Download HOW TO DRAW A HEXAGON 1. Introduction

HOW TO DRAW A HEXAGON
ANDREAS E. SCHROTH
Abstract. Pictures of the G2 (2) hexagon and its dual are presented. A way
to obtain these pictures is discussed.
1. Introduction
This paper presents pictures of the two classical generalised hexagons over the
eld with two elements. Information on these generalised hexagons is employed
so that the pictures make some properties of these hexagons obvious. Familiarity
with the theory of generalised polygons is not assumed. Perhaps this paper will
convince the uninitiated reader that even new and seemingly abstract geometrical
structures sometimes have nice visual presentations.
Generalised polygons were introduced by Tits in [7]. They serve, among other
things, as a geometric realisation of certain groups. Generalised triangles are essentially projective planes, generalised quadrangles are also known as polar spaces
of rank 2. Classical (thick) generalised n-gons, n > 2, exist only for n 2 f3; 4; 6; 8g.
By [3] also thick nite generalised n-gons, n > 2, exist only for n 2 f3; 4; 6; 8g.
If the pictures and discussions presented here whets your appetite for generalised
polygons, I recommend [6] and [9] for further reading.
There is a well-known picture of the smallest projective plane, also known as
the Fano plane (Figure 1). A picture of the smallest generalised quadrangle appeared rst on the cover of a book on generalised triangles [4] and reminds of an
ordinary pentagon (Figure 2). The picture is due to S. Payne, who calls it doily for
obvious reasons. More stunning pictures of geometries may be found in Polster's
Geometrical Picture Book [5].
Figure 1
Figure 2
1991 Mathematics Subject Classication. 51E12.
Key words and phrases. generalised hexagon, generalised polygon, coordinatisation, models of
geometries.
1
2
ANDREAS E. SCHROTH
The classical generalised hexagons over GF(2) are the G2 (2) hexagon and its
dual. Both are hexagons of order 2. There are 63 points and 63 lines. Moreover,
there are explicit descriptions of these generalised hexagons. So, in principle, it is
possible to produce some pictures of these hexagons. However, just placing and
labelling 63 dots on a sheet of paper and connecting collinear ones by a curve will
most likely produce a mess. To obtain a nice picture one has to know and employ
structural properties of the hexagons. The main property used here is the fact
that the cyclic group of order 7 acts on the G2 (2) hexagon and hence also on its
dual. Thus, the 63 points split into nine orbits with seven points each. Likewise
for lines. Also, three orbits of lines form dierent types of ordinary 7-gons. This
takes care of 6 orbits of points. There are lines connecting vertices of one kind of
7-gons with midpoints of another type of 7-gons. This gives three more orbits of
lines. In the G2 (2) hexagon this also gives an additional orbit of points. In the
G2 (2) hexagon the remaining two orbits of points and three orbits of lines form
the incidence graph of the Fano plane. The midpoints of the edges of this graph
are glued to midpoints of the three types of ordinary 7-gons. In the dual hexagon
the orbits not involved in the three ordinary 7-gons do not form easily recognisable
gures.
To nd a cyclic action of order seven and to work out the incidences two dierent
coordinatisations of the G2 (2) hexagon are used. The rst coordinatisation, due
to Tits [7], is good for calculations but less practical to detect incidences. The
second coordinatisation, introduced by De Smet and Van Maldeghem in [2], is ne
to detect incidences.
2. Preliminaries
A point-line geometry I = (P ; L; I) consists of a point set P , a line set L and an
incidence relation I between points and lines. Often lines are considered as sets of
points. Then incidence is given by inclusion and will not be mentioned explicitly.
The elements of the union V = P [ L are also called vertices.
A chain (v0 ; v1 ; : : : ; vn ), vi 2 V , is an ordered set of vertices such that vi?1 I vi
for 1 i n. The number of vertices in a chain diminished by 1 is called the
length of the chain. A chain (v0 ; : : : ; vn ) connects two vertices p and q if p = v0
and q = vn . Two vertices p and q are at distance k, if there is a chain of length k
connecting p to q and if the length of every chain connecting p to q is at least k.
Two vertices that cannot be connected by a chain are at innite distance.
