Download fpp revised

FINITE PROJECTIVE PLANES
DANNY KALMANOVICH
Undergraduate Thesis
February 2009
Department of Mathematics
Ben-Gurion University of the Negev
1. Introduction
This paper is a short survey of the first few small examples of finite projective
planes and some basic general results about the existence and explicit construction
of these objects. The second section will present the main concepts that will be
used throughout the paper. Sections 3 and 4 will discuss general existence and
explicit constructions in some cases. The following sections will present the planes
of order 2,3,4 and 5. Each such presentation will emphasize certain properties of
projective planes and some approaches to the construction of projective planes.
Some basic understanding of design theory, graph theory and automorphism groups
of these objects is assumed. The sources for this survey are cited within the paper
in the context of its subject except [6] which is a more intuitive and in a sense naive
presentation of a few projective planes.
2. Main Concepts
2.1. Projective planes. A projective plane is an incidence structure S =
(P, L) such that the following axioms hold:
(1) Any two distinct points p, q of P are incident with exactly one line denoted
pq.
(2) Any two distinct lines l, m of L intersect at exactly one point denoted l∩m.
(3) There exists a set of four points, no three of which are collinear, such a set
is called a quadrangle.
Theorem 1. A finite projective plane is a BIBD with parameters:
• v = b = n2 + n + 1
• k =r =n+1
• λ=1
In this case we say it is a projective plane of order n.
1
2
DANNY KALMANOVICH
2.2. Some substructures in projective planes.
Definition 1. Let S = (P, L) be a projective plane. Let A ⊂ P .
Then A is called an arc if it does not contain 3 collinear points.
Proposition 1. Let A be an arc in a projective plane of order n, then:
• If n is odd then |A| ≤ n + 1.
• If n is even then |A| ≤ n + 2.
In the first case, where n is odd, an arc of size n + 1 is called an oval.
In the second case, where n is even, an arc of size n + 2 is called a hyperoval.
For a general introduction and some more basic facts see [4].
2.3. Additional objects and notions.
Consider K6 the complete graph on 6 vertices.
Definition 2.
• A 1-factor of K6 is a set of 3 pairwise disjoint edges of K6 .
For example, f1 is a 1-factor of K6 :
1
•
•
6
2
•33
33
33
•3
•
5
•
4
f1
• A 1-factorization of K6 is a partition of all 15 edges of K6 into 5 factors.
For example, F = {f1 , f2 , f3 , f4 , f5 } is a 1-factorization of K6 :
1
•
6•
2
•33
33
33
•3
•
5
•
4
f1
1
•
2
•
6 •KKK
ss• 3
KK
KK ssss
Ks
ss KK
•s
•
5
4
f2
1
2
•
ss•
s
s
ss
ss
s
6 •s
•3
•
4
•
5
f3
FINITE PROJECTIVE PLANES
1
2
•KKK •
KK
KK
KK
K
•3
6 •33
33
33
•
•
5
4
f4
6•
1
2
•//
•
// // /
///
//
•
•
5
4
f5
3
•3
The names 1-factor and 1-factorization come from the fact that the spanning
subgraph of a 1-factor is a regular graph of valency 1, that is, a perfect matching
on the vertices of K6 .
Proposition 2. In K6 :
(1) There are 15 1-factors.
(2) There are 6 different 1-factorizations.
(3) Two disjoint 1-factors are contained in a unique 1-factorization.
For proof of this proposition see [3].
Proposition 3. There is only one 1-factorization of K6 (up to a permutation of
{1, 2, 3, 4, 5, 6}).
Proof. As the first factor we choose
1
•
6•
2
•33
33
33
•3
•
5
•
4
f1
The factor f2 containing the edge {1, 2} contains {3, 4} or {3, 5}, we may choose
{3, 5}:
1
•
2
•
6 •KKK
ss• 3
KK
KK ssss
Ks
ss KK
•s
•
5
4
f2
4
DANNY KALMANOVICH
Now the remaining 3 factors are determined uniquely:
6•
1
2
•
s•
s
s
ss
ss
s
ss
•3
•
4
•
5
f3
1
2
•KKK •
KK
KK
KK
K
•3
6 •33
33
33
•
•
5
4
f4
6•
1
2
•//
•
// // //
///
/
•
•
5
4
f5
•3
It is easy to see that if in the second choice that we made we were to choose the edge
{3, 4}, the obtained unique 1-factorization F 0 is mapped to F by the permutation
(4, 5).
Thus, there is essentially only one 1-factorization of K6 .
Considering that fact, we will be able to work with a suitable concrete 1-factorization
of K6 .
3. The Bruck-Ryser Theorem
This section will discuss a very important result giving restrictions on the set of
parameters of projective planes.
We will give a formulation of this theorem, and a general outline of its proof.
Also, an equivalent formulation will be given.
