*(*de Gruyter 2003

## Characterization of Grassmannians by one class of singular subspaces

E. E. Shult

(Communicated by W. M. Kantor)

Abstract.Spaces related to Grassmann spaces are characterized in terms of the relation of any point to members of some class of singular subspaces of a parapolar space. It is not assumed that all singular subspaces have ﬁnite projective rank—only that at least one subspace in the speciﬁed class does so.

1 Introduction

The classical point-line geometries take as their points the coset space of a parabolic subgroup of a group of Lie type. In the case of Lie groups, these are most of the ruled manifolds that have concerned analysts, topologists and physicists for over a century.

In general, however, they are deﬁned relative to arbitrary ﬁelds and division rings where they are studied from the points of view of algebraic geometry and incidence geometry.

The point of view of incidence geometry begins with a geometryðP;LÞof points and lines, and seeks to characterize a classical parabolic coset-space, by relatively simple axioms on these points and lines. The famous Veblen-Young theorem [18]

characterizing all projective spaces of rank at least three is an example of this sort;

so also is the characterization of all nondegenerate polar spaces of rank at least three (combined results of [17, 4, 12, 11]).

These two geometries make their appearance as proper convex subspaces of the remaining Lie-incidence geometries in the roles of singular subspaces and symplecta, respectively, and this fact has motivated the deﬁnition of parapolar spaceas well as the advancement of these spaces as a natural stage on which to characterize most of the remaining coset geometries for groups of spherical Lie type.

Of course the parapolar concept arose ﬁrst with the early papers of Cooperstein [9, 10] and evolved to more adaptable forms in the work of F. Buekenhout [3, 2] and unpublished notes of A. Cohen. The ﬁrst major characterization theorems depended on two break-through papers of A. Cohen [5, 6] and appeared in [8]. This theorem characterized at least one coset geometry for each group of exceptional Lie type but

did so on the basis of a very restricted relation between a symplecton and any exterior point.

The present paper is part of a program to follow the other alternative: to charac-
terize parapolar spaces by the relation of a point to a maximal singular subspace
belonging to a limited class of such spaces. The basic hypothesis is that the set of
points x^{?}VM collinear with a point x exterior to a maximal singular space M is
either empty, a single point, or has projective rankdd2. So far, the Grassmann and
half-spin geometries have been so characterized [15, 16], but the former character-
ization requires the hypothesis asMranges overallmaximal singular subspaces and
when all of these haveﬁnitesingular rank. Thus, as things stand, the characterization
of Grassmannians in [15] rests upon an unreasonably strong hypothesis not at all in
line with the second result [16] on half-spin geometries. The purpose of this paper is to
achieve a singular characterization of Grassmannians that can be a companion to the
half-spin characterization of [16]—that is, it must have these features: (1) the hypoth-
esis is only on a classMof maximal singular subspaces rich enough to cover every line,
and (2) it is not assumed that all singular subspaces possess ﬁnite projective rank.

Why study singular characterizations? Singular subspaces can be recognized in the point-residuals of any gamma space, while symplecta might not. In forthcom- ing work with S. Onefrei, the existing singular characterizations are used to obtain singular characterizations of several exceptional coset geometries—for exampleE6;4, and E7;7, where the points are the cosets of the maximal parabolic groups cor- responding to the node at the end of the shortest and middle-length arms of the respective Dynkin diagrams.

2 The basic axioms and results 2.1 The hypotheses.We assume:

(D) 1. G¼ ðP;LÞis a parapolar space.

2. There exists a classMof maximal singular subspaces ofGsuch that (a) Every line LALlies in a member ofM.

(b) Given a non-incident pairðp;MÞAPM,the set p^{?}VM is empty,or is
a line.

We now describe the properties of a point-residual of a point-line geometryGsat- isfying the hypothesis (D) above.

(E) 1. Gis a strong parapolar space.

2. Gcontains a classMof maximal singular subspaces with these properties:

(a) Every member ofMis a projective space (not necessarily of ﬁnite rank).

(b) Every point lies in a member ofM.

(c) IfMAM, andxis a point not in M, thenxis collinear with a unique point pMðxÞ. (ThuspM :PM!M is a well deﬁned mapping—the projection onto M.)

3. Any symplecton is classical.

Remark. Conclusion 3 comes from the fact that theGdiscussed above is the point- residual of a strong parapolar space of symplectic rank at least three. In particicular the point-diameter is only two.

Theorem 1. SupposeG¼ ðP;LÞis a point-line geometry satisfying the hypothesis(E) above.Then one of the following holds:

1. Gis a generalized quadrangle.

2. Gis the product geometry AB of two maximal singular subspaces A and B.

3. Some symplecton is not a grid,and the members ofMpartition the points.Any line which is not contained in a element of M is a maximal singular subspace.No line intersects all members ofM.

Theorem 2.Assume Gsatisﬁes hypothesis(D)where some maximal singular subspace ofMhas ﬁnite projective rank.Suppose further that some line lies on at least two mem- bers ofM.Then one of the following holds:

1. Gis a non-degenerate polar space of rank three.

2. G is the Grassmannian of all d-subspaces of a projective space PðVÞ, possibly of inﬁnite rank,and d is a ﬁnite integer greater than1.

3. All maximal singular subspaces of G have ﬁnite rank and G is isomorphic to the factor geometry A2n1;nðDÞ=hsi, where s is a polarity of Witt index at most n5.

Remark.If one wishes to omit the assumption that at least one line lies in at least two members ofM, the conclusion of Theorem 2 must be altered to include a third pos- siblity that is described in Theorem 22 in the last section of this paper. The above Theorem 2 is thus an immediate corollary of Theorem 22. No example of this third alternative to Theorem 22 is known to the author.

3 Review of basic concepts

3.1 The basic glossary.A point-line geometry is simply a pair of setsðP;LÞwith a symmetric incidence relation such that every member of Lis incident with at least two members ofP. The elements ofParepointswhile the elements ofLare called lines. The set of points incident with a given lineLALis called thepoint-shadow of L. A line incident with at least three points is said to bethick.

A subset S ofP is called a subspace if and only if the point-shadow of any line intersects it at zero, exactly one, or all of its points. The intersection of an arbitrary collection of subspaces is a subspace (which might be empty). The intersection of all subspaces ofGcontaining a subsetXofPis calledthe subspace generated by Xand is denotedhXi

G.

Two points arecollinearif and only if there exists a line incident with both of them.

Thepoint-collinearity graphofG¼ ðP;LÞis the graphDðGÞ(or justDifGis under- stood) whose vertex set is Pand whose edges are pairs of distinct points which are collinear. The geometryG is said to beconnected if and only if the graphDis con- nected.

A subspaceSis said to be asingular if any two of its points are collinear. If p is
a point, the symbol p^{?} denotes the set of all points which are either collinear with,
or equal to p (this notation goes back to D. Higman and is standard). A point-line
geometryG¼ ðP;LÞis called agamma spaceif and only if p^{?}is always a subspace,
for each point p. This is equivalent to saying that for every point p and lineLnot
incident with p, ifpis collinear with at least two distinct points of the point-shadow
ofL, then pis collinear with all points of that point-shadow.

If distinct lines always possess distinct point shadows, one may simply identify a line L with its own point-shadow. This certainly occurs whenðP;LÞ is apartial linear space, that is point-line geometry in which two distinct points are incident with at most one line. A partial linear space that is a singular space is called alinear space. (It is a standard convention to denote the unique line on two distinct collinear pointsxandyof a partial linear space by the symbolxy.) A linear spaceðP;LÞwith all lines thick is called a projective plane if and only if any two distinct lines in it intersect at a point. A linear space ðP;LÞwith all lines thick is called a projective space if and only if any two intersecting lines generate a projective plane. In this case, as is well known, the poset of all subspaces is the poset of ﬂats of a matroid, and so the cardinality of any two minimal generating sets is the same—a number which when diminished by one, is called the projective dimensionorrankof the projective space.

