INCIDENCE GEOMETRIES CHAPTER 4. Contents 1.Motivation 2.Incidence Geometries 3.Incidence Geometry...
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INCIDENCE GEOMETRIES
CHAPTER 4
Contents
1. Motivation
2. Incidence Geometries
3. Incidence Geometry Constructions
4. Residuals, Truncations - Sections, Shadow Spaces
5. Incidence Structures and Combinatorial Configurations
6. Substructures, Symmetry and Duality
7. Haar Graphs and Cyclic Configurations
8. Algebraic Structures
9. Euclidean Plane, Affine Plane, Projective Plane
10. Point Configurations, Line Arrangements and Polarity
1. Pappus and Desarguers Theorem
2. Existence and Countnig3. Coordinatization4. Combinatorial Configurations
on Surfaces5. Generalized Polygons6. Cages and Combinatorial
Configurations7. A Case Study – The Gray Graph8. Another Case Study - Tennis
Doubles9. Martinetti-Boben Theorem
1. Motivation
Motivation
• When Slovenia joined the European Union it obtained 7 seats in the Parliament of the European Union. In 2004 the first elections to the European Parliament in Slovenia were held.
• There were 13 political parties (7 parliamentary parties: 1, 2, 3, 4, 5, 6, 7, and 6 non-parliamentary parties: A, B, C, D, E, F) competing for these seats. TV Slovenia decided to cover the campain by hosting political parties in 6 TV shows: a,b,c,d,e,f.
• TV asked mathematicians to help them select the guests in a fair way.
Motivation
• With a little help from mathematicians TV came up with the following schedule.
a b c d e f
A B C D E F
1 2 3 1 2 3
4 6 4 7 5 6
5 7
Example – TV coverage of EU parliamentary elections in Slovenia
TV Shows Parties
aA 1 4 5
bB 2 6
cC 3 4
dD 1 7
eE 2 5
fF 3 6 7
Model
• We can model the above schedule as follows:
• Let P = {1,2,3,4,5,6,7,A,B,C,D,E,F}• Let L = {a,b,c,d,e,f}• Let I ½ P £ L such that• (p,L) 2 I if and only if political party p
appears in the show L.• I = {(A,a), (1,a), (4,a), (5,a), ... }
Incidence structure
• An incidence structure C is a triple– C = (P,L,I) where
• P is the set of points,
• L is the set of blocks or lines
– I P L is an incidence relation.– Elements from I are called flags.
Levi Graph
• The bipartite incidence graph (C) with black vertices P, white vertices L and edges I is known as the Levi graph of the structure C.
Levi graph for the Election structure
• On the left there is the Levi graph for the incidence structure of the media coverage of the European Union Parliament elections in Slovenia.
• Each parliamentary party appears twice and each non-parliamentary party appears once. (check valence!)
a
e
c
bE
f
B
A
F
C
154
6 73
2D
d
Menger graph
• Given an incidence structure C = (P,L,I) we say that two points p and q are collinear, if there is a line L that contains both of them.
• Menger graph M(C).• Vertices P
• p ~ q if and only if p and q are collinear.
Menger Graph from Levi Graph
• There is a simple procedure for computing M from L. Take the pure graph power L(2). It is obtained from L by taking the same vertex set and making two vertices adjacent in L(2) if and only if they are at distance two in L. Since L is bipartite L(2). has (at least) two components. The one defined on the black vertices (corresponding to points of the incidence structure) is Menger graph M. The other one is called dual Menger graph.
Menger graph for the Election structure
• On the left there is the Menger graph for the incidence structure of the media coverage of the European Union Parliament elections in Slovenia.
E
B
A
F
C
1
5
4
6
73
2 D
Configuration Graph
The configuration graph K is the complement of the Menger graph. The dual configuration graph is the complement of the dual Menger graph.
Dual Configuration graph for the Election structure
• On the left there is the dual configuration graph for the incidence structure of the media coverage of the European Union Parliament elections in Slovenia.
a
e
c
b
f
d
Dual Configuration graph for the Election structure
• The Hamilton path abcdef in the dual configuration graph guarantees that no political party appears in two consecutive TV shows. a
e
c
b
f
d
Examples
• 1. Each graph G = (V,E) is an incidence structure: P = V, L = E, (x,e) 2 I if and only if x is an endvertex of e.
• 2. Any family of sets F µ P(X) is an incidence structure. P = X, L = F, I = 2.
• 3. A line arrangement L = {l1, l2, ..., ln} consisting of a finite number of n distinct lines in the Euclidean plane E2 defines an incidence structure. Let V denote the set of points from E2 that are contained in at least two lines from L. Then: P = V, L = L and I is the point-line incidence in E2.
Exercises 1
• N1. Draw the Levi graph of the incidence structure defined by the complete bipartite graph K3,3.
• N2. Draw the Levi graph of the incidence structure defined by the power set P({a,b,c}).
• N3. Determine the Levi graph of the incidence structure, defined by an arrangement of three lines forming a triangle in E2.
• N4 Determine the Levi graph of the line arrangement on the left.
2. Incidence Geometry
Incidence geometry
• An incidence geometry (G,c) of rank k is a graph G with a proper vertrex coloring c, where k colors are used.
• Sometimes we denote the geometry by (G,~,T,c). Here c:V(G) ! T is the coloring and |T| = k is the number of colors, also known as the rank of G. The relation ~ is called the incidence.
• T is the set of types. Note that only objects of different types may be incident.
Morphisms or representations
• Given two incidence geometries (G,~,c,T) and (G',~,c',T') a pair (f,g) of mappings
• f: G ! G' and • g: T ! T' is called a morphism of geometries
(or representation) if the following is true:• for any v 2 V(G) c'(f(v)) = g(c(v)).• for any u,v 2 V(G): if u ~ v then either f(u) = f(v) or
f(u) ~ f(v).
Special morphisms• Some morphisms have nice properties and deserve special attention.• We call a representation dimension-preserving if
• for any u,v 2 V(G): if u ~ v then f(u) ~ f(v).
• We call a representation faithful or strong if • for any u,v 2 V(G): u ~ v if and only if f(u) ~ f(v).
• A faithful representation in which both f and g are injective is called realization.
• A morphism is an isomorphism if both f and g are bijections and the inverse pair (f-1,g-1´) is a morphism too.
• The image of a representation is geometry. The image of a realization is isomorphic to the original.
• For stirng geometries we seek representations and realizations in sets. Vertices are mapped to the elements (or singletons) of S and the faces to subsets of S. The incidence ui ~ uj, i < j is represented by inclusions S(u i) µ S(uj).
Automorphisms
• There are two types of automorphisms in a geometry (G,~,c,T). Aut0 G contains type-preserving automorphisms. (g = id). Aut G contains all (extended) automorphisms.
• In the case of string geometries we want the linear order on T to be respected (or reversed). In the case of extended automorphisms we speak of dualities that map faces of rank r to rank n-r.
Examples
• 1. Each incidence structure is a rank 2 geometry. (Actualy, look at its Levi graph.)
• 2. Each 3 dimensional polyhedron is a rank 3 geometry. There are three types of objects: vertices, edges and faces with obvious geometric incidence.
• 3. Each (abstract) simplicial complex is an incidence geometry. Incidence is defined by inclusion of simplices.
• 4. Any complete multipartite graph is a geometry. Take for instance K2,2,2, K2,2,2,2, K2,2, ..., 2. The vertex coloring defining the geometry in each case is obvious.
Pasini Geometry
• Pasini defines incidence geometry (that we call
Pasini geometry) in a more restrictive way.– For k=1, the graph must contain at least two vertices: |
V(G)|>1.
– For k>1:• G has to be connected,
• For each x V(G) the (k-1)-colored graph (Gx,c), called residuum, induced on the neigbors of x is a Pasini geometry of rank (k-1).
String geometries
• A geometry G over the set of types T = {-1,0,1, ..., n} is called a string geometry if the following (1-2) is true (the elements of G are called faces, faces of type 0 are called vertices (or points), faces of type 1 are called edges (or lines), faces of type n-1 are called facets.). It is called pure string geometry if (1-3) is true.
1. There are exactly two improper faces u-1 2 V(G) of type -1 and one element un 2 V(G) of type n (both incident with every other face). The rest are called proper faces.
2. If ui, uj, uk are elements of respective types i < j < k and ui ~ uj, uj ~ uk, then ui ~ uk.
3. Every collection of mutually incident faces U can be extended to a sequence of (n+2) mutually incident faces. (In other words: all chambers have rank n+2.)
Incidence geometries of rank 2
• Incidence geometries of rank 2 are simply bipartite graphs with a given black and white vertex coloring.
• Rank 2 Pasini geometries are in addition connected and the valence of each vertex is at least 2: (G) >1.
Example of Rank 2 Geometry
• Graph H on the left is known as the Heawood graph.
• H is connected• H is trivalent: (H) =
(H) = 3.• H is bipartite.• H is a Pasini
geometry.
Another View• The geometry of the
Heawood graph H has another interpretation.
• Rank = 2. There are two types of objects in Euclidean plane, say, points and curves.
• There are 7 points, 7 curves, 3 points on a curve, 3 curves through a point.
• The corresponding Levi graph is H!
In other words ...
• The Heawood graph (with a given black and white coloring) is the same thing as the Fano plane (73), the smallest finite projective plane.
• Any incidence geometry can be interpeted in terms of abstract points, lines.
• If we want to distinguish the geometry (interpretation) from the associated graph we refer to the latter as the Levi graph of the corresponding geometry.
Simplest Rank 2 Pasini Geometries
• “Simplest” geometries of rank 2 in the sense of Pasini are even cycles. For instance the Levi graph C6 corresponds to the triangle.
Cycle(Levi Graph)
Triangle(Geometry)
Rank 3
• Incidence geometries of rank 3 are exactly 3-colored graphs.
• Pasini geometries of rank 3 are much more restricted. Currently we are interested in those geometries whose residua are even cycles.
• Such geometries correspond to Eulerian surface triangulations with a given vertex 3-coloring.
Flag System as Geometries
• Any flag system µ V £ E £ F defines a rank 3 geometry on X = V t E t F. There are three types of elements and two distinct elements of X are incicent if and only if they belong to the same flag of .
