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

• C4 over C3 is no good. However, C6 over C3 is Ok.

Fibers and Sheets.

• We say that C6 is a twosheeted cover over C3. Red vertices are in the same fiber. Similarly, the dotted lines belong to the saem fiber.

• Graph mapping f: C6 C3 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 Q3 is a two fold cover over complete graph K4.

• The vertex fibers are composed of pairs of antipodal vertices.

Covers over Pregraphs

• Graph K4 can be understood as a four-fold cover over 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) 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 Z2

on each of its edges. It can also be described as the tensor product KC(X) = X £ K2.

Homework

• H1: Prove that Kronecker cover is bipartite.• H2: Prove that generalized Petersen graph G(10,2)

is a twofold cover over the Petersen graph G(5,2).• H3: Determine the Kronecker cover over G(5,2).

• H4: 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 over 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

• N1: Prove that each double sheeted cover is regular.

• N2: Find an example of a three sheeted cover that is not regular.

• N3: Express the graph on the left as a 6-fold cover over 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 over n is bipartite too. Dipole 3 jeis cubic, sometimes called the theta graph .

Cyclic cover over a dipole – Haar graph H(n).

• H(37) is determined by number 37, actually by its binary representation (1 0 0 1 0 1).

• k = 6 is the length of the sequence, hence group Z6.

• (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).

0 3 5

Z6

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

over .– 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 over 1-cage.

11

Z2

K4, the 3-cage

• K4 is a Z4 covering over the 1-cage.

• In general, we obtain a Z2n 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?

2

Z4

1

0

1

2

3

K3,3, the 4-cage

• K3,3 is a Z6 covering over the 1-cage.

• It can also be seen as a Z3 covering over the 2-cage .

• Exercise: Express K3,3 as a covering graph over . Dtermine a natural number n, such that K3,3 is a Haar graph H(n).

3

Z6

1

0

1

2

3

4

5

The Handcuff Graph G(1,1)

• 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.1

0

Z2

I graphs I(n,i,j) and Generalized Petersen graphs G(n,k)

• 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.

i0

Zn

j

The 6-cage

• The 6-cage is the Heawood graph on 14 vertices. It is a 7-fold cyclic cover over the graph. But it is also a dihedral cover over the 1-cage.

• Let the presentaion of Dn be given as follows: Dn = <a,b|an,b2, ab=ba-1>

• Then the Heawood is a covering described on the left.

ba

D7

Exercises

• N1. Express the 7-cage as a covering graph.

• N2. 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(1,1,1,1,1).

Coverings over (3,1)-cubic graphs

• 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 I-graphs, Y-graphs and H-graphs: I(n,i,j),Y(n,i,j,k) and H(n,i,j,k,l).

Zni

i

i

j

j

j

k

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 over 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 = {B1, ..., 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• Theorem. Given any two finite graphs G and H, the

following statements are equivalent: 1. G and H have the same universal cover,2. G and H have a common finite cover,3. G and H have a common cover,4. G and H have the same minimal valence refinement.5. 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 K2,2,2.

• The construction resembles truncation.

• Each vertex is truncated and an inverse figure is placed in the space provided for it.

• Theorem: If G is planar, then THE(G) is planar.

Homework

• H1. Given connected graph G with n vertices and e edges and with valence sequence (d1, d2, ..., dn). Determine the parameters for THE(G).

• H2. 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 G\T 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 H1(G,Z). We may replace Z by Zk and obtain the first Zk homology group Zk

m-n+1.

Pseudohomological Covers

• Idea: Let G be a graph and T its spanning tree and with the edges H = {h1,h2,...,hm-n+1} = E(G)\E(T). Let (H) be a group with m-n+1 interchangeable generators H. The pseudohomological -cover HOM(G,) 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,) 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,Z2,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,Z2,T) is connected if and only if G is not a tree.

Example

• The two voltage graphs on the left determine different pseudohomological Z2 covers.

• Cov(G,2) is bipartite and Cov(G,1) is not.

1

1

1 1

0

00

0

0

0

Z2

1

2

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 G\T. Define the voltage assignment such that each non-tree edge gets the voltage ei = (0,0,..,0,1,0,...,0) 2 Zp

k.• 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 = {h1, h2, ..., 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 Z2k.

Exercises

• N1. 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

• Let G be a graph with a given cycle basis C1, C2, ..., Ck. Direct each cycle and assign to each edge of Ci the voltage ei 2 Zn

k. 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.

(0,1)

(1,0)

(0,1)

(1,1) (1,0)

Z22

Least Common Cover

• Theorem: There exist finite connected graphs H1, H2, G1, G2 such that G1 and G2 are both double covers of H1 and H2.

• 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.

H1 and H2

• Define H1 and H2 as follows:

• H1 = G + X + X and H2 = G + 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 G1 = G(10,3) ++ G(10,3) ++ G(10,3) and G2 = G(10,3) ++ G(10,3) ++ 2X.

• G1 and G2 are distinct. They are both covers of H1 and H2.