Reconstructing triangulations of 3-manifolds from their intersection matrix
aa r X i v : . [ m a t h . G T ] F e b RECONSTRUCTING TRIANGULATIONS OF3-MANIFOLDS FROM THEIR INTERSECTIONMATRIX
Jorge L. Arochaemail [email protected] Fern´andez-Hidalgoemail [email protected]
Abstract
The intersection matrix of a simplicial complex has entries equal to the rankof the intersection of its facets. In [1] the authors prove the intersection matrix isenough to determine a triangulation of a surface up to isomorphism. In this work weshow the intersection matrix is enough to determine the triangulation of a 3-manifoldup to isomorphism.
Definition 1.1 (simplicial complex) . An abstract simplicial complex is a pair ( V, ∆)such that, V is a finite set, ∆ ⊆ V and the following condition is satisfied, if X ∈ ∆and Y ⊆ X then Y ∈ ∆. We call an element of ∆ a simplex. If a simplex has i elements we shall say it has rank i , dimension i − i − i -simplices with ∆ i . The maximal elements of ∆ are calledfacets. Definition 1.2 (isomorphism of simplicial complices) . Given two simplicial com-plices ( V, ∆) and ( V ′ , ∆ ′ ) we say a map ϕ : V → V ′ is an isomorphism if ϕ is bijectiveand X ∈ ∆ if and only if ϕ ( X ) ∈ ∆ ′ .In this work we focus on simplicial complices with associated geometric realiza-tions where the underlying space is a 3-dimensional manifold (we call such objectstriangulations of 3-dimensional manifolds). For our purposes it is convenient towork with the underlying abstract simplicial complex and only use the geometricrealization to give conditions the abstract simplicial complex must satisfy. One suchcondition is the following, given in terms of the neighbourhood of vertices. Definition 1.3 (neighbourhood of a vertex) . Let ( V, ∆) be a simplicial complexof dimension 3 and let v ∈ V . We define the neighbourhood of v as the simplicialcomplex of dimension 2 spanned by the set of facets ∆ ′ = { X − { v }| v ∈ X ∈ ∆ } . bservation 1.4. If ( V, ∆) is a triangulation of -manifold the neighbourhood of v must span a space homeomorphic to a sphere of dimension . The intersection matrix of a simplicial complex is the matrix such that entry i, j is equal to the rank of the intersection of the i ’th facet and the j ’th facet.We use the following definition of an intersection preserving map which capturesthe notion of two simplicial complices having the same intersection matrix. Definition 1.5 (intersection preserving map) . Given two simplicial complices ( V, ∆)and ( V ′ , ∆ ′ ) of dimension 3 with corresponding sets of facets ∆ and ∆ ′ we say amap f : ∆ → ∆ ′ is an intersection preserving map if f is bijective and | X ∩ Y | = | f ( X ) ∩ f ( Y ) | for all X, Y ∈ ∆ .We say f extends to an isomorphism if there exists an isomorphism ϕ : V → V ′ such that ϕ ( X ) = f ( X ) for all X ∈ ∆ .The main result we shall prove is the following, Theorem 1.6.
Let ( V, ∆) and ( V ′ , ∆ ′ ) be two triangulations of -dimensional man-ifolds and f : ∆ → ∆ ′ an intersection preserving map, then f extends to anisomorphism. Section 2 proves some results about extensions of intersection preserving maps,and provides sufficient conditions for such an extension to exist. Sections 3 and 4show these conditions hold in our case. Section 3 is dedicated to introducing andclassifyng 3-cyclic shells, and in section 4 we use this classification to obtain thedesired result.
Let us first introduce a ”reasonable” way of extending a map between facets to allsimplices.
Definition 2.1.
Let ( V, ∆) and ( V ′ , ∆ ′ ) be simplicial complices of dimension 3 andlet f : ∆ → ∆ ′ be any map. We define F : ∆ → ∆ ′ via F ( X ) = \ Y ∈ ∆ | X ⊆ Y f ( Y ) . Observation 2.2.
