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Infinite (Af, Af∗)-geometries
IOURNAL
OF COMBINATORIAL
THEORY,
%rkS
A 55, 133-139
( 1990)
Note Infinite
(Af, /If*)
CHRISTIANE UnioersitP
Libre
-Geometries
LEFEVRE-PERCSY
de Bru.~elles. Campus de la Plaine B-1050 Bruxelles, Belgium
Communicated
by Francis
Received
February
C.P. .?16.
Buekenhout 27, 1989
This paper extends to the infinite case a previous classification result on finite rank 3 geometries with afline planes and dual afflne point residues, by giving a new (” 1990 Academic Press, Inc. proof of it, free of any finiteness assumption.
1. INTRODUCTION Let r be a rank 3 geometry, assumed to be connected and firm in the sense of Buekenhout [Z], the three types of elements being called points, lines, and planes. Suppose that f admits the Buekenhout diagram .4f 0 0
af* 1
0 2
where Af means afline plane and Af* dual affine plane. Such a geometry is called an (Af, Af*)-geometry. Van Nypelseer and I have studied in [3] finite (Af, Af*)-geometries satisfying the following axiom: [A] any two points off are incident with at most one line. It was proved that they belong to two known classes of examples, denoted by A, and A,, and obtained from a 3-dimensional affine space A, either by deleting a point 0 and all lines and planes through 0, or by deleting a direction cc of lines and all planes parallel to that direction. The aim of the present note is to extend this result to the infinite case by giving a new proof of it, free of any finiteness assumption. THEOREM. Let r be an (Af, Af*)-geometry satisfying [A]. Then r is isomorphic to one of the (possibly infinite) geometries A,, and A,.
133 OQ97-3165190
$3.00
Copyright ~0 1990 by Academic Press, Inc. All rights of reproduction in any form reserved.
134
CHRISTIANE LEFEVRE-PERCSY
2. PRELIMINARY
RESULTS AND NOTATIONS
The proof of certain preliminary results of [S] does not use the finiteness assumption; we recall those results we shall need in the sequel. The symbol r always denotes an (Af, #*)-geometry satisfying [A]. PROPOSITION 1 [3, Proposition 11. r is isomorphic to the geometry of all points qf r provided with the set qf O-shadowsof all flags of r ordered by inclusion.
Thanks to this proposition, we use as in [3] expressions like “a point belongs to a line”, “ a line is contained in a plane”, etc. We will say that two distinct points a and b are adjacent (denoted by a-b) if there is a line of r on a and 6; this line (denoted by ah) is unique if [A] is satisfied. The non-adjacency relation is denoted by 7L. Note that a point is always assumed to be non-adjacent to itself. PROPOSITION 2 [3, Lemma
11. Zf r has a line with ,finite cardinality q
then each Iine of r has q points. PROPOSITION
3 [3, Lemma 23. 7L is an equivalence relation.
Notations 1 [3]. If p is a point of r, we will denote by C(p) the equivalence class of p for the relation 7L. Let P,(p), for i EZ(p), be the classes of parallel points in the residue of p (which is a dual affine plane: two points of this residue are parallel if they are non-collinear). Remark that 1Pi(p)1 3 2. We will denote by S;(p) the union of C(p) and all points on the lines of r belonging to P,(p).
3. PROOF OF THE THEOREM We shall study the structure of the linear space L, defined as follows. DEFINITION 1. Let L be the linear space whose points are the points of r and whose lines are the lines of r and the sets C(p), which will be called ideal lines of L.
The next two propositions are stated in [3] in the finite context. So we have to restate them here, together with their proof. PROPOSITION 4. For every point p’ E Si (p), there exists an index j EI( p’) such that Si(p’) = S;(p).
