# Three-Dimensional Pseudomanifolds  on Eight Vertices

Authors: Basudeb Datta; Nandini Nilakantan
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

A normal pseudomanifold is a pseudomanifold in which the links of simplices are also
pseudomanifolds. So, a normal 2-pseudomanifold triangulates a connected closed 2-manifold. But,
normald-pseudomanifolds form a broader class than triangulations of connected closed d-manifolds
for d≥3. Here, we classify all the 8-vertex neighbourly normal 3-pseudomanifolds. This gives a
classification of all the 8-vertex normal 3-pseudomanifolds. There are 74 such 3-pseudomanifolds,
39 of which triangulate the 3-sphere and other 35 are not combinatorial 3-manifolds. These 35
triangulate six distinct topological spaces. As a preliminary result, we show that any 8-vertex 3-pseudomanifold is equivalent by proper bistellar moves to an 8-vertex neighbourly 3-pseudomanifold.
This result is the best possible since there exists a 9-vertex nonneighbourly 3-pseudomanifold which
does not allow any proper bistellar moves.


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## Body

## 1. Introduction

Recall that asimplicial complex is a collection of
nonempty finite sets (sets of vertices) such that every nonempty subset
of an element is also an element. For i≥0,
the elements of size i+1 are called the i-simplices (or i-faces) of the complex.A simplicial complex is usually thought of as a
prescription for construction of a topological space by pasting geometric
simplices. The space thus obtained from a simplicial complexK is called the geometric carrier of K and is denoted by |K|.
We also say that Ktriangulates|K|.
A combinatorial2-manifold (resp., combinatorial2-sphere) is a simplicial complex which
triangulates a closed surface (resp., the 2-sphere S2).For a simplicial complexK,
the maximum of k such that K has a k-simplex, is called the dimension of K.
A d-dimensional simplicial complex K is called pure if each simplex of K is contained in a d-simplex of K.
A d-simplex in a pure d-dimensional simplicial complex is called a facet. A d-dimensional pure simplicial complex K is called a weak pseudomanifold if each (d−1)-simplex of K is contained in exactly two facets of K.With a pure simplicial complexK of dimension d≥1,
we associate a graph Λ(K) as follows. The vertices of Λ(K) are the facets of K and two vertices of Λ(K) are adjacent if the corresponding facets
intersect in a (d−1)-simplex of K.
If Λ(K) is connected, then K is called strongly connected. A strongly connected
weak pseudomanifold is called a pseudomanifold. Thus, for a d-pseudomanifold K,Λ(K) is a connected (d+1)-regular graph. This implies that K has no proper subcomplex which is also a d-pseudomanifold; (or else, the facets of such
a subcomplex would provide a disconnection of Λ(X)).For any setV with #(V)=d+2 (d≥0), let K be the simplicial complex whose simplexes are
all the nonempty proper subsets of V.
Then K is a d-pseudomanifold and triangulates the d-sphere Sd.
This d-pseudomanifold K is called the standardd-sphere and is denoted by Sd+2d(V) (or Sd+2d). By convention, S20 is the only 0-pseudomanifold.Ifσ is a face of a simplicial complex K,
then the link of σ in K,
denoted by lkK(σ) (or lk(σ)), is by definition the simplicial complex
whose faces are the faces τ of K such that τ is disjoint from σ and σ∪τ is a face of K.
Clearly, the link of an i-face in a weak d-pseudomanifold is a weak (d−i−1)-pseudomanifold. For d≥1,
a connected weak d-pseudomanifold is said to be a normald-pseudomanifold if the links of all the
simplices of dimension ≤d−2 are connected. Thus, any connected
triangulated d-manifold (triangulation of a closed d-manifold) is a normal d-pseudomanifold. Clearly, the normal
2-pseudomanifolds are just the connected combinatorial 2-manifolds; but, normal d-pseudomanifolds form a broader class than
connected triangulated d-manifolds for d≥3.Observe that ifX is a normal pseudomanifold, then X is a pseudomanifold. (If Λ(X) is not connected, then, since X is connected, Λ(X) has two components G1 and G2 and two intersecting facets σ1, σ2 such that σi∈Gi,i=1,2.
Choose σ1,σ2 among all such pairs such that dim⁡(σ1∩σ2) is maximum. Then dim⁡(σ1∩σ2)≤d−2 and lkX(σ1∩σ2) is not connected, a contradiction.) Notice
that all the links of positive dimensions (i.e., the links of simplices of
dimension ≤d−2) in a normal d-pseudomanifold are normal pseudomanifolds.
Thus, if K is a normal 3-pseudomanifold, then the link of
a vertex in K is a combinatorial 2-manifold. A vertex v of a normal 3-pseudomanifold K is called singular if the link of v in K is not a 2-sphere. The set of singular
vertices is denoted by SV(K).
Clearly, the space |K|∖SV(K) is a pl 3-manifold. If SV(K)=∅ (i.e., the link of each vertex is a 2-sphere),
then K is called a combinatorial3-manifold. A combinatorial3-sphere is a combinatorial 3-manifold
which triangulates the topological 3-sphere S3.LetM be a weak d-pseudomanifold. If α is a (d−i)-face of M,0<i≤d, such that lkM(α)=Si+1i−1(β) and β is not a face of M (such a face α is said to be a removable face of M), then consider the weak d-pseudomanifold (denoted by κα(M)) whose facet-set is {σ:σafacetofM,α⊈σ}∪{β∪α∖{v}:v∈α}.
The operation κα:M↦κα(M) is called a bistellari-move. For 0<i<d,
a bistellar i-move is called a proper bistellar move. If κα is a proper bistellar i-move and lkM(α)=Si+1i−1(β),
then β is a removable i-face of κα(M) (with lkκα(M)(β)=Sd−i+1d−i−1(α)) and κβ:κα(M)↦M is an bistellar (d−i)-move. For a vertex u,
if lkM(u)=Sd+1d−1(β),
then the bistellar d-move κ{u}:M↦κ{u}(M)=N deletes the vertex u (we also say that N is obtained from M by collapsing the vertex u). The operation κβ:N↦M is called a bistellar 0-move (we also say that M is obtained from N by starring the vertex u in the facet β of N). The 10-vertex combinatorial 3-manifold A103 in Example 3.15 is not neighbourly and does not
allow any bistellar 1-move. In [1], Bagchi and Datta have shown that if the number of
vertices in a nonneighbourly combinatorial 3-manifold is at most 9, then the
3-manifold admits a bistellar 1-move. Existence of the 9-vertex
3-pseudomanifold B93 in Example 3.16 shows that Bagchi and Datta's
result is not true for 9-vertex 3-pseudomanifolds. Here we prove the following
theorem.Theorem 1.1.
IfM is an 8-vertex 3-pseudomanifold, then there exists a sequence
of bistellar 1-moves κA1,…,κAm,
for some m≥0,
such that κAm(⋯(κA1(M))) is a neighbourly 3-pseudomanifold.In [2], Altshuler has shown that every combinatorial
3-manifold with at most 8 vertices is a combinatorial 3-sphere. In [3], Grünbaum and Sreedharan
have shown that there are exactly 37 polytopal 3-spheres on 8 vertices (namely, S8,13,…,S8,373 in Examples 3.1 and 3.3). They have also
constructed the nonpolytopal sphere S8,383.
In [4], Barnette
proved that there is only one more nonpolytopal 8-vertex 3-sphere (namely, S8,393). In [5], Emch constructed an 8-vertex normal 3-pseudomanifold
(namely, N1 in Example 3.5) as a block design. This is not a
combinatorial 3-manifold and its automorphism group is PGL(2,7) (cf. [6]). In [7], Altshuler has constructed another 8-vertex normal
3-pseudomanifold (namely, N5 in Example 3.5). In [8], Lutz has shown that there
exist exactly three 8-vertex normal 3-pseudomanifolds which are not
combinatorial 3-manifolds (namely, N1,N5 and N6 in Example 3.5) with vertex-transitive
automorphism groups. Here we prove the following theorem.Theorem 1.2.
LetS8,353,…,S8,383,N1,…,N15  be as in Examples 3.1 and 3.5.
(i)
ThenS8,i3≇S8,j3,Nk≇Nl, and S8,m3≇Nn for 35≤i<j≤38,1≤k<l≤15,35≤m≤38, and 1≤n≤15.(ii)
IfM is an 8-vertex neighbourly normal 3-pseudomanifold, then M is isomorphic to one of S8,353,…,S8,383,N1,…,N15.Corollary 1.3.
There are exactly39 combinatorial 3-manifolds on 8 vertices, all of which are combinatorial 3-spheres.Corollary 1.4.
There are exactly35 normal 3-pseudomanifolds on 8 vertices which are not combinatorial 3-manifolds. These are N1,…,N35 defined in Examples 3.5 and 3.8.The topological properties of these normal3-pseudomanifolds are given in Section 3.


## 2. Preliminaries

All the simplicial complexes considered in this paper
are finite (i.e., with finite vertex-set). The vertex-set of a simplicial
complexK is denoted by V(K).