Let ( V, ∆) be a triangulation of a -manifold, then for all v ∈ V the intersection of all facets containing v is { v } .Proof. Let v ∈ V , one has \ X ∈ ∆ | v ∈ X X = \ X ∈ ∆ | v ∈ X X This is because every triangle is contained in exactly two tetrahedra(since thesimplicial complex is a triangulation), so every intersacand of the RHS is the inter-section of two intersecands of the LHS. e can see \ X ∈ ∆ | v ∈ X X = { v } by only considering triangles contained in one tetrahedra X .The following lemma shows F can be the only extension of an intersection pre-serving map between 3-manifolds. Lemma 2.3.
Let ( V, ∆) and ( V ′ , ∆ ′ ) be triangulations of -manifolds and let f :∆ → ∆ ′ be a bijection between facets which is extended by the isomorphism ϕ : V → V ′ . Then for each v ∈ V the set F ( { v } ) is equal to { ϕ ( v ) } .Proof. Let v ∈ V . Since f ( X ) = ϕ ( X ) for all X ∈ ∆ we have, \ X ∈ ∆ | v ∈ X f ( X ) = \ X ∈ ∆ | v ∈ X ϕ ( X ) . Because ϕ is bijective and \ X ∈ ∆ | v ∈ X X = { v } we obtain, \ X ∈ ∆ | v ∈ X ϕ ( X ) = { ϕ ( v ) } We now show whenever this map and the inverse map can be defined, the resultingmaps are isomorphisms.
Theorem 2.4.
Let ( V, ∆) and ( V ′ , ∆ ′ ) be triangulations of -manifolds and f :∆ → ∆ ′ be a bijection between facets such that for all v ∈ V and v ′ ∈ V ′ thesets F ( { v } ) and F − ( { v ′ } ) are singletons. Then the map ϕ : V → V ′ defined by { ϕ ( v ) } = F ( v ) is an isomorphism and the map σ : V ′ → V defined by { σ ( v ′ ) } = F − ( { v ′ } ) is its inverse.Proof. We first show if v ∈ X ∈ ∆ then ϕ ( v ) ∈ f ( X ). One can see this by noticing { ϕ { v }} = \ Y ∈ ∆ | v ∈ Y f ( Y ) ⊆ f ( X )One can apply the previous argument once more (but in the opposite direction)using the element ϕ ( v ) ∈ f ( X ) ∈ ∆ ′ to obtain σ ( ϕ ( v )) ∈ X for all v ∈ X ∈ ∆ .This implies σ ( ϕ ( v )) ∈ \ X ∈ ∆ | v ∈ X X = { v } . Therefore ϕ is bijective and σ is itsinverse.It only remains to be proved that ϕ ( X ) = f ( X ) for all X ∈ ∆ . We alreadyhave ϕ ( X ) ⊆ f ( X ) and the equality is derived from the fact that ϕ is injective and X, f ( X ) are two finite sets of the same size. he following lemma will be used to prove theorem 1.6. In section 4 we showthe conditions of the lemma are met. Lemma 2.5.
Let ( V, ∆) and ( V ′ , ∆ ′ ) be triangulations of -manifolds and let f :∆ → ∆ ′ be an intersection preserving map such that F induces a bijection between ∆ i and ∆ ′ i for ≤ i ≤ . Then the map ϕ : V → V ′ given by { ϕ ( v ) } = F ( v ) is welldefined. This proof will be slightly long and is composed of various lemmas.
Lemma 2.6.
For all v ∈ V we have \ X ∈ ∆ | v ∈ X f ( X ) = \ X ∈ ∆ | v ∈ X F ( X ) .Proof. For the ⊇ containment we notice if X ∈ ∆ contains v then there is Y ∈ ∆ containing v ,such that Y ⊆ X , it follows f ( Y ) ⊆ F ( X ).For the ⊆ containment we use 2.5 to see every intersecand in the RHS is theintersection of two elements of the LHS. Lemma 2.7.
Let v ∈ V and X ∈ ∆ with v ∈ X . The set \ Y ∈ ∆ | v ∈ Y ⊆ X F ( Y ) containsexactly one element.Proof. This is the intersection of 3 triangles contained in the tetrahedra F ( X ).To see this is the case we note if Y contains X then F ( Y ) is contained in F ( X ),and because F is a bijection between triangles we can be sure there are 3 distinctintersecands. Lemma 2.8.