A generalised hexagon is a point-line geometry H = (P ; L) satisfying the following axioms:
1. The distance between two vertices is at most six.
2. For any two vertices at distance k < 6 there exists a unique connecting chain
of length k.
3. Every vertex is incident with at least three vertices.
The dual of an incidence geometry is obtained by interchanging the r^oles of
points and lines. Since the denition of a generalised hexagon is given entirely in
terms of vertices, the dual of a generalised hexagon is again a generalised hexagon.
HOW TO DRAW A HEXAGON
3
An ordinary hexagon satises the rst two axioms but not the third. Instead
every vertex is incident with precisely two vertices.
Two points p; q are said to be collinear if they can be joined by a line. The line
connecting two collinear points p and q is denoted p _ q. The set of lines incident
with a point p is denoted Lp.
A nite generalised hexagon is a generalised hexagon with only nitely many
vertices. One readily veries that for a nite generalised hexagon there are parameters s and t such that every line contains s +1 points and every point lies on t +1
lines. These parameters are also called the order of the generalised hexagon.
Suppose p is a point of a nite generalised hexagon of order (s; t). One sees that
the set of points at distance 6, 4 or 2 from p consists of s3 t2 , (t + 1)ts2 or (t + 1)s
points, respectively. Thus, all in all, there are s3 t2 + s2 t2 + s2 t + st + s + 1 points.
A generalised hexagon with a minimal number of points has parameters s = t = 2.
Thus there are 63 points. By duality there are also 63 lines.
3. The classical Hexagon
According to [1] there are only two generalised hexagons with parameters 2.
They are dual to each other and are related to the Chevalley group G2 (2). We
will call one of these hexagons the G2 (2) hexagon.
The G2 (2) hexagon can be described explicitly in terms of coordinates in several
ways. In this paper we will use two dierent coordinatisations. The rst coordinatisation, introduced by Tits in [7], views the points of the G2 (2) hexagon as the
points of the quadric Q(6; 2) represented in the projective space PG(6; 2). Thus
P = f(x0 ; x1; x2 ; x3 ; x4 ; x5 ; x6) j x0x4 + x1 x5 + x2 x6 = x23 g:
Lines of the G2 (2) hexagon are lines on this quadric whose Grassmanian coordinates satisfy
p12 = p34; p20 = p35 ; p01 = p36; p03 = p56 ; p13 = p64; p23 = p45 :
The advantage of this coordinatisation is that automorphisms can be described in
terms of matrices. The disadvantage is that collinearity is not easily detected.
The second coordinatisation is due to De Smet and Van Maldeghem and was
introduced in [2]. We will restrict to the case where the eld is GF(2). There is
one special point denoted (1). All other points are of the form (a0 ; : : : ; ak ), where
0 k 5 and ai 2 GF(2). Lines are denoted in the same way with the dierence
that square brackets are used instead of brackets. Incidence is given by
[k; b; k0 ; b0 ; k00 ] I (k; b; k0 ; b0 ) I [k; b; k0 ] I (k; b) I [k] I (1)
I [1] I (a) I [a; l] I (a; l; a0 ) I [a; l; a0 ; l0 ] I (a; l; a0 ; l0 ; a00 )
4
ANDREAS E. SCHROTH
DS-VM
(1)
(a)
(k; b)
(a; l; a0 )
(k; b; k0 ; b0 )
(a; l; a0 ; l0 ; a00 )
Tits
(1; 0; 0; 0; 0; 0; 0)
(a; 0; 0; 0; 0; 0; 1)
(b; 0; 0; 0; 0; 1; k)
(l + aa0 ; 1; 0; a; 0; a; a0 )
(k0 + bb0 ; k; 1; b; 0; b0 ; b + b0 k)
(al0 + a0 + a00 l + aa0 a00 ; a00 ; a; a0 + aa00 ; 1; l + aa00 ; l0 + a0 a00 )
Table 1: Translation between the two coordinatisations
and
8 a00 = ak + b
>
< 0 = ak + b0
(a; l; a0 ; l0 ; a00 ) I [k; b; k0 ; b0 ; k00 ] () > la =
+ k00 + aa00 + aa0
: l0 = ak
+ k0 + kk00 + akb + bb0 + ab
8 b = ak
+ a00
>
< b0 = ak
ak + a0
() > k00 = ak
0
: k0 = ak ++ ll0++abkl ++abaa00k + a0a00 + aa00
where all calculations are carried out in GF(2). In this coordinatisation it is not
too dicult to determine whether two points are collinear. There is however no
handy way to describe automorphisms.
