Two-dimensional weak pseudomanifolds on eight vertices

Proc. Indian Acad. Sci. (Math. Sci.) Vol. 112, No. 2, May 2002, pp. 257–281.© Printed in India

Two-dimensional weak pseudomanifolds on eight vertices

BASUDEB DATTA and NANDINI NILAKANTAN

Department of Mathematics, Indian Institute of Science, Bangalore 560 012, IndiaE-mail: [email protected]; [email protected]

MS received 20 September 2001

Abstract. We explicitly determine all the two-dimensional weak pseudomanifolds on8 vertices. We prove that there are (up to isomorphism) exactly 95 such weak pseudo-manifolds, 44 of which are combinatorial 2-manifolds. These 95 weak pseudomanifoldstriangulate 16 topological spaces. As a consequence, we prove that there are exactlythree 8-vertex two-dimensional orientable pseudomanifolds which allow degree threemaps to the 4-vertex 2-sphere.

Keywords. Two-dimensional complexes; pseudomanifolds; degree of a map.

1. Introduction

Recall that asimplicial complex(in short,complex) is a collection of non-empty finite setssuch that every non-empty subset of an element is also an element. Fori ≥ 0, the elementsof sizei + 1 are called thei-simplicesof the complex. Fori = 1, 2, thei-simplices arealso called theedgesand trianglesof the complex, respectively. For a complexX, themaximum ofk such thatX has ak-simplex is called thedimensionof X. The union ofall the simplices of a complexX is called thevertex-setof X and is denoted byV (X).Elements ofV (X) are calledverticesofX. A complexX is calledfinite if V (X) is a finiteset. Ak-simplex{v0, . . . , vk} of a complex is also denoted byv0 · · · vk.

If X1 andX2 are two complexes, then a simplicial map fromX1 to X2 is a mapϕ :V (X1) → V (X2) such thatσ ∈ X1 impliesϕ(σ) ∈ X2. A bijectionπ : V (X1) → V (X2)

is called anisomorphismif bothπ andπ−1 are simplicial. Two complexesX1, X2 are called(simplicially) isomorphicwhen such an isomorphism exists. We identify two complexes ifthey are isomorphic. An isomorphism from a complexX to itself is called anautomorphismof X. All the automorphisms ofX form a group, which is denoted by Aut(X). Twosimplicial mapsf, g : X1 → X2 are said to beequivalent(denoted byf ∼= g) if thereexistϕ ∈ Aut(X1) andψ ∈ Aut(X2) such thatψ ◦ f ◦ ϕ = g.

A d-dimensional simplicial complexX is called ad-dimensionalweak pseudomanifoldif each simplex ofX is contained in ad-simplex ofX and each(d − 1)-simplex ofX iscontained in exactly twod-simplices ofX. A d-dimensional weak pseudomanifoldX iscalled apseudomanifold(without boundary) if for any pairσ , λ of d-simplices ofX, thereexists a sequenceτ1, . . . , τn of d-simplices ofX, such thatσ = τ1, λ = τn andτi ∩ τi+1is a(d−1)-simplex ofX for 1 ≤ i ≤ n−1. (P1 andP2, given in §2, are pseudomanifoldsbutQ1 andQ2 are not.)

A simplicial complex is usually thought of as a prescription for the construction of atopological space by pasting together geometric simplices (see §3 for finite complexes,

257

258 Basudeb Datta and Nandini Nilakantan

[8] and [11] for the general case). The space thus obtained from a complexK is called thegeometric carrierof K and is denoted by|K|. We also say thatK triangulates|K|.

For any setV ond+2 (≥ 2) elements, letK be the simplicial complex whose simplicesare all the non-empty proper subsets ofV . Then|K| is homeomorphic to the sphereSd .This complex is called thestandardd-sphereand is denoted bySdd+2(V ) or simply bySdd+2. A finite complexX is called acombinatoriald-sphere, if |X| is PL-homeomorphicto |Sdd+2| [10]. Clearly, a finite one-dimensional complex is a combinatorial 1-sphere ifand only if it is a pseudomanifold. Forn ≥ 3, the combinatorial 1-sphere withn verticesis unique and is denoted byCn. The complexCn is also called ann-cycle. An n-cycle withedgesv1v2, . . . , vn−1vn, vnv1 is also denoted byCn(v1, . . . , vn).

If v is a vertex of a simplicial complexX, then thelink of v in X, denoted by LkX(v),is the complex whose simplices are those simplicesτ of X such thatv 6∈ τ and{v} ∪ τis a simplex ofX. The number of vertices in the link ofv is called thedegreeof v and isdenoted by deg(v). Clearly, the link of a vertex in ad-dimensional weak pseudomanifoldis a(d − 1)-dimensional weak pseudomanifold.

A finite simplicial complexX is called acombinatoriald-manifold if |X| is a d-dimensional PL-manifold (without boundary), i.e., LkX(v) is a combinatorial(d − 1)-sphere for each vertexv in X [7,10]. So,X is a combinatorial 2-manifold if the link ofeach vertex is a cycle. We also know (e.g., see [7]) that a finite simplicial complexK is acombinatorial 2-manifold if and only if|K| is a two-dimensional topological manifold.

A vertex of a finite two-dimensional weak pseudomanifold is calledsingular if its linkis not a cycle (and hence consists of more than one cycle). So, a two-dimensional weakpseudomanifold is not a combinatorial manifold if and only if it contains a singular vertex.(In each ofP1, P2,Q1 andQ2, 7 is a singular vertex.)

A combinatorial 2-manifoldX is calledd-equivelarif each vertex ofX has degreed.A combinatorial 2-manifold is calledequivelarif it is d-equivelar for somed.

If the number ofi-simplices of ad-dimensional finite complexX is fi(X) (0 ≤ i ≤ d),then the numberχ(X) := ∑d

i=0(−1)ifi(X) is called theEuler characteristicof X.If K is ad-dimensional oriented pseudomanifold, then [zK ] generatesHd(K,Z), where

zK := ∑σd∈K 1·σd (summation is taken over all the positively orientedd-simplices). Let

K andL be two orientedd-dimensional pseudomanifolds. Ifϕ : K → L is a simplicialmap thenϕ∗

d : Hd(K,Z) → Hd(L,Z) is a homomorphism and hence there existsm ∈ Z

such thatϕ∗d ([zK ]) = m[zL]. Thism is called thedegreeof ϕ and is denoted by deg(ϕ)

[11].LetK be a two-dimensional pseudomanifold andϕ : K → S2

4 be a simplicial map. Ifϕ(u) 6= ϕ(v) for each edgeuv of K thenϕ is called a 4-coloring [12].

It is known (e.g., see [3,5]) that if the number of vertices of a two-dimensional weakpseudomanifoldM is at most 6 thenM is a combinatorial 2-manifold andM is isomorphicto S1, . . . , S4 orR1 (given in §2).

In [5], we have seen that there are exactly nine 7-vertex combinatorial 2-manifoldsand four 7-vertex two-dimensional weak pseudomanifolds which are not combinatorial2-manifolds. Among the four non-manifolds two are pseudomanifolds, which triangulatethepinched sphere(the space obtained by identifying two points ofS2).

In [6], we have determined all the equivelar combinatorial 2-manifolds on at most 11vertices. There are 27 such equivelar combinatorial 2-manifolds.

Altshuler and Steinberg [2] showed that there are fourteen 8-vertex combinatorial 2-spheres. Cervone [4] showed that there are exactly six 8-vertex combinatorial 2-manifolds,which triangulate the Klein bottle. It is known (e.g., see [7,9]) that there does not exist

Two-dimensional weak pseudomanifolds 259

any 8-vertex combinatorial 2-manifold of Euler characteristic−1. Here, we classify allthe two-dimensional weak pseudomanifolds on 8 vertices. More explicitly, we prove:

Theorem 1.1. There are exactly44 distinct combinatorial2-manifolds on8 vertices,namely,S10, . . . , S23, R5, . . . , R20, T2, . . . , T8,K1, . . . , K6 andD (given in§2).

Theorem 1.2. There are exactly51distinct8-vertex two-dimensional weak pseudomani-folds which are not combinatorial2-manifolds, namely,P3, . . . , P39,Q3, . . . ,Q16 (givenin §2).

COROLLARY 1.3

LetM be a two-dimensional weak pseudomanifold onn (≤ 8) vertices.

(i) If |M| is a manifold then|M| is homeomorphic to the2-sphere(S2), the real projectiveplane(RP 2), the torus(S1 × S1), the Klein bottle(K) or the space consisting of twodisjoint2-spheres.

(ii) If |M| is not a manifold then|M| is homeomorphic to the pinched sphere(P ), RP 2#P ,P#P , RP 2#P#P , K#P , (S1 × S1)#P , the union of twoS2’s having one, two, threeor four points in common or the union ofS2 andRP 2 having three points in common(given at the end of§2). (Here,A#B denotes the connected sum ofA andB).

If ϕ : K → S24 is a simplicial map, whereK is an oriented 8-vertex two-dimensional

pseudomanifold, thenf2(K) ≤ 18 and hence deg(ϕ) ≤ 4. Here we prove:

Theorem 1.4. Let ϕ : K2n → S2

4 be a simplicial map, whereK2n is a two-dimensional

oriented pseudomanifold onn vertices. Letf , g andh be as in Example2.1. If n ≤ 8 thendeg(ϕ) ≤ 3 (and hence≥ −3). Equality is attained here if and only ifϕ is equivalent tof , g or h.