We remark that a gamma space, all of whose singular subspaces are linear spaces, must be a partial linear space.

A subspaceSofG¼ ðP;LÞis said to beconvexif and only if, for any two of its points—sayx and y—the intermediate vertices of any path of minimal length con- nectingxto yin the point-collinearity graphDare also points ofS.

The intersection of any collection of convex subspaces is also a convex subspace.

The intersection of all convex subspaces ofG¼ ðP;LÞcontaining a subsetXofP, is a convex subspace called theconvex closure(inGÞof X, and is denotedhhXii

G. 3.2 Product geometries. Now suppose Gi¼ ðPi;LiÞ is a point-line geometry for i¼1;2. We wish to describe the product geometry G1G2. Its set of points is the Cartesian productP1P2. There are two sorts of lines. Avertical lineis one whose point shadow has the formx1L2:¼ fðx1;y2Þ jy2AL2gwhere L2AL2. Similarly, a horizontal line is one whose point shadow has the form L1y2:¼ fðx1;y2Þ j x1AL1g where L1 is a line of L1. Then the product geometry, G1G2 becomes a point-line geometry ðP1P2;LVULHÞwhere LV and theLH are respectively all the vertical and horizontal lines. It is usually not necessary to mention these sets explicitly since they are completely determined by the quartetteðP1;L1;P2;L2Þ.

Suppose now,S_{1}andS_{2}are subspaces of a point-line geometryG¼ ðP;LÞ. If we
writeG¼S_{1}S_{2} we intend to assert that Gis isomorphic to the product geometry
S_{1}S_{2} where the symbolS_{i}is interpreted to be the point-line geometryðSi;LðSiÞÞ

where LðSiÞ denotes those lines of L whose point-shadow lies entirely inside the subspaceSi,i¼1;2.

3.3 Point residuals. Suppose G¼ ðP;LÞ is a gamma space all of whose singular subspaces are projective spaces. Then, of course,Gis a partial linear space. Given a point p, letLp andPpbe the collections of all lines and all projective planes, respec- tively, which are incident with point p. We say thatLALpis incident withpAPpif and only if LJp. With respect to this incidence relation, the geometry ResðpÞ ¼ ðLp;PpÞis a also a gamma space of ‘‘points’’ and ‘‘lines’’ whose singular subspaces are projective spaces. The geometry ResðpÞis called theresidue at point p, or more generally, apoint-residue.

3.4 Symplecta. Anon-degenerate polar spaceis a point-line geometry ðP;LÞ with these properties: (1) all lines are thick, (2) no point is collinear with all remaining points, and (3) given a point p and a lineL not incident with p, p is collinear with exactly one or collinear with all of the points of the point-shadow of L. Obviously property (3) makes polar spaces a species of gamma space. If it never happens that some point p is collinear with more than one point of a line not incident with it, then the polar space is called anon-degenerate polar space of rank2 or ageneralized quadrangle with thick lines. For the rest of this paper, we use the term polar space to mean a non-degenerate polar space, and the term generalized quadrangle(or just

‘‘quadrangle’’ if the context is clear) to mean generalized quadrangle with thick lines.

A generalized quadrangle in which each point is on just two lines is called a grid.

Such a quadrangle is a productLN of two lines.

It is well known that if a polar space is not a generalized quadrangle, then it is a partial linear gamma space whose singular subspaces are all projective spaces.

If one of these maximal singular subspaces has ﬁnite projective rankdd2, then all of its maximal singular subspaces possess the same ﬁnite projective rank d, and we say thatthe polar space has rank dþ1. Natural case-divisions make it convenient to distinguish those rankd polar spaces in which any second-maximal singular sub- space lies in just two maximal singular subspaces. We call these oriﬂamme polar spaces.

If there exists a maximal singular subspace of inﬁnite projective rank, then all maximal singular subspaces have inﬁnite projective rank, but these ranks need not be the same cardinality. In this case we simply say thatthe polar space has inﬁnite rank, and no particular inﬁnite cardinal is speciﬁed as the rank.

A convex subspaceSofGsuch thatS, together with the lines contained in it forms a polar space, is called a symplecton. For example, if G¼AB, where A and B are singular subspaces ofG, then the convex closurehhx;yiiof two pointsxand y which are not collinear is a convex subspace that is a grid, and hence is a symplecton.

A fundamental fact that we shall invoke many times is this:

If S is a symplecton of the gamma spaceG¼ ðP;LÞ,and x is a point not in S,then
x^{?}VS is a singular subspace:

3.5 Parapolar spaces. In this paper we adopt a deﬁnition of ‘‘parapolar space’’

equivalent to that introduced in A. Cohen’s survey article in the Handbook for Incidence Geometry [7]. The reader should be warned that this deﬁnition is more general than that given in the literature preceeding theHandbook.1

Aparapolar spaceis a connected point-line geometry with these properties:

(1) Iffx;ygis a non-collinear pair of distinct points, then either
(a) x^{?}Vy^{?}¼q,

(b) jx^{?}Vy^{?}j ¼1, or

(c) hhx;yiiis a symplecton.

(In this case the pairðx;yÞis called apolar pair.) (2) Every line lies in at least one symplecton.

It easily follows that a parapolar space is a partial linear gamma space with every 4-circuit in a unique symplecton—which together with (2) is the deﬁnition of [7, page 688].

A parapolar spaceGis said to havesymplectic rank(at least)kif and only if every
symplecton ofGhas rank (at least)kas a polar space.2Some perfectly natural par-
apolar spaces (for example the Lie incidence geometries of typeC_{n;2},nd5) possess
symplecta of two di¤erent polar ranks.

A standard result is the following:

Lemma 3.SupposeGis a parapolar space of symplectic rank at least three.Then the following holds.

1. Any singular subspace generated by a point and a line lies in a symplecton.

2. All singular subspaces ofGare projective spaces.

Proof. SupposeA¼hp;Li_{G}is a singular subspace of Ggenerated by a point p and
a line L. We must show thatAlies in a symplecton. By property (2) of a parapolar
space, their exists a symplectonScontainingL. IfScontains p we are done, so we
may assume p is not in S and so p^{?}VS is a singular subspace ofS. On the other
hand, sinceSis a polar space of rank at least three,L^{?}VSis not a singular subspace.

Thus there is a point xAL^{?}VSp^{?}VS. Then fp;xg is a pair of non-collinear
points with a lineLinp^{?}Vx^{?}. ThusR:¼hhp;xiiis a symplecton containing pand
Land so containsA. So part 1 is proved.

Part 2 follows from the fact that all singular subspaces of a polar space are pro- jective spaces—in particular the subspaceAof the previous paragraph is a projective plane. From the general choice of pandLit follows that all singular subspaces are projective spaces.

1The shift in deﬁnitions can be justiﬁed on utilitarian grounds. Under the old deﬁnition, there was no real name for the geometry of the point-residual of a parapolar space with sym- plecta of rank three. This was a disadvantage since (like the current paper) most of the argu- ments take place at this level.

2In [7] the symplectic rank is called thepolar rank, which to many students is the rank of a polar space, not a parapolar space.

The conclusions of Lemma 3 fail dramatically when there are symplecta of rank two—that is convex subspaces which are generalized quadrangles. For example, ifA andBare two arbitrary linear spaces, each containing at least a line, then the product geometryABis a parapolar space of symplectic rank two with singular subspaces isomorphic toAand toB.

A parapolar space is called a strong parapolar space if the alternative 1(c) of the deﬁnition of parapolar space never occurs—that is,everypair of points at distance 2 in the collinearity graph is covered by a symplecton.

A particular example of a strong parapolar space of rank exactly three is the object
being characterized by the Theorem 2 at the beginning of this paper. Let Vbe any
right vector space over a division ring D, and let d be a positive integer properly
bounded by the dimension ofV, if the latter is ﬁnite. LetV_{d} be the full collection of
all d-dimensional subspaces of V. LetVd1;dþ1 be the full collection of pairs ðA;BÞ
whereAandBare subspaces of dimensionsd1 anddþ1 respectively andAJB.