INFINITE (AA Af * )-GEOMETRIES
135
Proqf Suppose first p’ mp. Let P,( p’) be the class of parallel points of the residue of p’ which contains the line pp’; let Si(p’) be the corresponding set (see Notations above). If we show that Sj(p) is contained in S,(p’), then, as the roles of p and p’ are symmetric, we also show that S,(p’) is contained in S,(p) and the proposition is proved. Let s be any point of the line p’x is not coplanar with the line p’p; in other S,(P). If x-p’, words, p’x E Pi( p’) and so x E S, (p’). If x 7L p’, then .YE S,( p’) by definition. To finish the proof, suppose now p’ + p. Consider p” in S,(p) with p” - p; necessarily p” wp’. Then, by application of the preceding result for p, p” and p”, p’, there exist j and k such that S,(p) = S,(p”) = S,(p’). PROPOSITION 5. For every point p of r and every i in I(p), the set S,(p) is a linear subspaceof L.
Proof: Let p, and p2 be two points in S,(p). If p, and pz are both in C(p), the unique line of L through p, and pz is C(p), which is contained in S,(p). If one of the points is adjacent to p (say p1 -p) then, by Proposition 4, the line of L through p, and pz is in S,(p,) = S,(p) and the proof
is finished. This last result motivates the following definitions. DEFINITIONS 2. The sets S,(p), where p is a point of I- and i an index in Z(p), are called ideal planes of L. The linear space L provided with the planes of r and these ideal planes Si(p) becomes a rank 3 geometry which we will denote abusively by L. The residue of a point in this rank 3 geometry L is denoted by L,. The points of L, corresponding to the ideal lines through p are called ideal points of L,; the lines of L, corresponding to the ideal planes on p are called ideal lines of L,. PROPOSITION 6. L, is the projective pfane which is the completion of the residue of p in r.
Proof. The residue of p in r is a dual afIine plane. There is exactly one ideal line of L through p. Then, the proposition is an immediate consequence of the definition of the ideal planes S,(p) through p.
This proposition
has the following easy corollaries:
COROLLARY 1. Any three non-collinear points of L are contained either in exactly one plane of r or in exactly one ideal plane of L. COROLLARY 2. Two distinct ideal planes of L having a point in common intersect in an ideal line.
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CHRISTIANE LEFEVRE-PERCSY
COROLLARY 3. A plane of r and an ideal plane having a point in common intersect along a line of r.
A natural question arises: what is the structure of the ideal planes of L? LEMMA 1. Consider two ideal lines C(p,) and C(pz). Let S=Si(p,)= Sj(p2) be the ideal plane containing C(p,) and C(pz) (see Proposition 4). Supposer is a point qf r not in S. Then the class P,(r) containing the lines t-x1, for x, E C(p, ), and the class P?(r) containing the lines rxz, for x2 E C(p2), are distinct.
Proof Suppose, on the contrary, that P,(r) = PJr). This means that there is no plane of r containing rp, and rp,. So the lines p,r and p1 pz are also not coplanar in r. Consequently, r must belong to Si(p,) = S, a contradiction. LEMMA 2. Let S be an ideal plane of L and p a point not in S. Then S is isomorphic to a substructure &$of points and lines of the projective plane L,. Furthermore, each ideal line of S corresponds to an ideal line of L,.
Proof For every point I in S, the line px intersects S exactly in x, because S is in a linear subspaceof L (Proposition 5) and p is not in S. SO there is an injection of the points of S into L,. Thanks to Corollary 1, each line L of S is contained in a unique plane through p. This plane is ideal if and only if L is ideal. Furthermore, Corollaries 2 and 3 shows that two distinct lines of S determine two distinct planes through p; hence, there is an injection of the lines of S into the set of lines of L,, ideal lines of S corresponding to ideal lines of L,. Since the incidence between points and lines of S corresponds to the inclusion of lines into planes through p, the lemma is proved. LEMMA