Self-avoiding maps
• Recall that a map is self-avoiding if and only if neither the skeleton of the map nor the skeleton of its dual has a loop.
Self-avoiding maps as Geometries of rank 4
• Consider a generalized flag system µ V £ E £ F £ P that defines a rank 4 geometry on X = V t E t F t P.
• There are four types of elements and two distinct elements of X are incident if and only if they belong to the same flag of .
• We may take any self-avoiding map M and the four involutions 0,1,2 and 3 and define a geometry as above.
Exercises 2• N1. Prove that the Petrie dual of a self-
avoiding map is self-avoiding.• N2. Prove that any operation Du,Tr,Me,Su1, ...
of a self-avoiding map is self-avoiding.• N3. Prove that BS of any map is self-avoiding.• N4. Show that any self-avoiding map may be
considered as a geometry of rank 4 (add the fourth involution).
Homework 2
• H1 Describe the rank 4 geometry of the projective planar map on the left.
3. Incidence Geometry Constructions
Geometries from Groups
• Let G be a group and let {G1,G2,...,Gk} be a family of subgroups of G.
• Form the cosets xGt, t 2 {1,2, ..., k}.
• An incidence geometry of rank k is obtained as follows:
• Elements of type t 2 {1,2,...,k} are the cosets xGt.
• Two cosets are incident: xGt ~ yGs if and only if xGt Å yGs ;.
Q – The Quaternion Units
QQ 11 -1-1 ii -i-i jj -j-j kk -k-k
11 1 -1 i -i j -j k -k
-1-1 -1 1 -i i -j j -k k
ii i -i -1 1 k -k -j j
-i-i -i i 1 -1 -k k j -j
jj j -j k -k -1 1 i -i
-j-j -j j -k k 1 -1 -i i
kk k -k j -j -i i -1 1
-k-k -k k -j j i -i 1 -1
Geometry from Quaternions
• Example: Q = {+1,-1,+i,-i,+j,-j,+k,-k}.• Gi = {+1,-1,+i,-i}, Gj = {+1,-1,+j,-j}, Gk ={+1,-
1,+k,-k}.
Quaternions - Continiuation
• The Levi graph is an octahedron.
• Labels on the left:• i = {+1,-1,+i,-i}
• j,k = {+j,-j,+k,-k}, etc.
i
jk
j,k
i,k i,j
Quaternions– Examle of Rank 4 Geometry.
• Levi graph was an octahedron.
• Notation:• i = {+1,-1,+i,-i}
• j,k = {+j,-j,+k,-k}, etc.
• If we add the sugroup G0 = {+1,-1}, four more cosets are obtained:
• Additional notation:• 1 = {+1,-1},i’={+i,-i}, etc.
i
kj
j,k
i,k i,j
11
i’i’
k’k’ j’j’
Reye’s Configuration
• Reye’s configuration of points, lines and planes in 3-dimensional projective space consists of
• 8 + 1 + 3 = 12 points (3 at infinity)
• 12 + 4 = 16 lines
• 6 + 6 = 12 planes.
P=12 L=16
P=12 - 4 6
L=16 3 - 3
6 4 -
Theodor Reye
• Theodor Reye (1838 - 1919), German Geometer.
• Known for his book Geometrie der Lage (1866 and 1868).
• Published his famous configuration in 1878.
• Posed “the problem of configurations.”
Centers of Similitude• We are interested in
tangents common to two circles in the plane.
• The two intersections are called the centers of similitudes of the two circles. The blue center is called the internal, the red one is the external center.
• If the radii are the same, the external center is at infinity.
Reye’s Configuration -Revisited
• Reye’s configuration can be obtained from centers of similitudes of four spheres in three space (see Hilbert ...)
• Each plane contains a complete quadrangle.
• There are 2 C(4,2) = 2 4 3/2 = 12 points.
Exercises 3-1
• N1. Consider the geometry defined by Z3 and Z5 in Z15. Draw its Levi graph.
• N2. Draw the Levi graph of the geometry defined by all non-trivial subgroups of the symmetric group S3.
• N3. Draw the Levi graph of the geometry defined by all non-trivial subgroups of the group Z2
3.
Exercises 3-2
• N4. Let there be three circles in a plane giving rise to 3 internal and 3 external centers of similitude. Prove that the three external centers of similitude are colinear.
4. Residuals, Truncations - Sections, Shadow Spaces
Residual geometry
• Each incidence geometry• =(, ~, T,c)• (,~) a simple graph• c, proper vertex coloring,• T collection of colors.• c: V( ! T
• Each element x 2 V( determines a residual geometry x. defined by an induced graph defined on the neighborhood of x in .
x
x
Flags and Residuals
• In an incidence geometry a clique on m vertices (complete subgraph) is called a flag of rank m.
• Residuum can be definied for each flag F ½ V(). (F) = Å{(x) = x |x 2 F}.
Chambers and Walls
• A maximal flag (flag of rank |T|} is called a chamber. A flag of rank |T|-1 is called a wall.
• To each geometry we can associate the chamber graph:
• Vertices: chambers• Two chambers are adjacent if and only if they share a common
wall.
• (See Egon Shulte, ..., Tits systems)
The 4-Dimensional Cube Q4.
0000
1000
0010
0100
0001
Hypercube• The graph with one vertex
for each n-digit binary sequence and an edge joining vertices that correspond to sequences that differ in just one position is called an n-dimensional cube or hypercube.
• v = 2n
• e = n 2n-1
4-dimensional Cube.
0000
1000
0010
0100
0001
1100
1110
0110
01110011
1001 1101
11111011
1010
4-dimensional Cube and a Famous Painting by Salvador Dali
• Salvador Dali (1904 – 1998) produced, in 1954, the Crucifixion (Metropolitan Museum of Art, New York) in which the cross is a 3-dimensional net of a 4-dimensional hypercube.
4-dimensional Cube and a Famous Painting by Salvador Dali
• Salvador Dali (1904 – 1998) produced in 1954, the Crucifixion (Metropolitan Museum of Art, New York) in which the cross is a 3-dimensional net of a 4-dimensional hypercube.
The Geometry of Q4.
• Vertices (Q0) of Q4: 16
• Edges (Q1)of Q4: 32
• Squares (Q2) of Q4: 24
• Cubes (Q3) of Q4: 8
• Total: 80
• The Levi graph of Q4 has 80 vertices and is colored with 4 colors.
Residual geometries of Q4.
V E S Q3.
(V) - 4 6 4
(E) 2 - 3 3
(S) 4 4 - 2
(Q3) 8 12 6 -
Truncations or Sections
• Given a geometry G = (V,~,T,c) and a subset of types J µ T, define a J-section G/J of G as the geometry H = (U,~,J,c), where U = {v 2 V| c(v) 2 J} and H is the induced subgraph of G.
Quaternions– Example of Rank 4 Geometry - Section
i
k
j
j,k
i,k i,j
11
i’i’
k’k’ j’j’
i
jk
j,k
i,k i,j
Rank 4 geometry Rank 3 section
Shadow Spaces
• Given a geometry G = (V,~,T,c) and J µ T we may define an incidence structure Spa(G,J) whose points are J-flags and the blocks are composed of those sets of J-flags that belong to the residual geometry G(F) for some flag F from the original geometry G.
Shadow Spaces - An Example
• Let us denote the types• I = {g,r,b}.
• Let J = {r,b}. There are three J-flags: 26, 45 and 56. The set system for the shadow space:
• {{45},{26},{45,56},{26,56}}.
• For J = {g,b} we get three flags:• {16,14,34}
• The set system for the shadow space:
• {{16},{34},{14,16},{14,34}}
1 2
34
5
6
Shadow spaces of Maps
• For maps as rank 3 geometries the notion of shadow spaces gives rise to an interesting interpretation. There are three types of objects {v,e,f}.
• Hence, there are 7 types of shadow spaces:
• {v} - primal: id• {e} - medial: Me• {f} - dual: Du• {v,e} - truncation: Tr • {v,f} - Me Me• {e,f} - leapfrog: Le• {v,e,f}- Co
Shadows - Example• Our map is a prism. All flags (structured by type):
• ;,• 1,2,3,4,5,6
• a,b,c,d,e,f,g,h,i
• A,B,C,D,E
• 1a,1d,1e,2a,2b,2f,3e,3f,3i,4c,4d,4h,5b,5c,5g,6g,6h,6i
• 1A,1B,1C,2A,2B,2D,3B,3C,3D,4A,4C,4E,5A,5D,5E,6C,6D,6E
• aA,aB,bA,bD,cA,cE,dA,dC,eB,eC,fB,fD,gD,gE,hC,hE,iC,iD
• 1aA,1aB,1dA,1dC,1eB,1eC,2aA,2aB,2bA,2bD,2fB,2fD,3e
B,3eC,3fB,3fD,3iC,3iD,4cA,3cE,4dA,4dC,4hC,4hE,5bA,5bD,5cA,5cE,5gD,5gE,6gD,6gE,6hC,6hE,6iC,8iD1 2
3
4
6
5
a
b
c
d
e f
gh
A
B
C D
E
i
Shadows - Example - Primal• Our map is a prism. T ={v,e,f}:
• J = {v}
• J-flags: 1, 2, 3, 4, 5, 6
• Sets: 12, 13, 14, 23, 25, 36, 45, 46, 56, 123, 456, 1245, 1346, 2356.
1 2
3
4
6
5
a
b
c
d
e f
gh
A
B
C D
E
i
Shadows - Example - Dual• Our map is a prism. T = {v,e,f}:
• J = {f}
• Flags: A,B,C,D,E
• Sets: AB, AC, AD, AE, BC, BD, CD, CE, DE, ABC, ABD, BCD, CDE, ACE, ADE.
1 2
3
4
6
5
a
b
c
d
e f
gh
A
B
C D
E
i
Shadows - Example - Medial• Our map is a prism. T = {v,e,f}:
• J = {e}
• Flags: a,b,c,d,e,f,g,h,i
• Sets: ae,ab,ad,af,bc,bf,bg,cd,cg,ch,de,dh,ef,ei,fi,gh,gi,hi, aef, bfgi, dehi, abcd,cgh.