We say that ad-spaceCisincidentwith such a pairðA;BÞif and only ifAJCJB.

Then the point-line geometry G¼ ðVd;Vd1;dþ1Þ subject to the described incidence is called the Grassmannian of d-subspaces of V and is denoted AdðVÞ, or, if V has ﬁnite dimensionnþ1, it is denotedAn;d orAn;dðDÞif the division ringDneeds to be emphasized. When d ¼1 it is a projective space. If 1<d <dimV1, the Grass- mannian is a strong parapolar space of symplectic rank three.

3.6 Two cited results.The following Lemma is due to A. Cohen. Although it might be described as a technical result, it is a very important one for virtually every purely local characterization of the classical Lie incidence geometries ultimately depends on this lemma.

The polar spaces of rank three were completely classiﬁed by J. Tits in [17]. Any generalized quadrangle that is isomorphic to a point residue in such a polar space is said to be aclassical quadrangle.

Lemma 4. Suppose G¼ ðP;LÞis a generalized quadrangle whose point-set is parti- tioned by a subcollectionSof the line set(such a collection of linesSis usually called aline spread)with these properties:

1. Given any two distinct lines L and N of the spread,the subspacehL;Ni_{G} that they
generate is a grid, with all members GðL;NÞof the parallel class of the grid con-
taining L and N belonging to the line spreadS.

2. Let Gbe the collection of all subsets of Sof the form GðL;NÞfor distinct lines L and N belonging toS.ThenðS;GÞis a projective plane.

ThenGis not a classical generalized quadrangle.

This Lemma is proved in Cohen’s fundamental paper [6].

We also require the following characterization of the Grassmannian ofd-subspaces of a vector space.

Theorem 5.(Shult [14, Theorem 6.1, pp 173–4], Bichara and Tallini [1])SupposeGis
a strong parapolar space of symplectic rank three.Suppose the full collectionM^{}of all
maximal singular subspaces of G is partitioned into two subcollections, M^{}¼PþS
with these properties:

1. Any subspace inPintersects any subspace ofSin a line or the empty set.

2. Every line lies in exactly one member ofPand exactly one member ofS.

3. Some singular subspace inPcontains a ﬁnite unreﬁnable chain of subspaces.

ThenGis the Grassmannian A_{d}ðVÞof d-subspaces of a vector space V over some divi-
sion ring,where d is ﬁnite,but the dimension of V need not be.

Proof.This is a corollary of Theorem 6.1 of [14, page 173]. We verify the hypotheses of that theorem. Condition (T1) follows from the fact that G is a gamma space.

Conditions (T2)(i)–(ii) are simply hypotheses 1. and 2. given above. The intersection
property (T3) is an easy consequence of the parapolar hypothesis and the fact that
symplecta with hypothesis 2 are of typeD_{3}and any two maximal singular subspaces
of a given class of such a symplecton always meet at a point. Finally, condition (T4)
restates condition 3 above. The conclusion now follows from the cited Theorem 6.1.3
Corollary 6. Suppose G is a strong parapolar space each of whose point-residuals
ResGðpÞis a product geometry ApBp where Ap and Bp are projective spaces of pos-
itive rank,at least one of which has ﬁnite rank.Then one of the following holds:

1. GFA_{d}ðVÞfor some vector space V and integer d>2.

2. Gis the quotient geometry A_{2n1;}_{n}=hsiwheresis induced by a polarity ofPðVÞ ¼
PG2n1;1ðDÞof Witt index at most n5.

Remark. This would really be a Corollary of the beautiful theorem of A. Cohen [6]

were it not for the small detail that the latter requiresGto have ﬁnite singular rank—

that is,allsingular subspaces are projective spaces of ﬁnite rank (see [7, 6.3, p. 718]).

We avoid this by using the preceding theorem where ﬁniteness is required of only some maximal singular subspaces. The proof uses Cohen’s two-fold covering con- struction [6].

Proof. Here, every point-residue ResGðpÞ is a product geometryApBp, where Ap

andBpare two distinct maximal singular subspaces ofGwhich intersect at a line on p. In this case there are two classes of maximal singular subspacesAp andBp which contain p. Members of the same class pairwise intersect at exactlyfpg, while mem- bers of di¤erent classes intersect at some line on p.

Thus each line LALlies in exactly two maximal singular subspacesA_{L} and B_{L}
and at least one of these two is a member ofM, and so has ﬁnite projective rank.

Following the construction in Cohen [6], we now form a new geometryGG^ ¼ ðPP;^ LLÞ,^
3Note that the parameternappearing inA_{n;d}in the conclusion of Theorem 6.1 is any car-
dinal number not exceeded byd; there is no assumption there that it is ﬁnite.

where PP^is the collection of all pairs ðp;XpÞwhere Xp is either Ap orBp—that is, one of the two classes of maximal singular subspaces on p. For each lineLofLlet AL andBL be the two maximal singular subspaces containing L. LetLL^ be the set of all pairsðL;XLÞwhereXLAfAL;BLg. The pairðp;XpÞis said to be incident with ðL;XLÞ if and only if (1) p is incident with L in G and (2) XLAXp. There is an obvious geometry morphism f : ^GG!Gwhich takes pairðp;XpÞto pointpand takes pairðL;XLÞto lineL. The point-mappings and line-mappings are both onto and all ﬁbers have cardinality two.

Now each line LL^ ¼ ðL;ALÞofGG^ is the intersection of two maximal singular sub-
spaces: (1) The ﬁrst is the set M_{1}ðLLÞ^ :¼ fðx;XxÞ jxAA_{L};A_{L}AXxg. (2) The second
is M_{2}ðLLÞ^ :¼ fðx;XxÞ jxAB_{L};B_{L}BXxg. The intersection of these is LL^ and the two
spaces comprise all maximal singular spaces of GG^ containingLL. Let^ Mi:¼ fMiðLLÞ j^
L^

LALLg^ fori¼1;2. As is evident from their deﬁnition, the two classes contain no space in common. In fact the two collections M1 andM2 obey the hypotheses onS andPof Theorem 5, the intersection property resulting from the fact that the sym- plecta of Glift toGG^ to make any connected component of the latter a strong para- polar space. (These details are in Cohen’s fundamental paper [6].) It now follows from Theorem 5 that any connected component of GG^ is isomorphic to the Grass- mannianAdðVÞofd-spaces of a vector spaceV.

If the geometryGG^ is not connected, then each ﬁber of a point or line contains one object of each component geometry, and so the restriction of the morphism f : ^GG!G to one of the two connected components produces an isomorphismGFAdðVÞ. On the other hand, if Gis connected then all maximal singular subspaces have the same projective dimension, say n1, and f : ^GG¼AnðVÞ !G is a two-fold covering deﬁned by a deck-transformation sof degree 2 exchanging the two classesM1 and M2. Thussis induced by a polarity. The condition that shave Witt index at most n5 results from the fact that the imageGis a parapolar space.

4 Immediate consequences of the Hypothesis (D)

Lemma 7.The following statements hold:

1. If S is a symplecton and MAMmeets S in at least a line,then SVM is a maximal singular subspace of S and is a plane.

2. Gis a strong parapolar space of symplectic rank exactly three.

Proof. Before proving the rest, let us ﬁrst see why G is a strong parapolar space.

Supposexandyare a pair of non-collinear points collinear with a common pointv.

By hypothesis 2(a) there is a maximal singular subspace MAM containing the line
xv. Now y^{?}VM contains the pointvand so by hypothesis 2(b) y^{?}VM contains a
lineA. But asMis singular,x^{?}Vy^{?}contains the lineA, and soðx;yÞis a polar pair.

Thus all distance two point-pairs ofPare polar pairs, whenceGis a strong parapolar space.

SupposeS is a symplecton ofG,Lis a line in S, andMis a member ofM con-

tainingL(such anMexists by hypothesis). ThenSVMis a singular subspace of the
polar space S and so, for a point sASL^{?},s^{?}VðSVMÞ ¼H is a hyperplane of
SVMnot containingL. SinceLis a line andHis a hyperplane of the same singular
space,HVLis non-empty.