3. Let p, S, ,f?as above. Then L, - 3 contains at least one line.
Proqf: Let Y be a point of S. Since p $ S, the ideal line C(p) does not meet S and so px is a line of r. If Li is a line of S on x, which is not an ideal line, there is a plane n, of r containing p and Li (otherwise there would be an ideal plane on p and Li and so Lj would be ideal). Consequently, the plane 71,is an aftine plane and so there is a line Ii through p which is parallel to L, in ni. As 1P,(x) 13 2, there are at least two distinct lines L1, L, on x and so there are at least two distinct lines I,, I, through p. The lines I, and I, are contained in a plane a through p. If GIis disjoint from S, the lemma is proved. Thus suppose, by way of contradiction, that a intersects S. This cannot occur if CIis a plane of r, because in that case, c( is an afline plane and so the line an S intersect one of the I, and I,, contradicting the fact that I, and I2 do not meet S. Consequently, suppose
INFINITE (.4j; A~*)-GE~METRIES
137
further that c( is an ideal plane intersecting S. If L’ E c(n S and if M is a line of S through z, different from C(z), then there is a line m on p parallel to M. Since the plane containing h4 and m intersects a in pz, it is clear that m is not in c(. Consequently, there exists a plane 0, (resp. 0,) of f containing m and I, (resp. I,), otherwise m and 1, (resp. 1,) would be in the same ideal plane c( through p. As was shown above, /I, (resp. PJ must be disjoint from S and the lemma is proved. LEMMA 4.
Let p, S, 3 be as above. If L, - 3 contains an ideal line L,
then L, - S is L. Proof Suppose there exists a point aE L, - s- L. This point a is a line A on p different from C(p), for C(p) is contained in the ideal plane corresponding to L. Choose a point b on L, such that the corresponding line B is not C(p). Then the plane (A, B) on p is a plane of f and must be disjoint from S (for a $3). Consequently the line ab is disjoint from 3. As this holds for every point b on L such that B # C(p), this implies that all points of 3 would be on the line joining b to the point of L, corresponding to C(p); this is impossible, because S has more than one line! LEMMA 5. Let p, S, 3 be as above. Let c be the point of Lp corresponding to C(p). If L, - 3 contains a line L which is not ideal, then L, - 3 is the union of L and c. Proof It is clear that the point c is not in 3. Suppose there exists a point a # c in L, - 3 - L and choose b E L. If the corresponding lines A and B through p are in a plane of r, i.e. if they are not in a same ideal plane through p, then the plane (A, B) is disjoint from S (for a$ 3). As this holds for every point b on L such that CE ab, all points of 3 must be on the line joining a to the point c, a contradiction.
Lemmas 2,4 and 5 imply the next proposition. PROPOSITION 7. Let S be an ideal plane of L. Then the only possible structures for S are the following: (i) S is an affine plane, obtained from the projective plane L,, by deleting a projective line. Furthermore, the ideal lines of S are the elements of a direction of parallel lines in S.
(ii) S is an affine plane minus one point, obtained from the projective plane L, b-y deleting a projective line and a point not on it. Furthermore, the ideal lines of S are the lines through the deleted point. COROLLARY 4. Any three non-collinear points of L generate a subspace of L which is a plane of I or an ideal plane of L.
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CHRISTIANE
LEFEVRE-PERCSY
ProoJ: This follows from Corollary 1, from Proposition fact that the planes of r are alIme planes.
7, and from the
Note that Proposition 7 implies that all projective planes L,, for p $ S, are isomorphic, since S cannot have structure (i) in some L, and structure (ii) in another L,. Actually, we prove that all ideal planes of L have the same structure. PROPOSITION 8. If L has ati ideal plane of type (i), then all ideal planes of L are of type (i).
