Connectedness of Certain Graph Coloring Complexes
aa r X i v : . [ m a t h . C O ] F e b Connectedness of Certain Graph Coloring Complexes
Nandini Nilakantan ∗ , Samir Shukla † August 27, 2018
Abstract
In this article, we consider the bipartite graphs K × K n . We prove that the con-nectedness of the complex Hom( K × K n , K m ) is m − n − m ≥ n and m − G, K m ) is exactly m − d − m ≥ n , where d is the maximal degree of the graph G . Keywords : Hom complexes, Exponential graphs, Discrete Morse theory.2000
Mathematics Subject Classification primary 05C15, secondary 57M15
A classical problem in graph theory is the determination of the chromatic number of agraph which finds application in several fields. In 1955, Kneser posed the Kneser conjecturewhich dealt with the computation of the chromatic number of a certain class of graphs [11].This conjecture was proved in 1975 by Lova´sz [15]. Lova´sz constructed a simplicial complexcalled the neighborhood complex N ( G ) of a graph G to prove the Kneser conjecture.In [1], Lova´sz generalised the neighborhood complex by the prodsimplicial complex calledthe Hom complex , denoted by Hom(
G, H ) for graphs G and H . In particular, Hom( K , G )(where K denotes a complete graph with 2 vertices) and N ( G ) are homotopy equivalent.The idea was to be able to estimate the chromatic number of an arbitrary graph G byunderstanding the connectivity of the Hom complex from some standard graph into G . If H is the complete graph K n , then the complexes Hom( G, K n ) are highly connected. Thesecomplexes Hom( G, K n ) can be considered to be the spaces of n - colorings of the graph G .The complexes Hom( G, H ) are not very well understood.Babson and Kozlov made the following conjecture in [1].
Conjecture 1.1.
For a graph G with maximal degree d , Hom( G, K n ) is at least n − d − G is an odd cycle, Hom( G, K n ) is n − n ≥ . In [13], it is provedthat for even cycles C m , Hom( C m , K n ) is n − n ≥ . In [17], Malenproved a strong generalisation of [4], by showing that Hom(
G, K n ) is at least m − D − D = max H ⊂ G δ ( H ), δ ( H ) being the minimal degree of the graph H , and the ∗ Department of Mathematics and Statistics, IIT Kanpur, Kanpur-208016, India. [email protected]. † Department of Mathematics and Statistics, IIT Kanpur, Kanpur-208016, India. [email protected]. H of G . Since D ≤ d , this also impliesConjecture 1.1.It is natural to ask whether it is possible to classify the class of graphs G for which theHom complexes Hom( G, K n ) are exactly n − D − K × K n , which are n − D = d = n −
1. SinceHom( K × K n , K m ) ≃ Hom( K , K K n m ), where K K n m is an exponential graph, it is sufficientto determine the connectedness of Hom( K , K K n m ) which is the same as the connectednessof the neighborhood complex N ( K K n m ). The main result of this article is Theorem 1.1.
Let m, n ≥ . Thenconn ( Hom ( K × K n , K m )) = ( m − n − if m ≥ nm − otherwise . In the special case, where m = n + 1, in [18], it is shown that Theorem 1.2.
Let n ≥ and p = n !( n − n . Then H k ( N ( K K n n +1 ); Z ) = Z if k = 0 , or n − Z p − n !+12 if k = 20 otherwise . A graph G is a pair ( V ( G ) , E ( G )), where V ( G ) is the set of vertices of G and E ( G ) ⊂ V ( G ) × V ( G ) the set of edges. If ( x, y ) ∈ E ( G ), it is also denoted by x ∼ y . Here, x is saidto be adjacent to y . The degree of a vertex v is defined by deg( v ) = |{ y ∈ V ( G ) | x ∼ y }| .Here | X | represents the cardinality of the set X .A bipartite graph is a graph G with subsets X and Y of V ( G ) such that V ( G ) = X ⊔ Y and ( v, w ) / ∈ E ( G ) if { v, w } ⊆ X or { v, w } ⊆ Y . Examples of bipartite graphs includethe even cycles C n where V ( C n ) = { , , . . . , n } and E ( C n ) = { ( i, i + 1) | ≤ i ≤ n − } ∪ { (1 , n ) } . In this case X = { , , , . . . , n − } and Y = { , , , . . . , n } .A graph homomorphism from a graph G to graph H is a function φ : V ( G ) → V ( H )such that, ( v, w ) ∈ E ( G ) = ⇒ ( φ ( v ) , φ ( w )) ∈ E ( H ) . A finite abstract simplicial complex X is a collection of finite sets such that if τ ∈ X and σ ⊂ τ , then σ ∈ X . The elements of X are called simplices of X . If σ ∈ X and | σ | = k + 1,then σ is said to be k − dimensional . A k − k simplex σ is calleda facet of σ . A prodsimplicial complex is a polyhedral complex each of whose cells is a directproduct of simplices ([14]).Let v be a vertex of the graph G . The neighborhood of v is defined by N ( v ) = { w ∈ V ( G ) | ( v, w ) ∈ E ( G ) } . If A ⊂ V ( G ), the neighborhood of A is defined as N ( A ) = { x ∈ V ( G ) | ( x, a ) ∈ E ( G ) ∀ a ∈ A } .The neighborhood complex , N ( G ) of a graph G is the abstract simplicial complex whosevertices are all the non isolated vertices of G and whose simplices are those subsets of V ( G )which have a common neighbor.Consider distinct vertices u and v in G such that N ( u ) ⊂ N ( v ). The subgraph G \ { u } of G , where V ( G \ { u } ) = V ( G ) \ { u } and E ( G \ { u } ) = E ( G ) \ { ( v, w ) ∈ E ( G ) | at least one of v or w is u } , is called a fold of G .2or any two graphs G and H , Hom( G, H ) is the polyhedral complex whose cells areindexed by all functions η : V ( G ) → V ( H ) \ {∅} , such that if ( v, w ) ∈ V ( G ), then η ( v ) × η ( w ) ⊂ E ( H ).Elements of Hom( G, H ) are called cells and are denoted by ( η ( v ) , . . . , η ( v k )), where V ( G ) = { v , . . . , v k } . A cell ( A , . . . , A k ) is called a face of B = ( B , . . . B k ), if A i ⊂ B i ∀ ≤ i ≤ k . The Hom complex is often referred to as a topological space. Here, we arereferring to the geometric realisation of the order complex of the poset. The simplicialcomplex whose facets are the chains of the Poset P is called the order complex of P .A topological space X is said to be n - connected if π ∗ ( X ) = 0 for all ∗ ≤ n . By convention, π ( X ) = 0 means X is connected. The connectivity of a topological space X is denoted byconn( X ), i.e. , conn( X ) is the largest integer m such that X is m -connected. If X is a nonempty, disconnected space, it is said to be − −∞ connected.We now review some of the constructions related to the existence of an internal hom which is related to the categorical product. All these details can be found in [8, 9, 16]. • The categorical product of two graphs G and H , denoted by G × H is the graph where V ( G × H ) = V ( G ) × V ( H ) and ( g, h ) ∼ ( g ′ , h ′ ) in G × H if g ∼ g ′ and h ∼ h ′ in G and H respectively. • If G and H are two graphs, then the exponential graph H G is defined to be the graphwhere V ( H G ) contains all the set maps from V ( G ) to V ( H ). Any two vertices f and f ′ in V ( H G ) are said to be adjacent, if v ∼ v ′ in G implies that f ( v ) ∼ f ′ ( v ′ ) in H .Using tools from poset topology ([3]), it can be shown that given a poset P and a posetmap c : P → P such that c ◦ c = c and c ( x ) ≥ x ∀ x ∈ P , there is a strong deformationretract induced by c : P → c ( P ) on the relevant spaces. Here, c is called the closure map .From [5, Proposition 3.5] we have a relationship between the exponential graph and thecategorical product in the Hom-complex. Proposition 2.1.
Let G , H and K be graphs. Then Hom ( G × H, K ) can be included inHom ( G, K H ) so that Hom ( G × H, K ) is the image of the closure map on Hom ( G, K H ) . Inparticular, there is a strong deformation retract | Hom ( G × H, K ) | ֒ → | Hom ( G, K H ) | . From [1, Proposition 5.1] we have the following result which allows us to replace a graphby its fold in the Hom complex.
Proposition 2.2.
Let G and H be graphs such that u, v are distinct vertices of G and N ( u ) ⊂ N ( v ) . The inclusion i : G \ { u } ֒ → G respectively, the homomorphism φ : G → G \ { u } which maps u to v and fixes all the other vertices, induces the homotopy equivalence i H : Hom ( G, H ) → Hom ( G \ { u } , H ) , respectively φ H : Hom ( G \ { u } , H ) → Hom ( G, H ) . We introduce some tools from Discrete Morse Theory which have been used in this article.R. Forman in [6] introduced what has now become a standard tool in Topological Combi-natorics, Discrete Morse Theory. The principal idea of Discrete Morse Theory (simplicial)is to pair simplices in a complex in such a way that they can be cancelled by elementarycollapses. This will reduce the original complex to a homotopy equivalent complex, which isnot necessarily simplicial, but which has fewer cells. More details of discrete Morse theorycan be found in [10] and [14]. 3 efinition 3.1. A partial matching in a Poset P is a subset M of P × P such that • ( a, b ) ∈ M implies b ≻ a , i.e. a < b and c such that a < c < b . • Each element in P belongs to at most one element in M .In other words, if M is a partial matching on a Poset P then, there exists A ⊂ P andan injective map f : A → P \ A such that f ( x ) ≻ x for all x ∈ A . Definition 3.2. An acyclic matching is a partial matching M on the Poset P such thatthere does not exist a cycle f ( x ) ≻ x ≺ f ( x ) ≻ x ≺ f ( x ) ≻ x . . . f ( x t ) ≻ x t ≺ f ( x ) , t ≥ . If we have an acyclic partial matching on P , those elements of P which do not belongto the matching are said to be critical . To obtain the desired homotopy equivalence, thefollowing result is used. Theorem 3.1. (Main theorem of Discrete Morse Theory)[6]Let X be a simplicial complex and let A be an acyclic matching such that the empty setis not critical. Then, X is homotopy equivalent to a cell complex which has a d -dimensionalcell for each d -dimensional critical face of X together with an additional -cell. To prove Theorem 1.1, we first construct an acyclic matching on the face poset of N ( K K n m ).We then consider the corresponding Morse Complex to determine the connectedness ofHom( K × K n , K m ). In this article n ≥ , m − n = p ≥ k ] denotes the set { , , . . . , k } .Any vertex in the exponential graph K K n m is a set map f : [ n ] → [ m ]. Lemma 4.1.
The graph K K n m can be folded onto the graph G , where the vertices f ∈ V ( G ) have images of cardinality either 1 or n . Proof.
