aa r X i v : . [ m a t h . C O ] M a r Tensor Product of Polygonal Cell Complexes
Yu-Yen ChienMathematics DivisionNational Center for Theoretical Science
Taiwan [email protected]
September 18, 2018
Abstract
We introduce the tensor product of polygonal cell complexes, which interactsnicely with the tensor product of link graphs of complexes. We also develop theunique factorization property of polygonal cell complexes with respect to the tensorproduct, and study the symmetries of tensor products of polygonal cell complexes. A polygonal cell complex is a 2-dimensional CW-complex with polygons as 2-cells,namely a graph with polygons attached. To be precise, we take a rather formal definition:a polygonal cell complex is a 2-dimensional CW-complex satisfying:(1) Each 1-cell is an interval of length 1, and each 2-cell is a disc of positive integralcircumference.(2) For a 2-cell of circumference n , the attaching map sends exactly n points evenlydistributed on the boundary to the 0-skeleton.(3) For each boundary segment between the points described in (2), the attaching mapsends the segment isometrically onto an open 1-cell.Intuitively speaking, we can think of each 2-cell as a regular polygon, and the attachingmap glues vertices to vertices, and edges to edges. Those 2-cells act like faces of apolyhedron, and we will use the word face to denote a 2-cell alternatively. Note that theattaching map of a face may not be injective, and a polygonal cell complex can be quitedifferent from polyhedra. A polygonal complex , which simulates polyhedra better, isa polygonal cell complex satisfying:(i) The attaching map of each cell is injective.(ii) The intersection of any two closed cell is either empty or exactly one closed cell.Unless otherwise specified, when we use the word complex, it means polygonal cellcomplex, which may or may not be a polygonal complex.1 vvee ef Figure 1: a flag of a dunce hatHere is a concise way to describe the local structure of complexes. For a polygonalcell complex X , the link of X at a vertex v is a graph L ( X, v ) with vertices indexed byends of edges attached to v , and edges indexed by corners of faces attached to v . Twovertices v and v in L ( X, v ) are joined by an edge e if and only if the corresponding endsof v and v are joined by the corresponding corner of e . Basically a link describes theincidence relation of edges and faces at a vertex. Note that L ( X, v ) can also be identifiedas the set { x ∈ X | d ( x, v ) = δ } , where d is the distance function in X and δ is somepositive number less than 1/2.Take the dunce hat in Figure 1 as an example. Although there is only one edge in thecomplex, this edge has two ends attached to v , and therefore contributes two vertices tothe link at v . Notice that the top corner of the face joins these two ends, and correspondsto an edge joining two vertices in the link at v . The left corner of the face joins the sameend of the edge, and hence corresponds to a loop in the link, while the right corner of theface also corresponds to a loop at the other vertex. Therefore the link at v is a graph withtwo vertices e and e , one edge joining e and e , and two loops at e and e respectively.For polyhedra, a flag is an incident triple of face, edge, and vertex. Such definitionneeds to be modified for polygonal cell complexes. Take Figure 1 as an example again.It has only one vertex, one edge, and one face, but we would like it to have six flags justas a usual triangle. In a polygon, each flag corresponds to a triangle in its barycentricsubdivision. We can use this as an alternative definition of a flag, and this definitionworks for polygonal cell complexes as well. As Figure 1 shows, the shaded area is a flagof the dunce hat, and a dunce hat has six flags.Highly symmetric polygonal complexes have been studied in [1, 2, 9, 10]. In particular,simply-connected flag-transitive polygonal complexes with complete graphs as links areclassified in [2]. The main motivation of this paper is to use these flag-transitive complexesto generate more flag-transitive complexes. More specifically, we would like to developa product of complexes which preserves flag-transitivity, and the link of the product issome graph product of the links of factors. Suppose that • is certain type of graph product such that V (Γ • Γ ′ ) = V (Γ) × V (Γ ′ ), andwe want to define a complex product ∗ with the following property: for any complexes X and X ′ , and for any vertices v ∈ X and v ′ ∈ X ′ , we have L ( X, v ) • L ( X ′ , v ′ ) ∼ = L ( X ∗ X ′ , ( v, v ′ )) . V ( X ∗ X ′ ) = V ( X ) × V ( X ′ ). The above propertyprovides sufficient information about how the complex product ∗ shall be defined. If weassume the 1-skeletons of X and X ′ are simple graphs, by considering the vertex sets oftwo link graphs in the equation, we have { neighbours of v in X } × { neighbours of v ′ in X ′ } = { neighbours of ( v, v ′ ) in X ∗ X ′ } , which can be interpreted as two vertices ( v, v ′ ) and ( u, u ′ ) are adjacent in X ∗ X ′ if andonly if v is adjacent to u in X and v ′ is adjacent to u ′ in X ′ . This is essentially thedefinition of the direct product of simple graphs. Since the 1-skeletons of complexes arenot necessarily simple, we shall generalize the direct product to suit arbitrary graphs. Definition 2.1.
Suppose that Γ and Γ ′ are two arbitrary graphs with edge sets E (Γ) = { e α | α ∈ A } and E (Γ ′ ) = { e β | β ∈ B } . The tensor product of Γ and Γ ′ , denoted byΓ ⊗ Γ ′ , is a graph with vertex set V (Γ ⊗ Γ ′ ) = V (Γ) × V (Γ ′ ), and edge set E (Γ ⊗ Γ ′ ) = (cid:8) e δα,β | α ∈ A, β ∈ B, δ ∈ { , } (cid:9) , where e δα,β is an edge joining ( v , v ′ δ ) and ( v , v ′ − δ ), given e α joins v and v in Γ, and e β joins v ′ and v ′ in Γ ′ .Note that for simple graphs, the tensor product defined above is exactly the directproduct of graphs. Like direct product, each pair of edges from two factors generatestwo edges in the tensor product, even when loops are involved, as illustrated in Figures2 and 3. In some literatures such as [4], direct product is defined over graphs withoutparallel edges but admitting loops. In such definition, a loop serves as the identity ofdirect product. In particular a loop times an edge is an edge, and a loop times a loop isagain a loop, while in our definition a loop times an edge is two parallel edges, and a looptimes a loop creates two loops around the same vertex. Since we will need such directproduct later, we take a different name and symbol for our generalized product.There are some reasons to define tensor product in this manner. First, note the numberof vertices in L ( X, v ) is exactly the valency of v in X , where a loop at v contributes 2 tothe number. Assuming L ( X, v ) • L ( X ′ , v ′ ) ∼ = L ( X ∗ X ′ , ( v, v ′ )), this implies d X ( v ) · d X ′ ( v ′ ) = d X ∗ X ′ (( v, v ′ )) , which is true for the tensor product, but not for the direct product admitting loops.Secondly, when we glue a face along a loop, the orientation of gluing matters, and the3 =Figure 3: tensor product of two loopstensor product can keep track of such orientations. In Definition 2.1, when e α or e β is aloop, we shall think of it as an edge joining two different ends of the loop, say + and − ,and label two ends of e δα,β by + and − accordingly. We can then lift any given orientationof a loop in a factor to edges generated by this loop in the product, as illustrated inFigures 2 and 3. This also allows us to define projections unambiguously. Note that wedo not assume graphs to be directed. We just distinguish two ends of each loop. Definition 2.2.
Assume the notation of Definition 2.1. The projection from Γ ⊗ Γ ′ to Γ,denoted by π Γ , is a continuous function such that π Γ maps ( v, v ′ ) ∈ V (Γ ⊗ Γ ′ ) to v ∈ V (Γ),and e δα,β ∈ E (Γ ⊗ Γ ′ ) to e α ∈ E (Γ) isometrically between endpoints. The projection π Γ ′ from Γ ⊗ Γ ′ to Γ ′ is likewise defined.The projections defined above are graph homomorphisms in the following sense. Definition 2.3.
Let Γ and Γ ′ be two arbitrary graphs. A continuous function ϕ from Γto Γ ′ is a homomorphism if ϕ maps each vertex of Γ to a vertex of Γ ′ , and each openedge of Γ isometrically onto an open edge of Γ ′ . Remark.
In the above definition, the continuity of ϕ is essentially saying that a homo-morphism maps incident vertices and edges to incident vertices and edges. Meanwhile,the isometric condition helps to choose a representative from all homotopic maps.Note that the composition of two graph homomorphisms is again a graph homomor-phism. Together with the trivial automorphisms, the class of graphs forms a category.The following proposition shows that the tensor product defined above is actually thecategorical product of this category. Proposition 2.4.
Let Γ and Γ ′ be two arbitrary graphs. Suppose that Γ is a graphwith two homomorphisms ϕ : Γ → Γ and ϕ ′ : Γ → Γ ′ . Then there exists a uniquehomomorphism ψ : Γ → Γ ⊗ Γ ′ such that ϕ = π Γ ◦ ψ and ϕ ′ = π Γ ′ ◦ ψ . In other words,there exists a unique ψ such that the diagram in Figure 4 commutes. Proof.
Assume that there exists a continuous function ψ : Γ → Γ ⊗ Γ ′ such that ϕ = π Γ ◦ ψ and ϕ ′ = π Γ ′ ◦ ψ . Then ∀ v ∈ V (Γ ), we have ϕ ( v ) = π Γ ◦ ψ ( v ) and ϕ ′ ( v ) = π Γ ′ ◦ ψ ( v ). ByDefinition 2.2, we know that ψ ( v ) = ( ϕ ( v ) , ϕ ′ ( v )).Suppose that e is an open edge joining v and u in Γ , and we denote ϕ ( e ) and ϕ ′ ( e ) by e α and e β respectively. By the continuity of ψ , ψ ( e ) is an open path connecting ( ϕ ( v ) , ϕ ′ ( v ))and ( ϕ ( u ) , ϕ ′ ( u )). Notice that e α = ϕ ( e ) = π Γ ◦ ψ ( e ) and e β = ϕ ′ ( e ) = π Γ ′ ◦ ψ ( e ). ByDefinition 2.2, we know ψ ( e ) is either e α,β or e α,β , determined by endpoints ( ϕ ( v ) , ϕ ′ ( v ))and ( ϕ ( u ) , ϕ ′ ( u )). In case e α or e β is a loop, by keeping track of ends of the loop, ψ ( e ) isalso uniquely determined. Moreover, the local isometry over open edges of ϕ and π Γ forces ψ to map e isometrically to ψ ( e ). Note that we have explicitly constructed a continuous4 ⊗ Γ ′ Γ Γ ′ Γ π Γ π Γ ′ ϕ ϕ ′ ∃ ! ψ Figure 4: universal property of graph tensor product ψ satisfying our initial assumption. We have also shown that ψ is uniquely determined,and actually a homomorphism, which finishes the proof.For any two graphs Γ and Γ ′ , we denote the set of all homomorphisms from Γ to Γ ′ by Hom(Γ , Γ ′ ). We have the following corollary about the number of homomorphisms. Corollary 2.5.
