390 likes | 578 Views
Maps. Graphs on Surfaces. We are mainly interested in embeddings of graphs on surfaces: h : G ! S. An embedding should be differentiated from immersion. On the left we see some forbidden cases for embeddings. Cellular (or 2-cell) embedding.
E N D
Graphs on Surfaces • We are mainly interested in embeddings of graphs on surfaces: • h: G ! S. • An embedding should be differentiated from immersion. • On the left we see some forbidden cases for embeddings.
Cellular (or 2-cell) embedding • Embedding h:G ! S is cellular (or 2-cell), if S \ h(G) is a union of open disks. • A 2-cell embedding is strong (or proper) if the closure of each open disk is a closed disk. • Proposition: Only connected graphs admit 2-cell embeddings.. • On the left we see two embeddings of K4 in torus S1. The first one is cellular, the second ons is not!
2-Cell Embeddings and Maps • 2-cell embeddings of graphs are also known as maps. There is a subtile difference in the point of view. • In the former the emphasis is given to the graph while in the latter the emphasis is in the map, a structure, composed of vertices, edges and faces. Examples of maps include surfaces of polyhedra. • Maps include different, equivalent, cryptomorphic purely combinatorial definitions that can be used as a foundation of a theory of maps that is independent of topology.
Genus of a Graph • Let g(G) denote the genus of a graph G. This parameter denotes the minimal integer k, such that G admits an embedding into an orientable surface of genus k. • Note: (G) = 0 if and only if G is planar.
Euler Characteristics • To each closed surface S we associate a number (S) called Euler characteristics of S. • (Sg) = 2 – 2g, for orientable surface of genus g. • (Nk) = 2 – k, for non-orientable surface of crosscap number (non-orientable genus) k.
Euler Formula • Let G be a graph with v vertices, e edges cellularly embedded in surface S with f faces. Then • v – e + f = (S).
Rotation Scheme • Let G be a connected graph with the vertex set V, with arcs S and edges E. For each v 2 V define the set: S[v] = {s 2 S| i(s) = v}. Let and be mappings: • r: S ! S • l: S !{-1,+1}. • with the property: • Permutation r acts cyclically on S[v], for each v 2 V. • l(s) = l-1(r(s)), for each s 2 S. [Hence is a voltage assignment. In our case: l(s) = l(r(s))]. • The triple (G,r,l) is a called a rotation scheme, defining a 2-cell embedding of G into some surface.
Interpretation of Rotation Scheme • We follow arcs starting at s0 until we return to the initial arc. • s à s0, • s Ã(s). • positive à True. • While s s0 do • If positive then • If (s) = 1 then • s Ã(s) • else • positive à False; • s Ã(s)-1 • else • If (s) = 1 then • s Ã(s)-1 • else • positive à False; • s Ã(s) r(s) r2(s) s r(s) r3(s) r4(s) r(r(s)) r(s) r2(s) s r(s) r3(s) r4(s) r(r(s))
Rotation Scheme and Rotation Projection • Rotation scheme can be represented by rotation projection. • Rotation r can be reconstructed from the bottom drawing. Each arc s carries l(s) = 1.
Example • On the left we see the rotation projection of K4. The faces are triangles. • There is no cycle with an odd nunber of “crosses”. • V – E + F = 4 – 6 + 4 = 2. • The surface is a sphere! • Exercise: Analyse the faces of the embedding if all crosses are removed from the figure on the left.
Main Fact • Theorem: Any 2-cell embedding of a graph G into a surface S can be described by a rotation scheme (G,,). Furthermore, by face tracing algorithm the number of faces F can be computed yielding (S). Finally, S is non-orientable if and only if G contains a cycle • C = (e1,e2, ... , ek) such that • (C) := (e1) (e2) ... (ek) = -1
Combinatorial Theory of Maps • There are several cryptomorphic definitions of maps (graphs on surfaces.) • Rotation schemes represent such a tool. • Note that we start with a graph G and add additional information (G,,) in order to describe its 2-cell embedding. In some closed surface. • We may also start directly from maps or polyhedra.
f v e Flag Systems • Let V,E,F be disjoint (finite) sets. • Fµ V £ E £ F is a flag system. Here: • V vertex set, • E edge set • F face set. • A face that is a polygon with d sides, (a d-gon), consists of 2d flags (see figure on the left!)
Flag Systems are General • Using flag systems we can describe general complexes such as books. • Note the a 3-book contains a non-orientable Möbious strip.
Flag systems from 2-cell embeddings • To a 2-cell embedding we associate a flag system as follows. Let V be the set of vertices, E, the set of edges and F the set of faces of the embedding. Define • µ V £ E £ F as follows: • (v,e,f) 2 if and only if v, e, and f are pairwise incident.
