Covering Graphs Motivation Suppose you are taken to

Covering Graphs • Motivation: • Suppose you are taken to two different labyrinths. Is it possible to tell they are distinct just by walking around? • Let us call the first graph maze X, and the second one Y.

Question • Is it possible to distinguish between the two mazes? • Answer: Yes, we can. In the upper maze there are two adjacent trivalent vertices. This is not the case in the lower maze.

Local Isomorphism • On the other hand we cannot distinguish (locally) between the upper and lower graph. • To each walk upstairs we can associate a walk downstairs.

One More Example • C 4 over C 3 is no good. However, C 6 over C 3 is Ok.

Fibers and Sheets. • We say that C 6 is a twosheeted cover C 3. Red vertices are in the same fiber. Similarly, the dotted lines belong to the saem fiber. • Graph mapping f: C 6 C 3 is called covering projection. • Preimage of a vertex f-1(v) (or an edge f-1(e)) is called a fiber. • The cardinality of a fiber is constant. k =|f-1(v)| is called the number of sheets.

One More Example • The cube graph Q 3 is a two fold cover complete graph K 4. • The vertex fibers are composed of pairs of antipodal vertices.

Covers over Pregraphs • Graph K 4 can be understood as a fourfold cover a pregraph on one vertex (one loop and one half -edge).

Voltage Graphs • X = (V, S, i, r) – connected (pre)graph. • ( , A) – permutation group acting on space A. • : S – voltage assignment. • Condition: for each s 2 S we have [s] [r(s)] = id.

Voltage Graph Determines a Covering Graph • Each voltage graph (X, , A, ) determines a covering graph Y and the covering projection f: Y X as follows: • Covering graph Y = (V(Y), S(Y), i, r) • • V(Y) : = V(X) x A S(Y) : = S(X) x A i: S(Y) V(Y): i(s, a) : = (i(s), a). r: S(Y): r(s, a) : = (r(s), [s](a)). • Covering projection f • f: V(Y) V(X): f(x, a) : = x. • f: S(Y) S(X): f(s, a) : = s. • Sometimes we denote the covering graph Y by Cov(X; ).

(Rhetorical) Questions • “Different” voltage graphs may give rise to the “same” cover. What does it mean the “same” and how do we obtain all “different” voltage graphs? • The voltage graph is determined in essence by the abstract group. What is the role of permutation group? • How do we ensure that if X is connected then Y is connected, too?

Kronecker Cover • Let X be a graph. The canonical double cover or Kronecker cover: KC(X) is a twofold cover that is defined by a voltage graph that has nontrivial voltage from Z 2 on each of its edges. It can also be described as the tensor product KC(X) = X £ K 2.

Homework • H 1: Prove that Kronecker cover is bipartite. • H 2: Prove that generalized Petersen graph G(10, 2) is a twofold cover the Petersen graph G(5, 2). • H 3: Determine the Kronecker cover G(5, 2). • H 4: Determine a Zn covering over the handcuff graph G(1, 1), that is not a generalized Petersen graph G(n, r).

Regular Covers • Let Y be a cover X. We are interested in fiber preserving elements of Aut Y (covering transformations). • Let Aut(Y, X) · Aut Y be the group of covering transformations. • The cover Y is regular, if Aut(Y, X) acts transitively on each fiber. • Regular covers are denoted by voltage graphs, where permutation group ( , A) acts regularly on itself by left or right translations: ( , ).

Exercises • N 1: Prove that each double sheeted cover is regular. • N 2: Find an example of a three sheeted cover that is not regular. • N 3: Express the graph on the left as a 6 -fold cover a pregraph on a single vertex.

Dipole n • Dipole n has two vertices joined by n parallel edges. We may call one vertex black, the other white. On the left we see 5. • Each dipole is bipartite, that is why each cover n is bipartite too. Dipole 3 jeis cubic, sometimes called theta graph .

Cyclic cover a dipole – Haar graph H(n). 0 3 5 Z 6 • H(37) is determined by number 37, actually by its binary representation (1 0 0 1). • k = 6 is the length of the sequence, hence group Z 6. • (0 1 2 3 4 5) – positions of “ 1”. • Positions of “ 1”s: 0, 3 in 5. {0, 3, 5} are the voltages on . The corresponding covering graph is H(37).

Exercises • Graph on the left is called the Heawood graph H. Prove: – H is bipartite. – H is a Haar graph (Determine n, such that H = H(n)) – Express H as a cyclic cover . – Show that there are no cycles of lenght < 6 in H. – Show that H is the smallest cubic graph with no cycles of length < 6.

Cages as Covering Graphs • A g-cage is a cubic graph of girth g that has the least number of vertices. • Small cages can be readily described as covering graphs.