Now as M is a singular subspace, s^{?}VMSJL^{?}Vs^{?}JS (the second con-
tainment follows from the convexity ofS). This asserts that a set disjoint from Sis
contained inS, and so the former is empty. Thus

s^{?}VM¼s^{?}VðSVMÞ ¼H:

Since this is a non-empty set, hypothesis 2(b) forces the left-most term to be exactly a line. ThusSVM is a plane.

IfSVMwere not a maximal singular subspace ofSthen we could ﬁnd a point in
ðSVMÞ^{?}VSMand that would also contradict part 2 (b) of the Hypothesis. Thus
we see thatMVSis a plane as well as a maximal singular subspace ofS, so all parts
of the Lemma have been proved.

Lemma 8.Suppose P is a plane and M is an element ofMwhich meets P in exactly a
point p.Suppose P^{?}VM is not a line.Then there is a plane Q of M on p and a bijection

l:ðLðPÞÞ_{p}! ðLðQÞÞ_{p};

from the lines of P on point p to the lines of Q on point p taking each line L to the line
L^{?}VM.

Proof. We know how l is deﬁned. SupposeP^{?}VM is not a line. Then of course it
is just the pointp. Thus for two distinct linesLandNbelonging to the planePand
meeting at point p, L^{?}VM:¼L^{0} and N^{?}VM:¼N^{0} are distinct. Moreover, the
intersection of the perps of L^{0} and Nshare the line L and so their convex closure
S:¼hhN;L^{0}iiis a symplecton which contains planePand by Lemma 7 meetsMat
a planeQcontaining lineL^{0}. SincePandQare two planes of a non-degenerate rank
three polar spaceS, we see that the lines ofPonp are mapped bijectively onto the
lines ofQon pointp.

5 The Hypothesis (E)

Remark.This section, as well as Sections 6, 7, 8, 9, and 10 which follow, all assume hypothesis (E).

Lemma 9.The following are easy consequences of Hypothesis(E):

1. Every symplecton meets a member ofMat the empty set,or at a line.

2. If L is any line disjoint from MAM, then the mapping L!M induced by the projection mapping,has as its image,either a single point,or a line of M.

3. Every symplecton is a generalized quadrangle,consequentlyGis a strong parapolar space of symplectic rank exactly two.

Proof.Of course, this is just a re-hash of Lemma 7, localized at a point, but lets prove
it from (E) alone. Suppose the symplecton Sintersected the subspaceMofMnon-
trivially. IfSVM is not a maximal singular subspace ofS, then there exists a point
pAðSVMÞ^{?}S, at which point assumption E2(b) forcesSVMto contain exactly
one point, say m. Choose yASm^{?}. Again by part E2(b), y is collinear with a
unique point my of M. Then myAm^{?}Vy^{?}JS. But that contradicts SVM being
a point.

So we may assume that SVM is a maximal singular subspace ofS. Then choos-
ingxASM, we see that E2(b) implies x^{?}VðSVMÞis a point. But the latter is a
hyperplane ofSVM. Thus the maximal singular subspace SVM ofSis a line, and
soSis a generalized quadrangle.

Now this argument works for any symplecton Ssince any of its points lies in an element ofM. Parts 1 and 3 of the Lemma have been proved.

SupposeLis a line disjoint from a member ofMand let f :L!M be the map- ping which maps each point ofLto the unique point ofMwith which it is collinear.

Choose xALand let Rbe the unique symplecton containing fðxÞandL. Then by
Part 1,RmeetsMat a lineL^{0}oppositeLinR. ThenL^{0}¼ fðLÞ, so fðLÞis a line as
required.

6 Fibred symplecta

We assume hypotheses (E). We suppose MAMand thatSis a symplecton sharing
no point with M. Then the restriction of the projection mapping onto M produces
a mapping f :S!M which takes a line of Seither to a line of M, or to a point
ofM. Let fðSÞbe the collection of all image points—that is, fðSÞ ¼ fmAMjm^{?}V
S0 qg.

Lemma 10.Suppose mA fðSÞ.Then its ﬁbre f^{1}ðmÞis either a single point or is a line
of S.

Proof. By convexity ofS, the ﬁbre f^{1}ðmÞis a singular subspace of S. SinceSis a
generalized quadrangle, this ﬁbre is either a point or is a line.

Lemma 11. Let F be the set of points in M whose ﬁbres are lines of S.Then F is a
subspace of M.Moreover,if E is a line contained in F,then f^{1}ðEÞis a subquadrangle
of S which is a grid.Then ﬁbers of the points of E form one of the two line-spreads of
this grid.

Proof.Supposea^{0}andb^{0}are two distinct points ofF. LetAandBbe their respective
ﬁbres. ThenAandBare opposite lines ofS. Then for each pointxofA, there is a
unique line T_{x}onxwhich intersectsBat the unique pointbðxÞofx^{?}VB. Then this
transverse line maps by f to the unique line E¼a^{0}b^{0} of M. Thus, for each point
mAL^{0}, the ﬁbre DðmÞ:¼ f^{1}ðmÞmust intersect each transverse line T_{x}. This gives
L^{0}JF.

On the other hand ifzADðmÞ, then eitherDðmÞ ¼AandzAT_{z}, orDðmÞis a line
oppositeA, so, iffxg ¼z^{?}VA, thenx^{?}VDðmÞcontains bothzand the unique point
of TxVDðmÞwhile being a singleton set. This forces zATx. Thus the ﬁbre f^{1}ðmÞ
cannot contain a point not in6

xAATx.

Thus we see that the ﬁbres of points of L^{0} form a line-spread of G:¼6

xAATx

while the systemfTxjxAAgof transverse lines forAandBform another such sys- tem. Since the lines of each system are pairwise opposite, no further collinearities exist among the points. It follows thatGis a subspace ofSwhich is a grid with these two systems of line-spreads.

6.1 A particular situation. We suppose f is not injective. Then there are dis-
tinct points a and b of S collinear with a common point fðaÞ ¼ fðbÞ ¼a^{0}. Since
f^{1}ða^{0}Þ:¼a^{0?}VS is a clique with at least two points, and since lines are maximal
singular subspaces ofS, we see that the ﬁbre f^{1}ða^{0}Þis a lineA. Then any other line
Lon pointais mapped bijectively to a line fðLÞ:¼L^{0}ofM. Next supposeBis a line
meetingLat a pointbdistinct fromA, chosen so that its image fðBÞ:¼B^{0} is also a
line of M. Then B^{0} meetsL^{0} at the pointb^{0}¼ fðbÞ. ThenhL^{0};B^{0}i

M is a projective planep.

NowAandBare opposite lines of the symplectonS. Thus each pointai of lineA
is collinear with a unique point bi of B, and we denote the full collection of lines
fTi:¼aibigbyTand call them thetransversal lines for A and B. It is clear then that
these transversal lines are mapped onto the full pencil of lines ofpon pointa^{0}, so in
factpJfðSÞ. This situation is illustrated in Figure 1.

Let R:¼6ff^{1}ðpÞ jpApg. ClearlyRis a subspace ofSsincepis a subspace of
M. Select any point u on line L so thatu is distinct from both a and b. Let T_{i} be
a transversal from AtoBwhich is distinct from (and hence opposite to) L. Thenu
is collinear with a unique point v ofT_{i} andvis not in B. It follows that line uv is
opposite lineB.

Now fðuÞand fðvÞare points ofp ðB^{0}Ufa^{0}gÞ, on di¤erent members of the line
pencil on a^{0}. Then f maps the line uv of Sto the line fðuÞfðvÞ of p. Since p is a

Figure 1. A special situation.

projective plane, fðuÞfðvÞmeets lineB^{0}¼ fðBÞat a point fðbiÞfor someb_{i}AB. This
means uv contains a pointwwith fðwÞ ¼ fðbiÞ. Sinceuv andB are opposite lines,
w0biand so are distinct points belonging to the ﬁbreC:¼ f^{1}ðfðbiÞÞ. By Lemma
11 above,G¼ f^{1}ða^{0}fðbiÞÞ ¼hA;Ci_{S} is a grid.

