SSECTION PROBLEMS FOR GRAPH CONFIGURATION SPACES
ALEXANDER BAUMAN
Abstract.
We consider, for a finite graph G , when the surjective map Conf n +1 ( G ) → Conf n ( G ) of configuration spaces admits a section. We study when the answer depends onlyon the homotopy type of G , and give a complete answer. We also provide basic techniquesfor construction of sections and show that the induced map on fundamental groups is alwayssurjective. Introduction
By a graph , we mean a finite connected 1-dimensional cell complex. If n is a naturalnumber, then define the n -th ordered configuration space of G to be the set Conf n ( G ) = { ( x , . . . , x n ) : x i (cid:54) = x j for i (cid:54) = j } equipped with the subspace topology in the product space G n . Throughout, fix a graph G and a natural number n .Consider the “forgetting map” φ n +1 : Conf n +1 ( G ) → Conf n ( G ) defined by: ( x , . . . , x n +1 ) (cid:55)→ ( x , . . . , x n ) . Our main question is whether or not this surjection is split in the category of topologicalspaces, that is, whether or not there exists some function s n : Conf n ( G ) → Conf n +1 ( G ) such that φ n +1 ◦ s n = id Conf n ( G ) . The map s n is called a section . We will note that the data of a section is precisely the dataof a map that continuously adds a point to each configuration of Conf n ( G ) .We now state our main results, in terms of the Euler characteristic χ of graphs: Theorem 1.1.
When χ ( G ) < and n ≥ − χ ( G ) , then no section exists. Theorem 1.2.
When χ ( G ) = 0 , then a section exists for all n . Theorem 1.3.
When χ ( G ) = 1 , then a section exists if and only if n ≥ . We also show that when χ ( G ) < and n < − χ ( G ) , then a section may or may notexist by providing examples of both cases, which shows that it isn’t possible to improve ourresults without appealing to more than the value of χ ( G ) .The related problem of finding a section to the forgetting map Conf n +1 ( M ) → Conf n ( M ) where M is a manifold with dimension greater than 1 has been studied, for instance in[CGL20] and [Che19]. This problem is somewhat more amenable to sophisticated argu-ments from algebraic topology since the forgetting map is a fibration, and the homology of Conf n ( M ) is significantly better understood than the homology of Conf n ( G ) . The forgettingmap φ n +1 of graph configuration spaces is not a fibration, as the topology of G \ {∗} depends a r X i v : . [ m a t h . G T ] N ov ALEXANDER BAUMAN on our choice of point ∗ . As a result, our methods are primarily geometric, and make littleuse of algebraic topology.The paper proceeds as follows: In Section 2, we introduce notation and terminology. InSection 3, we study the map on fundamental groups induced by φ n +1 , and show that it isalways surjective and often split. In Section 4, we present elementary examples of sections.In Section 5, we introduce and develop the idea of connectable components , a technical tooluseful for the proof of Theorem 1.1. Section 6 is devoted to the proof of Theorem 1.1. InSection 7, we introduce a combinatorial model for Conf n ( G ) which is useful for proving moreadvanced existence results. In Section 8, we use this combinatorial model to demonstratemore sophisticated examples of sections. Acknowledgement.
The author would like to thank Bena Tshishiku for introducing him tograph configuration spaces, and for meeting with him regularly to discuss ideas and reviewdrafts of this paper. 2.
Notation
We assume that each graph is equipped with its natural cellular structure such that when G (cid:54)∼ = S , every vertex has degree (cid:54) = 2 , and when G ∼ = S , then G only has one vertex. A branched vertex is a vertex with degree ≥ . An edge e is a loop if it contains a single vertex.If we fix a homeomorphism between each edge and I = [0 , (or S in the case that the edgeis a loop) we obtain a metric d on G by taking these homeomorphisms to be local isometriesand using the shortest path metric.A subset of a graph is called a subgraph if it is closed and connected (this is not thestandard definition, but it will prove more useful and flexible for us). A graph is a tree if itis simply connected. A maximal subtree of a graph is a subgraph which is a tree, and whichcontains every vertex. It is a basic fact of graph theory that every graph contains a maximalsubtree.We call the elements of Conf n ( G ) configurations , and the entries in a configuration tokens .The integers , . . . , n are called indices . If p, q are distinct points in G which lie on a commonedge, then ( p, q ) (resp. [ p, q ] ) is the open (resp. closed) interval they bound. In the case, forexample, where G ∼ = S or p, q are vertices which are endpoints of multiple edges, there isambiguity in this definition, which we ignore, but the particular choice of edge will be clearfrom context in all cases we consider. If x = ( x , . . . , x n ) is a configuration, then we abusenotation, and denote by G \ x the complement G \ { x , . . . , x n } .3. The Group Homomorphism φ n +1 , ∗ In this section, we discuss the group-theoretic analogue of the main question. As we willsee it is seldom useful for showing the nonexistence of sections.Define the n -th pure braid group of G to be the fundamental group P B n ( G ) = π (Conf n ( G )) . We denote by φ n +1 , ∗ the induced map of φ n +1 on fundamental groups. If there exists asection s n , then we can deduce that the sequence:(1) → ker( φ n +1 , ∗ ) → P B n +1 ( G ) φ n +1 , ∗ −−−−→ P B n ( G ) → ECTION PROBLEMS FOR GRAPH CONFIGURATION SPACES 3 is a split exact sequence of groups. The converse, however, is not necessarily true, that is,the sequence (1) might be split exact even when s n does not exist. In this section we showthat φ n +1 , ∗ is always surjective and often split. Theorem 3.1. φ n +1 , ∗ is surjective.Proof. Let x = ( x , . . . , x n ) be a base configuration for Conf n ( G ) such that x lies in theinterior of an edge e which contains a vertex v , such that none of x , . . . , x n lie in between x and v . Suppose α = ( α , . . . , α n ) is a based loop in Conf n ( G ) . Up to homotopy, we canassume that the speed of α is piecewise constant, and that α never changes speed at abranched vertex or turns around on a dime at a branched vertex.Select ε > small enough that d ( α ( t ) , α i ( t )) ≥ ε for all i = 2 , . . . , n and all t ∈ I , andsuch that α always changes speed at least ε away from a branched vertex. We will definea function f : I → G such that d ( f ( t ) , α ( t )) = ε for all t . Choose f (0) arbitrarily out ofthe two possible options. Suppose α has constant speed on the interval [ t , t ] and supposethat f ( t ) is already defined. We simply define f to have the same speed and directionas α on this interval, so that f is either “leading” or “following” α . This gives a path ˜ α = ( α , . . . , α n , f ) in Conf n +1 ( G ) satisfying φ n +1 ◦ ˜ α = α . If ˜ α is a loop, then we are done.If not, then f (0) , f (1) are the two points with distance ε from f . We define a path β in Conf n +1 ( G ) such that ˜ α ∗ β exists and is a loop, and such that φ n +1 ◦ β is a trivial loop in Conf n ( G ) . Then, we will have φ n +1 ◦ ( ˜ α ∗ β ) = ( φ n +1 ◦ ˜ α ) ∗ ( φ n +1 ◦ β ) (cid:39) α , so we will bedone. We construct β by moving the first and ( n + 1) ’st tokens through a branched vertex,switching their positions as shown in Figure 1. β n +1 (0) β (0) β n +1 (1 / β (1 / β (1) β n +1 (1) Figure 1.
The path β (cid:3) The following proposition, also proven in [CD14] gives a sufficient condition for φ n +1 , ∗ tosplit. Recall that a vertex v of G is free if it has degree 1. Proposition 3.2.
Suppose G has at least one free vertex. Then φ n +1 , ∗ is a split surjectionof groups. ALEXANDER BAUMAN
Proof.