Consider the vertex f such that 1 < | Im f | < n . Since f is not injective there existdistinct i, j ∈ [ n ] such that f ( i ) = f ( j ) = α . Consider ˜ f ∈ V ( K K n m ) such that ˜ f ([ n ]) = α .By the definition of the exponential graph, any neighbor h of f will not have α in itsimage and therefore h will be a neighbor of ˜ f thereby showing that N ( f ) ⊆ N ( ˜ f ). K K n m can be folded to the subgraph K K n m \ { f } of K K n m . Repeating the above argument for allnoninjective, non constant maps from [ n ] to [ m ], K K n m can be folded to the graph G whosevertices are either constant maps or injective maps from [ n ] to [ m ].From Proposition 2.2, we observe that N ( K K n m ) ≃ N ( G ). Hence, it is sufficient to studythe homotopy type of N ( G ).We now fix the following notations.If f ∈ V ( G ) and f ([ n ]) = { x } , then f is denoted by < x > . In the other cases the string a a . . . a n denotes the vertex f where a i = f ( i ), 1 ≤ i ≤ n. Hence, if the notation a a . . . a n is used, it is understood that for 1 ≤ i < j ≤ n , a i = a j . Let σ = { f , . . . , f q } ⊂ V ( G ).Then X σi = { f j ( i ) | ≤ j ≤ q } , A σi = [ m ] \ S j = i X σj and X σ = n S i =1 X σi .4 .1 Construction of an acyclic matching Let ( P, ⊂ ) be the face poset of N ( G ), S = { σ ∈ P | < > / ∈ σ, σ ∪ < > ∈ P } and themap µ : S → P \ S be defined by µ ( σ ) = σ ∪ < > .Let S ′ = P \ { S , µ ( S ) } and for 2 ≤ i ≤ m, define S i = { σ ∈ S ′ i − | < i > / ∈ σ, σ ∪ < i > ∈ S ′ i − } and µ i : S i → S ′ i − \ S i by µ i ( σ ) = σ ∪ < i >, where S ′ i = S ′ i − \ { S i , µ i ( S i ) } . Since S i ∩ S j = ∅ ∀ ≤ i = j ≤ m , if σ ∈ S = m S i =1 S i , then σ belongs to exactly one S i . Define µ : S → P \ S by µ ( σ ) = µ i ( σ ) where σ ∈ S i . By the construction, µ is an injective mapand hence µ is a well defined partial matching on the poset P . Proposition 4.2. µ is an acyclic matching on P . Proof.
Assume that µ is not an acyclic matching, i.e. , there exist distinct simplices σ , . . . ,σ t in S such that µ ( σ i ) ≻ σ i +1(mod t ) , ≤ i ≤ t . Claim 1. σ i +1(mod t ) = µ ( σ i ) \ { f } , where | Im f | = 1. µ ( σ i ) ≻ σ i +1(mod t ) implies that σ i +1 = µ ( σ i ) \ { f } , for some f ∈ V ( G ). If | Im f | = n , then f / ∈ σ i +1(mod t ) ⇒ f / ∈ µ ( σ i +1(mod t ) ) ⇒ f / ∈ σ i +2(mod t ) . Hence, f / ∈ µ ( σ i ), acontradiction. So f = < j > for some j ∈ { , . . . , m } .Let k ∈ { , . . . , m } be the least integer such that { σ , . . . , σ t } ∩ S k = ∅ . Without loss ofgenerality assume that σ ∈ S k , i.e. < k > / ∈ σ and µ ( σ ) = σ ∪ < k > . Since σ = σ ,from Claim 1, σ = µ ( σ ) \ { < i > } , i = k . Therefore, < k > ∈ σ . If < k > ∈ σ i , ≤ i ≤ t, then < k > ∈ µ ( σ t ). µ ( σ ) \ { < k > } = σ = µ ( σ t ) \ { < k > } ⇒ µ ( σ ) = µ ( σ t ), acontradiction (since σ = σ t and µ is injective).Let l ∈ [ m ] be such that < k > / ∈ σ l and < k > ∈ σ j , ≤ j < l . Here, < k > ∈ µ ( σ l − )and therefore, σ l = µ ( σ l − ) \ { < k > } . Since σ l , µ ( σ l − ) / ∈ S j , µ j ( S j ) ∀ j < k , we see that σ l , µ ( σ l − ) ∈ S ′ k − , which implies that σ l ∈ S k (since µ ( σ l − ) = σ l ∪ < k > ). Here, σ l = σ l − and µ ( σ l ) = µ ( σ l − ), a contradiction. Thus our assumption that µ ( σ i ) ≻ σ i +1(mod t ) , ≤ i ≤ t is incorrect i.e. there exists no such cycle in P . Hence, µ is an acyclic matching. A subset σ of V ( G ) is a simplex in N ( G ) if and only if A σi = ∅ ∀ i ∈ [ n ]. In this section,we characterize the critical cells of N ( G ). If there exists z . . . z n ∈ N ( σ ) such that 1 / ∈{ z , . . . , z n } , then σ ∪ < > ∈ N ( G ), which implies that σ is not a critical cell. Further, if σ ∈ N ( < x > ) for some x = 1, then < > ∈ N ( σ ) and σ ∈ S . Therefore, for σ to be acritical cell, either N ( σ ) = < > or σ ⊂ N ( z . . . z n ) with 1 ∈ { z , z , . . . z n } . Case 1. σ ⊂ N ( z . . . z n ), where z k = 1 for some k ∈ { , , . . . , n } .In this case, clearly, σ = { < x >, . . . , < x l > } ∪ τ , where τ = { f , . . . , f q } with | Im f i | = n ∀ i ∈ { , . . . , q } and x , . . . x l ∈ [ m ] \ { z , . . . , z n } . Since σ ∼ z . . . z n and < z i > is not a neighbor of z . . . z n for all i ∈ [ n ] we see that < z i > / ∈ σ . If z r ∈ X σ for some r ∈ [ n ], then z r ∈ X τ , which implies that there exists f ∈ τ such that z r ∈ Im f . Since f ∼ z . . . z n , f ( i ) = z r ∀ i = r , which implies that f ( r ) = z r . Proposition 4.3.