Finally, a more general formulation is available about symmetric 2−(v, k, λ) designs,
which imply the formulation about projective planes as a corollary - this will be
discussed briefly.
3.1. Formulation.
Theorem 2. If a projective plane of order n exists, and n ≡ 1 or 2 (mod 4), then
n is a sum of two squares.
3.2. Proof outline.
The proof of the Bruck-Ryser theorem uses some Number Theory facts.
3.2.1. Necessary Number Theory facts.
Fact 1. The “four-square identity”:
(a21 + a22 + a23 + a24 )(b21 + b22 + b23 + b24 ) = c21 + c22 + c23 + c24
where:
c1 = a1 b1 − a2 b2 − a3 b3 − a4 b4 ,
c2 = a1 b2 + a2 b1 + a3 b4 − a4 b3 ,
c3 = a1 b3 + a3 b1 + a4 b2 − a2 b4 ,
c4 = a1 b4 + a4 b1 + a2 b3 − a3 b2 .
Fact 2. Every positive integer is the sum of four integer squares.
FINITE PROJECTIVE PLANES
5
Fact 3. For any integer n, if the equation x2 + y 2 = nz 2 has an integer solution
with x, y, z not all zero, then n is the sum of two integer squares.
3.2.2. Outline of the proof of Theorem 2.
We have a projective plane of order n where n ≡ 1 or 2 (mod 4), that is, the
number of points is N = n2 + n + 1 and N ≡ 3 (mod 4).
(1) The first step is to define the incidence matrix A of the plane.
Then the matrix AAT has entry (i, j) equal to the number of lines containing both the ith and the j th points,
this entry is equal to n + 1 if i = j, and 1 otherwise.
That is translated to the matrix identity:
AAT = nI + J.
(2) Let x1 , . . . , xN be indeterminants. Let x = (x1 , . . . , xN ) and xA = z.
then the above identity implies:
zz T = xAAT xT = nxxT + xJxT
which means:
2
z12 + . . . + zN
= n(x21 + . . . + x2N ) + w2 ,
where: w = x1 + . . . + xN .
(3) Now, we take a new indeterminate xN +1 and add nx2N +1 to both sides of
the equation. Note that N + 1 is divisible by 4.
By Fact 2 we write n as the sum of four squares: n = a21 + a22 + a23 + a24 ,
and use Fact 1 to write
2
2
n(x24i+1 + . . . + x24i+4 ) = y4i+1
+ . . . + y4i+4
,
where the y’s are linear combinations of the x’s.
we have:
2
2
2
z12 + . . . + zN
+ nx2N +1 = y12 + . . . + yN
+1 + w .
(4) In this step, we make some adjustments such that each xi is expressed as
a rational linear combination of the other x’s.
We start with x1 , it has a non-zero coefficient in at least one y and one z,
assume these are y1 and z1 .
If it has different coefficients in these two expressions, we impose the condition: y1 = z1 , otherwise, we impose: y1 = −z1 .
Here is an implicit example:
z1 = a11 x1 + a21 x2 + . . . + aN 1 xN ,
here a11 , a21 , . . . , aN 1 is the first column of A.
y1 = a1 x1 − a2 x2 − a3 x3 − a4 x4 , here n = a21 + a22 + a23 + a24 .
+a21
N1
If a1 6= a11 then we obtain: x1 = aa12 −a
x2 + . . . + a1a−a
xN .
11
11
aN 1
a2 −a21
If a1 = a11 then we obtain: x1 = a1 +a11 x2 + . . . − a1 +a11 xN .
In either case, x1 is expressed as a rational linear combination of the other
x’s and y12 = z12 , therefore we can cancel these terms.
Repeating this process for i = 2, 3, . . . N , we finally obtain:
2
2
nx2N +1 = yN
+1 + w ,
where yN +1 is a rational multiple of xN +1 .
6
DANNY KALMANOVICH
(5) In the final step, we can choose an integer value of xN +1 such that yN +1
and w are also integers, and we obtain a non-zero integer solution to the
above equation. now, by Fact 3 n is a sum of two integer squares.
3.3. An equivalent formulation.
Here is an equivalent formulation of the Bruck-Ryser theorem:
Theorem 3. If a projective plane of order n exists, and n ≡ 1 or 2 (mod 4), then
the square-free part of n has no prime divisors of the form 3 (mod 4).
This is due to another Number Theory fact: n is the sum of two integer squares
iff the square-free part of n has no prime divisors of the form 3 (mod 4).
3.4. Corollaries. Here are some simple corollaries of the Bruck-Ryser theorem:
• among the first few numbers of the form 1 (mod 4) which are:
5, 9, 13, 17, 21, 25, 29, 33
there are no projective planes of orders 21 and 33;
• among the first few numbers of the form 2 (mod 4) which are:
6, 10, 14, 18, 22, 26, 30, 34
there are no projective planes of orders 6, 14, 22 and 30.