1 2
3
4
6
5
a
b
c
d
e f
gh
A
B
C D
E
i
Shadows - Example - Truncation• Our map is a prism. T = {v,e,f}:
• J = {v,e}
• Flags: 1a,1d,1e,2a,2b,2f,3e,3f,3i,4c,4d,4h,5b,5c,5g,6g,6h,6i
• Sets: 1a,1d,1e,2a,2b,2f,3e,3f,3i,4c,4d,4h,5b,5c,5g,6g,6h,6i
• ...
1 2
3
4
6
5
a
b
c
d
e f
gh
A
B
C D
E
i
Posets
• Let (P,·) be a poset. We assume that we add two special (called trivial) elements, 0, and 1, such that for each x 2 P, we have 0 · x · 1.
Ranked Posets
• Note that a ranked poset (P,·) of rank n has the property that there exists a rank function r:P ! {-1,0,1,...,n}, r(0) = -1, r(1) = n and if y covers x then r(y) = r(x) +1. (All linear chains have the same length)
• If we are given a poset (P, ·) with rank function r, then such a poset defines a natural incidence geometry.
• V() = P.• x ~ y if and only if x < y. • c(x) := r(x). Vertex color is just the rank.
• The corresponding geometry is a string geometry. There is a one-to-one correspondence between the two concepts.
Intervals in Posets
• Let (P,·) be a poset.
• Then I(x,z) = {y| x · y · z} is called the interval between x and z.
• Note that I(x,z) is empty if and only if x £ z.
• I(x,z) is also a ranked poset with 0 and 1.
Connected Posets.
• A ranked poset (P,·) wih 0 and 1 is called connected, if either rank(P) = 1 or for any two non-trivial elements x and y there exists a sequence x = z0, z1, ..., zm = y, such that there is a path avoiding 0 and 1 in the Levi graph from x to y and the rank function is changed by § 1 at each step of the path.
Abstract Polytopes
• Peter McMullen and Egon Schulte define abstract polytopes as special ranked posets.
• Their definition is equivalent to the following:• (P,·) is a ranked poset with 0 and 1 (minimal and maximal
element)• For any two elements x and z, such that r(z) = r(x)+2, x < z there
exist exactly two elements y1, y2 such that x < y1 < z, x < y2 < z.• Each section is connected.
• Note that abstract poytopes are a special case of posets but they form also a generalization of the convex polytopes.
Convex vs abstract polytopes
• To each convex polytope we may associate an abstract polytope. For instance, the tetrahedron:
• 0
• 4 vertices: v1, v2, v3, v4.
• 6 edges: e1, e2, ..., e6,
• 4 faces: t1,t2,t3, t4
• 1
• e1 = v1v2, e2 = v1v3, e3 = v1v4, e4 = v2v3, e5 = v2v4, e6 = v3v4.
• t4 = v1v2v3, t1 = v2v3v4, t3 = v1v2v4, t2 = v1v3v4.
The Poset
• In the Hasse diagram we have the following local picture:
v2v1 v3 v4
e2e1e3 e4 e5
e6
t2t1 t3 t4
1
0
Diagram geometries
• For any incidence geometry G(V,~,T,c) we usually study for each pair i,j 2 T the section (truncation) of rank two: G/(i,j). We deliberatly make distinction between G/(i,j) and its dual G/(j,i). Sometimes each connected component of G/(i,j) has the same structure. This is indicated by a diagram. A diagram in an edge-labeled graph on the vertex set T, where the lables indicate the structure of each section.
String diagram geometries
• The edge between i an j is omitted if and only if G/(i,j) is a generalized digon. This means that each connected component is a complete bipartite graph.
• G is called a string diagram geometry if the corresponding diagram has a shape of a path (or union of paths).
• Example: Each abstract polytope is a string diagram geometry.
The Grassmann graph
• Let G(V,~,T,c) be an incidence geometry and let i 2 T be a type. Then we define the Grassmann graph G(i) to on the vertex set V(i) = {v 2 V| c(v) = i} and two vertices u and v are adjacent in G(i) if an only if for each j i there exists an w 2 V(j) such that u ~ w and w ~ v (in the original geometry.)
• Example: For instance, in the case of rank two geometries, the Grassmann graphs are exactly the Menger graph and the dual Menger graph.
Exercises 4-1• N1. Our map is a prism. I
= {v,e,f}:• For each set of type
• J = {v,f}
• J = {e,f}
• J = {v,e,f}
determine the shadow space.
1 2
3
4
6
5
a
b
c
d
e f
gh
A
B
C D
E
i
Exercises 4-2
• N2. Repeat the analysis of previous two slides for the simplex K5.
• N3. Repeat the analysis of the previous two slides for the generalized octahedron K2,2,2,2.
Exercises 4-3
• N4: Determine all residual geometries of Reye’s configuration
• N5: Determine all residual geometries of Q4.• N6: Determine all residual geometries of the
Platonic solids.• N7: Determine the Levi graph of the geometry for
the group Z2 £ Z2 £ Z2, with three cyclic subgroups, generated by 100, 010, 001, respectively.
Exercises 4-4
• N18: Determine the posets and Levi graphs of each of the polytopes on the left.
• Solution for the haxagonal pyramid:
• 0
• 7 vertices: v0, v1, v2, ..., v6.
• 12 edges: e1, e2, ..., e6, f1, f2, ..., f6
• 7 faces: h,t1,t2,t3,.., t6
• 1
• e1 = v1v2, e2 = v2v3, e3 = v3v4, e4 = v4v5, e5 = v5v6, e6 = v6v1, f1 = v1v0, f2 = v2v0,f3 = v3v0, f4 = v4v0, f5=v5v0, f6 = v6v0.
• h = v1v2v3v4v5v6,
• t1 = v1v2v0, t2 = v2v3v0, t3 = v3v4v0, t4 = v4v5v0, t5
= v5v6v0, t6 = v6v1v0,
5. Incidence Structures
Incidence structure
• An incidence structure C is a triple– C = (P,L,I) where
• P is the set of points,• L is the set of blocks or lines
– I P L is an incidence relation.– Elements from I are called flags.
• The bipartite incidence graph (C) with black vertices P, white vertices L and edges I is known as the Levi graph of the structure C.
(Combinatorial) Configuration
• A (vr,bk) configuration is an incidence structure C = (P,L,I) of points and lines, such that
• v = |P|
• b = |L|
• Each point lies on r lines.
• Each line contains k points.
• Two lines intersect in at most one point.
• Warning: Levi graph is semiregular of girth 6
Symmetric configurations
• A (vr,bk) configuration is symmetric, if • v = b (this is equivalent to r = k).
• A (vk,vk) configuration is usually denoted by (vk).
Small Configurations
• Triangle, the only (32) configuration.
• Pasch configuration (62,43) and its dual Perfect Quadrangle (43,62) have the same Levi graph.
6. Substructures, Symmetry and Duality
Substructures
• An incidence structure C’ = (P’, L’,I’) is a substructure of an incidence structure C = (P, L,I), C’ µ C, if P’ µ P, L’ µ L and I’ µ I.
Duality
Each incidence structure C = (P,L,I) gives rise to a dual structure Cd = (L,P,Id) with the role of points and lines reversed and keeping the incidence.
• The structures C and Cd share the same Levi graph with the roles of black and white vertices reversed.
Self-Duality and Automorphisms
• If C is isomorphic to its dual Cd , it is said that C is selfdual, the corresponding isomorphism is called a (combinatorial) duality.
• A duality of order 2 is called (combinatorial) polarity.
• An isomorphism mapping C to itself is called an automorphism or (combinatorial) collinearity.
Automorphisms and Antiautomorphisms
• Automorphisms of the incidence structure C form a grup that is called the group of automorphisms and is denoted by Aut0C.
• If automorphisms and dualities (antiautomorphisms) are considered together as permutations, acting on the disjoint union P L, we obtain the extended group of automorphism Aut C.
• Warning: If C is disconnected there may be mixed automorphisms.
Graphs and Configurations
• The Levi graph of a configuration is bipartite and carries complete information about the configuration.
• Assume that C is connected. The extended group of automorphisms AutC coincides with the group of automorphisms of the Levi graph L ignoring the vertex coloring, while Aut0C stabilises both colors.
Examples
• 1. Each graph G = (V,E) is an incidence structure: P = V, L = E, (x,e) 2 I if and only if x is an endvertex of e.
• 2. Any family of sets F µ P(X) is an incidence structure. P = X, L = F, I = 2.
• 3. A line arrangement L = {l1, l2, ..., ln} consisting of a finite number of n distinct lines in the Euclidean plane E2 defines an incidence structure. Let V denote the set of points from E2 that are contained in at least two lines from L. Then: P = V, L = L and I is the point-line incidence in E2.
Exercises 6
• N1: Draw the Levi graph of the incidence structure defined by the complete bipartite graph K3,3.
• N2: Draw the Levi graph of the incidence structure defined by the powerset P({a,b,c}).
• N3: Determine the Levi graph of the incidence structure, defined by an arrangemnet of three lines forming a triangle in E2.
7. Haar Graphs and Cyclic Configurations
Haar graph of a natural number
Let us write n in binary:
n = bk-12k-1 + bk-2 2k-2 + ...+ b12 + b0
where B(n) = (bk-1, bk-2, ..., b1, b0), bk-1= 1are binary digits of n. Graph H(n) = H(k; n), called the Haar graph of the natural number n, has vertex set ui, vi, i=0,1,...,k-1. Vertex ui is adjacent to vi+j, if and only if bj = 1 (arithmetic is mod k).
Remark
When defininig H(n) we assumed that k is the number of binary digits of n. In general, for H(k;n) one can take k to be greater than the number of binary digits. In such a case a different graph is obtained!
Example
Determine H(37).
Binary digits:• B(37) = {1,0,0,1,0,1}• k = 6.• H(37) = H(6;37) is
depicted on the left!
Dipoles n
• The dipole n has two vertices, joined by n parallel edges. If we want to distinguish the two vertices, we call one black, the other one white. On the left we see 5.
• Each dipole is a bipartite graph. Therefore each of its covering graphs is a bipartite graph.