For any graphs Γ, Γ , Γ , we have | Hom(Γ , Γ ⊗ Γ ) | = | Hom(Γ , Γ )) | · | Hom(Γ , Γ ) | . Proof.
An immediate consequence of Proposition 2.4.Note that for any graph Γ, there is a homomorphism from Γ to a loop. Since wedistinguish the orientations when we map an edge to a loop, there are actually 2 n suchhomomorphisms, where n is the number of edges of Γ. In particular, a loop is not theterminal object in the category of arbitrary graphs. Corollary 2.6.
Let Γ and Γ ′ be two graphs, P be a path in Γ of length n from v to u , and P ′ be a path in Γ ′ of length n from v ′ to u ′ . Then in Γ ⊗ Γ ′ , there exists aunique path, denoted by ( P, P ′ ) ⊗ , from ( v, v ′ ) to ( u, u ′ ) such that π Γ (( P, P ′ ) ⊗ ) = P and π Γ ′ (( P, P ′ ) ⊗ ) = P ′ . Proof.
Let I be a graph which is a path of length n . We can give I a specific orientationfrom one end to the other. Then there is a natural homomorphism ϕ from I to P , aswell as one ϕ ′ from I to P ′ . By Proposition 2.4, there exists a unique homomorphism ψ : I → Γ ⊗ Γ ′ such that ϕ = π Γ ◦ ψ and ϕ ′ = π Γ ′ ◦ ψ . Hence we have P = ϕ ( I ) = π Γ ◦ ψ ( I )and P ′ = ϕ ′ ( I ) = π Γ ′ ◦ ψ ( I ). Note that ψ ( I ) satisfies the conditions of ( P, P ′ ) ⊗ , and theuniqueness of ( P, P ′ ) ⊗ follows the uniqueness of ψ . Remark.
For simple graphs, this result is straightforward from the definition of tensorproduct. This corollary clarifies the case when P or P ′ contains a loop, where the orien-tation going through the loop will determine the edge to choose in ( P, P ′ ) ⊗ . To define our complex product more concisely, we would like to extend the notation( , ) ⊗ above. Let Γ and Γ be two graphs, C be a cycle of length n in Γ , and C bea cycle of length m in Γ . Both C and C are assigned initial vertices and orientations.5pecifically, C is ( v , e , v , e , . . . , e m − , v m = v ), where v i ∈ V (Γ ) and e j ∈ E (Γ ).Then for i ∈ { , , . . . , m − } we define( C , C ) i δ ⊗ := ( [ n, m ] n C , [ n, m ] m C i δ ) ⊗ , a cycle of length [ n, m ] in Γ ⊗ Γ , where [ n, m ] is the least common multiple of n and m , kC j is the cycle repeating C j k times, and C i is the same cycle as C , but starting at v i ,while C i is the reversed cycle of C starting at v i . Definition 3.1.
Let X and Y be two polygonal cell complexes with face sets F ( X ) = { f α | α ∈ A } and F ( Y ) = { f β | α ∈ B } . We denote the boundary length of f α and f β by n α and n β respectively, and let ( n α , n β ) denote the greatest common divisor of n α and n β . The tensor product of X and Y , denoted by X ⊗ Y , is a polygonal cell complexwith 1-skeleton X ⊗ Y , the tensor product of the 1-skeletons of X and Y , and face set F ( X ⊗ Y ) = (cid:8) f i δ α,β | α ∈ A, β ∈ B, i ∈ { , , . . . , ( n α , n β ) − } , δ ∈ { , } (cid:9) , where f i δ α,β is a face attached along ( C α , C β ) i δ ⊗ , while C α is the cycle along which f α isattached in X , and C β is the cycle along which f β is attached in Y . Remark.
We will use the jargon that f i δ α,β is generated by f α and f β , especially when facesare not clearly indexed. In the above definition, note that ( C α , C β ) i δ ⊗ and ( C α , C β ) i +( n α ,n β ) δ ⊗ are identical cycles with different starting vertices. To let a pair of corners of f α and f β contribute to exactly one face corner in X ⊗ Y , we only choose i ∈ { , , . . . , ( n α , n β ) − } .Here we discard repeated corner pairs, not faces in X ⊗ Y attached along the samecycle. For example, let X and Y be 15-gons wrapped around a cycle of length 3 and 5respectively. Note that the tensor product of a triangle and a pentagon is not the sameas X ⊗ Y . The former has only 2 · (3 ,
5) = 2 faces, while X ⊗ Y has 2 · (15 ,
15) = 30 facesin two groups, each of which has 15 faces with cyclically identical attaching maps.In the example of a triangle tensor a pentagon, the only two faces meet at every vertexin the product. In general, when n α = n β , two faces f i α,β and f i α,β meet at more thanone vertex. Therefore the tensor product of two polygonal complexes is not necessarilypolygonal. How about the case when n α = n β ? For n α even, note that f i α,β and f i α,β havetwo vertices (0 ,
0) and ( n α , n α ) in common, and the tensor product is not polygonal. Forodd cases, we have the following result. Proposition 3.2.
Suppose that X and Y are polygonal complexes with all faces of thesame odd length n . Then the tensor product X ⊗ Y is a polygonal complex. Proof.
Since X and Y are polygonal complexes, we know that X and Y are simplegraphs, and hence the 1-skeleton of X ⊗ Y , namely X ⊗ Y , is a simple graph as well.Consider the boundary of an arbitrary face f i δ α,β in X ⊗ Y , namely ( C α , C β ) i δ ⊗ . Note that C α and C β are both simple closed cycles of the same length n , as they are boundaries offaces of polygonal complexes. Therefore ( C α , C β ) i δ ⊗ is a simple closed cycle of length n . Inbrief, every face of X ⊗ Y is attached along a simple closed cycle.Now all we have to show is that the intersection of two faces in X ⊗ Y is either empty,a vertex, or an edge in X ⊗ Y . Suppose that there exist two faces f i δ α,β and f j δ ′ α ′ ,β ′ in X ⊗ Y f i δ α,β and f j δ ′ α ′ ,β ′ is neither empty, a vertex, nor an edge. Forthe case of n = 3, it is not hard to see that f i δ α,β and f j δ ′ α ′ ,β ′ share the same boundary, andin fact are the same face by the polygonality of X and Y . For the case of odd n >
3, notethat f i δ α,β and f j δ ′ α ′ ,β ′ share two vertices which are not consecutive on the boundary of faces.By the polygonality of X and Y , this implies that f α = f α ′ and f β = f β ′ . Consider theboundaries of f i δ α,β and f j δ ′ α,β , namely ( C α , C β ) i δ ⊗ and ( C α , C β ) j δ ′ ⊗ . When δ = δ ′ and i = j ,( C α , C β ) i δ ⊗ and ( C α , C β ) j δ ′ ⊗ have no vertex in common. When δ = δ ′ , notice that a commonvertex of ( C α , C β ) i δ ⊗ and ( C α , C β ) j δ ′ ⊗ corresponds to an integer m such that j + m ≡ i − m mod n ⇔ m = i − j mod n, which has a unique solution when n is odd. In other words, when δ = δ ′ , ( C α , C β ) i δ ⊗ and( C α , C β ) j δ ′ ⊗ intersect at exactly one vertex. Since f i δ α,β and f j δ ′ α,β share two vertices, we canconclude that δ = δ ′ and i = j . This finishes the proof.The complex tensor product does not preserve simple connectedness either. Proposition 3.3.
Let X and Y be an n -gon and m -gon respectively, where n and m aretwo positive integers. Then X ⊗ Y is simply-connected if and only if n = m = 1. Proof.
When n = m = 1, the 1-skeleton of X ⊗ Y is a vertex with two loops, as illustratedin Figure 3, and X ⊗ Y has two faces attached along these two loops respectively. In thiscase, X ⊗ Y is actually contractible, and of course simply-connected.Now suppose that n and m are not both equal to 1. Without loss of generality, wecan assume n ≥
2. Note that X has n vertices, n edges, and 1 face, whereas Y has m vertices, m edges, and 1 face. By Definition 3.1, the complex X ⊗ Y has nm vertices,2 nm edges, and 2( n, m ) faces. Therefore X ⊗ Y has Euler characteristic χ ( X ⊗ Y ) = nm − nm + 2( n, m ) = − nm + 2( n, m ) ≤ − m + 2 m = 0 . By the following lemma, we know X ⊗ Y is not simply-connected. Lemma 3.4.
Suppose X is a finite simply-connected polygonal cell complex. Then theEuler characteristic of X is at least 1. Proof.
Suppose X has v vertices, e edges, and f faces. First we find an arbitrary spanningtree T for the 1 skeleton of X , and then contract T to get a new complex X ′ , which is alsosimply-connected. Note that T has v − X ′ has 1 vertex, e − v + 1edges, and f faces. The fundamental group π ( X ′ ), a trivial group, can be presentedas a group with e − v + 1 generators and f relators. Consider the abelianization of π ( X ′ ), which is again trivial. Then the presentation can be expressed as f homogeneousequations of e − v + 1 unknowns over Z . To have only trivial solution, the number ofequations needs to be at least the number of unknowns. So we have f ≥ e − v + 1, andtherefore v − e + f ≥ Remark.
Let X and Y be two arbitrary complexes, and C be a cycle along the 1-skeletonof X ⊗ Y . This proposition shows that the contractibility of π X ( C ) and π Y ( C ) does notguarantee the contractibility of C . Conversely, when C is contractible in X ⊗ Y , can we7 bca b c ϕ c ab Figure 5: a complex homomorphism from a hexagon to a triangleconclude that π X ( C ) and π Y ( C ) are contractible? The answer is positive. We can finda series { C j } of homotopic cycles of C such that C = C , C n is a vertex, and each C j morphs through a single face f i δ α,β to obtain C j +1 . Note that π X ( C j ) can morph through asingle face f α to obtain π X ( C j +1 ), even when the length of f α properly divides the lengthof f i δ α,β . Therefore π X ( C ) = π X ( C ) is homotopic to π X ( C n ), which is a vertex.In the above remark, we actually abuse the notation π X , as we have not yet definedprojection maps for complex tensor products. To define such projection maps, first weintroduce some terminology. Let X and Y be an n -gon and m -gon with centre O X and O Y respectively. A function ρ : X → Y is radial if ρ sends O X to O Y , ∂X to ∂Y , andfor every point P ∈ ∂X , every real number t ∈ [0 , ρ (cid:0) t · O X + (1 − t ) P (cid:1) = t · O Y + (1 − t ) ρ ( P ) . Definition 3.5.