We identify the 0-faces of a complex with the vertices. The 1-faces of a complex K are also called the edges of K.IfK,L are two simplicial complexes, then an isomorphism from K to L is a bijection π:V(K)→V(L) such that for σ⊆V(K),σ is a face of K if and only if π(σ) is a face of L.
Two complexes K, L are called isomorphic when such an isomorphism
exists. We identify two complexes if they are isomorphic. An isomorphism from a
complex K to itself is called an automorphism of K.
All the automorphisms of K form a group under composition, which is
denoted by Aut(K).For a faceσ in a simplicial complex K,
the number of vertices in lkK(σ) is called the degree of σ in K and is denoted by deg⁡K(σ) (or by deg⁡(σ)). If every pair of vertices of a simplicial
complex K form an edge, then K is called neighbourly. For a simplicial complex K,
if U⊆V(K),
then K[U] denotes the induced complex of K on the vertex-set U.If the number ofi-faces of a d-dimensional simplicial complex K is fi(K)(0≤i≤d), then the number χ(K):=∑i=0d(−1)ifi(K) is called the Euler characteristic of K.Agraph is a simplicial complex of dimension ≤1.
A finite 1-pseudomanifold is called a cycle. An n-cycle is a cycle on n vertices and is denoted by Cn (or by Cn(a1,…,an) if the edges are a1a2,…,an−1an,ana1).For a simplicial complexK,
the graph consisting of the edges and vertices of K is called the edge-graph of K and is denoted by EG(K).
The complement of EG(K) is called the nonedge graph of K and is denoted by NEG(K).
For a weak 3-pseudomanifold M and an integer n≥3,
we define the graph Gn(M) as follows. The vertices of Gn(M) are the vertices of M.
Two vertices u and v form an edge in Gn(M) if uv is an edge of degree n in M.
Clearly, if M and N are isomorphic, then Gn(M) and Gn(N) are isomorphic for each n.IfM is a weak 3-pseudomanifold and κα:M↦κα(M)=N is a bistellar 1-move, then, from the
definition, (f0(N),f1(N),f2(N),f3(N))=(f0(M),f1(M)+1,f2(M)+2,f3(M)+1) and deg⁡N(v)≥deg⁡M(v) for any vertex v.
If κα:M↦κα(M)=L is a bistellar 3-move, then (f0(L),f1(L),f2(L),f3(L))=(f0(M)−1,f1(M)−4,f2(M)−6,f3(M)−3).Consider the binary relation “≤” on the set of weak 3-pseudomanifolds as M≤N if there exists a finite sequence of bistellar 1-moves κα1,…,καm,
for some m≥0,
such that N=καm(⋯κα1(M)).
Clearly, this ≤ is a partial order relation.Two weakd-pseudomanifolds M and N are bistellar equivalent (denoted by M∼N) if there exists a finite sequence of bistellar
operations leading from M to N.
If there exists a finite sequence of proper bistellar operations leading from M to N,
then we say M and N are properly bistellar equivalent and we
denote this by M≈N.
Clearly, “∼" and “≈" are equivalence relations on the set of
pseudomanifolds. It is easy to see that M∼N implies that |M| and |N| are pl homeomorphic.For two simplicial complexesX and Y with disjoint vertex sets, the simplicial
complex X∗Y:=X∪Y∪{σ∪τ:σ∈X,τ∈Y} is called the join of X and Y.LetK be an n-vertex (weak) d-pseudomanifold. If u is a vertex of K and v is not a vertex of K,
then consider the simplicial complex ΣuvK on the vertex set V(K)∪{v} whose set of facets is {σ∪{u}:σ is a facet of K and u∉σ}∪{τ∪{v}:τ is a facet of K}.
Then ΣuvK is a (weak) (d+1)-pseudomanifold and |ΣuvK| is the topological suspension S|K| of |K| (cf. [9]). It is easy to see that the links of u and v in ΣuvK are isomorphic to K.
This ΣuvK is called the one-point suspension of K.For twod-pseudomanifolds X and Y,
a simplicial map f:X→Y is called a k-fold branched covering (with discrete
branch locus) if |f|||X|∖f−1(U):|X|∖f−1(U)→|Y|∖U is a k-fold covering for some U⊆V(Y).
(We say that X is a branched cover of Y and Y is a branched quotient of X.)
The smallest such U (so that |f|||X|∖f−1(U):|X|∖f−1(U)→|Y|∖U is a covering) is called the branch locus. If N is a k-fold branched quotient of M and N is obtained from N by collapsing a vertex (resp., starring
a vertex in a facet), then N is the branched quotient of M,
where M can be obtained from M by collapsing k vertices (resp., starring k vertices in k facets). For proper bistellar moves we have
the following lemma.Lemma 2.1.
LetM and N be two d-pseudomanifolds and f:M→N be a k-fold branched covering. For 1≤l<d−1,
if α is a removable l-face, then f−1(α) consists of k removable l-faces α1,…,αk(say) and καk(⋯(κα1(M))) is a k-fold branched cover of κα(N).Proof.
LetlkN(α)=Sd−l+1d−l−1(β).
Since the dimension of α is >0,f−1(α) consists of kl-faces, α1,…,αk (say) of M.
Let lkM(αi)=Sd−l+1d−l−1(βi) and Mi:=M[αi∪βi] for 1≤i≤k.
Since f is simplicial, βi is not a face of M and hence αi is removable for each i.
Since 0<l<d−1,
it follows that Mi is neighbourly. For i≠j,
if x≠y∈V(Mi)∩V(Mj),
then xy is an edge in Mi∩Mj and hence the number of edges in f−1(f(x)f(y)) is less than k,
a contradiction. So, #(V(Mi)∩V(Mj))≤1 for i≠j.
This implies that βi is not a face in καj(M) and hence αi is removable in καj(M) for i≠j.
The result now follows.Remark3.14 shows that Lemma 2.1 is not true for l=d−1 (i.e., for bistellar 1-moves) in general.Example 2.2.
In Figure2, we present some weak 2-pseudomanifolds on at most seven
vertices. The degree sequences are
presented parenthetically below the figures. Each of S1,…,S9 triangulates the 2-sphere, each of R1,…,R4 triangulates the real projective plane and T triangulates the torus. Observe that P1,P2 are not pseudomanifolds.
We know that ifK is a weak 2-pseudomanifold with at most six
vertices, then K is isomorphic to S1,…,S4 or R1 (cf. [9]). In [10], we have seen the following.Figure 1(a)(b)(c)(d)Proposition 2.3.
There are exactly13 distinct 2-dimensional weak pseudomanifolds on 7 vertices, namely, S5,…,S9,R2,…,R4,T,P1,…,P3,
and P4.


## 3. Examples

We identify a weak pseudomanifold with the set of
facets in it.Example 3.1.
These four neighbourly 8-vertex combinatorial
3-manifolds were found by Grünbaum and Sreedharan (in [3], these are denoted by P358,P368,P378 and ℳ,
resp.). It follows from Lemma  3.4 that these are combinatorial
3-spheres. It was shown in [3] that the first three of these are polytopal 3-spheres
and the last one is a nonpolytopal sphere: (3.1)S8,353={1234,1267,1256,1245,2345,2356,2367,3467,3456,4567,1238,1278,2378,1348,3478,1458,4578,1568,1678,5678},S8,363={1234,1256,1245,1567,2345,2356,2367,3467,3456,4567,1268,1678,2678,1238,2378,1348,3478,1458,1578,4578},S8,373={1234,1256,1245,1457,2345,2356,2367,3467,3456,4567,1568,1578,5678,1268,2678,1238,2378,1348,1478,3478},S8,383={1234,1237,1267,1347,1567,2345,2367,3467,3456,4567,2358,2368,3568,1268,1568,1248,2458,1478,1578,4578}.Lemma 3.2.
S8,i3≇S8,j3 for 35≤i<j≤38.Proof.
Observe thatG6(S8,353)=C8(1,2,…,8),G6(S8,363)=(V,{23,34,45,67,78,81}), G6(S8,373)=(V,{23,34,56,78,81}),
and G6(S8,383)=(V,{17,23,58}),
where V={1,…,8}.
Since K≅L implies G6(K)≅G6(L),S8,i3≇S8,j3,
for 35≤i<j≤38.Example 3.3.
Some nonneighbourly 8-vertex combinatorial
3-manifolds. It follows from Lemma3.4 that these are combinatorial 3-spheres.
For 1≤i≤34,
the sphere S8,i3 is isomorphic to the polytopal sphere Pi8 in [3] and the sphere S8,393 is isomorphic to the nonpolytopal sphere found
by Barnette in [4]. We
consecutively define (3.2)S8,393=κ46(S8,383),S8,333=κ27(S8,373),S8,323=κ48(S8,373),S8,313=κ58(S8,373),S8,303=κ24(S8,373),S8,293=κ27(S8,313),S8,283=κ24(S8,313),S8,273=κ13(S8,313),S8,253=κ57(S8,313),S8,243=κ48(S8,313),S8,233=κ35(S8,313),S8,263=κ46(S8,273),S8,223=κ24(S8,253),S8,213=κ68(S8,253),S8,203=κ48(S8,253),S8,193=κ17(S8,253),S8,183=κ27(S8,253),S8,123=κ15(S8,253),S8,113=κ35(S8,253),S8,173=κ24(S8,193),S8,343=κ27(S8,263)=S30(1,3)∗S30(2,7)∗S30(4,6)∗S30(5,8),S8,163=κ13(S8,193),S8,153=κ28(S8,183),S8,143=κ47(S8,203),S8,103=κ15(S8,193),S8,93=κ35(S8,193),S8,83=κ47(S8,193),S8,133=κ38(S8,163),S8,73=κ24(S8,83),S8,63=κ35(S8,83),S8,53=κ48(S8,83),S8,43=κ15(S8,83),S8,33=κ48(S8,43),S8,23=κ48(S8,63),S8,13=κ16(S8,43).Lemma 3.4.