Let p be a free vertex. There exists an isotopy h t : G → G which fixes thecomplement of a neighborhood of p and such that p / ∈ h ( G ) (imagine “pushing in” theedge containing p by some small amount). Then, define g : Conf n ( G ) → Conf n +1 ( G ) by g ( x ) = ( h ( x ) , . . . , h ( x n ) , p ) . We see that φ n +1 ◦ g (cid:39) id Conf n ( G ) by applying h t to each com-ponent, and passing to the induced maps on fundamental groups gives the desired result. (cid:3) The spaces
Conf n ( G ) are aspherical, as proven in Theorem 3.1 of [Ghr01] and since thereis a natural bijection between homotopy classes of maps between aspherical spaces andhomomorphisms of their fundamental groups by Proposition 1B.9 of [Hat02], φ n +1 , ∗ is a splitsurjection of groups if and only if φ n +1 is split up to homotopy.We could not show in any examples that φ n +1 , ∗ does not split. This may be an interestingquestion for further study. 4. Basic Examples of Sections
Here we present three elementary cases where sections always exist. First we discussidentifying functions:Note that the data of a section is exactly the data of a function that continuously addsa token to each configuration of
Conf n ( G ) . In other words, this is equivalent to asking ifthere exists a map f : Conf n ( G ) → G such that for any x = ( x , . . . , x n ) ∈ Conf n ( G ) andany index j , we have that f ( x ) (cid:54) = x j . We call any function of this form an identifyingfunction . Note that identifying functions are in one-to-one correspondence with sections.For notational simplicity we will prefer to work with them henceforth. Proposition 4.1.
Suppose n = 1 . If χ ( G ) ≤ then G admits an identifying function.Proof. If χ ( G ) ≤ , then there exists an embedded circle S ⊂ G and a retract r : G → S .Define f : G → G by f ( x ) = − r ( x ) , the point on S antipodal to r ( x ) . Then f is anidentifying function. (cid:3) Proposition 4.2. If n ≥ and χ ( G ) = 1 , then G admits an identifying function.Proof. Note that χ ( G ) = 1 implies that G is a tree. Therefore there exists a unique embeddedline segment between x , x for each configuration x = ( x , . . . , x n ) . Let δ ( x ) = min( { d ( x , x k ) : k (cid:54) = 1 } ∪ { } ) / . Then let f ( x ) be the point on this embedded line segment with distance δ ( x ) away from x .Then f is an identifying function. (cid:3) Proposition 4.2 is one half of Theorem 1.3. We will prove the other direction in Proposition5.12. We conclude this section by proving Theorem 1.2.
Proof of Theorem 1.2.
The case n = 1 follows from Proposition 4.1, so assume n ≥ . Ourapproach will be to orient each edge, and then add a point near x with respect to thisorientation. There is a unique embedded circle S ⊂ G . Orient this circle, and orient everyother edge so that it’s “pointing towards” S (Note the condition χ ( G ) = 0 implies that G ishomeomorphic to a circle with trees attached, where each tree is attached at a single point).This gives us a unique “forwards direction” at each point on G . Let δ ( x ) = min { d ( x , x k ) : k (cid:54) = 1 } / , and let f ( x ) be the point a distance δ ( x ) away from x in the “forward direction.” Then f is an identifying function. (cid:3) ECTION PROBLEMS FOR GRAPH CONFIGURATION SPACES 5 Connectable Components and Chasing
In this section we develop tools which can be used to prove the nonexistence of sections.Consider the configurations x, x (cid:48) on the graph G , as depicted in Figure 2. x f ( x ) x x x (cid:48) x (cid:48) x (cid:48) Figure 2.
The configurations x, x (cid:48)
As we can see, G = Θ , the graph homeomorphic to a theta, n = 3 , x and x (cid:48) are thepictured configurations, and Y , Y , Y are the connected components of G \ x (cid:48) . Suppose that f is an identifying function on G , with f ( x ) as shown on the left. We are interested inknowing which of Y , Y , Y contains f ( x (cid:48) ) . Intuitively, we could imagine that if we move x up along the central edge then we “force” f ( x ) to choose between either Y or Y , and thatit is impossible for f ( x (cid:48) ) to lie in Y .Now consider the configuration x (cid:48)(cid:48) from Figure 3 x f ( x ) x x x (cid:48)(cid:48) x (cid:48)(cid:48) x (cid:48)(cid:48) Figure 3.
The configuration x (cid:48)(cid:48) Note that x (cid:48)(cid:48) differs from x only by moving tokens within edges, without ever enteringvertices. Therefore, we would intuitively expect that f ( x (cid:48)(cid:48) ) ∈ Z . These facts are notdifficult to prove individually using basic concepts from point-set topology, but we wouldlike generalizations of them that hold for every graph and any n . The main idea is thattokens entering branched vertices “slice up” the complement into more components, andalong paths where no token enters a branched vertex, the components of the complement donot increase in number.5.1. Connectable Components.
Suppose α = ( α , . . . , α n ) is a path in Conf n ( G ) . Wewill define a relation between the components of G \ α (0) and the components of G \ α (1) .First, we must define locally consistent systems of components along α . Definition 5.1.
Suppose α = ( α , . . . , α n ) is a path in Conf n ( G ) , and suppose that for each t ∈ I we have a connected component X t of G \ α ( t ) . The assignment t (cid:55)→ X t is called a system of components along α . Such a system is locally consistent if for each t ∈ I and eachpoint p ∈ X t , there exists an ε > such that p ∈ X s for all s ∈ ( t − ε, t + ε ) .We now define the relation on the components of G \ α (0) and the components of G \ α (1) where α is a path in Conf n ( G ) . ALEXANDER BAUMAN
Definition 5.2.
Let α be a path in Conf n ( G ) . If X is a component of G \ α (0) and Y is acomponent of G \ α (1) , then we say that X and Y are connectable by α , or write X (cid:32) α Y if there exists a locally consistent system X t of components along α such that X = X and X = Y . We say that the system X t connects X to Y (along α ).We also must consider globally consistent systems of components defined on subsets of Conf n ( G ) . Definition 5.3.
Suppose A ⊂ Conf n ( G ) . A (globally) consistent system of components on A consists of a connected component X x of G \ x , for each x ∈ A , such that for any path α : I → A , the assignment t (cid:55)→ X f ( α ( t )) is a locally consistent system of components along α . The next proposition assures us of the essential fact that identifying functions give rise toconsistent systems of components. Proposition 5.4.
Suppose f is an identifying function. Let X x be the component of G \ x containing f ( x ) . Then x (cid:55)→ X x is a consistent system of components on Conf n ( G ) .Proof. Let α = ( α , . . . , α n ) be path in Conf n ( G ) , and let X t = X α ( t ) . It suffices to show thatthe assignment t (cid:55)→ X t is a locally consistent choice of components along α . Select t ∈ I and p ∈ X t . Let K be a line segment in X t connecting p to f ( α ( t )) , and let d ( α ( t ) , K ) = δ > .Let φ : I → R be the continuous function defined by φ ( s ) = d ( α ( s ) , K ∪ f ( α ([ s, t ]))) (where we instead write [ t, s ] when t < s ). Since φ ( t ) = δ > , φ is positive in someneighborhood ( t − ε, t + ε ) of t . For any s ∈ ( t − ε, t + ε ) , we can see that the connectedset K ∪ f ( α ([ s, t ])) contains p and f ( α ( s )) , and is disjoint from α ( s ) . Therefore, it lies in asingle component of G \ α ( s ) , which is X s , as desired. (cid:3) Remark.
Our main method of disproving the existence of sections is to show that consistentsystems of components do not exist on
Conf n ( G ) .In this section, we will prove two important facts that will allow us to compute therelation (cid:32) α for two classes of paths. These facts formalize the intuitive ideas discussed inthe opening portion of this section. Given an identifying function f , these facts will allowus to determine by Proposition 5.4 the component of G \ x containing f ( x ) by moving alongpaths in Conf n ( G ) . Definition 5.5.
