Spring Semester 2015–2016 MATH41072/MATH61072 Algebraic Topology Supplementary read- ing §B. Graphs on Surfaces

B.1 Definition. A graph G is a 1-dimensional simplicial complex (see Definition 4.3) which can be thought of as a finite set V = V (G) of vertices and a set E = E(G) of pairs of vertices called edges (see Definition 4.12). The degree of a vertex is the number of edges containing that vertex. A graph G is connected if for each pair of vertices v, v0 ∈ V (G) we can find a sequence of edges {v0, v1}, {v1, v2},... {vn−1, vn} ∈ E(G) such that v0 = v 0 and vn = v (cf. Definition 2.3(b)).

B.2 Examples. (a) The complete graph on n vertices Kn has all possible n
2 edges.

(b) The complete bipartate graph Km,n has a vertex set V = V1 t V2 where V1 consists of m vertices and V2 consists of n vertices and the edge set is  E = {v1, v2} | v1 ∈ V1, v2 ∈ V2 .

B.3 Definition. A graph G can be embedded in a surface S if the geo- metric realization of G (thought of as a 1-dimensional simplicial complex) is homeomorphic to a subspace of S. A graph which embeds in S2 is called a planar graph (since embeddability 2 2 in S is clearly equivalent to embeddability in R .

B.4 Remark. It is a consequence of the Triangulation Theorem (Theo- rem 2.7) that G may be embedded in S if and only if there is a triangulation of S, i.e. a simplicial surface K such that |K| =∼ S such that G is a subcom- plex of K. We shall take this result for granted in what follows. In considering embedding questions ‘free edges’ for which one of the vertices has degree 1 can always be easily added to an embedding so in what follows we only need to consider graphs without such edges.

2 B.5 Examples. (a) K4 embeds in S .

v2

v3 v1 v4

1 (b) K6 embeds in P1, the projective plane (represented as a hexagon with opposite sides identified as indicated).

v2 v1

v5 v4

v3 v3

v6

v1 v2

(c) K7 embeds in T1, the torus (represented as a square with opposite sides identified as usual).

v1 v2 v3 v1

v4 v6 v4

v5 v7 v5

v1 v2 v3 v1

B.6 Definition. The (oriented) genus γ(G) of a graph G is the minimal genus of an oriented surface in which G embeds, i.e. γ(G) = k when G embeds in Tk but not in Tk−1. So a planar graph has genus 0.

B.7 Definition. Given an embedding of a graph G in a surface S the path-components of the complement S \ G are called the regions of the embedding.

B.8 Proposition. Given an embedding of a connected graph G with no free edges and with v vertices and e edges in a surface S, the number of regions r of the embedding satisfies r 6 2e/3. Proof. Having done this, each region is bounded by at least three edges. Each edge is on the boundary of at most two regions. Hence e > 3r/2 which gives the result. 

2 B.9 Definition. Given an embedding of a graph in a surface such that each of the regions is homeomorphic to an open disc then we say that the embedding is a 2-cell embedding.

B.10 Remark. If you think of the surface as a sphere with handles and cross-caps attached (see Remarks 1.22) then you may have one or both ends of a handle attached within a region or you may have cross-cap within a region. In either case the embedding would not be a 2-cell embedding.

B.11 Theorem. Suppose that a connected graph G with v vertices and e edges has a 2-cell embedding with r regions in a surface S with Euler characteristic χ(S). Then v − e + r = χ(S).

Proof. We may assume that G has no free edges since removing a free edge reduces both v and e by 1 and so does not change v − e + r. We can extend the embedded graph to a triangulation of S as follows. Each region of the embedding together with the adjacent edges and vertices is a topological polygon. We can triangulate this polygon by adding a new vertex in the interior of the region and joining this by edges to the vertices of the polygon. Thus if the polygon has k edges and vertices, we will add one vertex, add k edges and divide the polygon into k triangular regions. So for the new embedded graph G0 we obtain by the process v0 − e0 + r0 = (v + 1) − (e + k) + (f + k − 1) = v − e + r. Applying process to each region of the 2-cell embedding we obtain a triangulation of the surface and so v − e + r = χ(S) from the definition of the Euler characteristic. 