(a)S8,i3≈S8,j3,
for 1≤i,j≤39, (b)S8,m3 is a combinatorial 3-sphere for 1≤m≤39,
and (c)S8,k3≇S8,l3 for 1≤k<l≤39.Proof.
For0≤i≤6,
let 𝒮i denote the set of S8,j3's with i nonedges. Then 𝒮0={S8,353,S8,363,S8,373,S8,383},𝒮1={S8,303,S8,313,S8,323,S8,333,S8,393},𝒮2={S8,233,S8,243,S8,253,S8,273,S8,283,S8,293},𝒮3={S8,113,S8,123,S8,183,S8,193,S8,203,S8,213,S8,223,S8,263},𝒮4={S8,83,S8,93,S8,103,S8,143,S8,153,S8,163,S8,173,S8,343},𝒮5={S8,43,S8,53,S8,63,S8,73,S8,133}, and 𝒮6={S8,13,S8,23,S8,33}.
From the proof of Lemma4.7, S8,353≈S8,303≈S8,363≈S8,303≈S8,373≈S8,323≈S8,383.
Thus, S8,i3≈S8,j3 for 35≤i,j≤38.
Now, if S8,i3∈𝒮2∪𝒮3∪𝒮4∪𝒮5∪𝒮6,
then, from the definition of S8,i3,S8,i3≈S8,313≈S8,373.
This proves part (a).
SinceS8,343 is a join of spheres, S8,343 is a combinatorial 3-sphere. Clearly, if M≈N and M is a combinatorial 3-sphere, then N is so. Part (b) now follows from part (a).
Since the nonedge graphs of the members of𝒮6 (resp., 𝒮5) are pairwise nonisomorphic, the members of 𝒮6 (resp., 𝒮5) are pairwise nonisomorphic.
ForS8,i3,S8,j3∈𝒮4(i<j) and NEG(S8,i3)≅NEG(S8,j3) imply (i,j)=(8,9) or (14,15).
Since M≅N implies G6(M)≅G6(N) and G6(S8,83)≇G6(S8,93),G6(S8,143)≇G6(S8,153), the members of 𝒮4 are pairwise nonisomorphic.
ForS8,i3≠S8,j3∈𝒮3 and NEG(S8,i3)≅NEG(S8,j3) imply {i,j}={11,12} or 18≤i≠j≤21. Let
∑1={S8,113,S8,123},∑2={S8,183,S8,193,S8,203,S8,213},∑3={S8,223}and∑4={S8,263}.⁢ Since the nonedge graph of a member in Σi is nonisomorphic to the nonedge graph of a
member of Σj for i≠j,
a member of Σi is nonisomorphic to a member of Σj.
Observe that G6(S8,113)≇G6(S8,123) and for 18≤i<j≤21,G6(S8,i3)≅G6(S8,j3) implies (i,j)=(18,19).
Since G3(S8,183)≇G3(S8,193),
the members of 𝒮3 are pairwise nonisomorphic.
SinceG3(S8,i3)≇G3(S8,j3) for S8,i3≠S8,j3∈𝒮2,
the members of 𝒮2 are pairwise nonisomorphic. By the same
reasoning, the members of 𝒮1 are pairwise nonisomorphic.
By Lemma3.2, the members of 𝒮0 are pairwise nonisomorphic. Since a member of 𝒮i is nonisomorphic to a member of 𝒮j for i≠j,
the above imply part (c).Example 3.5.
Some 8-vertex neighbourly normal
3-pseudomanifolds:(3.3)N1={1248,1268,1348,1378,1568,1578,2358,2378,2458,2678,3468,3568,4578,4678,1247,1257,1367,1467,2347,2567,3457,3567,1236,2346,1345,1235,1456,2456},N2={1248,2458,2358,3568,3468,4678,4578,1578,1568,1268,2678,2378,1378,1348,1247,2457,2357,3567,3467,1567,1267,1347}=Σ78T,N3={1248,1268,1348,1378,1568,1578,2358,2378,2458,2678,3468,3568,4578,4678,1234,2347,2456,2467,3456,3457,1235,1256,1357},N4={1248,1268,1348,1378,1568,1578,2358,2378,2458,2678,3468,3568,4578,4678,1245,1256,2356,2367,3467,1347,1457},N5={1258,1268,1358,1378,1468,1478,2368,2378,2458,2478,3458,3468,1257,1267,1367,1457,2357,2467,3457,3467,2356,2456,1356,1456},N6={1358,1378,1468,1478,1568,2368,2378,2458,2478,2568,3458,3468,1235,1245,1457,1567,2357,2567,3457,1236,1246,1367,2467,3467},N7={1268,1258,1358,1378,1478,1468,2378,2368,2458,2478,3468,3458,1356,1367,2357,2356,3467,3457,1256,1467,2457},N8=κ348(κ238(κ56(κ67(N7)))),N9=κ235(κ67(N7)),N10=κ148(κ67(N7)),N11=κ348(κ56(N10)),N12=κ457(κ23(N9)),N13=κ567(κ23(N9)),N14=κ138(κ57(N8))≅Σ78R2,N15=κ158(κ23(N9)). All the vertices of N1 are singular and their links are isomorphic to
the 7-vertex torus T.
There are two singular vertices in N2 and their links are isomorphic to T.
The singular vertices in N3 are 8, 3, 4, 2, 5 and their links are
isomorphic to T,R2,R2,R3, and R3,
respectively. There is only one singular vertex in N4 whose link is isomorphic to T.
All the vertices of N5 (resp., N6) are singular and their links are isomorphic
to R4 (resp., R3). Each of N7,…,N15 has exactly two singular vertices and their
links are 7-vertex ℝP2's. Thus, each Ni is a normal 3-pseudomanifold.
It follows from the definition thatNi≈Nj for 7≤i,j≤15.
Here we prove the following lemmas.Lemma 3.6.
(a) The geometric carriers ofN1,N2,N3,N4,N5,
and N7 are distinct (non-homeomorphic),  (b)Ni≉Nj for 1≤i<j≤7, (c)N5∼N6.Proof.
For a normal 3-pseudomanifoldX,
let ns(X) denote the number of singular vertices.
Clearly, if M and N are two normal 3-pseudomanifolds with
homeomorphic geometric carriers, then (ns(M),χ(M))=(ns(N),χ(N)).
Now, (ns(N1),χ(N1))=(8,8),(ns(N2),χ(N2))=(2,2),(ns(N3),χ(N3))=(5,3), (ns(N4), χ(N4))=(1,1), (ns(N5),χ(N5))=(8,4),(ns(N7),χ(N7))=(2,1).
This proves part (a).
Part (b) follows from the fact thatNi is neighbourly and has no removable edge and,
hence, there is no proper bistellar move from Ni for 1≤i≤6.
LetN5′ be obtained from N5 by starring a new vertex 0 in the facet 1358.
Let N5′′=κ{0}(κ08(κ156(κ07(κ03(κ035(κ68(κ02(κ268(κ13(κ135(κ138(κ158(N5′))))))))))))),
then N5′′ is isomorphic to N6 via the map (2,3)(5,8).
This proves part (c).Lemma 3.7.
Nk≇Nl for 1≤k<l≤15.Proof.
Letns be as above. Clearly, if M and N are two isomorphic 3-pseudomanifolds, then (ns(M),f3(M))=(ns(N),f3(N)).
Now, (ns(N1),f3(N1))=(8,28),(ns(N2),f3(N2))=(2,22),(ns(N3),f3(N3))=(5,23),(ns(N4),f3(N4))=(1,21),(ns(N5),f3(N5))=(ns(N6),f3(N6))=(8,24),
and (ns(Ni),f3(Ni))=(2,21) for 7≤i≤15.
Since the links of each vertex in N5 is isomorphic to R4 and the links of each vertex in N6 is isomorphic to R3,
it follows that N5≇N6.
Thus, Ni≇Nj for 1≤i≤6, 1≤j≤15, i≠j.
Observe that the singular vertices inNi are 3 and 8 for 7≤i≤15.