Remark1.5. Observe thatP3, . . . , P39 (in Theorem 1.2) are pseudomanifolds, whereasQ3, . . . ,Q16 are not pseudomanifolds. Among the pseudomanifolds,P3, . . . , P20,P28, . . . , P36 andP39 are orientable andP21, . . . , P27, P37 andP38 are non-orientable.

Remark1.6. If M is an 8-vertex two-dimensional weak pseudomanifold, then it is easyto see thatχ(M) lies between−1 and 4. IfM is a combinatorial 2-manifold then, fromTheorem 1.1,χ(M) is 0, 1, 2 or 4. However, by Theorem 1.2, there exist weak pseudo-manifolds with Euler characteristic−1, . . . ,3.

Remark1.7. Theorem 1.1, Theorem 1.2 and Proposition 3.1 together with the resultsin [3], which determine all thed-dimensional weak pseudomanifolds on at mostd + 4vertices, classify all thed-dimensional (d 6= 3) weak pseudomanifolds on less than orequal to 8 vertices. Moreover, Theorem 1.1 together with Proposition 3.1, the results in[1], which classify all the combinatorial 3-manifolds on at most 8 vertices and the resultsin [3] classify all the combinatorial manifolds on less than or equal to 8 vertices.

In §2 we present all the two-dimensional weak pseudomanifolds on at most 8 vertices.In §3 we give some definitions, constructions and results which we shall need later. In§4 we consider combinatorial manifolds and prove Theorem 1.1. In §5 we consider weakpseudomanifolds which are not combinatorial manifolds and prove Theorem 1.2. In §6 weprove Corollary 1.3 and Theorem 1.4.


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2. Examples

In Theorems 1.1 and 1.2 we have stated that there are 95 two-dimensional weak pseudo-manifolds on 8 vertices. In this section we present all these 95 weak pseudomanifolds.We also present all the 18 two-dimensional weak pseudomanifolds on less than or equalto 7 vertices. The degree sequences are presented parenthetically below the figures. For0 ≤ i ≤ 7, i in the figures represents the vertexvi . At the end of this section, we present


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the geometric carriers of all the weak pseudomanifolds on 8 vertices. At the beginning wepresent three degree 3 maps (which we have mentioned in Theorem 1.4) to the 4-vertex2-sphere.

Example2.1. LetS1 = S24({a, b, c, d})with the positively oriented 2-simplicesabc, acd,

adb, bdc. Let S15, P34 andP35 be as given below:


7

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(a) Consider the orientation onS15 given by the positively oriented 2-simplices 176, 160,064, 104, 143, 132, 234, 524, 546, 562, 267, 127. Letf : S15 → S1 be the simplicialmap given byf (1) = f (5) = a, f (0) = f (2) = b, f (4) = f (7) = c andf (3) =f (6) = d.


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(b) Consider the orientation onP34 given by the positively oriented 2-simplices 475, 467,765, 056, 015, 061, 162, 426, 024, 043, 453, 135, 173, 037, 072, 127. Letg : P34 → S1be the simplicial map given byg(0) = g(4) = a, g(1) = g(7) = b, g(2) = g(5) = c

andg(3) = g(6) = d.(c) Consider the orientation onP35 given by the positively oriented 2-simplices 523, 537,

572, 274, 704, 760, 673, 163, 130, 203, 102, 124, 146, 564, 506, 540. Leth : P35 → S1be the simplicial map given byh(1) = h(5) = a, h(2) = h(6) = b, h(3) = h(4) = c

andh(0) = h(7) = d.


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Then deg(f ) = deg(g) = deg(h) = 3. The mapf is a 4-coloring butg andh are not.

3. Preliminaries

For a finite simplicial complexX, if ni (> 0) is the number of vertices of degreedi andd1 > d2 > · · · , thendn1

1 . . . . .dnkk is called thedegree sequenceof X, where

∑ki=1 ni is

equal to the number of vertices ofX.


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If mi (> 0) is the number of singular vertices of degreeci in a two-dimensional weakpseudomanifoldX and c1 > c2 > · · · then cm1

1 . . . . .cmkk is called thesingular degree

sequenceof X, where∑ki=1mi is equal to the number of singular vertices ofX.

If X is a finite simplicial complex then one defines a geometric realization ofX asfollows: Let V (X) = {v1, . . . , vn}. We choose a set ofn points{x1, . . . , xn} in R

N (forsomeN ) in such a way that a subsetS = {xj1 . . . , xji+1} of i + 1 points is affinelyindependent ifσ = vj1 · · · vji+1 is a simplex ofX. The convex set spanned byS is calledthe geometric carrierof σ or the geometric simplexcorresponding toσ and denoted


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24

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7

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3

6

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1

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75

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23

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76

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123

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74

1

1

23

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7

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4 22

by |σ |. SinceX is finite we can chooseN so thatσ ∩ γ = ∅ implies |σ | ∩ |γ | = ∅.The setX := {|σ | : σ ∈ X, σ ∩ γ = ∅ ⇒ |σ | ∩ |γ | = ∅} is called ageometricsimplicial complexcorresponding toX or ageometric realizationof X. The topologicalspace|X| := ∪σ∈X|σ | is called ageometric carrierof X. Clearly, if two finite complexeshave a common geometric realization, then they are isomorphic and isomorphic finitecomplexes have homeomorphic geometric carriers [8]. We identify a complex with itsgeometric realization.


S2 S2

P RP2# P # P

RP2

RP2

RP2 # P # PP

# P K # P

K

S

S

2

2

S x S1 1

( )

S2RP2

1 0

567

4

5

76 2

1

2

03

123

765

40

3

12

723

5

1

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07

645

1

Q 14(7.5 )

44Q 15

(6 )8

16Q (7.6.3)43

S2S2

S2S2

If M is ann-vertex two-dimensional weak pseudomanifold withf1 edges andf2 trian-gles then 3f2 = 2f1. Therefore,

χ(M) = n− f1 + f2 = n− f1

3= n− f2

2. (1)


In [5], we have seen the following:

PROPOSITION 3.1

There are exactly13distinct two-dimensional weak pseudomanifolds on7vertices, namely,T1, R2, R3, R4, S5, . . . , S9, P1, P2,Q1 andQ2.

Altshuler and Steinberg [2] showed the following:

PROPOSITION 3.2

There are exactly14distinct combinatorial2-spheres on 8 vertices, namely,S10, . . . , S23.

If X1, X2 are two simplicial complexes with disjoint vertex sets, then theirjoinX1 ∗X2is the complex whose simplices are those ofX1 andX2 and the unions of simplices ofX1 with simplices ofX2. If bothX1 andX2 are pseudomanifolds then so isX1 ∗ X2 [3].Observe thatS0

2({a, b}) ∗ S02({c, d}) = C4(a, c, b, d).

If M is a two-dimensional simplicial complex andτ=abc is a 2-simplex then(M\τ)∪(∂τ ∗ v) denotes the two-dimensional complex whose 2-simplices areabv, acv, bcv andthose ofM other thanτ , wherev 6∈ V (M). This complex is said to be obtained fromMby starring the vertexv in τ . Observe thatR2 is obtained fromR1 by starring the vertex0 in 123. Similarly, ifab is an edge (contained inabc andabd) andv /∈ V (M) then(M! \{ab})∪ (C4(a, c, b, d)∗v) denotes the two-dimensional complex whose 2-simplicesareacv, adv, bcv, bdv and those ofM other thanabc andabd. This complex is said tobe obtained fromM by starring the vertexv in the edgeab. The complexR3 is obtainedfromR1 by starring the vertex 0 in the edge 12.

LetM, N be two simplicial complexes withσ1, . . . , σm ∈ M andτ1, . . . , τm ∈ N . Wesay (M, σ1, . . . , σm) and (N, τ1, . . . , τm) are isomorphic (denoted by(M, σ1, . . . , σm)∼= (N, τ1, . . . , τm)) if there exists an isomorphismϕ fromM to N such thatϕ(σi) = τifor 1 ≤ i ≤ m.

LetM be ad-dimensional weak pseudomanifold. Letu andv be two distinct verticesof M such thatuv is not an edge. IfV (Lk(u)) ∩ V (Lk(v)) = ∅ then define the complexM = {τ ∈ M : u 6∈ τ, v 6∈ τ }∪ {(τ\{u})∪{w} : u ∈ τ }∪ {(τ\{v})∪{w} : v ∈ τ }. ThisMis called the simplicial complex obtained fromM by identifyingu andv (to a new vertexw). Observe thatP1 is obtained fromS20 by identifying vertices 0 and 3 ofS20 andP2 isis obtained fromS16 by identifying vertices 0 and 5 ofS16.

If τ1 = abc andτ2 = xyz are two disjoint 2-simplices of a two-dimensional simplicialcomplexM andv 6∈ V (M) then(M \{τ1, τ2}) ∪ ((∂τ1 ∪ ∂τ2) ∗ v) denotes the complexwhose 2-simplices areabv, acv, bcv, xyv, yzv, xzv and those ofM other thanτ1 andτ2.Observe that we getP33 from P2 andP39 from T1 by this process.

From these definitions one gets the following:

PROPOSITION 3.3

Let M be obtained fromM by identifying two verticesu andv.