Assume the notation of Definition 3.1. The projection from X ⊗ Y to X , denoted by π X , is a continuous function such that π X restricted to X ⊗ Y is exactly π X , the projection of the graph tensor product, and π X maps f i δ α,β ∈ F ( X ⊗ Y ) radiallyto f α ∈ F ( X ). The projection π Y from X ⊗ Y to Y is likewise defined.The projection maps defined above are complex homomorphisms in the following sense. Definition 3.6.
Let X and Y be two polygonal cell complexes. A continuous function ϕ from X to Y is a homomorphism if ϕ restricted to X is a graph homomorphism to Y , and ϕ maps each face of X radially to a face of Y and each open face corner (ignoringthe boundary) of X homeomorphically to an open face corner of Y . Remark.
In the above definition, the continuity of ϕ is essentially saying that a complexhomomorphism maps incident cells to incident cells. Similar to the isometric conditionin graph homomorphism, the radial condition is imposed to rule out homotopic complexhomomorphisms. Most important of all, the homeomorphic corner condition forces a faceof X to wrap around a face f of Y along the direction of the attaching map of f , possiblymore than once. In particular, a face of length n can only be mapped to a face of lengthdividing n . Figure 5 illustrates such phenomenon, where corners are mapped to a cornerwith the same label. The projection π X of complex tensor product mapping f i δ α,β to f α isalso a typical example.Note that the composition of two complex homomorphisms is again a complex homo-morphism. Together with the trivial automorphisms, the class of polygonal cell complexesforms a category. The following proposition shows that the complex tensor product de-fined above is actually the categorical product of this category.8 ⊗ YX YZπ X π Y ϕ X ϕ Y ∃ ! ψ Figure 6: universal property of complex tensor product
Proposition 3.7.
Let X and Y be two polygonal cell complexes. Suppose that Z is acomplex with two homomorphisms ϕ X : Z → X and ϕ Y : Z → Y . Then there exists aunique homomorphism ψ : Z → X ⊗ Y such that ϕ X = π X ◦ ψ and ϕ Y = π Y ◦ ψ . In otherwords, there exists a unique ψ such that the diagram in Figure 6 commutes. Proof.
Assume that there exists a continuous function ψ : Z → X ⊗ Y such that ϕ X = π X ◦ ψ and ϕ Y = π Y ◦ ψ . Note that ϕ X , ϕ Y , π X , and π Y restricted to the 1-skeletons oftheir domains are all graph homomorphisms. By Proposition 2.4, the restriction of ψ to Z is a uniquely determined graph homomorphism to X ⊗ Y .Suppose that f is a face in Z , ϕ X ( f ) wraps around a face f α in X , and ϕ Y ( f ) wrapsaround a face f β in X . Then ϕ X ( f ) = π X ◦ ψ ( f ) wraps around f α , and ϕ Y ( f ) = π Y ◦ ψ ( f )wraps around f β . By Definition 3.5, ψ ( f ) must wrap around f i δ α,β for some i and δ . Let c bea corner of f . Then we must have ϕ X ( c ) = π X ◦ ψ ( c ) and ϕ Y ( c ) = π Y ◦ ψ ( c ). By the remarkafter Definition 3.1, this pair of corners ( ϕ X ( c ) , ϕ Y ( c )), orientation included, appears inexactly one f i δ α,β . Therefore i and δ are uniquely determined, and ψ ( f ) wraps around this f i δ α,β . Moreover, the radiality of ϕ X and π X forces ψ to map f radially to f i δ α,β . Note thatwe have explicitly constructed a continuous ψ satisfying our initial assumption. We havealso shown that ψ is uniquely determined, and actually a complex homomorphism, whichfinishes the proof. Remark.
For any two complexes X and Y , we denote the set of all complex homomor-phisms from X to Y by Hom( X, Y ). Similarly to Corollary 2.5, we have | Hom(
Z, X ⊗ Y ) | = | Hom(
Z, X ) | · | Hom(
Z, Y ) | . As we mentioned earlier, for any graph Γ, there is a homomorphism from Γ to aloop. It is reasonable to ask the following question: for any complex X , is there always ahomomorphism from X to a 1-gon? The answer is negative. Take Figure 7 as an example.Once the image of the leftmost edge is determined, it determines the image of all otheredges. If we identify the leftmost and the rightmost edges with a twist, i.e. making it aMobius strip, then there is no way to have a homomorphism. Note that this question isnot related to orientability. If the complex is a strip with 3 squares, then the Mobius casehas a homomorphism, while the orientable case does not. Proposition 3.8.
Let X and Y be two polygonal cell complexes, and ϕ : X → Y be acomplex homomorphism mapping a vertex v ∈ V ( X ) to u ∈ V ( Y ). Then ϕ induces agraph homomorphism L ( ϕ ) from L ( X, v ) to L ( Y, u ). Moreover, let Z be another complexand ρ : Y → Z be a complex homomorphism mapping u to w ∈ V ( Z ). Then we have L ( ρ ◦ ϕ ) = L ( ρ ) ◦ L ( ϕ ), as illustrated in Figure 8.9igure 7: a homomorphism to a 1-gon X L ( X, v ) Y Zϕ ρρ ◦ ϕ L ( Y, u ) L ( Z, w ) L ( ϕ ) L ( ρ ) L ( ρ ◦ ϕ ) L Figure 8: functoriality of L Proof.
By definition, L ( X, v ) has vertices corresponding to edge ends around v in X ,and edges corresponding to face corners at v in X . Since ϕ restricted to X is a graphhomomorphism, ϕ maps an edge end around v in X to an edge end around u in Y . Inaddition, by the homeomorphic condition in Definition 3.6, ϕ maps a face corner at v joining two edge ends around v homeomorphically to a face corner at u joining two edgeends around u . Therefore ϕ induces a graph homomorphism L ( ϕ ) from L ( X, v ) to L ( Y, u ).Once these induced graph homomorphisms between link graphs are defined, the equality L ( ρ ◦ ϕ ) = L ( ρ ) ◦ L ( ϕ ) follows immediately. Remark.
To each polygonal cell complex, we can assign a distinguished vertex to bethe basepoint. Together with basepoint-preserving homomorphisms, the class of pointedpolygonal cell complexes also forms a category. The above proposition is essentially sayingthat L is a functor from this category to the category of graphs.Now we move back to the main purpose of this chapter: to develop a complex productinteracting nicely with some product of link graphs. From the above discussion, weknow that the complex tensor product arises naturally in the category of polygonal cellcomplexes. Does this natural categorical product fulfill the main job? Yes, it does. Theorem 3.9.
Suppose that X and Y are two polygonal cell complexes, and v and u aretwo vertices in X and Y respectively. Then we have L ( X, v ) ⊗ L ( Y, u ) ∼ = L ( X ⊗ Y, ( v, u )) . Proof.
We can identify edge ends incident to a vertex as paths of length 1 leaving thevertex, since a loop contributes to two edge ends as well as two such paths, which we call1-paths for short. By Corollary 2.6, there is a bijection between 1-paths leaving ( v, u ) in X ⊗ Y and pairs of 1-path leaving v in X and 1-path leaving u in Y . Therefore we canindex 1-paths leaving ( v, u ) in X ⊗ Y by such 1-path pairs in X and Y .10 e α e α u e β e β ⊗ = ( v, u )( e α , e β )( e α , e β )( e α , e β ) ( e α , e β )Figure 9: well linked tensor product v ′ e ′ α e ′ α u ′ e ′ β e ′ β ⊗ = ( v ′ , u ′ )( e ′ α , e ′ β )( e ′ α , e ′ β )( e ′ α , e ′ β ) ( e ′ α , e ′ β )Figure 10: automorphic image of Figure 9Suppose that f α ∈ F ( X ) has a corner c α at ( e α , v, e α ), and f β ∈ F ( Y ) has a cor-ner c β at ( e β , u, e β ), as illustrated in Figure 9. These e ∗ ’s should be understood as1-paths. By the remark after Definition 3.1, the pairing of these two corners appearsexactly once in f i α,β and f j α,β respectively, forming corners (( e α , e β ) , ( v, u ) , ( e α , e β )) and(( e α , e β ) , ( v, u ) , ( e α , e β )) in X ⊗ Y . Note that by taking projection maps, we know thatany face corner at ( v, u ) comes from some pairing of corners at v and u .Now we translate the above statements in terms of corresponding link graphs. Firstof all, we have V (cid:0) L ( X, v ) (cid:1) × V (cid:0) L ( Y, u ) (cid:1) ∼ = V (cid:0) L ( X ⊗ Y, ( v, u )) (cid:1) . Secondly, the corner c α is an edge joining vertices e α and e α in L ( X, v ), and c β is an edge joining vertices e β and e β in L ( Y, u ). Notice that the edge pair ( c α , c β ) contributes to one edge joining( e α , e β ) and ( e α , e β ), and one edge joining ( e α , e β ) and ( e α , e β ) in L ( X ⊗ Y, ( v, u )).Meanwhile, taking all possible pairings of edges exhausts all edges in L ( X ⊗ Y, ( v, u )). ByDefinition 2.1, this is exactly saying that L ( X, v ) ⊗ L ( Y, u ) ∼ = L ( X ⊗ Y, ( v, u )). Remark.
In the terminology of category theory, this theorem is essentially saying thatthe functor L from the category of pointed complexes to the category of graphs preservescategorical products, which is not always true for an arbitrary functor.As indicated in Propositions 3.2 and 3.3, the complex tensor product does not nec-essarily preserve polygonality and simple connectedness. Fortunately, complex tensorproduct does preserve the most important property for our purpose. Theorem 3.10.
Let X and Y be any two flag-transitive polygonal cell complexes. Thenthe complex tensor product X ⊗ Y is flag-transitive. Proof.