Moreover, (i) lkN7(3)≅lkN7(8)≅R4,
(ii) lkN8(3)≅R4 and lkN8(8)≅R3,
(iii) lkN9(3)≅R2 and lkN9(8)≅R4,
(iv) lkN10(3)≅lkN10(8)≅R3 and deg⁡N10(38)=6,
(v) lkN11(3)≅lkN11(8)≅R3 and deg⁡N11(38)=5,
(vi) lkN12(3)≅R2,lkN12(8)≅R3 and G3(N12)=(V,{32,21,17,75,54,46}),
(vii) lkN13(3)≅R2,lkN13(8)≅R3 and G3(N13)=(V,{32,21, 17,75,56,67,64,42}),
(viii) lkN14(3)≅lkN14(8)≅R2 and deg⁡N14(38)=3.
(xi) lkN15(3)≅lkN15(8)≅R2 and deg⁡N15(38)=6.
These imply that there is no isomorphism between Ni and Nj for 7≤i<j≤15.
This completes the proof.Example 3.8.
Some 8-vertex nonneighbourly normal
3-pseudomanifolds:(3.4)N16=κ67(N7),N17=κ24(N8),N18=κ238(κ56(κ67(N7))),N19=κ57(N8),N20=κ56(N10),N21=κ12(N9),N22=κ14(N11),N23=κ23(N9),N24=κ38(N14),N25=κ56(N16),N26=κ12(N16),N27=κ56(N17),N28=κ57(N18),N29=κ15(N18),N30=κ12(N23),N31=κ24(N22),N32=κ24(N26),N33=κ57(N25),N34=κ45(N28),N35=κ58(N29).Lemma 3.9.
(a)Ni≇Nj for 1≤i<j≤35 and (b)Nk≈Nl for 7≤k,l≤35.Proof.
For0≤i≤3,
let 𝒩i denote the set of 3-pseudomanifolds defined in
Examples 3.5 and 3.8 with i nonedges. Then 𝒩0={N1,…,N15},𝒩1={N16,…,N24},𝒩2={N25,…,N31},
and 𝒩3={N32,…,N35}.
The singular vertices in Ni are 3 and 8 for 7≤i≤35.
By Lemma3.7, the members of 𝒩0 are pairwise nonisomorphic.
Observe that (i)lkN16(3)≅R4 and lkN16(8)≅R3,
(ii) lkN17(3)≅lkN17(8)≅R4,
(iii) lkN18(3)≅lkN18(8)≅R3 and G6(N18)=(V,{73,31,18,84}),
(iv) lkN19(3)≅lkN19(8)≅R3 and G6(N19)=(V,{63,31,18,86}),
(v) lkN20(3)≅lkN20(8)≅R3 and G6(N20)=(V,{74,28,83,31}),
(vi) lkN21(3)≅R2,lkN21(8)≅R3 and G6(N21)=(V,{48,83,37,36}),
(vii) lkN22(3)≅R2,lkN22(8)≅R3 and G6(N22)=(V,{28,86,63,37,38}),
(viii) lkN23(3)≅R1 and lkN23(8)≅R3,
(ix) lkN24(3)≅lkN24(8)≅R1.
These imply that there is no isomorphism between any two members of 𝒩1.
Observe that (i)lkN25(3)≅R3 and lkN25(8)≅R4,
(ii) lkN26(3)≅lkN26(8)≅R3 and G6(N26)=(V,{53,38,84}),
(iii) lkN27(3)≅lkN27(8)≅R3,G6(N27)=(V,{78,81,13,37}) and NEG(N27)={24,56},
(iv) lkN28(3)≅lkN28(8)≅R3,G6(N28)=(V,{18,84,43,31}) and NEG(N28)={75,56},
(v) lkN29(3)≅R3 and lkN29(8)≅R2,
(vi) lkN30(3)≅R1 and lkN30(8)≅R3,
(vii) lkN31(3)≅lkN31(8)≅R2.
These imply that there is no isomorphism between any two members of 𝒩2.
Observe that (i)lkN32(3)≅lkN32(8)≅R3,
(ii) lkN33(3)≅lkN33(8)≅R4,
(iii) lkN34(3)≅lkN34(8)≅R2,
(iv) lkN35(3)≅R2 and lkN35(8)≅R1.
These imply that there is no isomorphism between any two members of 𝒩3.
Since a member of𝒩i is nonisomorphic to a member of 𝒩j for i≠j,
the above imply part (a).
Part (b) follows from the definition of Nk for 8≤k≤35.The 3-dimensionalKummer varietyK3 is the torus S1×S1×S1 modulo the involution σ:x↦−x.
It has 8 singular points corresponding to 8 elements of order 2 in the abelian
group S1×S1×S1.
In [11], Kühnel showed
that N5 triangulates K3.
For a topological space X,C(X) denotes a cone with base X.
Let H=D2×S1 denote the solid torus. As a consequence of
the above lemmas we get.Corollary 3.10.
All the8-vertex normal 3-pseudomanifolds triangulate seven distinct
topological spaces, namely, |S8,j3|=S3 for 1≤j≤38,|N1|,|N2|=S(S1×S1),|N3|,|N4|=H∪(C(∂H)),|N5|=|N6|=K3,and |Ni|=S(ℝP2) for 7≤i≤35.Proof.
LetK be an 8-vertex normal 3-pseudomanifold. If K is a combinatorial 3-sphere, then it triangulates the 3-sphere S3.
IfK is not a combinatorial 3-sphere, then, by Lemma 3.9(b), |K| is (pl) homeomorphic to |N1|,…,|N6|,
or |N7|.
Since N2=Σ78T,|N2| is homeomorphic to the suspension S(S1×S1).
In N4,
the facets not containing the vertex 8 form a solid torus whose boundary is the
link of 8. This implies that |N4|=H∪(C(∂H)).
It follows from Lemma 3.6(c) that |N6| is (pl) homeomorphic to |N5|=K3.
Since N24 is isomorphic to the suspension S20∗R1, |N24|=S(ℝP2).
Therefore, by Lemma 3.9(b), |Ni| is (pl) homeomorphic to |N24|=S(ℝP2) for 7≤i≤35.
The result now follows from Lemma 3.6(a).A 3-dimensionalpseudocomplexK is an ordered pair (Δ,Φ),
where Δ is a finite collection of disjoint tetrahedra
and Φ is a family of affine isomorphisms between
pairs of 2-faces of the tetrahedra in Δ.
Let |K| denote the quotient space obtained from the
disjoint union ⊔σ∈Δσ by setting x=φ(x) for φ∈Φ.
The quotient of a tetrahedron σ∈Δ in |K| is called a 3-simplex in |K| and is denoted by |σ|.
Similarly, the quotient of 2-faces, edges, and vertices of tetrahedra are
called 2-simplices, edges, and vertices in |K|,
respectively. If |K| is homeomorphic to a topological space X,
then K is called a pseudotriangulation of X.
A 3-dimensional pseudocomplex K=(Δ,Φ) is said to be regular if the following hold: (i) each
3-simplex in |K| has four distinct vertices, and (ii) for 2≤i≤3,
no two distinct i-simplices in |K| have the same set of vertices. So, for 2≤i≤3,
an i-simplex α in |K| is uniquely determined by its vertices and
denoted by u1⋯ui+1,
where u1,…,ui+1 are vertices of α.
(But, the edges in |K| may not form a simple graph.) So, we can
identify a regular pseudocomplex K=(Δ,Φ) with 𝒦:={|σ|:σ∈Δ}.
Simplices and edges in |K| are said to be simplices and edges of 𝒦.
Clearly, a pure 3-dimensional simplicial complex is a regular pseudocomplex.Letℳ be a regular pseudotriangulation of X and abcd,abce be two 3-simplices in ℳ.
If ade,bde,cde are not 2-simplices in ℳ,
then 𝒩:=(ℳ∖{abcd,abce})∪{abde,acde,bcde} is also a regular pseudotriangulation of X.
We say that 𝒩 is obtained from ℳ by the generalized bistellar1-moveκabc.
If there is no edge between d and e in ℳ,
then κF is called a bistellar1-move. If there exist 3-simplices of the
form xyuv,xzuv,yzuv in a regular pseudotriangulation 𝒫 of Y and xyz is not a 2-simplex, then 𝒬:=(𝒫∖{xyuv,xzuv,yzuv})∪{xyzu,xyzv} is also a regular pseudotriangulation of Y.
We say that 𝒬 is obtained from 𝒫 by the generalized bistellar2-moveκE,
where E is the common edge in xyuv,xzuv, and yzuv.
If E is the only edge between u and v in 𝒫, then κE is called a bistellar2-move.LetM be a pseudotriangulation of a closed
3-manifold and N a 3-pseudomanifold. A simplicial map f:M→N is said to be a k-fold branched covering (with discrete
branch locus) if there exists U⊆V(N) such that |f|||M|∖f−1(U):|M|∖f−1(U)→|N|∖U is a k-fold covering. The smallest such U (so that |f|||M|∖f−1(U):|M|∖f−1(U)→|N|∖U is a covering) is called the branch locus. It is known that N1 can be regarded as a branched quotient of a
regular hyperbolic tessellation (cf. [6]). In [11], Kühnel has shown that N5 is a 2-fold branched quotient of a
pseudotriangulation of the 3-dimensional torus. Here we prove the following
theorem.Theorem 3.11.
(a)N24 is a 2-fold branched quotient of a 14-vertex combinatorial 3-sphere.
