CiteSeerX — Rectangular Structures From Affine Planes Over Fields

Rectangular structures from ane planes over elds

Tim Boykett

Time's Up, Industriezeile 33b, A-4020 Linz, Austria Mathematik, Uni Linz, A-4040 Linz, Austria

[email protected]

Jan 1998

Abstract

A method for constructing rectangular structures, and thus reversible one-dimensional cellular automata from ane planes is presented. The construction of such examples from ane planes over nite elds is then investigated. Explicit formulae are given for the semicentral bigroupoid operations, and the automorphism group of the rectangular structure is determined.

1 Introduction

Denition 1

A Semicentral Bigroupoid is a (2

;

2)-algebra (

S;

^

;

^) satisfying the following axioms:

(

a

^

b

)^(

b

^

c

) =

b

(1)

(

a

^

b

)^(

b

^

c

) =

b

(2)

The name comes from the fact that this algebra is a generalisation of a central groupoid [5, 7]. Often we will omit the ^ and represent this operation by juxtaposition. Note that by the symmetry of the axioms, if (

S;

^

;

^) is a semicentral bigroupoid, then so is (

S;

^

;

^), the dual.

Example 1

Let

A

,

B

be two sets, and let

Q

=

A

^

B

. Dene

(

a

¹

;b

¹)^(

a

²

;b

²) = (

a

²

;b

¹) (3) (

a

¹

;b

¹)^(

a

²

;b

²) = (

a

¹

;b

²) (4) Then(

Q;

^

;

^) is a semicentral bigroupoid.

Semicentral bigroupoids are equivalent to reversible one dimensional cellular automata, see e.g. [1].

Denition 2

A Rectangular Structure on a set

S

, called the base set, is a collection ^R of ordered pairs of subsets, called rectangles, of

S

, such that

8(

s;t

)²

S

²⁹!

R

^2R such that (

s;t

)²

R

(5)

8

R;Q

^2R

;

^j

R

^1\

Q

^2j= 1

:

(6)

where we identify

R

= (

R

¹

;R

²) =

R

^1

R

².

We say two rectangular structures are isomorphic if there is an invertible map between the base sets that preserves rectangles.

Take two sets

A;B

. Dene

S

=

A

^

B

, and for all (

a;b

)²

S

dene

R

⁽a;b⁾= (^f

a

^g

B;A

^f

b

^g). Then

R=^f

R

⁽a;b^)j

a

²

A;b

²

B

^g (7) is a rectangular structure on

S

. For any ((

a;b

)

;

(

c;d

)) ²

S

², ((

a;b

)

;

(

c;d

)) ²

R

⁽_a;d⁾, and this is obviously unique. Let

R

=

R

⁽_a;b⁾,

Q

=

R

⁽_c;d⁾be two rectangles in^R. Then

R

¹ =^f

a

^g

B

and

Q

² =

A

^f

d

^g, so ^j

R

^1\

Q

^2j=^jff

a

^gf

d

^ggj= 1. Since this is a very simple construction, We will refer to it as a vanilla rectangular structure on

A;B

, in the same sense that vanilla icecream is the simplest.

Dene the map:

d

:^{R !}

S

(8)

R

^7!

r

where^f

r

^g=

R

^1\

R

² (9) This map is well dened since for every

R

^2R,^j

R

^1\

R

^2j= 1 by (6) above, so

R

^1\

R

²=^f

r

^gfor some

r

.

From a rectangular structure ^R, using the bijection

d

from equation (8) above and denoting by

R

(

s;t

) the unique rectangle on the pair (

s;t

) guaranteed by (5), dene

:

S

^

S

^!

S

(10)

(

s;t

) ^7!

u

where^f

u

^g= (

d

^?1(

s

))^2\(

d

^?1(

t

))¹

:

S

^

S

^!

S

(11)

(

s;t

) ^7!

d

(

R

(

s;t

)) as binary operations on

S

.