• In particular 3 is a cubic graph also known as the theta graph .
Cyclic covers over a dipole• Each Haar graph is a cyclic cover over a
dipole. One can use the following recipe:• H(37) is determined by a natural number
37, or, equivalently by a binary sequence:(1 0 0 1 0 1).
• The length is k=6, therefore the group Z6. • The indices are written below:• (1 0 0 1 0 1)• (0 1 2 3 4 5)• The “1”s appear in positions: 0, 3 in 5.
These numbers are used as voltages for H(37).
0 3 5
Z6
Connected Haar graphs
• Graph G is connected if there is a path between any two of its vertices.
• There exist disconnected Haar graphs, for instance H(10).
• Define n to be connected, if the corresponding Haar graph H(n) is connected.
• Disconnected numbers: 2,4,8,10,16,32,34,36,40,42,64...
The Mark Watkins Graph
• The cubic Haar graph H(536870930) has an interesting property. 536870930 is the smallest connected number that is cyclically equivalent to no odd number.
• Recall that two sets S,T µ Zn are cyclically equivalent if there exists a 2 Zn
* and b 2 Zn such that S = aT + b (mod n).
Girth of Connected Haar graphs
• K2 is the only connected 1-valent Haar graph.
• Even cycles C2n are connected 2-valent Haar graphs.
• Theorem: Let H be a connected Haar graph of valence d > 2. Then either girth(H) = 4 or girth(H) = 6.
Cyclic Configurations
• A symmetric (vr) configuration determined by its first column s of the configuration table where each additional column is obtained from s by addition (mod m) is called a cyclic configuration Cyc(m;s).
• The left figure depicts a cyclic Fano configuration Cyc(7;1,2,4) = Cyc(7;0,1,3).
a b c d e f gk 1 2 3 4 5 6 0
k+1 2 3 4 5 6 0 1k+3 4 5 6 0 1 2 3
Connection to Haar graphs
• Theorem: A symmetric configuration (vr), r ¸ 1 is cyclic, if and only if its Levi graph is a Haar graph with girth 4.
• Corollary: Each cyclic configuration is point- and line-transitive and combinatorially self-dual.
• Corollary: Each cyclic configuration (vr), r > 2 contains a triangle.
• Question: Does there exist a cyclic configuration that is not combinatorially self-polar?
Problem
• Study cyclic configurations with respect to flag orbits.
• Example: On the left we see the smallest 0-symmetric graph Haar(261) on 18 vertices. It is the Levi graph of the cyclic (93) configuration having 3 flag orbits.
Exercises 7-1• The graph on the left is the so-
called Heawood graph H. Prove:– N1: H is bipartite– N2: H is a Haar graph. (Find n!)– N3: Determine H as a cyclic
cover over 3..– N4: Prove that H has no cycle of
length < 6.– N5: Prove that H is the smallest
cubic graph of girth 6.– N6: Find a hexagonal torus
embedding of H .– N7: Determine the dual of the
embedded H.
Exercises 7-2• N8: Prove that each 2m is a disconnected number.
• N9: Show that the Möbius-Kantor graph G(8,3) is a Haar graph of some number. Which number is that?
• N10: (*) Determine all generalized Petersen graphs that are Haar graphs of some natural number.
• N11: Show that some Haar graphs are circulants.
• N12: Show that some Haar graphs are non-circulants.
Exercises 7-3
• N13: Prove that each Haar graph is vertex transitive.
• N14: Prove that each Haar graph is a Cayley graph for a dihedral group.
• N15: Prove that there exist bipartite Cayley graphs of dihedral groups that are not Haar graphs (such as the graph on the left).
Exercises 7-4• N16: The numbers n and m are cyclically equivalent, if the
binary string of the first number can be cyclically transformed to the binary string of the second number. This means that the string can be cyclically permuted, mirrored or multiplied by a number relatively prime with the string length.
• N17: The numbers n and m are Haar equivalent, if their Haar graphs are isomorphic: H(n) = H(m).
• N18: Prove that cyclic equivalence implies Haar equivalence.• N19: Determine all numbers that are cyclically equivalent to
69.• N20: Use a computer to show that 137331 and 143559 are
Haar equivalent, but are not cyclically equivalent.
Exercises 7-5
• N21: Show that each Haar graph of an odd number H(2n+1) is hamiltonian and therefore connected.
Homework 7
• Use Vega to explore the edge-orbits of cyclic Haar graphs.
• H1. Find an example of a cubic Haar graph that has 1,2, or 3 edge orbits.
• H2. Find an example of a quartic Haar graph that has 1, 2, 3, or 4 edge orbits. Study the graphs with 2 edge orbits.
8. Algebraic Structures
Real Numbers R.
• Let us review the structure of the set of real numbers (real line) R.
• In particular, consider addition + and multiplication £.
• (R,+) forms an abelian group.
• (R,£) does not form a group. Why?
• (R,+,£) forms a (commutative) field.
Real Numbers R. - Exercises
• N43: Write down the axioms for a group, abelian group, a ring and a field.
• N44: What algebraic structure is associated with the integers (Z,+,£)?
• N45: Draw a line and represent the numbers R. Mark 0, 1, 2, -1, ½, .
A Skew Field K• A skew field is a set K endowed with two constants 0 and 1, two unary operations• -: K ! K,• ‘: K ! K, • and with two binary operations:• +: K £ K ! K,• : K £ K ! K,• satisfying the following axioms:• (x + y) + z = x + (y +z) [associativity]• x + 0 = 0 + x = x [neutral element]• x + (-x) = 0 [inverse]• x + y = y + x [commutativity]• (x y) z = x (y z). [associativity]• (x 1) = (1 x) = x [unit]• (x x’) = (x’ x) = 1, for x 0. [inverse]• (x + y) z = x z + y z. [left distributivity]• x (y + z) = x y + y z. [right distributivity]• A (commutative) field satisfies also:• x y = y x.
Examples of fields and skew fields
• Reals R• Rational numbers Q• Complex numbers C• Quaterions H (non-commutative!! Will consider
briefly later!)• Residues mod prime p: Fp
• Residues mod prime power q = pk: Fq (more complicated, need irreducible poynomials!!Will consider briefly later!)
Complex numbers C
• = a + bi 2 C• * = a – bi
• = c + di 2 C• = (ac –bd) + (bc + ad)i
• 0, / = [(ac + bd) + (bc – ad)i]/[c2 + d2]
• -1 = (a –bi)/(a2 + b2)
Quaternions H.
• Quaternions form a non-commutative field.• General form:• q = x + y i + z j + w k., x,y,z,w 2 R.• i 2 = j 2 = k 2 =-1.
• q = x + y i + z j + w k.• q’ = x’ + y’ i + z’ j + w’ k.
• q + q’ = (x + x’) + (y + y’) i + (z + z’) j + (w + w’) k.• How to define q .q’ ?• i.j = k, j.k = i, k.i = j, j.i = -k, k.j = -i, i.k = -j.• q.q’ = (x + y i + z j + w k)(x’ + y’ i + z’ j + w’ k)
Quaternions H. - Exercises
• N46: There is only one way to complete the definition of multiplication and respect distributivity!
• N47: Represent quaternions by complex matrices (matrix addition and matrix multiplication)! Hint: q = [ ; -* *]. (We are using Matlab notation).
-* *
Residues mod n: Zn.
• Two views:
• Zn = {0,1,..,n-1}
• Define ~ on Z:
• x ~ y $ x = y + cn
• Zn = Z/~
• (Zn,+) is an abelian group, namely a cyclic group. Here + is taken mod n!!!
Example (Z6, +).
+ 0 1 2 3 4 5
0 0 1 2 3 4 5
1 1 2 3 4 5 0
2 2 3 4 5 0 1
3 3 4 5 0 1 2
4 4 5 0 1 2 3
5 5 0 1 2 3 4
Example (Z6, £).
£ 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 1 2 3 4 5
2 0 2 4 0 2 4
3 0 3 0 3 0 3
4 0 4 2 0 4 2
5 0 5 4 3 2 4
Example (Z6\{0}, £).
£ 1 2 3 4 5
1 1 2 3 4 5
2 2 4 0 2 4
3 3 0 3 0 3
4 4 2 0 4 2
5 5 4 3 2 4
It is not a group!!!
For p prime, (Zp\{0}, £) forms a group: (Zp, +,£) = Fp.
Vector space V over a field K
• +: V £ V ! V (vector addition)
• .: K £ V ! V (scalar multiple)
• (V,+) abelian group
• ( + )x = x + x
• 1.x = x
• ( ).x = ( x)
• .(x +y) = .x + .y
9. Euclidean Plane, Affine Plane, Projective Plane
Euclidean plane E2 and real plane R2
• R2 = {(x,y)| x,y 2 R}
• R2 is a vector space over R. The elements of R2 are ordered pairs of reals.
• (x,y) + (x’,y’) = (x+x’,y+y’)
• (x,y) = ( x, y)
• We may visualize R2 as an Euclidean plane (with the origin O).
Subspaces
• One-dimensional (vector) subspaces are lines through the origin. (y = ax)
• One-dimensional affine subspaces are lines. (y = ax + b)
o
y = ax y = ax + b
Three important results
• Thm1: Through any pair of distinct points passes exactly one affine line.
• Thm2: Through any point P there is exactly one affine line l’ that is parallel to a given affine line l.
• Thm3: There are at least three points not on the same affine line.
• Note: parallel = not intersecting or identical!
Affine Plane
• Axioms:• A1: Through any pair of distinct points passes
exactly one line.• A2: Through any point P there is exactly one line
l’ that is parallel to a given line l.• A3: There are at least three points not on the same
line. • Note: parallel = not intersecting or identical!
Examples
• Each affine plane is an incidence structure C = (P,L,I) of points and lines.
• Let K be a field, then K2 has a structure of an affine plane.
• K = Fp.
• Determine the number of points and lines in the affine plane A2(p) = Fp
2.
Parallel Lines
• Parallel lines l || m define an equivalence relation on the set of lines.
1. l || l2. l || m ) m || l 3. l || m, m || n ) l || n.
A pencil of parallel lines
• An equivalence class of parallel lines is called a pencil of parallel lines.