We say that a path α : I → Conf n ( G ) is a Type I path if no token ever enters a vertex. That is, if α i ( t ) is not a vertex, then for all s > t , α i ( s ) is also not a vertex.We say that a path is a Type II path if each coordinate of α except one is constant, and thenonconstant coordinate moves directly at constant speed from the interior of an edge to avertex. A path that is either Type I or Type II is called basic . Remark. If α is any path in Conf n ( G ) such that at least two tokens never enter a branchedvertex simultaneously, then α is homotopic to a concatenation of basic paths via a homotopy h t such that the topology of G \ h t ( s ) depends only on s , up to reparameterization. If twotokens simultaneously enter branched vertices, then we can slightly perturb α such that thetokens enter the branched vertices at different times, which does not change the relation (cid:32) α . If α, β are paths such that the concatenation α ∗ β is defined, and X is a component of G \ α (0) and Z is a component of β (1) , then it is clear that X (cid:32) α ∗ β Z if and only if there is ECTION PROBLEMS FOR GRAPH CONFIGURATION SPACES 7 some component Y of G \ α (1) such that X (cid:32) α Y and Y (cid:32) β Z . Therefore, if we computethe relation (cid:32) α for basic paths α , we can determine it for all paths.Our current goal is to determine exactly which components are connectable by basic paths.5.2. Type I Paths.Proposition 5.6. If α is a Type I path, and X is a component of G \ α (0) , then there is atmost one component Y of G \ α (1) such that X (cid:32) α Y .Proof. Suppose
Y, Z are distinct components of G \ α (1) , and that X (cid:32) α Y and X (cid:32) α Z .Suppose that Y t , Z t are locally consistent systems of components along α connecting X to Y, Z respectively. Let s = inf { t ∈ I : Y t (cid:54) = Z t } . If Y s = Z s , then for any p ∈ Y t , we can take arbitrarily small ε > such that Y s + ε (cid:54) = Z s + ε and p ∈ Y s + ε . However, this implies that p / ∈ Z s + ε for arbitrarily small ε , which contradictsthe fact that Z t is a locally consistent system of components. Therefore, Y s (cid:54) = Z s .Now, select p ∈ Y s , q ∈ Z s . There exists an ε > such that for all r ∈ [ s − ε, s ) , we have p, q ∈ Y r . Let K be an embedded line segment in Y s − ε connecting p to q . Since Y s (cid:54) = Z s ,there is some r ∈ ( s − ε, s ] such that for some i = 1 , . . . , n we have α i ( r ) ∈ K . We can in factassume that α i ( r ) ∈ ∂K (where ∂ refers to the topological boundary as a subspace of G ) bypossibly taking a smaller value of r . However, by our choice of ε , α i ( r ) (cid:54) = p, q , so α i ( r ) = v for some branched vertex contained in K , since these are the only other boundary points of K . This contradicts the fact that α is a Type I path, and we are done. (cid:3) We now describe explicitly, for any type I path α and any component X of G \ α (0) , theunique component Y of G \ α (1) such that X (cid:32) α Y . Proposition 5.7.
Suppose α = ( α , . . . , α n ) is a Type I path, and that X is a component of G \ α (0) . Define a component Y of G \ α (1) as follows:(i) If X contains a vertex v , then Y is the component containing v . X (ii) If X is an open interval ( α i (0) , α j (0)) in an edge e and α i ( t ) , α j ( t ) ∈ e for all t , then Y = ( α i (1) , α j (1)) ⊂ e . x i X x j (iii) If X is an open interval ( α i (0) , α j (0)) in an edge e and α i (0) = v is an endpoint of e , and α i ( t ) / ∈ e for some t , then Y is the component containing v . ALEXANDER BAUMAN x j X x i x i X (iv) If X is a loop e without its branched vertex v = α i (0) , then if α i is constant, Y = X ,and otherwise, Y is the component containing v .Then, X (cid:32) α Y .Proof. For any t ∈ I , consider the restricted path α | [0 ,t ] . Since any restriction of a Type Ipath remains Type I, the statement of the proposition gives us a component Y t of G \ α ( t ) .We claim this is a locally consistent system of components along α . We proceed by provingeach of the cases separately.(i) Select t ∈ I and p ∈ Y t . Since α is Type I, v ∈ Y t , so there exists a compactconnected K ⊂ Y t containing p and v . Let ζ = d ( K, α ( t )) , and select ε such that forall s ∈ ( t − ε, t + ε ) and all i = 1 , , . . . , n , we have d ( α i ( t ) , α i ( s )) < ζ . Since v ∈ K ,we have K ⊂ Y s for all s ∈ ( t − ε, t + ε ) , so p ∈ Y s , as desired.(ii) Select t ∈ I and p ∈ Y t . Let ζ = min { d ( α i ( t ) , p ) , d ( α j ( t ) , p ) } , and select ε such that d ( α η ( t ) , α η ( s )) < ζ for η = i, j and s ∈ ( t − ε, t + ε ) . Then p ∈ ( α i ( s ) , α j ( s )) = Y s for all s ∈ ( t − ε, t + ε ) , as desired.(iii) Select t ∈ I . If α i ( t + ε ) = v for some ε > , then apply (ii) to the restricted path α | [0 ,t + ε ] . If α i ( t − ε ) / ∈ e for some ε > , then apply (i) to the restricted path α | [ t − ε, .It suffices to consider the remaining case, where α i ( t ) = v but α i ( t + ε ) / ∈ e for all ε > . Select p ∈ Y t . As before, we can set ζ = min { d ( α i ( t ) , p ) , d ( α j ( t ) , p ) } , and select ε such that for all s ∈ ( t − ε, t + ε ) we have d ( α η ( s ) , α η ( t )) < ζ for η = i, j .If s ≤ t , then Y s = ( α i ( s ) , α j ( s )) , which contains p . If s > t , then Y s contains v .Since α i ( s ) / ∈ e , [ v, p ] is disjoint from α ( s ) , so p ∈ Y s .(iv) We can modify the cellular structure of G such that G has no loops, perhaps byadding a vertex to each loop. Since none of our techniques depend on the vertices of G being branched, we can reduce to part (i). (cid:3) ECTION PROBLEMS FOR GRAPH CONFIGURATION SPACES 9
Type II Paths.
We now turn to Type II paths. We begin with a lemma.
Lemma 5.8.
Suppose α is a path in Conf n ( G ) , and X t is a system of components along α . If X t ⊂ X s whenever t < s , then X t is the only locally consistent system of componentsalong α starting at X .Proof. We can easily observe that X t is locally consistent, so it suffices to show that this isthe unique locally consistent system of components along α starting at X . Suppose Y t isanother such system, and let s = inf { t ∈ I : X t (cid:54) = Y t } . By the argument from the proof of Proposition 5.6, we can see that in fact X s (cid:54) = Y s . Select p ∈ Y s , and ε such that p ∈ Y r for all r ∈ [ s − ε, s ] . Since Y r = X r by assumption, and X r ⊂ X s , we have p ∈ X s , a contradiction. (cid:3) We can now compute the relation (cid:32) α when α is a Type II path. Proposition 5.9.
Let α = ( α , . . . , α n ) be a Type II path where α i is the unique nonconstantcoordinate, and therefore has constant speed, and α i (1) = v for some vertex v . Suppose thatthe components of G \ α (0) , G \ α (1) near v are labelled as in the picture below (note that X , X and Y , . . . , Y k are not necessarily distinct). If X = X , and X , X are distinct, then Y = Y , and if X = X then Y = Y j for some j ≥ . If X = Z for some other component Z , then Y = Z . X X Y Y Y Y Y Proof.