3.5. A more general form of the Bruck-Ryser theorem.
3.5.1. Formulation.
Theorem 4. If v, k, λ are integers satisfying (v − 1)λ = k(k − 1) then for the
existence of a symmetric 2 − (v, k, λ) design it is necessary that:
(1) if v is even then k − λ is a square;
v−1
(2) if v is odd then z 2 = (k − λ)x2 + (−1) 2 λy 2 has a non-trivial solution in
integers x, y and z.
3.5.2. The way to the more general proof.
The proof of this, more general result, is very similar to the proof in section 2.
The main differences are that here we have to consider cases:
(1) where v ≡ 0 or 2 (mod 4)
(2) where v ≡ 1 or 3 (mod 4)
The case where v ≡ 3 (mod 4) is the same as in section 2,
the case where v ≡ 1 (mod 4) is very similar with the difference that we do not
need to add an extra indeterminant.
Section 3 is based mostly on [5] and also [4].
4. the model of a finite projective plane via a difference set
In this section we will present a way of constructing a model of a finite projective
plane with the use of a difference set.
First, the definition of a difference set will be given, illustrated by a simple example,
then we will focus on the connection with symmetric (v, k, λ) BIBD,
and in particular with finite projective planes.
Next, the main result of this section, Singer’s Theorem will be stated,
FINITE PROJECTIVE PLANES
7
And finally, as a result of Singer’s Theorem, some small order examples will be
presented in an evident simple form.
4.1. Definition of a difference set.
Definition 3. Let (H, +) be a group of order v, let D ⊆ H where |D| = k.
Then D is called a (v, k, λ) difference set if the set of differences {di − dj |di , dj ∈ D, i 6= j},
contains every element h 6= 0 of H exactly λ times.
Example 1. Consider (Z7 , +). Let D = {1, 2, 4},
then D ⊆ Z7 is a (7, 3, 1) difference set.
Indeed, the set of differences contains every element h 6= 0 of Z7 exactly once:
{di −dj |di , dj ∈ D} = {1−2 = 6, 1−4 = 4, 2−1 = 1, 2−4 = 5, 4−1 = 3, 4−2 = 2}.
Definition 4. Let H be a group, D ⊆ H, h ∈ H,
then D + h = {d + h|d ∈ D} is the translate of D by h.
Example 2. Consider again (Z7 , +). Let D = {1, 2, 4}, h = 3,
then the translate of D by h is:
D + h = {d + h|d ∈ D} = {1 + 3 = 4, 2 + 3 = 5, 4 + 3 = 0} = {0, 4, 5}.
Now we are ready to construct a symmetric (v, k, λ) BIBD:
Let (H, +) be a group, |H| = v. Let D ⊆ H, be a (v, k, λ) difference set,
denote the set of translates of D by B = {D + h|h ∈ H},
then S = (H, B) is a symmetric (v, k, λ) BIBD.
A counting argument shows that |B| = v: if h1 6= h2 and D + h1 = D + h2
then counting in two ways provides a contradiction, thus we have |H| = v = |B|
i.e. a symmetric design, and the size of each block is |D| = k.
Now, let x, y ∈ H be two distinct points of S, then x, y ∈ D + h iff x − h, y − h ∈ D,
since x − y 6= 0 there are exactly λ distinct pairs ci , di ∈ D, i = 1, 2, . . . , λ such that
x − y = ci − di . Denote fi = −di + y = −ci + x, then:
ci = x − fi
di = y − fi
that is:
x ∈ D + fi
y ∈ D + fi
for i = 1, 2, . . . , λ. These are λ blocks which contain the two given points.
These are the only blocks that contain both x and y.
In particular, if we take a group H with a (n2 + n + 1, n + 1, 1) difference set,
we can construct a projective plane of order n.
Next we will show the smallest such example, the Fano plane:
4.2. Example of The Fano Plane via a difference set.
Example 3. Consider (Z7 , +).
As we have seen in Example 1, D = {1, 2, 4} is a (7, 3, 1) difference set.
We construct the set B of all translates of D:
B = {D + 0 = {1, 2, 4} ,
8
DANNY KALMANOVICH
D + 1 = {2, 3, 5} ,
D + 2 = {3, 4, 6} ,
D + 3 = {4, 5, 0} ,
D + 4 = {5, 6, 1} ,
D + 5 = {6, 0, 2} ,
D + 6 = {0, 1, 3}}
We obtain a symmetric-(7, 3, 1)-BIBD.
In other words, this is the projective plane of order 2, aka The Fano Plane:
0
•///
///
//
/2
4 J
• JJJ ttt•///
t
t•J3JJJ ///
tttt
JJJ /
t
JJJ//
tttt
•
•
•
5
6
1
We now proceed to the main result of this section, this is a theorem which states
that such a construction can always be done providing that n is a prime power.