The 1-skeleton of a flag system. • Given a flag system µ V £ E £ F, we may study its projection to the first two factors: • A = {(v,e)| (v,e,f) 2}. • Define: • i:A ! V by i: (v,e) v and • Ve = {v 2 V| (v,e) 2 A}. • Assume |Ve| · 2, for each e 2 E. • We may define r:A ! A by: • r(v,e) = (w,e) if Ve = {v,w} and • r(v,e) = (v,e) if Ve = {v}. • The quadruple (V,A,i,r) is a pre-graph. It is called the 1-skeleton of . • Given there is an easy test whether the 1-skeleton is indeed a graph: for each e 2 E we must indeed have |Ve| = 2.
1-co-skeleton • If we replace the role of V and F in a flag system µ V £ E £ F we obtain a 1-co-skeleton. • We say that the skeleton and co-skeleton are dual graphs.
Homework H1: If one of 1-skeleton is a graph is the 1-co-skeleton a graph too? Prove or find a counterexample.
Exercises • N1. Determine the flag system describing the four-sided pyramid. • N2. Determine the 1-skeleton and 1-co-skeleton for N1. • N3. Define the notion of automorphism of a flag system . For the case N1 find the orbits of Aut .
When does a flag system define a surface? • As we have seen in the case of a book we may have an edge belonging to more than two faces. This clearly violates the rule that each point on a surface has a neighborhood homeomorphic to an open disk. • Therefore a necessary condition is: • Each for each flag (v,e,f) 2 there must exist a unique triple (v’,e’,f’) 2 V £ E £ F with v’ v, e’ e, f’ f such that (v’,e,f), (v,e’,f),(v,e,f’) 2. • Another obvious condition is that the 1-skeleton must be connected. • However, a flag system satisfying these two conditions may still represent more general spaces than surfaces. • It may represent a pseudosurface. • Let us define: • v = {(f,e)| (v,e,f) 2}. • e = {(v,f)|(v,e,f) 2}. • f = {(v,e}| (v,e,f) 2}. • Each of the three structures defined above can be represented as graph. More presicely, each of them is regular 2-valent graph. • is a surface if and only if each graph v, e and f is connected.
Limits of flag systems • Unfortunately, there are connected graphs whose 2-cell embeddings cannot be represnted by flag systems. • Proposition. Let G be a connected graph. If G contains a loop or a bridge no 2-cell embedding of G can be described by flag systems. • [A bridge is an edge whose removal disconnects the graph.]
Some limits of flag systems • On the left we see K4 embedded in torus with one 4-gon and one 8-con. • Green and red flag have all three matching components equal. • This map cannot be described by flag systems.
Self-avoiding maps • Theorem: A 2-cell embedding of G in some surface can be described by a flag system if an only if neither G nor its dual contains a loop. • A map that satisifies the conditions of this theorem will be called self-avoiding.
Flags, from a different view-point. • Let us forget about V,E, F for a moment. Let the set of flags F be given. • For instance, on the left, we see them as triangles. • Define the flag graph G(F): • V() = . • f ~ f’ if and only if triangles have a common side.
From flags to flag graph. First the vertices.
From flags to flag graph. First the vertices. Next: three kinds of new edges: along the edges across the edges. across the angles.
Flag graphs for 2-cell embedded graphs. Flag graph is: - connected - trivalent - contains a 2-factor of form m C4.
Flag graphs for 2-cell embedded graphs. A practical guide to the construction. The first step when rectangles are placed on each edge is shown.
Yet another view to flag graphs. • We may start with three involutions: • t0, t1, t2 : F!F • 02 = 12 = 22 = 1, each fixed-poit free. • t0t2 = t2t0, also fixed-point free. • Each invoultion corresponds to a 1-factor. Together they define a cubic graph: the flag graph (). • The group <0,1,2>, called monodromy group must act transitively on . [This is eaquivalent to saying that () is connected.] • These axioms define a (combinatorial) map on a surface.
Combinatorial Map. • Combinatorial map is defined by three involutions satisfying the axioms from the previous slide. • Orbits of <2,1> acting on define V. • Orbits of <0,2> acting on define E. • Orbits of <0,1> acting on define F.
Orientable Map • Theorem: A map is orientable if and only if the flag graph is bipartite.
Unique Embedding • Theorem (Whitney): Each 3-connected planar graph admits a unique embedding in the sphere. • Theorem (Mani). Let Aut G be the group of automorphism of a 3-connectede planar graph G and let Aut M be the group of automorphisms of the corresponding map. Then Aut G = Aut M.
Example - Exercises • On the left there is an embedding of Q3 on torus. • N1: Determine the rotation scheme for this embedding. • N2: Determine the flag graph for this embedding.
Example - Exercises • On the left there is a different embedding of Q3 on torus. • N1: Determine the rotation scheme for this embedding. • N2: Determine the flag graph for this embedding. • .
Levi graph of a map • Levi graph of a map M has the vertex set: • VM t EM t FM, • Edges are determined by the sides of flags (as triangles). • WARNING: The graph on the left is not simple!!
Characterisation • Theorem: Levi graph of a map is simple if neither 1-skeleton nor 1-co-skeleton has a loop. • Definition: A map M is simple,if and only if its Levi graph is simple.
Homework • H1: Given Flag graph of a map M, determine whether M is simple! (Prove previous theorem)