Let v ∈ V and X ∈ ∆ with v ∈ X . The set \ Y ∈ ∆ | v ∈ Y ⊆ X F ( Y ) containsexactly one element.Proof. This is the intersection of 2 edges contained in the triangle F ( X ). To see thisis the case we note if Y contains X then F ( Y ) is contained in F ( X ), and because F is a bijection between edges we can be sure there are 2 distinct intersecands. Lemma 2.9.
Let v ∈ V , X ∈ ∆ and Y ∈ ∆ with v ∈ Y ⊆ X , then we have \ Z ∈ ∆ | v ∈ Z ⊆ X F ( Z ) = \ Z ∈ ∆ | v ∈ Z ⊆ Y F ( Z ) (2.1) Proof.
We first show if Z ∈ ∆ and v ∈ Z ⊆ Y then, F ( Z ) = \ W ∈ ∆ | Z ⊆ W ⊆ X F ( W )To see this we first notice the ⊆ relation holds, because F ( Z ) ⊆ F ( W ) for bothtriangles W in the intersection. To see equality holds we notice there are two distinctintersecands in the right hand side, so the intersection is an edge. e have thus shown every intersecand in the RHS of 2.1 is the intersection oftwo intersecands of the LHS. It follows we have the ⊆ relation in 2.1. Because bothsets have exactly one element (by 2.7 and 2.8) it follows they are equal. Corollary 2.10. If X and Y are two tetrahedra and | X ∩ Y | = 3 then \ Z ∈ ∆ | v ∈ Z ⊆ X F ( Z ) = \ Z ∈ ∆ | v ∈ Z ⊆ Y F ( Z ) Proof.
Use 2.9 two times with the triangle X ∩ Y .Because of 1.4 we have the following observation. Observation 2.11.
Let
X, Y be two tetrahedra containing v . Then there is a finitesequence of tetrahedra X = X , . . . X k − = Y such that | X i ∩ X i +1 | = 3 for all ≤ i < k − and v ∈ X i for all ≤ i ≤ k − . Corollary 2.12. If X and Y are two tetrahedra containing v then \ Z ∈ ∆ | v ∈ Z ⊆ X F ( Z ) = \ Z ∈ ∆ | v ∈ Z ⊆ Y F ( Z )We are now finally able to prove F ( { v } ) is a singleton for each v ∈ V . We mustonly verify the following sequence of equalities, where the first equality is given by2.6. F ( { v } ) = \ X ∈ ∆ | v ∈ X F ( X ) = \ X ∈ ∆ | v ∈ X ( \ Y ∈ ∆ | v ∈ Y ⊆ X F ( Y )) . Finally we note by 2.12 this last intersection contains only one distinct set, whichhas only one element. This completes the proof of 2.5.
The main goal of this section is to classify all 3-cyclic shells. We begin with thedefinition of an n -cyclic shell. Definition 3.1 ( n -cyclic shell) . Let n ≥
2. An n -cyclic shell is a simplicial complexof dimension n with k ≥ { H , H . . . H k − } , such that for all 0 ≤ i < j ≤ k − | T i ∩ T j | = n if j = i + 1 or i = 0 , j = k −
1, and | T i ∩ T j | = n − Observation 3.2.