This is the classic Singer’s Theorem:
4.3. Singer’s Theorem.
Theorem 5. Let n be a prime power, then for any positive integer d there exists a
d+1
d
d+1
−1 nd−1 −1
( n n−1−1 , nn−1
, n−1 ) difference set of the group (Zv , +), where v = n n−1−1 .
For the proof of Singer’s theorem see [8].
In particular, when d = 2, for any prime power n, Singer’s Theorem
guarantees that a finite projective plane of order n exists.
Now, the construction of an appropriate difference set is the only missing piece:
4.3.1. How to construct an appropriate difference set.
We start by constructing the field with v elements as an extension of a field with p
elements (here p is a prime),then we make points of this field (considered as a vector
space over the field with p elements) correspond to points of P G(d, n). Explicitly:
• We choose an irreducible polynomial of degree d + 1 = 3 over Fn .
• Assume x is a root of this polynomial, this gives an expression of x3 in
terms of 1, x, x2 . Furthermore, all powers of x can now be expressed in
terms of 1, x, x2 and made to correspond to points of P G(2, n).
• Now we choose any line, for example x1 = 0, and the set of points of this
line is the desired difference set.
In the following example we will construct a model of the finite projective plane of
order 3 using this method:
FINITE PROJECTIVE PLANES
9
Example 4. By Singer’s Theorem we will be working over the group (Z13 , +).
Now, we choose an irreducible polynomial of degree 3 over F3 ,
for example: f (y) = y 3 − y + 1.
Assume x is a root of f (y), then x3 = 2 + x,
and the powers of x can be expressed in terms of 1, x, x2 and correspond to points
of P G(2, 3) in a natural way,
for example: x7 = x3 ·x3 ·x = (2 + x)·(2 + x)·x = 4 + 4x + x2 ·x = 1 + x + x2 ·
x = x + x2 + x3 = x + x2 + (2 + x) = 2 + 2x + x2 .
Thus: x7 = 2 + 2x + x2 ←→ (2, 2, 1).
Now we choose a line: for example, we choose the line x1 = 0 which contains the
points:
(0, 1, 0), (0, 0, 1), (0, 2, 1), (0, 1, 1)
These points correspond to:
x1 , x2 , x4 , x10
That is, we obtain a (13, 4, 1)-difference set D = {1, 2, 4, 10} for (Z13 , +).
4.3.2. Difference sets for some planes of small order.
In the same manner we obtain models of projective planes of a prime power order.
Here are some small order examples:
Example 5. D = {1, 2, 5, 15, 17} is a (21, 5, 1)-difference set of (Z21 , +).
This gives the projective plane of order 4.
Example 6. D = {1, 2, 4, 9, 13, 19} is a (31, 6, 1)-difference set of (Z31 , +).
This gives the projective plane of order 5.
Example 7. D = {1, 2, 4, 14, 33, 37, 44, 53} is a (57, 8, 1)-difference set of (Z57 , +).
This gives the projective plane of order 7.
5. Projective Planes of orders 2 and 3
In this section the planes of order 2 and 3 will be considered with more detail.
Particularly, we will focus on their uniqueness.
5.1. Uniqueness of the projective plane of order 2.
Theorem 6. The Projective Plane of order 2 is unique up to isomorphism.
Proof. Let S be a projective plane of order 2.
Assume P = {0, 1, 2, 3, 4, 5, 6} is the point set of S.
Consider a quadrangle in S, label the points of this quadrangle {0, 3, 5, 6}:
5
0
•///
//
//
//
//
//
•J3JJ //
t
t
JJJ //
tttt
JJJ/
tttt
J/
•
•
6
10
DANNY KALMANOVICH
Now we can label the 3 remaining intersection points of the 6 lines of the quadrangle.
We label the intersection of {0, 3} ∩ {5, 6} by 1, {0, 6} ∩ {3, 5} by 2, {0, 5} ∩ {3, 6}
by 4:
0
•///
///
//
/2
4 J
•
J
t
JJ tt •///
t
t•J3JJJ ///
tttt
JJJ /
t
JJJ//
tttt
•
•
•
5
6
1
At this stage, we must connect {1, 2, 4} by a line:
0
•///
///
//
/2
4 J
•
J
t
JJ tt •///
t
t•J3JJJ ///
tttt
JJJ /
t
JJJ//
tttt
•
•
•
5
6
1
Thus we have created all 7 lines and this is the only way that this can be done.
Clearly this projective plane is isomorphic to the Fano plane,
i.e. the unique projective plane of order 2.
5.2. The Projective Plane of order 3 via a difference set.
Consider (Z13 , +). Let D = {0, 1, 3, 9}, then D is a (13, 4, 1)-difference set.