First, suppose that X = X and that X , X are distinct. Then, we can define asystem of components Z t , such that Z t is the component of G \ α ( t ) containing X . ByLemma 5.8, this is the unique locally consistent system of components starting at X , so Y = Y .Now suppose that X = X , and that Z t is a locally consistent system of componentsstarting at X . It suffices to show that Z ⊂ Y ∪ · · · ∪ Y k . Select p ∈ Z , and ε such that p ∈ Z s whenever s ∈ [1 − ε, . For any s < , we note thatthe restriction α | [0 ,s ] is a Type I path, which implies that Z s is the component of G \ α ( s ) containing v by Proposition 5.7(i). When Y = Y j for some j ≥ , this implies that Z s = ( Y ∪ · · · ∪ Y k ) \ { α i ( s ) } , and since p ∈ Z − ε , this implies that p is in Y j for some j ≥ , as desired. When Y (cid:54) = Y j for all j ≥ , then Z s = Y ∪ · · · ∪ Y k ∪ [ α i ( s ) , v ] . If we can show p / ∈ [ α i (1 − ε ) , v ] , then we are done. However, this is impossible, sinceotherwise there would be some s ∈ (1 − ε, such that α i ( s ) = p , which contradicts ourchoice of ε . This completes the proof. (cid:3) Propositions 5.7 and 5.9 are generally used in conjunction with Proposition 5.4. If wehave a basic path α , and an identifying function f with f ( α (0)) ∈ X for some component X of G \ α (0) , then we obtain by Proposition 5.4 a locally consistent choice of componentsalong α . We can then use either Proposition 5.7 or Proposition 5.9 to determine the possiblecomponents of G \ α (1) which can contain f ( α (1)) .5.4. Applications.
Our first important application of Propositions 5.7 and 5.9 is the “chas-ing” lemma:
Lemma 5.10 (Chasing) . Suppose f is an identifying function, and x = ( x , . . . , x n ) is aconfiguration. Let X denote the connected component of G \ x containing f ( x ) , and supposethat X ∪ { x } is simply connected. Then there exists some point p ∈ X and an integer ≤ j ≤ n such that if x (cid:48) = ( p, x , . . . , x n ) , then f ( x (cid:48) ) ∈ ( x j , p ) .Proof. We induct on the number of vertices k in X . If k = 0 , then X is an interval, so wecan merely take p = x .Suppose we know the result for all (cid:96) < k . Since X ∪ { x } remains connected when x is removed, x must have degree 1 in the closure of X , since any point which is not a freevertex disconnects a simply connected 1-complex when removed. Consider the Type II path α in Conf n ( G ) such that α (0) = x and α moves along the unique edge in the closure of X which is connected to x . Let v = α (1) , and let X (cid:48) be the connected component of G \ α (1) containing f ( α (1)) . By Proposition 5.9, X (cid:48) borders v and x / ∈ X (cid:48) . Therefore, X (cid:48) ∪ { v } issimply connected, since it is a connected subset of a tree. Since v / ∈ X (cid:48) and X (cid:48) ⊂ X , X (cid:48) contains fewer than k vertices, so we are done by the inductive hypothesis. (cid:3) Of course, the chasing lemma applies equally well when x is replaced by any other token.The chasing lemma, while simple, is an important aspect of the proof of the main theorem.When we apply it (with x i taking the role of x ) we say that we are “chasing with x i .”The following corollary is worth noting separately: Corollary 5.11. If f is an identifying function on G , and T ⊂ G is a subtree of G whichis connected to the rest of G by a single point p , then for any configuration of the form x = ( p, x , . . . , x n ) with x , . . . , x n / ∈ T , we have f ( x ) / ∈ T .Proof. Suppose that x = ( p, x , . . . , x n ) is such that x , . . . , x n / ∈ T , and f ( x ) ∈ T . Then,applying Lemma 5.10, there is some i = 2 , . . . , n and p (cid:48) ∈ T such that if x (cid:48) = ( p (cid:48) , x , . . . , x n ) ,then f ( x (cid:48) ) ∈ ( p (cid:48) , x i ) . If p (cid:48) (cid:54) = p , then this is a contradiction since x i / ∈ T , and if there isany interval where exactly one endpoint is in T , then that endpoint must be in ∂T , andtherefore be equal to p . However, if p (cid:48) = p , then x (cid:48) = x , and we have f ( x ) / ∈ T , which is acontradiction, so we are done. (cid:3) We can now prove the second half of Theorem 1.3.
Proposition 5.12.
Suppose χ ( G ) = 1 and n = 1 . Then G does not admit a section. ECTION PROBLEMS FOR GRAPH CONFIGURATION SPACES 11
Proof.
Suppose f : G → G is an identifying function. Given any point p ∈ G , p separates G into subtrees T , . . . , T k . However, by Corollary 5.11, f ( p ) can be contained in none of them,which is a contradiction. (cid:3) Note that Proposition 5.12 is a generalization of the 1-dimensional Brouwer fixed-pointtheorem, as it says that for any tree T (not just the interval I ), any continuous map T → T has a fixed point. Remark.
We briefly return to our analysis of the map φ n +1 , ∗ on fundamental groups. Sup-pose that X x is a consistent system of components on Conf n ( G ) . We can associate to thisa subgroup H < P B n +1 ( G ) . An element [ α ] ∈ P B n +1 ( G ) lies in H if and only if it can berepresented by some path α = ( α , . . . , α n +1 ) such that α n +1 ( t ) ∈ X ( α ( t ) ,...,α n ( t )) for all t . Onecan verify that this is a subgroup. Suppose that s : Conf n ( G ) → Conf n +1 ( G ) is a section,and let s ∗ be the induced map on fundamental groups. Then, if the consistent system ofcomponents on Conf n ( G ) associated to s is X x , then the image of s ∗ must be contained in H as constructed above. This is because for any based loop β in Conf n ( G ) , the composedmap s ◦ β represents an element of H . This allows us to produce another algebraic necessityfor the existence of a section: If a section exists then there exists some consistent choice ofcomponents X x such that the restricted map: H (cid:44) → P B n +1 ( G ) → P B n ( G ) admits a section. 6. Proof of Theorem 1.1
In this section, we prove Theorem 1.1 and some necessary lemmas.6.1.
Distinguished Pairs.
First we define and discuss the basic properties of distinguishedpairs, which occur when, for a configuration x and identifying function f , f ( x ) lies betweentwo tokens of x . We will show that the existence of such pairs allow us to draw broadconclusions about the existence of sections, culminating in the proof of Theorem 1.1. Recallthe open interval ( p, q ) notation introduced at the start of Section 4. Definition 6.1.
Suppose G (cid:54)∼ = S , and e is an oriented edge such that G remains connectedwhen a single interior point of e is removed, and suppose that f is an identifying functionon G . If i, j are a pair of distinct indices, we say that ( i, j ) forms a distinguished pair (of f )on e if there exists a configuration x = ( x , . . . , x n ) such that:(a) x i , x j ∈ e ,(b) x i lies before x j on e with respect to the orientation on e ,(c) x k / ∈ ( x i , x j ) for all k ,(d) f ( x ) ∈ ( x i , x j ) .We then say that the configuration x is a witness to the distinguished pair ( i, j ) . x i f ( x ) x j Figure 4.
A witness to a distinguished pair ( i, j ) Note that if f is an identifying function and x = ( x , . . . , x n ) is a configuration whichsatisfies the hypotheses of Lemma 5.10 with respect to f , and additionally G \ { x (cid:96) } isconnected for each (cid:96) ≥ , then the conclusion of Lemma 5.10 in fact produces a distinguishedpair containing .Next we state and prove the most important fact about distinguished pairs. Proposition 6.2. If ( i, j ) is a distinguished pair on e , and y = ( y , . . . , y n ) is a configurationwhich satisfies the properties (a)-(c) in Definition 6.1 with respect to ( i, j ) for f , then y alsosatisfies property (d), that is, it is a witness to ( i, j ) .Proof. Let x = ( x , . . . , x n ) to ( i, j ) be a witness to ( i, j ) , and let y = ( y , . . . , y ) be as inthe proposition statement. Our goal is to construct a path α = ( α , . . . , α n ) in Conf n ( G ) connecting x, y such that α i , α j never leave e , which is a concatenation of basic paths. Thiswill imply by Propositions 5.7 and 5.9 that f ( y ) ∈ ( y i , y j ) .First consider the case where none of the x (cid:96) with (cid:96) (cid:54) = i, j lie in e , and y i , y j lie in theinterior of e . Then, we can take α to first move the i ’th and j ’th tokens to y i , y j respectively,and then move each of the remaining n − tokens of x to the corresponding tokens of y .Since Conf n − ( G \ [ x i , x j ]) is connected by Theorem 2.7 of [Abr00], this is possible.If (say) y i is at an endpoint of e , then let y (cid:48) i be a point in e which is very close to y i . Wecan modify the path α above by initially moving x i to y (cid:48) i , and then moving y (cid:48) i to y i once theremaining n − tokens agree with y .If some of the x (cid:96) lie in e for (cid:96) (cid:54) = i, j , then we can first apply a path which moves each ofthese x (cid:96) outside of e , and then reduce to the previous case.All these paths can be taken to be concatenations of basic paths, so we are done. (cid:3) Proposition 6.2 gives rise to the very useful observation:
Corollary 6.3. If ( i, j ) is a distinguished pair on e and ( k, (cid:96) ) is a distinguished pair on e (cid:48) ,then { i, j } ∩ { k, (cid:96) } (cid:54) = ∅ Proof.