As an example consider the vanilla rectangular structure on

A;B

. In this case

d

(

R

⁽_a;b⁾) = (

a;b

) (12)

R

((

a;b

)

;

(

c;d

)) =

R

⁽_a;d⁾ (13) thus

(

a;b

)^(

c;d

) = (

c;b

) (14) (

a;b

)^(

c;d

) = (

a;d

) (15) and we have a semicentral bigroupoid isomorphic to that dened in Example 1.

In general

Proposition 2

The algebra (

S;

^

;

^), with operations dened as in (10),(11) above, is an idempotent semicentral bigroupoid.

Proof: This is pure calculation.

(

a

^

b

)^(

b

^

c

) = (

d

^?1(

a

)^2\

d

^?1(

b

)¹)^(

d

^?1(

b

)^2\

d

^?1(

c

)¹) (16)

=

k

^

l

for some

k;l

²

S

(17)

=

d

(

R

(

k;l

)) (18)

But

k

²

d

^?1(

b

)¹ and

l

²

d

^?1(

b

)². Let

B

=

d

^?1(

b

). Then (

k;l

) ²

B

, thus

R

(

k;l

) =

B

, thus

k

^

l

=

d

(

B

) =

b

.

Now for the dual.

(

a

^

b

)^(

b

^

c

) =

d

(

R

(

a;b

))^

d

(

R

(

b;c

)) (19)

=

d

^?1(

d

(

R

(

a;b

)))^2\

d

^?1(

d

(

R

(

b;c

)))¹ (20)

=

R

(

a;b

)^2\

R

(

b;c

)¹ (21) Since

b

²

R

(

a;b

)²,

b

²

R

(

b;c

)¹ and their intersection is unique,

R

(

a;b

)^2\

R

(

b;c

)¹ =

b:

(22) Thus the two axioms of a semicentral bigroupoid are satised. Since

a

^

a

=

d

(

R

(

a;a

)) =

a

for all

a

²

S

we see that the ^ operation is idempotent, thus both operations are idempotent by the simple note that

a

^

a

= (

a

^

a

)^(

a

^

a

) =

a

(23) In general a complete equivalence exists between (idempotent) semicentral2

bigroupoids and rectangular structures, though we do not need that here. See e.g. [1, 2].

From such an idempotent semicentral bigroupoid, we can dene a collec- tion of others using a technique called lifting, then determine the exact num- ber of mutually non{isomorphic semicentral bigroupoids derivable from a given rectangular structure using knowledge about the automorphism group of the rectangular structure [2, Section 10.1].

This paper starts o looking at a special class of rectangular structures, then looks at a particular method of constructing them from ane planes. The following section looks at detail at the case of examples constructed from ane planes over nite elds, nding there are only two classes of examples, only one of which is interesting. The automorphism group of the rectangular structure is then constructed and investigated, and found to be very nicely structured in general. We nish by determining the automorphism group of the rectangular structures in general.

2 Partitioned rectangular structures

This section investigates a construction method for rectangular structures pro- posed by Tim Penttila [8] after noting certain regularities in the examples of rectangular structures of order 4 and 6. We use orthogonal partitions of a set

S

to construct examples.

We refer to this class as partitioned rectangular structures.

Denition 3

A pair of partitions  =^f

P

¹

;P

²

;:::;P

n^g

;

 =^f

T

¹

;T

²

;:::;T

m^g

of a set

S

are called orthogonal if for all

i;j

, ^j

P

i^\

T

j^j= 1.

Lemma 3

If  = ^f

P

¹

;P

²

;:::;P

n^g

;

 =^f

T

¹

;T

²

;:::;T

m^g are othogonal parti- tions of the set

S

, then ^j

P

i^j=

m

, ^j

T

j^j=

n

for all

i;j

, and ^j

S

^j=

mn

.

Proof: Fix

i

and look at



:^f1

;::: ;m

^g!

P

i

; j

^7!

P

i^\

T

j. Suppose

P

i^\

T

j =

P

i^\

T

k =^f

a

^g for some

j

⁶=

k

. Then

a

²

T

j ^\

T

k ⁶=^; so  is not a partition.

This is a contradiction, so we know that



is one to one. For every

a

²

P

i we know that

a

²

S

so there is some

j

such that

a

²

T

j, thus



(

j

) = ^f

a

^g, so we know that



is onto, thus an bijection and thus^j

P

i^j=

m

, independently of

i

.