(b) For7≤i≤35,Ni is a 2-fold branched quotient of a 14-vertex regular pseudotriangulation of the 3-sphere.Lemma 3.12.
LetM be a regular pseudotriangulation of a 3-manifold and N be a normal 3-pseudomanifold. Let f:M→N be a k-fold branched covering with at most two
vertices in the branch locus. If κe:N↦N is a bistellar 2-move, then there exist k generalized bistellar 2-moves κe1,…,κek such that κek(⋯(κe1(M))) is a k-fold branched cover of N.Proof.
LetlkN(e)=S31({x,y,z}).
Let f−1(e) consist of the edges e1,…,ek.
Let the end points of ei be ui, vi,
the 3-simplices containing ei be uivixiyi,uivixizi,uiviyizi,
and f(xi)=x,f(yi)=y,f(zi)=z for 1≤i≤k.
Since xyz is not a simplex in N,
it follows that xiyizi is not a 2-simplex in M.
Let Mi be the pseudocomplex consists of uivixiyi,uivixizi, and uiviyizi.
Since the number of vertices in the branched locus is at most 2, it follows
that the number of vertices common in Mi and Mj is at most 2 for i≠j.
In particular, #({xi,yi,zi}∩{xj,yj,zj})≤2.
Therefore, xjyjzj is not a 2-simplex in κei(M).
So, we can perform generalized bistellar 2-move κej on κei(M)=(M∖Mi)∪{xiyiziui,xiyizivi} for i≠j.
Clearly, M:=κek(⋯κe1(M)) is a k-fold branched cover of N (via the map f,
where f(w)=f(w) for w∈V(M)=V(M) and f(xiyiziui)=xyzu and f(xiyizivi)=xyzv).Lemma 3.13.
LetM be a regular pseudotriangulation of a 3-manifold and N be a normal 3-pseudomanifold. Let f:M→N be a k-fold branched covering with at most two
vertices in the branch locus. If κF:N↦N is a bistellar 1-move, then there exist k generalized bistellar 1-moves κF1,…,κFk such that κFk(⋯(κF1(M))) is a k-fold branched cover of N.Proof.
LetF=xyz and lkN(F)={u,v}.
Let f−1(F) consist of the 2-simplices F1,…,Fk.
Let Fi=xiyizi and the 3-simplices containing Fi be xiyiziui and xiyizivi and f(xi,yi,zi,ui,vi)=(x,y,z,u,v) for 1≤i≤k.
Since f is simplicial, it follows that xiuivi,yiuivi, and ziuivi are not 2-simplices in M.
Let Mi be pseudocomplex {xiyiziui,xiyizivi}.
Since the number of vertices in the branched locus is at most 2, it follows
that xjujvj,yjujvj, and zjujvj are not 2-simplices in κFi(M) for i≠j.
Then (by the similar arguments as in the proof of Lemma 3.12) κFk(⋯κF1(M)) is a k-fold branched cover of N.Proof  of Theorem3.11.
Ifℐ denotes the boundary of the icosahedron, then
there exists a simplicial 2-fold covering f:ℐ→R1.
Consider the simplicial map f:S20({a,b})∗ℐ→S20({c,d})∗R1 given by f(a)=c,f(b)=d and f(u)=f(u) for u∈V(ℐ).
Then f is a 2-fold branched covering with branch
locus {c,d}.
Since N24 is isomorphic to the suspension S20∗R1,
it follows that N24 is a 2-fold branched quotient of the 14-vertex
combinatorial 3-sphere S20({a,b})∗ℐ (with branch locus {3,8}). This proves part (a).
The result now follows from Lemmas3.9(a),
3.12, and 3.13. (In fact, to obtain a 2-fold branched cover N14 of N14 from R1∗S20,
one needs one bistellar 1-move and then one generalized bistellar 1-move; and
all other moves required in the proof are bistellar moves on regular
pseudotriangulations of S3.)Remark 3.14.
The combinatorial3-sphere R1∗S20 is a 2-fold branched cover of N24 and N14  can be obtained from N24 by a bistellar 1-move. Now, if f:M→N14 is a 2-fold branched covering and M is a combinatorial 3-manifold, then (since lkN14(8) is a 7-vertex triangulated ℝP2) the link of any vertex in f−1(8) is a 14-vertex triangulated S2 and hence f0(M)>14.
(Similarly, for i≠24,
if Ni is a branched quotient of a combinatorial
3-manifold M,
then f0(M)>14.)
So, there does not exist a combinatorial 3-sphere M which is a branched cover of N14 and which can be obtained from R1∗S20 by proper bistellar moves.In [7], Altshuler observed that N1 is orientable and |N1| is simply connected. In [8], Lutz showed that (H1(N1),H2(N1),H3(N1))=(0,ℤ8,ℤ). The normal 3-pseudomanifold N3 is the only among all the 35 which has
singular vertices of different types, namely, one singular vertex whose link is
a triangulated torus and four singular vertices whose links are triangulated
real projective planes. Using polymake [12], we find that (H1(N3),H2(N3),H3(N3))=(0,ℤ2⊕ℤ2,0). We summarized all the findings about N1,…,N35 in Table 1.Table 1
8-vertex normal
3-pseudomanifolds which are not combinatorial 3-manifolds.
Xf-vector (f1,f2,f3)χ(X)ns(X)links of singular verticesGeometric carriers, Homology(H1,H2,H3)N1(28,56,28)88all areT|N1| is simply connected, (H1,H2,H3)=(0,ℤ8,ℤ)N2(28,44,22)22both areT|N2|=S(S1×S1)N3(28,46,23)35T,R2,R2,R3,R3(H1,H2,H3)=(0,ℤ2⊕ℤ2,0)N4(28,42,21)11T|N4|=H∪(C(∂H))N5(28,48,24)48all areR4|N5|=K3N6,,,,,,all areR3|N6|=K3N7(28,42,21)12both areR4|N7|=S(ℝP2)Ni, 8≤i≤15,,,,,,both are in{R1,…,R4}|Ni|=S(ℝP2)Ni,16≤i≤24(27,40,20),,,,,,,,Ni,25≤i≤31(26,38,19),,,,,,,,Ni,32≤i≤35(25,36,18),,,,,,,,[HereK3 is the 3-dimensional Kummer variety, H=D2×S1 is the solid torus, S(Y) is the topological suspension of Y,
and ns(X) is the number of singular vertices in X.]Example 3.15.
Ford≥2,
let (3.5)K2d+3d={vi⋯vj−1vj+1⋯vi+d+1:i+1≤j≤i+d,1≤i≤2d+3} (additions in the suffixes are
modulo 2d+3). It was shown in [13] the following : (i) K2d+3d is a triangulated d-manifold for all d≥2,
(ii) K2d+3d triangulates Sd−1×S1 for d even, and triangulates the twisted product Sd−1×−S1 (the twisted Sd−1-bundle over S1) for d odd. For d≥3, K2d+3d is the unique nonsimply connected (2d+3)-vertex triangulated d-manifold (cf. [14]). The combinatorial
3-manifolds K93 was first constructed by Walkup in [15].
FromK93,
we construct the following 10-vertex combinatorial 3-manifold: (3.6)A103:=(K93∖{v1v2v3v5,v2v3v5v6,v3v5v6v7,v3v4v6v7,v4v6v7v8})∪{v0v1v2v3,v0v1v2v5,v0v1v3v5,v0v2v3v6,v0v2v5v6,v0v3v5v7,v0v5v6v7,v0v3v4v6,v0v3v4v7,v0v4v6v8,v0v4v7v8,v0v6v7v8}. [Geometrically, first we remove
a pl 3-ball consisting of five 3-simplices from |K93|.
This gives a pl 3-manifold with boundary and the boundary is a 2-sphere. Then
we add a cone with base this boundary and vertex v0.
So, the new polyhedron |A103| is pl homeomorphic to |K93|.
This implies that the simplicial complex A103 is a combinatorial 3-manifold.]
The only nonedge inA103 is v0v9 and there is no common 2-face in the links of v0 and v9 in A103.
So, A103 does not allow any bistellar 1-move. So, A103 is a 10-vertex nonneighbourly combinatorial
3-manifold which does not admit any bistellar 1-move.
Similarly, fromK114,
we construct the following 12-vertex triangulated 4-manifold: (3.7)A124:=(K114∖{v1v2v3v4v6,v2v3v4v6v7,v3v4v6v7v8,v4v6v7v8v9,v4v5v7v8v9,v5v7v8v9v10})∪{v0v1v2v3v4,v0v1v2v3v6,v0v1v2v4v6,v0v1v3v4v6,v0v2v3v4v7,v0v2v3v6v7,v0v2v4v6v7,v0v3v4v6v8,v0v3v4v7v8,v0v3v6v7v8,v0v4v6v7v9,v0v4v6v8v9,v0v4v7v8v9,v0v4v5v7v9,v0v4v5v8v9,v0v4v7v8v9,v0v5v7v8v10,v0v5v7v9v10,v0v5v8v9v10}. The only nonedge in A124 is v0v11 and there is no common 2-face in the links of v0 and v11 in A124.
So, A124 does not allow any bistellar 1-move. So, A124 is a 12-vertex nonneighbourly triangulated
4-manifold which does not admit any bistellar 1-move.