1 -Cage • Usually we consider only simple graphs. For our purposes it makes sense to define also a 1 -cage as a pregraph on the left. • 1 -cage is the unique smallest cubic pregraph.

2 -Cage • The only 2 -cage is the graph. • We may view 2 -cage, as the Kronecker cover 1 -cage. 1 1 Z 2

3 K 4, the 3 -cage 2 1 0 2 Z 4 1 • K 4 is a Z 4 covering over the 1 -cage. • In general, we obtain a Z 2 n covering over the 1 -cage by assigning voltage 1 to the loop and voltage n to the half-edge. • Exercise: What is the covering graph in such a case?

K 3, 3, the 4 -cage 5 4 3 2 1 0 3 Z 6 1 • K 3, 3 is a Z 6 covering over the 1 -cage. • It can also be seen as a Z 3 covering over the 2 -cage . • Exercise: Express K 3, 3 as a covering graph over . Dtermine a natural number n, such that K 3, 3 is a Haar graph H(n).

The Handcuff Graph G(1, 1) 1 0 Z 2 • By changing the voltage on the loop of the 1 -cage we obtain a double cover G(1, 1), the smallest generalized Petersen graph, known as the Handcuff graph.

I graphs I(n, i, j) and Generalized Petersen graphs G(n, k) 0 i Zn • Cyclic covers over the handcuff graph are called I -graphs. Each I-graph can be described by three parameters I(n, i, j) with i · j. In case i = 1 we call I(n, i, k) = G(n, k), the generalized Petersen graph. • In particular, I(5, 1, 2) is the 5 -cage. j

The 6 -cage b a D 7 • The 6 -cage is the Heawood graph on 14 vertices. It is a 7 -fold cyclic cover the graph. But it is also a dihedral cover the 1 cage. • Let the presentaion of Dn be given as follows: Dn = <a, b|an, b 2, ab=ba-1> • Then the Heawood is a covering described on the left.

Exercises • N 1. Express the 7 -cage as a covering graph. • N 2. Express the 8 -cage as a covering graph.

(3, 1)-trees • A (3, 1)-tree is a tree whose vertices have valence 3 and 1 only. • On the left we see the smallest (3, 1)-trees I, Y and H.

(3, 1)-cubic graphs • A (3, 1)-cubic graph is obtained from a (3, 1)tree by adding a loop at each vertex of valence 1. • On the left we see the smallest (3, 1)-cubic graphs I(1, 1, 1), Y(1, 1, 1, 1) and H(1, 1, 1).

Coverings over (3, 1)-cubic graphs j i Zn j i k i j k l • By putting 0 on the tree edges and appropriate voltages on the loops of (3, 1)-cubic graph we obtain their Zn coverings. • In the case of the graphs on the left we obtain the Igraphs, Y-graphs and Hgraphs: I(n, i, j), Y(n, i, j, k) and H(n, i, j, k, l).

Covers Determined by Graphs • We know already that there exists a cover, namely Kronecker cover, that depends only on X itself and the voltage assignment plays a minor role. • Now we will present some covers that have a similar property.

Coverings and Trees • Let X be a connected graph and let Cov(X) denote all connected covers over X: • Cov(X) = {(Y, )| Y connected and : Y ! X, covering projection}. For each connected X we have (X, id) 2 Cov(X). • Proposition: For a connected X we have Cov(X) = {(X, id)} if and only if X is a tree. • This fact holds both for finite and locally finite trees.

Universal cover • • Let X, Y and Z be connected graphs and let : Y ! X and : Z ! Y be covering projections. On the other hand, we may consider the class Cov(X) of all coverings over X. We may introduce a partial order in Cov(X). (Y, ) < (Z, ) if there exists a covering projection (Z, ) 2 Cov(Y) so that = . Proposition: Any connected finite or locally finite graph X can be covered by some tree T; : T ! X. Proposition: Any connected finite or locally finite graph X can be covered by at most one tree T. Proposition: Let : T ! X be a covering projection form a tree to a connected graph X. Then for each covering : Y ! X there exists a covering : T ! Y such that = . Corollary: For each connected X the poset Cov(X) has a maximal element (T, ) where T is a tree. The maximal element (T, ) 2 Cov(X) is called the universal covering of X.

Construction of Universal Cover • There is a simple construction of the universal covering projection. • Let X be a connected graph and let T µ X be a spanning tree. Furthermore, let S = E(G) E(T) be the set of edges not in tree T. • Consider S to be the set of generators for a free group F(S) and F(S) to be the voltage group. • Let us assing voltages on E(G) as follows: • If e 2 E(T) the voltage on e is identity. • If e 2 S the voltage is the corresponding generator (or its inverse) • Note: The construction does not depend on the choice of direction of edges. • Proposition: The described construction gives rise to the universal cover.