NowRcontainshA;Bi_{S}and so properly contains the gridG. Then any point ofG
lies on a line ofRwhich is not inG, and these lines map onto lines ofp. Let us choose
distinct points x and y on a line T transversal to two of the line-ﬁbres of G. Let
Lx and Ly be lines of Rwhich lie on x andy, respectively, but are not lines of G.

LxVLy ¼q since any point of their intersection would lie outside T while being
collinear with distinct pointsxandyofT. Then fðxÞ0fðyÞand fðLxÞand fðLyÞ
are lines which intersectL^{0}¼a^{0}fðbiÞat distinct points fðxÞand fðyÞand so intersect
each other at a point pApL^{0}. But that means f^{1}ðpÞcontains a point ofL_{x}and a
point ofL_{y} and since these are disjoint lines we see that pAFVp.

It now follows from Lemmas 10 and 11 that Ris a generalized quadrangle con- taining a line spreadF(the ﬁbres of f :R!p), any two of which generate a grid, and, letting G be the collection of grids formed in this way, the incidence system ðF;GÞis isomorphic toðPp;LpÞ, the points and lines of the projective planep.

By E(4)Sis a classical quadrangle. Now Lemma 4 of Arjeh Cohen [6] shows that this is impossible.

We have proved the following:

Theorem 12.Suppose S is a symplecton disjoint from a subspace M ofM.Then either 1. the projection into M induces a projective embedding f :S!M,or

2. S is a grid and the projection on M induces a mapping f :S!M onto a line L of M.The ﬁbres of the points of L form one of the line-spreads of the grid S.

Corollary 13.Suppose S is a symplecton which is not a grid.Supposepis a plane which
meets S at a line.Then for any point xApS,x lies in a unique member M_{x}ofMand
the planepitself lies in M_{x}.

Proof. Suppose pVS is a line L and choose x in pL. Suppose x belongs to a
singular subspace M^{0} of M which does not contain p. Then pVM^{0}¼ fxg. Also,
convexity of S forces M^{0}VS¼q. But now projection on M^{0} forces a mapping
f :S!M^{0} which possesses a non-trivial ﬁbre. By the Theorem 12, Sis a grid, an
absurdity. So no suchM^{0}exists.

But by hypothesis xlies in some memberMx ofM, and so we see that we must havepJMxfor any such singular space. This fact forces the uniqueness ofMx.

A line is called anM-lineif it is contained in one of the singular subspaces ofM.

By axiom E2(a), theM-lines cover all the points.

Corollary 14.Suppose S is a symplecton which is not a grid.Then every line of S is either anM-line or is already a maximal singular subspace ofG.

Proof.This is immediate from Corollary 13.

7 Unﬁbered symplecta

7.1 *M-projections which embedd a symplecton.*LetSbe a symplecton. We letN^{2}ðSÞ
denote the set of points xAPS for which x^{?}VS is a line. Similarly we write
N^{1}ðSÞfor the set of pointsxAPS for whichx^{?}VS is a single point, and ﬁnally
writeN^{0}ðSÞfor the set of pointsxAPSfor whichx^{?}VS¼q. Then we have the
following partition of points:

P¼SþN^{2}ðSÞ þN^{1}ðSÞ þN^{0}ðSÞ:

The conclusion of Corollary 13 motivates another deﬁnition. The set U of all points pAPwhich lie in a unique member ofMwill be calledthe uniqueness set.

Theorem 15. Suppose MAM and S is a symplecton which shares no point with M.

Suppose further that the projection onto M induces an embedding f :S!M as in the
ﬁrst case of Theorem12.Choose any pairðx;yÞof non-collinear points of S,let M_{y}be
a member Mcontaining y,let Ly be the line MyVS and let t be the unique point of
x^{?}VLy.Let R be the unique symplecton containing t and fðxÞ.Then R is not a grid.

The following uniqueness results hold:

1. The subspace My is the unique element ofMcontaining the point y.

2. Since y is arbitrarily chosen in S, we see that every point of S lies in a unique member ofM.

3. Every point of M lies in a unique member ofM.

4. Every line of S is either anM-line or is itelf a maximal singular space.

Proof.Letx, y,M_{y},L_{y} andtbe as chosen. Then fðxÞis the unique point ofMcol-
linear withx. We haveMVM_{y}¼qsince otherwise fðLyÞis a single point, against
f being an embedding. Then fðxÞis collinear to a unique vertexzinM_{y}L_{y}. Now
ðx;fðxÞ;z;t;xÞis a 4-circuit lying in a symplectonR. NowRmeetsSat linextandR
meetsMyat linetz. Also, sinceSVM is non-empty, it too is a line—in fact it is the
line fðxÞfðtÞ. The conﬁguration is illustrated in Figure 2.

Now we see that the point fðxÞsits on three distinct lines ofR: fðxÞx, fðxÞz, and fðxÞfðtÞ. ThusRcannot be a grid.

Now we note that yAMyRandMymeetsRat a line. Thus by Corollary 13 y lies in a unique member ofM.

Similarly every point of M ðMVRÞ lies in a unique member of M. But R is
(now, uniquely) determined by the choice ofðx;yÞ. But if we chooseuASVy^{?}t^{?},
then replacement of the pair ðx;yÞ by ðu;tÞ in the construction produces a new
symplecton R^{0} which is not a grid, and which meets Mat the line fðuÞfðyÞ. Note
that lineuyis opposite linextinSand so fðtÞfðxÞand fðuÞfðyÞare disjoint lines in
Msince f :S!Mis an embedding. But as before all points of line fðzÞfðtÞlie in a
unique member ofM, so all points ofMhave this uniqueness property.

There is more. Letx, y,u,tbe as in the previous paragraph. NowM_{y}VR¼tzand
M_{y}VR^{0}¼yz^{0}wherez^{0}is the unique point ofM_{y} collinear with fðuÞ. These lines are

disjoint since fðuÞfðyÞand fðtÞfðxÞare disjoint lines ofM(as observed in the pre-
vious paragraph) and the perpendicular relation produces an isomorphismM_{y}!M.

But since the points outside either of these lines are uniqueness points, we have
M_{y}JU.

Finally, supposepis a plane not in an element ofMmeetingSat a lineN. ThenN
is not anM-line. Choose distinct pointsxandtof lineNand letM_{t}be a memberM
on pointt. ThenM_{t}VSis a lineL_{t}ont. ClearlyL_{t}0N. Now chooseyinL_{t}distinct
fromt. Then yis not collinear with x. If we rename things, writing M_{y} forM_{t} and
Ly forLt we have exactly the construction of the symplectonRat the beginning of
the Theorem. NowRis not a grid, but sits on the lineN¼xtof the planep. This is
contrary to Corollary 5a. Thus no suchpexists. It follows that all lines ofSwhich are
notM-lines are already maximal singular subspaces. The proof is complete.

Corollary 16. Suppose the symplecton S is not a grid and MVS¼q, for some MAM.Then every point ofPlies in a unique member ofM.

Proof. Let U be the collection of uniqueness points—those points which lie in a unique member ofM. Our objective is to prove thatU¼P.

By Theorem 12, since S is not a grid, the mapping f :S!M induced by the
projection into M is an embedding. By Theorem 15 part 2, SJU. By Corollary
14, since Sis not a grid, every point ofN^{2}ðUÞis a uniqueness point. Now suppose
yAN^{1}ðSÞUN^{0}ðSÞ. Then by our hypothesis, ylies in a singular subspace M_{y}AM.

ThenMyVM¼q. Then, asSis not a grid, one obtains an embedding fy:S!My. Upon replacing MbyMy in Theorem 15 part 3, we see thatMyJU and in partic- ular yAU.