(a) If M is a (weak) pseudomanifold, then so isM.(b) If N is obtained fromN by identifying two verticesu1 and v1 and (N, {u1}, {v1})∼= (M, {u}, {v}), thenN ∼= M.(c) If M is a two-dimensional pseudomanifold and bothu and v are non-singular then

|M| is homeomorphic to the connected sum of|M| and the pinched sphere.


PROPOSITION 3.4

LetM be a two-dimensional simplicial complex.

(a) Let τ be ani-simplex(1 ≤ i ≤ 2) andM be obtained fromM by starring a vertex inτ . If M is a weak pseudomanifold, pseudomanifold or combinatorial2-manifold thenso isM with the same geometric carrier.

(b) Letτ1, τ2 are disjoint 2-simplices ofM andu, v,w 6∈ V (M). LetM := (M\{τ1, τ2})∪((∂τ1 ∪ ∂τ2) ∗w). LetN be the complex obtained fromM by starringu in τ1 andv inτ2. LetN be obtained fromN by identifyingu andv. IfM is a(weak) pseudomanifoldthen so isN and N ∼= M (and hence|M| is homeomorphic to the connected sum of|M| and the pinched sphere wheneverM is a pseudomanifold).

PROPOSITION 3.5

LetM1 andM2 be two-dimensional weak pseudomanifolds.

(a) Let Mj be obtained fromMj by starring a vertex on ani-simplex(1 ≤ i ≤ 2) σj forj = 1, 2. If (M1, σ1) ∼= (M2, σ2), thenM1 ∼= M2.

(b) If (M1, σ1, τ1) ∼= (M2, σ2, τ2) and u1, u2 /∈ V (M1) ∪ V (M2), whereσj , τj aredisjoint 2-simplices ofMj for j = 1, 2, then(M1\{σ1, τ1}) ∪ ((∂σ1 ∪ ∂τ1) ∗ u1) ∼=(M2\{σ2, τ2}) ∪ ((∂σ2 ∪ ∂τ2) ∗ u2).

PROPOSITION 3.6

Letϕ : K → L be a simplicial map of degreed > 0, whereK andL are two-dimensionaloriented pseudomanifolds. For a vertexv of L, let Sv := {σ ∈ K : ϕ(σ) is a 2-simplexcontainingv}.(a) If σ is a 2-simplex ofL thenϕ−1(σ ) contains at leastd simplices.(b) If for some 2-simplexσ ofL, ϕ−1(σ ) containsd or d + 1 2-simplices, sayσ1, . . . , σd

(or σd+1), thenσi andσj have at most one vertex in common fori 6= j .(c) If for some vertexv of degreec of L, Sv contains the 2-simplicesτi ∪ {ui}, τi ∪ {vi},

whereϕ(ui) = ϕ(vi) = v for 1 ≤ i ≤ p, then#(Sv) ≥ cd + 2p.

Proof. (a) follows from the definition of the degree of a simplicial map.If possible letϕ(uvx) = ϕ(uvy) = σ . If ϕ2 : C2(K) → C2(L) is the homomorphism

induced byϕ then, for any orientations ofK andL,ϕ2(+uvx) = −ϕ2(+uvy) inC2(L). So,if m is the coefficient of+σ in ϕ2(C2(K)) then|m| ≤ d+1−2 and hence deg(ϕ) ≤ d−1,a contradiction. This proves (b).

By the same argument as in (b),ϕ2(+τi ∪ {ui}) = −ϕ2(+τi ∪ {vi}), for 1 ≤ i ≤ p.Therefore, for each 2-simplexσ containingv (as the degree ofϕ is d) #(ϕ−1

2 (σ )\{τi ∪{ui}, τi ∪ {vi} : 1 ≤ i ≤ p}) ≥ d. This proves (c). 2

For a simplicial mapϕ : K → L, ad-simplexσ is said to becollapsingif ϕ(σ) is not ad-simplex. Letϕ : K → S2

4 be a simplicial map, whereK is anm-vertex two-dimensionalweak pseudomanifold andS2

4 is the standard 2-sphere (with 2-simplicesσ1, . . . , σ4). Themapϕ is said to be oftype (n1, n2, n3, n4) if ni is the number of triangles (inK) withimageσi (1 ≤ i ≤ 4) andn1 ≥ n2 ≥ n3 ≥ n4.


4. Proof of Theorem 1.1

Throughout this sectionM is an 8-vertex combinatorial 2-manifold.The first lemma follows from the description ofR2, R3 andR4 in §2.

Lemma4.1. If R2, R3 andR4 are as in§2 then we have the following:

(i) (R2, v0v1v2)α1∼= (R2, v0v1v3)

α2∼= (R2, v0v2v3).

(ii) (R2, v1v3v5)α2∼= (R2, v2v3v6)

α3∼= (R2, v1v2v4).

(iii) (R2, σ ) ∼= (R2, v3v4v5), for σ = v3v4v6, v2v5v6, v2v4v5, v1v4v6, v1v5v6.

(iv) (R3, v0v1v3)α2∼= (R3, v0v2v3)

α4∼= (R3, v0v2v4)α2∼= (R3, v0v1v4).

(v) (R3, v1v3v5)α2∼= (R3, v2v3v6)

α4∼= (R3, v2v4v5)α2∼= (R3, v1v4v6).

(vi) (R3, v3v4v5)α2∼= (R3, v3v4v6), (R3, v1v5v6)

α2∼= (R3, v2v5v6).

(vii) (R4, σ1) ∼= (R4, σ2) for any 2 trianglesσ1, σ2 ofR4,

whereα1, . . . , α8 : {v0, . . . , v6} → {v0, . . . , v6} are the permutations given byα1 =(v2, v3)(v4, v5), α2 = (v1, v2)(v5, v6), α3 = (v1, v3)(v4, v6), α4 = (v3, v4)(v5, v6), α5 =(v2, v3)(v0, v6), α6 = (v2, v4)(v0, v5), α7 = (v1, v4)(v0, v6) andα8 = (v1, v3)(v0, v5).

Here(Ri, τ )αl∼= (Rj , σ ) means(Ri, τ ) and(Rj , σ ) are isomorphic via the mapαl .

Lemma4.2. If χ(M) = 1 andM has a vertex of degree3, then M is isomorphic toR5,. . . , R11 or R12 defined above.

Proof. Let v7 be a vertex of degree 3 ofM and let Lk(v7) = C3(a, b, c). SinceM 6∼= S24,

abc is not a simplex. LetM = (M \ {v7}) ∪ {τ }. Then M is a 7-vertex combinato-rial 2-manifold withχ(M) = χ(M) = 1. Hence, by Lemma 4.1,(M, τ ) is isomorphicto (R2, v0v1v2), (R2, v2v3v6), (R2, v1v5v6), (R3, v0v1v3), (R3, v2v3v6), (R3, v3v4v5),(R3, v1v5v6) or (R4, v0v1v3).

If (M, τ ) is isomorphic to(R2, v0v1v2) then, by Proposition 3.5(a),M = (M \{τ }) ∪(∂τ ∗ v7) is isomorphic toR2,1, whereR2,1 = (R2\{v0v1v2})∪ ({v0v1, v1v2, v0v2} ∗ v7).

SinceR2,1 isR5,M is isomorphic toR5.Similarly in the other casesM is isomorphic to one of the following:R2,2 := (R2\

{v2v3v6}) ∪ ({v2v3, v3v6, v2v6} ∗ v7), R2,3 := (R2 \ {v1v5v6}) ∪ ({v1v5, v5v6, v1v6} ∗v7), R3,1 := (R3 \ {v0v1v3}) ∪ ({v0v1, v1v3, v0v3} ∗ v7), R3,2 := (R3 \ {v2v3v6}) ∪({v2v3, v3v6, v2v6}∗v7),R3,3 := (R3\{v3v4v5})∪({v3v4, v4v5, v3v5}∗v7),R3,4 := (R3\{v1v5v6})∪ ({v1v5, v5v6, v1v6} ∗ v7),R4,1 := (R4 \ {v0v1v3})∪ ({v0v1, v1v3, v0v3} ∗ v7).

Now observe thatR2,2 = R6, R2,3 = R7, R3,1 = R8, R3,2 = R9, R3,3 = R10, R3,4 =R11 andR4,1 = R12. This proves the lemma. 2

Lemma4.3. If χ(M)=1 and there exists a vertex of degree7 ofM and no vertex of degree3, thenM is isomorphic toR13, . . . , R16 or R17.

Proof. By (1), the number of 2-simplices ofM is 14. Letv0 be a vertex of degree 7 andLk(v0) = C7(v1, . . . , v7).

Claim. Any 2-simplex not containingv0 contains exactly one edge fromC7(v1, . . . , v7).


Since the degree of each vertex is more than 3,vivi+1vi+2 is not a simplex for 1≤ i ≤ 7(addition in the subscript is modulo 7). So, no 2-simplex contains more than one edge fromC7(v1, . . . , v7). Let abc be a simplex not containingv0. If none ofab, bc or ca is fromC7(v1, . . . , v7) then the number of 2-simplices is more than 14 (7 throughv0, 7 throughthe edges inC7(v1, . . . , v7) andabc), a contradiction. These prove the claim.

First consider the case when there exists no triangle of the formvivi+1vi+3 orvivi+1vi+5.In this case, by the claim, the other triangles arev1v2v5, v2v3v6, v3v4v7, v1v4v5, v2v5v6,v3v6v7 andv1v4v7. ThenM isR17.