The projective plane of order 3 is S = (P, L) where:
P = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
L = {{ 0, 1, 3, 9 },
{ 1, 2, 4, 10},
{ 2, 3, 5, 11},
{ 3, 4, 6, 12},
{ 4, 5, 7, 0 },
{ 5, 6, 8, 1 },
{ 6, 7, 9, 2 },
{ 7, 8, 10, 3},
{ 8, 9, 11, 4},
{9, 10, 12, 5},
{10, 11, 0, 6},
{11, 12, 1, 7},
{ 12, 0, 2, 8}}
FINITE PROJECTIVE PLANES
11
5.3. Uniqueness of the projective plane of order 3.
The following theorem’s proof is of very similar nature as the proof of the uniqueness of the projective plane of order 2. In this proof there is no graphical presentation of the situation due to the many graphical details such a presentation requires.
Theorem 7. The projective plane of order 3 is unique up to isomorphism.
Proof. Let T be a projective plane of order 3.
Assume P = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} is the point set of T .
Consider a quadrangle in T , that is,
a set of 4 points such that no 3 of them are collinear.
We label them: {0, 1, 2, 5}
The line containing the points {0, 1} must cross the line containing {2, 5} we label
this crossing point by 3, also, each line contains 4 points,
and so we obtain the lines: {0, 1, 3, 9} and {2, 3, 5, 11}.
Using a similar consideration about {1, 2} and {0, 5}
we obtain the lines: {1, 2, 4, 10} and {4, 5, 7, 0}.
Again, by the same argument we obtain for {0, 2} and {1, 5}
the lines: {12, 0, 2, 8} and {5, 6, 8, 1}.
At this stage we have all 13 points of the plane, and 6 lines.
Each of the 4 points of the quadrangle is contained in 3 lines.
Each of the 3 points of {3, 4, 8} is contained in 2 lines,
and the remaining 6 points are each on one line.
For each point of the quadrangle there are exactly 3 points of P that are not
collinear with it, thus, the last line through each of the points of the quadrangle is
determined and we obtain the lines:
{10, 11, 0, 6}, {11, 12, 1, 7}, {6, 7, 9, 2} and {9, 10, 12, 5}.
At this stage, each point of {6, 7, 9, 10, 11, 12} has 3 lines containing it,
and each point of {3, 4, 8} has 2 lines containing it.
Again, for each point of {6, 7, 9, 10, 11, 12} there are exactly 3 points of P that are
not collinear with it, thus, the last line through each of these points is determined
and we obtain the remaining 3 lines:
{3, 4, 6, 12}, {7, 8, 10, 3} and {8, 9, 11, 4}.
Clearly T is isomorphic to S from the previous section,
i.e. The projective plane of order 3 is unique.
For similar proofs about the planes of order 4 and 5 see [7].
12
DANNY KALMANOVICH
6. The Projective Plane of order 4
In this section the plane of order 4 will be presented using some substructures
contained in the plane: We will show a way to start from an arbitrary quadrangle
and extend it uniquely to a hyperoval. In the next step, we take a hyperoval, obtained from a quadrangle, and uniquely extend it to the whole plane.
During this presentation we will also see some nice connections between the projective plane of order 4 and the projective plane of order 2.
We begin by stating a few properties of hyperovals:
6.1. Some simple properties of hyperovals.
• No 3 collinear points.
• Any line of the plane either meets the hyperoval at exactly 2 points, or 0
points.
6.2. A hyperoval in a projective plane of order 4.
Let S = (P, L) be a projective plane of order 4.
Consider a quadrangle Q = {p1 , p2 , p3 , p4 } of S.
Denote by:
d 1 = p1 p2 ∩ p3 p4
d 2 = p2 p3 ∩ p1 p4
d 3 = p1 p3 ∩ p2 p4
the diagonal points of Q.
Lemma 1. The diagonal points of a quadrangle Q of S are collinear.
Proof. Assume that the diagonal points of Q are not collinear.
There are 9 lines that intersect Q ∪ {d1 , d2 , d3 } in at least two points:
6 diagonals, namely, the lines pi pj , 1 ≤ i, j ≤ 4, i 6= j,
3 lines through the diagonal points, namely, the lines di dj , 1 ≤ i, j ≤ 3, i 6= j.
Moreover, there are 11 lines having just one intersection point with Q ∪ {d1 , d2 , d3 }:
2 for each point of Q, 1 for each diagonal point of Q.
Since 9 + 11 < 21, there exists a line l having no intersection with Q ∪ {d1 , d2 , d3 }
therefore l intersects the 6 diagonals of Q at 6 different points.
This is a contradiction to the fact that any line has 5 points on it.
Denote by d the line containing the diagonal points of Q. By the lemma,
adding the two remaining points p, q on d to the points of Q we obtain 6 points
{p1 , p2 , p3 , p4 , p, q} no 3 of which are collinear, namely, a hyperoval in S.