There is a special family of n -cyclic shells we denote nCW k , thesecontain facets H i = X ∪ { v i , v i +1 } for ≤ i ≤ k − and H k − = X ∪ { v k − , v } ,where X is a n − simplex. We denote such n -cyclic shells by nCW k . One can seethe -cyclic shells are precisely the simplicial complices with the same intersectionmatrices as CW k .In particular we note the neighbourhood of an edge { t, b } in the triangulationof a -manifold is spanned by facets H i = { v i , v i +1 , t, b } for ≤ i ≤ k − and H k − = { v k − , v , t, b } . n order to classify all 3-cyclic shells we follow a procedure similar to the onefollowed to classify 2-cyclic shells in [1] (note these objects are simply called cyclicshells in [1]). With this in mind we introduce the 3-lineal shells. Definition 3.3 ( n -lineal shell) . An n -lineal shell is a simplicial complex of dimension n spanned by a set of k ≥ { H , H . . . H k − } such that for all 0 ≤ i < j ≤ k − | T i ∩ T j | = n if j = i + 1 and | T i ∩ T j | = n − n -lineal shells which we denote nLW k , these arespanned by k facets H i = { v i , v i +1 }∪ X for 0 ≤ i ≤ k −
1, where X is an n − v i . Notice that if H , H , . . . H k − spans an n -lineal shellthen for every 0 ≤ i < j ≤ k − H i , H i +1 . . . H j also span an n -lineal shell.This allows us to classify the 3-lineal shells starting from the small ones and gluingnew facets one by one onto one of the 3 ”unused” faces of H or H k − . In doing sowe will need to deal with a large number of cases, but the arguments used in eachcase are simple.Given a 2-cyclic shell ( V, ∆) and a new vertex u not in V we can consider thesimplicial complex ( V ∪ { u } , ∆ ′ ) of dimension 3 in which the facets are ∆ ′ = { X ∪{ u }| X ∈ ∆ } .One can see ( V ∪ { u } , ∆ ′ ) is a 3-cyclic shell, we refer to this process as lifting a2-shell. We denote a lifted 2-shell by using the same name as in [1] with a 3 to theleft. We now make the definition of a ”lifted” 3-lineal and 3-cyclic shell precise. Definition 3.4.
We say a 3-lineal shell ( V, ∆) (respectively 3-cyclic shell) is a lifted3-lineal shell (respectively 3-cyclic shell) if there exists a vertex α contained in all itsfacets. In this case we notice the 2-dimensional simplicial complex spanned by allthe 2-simplices that do not contain α is a 2-lineal shell (respectively 2-cyclic shell),we say ( V, ∆) is obtained by lifitng this 2-lineal shell (respectively 2-cyclic shell).We now include the classification of 2-cyclic and 2-lineal shells found in [1]. Theclassification of 2-shells (and thus of lifted 3-shells) will reduce computations whenwe classify the 3-shells. Theorem 3.5.
The lineal -shells are LW k as well as the following,name vertex set facet set LE { v, a, b, b ′ , a ′ } { v, a, b } , { a, b, b ′ } , { a ′ , b, b ′ } , { v, a ′ , b ′ } LE same as LE and c same as LE and { b ′ , v, c } LE same as LE and d same as LE and { v, d, b } Theorem 3.6.
The cyclic -shells are CW k as well as the following,name vertex set facet set CE same as LE same as LE and { v, a, a ′ } CE same as LE same as LE and { v, b, c } a b b ′ a ′ va CE vb a b ′ a ′ c bv CE Figure 1: The non-orientable simplicial complices CE and CE We now begin with the classification of 3-lineal shells up to isomorfism. Easyarguments show that 3 LW and 3 LW are the only 3-lineal shells with 2 and 3 facets.We continue by analyzing the cases with 4 facets. Every 3-cyclic shell with 4 facets can be obtained by gluing a new facet onto 3 LW .Because of symmetry we can assume it shares 3 vertices with H = { v , v , t, b } and contains t . There are 2 cases (from here on out α will always be the symbolused for the new vertex, we will initially assume α is a new vertex, and then con-sider the possibility of identifying α with a preexisting vertex, we will also denote thenew facet by H k or H − depending on whether it shares 3 vertices with H k − or H ) : H = { v , t, b, α } , this yields 3 LW H = { v , v , t, α } , we must identify α with a vertex of H , if α = v or α = b then | H ∩ H | = 1, so α = v , this yields a 3-lineal shell we shall denote 3 LE . First we try to extend 3 LW . Because of symmetry we can assume H shares 3vertices with H = { v , v , t, b } and contains t , there are 2 cases: H = { v , t, b, α } , this yields 3 LW H = { v , v , t, α } , we need | H ∩ H | = 2 and | H ∩ H | = 2 so α must be equalto the only common vertex of H and H that isn’t t or b , that is α = v . This yieldsa 3-lineal shell we shall denote 3 LE We now try to extend 3 LE , we will relable it to have vertices { v, a, b, b ′ , a ′ , z } andfacets H = { v, a, b, z } , H = { a, b, b ′ , z } , H = { b, b ′ , a ′ , z } , H = { b ′ , a ′ , v, z } . Noticethe permutation ( a, a ′ )( b, b ′ ) is an automorphism, so we can assume H shares 3vertices with { b ′ , a ′ , v, z } , there are 3 cases: H = { v, z, b ′ , α } , yields 3 LE ( to see this notice H , H , H , H span a complexisomorphic to 3 LW , so this shell already appeared previously). = { v, a ′ , z, α } , the rank of every intersection is correct except for H ∩ H , sowe must identify α with a vertex exclusive to H , none exist. H = { v, a ′ , b ′ , α } , We must have | H ∩ H | = 2 and | H ∩ H | = 2, so α = a, b or z . The last two options imply | H , H | = 3. So we must have α = a , this yieldsa 3-lineal shell we shall denote LS . Lemma 3.7.