In case X or Y has no faces, then X ⊗ Y is simply a graph, and the flag-transitivityfollows easily from the definition of graph tensor product. Hereafter we assume that both X and Y have at least one face. 11et (( e α , e β ) , ( v, u ) , ( e α , e β )) be a face corner in X ⊗ Y , which projects to a cor-ner ( e α , v, e α ) in X and a corner ( e β , u, e β ) in Y , as illustrated in Figure 9. Let(( e ′ α , e ′ β ) , ( v ′ , u ′ ) , ( e ′ α , e ′ β )) be another face corner in X ⊗ Y , which projects to a cor-ner ( e ′ α , v ′ , e ′ α ) in X and a corner ( e ′ β , u ′ , e ′ β ) in Y , as illustrated in Figure 10. Since X and Y are flag-transitive, there exist ρ ∈ Aut( X ) mapping ( e α , v, e α ) to ( e ′ α , v ′ , e ′ α )and σ ∈ Aut( Y ) mapping ( e β , u, e β ) to ( e ′ β , u ′ , e ′ β ). Comparing Figures 9 and 10,note that ( ρ, σ ) gives an automorphism of X ⊗ Y mapping (( e α , e β ) , ( v, u ) , ( e α , e β ))to (( e ′ α , e ′ β ) , ( v ′ , u ′ ) , ( e ′ α , e ′ β )). The above discussion shows that Aut( X ⊗ Y ) acts tran-sitively on face corners with orientations, and therefore transitively on half-corners. Inother words, Aut( X ⊗ Y ) acts transitively on flags. Remark.
In Figure 9, flipping both corners in X and Y will flip both corners in X ⊗ Y ,whereas flipping only one corner in either X or Y will swap two corners in X ⊗ Y . In the proof of Theorem 3.10, the key fact we used is the following relation:Aut( X ) × Aut( Y ) ≤ Aut( X ⊗ Y ) . Is it possible that these two groups are actually isomorphic? When X an Y are isomorphic,we can swap X and Y to obtain an extra automorphism, since the complex tensor productis commutative up to isomorphism. In addition to swapping, the following propositiongives more extra automorphisms in a less obvious way. Proposition 4.1.
Let X , Y , and Z be polygonal cell complexes. Then we have( X ⊗ Y ) ⊗ Z ∼ = X ⊗ ( Y ⊗ Z ) . In other words, complex tensor product is associative up to isomorphism.
Proof.
A categorical result of the universal property in Proposition 3.7. See [6].The associativity of the complex tensor product complicates Aut( X ⊗ Y ). For example,if Y can be factorized into X ⊗ Z , then X ⊗ Y ∼ = X ⊗ ( X ⊗ Z ) has an automorphismswapping the two copies of X . Hence the symmetry of the product of complexes is alsorelated to the factoring of complexes. In response to associativity, we modify the originalquestion as follows: for complexes X i which are irreducible with respect to complex tensorproduct, is the automorphism group Aut( ⊗ X i ) generated by automorphisms of X i ’s,together with permutations of isomorphic factors? By a Cartesian automorphism , wemean an element in the subgroup of Aut( ⊗ X i ) generated in the above manner.There have been lots of studies about the symmetry of different products of graphs.One of the major goals of this chapter is to apply the theory of the graph direct productto the complex tensor product. Hence we first introduce related theorems about the graphdirect product. The book [4] by Hammack, Imrich, and Klavˇzar offers a comprehensivesurvey of products of graphs, and we shall follow their approach and terminology here.We briefly mentioned the direct product of graphs in Chapter 2. Here we give thedefinition again, with an emphasis on the possible presence of loops. We say that a graphΓ is a simple graph with loops admitted if for any u, v ∈ V (Γ), there is at most one12igure 11: direct product of graphs in S edge joining u and v , including the case u = v . In particular, there is at most one loopat a vertex. For convenience, we use S to denote the class of simple graphs, and S todenote the class of simple graphs with loops admitted. Definition 4.2.
Let Γ and Γ ′ be two graphs in S . The direct product of Γ and Γ ′ ,denoted by Γ × Γ ′ , is a graph in S with vertex set V (Γ × Γ ′ ) = V (Γ) × V (Γ ′ ). Thereis an edge joining two vertices ( v, v ′ ) and ( u, u ′ ) in Γ × Γ ′ if and only if there is an edgejoining v and u in Γ, and there is an edge joining v ′ and u ′ in Γ ′ .Note in the above definition, v and v ′ could be the same vertex, as well as u and u ′ .Figure 11 illustrates the direct product of two graphs in S . Under this definition, noticethat a loop L serves as the identity element of direct product of graphs. In other words,for any simple graph Γ with loops admitted, we always have L × Γ ∼ = Γ × L ∼ = Γ . Also note that the direct product of two edges is again two edges, laid out as a crossin the figure, which is part of the reason why graph theorists choose the symbol “ × ” [4].Therefore the direct product of two connected graphs is not necessarily connected. Thefollowing theorem is known as Weichsel’s Theorem [4]. Theorem 4.3.
Suppose that Γ and Γ ′ are two connected simple graphs with at least twovertices. If Γ and Γ ′ are both bipartite, then Γ × Γ ′ has exactly two components. If atleast one of Γ and Γ ′ is not bipartite, then Γ × Γ ′ is connected. Proof.
The first part of the theorem is straightforward. For the second part, note thata simple graph is not bipartite if and only if there is an odd cycle in the graph. Byexploiting such a cycle properly, the second part of the theorem follows. For a detailedproof, please refer to Theorem 5.9 in [4].A graph Γ is prime if Γ has more than one vertex, and Γ ∼ = Γ × Γ implies that eitherΓ or Γ is a loop. Note that the idea of being prime depends on the class of graphs weare talking about. For example, let Γ be a path of length 3, which has 4 vertices. ThenΓ is prime in S , as the only possible factoring is the product of two edges, which is thedisjoint union of two edges. And the statement that Γ ∼ = Γ × Γ implies either Γ or Γ is a loop is still logically true. However, Γ can be factorized in S as the graph on theleft of Figure 11 times one edge in the bottom, and hence Γ is not prime in S .13igure 12: vertices with the same set of neighboursConsider the question of factoring a graph into the product of prime graphs. Fora finite graph, such a prime factorization always exists, since the number of verticesof factors decreases as the factoring goes. However, such a prime factorization is notnecessarily unique, and it depends on the graph itself and the class of graphs where wedo the factoring. For example, a path of length 3 together with associativity can be usedto create graphs with non-unique prime factorizations in S . There are also graphs withnon-unique prime factorizations in S , an example of which can be found in [4]. Thefollowing theorem of unique prime factorization is due to McKenzie [8]. Theorem 4.4.
Suppose that Γ ∈ S is a finite connected non-bipartite graph with morethan one vertex. Then Γ has a unique factorization into primes in S .The next question is about the automorphism group of direct product, which hopefullyhas only these Cartesian automorphisms with respect to the product. Note that a pairof vertices with the same set of neighbours creates pairs of vertices with the same setof neighbours in the direct product, and results in lots of non-Cartesian automorphisms.This phenomenon is illustrated in Figure 12 , where a vertex with a loop should have itselfas a neighbour. We say that a graph is R -thin if there are no vertices with the same setof neighbours. In addition to R -thinness, the disconnectedness due to Theorem 4.3 alsocreates non-Cartesian automorphisms. Even when the direct product is connected, theremight still be some exotic automorphisms. The following theorem is due to D¨orfler [3]. Theorem 4.5.
Suppose that Γ ∈ S is a finite connected non-bipartite R -thin graphwith a prime factorization Γ = Γ × Γ × · · · × Γ n in S . Then Aut(Γ) is generated byautomorphisms of prime factors and permutations of isomorphic factors.We would like to use Theorems 4.4 and 4.5 to develop similar results for the complextensor product. The first problem we immediately encounter is that, for the complextensor product, we obtain the 1-skeleton of the product through the graph tensor prod-uct, which is not exactly the same as the direct product of graphs. Fortunately, such adifference does not really take place in graphs with higher symmetries. Proposition 4.6.
Let Γ ∈ S be a finite connected non-bipartite R -thin graph with morethan one vertex, and Γ = Γ × Γ × · · · × Γ n be the unique prime factorization in S . IfΓ is edge-transitive, then Γ and each prime factor Γ i are in S .14 roof. Since Γ has more than one vertex, the connectedness of Γ implies that Γ has anon-loop edge. By the edge-transitivity of Γ, we know Γ has no loop, and hence is in S .If each factor Γ i has a loop, then the product Γ will have a loop, which is not true. Ifeach factor Γ i is loop-free, then we have finished the proof. Hence we can assume there isat least one factor with a loop, and at least one factor without a loop.Let Γ α be the direct product of all factors with a loop, and Γ β be the direct product ofall factors without a loop. Then we have Γ = Γ α × Γ β . Note that permuting isomorphicfactors of Γ does not involve permuting factors of Γ α with factors of Γ β . By Theorem4.5, we have Aut(Γ) = Aut(Γ α ) × Aut(Γ β ). Since a prime factor has more than onevertex, Γ α and Γ β both have more than one vertex. Since Γ is connected, Γ α and Γ β areboth connected. Hence Γ α has a loop at some vertex v and a non-loop edge joining twovertices v α and v ′ α , while Γ β has a non loop edge joining two vertices v β and v ′ β . Then inΓ = Γ α × Γ β , there is an edge joining ( v, v β ) and ( v, v ′ β ), and another edge joining ( v α , v β )and ( v ′ α , v ′ β ). Notice that Aut(Γ) = Aut(Γ α ) × Aut(Γ β ) can not send the first edge to thesecond one, contradicting the assumption that Γ is edge-transitive. Remark.
To visually interpret the last few lines of the proof, it says that a Cartesianautomorphism can not permute horizontal edges with slant edges in Figure 12.Now we move on to the factorization of polygonal cell complexes. First consider thefollowing example. Let X and Y be a triangle and a pentagon respectively, X ′ be a cycleof length 3 with two triangles attached, and Y ′ be a cycle of length 5 with two pentagonsattached. Since the numbers of vertices of these complexes are prime, the only possibleway to factorize them is to have a factor of one vertex with at least a loop and a face,which creates double edges in the product. Hence we know these complexes can not befactorized further, and we have non-unique factorizations X ⊗ Y ′ ∼ = X ′ ⊗ Y .Here we give another example of non-unique factorization. Let X be a triangle, and Y ′ be a (7 · · X ⊗ Y ′ has two faces of length 3 · ·
7, wrapped around two cycles oflength 3 · · X ′ wrapped around a cycle of length 3,and a pentagon Y . It is easy to see that X ⊗ Y ′ ∼ = X ′ ⊗ Y , and these complexes can notbe factorized further. To avoid these non-uniquely factorized situations, we restrict ourdiscussion to the factorization of simple complexes. Definition 4.7.