Examples • Example: The universal cover any regular k-valent graph is a regular infinte tree T(1, k).

Valence Partition and Valence Refinement • Let G be a graph and let B = {B 1, . . . , Bk} be a partition of its vertex set V(G) for which there are constants rij, 1 · i, j · k such that for each v 2 Bi there are rij edges linking v to the vertices in Bj. Let R = [rij] be the corresponding k £ k matrix, Then B is called valence partition and R is called valence refinement. If k is minimal, then B is called minimal valence partition and R is called minimal valence refinement. • Two refinements R and R’ are considered the same if one can be transformed to the other one by simultaneous permutation of rows and columns. • A refinement is uniform, if each row is constant.

Construction • • • Given graphs G and G’ with a common refinement. Let mij denote the number of arcs in G of type i ! j. Let ni denote the number of vertices in G of type i. Let bij = lcm(mij)/mij. (If mij = 0 , let bij undefined). Let ai = lcm(mij)/ni. Note that bij and ai depend only on the common matrix R and are the same for both graphs G and G’. Let l(e) or l(e’) be a linear order given to all type i ! j arcs with a common initial vertex i(e) (or i(e’)). Let V(H) = {(i, v, v’, p)|v and v’ of type i, p 2 Zai} Let S(H) = {(i, j, e, e’, q)|e and e’ of type i ! j, q 2 Zbij} r(i, j, e, e’, q) : = (j, i, r(e), r(e’), q) i(i, j, e, e’, q) : = (i, i(e), i(e’), q rij + l(e)-l(e’)} H is a common cover of G and G’.

Computing Minimal Valence Refinement • Let r[u, B] denote the number of edges linking u to the vertices in B. • Algorithm [F. T. Leighton, Finite Common Coverings of Graphs, JCT(B) 33 1982, 231 -238. ] • Step 1. Place two vertices in the same block if and only if they have the same valence. • Step 2. While there exist two blocks B and B’ and two distinct vertices u, v in B with r[u, B’] r[v, B’] repeat the following: • Partition the block B into subblocks in such a way that two vertices u, b of B remain in the same block if and only if r[u, B’] = r[v, B’] for each B’ of the previous partition. • Step 3. From minimal valence partition B compute the minimal vertex refinement R. • Note: We may maintain R during the run of the algorithm as a matrix whose elements are sets of numbers.

Comon Cover • 1. 2. 3. 4. 5. • Theorem. Given any two finite graphs G and H, the following statements are equivalent: G and H have the same universal cover, G and H have a common finite cover, G and H have a common cover, G and H have the same minimal valence refinement. G and H have the same some valence refinement. Homework. Find the result in the literature and construct a finite comon cover of G(5, 2) and G(6, 2).

Petersen graph • An unusual drawing of Petersen graph.

Petersen graph G(5, 2) and graph X.

Kronecker Cover - Revisited • Kronecker cover KC(G) is an example of covers, determined by the graph itself. • Exercise. Show that G(5, 2) and X have the same Kronecker cover.

THE covering graph • Let G be a graph with the vertex set V. By THE(G) we denote the following covering graph. • To each edge e = uv we assing transposition e = (u, v) 2 Sym(V). The resulting covering graph has two components, one being isomorphic to G. The other componet is called THE covering graph.

Examples • On the left we see The covering graph of K 2, 2, 2. • The construction resembles truncation. • Each vertex is truncated an inverse figure is placed in the space provided for it. • Theorem: If G is planar, then THE(G) is planar.

Homework • H 1. Given connected graph G with n vertices and e edges and with valence sequence (d 1, d 2, . . . , dn). Determine the parameters for THE(G). • H 2. Determine all connected graphs G for which girth(G) girth(THE(G)).

The fundamental group of a graph. • Let G be a connected graph rooted at r 2 V(G) and let denote the collection of closed walks rooted at r. • Let and be two closed walks rooted at r. The compositum is also a closed walk rooted at r. • We may also define -1 as the inverse walk. • Finally, we need equality (equivalence). • 1 2 ~ 1 e e-1 2. • (G, r) : = /~ is a group, called the fundamental group of G (first homotopy group). • Fact: (G, r) is a free group generated with m-n+1 generators.

The first Homology group of a graph • Let G be a connected graph and T one of its spanning trees. Each edge h 2 GT of the co-tree defines a unique cycle C(h) µ E(G). • The charactersitic vector h 2 {0, 1}m, h(e) = 1, if e 2 C(h) and h(e) = 0, represents C(h). The set of all charactersitic vectors spans a m-n+1 dimensional Z-module in Zm. This can be also viewed as a free abelian group isomorphic to Zm-n+1. • This group is called the first homology group H 1(G, Z). We may replace Z by Zk and obtain the first Zk homology group Zkm-n+1.

