operatorname{Link}_{K}\alpha=\partial\beta$, we have that
$\alpha\cap\beta=\emptyset$. We write $\alpha=(v_{0},\dots,v_{n-h})$ and
$\beta=(v_{n-h+1},\dots,v_{n+1})$. Then there are the following $h+1$
simplices of dimension $n$ in $K$

$F_{i}\,\mathbin{\raisebox{0.3pt}{:}\\!=}\,\alpha\cup\beta\setminus\\{v_{n+1-i}\\},\quad
i=0,\dots,h,$

where $\alpha=\bigcap_{i=0}^{h}F_{i}$, and there are the following $n-h+1$
simplices of dimension $n$ in ${\operatorname{{\bf bm}}}_{\alpha}K$

$H_{j}\,\mathbin{\raisebox{0.3pt}{:}\\!=}\,\alpha\cup\beta\setminus\\{v_{j}\\},\quad
j=0,\dots,n-h,$

where $\beta=\bigcap_{j=0}^{n-h}H_{j}$. Since ${\operatorname{{\bf
bm}}}_{\alpha}K=(K\setminus(\alpha\ast\partial\beta))\cup(\partial\alpha\ast\beta)$,
we see that

$\mathcal{F}_{n}^{K}\setminus\\{F_{0},\dots,F_{h}\\}\,=\,\mathcal{F}_{n}^{{\operatorname{{\bf
bm}}}_{\alpha}K}\setminus\\{H_{0},\dots,H_{n-h}\\}.$

Moreover, it follows from $f_{n}({\operatorname{{\bf
bm}}}_{\alpha}K)=f_{n}(K)$ that $h+1=n-h+1,$ so $n=2h$. ∎

We also see from the proof of Proposition 2.3 that $f_{n}({\operatorname{{\bf
bm}}}_{\alpha}K)=f_{n}(K)$ if and only if $n=2h$. Using a similar argument to
the proof of Proposition 2.3, we also find that $f_{n-1}({\operatorname{{\bf
bm}}}_{\alpha}K)=f_{n-1}(K)$ if and only if $n=2h$. Thus we have:

######  Corollary 2.4.

Let $K$ be a triangulated manifold of dimension $n$ and assume that
$(\alpha,\beta)$ is a bistellar pair of type $h$ in $K$. Then the following
statements are equivalent:

  1. (1)

$f({\operatorname{{\bf bm}}}_{\alpha}K)=f(K)$;

  2. (2)

$f_{n}({\operatorname{{\bf bm}}}_{\alpha}K)=f_{n}(K)$;

  3. (3)

$f_{n-1}({\operatorname{{\bf bm}}}_{\alpha}K)=f_{n-1}(K)$;

  4. (4)

$n=2h$.

In addition, we also see from the proof of Proposition 2.3 that the bistellar
$h$-move ${\operatorname{{\bf bm}}}_{\alpha}$ on $K$ only changes the
subcomplex
$\Lambda_{\alpha}\mathbin{\raisebox{0.3pt}{:}\\!=}\alpha\ast\partial\beta$ of
$K$, determined by the $h+1$ simplices of dimension $n$ in ${\bf
F}_{\alpha}\mathbin{\raisebox{0.3pt}{:}\\!=}\\{F_{0},\dots,F_{h}\\}$, which is
exactly transformed into the subcomplex ${\operatorname{{\bf
bm}}}_{\alpha}\Lambda_{\alpha}\,=\,\Lambda_{\beta}\,\mathbin{\raisebox{0.3pt}{:}\\!=}\,\partial\alpha\ast\beta$
of ${\operatorname{{\bf bm}}}_{\alpha}K$, determined by the $n-h+1$ simplices
of dimension $n$ in ${\bf
H}_{\beta}\mathbin{\raisebox{0.3pt}{:}\\!=}\\{H_{0},\dots,H_{n-h}\\}$.

######  Corollary 2.5.

The subcomplexes $\Lambda_{\alpha}$ and $\Lambda_{\beta}$ are combinatorially
equivalent if and only if $n=2h$.

######  Remark 2.6.

If $n=2h$, since $\dim\alpha=\dim\beta=h$, an easy observation shows that for
each $0\leq i<h$,
$\mathcal{F}_{i}^{\Lambda_{\alpha}}=\mathcal{F}_{i}^{\Lambda_{\beta}}$, and
for each $h\leq i\leq n$,
$\mathcal{F}_{i}^{\Lambda_{\alpha}}\not=\mathcal{F}_{i}^{\Lambda_{\beta}}$ but
that
$|\mathcal{F}_{i}^{\Lambda_{\alpha}}|=|\mathcal{F}_{i}^{\Lambda_{\beta}}|$.

###  2.2. PL Manifolds and Pachner’s Theorem

Two triangulated manifolds $K_{1}$ and $K_{2}$ are said to be _PL
homeomorphic_ if there exist subdivisions $K^{\prime}_{1}$ and
$K^{\prime}_{2}$ of $K_{1}$ and $K_{2}$ such that $K^{\prime}_{1}$ and
$K^{\prime}_{2}$ are combinatorially equivalent.

A _PL manifold_ of dimension $n$ is a special triangulated manifold $K$ of
dimension $n$ such that $\operatorname{Link}_{K}\alpha$ is a _PL sphere_ of
dimension $n-|\alpha|$ for every nonempty simplex $\alpha\in K$, where a PL
sphere is a triangulated sphere which is PL homeomorphic to the boundary of a
simplex.

The following theorem, due to Pachner, gives a description of PL
homeomorphisms in terms of bistellar moves.

######  Theorem 2.7 (Pachner [11, 12]).

Two PL manifolds $K_{1},K_{2}$ are PL homeomorphic if and only if $K_{1}$ can
be transformed into $K_{2}$ by a finite sequence of bistellar moves.

It is well-known that in dimensions $\leq 3$, the categories of PL,
topological and smooth manifolds are the same. However, the situation in
dimensions $>3$ is quite different. A Theorem of Whitney tells us that a
smooth manifold of any dimension always admits a PL structure. In dimension 4,
there exist topological manifolds that admit no PL structure, and even worse,
admit no triangulation. However, the categories of PL and smooth manifolds in
dimension 4 agree. Namely, there is exactly one smooth structure on every PL
4-manifold. This means that the classification of smooth 4-manifolds is
equivalent to the classification of PL 4-manifolds. But so far, the
classification problem is still quite difficult and remains wide open.

##  3\. Exchange Matrices

Let $\mathbb{TM}_{n}$ denote the set of all closed oriented triangulated
manifolds of dimension $n>1$. For each closed oriented triangulated manifold
$K$, since $K$ is compact, it has finitely many vertices as a simplicial
complex. Without the loss of generality, we assume that $K$ is a simplicial
complex on vertex set $[m]=\\{1,\dots,m\\}$, so $K$ is a subcomplex of
$2^{[m]}$, where $2^{[m]}$ is the $(m-1)$-dimensional simplicial complex whose
geometric realization is a ball of dimension $m-1$.

###  3.1. Pair Ordering

Using the ordering $1<\dots<m$ on the vertices, we may impose the
lexographical ordering $<$ on $2^{[m]}$. For example, when $m=3$, we have that

$\emptyset\,<\,\\{1\\}\,<\,\\{2\\}\,<\,\\{3\\}\,<\,\\{1,2\\}\,<\,\\{1,3\\}\,<\,\\{2,3\\}\,<\,\\{1,2,3\\}.$

By convention we will always assume a natural lexographical ordering $<$ on
all faces of $2^{[m]}$. For each nonempty subset $\alpha$ in $[m]$, clearly
$2^{\alpha}$ admits a lexographical ordering $<$ which agrees with that on
$2^{[m]}$.

For each simplex $\alpha$ with vertices
$\\{\alpha_{0},\alpha_{1},\dots,\alpha_{n}\\}\subseteq[m]$, by
$(\alpha_{0},\alpha_{1},\dots,\alpha_{n})$ we denote the simplex $\alpha$ with
its orientation determined by the order of its vertices. By
$-(\alpha_{0},\alpha_{1},\dots,\alpha_{n})$ we denote the oriented simplex
$(\alpha_{0},\alpha_{1},\dots,\alpha_{n})$ with the opposite orientation. We
can also give a _canonical orientation_ of $\alpha$ in such a way that the
orientation of $\alpha$ agrees with the lexographical ordering of
$\\{\alpha_{0},\alpha_{1},\dots,\alpha_{n}\\}$. For example, if
$\alpha_{0}<\alpha_{1}<\dots<\alpha_{n}$, then the corresponding oriented
simplex $(\alpha_{0},\alpha_{1},\dots,\alpha_{n})$ has the canonical
orientation. Clearly, each simplex in $2^{[m]}$ can be naturally equipped with
the canonical orientation.

Let $\alpha=(\alpha_{0},\alpha_{1},\dots,\alpha_{n})$ be an oriented simplex
in $2^{[m]}$ where $n\geq 0$. The boundary operator $\partial$ on $\alpha$ is
defined as follows:

(3.1)
$\partial(\alpha_{0},\alpha_{1},\dots,\alpha_{n})\,=\,\sum_{i=0}^{n}(-1)^{i}(\alpha_{0},\dots,\widehat{\alpha_{i}},\dots,\alpha_{n}).$

We may generalize the formula (3.1) as follows: for any $k$ with $1\leq k\leq
n+1$, define

$\partial^{(k)}(\alpha_{0},\alpha_{1},\dots,\alpha_{n})\,=\,\sum_{0\leq
i_{1}<\dots<i_{k}\leq
n}(-1)^{i_{1}+\dots+i_{k}}\left(\dots,\widehat{\alpha_{i_{1}}},\dots,\widehat{\alpha_{i_{k}}},\dots\right).$

Obviously $\partial^{(1)}$ is exactly the boundary operator $\partial$ as
mentioned in (3.1).

With this understanding, now let us define a _pair ordering_ $\prec$ of all
$d$-faces in a simplex $\alpha=(\alpha_{0},\alpha_{1},\dots,\alpha_{n})$,
where $0\leq d\leq n$.

Note that when $n=0$ or $d=n$, the collection of $d$-faces
$\mathcal{F}^{\alpha}_{d}$ of $\alpha$ only contains a single face, so in this
case, we don’t need to consider the pair ordering. Without the loss of
generality, we assume that $n\geq 1$ and $0\leq d<n$. In addition, we assume
that all faces of dimension less than $n$ are always equipped with the
canonical orientation.

######  Definition 3.1.

Take two $d$-faces $f$ and $g$ in $\mathcal{F}_{d}^{\alpha}$ such that $f<g$.
If $f\cap g=\emptyset$, we say that there is no pair ordering between $f$ and
$g$. Assume that $f\cap g\neq\emptyset$ and $|f\cap g|=k+1$. Consider the
coefficient $c_{fg}$ of the face $(f\cup g)\setminus(f\cap g)$, with its
canonical orientation, in the expression of $\partial^{(n-2(d-k)+1)}\alpha$.
Then the _pair ordering_ between $f$ and $g$ is defined as follows:

$\displaystyle g\prec f,\quad$ if $c_{fg}=+1$; $\displaystyle f\prec g,\quad$
if $c_{fg}=-1$.

If $d=0$, then there is no pair ordering on $\mathcal{F}_{0}^{\alpha}$ since
any two $0$-faces in $\mathcal{F}_{0}^{\alpha}$ have empty intersection. If
$\dim\alpha=n=2h$, then any two faces in $\mathcal{F}^{\alpha}_{h}$ have a
nonempty intersection because any middle-dimensional face $f$ in $\alpha$
contains $h+1$ vertices.

######  Example 3.2.

For example, when $n=2$, take $\alpha=(1,2,3)$. In this case, we have that
$\mathcal{F}_{1}^{\alpha}=\\{(12),(13),(23)\\}$ with the lexographical
ordering $(12)<(13)<(23)$. From the formula
$\partial^{(1)}(1,2,3)=(23)-(13)+(12)$, we can obtain the pair orderings for
all faces in $\mathcal{F}_{1}^{\alpha}$ as follows:

$(2,3)\prec(1,3);\quad(1,3)\prec(1,2);\quad(1,2)\prec(2,3).$

######  Example 3.3.

When $n=3$ we take $\alpha=(1,2,3,4)$. On $\mathcal{F}_{2}^{\alpha}$, one has
the lexographical ordering

$(1,2,3)\,<\,(1,2,4)\,<\,(1,3,4)\,<\,(2,3,4).$

From the formula

$\displaystyle\partial^{(2)}(1,2,3,4)\,=\,-(3,4)+(2,4)-(2,3)-(1,4)+(1,3)-(1,2)$

we can obtain the pair orderings for all 2-faces of $\alpha$ as follows:

$\displaystyle(1,3,4)\prec(2,3,4);\quad(1,2,3)\prec(2,3,4);\quad(2,3,4)\prec(1,2,4);$
$\displaystyle(1,2,4)\prec(1,3,4);\quad(1,3,4)\prec(1,2,3);\quad(1,2,3)\prec(1,2,4).$

If we take $\alpha=-(1,2,3,4)$, then all 2-faces of $\alpha$ have the reverse
pair orderings.

###  3.2. Exchange Matrices

Throughout the following, we use a convention of orientations and orderings on
all simplices in each closed oriented triangulated manifold $K$ (as a
subcomplex of $2^{[m]}$) of $\mathbb{TM}_{n}$. This convention is called the
_orientation-ordering convention_ , which is stated as follows:

  1. (1)

we orient all $n$-dimensional simplices in $K$ such that their sum is a cycle
chain in the chain group $C_{n}(K)$;

  2. (2)

all other simplices of dimension less than $n$ are equipped with the canonical
orientation as defined before;

  3. (3)

there is a natural lexographical ordering $<$ on all simplices in $K\subseteq
2^{[m]}$.

Given a $K\in\mathbb{TM}_{n}$, following the idea of [5, Definition 4.1], we
are going to define a skew-symmetric integer square matrix $B(K)$ of size
$f_{n-1}(K)=|\mathcal{F}(K)|$ with entries corresponding to all
$(n-1)$-dimensional simplices in $K$.

######  Definition 3.4.

Take any oriented $n$-simplex $\alpha\in\mathcal{F}_{n}^{K}$. We first define
a skew-symmetric integer square matrix
$B^{\alpha}(K)=(b^{\alpha}_{fg})_{f,g\in\mathcal{F}(K)}$ of size $f_{n-1}(K)$
by setting

$b^{\alpha}_{fg}\,=\ \begin{cases}+1,&\text{if $f,g\in\mathcal{F}(\alpha)$
such that $g\prec f$ in $\mathcal{F}(\alpha)$};\\\ -1,&\text{if
$f,g\in\mathcal{F}(\alpha)$ such that $f\prec g$ in $\mathcal{F}(\alpha)$};\\\
\phantom{+}0,&\text{otherwise.}\end{cases}$

Then the matrix $B(K)$ is defined by

$B(K)\,=\,\sum_{\alpha\in\mathcal{F}_{n}^{K}}B^{\alpha}(K).$

Note that the entries of the matrix $B(K)$ belong to $\\{-1,0,+1\\}$.
Definition 3.4 is a generalization of [5, Definition 4.1].

######  Remark 3.5.

For an oriented $n$-simplex $\alpha\in\mathcal{F}_{n}^{K}$, if we define
$c_{gf}\mathbin{\raisebox{0.3pt}{:}\\!=}-c_{fg}$, for $f<g$, then we see that
$b_{fg}=c_{fg}$, for all $f,g\in\mathcal{F}(\alpha)$.

######  Example 3.6.

Let $K$ be the boundary complex of $2^{[5]}$, which is 3-dimensional. Orient
the five 3-simplices of $K$ by

$\partial^{(1)}(1,2,3,4,5)\,=\,(2,3,4,5)-(1,3,4,5)+(1,2,4,5)-(1,2,3,5)+(1,2,3,4).$

So $K$ has the following five oriented 3-simplices

$\displaystyle\alpha_{1}\,=\,(2,3,4,5),\ \alpha_{2}\,=\,-(1,3,4,5),\
\alpha_{3}\,=\,(1,2,4,5),\alpha_{4}\,=\,-(1,2,3,5),\ \alpha_{5}\,=\,(1,2,3,4)$

such that the sum $\alpha_{1}+\dots+\alpha_{5}$ is a cycle chain. Also, $K$
contains ten 2-simplices

(3.2) $\displaystyle(1,2,3)\,<\,(1,2,4)\,<\,(1,2,5)\,<\,(1,3,4)\,<\,(1,3,5)$
$\displaystyle<\,$
$\displaystyle(1,4,5)\,<\,(2,3,4)\,<\,(2,3,5)\,<\,(2,4,5)\,<\,(3,4,5)$

ordered by the lexographical ordering, so $B(K)$ is a skew-symmetric integer
square matrix of size 10, with all rows and columns ordered by (3.2). First we
can write out $B^{\alpha_{i}}(K)$ by using Example 3.3, and then obtain

$B(K)\,=\,\left(\begin{array}[]{cccccccccc}0&-1&+1&+1&-1&0&-1&+1&0&0\\\
+1&0&-1&-1&0&+1&+1&0&-1&0\\\ -1&+1&0&0&+1&-1&0&-1&+1&0\\\
-1&+1&0&0&+1&-1&-1&0&0&+1\\\ +1&0&-1&-1&0&+1&0&+1&0&-1\\\
0&-1&+1&+1&-1&0&0&0&-1&+1\\\ +1&-1&0&+1&0&0&0&-1&+1&-1\\\
-1&0&+1&0&-1&0&+1&0&-1&+1\\\ 0&+1&-1&0&0&+1&-1&+1&0&-1\\\
0&0&0&-1&+1&-1&+1&-1&+1&0\end{array}\right).$

##  4\. Matrix Mutations under Bistellar Moves

Given a closed oriented triangulated manifold $K$ of $\mathbb{TM}_{n}$ we
regard $K$ as a subcomplex of $2^{[m]}$ where $m$ is the number of vertices of
$K$.