A polygonal cell complex X is a simple complex if X has at least oneface, X has no pairs of faces attached along the same cycle, and the attaching map ofeach face does not wrap around a cycle more than once. A polygonal cell complex X isa prime complex if there do not exist complexes X and X such that X = X ⊗ X . Remark.
Figure 13 above is a simple complex with two 1-gons. If we add another 2-gonattached along two different loops, the resulting complex is still a simple complex, as theboundary cycles of theses faces are not exactly the same.To factorize a complex X , our general setting is as follows. We assume that we knowa factorization of the 1-skeleton X = Γ ⊗ Γ , and try to find a complex factorization X = X ⊗ X such that X = Γ and X = Γ . A natural thought is to project the faces of X down to Γ and Γ to be faces. Consider the complex tensor product of a triangle anda pentagon, which is a complex with two 15-gons. Note that when we project these two15-gons back to the 1-skeletons of factors, what we obtain are 15-gons wrapped aroundcycles of length 3 and 5 respectively, not the original faces.15igure 13: a simple complex with two 1-gons Definition 4.8.
Let X be a polygonal cell complex, f be a face of X attached alonga cycle C f , and Γ and Γ be two graphs such that X = Γ ⊗ Γ . The reductiveprojection of f to Γ i , denoted by π Γ i ( f ), is a face attached along the reduced cycle of π Γ i ( C f ) in Γ i , namely the shortest cycle C such that repeating C gives π Γ i ( C f ). Remark.
In exactly the same way, we can define π Γ i ( f ) for the case X = ⊗ ni =1 Γ i . Notethat when X = Γ ⊗ Γ ⊗ Γ , we have π Γ ( f ) = π Γ ( π Γ ⊗ Γ ( f )) = π Γ ( π Γ ⊗ Γ ( f )). Proposition 4.9.
Let X be a simple complex, and Γ and Γ be two graphs such that X = Γ ⊗ Γ . If there exist two complexes X and X with 1-skeletons Γ and Γ respectively such that X = X ⊗ X , then X and X are simple complexes whose facesare precisely the reductive projections of faces of X . Proof.
Suppose that such complexes X and X exist. Let f be a face of X attachedalong a cycle C f of length n , and let C j of length n j be the reduced cycle of π Γ j ( C f ) in X j for j ∈ { , } . Note that f is generated by a face f of X attached along m C , andby a face f of X attached along m C , where m i C i is the cycle made by repeating C i for m i times. By Definition 3.1, f and f generate faces attached along ( m C , m C ) i δ ⊗ ,where i ∈ { , , . . . , ( m n , m n ) − } and δ ∈ { , } . By the Euclidean algorithm, wecan find an integer k > k ≡ n and k ≡ ( n , n ) mod n . Note thatin k steps along ( m C , m C ) δ ⊗ , we can walk from the starting vertex of ( m C , m C ) δ ⊗ to the starting vertex of ( m C , m C ) ( n ,n ) δ ⊗ , so these two cycles are identical. Since X is simple, there are no pairs of faces attached along the same cycle in X . Therefore wehave ( n , n ) ≥ ( m n , m n ) ≥ ( n , n ) . Now consider the length of the face f , which is n = [ m n , m n ] = m n · m n ( m n , m n ) = m m · n n ( n , n ) = m m · [ n , n ] . This shows that f is attached along some cycle ( C , C ) i δ ⊗ of length [ n , n ] for m m rounds, and the simplicity of X implies that m = m = 1. In other words, X i must havethe reductive projection π Γ i ( f ) of f as its face. Note that different faces of X might havethe same reductive projection in X i , and we have to discard duplicated ones. Otherwiseduplicated faces in X i will generate duplicated faces in X , violating the simplicity of X .Conversely, any faces f of X and f of X are the reductive projections of the faces in X they generate. Hence X and X are the simple complexes with exactly those facesfrom the reductive projections of faces of X . Proposition 4.10.
Let X , X , and X be polygonal cell complexes such that X = X ⊗ X . Then X is a simple complex if and only if X and X are simple complexes. Proof.
Proposition 4.9 takes care of the only if part, and here we prove the if part. Supposethat X has an n -gon f attached along a cycle for m rounds. Since X and X are simple, f f to X and X , which are of length l and l respectively. Note that l and l both divide nm . Then the two reductive projectionsgenerate faces of length n = [ l , l ] ≤ nm . Hence we can conclude that m = 1. If thereis another face f ′ in X attached along the same cycle with f , then f ′ is also generatedby the reductive projections of f . If we can show a face in X and a face in X do notgenerate duplicated faces in X , then this implies X is a simple complex.Suppose that a face f of X has vertices v , v , . . . , v p − , v in order, and a face f of X has vertices u , u , . . . , u q − , u in order. By the remark after Definition 3.1, everypair of corners of f and f appears exactly once in the faces generated by f and f . Iftwo faces generated by f and f are attached along the same cycle in X , there must betwo pairs of corners of f and f forming the same corner in X . In particular, we canfind ( v i , u i ′ ) = ( v j , u j ′ ) such that i = j or i ′ = j ′ . When i = j , we have v i = v j and v i + k = v j + k for any integer k mod p . This implies that f wraps around a cycle morethan once, violating the simplicity of X . Similarly i ′ = j ′ contradicts the simplicity of X . The contradiction results from the assumption that two faces generated by f and f are attached along the same cycle in X . Hence we know that f and f does not generateduplicated faces, and the simplicity of X follows. Proposition 4.11.
Let X be a simple complex, and Γ and Γ be two graphs such that X = Γ ⊗ Γ . Then the following two statements are equivalent:(1) There exist two complexes X and X such that X i = Γ i and X = X ⊗ X .(2) For any faces f and f of X , X contains all faces generated by π Γ ( f ) and π Γ ( f ). Proof.
Assume (1). By Proposition 4.9, X and X are the simple complexes with exactlythose reductive projections of X as faces. For any faces f and f of X , π Γ ( f ) is a faceof X , and π Γ ( f ) is a face of X . Since X = X ⊗ X , X contains all faces generated by π Γ ( f ) and π Γ ( f ). Hence (1) implies (2).Assume (2). First we show that a face f of X can be generated by π Γ ( f ) and π Γ ( f ).Let C f , C , and C be the boundary cycles of f , π Γ ( f ), and π Γ ( f ) respectively. ByDefinition 4.8, we can assume that π Γ j ( C f ) = n j C j for j ∈ { , } , namely repeating C j for n j times gives π Γ j ( C f ). Note that f is attached along some cycle ( n C , n C ) i δ ⊗ , whichcan be rewritten as ( n , n )( n ( n ,n ) C , n ( n ,n ) C ) i δ ⊗ . Since the simple complex X has no faceattached around a cycle more than once, we know that ( n , n ) = 1, and thereforelength C f = n · (length C ) = n · (length C ) = [length C , length C ] . This shows that π Γ ( f ) and π Γ ( f ) can generate the face f . Now let X and X bethe simple complexes with exactly those faces from the reductive projections of X . ByProposition 4.10, X ⊗ X is a simple complex, and in particular X ⊗ X has no duplicatedfaces. By the assumption of (2), X contains all the faces of X ⊗ X . Conversely, anyface f of X is a face of X ⊗ X , since f can be generated by π Γ ( f ) and π Γ ( f ). Thenwe have X = X ⊗ X , and hence (2) implies (1).Although we already know the associativity of complex tensor product through theuniversal property, it will be helpful to understand how faces are formed in the productof more than two complexes. First let us review the product of two complexes. Let f α be a face of length n α attached along a cycle C α in X , and f β be a face of length n β
65 6 Figure 14: a face generated by 3 faces in complex tensor productattached along a cycle C β in Y . By Definition 3.1, f α and f β generate faces f i δ α,β of length[ n α , n β ] attached along ( C α , C β ) i δ ⊗ , i ∈ { , , . . . , ( n α , n β ) − } , δ ∈ { , } . To explain theboundary cycle of f i δ α,β in plain language, basically we pick a pair of corners of f α and f β to start, and go around C α and C β in two coordinates respectively until we return to thestarting pair of corners. Note that the index i is chosen in such a way that each pair ofcorners appears exactly once among all faces generated by f α and f β .A good way to visualize this is a slot machine of two reels of length [ n α , n β ], cyclicallylabeled by the vertices of f α and f β respectively. Faces generated by f α and f β have a one-to-one correspondence with different combinations of two reels, with flipping allowed forthe second reel. From this aspect, it is easy to see that for face f j of length n j in complex X j , j ∈ { , , . . . , m } , f , f , . . . , f m generate faces in ⊗ mj =1 X j of length [ n , n , . . . , n m ]such that each m -tuple of corners appears exactly once among all generated faces. Facesgenerated by f , f , . . . , f m have a one-to-one correspondence with different combinationsof m reels of length [ n , n , . . . , n m ], cyclically labeled by the vertices of f j respectively,with flipping allowed from the second reel on. Figure 14 illustrates how a face is generatedby the complex tensor product of 3 faces from such an aspect. Theorem 4.12.
Let X be a simple polygonal cell complex. If the 1-skeleton of X is afinite simple connected non-bipartite R -thin edge-transitive graph with more than onevertex, then X has a unique factorization into prime complexes. Proof.
