305
I nt. J. M. Math.
Vol. 4 No. 2 (19811 305-319
ON RANK 4 PROJECTIVE PLANES
O. BACHNLANN
Dpartement de mathmatiques
Ecole polytechnique fdrale
CH-1007 Lausanne, Suisse
(Received October 4, 1979)
Let a finite projective plane be called rank m plane if it admits a
ABSTRACT.
collineation group G of rank m, let it be called strong rank m plane if moreover
Gp
G
1
for some point-line pair (P,I).
It is well known that every rank 2 plane
It is conjectured that the only
is desarguesian (Theorem of Ostrom and Wagner).
rank 3 plane is the plane of order 2.
plane is the plane of order 2.
By [i] and [7] the only strong rank 3
In this paper it is proved that no strong rank 4
plane exists.
KEY WORDS AND PHRASES.
Projec//ve p/anew, rank 4 gscoap.
980 ATHENATIC$ SUBJECT CLASSIFICATION CODE:
1
50
Geome.
INTRODUCTION
In
[6]
Kallaher gives restrictions for the order n of a finite rank 3 pro-
306
O. BACHMANN
jective plane and conjectures that no such plane exists if n # 2.
Let a finite
projective plane be called a strong rank m projective plane if it admits a rank
m collineation group G such that
Bachmann
[I]
and Kantor
[7]
Gp
G
for some point-line pair (P,1).
I
no strong rank 3 projective plane of order n
By
# 2
exists. If the conjecture is true that for projective designs the representations
on the points and on the blocks of an arbitrary transitive collineation group are
[2],
similar (see Dembowski
p. 78), then every rank m projective plane is a strong
rank m p lane.
We shall prove in this article the following
No strong rank 4 projective plane exists.
THEOREM:
To prove the Theorem we first divide the strong rank 4 planes into 3 classes
(see Lemma 2 and 3).
C of trace 0 (see
Then we associate with each such plane (O
[3]).
,
)-matrices A and
Finally we show that for each class the trace condition
contradicts the integrality of the multiplicities of the eigenvalues of A or C.
We shall use the following notations, definitions and basic results (see
Dembowski
[2] )
A collineation group of a projective plane has equally many point orbits and
line orbits.
The rank of a transitive permutation group is the number of orbits
of the stabilizer of one of the permuted elements.
transitive collineation group of a projective plane,
ranks are equal (Kantor
[8]). A
If G is a (point or line)
then the point and line
rank m projective plane is a projective plane
which admits a transitive collineation group whose (point or line ) rank is m
(m 2).
them.
The lines (points) are identified with the set of points (lines) on
We write
P E i
G
if and only if
P
i Y for all y
G.
307
RA 4 PROJECTIVE PLANES
Let
4
-
PROOF OF THE THEOREM.
2.
(P,L,) be a projective plane of finite order n and let G be a rank
collineation group of
It is easily seen that
pO
PYO
and
write i.
1
P..
for
1
Clearly
pd
i
pyO
I
pay
For
PP
Gp
P
G
P---- L is defined by
O
-1
iY
I
G.
for some
O
If
i
we
and
O
iY
Y
-1
IL
P
for all
{P),
4 orbits
has exactly
(Po’lo)"
for some point-line pair
IO
A bijective map
n .3.
if and only if
i
Gp O
such that
A(P),
r(P),
[.
(P).
G
(i)
We choose
the notation in such a way that
LEMM 2.1:
If
I(A’)2(B’)I
PROOF:
If
A
61, /2’
if
A
J3 e [L, r ,][}
3(B)
A’6
and
A3(B), A’B’),
then
IAI(A) A2(B)[
3(B’).
then for some
e__G,
7aG
whence by (2)
(A) 2(B)I
L 2.2:
orbits, say
and
P3
Suppose that
d(Po)
(Po )
lo-{Po}
Po
i
Then
O
and
i
GPo2
can be chosen such that
1
O
-(Po}
and
Po-(lo}
Gp o
are
PGP2
U(P)"
Po -{I with
Pifi Io; Po’ P2’ P3 e 12; P2
(Fig. i).