Similarly^j

T

j^j=

n

for all

j

.

Since every element of

S

appears in exactly one

P

i, we know that ^j

S

^j =

n

^j

P

i^j=

mn

. ²

Proposition 4

Let  be a partition of

S

. For all

P

², let P be a partition of

S

orthogonal to . Then the set ^R = ^f(

P;Q

)^j

P

² 

; Q

² P^g is a rectangular structure, as is^R=^f(

Q;P

)^j

P

²

; Q

²P^g.

Proof: Consider the case ^R =^f(

P;Q

)^j

P

²

; Q

²P^g. Take an ordered pair (

a;b

) ²

S

². Since  is a partition, there is a unique

P

²  with

a

²

P

. Since P is a partition, there is then some unique

Q

²P with

b

²

Q

. Thus there is a unique (

P;Q

)^2R such that (

a;b

) ²(

P;Q

), thus satisfying the rst rectangular structure axiom (5).

Take (

P;Q

)

;

(

R;T

) ^{2 R}. Then

T

² R, R orthogonal to  implies that

j

T

^\

P

^j= 1. Thus equation (6) is satised, and we have a rectangular structure.

The second construction is proven analogously. ² If we have two orthogonal partitions 

;

 of

S

, and take P =  for all

P

, then we obtain a vanilla rectangular structure.

Partitioned rectangular structures are somewhat simpler in structure than those that are not, they are also somewhat simpler to construct. A straight- forward analysis shows that the Fredkin construction for reversible cellular au- tomata (see e.g. [9] section 5.4) gives a rectangular structure that is partitioned.

Note that in general a rectangular structure is not partitioned, although the numbers of partitioned examples outweighs the nonpartitioned examples. The smallest non{partitioned example is of order 6 and can be found in detail in the appendices of [2].

In the following section we will look at a construction of partitioned rect- angular structure from an ane plane.

3 Ane plane construction

In this section we will investigate a method to construct partitioned rectangular structures from an ane plane. Because all parallel classes of lines in an ane plane are mutually orthogonal, we can use them to construct examples. Taking one parallel class as the primary partition  with

S

being the point set of the ane plane, for every line

L

² select any other parallel class as the orthogonal



P

o





?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

P































L

Figure 1: Partitioned Rectangular Structure from an ane plane.

partition L. The following section looks at one particular method to select the secondary partitions L.

Take an ane plane of order

n

with incidence relation

aIL

if point

a

is incident with line

L

. Let [

M

] represent the set of lines parallel to a line

M

. Take  the be some parallel class of lines. Select some point

o

and some line

L

not in . For each

P

²,

o

ⁿ

IP

, take P to be the class of lines parallel to the line through

o

and the intersection of

P

and

L

, P = [

o

(

PL

)]. For the

P

² with

oIP

, P = [

L

]. This gives a suitable 

;

P for a partitioned rectangular structure. See Figure 1 for a diagram.

Proposition 5

The partitions  and P above are orthogonal.

Proof:  is a set of parallel lines, thus it is a partition. For any

P

²,

o

ⁿ

IP

, there is a unique point

p

=

P

^\

L

, as

L

is not in  and

op

is a line also not in

. This each line in the parallel class of

op

, i.e. P, intersects with each line in

 exactly once, thus P is orthogonal to .

If

oIP

then P is the parallel class including

L

, which is not the parallel class , thus each line in

Pi

P intersects each line in  exactly once, and  and

P are orthogonal.

2

Proposition 6

The following are equivalent:

1. The resulting rectangular structure is vanilla.

2. P = Q for all

P;Q

² 3.

oIL

4. P = Q for some

P;Q

²,

P

⁶=

Q

Proof: If the resulting rectangular structure is vanilla, then all the secondary partitions are equal, thus P = Q for all

P;Q

². Similarly if P = Q for all

P;Q

² then the resulting rectangular structure is vanilla. Thus 1^, 2.

Obviously 2⁾4.