• Thm. Each pencil of parallel lines defines an equivalence relation on the set of lines.
Ideal points and Ideal line
• Each pencil of parallel lines defines a new point, called an ideal point (or a point at inifinity.) New point is incident with each line of the pencil.
• In addition we add a new ideal line (or line at infinity)
Extended Plane
• Let A be an arbitrary affine plane. The incidence structure obtained from A by adding ideal points and ideal lines is called the extended plane and is denoted by P(A).
• Theorem. Let C be an extended plane obtained from any affine plane. The following holds:
• T1. For any two distinct points P and Q there exists a unique line l connecting them.
• T2. For any two distinct lines l and m there exists a unique point P in their intersection.
• T3. There exist at least four points P,Q,R,S such that no three of them are colinear.
Projective Plane
• Axioms for the Projective Plane. Let C be an incidence structure of points and lines that satisfies the following axioms:
• P1. For any two distinct points P and Q there exists a unique line l connecting them.
• P2. For any two distinct lines l and m there exists a unique point P in their intersesction.
• P3. There exist at least four points P,Q,R,S such that no three of them are colinear.
Linear Transformations
• In a vector space the important mappings are linear transformations:
• L( x + y) = L(x) + L(y). L-1 exists.
• L can be represented by a nonsingular square matrix.
Semi Linear Transformations
• A semi linear transformation is more general:
• L(x + y) = f() L(x) + f() L(y). L-1 exists, f: K ! K is an automorphism of K.
Affine Transformations
• In an affine plane the important mappings are affine transformations (=affinities).
• An affine transformation maps sets of collinear points to collinear points.
• Each affine transformation is of the form A(x) + c, where A is a semilinear transformation.
Projective plane from R3
• Consider the incidence structure defined by 1-dimensional and 2-dimensional subspaces of R3 where the incidence is defined by inclusion.
• Call 1-dimensional subspaces points and 2-dimensional subspaces lines.
Homogeneous Coordinates
• Let (a,b,c) (0,0,0) be a point in R3. There is exactly one line through the origin passing through (a,b,c). Hence a projective point can be represented by (a,b,c). However, for any 0 the same projective point can be represented by ( a, b, c).
• That is why (a,b,c) are called homogeneous coordinates.
Unit sphere model
• Take a unit sphere in R3.
• Let pairs of antipodal points be projective points.
• Let big circles be projective lines.
• Prove that this system is a model for a projective plane.
Exercises 9-1
• N1. Conditions 1. and 2. are true for any incidence structure. (Prove it!)
• N2: Prove condition 3 for affine planes and find a counter-example for general incidence structure.
• N3. Prove that this structure satisfies all three axioms for the projective plane.
• N4: Prove that in R, Q, Fp, (p- prime) there are no nontrivial automorphisms.
Exercises 9-2
• N5: Prove that z z* (conjugate) is an automorphism of C.
• N6: Go to the library or the internet and find a reference to the group of authomorphisms of the complex numbers C and the quaternions H.
• N7: Determine the size of the group of automorphisms of Fq, for q = pk, a power of a prime.
10. Point Configurations, Line Arrangements, Polarity
Point Configuration
• A point configuration in R2 is a collection of points affinely spanning R2.
• In other words: not all points are collinear.
Line Arrangement
• A line arrangement is a partitioning of the plane R2 into connected regions (cells, edges, and vertices) induced by a finite set of lines.
Area of a Triangle
Area of the green trapezoid:
A12= (1/2)(y2 +y1) (x2 – x1)
In the same way:A23= (1/2)(y2 +y3) (x3 – x2)
A13= (1/2)(y3 +y1) (x3 – x1)
Area of the triangle:
T = A12 + A23 – A13.
P2(x2,y2)y2
x2
P1(x1,y1)
x1
y1
O
P3(x3,y3)
x3
y3
Area of a Triangle
P2(x2,y2)y2
x2
P1(x1,y1)
x1
y1
O
P3(x3,y3)
x3
y3
Triple of Collinear Points
P2(x2,y2)y2
x2
P1(x1,y1)
x1
y1
O
P3(x3,y3)
x3
y3
The points P1(x1,y1), P2(x2,y2), P3(x3,y3), are collinear if and only if T = 0.
Point Configurations – Line Arrangements
• Each point configuration S gives rise to a line arrangement A(S). The lines are determined by all pairs of points.
• Another line arrangement A3(S) is determined by triples of collinear points.
Polarity with Respect to a Circle
• Let us consider the extended plane and a circle K in it. There is a mapping from points to lines (and vice versa). : p P.
• p – polar• P – pole
Pp
Pp
P
p
Polarity with respect to the unit circle
• Given P(a,b) the equation of the polar is• p: y = (-a/b)x + (1/b)• p: by + ax = 1• In general: Circle K(p,q;r)• p: y(b-q) + x(a-p) = p(a-p) + q(b-q) + r2.• Given• p: y = kx + n• P(a,b)• a = -k/n• b = 1/n• In general:• a = p-kr2/(kp + n – q) • b = q+ r2/(kp + n –q)
Natural Parameters p,q,r
• For a given point configuration S the center of the circle(p,q) is determined as the barycenter of S while the radius is given as the average distance from the center.
Polarity in General
• A general polarity is defined with respect to a conic section (ellipse, hyperbola, or parabola).
Polar Duality of Vectors and Central Planes in R3.
• A polar duality is a mapping associating a vector v 2 R3 with an oriented central plane having v as its normal vector and vice versa.
A Standard Affine Polar-Duality
• A standard affine polar duality is a mapping between non-vertical lines and points of R2 associating the non-vertical line y = ax + b with the point (a,-b) and vice versa.
Polar Duality of Points and Lines in the Affine Space.
• General rule: Take a polar-duality of vectors and central planes and consider the intersetion with some affine plane in R3 .
Homogeneous Coordinates
• Take the affine plane z = 1. A point with Euclidean coordinates (x,y) can be assigned the homogeneous coordinates (x,y,1). Ideal points get homogeneous coordinates (x,y,0).
(x0,y0)
(x0,y0,1)
(z0x0,z0y0,z0)
Equation of a plane through the origin
• Recall general plane:• ax + by + cz = d.
• Equation of a plane through the origin:• ax + by + cz = 0-
• Another meaning:• (x,y,z) homogeneous coordinates of a projective
point
• [a,b,c] homogeneous coordinates of a projective line.
Point on a Line
• Let (a,b,c) be homogeneous coordinates of a point P and let [A,B,C] be homogeneous coordinates of a line p.
• Then P lies on p if and only if aA + bB + cC = 0.
• Let P(a,b,c) and P’(a’,b’,c’). The equation of a line through P Æ P’. is defined by the cross product [A,B,C] = (a,b,c) £ (a’,b’,c’).
• Similarly we get the intersection of two lines.
Example
• Polarity of a point configuration consisting of the points of a 10 £ 10 grid.
• Parameters of the circle are determined automatically.
Star Polygons (n/k).
• By (n/k) we denote star polygons.
• Note that each of them defines an incidence structure. in which the points are the vertices and intersections while the lines are the edges of a polygon.
3/1 4/1 5/1
5/2 6/1 6/2
7/1 7/2 7/3
Fano Plane
• We obtain the Fano plane from F2
3. There are obviously 7 (non-zero) points: Any pair of points defines a unique line that contians exactly one additional point.
0 0 1 0 1 1 1
0 1 0 1 0 1 1
1 0 0 1 1 0 1
Exercises 10-1
• A polarity maps a point configuration to a line arrangement and vice versa.
• N1:Take an equilateral triangle ABC with sides a,b,c. Find a polarity, such that a A, b B and c C.
• N2: Determine the polar figure of point configuration determined by the vertices of a regular n-gon with respect to its inscribed circle.
• N3: Determine the polar of an ideal point and the pole of the ideal line.
Exercises 10-2
• N4: Determine the number of points and lines of the incidence structure defined by the star polygon 5/2.
• N5(*): Determine the number of points and lines of the incidence structure determined by the star polygon n/k.
Homework 10
• H1: Prove that the number of points on each line of any finite projective plane P is constant, say q+1. [Then q is called the order of P.]
• H2: Which axioms of a projective planes are valid for a near-pencil N(n)?
11. Pappus and Desargues Theorem
Pappus Theorem
• Let A, B, C be three collinear points and let A', B' , C' be another triple of collinear points. Let A'' be the intersection of (BC') and (B'C), B'' the intersection (A,C') and (A'C), C'' the intersetion of (AB') and (A'B). Then the points A'', B'' and C'' are collinear.
AB
C
C'' B'' A''
A' B'C'
Desargues Theorem
• Let ABC and A'B'C' be two triangles. Let A'' be the intersection of BC and B'C', let B'' be the intersection of AC and A'C' and C'' be the intersection of AB and A'B'. The lines AA',BB' and CC' intersect in a common point O if and only if A'', B'' and C'' are collinear.
O
A
A'
B'
C'CB
A''
C''
B''
Ternary ring coordinatization.
• Ternary operation, desrcibed in geometric terms.
• Properties:• (a) x*0*b = 0*x*b=b• (b)x*1*0 = 1 * x * 0 = x• (c) Given x,y,a, there is a unique b
such that y = x*a*b• (d) Given x,x’,y,y’ with x x’ there
is a unique ordered pair (a,b) such that y = x*a*b and y’=x’*a*b.
• (e) Given a,a’,b,b’ with a a’, there is a unique x such that x*a*b=x*a’*b’.
[0,0] [1,0]
[1,b][0,b]
[a,0]
[0,a*b*c]
[0,c]
[b]
[b]
Pappian and Desarguesian Projective Planes
• Thm. A projective plane is desarguesian if and only if the ternary ring is a field or a sqew-field.
• Thm. A projective plane is pappian if an only if the ternary ring is a field.
Non-Desarguesian Projective Plane
• F.R.Multon (1902)
• Points: points in the real projective plane.
• Lines: • y = mx+n, m· 0.
• y = mx + n, x¸(-n/m), m¸0
• y = (m/2)x + n), m¸0,y·0.
• Line at infinity contains points [m].
Exercises 11
• N1(*): Prove the Pappus theorem in the Euclidean plane.