The case described by Lemma 2 will be called case I.
lo-{Po}is not a
lo-{Po}ffi (Po)U r(Po). Then
pd. This leads to the
with i
PROOF:
say
P
GP
If
GPo
orbit, then it is the union of 2 orbits,
Po-iloiS
a linQ orbit
contradiction
i
GP
and
][(Po)
O. BACHMANN
308
GPo
1Gpo
Hence we may assume that
GPo
Gpod
2
lo-{Po}
A(Po)
Dually:
Po-{io] where
-2
P
Po
r (Po)
(note
that
For any poin Q
P{
Let
A(Po).
[((11’ 12 )d
P1 (Po )
l
Io )
P2 e
that
P3
P3
P16 Qo
exists then
P
t r6G be such that
TT
P PI.
put
GPo’P2
and hence
It rins to prove
such
P2 e 1{
If
then
P2 la
P2P’P2
this plies
P1
for some point
1
I.
6 (Po)
12 {Po P2}
for all Q 6
i
en
Po"
exists such that
12
P1 13"
It follows that
Gpo p2 GpoP 1"
and hence
I O and therefore
(7 7o) -I
Ceo
If no
Po lo’ P0
# P O end
GPo
Hence
Further
I
P2
for otherwise
Z’
P2
6
l’
po
"To
for all
i
for some
Yl,r’6 Gpo
for some
oGpo
(4)
which cannot occur.
Yo 6 Gpo
if and only if
IZ-{Po}
prove (5) notsthat by (4) through any point of
one and hence exactly one line of
Let’s apply (5)
to
GpI
Gp
ii o.
in place of
(3) and
Gpo:
P2 I t
oGP2.
goes at least
then imply (5).
GY
(5)
,_iGp Y
poG
y
[Po, Pi};
(Po,
Pof
hence
6
1Zcfor
Po; (P)
’16Gp
some
pP1for
R
Of the 3 orbits
(Poll)G, (Po,12)G,
I
G
only one consists of
II) belon
and (E,
Hence there exists
LEMMA 2.3:
A (Po)
(Po)
P3
P
G
(Po, r)
12
(Po),
P2
T16Gpo"
rforsome
Let
iuced by G on
(Pi, r)
and
and then also
R
This contricts
Gp
Po
"
P1 13"
such that
Po io"
Io
S
for some
’"
(PI’ io)
Thus
Gp
if and only if
to the se G-orbit.
Suppose that
io"
flabs.
S
12-Po
.ny R6
It follows that
(Po ll)
say
PI’ P2
Po
YEG such that
where
309
PROJECTIVE PLANES
RANK
Then
io
and dually
F(Po)’
P3
6
][(Po)
Po
are
Gpo-orbits
can be chosen such that
Po’ P2’ P3 i; P 12, 13; P2 i;. P3 12
or
Po’ P1 "P’ 3 61 ;2 PI IO; (Po ) 12 P2Y} for someYoGpo
P3 iI, 13. In both cases n4.
either
P
ii;
PI’ P2’
The 2 cases described by Lena 3
will be called case III
II2
(Fig. 2).
It is easily seen that
PROOF:
and
Po
(Po)
"P2’ P3
6
(P3’ 12)
II
(PI’
[Po, Io
lo)G’
P
11
so
i
Gpo-
orbits; say
P
Let
i.
Case IIi:
P
Case I12:
Po II"
Figure I
Clearly
are
A(Po)"
We
have to distinguish 2 cases:
Case I
CASE Ill:
resp. case
P
If
P3 6 12"
and
r(po
P2 13
then
pPo,
(P2’ 13)
o
i_
(Po)
6
(Po’
pPo
for some
II)G’ hence
O. BACHHANN
310
Case IIi
Case ZT2
Figure 2
Analogously
P2 13
P2 13
if
P3 6 12"
Thus
if and only if
(6)
P3 6 12"
Similarly one proves
PIe
If
then, by (6), we can choose
n > 3
Let’s show
that
n > 3
(Fig. 3).