Let P = Q for some

P;Q

² ,

P

⁶=

Q

. Then [

o

(

PL

)] = [

o

(

QL

)], but these two lines are incident with

o

, so

o

(

PL

) =

o

(

QL

) =

L

⁰ for some

L

⁰. Since

P;Q

are parallel and distinct,

PL

and

QL

are distinct and

PL

and

QL

are both incident with

L

and with

L

⁰, so

L

=

L

⁰. Thus

oIL

, so 4⁾3.

If

oIL

, then P includes

L

for all

P

² , so P = Q for all

P;Q

² .

Thus 3⁾2. ²

Corollary 7

Either P = Q for all

P;Q

² or P ⁶= Q for all

P;Q

².

In general, given an ane plane we can generate at least two nonisomorphic rectangular structures from it using this scheme, one being vanilla and the other not.In order to specify an ane plane generated rectangular structure, we need only a pair of incident lines

P;L

and a point

oIP

. The

P

denes the parallel class  = [

P

],

o

and

L

carry their previous meanings.

Given two such congurations (

o;P;L

) and (

o

⁰

;P

⁰

;L

⁰), if there is a collin- eation of the ane plane carrying one conguration to the other, then this collineation is an isomorphism of the derived rectangular structures, as it is a mapping of the points which takes lines to lines and therefore does not disrupt the resulting structure.

Then the above comment about at least two classes of examples amounts to saying that the set of congurations^f(

o;P;L

)^j

oIP; P

not parallel to

L

^ghas at least two orbits under the collineation group, one with

oIL

and one without.

Below we will see that for ane planes over elds, these are the only two orbits. In general even if there are more orbits of the set of congurations with

o

ⁿ

IL

under the collineation group, the resulting rectangular structures may be isomorphic across orbit boundaries.

The construction above is a specic case of a collection of construction meth- ods. As mentioned above, we need only select some parallel class as the primary partition  and then a collection of parallel classes as the secondary partitions

L. We do not even need the completeness of an ane plane, in general a

k

{net will suce [4, p. 251]. Since a

k

{net is equivalent to a collection of mu- tually orthogonal latin squares [4, p. 269], we can choose to look at mutually orthogonal latin squares. Even this is too much, in general given a latin square and a collection of orthogonal mates [4, p. 155] we can construct rectangular structures.

4 Field case

In this section we will be concentrating upon the ane plane over eld case.

Note that transitive on triangles means transitive on non{collinear triples of points as triples, not as sets.

Proposition 8

Up to isomorphism there are only two types of congurations in an ane plane over a eld.

Proof: The result follows using the fact that the collineation group on an ane plane over a nite eld is transitive on triangles.

Given two nonvanilla congurations (

o;P;L

) and (

o

⁰

;P

⁰

;L

⁰), construct the two triples of points (

o;PL;l

) with

lIL

not on

P

and (

o

⁰

;P

⁰

L

⁰

;l

⁰) with

l

⁰

IL

⁰ not on

P

⁰. By transitivity on triangles we nd a collineation



of the ane plane with (

o;PL;l

) = (

o

⁰

;P

⁰

L

⁰

;l

⁰), so (

o;P;L

) = (

o

⁰

;P

⁰

;L

⁰).

Similarly take two vanilla congurations (

o;P;L

) and (

o

⁰

;P

⁰

;L

⁰) with

oIP

and

o

⁰

IP

⁰. Construct the two triples of points (

o;p;l

) with

lIL

not on

P

and

pIP

not on

L

and (

o

⁰

;p

⁰

;l

⁰) with

l

⁰

IL

⁰ not on

P

⁰ and

p

⁰

IP

⁰ not on

L

⁰. By transitivity on triangles we nd a collineation



of the ane plane with (

o;p;l

) = (

o

⁰

;p

⁰

;l

⁰), so (

o;P;L

) = (

o

⁰

;P

⁰

;L

⁰). ²

This result has a converse.

Proposition 9

If there are only two orbits in the set of congurations under the action of the collineation group, then the ane plane is over a eld.