Throughout this section, we assume that we can perform a bistellar $h$-move
${\operatorname{{\bf bm}}}_{\alpha}$ on $K$ at an $(n-h)$-dimensional simplex
$\alpha$, that is, there exists a bistellar pair $(\alpha,\beta)$ such that
$\operatorname{Link}_{K}\alpha=\partial\beta$ but $\beta\not\in K$. We write

$L\,\mathbin{\raisebox{0.3pt}{:}\\!=}\,{\bf bm}_{\alpha}K.$

By Definition 3.4, we see that

$B(K)\,=\,(b_{fg}^{K})_{f,g\in\mathcal{F}(K)},\quad\text{where we write
$\mathcal{F}(K)\mathbin{\raisebox{0.3pt}{:}\\!=}\mathcal{F}_{n-1}^{K}$,}$

is of size $f_{n-1}(K)$ and $B(L)=(b_{fg}^{L})_{f,g\in\mathcal{F}(L)}$ is of
size $f_{n-1}(L)$. On the other hand, we know from Corollary 2.4 that
$f_{n-1}(K)=f_{n-1}(L)$ if and only if $n=2h$. Thus, we will now concentrate
on the case $f_{n-1}(K)=f_{n-1}(L)$ and consider how $B(K)$ changes into
$B(L)=B({\operatorname{{\bf bm}}}_{\alpha}K)$ under the bistellar move
${\operatorname{{\bf bm}}}_{\alpha}$.

###  4.1. Defining the Matrix Mutation

In the case of $f_{n-1}(K)=f_{n-1}(L)$, we have $n=2h$, and then
$\dim\alpha=\dim\beta=h$. Without loss of generality, assume that

$\alpha\,=\,(v_{0},v_{1},\dots,v_{h})\quad\text{and}\quad\beta\,=\,(v_{h+1},v_{h+2},\dots,v_{n+1}),$

and the oriented $n$-simplex $(v_{0},v_{1},\dots,v_{n})\in K$ is compatible
with the given orientations of the $n$-simplices of $K$. Then ${{\bf
F}_{\alpha}}=\mathcal{F}_{n}^{\Lambda_{\alpha}}$ must consist of the following
$h+1$ oriented simplices of dimension $n$:

$F_{i}\,=\,(-1)^{i}(v_{0},v_{1},\dots,v_{h},v_{h+1},\dots,\widehat{v_{n+1-i}},\dots,v_{n+1}),\quad\text{for
$i=0,\dots,h$,}$

such that $\alpha$ is the intersection of the $F_{i}$, for $i=0,\dots,h$.
Since ${\operatorname{{\bf bm}}}_{\alpha}$ only changes $\Lambda_{\alpha}$
into ${\operatorname{{\bf bm}}}_{\alpha}\Lambda_{\alpha}=\Lambda_{\beta}$ the
unchanged $n$-simplices

$\mathcal{F}_{n}^{K}\setminus{\bf
F}_{\alpha}\,=\,\mathcal{F}_{n}^{L}\setminus{\bf H}_{\beta},$

will keep their orientations, and the $h+1$ oriented simplices of dimension
$n$ in ${{\bf H}_{\beta}}=\mathcal{F}_{n}^{\Lambda_{\beta}}$ will be

$H_{i}\,=\,(-1)^{i}(v_{0},v_{1},\dots,\widehat{v_{i}},\dots,v_{h},v_{h+1},\dots,v_{n+1}),\quad\text{for
$i=0,\dots,h$,}$

such that $\beta$ is the intersection of the $H_{i}$, for $i=0,\dots,h$. This
makes sure that $L={\operatorname{{\bf bm}}}_{\alpha}K$ satisfies the
orientation-ordering convention.

######  Lemma 4.1.

For any $F\in\mathcal{F}_{n}^{K}\setminus{\bf F}_{\alpha}$ and any
$f,g\in\mathcal{F}(F)$, we have that $b_{fg}^{K}=b^{L}_{fg}$. Furthermore,

$\sum_{F\in\mathcal{F}_{n}^{K}\setminus{\bf
F}_{\alpha}}B^{F}(K)\,=\,\sum_{F\in\mathcal{F}_{n}^{L}\setminus{\bf
H}_{\beta}}B^{F}(L).$

###### Proof.

This is because $\mathcal{F}_{n}^{K}\setminus{\bf
F}_{\alpha}=\mathcal{F}_{n}^{L}\setminus{\bf H}_{\beta}$. ∎

By Lemma 4.1, it remains to study how $\sum_{F\in{\bf F}_{\alpha}}B^{F}(K)$
changes into $\sum_{F\in{\bf H}_{\beta}}B^{F}(L).$ To do this, we need to
further analyze the structures of both subcomplexes $\Lambda_{\alpha}$ and
$\Lambda_{\beta}$. An easy observation shows that both $\Lambda_{\alpha}$ and
$\Lambda_{\beta}$ have $\binom{h+1}{1}\binom{h+1}{1}+\binom{h+1}{2}$ simplices
of dimension $n-1$, and there are $\binom{h+1}{1}\binom{h+1}{1}$ simplices of
dimension $n-1$ that lie in both $\Lambda_{\alpha}$ and $\Lambda_{\beta}$. Let

$\displaystyle\mathcal{D}_{\alpha}\,\mathbin{\raisebox{0.3pt}{:}\\!=}\,$
$\displaystyle\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{F}(\Lambda_{\beta})$
$\displaystyle\,=\,$
$\displaystyle\\{(v_{0},v_{1},\dots,v_{h},v_{h+1},\dots,\widehat{v_{h+1+i}},\dots,\widehat{v_{h+1+j}},\dots,v_{n+1})\mid
0\leq i<j\leq h\\}$

which consists of $\binom{h+1}{2}$ simplices of dimension $n-1$ in
$\Lambda_{\alpha}$. Similarly,

$\displaystyle\mathcal{D}_{\beta}\,\mathbin{\raisebox{0.3pt}{:}\\!=}\,$
$\displaystyle\mathcal{F}(\Lambda_{\beta})\setminus\mathcal{F}(\Lambda_{\alpha})$
$\displaystyle\,=\,$
$\displaystyle\\{(v_{0},v_{1},\dots,\widehat{v_{i}},\dots,\widehat{v_{j}},\dots,v_{h},v_{h+1},\dots,v_{n+1})\mid
0\leq i<j\leq h\\}$

consists of $\binom{h+1}{2}$ simplices of dimension $n-1$ in
$\Lambda_{\beta}$. Obviously,
$\mathcal{D}_{\alpha}\cap\mathcal{D}_{\beta}=\emptyset$. Then we have that

(4.1)
$\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}\,=\,\mathcal{F}(\Lambda_{\beta})\setminus\mathcal{D}_{\beta}$

which contains exactly $\binom{h+1}{1}\binom{h+1}{1}$ $(n-1)$-simplices.

######  Lemma 4.2.

Suppose $g\notin\mathcal{F}(\Lambda_{\alpha})$. Then for any
$f\in\mathcal{D}_{\alpha}$, $b_{fg}^{K}=0$ in $B(K)$, and for any
$f\in\mathcal{D}_{\beta}$, $b_{fg}^{L}=0$ in $B(L)$.

###### Proof.

This is because each $(n-1)$-simplex in $\mathcal{D}_{\alpha}$ is not a face
of any $n$-simplex in $\mathcal{F}_{n}^{K}\setminus{{\bf F}_{\alpha}}$ and
each $(n-1)$-simplex in $\mathcal{D}_{\beta}$ is not a face of any $n$-simplex
in $\mathcal{F}_{n}^{L}\setminus{{\bf H}_{\beta}}$. ∎

By Corollary 2.5, $\Lambda_{\alpha}$ and $\Lambda_{\beta}$ are combinatorially
equivalent.

######  Lemma 4.3.

Let $\sigma$ be a permutation on $[n+2]$ such that $\sigma(\alpha)=\beta$.
Then $\sigma$ induces a combinatorial equivalence between $\Lambda_{\alpha}$
and $\Lambda_{\beta}$.

###### Proof.

This is because $\Lambda_{\alpha}$ and $\Lambda_{\beta}$ have the same vertex
set $[n+2]$, and $\alpha=\bigcap_{i=0}^{h}F_{i}$ and
$\beta=\bigcap_{i=0}^{h}H_{i}$. ∎

######  Remark 4.4.

If $\sigma$ is a permutation on $[n+2]$ such that $\sigma(\alpha)=\beta$, then
it is easy to see that $\sigma$ not only gives a bijection from
$\mathcal{F}(\Lambda_{\alpha})\to\mathcal{F}(\Lambda_{\beta})$, but is also
invariant on $\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}$ and
maps $\mathcal{D}_{\alpha}$ onto $\mathcal{D}_{\beta}$ bijectively. Note that
the permutation $\sigma$ on $[n+2]$ with $\sigma(\alpha)=\beta$ is, of course,
not unique.

We fix the following permutation in the symmetric group $\mathcal{S}_{n+2}$:

$\sigma\,=\,\left(\begin{array}[]{ccccc}1&2&\cdots&n+1&n+2\\\
n+2&n+1&\cdots&2&1\\\ \end{array}\right)\,\in\,\mathcal{S}_{n+2}.$

Note that $\sigma(\alpha)=\beta$.

######  Definition 4.5.

We define the _matrix mutation_

$\mu_{\alpha}(B(K))\,=\,(\bar{b}_{fg})_{f,g\in\mathcal{F}(L)}$

along $\alpha$ as follows:

$\bar{b}_{fg}\,=\,\begin{cases}-b^{K}_{\sigma(f)\sigma(g)},&\text{if
$f,g\in\mathcal{F}(\Lambda_{\beta})$;}\\\
\phantom{-}b^{K}_{fg},&\text{otherwise.}\end{cases}$

######  Remark 4.6.

The permutation $\sigma=\left(\begin{array}[]{ccccc}1&2&\cdots&n+1&n+2\\\
n+2&n+1&\cdots&2&1\\\ \end{array}\right)$ in $\mathcal{S}_{n+2}$ is of order
2.

######  Proposition 4.7.

If $L=\operatorname{{\bf bm}}_{\alpha}K$, where $K\in\mathbb{TM}_{2h}$ and
$\alpha\in\mathcal{F}_{h}^{K}$, then

$\mu_{\alpha}(B(K))\,=\,B(L).$

###### Proof.

Suppose that $f<g$ in $\mathcal{F}(\Lambda_{\alpha})$ and that $f\cup
g=F_{i}$, for some $i=0,\dots,h$. Then $\sigma(g)<\sigma(f)$ and
$\sigma(F_{i})=H_{i}$. The coefficient $c_{fg}$ appears in
$\partial^{(n-1)}F_{i}$. If $v_{i}\in f\cap g$ appears in an even position in
$F_{i}$, then so will $\sigma(v_{i})$ in $H_{i}$ (and similarly for odd
positions) since $\dim F_{i}=n$ is even. Therefore,
$c_{fg}=c_{\sigma(g)\sigma(f)}$ which gives the result.

If either $f$ or $g$ is not in $\mathcal{F}(\Lambda_{\alpha})$, then this
follows from Lemma 4.1. ∎

######  Remark 4.8.

Since ${\operatorname{{\bf bm}}}_{\beta}L={\operatorname{{\bf
bm}}}_{\beta}{\operatorname{{\bf bm}}}_{\alpha}K=K$, it is easy to see that
$\mu_{\alpha}$ has an inverse $\mu_{\beta}$, that is,

$\mu_{\beta}(\mu_{\alpha}(B(K)))\,=\,\mu_{\beta}(B(L))\,=\,B(K).$

###  4.2. Examples

Let us now look at the special cases of $h=1$ and $2$.

#### The case $h=1$

When $h=1$, we have that $\alpha=(1,2)$ and $\beta=(3,4)$, so

${{\bf F}_{\alpha}}\,=\,\\{F_{0}=(1,2,3),\ F_{1}=-(1,2,4)\\},\quad{{\bf
H}_{\beta}}=\\{H_{0}=(2,3,4),\ H_{1}=-(1,3,4)\\}$

and

$\mathcal{F}(\Lambda_{\alpha})\,=\,\\{(1,2),(1,3),(1,4),(2,3),(2,4)\\},\quad\mathcal{F}(\Lambda_{\beta})\,=\,\\{(1,3),(1,4),(2,3),(2,4),(3,4)\\}.$

Also, $\mathcal{D}_{\alpha}=\\{(1,2)\\}$ and
$\mathcal{D}_{\beta}=\\{(3,4)\\}$. Furthermore, $B(\Lambda_{\alpha})$ and
$B(\Lambda_{\beta})$ determine nontrivial $5\times 5$ matrices. By a direct
calculation, we have that

$B(\Lambda_{\alpha})\,=\,\left(\begin{array}[]{ccccc}0&1&-1&-1&1\\\
-1&0&0&1&0\\\ 1&0&0&0&-1\\\ 1&-1&0&0&0\\\ -1&0&1&0&0\\\ \end{array}\right)$

with all rows and columns ordered by $(1,2),(1,3),(1,4),(2,3),(2,4)$, and

$B(\Lambda_{\beta})\,=\,\left(\begin{array}[]{ccccc}0&-1&1&1&-1\\\
1&0&-1&0&0\\\ -1&1&0&0&0\\\ -1&0&0&0&1\\\ 1&0&0&-1&0\\\ \end{array}\right)$

with all rows and columns ordered by $(3,4),(1,3),(1,4),(2,3),(2,4)$.

Here $\sigma=\left(\begin{array}[]{ccccc}1&2&3&4\\\ 4&3&2&1\\\
\end{array}\right)$, and $\sigma$ maps $\mathcal{F}(\Lambda_{\alpha})$
bijectively onto $\mathcal{F}(\Lambda_{\beta})$, as shown in the following
table:

$f$ | (1,2) | (1,3) | (1,4) | (2,3) | (2,4)
---|---|---|---|---|---
$\sigma(f)$ | (3,4) | (2,4) | (1,4) | (2,3) | (1,3)

Then we can read out that

$b_{\sigma(f)\sigma(g)}^{L}\,=\,-b_{fg}^{K}.$

#### The case $h=2$

When $h=2$, we have that $\alpha=(1,2,3)$ and $\beta=(4,5,6)$, so

${{\bf F}_{\alpha}}\,=\,\\{F_{0}=(1,2,3,4,5),\ F_{1}=-(1,2,3,4,6),\
F_{2}=(1,2,3,5,6)\\}$

and

${{\bf H}_{\beta}}\,=\,\\{H_{0}=(2,3,4,5,6),\ H_{1}=-(1,3,4,5,6),\
H_{2}=(1,2,4,5,6)\\}.$

Furthermore, $\mathcal{F}(\Lambda_{\alpha})$ consists of twelve 3-simplices

$\displaystyle(1,2,3,4),(1,2,3,5),(1,2,3,6),(1,2,4,5),(1,2,4,6),(1,2,5,6),$
$\displaystyle(1,3,4,5),(1,3,4,6),(1,3,5,6),(2,3,4,5),(2,3,4,6),(2,3,5,6)$

in $\Lambda_{\alpha}\subset K$, and $\mathcal{F}(\Lambda_{\beta})$ consists of
twelve 3-simplices

$\displaystyle(1,4,5,6),(2,4,5,6),(3,4,5,6),(1,2,4,5),(1,2,4,6),(1,2,5,6),$
$\displaystyle(1,3,4,5),(1,3,4,6),(1,3,5,6),(2,3,4,5),(2,3,4,6),(2,3,5,6)$

in $\Lambda_{\beta}\subset L$. Also,

$\displaystyle\mathcal{D}_{\alpha}\,=\,\\{(1,2,3,4),(1,2,3,5),(1,2,3,6)\\}\quad\text{and}\quad\mathcal{D}_{\beta}\,=\,\\{(1,4,5,6),(2,4,5,6),(3,4,5,6)\\}.$

In this case,
$\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}=\mathcal{F}(\Lambda_{\beta})\setminus\mathcal{D}_{\beta}$
contains nine $3$-simplices

$\displaystyle(1,2,4,5),(1,2,4,6),(1,2,5,6),(1,3,4,5),(1,3,4,6),(1,3,5,6),(2,3,4,5),(2,3,4,6),(2,3,5,6)$

in $\Lambda_{\alpha}$ and $\Lambda_{\beta}$. Both $B(\Lambda_{\alpha})$ and
$B(\Lambda_{\beta})$ determine $12\times 12$ skew-symmetric matrices. A direct
calculation shows that

$B(\Lambda_{\alpha})\,=\,\left(\begin{array}[]{cccccccccccc}0&-1&1&1&-1&0&-1&1&0&1&-1&0\\\
1&0&-1&-1&0&1&1&0&-1&-1&0&1\\\ -1&1&0&0&1&-1&0&-1&1&0&1&-1\\\
-1&1&0&0&0&0&-1&0&0&1&0&0\\\ 1&0&-1&0&0&0&0&1&0&0&-1&0\\\
0&-1&1&0&0&0&0&0&-1&0&0&1\\\ 1&-1&0&1&0&0&0&0&0&-1&0&0\\\
-1&0&1&0&-1&0&0&0&0&0&1&0\\\ 0&1&-1&0&0&1&0&0&0&0&0&-1\\\
-1&1&0&-1&0&0&1&0&0&0&0&0\\\ 1&0&-1&0&1&0&0&-1&0&0&0&0\\\
0&-1&1&0&0&-1&0&0&1&0&0&0\\\ \end{array}\right)$

with all rows and columns ordered by $(1,2,3,4)$, $(1,2,3,5)$, $(1,2,3,6)$,
$(1,2,4,5)$, $(1,2,4,6)$, $(1,2,5,6)$, $(1,3,4,5)$, $(1,3,4,6)$, $(1,3,5,6)$,
$(2,3,4,5)$, $(2,3,4,6)$, $(2,3,5,6)$, and

$B(\Lambda_{\beta})\,=\,\left(\begin{array}[]{cccccccccccc}0&-1&1&1&-1&1&-1&1&-1&0&0&0\\\
1&0&-1&-1&1&-1&0&0&0&1&-1&1\\\ -1&1&0&0&0&0&1&-1&1&-1&1&-1\\\
-1&1&0&0&-1&1&0&0&0&0&0&0\\\ 1&-1&0&1&0&-1&0&0&0&0&0&0\\\
-1&1&0&-1&1&0&0&0&0&0&0&0\\\ 1&0&-1&0&0&0&0&1&-1&0&0&0\\\
-1&0&1&0&0&0&-1&0&1&0&0&0\\\ 1&0&-1&0&0&0&1&-1&0&0&0&0\\\
0&-1&1&0&0&0&0&0&0&0&-1&1\\\ 0&1&-1&0&0&0&0&0&0&1&0&-1\\\
0&-1&1&0&0&0&0&0&0&-1&1&0\\\ \end{array}\right)$

with all rows and columns ordered by $(1,4,5,6)$, $(2,4,5,6)$, $(3,4,5,6)$,
$(1,2,4,5)$, $(1,2,4,6)$, $(1,2,5,6)$, $(1,3,4,5)$, $(1,3,4,6)$, $(1,3,5,6)$,
$(2,3,4,5)$, $(2,3,4,6)$, $(2,3,5,6)$.