(7)
12, 13
P2’ P3
such that
Suppose that
n- 3.
Then, since
P2 13 P3 12"
Put
P4 I i.
Po 11
and
P2 6 1o,
14 POP1" Let P56 I.o-(P1, P}.
Then PoE
15 and then 15 12
(POP5 f 13)P 4/ 12 Denote this
point by T. Clearly P2Ps 12
-I
T. Since (P2P5)
e 12 15 we
obtain the contradiction
-I
Figure 3
(P2Ps)
[
P2P5
RANK 4 PROJECTIVE PLANES
CASE 112:
Gp
o’
GPo ’P
PI
11 2/
Po’ PI’ P3 12
U
We may assume that
IP 2
n>3.
We first assume that
2
).
P2 e ,(Po
where
let
12
Po’ PI’ pO}
n>3,
o
11/ 1 2 P2
1 (
2
r (Po)
{
and hence, since
The only G-orbit of
G
n+l, hence
Ii 2
x_L
with som,
To
Gp
7o
o
G ’PI
GPl
consisting of flags is (P
and since
O’
12)
G
(PI’ 12)
(P2’ 11)’ (P2’ 13)’
Hence
1
-PI’ P72}
2
6
(P2’ io )G"
Gll GP1
which is impossible.
P1 13"
If n=3 then
To
(P2’ 12
wisw
is transitive on
’PI
O
then implies that
(Po’ 12 )G" Po
(Po 12 )G in particular P2 i O’ iI 13
If
P1 6 1 3 then (PI’ 13) fi (Po’ 12 )G and hence (P3’ ii)
Since also (Po’ ii) E (P2’
io )G we have pl
o
P3 for some 7 1
This implies that
P3
Then
is invariant under
P2
(P3’ 12)’ (PI’ io)
(Po II) (P2 io)
Hence
Then
Gp
I;
r(P o)
311
TO
E
1’o
(P2
2G
o
12)
GPo, P2
2
70
P (P)
[9] ).
If
Hence
Q
[(P2P
4
4,2for if T
Yo
P2 P2
2
1
O
then
since other-
which contradicts
3
o
This completes the proof of the Lemma.
P2 P2
Let us now associate with
If
is of order
which is impossible. Moreover
I. It follows that
o P2P27
(P2P2)
7o.
To
PI’ P2 P3} ’o
p[Po’
12
(G,]) 3 (O,)-matrices.
is a G-orbit then let P’(P) denote the paired orbit (see Wielandt
Q6@(P)
then
e eP’(Q),
Q=P
for
some’EG
and
QY (p(p))7
#
(pT)
p (Q).
i.e.
Q 6P(P)
implies that
P
e’ (Q).
(8)
312
O. BACHMANN
This implies that in
...,
Case 112:
Case III:
Case I:
A’(P)
F(P)
A’ (P)
A(P)
r’(P)
(P)
r’ (r)
IT(P)
resp.
,(r)
(r)
II (e)
Y(P)
resp.
A,(P)
r(P)
r(P)
(P)
(P)
(P)
n,(p)
(p).
v). Let A be the (O ,l) -matrix with rows enumerated by the points
lumns by
(Pk)
and such that
B, C be the analogous matrices
Pk
and co-
if and only if PkEA(Pi).
(Pk’ (Pi ))
with r(Pk) resp. K(Pk in place of (Pk ).
Let
We have in
case III:
case I:
At
t
B, C
C
At
t
A, B
C
t
t
A, B
t
B, C
A
Let
k
=IA(P)I
i
if
Ja(P) a(Q)J
v
if r’ (P)-n(P)
C
A
if r’(P)=r(P)
m--IH(P)I
rP)l
A (Q)I
A.(P)
case 112:
r(P)
Q
I ,
if
Q 6
fH(p)
A(P)
A straightforward calculation shows that
I + A + B + C
t
A A
C
t
C
J
the
k I + AA
m I
vxv-matrix with l’s in every entry
+B
+C
+p’A +V’B +MC
t
t
B, C
C
RANK 4 PROJECTIVE PLANES
313
AJkJ
BJ-IJ
CJmJ.