Now, using a hyperoval, we proceed to the construction of the whole plane.
6.3. A model of the projective plane of order 4.
As we have seen, given a quadrangle we can construct a unique hyperoval containing it. Next, we will show that this hyperoval can be uniquely extended to the
whole plane.
Proposition 4. Let H = {A, B, C, D, E, F } be a hyperoval.
Then the lines AB, CD and EF meet at a point P.
FINITE PROJECTIVE PLANES
13
Proof. Denote P = AB ∩ CD.
By the fact that each line meets H at 0 or 2 points, the line P E must contain
F.
This proposition implies that every 1-factor of H (here we think of H as K6 )
has a corresponding point p outside of H. Now, given a hyperoval H, the number
of 1-factors of H is 15. The corresponding 15 points are all different, because if a
point p is a crossing point of 2 different 1-factors then p has at least 6 different lines
through it. This is a contradiction.
Therefore, we can label all 21 points:
15 points outside of H we label by the corresponding unique 1-factor of H,
6 points of H - labeled naturally {A, B, C, D, E, F }.
Also, we can label 15 lines which intersect H with the 15 edges of H (here, again
we think of H as K6 ).
Now, we need only to find a way to label the lines outside of H:
Let l be such a line, then each point of l corresponds to some 1-factor,
By Proposition 2 the corresponding 5 different 1-factors together form a unique
1-factorization of H (as K6 ).
So we can label l with this unique 1-factorization of H.
By Proposition 2 there are exactly 6 different 1-factorizations each of them corresponding to a unique line outside of H.
So we can finally label the remaining 6 lines and obtain the projective plane of
order 4.
For a more detailed presentation see [1].
7. the projective plane of order 5
In this section the projective plane of order 5 will be presented, The construction
of the plane of order 5 will again be accomplished with the aid of K6 .
We consider the notions of a 1-factor and a 1-factorization, this time with more
sophistication and delicacy.
7.1. The complete graph K6 and its 1-factors and 1-factorizations.
Using Proposition 3 we may consider the concrete 1-factorization F = {f1 , f2 , f3 , f4 , f5 }
of K6 :
1
•
6•
2
•33
33
33
•3
•
5
•
4
f1
1
•
2
•
6 •KKK
ss• 3
KK
KK ssss
Ks
ss KK
•s
•
5
4
f2
1
2
•
ss•
s
s
ss
ss
s
6 •s
•3
•
4
•
5
f3
14
DANNY KALMANOVICH
1
2
•KKK •
KK
KK
KK
K
•3
6 •33
33
33
•
•
5
4
f4
6•
1
2
•//
•
// // //
///
/
•
•
5
4
f5
•3
Denote by F 0 = {f6 , . . . , f15 } the remaining 10 1-factors of K6 :
1
•
2
•
6 •33
33
•
5
•3
•
4
1
•
6•
•3
•
5
f6
1
2
•
•
•3
6•
•
•
5
4
f11
1
•
2
•
•
4
f7
6
1
•
•
2
•
s• 3
ss
ss
s
•s •
5
4
f12
2
•33
33
K
•3
6 • KK
KK
KK
•
•
5
4
f8
1
2
•// •33
33
//
/
/
•3
6 •33
33 ///
•
•
5
4
f9
1
2
•KKK •
K
KKK
K
•3
6 • KK KK
KK
•
•
5
4
f13
1
2
•// ss•
s/s
ss //
s
// ss• 3
6•
s
ss//
s
s
•
•
5
4
f14
Now we turn to the construction of the projective plane of order 5 in terms of K6 ,
F and F 0 :
7.2. A model of the projective plane of order 5.
The points are of three types:
- 6 vertices of K6 - denoted by vi , i ∈ {1, 2, . . . , 6}.
- 10 1-factors forming F 0 - denoted by fi , i ∈ {6, 7, . . . , 15}.
- 15 pairs of edges of 1-factors of F - denoted by {l, l0 }.
Also, there are three types of lines:
- 6 vertices of K6 - denoted by l(vi ), i ∈ {1, 2, . . . 6}.
- 10 1-factors forming F 0 - denoted by l(fi ), i ∈ {6, 7, . . . , 15}.
- 15 edges of K6 - denoted by li , i ∈ {1, 2, . . . , 15}.
Now we define the incidence rules:
The point vi is incident with the line l(vi ) and with the line li iff vi ∈ li .
The point fi is incident with the line li iff li ∈ fi and with the line l(fj ) iff
|fi ∩ fj | = 1.