By Theorem 4.4, since X ∈ S ⊂ S is a finite connected non-bipartite graph withmore than one vertex, X has a unique factorization X = Γ × Γ × · · · × Γ n into primesin S with respect to direct product of graphs. By Proposition 4.6, the edge-transitivityof X implies that each prime factor Γ i is in fact a simple graph. On the other hand, ifwe factorize X with respect to graph tensor product, each factor would also be a simplegraph with more than one vertex, because a loop creates double edges in the product,and a single vertex breaks the connectivity of the product. Note that direct productand graph tensor product coincide in S . Hence we know X has a unique factorization X = Γ ⊗ Γ ⊗ · · · ⊗ Γ n into primes in S with respect to graph tensor product.Now we consider the factorization of the complex X . Note that we can always obtain aprime factorization of X , since the number of vertices of factors decreases as the factoringgoes. Suppose X has two factorizations A and B , and X is a prime factor of X in A with 1-skeleton Γ ⊗ Γ . By Proposition 4.11, there exist two faces f and f such that X lacks certain face generated by π Γ ( f ) and π Γ ( f ). In other words, there is certain18air of corners of π Γ ( f ) and π Γ ( f ) missing in the faces of X , and hence such pairwill be absent in the n -tuples representing face corners of X . By Proposition 4.9, we canfind faces f and f of X such that π Γ ⊗ Γ ( f ) = f and π Γ ⊗ Γ ( f ) = f , and we have π Γ ( f ) = π Γ ( f ) and π Γ ( f ) = π Γ ( f ). If Γ and Γ belong to different prime factors X and X in B , we can reductively project f i to X i to obtain a face f ′ i of X i , i ∈ { , } .Then we have π Γ ( f ′ ) = π Γ ( f ) = π Γ ( f ) and π Γ ( f ′ ) = π Γ ( f ) = π Γ ( f ). Notice that f ′ and f ′ generate all possible pairs of corners of π Γ ( f ) and π Γ ( f ) in X ⊗ X andhence in X , a contradiction. So Γ and Γ belong to the same prime factor in B .The above argument can be applied to the case when the 1-skeleton of X is the graphtensor product of more than two prime graphs, simply by splitting prime graph factorsinto two groups. It follows that every prime 1-skeleton factor of X belongs to the sameprime complex X ′ in B . Conversely, every prime 1-skeleton factor of X ′ belongs to X ,and hence X and X ′ are actually the same. In case X has a prime 1-skeleton Γ j , then Γ j belongs to some X ′ in B with a prime 1-skeleton, otherwise the prime 1-skeleton factorsof X ′ belong to at least two complexes in A . In conclusion, we know two factorizations A and B are identical, and X has a unique factorization into prime complexes. Theorem 4.13.
Suppose that X is a simple polygonal cell complex, and its 1-skeleton isa finite simple connected non-bipartite edge-transitive R -thin graph with more than onevertex. Let X = X ⊗ X ⊗ · · · ⊗ X n be a prime factorization of X . Then Aut( X ) isgenerated by automorphisms of prime factors and permutations of isomorphic factors. Proof.
Since X has no faces attached along the same cycle, an automorphism of X iscompletely determined by its action on the 1-skeleton X , and we can identify Aut( X ) asa subgroup of Aut( X ). To understand Aut( X ), by the argument in the proof of Theorem4.12, we know X has a unique factorization X = Γ × Γ × · · · × Γ m = Γ ⊗ Γ ⊗ · · · ⊗ Γ m into primes in S . By Theorem 4.5, the extra R -thin condition on X implies that Aut( X )is generated by automorphisms of Γ i ’s and permutations of isomorphic Γ j ’s.Let ϕ be an arbitrary automorphism of X , which can be represented as some ρ ∈× mi =1 Aut(Γ i ) followed by a permutation of Γ j ’s. This implies that for any face f of Xϕ (cid:0) π ⊗ i ∈ I Γ i ( f ) (cid:1) = π ϕ ( ⊗ i ∈ I Γ i ) (cid:0) ϕ ( f ) (cid:1) = π ⊗ i ∈ I ϕ (Γ i ) (cid:0) ϕ ( f ) (cid:1) , where I is an arbitrary non-empty subset of { , , . . . , m } . Suppose that X has 1-skeleton X = ⊗ i ∈ I Γ i for some I ⊂ { , , . . . , m } . We claim that ∀ i ∈ I , ϕ (Γ i ) belongs to the sameprime factor X k of X . If not, then we can find I ⊔ I = I such that ∀ i ∈ I , ∀ j ∈ I , ϕ (Γ i )and ϕ (Γ j ) belong to different prime factors of X . Let Γ α = ⊗ i ∈ I Γ i and Γ β = ⊗ j ∈ I Γ j ,and hence we have X = Γ α ⊗ Γ β . Since X is prime, by Proposition 4.11, we can findfaces f and f of X such that X lacks certain face generated by π Γ α ( f ) and π Γ β ( f ).By Proposition 4.9, we can find faces f and f of X such that π Γ α ⊗ Γ β ( f ) = f and π Γ α ⊗ Γ β ( f ) = f . Then the complex X lacks certain corner combination of π Γ α ( f ) and π Γ β ( f ) in the m -tuples representing face corners of X . By taking the automorphism ϕ , thecomplex X lacks certain corner combination of π ⊗ i ∈ I ϕ (Γ i ) ( ϕ ( f )) and π ⊗ j ∈ I ϕ (Γ j ) ( ϕ ( f )),which is impossible because ϕ (Γ i ) and ϕ (Γ j ) belong to different prime factors of X , andtaking complex tensor product of these factors generates all the corner combinations.Hence for every 1-skeleton factor Γ i of X , ϕ (Γ i ) belongs to the same prime factor X k of X . By considering ϕ − , we know that X k has exactly these ϕ (Γ i )’s as 1-skeleton19actors. Moreover, ϕ (cid:0) π ⊗ i ∈ I Γ i ( f ) (cid:1) = π ⊗ i ∈ I ϕ (Γ i ) (cid:0) ϕ ( f ) (cid:1) implies that ϕ induces an isomor-phism from X to X k . This shows that every ϕ ∈ Aut( X ) can be represented as some σ ∈ × ni =1 Aut( X i ) followed by a permutation of X j ’s, and the theorem holds. Remark.
Let e X be the disjoint union of prime factors of X . Then the above theoremimplies that Aut( X ) ∼ = Aut( e X ), which is a convenient way to describe Aut( X ).The following corollary is a partial converse of Theorem 3.10. Corollary 4.14.
Suppose that X is a simple polygonal cell complex, and its 1-skeleton isa finite simple connected non-bipartite edge-transitive R -thin graph with more than onevertex. If X is flag-transitive, then any factor of X is flag-transitive. Proof.
Note that it suffices to show that any prime factor of X is flag-transitive. Thenby Theorem 4.12 and Theorem 3.10, any factor of X is a complex tensor product offlag-transitive prime factors of X , and hence is flag-transitive.By Theorem 4.12, X has a unique prime factorization X = X ⊗ X ⊗ · · · ⊗ X n .Suppose that one of the prime factors is not flag-transitive, without loss of generality say X , and X i is isomorphic to X if and only of 1 ≤ i ≤ m for some integer m ≤ n . Since X is not flag-transitive, there exist two oriented face corners ( e , v , e ) and ( e ′ , v ′ , e ′ )in X such that Aut( X ) can not map one corner to the other. For each j such that m + 1 ≤ j ≤ n , we pick an arbitrary corner ( e j , v j , e j ) of X j . Consider the following twocorners of X :(( e , . . . , e , e m +1 , . . . , e n ) , ( v , . . . , v , v m +1 , . . . , v n ) , ( e , . . . , e , e m +1 , . . . , e n )) and(( e ′ , . . . , e ′ , e m +1 , . . . , e n ) , ( v ′ , . . . , v ′ , v m +1 , . . . , v n ) , ( e ′ , . . . , e ′ , e m +1 , . . . , e n )) . By Theorem 4.13, Aut( X ) is generated by automorphisms of prime factors and permuta-tion of isomorphic factors. In particular, it is impossible for Aut( X ) to map one of theabove corners to the other, contradicting to the flag-transitivity of X . Therefore we canconclude that any prime factor of X is flag-transitive.The corollary below answers the question we posed in the beginning of the chapter. Corollary 4.15.
For i ∈ { , , . . . , n } , let X i be a simple prime complex with a finitesimple connected non-bipartite symmetric R -thin 1-skeleton having more than one ver-tex. Then the complex tensor product X = ⊗ ni =1 X i has automorphism group Aut( X )generated by Aut( X i )’s and permutations of isomorphic X j ’s. Proof.
By Proposition 4.10, we know X is a simple complex. By the definition of graphtensor product, we know X is a finite simple graph. Note that a simple graph is non-bipartite if and only if there is a cycle of odd length. Then the graph tensor product of twonon-bipartite graphs contains a cycle of odd length and hence is non-bipartite. Inductionshows that X is non-bipartite, and by Theorem 4.3 we know that X is connected. Bythe special case of Theorem 3.10 (complexes without faces), we know X is symmetricand hence edge-transitive. Note that for two graphs Γ and Γ , the set of neighbours ofa vertex ( u, v ) ∈ V (Γ ⊗ Γ ) is the direct product of the set of neighbours of u in Γ with the set of neighbours of v in Γ . This implies the graph tensor product of R -thingraphs is a R -thin graph. To summarize, we know X is a simple complex with a primefactorization X = ⊗ ni =1 X i , and its 1-skeleton X is a finite simple connected non-bipartiteedge-transitive R -thin graph with more than one vertex. By Theorem 4.13, we know thatAut( X ) is as described in the corollary. 201 ,
1) (0 , ,
2) ( − , −
1) (1 , −
1) ( − , , , Remark.
The tensor products of edge-transitive graphs are not necessarily edge-transitive.Therefore we require each X i to be symmetric to ensure the edge-transitivity of X .Note that when a complex has a face of odd length, then the 1-skeleton of the complexis non-bipartite, and Corollary 4.15 has a chance to work. In the next chapter, we willinvestigate the automorphism group of the tensor product of complexes with only facesof even lengths from a different aspect. In this chapter we investigate the tensor product of complexes with only faces of evenlengths, and our goal is to develop results similar to Corollary 4.15, which basically says anautomorphism of certain complex tensor products must be of Cartesian type. Note thatwhen there is more than one bipartite factor, Theorem 4.3 implies that the complex tensorproduct is disconnected, and the product is likely to have non-Cartesian automorphismsfrom the direct product of automorphism groups of components. Hence in such a context,the proper question to pose should be as follows: for complexes X i with only faces of evenlengths, is the automorphism group of a component of ⊗ X i generated by automorphismsof X i ’s together with permutations of isomorphic factors?For graph tensor products, the connectedness of the product does not guarantee theabsence of non-Cartesian automorphisms. For complex tensor products, we hope thatthe extra face structure helps to eliminate non-Cartesian automorphisms. For example,let us look at the complex tensor product of two squares, which has two isomorphiccomponents. We denote vertices of a square by 0 , , , − · ·
4! automorphisms, and not all of themgive a complex automorphism due to the extra face structure.Figure 15 also reveals an important fact of the tensor product of complexes with onlyfaces of even lengths: a face is antipodally attached to another face generated by thesame pair of faces, and through such antipodally attached relation we can find all otherfaces generated by the same pair of faces in that component. Such face blocks (definedin Definition 5.4) help to determine the Cartesian structure of a complex tensor product,and if we can show a generic face block has only Cartesian automorphisms, then we have a21hance to force a complex automorphism stabilizing a face block to be of Cartesian type.To simplify the problem, we restrict our discussion to the tensor product of complexeswith faces of the same even length, and the first step is to establish the Cartesian resultfor the tensor product of 2 n -gons. The following lemma is a useful tool for this purpose. Lemma 5.1.