Now we determine the eigenvalues of A in case 111 and of C in the cases I
and II 2.
CASE III:
k
n + I
i
n
n
m
k + i + m +
I
A
It follows that
(A
n2+
v
2
AtA
n +
1,
k
We have
where
n
3(n
+ I)
where
n
n + I,
n2+
n
Gp o p1
IP3
3
n-
3
IP 2
2
I,
I.
k
n I + J
hence
O. This gives the eigenvalues of A:
+f’’"
A2,3
CASE I:
+ I)
(n + I) I + A + B + C
(n + I) I)(A2- n I)
1
2(n
i
n,
’
I]I(Po)
v’
II[(Po)
mffin
q
1), k + I +m + I v
(n
n2+ n + I.
X(P3)
Let’s calculate ’:
I)
n(n
(note that
n
Ill(P3)] JII(P3 )6 A(Po)I
r(P3)
IA (Po)#
Gp
P2
#A
3
Po
and hence
(Po) n (P3)I
+
.
+
II
Ill (P3)
+Ill (P3) II (Po)[ +I
/
(9)
(P3)).
IA (Po) /’)- I’(P3)
+
IA (Po) 61 " (P3)I
Io)
Clearly
(II)
A
(P3) q ](Po)l
2.
(12)
O. BACHMANN
314
by Lemma
P3 6 E(Po hence
Gp
o
I implies
n
Thus IF o’P21
3
P
IA(p3) r(eo) e2ll for some To6G
G o
..oGp o ,p 2 we
To
that
11oGPo,P21 n I. Hence II Po,P21 n -i. Since P1 I
Gp
Since both groups are conjugate
then have
P1 ’P2 PI’ i.e.
P el(P
PROOF of (12):
(see the proof of
Moreover
P2
o’
and
O
GPo’P2GPo’P1
GPoPI GPo ’P2" T’
o6Gp
P2 13’ P2
lo
P2
3)
"
2) this gives
Lemma,,
To 6Gp
{
IP20 GPo ’P21
for some
{Po’ PI}’
O
if d only if
Thus
Gp o’2
p
Gp
P2 o{ 113 (
o
n-
I;
is transitive on
by the above,
Since
To
113
hence
]/’
o
io
G
P2Pol
2. This
proves (12).
Equations (I0), (II), (12) imply
H(P3)
[H(p3) (Po)
To determine
n
By (12)
Gp
Ir(Po)
l(Po)
Gp
[r(P o)
Gp
A(P3)
Gp
P2 o fl P2 3 ]12 o 12 3
A(P3)I
2.
f
A(Po)I
n
(13)
3.
we use
l(Po N ](P3)I + Ir(P o)
Gp
Since r(P 3)
P2 3, r(P o)
+
]I(P3)[
(14)
Y(P3)
1. It follows that
I](P3) 1(Po)l
n- 3.
(15)
Equations (9), (13) and (15) imply that
’
Analogously we calculate
n2- 3n + 5.
’ ’
and
(16)
RANK 4 PROJECTIVE PLANES
r(P o)
Gp
112 o
i
PI
IH(P1
O. Hence
I
’
n--
IA(Po)
A(P2)
Ii I
(note that
Gp
P21
eI
21
Ir(eo)
n,
Gp
Gp
I12 o O i I 21
Ir(Po)
O. Hence
(q
1(Po)l + IH(P 2) H(Po)I"
IA(P o) ctr(P2)[ + A(Po C If(P2)
A(P2)
O
l(eo)l
[H(P2
(n
l)(n
2)(A + B) + (n 2- 3n + 5) C
l)(n
and then
(C
n(n- ) I)(C 2- 3C
The eigenvalues of C are
CASE 112:
k
n +
i
By the proof of Lemma 3
(n + l)(n
][(Po)
[(P3)I
2)
!. In n + 1
clearly
l(Po
I,
m
Gp
Gp
o
r(e 2)
el 21
Gp
el
III
o/q i
IP(P o) (ql(P2)I
and
Gp
N
1
2
I
2). Hence IH(P 2)
A(P2)I -II(Po
Ir(Po)
(I
A(eo)
r(P2)l
Gp
Pl 21
1
Gp
e 2.o
I. It follows that
n
2).