Now, assume that there exists a simplex of the formvivi+1vi+3 or vivi+1vi+5. We mayassume thatv1v2v4 is a simplex (the case whenv1v2v6 is a simplex gives isomorphiccomplexes). Then, by repeated use of the claim, the following are the possibilities for theremaining 6 triangles.

(1) v2v4v5, v2v3v5, v3v5v6, v3v4v6, v4v6v7, v1v4v7,(2) v2v4v5, v2v3v5, v3v5v6, v3v6v7, v1v3v7, v1v3v4,(3) v2v4v5, v2v3v5, v3v5v6, v3v6v7, v3v4v7, v1v4v7,(4) v2v4v5, v2v5v6, v2v3v6, v3v4v6, v4v6v7, v1v4v7,(5) v2v4v5, v2v5v6, v2v3v6, v3v6v7, v3v4v7, v1v4v7,(6) v2v4v5, v2v5v6, v2v3v6, v3v6v7, v1v3v7, v1v3v4,(7) v2v4v5, v2v5v6, v2v6v7, v2v3v7, v3v4v7, v1v4v7,(8) v2v4v5, v2v5v6, v2v6v7, v2v3v7, v1v3v7, v1v3v4.

For 1 ≤ i ≤ 8, let Mi be the combinatorial 2-manifold arising in case (i) above. Letα1, α2, α3 : {v0, . . . , v7} → {v0, . . . , v7} be the permutations given byα1 = (v1, v6)

(v2, v5)(v3, v4), α2 = (v1, v5)(v2, v4)(v6, v7) andα3 = (v1, v4)(v2, v3)(v5, v7). Then

M8

α3∼= M2

α1∼= M1 = R13, M6

α2∼= M3 = R14, M7

α2∼= M4 = R15, M5 = R16.

So, in this case,M is isomorphic toR13, R14, R15 orR16. This proves the lemma. 2

Lemma4.4. If χ(M) = 1 and3 < deg(v) < 7, for each vertexv, thenM is isomorphictoR18, R19 or R20.

Proof. Since (by (1)) the number of edges ofM is 21, there exists a vertex (sayv0) ofdegree 6. Let Lk(v0) = C6(v1, . . . , v6). Let v7 be the remaining vertex. If Lk(v0) andLk(v7) have no common edge then number of 2-simplices is more than 14 (6 throughv0, 6more through the edges inC6(v1, . . . , v6) and at least 4 more throughv7), a contradiction.Without loss of generality, we can assume thatv4v5v7 is a simplex.

Let vv1v2 (6= v0v1v2) be the other simplex containingv1v2. Since the degree of eachvertex is more than 3,v 6∈ {v3, v6}. Hence, without loss of generality, we can assume thatv = v4 or v7.

Case I.v1v2v4 is a simplex. Asv2v3v4 is not a simplex, Lk(v4) = C6(v3, v0, v5, v7, v2, v1).Sincev1v5v6 is not a simplex, Lk(v1) isC5(v6, v0, v2, v4, v3) orC6(v6, v0, v2, v4, v3, v7).

Subcase I.1. Lk(v1) = C5(v6, v0, v2, v4, v3). Since deg(v7) > 3, v2v5 /∈ Lk(v7). There-forev2v3 or v2v6 is in Lk(v7).

In the first case, from the links ofv3 andv7, v3v6v7 andv5v6v7 are simplices. ThenMis isomorphic, via the map(v0, v6, v2, v1, v5, v7)(v3, v4), toR20.


In the second case,v2v3v6 is a simplex throughv2 and hencev5v6v7 is a simplex. ThenM is isomorphic, via the map(v0, v6, v3, v5, v7)(v1, v4, v2), toR19.

Subcase I.2. Lk(v1) = C6(v6, v0, v2, v4, v3, v7). Then Lk(v7) isC6(v2, v4, v5, v3, v1, v6)

or C6(v2, v4, v5, v6, v1, v3). If Lk (v7)=C6(v2, v4, v5, v3, v1, v6) then Lk(v2) = C6(v1,

v0, v3, v6, v7, v4). These give more than 14 triangles, which is not possible. So, Lk(v7) =C6(v2, v4, v5, v6, v1, v3). ThenM is isomorphic, via the map(v0, v2, v5)(v7, v3, v4, v1, v6),toR19.

Case II. v1v2v7 is a simplex. Since the degree of each vertex is more than 3, the 2-simplex(6= v0v3v4) containingv3v4 is v1v3v4, v3v4v6 or v3v4v7.

In the first case, by similar argument as above,M is isomorphic toR18 orR20.In each of the other two cases,M is isomorphic toR18, R19 orR20. This completes the

proof of Lemma 4.4. 2

Lemma4.5. If χ(M) = 0 and there exists a vertex of degree3 inM, thenM is isomorphicto T3.

Proof. Let deg(v0) = 3 and let Lk(v0) = C3(v1, v2, v3). SinceM 6∼= S24, v1v2v3 is not a

simplex. LetM be the complex on the vertex setV (M)\{v0} and whose 2-simplices arev1v2v3 and those ofM which do not containv0. ThenM is a 7 vertex 2-manifold whichhas 14 triangles and hence 21 edges. From Proposition 3.1,M is isomorphic toT1. Then,by Proposition 3.5(a),M is isomorphic toT3. 2

Lemma4.6. If χ(M) = 0 andM has no vertex of degree7, thenM is isomorphic toT2.

Proof. Since the degree of each vertex is less than 7, the degree of each vertex has to be6. Letv0 be a vertex whose link isC6(v1, . . . , v6). Let v7 be the remaining vertex. Thenv1, . . . , v6 ∈ Lk(v7). Since the degree of each vertex is more than 3,vivi+1vi+2 is nota simplex inM for 1 ≤ i ≤ 6 (addition in the subscripts are modulo 6). Since|M| isnot homeomorphic toS2, Lk(v7) 6= C6(v1, . . . , v6). Without loss of generality, assumev1v2 /∈ Lk(v7). Then eitherv1v2v4 or v1v2v5 is the other 2-simplex containingv1v2.Assume, without loss of generality, thatv1v2v4 is a simplex. So,v3v0, v0v5 andv1v2 areedges in the link ofv4.

Claim. Neitherv1v3 norv2v5 is an edge in the link ofv4.

If v2v5 ∈ Lk(v4) then Lk(v4) = C6(v3, v0, v5, v2, v1, v7). Sincev1v5v6 is not a simplex,Lk(v1) = C6(v6, v0, v2, v4, v7, v3) and hence Lk(v3) = C6(v2, v0, v4, v7, v1, v6). Theseimply deg(v2) = 7, a contradiction.

If v1v3 ∈ Lk(v4) then Lk(v1) = C6(v6, v0, v2, v4, v3, v7), Lk(v4) = C6(v3, v0, v5, v7,v2, v1) and Lk(v3) = C6(v2, v0, v4, v1, v7, v), wherev = v5 or v6. If v = v5 then,Lk(v5) = C6(v4, v0, v6, v2, v3, v7). These imply,v0v2, v1v2, v3v2, . . . , v7v2 are edges inM and hence, deg(v2) = 7, a contradiction. Ifv = v6 thenC3(v1, v3, v6) ⊆ Lk(v7), acontradiction. This proves the claim.

Sincev2v3v4 is not a simplex, by the claim, Lk(v4) = C6(v3, v0, v5, v1, v2, v7) andLk(v1) = C6(v6, v0, v2, v4, v5, v7). Then Lk(v5) = C6(v6, v0, v4, v1, v7, y), wherey =v2 or v3. If y = v2 then deg(v2) = 7, a contradiction. So,y = v3. These imply Lk(v2) =C6(v1, v0, v3, v6, v7, v4). ThenM is isomorphic, via the map(v1, v4, v5, v3, v7)(v2, v6),to T2. 2


Lemma4.7. If χ(M) = 0 and there exists a vertex of degree7 in M and no vertex ofdegree3, thenM is isomorphic toT4, . . . , T8,K1, . . . , K5 or K6.

Proof. Let deg(v0) = 7 and let Lk(v0) = C7(v1, . . . , v7). Since the degree of each vertexis more than 3 andM 6= R17, there existsi ∈ {1, . . . ,7} such thatvivi+1vi+3 orvivi+1vi+5is a simplex (additions in the subscripts are modulo 7). We may assume thatv1v2v4 is asimplex (cases whenv1v2v6 is a simplex give isomorphic complexes). Then the 2-simplex(6= v0v1v7) containingv1v7 is v1v7v3, v1v7v4 or v1v7v5.

Case I. v1v4v7 is a simplex. Ifv4v7v6 is a simplex then Lk(v7) = C4(v1, v0, v6, v4),Lk(v4) = C7(v3, v0, v5, v2, v1, v7, v6) and Lk(v1) = C4(v2, v0, v7, v4). Now, Lk(v2) isC5(v3, v0, v1, v4, v5) or C6(v3, v0, v1, v4, v5, v6). In the first case, Lk(v3) = C5(v2, v0,

v4, v6, v5). Then the links of the remaining vertices are complete and the degree sequenceof M is 72 · 54 · 42 and henceχ(M) = −1, a contradiction. In the second case, the linksof the remaining vertices are complete and the degree sequence ofM is 72 · 62 · 44, acontradiction again. So, the 2-simplex (6= v0v4v7) containingv4v7 is v3v4v7 or v4v5v7.