ProoJ: Let S be an ideal plane of type (i). This means that, through every point p of L - S, there is exactly one ideal plane disjoint from S. So there is a partition of L into ideal planes and each of them must be of type (i). Now suppose there is an ideal plane S’ of type (ii). Then, necessarily, S’ meets S and S n S’ is an ideal line (see Corollary 2). Since S’ is of type (ii), there is a plane n of r disjoint from S’. This plane rr must intersect S (for S is of type (i)) and Sn rr is a line disjoint from Sn S’. As S is an affme plane, this means that S n ‘/I and S n S’ are parallel lines. But all lines of S that are parallel to the ideal line Sn S’ are ideal lines (see the description of case (i)). And so arises a contradiction: the intersection of S with the plane rc of r must be a line of ZY
We are now able to finish the proof of the theorem. Conclusion. Thanks to Corollary 4, the two above propositions allow us to determine the structure of the linear subspaces of L generated by three non-collinear points: if the ideal planes of L are of tye (i), then all 2-dimensional subspaces of L are affme planes; if the ideal planes of L are of type (ii), then the 2-dimensional subspaces of L are either alline planes or affine planes minus one point. In the last case, complete L by a new point 0, which is decided to be incident with all ideal lines and ideal planes of L; we get a new linear space L whose planes are all affme. Now we can apply Buekenhout’s characterization of afline spaces [l] and claim that either L or i. is an affine space, whenever the lines of L of L have at least four points. Since the residue of a point in L or L is a projective plane, this implies that L or L is a 3-dimensional afline space. Consequently, the geometry r is isomorphic to one of the geometries A, and A, defined at the beginning. To finish the proof, suppose that there is a line of L or L having less than four points. Then Proposition 1 shows that all lines of r have the same cardinality q < 4 and we can refer to the last paragrah of the proof of the finite case [3, “Proof of the theorem”].
INFINITE ( Af
Af * )-OE~METRIES
139
REFERENCES 1. F.
BLJEKENHOUT, Une caracttrisation des espaces affns baste sur la notion de droite. Math. Z. 111 (1969), 367-371. 2. F. BUEKENHOUT, The basic diagram of a geometry, in “Geometries and Groups” (M. Aigner and D. Jungnickel, Eds.), pp. l-29, Springer Verlag, Berlin, 1981. 3. C. LEFEVRE-PERCSY AND L. VAN NYPELSEER. Finite rank 3 geometries with afline planes and dual afline point residues. Discrete Math., to appear.
OF COMBINATORIAL
THEORY,
%rkS
A 55, 133-139
( 1990)
Note Infinite
(Af, /If*)
CHRISTIANE UnioersitP
Libre
-Geometries
LEFEVRE-PERCSY
de Bru.~elles. Campus de la Plaine B-1050 Bruxelles, Belgium
Communicated
by Francis
Received
February
C.P. .?16.
Buekenhout 27, 1989
This paper extends to the infinite case a previous classification result on finite rank 3 geometries with afline planes and dual afflne point residues, by giving a new (” 1990 Academic Press, Inc. proof of it, free of any finiteness assumption.
1. INTRODUCTION Let r be a rank 3 geometry, assumed to be connected and firm in the sense of Buekenhout [Z], the three types of elements being called points, lines, and planes. Suppose that f admits the Buekenhout diagram .4f 0 0
af* 1
0 2
where Af means afline plane and Af* dual affine plane. Such a geometry is called an (Af, Af*)-geometry. Van Nypelseer and I have studied in [3] finite (Af, Af*)-geometries satisfying the following axiom: [A] any two points off are incident with at most one line. It was proved that they belong to two known classes of examples, denoted by A, and A,, and obtained from a 3-dimensional affine space A, either by deleting a point 0 and all lines and planes through 0, or by deleting a direction cc of lines and all planes parallel to that direction. The aim of the present note is to extend this result to the infinite case by giving a new proof of it, free of any finiteness assumption. THEOREM. Let r be an (Af, Af*)-geometry satisfying [A]. Then r is isomorphic to one of the (possibly infinite) geometries A,, and A,.
133 OQ97-3165190
$3.00
Copyright ~0 1990 by Academic Press, Inc. All rights of reproduction in any form reserved.