The permutation $\sigma=\left(\begin{array}[]{cccccc}1&2&3&4&5&6\\\
6&5&4&3&2&1\\\ \end{array}\right)$ then maps $\mathcal{F}(\Lambda_{\alpha})$
bijectively to $\mathcal{F}(\Lambda_{\beta})$ similarly to the previous case.
Then we can easily read out that

$b_{\sigma(f)\sigma(g)}^{L}\,=\,-b_{fg}^{K},\quad\text{for
$f,g\in\mathcal{F}(\Lambda_{\alpha})$.}$

##  5\. Bistellar Cluster Algebras

This section is dedicated to defining a class of algebras from closed oriented
triangulated manifolds, which we call _bistellar cluster algebras_.

Given a closed oriented triangulated manifold $K\in\mathbb{TM}_{n}$, assume
that $K$ is a simplicial complex on the vertex set $[m]$, so $K$ is a
subcomplex of $2^{[m]}$. We shall mainly be concerned with the case in which
$n$ is even and study the algebraic behaviour of middle-dimensional bistellar
moves on $K$. We write $n=2h$ and assume that we can perform a bistellar
$h$-move on $\alpha\in K$ and set
$L\mathbin{\raisebox{0.3pt}{:}\\!=}{\operatorname{{\bf bm}}}_{\alpha}K$.

Now we are going to construct an algebra $\mathcal{A}_{K}$ associated to $K$.
Our construction differs from the one of a classical cluster algebra (see [6,
5]), but the basic idea of our construction follows closely.

###  5.1. Seeds and Bistellar Seed Mutations

Let $(\mathbb{P},\oplus,\cdot)$ be a semifield, that is, $(\mathbb{P},\oplus)$
is a commutative semigroup and $(\mathbb{P},\cdot)$ is an abelian group such
that multiplication distributes over the auxiliary addition $\oplus$. By
$\mathbb{Z}\mathbb{P}$ we denote the group ring of $(\mathbb{P},\cdot)$ with
integer coefficients.

######  Definition 5.1.

Let $K$ be a closed oriented triangulated manifold in $\mathbb{TM}_{n}$. Let
$\mathbb{F}\mathcal{X}(K)$ denote the field of rational functions over
$\mathbb{ZP}$ with independent variables given by $\mathcal{X}(K)$, where
$\mathcal{X}(K)=\\{x_{f}\mid f\in\mathcal{F}(K)\\}$ is called a _cluster_.

Associated to $K$, we get a triple $\Sigma_{K}=(\mathcal{X}(K),{\bf
p}_{K},B(K))$, called a _seed_ , where

  1. (1)

$\mathcal{X}(K)$ is the cluster corresponding to $K$, as defined above;

  2. (2)

the coefficient tuple ${\bf p}_{K}=(p^{\pm}_{K,x})_{x\in\mathcal{X}(K)}$ is a
$2d$-tuple of elements of $\mathbb{P}$ satisfying the normalization condition
$p_{K,x}^{+}\oplus p_{K,x}^{-}=1$ for all $x\in\mathcal{X}(K)$, where
$d=f_{n-1}(K)$;

  3. (3)

$B(K)$ is the exchange matrix of $K$.

First let us analyse the behaviour of the seeds under the actions of bistellar
$h$-moves. Let $(\alpha,\beta)$ be a bistellar pair in $K$ of type $h$. Set
$\mathcal{X}(\mathcal{D}_{\alpha})=\\{x_{f}\mid f\in\mathcal{D}_{\alpha}\\}$
and $\mathcal{X}(\mathcal{D}_{\beta})=\\{x_{f}\mid
f\in\mathcal{D}_{\beta}\\}$. Then $\mathcal{X}(\mathcal{D}_{\alpha})$ and
$\mathcal{X}(\mathcal{D}_{\beta})$ are subsets of $\mathcal{X}(K)$ and
$\mathcal{X}(L)$, respectively. Note that by Corollary 2.4, we have
$f_{n-1}(L)=f_{n-1}(K)=d$, so $|\mathcal{X}(K)|=|\mathcal{X}(L)|$.

######  Definition 5.2.

Elements in $\mathcal{X}(\mathcal{D}_{\alpha})$ and
$\mathcal{X}(\mathcal{D}_{\beta})$ are said to be _exchangeable bistellar
cluster variables_ in $\mathcal{X}(K)$ and $\mathcal{X}(L)$, respectively.

With this convention, for the bistellar pair $(\alpha,\beta)$ in $K$, we have
the following relations between the seeds $\Sigma_{K}=(\mathcal{X}(K),{\bf
p}_{K},B(K))$ and $\Sigma_{L}=(\mathcal{X}(L),{\bf p}_{L},B(L))$ as follows:

  1. (1)

Firstly, we see that between the two clusters $\mathcal{X}(K)$ and
$\mathcal{X}(L)$ we have the following:

$\mathcal{X}(L)\,=\,(\mathcal{X}(K)\setminus\mathcal{X}(\mathcal{D}_{\alpha}))\cup\mathcal{X}(\mathcal{D}_{\beta})\quad\text{and}\quad\mathcal{X}(K)\,=\,(\mathcal{X}(L)\setminus\mathcal{X}(\mathcal{D}_{\beta}))\cup\mathcal{X}(\mathcal{D}_{\alpha}),$

which exactly corresponds to

$L\,=\,{\operatorname{{\bf
bm}}}_{\alpha}K\,=\,(K\setminus\Lambda_{\alpha})\cup\Lambda_{\beta}\quad\text{and}\quad
K\,=\,\operatorname{{\bf
bm}}_{\beta}L\,=\,(L\setminus\Lambda_{\beta})\cup\Lambda_{\alpha}$

respectively. Since
$\sigma(\mathcal{F}(\Lambda_{\alpha}))=\mathcal{F}(\Lambda_{\beta})$, we can
use $\sigma$ to define a bijection
$\varphi^{K}_{\alpha}\colon\mathcal{X}(K)\to\mathcal{X}(L)$ by

(5.1) $\varphi^{K}_{\alpha}(x_{f})\,=\,\begin{cases}x_{\sigma(f)},&\text{if
$f\in\mathcal{F}(\Lambda_{\alpha})$;}\\\ x_{f},&\text{if
$f\notin\mathcal{F}(\Lambda_{\alpha})$,}\end{cases}$

satisfying
$\varphi^{K}_{\alpha}(\mathcal{X}(\mathcal{D}_{\alpha}))=\mathcal{X}(\mathcal{D}_{\beta})$
since $\sigma(\mathcal{D}_{\alpha})=\mathcal{D}_{\beta}$. Conversely, since
$\mathcal{F}(K)\setminus\mathcal{F}(\Lambda_{\alpha})=\mathcal{F}(L)\setminus\mathcal{F}(\Lambda_{\beta})$
and $\sigma$ is an involution, we can also use $\sigma$ to define a bijection
$\varphi^{L}_{\beta}\colon\mathcal{X}(L)\rightarrow\mathcal{X}(K)$ similarly
satisfying
$\varphi^{L}_{\beta}(\mathcal{X}(\mathcal{D}_{\beta}))=\mathcal{X}(\mathcal{D}_{\alpha})$.
Clearly, $\varphi^{L}_{\beta}$ is the inverse of $\varphi^{K}_{\alpha}$. So
$\varphi^{L}_{\beta}(\varphi^{K}_{\alpha}(\mathcal{X}(K)))\,=\,\mathcal{X}(K)$.

  2. (2)

Secondly, $\varphi^{K}_{\alpha}$ and $\varphi^{L}_{\beta}$ determine two
bijections

$c_{K}\colon{\bf p}_{K}\longrightarrow{\bf p}_{L}\quad\text{given by
$p^{\pm}_{K,x}\longmapsto p^{\pm}_{L,\varphi^{K}_{\alpha}(x)}$}$

and

$c_{L}\colon{\bf p}_{L}\longrightarrow{\bf p}_{K}\quad\text{given by
$p^{\pm}_{L,x}\longmapsto p^{\pm}_{K,\varphi^{L}_{\beta}(x)}$}.$

Obviously, $c_{L}(c_{K}({\bf p}_{K}))={\bf p}_{K}$.

  3. (3)

Finally, by Proposition 4.7, the matrix mutation $\mu_{\alpha}$ satisfies the
property that $\mu_{\alpha}(B(K))=B(L)$. We have, for each $b_{fg}^{K}\in
B(K)$,

(5.2)
$\mu_{\alpha}(b_{fg}^{K})\,=\,\begin{cases}-b_{fg}^{K}=b_{\sigma(f)\sigma(g)}^{L},&\text{if
$f,g\in\mathcal{F}(\Lambda_{\alpha})$};\\\
\phantom{+}b_{fg}^{K}=b_{fg}^{L},&\text{otherwise}.\end{cases}$

Since ${\operatorname{{\bf bm}}}_{\beta}L={\operatorname{{\bf
bm}}}_{\beta}{\operatorname{{\bf bm}}}_{\alpha}K=K$ and $\sigma$ is an
involution, we have that $\mu_{\beta}(B(L)=B(K)$, so

$\mu_{\beta}(\mu_{\alpha}(B(K)))\,=\,B(K).$

We now set up some notation. Suppose that $K\in\mathbb{TM}_{n}$ with bistellar
pair $(\alpha,\beta)$ of type $h$ with $n=2h$. Then for $f\in\mathcal{F}(K)$
we write

$u_{f}^{K}\,\mathbin{\raisebox{0.3pt}{:}\\!=}\,p_{K,x_{f}}^{+}/p_{K,x_{f}}^{-}.$

It then follows from $p^{+}_{K,x_{f}}\oplus p^{-}_{K,x_{f}}=1$ that

$p^{+}_{K,x_{f}}\,=\,\frac{u_{f}^{K}}{1\oplus u_{f}^{K}}\quad\text{and}\quad
p^{-}_{K,x_{f}}\,=\,\frac{1}{1\oplus u_{f}^{K}}.$

We also write

$\pi_{\alpha,f}^{K}\,\mathbin{\raisebox{0.3pt}{:}\\!=}\,\prod\limits_{g\in\mathcal{D}_{\alpha}}(p^{+}_{K,x_{g}})^{[b^{K}_{gf}]_{+}}(p^{-}_{K,x_{g}})^{-[-b^{K}_{gf}]_{+}},$

where $[b]_{+}\mathbin{\raisebox{0.3pt}{:}\\!=}\max\\{b,0\\}$.

######  Definition 5.3.

With the above understanding, we define the _bistellar seed mutation_

$\Phi_{\alpha}\colon\Sigma_{K}\,=\,(\mathcal{X}(K),{\bf
p}_{K},B(K))\longrightarrow\Sigma_{L}\,=\,(\mathcal{X}(L),{\bf p}_{L},B(L))$

as follows:

  1. (1)

On $\mathcal{X}(K)$,
$\varphi^{K}_{\alpha}(\mathcal{X}(K))=\mathcal{X}(L)=(\mathcal{X}(K)\setminus\mathcal{X}(\mathcal{D}_{\alpha}))\cup\mathcal{X}(\mathcal{D}_{\beta})$
such that there are _exchange relations_ between exchangeable bistellar
cluster variables of $\mathcal{X}(\mathcal{D}_{\alpha})$ and
$\mathcal{X}(\mathcal{D}_{\beta})$ as follows: for $f\in\mathcal{D}_{\alpha}$,

(5.3)
$x_{f}x_{\sigma(f)}\,=\,\frac{1}{D^{K}_{\alpha,f}}\left(p^{+}_{K,x_{f}}\operatorname{lcm}\left(\prod_{g\in\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}}x_{g}^{[b_{fg}^{K}]_{+}},\prod_{g\in\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}}x_{\sigma(g)}^{[b_{fg}^{K}]_{+}}\right)\right.\\\
+\left.p^{-}_{K,x_{f}}\operatorname{lcm}\left(\prod\limits_{g\in\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}}x_{g}^{[-b_{fg}^{K}]_{+}},\prod\limits_{g\in\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}}x_{\sigma(g)}^{[-b_{fg}^{K}]_{+}}\right)\right)$

and

(5.4)
$D^{K}_{\alpha,f}\,\mathbin{\raisebox{0.3pt}{:}\\!=}\,\prod_{\begin{subarray}{c}g\in\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}\\\
b_{fg}^{K}>0,\
b_{f\sigma(g)}^{K}<0\end{subarray}}x_{g}^{b_{fg}^{K}}x_{\sigma(g)}^{b_{fg}^{K}}\,=\,\prod_{\begin{subarray}{c}g\in\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}\\\
b_{fg}^{K}<0,\
b_{f\sigma(g)}^{K}>0\end{subarray}}x_{g}^{-b_{fg}^{K}}x_{\sigma(g)}^{-b_{fg}^{K}},$

which follows from
$\sigma(\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha})=\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}$
and $\sigma$ being of order 2.

There are $\binom{h+1}{2}$ exchange relations since
$|\mathcal{X}(\mathcal{D}_{\alpha})|=|\mathcal{X}(\mathcal{D}_{\beta})|=\binom{h+1}{2}$.

  2. (2)

On ${\bf p}_{K}$, $c_{K}({\bf p}_{K})={\bf
p}_{L}=(p^{\pm}_{L,x_{f}})_{f\in\mathcal{F}(L)}$ is uniquely determined by the
normalization condition $p^{+}_{L,x_{f}}\oplus p^{-}_{L,x_{f}}=1$ together
with

$u^{L}_{f}\,=\,\begin{cases}\left(u^{K}_{\sigma(f)}\right)^{-1},&\text{ if
$f\in\mathcal{D}_{\beta}$;}\\\
\pi_{\alpha,\sigma(f)}^{K}u^{K}_{\sigma(f)},&\text{if
$f\in\mathcal{F}(\Lambda_{\beta})\setminus\mathcal{D}_{\beta}$};\\\
u^{K}_{f},&\text{ otherwise.}\end{cases}$

  3. (3)

On $B(K)$, $\mu_{\alpha}(B(K))=B(L)$.

######  Remark 5.4.

Since ${\operatorname{{\bf bm}}}_{\beta}L={\operatorname{{\bf
bm}}}_{\beta}{\operatorname{{\bf bm}}}_{\alpha}K=K$, in a similar way as above
we also can define the bistellar seed mutation

$\Phi_{\beta}\colon\Sigma_{L}\,=\,(\mathcal{X}(L),{\bf
p}_{L},B(L))\longrightarrow\Sigma_{K}\,=\,(\mathcal{X}(K),{\bf p}_{K},B(K)).$

From Definition 5.3(1), we write

$M^{+}_{K,\alpha,f}\,=\,\frac{p^{+}_{K,x_{f}}}{D^{K}_{\alpha,f}}\operatorname{lcm}\left(\prod_{g\in\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}}x_{g}^{[b_{fg}^{K}]_{+}},\prod_{g\in\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}}x_{\sigma(g)}^{[b_{fg}^{K}]_{+}}\right)$

and

$M^{-}_{K,\alpha,f}\,=\,\frac{p^{-}_{K,x_{f}}}{D^{K}_{\alpha,f}}\operatorname{lcm}\left(\prod_{g\in\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}}x_{g}^{[-b_{fg}^{K}]_{+}},\prod_{g\in\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}}x_{\sigma(g)}^{[-b_{fg}^{K}]_{+}}\right),$

so $x_{f}x_{\sigma(f)}=M^{+}+M^{-}$.