Suppose ( i, j ) , ( k, (cid:96) ) , are distinguished pairs such that { i, j } ∩ { k, (cid:96) } = ∅ . Then, wecan find a configuration x = ( x , . . . , x n ) satisfying the hypotheses for Proposition (6.2) withrespect to both distinguished pairs. This necessitates f ( x ) ∈ ( x i , x j ) and f ( x ) ∈ ( x k , x (cid:96) ) ,which is a contradiction. (cid:3) The Case n > . We now present the proof of the main theorem in the case n > :Suppose f is an identifying function, and consider the statements below:(A) Every index belongs to some distinguished pair of f .(B) No index belongs to every distinguished pair of f . Lemma 6.4.
When n > , the statements (A), (B) contradict.Proof. Suppose that statements (A), (B) are both true, and suppose without loss of gen-erality that (1 , is a distinguished pair. Then is contained in some distinguished pair, ( i, . If i (cid:54) = 1 , , then this contradicts Corollary 6.3, so suppose i = 1 . Then, (1 , mustbe a distinguished pair, since if is in any other distinguished pair, we can produce a coun-terexample to Corollary 6.3. By (B), there is some distinguished pair ( i, j ) with i, j (cid:54) = 1 .This contradicts Corollary 6.3 applied to either the pairs ( i, j ) and (1 , , the pairs ( i, j ) and (1 , , or the pairs ( i, j ) and (1 , , depending on i, j . (cid:3) It therefore suffices to establish that (A) and (B) hold when n ≥ − χ ( G ) and χ ( G ) < . ECTION PROBLEMS FOR GRAPH CONFIGURATION SPACES 13
Proof of (A) when n ≥ − χ ( G ) . Let T be a maximal subtree of G , let k = 1 − χ ( G ) , and let e , e , . . . , e k be the edges in the complement of T . Let A , A be closed intervals containedin the interiors of e , e respectively, and let γ : I → T ∪ e ∪ e be an embedded path connecting an endpoint of A to an endpoint of A , such that theintersection of the image of γ with A ∪ A is exactly these two endpoints. We can assume that T is not contained in the image of γ (if not, enlarge T slightly so that it is not homeomorphicto an interval). The situation is summarized in Figure 5. e e e e e e γA A T Figure 5.
A graph G with labelled edges, with labelled A , A and γ .We will prove that is contained in some distinguished pair. Let x ∈ G be a point in T not contained in the image of γ . Let x , . . . , x k − be points in the interiors of e , . . . , e k respectively. Let x k , . . . , x n − be points spaced out equally in A in order such that x n − ison the endpoint contained in the image of γ , and let x n be the endpoint of A not containedin the image of γ . Let x = ( x , . . . , x n ) . The configuration x is pictured in Figure 6.The connected components of ( G \ x ) ∪ { x } are the intervals ( x k , x k +1 ) , . . . , ( x n − , x n − ) ,and a large simply connected component X which contains x . If f ( x ) ∈ X , then let ˜ X bethe component of X \ { x } containing f ( x ) . Note that ˜ X ∪ { x } is simply connected, so wecan apply Lemma 5.10 to produce a distinguished pair containing 1, as desired.Otherwise, suppose that f ( x ) ∈ ( x n − , x n − ) (the cases when f ( x ) lies in a different ofthese intervals is similar, by renumbering the indices and using Proposition 6.2). Let α bethe path in Conf n ( G ) which moves x n − along γ , and fixes each of the other tokens. Callthe ending configuration x (cid:48) = ( x , . . . , x n − , γ (1) , x n ) , and let X (cid:48) be the big simply connected component in ( G \ x (cid:48) ) ∪ { x } . For each t , let X t be thecomponent of G \ β ( t ) which contains ( x n − , x n − ) . Note that X t is a system of componentswhich satisfies the hypotheses of Lemma 5.8, which implies that f ( x (cid:48) ) ∈ X = X (cid:48) . As before,we can apply Lemma 5.10 to x (cid:48) to produce a distinguished pair containing . (cid:3) Proof of (B) when n ≥ − χ ( G ) . Suppose for sake of contradiction that is in every dis-tinguished pair, and that (1 , is a distinguished pair in e (a distinguished pair exists by x x x x x x x x n − x n Figure 6.
The configuration x (A)). Recall that this equips e with an specified orientation. Let T be a maximal subtreenot containing e , and let e be some other edge in G \ T . Suppose further that T “peeksin” to each edge, such that each point in ∂T is contained in the closure of exactly one edgein G \ T . Figure 7 show an example of one of these constructions. e e T Figure 7.
Let e , . . . , e k be the other edges not contained in T , where k = 1 − χ ( G ) , as before. Let H be the subgraph formed from T ∪ e ∪ e by removing free edges until every vertex has degreeat least . Since H is a deformation retract of T ∪ e ∪ e , we can see that χ ( H ) = − , and H has no free edges. Therefore H is homeomorphic to either the figure eight graph ( ∞ ) , thetheta graph (Θ) , or the dumbbell graph D . The three possibilities are pictured in Figure 8.Define a configuration x = ( x , . . . , x n ) as follows: Let x , x , x , . . . , x k +1 be points inthe interior of e , e , . . . , e k , respectively. In the interior of e , place x , x k +2 , . . . , x n equallyspaced apart in e , such that x is farthest back on e with respect to the orientation on e ,and let A = [ x , x n ] . Finally, place x on a branched vertex in T ∩ H as follows:(i) When H = ∞ , then x is the unique branched vertex of H . ECTION PROBLEMS FOR GRAPH CONFIGURATION SPACES 15 e e e e e e Figure 8.
Neighborhoods of the possible subgraphs H (ii) When H = Θ , then x is the branched vertex of H closer to the back end of e , suchthat when x slides into e , we obtain a witness to (1 , .(iii) When H = D , then x is the branched vertex of H lying closer to e Further, let B = G \ ( x ∪ A ) , and label the components of G \ B by X, Y, Z (and X (cid:48) when H = ∞ ) as indicated in Figure 9.When we slide x into the back of e , we obtain a witness to (1 , by Proposition 6.2, andthe reverse path gives a locally system of components by Lemma 5.8, which implies exactlythat f ( x ) ∈ X .Let x (cid:48) = ( x , x , x , x , . . . , x n ) be the configuration which is x except with x and x swapped. We will argue that f ( x (cid:48) ) cannot lie in any component of B . If f ( x (cid:48) ) ∈ X (or X (cid:48) ),then applying Lemma 5.10 with x produces a distinguished pair which does not contain .Therefore we must have f ( x (cid:48) ) ∈ Y or f ( x (cid:48) ) ∈ Z .We will define two paths β = ( β , . . . , β n ) and γ = ( γ , . . . , γ n ) in Conf n ( G ) from x to x (cid:48) which will allow us to see that neither case is possible. Take each of these paths to beconcatenations of basic paths which fix each token except for the first and third, and suchthat β moves x , x around the right loop in H a half-turn clockwise , and γ moves x , x around the right loop in H a half-turn counter-clockwise .Note that for each t , there are exactly two components of G \ β ( t ) which border both β ( t ) and β ( t ) . Since β , β both move clockwise, β , β are each moving “towards” a specificwell-defined component at each t , and they likewise are each moving “away” from a specificwell-defined component at each time. Further, for t such that β ( t ) (cid:54) = x , the path β ismoving “away” from the component that β is moving “towards.” Our immediate goal is to Here “clockwise” and “counter-clockwise” are with respect to the orientations implied by the figure Ax x n − x XX (cid:48) x YZ Ax x n − x x X Y ZA x x n − x x X YZ
Figure 9.