134
CHRISTIANE LEFEVRE-PERCSY
2. PRELIMINARY
RESULTS AND NOTATIONS
The proof of certain preliminary results of [S] does not use the finiteness assumption; we recall those results we shall need in the sequel. The symbol r always denotes an (Af, #*)-geometry satisfying [A]. PROPOSITION 1 [3, Proposition 11. r is isomorphic to the geometry of all points qf r provided with the set qf O-shadowsof all flags of r ordered by inclusion.
Thanks to this proposition, we use as in [3] expressions like “a point belongs to a line”, “ a line is contained in a plane”, etc. We will say that two distinct points a and b are adjacent (denoted by a-b) if there is a line of r on a and 6; this line (denoted by ah) is unique if [A] is satisfied. The non-adjacency relation is denoted by 7L. Note that a point is always assumed to be non-adjacent to itself. PROPOSITION 2 [3, Lemma
11. Zf r has a line with ,finite cardinality q
then each Iine of r has q points. PROPOSITION
3 [3, Lemma 23. 7L is an equivalence relation.
Notations 1 [3]. If p is a point of r, we will denote by C(p) the equivalence class of p for the relation 7L. Let P,(p), for i EZ(p), be the classes of parallel points in the residue of p (which is a dual affine plane: two points of this residue are parallel if they are non-collinear). Remark that 1Pi(p)1 3 2. We will denote by S;(p) the union of C(p) and all points on the lines of r belonging to P,(p).
3. PROOF OF THE THEOREM We shall study the structure of the linear space L, defined as follows. DEFINITION 1. Let L be the linear space whose points are the points of r and whose lines are the lines of r and the sets C(p), which will be called ideal lines of L.
The next two propositions are stated in [3] in the finite context. So we have to restate them here, together with their proof. PROPOSITION 4. For every point p’ E Si (p), there exists an index j EI( p’) such that Si(p’) = S;(p).
INFINITE (AA Af * )-GEOMETRIES
135
Proqf Suppose first p’ mp. Let P,( p’) be the class of parallel points of the residue of p’ which contains the line pp’; let Si(p’) be the corresponding set (see Notations above). If we show that Sj(p) is contained in S,(p’), then, as the roles of p and p’ are symmetric, we also show that S,(p’) is contained in S,(p) and the proposition is proved. Let s be any point of the line p’x is not coplanar with the line p’p; in other S,(P). If x-p’, words, p’x E Pi( p’) and so x E S, (p’). If x 7L p’, then .YE S,( p’) by definition. To finish the proof, suppose now p’ + p. Consider p” in S,(p) with p” - p; necessarily p” wp’. Then, by application of the preceding result for p, p” and p”, p’, there exist j and k such that S,(p) = S,(p”) = S,(p’). PROPOSITION 5. For every point p of r and every i in I(p), the set S,(p) is a linear subspaceof L.
Proof: Let p, and p2 be two points in S,(p). If p, and pz are both in C(p), the unique line of L through p, and pz is C(p), which is contained in S,(p). If one of the points is adjacent to p (say p1 -p) then, by Proposition 4, the line of L through p, and pz is in S,(p,) = S,(p) and the proof
is finished. This last result motivates the following definitions. DEFINITIONS 2. The sets S,(p), where p is a point of I- and i an index in Z(p), are called ideal planes of L. The linear space L provided with the planes of r and these ideal planes Si(p) becomes a rank 3 geometry which we will denote abusively by L. The residue of a point in this rank 3 geometry L is denoted by L,. The points of L, corresponding to the ideal lines through p are called ideal points of L,; the lines of L, corresponding to the ideal planes on p are called ideal lines of L,. PROPOSITION 6. L, is the projective pfane which is the completion of the residue of p in r.
Proof. The residue of p in r is a dual afIine plane. There is exactly one ideal line of L through p. Then, the proposition is an immediate consequence of the definition of the ideal planes S,(p) through p.
This proposition
has the following easy corollaries:
COROLLARY 1. Any three non-collinear points of L are contained either in exactly one plane of r or in exactly one ideal plane of L. COROLLARY 2. Two distinct ideal planes of L having a point in common intersect in an ideal line.