Proof: If there are only two orbits, then for any pair of pairs of points (

o;p

) and (

o

⁰

;p

⁰) there are two congurations (

o;P;L

) with

P

=

op

and

pIL

and (

o

⁰

;P

⁰

;L

⁰) with

P

⁰ =

o

⁰

p

⁰ and

p

⁰

IL

⁰. Then there is a collineation of the ane plane mapping one conguration to the other, thus the collineation group is 2{transitive on the plane.

By [6] there are only three possibilities for ane planes with 2{transitive automorphism groups. There are the ane planes over elds, the ane plane over the exceptional neareld of order 9 and the Hering ane plane of order 27 described in [3, p. 236]. In both these special cases the automorphism groups are not transitive on congurations, thus the only possibilities are the ane

planes over elds. ²

Let

K

=

GF

(

q

) be the nite eld over which we construct the ane plane, and let

K

^ be the nonzero elements of

K

. Since we have a lot of freedom to choose the lines

P

and

L

and the point

o

, we select the two cases:



L

=^f(0

;y

)^j

y

²

K

^g,

P

=^f(

x;

0)^j

x

²

K

^g,

o

= (0

;

0). This is the vanilla conguration.



L

=^f(0

;y

)^j

y

²

K

^g,

P

=^f(

x;

0)^j

x

²

K

^g,

o

= (1

;

0).

The vanilla conguration leads to a structure nearly devoid of interest, so we will ignore it in the sequel.

We will use the terminology

P

a to mean the horizontal line^f(

x;a

)^j

x

²

K

^g. Then we nd the partitions as follows.

 = ^f

P

a^j

a

²

K

^g= [

P

⁰] (24)

P⁰ = ^ff(

a;y

)^j

y

²

K

^gj

a

²

K

^g(vertical lines) (25)

= [

L

] (26)

P^a = [^f(

x;a

(1^?

x

))^j

x

²

K

^g] (27)

For ease of notation dene

Q

⁰;k = ^f(

k;y

)^j

y

²

K

^g (28) (

a

⁶= 0)

Q

a;k = ^f(

x;a

(1^?

x

) +

k

)^j

x

²

K

^g (29) A rectangle is then a pair (

P

a

;Q

a;k) for some

a;k

²

K

. Dene the rectan- gular structure^R=^f(

P

a

;Q

a;k)^j

a;k

²

K

^g.

The diagonal map is the dened on a rectangle (

P

a

;Q

a;k) by

d

((

P

⁰

;Q

⁰;k)) = (

k;

0) (30) (

a

⁶= 0)

d

((

P

a

;Q

a;k)) = (

k

a;a

^{) (31)}

and the inverse

d

^?1(

a;

0) = (

P

⁰

;Q

⁰;a) (32) (

b

⁶= 0)

d

^?1(

a;b

) = (

P

b

;Q

b;ab) (33) which then leads to the following denition of an idempotent semicentral bi- groupoid

(

a;

0)^(

c;d

) = (

a;d

) (34) (

b

⁶= 0) (

a;b

)^(

c;d

) = (

a

+ 1^?

d

b;d

^{) (35)}

(

a;

0)^(

c;d

) = (

c;

0) (36) (

b

⁶= 0) (

a;b

)^(

c;d

) = (

c

^?1 +

d

b;b

^{) (37)}

5 Automorphism group

The number of nonisomorphic semicentral bigroupoids that can be dened from this idempotent semicentral bigroupoid is determined by the automorphism group of the semicentral bigroupoid. In [2] it is shown that this automorphism group can be easily determined by intersecting the automorphism groups of two graphs associated with the algebra. In this section we will look at these graphs and their automorphism groups.

Both graphs are dened with the vertex set being the point set of the under- lying rectangular structure, in this case

K

^

K

. When looking at the idempotent semicentral bigroupoid one can derive the associated graphs directly from the rectangular structure. We will refer to the two graphs as the red graph and the blue graph for ease of notation. Let^Rbe the rectangular structure dened above.

The red graph

G

r has edges

f(

a;b

)^j

a

²

d

(

R

)

; b

²

R

²

; R

^2Rg (38) or more explicitly

(

k;

0) ^! (

k;y

)

y

²

K

(39)

(

b

⁶= 0) (

a;b

) ^! (

x;b

(

a

+ 1^?

x

))

x

²

K

(40)

Lemma 10

In

G

r there are edges (

a;b

) ^! (

a

+ 1

;

0) and these are the only non{loop edges going to nodes of the form (

x;

0).