Luckily there is a translation between the two coordinatisations. Table 1 states
the translation for points if the eld is GF(2). We will mainly work with points,
so we do not need a translation table for lines. For more details on these coordinatisations see [2].
4. An automorphism of order seven
From now on we work entirely in the G2 (2) hexagon. All indices of points and
lines are taken in Z7. Our rst aim is to nd an automorphism of order seven
in the hexagon. Then the 63 points split into nine orbits of seven points each.
Analogously for the lines. In the Tits coordinatisation every automorphism can
be represented by a matrix ([7]).
An ordinary ordered 7-gon is an ordered set of points (a0 ; : : : ; a6 ) such that ak is
collinear with but dierent from ak+1 (see Figure 3). By [8] we know that the group
of automorphisms acts transitively on ordinary ordered 7-gons. In particular, if
(a0 ; : : : ; a6 ) forms an ordinary ordered 7-gon, then there is an automorphism such that ai = ai+1 . As remarked in [8], the order of the group of automorphisms
of the G2 (2) hexagon equals the number of ordered 7-gons. Thus the group of
automorphisms acts sharply transitive on the set of ordered 7-gons. This implies,
that any such has order seven. Our aim is to rst present such a 7-gon and then
to determine .
HOW TO DRAW A HEXAGON
a0
a1
a2
b5
b4 b3
c0
b2
b6 b b1
0
a3
a4
Figure 3
a6
a5
c1
c2
d1 d0 d6
d2
d5
d 3d 4
c3
c4
Figure 4
c6
c5
5
e0
e6
e1
e5
e2
e3
e4
Figure 5
Here is an ordinary 7-gon. It is obtained by gradually moving away from the
point (1) and then coming back via a dierent path.
a0 := (1)
(1; 0; 0; 0; 0; 0; 0);
a1 := (0)
(0; 0; 0; 0; 0; 0; 1);
a2 := (0; 0; 0) (0; 1; 0; 0; 0; 0; 0);
a3 := (0; 0; 0; 0; 0) (0; 0; 0; 0; 1; 0; 0);
a4 := (1; 1; 1; 1; 1) (0; 1; 1; 0; 1; 0; 0);
a5 := (0; 1; 1; 1) (0; 0; 1; 1; 0; 1; 1);
a6 := (0; 1)
(1; 0; 0; 0; 0; 1; 0):
The connecting lines are:
A4 := a0 _ a1 = [1];
A5 := a1 _ a2 = [0; 0];
A6 := a2 _ a3 = [0; 0; 0; 0];
A0 := a3 _ a4 = [1; 0; 0; 0; 0];
A1 := a4 _ a5 = [0; 1; 1; 1; 1];
A2 := a5 _ a6 = [0; 1; 1];
A3 := a6 _ a0 = [0]:
In the Tits coordinatisation the third point on every line is the sum of the other
two points. Thus, the third point on Ak besides ak+3 and ak?3 is bk := ak?3 + ak+3.
Table 3 lists the points bi .
The points ak , 0 k 6, form a basis of the GF(2)7 . Thus there is a unique
linear map of GF(2)7 with ak = ak+1 , or equivalently, a0 k = ak . By linearity
we also have b0 k = bk and A0 k = Ak . By elementary linear algebra one obtains:
00 0 0 0 0 1 01
B
0 0 1 0 1 0 1C
B
C
B
0 0 0 0 1 0 0C
B
C
C
0
0
1
1
0
0
0
=B
:
B
C
B
C
0
1
0
0
1
0
0
B
@0 0 1 0 0 0 0 C
A
1 0 1 0 0 0 0
6
ANDREAS E. SCHROTH
5. Working out the incidence
We now have to nd the other seven orbits of points and eight orbits of lines.
Indices of points and lines are such that vk = v0 k .
To obtain the other orbits we gradually move away from the point (1). The
lines through a0 = (1) are A3 = [0], A4 = [1] and B0 := [1] = f(1); (1; 0); (1; 1)g.