Suppose on a real line, someone wants to take d steps to walk from aninteger d − k to 0, where ⌊ d ⌋ ≥ k ≥ (cid:0) d − k (cid:1) ways to arrive from 1, and (cid:0) d − k − (cid:1) ways to arrive from − (cid:0) d − k (cid:1) / (cid:0) d − k − (cid:1) is greater than or equal to 1, with equality if and only if d − k = 0.Moreover, when d is fixed and k is increasing, the ratio is decreasing. Proof.
Suppose this person takes x steps of minus 1 and y steps of plus 1 to arrive at 0.Then we have x + y = d and − x + y = − d + 2 k , and therefore x = d − k and y = k .By ordering two types of steps arbitrarily, we can obtain all different ways to arrive at 0.To arrive from 1, the last step must be minus 1, and there are (cid:0) d − k (cid:1) such combinations.To arrive from −
1, the last step must be plus 1, and there are (cid:0) d − k − (cid:1) such combinations.When d is odd, we have d − ≥ k and hence (cid:0) d − k (cid:1) > (cid:0) d − k − (cid:1) . When d is even, we have d ≥ k which implies d − > k − (cid:0) d − k (cid:1) ≥ (cid:0) d − k − (cid:1) , with equality if and only if k + ( k −
1) = d −
1, namely d − k = 0. To show that the ratio (cid:0) d − k (cid:1) / (cid:0) d − k − (cid:1) decreases as k increases, we simply have to verify the following inequality: (cid:0) d − k (cid:1) / (cid:0) d − k − (cid:1) > (cid:0) d − k +1 (cid:1) / (cid:0) d − k (cid:1) ⇔ (cid:0) d − k (cid:1)(cid:0) d − k (cid:1) > (cid:0) d − k +1 (cid:1)(cid:0) d − k − (cid:1) ⇔ ( d − ... ( d − k ) k ! ( d − ... ( d − k ) k ! > ( d − ... ( d − k − k +1)! ( d − ... ( d − k +1)( k − ⇔ d − kk > d − k − k +1 ⇔ dk − > dk +1 − ⇔ k + 1 > k, which is obviously true. Proposition 5.2.
For i ∈ { , , . . . , m } , let C i be a graph which is a cycle of length 2 n ,where n is an integer at least 3. Then the automorphism group of a component of ⊗ mi =1 C i can be generated by elements of Aut( C i )’s together with permutations of C i ’s. Proof.
We denote vertices of C i by 0 , , . . . , n − , n, − ( n − , − ( n − , . . . , − ⊗ mi =1 C i containing the vertex v = (0 , , . . . , × mi =1 Aut( C i ) acts transitively on vertices of ⊗ mi =1 C i . Therefore to prove this proposition,it suffices to show that the v -stabilizer G v of Aut(Γ) can be generated by elements ofAut( C i )’s together with permutations of C i ’s. Notice that there are 2 m · m ! Cartesianautomorphisms of Γ fixing v , generated by the reflection fixing 0 in each C i and allpermutations of m factors. If we can show | G v | ≤ m · m !, then the proposition follows.First we show that Γ is a rigid graph. Namely we want to show that if ϕ ∈ G v fixesall neighbours of v , then ϕ must be trivial. Note that two vertices ( b , b , . . . , b m ) and( c , c , . . . , c m ) are adjacent if and only if b i − c i ≡ ± n for all i , and therefore V (Γ) ⊆ V ∗ = { ( a , a , . . . , a m ) ∈ V ( ⊗ mi =1 C i ) | a ≡ a ≡ · · · ≡ a m mod 2 } . u = ( a , a , . . . , a m ) ∈ V ∗ , there is a path of length d = max {| a | , | a | , . . . , | a m |} from u to v , because we can reach 0 in d steps in the coordinates with absolute value d ,and we can also reach 0 in d steps in the other coordinates by walking back and forth aseach coordinate has the same parity. Hence V (Γ) = V ∗ , and d ( u, v ) = d follows easily.Note that the number of geodesics from u to v is the product of the number of waysin each coordinate to walk to 0 in d steps. Look at the i -th coordinate of v . For now weassume that a i ≥
0, and let k i be the integer such that a i = d − k i . If n > a i >
0, wehave ⌊ d ⌋ ≥ k i ≥
0, and walking to 0 in d steps is equivalent to the setting of Lemma 5.1.By the lemma, the ratio of numbers of u − v geodesics arriving from 1 and from − i -th coordinate is (cid:0) d − k i (cid:1) / (cid:0) d − k i − (cid:1) >
1. Since the automorphism ϕ fixes ( ± , ± , . . . , ± ϕ . Again by the Lemma, k i must remain the same to keep this ratio, and hence the i -th coordinate of ϕ ( u ) must be a i . If a i = 0, then u has a neighbour w with the i -th coordinate 1. Note that ϕ ( u ) isadjacent to ϕ ( w ) with the i -th coordinate 1, and the i -th coordinate of ϕ ( u ) is either 0 or2. In the latter case, since n > >
0, by taking ϕ − the above argument implies a i = 2,a contradiction. Hence the i -th coordinate of ϕ ( u ) is 0. Similarly if a i = n , then the i -thcoordinate of ϕ ( u ) is n . For negative a i , by applying the mirror version of Lemma 5.1, weknow that the i -th coordinate of ϕ ( u ) is a i . Note that the above result is true for everycoordinate. Hence ϕ ( u ) = u for every u ∈ V (Γ), and ϕ is trivial.Now look at the local structure around v . Note that two neighbours of v taking differ-ent values in k coordinates have 2 m − k common neighbours. In particular, two neighboursof v differ in exactly one coordinate if and only of they have 2 m − common neighbours.Hence among the neighbours of v , the relation of differing in exactly one coordinate ispreserved under G v . If we draw an auxiliary edge between any two such neighbours of v ,then the 2 m neighbours of v plus these auxiliary edges form a hypercube Q m preservedunder G v . Since Γ is rigid, an automorphism of G v is completely determined by its actionon the neighbours of v , which also induces an automorphism of the auxiliary Q m . As aresult, we have | G v | ≤ Aut( Q m ) = 2! · m !, which finishes the proof. Remark.
Let H be the subgroup of × mi =1 Z n generated by S = { ( ± , ± , . . . , ± } . Notethat the component Γ in the above proof is actually isomorphic to the Cayley graph of H with respect to the generating set S . Corollary 5.3.
Suppose that X i is a 2 n -gon for i ∈ { , , . . . , m } , where n is an integerat least 3. Then the automorphism group of a component of ⊗ mi =1 X i can be generated byelements of Aut( X i )’s together with permutations of X i ’s. Proof.
Note that a 2 n -gon has the same automorphism group as its 1-skeleton, and ⊗ mi =1 X i has the same Cartesian automorphisms as ⊗ mi =1 X i . Hence a vertex stabilizer G v of acomponent X of ⊗ mi =1 X i has 2 m · m ! Cartesian automorphisms, and | G v | is at most thecardinality of the stabilizer of v in X , which is 2 m · m ! by Proposition 5.2. Remark.
We do need the condition n ≥ n = 2,Figure 15 illustrates a component of the tensor product of two squares. Its 1-skeleton isthe complete bipartite graph K , with lots of non-Cartesian automorphisms. With theface structure, there are much fewer complex automorphisms, but swapping (0 ,
2) and(2 ,
0) still gives a non-Cartesian complex automorphism.23 uv v
Figure 16: a non-elementary complexNow we formally define the face blocks mentioned in the beginning of the chapter.An intuitive definition of a face block in a complex tensor product ⊗ mi =1 X i would be anyconnected component in ⊗ mi =1 f i , where each f i is a face of X i . Note that if each f i isan even gon attached injectively, then ⊗ mi =1 f i has 2 m − components, and hence 2 m − faceblocks. If these f i ’s are attached non-injectively, then the above face blocks could haveextra incidence relations, and we might end up having fewer components. We would liketo define a face block regardless of attaching maps, so we take the following definition. Definition 5.4.
For i ∈ { , , . . . , m } , let X i be a polygonal cell complex with only facesof even length 2 n ≥
2. Let f i be a face of X i with corners labeled by 0 , , . . . , n − face block generated by f , f , . . . , f m is a subcollection of faces generatedby f , f , . . . , f m such that two faces f a and f b are in the same face block if and only if acorner of f a with label ( a , a , . . . , a m ) and a corner f b with label ( b , b , . . . , b m ) have a − b ≡ a − b ≡ · · · ≡ a m − b m mod 2 . Remark.
It is easy to see that a face block is well-defined no matter how faces are cyclicallylabeled and no matter which corners are chosen to verify the above criterion. In generalit is not obvious whether or not two faces are in the same face block of a complex tensorproduct without knowing the tensor product structure. In the tensor product of thefollowing class of complexes, recognizing a face block is much easier.
Definition 5.5.
A connected polygonal cell complex X is an elementary complex if X satisfies the following three conditions:(1) Every face of X is of the same even length ≥ Remark.
Condition (3) basically says no two faces can be attached antipodally, and in aface different pairs of antipodal corners are not attached to the same pair of vertices. Forexample, the complex in Figure 16 is not an elementary complex.
Proposition 5.6.