The components
X, Y, Z , (and X (cid:48) ) of G \ B .show that f ( x (cid:48) ) = f ( β (1)) must lie in the component of G \ β (1) which β is moving awayfrom. We show this inductively on the basic pieces of β .Consider a Type I piece of β . We can see that f must remain in the component β ismoving away from by Propositions 5.6 and 5.7.Consider a Type II piece of β where only β moves to a vertex v . Since the componentthat β is moving away from is distinct from the components bordering v , we can deduce byProposition 5.9 that f must remain in the component which f is moving away from at theend of this basic piece.Finally, consider a Type II piece of β where β moves, and suppose that f takes valuesin a component of G \ β which borders β . Since we inductively assume that f lies in thecomponent of G \ β which β is moving away from, f must lie on the component of G \ β that β is moving towards. Therefore, once β reaches the end of this Type II piece at sometime t , f must lie in some component W of G \ β ( t ) which borders β ( t ) but is not behind β at t by Proposition 5.9. If W does not border β ( t ) , then by our construction of theconfiguration x , W ∪ { β ( t ) } is simply connected, so we can apply Lemma 5.10 to producea distinguished pair which does not contain . Therefore, W must border β ( t ) , so it mustin fact be the component which β is moving towards, and therefore the component which β is moving away from.We have shown that f ( x (cid:48) ) is in the component which β is moving away from. Since β moves clockwise, we must have f ( x (cid:48) ) ∈ Y . However, if we apply the exact same argumentwith γ , we get f ( x (cid:48) ) ∈ Z , which is a contradiction. (cid:3) ECTION PROBLEMS FOR GRAPH CONFIGURATION SPACES 17
The Case n = 3 . When n = 3 and χ ( G ) = − , the statements (A) and (B) do notcontradict Corollary 6.3, so we must rely on a different technique. We first introduce aproposition which simplifies the proof. Proposition 6.5.
Suppose H ⊂ G is a deformation retract of G , and suppose there existsan identifying function on G . Then there exists an identifying function on H .Proof. Since H is a deformation retract of G , G can be obtained from H by attaching treesto H at single points. Therefore, there exists a retract r : G → H which collapses each ofthese trees to a single point. We define g : Conf n ( H ) → H by g ( x ) = r ( f ( x )) . We can seethat g is continuous, and we can show that it is also an identifying function. Suppose that x = ( x , . . . , x i ) is a configuration, and suppose x i = r ( f ( x )) . Then f ( x ) ∈ T for some tree T attached to G . Corollary 5.11 produces a contradiction. (cid:3) Finally, we present another consequence of Proposition 6.2 which will be useful to us.
Proposition 6.6. If ( i, j ) and ( k, i ) are distinguished pairs on e , then j = k .Proof. Suppose otherwise, then we can construct a configuration x = ( x , . . . , x n ) which is asimultaneous witness to ( i, j ) and ( k, i ) , by picking x i , x j , x k all lying on e with x k < x i < x j with respect to the orientation on e , and no other tokens contained in the interval [ x k , x j ] . ByProposition 6.2, such a configuration is a witness to both distinguished pairs, necessitating f ( x ) ∈ ( x i , x j ) and f ( x ) ∈ ( x k , x i ) , which is a contradiction. (cid:3) We can now present the final portion of the proof of Theorem 1.1.
Proof when n = 3 and χ ( G ) = − . It suffices to consider this in the case where G has nofree vertices by Prop 6.5. The only such graphs with χ ( G ) = − are the graphs ∞ , Θ , and D as before. Suppose for sake of contradiction that f is an identifying function on G .We will define two functions Hor : Σ → {± } , Ver : Σ → {± } . For a permutation σ ∈ Σ , consider the configuration x σ depicted in Figure 10. x σ, x σ, x σ, x σ, x σ, x σ, x σ, x σ, x σ, Figure 10.
The configurations x σ Denote by x σ,i the i ’th token of x σ . We consider the components of G \ x σ to be either onthe “top” or “bottom,” (resp. “left,” “right”) corresponding to their depictions in the figure.When G = ∞ , Θ , then we say Ver( σ ) = +1 if f ( x σ ) lies in one of the top components of G \ x σ , and Ver( σ ) = − otherwise. Similarly, when G = ∞ , D we say Hor( σ ) = +1 if f ( x σ ) lies in one of the right components of G \ x σ and Hor( σ ) = − otherwise. To compute Ver for G = D , define x (cid:48) σ to be the configuration obtained by moving along a Type II pathwhich moves x σ, towards f ( x σ ) . By Proposition 5.9, f ( x (cid:48) σ ) lies in either the top or bottomcomponent of G \ x (cid:48) σ contained in the loop containing x σ, . We say that Ver( σ ) = +1 if f ( x (cid:48) σ ) is in the top component of G \ x (cid:48) σ in this loop, and Ver( σ ) = − if f ( x (cid:48) σ ) is in thebottom component of G . Similarly, for G = Θ , define x (cid:48) σ to be the configuration obtained by moving along a Type II path which moves x σ, towards f ( x σ ) . As before, we can concludeby Proposition 5.9 that f ( x (cid:48) σ ) lies on one of the components of G \ x (cid:48) σ in the direction that x σ, moved. We can therefore define Hor( σ ) = +1 if f ( x (cid:48) σ ) ∈ ( x (cid:48) σ, , x (cid:48) σ, ) , and Hor( σ ) = − otherwise. Figure 11 shows a particular example of this construction. x (cid:48) σ, x (cid:48) σ, f ( x (cid:48) σ ) x (cid:48) σ, x (cid:48) σ, x (cid:48) σ, f ( x (cid:48) σ ) x (cid:48) σ, Figure 11.
Label the edges containing x σ, and x σ, by e , e respectively. Note that Ver and
Hor associate a distinguished pair to each σ which contains σ (2) . Explicitly, this association isgiven by: σ (cid:55)→ ( σ (2) , σ (2 + Hor( σ ))) on Ver( σ ) · e σ ) , where e , e are equipped with appropriate orientation, and where {± } acts on the ori-ented edges of G by changing orientation. Our next task is to determine how Ver and
Hor interact with the multiplication in Σ . By possibly applying a symmetry to G , assume that Hor(id) = Ver(id) = +1 . We make the following claims for all σ ∈ Σ : Hor( σ ) = +1 , Ver((12) σ ) = Ver( σ ) , Ver((23) σ ) = − Ver( σ ) . If we can prove these claims, then the remainder of the proof is simple, since we can write:
Ver(id) = Ver((123)(123)(123)) = Ver((12)(23)(12)(23)(12)(23)) = − Ver(id) , a contradiction.All that remains is to prove the claim. We will prove this only in the case G = ∞ , butthe other cases are not so different.We will show that Hor((12)) = Hor((23)) = Ver((12)) = +1 , and
Ver((23)) = − . Thiswill imply the rest of the claim by replacing f with the identifying function f σ , defined by ( x , x , x ) (cid:55)→ f ( x σ (1) , x σ (2) , x σ (3) ) . Suppose the components of G \ x id are labelled as in Figure 12 X X X X Figure 12.
