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As Francis Buekenhout defined it in [Bue79], the diagram of a firm, residually connected, flag--transitive
geometry Γ is a complete graph K, whose vertices are
the elements of the set of types I of Γ, provided with
some additional structure which is further described as follows.
To each vertex i ∈I, we attach the order si which is | ΓF | - 1 where F is any flag of type I\{
i }, and the number ni of elements of type i of Γ.
To every edge { i, j } of K, we associate three
positive integers dij, gij and dji where gij
(the gonality) is equal to half the girth of the incidence graph
of a residue ΓF of type { i, j }, and dij
(resp. dji), the i--diameter (resp. j--diameter) is
the greatest distance from some fixed i--element (resp. j--element)
to any other element in ΓF.
On a picture of the diagram, this structure will often be depicted as
follows.
i d_ij g_ij d_ji j
O--------------------O
s_i-1 s_j-1
n_i n_j
Moreover, when gij = dij = dji = n != 2, we only write
gij and if n = 2, we do not draw the edge { i, j }.
Diagram(C) : CosetGeom -> GrphUnd, GrphVertSet, GrphEdgeSet
Given a firm, residually connected and flag--transitive incidence
geometry D or a coset geometry C, return a complete graph K whose
vertices and edges are labelled in the following way. Every vertex i
of K is labelled with a sequence of two positive integers, the first
one being si and the second one being the number of elements of
type i in D. Every edge { i, j } is labelled with a
sequence of three positive integers which are respectively dij,
gij and dji.
Observe that both these functions do the same but the second one
works much faster than the first one since it uses groups to compute
the parameters of the diagram. So when the user has to compute the
diagram of an incidence geometry, it is strongly advised that he first
converts it into a coset geometry and then computes the diagram of
the corresponding coset geometry.
Back to the Petersen graph : let us now explore a little more the first
incidence geometry we constructed. Suppose that ig is the rank two
incidence geometry corresponding to the Petersen graph. We may test
whether it is firm, residually connected and flag--transitive.
> IsFirm(ig);
true
> IsRC(ig);
true
> IsFTGeometry(ig);
true
Since it satisfies these three properties, we can compute its diagram.
> d, v, e := Diagram(ig);
> d;
Graph
Vertex Neighbours
1 2 ;
2 1 ;
> for x in v do print x, Label(x);end for;
1 [ 1, 10 ]
2 [ 2, 15 ]
> for x in e do print x, Label(x);end for;
{1, 2} [ 5, 5, 6 ]
We thus see that the diagram of ig can be drawn as follows.
1 5 5 6 2
O-------------------O
1 2
10 15
Back to the cube: let ig be the rank three incidence geometry of the cube as constructed above and compute its diagram.
> cube := IncidenceGeometry(g);
> d, v, e := Diagram(cube);
> d;
Graph
Vertex Neighbours
1 2 3 ;
2 1 3 ;
3 1 2 ;
> for x in v do x, Label(x); end for;
1 [ 1, 8 ]
2 [ 1, 12 ]
3 [ 1, 6 ]
> for x in e do x, Label(x); end for;
{1, 2} [ 4, 4, 4 ]
{1, 3} [ 2, 2, 2 ]
{2, 3} [ 3, 3, 3 ]
So the diagram can be depicted as follows.
1 4 2 3 3
O-------------------O-------------------O
1 1 1
8 12 6
Back to the Hoffman-Singleton graph:
> ig := IncidenceGeometry(HoffmanSingletonGraph());
> d, v, e := Diagram(ig); d;
Graph
Vertex Neighbours
1 2 ;
2 1 ;
> for x in v do print x, Label(x); end for;
1 [ 1, 50 ]
2 [ 6, 175 ]
> for x in e do print x, Label(x); end for;
{1, 2} [ 5, 5, 6 ]
So the diagram can be depicted as follows.
1 5 5 6 2
O-------------------O
1 6
50 175
Back to Neumaier's geometry : let ig be the rank four incidence
geometry of Neumaier, as described above and compute its diagram.
> d, v, e := Diagram(neumaier);
> d;
Graph
Vertex Neighbours
1 2 3 4 ;
2 1 3 4 ;
3 1 2 4 ;
4 1 2 3 ;
> for x in v do print x, Label(x); end for;
1 [ 2, 50 ]
2 [ 2, 50 ]
3 [ 2, 50 ]
4 [ 2, 50 ]
> for x in e do print x, Label(x); end for;
{1, 2} [ 4, 4, 4 ]
{1, 3} [ 2, 2, 2 ]
{1, 4} [ 3, 3, 3 ]
{2, 3} [ 3, 3, 3 ]
{2, 4} [ 2, 2, 2 ]
{3, 4} [ 4, 4, 4 ]
So the diagram can be depicted as follows.
. 1 4 2
. 2 O-----------------O 2
. 50 | | 50
. | |
. | |
. 3 | | 3
. | |
. | |
. | 4 |
. 3 O-----------------O 4
. 2 2
. 50 50
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