The discoveries of the previous paragraph can be summarized by asserting
P¼SþN^{2}ðSÞ þN^{1}ðSÞ þN^{0}ðSÞJU;

which we were to prove.

*R*

*S*
*L*_{y}

*x*

*f*(x)

*M*
*z*
*M*_{y}

*t*
*y*

Figure 2. The conﬁguration of Theorem 15.

8 The case that some member ofM is a line

Theorem 17.If a line L is a member ofM,then one of the following holds:

1. Gis itself a generalized quadrangle.

2. G is a product geometry LP, where P is a maximal singular subspace. Every symplecton ofGis a grid.

Proof. SupposeLis a line inM. SinceGis a strong parapolar space of rank two,L lies in a symplecton S. IfS¼P, the ﬁrst conclusion holds and we are done. So we assumeS0P.

Choose yAPS. Then y^{?}VLis a single point, say p. We claim that inS, there
is only one further line on pbesidesLand that line is y^{?}VS. If not there would be
a line N ofS on p not in y^{?}. We could then form the symplecton R:¼hhy;Nii
and choose a pointzARp^{?}. Thenzis collinear with a pointqofLdistinct from p.

Then qJp^{?}Vz^{?}JR, and so lies in RVL¼RVSVL¼NVL¼ fpg, an absur-
dity.

Thus, for each yAp^{?}Swe have y^{?}VS¼N, for any lineNinSwhich contains
p and is distinct from L. Since y^{?}VS is a clique, the line Nis unique. Thus Sis a
grid. But then we have p^{?}SJN^{?}, which must be a singular subspace (otherwise
Gwould not have symplectic rank two). Thus, for any pof the lineL, we have

p^{?}¼LUAp; and LVAp¼ fpg;

where Ap¼N_{p}^{?} andNp is the unique line of the gridSon psuch thatNp is distinct
from the lineL. Clearly eachA_{p} is a singular subspace which is not a line. Moreover
we have a partition into maximal singular subspaces:

P¼*]*

fApjpALg: ð1Þ

Now choose any pointxinP. IfxALset pðxÞ ¼xand note thatLis the unique
line on x not in ApðxÞ. If xAPL, let pðxÞ be the unique point of x^{?}VL. Then
xAApðxÞ. We claim that there is a unique lineLx on xwhich is not in ApðxÞ. First
there is at least one such line, since the symplectonRx:¼hhxpðxÞ;Liicontains one.

On the other hand, ifL^{0}were such a line, then the symplectonhhxpðxÞ;L^{0}iiwould
intersectLnon-trivially and hence would containL(Lemma 9, part 1), forcing it to
coincide with theRx. ButRxsatisﬁes the hypothesis that we had forSabove, and so
Rxis a grid. That means there is only one lineLxonxwhich does not lie in the sin-
gular spaceA_{pðxÞ}, and that line is oppositeL.

Thus all lines ofGwhich are not in one of the singular subspacesA_{p}form a spread
of lines transverse to the componentsA_{p} of the partition in equation (1). Every sym-
plecton on such a transverse line is a grid with its intersections with the A_{p} form-
ing a spread. It follows that for any two distinct points x and y of the line L, the
point-bijection A_{x}!A_{y} induced by the system of transverse lines takes lines ofA_{x}

to linesA_{y}. This is the last step needed to conclude thatG is the product geometry
LAp.

9 Symplecta disjoint from no member ofM

Theorem 18.Suppose S is a symplecton that is not disjoint from any singular subspace ofM.Then either(a)Gis a generalized quadrangle, (b)S is a grid,or(c)all points are uniqueness points.

Proof.We suppose thatSis a symplecton disjoint from no member ofM. By way of
contradiction we assume thatS is not a grid and that M_{1} andM_{2} are two distinct
members ofMwhich meet at a point pin the symplectonS.

If either M_{i} were a line, we could apply Theorem 17 to conclude that Gis a gen-
eralized quadrangle, or thatG¼LPfor some singular subspaceP. But in the lat-
ter caseSwould be a grid, contrary to assumption. SoGis a generalized quadrangle,
which is one of our conclusions. Thus we may assume that neither M1 norM2 are
lines.

Fori¼1;2, letLi:¼SVMi and supposeNiis a line ofMion pdistinct from the
lineLi(this is possible sinceMiis not a line). LetR:¼hhN1;N2ii, the unique sym-
plecton on theNi. Choose a pointzARp^{?}and letMzbe a member ofMon point
z. Now by our hypothesis,Mzcannot be disjoint fromS, and soMzVSis anM-line
AzofSwhich is not no point p. Then the unique pointazof p^{?}VAzis collinear with
both pandzand so belongs toR—that isRVS¼paz. Alsoazis on neither lineLi

since the unique point mi ofz^{?}VMi lies inNi fpg, i¼1;2. Thus the three lines
za_{z},zm_{1}andzm_{2}on pointzare all distinct and belong toR. It follows thatRis not a
grid.

Now choose any pointwASp^{?}and letM_{w}be a member ofMcontaining point
w. IfM_{w}VR¼q, then Corollary 16 (applied with the non-gridRand singular space
M_{y} replacingS andMrespectively) would not allow M_{1} and M_{2} to intersect at p.

Thus we must conclude thatM_{w}VRis a lineBofRnot containingp. Then the point
tofp^{?}VBlies inRand also lies in p^{?}Vw^{?}JS, and so is a point ofRVS¼pazas
well as a point of theM-lineLw¼MwVS.

But in this conﬁguration we can replaceRby a new symplectonR^{0}:¼hhN1;N_{2}^{0}ii
whereN_{2}^{0}is another line ofM2on pdistinct from bothN2andL2. ThenR^{0}VR¼N1.
But just as we argued forRabove,R^{0}is not a grid, and meets Sat a line. We note
that the two lines RVS and R^{0}VS are distinct since RVR^{0}¼N1. But in the last
line of the previous paragraph we saw that theM-lineLw¼MwVSintersectedRVS
at a point (we called t). So similarly, Lw intersects line R^{0}VS at a point sdistinct
fromt. The gamma space property ofSthen forcesLwJp^{?}, which is absurd since
wAL_{w}p^{?}.

Thus, if Sis not a grid, we have shown that all points ofSare uniqueness points.

Now consider any point yofPS. Then by assumption, y^{?}VSis a lineL_{y}. But if
M_{y} is any element ofM on y, then by assumption,M_{y}VSis a line, and so must be
L_{y}. Thus any element ofMon ycontains the planehy;L_{y}i, and this forces it to be
M_{y}. Thus yis also a uniqueness point.

10 When all symplecta are grids In this section we add to (E) the extra hypothesis:

(G) Every symplecton ofGis a grid.

Theorem 19.If hypothesis(G)is assumed,thenGFMA where MAMand A is a singular subspace.

Proof.By way of contradiction suppose the conclusion is false. Then by Theorem 11, we may assume that no member ofMis a line.

Choose any singular subspaceMofM. Suppose, for some point pinM,L1andL2

are two distinct lines on pwhich are not inM. IfL1is not inL^{?}_{2}, then the symplecton
R on L1 andL2 must meetMat a third line (see Lemma 7), and so is not a grid,
against (G). Thus, always we must have

Step 1: For any point p in a singular subspace M ofM,there exists another maximal singular subspace AðpÞsuch that

p^{?}¼MUAðpÞ where MVAðpÞ ¼ fpg:

Suppose, for the moment that pandMare ﬁxed as in Step 1. Choose any point y
not in p^{?}. Then by hypothesis (E), yis collinear with a unique point y1ofM fpg.

Then there is a symplecton Gwhich is a grid on the intersecting lines yy1 and y1p.

Then plies on a unique lineLof this grid distinct from the line py1. Then by Step 1,
Lis a line of the singular spaceAðpÞ. Moreover yis collinear with a unique point y_{2}
of lineL, since all of this occurs within the gridG.

Now supposeywere collinear with another point y_{2}^{0} ofAðpÞ. Then y_{2}^{0} would lie in
y^{?}Vp^{?}JG, and so would lie inGVAðpÞ ¼L. But that forces y_{2}^{0} ¼y_{2}. Thus y_{2} is
the unique point ofAðpÞwhich is collinear with such a pointy.