136
CHRISTIANE LEFEVRE-PERCSY
COROLLARY 3. A plane of r and an ideal plane having a point in common intersect along a line of r.
A natural question arises: what is the structure of the ideal planes of L? LEMMA 1. Consider two ideal lines C(p,) and C(pz). Let S=Si(p,)= Sj(p2) be the ideal plane containing C(p,) and C(pz) (see Proposition 4). Supposer is a point qf r not in S. Then the class P,(r) containing the lines t-x1, for x, E C(p, ), and the class P?(r) containing the lines rxz, for x2 E C(p2), are distinct.
Proof Suppose, on the contrary, that P,(r) = PJr). This means that there is no plane of r containing rp, and rp,. So the lines p,r and p1 pz are also not coplanar in r. Consequently, r must belong to Si(p,) = S, a contradiction. LEMMA 2. Let S be an ideal plane of L and p a point not in S. Then S is isomorphic to a substructure &$of points and lines of the projective plane L,. Furthermore, each ideal line of S corresponds to an ideal line of L,.
Proof For every point I in S, the line px intersects S exactly in x, because S is in a linear subspaceof L (Proposition 5) and p is not in S. SO there is an injection of the points of S into L,. Thanks to Corollary 1, each line L of S is contained in a unique plane through p. This plane is ideal if and only if L is ideal. Furthermore, Corollaries 2 and 3 shows that two distinct lines of S determine two distinct planes through p; hence, there is an injection of the lines of S into the set of lines of L,, ideal lines of S corresponding to ideal lines of L,. Since the incidence between points and lines of S corresponds to the inclusion of lines into planes through p, the lemma is proved. LEMMA
3. Let p, S, ,f?as above. Then L, - 3 contains at least one line.
Proqf: Let Y be a point of S. Since p $ S, the ideal line C(p) does not meet S and so px is a line of r. If Li is a line of S on x, which is not an ideal line, there is a plane n, of r containing p and Li (otherwise there would be an ideal plane on p and Li and so Lj would be ideal). Consequently, the plane 71,is an aftine plane and so there is a line Ii through p which is parallel to L, in ni. As 1P,(x) 13 2, there are at least two distinct lines L1, L, on x and so there are at least two distinct lines I,, I, through p. The lines I, and I, are contained in a plane a through p. If GIis disjoint from S, the lemma is proved. Thus suppose, by way of contradiction, that a intersects S. This cannot occur if CIis a plane of r, because in that case, c( is an afline plane and so the line an S intersect one of the I, and I,, contradicting the fact that I, and I2 do not meet S. Consequently, suppose
INFINITE (.4j; A~*)-GE~METRIES
137
further that c( is an ideal plane intersecting S. If L’ E c(n S and if M is a line of S through z, different from C(z), then there is a line m on p parallel to M. Since the plane containing h4 and m intersects a in pz, it is clear that m is not in c(. Consequently, there exists a plane 0, (resp. 0,) of f containing m and I, (resp. I,), otherwise m and 1, (resp. 1,) would be in the same ideal plane c( through p. As was shown above, /I, (resp. PJ must be disjoint from S and the lemma is proved. LEMMA 4.
Let p, S, 3 be as above. If L, - 3 contains an ideal line L,
then L, - S is L. Proof Suppose there exists a point aE L, - s- L. This point a is a line A on p different from C(p), for C(p) is contained in the ideal plane corresponding to L. Choose a point b on L, such that the corresponding line B is not C(p). Then the plane (A, B) on p is a plane of f and must be disjoint from S (for a $3). Consequently the line ab is disjoint from 3. As this holds for every point b on L such that B # C(p), this implies that all points of 3 would be on the line joining b to the point of L, corresponding to C(p); this is impossible, because S has more than one line! LEMMA 5. Let p, S, 3 be as above. Let c be the point of Lp corresponding to C(p). If L, - 3 contains a line L which is not ideal, then L, - 3 is the union of L and c. Proof It is clear that the point c is not in 3. Suppose there exists a point a # c in L, - 3 - L and choose b E L. If the corresponding lines A and B through p are in a plane of r, i.e. if they are not in a same ideal plane through p, then the plane (A, B) is disjoint from S (for a$ 3). As this holds for every point b on L such that CE ab, all points of 3 must be on the line joining a to the point c, a contradiction.