For x ∈ [ m ] with x = z i , ∀ i ∈ [ n ] and < x > / ∈ σ , define a = min { x, z r } if x ∈ X τr \ S r = j X τj x if x ∈ X τi ∩ X τj , i = j. et η be the set { σ ∪ < a >, . . . , < a s > } \ { < b >, . . . , < b t > } , s, t ≥ , a i , b j < a, ≤ i ≤ s, ≤ j ≤ t . Then(i) η ∈ N ( G ) ⇐⇒ η ∪ < a > ∈ N ( G ) . (ii) η ∈ S ⇐⇒ η ∪ < a > ∈ S . Proof. If η ∪ < a > ∈ N ( G ) or S , then η ∈ N ( G ) or S , respectively. Case (i) a = min { x, z r } .Since a ≤ x, z r ∈ X τr \ S r = j X τj and < x >, < z r > / ∈ σ , we observe that x, z r ∈ X ηr \ S r = j X ηj .Any neighbor f of η has the property that f ( i ) = x, z r ( i.e. f ( i ) = a ) if i = r . If < y > ∼ η, i.e. η ∈ N ( G ), then y = a and < y > ∼ < a > implying that < y > ∼ η ∪ < a > and thus η ∪ < a > ∈ N ( G ).If η ∈ S and < y > ∼ η ∪ < > , then < y > ∼ < a > and < > . Therefore η ∪ < a > ∪ < > ∈ N ( G ) and η ∪ < a > ∈ S . Let x r = max { x, z r } . Clearly, a, x r ∈ X τr \ S i = r X τi . For any neighbor y = y . . . y n of η , y i = x r , a ∀ i = r and so y . . . y r − x r y r +1 . . . y n ∼ η and < a > . Therefore η ∪ < a > ∈ N ( G ) if η ∈ N ( G ).If η ∈ S and y ∼ η , then y i = 1 ∀ i ∈ [ n ]. Since x r = 1 , y . . . y r − x r y r +1 . . . y n ∼ η, < a > and < > . Thus, η ∪ < a > ∈ S . Case (ii) x = a ∈ X τi ∩ X τj , i = j .In this case, a = z t and a / ∈ A ηt ∀ t ∈ [ n ]. Since A η ∪ t = A ηt \ { a } and a / ∈ A ηt , weget A η ∪ t = A ηt . If f ∈ N ( η ), then ∀ t ∈ [ n ], f ( t ) ∈ A ηt = A η ∪ t implying that f ∈ N ( η ∪ < a > ). N ( η ∪ < a > ) ⊂ N ( η ) always, and therefore, N ( η ) = N ( η ∪ < a > ). ( i )and ( ii ) now follow. Proposition 4.4.
Let a be as defined in Proposition 4.3, σ / ∈ S t ∪ µ t ( S t ) ∀ t < a and σ ∪ < a > ∈ S i ∪ µ i ( S i ) , i < a . Then, there exists s ∈ { , . . . , m } and a cell ξ s = σ ∪ ( { < a >, < b >, . . . , < b r > } \ { < l >, . . . , < l t > } if s is even { < b >, . . . , < b r > } \ { < l >, . . . , < l t > } if s is odd , where r, t ≥ and b , . . . , b r , l , . . . , l t < a such that ξ s ∈ S ∪ µ ( S ) but ξ s \ { < a > } (if < a > ∈ ξ s ) or ξ s ∪ < a > ( if < a > / ∈ ξ s ) / ∈ S ∪ µ ( S ) . Proof.
Clearly, i > σ ∪ < a > ∈ S ∪ µ ( S ) implies that σ ∈ S ∪ µ ( S ), acontradiction. Define the cell ξ = ( σ \ { < i > } if < i > ∈ σσ ∪ < i > if < i > / ∈ σ. Since, σ ∪ < a > ∈ S i ∪ µ i ( S i ), ξ ∪ < a > is always a simplex, thereby showingthat ξ ∈ N ( G ). If ξ ∈ S i ∪ µ i ( S i ), i.e. ξ / ∈ S j ∪ µ j ( S j ) ∀ j < i , then σ willbelong to S i ∪ µ i ( S i ), which contradicts the hypothesis. Therefore, ∃ i < i such that ξ ∈ S i ∪ µ i ( S i ). By Proposition 4.3(i), ξ ∪ < a > , which is { σ ∪ < a > } \ { < i > } or σ ∪ < a > ∪ < i > , belongs to S i ∪ µ i ( S i ) . So, if i = 1 the result holds. Let i > ξ = ( { ξ ∪ < a > } \ { < i > } if < i > ∈ σξ ∪ < a > ∪ < i > if < i > / ∈ σ.