Then it is easy to see, using equation (5.4), that:

######  Lemma 5.5.

The two monomials $M^{+}_{K,\alpha,f}$ and $M^{-}_{K,\alpha,f}$ have no common
divisor.

######  Lemma 5.6.

Following from Definition 5.3(2) we have the following relations:

$p^{+}_{L,x_{f}}\,=\,\begin{cases}p^{-}_{K,\sigma(f)},&\text{if
$f\in\mathcal{D}_{\beta}$};\\\
\frac{\pi_{\alpha,\sigma(f)}^{K}u^{K}_{\sigma(f)}}{1\oplus\pi_{\alpha,\sigma(f)}^{K}u^{K}_{\sigma(f)}},&\text{if
$f\in\mathcal{F}(\Lambda_{\beta})\setminus\mathcal{D}_{\beta}$};\\\
p^{+}_{K,x_{f}},&\text{otherwise},\end{cases}\quad\text{and}\quad
p^{-}_{L,x_{f}}\,=\,\begin{cases}p^{+}_{K,\sigma(f)},&\text{if
$f\in\mathcal{D}_{\beta}$};\\\
\frac{1}{1\oplus\pi_{\alpha,\sigma(f)}^{K}u^{K}_{\sigma(f)}},&\text{if
$f\in\mathcal{F}(\Lambda_{\beta})\setminus\mathcal{D}_{\beta}$};\\\
p^{-}_{K,x_{f}},&\text{otherwise}.\end{cases}$

###### Proof.

If $f\in\mathcal{D}_{\beta}$, since
$p^{+}_{L,x_{f}}/p^{-}_{L,x_{f}}=p^{-}_{K,x_{\sigma(f)}}/p^{+}_{K,x_{\sigma(f)}}$
and $p^{+}_{K,x_{\sigma(f)}}\oplus p^{-}_{K,x_{\sigma(f)}}=1$, then
$p^{+}_{K,x_{\sigma(f)}}=\frac{1}{1\oplus u_{f}^{L}}$ and
$p^{-}_{K,x_{\sigma(f)}}=\frac{u_{f}^{L}}{1\oplus u_{f}^{L}}$, so
$p^{\pm}_{L,x_{f}}=p^{\mp}_{K,x_{\sigma(f)}}$ as required.

If $f\in\mathcal{F}(L)\setminus\mathcal{F}(\Lambda_{\beta})$, since
$p^{+}_{L,x_{f}}/p^{-}_{L,x_{f}}=p^{+}_{K,x_{\sigma(f)}}/p^{-}_{K,x_{\sigma(f)}}$,
then clearly $p^{\pm}_{L,x_{f}}=p^{\pm}_{K,x_{f}}$.

If $f\in\mathcal{F}(\Lambda_{\beta})\setminus\mathcal{D}_{\beta}$, by (5.2),
by direct calculation

$p^{+}_{L,x_{f}}\,=\,\frac{u^{L}_{f}}{1\oplus
u^{L}_{f}}\,=\,\frac{\pi^{K}_{\alpha,\sigma(f)}u^{K}_{\sigma(f)}}{1\oplus\pi^{K}_{\alpha,\sigma(f)}u^{K}_{\sigma(f)}}\quad\text{and}\quad
p^{-}_{L,x_{f}}\,=\,\frac{1}{1\oplus
u^{L}_{f}}\,=\,\frac{1}{1\oplus\pi^{K}_{\alpha,\sigma(f)}u^{K}_{\sigma(f)}}.$

∎

######  Remark 5.7.

We also have that

$\displaystyle\pi_{\alpha,\sigma(f)}^{K}\,$
$\displaystyle=\,\prod\limits_{g\in\mathcal{D}_{\alpha}}(p^{+}_{K,x_{g}})^{[b^{K}_{g\sigma(f)}]_{+}}(p^{-}_{K,x_{g}})^{-[-b^{K}_{g\sigma(f)}]_{+}}\,=\,\prod\limits_{g\in\mathcal{D}_{\beta}}(p^{+}_{K,x_{\sigma(g)}})^{[b^{K}_{\sigma(g)\sigma(f)}]_{+}}(p^{-}_{K,x_{\sigma(g)}})^{-[-b^{K}_{\sigma(g)\sigma(f)}]_{+}}$
$\displaystyle=\,\prod\limits_{g\in\mathcal{D}_{\beta}}(p^{-}_{L,x_{g}})^{[-b^{L}_{gf}]_{+}}(p^{+}_{L,x_{g}})^{-[b^{L}_{gf}]_{+}}\,=\,\left(\prod\limits_{g\in\mathcal{D}_{\beta}}(p^{+}_{L,x_{g}})^{[b^{L}_{gf}]_{+}}(p^{-}_{L,x_{g}})^{-[-b^{L}_{gf}]_{+}}\right)^{-1}\,=\,\left(\pi^{L}_{\beta,f}\right)^{-1}.$

######  Lemma 5.8.

For $f\in\mathcal{D}_{\alpha}$,
$M^{\pm}_{K,\alpha,f}=M^{\mp}_{L,\beta,\sigma(f)}$. Furthermore,

$M^{+}_{L,\beta,\sigma(f)}+M^{-}_{L,\beta,\sigma(f)}\,=\,M^{-}_{K,\alpha,f}+M^{+}_{K,\alpha,f}\,=\,x_{f}x_{\sigma(f)}.$

###### Proof.

For $f,g\in\mathcal{F}(\Lambda_{\alpha})$, we have by (5.2) that
$b_{\sigma(f)\sigma(g)}^{L}=-b^{K}_{fg}$. For $f\in\mathcal{D}_{\alpha}$, by
Lemma 5.6 we have that $p^{\pm}_{L,x_{\sigma(f)}}=p^{\mp}_{K,x_{f}}$. We note
that $\sigma$ is an involution,
$\sigma(\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha})=\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}=\mathcal{F}(\Lambda_{\beta})\setminus\mathcal{D}_{\beta}$
and $\sigma(\mathcal{D}_{\alpha})=\mathcal{D}_{\beta}$. Thus, for
$f\in\mathcal{D}_{\alpha}$,

$\displaystyle M^{\mp}_{L,\beta,\sigma(f)}\,=\,$ $\displaystyle\
\frac{p^{\mp}_{L,x_{\sigma(f)}}}{D^{L}_{\beta,\sigma(f)}}\operatorname{lcm}\left(\prod\limits_{g\in\mathcal{F}(\Lambda_{\beta})\setminus\mathcal{D}_{\beta}}x_{g}^{[\mp
b^{L}_{\sigma(f)g}]_{+}},\prod\limits_{g\in\mathcal{F}(\Lambda_{\beta})\setminus\mathcal{D}_{\beta}}x_{\sigma(g)}^{[\mp
b^{L}_{\sigma(f)g}]_{+}}\right)$ $\displaystyle\,=\,$
$\displaystyle\frac{p^{\pm}_{K,x_{f}}}{D^{L}_{\beta,\sigma(f)}}\operatorname{lcm}\left(\prod\limits_{g\in\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}}x_{\sigma(g)}^{[\mp
b^{L}_{\sigma(f)\sigma(g)}]_{+}},\prod\limits_{g\in\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}}x_{\sigma^{2}(g)}^{[\mp
b^{L}_{\sigma(f)\sigma(g)}]_{+}}\right)$ $\displaystyle\,=\,$
$\displaystyle\frac{p^{\pm}_{K,x_{f}}}{D^{L}_{\beta,\sigma(f)}}\operatorname{lcm}\left(\prod\limits_{g\in\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}}x_{g}^{[\pm
b^{K}_{fg}]_{+}},\prod\limits_{g\in\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}}x_{\sigma(g)}^{[\pm
b^{K}_{fg}]_{+}}\right)$

where

$D^{L}_{\beta,\sigma(f)}\,=\,\prod_{\begin{subarray}{c}g\in\mathcal{F}(\Lambda_{\beta})\setminus\mathcal{D}_{\beta}\\\
b_{\sigma(f)g}^{L}>0,\
b_{\sigma(f)\sigma(g)}^{L}<0\end{subarray}}x_{g}^{b_{\sigma(f)g}^{L}}x_{\sigma(g)}^{b_{\sigma(f)g}^{L}}\,=\,\prod_{\begin{subarray}{c}g\in\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}\\\
-b_{fg}^{K}>0,\
-b_{f\sigma(g)}^{K}<0\end{subarray}}x_{\sigma(g)}^{-b_{fg}^{K}}x_{g}^{-b_{fg}^{K}}\,=\,D^{K}_{\alpha,f}$

using equation (5.4). Thus, we have that
$M^{\pm}_{K,\alpha,f}=M^{\mp}_{L,\beta,\sigma(f)}$. ∎

The above arguments imply that:

######  Proposition 5.9.

The bistellar seed mutation
$\Phi_{\alpha}\colon\Sigma_{K}\longrightarrow\Sigma_{L}$ is invertible, and
its inverse is $\Phi_{\beta}$.

The seed mutation $\Phi_{\alpha}$ also determines a natural
$\mathbb{ZP}$-linear field isomorphism
$\mathcal{L}_{(K,L)}\colon\mathbb{F}\mathcal{X}(K)\longrightarrow\mathbb{F}\mathcal{X}(L)$
given by

(5.5)
$\mathcal{L}_{(K,L)}(x_{f})\,=\,\varphi^{K}_{\alpha}(x_{f})\,=\,\begin{cases}x_{f}&\text{if
$f\in\mathcal{F}(K)\setminus\mathcal{F}(\Lambda_{\alpha})$;}\\\
x_{\sigma(f)}&\text{if
$f\in\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}$;}\\\
x_{\sigma(f)}=\frac{M^{+}_{K,\alpha,f}+M^{-}_{K,\alpha,f}}{x_{f}}&\text{if
$f\in\mathcal{D}_{\alpha}$.}\end{cases}$

Clearly, $\mathcal{L}_{(K,L)}$ is invertible and
$\mathcal{L}_{(L,K)}=\mathcal{L}_{(K,L)}^{-1}$. Since
$\mathcal{X}(K)\setminus\mathcal{X}(\mathcal{D}_{\alpha})=\mathcal{X}(L)\setminus\mathcal{X}(\mathcal{D}_{\beta})$
and

$\frac{M^{+}_{K,\alpha,f}+M^{-}_{K,\alpha,f}}{x_{f}}\,\in\,\mathbb{F}\mathcal{X}(L)$

for $f\in\mathcal{D}_{\alpha}$, this gives:

######  Corollary 5.10.

Under the action of the seed mutation $\Phi_{\alpha}$ (or $\Phi_{\beta}$),
$\mathbb{F}\mathcal{X}(K)$ can be identified with $\mathbb{F}\mathcal{X}(L)$.

We can now actually define the required bistellar cluster algebra
$\mathcal{A}_{K}$ from the initial seed $\Sigma_{K}$ via bistellar seed
mutations by performing all possible bistellar $h$-moves on $K$, as in the
definition of a classical cluster algebra. To understand the structure of
$\mathcal{A}_{K}$ more clearly, we would like to give more details in the next
subsection.

###  5.2. Bistellar Exchange Graphs and Bistellar Cluster Algebras

We define an equivalence relation on $\mathbb{TM}_{2h}$ by saying that
$K_{1}\sim K_{2}$ if and only if $K_{2}$ can be obtained from $K_{1}$ by a
finite sequence of bistellar $h$-moves (including the empty sequence). This is
clearly an equivalence relation and we denote by $[K]$ the equivalence class
containing $K$.

Note that it is possible that $[K]$ only contains $K$ itself. For example, if
$K$ is the boundary complex of an $(2h+1)$-simplex, then we cannot perform any
bistellar $h$-moves on $K$, so $[K]=\\{K\\}$. If $|[K]|>1$, then by Corollary
2.4, any triangulated manifold $L\in[K]$ is also a subcomplex of of $2^{[m]}$,
so $[K]$ is finite. Throughout the following, we assume that $|[K]|>1$.

######  Definition 5.11.

A finite sequence of pairs
$(\alpha_{1},\beta_{1}),\dots,(\alpha_{l},\beta_{l})$ in
$\mathcal{F}_{h}^{2^{[m]}}$ is called a _bistellar sequence_ of $[K]$ if there
exists $L^{\prime},L^{\prime\prime}\in[K]$ such that
$L^{\prime\prime}={{\operatorname{{\bf
bm}}}_{\alpha_{l}}\cdots{\operatorname{{\bf bm}}}_{\alpha_{1}}L^{\prime}}$.

A pair $(\alpha,\beta)$ in $\mathcal{F}_{h}^{2^{[m]}}$ is said to be a
_bistellar pair_ of $[K]$ if there is a bistellar sequence of $[K]$,
$(\alpha_{1},\beta_{1}),\dots,(\alpha_{l},\beta_{l})$, such that
$(\alpha,\beta)$ is one of
$(\alpha_{1},\beta_{1}),\dots,(\alpha_{l},\beta_{l})$.

Obviously, if $(\alpha_{1},\beta_{1}),\dots,(\alpha_{l},\beta_{l})$ is a
bistellar sequence of $[K]$, then so is its reverse.

By $\mathcal{S}_{bp}^{[K]}$ we denote the set of all bistellar pairs of $[K]$,
and for each $L\in[K]$, by $\mathcal{S}_{bp}^{L}$ we denote the subset of
$\mathcal{S}_{bp}^{[K]}$ consisting of bistellar pairs in $L$. Obviously, each
$\mathcal{S}_{bp}^{L}$ is nonempty. On the other hand, given a pair
$(\alpha,\beta)\in\mathcal{S}_{bp}^{[K]}$, it is easy to see that there must
be a triangulated manifold $L\in[K]$ such that $(\alpha,\beta)$ is a bistellar
pair in $L$, that is, $(\alpha,\beta)\in\mathcal{S}_{bp}^{L}$. Thus,

$\mathcal{S}_{bp}^{[K]}\,=\,\bigcup_{L\in[K]}\mathcal{S}_{bp}^{L}.$

Note that generally, for two $L,L^{\prime}\in[K]$, the intersection
$\mathcal{S}_{bp}^{L}\cap\mathcal{S}_{bp}^{L^{\prime}}$ may not be empty.

######  Lemma 5.12.

The set of bistellar pairs $\mathcal{S}_{bp}^{[K]}$ has the following
properties:

  1. (1)

$\mathcal{S}_{bp}^{[K]}$ is a finite set.

  2. (2)

$\mathcal{S}_{bp}^{[K]}$ is nonempty if and only if $|[K]|>1$.

  3. (3)

$(\alpha,\beta)\in\mathcal{S}_{bp}^{[K]}$ if and only if
$(\beta,\alpha)\in\mathcal{S}_{bp}^{[K]}$.

  4. (4)

Let $L\in[K]$ and $(\alpha,\beta)\in\mathcal{S}_{bp}^{[K]}$. Then
$(\alpha,\beta)\in\mathcal{S}_{bp}^{L}$ if and only if
$(\beta,\alpha)\in\mathcal{S}_{bp}^{{\operatorname{{\bf bm}}}_{\alpha}L}$.

  5. (5)

If $(\alpha,\beta),(\alpha^{\prime},\beta^{\prime})\in\mathcal{S}_{bp}^{L}$
are distinct pairs, then
$\mathcal{D}_{\alpha}\cap\mathcal{D}_{\alpha^{\prime}}=\emptyset$.

###### Proof.

Properties (1)–(4) are immediate. It only remains to show that the property
(5) holds.

Assume that
$\mathcal{D}_{\alpha}\cap\mathcal{D}_{\alpha^{\prime}}\not=\emptyset$. Then
there is at least one $(n-1)$-simplex
$f\in\mathcal{D}_{\alpha}\cap\mathcal{D}_{\alpha^{\prime}}$ such that

  1. (a)

$\alpha\subseteq f$ and $\alpha^{\prime}\subseteq f$ (since
$\alpha=\bigcap_{I\in\mathcal{D}_{\alpha}}I$ and
$\alpha^{\prime}=\bigcap_{J\in\mathcal{D}_{\alpha^{\prime}}}J$); and
furthermore

  2. (b)

there are two vertices $v,w\in\beta$ and two vertices
$v^{\prime},w^{\prime}\in\beta^{\prime}$ such that

$f\,=\,(\alpha\cup\beta)\setminus\\{v,w\\}\,=\,(\alpha^{\prime}\cup\beta^{\prime})\setminus\\{v^{\prime},w^{\prime}\\}.$

Since $(\alpha,\beta)$ and $(\alpha^{\prime},\beta^{\prime})$ are different,
at most one of $v^{\prime}$ and $w^{\prime}$ is in $\beta$. Without the loss
of generality, assume that $v^{\prime}\not\in\beta$. Since
$F=(\alpha^{\prime}\cup\beta^{\prime})\setminus\\{w^{\prime}\\}$ is an
$n$-simplex of $L$ and $\alpha\subset f\subset F$, we have that
$F\setminus\alpha$ is a simplex in
$\operatorname{Link}_{L}\alpha=\partial\beta$, so $v^{\prime}\in F$ is a
$0$-simplex of $\partial\beta$, that is, $v^{\prime}\in\beta$. This is a
contradiction. Thus
$\mathcal{D}_{\alpha}\cap\mathcal{D}_{\alpha^{\prime}}=\emptyset$ as desired.
∎

######  Example 5.13.

Let $K$ be a triangulation with five vertices of the $2$-sphere, as shown in
Figure 1.