Proof: The rst part follows directly from inserting

x

=

a

+1 in (40) above.

The second claim can be seen by setting the second coordinate of the to node in (39) to zero for a loop edge. Set the second coordinate of the to node in (40) to zero then

b

⁶= 0 implies that

x

=

a

+ 1 completing the proof. ²

The blue graph

G

b has edges

f(

a;b

)^j

a

²

R

¹

; b

²

d

(

R

)

; R

^2Rg (41) or more explicitly

(

a;b

) ^! (

x;b

)

x

²

K

(42)

That is, the blue graph consists of a set of

q

{cliques on the points sharing the same

y

{value.

Since the blue graph is so well{behaved, we know that if



²

Aut

(

G

b), then is can be rewritten as



(

a;b

) = (



¹(

a;b

)

;

²(

b

)) (43) where



² is a permutation of

K

.

We are interested in

G

=

Aut

(^R) =

Aut

(

G

r)^\

Aut

(

G

b). The next result looks at the part of

G

that xes the baseline

P

⁰ as a set (not pointwise).

Proposition 11 G

P⁰ =^f



: (

a;b

)^7!(

a

+

k

⁰

;k

¹

b

)^j

k

⁰

;k

^{1 2}

K; k

^{1 6}= 0^g

Proof: Let



²

G

x the line

P

⁰, thus



² xes 0. Dene



³ :

K

^!

K

by



(

x;

0) = (



³(

x

)

;

0).

Note that there are red edges (

a;

0) ^! (

a;b

) thus



(

a;

0) = (



³(

a

)

;

0) ^!



(

a;b

) = (



¹(

a;b

)

;

²(

b

)) which must be of the form (39) thus



¹(

a;b

) =



³(

a

).

Thus



(

a;b

) = (



³(

a

)

;

²(

b

)).

Take the red edge (0

;a

) ^! (1

;

0) with

a

⁶= 0 and its image under



, (



³(0)

;

²(

a

))^!(



³(1)

;

²(0)) = (



³(1)

;

0) By Lemma 10 we know that



³(1) =



³(0) + 1. Similarly taking the red edge (

i;a

) ^!(

i

+ 1

;

0),

a

⁶= 0, and its im- age (



³(

i

)

;

²(

a

)) ^! (



³(

i

+ 1)

;

0) we know that



³(

i

+ 1) =



³(

i

) + 1, thus



³(

k

) =



³(0) +

k

so



³ is dened purely in terms of addition and the value of



³(0).

We know that

G

r contains the edges (^?1

;

1) ^!(^?

x;x

) for all

x

⁶= 0. This edge has image (



³(^?1)

;

²(1))^!(



³(^?

x

)

;

²(

x

)) under



. By the denition of the red graph

G

r, we know that the red edges leaving (



³(^?1)

;

²(1)) have the form (



³(^?1)

;

²(1))^!(

x;

²(1)(



³(^?1)+1^?

x

)). Dene 

x

=



³(^?

x

) =

k

^0?

x

with

k

⁰ =



³(0).



²(

x

) =



²(1)(



³(^?1) + 1^?

x

) (44)

=



²(1)((^?1 +

k

⁰) + 1^?(

k

^0?

x

)) (45)

=



²(1)(

x

) (46)

thus



²(

x

) =



²(1)

x

=

k

¹

x

with

k

¹=



²(1)⁶= 0.

Thus we see that

G

P^{0 f}



: (

a;b

)^7!(

a

+

k

⁰

;k

¹

b

)^j

k

⁰

;k

¹²

K; k

^{1 6}= 0^g. Now for the reverse inclusion. Dene



(

a;b

) = (

a

+

k

⁰

;k

¹

b

),

k

⁰

;k

^{1 2}

K; k

¹⁶= 0. This map preserves the line

P

⁰. We must show that this is an automorphism of both the red and the blue graphs.