Subcase I.1. v3v4v7 is a simplex. Sincev4v5v6 is not a simplex, Lk(v4) = C6(v3, v0, v5,v2, v1, v7) and hence Lk(v1) = C4(v7, v0, v2, v4). Then the edgev5v6 (of Lk(v0)) is eitherin Lk(v2) or in Lk(v3).

If v5v6 is in Lk(v2), then Lk(v2) isC7(v3, v0, v1, v4, v5, v6, v7) orC6(v3, v0, v1, v4, v5,v6). In the first case, the links of the remaining vertices are complete. These imply thatf2(M)=14 and henceχ(M)=−1, a contradiction. In the second case, Lk(v6)=C5(v7,

v0, v5, v2, v3). Then no more 2-simplices are possible. These imply thatf2(M) = 14, acontradiction again. So,v5v6 is in Lk(v3).

If v2v3v5 is a simplex then Lk(v3) = C6(v2, v0, v4, v7, v6, v5). These complete all thelinks of the remaining vertices. In this case,f2(M) = 14, a contradiction. Hence, the 2-simplex (6= v0v2v3) containingv2v3 is v2v3v6. Then Lk(v3) = C6(v2, v0, v4, v7, v5, v6).Now, Lk(v2) and Lk(v7) show that the remaining two simplices arev2v6v7 andv2v5v7.ThenM is isomorphic, via the map(v0, v4, v2)(v1, v3, v5), toK3.

Subcase I.2. v4v5v7 is a simplex. To complete Lk(v4), v2v4v6 and v3v4v6 have to besimplices. Then the other triangle containingv5v6 is eitherv2v5v6 v3v5v6.

In the first case, using the similar method as above,M is isomorphic toT4 or T7.In the second case,M is isomorphic toK3.

Case II. v1v3v7 is a simplex. Sincev6v7 is an edge, one ofv2v6v7, v3v6v7 or v4v6v7 is atriangle.

In the first case,M is isomorphic toK1, . . . , K4, T4 or T5.In the second case,M is isomorphic toT5, T6, T7,K1, . . . , K4 orK6.In the third case,M is isomorphic toT4 orK3.

Case III. v1v5v7 is a simplex. Then one ofv1v5v6, v2v5v6 orv3v5v6 is a triangle containingv5v6.

In the first case,M is isomorphic toK1, . . . , K6.In the second case,M is isomorphic toT5, . . . , T8,K2,K3 orK4,In the third case,M is isomorphic toK1, . . . , K5, T5 or T8. This completes the proof of

Lemma 4.7. 2

Lemma4.8. The combinatorial2-manifolds mentioned in Theorem1.1are pairwise non-isomorphic.


Proof. For 2≤ i < j ≤ 8, Ti ∼= Tj implies that the degree sequence ofTi is the same asthe degree sequence ofTj . This implies that(i, j) = (5, 8). If ϕ is an isomorphism fromT5to T8, thenϕ(v1v2) = v2v5 (since degT5

(vj ) = degT8(ϕ(vj )). This impliesϕ(v1v2v4) =

v2v5v0 or v2v5v3. Thenϕ(v4) = v0 or v3. But degT5(v4) = 6 and degT8

(v0) = 5 =degT8

(v3), a contradiction. So,Ti 6∼= Tj for 2 ≤ i 6= j ≤ 8.Similarly, for 1 ≤ i < j ≤ 6,Ki ∼= Kj implies that the degree sequence ofKi is the

same as the degree sequence ofKj and hence(i, j) = (2, 4). If α is an isomorphism fromK2 toK4, thenα({v0, v4, v5}) = {v1, v4, v5} (sinceα takes vertices of degree 7 inK2 tovertices of degree 7 ofK4). Now, v0v4v5 ∈ K2 whereasv1v4v5 6∈ K4. Hence,K2 6∼= K4.So,Ki 6∼= Kj for 1 ≤ i 6= j ≤ 6.

Repeating the above argument we find that for 5≤ i < j ≤ 20,Ri ∼= Rj implies(i, j) = (5, 10), (14, 16) or (18, 19).

If possible, letβ1 : R5 → R10 be an isomorphism. Since deg(β1(v)) = deg(v) for eachvertexv in R5, β1(v7) = v7 andβ1(v0) = v0. Now,v0v7 is an edge inR5 whereasv0v7 isnot an edge inR10. So,R5 6∼= R10.

If possible, letβ2 : R14 → R16 be an isomorphism. Since deg(β2(v)) = deg(v) foreach vertexv in R14, β2(v0) = v0 andβ1({v3, v4}) = {v2, v4}. Now, v0v3v4 is a simplexin R14 whereasv0v2v4 is not a simplex inR16. Thus,R14 6∼= R16.

In R19, both the degree 4 vertices form an edge but that is not the case inR18. Thus,R18 6∼= R19. So,Ri 6∼= Rj for 5 ≤ i 6= j ≤ 20.

If, for 10 ≤ i < j ≤ 23,Si ∼= Sj then the degree sequence ofSi is equal to the degreesequence ofSj . This implies thati = 16 andj = 18. Now, ifγ is an isomorphism betweenS16 andS18, thenγ takes vertices of degree 3 inS16 to vertices of degree 3 ofS18. Observethat the links of the vertices of degree 3 ofS16 have a vertex of degree 4 whereas those ofS18 have no vertex of degree 4. This shows that such aγ does not exist. Hence,Si 6∼= Sjfor 10 ≤ i 6= j ≤ 23.

The lemma now follows from the fact thatSi (10 ≤ i ≤ 23) triangulatesS2, Rj(5 ≤ j ≤ 20) triangulatesRP 2, Tk (2 ≤ k ≤ 8) triangulatesS1 × S1, Kp (1 ≤ p ≤ 6)triangulates the Klein bottle and|D| is disconnected. 2

Proof of Theorem1.1. LetM be an 8-vertex combinatorial 2-manifold. Then3×82 ≤

f1(M) ≤(

82

), χ(M) = 8 − f1(M) + f2(M) and 3f2(M) = 2f1(M). This implies

12 ≤ f1(M) ≤ 27 and hence−1 ≤ χ(M) ≤ 4. But it is known (e.g., see [7,9]) that theredoes not exist any 8-vertex combinatorial 2-manifoldM with χ(M) = −1. If χ(M) ≥ 3then|M| is disconnected and since the number of vertices is 8,M is isomorphic toD. So,if M 6= D then 0≤ χ(M) ≤ 2. The theorem now follows from Lemmas 4.2,. . . , 4.8 andProposition 3.2. 2

5. Proof of Theorem 1.2

Throughout this sectionM is an 8-vertex two-dimensional weak pseudomanifold whichis not a combinatorial 2-manifold.

Lemma5.1. If M has a vertex, sayv0, whose link is of the formC3(v1, v2, v3) tC4(v4, v5, v6, v7), wherev1v2v3 is a simplex, thenM is isomorphic toQ3,Q5,Q6,Q12or Q16.


Proof. Here we have two cases: (I) At least one ofv4v6 or v5v7 is an edge ofM and (II)Neitherv4v6 norv5v7 is an edge ofM.

Case I. We can assume, without loss of generality, thatv4v6 is an edge ofM.

Subcase I.1. The triangles throughv4v6 arev4v5v6 andv4v6v7. Then,M isQ3.

Subcase I.2. Exactly one ofv4v5v6 or v4v6v7 is a simplex. Assume, without loss of gener-ality, thatv4v5v6 is a simplex. Then, the other triangle throughv4v6 must be one ofv1v4v6,v2v4v6 or v3v4v6. Assume, without loss of generality, thatv1v4v6 is a simplex inM. Thenv1v6v7 andv1v4v7 have to be simplices. HereM isQ5.

Subcase I.3. Neitherv4v5v6 norv4v6v7 is a simplex ofM. The triangles throughv4v6 areof the formv1v4v6, v2v4v6 or v3v4v6. Assume, without loss of generality, thatv1v4v6 andv2v4v6 are simplices ofM. The triangle (6= v1v4v6) havingv1v4 as an edge is eitherv1v4v5or v1v4v7. We can assume, without loss of generality, thatv1v4v5 is a simplex. Then, theother triangle throughv1v5 is v1v5v6 or v1v5v7.

In the first case (by considering Lk(v6) and Lk(v4)), v2v6v7 andv2v4v7 are simplices.HereM isQ12.

In the second case, Lk(v4) = C6(v5, v0, v7, v2, v6, v1) and Lk(v1) = C3(v0, v2, v3) tC4(v6, v4, v5, v7) and hence Lk(v6) = C6(v5, v0, v7, v1, v4, v2). The link ofv7 shows thatv2v5v7 is a simplex. ThenM isQ16.

Case II. Assume, without loss of generality, that the second 2-simplex containingv4v5is v1v4v5. Then (since,v4v6 andv5v7 are non-edges), Lk(v1) = C3(v0, v2, v3) t C4(v4,

v5, v6, v7). In this case,M isQ6. 2

Lemma5.2. If M has a vertex, sayv0, whose link is of the formC3(v1, v2, v3) tC4(v4, v5, v6, v7), wherev4v6 andv5v7 are edges butv1v2v3 is not a simplex, thenM isisomorphic toP29, . . . , P32, P35, P37, P38,Q11 or Q14.

Proof. We have three cases: (I) Bothv4v5v6 andv4v6v7 are simplices, (II) exactly one ofv4v5v6 andv4v6v7 is a simplex and (III) neitherv4v5v6 norv4v6v7 is a simplex.