.............................................................................$3$$4$$5$$1$$2$
Figure 1. Triangulation of the 2-sphere with five vertices

Then there are ten elements in $[K]$ and thirty bistellar pairs in
$\mathcal{S}_{bp}^{[K]}$, which are described with their bistellar moves in
Figure 2.

................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................$2$$5$$2$$3$$5$$1$$K_{32}$$K_{12}$$K_{21}$$K_{22}$$K_{1}$$K_{11}$$K_{3}$$K_{31}$$4$$5$$1$$3$$K_{2}$${\bf
bm}_{(2,4)}$${\bf bm}_{(3,5)}$${\bf bm}_{(1,4)}$${\bf bm}_{(3,5)}$${\bf
bm}_{(2,5)}$${\bf bm}_{(1,4)}$${\bf bm}_{(3,5)}$${\bf bm}_{(1,2)}$${\bf
bm}_{(1,2)}$${\bf bm}_{(4,5)}$${\bf bm}_{(2,3)}$${\bf bm}_{(4,5)}$

${\bf bm}_{(1,2)}$

${\bf bm}_{(3,4)}$

${\bf bm}_{(4,5)}$

${\bf bm}_{(1,3)}$

${\bf bm}_{(2,3)}$

${\bf bm}_{(1,5)}$

${\bf bm}_{(1,5)}$${\bf bm}_{(3,4)}$

${\bf bm}_{(3,4)}$

${\bf bm}_{(2,5)}$${\bf bm}_{(2,4)}$${\bf bm}_{(1,3)}$

${\bf bm}_{(1,5)}$

${\bf bm}_{(2,5)}$

${\bf bm}_{(1,3)}$

${\bf bm}_{(2,4)}$

${\bf bm}_{(1,4)}$${\bf
bm}_{(2,3)}$.$4$$1$$2$$3$$5$$4$$3$$1$$2$$5$$4$$1$$3$$2$$5$$K$$2$$3$$4$$4$$4$$1$$3$$4$$3$$2$$4$$2$$1$$3$$5$$1$$5$$4$$2$$5$$3$$1$$1$$2$$5$
Figure 2. Bistellar pairs and moves in $[K]$.

######  Definition 5.14.

The _bistellar exchange graph_ $\mathbb{G}_{[K]}$ of $[K]$ is defined as
follows:

  1. (1)

The vertex set of $\mathbb{G}_{[K]}$ is $[K]$.

  2. (2)

Each edge corresponds to a bistellar $h$-move and its inverse. In other words,
for $(\alpha,\beta)\in\mathcal{S}_{bp}^{[K]}$, if $(\alpha,\beta)$ is a
bistellar pair in $L\in[K]$, then both $(\alpha,\beta)$ and $(\beta,\alpha)$
define an edge between $L$ and ${\operatorname{{\bf bm}}}_{\alpha}L$.

The bistellar exchange graph $\mathbb{G}_{[K]}$ of $[K]$ indicates how all
triangulated manifolds in $[K]$ exchange amongst themselves by performing
bistellar $h$-moves.

######  Example 5.15.

The bistellar exchange graph of $[K]$ in Example 5.13 is a trivalent graph
with ten vertices shown in Figure 3.

$K_{11}$$K_{1}$$K_{3}$$K_{32}$$K_{12}$$K_{21}$$K$$K_{2}$$K_{22}$$K_{31}$
Figure 3. Bistellar exchange graph.

######  Proposition 5.16.

The bistellar exchange graph $\mathbb{G}_{[K]}$ is connected.

###### Proof.

This immediately follows from $[K]$ being an equivalence class. ∎

######  Remark 5.17.

Generally, the bistellar exchange graph $\mathbb{G}_{[K]}$ may not be regular.

Now, to each vertex $L\in\mathbb{G}_{[K]}$ (that is, a triangulated manifold
in $[K]$), we can associate two pieces of data: a bistellar seed
$\Sigma_{L}=(\mathcal{X}(L),{\bf p}_{L},B(L))$ and a field
$\mathbb{F}\mathcal{X}(L)$. To each edge $\overline{LL^{\prime}}$ in
$\mathbb{G}_{[K]}$ (that is, there is a bistellar pair
$(\alpha,\beta)\in\mathcal{S}_{bp}^{[K]}$ such that
$(\alpha,\beta)\in\mathcal{S}_{bp}^{L}$ and
$(\beta,\alpha)\in\mathcal{S}_{bp}^{L^{\prime}}$, so
$L^{\prime}={\operatorname{{\bf bm}}}_{\alpha}L$ and $L={\operatorname{{\bf
bm}}}_{\beta}L^{\prime}$), we can associate the seed mutation
$\Phi_{\alpha}\colon\Sigma_{L}\to\Sigma_{L^{\prime}}$ and its inverse
$\Phi_{\beta}\colon\Sigma_{L^{\prime}}\to\Sigma_{L}$. We know from (5.5) that
the seed mutation $\Phi_{\alpha}$ determines a $\mathbb{ZP}$-linear field
isomorphism
$\mathcal{L}_{(L,L^{\prime})}\colon\mathbb{F}\mathcal{X}(L)\to\mathbb{F}\mathcal{X}(L^{\prime})$
given by

$\mathcal{L}_{(L,L^{\prime})}(x_{f})\,=\,\varphi^{L}_{\alpha}(x_{f})\,=\,\begin{cases}x_{f},&\text{if
$f\in\mathcal{F}(L)\setminus\mathcal{F}(\Lambda_{\alpha})$;}\\\
x_{\sigma(f)},&\text{if
$f\in\mathcal{F}(\Lambda_{\alpha})\setminus\mathcal{D}_{\alpha}$;}\\\
x_{\sigma(f)}=\frac{M^{+}_{L,\alpha,f}+M^{-}_{L,\alpha,f}}{x_{f}},&\text{if
$f\in\mathcal{D}_{\alpha}$,}\end{cases}$

and the isomorphism $\mathcal{L}_{(L^{\prime},L)}$ determined by
$\Phi_{\beta}$ is exactly $\mathcal{L}_{(L,L^{\prime})}^{-1}$. By Corollary
5.10, $\mathbb{F}\mathcal{X}(L)$ can be identified with
$\mathbb{F}\mathcal{X}(L^{\prime})$. Since $\mathbb{G}_{[K]}$ is connected,
all the fields $\mathbb{F}\mathcal{X}(L)$, for $L\in[K]$, can be identified
with each other, and then they can be regarded as a single field $\mathbb{F}$
such that:

  1. (1)

$\mathbb{F}$ contains all the elements in
$\mathcal{X}_{[K]}=\bigcup_{L\in[K]}\mathcal{X}(L)$;

  2. (2)

there are the exchange relations (5.3) in $\mathbb{F}$ for exchangeable
bistellar variables

$\mathcal{EX}_{[K]}\,=\,\left\\{x_{f}\mid
f\in\bigcup_{(\alpha,\beta)\in\mathcal{S}_{bp}^{[K]}}\mathcal{D}_{\alpha}\cup\mathcal{D}_{\beta}\right\\}.$

######  Definition 5.18.

The _bistellar cluster algebra_ $\mathcal{A}_{[K]}$ of $[K]$ is the subalgebra
of $\mathbb{F}$, generated by all cluster variables
$x_{f}\in\mathcal{X}_{[K]}$ with the exchange relations given by (5.3).

Definition 5.18 is still suitable for the case of $|[K]|=1$. In this case,
$\mathcal{A}_{[K]}$ is just a polynomial ring generated by all variables
$x_{f}\in\mathcal{X}(K)$.

######  Remark 5.19.

Since $\mathbb{G}_{[K]}$ is connected, all exchangeable bistellar variables
$x_{f}\in\mathcal{EX}_{[K]}$ can be reduced to being rational functions in
$\mathbb{F}\mathcal{X}(K)$ by using the exchange relations. So the bistellar
cluster algebra $\mathcal{A}_{[K]}$ can be regarded as a subalgebra of
$\mathbb{F}\mathcal{X}(K)$ with $\Sigma_{L}=(\mathcal{X}(K),{\bf p}_{K},B(K))$
as an initial bistellar seed.

######  Definition 5.20.

For each $L\in[K]$, $\mathcal{A}_{L}$ is defined as the subalgebra in
$\mathbb{F}\mathcal{X}(L)$ with $\Sigma_{L}=(\mathcal{X}(L),{\bf p}_{L},B(L))$
as an initial bistellar seed.

######  Theorem 5.21.

Let $K,K^{\prime}\in\mathbb{TM}_{2h}$ be two triangulated manifolds. If
$K^{\prime}$ is obtained from $K$ by a finite sequence of bistellar $h$-moves,
then $\mathcal{A}_{K}\cong\mathcal{A}_{K^{\prime}}$.

###### Proof.

This is because $[K]=[K^{\prime}]$. ∎

###  5.3. A Simple 2-dimensional Example

Let $K$ be the triangulation of the 2-sphere with five vertices in Example
5.13. So there are ten elements in $[K]$ and thirty bistellar pairs in
$\mathcal{S}_{bp}^{[K]}$, as shown in Example 5.13. First we see easily that

  1. (1)

The union $\bigcup_{L\in[K]}\mathcal{F}(L)$ is equal to
$\mathcal{F}_{1}^{2^{[5]}}$. So $\mathcal{X}_{[K]}$ consists of ten elements.
More precisely,

$\mathcal{X}_{[K]}\,=\,\\{x_{(1,2)},x_{(1,3)},x_{(1,4)},x_{(1,5)},x_{(2,3)},x_{(2,4)},x_{(2,5)},x_{(3,4)},x_{(3,5)},x_{(4,5)}\\}.$

  2. (2)

The set $\mathcal{EX}_{[K]}$ of all exchangeable bistellar variables is the
same as $\mathcal{X}_{[K]}$. So $\mathcal{A}_{[K]}$ is the subalgebra of
$\mathbb{F}$, generated by the ten elements in $\mathcal{X}_{[K]}$.

To determine the structure of $\mathcal{A}_{[K]}$, let us look at the exchange
relations between exchangeable bistellar variables. Since
$\mathcal{S}_{bp}^{[K]}$ contains thirty bistellar pairs, there are 15
exchange relations. These exchange relations with the relations among the
${\bf p}_{L}$, for $L\in[K]$, are stated as follows:

  1. (1)

There is the following exchange relation between $\mathcal{X}(K)$ and
$\mathcal{X}(K_{1})$.

$\displaystyle x_{(1,2)}x_{(4,5)}\,=\,$ $\displaystyle
p_{K,x_{(1,2)}}^{+}x_{(1,4)}x_{(2,5)}+p_{K,x_{(1,2)}}^{-}x_{(1,5)}x_{(2,4)}$
$\displaystyle=\,$ $\displaystyle
p_{K_{1},x_{(4,5)}}^{-}x_{(1,4)}x_{(2,5)}+p_{K_{1},x_{(4,5)}}^{+}x_{(1,5)}x_{(2,4)}$

with $p_{K,x_{(1,2)}}^{\pm}=p_{K_{1},x_{(4,5)}}^{\mp}$.

  2. (2)

There is the following exchange relation between $\mathcal{X}(K)$ and
$\mathcal{X}(K_{2})$.

$\displaystyle x_{(1,3)}x_{(4,5)}\,=\,$ $\displaystyle
p_{K,x_{(1,3)}}^{+}x_{(1,4)}x_{(3,5)}+p_{K,x_{(1,3)}}^{-}x_{(1,5)}x_{(3,4)}$
$\displaystyle=\,$ $\displaystyle
p_{K_{2},x_{(4,5)}}^{-}x_{(1,4)}x_{(3,5)}+p_{K_{2},x_{(4,5)}}^{+}x_{(1,5)}x_{(3,4)}$

with $p_{K,x_{(1,3)}}^{\pm}=p_{K_{2},x_{(4,5)}}^{\mp}$.

  3. (3)

There is the following exchange relation between $\mathcal{X}(K)$ and
$\mathcal{X}(K_{3})$.

$\displaystyle x_{(2,3)}x_{(4,5)}\,=\,$ $\displaystyle
p_{K,x_{(2,3)}}^{+}x_{(2,4)}x_{(3,5)}+p_{K,x_{(2,3)}}^{-}x_{(2,5)}x_{(3,4)}$
$\displaystyle=\,$ $\displaystyle
p_{K_{3},x_{(4,5)}}^{-}x_{(2,4)}x_{(3,5)}+p_{K_{3},x_{(4,5)}}^{+}x_{(2,5)}x_{(3,4)}$

with $p_{K,x_{(2,3)}}^{\pm}=p_{K_{3},x_{(4,5)}}^{\mp}$.

  4. (4)

There is the following exchange relation between $\mathcal{X}(K_{1})$ and
$\mathcal{X}(K_{11})$.

$\displaystyle x_{(1,2)}x_{(3,5)}\,=\,$ $\displaystyle
p_{K_{11},x_{(1,2)}}^{+}x_{(1,3)}x_{(2,5)}+p_{K_{11},x_{(1,2)}}^{-}x_{(1,5)}x_{(2,3)}$
$\displaystyle=\,$ $\displaystyle
p_{K_{1},x_{(3,5)}}^{-}x_{(1,3)}x_{(2,5)}+p_{K_{1},x_{(3,5)}}^{+}x_{(1,5)}x_{(2,3)}$

with $p_{K_{11},x_{(1,2)}}^{\pm}=p_{K_{1},x_{(3,5)}}^{\mp}$.

  5. (5)

There is the following exchange relation between $\mathcal{X}(K_{1})$ and
$\mathcal{X}(K_{12})$.

$\displaystyle x_{(1,2)}x_{(3,4)}\,=\,$ $\displaystyle
p_{K_{12},x_{(1,2)}}^{+}x_{(1,3)}x_{(2,4)}+p_{K_{12},x_{(1,2)}}^{-}x_{(1,4)}x_{(2,3)}$
$\displaystyle=\,$ $\displaystyle
p_{K_{1},x_{(3,4)}}^{-}x_{(1,3)}x_{(2,4)}+p_{K_{1},x_{(3,4)}}^{+}x_{(1,4)}x_{(2,3)}$

with $p_{K_{12},x_{(1,2)}}^{\pm}=p_{K_{1},x_{(3,4)}}^{\mp}$.

  6. (6)

There is the following exchange relation between $\mathcal{X}(K_{2})$ and
$\mathcal{X}(K_{21})$.

$\displaystyle x_{(1,3)}x_{(2,5)}\,=\,$ $\displaystyle
p_{K_{21},x_{(1,3)}}^{+}x_{(1,2)}x_{(3,5)}+p_{K_{21},x_{(1,3)}}^{-}x_{(1,5)}x_{(2,3)}$
$\displaystyle=\,$ $\displaystyle
p_{K_{2},x_{(2,5)}}^{-}x_{(1,2)}x_{(3,5)}+p_{K_{2},x_{(2,5)}}^{+}x_{(1,5)}x_{(2,3)}$

with $p_{K_{21},x_{(1,3)}}^{\pm}=p_{K_{2},x_{(2,5)}}^{\mp}$.

  7. (7)

There is the following exchange relation between $\mathcal{X}(K_{2})$ and
$\mathcal{X}(K_{22})$.

$\displaystyle x_{(1,3)}x_{(2,4)}\,=\,$ $\displaystyle
p_{K_{22},x_{(1,3)}}^{+}x_{(1,2)}x_{(3,4)}+p_{K_{22},x_{(1,3)}}^{-}x_{(1,4)}x_{(2,3)}$
$\displaystyle=\,$ $\displaystyle
p_{K_{2},x_{(2,4)}}^{-}x_{(1,2)}x_{(3,4)}+p_{K_{2},x_{(2,4)}}^{+}x_{(1,4)}x_{(2,3)}$

with $p_{K_{22},x_{(1,3)}}^{\pm}=p_{K_{2},x_{(2,4)}}^{\mp}$.

  8. (8)

There is the following exchange relation between $\mathcal{X}(K_{3})$ and
$\mathcal{X}(K_{31})$.

$\displaystyle x_{(1,4)}x_{(2,3)}\,=\,$ $\displaystyle
p_{K_{3},x_{(1,4)}}^{+}x_{(1,2)}x_{(3,4)}+p_{K_{3},x_{(1,4)}}^{-}x_{(1,3)}x_{(2,4)}$
$\displaystyle=\,$ $\displaystyle
p_{K_{31},x_{(2,3)}}^{-}x_{(1,2)}x_{(3,4)}+p_{K_{31},x_{(2,3)}}^{+}x_{(1,3)}x_{(2,4)}$

with $p_{K_{31},x_{(2,3)}}^{\pm}=p_{K_{3},x_{(1,4)}}^{\mp}$.

  9. (9)

There is the following exchange relation between $\mathcal{X}(K_{3})$ and
$\mathcal{X}(K_{32})$.

$\displaystyle x_{(1,5)}x_{(2,3)}\,=\,$ $\displaystyle
p_{K_{3},x_{(1,5)}}^{+}x_{(1,2)}x_{(3,5)}+p_{K_{3},x_{(1,5)}}^{-}x_{(1,3)}x_{(2,5)}$
$\displaystyle=\,$ $\displaystyle
p_{K_{32},x_{(2,3)}}^{-}x_{(1,2)}x_{(3,5)}+p_{K_{32},x_{(2,3)}}^{+}x_{(1,3)}x_{(2,5)}$

with $p_{K_{32},x_{(2,3)}}^{\pm}=p_{K_{3},x_{(1,5)}}^{\mp}$.