By the same way, one can construct a(2d+4)-vertex nonneighbourly triangulated d-manifold A2d+4d (from K2d+3d) which does not admit any bistellar 1-move
for all d≥3.Example 3.16.
LetN3 be as in Example 3.5. Let M be obtained from N3 by starring two vertices u and v in the facets 1248 and 3568,
respectively, that is, M=κ1248(κ3568(N3)).
Then M is a 10-vertex normal 3-pseudomanifold. Let B93 be obtained from M by identifying the vertices u and v.
Let the new vertex be 9. Then (3.8)B93:=(N3∖{1248,3568})∪{1249,1289,1489,2489,3569,3589,3689,5689}. The degree 3 edges in B93 are 16, 17,
and 67;
but none of these edges is removable. So, no bistellar 2-moves are possible from B93.
The only nonedge in B93 is 79.
Since there is no common 2-face in the links of 7 and 9,
no bistellar 1-move is possible. So, B93 is a 9-vertex nonneighbourly 3-pseudomanifold which
does not admit any proper bistellar move.


## 4. Proofs

Forn≥4,
by an Sn2 we mean a combinatorial 2-sphere on n vertices. If κβ:M↦N is a bistellar 1-move, then deg⁡N(v)≥deg⁡M(v) for v∈V(M).
Here we prove the following.Lemma 4.1.
LetM be an n-vertex 3-pseudomanifold and u be a vertex of degree 4.
If n≥6,
then there exists a bistellar 1-move κβ:M↦N such that deg⁡N(u)=5.Proof.
LetlkM(u)=S42({a,b,c,d}) and β=abc.
Let lkM(β)={u,x}.
If x=d,
then the induced complex K=M[{u,a,b,c,d}] is a 3-pseudomanifold. Since n≥6, K is a proper subcomplex of M.
This is not possible. So, x≠d and hence ux is a nonedge in M.
Then κβ is a bistellar 1-move. Since ux is an edge in κβ(M), κβ is a required bistellar 1-move.Lemma 4.2.
LetM be an n-vertex 3-pseudomanifold and u be a vertex of degree 5.
If n≥7,
then there exists a bistellar 1-move κβ:M↦N such that deg⁡N(u)=6.Proof.
Sincedeg⁡M(u)=5,
the link of u in M is of the form S20({a,b})∗S31({x,y,z}) for some vertices a,b,x,y,z of M.
If both xyza and xuzb are facets, then the induced subcomplex M[{x,y,z,u,a,b}] is a 3-pseudomanifold. This is not possible
since n≥7.
So, without loss of generality, assume that xyza is not a facet. Again, if xyab,xzab,
and yzab all are facets, then the induced subcomplex M[{u,x,y,z,a,b}] is a 3-pseudomanifold, which is not possible.
So, assume that xyab is not a facet.
Consider the faceβ=xya.
Suppose lkM(β)={u,w}.
From the above, w∉{z,b}.
So, uw is a nonedge and hence κβ is a required bistellar 1-move.Lemma 4.3.
LetM be a nonneighbourly 8-vertex 3-pseudomanifold and u be a vertex of degree 6.
If the degree of each vertex is at least 6,
then there exists a bistellar 1-move κτ:M↦N such that deg⁡N(u)=7.Proof.
Letu be a vertex with deg⁡M(u)=6 and uv be a nonedge. Let L=lkM(u).
Claim 1.
There exists a2-face τ such that τ∪{u} and τ∪{v} are facets.
First consider the case when there exists a vertexw such that degL(w)=5.
Let lkL(w)(=lkM(uw))=C5(1,2,3,4,5).
LetK=lkM(w).
Since deg⁡(v)=6, vw is an edge. Thus K contains 7 vertices. If one of 12v,…,45v,51v is a 2-face, say 12v,
then 12wv and 12wu are facets. In this case, τ=12w serves the purpose. So, assume that 12v,…,45v,51v are nonfaces in K.
Then there are at least three 2-faces (not containing u) containing the edges 12,…,45,51 in K.
Also, there are at least three 2-faces containing v in K.
So, the number of 2-faces in K is at least 11.
This implies that deg⁡K(v)=3 or 4 and K is a 7-vertex ℝP2 or P4.
Since deg⁡K(u)=5,
it follows that K is isomorphic to R2,R3,or P4 (defined in Section 2). In each case, (since deg⁡K(u)=5,deg⁡K(v)=3 or 4, and uv is a nonedge) there exists an edge α in K such that α∪{u} and α∪{v} are 2-faces in K and hence τ=α∪{w} serves the purpose.
Now, assume thatL has no vertex of degree 5. Then L must be of the form S20({a1,a2})∗S20({b1,b2})∗S20({c1,c2}).
If possible, let aibjckv is not a facet for 1≤i,j,k≤2.
Consider the 2-face a1b1c1.
There exists a vertex x≠u such that a1b1c1x is a facet. Assume, without loss of
generality, that a1b1c1a2 is a facet. Since deg⁡(c1)>5 (resp., deg⁡(b1)>5), a1a2b2c1 (resp., a1a2b1c2) is not a facet. So, the facet (other than a1b2c1u) containing a1b2c1 must be a1b2c1c2.
Similarly, the facet (other than a1b1c2u) containing a1b1c2 must be a1b1b2c2.
Then a1b2c1c2,a1b1b2c2, and a1b2c2u are three facets containing a1b2c2,
a contradiction. This proves the claim.
By the claim, there exists a 2-simplexτ such that lkM(τ)={u,v}.
Since uv is a nonedge of M,κτ:M↦κτ(M)=N is a bistellar 1-move. Since uv is an edge in N,
it follows that deg⁡N(u)=7.Proof  of Theorem1.1.
LetM be an 8-vertex 3-pseudomanifold. Then, by Lemma 4.1, there
exist bistellar 1-moves κA1,…,κAk,
for some k≥0,
such that the degree of each vertex in κAk(⋯(κA1(M)) is at least 5. Therefore, by Lemma 4.2, there
exist bistellar 1-moves κAk+1,…,κAl,
for some l≥k,
such that the degree of each vertex in κAl(⋯κAk(⋯(κA1(M))) is at least 6. Then, by Lemma 4.3, there exist
bistellar 1-moves κAl+1,…,κAm,
for some m≥l,
such that the degree of each vertex in κAm(⋯κAl(⋯κAk(⋯(κA1(M)))) is 7. This proves the theorem.Lemma 4.4.
LetK be an 8-vertex combinatorial 3-manifold. If K is neighbourly, then K is isomorphic to S8,353,S8,363,S8,373,
or S8,383.Proof.
SinceK is a neighbourly combinatorial 3-manifold, by
Proposition 2.3, the link of any vertex is isomorphic to S5,…,S8,
or S9.
Claim 1.
The links of all the vertices
cannot be isomorphic toS9 (=S20∗C5).
Otherwise, letlk(8)=S20(6,7)∗C5(1,2,…,5).
Consider the vertex 2. Since the degree of 2 is 7, 1267 or 2367 is not a facet. Assume, without loss of
generality, that 1267 is not a facet. Again, if 1236 is a facet, then deg⁡lk(2)(6)=3 and hence lk(2)≇S9.
So, 1236 is not a facet. Similarly, 1256 is not a facet. Then the facet other than 1268 containing 126 must be 1246.
Similarly, 1247 is a facet. This implies that lk(2)=S20(6,7)∗C5(1,4,5,3,8).
Thus deg⁡(26)=5.
Similarly, deg⁡(16)=deg⁡(36)=deg⁡(46)=deg⁡(56)=5.
Then, the 7-vertex 2-sphere lk(6) contains five vertices of degree 5. This is
not possible. This proves the claim.Case 1.
Consider the case whenK has a vertex, (say 8) whose link is isomorphic to S8.
Assume, without loss of generality, that the facets containing the vertex 8 are 1238, 1268, 1348, 1458, 1568, 2348, 2478, 2678, 4578,
and 5678.
Since deg⁡(3)=7,1234∉K.
Hence the facet other than 1238 containing the face 123 is one of 1235, 1236,
or 1237.
If1236∈K,
then, clearly, deg⁡(17)=3 or 4.
If deg⁡(17)=4,
then on completing lk(1),
we see that 1457, 1567∈K,
thereby showing that deg⁡(5)=5,
an impossibility. Hence, deg⁡(17)=3 and, therefore, 1457∈K.
There are two possibilities for the completion of lk(1).
If 1347, 1356, 1357∈K,
from the links of 4 and 3,
we see that 2346,2467,3467,3567∈K.
Here, deg⁡(5)=6.
If 1346,1467,1567∈K, then deg⁡(5)=5.
Thus, 1236∉K.Subcase 1.1.
1235∈K.
Since deg⁡(1)=7,
either 1345 or 1256 is a facet. In the first case, 1257,1267,1567∈K.
Here, deg(6)=5,
a contradiction. So, 1256∈M and hence 1347,1357,1457∈K.
From the links of the vertices 1,4,7 and 5,
we see that 1256,2346,2467,3467,3567,2356∈K.