Now ylies in some member, sayM_{y}, ofM, and so applying Step 1 withM_{y}replac-
ingM and y replacing p, we see that y^{?}¼MyUA, the union of two maximal sin-
gular subspaces which intersect at point y. Since yy1andyy2are distinct lines of grid
Gon y, thenyy1lies in one of the maximal singular subspaces (MyorA) and yy2lies
in the other.

We have established

Step 2: Let p and M be as in Step1.If yAPp^{?},then y is collinear with a unique
point y1of M and with a unique point y2of AðpÞ.Then y^{?}is the union of two
maximal singular spaces;one is Aðy1Þand the other contains yy2.Thus every
line on point y is either in y^{?}_{1} or is in y^{?}_{2}.

We next show

Step 3: Let p,M, y, y_{1}and y_{2}be as in(Step2).If a point y^{0}is collinear with both y_{1}
and y_{2},it is either y or p.

Clearly, any such y^{0}lies in the gridGwhich contains them, and the result follows.

Now we can complete the proof of the Theorem. First we can uniquely assign
coordinates from MAðpÞ to each point at follows. If y is not in p^{?}, we assign
coordinatesðy1;y2Þtoy. Ifmis a point ofM, we assign coordinatesðm;pÞtom. Ifa
is a point ofAðpÞ, we assign the coordinatesðp;aÞtoa. Note that phas coordinates
ðp;pÞand by (Step 3) each point ofP receives a unique coordinate by this device.

Conversely if ðm;aÞ is arbitrarily chosen with neither coordinate equal to p, then there is a unique gridGðm;aÞonfm;agcontaining a unique point yAGðm;aÞ fpg collinear with both m and a. Thus introducing coordinates produces a complete bijectionP!MAðpÞ.

Now let us consider the collection of coordinates of the points on an arbitrary
line ofL. IfLis inMor inAðpÞ, then one of the coordinates is constantly p while
the other coordinates range through a line of M orAðpÞ, respectively. Similarly, if
L contains a point y¼ ðy_{1};y_{2}Þ not in p^{?}, then by Step 2, this line is either in y^{?}_{1}
or in y^{?}_{2} but not both. In the former case the left coordinates of the points of L
are constantly y_{1} while the right-hand coordinates range over the line py_{2} at which
the grid G^{0}¼hhy1;y2iiintersects AðpÞ. In the latter case the right coordinates of
all points of L are constantly y2 while the right coordinates range over the line

py1.

Thus all lines have the formmLwheremis a point ofMandLis a line ofAðpÞ, or else have the formNawhereNis a line ofMandais a point ofAðpÞ.

The geometry on MAðpÞwith this collection of lines is precisely the product geometry,MAðpÞ, and the desired isomorphism follows.

This contradicts the assumption the theorem was false, completing the proof.

11 The proof of Theorem 1

We assume Hypothesis (E). We are to prove the following conclusion:

(C) One of the following holds:

1. Gis a generalized quadrangle.

2. G¼AB, the product of two maximal singular subspaces.

3. G properly contains a symplecton which is not a grid. The members ofM
partition the points of G. Moreover, if M is the full collection of all maxi-
mal singular subspaces A of G with the property that jp^{?}VAj ¼1 for all
pAPA, then in this case, every line that is not an M-line, is itself a
maximal singular subspace ofG. Finally, no line that is not anM-line inter-
sects every member ofM.

As before, letUbe the set of points ofGwhich lie in a unique member ofM.

Step 1: IfP0U,then either(i)Gis a generalized quadrangle,or(ii)every symplecton ofGis a grid.

Proof.SupposeP0U. Suppose by way of contradiction thatGis not a generalized quadrangle and that there exists a symplectonSwhich is not a grid.

Now by Theorem 18, ifSis disjoint from no member ofM, then either (1)Gis a generalized quadrangle, (2)Sis a grid or (3)P¼U. But any of the conclusions (1), (2) or (3) goes against the suppositions of the previous paragraph.

Thus, we must assume that there is a singular subspace MAM such that SVM¼q. Then, sinceS is not a grid, restriction of the projection mapping into M produces an embedding f :S!M. In that case the hypotheses of Corollary 16 are in place, forcing us to conclude that P¼U, contrary to the hypothesis of the theorem.

This contradiction completes the proof of this Step 1.

Step 2: If P0U,then either(i) Gis a generalized quadrangle or (ii)Gis a product geometry AB of two singular subspaces A and B.

Proof. By Step 1, either (i) holds or all symplecta are grids. But by Theorem 19, the latter case forces conclusion (ii) above.

Now we can complete the proof of the theorem. Suppose, that the ﬁrst two con- clusions of (C) fail—that is,Gis not a generalized quadrangle nor is it a product of two maximal singular subspaces. Then, from contrapositive of the statement of Step 2,P¼U.

Let us assume now thatMis the full collection of all maximal singular subspaces
Awith the property that every point outsideAis collinear with exactly one point of
A. SinceGis not a generalized quadrangle, nor a product of two maximal singular
subspaces, the elements ofMpartition the points. IfLis not anM-line, thenLmust
be disjoint from some member ofM, otherwiseLAM, by maximality ofM, and so
G is a product of two singular spaces, contrary to assumption. Choose MAM so
that LVM¼q. Then the set of points of M which are collinear with a point of
L themselves form a line L^{0} of M. Choose point y in L, and let M_{y} be the (now
unique) member of Mcontaining y. SinceLis not anM-line,M_{y}VL¼ fyg. Since
M_{y}VM¼q, there is a bijection f :M_{y}!Mtaking each point ofM_{y}to the unique
point of M collinear with it. This mapping f is an isomorphism of linear spaces.

Note that fðyÞAL^{0}. Thus there is a line Nin My such that fðNÞ ¼L^{0}. Now let
R¼hhy;L^{0}ii, the symplecton on yandL^{0}. Now each point of NUL fygis col-
linear with yand a point ofL^{0} ffðyÞg, forcingNULJR. Thus ylies on at least
three lines ofR, namelyL,N, and yfðyÞ, and soRis not a grid.

If Lwere properly contained in a singular subspaceB, thenBBMasLis not an M-line. Then, choosing xABL, andMxAM containingx we see that the con- vexity ofRforcesRVMx¼q. But the mappingg:R!Mxinduced by projection onMxis not an embedding sincegmaps the lineLtox. By Theorem 12Rmust be a grid. This contradicts the conclusion of the previous paragraph.

Thus L cannot properly lie in another singular subspace—that is, it is itself a maximal singular subspace.

11.1 Local recognition of the three conclusions of Theorem 1.Our purpose here is to identify one of the three alternatives of conclusion (C) above, by a property of any one of its points. The reason for doing this will emerge in the next section.

Theorem 20.SupposeG¼ ðP;LÞsatisﬁes hypothesis(E)so that one of the three con- clusions listed in(C)above holds.

1. Suppose Gcontains a point x such that every maximal singular subspace on x is a line.ThenGis a generalized quadrangle.

2. Suppose Gcontains a point x that lies in exactly two maximal singular subspaces.

ThenGis a product geometry.

3. SupposeGcontains a point x which lies in at least three maximal singular subspaces, only one of which is a member ofM and it is not a line.ThenGcontains a proper symplecton which is not a grid,and the elements ofMpartition the points ofG.

Proof. Part 1. Suppose every line onxis a maximal singular subspace. Then in par- ticular, an elementMxAMcontainingxmust be a line. By Theorem 17Gis either a generalized quadrangle or a product geometryLAwhereLis anM-line andAis a maximal singular subspace. But in the latter case,xlies in a subspaceAxisomorphic toA, and by hypothesisAxis a line. ThusAis a line, andG¼LAis a grid. Thus in either case,Gis a generalized quadrangle.