ctc
C 2=
Equations (16), (17) and (18) imply that
(n
(17)
I(P 2)
+
and hence
II(P o )I
’
and thus
Gp
n
[H(P2)f 1"(Po)
Further
where
G
P
o
2).
l)(n
1(e2)I =[el
IA(Po)
and
0
I
n
IH(P2) --IH(P 2) f A(Po)
IA(Po) (A(Po)q A(P2) +
n(n- I)
In
(n
(eo)
IP2
(Pl)l
}r(P o)
the proof of (12) and
Y
(P)I
315
n(n
l)(n
I) I + (n
n(n
2(n
I) I)
n(n
I);
I) I +
2)(J- I) + 3C
O.
(3
12, 3
+_
Sn
+’I’
(n- 2)(n + I), k + i + m + I
)/2.
v
2
n + n + I.
n4. Let’s determine I’
I[(P3) l(P3)q Z(Po) + I(P3)/ r(po)I + l][(P3)/
IA(Po) [A(Po)(q A(P3) + IA(Po)/q I(P3)I + IA(Po) (
(P3)[
(Po)f] T’(P3)
I. Let’s show that
{(P3)[ l’(Po)
(Pa)/ I (Po)
2.
2
(19)
(20)
316
O. BAC’HNANN
PROOF of (19) and (20): By Lemma
For
Yo 6 Gp
o’ P
(IP21-
o
oGeo,e I
pO2
for otherwise
IIOGp
o o Yo o
P2 ii ii P2 P2
O
if and only if
GPo Pl
GP ’PI,
2(n
2)
I
n +
which leads to the contradiction
7’o Gpo,PI
iI
o (
I).
Further
P2
otherwise,Gp
of
P2
since
Gp
P2
iI U
P2P2
o GP
for some
o 75, every line of
GPo
ii
o and this would iply that a point of
-
21 +
(22)
would contain at least 3 points
Gp
P2
O
o
o
o
(llk]P 2 P2
exists. By (21)
’
70 70 7o 7o for some *
"7GP ’PI
eGpo Since
70
P2
613
P2
P2
P2
Gp
Ie2 o131 l](Po) (P3)l 2. GpThis proves (19).
o
o
Each of the n- 2 lines of
13 o -{P2 P2} through P2 contains exactly one
point of
P2 o
I Together with P P2 this gives n points of P2 o. It
follows that exactly one point of
P2 o { o] lies on iI. This proves (20).
and (221
By means of (19) we obtain
ill(P3 )( (Po) n 2.
In n + 1
[r(Po )] [](Po A(P3 )] + lr(Po ) F(P3)[ + lr(Po) (P3)[
Gp
Gp
Gp
Gp
lr(Po) (3 a(P3)l 2 by (19) and Ir(Po) iq T(P3)] [P2 o n P2 31
I. Hence [H(P3) r(Po)
n- 2. It follows that l’
n2- 3n + I.
u’ III(P o) q II(PI)
[H(PI)I III(P 1) (3 A(Po) [II(P 1) (Po)[ where
Ill(P1) (n + l)(n- 2), [II(PI) f% A(Po) -n- 2 and ill(PI) (] [[’(Po)[ lr(Po)
by (20)
[I’(PolO(Pa)l -Ir(Po) I’(P)I. Ir(P o) C (Pa)l
1. Hence In(P_)cl r(Po)
-2 and
IP2 o (I P2 11 112 o F 1
RANK 4 PROJECTIVE PLANES
17
(n- )(n- I).
{H(P2) r(Po)
Ill(P_)f
P3Po
through y
’
nt
2)(n
(n
on 1
goes aculy one 1ne of
o
C
2
CtC
GpO
l
o and one
Gp
o
2
o. Hence
1)(n
O. The eigenvalues of C are
:
REMARK: Let
(C
2) J whence
(n + 1)(n
(n + l)(n
)’i
2),
2)(A + B) +
l)(n
2) I + (n
(n + 1)(n
2) I + (n 2- 3n + 2)(A + B + C)
(n + 1)(n
(n
2(n- 2) I
’o
’o
(
).