Let
g0 := (1; 0) (0; 0; 0; 0; 0; 1; 1);
f0 := (1; 1) (1; 0; 0; 0; 0; 1; 1):
The orbits of g0 and f0 under are listed in Table 3. By construction we have
Bk = fak ; fk ; gk g.
The lines intersecting the line B0 besides A4 and A5 are:
D0 := [1; 0; 0] f(1; 0); (1; 0; 0; 0); (1; 0; 0; 1)g;
F0 := [1; 0; 1] f(1; 0); (1; 0; 1; 0); (1; 0; 1; 1)g;
I0 := [1; 1; 0] f(1; 1); (1; 1; 0; 0); (1; 1; 0; 1)g;
E0 := [1; 1; 1] f(1; 1); (1; 1; 1; 0); (1; 1; 1; 1)g:
Incident with one of these lines but not with B0 are the points:
(1; 0; 0; 0) (0; 1; 1; 0; 0; 0; 0) = b0 ;
(1; 0; 0; 1) (0; 1; 1; 0; 0; 1; 1) =: c0 ;
(1; 0; 1; 0) (1; 1; 1; 0; 0; 0; 0) =: e0 ;
(1; 0; 1; 1) (1; 1; 1; 0; 0; 1; 1) =: d0 ;
(1; 1; 0; 0) (0; 1; 1; 1; 0; 0; 1) =: x0 ;
(1; 1; 0; 1) (1; 1; 1; 1; 0; 1; 0) =: y0 ;
(1; 1; 1; 0) (0; 1; 1; 1; 0; 1; 0) =: u0 ;
(1; 1; 1; 1) (1; 1; 1; 1; 0; 0; 1) =: v0 :
Table 3 lists the orbits of c0 , d0 and e0 under . We nd u0 = e5 and v0 = e2 .
Hence Dk = fbk ; ck ; gk g, Fk = fdk ; ek ; gk g and Ek = fek+2 ; ek?2 ; fk g. The orbit of
e0 forms an ordinary 7-gon where ek is collinear with ek+3 . The connecting line
is Ek?2 . Figure 5 depicts such a 7-gon. We say such a 7-gon is of type 3.
We now look at the lines through d0 . These are
F0 = [1; 0; 1];
H0 := [1; 0; 1; 1; 0] f(1; 0; 1; 1); (1; 0; 0; 0; 1); (0; 0; 1; 1; 0)g;
C0 := [1; 0; 1; 1; 1] f(1; 0; 1; 1); (1; 1; 0; 1; 1); (0; 1; 1; 0; 0)g:
For the points on H0 or C0 dierent from g0 we have:
(1; 0; 0; 0; 1) (0; 1; 1; 1; 1; 1; 0) =: h0 ;
(1; 0; 1; 1; 0) (1; 0; 0; 1; 1; 0; 1) =: i0 ;
(1; 1; 0; 1; 1) (0; 1; 1; 1; 1; 0; 1) = c6 ;
(0; 1; 1; 0; 0) (1; 0; 0; 1; 1; 1; 0) = c1 :
Table 3 lists the orbits of h0 and i0 . We nd x0 = h4 and y0 = i3 . Thus Hk =
fdk ; hk ; ik g, Ik = ffk ; hk?3; ik+3 g and Ck = fck+1; ck?1 ; dk g. The last equation
shows that the orbit of c0 forms an ordinary 7-gon where ck is collinear with ck+2 .
HOW TO DRAW A HEXAGON
7
The connecting line is Ck+1 . Figure 4 depicts such a 7-gon. We say such a 7-gon
is of type 2.
By now we have all points but one orbit of lines is still missing. One of the
missing lines is incident with b0 . The lines through b0 are D0 = [1; 0; 0], A0 =
[1; 0; 0; 0; 0] and G0 := [1; 0; 0; 0; 1]. The points on G0 dierent from b0 are
(1; 0; 1; 0; 1) (0; 1; 1; 0; 1; 1; 1) = i2 ;
(0; 1; 0; 1; 0) (0; 0; 0; 0; 1; 1; 1) = h5 :
Thus Gk = fbk ; hk?2 ; ik+2 g.
