REU LECTURES ON COMBINATORIAL TOPOLOGY
JENS HARLANDER
1. Graphs
A (directed) graph Γ consists of a set of vertices V and a set of edges E ⊆ V × V . If
e = (v, w) is an edge, then v is the initial vertex of e and w is the terminal vertex of e.
The case v = w is allowed. A graph can be drawn in the plane in many different ways. If
a graph can be drawn in the plane without edge crossings, we say the graph is planar. It
is actually better to think of a planar graph as drawn on a 2-sphere S 2 . Not all graphs are
planar. We will look at details in a moment. Given a planar graph drawn on a 2-sphere,
then number of vertices − number of edges + number of faces = 2. It is remarkable that
this alternating sum, called the Euler characteristic, does not depend on how the graph is
drawn on the sphere. Try to prove this. Using the Euler characteristic it is not difficult to
give examples of graphs that are not planar.
The graph K3,3 has three upper vertices and three lower vertices, and every upper vertex
is connected by an edge to every lower vertex. This gives a total of nine edges. This graph
is not planar. Suppose we could draw the graph on a sphere. Then 6 − 9 + f = 2 where f
is the number of faces. So f = 5. Thus we have tessellated the sphere with 5 faces, each
face having at least 4 edges in its boundary because K3,3 does not contain reduced cycles
of length less than 4, each vertex having valency 3. Let v(Fi ) be the number of vertices
in
P5the boundary of the face Fi . Since the valency of every vertex is exactly 3, we have
i=1 v(Fi ) = 3 number of vertices = 18. This is a contradiction because v(Fi ) ≥ 4.
Theorem 1.1. (Kuratowski’s Theorem) A graph is planar if and only it does not contain
a subdivision of K3,3 or of K5 , the complete graph on 5 vertices.
A link diagram is a planar 4-regular planar graph with over- and under-crossing information provided at the vertices. A virtual link diagram is a 4-regular graph with over- and
under-crossing information provided at the vertices. Edge crossings that are not vertices
but come from the fact that we draw a (possibly) non-planar graph in the plane are typically circled. Reidemeister moves provide a good (why?) notion of equivalence on the set
of link diagrams. Generalized Reidemeister moves also exist in the world of virtual knot
diagrams.
Date: Summer 2012.
1
2
JENS HARLANDER
2. 2-Complexes
A 2-complex consists of a directed graph together with a set of closed edge paths. Along
these edge paths one can attach discs, also called 2-cells.. There are various 2-complexes
associated with a virtual link diagram, such as the Wirtinger complex and the Dehn complex. A tiny creature standing at a vertex in a 2-complex looking out towards the horizon
would see a graph, called the link at that vertex. Let us be more precise. Consider a vertex
v in a 2-complex K. If e is edge starting at v mark a point e− on e close to v. If e is an
edge ending at v mark a point e+ on e close to v. The points arising in this fashion will
make up the vertex set of the link at v. Now suppose d is a 2-cell with the vertex v in its
boundary. Two edges, say e1 and e2 in the boundary of d will contain the vertex v. Draw
a corner in d around v. That corner is a edge in the link at v. Best to draw a picture and
look at many examples.
Topology is the study of spaces and continuous maps between spaces, up to deformations.
Of central importance is the study of maps from spheres into spaces, so called homotopy
theory. Some of these notions can be captured combinatorially.
A map S 1 → Γ, where Γ is a graph, can be encoded by a circle diagram, that is,
a subdivided circle, the edges in the circle are directed and labeled by an edge in Γ.
We require that the edge sequence obtained by reading off the labels along the circle in
clockwise direction is a closed edge path in Γ. A circle diagram is called reduced if we
do not encounter a labeling sub-sequence ee−1 or e−1 e. Note that a graph admits only
reducible circle diagrams if and only if that graph is a tree.
A map S 2 → K, where K is a 2-complex, can be encoded by a spherical diagram, that
is a tessellated sphere. The edges in that tessellation are oriented and labeled by the edges
in the 1-skeleton of K (that is the graph underlying K). We require that reading off the
labels along the boundary of a face yields a cyclic permutation of the boundary of a 2cell of K or its inverse. A spherical diagram is called reduces if there is no edge in the
tessellation that bounds two faces with boundary words that are inverse to each other.