Part 2. Now supposexis a point on exactly two maximal singular subspaces. Now
one of these is a maximal singular subspaceM_{x}belonging to the special classM. The
other, we shall callA_{x}, which may or may not belong toM. Thusx^{?}¼M_{x}UA_{x}.

Now let ybe any point ofGnot inx^{?}. We claim that yis collinear with a unique
point y_{2} of A_{x}. By hypothesis (E), y is collinear with a point y_{1} of M_{x}, and
S:¼hhx;yii is a symplecton containing x. Then there is a line N of S which
contains x but does not lie in y^{?}_{1}. Then Nmust lie inAx and y is collinear with a
unique point y2 ofN. Suppose y_{2}^{0} were any point of y^{?}VAx. Now y_{2}^{0?}contains the
2-coclique fx;yg, and so belongs toS, and in factSVAx¼N. Thus y_{2}^{0} ¼y2. Thus
Axhas the property that every point ofPAxis collinear with exactly one point of
Ax. By maximality ofMwe see thatAxAM. Thusxlies in two members ofM, and
so the third conclusion of (C) is impossible. If the ﬁrst conclusion held, we would
haveS¼Pis a generalized quandrangle with a pointxon exactly two lines. In that
caseGis a grid and so is the product of two lines. Thus the second conclusion of (C)
must hold in any case.

Part 3. Here,xlies in at least three maximal singular subspaces, one of which is not a line. It follows that G cannot be a product geometry nor can it be a generalized quadrangle. Thus the third conclusion of (C) must hold.

12 Proof of Theorem 2

12.1 Uniformity of point residuals.

Lemma 21. We assume that G is a gamma space with all singular subspaces projec-
tive spaces.Assume that for every point p,the point-residual Res_{G}ðpÞ:¼ ðLp;P_{p}Þis a
strong parapolar space of symplectic rank exactly 2satisfying exactly one of the three
conclusions of conclusion(C)—the choice depending on the particular point p.

Let us deﬁne three sets of points:

1. X_{1}: the set of all points p which lie in some line L with the property that every
maximal singular subspace containing L is a plane.

2. X2:the set of all points p which lie in some line L which lies in exactly two maximal singular subspaces.

3. X2:the set of points p which lie on some line L such that (a) L lies in at least three maximal singular subspaces, (b) L lies in a unique member MLofM,and

(c) ML has projective rank at least three.

Then each set X_{i} is a union of connected components of the point-collinearity graph
D¼ ðP;@Þ.

Proof.Suppose pAX_{1}andqis another point collinear withp. plies on a lineLwith
every maximal singular subspace containing La plane. Now Res_{G}ðpÞsatisﬁes con-
clusion (C) with a ‘‘point’’Lhaving each maximal singular subspace containing it a

‘‘line’’. By Theorem 20, Res_{G}ðpÞis a generalized quadrangle all of whose ‘‘lines’’ are
maximal singular subspaces. Thus the line pqhas all the maximal singular subspaces
containing it planes. It follows thatqAX1.

Suppose pAX2 andqis another point collinear with p. plies on a lineLlying on exactly two maximal singular subspaces. Since ResGðpÞsatisﬁes conclusion (C) with a

‘‘point’’Lin just two maximal singular spaces, by Theorem 20, ResGðpÞis a product geometry, all of whose ‘‘points’’ lie in just two maximal singular subspaces. In par- ticular, the line pqis in exactly two maximal singular subspaces, and soqAX2.

Suppose pAX_{3} andqis another point collinear with p. plies on a lineLlying in
a unique singular subspace M_{L} of M, with all other singular subspaces on L, of
which there are at least two, being planes. By Theorem 20, every line on phas this
property—in particular line pqdoes. ThusqAX_{3}.

Thus for all i, ifpAX_{i}andq is a point collinear with p, thenqAX_{i}. The proof is
complete.

Theorem 22.SupposeGis a connected parapolar space satisfying condition(D).Then one of the following three conclusions holds:

1. Gis a non-degenerate polar space of rank exactly three.

2. G is either (1) the Grassmannian AdðVÞ of d-subspaces of a (possibly inﬁnite- dimensional) vector space V, where d is a positive integer or (2) is the quotient geometry A2n1;nðDÞ=hsi of the Grassmannian of n spaces of a 2n-dimensional vector space by a polaritysof Witt index at most n5.

3. In the geometry G,every line L lies in a unique member M_{L}ofMwhich is of ﬁnite
projective dimension d>2and in at least two other maximal singular subspaces—

all of these maximal singular subspaces being planes.Moreover,each point lies in a proper rank three symplecton which is not oriﬂamme. For each planep that is not contained in a member ofMand for each point p inp,there exists a member ofM meetingpexactly at point p.

Proof.SupposeGis a connected parapolar space satisfying condition (D). Then every point residual ResGðpÞ satisﬁes condition (E), and so, by Theorem 1, such a point- residual must satisfy one of the three conclusions of (C). This meansP¼X1þX2þ X3, where the Xi are deﬁned as in Lemma 21. By assumption, G has a connected point-collinearity graph, so we have three cases emanating fromP¼Xi,i¼1;2;3.

Case 1.P¼X1. Here, every point-residue ResGðpÞis a generalized quadrangle. It follows from the main result of [13] thatGis a polar space of rank three.

Case 2.P¼X2. Here, every point-residue ResGðpÞis a product geometryApBp, whereApandBpare two distinct maximal singular subspaces ofGwhich intersect at a line on p. Conclusion 2 now follows from Corollary 6.

Case 3.P¼X_{3}. Every point-residue Res_{G}ðpÞ has its ‘‘points’’Lp partitioned by
the elements of M on p. It also contains a proper symplecton which is not a grid,
each of its ‘‘lines’’ (elements of P_{p}) is either contained in an element of M, or is
a maximal singular subspace. Now suppose a plane p on p were contained in no
member ofM. Thenpcorresponds to a ‘‘line’’ of Res_{G}ðpÞ. If this ‘‘line’’ met every
member ofMpand a ‘‘point’’, then it could be adjoined to the collectionMpwithout
changing the hypothesis (E), and Theorem 17 would then force ResGðpÞ to be a
product geometry of the formLMpwhereLis a ‘‘line’’ andMpis a singular space.

But in that case, every symplecton of ResGðpÞwould be a grid contrary to the fact that it contains a proper symplecton which is not a grid. Thus there must be a mem- berMofMpwhich intersects the planepexactly at point p.

Now Theorem 2 is an immediate corollary of the above Theorem 22.

Acknowledgement. The author is grateful to Prof. Anna Kasikova for a number of insightful discussions concerning this problem.

References

[1] A. Bichara, G. Tallini, On a characterization of Grassmann space representing the h- dimensional subspaces in a projective space. In:Combinatorics ’81(Rome, 1981), volume 18 ofAnn. Discrete Math., 113–131, North-Holland 1983. MR 84h:51013 Zbl 0506.51019 [2] F. Buekenhout, An approach to building geometries based on points, lines and convexity.

European J. Combin.3(1982), 103–118. MR 83j:51011 Zbl 0504.51006

[3] F. Buekenhout, Cooperstein’s theory.Simon Stevin57(1983), 125–140. MR 85b:51002 Zbl 0548.51004

[4] F. Buekenhout, E. Shult, On the foundations of polar geometry.Geom. Dedicata3(1974), 155–170. MR 50 #3091 Zbl 0286.50004

[5] A. M. Cohen, An axiom system for metasymplectic spaces.Geom. Dedicata12 (1982), 417–433. MR 84c:51011 Zbl 0511.51009

[6] A. M. Cohen, On a theorem of Cooperstein.European J. Combin.4(1983), 107–126.

MR 84j:51011 Zbl 0524.51001

[7] A. M. Cohen, Point-line spaces related to buildings. In:Handbook of incidence geometry, 647–737, North-Holland 1995. MR 96k:51009 Zbl 0829.51004

[8] A. M. Cohen, B. N. Cooperstein, A characterization of some geometries of Lie type.

Geom. Dedicata15(1983), 73–105. MR 85c:51010 Zbl 0541.51010