Suppose it was no bijection. Then there exists

a;b;a

⁰

;b

^{0 2}

K

such that



(

a;b

) =



(

a

⁰

;b

⁰), but

a

+

k

⁰ =

a

⁰+

k

^{0 )}

a

=

a

⁰ and

k

¹

b

=

k

¹

b

^{0 )}

b

=

b

⁰ so this is not the case and



is a bijection.

In

G

b

a

^!

b

^,

a

²=

b

² (47)

,

k

¹

a

² =

k

¹

b

² (48)

, (



(

a

))² = (



(

b

))² (49)

,



(

a

)^!



(

b

) (50)

So



is an automorphism of

G

b. In

G

r we have two cases.

Case 1: The edges (

k;

0) ^! (

k;y

).



((

k;

0)) = (

k

+

k

⁰

;

0) and



((

k;y

)) = (

k

+

k

⁰

;k

¹

y

) and by the denition of

G

bthere is an edge (

k

+

k

⁰

;

0)^!(

k

+

k

⁰

;k

¹

y

).

Case 2: The edges (

a;b

)^!(

x;b

(

a

+ 1^?

x

) with

b

⁶= 0. Now



((

a;b

)) = (

a

+

k

⁰

;k

¹

b

) (51)



((

x;b

(

a

+ 1^?

x

))) = (

x

+

k

⁰

;k

¹

b

(

a

+ 1^?

x

)) (52) The general form of an edge from



((

a;b

)) is (

a

+

k

⁰

;k

¹

b

) ^! (

x;

(

k

¹

b

)((

a

+

k

⁰) + 1^?

x

)). Setting 

x

=

x

+

k

⁰, we nd that (

k

¹

b

)(

a

+

k

⁰+ 1^?(

x

+

k

⁰)) = (

k

¹

b

)(

a

+ 1^?

x

) =



((

x;b

(

a

+ 1^?

x

))) so



²

Aut

(

G

r) and we are done.

Looking at the ane plane over

GF

(3), one nds that the automorphism2

group of the two graphs is generated by the permutations

(4

;

7)(5

;

8)(6

;

9) (53)

(1

;

2

;

3)(4

;

5

;

6)(7

;

8

;

9) (54) (1

;

4

;

2

;

5

;

3

;

6)(7

;

9

;

8) (55) with

i

=

a

+

pb

+ 1 coding the point (

a;b

). This group does not x the line

P

⁰.

Proposition 12 Aut

(^R) =

G

P⁰ except for

q

= 2

;

3

;

4.

Proof: Given an automorphism



of ^Rnot xing

P

⁰, let

P

a=



(

P

⁰) and dene automorphism



: (

x;y

) ^7! (

x;cy

) for some

c

²

K; C

⁶= 0

;

1. This is an automorphism by Prop 11 above. Then



=



^?1



^



is an automorphism of

Rthat exchanges

P

⁰ and

P

k

; k

=



^?12 (

ca

).



((

x;

0)) =



^?1(



((



¹(

x;

0)

;a

))) (56)

=



^?1((



¹(

x;

0)

;ca

)) (57)

= (

m;

^?12 (

ca

))²

P

k for some

m

(58)



((

x;

^?12 (

ca

))) =



^?1(



((



¹(

x;

^?12 (

ca

))

;ca

))) (59)

=



^?1((



¹(

x;

^?12 (

ca

))

;a

)) (60)

= (

m;

^?12 (

a

)) (61)

= (

m;

0)²

P

⁰ for some

m

(62) Thus in general we can assume that we have



²

Aut

(^R) which exchanges

P

⁰ and

P

k for some

k

⁶= 0.

Look at the red edge (

i;

0) ^! (

i;k

). This is mapped to another edge by



,



(

i;

0) = (



¹(

i;

0)

;k

)^!(



¹(

i;k

)

;

0) thus by Lemma 10



¹(

i;k

) =



¹(

i;

0) + 1.

Similarly take red edge (

i;k

) ^! (

i

+ 1

;

0) and note that (



¹(

i;k

)

;

0) ^! (



¹(

i

+ 1

;

0)

;k

) and thus



¹(

i;k

) =



¹(

i

+ 1

;

0).