The point {l, l0 } is incident with the line l(vi ) iff vi ∈ l00 where l00 is the third edge
1
2
•KKsKs•
ss KKK
K
ss
s
•3
6•
•
5
•
4
f10
1
•
2
•
6•
•3
•
5
•
4
f15
FINITE PROJECTIVE PLANES
15
of the 1-factor containing {l, l0 }, and with the line li iff li ∈ {l, l0 }, finally the point
{l, l0 } is incident with the line l(fi ) iff l00 ∈ fi .
Finally, it is easy to verify that this is indeed a projective plane of order 5:
There are 31 points and 31 lines.
Let vi be a point of the first type, then it is incident with 1 line of the first type
(the line l(vi )) and with 5 lines of the third type (one for each edge through vi ).
Let fi be a point of the second type, then it is incident with 3 lines of the third
type (one for each edge of fi ) and with 3 lines of the second type (because every
edge is contained in exactly 2 1-factors of F 0 ).
Let {l, l0 } be a point of the third type, then it is incident with 2 lines of the first
type (2 endpoints of the edge l00 ) and with 2 lines of the third type (the lines li = l
and lj = l0 ) and with 2 lines of the second type (one for each 1-factor containing
l00 ).
Let l(vi ) be a line of the first type, then it is incident with 1 point of the first type
(the point vi ) and 5 points of the third type (one for each edge l00 through vi ).
Let l(fi ) be a line of the second type, then it is incident with 3 points of the second
type (one for each fj such that |fi ∩ fj | = 1) and 3 points of third type (one for
each 1-factor of F containing an edge of fi ).
Let li be a line of the third type, then it is incident with 2 points of the first type
(one for each endpoint of li ) and 2 points of the second type (one for each 1-factor
fi containing li ) and 2 points of the third type (because for the edge li there are
exactly 2 pairs {l, l0 } containing li in the unique 1-factor of F containing li ).
7.3. General idea of the proof of uniqueness.
We begin as always with the existence of a small substructure, this time it is an
oval or a hexagon - a set of 6 points such that no 3 are collinear.
Let P be a projective plane of order 5, let Q = {p1 , p2 , p3 , p4 } be a quadrangle in
P . Denote by d1 = p1 p2 ∩ p3 p4 , d2 = p1 p3 ∩ p2 p4 , d3 = p1 p4 ∩ p2 p3 the diagonal
points of Q. We have:
Lemma 2. The diagonal points of Q are not collinear.
Proof. Assume the contrary that d1 , d2 , d3 are collinear, then there are 7 lines which
have 3 points in common with Q ∪ {d1 , d2 , d3 }. Moreover, through any point of
Q ∪ {d1 , d2 , d3 } there are 6 − 3 lines that intersect Q ∪ {d1 , d2 , d3 } in just one point.
Therefore there are 7 + 7(6 − 3) = 28 lines which intersect Q ∪ {d1 , d2 , d3 } in at
least one point, hence there is a line l disjoint from Q ∪ {d1 , d2 , d3 }, since all lines
of Q ∪ {d1 , d2 , d3 } intersect in Q ∪ {d1 , d2 , d3 }, l meets these 7 lines in 7 distinct
points - this is a contradiction to the fact that l has just 6 points.
The 3 lines which contain two diagonal points of the quadrangle Q are said to
be the diagonal lines of Q.
Lemma 3. Let Q be a quadrangle of P and let q be a point such that Q ∪ {q} is a
5-gon then:
(a) Q lies on a diagonal line.
(b) Conversely, on every diagonal line there are exactly two candidates for q.
(c) Q is contained in precisely six 5-gons.
16
DANNY KALMANOVICH
Proof. (a) Since q is off the lines pi pj no line qpi contains a diagonal point. If q were
off any diagonal line, then the lines qd1 , qd2 , qd3 would be distinct and q would be
incident with 7 lines - a contradiction.
(b) The diagonal line D1 through d2 , d3 meets the lines p1 p2 and p3 p4 in two distinct
points, therefore there remains exactly two points on D1 which are not collinear
with two points of Q.
Now we can determine the hexagons which contain a given quadrangle:
Theorem 8. Let Q ∪ {q1 } = {p1 , p2 , p3 , p4 , q1 } be a 5-gon, and let D1 = d2 d3 be
the diagonal line of Q through q1 .
(a) If q2 is the second point on D1 such that Q ∪ {q2 } is a 5-gon then,
H = Q ∪ {q1 , q2 } is a hexagon, moreover H is the only hexagon through Q ∪ {q1 }.
(b) Any 5-gon lies in a unique hexagon.
(c) Any quadrangle is contained in precisely 3 hexagons.
Proof. (a) Since D1 intersects H only in q1 and q2 , H is a hexagon.
On the other hand, if q3 , . . . , q6 are the points on d2 and d3 such that Q ∪ {qj } is a
5-gon, then the sets Q ∪ {q1 , qj } are the only candidates other then H of hexagons
through Q ∪ {q1 }, however since the lines D1 , q1 d1 , q1 p1 , q1 p2 , q1 p3 and q1 p4 are
distinct, these are precisely the six lines through q1 . So, qj is on one of these lines.