6y Proposition 4.3, ξ ∈ N ( G ). Since ξ ∪ < a > / ∈ S i ∪ µ i ( S i ), there exists i < i such that ξ ∈ S i ∪ µ i ( S i ). Here, ξ \ { < a > } which is ξ \ { < i > } or ξ ∪ < i > ∈ S i ∪ µ i ( S i ), which implies that the result holds if i = 1. Inductively, assume that thereexists l, < l < m , where i l +1 < i l < . . . < i < a and ξ t ∈ S i t +1 ∪ µ i t +1 ( S i t +1 ), 1 ≤ t ≤ l such that ξ t = { ξ t − ∪ < a > } \ { < i t > } if < i t > ∈ ξ t − , t is even ξ t − ∪ < a > ∪ < i t > if < i t > / ∈ ξ t − , t is even ξ t − \ { < a >, < i t > } if < i t > ∈ ξ t − , t is odd { ξ t − ∪ < i t > } \ { < a > } if < i t > / ∈ ξ t − , t is odd . Since ξ l \ { < a > } (or ξ l ∪ < a > ) = ξ l − \ { < i l > } or ξ l − ∪ < i l > ∈ S i l ∪ µ i l ( S i l ) and i l > i l +1 , the result holds if i l +1 = 1. If i l +1 >
1, by induction the result follows.We first get some necessary conditions for σ to be a critical cell. Lemma 4.5.
Let σ be a critical cell. Then(i) X σ = [ m ] or [ m ] \ { } . (ii) x ∈ [ m ] \ { z , . . . , z n } ⇒ < x > ∈ σ . Proof.
A critical cell σ does not belong to S i ∪ µ i ( S i ) for all i ∈ { , . . . , m } . (i) If x ∈ [ m ] \ { } such that x / ∈ X σ , then σ, < > ∼ < x > , thereby implying that σ ∈ S . Hence [ m ] \ { } ⊂ X σ .(ii) Let x ∈ [ m ] \ { z , . . . , z n } such that < x > / ∈ σ . Since z . . . z n is a neighbor of both
Let σ be a critical cell. Then(i) X τr ∩ X τs = ∅ if r = s .(ii) x ∈ X τi ⇒ x ≥ z i , i ∈ { , . . . , n } .(iii) ∃ a ∈ X τk such that a < min { z , . . . , b z k , . . . , z n } . Proof. (i) Assume that x ∈ X τr ∩ X τs . For any f ∈ τ, f ( i ) = z j ∀ i = j, x ∈ [ m ] \ { z , . . . , z n } and x = 1. Therefore, < x > ∈ σ by Lemma 4.5. Since < x > / ∈ σ \ { < x > } , Proposition4.3 holds, when σ is replaced by σ \ { < x > } in the definition of η .Since σ / ∈ S i ∪ µ i ( S i ) ∀ i , we see that σ \ { < x > } / ∈ S x ∪ µ x ( S x ). Therefore thereexists i < x such that σ \ { < x > } ∈ S i ∪ µ i ( S i ). Considering σ \ { < x > } insteadof σ ∪ < a > in Proposition 4.4 and by an argument similar to that in the proof ofProposition 4.4, we see that there exists s ∈ { , . . . , m } and a cell ξ s such that7 s = σ ∪ ( { < b >, . . . , < b r > } \ { < x >, < l >, . . . , < l t > } if s is even { < b >, . . . , < b r > } \ { < l >, . . . , < l t > } if s is odd , where r, t ≥ b , . . . , b r , l , . . . , l t < x such that, ξ s ∈ S ∪ µ ( S ) but ξ s \ { < x > } (if < x > ∈ ξ s ) or ξ s ∪ < x > (if < x > / ∈ ξ s ) / ∈ S ∪ µ ( S ). But, this is not possibleby Proposition 4.3. Hence X τr ∩ X τs = ∅ .(ii) Let x ∈ X τi such that x < z i . As in the case ( i ), x ∈ [ m ] \ { z , . . . , z n } and hence < x > ∈ σ . By exactly the same proof as the one above, we get a contradiction andthus x ≥ z i .(iii) Let z = min { z , . . . , b z k , . . . , z n } and a = min { x | x ∈ X τk } . If X σ = [ m ] , then a = 1and the result follows. Let X σ = [ m ] \ { } and a ≥ z . σ, < z > ∼ < > implies that σ ∪ < z > ∈ N ( G ) and σ is a critical cell implies that σ ∪ < z > / ∈ S . If z = 2, then σ, σ ∪ < > / ∈ S and hence σ ∈ S , which is not possible. Therefore, z >
2. Choose t , 1 < t < z . Here, t = z i ∀ i and therefore t ∈ [ m ] \ { z , . . . , z n } and t / ∈ X τi ∀ i ∈ [ n ](by ( ii )). Since t / ∈ X σ \
1, this case is not possible. Hence, m / ∈ X σ \