The components of G \ x id ECTION PROBLEMS FOR GRAPH CONFIGURATION SPACES 19
By assumption, we have f ( x id ) ∈ X . Consider the path α between x id and x (23) whichrotates x id , and x id , a half turn clockwise around the loop on the right. By representing α as a concatenation of a Type I and Type II path, we can see using Propositions 5.7 and 5.9that f ( x (23) ) ∈ X , so Ver((23)) = − and Hor((23)) = +1 .Now, consider the path β from x id and x (12) which rotates x id , and x id , clockwise ahalf-turn around the right circle, and the path β which rotates them counter-clockwise ahalf-turn. By representing these as concatenations of a Type I and a Type II path, we canuse Propositions 5.7 and 5.9 to show that X (cid:54) (cid:32) β X , and X (cid:54) (cid:32) β X . Finally, Proposition6.6 shows that f ( x (12) ) / ∈ X , otherwise (2 , , (3 , would both be distinguished pairs in e .Therefore, f ( x (12) ) ∈ X , so Ver((12)) = Hor((12)) = +1 , thereby establishing the claim,and concluding the proof. (cid:3) A Combinatorial Model
In this section, we discuss K n ( G ) , a combinatorial approximation to the spaces Conf n ( G ) which was introduced in [Lü14]. These spaces are very useful for constructing sections in thecases when they exist, as we will show that partially defined identifying functions on K n ( G ) often extend to Conf n ( G ) .The outline of this section is as follows: we will first define a poset P n ( G ) , which is definedslightly differently from the definition given in [Lü14], but the two are easily seen to beequivalent. We will define K n ( G ) as the cube complex whose face poset is P n ( G ) , and statewithout proof the fact that K n ( G ) embeds as a deformation retract of Conf n ( G ) . We willconclude by discussing which identifying functions on K n ( G ) extend to Conf n ( G ) .Let E be the set of oriented edges of G , and let ¯ E ⊂ E be the set of positively orientededges, where we fix some positive orientation on each edge. Let B be the set of branchedvertices of G . For each e ∈ E , let v e be the endpoint of e in the direction that e is pointing,and let ¯ e be the element of ¯ E corresponding to e . Recall that an index denotes an integer , . . . , n . First we define the k -dimensional faces of P n ( G ) . Definition 7.1. A k -face F of P n ( G ) associates indices to the elements of ¯ E (cid:116) B (cid:116) E asfollows:(i) To each positively oriented edge ¯ e ∈ ¯ E , there is an ordered tuple of associated indices: F (¯ e ) = ( i ¯ e, , . . . , i ¯ e,(cid:96) ) . (ii) To each branched vertex v ∈ B , there is at most one associated index F ( v ) = i v .(iii) To exactly k oriented edges e ∈ E such that v e ∈ B , there is an associated index F ( e ) = i e .(iv) Each index occurs exactly once as an index associated to an element of ¯ E (cid:116) B (cid:116) E .(v) For each vertex v ∈ B , there is at most one index of the form F ( v ) or F ( e ) , where e ∈ E is an oriented edge with v e = v .This gives a definition of P n ( G ) as a graded set. We now can define the partial ordering. Definition 7.2. If F is a k -face, and e ∈ E is an oriented edge such that F ( e ) is defined,then we will define two ( k − -faces, F + e and F − e , which we will use to define the order on P n ( G ) . First, we define F + e as follows:(i) F + e (¯ e (cid:48) ) = F (¯ e (cid:48) ) for all ¯ e (cid:48) ∈ ¯ E .(ii) F + e ( v e ) = F ( e ) . For v ∈ B with v (cid:54) = v e , F + e ( v ) exists if and only if F ( v ) does, and ifthey are defined, the two are equal. (iii) F + e ( e ) is undefined, and for e (cid:48) ∈ E with e (cid:48) (cid:54) = e , F + e ( e (cid:48) ) is defined if and only if F ( e ) is, and if they are defined, the two are equal.Now we define F − e as follows:(i) F − e (¯ e ) is equal to F (¯ e ) with F ( e ) attached to the back when e = ¯ e (that is, when e has positive orientation), and F (¯ e ) with F ( e ) attached to the front when − e = ¯ e .For ¯ e (cid:48) ∈ ¯ E with ¯ e (cid:48) (cid:54) = ¯ e , define F − e (¯ e (cid:48) ) = F (¯ e (cid:48) ) .(ii) For v ∈ B , F − e ( v ) exists if and only if F ( v ) does, and when they both exist the twoare equal.(iii) F − e ( e ) is undefined, and for e (cid:48) ∈ E with e (cid:48) (cid:54) = e , F − e ( e (cid:48) ) exists if and only if F ( e (cid:48) ) does,and when they both exist the two are equal.One can verify that F + e and F − e are indeed ( k − -faces of P n ( G ) . We define the ordering (cid:31) on P n ( G ) to be generated by the relations F (cid:31) F + e and F (cid:31) F − e for each face F and eachoriented edge e such that F ( e ) exists.The important facts about P n ( G ) are summarized in the following theorem, whose state-ments are proven in [Lü14]. Theorem 7.3.
The graded poset P n ( G ) is the face poset of a unique (up to isomorphism)abstract cube complex K n ( G ) . There is an embedding i : K n ( G ) (cid:44) → Conf n ( G ) , and a de-formation retract h t : Conf n ( G ) → Conf n ( G ) of Conf n ( G ) onto the image of K n ( G ) in Conf n ( G ) . In addition, the following holds:(a) The embedding i takes vertices of K n ( G ) exactly to the configurations where the tokensare spread out evenly on each edge.(b) For any configuration x ∈ Conf n ( G ) , no tokens of x leave or enter any edges orvertices along the path t (cid:55)→ h t ( x ) . Alternatively, the paths t (cid:55)→ h t ( x ) and t (cid:55)→ h − t ( x ) are both Type I for every x .(c) Tokens of configurations in K n ( G ) are uniformly far from each other. That is, thereexists some ε > such that for all x = ( x , . . . , x n ) ∈ K n ( G ) , we have d ( x i , x j ) ≥ ε . Corollary 7.4. If G has k branched vertices, then Conf n ( G ) has the homotopy type of a min { k, n } -dimensional cell complex.Proof. Indeed, one can check that K n ( G ) is min { k, n } dimensional. Details can be found in[Lü14]. (cid:3) Corollary 7.5. If G has only one branched vertex, then the short exact sequence → ker( φ n +1 , ∗ ) → P B n +1 ( G ) φ n +1 , ∗ −−−−→ P B n ( G ) → of groups, discussed in Section 3, splits for every n .Proof. By Corollary 7.4,
Conf n ( G ) has the homotopy type of a graph, so P B n ( G ) is a freegroup, and the sequence splits. (cid:3) Corollary 7.5 provides, along with Proposition 3.2, a class of examples of graphs such thatthe group homomorphism φ n +1 , ∗ splits, although the map φ n +1 of topological spaces doesnot admit a section.Intuitively, we can imagine the vertices of K n ( G ) to be the configurations of Conf n ( G ) where the vertices in each edge are equally spaced, as suggested in Theorem 7.3(a). Givensuch a configuration x , we discuss how to define the associated vertex F of P n ( G ) . For a ECTION PROBLEMS FOR GRAPH CONFIGURATION SPACES 21 positively oriented edge ¯ e ∈ ¯ E , set F (¯ e ) to be the tuple of indices whose corresponding tokensof x lie in ¯ e , ordered consistently with the orientation on ¯ e . For a branched vertex v ∈ B ,set F ( v ) to be the index of the token of x , if it exists, occupying v .An edge in K n ( G ) corresponds to a continuous move between vertices of K n ( G ) whichmoves a single token from a vertex into the interior of an edge. Similarly, a k -dimensionalface corresponds to a set of k moves which can be performed consistently, that is, theyinvolve distinct vertices and tokens.In our construction of the poset P n ( G ) , for a fixed face F , the indices associated withpositively oriented edges ¯ e and vertices v correspond to tokens which are essentially fixed onthe image of F , and the location of the token corresponding to such an index is determined bywhich positively oriented edge or vertex it is associated to. Conversely, the indices associatedto oriented edges e correspond to tokens which can move from the interior of e to the vertex v e throughout F . For such an e , the associated face F + e corresponds to the boundary face of F where the moving token on e is instead taken to be stationary at v e , and F − e correspondsto the boundary face where the moving token on e is taken to be stationary on the interiorof e . Example 7.6.
Let G be the dumbbell graph pictured in Figure 13, with the given labelsand choices of positive orientations, and let n = 5 . e b e b e Figure 13.