Lemmas 2,4 and 5 imply the next proposition. PROPOSITION 7. Let S be an ideal plane of L. Then the only possible structures for S are the following: (i) S is an affine plane, obtained from the projective plane L,, by deleting a projective line. Furthermore, the ideal lines of S are the elements of a direction of parallel lines in S.
(ii) S is an affine plane minus one point, obtained from the projective plane L, b-y deleting a projective line and a point not on it. Furthermore, the ideal lines of S are the lines through the deleted point. COROLLARY 4. Any three non-collinear points of L generate a subspace of L which is a plane of I or an ideal plane of L.
138
CHRISTIANE
LEFEVRE-PERCSY
ProoJ: This follows from Corollary 1, from Proposition fact that the planes of r are alIme planes.
7, and from the
Note that Proposition 7 implies that all projective planes L,, for p $ S, are isomorphic, since S cannot have structure (i) in some L, and structure (ii) in another L,. Actually, we prove that all ideal planes of L have the same structure. PROPOSITION 8. If L has ati ideal plane of type (i), then all ideal planes of L are of type (i).
ProoJ: Let S be an ideal plane of type (i). This means that, through every point p of L - S, there is exactly one ideal plane disjoint from S. So there is a partition of L into ideal planes and each of them must be of type (i). Now suppose there is an ideal plane S’ of type (ii). Then, necessarily, S’ meets S and S n S’ is an ideal line (see Corollary 2). Since S’ is of type (ii), there is a plane n of r disjoint from S’. This plane rr must intersect S (for S is of type (i)) and Sn rr is a line disjoint from Sn S’. As S is an affme plane, this means that S n ‘/I and S n S’ are parallel lines. But all lines of S that are parallel to the ideal line Sn S’ are ideal lines (see the description of case (i)). And so arises a contradiction: the intersection of S with the plane rc of r must be a line of ZY
We are now able to finish the proof of the theorem. Conclusion. Thanks to Corollary 4, the two above propositions allow us to determine the structure of the linear subspaces of L generated by three non-collinear points: if the ideal planes of L are of tye (i), then all 2-dimensional subspaces of L are affme planes; if the ideal planes of L are of type (ii), then the 2-dimensional subspaces of L are either alline planes or affine planes minus one point. In the last case, complete L by a new point 0, which is decided to be incident with all ideal lines and ideal planes of L; we get a new linear space L whose planes are all affme. Now we can apply Buekenhout’s characterization of afline spaces [l] and claim that either L or i. is an affine space, whenever the lines of L of L have at least four points. Since the residue of a point in L or L is a projective plane, this implies that L or L is a 3-dimensional afline space. Consequently, the geometry r is isomorphic to one of the geometries A, and A, defined at the beginning. To finish the proof, suppose that there is a line of L or L having less than four points. Then Proposition 1 shows that all lines of r have the same cardinality q < 4 and we can refer to the last paragrah of the proof of the finite case [3, “Proof of the theorem”].
INFINITE ( Af
Af * )-OE~METRIES
139
REFERENCES 1. F.
BLJEKENHOUT, Une caracttrisation des espaces affns baste sur la notion de droite. Math. Z. 111 (1969), 367-371. 2. F. BUEKENHOUT, The basic diagram of a geometry, in “Geometries and Groups” (M. Aigner and D. Jungnickel, Eds.), pp. l-29, Springer Verlag, Berlin, 1981. 3. C. LEFEVRE-PERCSY AND L. VAN NYPELSEER. Finite rank 3 geometries with afline planes and dual afline point residues. Discrete Math., to appear.