Any element of C is different from 1 and therefore belongs to X σ . For j ∈ { , . . . , l } ,there exists at least one element y ∈ C such that y ∈ A αi j and therefore in X αi j . If t ∈ [ n ] \ { i , . . . , i l } , then z t ∈ A αt . Hence, A αt = ∅ ∀ t ∈ [ n ] and therefore α ∈ N ( G ) . (i) Let z = 1. If t ∈ [ n ] \ { i , . . . , i l } , then z t = 1 and z t ∈ A α ∪ < >t , which implies that A α ∪ < >t = ∅ . Since T ∩ { } = ∅ , y ∈ A αi j implies that y ∈ A α ∪ < >i j ∀ j ∈ { , . . . , l } . Thus, α ∪ < > ∈ N ( G ) and therefore α ∈ S ∪ µ ( S ).(ii) Let z > . Consider the cell η = { σ ∪ { < a i >, ≤ i ≤ s }} \ { < b j >, ≤ j ≤ r } , s, r ≥ , a i , b j < z . Let D ′ = { < x > | x ∈ D } , C ′ = { < y > | y ∈ C } and γ = ( { η ∪ D ′ } \ C ′ if < z > / ∈ D ′ { η ∪ D ′ } \ { < z >, C ′ } if < z > ∈ D ′ , Proposition 4.12. η ∈ N ( G ) ⇔ γ, γ ∪ < z > ∈ N ( G ) . In particular, η ∈ S ⇔ γ, γ ∪ < z > ∈ S . Since σ is critical, by Theorem 4.8 ( iii ) y ∈ C has the property y > z ,. If j ∈ { , . . . , l } , then there exists y = 1 , z in C such that y belongs to A γi j and therefore also to A γ ∪ < >i j , A γ ∪
1. Define the cell11 = ( { ξ ∪ D ′ } \ { C ′ , < i > } if < i > ∈ σξ ∪ D ′ ∪ < i > if < i > / ∈ σ. By Proposition 4.12, ξ ∈ N ( G ). Since { ξ ∪ D ′ } \ C ′ / ∈ S i ∪ µ i ( S i ), there exists i < i such that ξ ∈ S i ∪ µ i ( S i ). Here, { ξ ∪ C ′ } \ D ′ which is ξ \ { < i > } or ξ ∪ < i > ∈ S i ∪ µ i ( S i ). If i = 1, then we get a contradiction by Proposition4.12. Inductively, assume that there exists l, < l < m , where i l +1 < i l < . . . < i < z and ξ t ∈ S i t +1 ∪ µ i t +1 ( S i t +1 ), 1 ≤ t ≤ l such that ξ t = { ξ t − ∪ D ′ } \ { C ′ , < i t > } if < i t > ∈ ξ t − , t is even { ξ t − ∪ D ′ ∪ < i t > } \ C ′ if < i t > / ∈ ξ t − , t is even { ξ t − ∪ C ′ } \ { D ′ , < i t > } if < i t > ∈ ξ t − , t is odd { ξ t − ∪ C ′ ∪ < i t > } \ D ′ if < i t > / ∈ ξ t − , t is odd . Since { ξ l ∪ C ′ } \ D ′ (or { ξ l ∪ D ′ } \ C ′ ) = ξ l − \ { < i l > } or ξ l − ∪ < i l > ∈ S i l ∪ µ i l ( S i l )and i l > i l +1 , if i l +1 = 1, using Proposition 4.12 we arrive at a contradiction.By induction, there exists 1 < s ≤ m , and a cell ξ s = σ ∪ ( { D ′ , h b i , . . . , h b r i} \ { C ′ , h l i , . . . , h l t i} if s is even { < b >, . . . , < b r > } \ {h l i , . . . , h l t i} if s is odd , where r, t ≥ b , . . . , b r , l , . . . , l t < z such that, ξ s ∈ S ∪ µ ( S ) but { ξ s ∪ C ′ } \ D ′ (if s is even) or { ξ s ∪ D ′ } \ C ′ (if s is odd) / ∈ S ∪ µ ( S ). But, this is not possible byProposition 4.12.Hence α / ∈ S i ∪ µ i ( S i ) ∀ i < z . Replacing α by α ∪ < z > (if < z > / ∈ α ) or by α \ { < z > } (if < z > ∈ α ) and applying an argument similar to the one above, wesee that α ∪ < z > (if < z > / ∈ α ) or α \ { < z > } (if < z > ∈ α ) / ∈ S i ∪ µ i ( S i ) ∀ i < z. Hence α ∈ S z ∪ µ z ( S z ). Lemma 4.13.
Let η = { < x >, . . . , < x p >, f , f } ⊂ N ( w . . . w n ) be a simplex, where w k = 1 , w i = min { w , . . . , c w k , . . . , w n } , f ( i ) = w i ∀ i = k and [ m ] \ { w , . . . , w n } = { x , . . . , x p } . If f ( k ) > w i , then η \ { f } ∈ S w i . Proof. f ( k ) > w i > f ( k ) ∈ { x , . . . , x p } and 1 / ∈ Im f . Therefore, { η ∪
No element of S ∪ µ ( S ) belongs to any alternating path between two criticalcells. Proof.
Let c = { τ, y , µ ( y ) , . . . , y t , µ ( y t ) , α } be an alternating path between the criticalcells τ and α . Let S = m S i =1 S i and µ : S → P \ S be the map such that µ | S i = µ i . Here,12 i ∈ S , µ ( y i ) ∈ µ ( S ) and S ∩ µ ( S ) = ∅ . If γ ∈ c and γ ∈ µ ( S ), then γ = µ ( y i ) for some i ∈ [ t ]. Since [ µ ( y i ) : y i ] = ± y i = µ ( y i ) \ { < > } . In the alternating path, y i +1 ≺ µ ( y i )and y i +1 = y i implies that < > ∈ µ ( y i ) \ { < > } . This implies that y i +1 ∈ µ ( S ), acontradiction. Therefore γ / ∈ c. If γ ∈ S , then µ ( γ ) ∈ c , a contradiction.We are now ready to prove the main result of this section. Theorem 4.15.
There exist two critical p -cells β and γ such that there exists exactly onealternating path from σ to each of β and γ . Further, there exists no alternating path from σ to any other critical p -cell. Proof.