Here, K≅S8,383 by the map (1,5,8,6)(2,7)(3,4).Subcase 1.2.
1237∈K.
By the same argument as in Case 1.1 (replace the vertex 1 by vertex 2), we
get 1267,2345,2357,2457∈K.
From lk(1) and lk(7),1346,1456,3456,1367,3567∈K.
Here, K≅S8,383 by the map (1,7,8,6)(2,5)(3,4).Case 2.
K has no vertex whose link is isomorphic to S8 but has a vertex whose link is isomorphic to S6.
Using the same method as in Case 1.1, we find that K≅S8,373.Case 3.
K has no vertex whose link is isomorphic to S8 or S6 but has a vertex whose link is isomorphic to S7.
Using the same method as in Case 1.1, we find that K≅S8,363.Case 4.
K has no vertex whose link is isomorphic to S6, S7,
or S8 but has a vertex (say 8) whose link is
isomorphic to S5.
The facets through 8 can be assumed to be 1238, 1278, 1348, 1458, 1568, 1678, 2348, 2458, 2568,
and 2678.
Clearly, 1234, 1267∉K.
If deg⁡(15)=6,
then from lk(1) and lk(5),
we see that 1235, 1345, 2345∈K,
thereby showing that deg⁡(3)=5.
Hence 1237∈K.
Now, we can assume, without loss of generality, that the facets required to
complete lk(1) are 1347, 1457,
and 1567.
Now, consider lk(2).
If deg⁡(27)=6,
then after completing the links of 2 and 7, we observe that deg⁡(4)=6.
Hence deg⁡(23)=6.
The links of 2, 7,
and 6 show that 2345, 2356, 2367, 3467, 4567,
and 3456∈K.
Here, K≅S8,353 by the map (2,3,4,5,6,7,8).
This completes the proof.Lemma 4.5.
LetK be an 8-vertex neighbourly normal 3-pseudomanifold. If K has one vertex whose link is the 7-vertex torus T,
then K is isomorphic to N1,N2,N3, or N4.Proof.
Let us assume thatV(K)={1,…,8} and the link of the vertex 8 is the 7-vertex
torus T.
So, the facets containing 8 are 1248, 1268, 1348, 1378, 1568, 1578, 2358, 2378, 2458, 2678, 3468, 3568, 4578,
and 4678.
We have the following cases.
Case 1.
There is a vertex (other than the
vertex 8), say 7, whose link is isomorphic toT.
Then lk(7) has no vertex of degree 3 and hence 2367,1457,1237,1357∉K.
This implies that the facet (other than 1378) containing 137 is 1367 or 1347.
In the first case, lk(17)=C6(5,8,3,6,4,2).
Thus, 1367,1467,1247,1257∈K.
Then, from the links of 67 and 37,
we get 2567,3567,2347,3457∈K.
Now, from lk(34),1346∉K.
Then, from the links of 36,34,23,14,
and 26,
we get 1236,2346,1345,1235,1456,2456∈K.
Here, K=N1.
In the second case,lk(37)=C6(2,8,1,4,6,5).
Thus, 1347,3467,3567,2357∈K.
Now, from the links of 47 and 67,
we get 1247,2457,1567,1267∈K.
Here, K=N2.Case 2.
There is a vertex whose link is a
7-vertexℝP2.
Claim 1.
There exists a vertex inK whose link is isomorphic to R2.
If there is vertex whose link is isomorphic toR2,
then we are done. Otherwise, since Aut(lk(8)) acts transitively on {1,…,7},
assume that lk(4)≅R3 (resp., R4). Since (1,2,5,7,6,3)∈Aut(lk(8)),
we may assume that the degree 4 vertex (resp., vertices) in lk(4) is 1 (resp., are 1, 5, 6). Then, from lk(4),1247,1347,2467∈K.
This implies that lk(7) is a nonsphere and deg⁡(67)=3.
Hence lk(7)≅R2.
This proves the claim.
By the claim, we can assume thatlk(4)≅R2.
Again, we may assume that the vertex 1 is of degree 3 in lk(4).
Then, from 1k(4),1234,2347,2456,2467,3456,3457∈K.
Considering the links of the edges 36, 26, 27, 25,
and 13,
we get 1256,1235,1357∈K.
Here, K=N3.Case 3.
Only singular vertex inK is 8.
So, the link of each vertex (other than vertex 8) is an S72 (a 7-vertex 2-sphere). Since 8 is a degree 6 vertex in lk(u),
it follows that lk(u) is isomorphic to one of S5,S6, or S7 (defined in Example 2.2) for any vertex u≠8.
If lk(1)≅S5,
then (since (3,4,2,6,5,7)∈Aut(lk(8))), we may assume that the other degree 6
vertex in lk(1) is 3. Then, from the links of 1 and 3, 1348, 1234, 1346 are facets containing 134,
a contradiction. If lk(1)≅S6,
then (since lk(18)=C6(3,4,2,6,5,7)) we may assume that the degree 5 vertices in lk(1) are 2, 3, and 5. Then lk(3) cannot be an S72,
a contradiction. So, lk(1)≅S7.
Since Aut(lk(8)) acts transitively on {1,…,7},
it follows that the link of each vertex is isomorphic to S7.
Sincelk(18)=C6(3,4,2,6,5,7) and (3,4,2,6,5,7)∈Aut(lk(8)),
we may assume that the degree 5 vertices in lk(1) are 4 and 5. Since lk(4)≅S7,
it follows that 1456∉K.
Then, from lk(1),1245,1256,1347,1457∈K.
Now, from the links of 4 and 5,
we get 3467,2356∈K.
Then, from lk(2),2367∈K.
Here K=N4.
This completes the proof.Lemma 4.6.
LetK be an 8-vertex neighbourly normal 3-pseudomanifold. If K is not a combinatorial 3-manifold and has no vertex whose link is
isomorphic to the 7-vertex torus T then K is isomorphic to N5,…,N14 or N15.
Proof.
Letns be the number of singular vertices in K.
Since K is neighbourly, by Proposition 2.3, the link
of any vertex is either a 7-vertex ℝP2 or a 7-vertex S2.
So, the number of facets through a singular (resp., nonsingular) vertex
is 12 (resp., 10). Let f3 be the number of facets of K.
Consider the set S={(v,σ):σ is a facet of K and v∈σ is a vertex }.
Then f3×4=#(S)=ns×12+(8−ns)×10=80+2ns.
This implies ns is even. Since K is not a combinatorial 3-manifold, it follows
that ns≠0 and hence ns≥2.
So, K has at least two vertices whose links are
isomorphic to R2,R3, or R4.
Case 1.
There exist (at least) two
vertices whose links are isomorphic toR4.
Assume that lkM(8)=R4.
Then 1258,1268,1358,1378,1468,1478,2368,2378,2458,2478,3458,3468∈K.
Since (1,3,4)(5,6,7),(1,2)(3,4)∈Aut(lk(8)),
we may assume that lk(3) or lk(7)≅R4.Subcase 1.1.
lk(7)≅R4.
Since lklk(7)(8)=C4(1,3,2,4),
it follows that 1, 2, 3, 4 are degree 5 vertices in lk(7).
Since (3,4)(5,6)∈Aut(lk(8)),
assume without loss that 136,145∈lk(7).
Then, from lk(7),
we get 1257,1267,1367,1457,2357,2467,3457,3467∈K.
This shows that lk(2) is an ℝP72.
Since 3457,3458∈K,
it follows that 2345∉K.
Then, from lk(2), 2356,2456∈K.
Then, from the links of 3 and 4, 1356,1456∈K.
Here K=N5.Subcase 1.2.
lk(7)≇R4.
So, lk(3)≅R4.
Since lklk(3)(8)=C6(1,7,2,6,4,5),
the degree 4 vertices in lk(3) are either 5,6,7,
or 1,2,4.
In the first case, on completion of lk(3),
we observe that 56, 67, 57 remain nonedges in K.
So, the degree 4 vertices in lk(3) are 1,2,
and 3.
Then 1356, 1367, 2356, 2357, 3457,
and 3467 are facets. Since lk(7)≇R4 and deg⁡(78)=4,
either lk(7)≅R3 or lk(7) is an S72.
In the former case, 2567 is a facet. This is not possible from lk(25).
So, lk(7) is an S72.
Then, from lk(7),1467,2457∈K.
Now, from lk(1),1256∈K.
Here, K=N7.Case 2.
Exactly one vertex whose link is
isomorphic toR4 and there exists a vertex whose link is
isomorphic to R3.
Using the same method as in Case 1, we find that K≅N8.Case 3.
Exactly one vertex whose link is
isomorphic toR4,
there is no vertex whose link is isomorphic to R3 and there exists (at least) a vertex whose
link is isomorphic to R2.
Using the same method as in Case 1, we find that K≅N9.Case 4.
There is no vertex whose link is
isomorphic toR4 and there exist (at least) two vertices whose
links are isomorphic to R3.
Assume that lkK(8)=R4,
so that deg⁡(78)=4.
Using the same method as in Case 1, we get the following: (i) if lkK(7)≅R3,
then K=N6 and (ii) if lkK(7)≇R3,
then K is isomorphic to N10 or N11.Case 5.