Thus we obtain



¹(

i;

0) =



¹(0

;

0) +

i

(63)



¹(

i;k

) =



¹(0

;

0) +

i

+ 1 (64) Now take some

a

⁶= 0

;k

and look at the red path (

x

¹

;a

)^!(

i;k

) ^!(

x

²

;a

).

Because



((

i;k

)) = (



¹(

i;k

)

;

0) we know that



¹(

x

¹

;a

) =



¹(

i;k

)^?1 =



¹(0

;

0) +

i

+ 1^?1 =



¹(0

;

0) +

i

(65) and similarly



¹(

x

²

;a

) =



¹(

i;k

) =



¹(0

;

0) +

i

+ 1 (66) Now, by the rules for edges of

G

rwe know that (

x

¹

;a

)^!(

i;k

) = (

x;a

(

x

¹+ 1^?

x

)), thus

i

=

x

¹+ 1^? _ka. Thus



¹(

x

¹

;a

) =



¹(0

;

0) +

x

¹+ 1^?

k

a

⁽⁶⁷⁾

Similarly by looking at (

i;k

)^!(

x

²

;a

) we nd

i

= _ak+

x

^2?1 so



¹(

x

²

;a

) =



¹(0

;

0) +

a

k

⁺

x

^2?1 + 1 (68)

Then equating (67) and (68) with

x

¹=

x

² we see that



¹(

x;a

) =



¹(0

;

0) +

x

+ 1^?

k

a

⁼



¹(0

;

0) +

a

k

⁺

x

^?1 + 1 (69)

)1^?

k

a

⁼

a

k

⁽⁷⁰⁾

)

ka

^?

k

² =

a

² (71)

)

a

^2?

ka

+

k

² = 0 (72)

Thus we need to have some

k

²

K

^such that for all

a

⁶= 0

;k; a

^2?

ka

+

k

² = 0.

This polynomial has at most 2 zeroes for any nonzero value of

k

, thus

q

^4.

Indeed for

q

= 2

;

3

;

4 we nd that

P

⁰ is not xed, and we are done. ²

6 Conclusion

In this paper we have seen an examples of a way to construct rectangular structures from ane planes, and we have analysed the construction from ane planes over elds to a large extent. Importantly, the automorphism group of the rectangular structure was able to be completely determined.

Generalisations of this work to other ane planes would be of interest, or to more general constructions using

k

{nets or latin squares as outlined above.

The determination of special properties of the semicentral bigroupoids and cellular automata constructed using these methods would also be of interest.

Acknowledgements

I would like to extend thanks to Tim Pentilla whose discussions regarding these and many other adeas have been most fruitful. I would also like to thank the UWA mathematics Department for being such gracious hosts during my stay here. I gratefully acknowledge nancial support from Linz University Hochschulfonds and Wirtschaftsfoerderung des Landes from the government of Upper Austria.

References

[1] Tim Boykett. Combinatorial construction of one dimensional reversible cel- lular automata. In G. Pilz, editor, Contributions to General Algebra 9, pages 81{90. Holder{Pichler{Tempsky, Teubner, 1995.

[2] Tim Boykett. Algebraic Aspects of Reversible Computation. PhD thesis, University of Western Australia, 1997.

[3] P. Dembowski. Finite Geometries, volume 7 of Ergebnisse der Mathematik und ihrer grenzegebiete. Springer Verlag, 1968.

[4] J. Denes and A.D. Keedwell. Latin Squares and their Applications. English Universities Press, London, 1974.

[5] Trevor Evans. Products of points { some simple algebras and their identities.

Am. Math. Monthly, pages 362{372, April 1967.

[6] W.M. Kantor. Homogeneous designs and geometric lattices. J. Comb. The- ory A, 38:66{74, 1985.

[7] Donald E. Knuth. Notes on central groupoids. Journal of Combinatorial Theory, 8:376{390, 1970.

[8] Tim Penttila. Personal communication, 1995.

[9] Tommaso Tooli and Norman H. Margolus. Invertible cellular automata: a review. Physica D, 45:229{253, 1993.