If a -linear shell with k facets contains LW then it is LW k Proof.
We just have to prove 3 LW k can only be extended to 3 LW k +1 for k ≥ H k shares 3 vertices with { v k − , v k , t, b } and t ∈ H k There are two cases: H k = { v k , t, b, α } , this yields 3 LW k +1 H k = { v k − , v k , t, α } , we have | H k ∩ H | = 1 and H k ∩ H | = 1, so z must be t or b , which is impossible. First we try to extend 3 LE , we present it with vertices { v, a, a ′ , b, b ′ , z, c } andfacets H = { v, a, b, z } , H = { a, b, b ′ , z } , H = { b, b ′ , a ′ , z } , H = { b ′ , a ′ , v, z } , H = { b ′ , v, c, z } . Notice that if z is in the new facet then the complex must be a lifted3-lineal shell, so this would only yield 3 LE , there are only two other options: H − = { v, a, b, α } , notice | H − ∩ H | = | H − ∩ H | = 1, so α = b ′ or α = z ,thefirst option forces | H − ∩ H | = 3 and the second H − = H . H = { b ′ , v, c, α } , notice | H ∩ H | = | H ∩ H | = 1, so α = b or α = z , the secondoption forces H = H and the first yields a 3-lineal shell we shall denote LF We now try to extend LS , we present it with vertices { v , v , v , v , t, b } andfacets H = { v , v , v , t } , H = { v , v , t, b } , H = { v , v , t, b } , H = { v , v , t, b } , H = { v , v , v , b } , since the permutation ( v , v )( v , v )( t, b ) is an automorphism we canassume H shares 3 vertices with H . There are 3 cases: H = { v , v , v , α } , every intersection has correct rank except H ∩ H and H ∩ H , so α must be a vertex exclusive to H and H , none exist. H = { v , v , b } , every intersection has correct rank except H ∩ H , so α mustbe a vertex exclusive to H , none exist. H = { v , v , b } , every intersection has correct rank except H ∩ H , so α mustbe a vertex exclusive to H , none exist .4 7 facets We first try to extend 3 LE , we present it with vertices { d, v, b, a, b ′ , a ′ , c, z } andfacets H = { d, v, b, z } , H = { v, b, a, z } , H = { a, b, b ′ , z } , H = { b, b ′ , a ′ , z } , H = { b ′ , a ′ , v, z } , H = { b ′ , v, c, z } , if z is in the new facet then the complex must bea lifted 3 lineal shell. Notice the permutation ( a, a ′ )( b, b ′ )( c, d ) is an automorphismof 3 LE , so there is only one case: H = { b ′ , v, c, α } , notice | H ∩ H | = | H ∩ H | = 1, so α = b , this yields a 3-linealshell we shall denote LF We now try to extend LF , we present it with vertices { a , a , a , a , a , a , a } and facets H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } . Notice the permutation(0 , , , (2 ,
4) is an automorphism, so there are 3 cases: H = { a , a , a , α } , notice | H ∩ H | = | H ∩ H | = 1, so α = a or a , in eithercase we would have | H ∩ H | = 3. H = { a , a , a , α } , this yields a 3-lineal shell , we can see it is isomorphic to LF via the isomorphism a v, a c, a b ′ , a a ′ , a z, a a, a b, α d . H = { a , a , a , α } , notice | H ∩ H | = | H ∩ H | = 1, so α = a or a , in eithercase we would have | H ∩ H | = 3. We now try to extend LF , we present it with vertices { a , a , a , a , a , a , a , a } and facets H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } .If we add a facet H − it must be { a , a , a , α } with α a new vertex (because thiswas the only way to extend LF ). This works and we call this 3-lineal shell LF .There are 3 options for H : H = { a , a , a , α } , notice | H ∩ H | = | H ∩ H | = 1, so α = a or a , if α = a then | H ∩ H | = 3 and if α = a then H = H H = { a , a , a , α } , notice | H ∩ H | = | H ∩ H | = 1, so α = a or a , if α = a then | H ∩ H | = 3 and if α = a then H = H H = { a , a , a , α } , notice | H ∩ H | = | H ∩ H | = 1 so α = a or a , if α = a then H = H and if α = a then | H ∩ H | = 3 .6 9 facets We present LF with vertices { a − , a , a , a , a , a , a , a , a } and facets H = { a , a , a , a − } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } . No-tice the bijection ( a − , a )( a , a )( a , a )( a , a ) is an isomorphism. So no new facetcan be added (because we already showed no new facet can be added to the otherend of LF ) . We have now proved the following theorem: Theorem 3.8.