B.12 Remark. By a similar argument we can show that, for any embed- ding of a graph with v vertices and e edges in a surface S (not necessarily a 2-cell embedding), v − e + r > χ(S). Where equality holds exactly in the case of a 2-cell embedding. The details are omitted. The basic idea is that the handles and cross-caps preventing the embedding from being 2-cell can be removed from S giving a new surface S0 in which the graph has a 2-cell embedding. The process of removing handles and cross-caps increases the Euler characteristic since it reduces the genus of the surface.

B.13 Corollary. If a graph G with v vertices and e edges embeds in a surface of Euler characteristic χ then χ 6 v−e/3 and so γ(G) > e/6−v/2+1.

Proof. Exercise. 

3 B.14 Example. For n > 3, γ(Kn) > (n − 3)(n − 4)/12.

Proof. Exercise. 

Minimal triangulations

B.15 Remark. It turns out that there is a connection between the embed- ding of complete graphs in surfaces and the problem of finding triangulations of a closed surface with the minimum number of triangles.

B.16 Theorem. If a triangulation of a closed surface S with Euler char- √ acteristic χ has v vertices then v > (7 + 49 − 24χ)/2. Proof. Suppose that the triangulation has e edges and f triangles. Then we know the following.

(i) v − e + f = χ (from Definition 3.1).

(ii) e 6 v(v − 1)/2 (since the maximum number of edges has every pair of vertices joined by an edge).

(iii) 2e = 3f (since each triangle has three edges and each edge is an edge of two triangles).

2 It follows that v − 7v + 6χ > 0 [Exercise]. √ √ From this it follows that v > (7 + 49 − 24χ)/2 or v 6 (7 − 49 − 24χ)/2 [Exercise]. The only possible solution of the second inequality (since v > 3 and χ 6 2) is χ = 2 and v = 3 which gives e = 3 and f = 2. This corresponds to two triangles with the same edges and vertices which would violate the intersection condition. So the result follows. 

√ B.17 Proposition. In Theorem B.16, if v = (7 + 49 − 24χ)/2 then e = v(v −1)/2 and so the 1-skeleton of the triangulation gives an embedding of Kv, the complete graph on v vertices.

Proof. Exercise. 

B.18 Remark. Corollary B.13 shows that, if the complete graph on v vertices embeds in a closed surface with Euler characteristic χ, then v2 − √ 7v + 6χ 6 0 which means that v 6 (7 + 49 − 24χ)/2. Details are left as an [Exercise]. The following was proved by Ringel and Youngs in 1968.

4 B.19 Theorem. The largest complete graph which may be embedded in a closed surface of Euler characteristic χ is the one with v = [(7 + √ 49 − 24χ)/2] (where [x] denotes the largest integer no greater than x) with the exception of the Klein bottle for which the largest embeddable complete graph is K6 rather than K7. [This number is known as the Heawood number because of Theorem B.23 below.]

√ B.20 Remarks. This means that in every case where v = (7+ 49 − 24χ)/2 is an integer (e.g. χ = 2, χ = 1, χ = 0, χ = −10) there is an embedding of the complete graph on v vertices in any surface of Euler characteristic χ which gives a minimal triangulation of the surface [see exercise] (with the exception of the Klein bottle P2 in which there is no embedding of K7 and for which the minimal triangulation has eight vertices.) The embeddings of 2 K4 in S , K6 in P1 and K7 in T1 giving minimal triangulations are given in Examples B.5.

The chromatic number of a surface

B.21 Remarks. One of the famous popular problems of mathematics was the ‘four colour conjecture’ which states that every map on the sphere (or the plane) may be coloured using only four colours so that regions with a common border have a different colour. This conjecture was first made in 1852 and over the years many false proofs were given. It is not difficult to prove that six colours are sufficient and in 1890 Heawood proved that five colours are sufficient. However, the result was only finally proved in 1976 by Appel and Haken (for some years afterwards letters from the University of Illinois were franked ‘four colors suffice’). The proof used computing techniques to check 1936 special configurations which it had been shown were sufficient. The remarkable thing is that the corresponding question for maps on surfaces other than the sphere is an easier problem. Notice that a map corresponds to a graph by putting one vertex in each region of the map and joining two vertices if their corresponding regions have a common boundary then we can reformulate the map colouring question in terms of graphs as follows. A free edge would only arise if one region was entirely within another. These are easy to deal with so we will assume that these do not arise in what follows.