• N2(*): Prove the Desargues theorem in the Eucliudean plane.
12. Existence and Counting of Combinatorial Configurations
Lineal Configurations
• In order to emphasise configurations as partial linear spaces we call them lineal configurations (= digon – free configurations).
Existence of Lineal Configurations
• Proposition: For each lineal (vr,bk) configuration (r ¸ k) the following is true:
• v.r = b.k• b ¸ v ¸ 1 + r(k – 1)
• Corollary: For symmetric (vk) configurations the following lower bound is obtained:
• v ¸ 1 + k(k-1) = 1 –k + k2
• In particular:• For k = 3 we have v ¸ 7,• For k = 4 we have v ¸ 13,• For k = 5 we have v ¸ 21.
Lower Bounds (Adapted from Grünbaum)
r\k 3 4 5 6 7
3 (73) (123,94) (203,125) (263,136) (353,157)
4 (94,123) (134) (204,165) ?(304,206)? ?(494,,287)?
5 (125,203) (165,204) (215) (305,256) ?(425,307)?
6 (136,263) ?(206,304)? (256,305) (316) X(496,427)X
7 (157,353) ?(287,494)? ?(307,425)? X(427,496)X X(437)X
Blocking Set
• A set of points B of an incidence structure is called a blocking set, if each line L contains two points x and y, such that:
• x 2 B and (x,L) 2 I,
• y B and (y,L) 2 I.
Notation
Counting (v3) Configurations
Counting Triangle-Free (v3) Configurations
13. Coordinatization
Coordinatization
Reconstruct an incidence structure from a matrix M:
• Columns are homogeneous coordinates in some field or sqew-field F.
• <ijk> = det (Mi Mj Mk)
• ijk form a line if and only if <ijk> = 0.
m11 m12 ... ... ... ... m1n
m21 m22 ... ... ... ... m2n
m31 m32 ... ... ... ... m3n
Fano plane (73).• We can reconstruct (73) from
the matrix M.• Columns are homogeneous
coordinates in F2.
• <ijk> = det (Mi Mj Mk)
• ijk form a line if and only if <ijk> = 0.
0 0 0 1 1 1 1
0 1 1 0 0 1 1
1 0 1 0 1 0 1
Möbius-Kantor Configuration – Revisited
• Möbius-Kantor configuration is the only (83) configuration. Its Levi graph is the generalized Petersen graph G(8,3).
• The configuration has no geometric realization with (real) points and lines in the Euclidean plane.
Affine plane of order 3
• (94,123) configuration is the affine plane of order 3.
• It contains the Pappus configuration.
• It contains also the Möbius-Kantor configuration.
2211 33
44 55 66
77 88 99
Complex Coordinatization of (94,123)
A Z3 coordinatization of (134) = PG(2,3)
A Z3 coordinatization of (123, 94)
• By removing one point from the projective plane we get the affine plane. (Its dual is (94,123))
Dual coordinates and dual lines
0 0 0 1 1 1 1 1 10 1 1 0 0 1 1 1 11 0 1 0 1 0 1 1 1
1 1 1 1 2 2 2 3 3 3 4 52 4 6 7 4 5 7 4 5 6 8 63 5 9 8 6 8 9 7 9 8 9 7
Möbius-Kantor Configuration – Revisited
• Möbius-Kantor configuration is coordinatizable over the complex field and over F3.
• Something is wrong here. I expected that one can change
• -1 ! 2• ! 2
• in the top matrix, and get the desired coordinatization. But columns 1 and 4 become identical.
0 -1 1 0 1 0 1
1 0 -1 1 0 0
-1 1 0 1 0 1
0 -1 1 0 1 0 1
1 0 -1 1 0 0
-1 1 0 1 0 1
Complex Coordinatization of (83)
• By removing one point (and 4 incident lines) we get (83) from (94,124).
Exercises 13-1
• N1: Determine the homogeneous coordinates of the 9 lines from the previous problem.
• N2: Write a computer program that will find the matrix for the polar.
• N3:Show that by deleting any column of the matrix for (94,123) a coordinatization of (83) is obtained.
• N4: Given the Levi graph G(8,3) of (83), determine the Levi graph of (94,123).
Exercises 13-2
• N5: Determine whether the incidence structure defined by 9 points and 8 lines on the left has a blocking set.
• N6: Prove that the game TIC-TAC-TOE has no possibility of a draw.
14. Combinatorial Configurations on Surfaces
Menger Graph on Torus
• On the left there is a hexagonal embedding of the Heawood graph in the torus. (Heawood = Levi graph of Fano)
• Its dual is a triangular embedding of K7 in S2.
Menger Graph on Torus
• Menger graph (of Fano) is K7 and has a triangular embedding in torus. (Consider only red vertices).
• Later we show how to generalize this construction.
Menger graph of Möbius-Kantor Configuration
• Menger graph of this configuration is depleated K8: DK8 = K8 – 4K2
• Vertices represent configuration points while triangles represent lines.
Möbius-Kantor Graph in Double Torus
The Möbius-Kantor graph is embedded in the double torus such that:
• The embedding is octagonal.
• The map is regular.
Möbius-Kantor Graph in Double Torus
• This embedding of the Möbius-Kantor graph gives rise to an embedding of the Menger graph DK8 in the same surface with 8 triangles and 6 quadrilaterals.
• By adding 4 missing edges we get an embedding of K8 in the double torus with all triangles, except for two quadrilaterals.
The Dual
• The dual graph is S[2]
(K4).• Let G be any graph.
Recall that S(G) is the subdivision graph.
• S[k](G) is obtained from S(G) by multiplying the original vertices of G k times.
Pappus configuration
• Pappus (93) configuration can be represented in the plane by exhibiting homogeneous coordinates for each point (a,b,c). Each line can be described in a similar way: [p,q,r] where the incidence is given by ap+bq+cr=0.
• This can be considered as an example of an orthogonal representation of (Levi) graphs where u~v implies(u) (v).
Pappus Graph on Torus
• Collection of hexagons:1. {10, 17, 18, 13, 12, 11}
2. {8, 15, 16, 17, 10, 9}
3. {7,12, 13, 14, 15, 8}
4. {4, 11, 12, 7, 6, 5}
5. {3,4, 5, 16, 15, 14}
6. {2, 9, 10, 11, 4, 3}
7. {1, 2, 3, 14, 13, 18}
8. {1, 18, 17, 16, 5, 6}
9. {1, 6, 7, 8, 9, 2}
• Euler formula:
• 18 - 27 + 9 = 0
• g = 1
Three (93) Configurations
Three (93) Configurations
• They are all combinatorially self-polar.
• Pappus (red)• Cyclic (green)• Non-cyclic (yellow?).
Three (93) Configurations
• List of faces:1. {5, 11, 14, 7, 15, 16, 12, 6}
2. {4, 10, 18, 17, 11, 5}
3. {3, 9, 17, 18, 12, 16, 8, 13, 10, 4}
4. {2, 8, 16, 15, 9, 3}
5. {1, 2, 3, 4, 5, 6}
6. {1, 6, 12, 18, 10, 13, 14, 11, 17, 9, 15, 7}
7. {1, 7, 14, 13, 8, 2}
• Euler formula: V = 18, E = 27, F = 7
• 18 - 27 + 7 = -2 = 2 - 2g.
• g = 2
Three (93) Configurations
• List of faces:1. {10, 16, 15, 11, 17, 18}
2. {8, 18, 17, 9, 14, 13}
3. {7, 15, 16, 12, 13, 14}
4. {4, 5, 6, 12, 16, 10}
5. {3, 9, 17, 11, 5, 4}
6. {2, 3, 4, 10, 18, 8}
7. {1, 2, 8, 13, 12, 6}
8. {1, 6, 5, 11, 15, 7}
9. {1, 7, 14, 9, 3, 2}
• g = 1.
Menger and Levi - Pappus
Menger and Levi – Non-Cyclic
Menger and Levi - Cyclic
Again - Shaken (coordinates slightly perturbed)
Menger and Its Complementof G(10,3)
Genus of G(10,3) is 2.
• List of faces:1. {6, 7, 17, 20, 13, 16}
2. {5, 6, 16, 19, 12, 15}
3. {4, 5, 15, 18, 11, 14}
4. {3, 13, 20, 10, 9, 8, 7, 6, 5, 4}
5. {2, 3, 4, 14, 17, 7, 8, 18, 15, 12}
6. {1, 2, 12, 19, 9, 10}
7. {1, 10, 20, 17, 14, 11}
8. {1, 11, 18, 8, 9, 19, 16, 13, 3, 2}• Euler formula: V - E + F = 2 - 2g.
• 20 - 30 + 8 = -2 = 2 - 2g.
• g = 2.
Clebsch hexagon
Clebsch hexagon – revisited
123
4
5
6
7
8
9
10
11
12
13
14
15
16
Clebsch graph
Hypercube Q4
Clebsch graph – revisited
• Connection to hypercube?
Exercises 14
• N1: Show that each (93) configuration is combinatorially self-polar.
• N2: Determine the groups of automorphisms and extended automorphisms.
• N3: Show that the genus of two configurations is 1 while the genus of the third one is 2. Make models!
• N4: Determine the three Menger graphs and their duals on the minimal surfaces.
• N5: Prove that the complements of the three Menger graphs are respectively C9, C6 [ C3, 3C3.
15. Generalized Polygons
Generalized Polygons
• A generalized polygon is a bipartite graph of diameter d and girth 2d.
• (From Godsil and Royle)
• Any Km,n is a generalized 2-gon.
Near-Pencil
• N(n) is a near-pencil (or degenerate projective plane) with n+1 points and n+1 lines with the incidence shown on the left.
N(4)
Projective Space PG(3,q)
• Let V = Fq4 be the
four-dimensional vector space over the field of order q and let PG(3,q) be the corresponding projective space.
• There are (q4 – 1)/(q-1) = (q + 1)(q2 + 1) projective points in PG(3,q).
The Matrix H
• The matrix H 2 Fq4 £ Fq
4 is defined below.
Totally Isotropic Subspace S of V.
• A subspace S ½ V is totally isotropic if uT H v = 0 for all u,v 2 S.