The -lineal shells are LE , LE , LE , LS , LF , LF , LF and LW k with k ≥ . -lineal shells In order to find all 3-cyclic shells we must analyze all possible ways to add a facet H k to a 3-lineal shell H , H , . . . H k − so that the facets H , H . . . H k span a 3-cyclicshell. In such a situation it is clear that both vertices of H ∩ H k − must be containedin H k , as well as one vertex from H \ H k − and one from H k − \ H . So we have re-duced the possibilities for H k to 4 facets. This observation clearly also implies that ifa lifted 3-lineal shell is closed, this results in a lifted 3-cyclic shell. So we must onlyconsider closing the following 3-lineal shells: LS , LF , LF , LF , LG , LG . Thisprocess just consists of revising each of the 4 × LS : we present it with vertices { v , v , v , v , t, b } and facets H = { v , v , v , t } , H = { v , v , t, b } , H = { v , v , t, b } , H = { v , v , t, b } , H = { v , v , v , b } H incorrect intersection { v , v , v , v } none { v , v , v , b } | H ∩ H | = 3 { v , v , t, v } | H ∩ H | = 3 { v , v , t, b } | H ∩ H | = 3We denote the resulting shell CS . LF : we present it with vertices { a , a , a , a , a , a , a } and facets H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } H incorrect intersection { a , a , a , a } | H ∩ H | = 3 { a , a , a , a } | H ∩ H | = 1 { a , a , a , a } | H ∩ H | = 3 { a , a , a , a } | H ∩ H | = 3 LF : we present it with vertices { a , a , a , a , a , a , a , a } and facets H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } , H = a , a , a , a } , H = { a , a , a , a } H incorrect intersection { a , a , a , a } | H ∩ H | = 3 { a , a , a , a } | H ∩ H | = 1 { a , a , a , a } | H ∩ H | = 3 { a , a , a , a } noneWe denote the resulting shell CF . LF : we present it with vertices { a − , a , a , a , a , a , a , a , a } and facets H = { a , a , a , a − } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } , H = { a , a , a , a } H incorrect intersection { a , a , a , a } | H ∩ H | = 3 { a , a , a , a } | H ∩ H | = 3 { a , a , a − , a } | H ∩ H | = 3 { a , a , a − , a } | H ∩ H | = 1 Theorem 3.9.
The -cyclic shells are CE , CE , CS , CF and CW k for k ≥ . We devote this section to proving the following theorem:
Theorem 4.1.