B.22 Definition. A colouring of a graph a function c: V (G) → A to a set A such that, if hv1, v2i is an edge of the graph then c(v1) 6= c(v2). The chromatic number k(G) of a graph G is the cardinality of the smallest set A

5 so that it has a colouring V (G) → A. [We are thinking here of A as a set of colours.] The chromatic number k(S) of a closed surface S as the maximum chromatic number of the graphs embeddable in S. Thus the Four Colour Theorem states that k(S2) = 4.

B.23 Theorem [Heawood (1890). If G may be embedded in a closed surface S of Euler characteristic χ < 2 then p k(G) 6 N = [(7+ 49 − 24χ)/2] where [x] denotes the integer part of the real number x.

So k(S) 6 N.

B.24 Remarks. (a) Theorem B.19 shows that if χ(S) = χ then k(S) > N (apart from S = P2), the number in the above theorem, since k(Kn) = n 2 and so k(S) = N apart from S = P2 (in fact k(P2) = 6) or (possibly) S = S for which Heawood’s proof of Theorem B.23 does not apply. (b) When S = S2 the number χ = 2 and so N = 4. So the corresponding result to Theorem B.21 in this case is the Four Colour Theorem. √ Proof of Theorem B.23. Put x = (7 + 49 − 24χ)/2 so that N = [x]. Then x(x − 7) + 6χ = 0. Suppose that a connected graph G with v vertices embeds in S. The degree of a vertex is the number of edges connected to this vertex. We prove that G must have a vertex of degree 6 N −1 so that theorem follows by induction on the number of vertices since, if we suppose that graphs of a smaller number of vertices have a chromatic number no greater than N then we can extend a colouring of the graph without this vertex to a colouring of the whole graph by selecting a colour for this vertex different from the colours of the adjacent vertices. This is certainly possible if v 6 N and so we may suppose that v > N which means that v > x. Now the average degree of the vertices of the graph is 2e/v since each edge has two vertices. It follows from Remark B.12 that 2e/v 6 6(1 − χ/v). [Exercise.] Now, if χ = 1 this implies that 2e/v < 6 and so G has a vertex of degree < 6. But in this case N = 6 and so the above inductive argument works. Otherwise, χ 6 0 and so 2e/v 6 6(1 − χ/v) 6 6(1 − χ/x) (since we are 2 supposing that v > x) = x − 1 (since x − 7x + 6χ = 0). Thus 2e/v 6 x − 1 and so G has a vertex of degree 6 x−1 and so of degree 6 N −1 as required to carry out the inductive argument. 

6 Exercises

1. Prove Corollary B.13.

2. Derive the result in Example B.14. Hence prove that γ(K7) = 1.

3, Prove that γ(K3,3) = 1. [Hint: To prove that K3,3 is not planar (the well-known ‘gas, water and electricity supply puzzle’) it is necessary to mod- ify the argument in the proof of Proposition B.8 to get an improved result for this graph.]

4. Complete the proof of Theorem B.16.

5. Prove Proposition B.17.

6. Complete the proof of Theorem B.23.

√ 7. For which values of χ is the Heawood number (7 + 49 − 24χ)/2 an integer?

8. Where does the proof of Theorem B.23 break down for S2? What does this proof tell you about the chromatic number of S2?

9. Verify that the triangulation in Example B.5(b) has K6 as its 1-skeleton and is unique (up to isomorphism) with this property.

10. Verify the claim of Remark B.18, i.e show that for the embedding of 2 √ the complete graph Kv one has v − 7v + 6χ 6 0 and v 6 (7 + 49 − 24χ)/2

11. To check the claim of Remark B.20

(a) Show that for the embedding of the complete graph Kv into a surface 2 of Euler characteristic χ fulfilling χ = 7v−v or equivalently v = (7 + √ 6 49 − 24χ)/2 the embedding fulfils the following necessary conditions to be the 1-skeleton of a triangulation:

• the embedding is 2-cell, • every region is adjacent to exactly three edges, • every edge lies in the boundary of exactly two regions.

7 (b) If the graph is the 1-skeleton of a triangulation, then this triangulation is minimal (with respect to the number of triangles).

8