Define the sets A i , B i , i ∈ { , } to be A i = T ∩ { Im f i \ { f i ( k ) }} and B i = { z , . . . , b z k , . . . , z n } \ Im f i . Since A i ⊂ T and B i ⊂ { z , . . . , z n } , A i ∩ B i = ∅ for 1 ≤ i ≤ . If z i / ∈ { f ( i ) , f ( i ) } for i = k , then z i / ∈ Im f , Im f , T and so z i / ∈ X σ . Since z i = 1 , < z i > ∼ σ, < > and thus σ ∈ S . Therefore, z i ∈ { f ( i ) , f ( i ) } ∀ i ∈ [ n ] \ { k } ,thus implying that B ∩ B = ∅ , B ⊂ Im f and B ⊂ Im f .Let x ∈ A ∩ A . For i ∈ [ n ] \ { k } , z i ∈ { f ( i ) , f ( i ) } and therefore, ∃ i = j ∈ [ n ]such that f ( i ) = f ( j ) = x . Here, X τi ∩ X τj = ∅ , a contradiction to 4.8 ( iv ) and hence A ∩ A = ∅ . Let W = { z | z = f ( i ) = f ( i ) } and T = T \ { A ∪ A } . Given a set B ⊂ B ∪ B ,define a corresponding set A ⊂ A ∪ A as follows. If z i ∈ B ∩ B , then z i = f ( i ) and hence, f ( i ) ∈ A . For z i ∈ B ∩ B , f ( i ) ∈ A . The number of elements in A and B are the same.Let the facet of σ in the alternating path c be y . We consider the following two cases. Case 1. A , A = ∅ .In this case, f ( i ) = f ( i ) = z i ∀ i = k and { f ( k ) , f ( k ) } ∩ T = ∅ . If an element x of T does not belong to X τ , then σ \ { < x > } ∼ < x > and so σ \ { < x > } ∈ S . If x ∈ X τ , then x ∈ { f ( k ) , f ( k ) } and σ \ { < x > } ⊂ N ( z . . . z k − xz k +1 . . . z n ) which implies σ \ { < x > } ∈ S and from 4.14, σ \ { < x > } / ∈ c . Therefore, y is either σ \ { f } or σ \ { f } .Each of X σ \{ f } and X σ \{ f } will be either [ m ] or [ m ] \ { } . Further, σ \ { f } and σ \ { f } satisfy the properties ( ii ) , ( iii ) and ( iv ) of Theorem 4.8. By Theorem 4.8 ( v ), at least oneof f ( k ) or f ( k ) < z i , where z i = min { z , . . . , b z k , . . . , z n } .If f ( k ) < z i , then σ \ { f } is a critical p -cell as it satisfies all the five conditions ofTheorem 4.8 and c = { σ, σ \ { f }} .Similarly, f ( k ) < z i implies σ \ { f } is critical and c = { σ, σ \ { f }} .Now, assume that f ( k ) < z i , f ( k ) > z i and y = σ \ { f } . Since f ( i ) = z i ∀ i = k and f ( k ) > z i , σ \ { f } ∈ S z i by Lemma 4.13. So µ ( y ) = { σ ∪ < z i > } \ { f } . If x ∈ [ m ] \ { z , . . . , z n , f ( k ) } , then µ ( y ) \ { < x > } ∼ < x > and hence belongs to S . µ ( y ) \ { < x > } = y if x = z i . { σ ∪ < z i > } \ { f , f } ∼ < z i > for i = i , k and so y has to be µ ( y ) \ { < x > } where x = f ( k ). By Theorem 4.8, y ⊂ N ( ˜ z . . . ˜ z n ), where˜ z k = f ( k ) , ˜ z i = 1 and ˜ z i = z i , i = i , k , is a critical cell and the alternating path in thiscase is { σ, σ \ { f } , { σ ∪ < z i > } \ { f } , { σ ∪ < z i > } \ { f , < f ( k ) > }} . If f ( k ) < z i , f ( k ) > z i and y = σ \ { f } , then the alternating path is { σ, σ \{ f } , { σ ∪ < z i > } \ { f } , { σ ∪ < z i > } \ { f , < f ( k ) > }} . Case 2.
At least one of A or A is non empty.We prove some results necessary for the construction of the alternating paths c from σ . Lemma 4.16.
Let α = { σ ∪ { < b > | b ∈ B }} \ { < a > | a ∈ A } , where B = ∅ . Then,(i) α \ { < x > } ∈ S ∀ x ∈ T . ii) If B ∩ B = ∅ , B (or B ∩ B = ∅ , B ), then α \ { f } , α \ { f } ∈ S .(iii) If B = B ∪ B and B , B = ∅ , then α \ { f } , α \ { f } ∈ S .(iv) If B = B (or B = B ), then α \ { f } ∈ S and α \ { f } / ∈ S (or α \ { f } / ∈ S and α \ { f } ∈ S .) Proof. (i) Let x ∈ T . If x / ∈ X τ , then x / ∈ X α \
Let B = B , B , η = { σ ∪ { < b i > | ≤ i ≤ q }} \ { < x i > | ≤ i ≤ q } ∈ S b and α = µ b ( η ) = η ∪ < b > ∈ µ b ( S b ) . If(i) α ∈ c and b / ∈ B , then the only facets y of α which can belong to c are of the form α \ { < x > } , where b x < b and x ∈ { A ∪ A } \ A .(ii) b = b , then no facet of α belongs to c and thus α / ∈ c . Proof.
By Lemma 4.16, ∀ x ∈ T , α \ { < x > } , α \ { f } and α \ { f } belong to S . For b ∈ B \ { b } , by Lemma 4.11, α \ { < b > } ∈ µ b ( S b ) and α \ { < b > } = η . Since α ∈ c implies that η ∈ c , the facet y = α \ { < b > } . For x ∈ A ∪ A such that b x > b , byLemma 4.11, α \ { < x > } ∈ µ b ( S b ). Thus, the only possible facets of α which can belongto c are of the type α \ { < x > } , where b x < b .If b = b , then no facet of α can belong to c . Since α ∈ µ b ( S b ), it is not a critical celland thus α / ∈ c .From the above Lemma, we conclude that no cell α = { σ ∪ { < b > | b ∈ B }} \ {
Let α = { σ ∪ { < b > | b ∈ B }} \ { < x > | x ∈ A } and b ∈ B . If α ∈ c ,then B = B . Proof.