  10. (10)

There is the following exchange relation between $\mathcal{X}(K_{11})$ and
$\mathcal{X}(K_{31})$.

$\displaystyle x_{(1,4)}x_{(3,5)}\,=\,$ $\displaystyle
p_{K_{11},x_{(1,4)}}^{+}x_{(1,3)}x_{(4,5)}+p_{K_{11},x_{(1,4)}}^{-}x_{(1,5)}x_{(3,4)}$
$\displaystyle=\,$ $\displaystyle
p_{K_{31},x_{(3,5)}}^{-}x_{(1,3)}x_{(4,5)}+p_{K_{31},x_{(3,5)}}^{+}x_{(1,5)}x_{(3,4)}$

with $p_{K_{11},x_{(1,4)}}^{\pm}=p_{K_{31},x_{(3,5)}}^{\mp}$.

  11. (11)

There is the following exchange relation between $\mathcal{X}(K_{11})$ and
$\mathcal{X}(K_{22})$.

$\displaystyle x_{(2,4)}x_{(3,5)}\,=\,$ $\displaystyle
p_{K_{11},x_{(2,4)}}^{+}x_{(2,3)}x_{(4,5)}+p_{K_{11},x_{(2,4)}}^{-}x_{(2,5)}x_{(3,4)}$
$\displaystyle=\,$ $\displaystyle
p_{K_{22},x_{(3,5)}}^{-}x_{(2,3)}x_{(4,5)}+p_{K_{22},x_{(3,5)}}^{+}x_{(2,5)}x_{(3,4)}$

with $p_{K_{11},x_{(2,4)}}^{\pm}=p_{K_{22},x_{(3,5)}}^{\mp}$.

  12. (12)

There is the following exchange relation between $\mathcal{X}(K_{12})$ and
$\mathcal{X}(K_{21})$.

$\displaystyle x_{(2,5)}x_{(3,4)}\,=\,$ $\displaystyle
p_{K_{12},x_{(2,5)}}^{+}x_{(2,3)}x_{(4,5)}+p_{K_{12},x_{(2,5)}}^{-}x_{(2,4)}x_{(3,5)}$
$\displaystyle=\,$ $\displaystyle
p_{K_{21},x_{(3,4)}}^{-}x_{(2,3)}x_{(4,5)}+p_{K_{21},x_{(3,4)}}^{+}x_{(2,4)}x_{(3,5)}$

with $p_{K_{12},x_{(2,5)}}^{\pm}=p_{K_{21},x_{(3,4)}}^{\mp}$.

  13. (13)

There is the following exchange relation between $\mathcal{X}(K_{12})$ and
$\mathcal{X}(K_{32})$.

$\displaystyle x_{(1,5)}x_{(3,4)}\,=\,$ $\displaystyle
p_{K_{12},x_{(1,5)}}^{+}x_{(1,3)}x_{(4,5)}+p_{K_{12},x_{(1,5)}}^{-}x_{(1,4)}x_{(3,5)}$
$\displaystyle=\,$ $\displaystyle
p_{K_{32},x_{(3,4)}}^{-}x_{(1,3)}x_{(4,5)}+p_{K_{32},x_{(3,4)}}^{+}x_{(1,4)}x_{(3,5)}$

with $p_{K_{12},x_{(1,5)}}^{\pm}=p_{K_{32},x_{(3,4)}}^{\mp}$.

  14. (14)

There is the following exchange relation between $\mathcal{X}(K_{21})$ and
$\mathcal{X}(K_{31})$.

$\displaystyle x_{(1,4)}x_{(2,5)}\,=\,$ $\displaystyle
p_{K_{21},x_{(1,4)}}^{+}x_{(1,2)}x_{(4,5)}+p_{K_{21},x_{(1,4)}}^{-}x_{(1,5)}x_{(2,4)}$
$\displaystyle=\,$ $\displaystyle
p_{K_{31},x_{(2,5)}}^{-}x_{(1,2)}x_{(4,5)}+p_{K_{31},x_{(2,5)}}^{+}x_{(1,5)}x_{(2,4)}$

with $p_{K_{21},x_{(1,4)}}^{\pm}=p_{K_{31},x_{(2,5)}}^{\mp}$.

  15. (15)

There is the following exchange relation between $\mathcal{X}(K_{22})$ and
$\mathcal{X}(K_{32})$.

$\displaystyle x_{(1,5)}x_{(2,4)}\,=\,$ $\displaystyle
p_{K_{22},x_{(1,5)}}^{+}x_{(1,2)}x_{(4,5)}+p_{K_{22},x_{(1,5)}}^{-}x_{(1,4)}x_{(2,5)}$
$\displaystyle=\,$ $\displaystyle
p_{K_{32},x_{(2,4)}}^{-}x_{(1,2)}x_{(4,5)}+p_{K_{32},x_{(2,4)}}^{+}x_{(1,4)}x_{(2,5)}$

with $p_{K_{22},x_{(1,5)}}^{\pm}=p_{K_{32},x_{(2,4)}}^{\mp}$.

######  Lemma 5.22.

We have the following relations:

(5.6) $\displaystyle x_{(1,2)}x_{(4,5)}\,$
$\displaystyle=\,x_{(1,4)}x_{(2,5)}\,=\,x_{(1,5)}x_{(2,4)}$ (5.7)
$\displaystyle x_{(1,3)}x_{(4,5)}\,$
$\displaystyle=\,x_{(1,4)}x_{(3,5)}\,=\,x_{(1,5)}x_{(3,4)}$ (5.8)
$\displaystyle x_{(2,3)}x_{(4,5)}\,$
$\displaystyle=\,x_{(2,4)}x_{(3,5)}\,=\,x_{(2,5)}x_{(3,4)}$ (5.9)
$\displaystyle x_{(1,2)}x_{(3,5)}\,$
$\displaystyle=\,x_{(1,3)}x_{(2,5)}\,=\,x_{(1,5)}x_{(2,3)}$ (5.10)
$\displaystyle x_{(1,2)}x_{(3,4)}\,$
$\displaystyle=\,x_{(1,3)}x_{(2,4)}\,=\,x_{(1,4)}x_{(2,3)}.$

###### Proof.

We only prove that (5.6) holds since the proofs of other cases (5.7)–(5.10)
are similar. Let

$p_{K,x_{(1,2)}}^{+}/p_{K,x_{(1,2)}}^{-}\,=\,u,\quad
p_{K_{21},x_{(1,4)}}^{+}/p_{K_{21},x_{(1,4)}}^{-}\,=\,v\quad\text{and}\quad
p_{K_{22},x_{(1,5)}}^{+}/p_{K_{22},x_{(1,5)}}^{-}\,=\,w.$

Since $p_{K,x_{(1,2)}}^{+}\oplus p_{K,x_{(1,2)}}^{-}=1$, one has that
$p_{K,x_{(1,2)}}^{+}=\frac{u}{1\oplus u}$ and
$p_{K,x_{(1,2)}}^{-}=\frac{1}{1\oplus u}$. Similarly, one has that
$p_{K_{21},x_{(1,4)}}^{+}=\frac{v}{1\oplus v}$ and
$p_{K_{21},x_{(1,4)}}^{-}=\frac{1}{1\oplus v}$, and
$p_{K_{22},x_{(1,5)}}^{+}=\frac{w}{1\oplus w}$ and
$p_{K_{22},x_{(1,5)}}^{-}=\frac{1}{1\oplus w}$. Thus we rewrite (1), (14) and
(15) as follows:

(5.11) $\displaystyle x_{(1,2)}x_{(4,5)}\,$ $\displaystyle=\,\frac{u}{1\oplus
u}x_{(1,4)}x_{(2,5)}+\frac{1}{1\oplus u}x_{(1,5)}x_{(2,4)}$ (5.12)
$\displaystyle x_{(1,4)}x_{(2,5)}\,$ $\displaystyle=\,\frac{v}{1\oplus
v}x_{(1,2)}x_{(4,5)}+\frac{1}{1\oplus v}x_{(1,5)}x_{(2,4)}$ (5.13)
$\displaystyle x_{(1,5)}x_{(2,4)}\,$ $\displaystyle=\,\frac{w}{1\oplus
w}x_{(1,2)}x_{(4,5)}+\frac{1}{1\oplus w}x_{(1,4)}x_{(2,5)}.$

From (5.11) and (5.12) we have that

$\displaystyle x_{(1,2)}x_{(4,5)}\,=$ $\displaystyle\,\frac{u}{1\oplus
u}\cdot\frac{v}{1\oplus v}x_{(1,2)}x_{(4,5)}+\left(\frac{u}{1\oplus
u}\cdot\frac{1}{1\oplus v}\oplus\frac{1}{1\oplus u}\right)x_{(1,5)}x_{(2,4)}$
$\displaystyle=$ $\displaystyle\,\frac{uv}{(1\oplus u)(1\oplus
v)}x_{(1,2)}x_{(4,5)}+\frac{1\oplus u\oplus v}{(1\oplus u)(1\oplus
v)}x_{(1,5)}x_{(2,4)}$

so $(1\oplus u\oplus v)x_{(1,2)}x_{(4,5)}=(1\oplus u\oplus
v)x_{(1,5)}x_{(2,4)}$. Thus, $x_{(1,2)}x_{(4,5)}=x_{(1,5)}x_{(2,4)}$. In a
similar way as above, one can obtain from (5.11) and (5.13) that
$x_{(1,2)}x_{(4,5)}=x_{(1,4)}x_{(2,5)}$. This shows that (5.6) holds. ∎

######  Remark 5.23.

The relation (5.7) is determined by (2), (10) and (13); (5.8) is determined by
(3), (11) and (12); (5.9) is determined by (4), (6) and (9); (5.10) is
determined by (5), (7) and (8). In addition, it is easy to see that (5.9) and
(5.10) can be induced from (5.6)–(5.8).

######  Proposition 5.24.

As a subalgebra of $\mathbb{F}$, $\mathcal{A}_{[K]}$ is the quotient algebra
of the polynomial algebra

$\mathbb{ZP}[x\mid
x\in\mathcal{X}_{[K]}]\,=\,\mathbb{ZP}[x_{(1,2)},x_{(1,3)},x_{(1,4)},x_{(1,5)},x_{(2,3)},x_{(2,4)},x_{(2,5)},x_{(3,4)},x_{(3,5)},x_{(4,5)}]$

as follows:

$\mathcal{A}_{[K]}\,=\,\mathbb{ZP}[x\mid x\in\mathcal{X}_{[K]}]/\mathcal{I},$

where $\mathcal{I}$ is the ideal determined by those relations (5.6)–(5.8).

As a subalgebra in $\mathbb{F}\mathcal{X}(K)$ with $(\mathcal{X}(K),{\bf
p}_{K},B(K))$ as an initial bistellar seed,
$\mathcal{A}_{K}\cong\mathcal{A}_{[K]}$ is generated by
$x_{(1,2)},x_{(1,3)},x_{(1,4)},x_{(1,5)},x_{(2,3)},x_{(2,4)},x_{(2,5)},x_{(3,4)},x_{(3,5)}$
and

$\frac{x_{(1,4)}x_{(2,5)}}{x_{(1,2)}}\,=\,\frac{x_{(1,5)}x_{(2,4)}}{x_{(1,2)}}\,=\,\frac{x_{(1,4)}x_{(3,5)}}{x_{(1,3)}}\,=\,\frac{x_{(1,5)}x_{(3,4)}}{x_{(1,3)}}\,=\,\frac{x_{(2,4)}x_{(3,5)}}{x_{(2,3)}}\,=\,\frac{x_{(2,5)}x_{(3,4)}}{x_{(2,3)}}.$

##  6\. Algebras as PL Invariants

In this section, our purpose is to construct PL invariants by using bistellar
cluster algebras. We shall work in the setting of PL manifolds. Let
$\mathbb{PL}_{n}$ be the set consisting of all $n$-dimensional PL manifolds in
$\mathbb{TM}_{n}$. We shall mainly be concerned with the case in which $n$ is
even, especially for $n=4$.

Write $n=2h$ and take a PL manifold $K\in\mathbb{PL}_{n}$. Let
$\mathfrak{S}_{K}$ be the set consisting of all PL manifolds that are PL
homeomorphic to $K$. Then we can write

$\mathfrak{S}_{K}\,=\,\bigcup_{m\geq m_{0}}\mathfrak{S}_{K}^{m},$

where $m_{0}$ is the minimal number of vertex numbers of all PL manifolds in
$\mathfrak{S}_{K}$, and $\mathfrak{S}_{K}^{m}$ consists of those PL manifolds
with $m$ vertices in $\mathfrak{S}_{K}$.

Consider the equivalence relation $\sim$ on $\mathfrak{S}_{K}$ as follows: for
$K_{1},K_{2}\in\mathfrak{S}_{K}$, $K_{1}\sim K_{2}$ if and only if one of
$K_{1}$ and $K_{2}$ can be transformed into the other by only doing a finite
sequence of bistellar $h$-moves (including the empty sequence). By
$\widetilde{\mathfrak{S}}_{K}$ we denote the quotient set
$\mathfrak{S}_{K}/\sim$. Since the bistellar $h$-moves do not change the
number of vertices, the equivalence relation $\sim$ is graded on
$\mathfrak{S}_{K}=\bigcup_{m\geq m_{0}}\mathfrak{S}_{K}^{m}$, so

$\widetilde{\mathfrak{S}}_{K}\,=\,\bigcup_{m\geq
m_{0}}\widetilde{\mathfrak{S}}_{K}^{m}.$

For a PL manifold $L$ in $\mathfrak{S}_{K}$, the equivalence class of $L$ in
$\widetilde{\mathfrak{S}}_{K}$ is denoted by $[L]$.

We define a binary relation $\preceq$ on $\widetilde{\mathfrak{S}}_{K}$ as
follows: for two elements $[K_{1}],[K_{2}]\in\widetilde{\mathfrak{S}}_{K}$ we
say

$\displaystyle[K_{1}]\preceq[K_{2}]\quad\Longleftrightarrow\quad$ $K_{2}$ is
obtained from $K_{1}$ by doing a finite sequence of bistellar $\displaystyle\
\text{ moves ${\operatorname{{\bf bm}}}_{\alpha_{1}},\dots,{\operatorname{{\bf
bm}}}_{\alpha_{l}}$, where $\dim\alpha_{i}\geq h$}.$

Note that for $[K^{\prime}]\in\widetilde{\mathfrak{S}}_{K}$, if
$|[K^{\prime}]|=1$, then we use the convention that
$[K^{\prime}]\preceq[K^{\prime}]$.

######  Lemma 6.1.

When $h=1,2$, $(\widetilde{\mathfrak{S}}_{K},\preceq)$ forms a directed set.

###### Proof.

When $h=1$, let $[L]\in\widetilde{\mathfrak{S}}_{K}$. Clearly,
$[L]\preceq[L]$. It is well-known that for any $m\geq m_{0}$ and any two
$L_{1},L_{2}\in\mathfrak{S}_{K}^{m}$, that $L_{1}$ can be obtained from
$L_{2}$ by performing a finite sequence of bistellar 1-moves (this can also be
seen by simplifying the case for $h=2$ below). So
$\widetilde{\mathfrak{S}}^{m}_{K}$ contains exactly one element. It then
follows easily from this that $(\widetilde{\mathfrak{S}}_{K},\preceq)$ forms a
directed set.

When $h=2$, let $[L]\in\widetilde{\mathfrak{S}}_{K}$. First, clearly one has
that $[L]\preceq[L]$. Let
$[L_{1}],[L_{2}],[L_{3}]\in\widetilde{\mathfrak{S}}_{K}$ be such that
$[L_{1}]\preceq[L_{2}]$ and $[L_{2}]\preceq[L_{3}]$. Then $L_{2}$ is obtained
from $L_{1}$ by performing finitely many bistellar $0$-moves, $1$-moves or
$2$-moves, and $L_{3}$ is obtained from $L_{2}$ similarly. So $L_{3}$ is
obtained from $L_{1}$ by performing finitely many bistellar $0$-moves,
$1$-moves or $2$-moves. Thus, $[L_{1}]\preceq[L_{3}]$. It remains to show that
for any two $[L_{1}],[L_{2}]\in\widetilde{\mathfrak{S}}_{K}$, there is an
$[L]\in\widetilde{\mathfrak{S}}_{K}$ such that $[L_{1}]\preceq[L]$ and
$[L_{2}]\preceq[L]$.

Clearly if either $L_{1}$ is obtained from $L_{2}$ by performing finitely many
bistellar $0$-moves, $1$-moves or $2$-moves or $L_{2}$ is obtained from
$L_{1}$ in a similar fashion, then we can choose $[L]$ as $[L_{1}]$ or
$[L_{2}]$ such that $[L_{1}]\preceq[L]$ and $[L_{2}]\preceq[L]$. Generally, by
Pachner’s Theorem [12], there always exists a finite sequence of bistellar
pairs $(\alpha_{1},\beta_{1}),\dots,(\alpha_{l},\beta_{l})$ such that

(6.1) ${\operatorname{{\bf bm}}}_{\alpha_{l}}\cdots{\operatorname{{\bf
bm}}}_{\alpha_{1}}L_{1}\,=\,L_{2},$

where $0\leq\dim\alpha_{i}\leq 4$. Such a sequence
$(\alpha_{1},\beta_{1}),\dots,(\alpha_{l},\beta_{l})$ is not unique in
general. To complete the proof, it suffices to show that there exists such a
sequence $(\alpha_{1},\beta_{1}),\dots,(\alpha_{l},\beta_{l})$, satisfying
(6.1) and the property that there is some $1\leq u\leq l$ such that for $1\leq
i\leq u$, $\dim\alpha_{i}\geq 2$ and for $u<i\leq l$, $\dim\alpha_{i}<2$, so
that we can obtain $L={\operatorname{{\bf
bm}}}_{\alpha_{u}}\cdots{\operatorname{{\bf
bm}}}_{\alpha_{1}}L_{1}={\operatorname{{\bf
bm}}}_{\beta_{u+1}}\cdots{\operatorname{{\bf bm}}}_{\beta_{l}}L_{2}$ as
desired.