Case I. The triangles throughv5v7 are of the formv1v5v7, v2v5v7 or v3v5v7. We canassume, without loss of generality, thatv1v5v7 andv2v5v7 are inM. Consider the trianglethroughv1v5 other thanv1v5v7. Clearly, it is eitherv1v2v5 or v1v3v5.

In the former case, to complete Lk(v1) and Lk(v3), v1v3v7 and v2v3v7 have to besimplices. HereM isQ11.

In the latter case,v2v3v5 has to be a simplex. This impliesv1v2v7 is also a simplex.ThenM is isomorphic, via the map(v5, v7), toQ11.

Case II. Assume, without loss of generality, thatv4v5v6 is a simplex. Then, the othertriangle throughv4v6 must be one ofv1v4v6, v2v4v6 or v3v4v6. Assume, without loss ofgenerality, thatv1v4v6 is inM. The triangles throughv5v7 are of the formv1v5v7, v2v5v7or v3v5v7. Without loss of generality we can assume that eitherv1v5v7 andv2v5v7 are inM or v2v5v7 andv3v5v7 are inM.

In the first case,M is P29, P30 or P31.In the second case,M is isomorphic toP32, P35 orQ14.

Case III. Without loss of generality, assume that the triangles throughv4v6 arev1v4v6 andv2v4v6. Then assume, without loss of generality, that the triangles throughv5v7 are eitherv1v5v7 andv2v5v7 or v2v5v7 andv3v5v7.


In the first case,M is isomorphic toP37.In the second case,M is isomorphic toP38. 2

Lemma5.3. If there exists a vertex, sayv0, whose link inM is of the formC3(v1, v2,v3) t C4(v4, v5, v6, v7), wherev1v2v3 andv4v6 are not simplices, thenM is isomorphicto P3, P4, P6, P9, P11, P13, P15, P22, P23, P25, P28, P30, P34,Q7,Q8 or Q11.

Proof. The complexM = (M \ {v0}) ∪ {v1v2v3, v4v5v6, v4v6v7}) is a 7-vertex two-dimensional weak pseudomanifold. From the classification of 7-vertex two-dimensionalweak pseudomanifolds, we observe that(M, v1v2v3, v4v5v6, v4v6v7) is (isomorphic to)(R2, v0v1v2, v3v4v5, v3v4v6), (R3, v0v1v3, v2v4v5, v2v5v6), (R3, v1v5v6, v0v2v3, v0v2v4),(S5, v2v3v4, v1v6v7, v1v5v6), (S6, v1v6v7, v2v3v4, v2v4v5), (S7, v2v4v5, v3v6v7, v1v3v7),(S7, v1v3v7, v4v5v6, v2v5v6), (S7, v1v3v7, v2v4v5, v2v5v6), (S7, v1v3v7, v2v4v5, v4v5v6),(S8, v1v2v3, v4v5v6, v5v6v7), (S8, v2v5v7, v1v3v4, v1v4v6), (S8, v5v6v7, v1v2v3, v1v3v4),(S9, v1v2v3, v4v5v6, v5v6v7), (Q1, v1v2v3, v4v5v7, v4v5v6), (Q1, v1v2v3, v4v6v7, v4v5v7),(Q2, v1v2v3, v4v5v7, v5v6v7), (P1, v1v5v6, v2v3v7, v2v3v4), (P2, v2v3v7, v1v5v6, v1v4v5)

or (P2, v2v3v4, v1v5v6, v5v6v7).First, assume that(M, v1v2v3, v4v5v6, v4v6v7) is (R2, v0v1v2, v3v4v5, v3v4v6). Let M

be the complex obtained fromM by starring a vertexu8 in v1v2v3 andu9 in the edgev4v6.Let R2 be the complex obtained fromR2 by starring a vertexv8 in v0v1v2 andv9 in theedgev3v4. Then, by Proposition 3.4(a),M ∼= R2.

The complex obtained fromM by identifyingu8 andu9 (to a new vertexv0) isM and thecomplex obtained fromR2 by identifyingv8 andv9 (to a new vertexv7) isP22. Therefore,by Proposition 3.3(b),M ∼= P22.

Similarly, if (M, v1v2v3, v4v5v6, v4v6v7) is one of (R3, v0v1v3, v2v4v5, v2v5v6),(R3, v1v5v6, v0v2v3, v0v2v4), (S5, v2v3v4, v1v6v7, v1v5v6), (S6, v1v6v7, v2v3v4, v2v4v5),(S7, v1v3v7, v4v5v6, v2v5v6), (S7, v1v3v7, v2v4v5, v2v5v6), (S7, v1v3v7, v2v4v5, v4v5v6),(S8, v2v5v7, v1v3v4, v1v4v6), (S8, v5v6v7, v1v2v3,v1v3v4), (Q1, v1v2v3, v4v5v7, v4v5v6),(Q1, v1v2v3, v4v6v7, v4v5v7), (P1, v1v5v6, v2v3v7, v2v3v4) or (P2, v2v3v4, v1v5v6,

v5v6v7), thenM isP23,P25,P6,P11,P4,P9,P3,P15,P13,Q7,Q8,P34 orP28 respectively.If (M, v1v2v3, v4v5v6, v4v6v7) is (S7, v2v4v5, v3v6v7, v1v3v7), (S8, v1v2v3, v4v5v6,

v5v6v7), (S9, v1v2v3, v4v5v6, v5v6v7), (Q2, v1v2v3, v4v5v7, v5v6v7) or (P2, v2v3v7,v1v5v6, v1v4v5) thenM is isomorphic toP6, P11, P15,Q11 or P30 respectively. 2

Lemma5.4. If M has no singular vertex of degree7, thenM is isomorphic toP5, P7, P8,P10, P12, P14, P16, . . . , P21, P24, P26, P27, P33, P36, P39,Q4,Q9,Q10,Q13 or Q15.

Proof. SinceM is not a combinatorial 2-manifold, there exists a vertex, sayv0, whoselink is of the formC3(v1, v2, v3) t C3(v4, v5, v6). SinceM 6∼= Q1, at most one ofv1v2v3or v4v5v6 can be a simplex. Letv7 be the remaining vertex ofM.

Case I. Exactly one ofv1v2v3 or v4v5v6 is a simplex. Assume, without loss of generality,that v1v2v3 is a simplex. Then the triangle (6= v0v4v5) throughv4v5 is v1v4v5, v2v4v5,v3v4v5 or v4v5v7. Without loss of generality, we can assume that eitherv1v4v5 or v4v5v7is a simplex.

In the first case,M is isomorphic toQ9,Q13 orQ15.In the second case,M is isomorphic toQ4,Q9 orQ13.

Case II. Neitherv1v2v3 norv4v5v6 is a simplex. ThenM = (M\{v0})∪{v1v2v3, v4v5v6} is a7-vertex two-dimensional weak pseudomanifold. SinceM has no singular vertex of degree


7, none ofv1, . . . , v6 is singular inM. From the classification of 7-vertex two-dimensionalweak pseudomanifolds we observe that(M, v1v2v3, v4v5v6) is (isomorphic to)(T1, v1v2v4, v3v5v6), (R2, v0v1v3, v2v5v6), (R3, v0v1v3, v2v4v5), (R3, v0v1v3, v2v5v6),(R4, v0v1v3, v2v4v5), (S5, v2v3v4, v1v6v7), (S5, v2v3v4, v1v5v6), (S6, v2v3v4, v1v6v7),(S7, v1v2v3, v4v5v6), (S7, v1v2v7, v3v4v6), (S7, v1v2v7, v4v5v6), (S7, v1v3v7, v4v5v6),(S8, v1v2v3, v4v5v6), (S8, v1v3v4, v5v6v7), (S8, v1v2v7, v4v5v6), (S9, v1v2v3, v4v5v6),(Q1, v1v2v3, v4v5v6), (P1, v1v2v6, v3v4v5) or (P2, v1v5v6, v2v3v4).

If (M, v1v2v3, v4v5v6) is (T1, v1v2v4, v3v5v6) then, by Proposition 3.5(b),M is P39.Similarly, if (M, v1v2v3, v4v5v6) is (R2, v0v1v3, v2v5v6), . . . , (S9, v1v2v3, v4v5v6),

(Q1, v1v2v3, v4v5v6), (P1, v1v2v6, v3v4v5) or (P2, v1v5v6, v2v3v4), thenM is P24, P26,P27,P21,P7,P5,P8,P12,P10,P14,P17,P18,P19,P16,P20,Q10,P36 orP33 respectively.2

Lemma5.5. The weak pseudomanifolds mentioned in Theorem1.2 are pairwise non-isomorphic.

Proof. First, we observe that all thePi ’s (1 ≤ i ≤ 39) are pseudomanifolds, while all theQi ’s (1 ≤ i ≤ 16) are not pseudomanifolds.

For 3 ≤ i < j ≤ 39,Pi ∼= Pj implies that the degree sequences ofPi andPj are thesame and the singular degree sequences ofPi andPj are the same and hence, from thedescription ofPi ’s in §2,(i, j) ∈ {(17, 18), (19, 20), (22, 25)}.

If there exists an isomorphismα : P17 → P18, thenα(v0) = v0 (since, these are the onlysingular vertices). InP17, each of theC3’s in the link ofv0 has a vertex of degree 4, whileonly oneC3 in Lk(v0) in P18 has a vertex of degree 4. Therefore,α is not an isomorphism.