For i ∈ { , , . . . , m } , let X i be an elementary complex with facesof even length 2 n ≥
2. Then in the complex tensor product ⊗ mi =1 X i , for any antipodalvertices u and v of a face in ⊗ mi =1 X i , there are exactly 2 m − faces having u and v asantipodal vertices, and these faces are in the same face block. Moreover, for any twofaces f and f ′ in the same face block, we can find a series of faces f , f , . . . , f k such that f = f , f k = f ′ , f i and f i +1 share antipodal vertices for i ∈ { , , . . . , k − } , and k ≤ n .24 roof. In ⊗ mi =1 X i , suppose that u = ( u , u , . . . , u m ) and v = ( v , v , . . . , v m ) are antipodalvertices of a face f generated by f , f , . . . , f m , where u i and v i are vertices of X i and f i is a face of X i for i ∈ { , , . . . , m } . Note that for each i ∈ { , , . . . , m } , projecting f to X i gives f i , and f i has u i and v i as antipodal vertices. Since X i is elementary, u i and v i are not the same vertex, and f i is the only face of X i having u i and v i as antipodalvertices, with a unique pair of antipodal corners attached to u i and v i . Hence any face in ⊗ mi =1 X i having u and v as antipodal vertices must be generated by f , f , . . . , f m in sucha way that the corresponding corners c i of the f i ’s at u i are combined together. With thecorner c of f fixed, flipping f i at c i for i ∈ { , , . . . , m } gives all 2 m − faces having u and v as antipodal vertices, and these faces are in the same face block.Now suppose that f and f ′ are two faces in the same face block B generated byfaces with corners labeled by 0 , , . . . , n − B according to such a corner labeling, and by following steps of ( ± , ± , . . . , ± v of f to reach any other vertex in B in n steps. In particular, thereis a unique vertex in B such that we need n steps to reach it from v . Since f ′ has morethan one vertex, we can start from v to reach a vertex u of f ′ in n − f and one step in f ′ if necessary, we can find a path from f to f ′ of lengthat most n + 1 such that the first and the last steps are in f and f ′ respectively. Notethat each ( ± , ± , . . . , ±
1) step determines a unique face in B , and hence the above pathdetermines a series of faces f , f , . . . , f k such that f = f , f k = f ′ , and k ≤ n . If f i and f i +1 are determined by the same ( ± , ± , . . . , ±
1) step, then f i and f i +1 are actuallythe same face, and we can remove one of them from the sequence. If f i and f i +1 aredetermined by different ( ± , ± , . . . , ±
1) steps, then f i and f i +1 are two different faceswith a common vertex with label ( a , a , . . . , a m ). Note that( a , a , . . . , a m ) + n ( ± , ± , . . . , ±
1) = ( a + n, a + n, . . . , a m + n ) mod 2 n, which is also a common vertex of f i and f i +1 , and therefore f i and f i +1 share antipodalvertices. The above argument is illustrated in Figure 15.Proposition 5.6 allows us to easily recognize a face block in a complex tensor product.If we impose the following conditions on each factor, then we can read the Cartesianstructure of a complex tensor product through the incidence relation of face blocks. Definition 5.7.
A connected polygonal cell complex X is an ordinary complex if everyface f of X is of the same even length 2 n ≥
4, and satisfies the following extra conditions:(1) If we label corners of f cyclically from 1 to 2 n , then any two corners with differentparities are not attached to the same vertex.(2) For any face f ′ incident to f , either f has only one corner meeting f ′ , or f has onlytwo consecutive corners meeting f ′ . Remark.
If the 1-skeleton of X is bipartite, then X satisfies (1) automatically. Alsonote that a polygonal complex satisfies both (1) and (2). The reader might have noticedthat (2) implies the condition (3) of an elementary complex. Since there are alternativeconditions serving our purpose as effectively as (2), we avoid defining ordinary complexesas a subclass of elementary complexes. 25 :Figure 17: how to avoid incidence Proposition 5.8.
For i ∈ { , , . . . , m } , suppose that X i is an ordinary complex withfaces of even length 2 n ≥
4. Let B be a face block generated by f , f , . . . , f m and B ′ bea face block generated by f ′ , f ′ , . . . , f ′ m , where f i and f ′ i are faces of X i . If B and B ′ areincident, then the following two statements are equivalent:(1) ∃ j such that f j is incident to f ′ j in X j , and ∀ i = j we have f i = f ′ i .(2) Every face of B is incident to a face of B ′ . Proof.
Assume (1). Without loss of generality, we can assume that j = 1. Since B and B ′ are incident, there is a face corner c of B meeting a face corner c ′ of B ′ . Suppose that c is the combination of corners c i of the f i ’s, and c ′ is the combination of corners c ′ i ofthe f ′ i ’s. Note that c of f meets c ′ of f ′ in X . Also note that for i = 1, c i and c ′ i arein the same face f , and they are either the same corner or different corners attached tothe same vertex. In particular, by condition (1) of Definition 5.7, c i and c ′ i have the sameparity under cyclic Z n labeling for i = 1. Let f be an arbitrary face of B generated bycombining c of f with corners c i of the f i ’s for i = 1. By Definition 5.4, c i has the sameparity as c i , and therefore has the same parity as c ′ i . Then again by Definition 5.4, theface f ′ generated by combining c ′ of f ′ with c i ’s of the f i ’s is a face of B ′ . It is obviousthat f is incident to f ′ . To summarize, given an arbitrary face f of B , we can find a face f ′ of B ′ incident to f . Hence (1) implies (2).Assume (2). If f i and f ′ i are disjoint, then B and B ′ are disjoint, which contradicts(2). Hence for each i ∈ { , , . . . , m } , f i and f ′ i are either incident or actually the same.Suppose that there is more than one j , say for j ∈ { , } , such that f j and f ′ j are incident.By condition (2) of Definition 5.7, f and f have either one corner or two consecutivecorners meeting f ′ and f ′ respectively. Pick two consecutive corners of f containing allcorners meeting f ′ and colour them blue. Similarly pick two consecutive corners of f containing all corners meeting f ′ and colour them red. Consider the faces generated by f , f , . . . , f m with the following corner combination: coloured corners of f and f areplaced at the opposite positions, as illustrated in Figure 17. Note that these faces aredisjoint with faces generated by f ′ , f ′ , . . . , f ′ m . If B does not contain any of these faces,we can flip two red corners of f to generate faces of B , and the resulting faces are stilldisjoint with faces generated by f ′ , f ′ , . . . , f ′ m . In other words, we can find a face of B incident to no face in B ′ , a contradiction. So there is at most one j such that f j and f ′ j are incident. Moreover, condition (1) of Definition 5.7 implies that different face blocksgenerated by f , f , . . . , f m are disjoint. Since B and B ′ are incident, we know that thereis exactly one j such that f j and f ′ j are incident. Hence (2) implies (1).26 f f f B , B , B , f B , B , B , Figure 18: Cartesian structure of face blocks
Remark.
Note that condition (2) of Definition 5.7 is only used for the argument illustratedin Figure 17. It is not hard to have alternative conditions serving this purpose, especiallywhen the length of faces is higher. We also want to point out that through finer exami-nation of incidence relation between face blocks, it is possible to obtain more informationsuch as how f j meets f ′ j in X j , perhaps under weaker conditions.With Propositions 5.6 and 5.8, in a tensor product X = X ⊗ X ⊗ · · · ⊗ X m whereeach X i is an elementary ordinary complex with only faces of even length 2 n ≥
4, we canrecognize face blocks and the Cartesian structure of X through the incidence relation onfaces, which is preserved under automorphisms of X . Now we define a graph Γ X to encodethe Cartesian structure of X . Let Γ X be a simple graph with vertex set × mi =1 F ( X i ), wherea vertex ( f , f , . . . , f m ) represents all faces of X generated by f , f , . . . , f m , such thattwo vertices are adjacent if and only if they take the same face in m − X i be a simple graph with vertexset F ( X i ), such that two vertices are adjacent if and only if the corresponding facesare incident in X i . Notice that Γ X = Γ X (cid:3) Γ X (cid:3) · · · (cid:3) Γ X i , where (cid:3) is the Cartesianproduct of graphs (see [4] for the definition). Figure 18 illustrates the case m = 2, where B i,j = ( f i , f j ) represents all faces generated by f i and f j . The following theorem due toImrich [5] and Miller [7] restricts the automorphism group of Γ X . Theorem 5.9.
Suppose that Γ is a finite simple connected graph with a factorizationΓ = Γ (cid:3) Γ (cid:3) · · · (cid:3) Γ m , where each Γ i is prime with respect to Cartesian product. Thenthe automorphism group of Γ is generated by automorphisms of prime factors and per-mutations of isomorphic factors.We can not guarantee Γ X i is prime, but at least X i is indeed a prime complex. Proposition 5.10.
Let Y be an elementary complex. Then Y is a prime with respect tocomplex tensor product, and Y is not a component of any complex tensor product. Proof.
Suppose that there exist complexes Y and Y such that Y is a component of Y ⊗ Y . Note that a face of Y is of even length, and must be generated by either twoeven faces or by one even and one odd face. In either case, by Definition 3.1, Y will havefaces antipodally attached together, violating that Y is elementary.Note that in Figure 18, each B i,j actually contains two face blocks generated by f i and f j , and in general each vertex of Γ X defined above contains 2 m − face blocks. Evenif we have some control over the automorphism group of Γ X , having multiple face blocks27 vv v v v v v v v v v v v u u u u u u f f f f Figure 19: tensor product of a hexagon with a 3-hexagon necklaceat one vertex of Γ X could lead to non-Cartesian automorphisms of X . Let us look atthe tensor product of a hexagon with a 3-hexagon necklace as illustrated in Figure 19,where v i is the vertex generated by u i and v , and coloured vertices in the product aregenerated by coloured u , u , and u . For brevity, half of the faces in the product areomitted. Consider the automorphism ρ of the product induced by fixing f , f , and f but flipping f (swapping the top and the bottom edges) in two factors. Then ρ fixes thefour face blocks on the left and right, and permutes vertices in each of the two middleblocks. In particular, we can permute vertices in a block while its two incident blocks arefixed. Therefore we can permute vertices in one middle block and fix all other five blocks.This gives a non-Cartesian automorphism.There are two main reasons why we have the above non-Cartesian automorphism.First, there is more than one face block generated by the same faces lying in the samecomponent of the product. Secondly, factors are not rigid enough, so the action on oneface block can not affect incident blocks, and can not be transmitted to blocks generatedby the same faces. We suspect that if either of these two reasons is absent, then eachcomponent of the product might have only Cartesian automorphisms. In particular, ifthe 1-skeleton of each factor is bipartite, then face blocks generated by the same faces arein different components. Also note that if a complex is a surface, it is rigid enough thatthe action on one face completely determines the whole automorphism. So far we do nothave a definite result yet, and hence we pose the following two conjectures. We hope toresolve these problems in the near future. Conjecture 5.11.
For i ∈ { , , . . . , m } , suppose that X i is an elementary ordinarycomplex with faces of the same even length 2 n ≥
6, and X i has bipartite 1-skeleton. Thenfor any component X of the complex tensor product ⊗ mi =1 X i , Aut( X ) can be generatedby automorphisms of X i ’s together with permutations of isomorphic factors. Conjecture 5.12.
For i ∈ { , , . . . , m } , suppose that X i is an elementary ordinarycomplex with faces of the same even length 2 n ≥
6, and X i has surface structure. Then28or any component X of the complex tensor product ⊗ mi =1 X i , Aut( X ) can be generatedby automorphisms of X i ’s together with permutations of isomorphic factors. The author appreciates the support of National Center for Theoretical Science, Taiwan,and would like to thank Ian Leary for introducing this topic and valuable advices.
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