For this we let $L\in[K]$, and then it suffices to show that for any two
bistellar pairs $(\alpha,\beta)$ with $\dim\alpha<2$ and
$(\alpha^{\prime},\beta^{\prime})$ with $\dim\alpha^{\prime}\geq 2$, if we can
perform the operation ${\operatorname{{\bf
bm}}}_{\alpha^{\prime}}{\operatorname{{\bf bm}}}_{\alpha}$ on $L$, then
${\operatorname{{\bf bm}}}_{\alpha^{\prime}}{\operatorname{{\bf
bm}}}_{\alpha}L$ can be written as ${\operatorname{{\bf
bm}}}_{\alpha_{l}}\cdots{\operatorname{{\bf
bm}}}_{\alpha_{u}}\cdots{\operatorname{{\bf bm}}}_{\alpha_{1}}L$ such that for
$1\leq i\leq u$, $\dim\alpha_{i}\geq 2$ and for $u<i\leq l$,
$\dim\alpha_{i}<2$. We divide this into two cases: (I) $\dim\alpha=0$; (II)
$\dim\alpha=1$.

  1. (I)

Without loss of generality, we assume that $\alpha=(1)$ and
$\beta=(2,3,4,5,6)$, so $\alpha\ast\partial\beta\subset L$. Then
$\beta=\partial\alpha\ast\beta\subset{\bf bm}_{\alpha}L$.

If $\dim\alpha^{\prime}=4$ and $\alpha^{\prime}=\beta$ (i.e.,
$|\alpha^{\prime}\cap\beta|=5$), then ${\bf bm}_{\alpha^{\prime}}{\bf
bm}_{\alpha}L=L$. In this case, first we do a bistellar 0-move at
$(1,2,3,4,5)$ on $L$, so that $(1,2,3,4,5)$ becomes a missing face of ${\bf
bm}_{(1,2,3,4,5)}L$, and let $(6^{\prime})\in{\bf bm}_{(1,2,3,4,5)}L$ such
that $\operatorname{Link}_{L}(1,2,3,4,5)=\partial(6^{\prime})$. Then we do the
inverse operation of ${\bf bm}_{(1,2,3,4,5)}$, so that

${\bf bm}_{\alpha^{\prime}}{\bf bm}_{\alpha}L\,=\,{\bf bm}_{(2,3,4,5,6)}{\bf
bm}_{(1)}L\,=\,L\,=\,{\bf bm}_{(6^{\prime})}{\bf bm}_{(1,2,3,4,5)}L.$

If $\dim\alpha^{\prime}=4$ and $|\alpha^{\prime}\cap\beta|\leq 4$, without
loss of generality, we may assume that $\alpha^{\prime}=(2,3,4,5,6^{\prime})$
or $(2,3,4,5^{\prime},6^{\prime})$ or $(2,3,4^{\prime},5^{\prime},6^{\prime})$
or $(2,3^{\prime},4^{\prime},5^{\prime},6^{\prime})$ or
$(2^{\prime},3^{\prime},4^{\prime},5^{\prime},6^{\prime})$. Clearly,
$\alpha^{\prime}$ is a simplex of $L$ so
$\operatorname{Link}_{L}\alpha^{\prime}$ is the boundary of a ghost vertex
$\beta^{\prime}=v$ (i.e., $\beta^{\prime}\not\in L$). Thus $\alpha\in{\bf
bm}_{\alpha^{\prime}}L$ and $\alpha^{\prime}\in{\bf bm}_{\alpha}L$. It is easy
to check that

${\bf bm}_{\alpha^{\prime}}{\bf bm}_{\alpha}L\,=\,{\bf bm}_{\alpha}{\bf
bm}_{\alpha^{\prime}}L$

as required.

If $\dim\alpha^{\prime}=3$ and $|\alpha^{\prime}\cap\beta|=4$, without loss of
generality, assume that $\alpha^{\prime}=(2,3,4,5)$, then we can write
$\beta^{\prime}=(6,6^{\prime})\not\in{\bf bm}_{\alpha}L$. We easily check that
${\bf bm}_{\alpha^{\prime}}{\bf bm}_{\alpha}L$ must contain the following four
4-simplices $(2,3,4,6,6^{\prime})$, $(2,3,5,6,6^{\prime})$,
$(2,4,5,6,6^{\prime})$ and $(3,4,5,6,6^{\prime})$. Now let us perform the
following bistellar moves on $L$.

    1. (a)

Firstly we see that $(1,6^{\prime})\not\in L$ and
$\operatorname{Link}_{L}(2,3,4,5)=\partial(1,6^{\prime})$, so we can do a
bistellar 1-move at $(2,3,4,5)$ on $L$.

    2. (b)

Secondly we have that $(6,6^{\prime})\not\in{\bf bm}_{(2,3,4,5)}L$ and
$\operatorname{Link}_{{\bf bm}_{(2,3,4,5)}L}(1,2,3,4)=\partial(6,6^{\prime})$,
so we can do a bistellar 1-move at $(1,2,3,4)$ on ${\bf bm}_{(2,3,4,5)}L$, and
in particular, we have $(2,3,4,6,6^{\prime})\in{\bf bm}_{(1,2,3,4)}{\bf
bm}_{(2,3,4,5)}L$.

    3. (c)

Thirdly we have that $(5,6,6^{\prime})\not\in{\bf bm}_{(1,2,3,4)}{\bf
bm}_{(2,3,4,5)}L$ and

$\operatorname{Link}_{{\bf bm}_{(1,2,3,4)}{\bf
bm}_{(2,3,4,5)}L}(1,3,4)\,=\,\partial(5,6,6^{\prime}).$

Then we perform a bistellar 2-move at $(1,3,4)$ on ${\bf bm}_{(1,2,3,4)}{\bf
bm}_{(2,3,4,5)}L$, so that $(3,4,5,6,6^{\prime})\in{\bf bm}_{(1,3,4)}{\bf
bm}_{(1,2,3,4)}{\bf bm}_{(2,3,4,5)}L$.

    4. (d)

Next it is easy to see that $(2,5,6,6^{\prime})\not\in{\bf bm}_{(1,3,4)}{\bf
bm}_{(1,2,3,4)}{\bf bm}_{(2,3,4,5)}L$ and

$\operatorname{Link}_{{\bf bm}_{(1,3,4)}{\bf bm}_{(1,2,3,4)}{\bf
bm}_{(2,3,4,5)}L}(1,4)\,=\,\partial(2,5,6,6^{\prime}).$

After we perform a bistellar 3-move at $(1,4)$ on ${\bf bm}_{(1,3,4)}{\bf
bm}_{(1,2,3,4)}{\bf bm}_{(2,3,4,5)}L$, we have

$(2,4,5,6,6^{\prime})\,\in\,{\bf bm}_{(1,4)}{\bf bm}_{(1,3,4)}{\bf
bm}_{(1,2,3,4)}{\bf bm}_{(2,3,4,5)}L.$

    5. (e)

Finally we can check that

$\operatorname{Link}_{{\bf bm}_{(1,4)}{\bf bm}_{(1,3,4)}{\bf
bm}_{(1,2,3,4)}{\bf bm}_{(2,3,4,5)}L}(1)\,=\,\partial(2,3,5,6,6^{\prime})$

but $(2,3,5,6,6^{\prime})\not\in{\bf bm}_{(1,4)}{\bf bm}_{(1,3,4)}{\bf
bm}_{(1,2,3,4)}{\bf bm}_{(2,3,4,5)}L$. After performing a bistellar 4-move at
$(1)$ on ${\bf bm}_{(1,4)}{\bf bm}_{(1,3,4)}{\bf bm}_{(1,2,3,4)}{\bf
bm}_{(2,3,4,5)}L$, we obtain that

$(2,3,5,6,6^{\prime})\in{\bf bm}_{(1)}{\bf bm}_{(1,4)}{\bf bm}_{(1,3,4)}{\bf
bm}_{(1,2,3,4)}{\bf bm}_{(2,3,4,5)}L.$

Moreover, an easy argument shows

${\bf bm}_{\alpha^{\prime}}{\bf bm}_{\alpha}L\,=\,{\bf bm}_{(1)}{\bf
bm}_{(1,4)}{\bf bm}_{(1,3,4)}{\bf bm}_{(1,2,3,4)}{\bf bm}_{(2,3,4,5)}L$

as required.

If $\dim\alpha^{\prime}=3$ and $|\alpha^{\prime}\cap\beta|\leq 3$, without
loss of generality, we may assume that $\alpha^{\prime}=(2,3,4,5^{\prime})$ or
$(2,3,4^{\prime},5^{\prime})$ or $(2,3^{\prime},4^{\prime},5^{\prime})$ or
$(2^{\prime},3^{\prime},4^{\prime},5^{\prime})$. Similarly to the argument of
the case $\dim\alpha^{\prime}=4$ and $|\alpha^{\prime}\cap\beta|\leq 4$, we
have the following equation

${\bf bm}_{\alpha^{\prime}}{\bf bm}_{\alpha}L\,=\,{\bf bm}_{\alpha}{\bf
bm}_{\alpha^{\prime}}L$

as required.

As in the above case, if $\dim\alpha^{\prime}=2$ and
$|\alpha^{\prime}\cap\beta|\leq 2$, then we easily see that

${\bf bm}_{\alpha^{\prime}}{\bf bm}_{\alpha}L\,=\,{\bf bm}_{\alpha}{\bf
bm}_{\alpha^{\prime}}L.$

If $\dim\alpha^{\prime}=2$ and $|\alpha^{\prime}\cap\beta|=3$, we may assume
that $\alpha^{\prime}=(2,3,4)$. We see that
$\operatorname{Link}_{L}\alpha^{\prime}$ is a square formed by four edges
$(1,5)$, $(1,6)$, $(5,6^{\prime})$ and $(6,6^{\prime})$. This implies that
$\beta^{\prime}=(5,6,6^{\prime})$, which is a 2-simplex of ${\bf
bm}_{\alpha^{\prime}}{\bf bm}_{\alpha}L$ but not a 2-simplex of $L$ and ${\bf
bm}_{\alpha}L$. A direct check gives that ${\bf bm}_{\alpha^{\prime}}{\bf
bm}_{\alpha}L$ must contain the following three 4-simplices
$(2,3,5,6,6^{\prime})$, $(2,4,5,6,6^{\prime})$ and $(3,4,5,6,6^{\prime})$. Now
for our purpose, we perform the following bistellar moves on $L$:

    1. (a)

Firstly we perform a bistellar 1-move at $(2,3,4,5)$ on $L$ since
$(1,6^{\prime})\not\in L$ and
$\operatorname{Link}_{L}(2,3,4,5)=\partial(1,6^{\prime})$.

    2. (b)

Secondly we can perform a bistellar 2-move at $(2,3,4)$ on ${\bf
bm}_{(2,3,4,5)}L$. This is because $(1,6,6^{\prime})\not\in{\bf
bm}_{(2,3,4,5)}L$ and $\operatorname{Link}_{{\bf
bm}_{(2,3,4,5)}L}(2,3,4)=\partial(1,6,6^{\prime})$.

    3. (c)

Thirdly we see that $(5,6,6^{\prime})\not\in{\bf bm}_{(2,3,4)}{\bf
bm}_{(2,3,4,5)}L$ and

$\operatorname{Link}_{{\bf bm}_{(2,3,4)}{\bf
bm}_{(2,3,4,5)}L}(1,3,4)\,=\,\partial(5,6,6^{\prime}).$

Thus we can perform a bistellar 2-move at $(1,3,4)$ on ${\bf bm}_{(2,3,4)}{\bf
bm}_{(2,3,4,5)}L$, so that $(3,4,5,6,6^{\prime})\in{\bf bm}_{(1,3,4)}{\bf
bm}_{(2,3,4)}{\bf bm}_{(2,3,4,5)}L$.

    4. (d)

Next we can check that $(2,5,6,6^{\prime})\not\in{\bf bm}_{(1,3,4)}{\bf
bm}_{(2,3,4)}{\bf bm}_{(2,3,4,5)}L$ and

$\operatorname{Link}_{{\bf bm}_{(1,3,4)}{\bf bm}_{(2,3,4)}{\bf
bm}_{(2,3,4,5)}L}(1,4)\,=\,\partial(2,5,6,6^{\prime}).$

After we perform a bistellar 3-move at $(1,4)$ on ${\bf bm}_{(1,3,4)}{\bf
bm}_{(2,3,4)}{\bf bm}_{(2,3,4,5)}L$, we have

$(2,4,5,6,6^{\prime})\,\in\,{\bf bm}_{(1,4)}{\bf bm}_{(1,3,4)}{\bf
bm}_{(2,3,4)}{\bf bm}_{(2,3,4,5)}L.$

    5. (e)

Finally we can see that $\operatorname{Link}_{{\bf bm}_{(1,4)}{\bf
bm}_{(1,3,4)}{\bf bm}_{(2,3,4)}{\bf
bm}_{(2,3,4,5)}L}(1)=\partial(2,3,5,6,6^{\prime})$ but

$(2,3,5,6,6^{\prime})\not\in{\bf bm}_{(1,4)}{\bf bm}_{(1,3,4)}{\bf
bm}_{(2,3,4)}{\bf bm}_{(2,3,4,5)}L.$

Thus, after performing a bistellar 4-move at $(1)$ on

${\bf bm}_{(1,4)}{\bf bm}_{(1,3,4)}{\bf bm}_{(2,3,4)}{\bf bm}_{(2,3,4,5)}L,$

we conclude that

$(2,3,5,6,6^{\prime})\,\in\,{\bf bm}_{(1)}{\bf bm}_{(1,4)}{\bf
bm}_{(1,3,4)}{\bf bm}_{(2,3,4)}{\bf bm}_{(2,3,4,5)}L.$

Together with the above arguments, we may obtain the required equation

${\bf bm}_{\alpha^{\prime}}{\bf bm}_{\alpha}L\,=\,{\bf bm}_{(1)}{\bf
bm}_{(1,4)}{\bf bm}_{(1,3,4)}{\bf bm}_{(2,3,4)}{\bf bm}_{(2,3,4,5)}.L$

  2. (II)

Without loss of generality, we assume that $\alpha=(1,2)$ and
$\beta=(3,4,5,6)$.

If $\dim\alpha^{\prime}=4$ and $|\alpha^{\prime}\cap(\alpha\cup\beta)|=5$,
then $\alpha^{\prime}\in{\bf H}_{\beta}=\\{(1,3,4,5,6),(2,3,4,5,6)\\}$.
Without loss of generality, we may assume that $\alpha^{\prime}=(1,2,3,4,5)$.
Let $\beta^{\prime}=(7)$, which is a ghost vertex of $L$ and ${\bf
bm}_{\alpha}L$. Then ${\bf bm}_{\alpha^{\prime}}{\bf bm}_{\alpha}L$ must
contain the following six 4-simplices $(2,3,4,5,6)$, $(3,4,5,6,7)$,
$(1,3,4,5,7)$, $(1,3,4,6,7)$, $(1,3,5,6,7)$ and $(1,4,5,6,7)$. In a similar
way to case (I), we perform the following bistellar moves on $L$:

    1. (a)

Firstly we perform a bistellar $0$-move at $(1,2,3,4,5)$ on $L$ since
$\operatorname{Link}_{L}(1,2,3,4,5)=\partial(7)$. Then we easily see that
$(1,3,4,5,7)\in{\bf bm}_{(1,2,3,4,5)}L$.

    2. (b)

Secondly we can perform a bistellar 1-move at $(1,2,3,4)$ on ${\bf
bm}_{(1,2,3,4,5)}L$ since $(6,7)\not\in{\bf bm}_{(1,2,3,4,5)}L$ and
$\operatorname{Link}_{{\bf bm}_{(1,2,3,4,5)}L}(1,2,3,4)=\partial(6,7)$, so
that we can obtain that

$(1,3,4,6,7)\,\in\,{\bf bm}_{(1,2,3,4)}{\bf bm}_{(1,2,3,4,5)}L.$

    3. (c)

Thirdly we see that $(5,6,7)\not\in{\bf bm}_{(1,2,3,4)}{\bf
bm}_{(1,2,3,4,5)}L$ and

$\operatorname{Link}_{{\bf bm}_{(1,2,3,4)}{\bf
bm}_{(1,2,3,4,5)}L}(1,2,3)\,=\,\partial(5,6,7).$

Thus we can perform a bistellar 2-move at $(1,2,3)$ on ${\bf
bm}_{(1,2,3,4)}{\bf bm}_{(1,2,3,4,5)}L$, so that $(1,3,5,6,7)\in{\bf
bm}_{(1,2,3)}{\bf bm}_{(1,2,3,4)}{\bf bm}_{(1,2,3,4,5)}L$.