A 2-complex is called diagramatically reducible, DR for short, if it only admits reducible
spherical diagrams. Here is a good exercise: show that a product of graphs is a 2-complex
that is DR. We will come back to this when we talked about curvature and DR tests.
Spherical pictures over Wirtinger 2-complexes of virtual knots can be represented by
labeled links.
There are important open questions concerning spherical diagrams. There are other
types of reductions one can do on spherical diagrams, so called bridge and diamond moves.
A 2-complex K is called aspherical if every spherical diagram over K can be reduced by all
allowed moves to the empty diagram. A famous open problem in low dimensional topology
is the Whitehead conjecture: A subcomplex of an aspherical 2-complex is aspherical. One
dimension lower this statement translates to the statement “a subgraph of a cycle free
graph is cycle free ”, which is obviously true. At the heart of Whitehead’s conjecture is the
REU LECTURES ON COMBINATORIAL TOPOLOGY
3
following open question: Is the Wirtinger complex of a long virtual knot aspherical? It is
known that not all Wirtinger complexes are DR. It is known that the Wirtinger complex of
a long alternating prime virtual knot can be deformed to a 2-complex that is DR. Here is a
key problems you will be working on for this REU: find ways to construct reduced diagrams
over Wirtinger complexes of long virtual knots. There are known techniques developed by
Rosebrock, but they seem to be special. There are sporadic spherical diagrams that have
been found by computer search, but the nature of these diagrams remain mysterious.
3. Curvature
Suppose S is a tessellated closed surface. Assign angles to the corners of the tiles (or
faces). We will define P
combinatorial curvature at vertices and at the faces. If v is a vertex
of S then κ(v) = 2π − αi , where the
Pαi are the angles assigned to the corners around the
vertex v. If d is a face, then κ(d) =
βi − (k − 2)π, where the βi are the interior angles
of the face d and k is the number of edges in the boundary of d.
Theorem 3.1. (combinatorial Gauss-Bonnet)
X
X
κ(v) +
κ(d) = 2πχ(S).
Proof. Exercise.
Recall the the Euler characteristic of S, χ(S), is the alternating sum number of vertices−
number of edges + number of faces. For example if S is the 2-sphere, then χ(S) = 2, and
if S is the torus, then χ(S) = 0. Using this result one can prove Stalling’s lemma.
Lemma 3.2. Let S be a tessellated 2-sphere. Give the 1-skeleton an orientation. Then
one of the following has to occur:
• S contains a face with consistently oriented boundary;
• S contains a sink vertex;
• S contains a source vertex.
Proof. We assign angles to the corners of the tessellation in the following way. Consider a
face d. Give a corner in d angle 0 if it is a corner with a source or sink vertex. Otherwise
P
assign the angle π. If the boundary of a face d id not consistently oriented, then
βi ≤
(k − 2)π, where the βi are the interior angles ofPd (which are π or 0.) Hence κ(d) ≤ 0. If v
is a vertexP
that is not a sink or a source, then
αi ≥ 2π. Hence κ(v) ≤ 0. It follows that
P
κ(v) + κ(d) ≤ 0, which is a contradiction because χ(S 2 ) = 2.
The weight test, an asphericity test, or rather a DR test for 2-complexes is based on
these ideas.
Theorem 3.3. Consider a 2-complex K. Assigne angles to the corners in the faces of K.
Suppose the following is true:
(1) κ(d) ≤ 0 for all faces d;
(2) if z = c1 ...cn is a reduced closed edge loop in
P the link of some vertex, and αi is the
angle assigned to the corner ci , then 2π − αi ≤ 0.
4
JENS HARLANDER
Then the 2-complex K is DR.
Proof. Consider a reduced spherical diagram S over K. The angle assignment for K implies
one on S. The two conditions in the theorem imply that the curvature at a vertex and
at a face is less or equal to zero. Hence, by the combinatorial Gauss-Bonnet, χ(S) ≤ 0, a
contradiction to the fact that S is a tessellated 2-sphere.
We need to look at many examples.