By Lemma 4.11, α ∈ µ b ( S b ) and is thus not a critical cell. If B ( B , then byLemma 4.17 ( i ), the facet of α ∈ c is of the form α \ { < x > } , where b x < b . Since b isthe least element of B , b x ∈ B , a contradiction. Thus B = B .Hence, we can conclude that y = σ \ { < x r > } or σ \ { < x l > } . Subcase 3. y = σ \ { < x r > } . 15sing Lemmas 4.17 and 4.18, y i = { σ ∪ { < b t > | r − i + 1 < t ≤ r }} \ { < x r >, . . . ,
3, using Theorem 4.8 and Lemma 4.9, we can concludethat C i = ∅ for 0 < i ≤ p −
1. Let C p = { α , . . . , α r } and C p +1 = { τ , . . . , τ r } . Let A = [ a ij ]be a matrix of order | C p | × | C p +1 | , where a ij = 1, if there exists an alternating path from τ j to α i and 0 if no such path exists. Using Theorem 4.15, each column of A contains exactlytwo non zero elements which are 1 (except, when { < >, . . . , < m > } ∈ C p +1 and in thiscase the column of A corresponding to this cell is zero, as there is no alternating path from { < >, . . . , < m > } to any critical cell). Theorem 4.19.
Let m − n = p ≥ . Then H p ( N ( G ); Z ) = 0 . Proof. M p ∼ = Z | C p | and M p +1 ∼ = Z | C p +1 | . Since each column of A is either zero (when { < >, . . . , < m > } ∈ C p +1 ) or contains exactly two non zero elements, both being 1,the column sum is zero (mod 2). Therefore, rank( A ) < | C p | . In particular, rank of theboundary map ∂ p +1 : Z | C p +1 | → Z | C p | is strictly less than | C p | .Since Hom( K × K n , K m ) ≃ Hom( K , K K n m ) ≃ N ( K K n m ) ≃ N ( G ) and the maximumdegree of K × K n is n −
1, conn(Hom( K × K n , K m )) = conn( N ( G )) ≥ m − n −
1. Hence N ( G ) is path connected and therefore H ( N ( G ); Z ) ∼ = Z .If p = 1, then since M ∼ = Z , Ker( ∂ ) ∼ = Z | C | , where ∂ : M −→ M . Hence H ( N ( G ); Z ) = 0. If p >
1, then C p − = ∅ implies that M p − = 0. Thus, Ker( ∂ p ) ∼ = Z | C p | ,where ∂ p : M p −→ M p − is the boundary map. Since rank( ∂ p +1 ) < | C p | , we see that H p ( N ( G ); Z ) = 0.We have developed all the necessary tools to prove the main result. We recall thefollowing results to prove Theorem 1.1. Proposition 4.20. (Theorem 3A.3, [7])If C is a chain complex of free abelian groups, then there exist short exact sequences −→ H n ( C ; Z ) ⊗ Z −→ H n ( C ; Z ) −→ Tor ( H n − ( C ; Z ) , Z ) −→ for all n and these sequences split. Proposition 4.21. (The Hurewicz Theorem)If a space X is ( n − connected, n ≥ , then ˜ H i ( X ; Z ) = 0 for i < n and π n ( X ) ∼ = H n ( X ; Z ) . Proof of Theorem 1.1. If n = 2, then Hom( K × K , K m ) ≃ Hom( K ⊔ K , K m ) ≃ Hom ( K , K m ) × Hom( K , K m ) ≃ S m − × S m − . Hence conn(Hom( K × K , K m )) = conn( S m − × S m − ) = m − n ≥
3. If m < n , 4.1 shows that K K n m can be folded to the graph G ′ , where V ( G ′ ) = { < x > | x ∈ [ m ] } . Then N ( < x > ) = { < y > | y ∈ [ m ] \ { x }} , for all < x > ∈ V ( G ′ )and therefore N ( G ′ ) is homotopic to the simplicial boundary of ( m − K × K n , K m ) ≃ N ( K K n m ) ≃ N ( G ′ ) ≃ S m − . Therefore conn(Hom( K × K n , K m )) = m − m = n , then for any f ∈ V ( K K n n ) with Im f = [ n ], N ( f ) = { f } . Since n ≥ N ( K K n n )is disconnected. 17ssume m − n = p ≥ . Since conn(Hom( K × K n , K m )) ≥ m − n −
1, ˜ H i (Hom( K × K n , K m ); Z ) = 0 ∀ ≤ i ≤ p − . Since Hom( K × K n , K m ) ≃ N ( G ) , ˜ H i ( N ( G ); Z ) =0 ∀ ≤ i ≤ p − H p (Hom( K × K n , K m ); Z ) = 0, by Theorem 4.19. By Proposition4.20, H p ( N ( G ); Z ) ∼ = H p ( N ( G ); Z ) ⊗ Z ⊕ Tor( H p − ( N ( G ); Z ) , Z ). Since p ≥ , N ( G )is path connected and hence Tor( H ( N ( G ); Z ) , Z ) ∼ = Tor( Z , Z ) = 0. Further, since forany 0 < q < p , ˜ H i ( N ( G ); Z ) = 0, H p ( N ( G ); Z ) = 0 implies H p ( N ( G ); Z ) = 0, which isa contradiction. Hence H p (Hom( K × K n , K m ); Z ) ∼ = H p ( N ( G ); Z ) = 0. If p = 1, thensince the abelinazitation of π (Hom( K × K n , K m )) is H (Hom ( K × K n , K m ); Z ) = 0, π (Hom( K × K n , K m )) = 0. For p >
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