    4. (d)

Next we can check that $(4,5,6,7)\not\in{\bf bm}_{(1,2,3)}{\bf
bm}_{(1,2,3,4)}{\bf bm}_{(1,2,3,4,5)}L$ and

$\operatorname{Link}_{{\bf bm}_{(1,2,3)}{\bf bm}_{(1,2,3,4)}{\bf
bm}_{(1,2,3,4,5)}L}(1,2)\,=\,\partial(4,5,6,7),$

we can perform a bistellar 3-move at $(1,2)$ on ${\bf bm}_{(1,2,3)}{\bf
bm}_{(1,2,3,4)}{\bf bm}_{(1,2,3,4,5)}L$, and then $(1,4,5,6,7)\in{\bf
bm}_{(1,2)}{\bf bm}_{(1,2,3)}{\bf bm}_{(1,2,3,4)}{\bf bm}_{(1,2,3,4,5)}L$.

    5. (e)

Finally we can see that $\operatorname{Link}_{{\bf bm}_{(1,2)}{\bf
bm}_{(1,2,3)}{\bf bm}_{(1,2,3,4)}{\bf
bm}_{(1,2,3,4,5)}L}(2,7)=\partial(3,4,5,6)$ but $(3,4,5,6)\not\in{\bf
bm}_{(1,2)}{\bf bm}_{(1,2,3)}{\bf bm}_{(1,2,3,4)}{\bf bm}_{(1,2,3,4,5)}L$.
Thus, after performing a bistellar 3-move at $(2,7)$ on ${\bf bm}_{(1,2)}{\bf
bm}_{(1,2,3)}{\bf bm}_{(1,2,3,4)}{\bf bm}_{(1,2,3,4,5)}L$, we conclude that

$(2,3,4,5,6),(3,4,5,6,7)\,\in\,{\bf bm}_{(2,7)}{\bf bm}_{(1,2)}{\bf
bm}_{(1,2,3)}{\bf bm}_{(1,2,3,4)}{\bf bm}_{(1,2,3,4,5)}L.$

The above arguments give

${\bf bm}_{\alpha^{\prime}}{\bf bm}_{\alpha}L\,=\,{\bf bm}_{(2,7)}{\bf
bm}_{(1,2)}{\bf bm}_{(1,2,3)}{\bf bm}_{(1,2,3,4)}{\bf bm}_{(1,2,3,4,5)}L.$

If $\dim\alpha^{\prime}=4$ and $|\alpha^{\prime}\cap(\alpha\cup\beta)|\leq 4$,
then an easy observation shows that

${\bf bm}_{\alpha^{\prime}}{\bf bm}_{\alpha}L\,=\,{\bf bm}_{\alpha}{\bf
bm}_{\alpha^{\prime}}L.$

Actually, we can also prove easily that if $\dim\alpha^{\prime}=i$ and
$|\alpha^{\prime}\cap(\alpha\cup\beta)|\leq i$ for $i=2,3$, then

${\bf bm}_{\alpha^{\prime}}{\bf bm}_{\alpha}L\,=\,{\bf bm}_{\alpha}{\bf
bm}_{\alpha^{\prime}}L.$

Thus it remains to show the case in which $\dim\alpha^{\prime}=i$ and
$|\alpha^{\prime}\cap(\alpha\cup\beta)|=i+1$ for $i=2,3$.

When $i=2$, without a loss, we may assume that $\alpha^{\prime}=(1,3,4)$. Then
$\beta^{\prime}=(5,6,6^{\prime})$, and further ${\bf bm}_{\alpha^{\prime}}{\bf
bm}_{\alpha}L$ must contain the following three $4$-simplices
$(1,3,5,6,6^{\prime})$, $(1,4,5,6,6^{\prime})$ and $(3,4,5,6,6^{\prime})$.
Next we perform the following bistellar moves on $L$:

    1. (a)

Firstly we perform a bistellar $1$-move at $(1,3,4,5)$ on $L$ since
$\operatorname{Link}_{L}(1,3,4,5)=\partial(2,6^{\prime})$.

    2. (b)

Secondly we perform a bistellar 2-move at $(1,3,4)$ on ${\bf bm}_{(1,3,4,5)}L$
since $(2,6,6^{\prime})\not\in{\bf bm}_{(1,3,4,5)}L$ and
$\operatorname{Link}_{{\bf bm}_{(1,3,4,5)}L}(1,3,4)=\partial(2,6,6^{\prime})$.

    3. (c)

Thirdly we see that $(5,6,6^{\prime})\not\in{\bf bm}_{(1,3,4)}{\bf
bm}_{(1,3,4,5)}L$ and

$\operatorname{Link}_{{\bf bm}_{(1,3,4)}{\bf
bm}_{(1,3,4,5)}L}(1,2,3)\,=\,\partial(5,6,6^{\prime}).$

Thus we can perform a bistellar 2-move at $(1,2,3)$ on ${\bf bm}_{(1,3,4)}{\bf
bm}_{(1,3,4,5)}L$, so that $(1,3,5,6,6^{\prime})\in{\bf bm}_{(1,2,3)}{\bf
bm}_{(1,3,4)}{\bf bm}_{(1,3,4,5)}L$.

    4. (d)

Next we can check that $(4,5,6,6^{\prime})\not\in{\bf bm}_{(1,2,3)}{\bf
bm}_{(1,3,4)}{\bf bm}_{(1,3,4,5)}L$ and

$\operatorname{Link}_{{\bf bm}_{(1,2,3)}{\bf bm}_{(1,3,4)}{\bf
bm}_{(1,3,4,5)}L}(1,2)\,=\,\partial(4,5,6,6^{\prime}),$

we can perform a bistellar 3-move at $(1,2)$ on ${\bf bm}_{(1,2,3)}{\bf
bm}_{(1,3,4)}{\bf bm}_{(1,3,4,5)}L$, and then

$(1,4,5,6,6^{\prime})\in{\bf bm}_{(1,2)}{\bf bm}_{(1,2,3)}{\bf
bm}_{(1,3,4)}{\bf bm}_{(1,3,4,5)}L.$

    5. (e)

Finally we can see that $\operatorname{Link}_{{\bf bm}_{(1,2)}{\bf
bm}_{(1,2,3)}{\bf bm}_{(1,3,4)}{\bf
bm}_{(1,3,4,5)}L}(2,6^{\prime})=\partial(3,4,5,6)$ but $(3,4,5,6)\not\in{\bf
bm}_{(1,2)}{\bf bm}_{(1,2,3)}{\bf bm}_{(1,3,4)}{\bf bm}_{(1,3,4,5)}L$. Thus,
after performing a bistellar 3-move at $(2,6^{\prime})$ on ${\bf
bm}_{(1,2)}{\bf bm}_{(1,2,3)}{\bf bm}_{(1,3,4)}{\bf bm}_{(1,3,4,5)}L$, we
conclude that

$(3,4,5,6,6^{\prime})\,\in\,{\bf bm}_{(2,6^{\prime})}{\bf bm}_{(1,2)}{\bf
bm}_{(1,2,3)}{\bf bm}_{(1,3,4)}{\bf bm}_{(1,3,4,5)}L.$

The above arguments give the following equality

${\bf bm}_{\alpha^{\prime}}{\bf bm}_{\alpha}L\,=\,{\bf
bm}_{(2,6^{\prime})}{\bf bm}_{(1,2)}{\bf bm}_{(1,2,3)}{\bf bm}_{(1,3,4)}{\bf
bm}_{(1,3,4,5)}L.$

When $i=3$, without loss of generality, we assume that
$\alpha^{\prime}=(1,3,4,5)$. Then $\beta^{\prime}=(6,6^{\prime})$ and ${\bf
bm}_{\alpha^{\prime}}{\bf bm}_{\alpha}L$ must contain the following three
$4$-simplices $(2,3,4,5,6)$, $(1,3,4,6,6^{\prime})$, $(1,3,5,6,6^{\prime})$,
$(1,4,5,6,6^{\prime})$ and $(3,4,5,6,6^{\prime})$. Now we perform the
following bistellar moves on $L$:

    1. (a)

Firstly we perform a bistellar $1$-move at $(1,3,4,5)$ on $L$ since
$\operatorname{Link}_{L}(1,3,4,5)=\partial(2,6^{\prime})$.

    2. (b)

Secondly we perform a bistellar 1-move at $(1,2,3,5)$ on ${\bf
bm}_{(1,3,4,5)}L$ since $(6,6^{\prime})\not\in{\bf bm}_{(1,3,4,5)}L$ and
$\operatorname{Link}_{{\bf bm}_{(1,3,4,5)}L}(1,2,3,5)=\partial(6,6^{\prime})$,
so that we can obtain that

$(1,3,5,6,6^{\prime})\in{\bf bm}_{(1,2,3,5)}{\bf bm}_{(1,3,4,5)}L.$

    3. (c)

Thirdly we see that $(4,6,6^{\prime})\not\in{\bf bm}_{(1,2,3,5)}{\bf
bm}_{(1,3,4,5)}L$ and

$\operatorname{Link}_{{\bf bm}_{(1,2,3,5)}{\bf
bm}_{(1,3,4,5)}L}(1,2,3)\,=\,\partial(4,6,6^{\prime}).$

Thus we can perform a bistellar 2-move at $(1,2,3)$ on ${\bf
bm}_{(1,2,3,5)}{\bf bm}_{(1,3,4,5)}L$, so that $(1,3,4,6,6^{\prime})\in{\bf
bm}_{(1,2,3)}{\bf bm}_{(1,2,3,5)}{\bf bm}_{(1,3,4,5)}L$.

    4. (d)

Next we can check that $(4,5,6,6^{\prime})\not\in{\bf bm}_{(1,2,3)}{\bf
bm}_{(1,2,3,5)}{\bf bm}_{(1,3,4,5)}L$ and

$\operatorname{Link}_{{\bf bm}_{(1,2,3)}{\bf bm}_{(1,2,3,5)}{\bf
bm}_{(1,3,4,5)}L}(1,2)\,=\,\partial(4,5,6,6^{\prime}),$

we can perform a bistellar 3-move at $(1,2)$ on ${\bf bm}_{(1,2,3)}{\bf
bm}_{(1,2,3,5)}{\bf bm}_{(1,3,4,5)}L$, and then

$(1,4,5,6,6^{\prime})\in{\bf bm}_{(1,2)}{\bf bm}_{(1,2,3)}{\bf
bm}_{(1,2,3,5)}{\bf bm}_{(1,3,4,5)}L.$

    5. (e)

Finally we see that $\operatorname{Link}_{{\bf bm}_{(1,2)}{\bf
bm}_{(1,2,3)}{\bf bm}_{(1,2,3,5)}{\bf
bm}_{(1,3,4,5)}L}(2,6^{\prime})=\partial(3,4,5,6)$ but $(3,4,5,6)\not\in{\bf
bm}_{(1,2)}{\bf bm}_{(1,2,3)}{\bf bm}_{(1,2,3,5)}{\bf bm}_{(1,3,4,5)}L$. Thus,
after performing a bistellar 3-move at $(2,6^{\prime})$ on ${\bf
bm}_{(1,2)}{\bf bm}_{(1,2,3)}{\bf bm}_{(1,2,3,5)}{\bf bm}_{(1,3,4,5)}L$, we
conclude that

$(2,3,4,5,6),(3,4,5,6,6^{\prime})\,\in\,{\bf bm}_{(2,6^{\prime})}{\bf
bm}_{(1,2)}{\bf bm}_{(1,2,3)}{\bf bm}_{(1,2,3,5)}{\bf bm}_{(1,3,4,5)}L.$

Moreover, the above arguments give the following equality

${\bf bm}_{\alpha^{\prime}}{\bf bm}_{\alpha}L\,=\,{\bf
bm}_{(2,6^{\prime})}{\bf bm}_{(1,2)}{\bf bm}_{(1,2,3)}{\bf bm}_{(1,2,3,5)}{\bf
bm}_{(1,3,4,5)}L.$

∎

######  Proposition 6.2.

For $h=1,2$, $\\{\mathcal{A}_{[L]}\\}_{[L]\in\widetilde{\mathfrak{S}}_{K}}$
forms a direct system.

###### Proof.

When $h=1$, the proof is not difficult. We would like to leave it as an
exercise to the reader.

When $h=2$, by Lemma 6.1 it suffices to show that for $[L]\preceq[L^{\prime}]$
in $\widetilde{\mathfrak{S}}_{K}$, there is a canonical embedding homomorphism
$e_{[L][L^{\prime}]}\colon\mathcal{A}_{[L]}\hookrightarrow\mathcal{A}_{[L^{\prime}]}$.
This is equivalent to showing that when $L^{\prime}={\operatorname{{\bf
bm}}}_{\alpha}L$ with $\dim\alpha=3,4$, such a homomorphism
$e_{[L][L^{\prime}]}$ exists. We will only prove the case of $\dim\alpha=3$
since the proof of the case $\dim\alpha=4$ is similar.

Let $\dim\alpha=3$. Without loss of generality, assume that $\alpha=(1,2,3,4)$
and $\beta=(5,6)$. Then

$\mathcal{X}(L^{\prime})\,=\,\left(\mathcal{X}(L\right)\setminus\\{x_{(1,2,3,4)}\\})\cup\\{x_{(1,2,5,6)},x_{(1,3,5,6)},x_{(1,4,5,6)},x_{(2,3,5,6)},x_{(2,4,5,6)},x_{(3,4,5,6)}\\}.$

Define $\varphi_{\alpha}\colon\mathcal{X}(L)\to\mathcal{X}(L^{\prime})$ by

$\varphi_{\alpha}(x)\,=\,\begin{cases}x,&\text{if
$x\in\mathcal{X}(L)\setminus\\{x_{(1,2,3,4)}\\}$;}\\\ x_{(1,2,5,6)},&\text{if
$x=x_{(1,2,3,4)}$.}\end{cases}$

Then it is easy to see that $\varphi_{\alpha}$ induces an embedding
homomorphism
$\widetilde{\varphi}_{\alpha}\colon\mathbb{F}\mathcal{X}(L)\to\mathbb{F}\mathcal{X}(L^{\prime})$.

######  Lemma 6.3.

Let $(\alpha^{\prime},\beta^{\prime})$ be a bistellar pair of type $2$ in $L$.
Then $(\alpha^{\prime},\beta^{\prime})$ must also be a bistellar pair of
${\operatorname{{\bf bm}}}_{\alpha}L$.

###### Proof.

We note that $\operatorname{Link}_{L}\alpha^{\prime}=\partial\beta^{\prime}$.
First we show that $\alpha^{\prime}\subset\alpha\cup\beta$ is impossible. In
fact, there are two possibilities: either $\alpha^{\prime}\subset\alpha$, or
$\alpha^{\prime}\nsubseteq\alpha$. If $\alpha^{\prime}\subset\alpha$, assume
that $\alpha^{\prime}=(1,2,3)$, then $\operatorname{Link}_{L}\alpha^{\prime}$
will not be the boundary of a missing face – so this case is impossible. If
$\alpha\nsubseteq\alpha$, assume that $\alpha^{\prime}=(1,2,5)$, then
$\beta^{\prime}=(3,4,7)$. Since what we have satisfies that
$\operatorname{Link}_{L}\alpha^{\prime}=\partial\beta^{\prime}$, this induces
that $(1,2,3,5,7),(1,2,4,5,7)\in L$. However,
$(1,2,3,5,7)\cap(1,2,4,5,7)=(1,2,5,7)$, which means that $(3,4)\not\in L$
since $L$ is a closed PL manifold. This is a contradiction. Therefore,
$\alpha^{\prime}\subset\alpha\cup\beta$ is impossible. Next, it is easy to see
that $\beta^{\prime}\subset\alpha\cup\beta$ is impossible.

Now if $|\beta^{\prime}\cap(\alpha\cup\beta)|\leq 2$, then clearly, whichever
$|\alpha^{\prime}\cap(\alpha\cup\beta)|=0$ or 1 or 2, we have that
$(\alpha^{\prime},\beta^{\prime})$ is a bistellar pair of ${\bf
bm}_{\alpha}L$. ∎

By the exchange relations (5.3) in Definition 5.3 and Lemma 6.3 we see that
all possible exchange relations in $\mathcal{A}_{[L]}$ are also the exchange
relations in $\mathcal{A}_{[{\bf bm}_{\alpha}L]}$. Furthermore, by Definition
5.18, the embedding homomorphism
$\widetilde{\varphi}_{\alpha}\colon\mathbb{F}\mathcal{X}(L)\to\mathbb{F}\mathcal{X}({\bf
bm}_{\alpha}L)$ induces a canonical embedding homomorphism $e_{[L][{\bf
bm}_{\alpha}L]}\colon\mathcal{A}_{[L]}\hookrightarrow\mathcal{A}_{[{\bf
bm}_{\alpha}L]}$. ∎

######  Definition 6.4.

Define

$\mathcal{A}_{K}^{PL}\,\mathbin{\raisebox{0.3pt}{:}\\!=}\,\lim_{[L]\in\widetilde{\mathfrak{S}}_{K}}\mathcal{A}_{[L]}.$

######  Theorem 6.5.

For $n=2,4$, assume that $K$ and $K^{\prime}$ in $\mathbb{PL}_{n}$ are PL
homeomorphic. Then $\mathcal{A}_{K}^{PL}\cong\mathcal{A}_{K^{\prime}}^{PL}$.

###### Proof.

This is because $K$ and $K^{\prime}$ determine the same directed set. ∎

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