• Each one-dimensional subspace S is totally isotropic: uT H u = 0.
• A two-dimensional subspace S, spanned by u and v: S = span{u,v} is totally isotropic if and only if uT H v = 0.
u?.
• For u 0 define u? = {v 2 V| uT H v = 0}.• Note that u? is a 3-dimensional subspace of V, containing u, that
is orthogonal to HTu.• In order to count the number of totally isotropic 2-dimensional
subspaces of V, we proceed as follows:• There are q4 – 1 non-zero vectors u 2 V. There are q3-q vectors v
2 u? – span{u}. Hence there are (q4 – 1)(q3 – q) pairs of vectors.• Each 2-dimensional subspace is spanned by (q2-1)(q2-q) pairs,
hence the number of totally isotropic 2-dimensional subspaces of V is given by: (q4 – 1)(q3 – q)/[(q2 –1)(q2 – q)] = (q2 + 1)(q + 1).
W(q)
• W(q) is the incidence structure of all totally isotropic points and totally isotropic lines in PG(3,q). It is a ((q2 + 1)(q + 1)q+1) configuration.
Generalized Quadrangle
• A generalized quadrangle is a partial linear space satisfying the following two conditions:
• Given any line L and a point p not on L there is a unique point p’ on L such that p and p’ are collinear.
• There are non-collinear points and non-concurrent lines.
L
p’
p
Tutte’s 8-Cage
• In 1947 Tutte gave a construction of the only 8-cage on 30 vertices.
Tutte’s 8-Cage – Construction (I)
• Take S(Q3). There are 6 pairs of antipodal new vertices of valence 2. These 6 pairs are naturally grouped into 3 quadruples – a quadruple represents a 1-factor.
Tutte’s 8-Cage – Construction (II)
• The tree on the left has 6 pairs of leaves and these 6 pairs are naturally grouped into 3 quadruples.
Tutte’s 8-Cage – Construction (III)
• By gluing appropriately the leaves of the tree on the left to the midpoints of the edges of the cube on the right one obtains Tutte’s 8-Cage.
• Cubic graph
• Bipartite graph
• Girth 8
• Diameter 4.
Question
• Q. If we subdivide the edges of K4 we may attach the tree on the left to it in such a way that we avoid quadrangles. What graph is produced in ths way?
Similar Question• Same for the S(K2,2,2) and
the tree. First layer antipodal edges, second layer main squares of the octahedron. Truncate vertices of valence 4.
• (What about S(Q4)?)
4-dimensional Cube Q4.
0000
1000
0010
0100
0001
1100
1110
0110
01110011
1001 1101
11111011
1010
W(2) and Q4.
• W(2) can be modelled on the vector space F2
4 (represented as hypercube).
• What are totally isotropic points (lines throug the origin) and lines (planes through the origin)?
0000
1000
0010
0100
0001
1100
1110
0110
0111
0011
1001 1101
11111011
1010
W(2)
• W(2) is a (153) configuration. Its Levi graph is Tutte’s 8-cage.
• W(2) admits geometric realization that is known as the Cremona-Richmond Configuration.
Cremona Richmond Configuration
• Cremona Richmond Configuration can be drawn by exhibiting pentagonal cyclic symmetry.
• It is the smallest triangle-free (v3) configuration.
Cremona-Richmond Configuration in Space
• Take the following points related to tetrahedon.• 4 vertices
• 6 midpoins of the edges.
• 4 centers of triangles
• 1 center of the tetrahedon
• The following lines:• 4 x 3 = 12 triangle hights
• 3 lines connecting antipodal midpoints of edges and the center
• The resulting structure is the Cremona-Richmond configuration.
Exercises 15-1
• N1: Prove that in PG(n,q) there are (qn+1 – 1)(qn+1 – q) ... (qn + 1 – qp)/ [(qp+1 – 1)(qp+1- q) ... (qp+1 – qp)] projective subspaces of dimension p.
Exercises 15-2
• N2: Study properties of W(3). By definition it is a (404) triangle-free configuration. What is its symmetry group?
• N3: Find one of its drawings.
• N4: Prove that the Levi graph of W(3) is semi-symmetric (= regular, edge-transitive but not vertex-transitive).
16. Cages and Configurations
The Balaban 10-cage• The Balaban 10-cage is presented on
the left. This is one of the three smallest cubic graphs of girth 10. It has 70 vertices, a symmetry is clearly visible.
• The cage has a Hamilton cycle. For instance one of its LCF codes is given here:
• [-9, -25, -19, 29, 13, 35, -13, -29, 19, 25, 9, -29, 29, 17, 33, 21, 9, -13, -31, -9, 25, 17, 9, -31, 27, -9, 17, -19, -29, 27, -17, -9, -29, 33, -25, 25, -21, 17, -17, 29, 35, -29, 17, -17, 21, -25, 25, -33, 29, 9, 17, -27, 29, 19, -17, 9, -27, 31, -9, -17, -25, 9, 31, 13, -9, -21, -33, -17, -29, 29]
The other two 10-cages• Besides the Balaban
cage there are two more 10-cages. The more symmetric one is drawn here.
• LCF:• [(-29, -19, -13, 13,
21, -27, 27, 33, -13, 13, 19, -21, -33,
29)5]
The third 10-cage
• The third 10-cage is the least symmetric.
• LCF:• [9, 25, 31, -17, 17, 33, 9,
-29, -15, -9, 9, 25, -25, 29, 17, -9, 9, -27, 35, -9, 9, -17, 21, 27, -29, -9, -25, 13, 19, -9, -33, -17, 19, -31, 27, 11, -25, 29, -33, 13, -13, 21, -29, -21, 25, 9, -11, -19, 29, 9, -27, -19, -13, -35, -9, 9, 17, 25, -9, 9, 27, -27, -21, 15, -9, 29, -29, 33, -9, -25].
10-cages
• All 10-cages are hamiltonian (see their LCF description).
• Respective automorphism group orders: • 80, 120, 24.
• Reference: T.Pisanski, M. Boben, D. Marušič, A. Orbanič, A. Graovec: The 10-cages and derived Configurations, Discrete Math. 275 (2003), 265--276.
17. A Case Study – The Gray Graph
The Gray Graph G
• The smallest known cubic edge- but not vertex-transitive graph has 54 vertices and is known as the Gray graph. It is denoted by G.
• Since its girth is 8, it is the Levi graph of two dual, smallest, triangle-free, point-, line- and flag-transitive, non-self-dual (273)-configurations.
The Gray Configuration
• Cyclic drawings of two dual Gray configurations.
• These drawings show the problem of a straight-line realizations of configurations. They both contain false incidences.
Gray Configuration Revisited
• There is a much better drawing of the Gray configuration available.
• Using this drawing it becomes clear that the Menger graph M of the Gray configuration is isomorphic to K3 K3 K3 .
Menger and Dual Menger Graph
• The representation of the configuration in the previous slide determines the choice between the Gray configuration (and Menger graph M = K3
3) and the dual Gray configuration and the dual Menger graph D.
The genus of a graph
• Let (G) denote the genus of the graph G. This parameter denotes the least integer k, such that G admits an embedding into an orientable surface of genus k.
The Genus of K3 K3 K3
• Several years ago it was shown that the genus of (K3 K3 K3 ) = 7. The genus embedding was constructed by Mohar, Pisanski, Škoviera and White. It is depicted in the figure.
b a
c
c'
b'
ac'
b'
c'
a'
b'
a'
b'
a'
c
c'
b
b
b'
a'
c
a
c+
c' b
a'
a
The Genus Embedding
• The genus embedding has some very nice features.– It contains a bipartite dual.– If we color the faces in two colors all 27
triangles get a single color.
• This means that points of the Gray configurations correspond to vertices while the lines correspond to the triangles.
The Gray Graph admits an embedding into a surface of
genus 7• If we keep the original
vertices and introduce the centers of triangles as new vertices with an old vertex v adjacent to a new vertex t if and only if v lies on the boundary of the triangle t, the resulting graph is the Gray graph.
• Hence the Gray graph fits onto the same surface!
The Gray Graph admits an embedding into a surface of
genus 7• If we keep the original
vertices and introduce the centers of triangles as new vertices with an old vertex v adjacent to a new vertex t if and only if v lies on the boundary of the triangle t, the resulting graph is the Gray graph.
• Hence the Gray graph fits onto the same surface!
The lower bound
• The upper bound for genus is 7. The lower bound 7 follows from the following:
• Proposition: Let L be the Levi graph and let M be the Menger graph of some (v3) configuration C, then (M) (L).
• Proof: Start with the genus embedding of L. By the reverse process depicted in the figure one can obtain the embedding of M in the same surface.
The dual Menger graph D
• There is just one unfinished case to consider. Namely, the dual Menger graph D can also be embedded into the surface of genus 7. It turns out that this graph is quite interesting. It is a Cayley graph of the semidirect product Z3 ⋉ Z9. It can be described as a Z9-covering graph over the base graph in the next slide.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
The Voltage Graph
• The Dual Menger graph is the Z9 covering graph over the voltage graph on the left.
• It can be viewed as the Cayley graph for the following presentation.
+4
+1
-1+4
-2
+2
+2
-4
+1
The Holt Graph
• The 4-valent Holt graph H is a spanning subgraph of on 27 vertices is the smallest 1/2-arc transitive graph. This means it is vertex- and edge- but not arc-transitive. It is depicted in the figure on the left.
The Holt Graph - again
• The 4-valent Holt graph H is an induced subgraph of the graph D. It is obtained from D by removing a suitable 2-factor composed of three 9-cycles. H is a Z9-covering graph over the green voltage graph on the left [with the 3 red loops removed.]
+4
+1
-1+4
-2
+2
+2
-4
+1
Some Presentations for Z3 ⋉ Z9
• Both H and D are Cayley graphs for the same group Z3 ⋉ Z9.
– Z3 ⋉ Z9 = <a, b | a9 = b3 = 1, b-1ab = a2>
– D = <x, y, z | x9 = y9 = z9 = 1, y-1xy = x2, y-1zy = z2, x-1yx = y2, x-1zx = z2, z-1xz = x2, z-
1yz = y2>– H is obtained from D by omitting any single
generator x,y,or z.