If β : P19 → P20 is an isomorphism, thenβ(v0) = v0 (since, these are the only singularvertices). We see that Lk(v0) in P19 has a vertex (namely,v3) of degree 4 whereas Lk(v0)

in P20 has no vertex of degree 4. So,P19 6∼= P20.If there exists an isomorphismψ : P22 → P25, thenψ(v7) = v7 (since, these are the

only singular vertices of degree 7). InP25 the link ofv7 has aC3 all of whose vertices havedegree 6 but the link ofv7 in P22 has no suchC3. So,P22 6∼= P25.

For 3≤ i < j ≤ 16,Qi∼= Qj implies that the degree sequences ofQi andQj are the

same and hence, from the description ofQi ’s in §2,(i, j) = (6, 7).Two degree 3 vertices inQ6 form an edge but that is not the case inQ7. Thus,Q6 6∼= Q7.

This completes the proof of the lemma. 2

Proof of Theorem1.2. LetM be a two-dimensional 8-vertex weak pseudomanifold whichis not a combinatorial 2-manifold. ThenM has a singular vertex.

First consider the case whenM has a singular vertex, sayv0, of degree 7. Then the linkof v0 is of the formC3 t C4. Let Lk(v0)= C3(v1, v2, v3) t C4(v4, v5, v6, v7).

If v1v2v3 ∈ M then, by Lemma 5.1,M is isomorphic toQ3,Q5,Q6,Q12 orQ16.If v1v2v3 is not a simplex then we have two cases, namely, (i) either bothv4v6 andv5v7

are edges ofM or (ii) at least one ofv4v6 or v5v7 is a non-edge ofM, say (without loss ofgenerality)v4v6 is not an edge. Then, by Lemmas 5.2 and 5.3,M is isomorphic toP3, P4,P6,P9,P11,P13,P15,P22,P23,P25,P28, . . . , P32,P34,P35,P37,P38,Q7,Q8,Q11 orQ14.

Finally, consider the case whenM has no singular vertex of degree 7. In this case, byLemma 5.4,M is isomorphic toP5, P7, P8, P10, P12, P14, P16, . . . , P21, P24, P26, P27,P33, P36, P39,Q4,Q9,Q10,Q13 orQ15. This completes the proof. 2


6. Applications

Proof of Corollary1.3. Observe thatS1, . . . , S23 triangulateS2, R1, . . . , R20 triangulateRP 2, T1, . . . , T8 triangulateS1 × S1,K1, . . . , K6 triangulate the Klein bottle (K) andDtriangulatesS2 t S2. These, Proposition 3.1 and Theorem 1.1 imply Corollary 1.3(i).

By Proposition 3.3(c),P1, . . . , P20 triangulate the pinched sphere (P ), P21, . . . , P27triangulateRP 2#P , P28, . . . , P36 triangulateP#P , P37 triangulatesRP 2#P#P , P38 tri-angulatesK#P andP39 triangulates(S1 × S1)#P .

Also (from the pictures in §2)Q1,Q3 andQ4 triangulate the union of twoS2’s havingone point in common,Q2, Q5, . . . ,Q10 triangulate the union of twoS2’s having twopoints in common,Q11,Q12 andQ13 triangulate the union of two 2-spheres having threepoints in common,Q14 andQ15 triangulate the union of two 2-spheres having four pointsin common,Q16 triangulates the union ofS2 andRP 2 having three points in common.These, Proposition 3.1 and Theorem 1.2 imply Corollary 1.3(ii). 2

Proof of Theorem1.4. Letϕ : K2n → S2

4 be a simplicial map, whereK2n is an oriented

n-vertex two-dimensional pseudomanifold andS24 is the 4-vertex 2-sphere with an orien-

tation.If n ≤ 7, then there exists a vertex, saya, of S2

4 whose inverse image contains less than2 vertices and hence there exists a triangle througha whose inverse image contains lessthan 3 triangles. Thus, deg(ϕ) < 3.

If n = 8, thenf2(K28) ≤ 18 and hence deg(ϕ) ≤ 18/4. Letϕ be of type(n1, n2, n3, n4).

Assume, deg(ϕ) ≥ 3. By the same argument as above, each vertex ofS24 has two inverse

images. Letϕ−1(a) = {a1, a2}, ϕ−1(b) = {b1, b2}, ϕ−1(c) = {c1, c2} andϕ−1(d) ={d1, d2}, wherea, b, c andd are the vertices ofS2

4. So, there does not exist any 2-simplexσ such thatϕ(σ) is a vertex and hence we have:

Claim 6.1. Each collapsing triangle contains exactly one collapsing edge. On the otherhand, both the triangles through a collapsing edge are collapsing.

It is also easy to see the following:

Claim 6.2. If Sa := {σ ∈ K28 : ϕ(σ) is a 2-simplex containinga}, then #(Sa) ≤ 12.

Further, ifa1a2 is an edge then #(Sa) ≤ (deg(a1)− 2)+ (deg(a2)− 2) ≤ 10.

If deg(ϕ) = 4, thenf2(K28) ≥ 16 and henceχ(K2

8) ≤ 0. If χ(K28) = −1 then by

Theorems 1.1 and 1.2,K28 = P37, P38 or P39. SinceP37 andP38 are non-orientable,

K28 = P39. Clearly, from the degree sequence ofP39, there exists an edgexy such that

ϕ(x) = ϕ(y). Then (by Claim 6.2) deg(ϕ) ≤ 10/3, a contradiction. So,χ(K28) = 0 and

hencef2(K28) = 16. Thenϕ is of type (4, 4, 4, 4) and hence there is no collapsing 2-

simplex and hence, by Claim 6.1, no collapsing edge. So, ifϕ(u) = ϕ(v), thenuv is a

non-edge and conversely (since the number of non-edges is

(82

)−24 = 4). These imply

that the 4 non-edges are disjoint and hence the degree sequence ofK28 is 68. Then, by

Theorems 1.1 and 1.2,K28 = T2 or P36. In both the casesv3v6 is a non-edge and hence

ϕ(v3) = ϕ(v6). If K28 = T2 thenϕ(v2v3v7) andϕ(v2v6v7) are the same 2-simplex, a

contradiction to Proposition 3.6(b). IfK28 = P36 thenϕ(v2v3v4) andϕ(v2v4v6) are the

same 2-simplex, a contradiction to Proposition 3.6(b). Thus deg(ϕ) ≤ 3. This proves thefirst part of the theorem.

Now assume deg(ϕ) = 3. In this case #(Sa) ≥ 9 for each vertexa of S24. Thus,

f2(K28) ≥ 12 and hence, by Theorems 1.1 and 1.2,−1 ≤ χ(K2

8) ≤ 2.


Case I. χ(K28) = −1. By Theorem 1.2 (asK2

8 is orientable),K28 = P39. Let ϕ(v0) = a.

If ϕ(v7) = a then (since #(Sa) ≥ 9) ϕ(v1), ϕ(v2), ϕ(v4) are distinct andϕ(v3), ϕ(v5),ϕ(v6) are distinct. Again,ϕ(v1) = ϕ(v3) or ϕ(v5) implies that there are 4 collapsing2-simplices throughv7. This implies, #(Sa) ≤ 8, a contradiction. So,ϕ(v1) = ϕ(v6).Similarly, ϕ(v2) = ϕ(v5) andϕ(v3) = ϕ(v4). Thenϕ−1(bcd) contains no 2-simplices, acontradiction. So,v7 6∈ ϕ−1(a). If v1 ∈ ϕ−1(a) thenϕ(v0v5v6) = ϕ(v1v5v6) and hence,by Proposition 3.6(c), #(Sa) ≥ 9 + 2 = 11. On the other hand #(Sa) ≤ (deg(v0)− 2)+(deg(v1) − 2) = 9, a contradiction. Similarly, forv2, v3, v4, v5 or v6 ∈ ϕ−1(a) we getcontradictions.

Case II. χ(K28) = 0. By Claim 6.1, the inverse image of each triangle ofS2

4 contains 3,5 or 7 triangles. Also, by Claim 6.2,(n1, n2, n3, n4) = (7, 3, 3, 3) or (5, 5, 3, 3) is notpossible. So,(n1, n2, n3, n4) is (5, 3, 3, 3) or (3, 3, 3, 3).

Subcase II.1. If (n1, n2, n3, n4)= (5, 3, 3, 3), then we have two collapsing triangles andhence, by Claim 6.1, exactly one collapsing edge. Assume (if necessary, by taking acomposition with an automorphism ofS2

4) that the number of triangles ofϕ−1(abc) is 5.Then #(Sa) ≥ 5 + 3 + 3 = 11 and hence, by Claim 6.2,a1a2 is not an edge. So, 11≤deg(a1)+ deg(a2) ≤ 12. Similarly,b1b2 andc1c2 are not edges, 11≤ deg(b1)+ deg(b2),deg(c1)+ deg(c2) ≤ 12 and #(Sb), #(Sc) ≥ 11. So,d1d2 is the collapsing edge. Withoutloss of generality, let deg(a1) = deg(b1) = deg(c1) = 6.