The equations for Hk , Ik and Gk show that the orbits of h0 and i0 may be
represented as a 14-gon, where ik is connected to hk by the line Hk , to hk+1 via
the line Ik?3 and to hk+3 via the line Gk?2. This is the incidence graph of the
Fano plane (Figure 6). The dual of this conguration is depicted in Figure 7.
i1
h1
i0 h0
i6
i5
h2
i2
h6
h3
i3 h4
i4
h5
Figure 6
G0
G1
G6
I0
H0
G2
G3
G5
G4
Figure 7
Table 2 summarises all incidences between points and lines.
Ak = fak+3; ak?3 ; bk g
Bk = fak ; fk ; gk g
Ck = fck+1 ; ck?1; dk g
Dk = fbk ; ck ; gk g
Ek = fek+2 ; ek?2; fk g
Fk = fdk ; ek ; gk g
Gk = fbk ; hk?2 ; ik+2g
Hk = fdk ; hk ; ik g
Ik = ffk ; hk?3; ik+3 g
Lak = fAk+3 ; Ak?3; Bk g
Lbk = fAk ; Dk ; Gk g
Lck = fCk+1; Ck?1 ; Dk g
Ldk = fCk ; Fk ; Hk g
Lek = fEk+2; Ek?2 ; Fk g
Lfk = fBk ; Ek ; Ik g
Lgk = fBk ; Dk ; Fk g
Lhk = fGk+2; Hk ; Ik+3 g
Lik = fGk?2; Hk ; Ik?3 g
Table 2: Lines and line pencils in the G2 (2) hexagon
8
ANDREAS E. SCHROTH
a0 = (1;0;0;0;0;0;0)
a1 = (0;0;0;0;0;0;1)
a2 = (0;1;0;0;0;0;0)
a3 = (0;0;0;0;1;0;0)
a4 = (0;1;1;0;1;0;0)
a5 = (0;0;1;1;0;1;1)
a6 = (1;0;0;0;0;1;0)
d0 = (1;1;1;0;0;1;1)
d1 = (1;0;0;1;1;1;1)
d2 = (1;0;1;1;1;0;0)
d3 = (0;0;1;0;1;1;0)
d4 = (1;0;1;1;1;1;0)
d5 = (1;0;1;0;1;1;1)
d6 = (1;1;1;1;1;1;1)
g0 = (0;0;0;0;0;1;1)
g1 = (1;1;0;0;0;0;1)
g2 = (0;1;0;0;1;0;1)
g3 = (0;0;1;0;0;0;0)
g4 = (0;1;0;1;0;1;1)
g5 = (1;1;0;1;1;0;1)
g6 = (0;0;1;1;0;0;1)
b0 = (0;1;1;0;0;0;0)
b1 = (0;1;0;1;1;1;1)
b2 = (1;0;1;1;0;0;1)
b3 = (0;0;0;0;0;1;0)
b4 = (1;0;0;0;0;0;1)
b5 = (0;1;0;0;0;0;1)
b6 = (0;1;0;0;1;0;0)
e0 = (1;1;1;0;0;0;0)
e1 = (0;1;0;1;1;1;0)
e2 = (1;1;1;1;0;0;1)
e3 = (0;0;0;0;1;1;0)
e4 = (1;1;1;0;1;0;1)
e5 = (0;1;1;1;0;1;0)
e6 = (1;1;0;0;1;1;0)
h0 = (0;1;1;1;1;1;0)
h1 = (1;0;1;0;0;1;0)
h2 = (1;1;0;1;0;1;1)
h3 = (1;1;0;1;1;0;0)
h4 = (0;1;1;1;0;0;1)
h5 = (0;0;0;0;1;1;1)
h6 = (1;0;1;0;1;0;1)
c0 = (0;1;1;0;0;1;1)
c1 = (1;0;0;1;1;1;0)
c2 = (1;1;1;1;1;0;0)
c3 = (0;0;1;0;0;1;0)
c4 = (1;1;0;1;0;1;0)
c5 = (1;0;0;1;1;0;0)
c6 = (0;1;1;1;1;0;1)
f0 = (1;0;0;0;0;1;1)
f1 = (1;1;0;0;0;0;0)
f2 = (0;0;0;0;1;0;1)
f3 = (0;0;1;0;1;0;0)
f4 = (0;0;1;1;1;1;1)
f5 = (1;1;1;0;1;1;0)
f6 = (1;0;1;1;0;1;1)
i0 = (1;0;0;1;1;0;1)
i1 = (0;0;1;1;1;0;1)
i2 = (0;1;1;0;1;1;1)
i3 = (1;1;1;1;0;1;0)
i4 = (1;1;0;0;1;1;1)
i5 = (1;0;1;0;0;0;0)
i6 = (0;1;0;1;0;1;0)
Table 3: Orbits of points under 6. Drawing the hexagons
All information needed to draw the G2 (2) hexagon and its dual is encoded in
Table 2. We know already that both hexagons contain disjoint ordinary 7-gons of
type 1, 2 and 3. Now, in the G2 (2) hexagon every vertex of a 7-gon of type i is
collinear with the opposite midpoint of the 7-gon of type i + 1. (The types are
enumerated modulo 3.) The connecting lines are Bk , Dk and Fk . In the dual
hexagon every vertex of a 7-gon of type i is collinear with the opposite midpoint of
the 7-gon of type i ? 1. The connecting lines are Lbk , Ldk and Lfk . This indicates
that the G2 (2) hexagon is not selfdual.