The Final Problems
• What is the genus of D? What is the genus of H? The genus of Z3 ⋉ Z9 is known. It is (Z3 ⋉ Z9) = 4. On the other hand we proved that D admits an embedding into the surface of genus 7. Hence– 4 (D) 7– 4 (H) 7
• In the first case one should improve the lower bound, in the second, the upper bound should be improved.
18. Another Case Study - Tennis Doubles
Story
• This problem was posed by an undergraduate student, Jure Kališnik, a tennis doubles fan, during the lectures on configurations that were held in Ljubljana in 2002.
Tennis Club Problem
• There are n players in a Tennis Club TC. The club owns two tennis courts. Occasionaly the club organizes a Tennis Doubles tournament with m rounds on both courts (altogether 2m games) along the following rules:
• TC1. Each player plays k games.
• TC2. During the tournamet each player meets any other player on a court at most once.
• TC3. No player may appear in the same round in both courts.
The Model
• We may model the tournament schedule by a configuration table. The table has 4 rows and 2m columns. The first m columns correspond to the games played on court A while the second m columns correspond to the games played on court B.
1. Each column has distinct marks. 2. Each pair of marks appears in the same column at most once.3. Each mark appears k times.4. 8m = nk5. The first m columns are permuted arbitrarily.
6. The second m columns are permuted in such a way that Ci Å Ci+m = ;, for i = 1..m.
The Model and Configurations
• We may model the schedule as a (nk,2m4) configuration. Clearly k = 8m/n.
• For k · 4 we use the inequality:• n ¸ 2m ¸ 1 + 4(k - 1)
• may obtain smallest cases:• k = 1, n = 8, m = 1 (81, 24)
• k = 2, n = 12, m = 3 (122, 64)
• k = 3, n = 16, m = 6 (163,124)
• k = 4, n = 14, m = 7 (144,144)
The Model and Configurations
• Again, k = 8m/n.• For k > 4, we use the inequality:
• 2m ¸ n ¸ 1 + 3k.
• may obtain smallest cases:• k = 5, n = 16, m = 10 (165, 204)
• k = 6, n = 20, m = 15 (206, 304)
• k = 7, n = 24, m = 21 (247, 424)
• k = 8, n = 25, m = 25 (258, 504)
• k = 9, n = 32, m = 36 (329, 724)
Tennis Court A
1 2 - 3 4
9 10 - 11 12
1 6 - 11 16
2 7 - 12 17
4 9 - 14 19
4 7 - 10 13
2 5 - 16 19
8 11 - 14 17
2 10 - 14 18
8 12 - 16 20
Tennis Court B
5 6 - 7 8
13 14 - 15 16
17 18 - 19 20
3 8 - 13 18
5 10 - 15 20
3 6 - 9 20
1 12 - 15 18
1 5 - 9 13
3 7 - 11 19
4 6 - 15 17
19. The Martinetti-Boben Theorem
Martinetti's "Theorem"• In 1887 V. Martinetti introduced a notion of reduction, and the notion
of irreducible configuration. Using his reduction, a (v3) configuration can be reduced to a ((v-1)3) configuration. He also determined the class of all irreducible configurations.
• Using Martinetti's idea each (v3) configuration can be obtained by a sequence of operations that are inverse to reductions.
• There are two problems with his approach. • Martinetti never made any clear distinctions between geometric and combinatorial
configurations. His method is purely combinatorial and gives no hint to what extent it applies to geometric configurations.
• The other problem lies in the fact that Martinetti made an error in the proof and omitted some irreducible configurations. So not only the proof but the theorem itself has to be modified. This was done by Marko Boben in his master’s thesis.
(v3) graphs
• A connected, cubic, bipartite graph with girth at least 6 is called a (v3) graph.
• A (v3) graph is a Levi graph of some connected (v3) configuration.
Martinetti reduction
• A Martinetti reduction on a Levi graph is depicted on the left. In the corresponding configuration a line and incident point is deleted.
• Attention: A Martinetti reduction can be applied in two ways by reattaching pending edges! It can only be applied if no quadrilaterals are formed.
Martinetti reduction and connectivity.
• The question is whether one can apply a Martinetti reduction on a connected graph and obtain a disconnected reduced graph.
• Here is a construction.
u
v
u'
v'
Martinetti reduction and connectivity.
• Take two Levi graphs and consider an edge uv in the first one and en edge u'v' in the second one.
• Subdivide each edge twice: u v1 u1 v and u' v'1 u'1 v' and identify the vertices u1 = u'1 and v1 = v'1
u
v
u'
v'
u'1
v1v'1
u1
Martinetti reduction and connectivity.
• Take two Levi graphs and consider an edge uv in the first one and en edge u'v' in the second one.
• Make double subdivision of each edge: u v1 u1 v and u' v'1 u'1 v' and identify the vertices u1 = u'1 and v1 = v'1
u
v
v1
u1
u'
v'
u'1
v'1
Conditions of Use
• We make the following assumptions on the use of Martinetti’s reduction:
1. It can be used only if no quadrilaterals are introduced.
2. It can only be used if the number of connected components is not increased.
Irreducible configurations and graphs.
• A (v3) combinatorial configuration (or its Levi graph) is irreducible, if we cannot apply Martinetti’s reduction. Other (v3) configurations and (v3) graphs are reducible.
Lemma
• Lemma: A (v3) graph G is irreducible if and only if for each edge e of G we have:
• e together with another edge forms the intersection of two 6-cycles or
• e lies on a path efg of lengh 3, that is the intersection of two 6-cycles.
Proof of Lemma. Part I.
• Let e be an arbitrary edge of an irreducible connected graph G. If it is reduced by Martinetti a 4-cycle is obtained.
• By inspection we see that G contains two adjacent 6-cycles both containing e, and the intersection is a path of length 2 or 3 [and e is not the central edge of the intersection].
e e
Proof of Lemma, Part II.
• If e does not lie on the intersection of two hexagons, the reduction is possible.
• Therefore, let e be an edge in the intersection of two 6-cycles.
• There are several cases to consider. If there is only one edge in the intersection, the graph is reducible. There may be two or three edges in the intersection.
e e
Proof of Lemma, Part II
• Let e be the edge on the intersection of two 6-cycles. In each case a four-cycle would be obtained in the reduction process.
e e
Graph t, Families t(n) and T(n)
Graph t = S(K_4) on the left is the basis for the construction of families t(n) and T(n) of graphs.
T(2) = t(4)
• T(2) = t(4)• T(n) has 20n vertices. Only
6 vertices of valence 2: three on the top, three on the bottom. If the three upper vertices are connected by the three lower vertices a cubic graph results. There are 6 ways of doing this but only three different graphs are obtained: T1(n), T2(n) and T3(n).
Family of graphs C(m)
• Graph C(m) has 6m vertices. It is obtained by connecting m 6-cycles as shown on the left.
• From C(m) three different graphs are obtained D(3m), D(3m+1) and D(3m+2). Each graph D(n) has 2n vertices.
C(3)
A
B
Cca
b
Family of graphs D(3m)
• From C(m) the graph D(3m) is constructed with no new vertices. Only three edges are added: A-a,B-b and C-c and the resulting graph D(3m) is cubic..
D(9)
A
B
Cca
b
Family of graphs D(3m+1)
• From C(m) the graph D(3m+1) is obtained by adding two new vertices: X and x. Five edges are added as shown in the figure.
D(10)
A
B
Cca
b
Family of graphs D(3m+2)
• Exercise: Find a paper and describe the family D(3m+2)
• In order to obtain D(3m+2) there are 4 new vertices added and a path of length 3.
Graphs D(n), n ¸ 7.
• The graphs D(n), n ¸ 7 are Zn covers over the dipole 3. with voltages 1,2,4. Hence D(n) = H(2n-1 + 5) are Haar graphs. It is not hard to verify that the LCF code LCF = [5,-5]n
generates them.
1 2 4
Zn
D(7)
• D(7) = Heawood graph.
D(8)
• D(8) = Möbius- Kantor graph, Levi graph of the unique (83) configuration.
D(9)
• D(9) = Levi graph of the unique cyclic (93) configuration.
D(10)
• D(10) = Levi graph of the unique cyclic (103) configuration.
The Martinetti-Boben Theorem
• Irreducible, connected (v3) graphs are:• The graph of the Pappus configuration.
• Graphs T1(n), T2(n), T3(n), for n ¸ 1
• Graphs D(n), for n ¸ 7.
• Remark: Martinetti overlooked the cases T2(n) and T3(n) and did not mention D(7) and D(8).
Boben reduction
• Recently Marko Boben introduced another reduction in order to further lower the number of irreducible (v3) graphs.
• Take two nonadjacent vertices of different colors. remove them and re-attach the neigubours of one to the neighbours of the second one. There are 6 possibilities.
Theorem
• Theorem: Irreducible, connected (v3) graphs are:
• Pappus graph.
• Heawood graph D(7).
• Remark: Both Martinetti and Boben reductions are allowed and the original condition for use is assumed for both of them.
New Conditions of Use
• We can make new assumtions on the use of Martinetti’s reduction:
1. It can be used only if the girth is not decreased.
2. It can only be used if the number of connected components is not increased.
• In this case we obtain irreducible triangle-free configurations, etc.
Some Questions
• This approach raises a number of interesting questions.
• Which geometric configurations are irreducible in the sense of Martinetti?
• Which triangle-free (combinatorial) configurations are irreducible in the sense of Martinetti?
• Which triangle-free (geometric) configurations are irreducible in the sense of Martinetti?
• Same questions if Boben reductions are added.
Flag sum of combinatorial configurations
• Take two incidence structures and a flag in each of them.
• In Pappus take the marked point and line. And in Fano, the same.
• Switch the adjacencies.
Point – Line Sum of Configurations
• Take two incidence structures and line L in the first one and a point p in the second. (Valence of L must be equal to the valence of p!)
• Delete L from the first and p from the second.
• Attach deficient lines from the second to the deficient points from the first.
• Note: we select the circle in the top Fano plane and the top point of the bottom Fano plane.
Chapter 4. Statistics Page
• Number of slides:276
• Number of sections:19
• Number of exercises:70
• Number of homeworks:3