Pseudohomological Covers • Idea: Let G be a graph and T its spanning tree and with the edges H = {h 1, h 2, . . . , hm-n+1} = E(G)E(T). Let (H) be a group with m-n+1 interchangeable generators H. The pseudohomological -cover HOM(G, , T) is determined by a voltage graph with (e) = id, for e 2 E(T) and (h) = h, for h 2 E(G)E(T). • Main Question. Is HOM(G, , T) independent of the choice of T and the selection of the generators or their inverses? If the answer is yes, the covering is called homological cover.

Pseudohomological 2 -cover • Let G be a graph and T its spanning tree. The pseudohomological 2 -cover HOM(G, Z 2, T) is determined by a voltage graph with (e) = 0, for e 2 E(T) and (e) = 1, for e E(T). • Theorem. If G is connected then HOM(G, Z 2, T) is connected if and only if G is not a tree.

Example 0 1 0 0 1 1 Z 2 0 1 0 2 • The two voltage graphs on the left determine different pseudohomological Z 2 covers. • Cov(G, 2) is bipartite and Cov(G, 1) is not.

Switching • Let (G, ) be a voltage graph. Let : V(G) ! be an arbitrary mapping, called switching, that assigns voltages to vertices. Define a new voltage assignment as follows: • (s) : = (i(s)) (s) (i(r(s))-1. • is well-defined. • Namely (r(s)) = (i(r(s))) (r(s)) (i(s))-1. • Hence (r(s))-1 = (i(s)) (r(s))-1 (i(r(s)))-1 = (i(s)) (s) (i(r(s)))-1 = (s). • Clearly for any switching the graphs Cov(G, ) and Cov(G, ) coincide. • Given (G, ) and any spanning tree T. There exists a switching such that the resulting is identity on T. • If, in addition, T is rooted at v, we may select (v) = id (or arbitrarily) and this determines switching completely.

Homological Elementary Abelian Covers • Let G be a graph with a spanning tree T. Let k = m-n+1 be the number of edges in GT. Define the voltage assignment such that each non-tree edge gets the voltage ei = (0, 0, . . , 0, 1, 0, . . . , 0) 2 Zpk. • Claim: If p is prime, then Cov(G, ) is independent of T. • Question: What happens in the case p is not prime?

Tree-To-Tree Switch • Let T and T’ be two spanning trees of G. Let H = {h 1, h 2, . . . , hk} be the co-tree edges of T. Let r be the root of G. For each vertex w 2 V(G) there is a unique path P(T’, w, r) on the three T’ from w to v. Let S(w) µ H be the collection of co-tree edges on this path. Let S(w) be the label given to w. Hence (w) = { hi| hi 2 S(w)}. • Claim: Starting with homological voltage assignment relative to T and applying the tree-to-tree switch , the voltages are given as follows: • The edges on T’ get voltage 0. • An edge e = uv on a co-tree T’ get the voltage: • k(e) = S(u) + S(v) if e 2 T. • k(e) = S(u) + S(v) + h(e) if e T. • Each co-tree edge e defines a cycle C(e). The net voltage on C(e) is equal to k(e). • The voltages k(e), for e T’ span the whole Z 2 k.

Exercises • N 1. Let Znk be an elementary abelian group. Let S be a set of generators with the following property. Each element is a 0 -1 vector. They generate the whole group. • Show that |S| = k. • Show that there is an automorphism of the group mapping S to the standard generating set.

Real Homological Cover (0, 1) (1, 0) Z 2 2 • Let G be a graph with a given cycle basis C 1, C 2, . . . , Ck. Direct each cycle and assign to each edge of Ci the voltage ei 2 Znk. The final voltage assignmnet is given by adding the partial voltages. • An example is given on the left. The cycle basis is determined by a spanning tree.

Least Common Cover • Theorem: There exist finite connected graphs H 1, H 2, G 1, G 2 such that G 1 and G 2 are both double covers of H 1 and H 2. • Proof. We start with graphs G = G(5, 2) and X that we know from earlier.

G+X and G + G • Given two graphs G and H we form G+H by adding an edge between them. • On the left we see G + X and G + G. • The resulting graph depends on the choice of the two vertices.

H 1 and H 2 • Define H 1 and H 2 as follows: • H 1 = G + X and H 2 = G + X.

Covers of G+H. • A double cover of G+H can be split into two double covers G* and H* and then joint them by a pair of edges. We denote the resulting graph by G* ++ H*. • For instance KC(G + X) = KC(G) ++ KC(X) = G(10, 3) ++ G(10, 3).

End of Proof • Let G 1 = G(10, 3) ++ G(10, 3) and G 2 = G(10, 3) ++ 2 X. • G 1 and G 2 are distinct. They are both covers of H 1 and H 2.