A vertex of K n ( G ) maps to a configuration like x , pictured in Figure 14, such that thevertices on each edge are evenly spaced apart. x x x x x Figure 14.
Let F x be the -face which maps to x . Explicitly, we can compute: F ( ¯ e ) = (3 , , ,F ( ¯ e ) = (4) ,F ( b ) = 2 , and every other value of F is trivial.We now turn to the question of extending identifying functions defined on K n ( G ) to Conf n ( G ) .The important fact is that under certain mild conditions, we can extend any identifyingfunction defined on K n ( G ) to all of Conf n ( G ) . Definition 7.7.
Suppose A ⊂ K n ( G ) is a subset containing the -skeleton of K n ( G ) , and f : A → G is a partially defined identifying function. We say that f is extendable if the following situation never occurs: Select a free edge e (edge connected to a free vertex) whichis connected to a branched vertex v . Then there exists a vertex F of K n ( G ) where F ( v ) exists, F (¯ e ) is empty, and f ( F ) ∈ e . Proposition 7.8.
A partially defined identifying function f : K n ( G ) → G extends to anidentifying function on Conf n ( G ) if and only if f is extendable.Proof. Suppose f is extendable, and let x = ( x , . . . , x n ) be a configuration, let y = ( y , . . . , y n ) = h ( x ) , and let p = f ( y ) . If p is a branched vertex of G , then define f ( x ) = p . Note that p (cid:54) = x , . . . , x n , since the path t (cid:55)→ h − t is Type I by Theorem 7.3(b). If p is not a branchedvertex of G , then p lies in some component A of G \ ( h ( x ) ∪ B ) . Note that A must be either:an open interval ( a , a ) or a half-open interval ( a , a ] , where a is a free vertex of G . Let V be the set of (branched or free) vertices of G . Define a function ψ : { y , . . . , y n } (cid:116) V → G by: ψ ( q ) = (cid:40) x j q = y j q q ∈ V .
We would like to define f ( x ) to lie in the open interval ( ψ ( a ) , ψ ( a )) , where we take a , a ∈ { y , . . . , y n } if possible.First we must check that ( ψ ( a ) , ψ ( a )) is nonempty. If ψ ( a ) = ψ ( a ) , then we must havethat a is a free vertex of G and a = y j for some j . Then, there is a path in Conf n ( G ) from y to a vertex F of K n ( G ) such that y j moves directly onto the nearest branched vertex v . By Lemma 5.8 and Proposition 5.9, if f extends, we must have f ( F ) ∈ [ a , v ) , whichcontradicts the fact that f is extendable. Therefore ( ψ ( a ) , ψ ( a )) is a nonempty interval.We define f ( x ) to lie between ψ ( a ) , ψ ( a ) in the same proportion that f ( y ) lies between a , a . This defines f on Conf n ( G ) , and it is easily verified to be a continuous identifyingfunction.Now suppose f is not extendable. Note that the situation in Definition 7.7 contradictsCorollary 5.11, so f does not extend to an identifying function on Conf n ( G ) . (cid:3) Corollary 7.9. If G has no free vertices, then every identifying function on K n ( G ) extendsto Conf n ( G ) . Sharpness of n ≥ − χ ( G ) In this section we use the methods of Section 7 to show that we cannot make any strongerclaims about the existence or nonexistence of sections based only on the homotopy type of agraph, that is, whenever n < − χ ( G ) and χ ( G ) < , it is impossible to determine whether G admits an identifying function. First, we prove an existence result. Proposition 8.1.
Suppose G contains a vertex w , such that G \ w contains k componentswhich are connected to w at least twice. Then, if n ≤ k , G admits an identifying function.Proof. First, we consider the graph H = (cid:87) k S k , a wedge of k circles, where k ≥ n . Let v bethe unique branched vertex, and let p , . . . , p k be the points such that p i is the point of S i antipodal to v . Let S be the set { , , . . . , k } , and let F be the class of functions ψ : S → N such that (cid:88) s ∈ S ψ ( s ) = n − . ECTION PROBLEMS FOR GRAPH CONFIGURATION SPACES 23
Let h : F → S be a function such that ψ (( h ( ψ ))) = 0 for all ψ ∈ F . Since k ≥ n , such afunction always exists.Select some small ε > , and let M ⊂ H n be the subset: M = { ( x , . . . , x n ) : x i ∈ H, there is at most one i such that d ( x i , v ) < ε } . (note that we do not require the x i to be distinct). We will define a Σ n -invariant identifyingfunction on M , that is, a function f : M → H such that f ( x , . . . , x n ) (cid:54) = x i for all x =( x , . . . , x n ) ∈ M , and i = 1 , . . . , n such that f ( σx ) = f ( x ) , where Σ n acts on M bypermuting indices. Let N ⊂ H be the ε -neighborhood of v , and let A = ( G \ N ) n and B = N × ( G \ N ) n − . Note that A and B are disjoint, and every element of M is in the Σ n -orbit of an element in A or B (but not both), so it suffices to define f on A ∪ B . On A ,we simply define f to be constantly equal to v . On B , we will define f separately on eachcomponent X of B . For each X , we can define a function ψ X : S → N by setting ψ X ( i ) tobe the number of tokens of an element of X contained in S i \ N . Clearly, ψ X ∈ F . Let π : B → N be the projection of B onto N . We can define a map λ : ¯ N → S h ( ψ X ) such that:(a) λ ( v ) = p h ( ψ X ) (b) For each free vertex u of ¯ N , λ ( u ) = v (c) λ has no fixed points, that is, λ ( q ) (cid:54) = q for all q ∈ N ∩ S h ( ψ X ) .By the definition of h , ψ X ( h ( ψ X )) = 0 , so S h ( ψ X ) contains no tokens of any element of X bythe definition of ψ X . Therefore, such a map exists. We define f = λ ◦ π on X . This is anidentifying function on X , and by applying this process to all components of B , we get anidentifying function on B . Further, by taking f to be Σ n invariant, we get an identifyingfunction on M , as desired, by noting that the local definitions of f glue continuously.Now, we turn to the task of defining an identifying function on G . By our assumptionson w , there exists an embedding i : H (cid:44) → G such that i ( v ) = w , and a retract r : G → H such that r − ( w ) = { v } . Note that r induces a map R : K n ( G ) → H n , by applying r to eachtoken of a configuration. By Theorem 7.3(c), there is some ε such that at most one tokenof any configuration of K n ( G ) is within ε of w . Taking sufficiently small ε , we can see thatthis implies that R ( K n ( G )) ⊂ M . Then, let g = i ◦ f ◦ R , where f is as constructed above.For any x ∈ K n ( G ) , note that none of the tokens of x lie in the image of the circle whichcontains g ( x ) , so g is a partially defined identifying function. Further, the image of H in G intersects no free edges, so g is extendable, and by Proposition 7.8, we conclude. (cid:3) Now we present a class of examples with n < − χ ( G ) where sections do not exist. Example 8.2 (Wedge of balloons) . Let B be the balloon graph, as pictured in Figure 15 Figure 15.
Suppose G = (cid:87) k B is a wedge of balloons, glued at the free vertices of each balloon, andsuppose that n ≤ k . If n ≥ , then statement (A) holds, and if n ≥ , statement (B) holds,and the proofs of these are quite similar to the proof of Theorem 1.1. The key observationis that whenever we have a configuration x with f ( x ) in a copy of B and a token (say x ) as depicted in Figure 16, there always exists a distinguished pair (1 , j ) in the circle in thatcopy of B (in fact for any j (cid:54) = 1 , possibly after changing orientation). x f ( x ) Figure 16.
To see why, note that we can slide any token x j into B , and continue through the pathdepicted by Figure 17. x f ( x ) x j x f ( x ) x j f ( x ) x Figure 17.
To prove (A) and (B), note that we can avoid placing tokens in any balloons we wouldlike, and copy the proof of Theorem 1.1, and the fact above when necessary.Therefore we have shown that when n < − χ ( G ) (and n ≥ ), we cannot say for surewhether or not G admits a section. References [Abr00] Aaron David Abrams.
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