Let ( V, ∆) and ( V ′ , ∆ ′ ) be triangulations of -manifolds. If f : ∆ → ∆ ′ is an intersection preserving map and the facets { H , H . . . H k − } span CW k then { f ( H ) , f ( H ) . . . f ( H k − ) } span CW k . Observation 4.2. If ( V, ∆) is a triangulation of a -manifold it cannot contain E or E , as the neighbourhood of the lifting vertex would contain a copy of E or E ,this is impossible as these are non-orientable and would contradict 1.4. With this in mind we will only have to prove two lemmas concerning CS and CF to complete the proof of 4.1. Surprisingly the proof of both lemmas are rathersimilar. Lemma 4.3. If f : ( V, ∆) → ( V ′ , ∆ ′ ) is an intersection preserving map between tri-angulations of -manifolds and the facets { H , H , H , H , H , H } span CW then { f ( H ) , f ( H ) . . . f ( H n − ) } cannot span CS . roof. Assume it does, let H = { a , a , t, b } , H = { a , a , t, b } , H = { a , a , t, b } , H = { a , a , t, b } , H = { a , a , t, b } , H = { a , a , t, b } and T = { b , b , b , b } , T = { b , b , b , b , } , T = { b , b , b , b } , T = { b , b , b , b } , T = { b , b , b , b } , T = { b , b , b , b } such that F ( H i ) = T i for all 0 ≤ i < b b b b b b T b b b b b b T b b b b b b T b b b b b b T b b b b b b T b b b b b b T Figure 2: The facets T i Let J = { a , a , t, α } be the only other facet of ( V, ∆) with { a , a , t } ⊆ J . Noticethat { b , b , b } ⊆ f ( J ) or { b , b , b } ⊆ f ( J ). This is because f ( J ) must sharethree vertices with T and cannot be one of the T i , as f is bijective.Case 1: { b , b , b } ⊆ f ( J ), then we have | f ( J ) ∩ T | , | f ( J ) ∩ T | ≥
2. This implies | J ∩ H | , | J ∩ H | ≥
2. Which is only possible if α = a .Case 2: { b , b , b } ⊆ f ( J ), then we have | f ( J ) ∩ T | , | f ( J ) ∩ T | ≥
2. This implies | J ∩ H | , | J ∩ H | ≥
2. Which is only possible if α = a .Hence α = a or a . We can define J i analogously for 0 ≤ i <
6. Explicitly J i is theonly other facet containing H i − { b } . Because of rotational simmetry we have thefollowing observation. Observation 4.4. J i = { a i , a i +1 , t, a i +3 } or J i = { a i , a i +1 , t, a i +4 } for ≤ i < (here we are using the notation a i = a i − for ≤ i < ). Corollary 4.5.
For each ≤ i < at least one of the triangles { a i , a i +1 , a i +3 } or { a i , a i +1 , a i +4 } appears in the boundary of t . Now notice that the triangles { a i , a i +1 , b } are all contained in the boundary of t andspan a surface homeomorphic to a disk whose boundary is the cycle a , a , a , a , a , a . herefore the remaining triangles in the boundary of t must also span a disk withthe same boundary.Applying 4.5 with i = 0 we get two cases,Case 1: Triangle { a , a , a } appears in the boundary of t . Using 4.5 with i = 1 we have at least one of { a , a , a } and { a , a , a } must appear. Noticethat the edge a a ∈ { a , a , a } crosses both a a ∈ { a , a , a } and a a ∈{ a , a , a } , this is a contradiction, as the triangles would not form a disk withboundary a , a , a , a , a , a .Case 2: Triangle { a , a , a } appears in the boundary of t . Using 4.5 with i = 2 we have at least one of { a , a , a } and { a , a , a } must appear. Noticethat the edge a a ∈ { a , a , a } crosses both a a ∈ { a , a , a } and a a ∈{ a , a , a } , this is a contradiction, as the triangles would not form a disk withboundary a , a , a , a , a , a . Lemma 4.6. If f : ( V, ∆) → ( V ′ , ∆ ′ ) is an intersection preserving map betweentriangulations of -manifolds and the facets { H , H , H , H , H , H , H , H } span CW then { f ( H ) , f ( H ) . . . f ( H n − ) } cannot span CF .Proof. Assume it does, let H = { a , a , t, b } , H = { a , a , t, b } , H = { a , a , t, b } , H = { a , a , t, b } , H = { a , a , t, b } , H = { a , a , t, b } , H = { a , a , t, b } and T = { b , b , b , b } , T = { b , b , b , b } , T = { b , b , b , b } , T = { b , b , b , b } , T = { b , b , b , b } , T = { b , b , b , b } , T = { b , b , b , b } , T = { b , b , b , b } such that f ( H i ) = T i for all 0 ≤ i < J = { a , a , t, α } be the only other facet in ( V, ∆) with { a , a , t } ⊆ J . Noticethat { b , b , b } ⊆ f ( J ) or { b , b , b } ⊆ f ( J ). This is because f ( J ) must sharethree vertices with T and cannot be one of the T i , as f is bijective.Case 1: { b , b , b } ⊆ f ( J ), then we have | f ( J ) ∩ T | , | f ( J ) ∩ T | ≥