Since qj ∈
/ d1 , q1 d1 , it is on a line q1 pi , therefore Q ∪ {qi , qj } cannot be a hexagon.
(b) A direct result of (a).
(c) By the previous lemma, every quadrangle is contained in six 5-gons. These six
5-gons are divided into three pairs - one for every diagonal line, such that each pair
is contained in one hexagon.
We are now ready to describe the points and lines of P in terms of a given
hexagon H:
There are three types of lines:
(1) l is called a secant if |l ∩ H| = 2.
(2) l is called a tangent if |l ∩ H| = 1.
(3) l is called a non-secant if |l ∩ H| = 0.
Since any point of H is on 5 secants, it is incident with exactly one tangent. Therefore the total number of tangents is 6.
Any point off H is incident with either 0 or 2 tangents - this is because any of the 5
points on a given tangent other then the point of tangency must have at least one
other tangent while there are only 5 other tangents.
There are also three types of points:
(1) points of H.
(2) p is called an outer point if p is incident with 2 tangents.
(3) p is called an inner point if p is incident with 0 tangents.
Since each of the 6 tangents has 5 outer points and every outer point is incident
with 2 tangents, the number of outer points is: 6 · 5/2 = 15, consequently, the
number of inner points is 31 − 6 − 15 = 10.
We are now ready to see that this model is isomorphic to the model from
section 7.2:
For an inner point i, we denote by F (i) the set of secants through i.
For an outer point o, let t1 and t2 be the tangents through o and let x1 = t1 ∩ H
FINITE PROJECTIVE PLANES
17
and x2 = t2 ∩ H, we denote by F (o) the set of secants through o together with the
line m(o) = x1 x2 .
So, if p is an inner or an outer point, F (p) is a set of 3 lines with the property that
every point of H is on exactly one line of F (p).
If we think of H as K6 , F (p) is a 1-factor of K6 . We call F (p) an inner 1-factor if
p is an inner point, and an outer 1-factor if p is an outer point.
Proposition 5.
• Every 1-factor of K6 is an inner 1-factor or an outer 1-factor.
• The outer 1-factors form a 1-factorization.
This proposition together with the following description of lines implies that the
model constructed from H is actually the same as the model constructed in the
previous section 7.2 and since there is only one 1-factorization of K6 - P is unique.
We now turn to describe the lines:
Let tx be a tangent (at the point x), every point on tx other then x is an outer
point o such that x ∈ m(o), that is, iff there is an outer 1-factor f = {l1 , l2 , l3 } with
o = l2 ∩ l3 and x ∈ l1 .
Let s be a secant, the inner points i on s are such that s ∈ F (i), the outer points
on s are s ∩ l2 and s ∩ l3 where f = {s, l2 , l3 } is the outer 1-factor through s.
Let l be a non-secant, it has 3 outer points o1 , o2 , o3 and 3 inner points i1 , i2 , i3 .
Since the tangents meet in pairs at the outer points on l, the set n(l) = {m(o1 ), m(o2 ), m(o3 )}
is an inner 1-factor, Therefore the outer points on l are obtained as follows: let
F = {s, l2 , l3 } be an outer 1-factor that has exactly one line s in common with n(l)
then l2 ∩ l3 is on l. For any inner point pj ∈ l, F (pj ) 6= n(l), therefore an inner
point i is on l iff |F (i) ∩ n(l)| = 1.
For a more detailed analysis of this presentation see [2].
References
[1] A. Beutelspacher: 21-6=15: a connection between two distinguished geometries, The
American Mathematical Monthly., Vol. 93, No. 1, Jan. 1986, pp. 29-41.
[2] A. Beutelspacher: a combinatorial approach to the projective plane of order five,
Journal of Geometry., Vol. 30 (1987), pp. 182-195.
[3] P. J. Cameron and J. H. van Lint, Designs, Graphs, Codes and their Links, Cambridge
University Press 1991, pp. 29-35, 81-88.
[4] P. J. Cameron, Combinatorics: Topics, Techniques, Algorithms, Cambridge University Press
1994, pp. 128-144.
[5] D. R. Hughes and F. C. Piper, Design theory, Cambridge University Press 1985, pp. 52-64,
93-98.
[6] C. R. Mac Innes: finite planes with less than eight points on a line, The American
Mathematical Monthly., Vol. 14, Vol. 14, pp. 171-174.
[7] R. G. Stanton: a combinatorial approach to small projective geometries, Bulletin of
the institute of combinatorics and its applications., Vol. 17 (1996), pp. 63-70.
[8] J. Singer: a theorem in finite projective geometry and some applications to number
theory, Trans. Amer. Math. Soc. 43 (1938), no. 3, pp. 377-385.