Since the sum of the degrees of all the vertices is 48, we get 12≤ deg(d1)+deg(d2) ≤ 14.If deg(d1) + deg(d2) = 12 then (by Claim 6.2) #(Sd) ≤ 12 − 4, a contradiction. Ifdeg(d1)+ deg(d2) = 14 then, deg(d1) = deg(d2) = 7. Without loss, we can assume thatdeg(a2) = deg(b2) = 5 and deg(c2) = 6. Letx, y be the vertices ofK2

8 such thatd1d2x

andd1d2y are 2-simplices. Ifx = a1, then #(Sa) ≤ 6 + 5 − 1 = 10, a contradiction.By a similar argument we see thatx, y /∈ {a1, a2, b1, b2}. Hencex, y ∈ {c1, c2}. Then#(Sc) ≤ 12− 2 = 10, a contradiction. Thus, deg(d1)+ deg(d2) = 13.

We can assume (if necessary, by taking composition with automorphisms ofS24 and

K28, i.e., up to an equivalence) that deg(d1) = 7, deg(d2) = 6, deg(b2) = deg(c2) = 6

and deg(a2) = 5. Hence there is no collapsing 2-simplex througha1 or a2. So, the degreesequence ofK2

8 is 7 · 66 · 5 and hence, by Theorems 1.1 and 1.2,K28 = P35 andd1 = v0,

a2 = v1. We can also assume that{b1, b2} = {v2, v6} and {c1, c2} = {v3, v6}. Then,{a1, d2} = {v5, v7}.

If (a1, d2) = (v7, v5) then {v1v2v4, v7v2v4, v1v3v6, v7v3v6} ⊆ Sa . This implies, byProposition 3.6(c), #(Sa) ≥ 13, a contradiction. So,(a1, d2) = (v5, v7). Then,ϕ is equiv-alent toh.

Subcase II.2. If (n1, n2, n3, n4) = (3, 3, 3, 3), then we have 4 collapsing triangles andhence, by Claim 6.1, exactly 2 collapsing edges.

Without loss of generality, we can assume thata1a2 andb1b2 are edges whereasc1c2andd1d2 are not. Then 9≤ deg(c1) + deg(c2), deg(d1) + deg(d2) ≤ 12. We see that(deg(a1) − 2) + (deg(a2) − 2)=#(Sa) = 9. Hence deg(a1) + deg(a2) = 13. Similarly,deg(b1) + deg(b2) = 13. We can assume that deg(a1) = deg(b1) = 7 and deg(a2) =deg(b2) = 6. Thus, deg(c1) + deg(c2) + deg(d1) + deg(d2) = 22. This shows that10 ≤ deg(c1)+ deg(c2), deg(d1)+ deg(d2) ≤ 12. Since deg(c1), deg(c2) ≤ 6, it is clearthat deg(c1), deg(c2) ≥ 4. Thus, there exists no vertex of degree 3. We may assume thatdeg(c1)+deg(c2) ≥ deg(d1)+deg(d2), deg(c1) ≥ deg(c2) and deg(d1) ≥ deg(d2). Thenwe have the following three possibilities for the degrees of the remaining four vertices.


(II.2.1) deg(c1) = deg(c2) = deg(d1) = 6 and deg(d2) = 4.(II.2.2) deg(c1) = deg(c2) = 6 and deg(d1) = deg(d2) = 5.(II.2.3) deg(c1) = deg(d1) = 6 and deg(c2) = deg(d2) = 5.

(II.2.1) In this case, from Theorems 1.1 and 1.2, we see thatK28 isT4. The vertex of degree

4 of T4 is v0. Henced2 = v0. We can assume, without loss of generality, thata1 = v7 andb1 = v2. Sincev4 andv5 are the only vertices of degree 6 which do not form an edge, we canassumec1 = v4 andc2 = v5. Thenϕ−1(acd) containsv0v4v7 andv0v5v7, a contradictionto Proposition 3.6(b) (since, #(ϕ−1(σ )) = 3 = deg(ϕ), for each 2-simplexσ in S2

4).

(II.2.2) In this case, the degree sequence ofK28 is 72 · 64 · 52 and hence, by Theorems 1.1

and 1.2,K28 is T5, T8,K1, P26, P32, P33 or P34. AsK1 andP26 are non-orientable,K2

8 isnotK1 or P26. Since the two degree 5 vertices ofT5 form an edge,K2

8 6= T5.If K2

8 is T8 then{d1, d2} = {v0, v3} and{a1, b1} = {v2, v5}. Thenϕ−1(abd) containsv0v2v5 andv2v3v5, a contradiction to Proposition 3.6(b).

If K28 = P32 then {d1, d2} = {v3, v6} and we can assumea1 = v0, b1 = v7. Then

{c1, c2} = {v1, v5} and{a2, b2} = {v2, v4}. If (a2, b2) = (v2, v4) thenv0v1v2, v0v2v3,v0v4v7, v2v4v7 are collapsing and hence #(Sa) ≤ 7, a contradiction. If(a2, b2) = (v4, v2)

thenv0v4v5, v0v4v7, v2v4v7, v2v5v7 are collapsing and hence #(Sa) ≤ 8, a contradiction.If K2

8 = P33 then{d1, d2} = {v3, v5} and{a1, b1} = {v1, v4}. Thenϕ−1(abd) containsv1v3v4 andv1v4v5, a contradiction to Proposition 3.6(b).

If K28 = P34 then{d1, d2} = {v2, v3} and{a1, b1} = {v0, v7}. Thenϕ−1(abd) contains

v0v2v7 andv0v3v7, a contradiction to Proposition 3.6(b).

(II.2.3) In this case also, the degree sequence ofK28 is 72 · 64 · 52 and hence (sinceK2

8 isorientable), by Theorems 1.1 and 1.2,K2

8 is T5, T8, P32, P33 or P34.If K2

8 = T5 then assume (up to an equivalence) that,a1 = v1, b1 = v2, c2 = v0,d2 = v5. Then,v4 = d1. These imply,ϕ(v0v2v4) = ϕ(v0v2v5) = bcd, a contradiction toProposition 3.6(b).

If K28 = T8 then assume, without loss of generality, thata1 = v2, b1 = v5, c2 = v0

andd2 = v3. Thenc1 = v1 andd1 = v6. So,{a2, b2} = {v4, v7}. If (a2, b2) = (v4, v7)

theϕ(v2v3v7) = ϕ(v2v6v7) = abd, a contradiction to Proposition 3.6(b). If(a2, b2) =(v7, v4) theϕ(v0v5v7) = ϕ(v1v5v7) = abc, a contradiction to Proposition 3.6(b).

If K28 = P32 then we can assume (up to an equivalence)a1 = v0, b1 = v7, c2 = v3,

d2 = v6 and hencec1 = v4 andd1 = v2. Thenϕ(v1v2v4) = ϕ(v1v4v6), a contradiction toProposition 3.6(b).

If K28 = P33 then assume, without loss of generality, thata1 = v1, b1 = v4, c2 = v3 and

d2 = v5. Thenc1 = v6 andd1 = v2. Clearly,(a2, b2) = (v0, v7) or (v7, v0). In the firstcaseϕ(v1v3v4) = ϕ(v0v3v4) = abc, a contradiction to Proposition 3.6(b). In the secondcase,ϕ(v1v4v5) = ϕ(v4v5v7) = abd, a contradiction to Proposition 3.6(b) again.

If K28 = P34 then assume (up to an equivalence) that,a1 = v0, b1 = v7, c2 = v2

andd2 = v3. Thenc1 = v5, d1 = v6 and {a2, b2} = {v1, v4}. If (a2, b2) = (v1, v4)

thenϕ(v0v3v7) = ϕ(v1v3v7) = abd, a contradiction to Proposition 3.6(b). So,(a2, b2) =(v4, v1). In this case,ϕ is equivalent tog.

Case III. χ(K28) = 2. In this casef2(K

28) = 12 and hence there is no collapsing 2-

simplex and(n1, n2, n3, n4) = (3, 3, 3, 3). So, deg(a1)+deg(a2) = deg(b1)+deg(b2) =deg(c1)+ deg(c2) = deg(d1)+ deg(d2) = 9. Further, by Claim 6.1,a1a2, b1b2, c1c2 andd1d2 are not edges.


Observe thatK28 cannot have a vertex of degree 4. (If possible let there exist a vertex,

saya1, of degree 4. Letb1c1, c1d1, d1x andxb1 be the edges in Lk(a1). Clearly,x 6= a2.If x = b2 or d2 then there is a collapsing 2-simplex, a contradiction. Ifx = c2 thenProposition 3.6(b) is contradicted.) Hence, by Theorems 1.1 and 1.2,K2

8 is S15 or S20.But, S20 has no pair of vertices, the sum of whose degrees is 9. So,K2

8 is S15.If deg(a1) ≥ deg(a2), then(a1, a2) ∈ {(v1, v5), (v2, v0), (v4, v7), (v6, v3)}. We observe

that (v1, v2)(v5, v0), (v1, v4)(v5, v7), (v1, v6)(v3, v5), (v2, v4)(v0, v7), (v2, v6)(v0, v3)

and (v4, v6)(v0, v3) are all automorphisms ofS15. So, we may assume that(a1, a2) =(v1, v5), (b1, b2) = (v2, v0), (c1, c2) = (v4, v7) and(d1, d2) = (v6, v3). Then (up to anequivalence),ϕ = f . The theorem now follows from Example 2.1. 2

Remark6.3. Some of the steps in the proofs of the lemmas in §4 and 5 are similar to theothers. Hence we have omitted these details for the sake of brevity. Complete proofs areavailable with the authors.

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Two-dimensional weak pseudomanifolds on eight vertices

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