2. This implies | J ∩ H | , | J ∩ H | ≥
2. Which is only possible if α = a .Case 2: { b , b , b } ⊆ f ( J ), then we have | f ( J ) ∩ T | , | f ( J ) ∩ T | ≥
2. This implies | J ∩ H | , | J ∩ H | ≥
2. Which is only possible if α = a .Hence α = a or a . We can define J i analogously for 0 ≤ i <
8. Explicitly J i is the only other facet containing H i − { b } . Notice that both complices have thedihedral symmetries of the squares a , a , a , a and b , b , b , b . These symmetriesdetermine the possiblilities for all J i . Observation 4.7. If J i = { a i , a i +1 , t, α } then α must be one of the two vertices in { a , a , a , a } opposite the odd vertex among a i , a i +1 (here we are using the notation b b b b b b b T b b b b b b b b T b b b b b b b b T b b b b b b b b T b b b b b b b b T b b b b b b b b T b b b b b b b b T b b b b b b b b T Figure 3: The facets T i a i = a i − for ≤ i < ). Corollary 4.8.
For each ≤ i < at least one of the triangles { a , a i +1 , α } and { a , a i +1 , α } appears in the boundary of t (where α and α are the two vertices in { a , a , a , a } opposite the odd vertex among a i , a i +1 . Now notice that the triangles { a i , a i +1 , b } are all contained in the boundary of t andspan a surface homeomorphic to a disk whose boundary is the cycle a , a , . . . , a .Therefore the remaining triangles in the boundary of t must also span a disk withthe same boundary.Applying 4.8 with i = 0 we get two cases,Case 1: Triangle { a , a , a } appears in the boundary of t . Using 4.8 with i = 2we have at least one of { a , a , a } or { a , a , a } must appear. Notice that the edge a a ∈ { a , a , a } crosses both a a ∈ { a , a , a } and a a ∈ { a , a , a } . This is acontradiction, as the triangles would not form a disk with boundary a , a , . . . , a .Case 2: Triangle { a , a , a } appears in the boundary of t . Using 4.8 with i = 7we have at least one of { a , a , a } or { a , a , a } must appear. Notice that the edge a a ∈ { a , a , a } crosses both a a ∈ { a , a , a } and a a ∈ { a , a , a } . This is acontradiction, as the triangles would not form a disk with boundary a , a , . . . , a .We finish this section by proving the conditions of 2.5 are met, thus showing theconditions for 2.4 are met and thus proving 1.6. emma 4.9. Let ( V, ∆) and ( V ′ , ∆ ′ ) be triangulations of -manifolds and f : ∆ → ∆ ′ an intersection preserving map. Then F induces bijections between ∆ i and ∆ ′ i for i ≥ .Proof. For i = 3 the claim is clear because F coincides with f .For i = 2 notice every 2-simplex is the intersection of two facets. Because f isintersection preserving we have that f induces a bijection between the pairs of facetswhose intersection is a 1-simplex.For i = 3 notice every 1-simplex is the intersection of all facets containing it,these facets span a complex isomorphic to CW k , on the other hand every complexisomorphic to CW k in ( V, ∆) corresponds to the simplex containing the two verticesthat belong to every facet. Because of 4.1 we have that f induces a bijection betweencomplices isomorphic to CW k , therefore F induces a bijection between ∆ and ∆ ′ . References [1] Arocha, J. L., Bracho, J., Garc´ıa-Col´ın, N., & Hubard, I. (2015). Reconstruct-ing Surface Triangulations by Their Intersection Matrices. Discussiones Math-ematicae Graph Theory, 35(3), 483-491.[1] Arocha, J. L., Bracho, J., Garc´ıa-Col´ın, N., & Hubard, I. (2015). Reconstruct-ing Surface Triangulations by Their Intersection Matrices. Discussiones Math-ematicae Graph Theory, 35(3), 483-491.