Disjoint circles: a classification
Authors: Ebert Gary L. 
Published: July 2010

For q a prime-power, let

IP

(
q
)

{\text {IP}}(q)

denote the miquelian inversive plane of order q. The classification of certain translation planes of order

q
2

{q^2}

, called subregular, has been reduced to the classification of sets of disjoint circles in

IP

(
q
)

{\text {IP}}(q)

. While R. H. Bruck has extensively studied triples of disjoint circles, this paper is concerned with sets of four or more circles in

IP

(
q
)

{\text {IP}}(q)

. In a previous paper, the author has shown (for odd q) that the number of quadruples of disjoint circles in

IP

(
q
)

{\text {IP}}(q)

is asymptotic to

q

12

/

1536

{q^{12}}/1536

. Hence a judicious approach to the classification problem is to study “interesting” quadruples. In general, let

C
1

,
…
,

C
n

{C_1}, \ldots ,{C_n}

be a nonlinear set of n disjoint circles in

IP

(
q
)

{\text {IP}}(q)

. Let H be the subgroup of the collineation group of

IP

(
q
)

{\text {IP}}(q)

composed of collineations that permute the

C
i

{C_i}

’s among themselves, and let K be that subgroup composed of collineations fixing each of the

C
i

′

s

{C_i}’s

. An interesting set of n disjoint circles would be one for which

K
=
1

K = 1

. It is shown that

K
=
1

K = 1

if and only if
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="StartLayout 1st Row with Label left-parenthesis asterisk right-parenthesis EndLabel StartLayout Enlarged left-brace 1st Row left-parenthesis i right-parenthesis there does not exist a circle upper D orthogonal to 2nd Row each of the given n circles comma and 3rd Row left-parenthesis ii right-parenthesis we do not have one circle in our set orthogonal 4th Row to each of the other n minus 1 circles period EndLayout EndLayout">

(

∗

)

{

(i)

there

does

not

exist

a

circle

D

orthogonal

to

each

of

the

given

n

circles,

and

(ii)

we

do

not

have

one

circle

in

our

set

orthogonal

to

each

of

the

other

n
−
1

circles

.

\begin{equation}\tag {$*$} \left \{ {\begin {array}{*{20}{c}} {{\text {(i)}}\;{\text {there}}\;{\text {does}}\;{\text {not}}\;{\text {exist}}\;{\text {a}}\;{\text {circle}}\;D\;{\text {orthogonal}}\;{\text {to}}} \\ {{\text {each}}\;{\text {of}}\;{\text {the}}\;{\text {given}}\;n\;{\text {circles,}}\;{\text {and}}} \\ {{\text {(ii)}}\;{\text {we}}\;{\text {do}}\;{\text {not}}\;{\text {have}}\;{\text {one}}\;{\text {circle}}\;{\text {in}}\;{\text {our}}\;{\text {set}}\;{\text {orthogonal}}} \\ {{\text {to}}\;{\text {each}}\;{\text {of}}\;{\text {the}}\;{\text {other}}\;n - 1\;{\text {circles}}{\text {.}}} \\ \end{array} } \right .\end{equation}

When

n
=
4

n = 4

and under mild restrictions on q, an algorithm is developed that finds all nonlinear quadruples of disjoint circles satisfying the orthogonality conditions

(
∗
)

( \ast )

and having nontrivial group H. Given such a quadruple, the algorithm determines exactly what group H is acting. It is also shown that most quadruples in

IP

(
q
)

{\text {IP}}(q)

, for large q, do indeed satisfy the conditions

(
∗
)

( \ast )

. In addition, the cases when

n
=
5
,
6
,

n = 5,6,

or 7 are explored to a lesser degree.