I follows that
(n2- 3n + ) C
by Lemma I
r(Pt)
2)
A2
C and
I)(C2+
C
(-I +
/8n
C2+ C
2(n
2) I)
q_5)/2
be the matrix representation of G obtained by
G--GL ()
v
associating with each T6G the corresponding permutation matrix (7) (the ordering
of
P is
the same as used in constructing the matrices A, B, C). By (2)
commutes with
A, B, C for all YG. Hence, by [9] Theorem 28.4, {I, A, B,
the basis of the commuting algebra
If(G)
of @. By
[9]
of
D.x
then
I, D 2,
D
1
f
I
and.
D3,
ilffi fi
C}
is
Theorem 29.5 W(G) is connu-
tative and hence, by [9] Theorem 29.4, the representation
constituents
()
has 4 irreducible
D 4, each with multiplicity I. If
f.l
is the degree
v
,
Let us finally show how the fact that A and C have trace 0 contradicts the
integrality of the mmltiplicities of
12’ 13"
318
O. BACHMANN
In the 3 cases
of
A2;
f
then v
AI appears with multiplicity I.
I is the multiplicity of
Sn
3"
This leads to
+ 1 )/2 + (n(n + I)
O
n(n- i) + f(3 +
O
(n + I) +
f+ (-)(n(n
O
(n + l)(n
2) + f(-I +
+ I)
8n
Let f denote the multiplicity
)/2
f)(3
f)
in case I,
in case
15)/2 + (n(n + I)
15’ )/2
8n
f)(-I
in case 112.
In any case this contradicts the fact that
n>2
In case III this is clear.
In case I suppose that a prime p divides
and then
that
p
8n
3n, i.e. p-- 3. This implies that
n(Sn + l)/Sn +
(3 2i-
I’
2
Xn + I, p2 n
19112].
4. Hence
Suppose that n
15
8n
38. Since
{172
tion group is of odd order (Hughes
n668, 19112}.
232
-
Then
3
n=-2
2i
are integers:
pn,
5n +
hence P
for some
i72
Then
p
&(17,
23
2
2
17-23 }, i.e.
n
e38,
68,
Sn 15’.
geo,P2 is transitive on
is even. This contradicts the fact that if
Suppose that
8n + I
I)(5-32i-I+ I)/82-3 i-I
In case 112 suppose that a prime p divides
P
/
fI
and
12-{P o, PI’
P2}
and
and
IG
rood 4, then the full collinea-
[5]).
2
Then n is not a square and n + n + I not a prime.
Hence, since G is flag-transitive, n is a prime power (Higman and Mc Laughlin [4])
which is absurd.
REFERENCES
I.
Bachmann, 0., On rank 3 projective
2.
Dembowski, P., Finite Geometries, New York, Springer, 1968.
3.
Higman, D.G., Finite permutation groups of rank 3, Math. Z. 86 (1964), 145-
desi.ns,
Mh. Math.
8__9 (1980), 175-183.
156.
4.
Higman, D.G. and J.E. Mc Laughlin, Geometric ABA-groups, IIi. J. Math.
(1961), 382-397.
5__
RANK 4 PROJECTIVE PLANES
3].9
5.
Hughes, D.R., Generalized incidence matrices over
Math. I (1957), 545-551.
6.
Kallaher, M.J., On rank 3 proective planes, Pacif. J. Math. 39 (1971), 207-
group algebras,
Iii. J.
214.
7.
Kantor, W.M., Moore geometries and rank 3 groups having -I, Quart. J. Math.
Oxford (2) 28 (1977), 309-328.
8.
Kantor, W.M.,
9.
Wielandt, H., Finite permutation groups, New York, Academic Press, 1964.
A....tomorPhism .roups
of designs, Math. Z. 09 (1969), 246-252.