There is no vertex whose link is
isomorphic toR4,
there exists exactly one vertex whose link is isomorphic to R3 and there exists (at least) a vertex whose
link is isomorphic to R2.
Using the same method as in Case 1, we find that K is isomorphic to N12 or N13.Case 6.
There is no vertex whose link is
isomorphic toR4 or R3 and there exist (at least) two vertices whose
links are isomorphic to R2.
Using the same method as in Case 1, we find that K is isomorphic to N14 or N15.
This completes the proof.Proof  of Theorem1.2.
SinceS8,m3's are combinatorial 3-manifolds and Nn's are not combinatorial 3-manifolds, S8,m3≇Nn for 35≤m≤38,1≤n≤15.
Part (a) now follows from Lemmas 3.2, 3.7. Part (b) follows from Lemmas 4.4, 4.5, and 4.6.Lemma 4.7.
Let𝒮0,…,𝒮6 be as in the proof of Lemma 3.4.
If a combinatorial 3-manifold K is obtained from a member of 𝒮j by a bistellar 2-move, then K is isomorphic to a member of 𝒮j+1 for 0≤j≤5.
Moreover, no bistellar 2-move is possible from a member of 𝒮6.Proof.
Recall that𝒮0={S8,353,S8,363,S8,373,S8,383}.
The removable edges in S8,373 are 13, 16, 17, 24, 27, 35, 46, 48,
and 58.
Since (1,4)(2,7)(3,8)∈Aut(S8,373),
up to isomorphisms, it is sufficient to consider the bistellar 2 -moves κ27,κ24,κ48,κ58, and κ46 only. Here S8,333:=κ27(S8,373),S8,303:=κ24(S8,373),S8,323:=κ48(S8,373),S8,313:=κ58(S8,373), and κ46(S8,373)≅S8,313 by the map (1,4,5)(2,7)(3,6,8).
The removable edges inS8,383 are 13, 38, 78, 27, 25, 15,
and 46.
Since (1,2,8)(7,3,5),(1,2)(3,7)(4,6)∈Aut(S8,383), it is sufficient to consider the bistellar 2-moves κ46 and κ78 only. Here S8,393:=κ46(S8,363) and κ78(S8,383)≅S8,323 by the map (1,7,8,4,6)(2,3).
The removable edges inS8,363 are 13,35,58,68,46,24,27, 17.
Since (1,5,6,2)(3,8,4,7) is an automorphism of S8,363,
it is sufficient to consider the bistellar 2-moves κ58 and κ68 only. Here κ58(S8,363)=S8,313 and κ68(S8,363)≅S8,303 by the map (1,6,4,8,2,5,7,3).
The removable edges inS8,353 are 13,35,57,71,24,46,68,
and 82.
Since (1,2,…,8),(1,8)(2,7)(3,6)(4,5)∈Aut(S8,353), it is sufficient to consider the bistellar 2-moves κ68 only. Here κ68(S8,353)≅S8,303 by the map (1,7,3)(2,8,4,5,6).
This proves the result for j=0.By the same arguments as in the case forj=0,
one proves for the cases for 1≤j≤5.
We summarize these cases in Figure 3 below. Last part follows from
the fact that none of S8,13,S8,33, or S8,33 has any removable edges.Figure 3
Hasse diagram of the poset of the 8-vertex
combinatorial 3-manifolds (the partial order relation is as defined in Section2).Lemma 4.8.
Let𝒩0,…,𝒩3 be as in the proof of Lemma 3.9.
If a 3-pseudomanifold K is obtained from a member of 𝒩j by a bistellar 2-move, then K is isomorphic to a member of 𝒩j+1 for 0≤j≤2.
Moreover, no bistellar 2-move is possible from a member of 𝒩3.Proof.
Recall that𝒩0={N1,…,N15}.
Since there are no degree 3 edges in N1, N2, N5,
and N6,
no bistellar 2-moves are possible from N1,N5,N6,
or N2.
The degree 3 edges in N3 (resp., in N4) are 14,16,17,36,67 (resp., 13,35,57,72,24,46,61). But, none of these edges is removable. So,
bistellar 2-moves are not possible from N3 or N4.
The removable edges inN7 are 12,14,24,56,57,
and 67.
Since (1,2)(6,7), (1,2)(5,6),
and (1,5)(2,6)(3,8)(4,7) are automorphisms of N7,
it follows that up to isomorphisms, we only have to consider the bistellar
2-move κ67.
Here, N16=κ67(N7).
The removable edges inN8 are 15,17,24,56,57,
and 67.
Since (1,6)(2,4),(1,6)(5,7),(2,4)(5,7)∈Aut(N8), we only consider the bistellar 2-moves κ24, κ56,
and κ57.
Here, N17=κ24(N8),N18=κ56(N8), and N19=κ57(N8).
The removable edges inN9 are 12,23,24,
and 67.
Since (1,4)(6,7)∈Aut(N9),
we consider only κ12,κ23,
and κ67.
Here, N21=κ12(N9),N23=κ23(N9), and κ67(N9)=N16.
The removable edges inN10 are 12,14,24,56,57,
and 67.
Since (1,7)(2,5)(3,8)(4,6),(1,4)(6,7)∈Aut(N10), we consider the bistellar 2-moves κ56 and κ57 only. Here, N20=κ56(N10) and κ67(N10)=N16.
The removable edges ofN11 are 14,24,56,57,
and 67.
Since (1,2)(5,6)(3,8)∈Aut(N11),
we only consider the bistellar 2-moves κ14,κ56, and κ67.
Here, N22=κ14(N11),κ56(N11)=N20, and κ67(N11)≅N18 (by the map (2,4)(5,7)).
The removable edges inN12 are 12, 23, 45,
and 57.
Here, κ12(N12)≅N22 (by the map (2,4,6)), κ23(N12)=N23,κ45(N12)≅N21 (by the map (1,6,5,2,7,4)(3,8)), and κ57(N12)≅N18 (by the map (1,6,7,4)).
The removable edges inN13 are 12,23,24,56,57,
and 67.
Since (1,4)(6,7)∈Aut(N13),
we only consider κ12,κ23,κ57,
and κ67.
Here, κ12(N13)≅N22 (by the map (2,7,5,4)), κ23(N13)=N23,κ57(N13)≅N18 (by the map (1,4)(6,7)), and κ67(N13)=N16.
The removable edges inN14 are 38,56,57,67.
Since (1,2,4)(5,6,7)(3,8)∈Aut(N14),
we only consider κ38 and κ57.
Here, N24=κ38(N14) and κ57(N14)=N19.
The removable edges inN15 are 15,23,24,58.
Since (1,7)(2,5)(3,8)(4,6)∈Aut(N15),
we only consider the bistellar 2-moves κ23 and κ24.
Here, κ23(N15)=N23 and κ24(N15)≅N21 (by the map (1,6,5,7,4)). This proves the result for j=0.By the same arguments as in the case forj=0,
one proves the same for other cases (namely, for j=1,2) as well. We summarize these cases in Figure 4. Last part
follows from the fact that, for Ni∈𝒩3,Ni has no removable edge.Figure 4
Hasse diagram of the poset of all the 3-pseudomanifoldsN7,…,N35.Proof  of Corollary1.3.
Let𝒮0,…,𝒮6 be as in the proof of Lemma 3.4.
Let M be an 8-vertex combinatorial 3-manifold. Then,
by Theorem 1.1, there exist bistellar 1-moves κA1,…,κAm,
for some m≥0,
such that M1:=κAm(⋯(κA1(M))) is a neighbourly 8-vertex 3-pseudomanifold.
Since bistellar moves send a combinatorial 3-manifold to a combinatorial
3-manifold, M1 is a combinatorial 3-manifold. Then, by Theorem
1.2, M1∈𝒮0.
In other words, M=κe1(⋯(κem(M1))),
where M1∈𝒮0 and κem:M1↦κem(M1), κei:κei+1(⋯(κem(M1)))↦κei(⋯(κem(M1))),
for 1≤i≤m−1,
are bistellar 2-moves. Therefore, by Lemma 4.7, M∈𝒮0∪⋯∪𝒮6.
The result now follows from Lemma 3.4.Proof  of Corollary1.4.
Let𝒩0,…,𝒩3 be as in the proof of Lemma 3.9.
Let M be an 8-vertex normal 3-pseudomanifold. Then,
by Theorem 1.1, there exist bistellar 1-moves κA1,…,κAm,
for some m≥0,
such that M1:=κAm(⋯(κA1(M))) is a neighbourly 3-pseudomanifold. Since
bistellar moves send a normal 3-pseudomanifold to a normal 3-pseudomanifold, M1 is normal. Hence, by Theorem 1.2, M1∈𝒩0.
In other words, M=κe1(⋯(κem(M1))),
where M1∈𝒩0 and κem:M1↦κem(M1), κei:κei+1(⋯(κem(M1)))↦κei(⋯(κem(M1))),
for 1≤i≤m−1,
are bistellar 2-moves. Therefore, by Lemma 4.8, M∈𝒩0∪𝒩1∪𝒩2∪𝒩3.
The result now follows from Lemma 3.9.



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