But also these lines connecting vertices to midpoints behave dierently in the
G2 (2) hexagon and its dual. In the G2 (2) hexagon the lines Bk , Dk and Fk
intersect. The common point is gk . In the dual the corresponding lines Lbk , Ldk
and Lfk do not intersect. However, the midpoints Bk , Dk and Fk are collinear.
The automorphism together with the reection where (vk ) = v?k if
k 2 fa; b; c; d; e; f; gg and (hk ) = i?k , (ik ) = h?k generate a subgroup of the
automorphism group isomorphic to the dihedral group of order 14. However, if
one draws the points and lines considered so far in such a fashion that the dihedral
group acts on the picture, then dierent lines intersect in intervals and can no
HOW TO DRAW A HEXAGON
9
longer be distinguished. More precisely, for any 0 k 6 the lines Bk , Dk and Fk
overlap. Thus, in the picture some of the `midpoints' are slightly o the middle.
To complete the picture of the G2 (2) hexagon the incidence graph of the Fano
plane is added in in such a fashion that the midpoints of the edges are glued to
midpoints of the 7-gons. To avoid overlapping lines the edges are bent. The picture
of the dual hexagon is completed analogously.
Figure 8: The G2 (2) hexagon
Figure 8 depicts the G2 (2) hexagon while Figure 9 depicts the dual hexagon. If
we use one more dimension, that is, if we construct spatial models of the hexagons,
we can avoid overlapping lines by placing the dierent 7-gons in dierent but parallel planes. This produces models upon which the dihedral group acts. Stereograms
of such models are presented on my homepage.
10
ANDREAS E. SCHROTH
Figure 9: The dual of the G2 (2) hexagon
Acknowledgements. I acknowledge the contribution made by one of the referees.
He or she made me realise the action of the dihedral group. This resulted in
improved pictures.
[1]
[2]
[3]
[4]
References
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W. Feit and G. Higman. The nonexistence of certain generalized polygons. J. Algebra, 1:114{
131, 1964.
S.E. Payne. Finite generalized quadrangles, a survey. In Proceedings of the International Conference on Projective Planes, pages 219{261. Washington State University Press, Pullman,
1973.
HOW TO DRAW A HEXAGON
11
[5] B. Polster. A Geometrical Picture Book. Springer, New York, 1998.
[6] J.A. Thas. Generalized polygons. In F. Buekenhout, editor, Handbook of Incidence Geometry:
Buildings and Foundations, pages 383{432. North-Holland, Amsterdam, 1995.
[7] J. Tits. Sur la trialite et certains groupes qui s'en deduisent. Publ. Math. I.H.E.S., 2:13{60,
1959.
[8] H. Van Maldeghem. A nite generalized hexagon admitting a group acting transitively on
ordered heptagons is classical. J. Combin. Theory Ser. A, 75(2):254{269, 1996.
[9] H. Van Maldeghem. Generalized Polygons. Birkhauser, Basel, 1998.
Institut fu r Analysis, TU Braunschweig, Pockelsstr. 14, D-38106 Braunschweig,
Germany,
E-mail address : [email protected]
URL: http://fb1.math.nat.tu-bs.de/~top/aschroth