(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the early papers of Cooperstein [9, 10] and evolved to more adaptable forms in the work of F. Buekenhout [3, 2] and unpublished notes of A. Cohen.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,S1andS2are subspaces of a point-line geometryG¼ ðP;LÞ. If we writeG¼S1S2 we intend to assert that Gis isomorphic to the product geometry S1S2 where the symbolSiis 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 typeCn;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;LiGis 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. LetVd 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;NiG 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 collectionMof all maximal singular subspaces of G is partitioned into two subcollections, M¼PþS 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 AdðVÞof 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 typeD3and 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. GFAdðVÞfor some vector space V and integer d>2.
2. Gis the quotient geometry A2n1;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 subspacesAL and BL 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 inAn;din 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 M1ðLLÞ^ :¼ fðx;XxÞ jxAAL;ALAXxg. (2) The second is M2ðLLÞ^ :¼ fðx;XxÞ jxABL;BLBXxg. 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:ðLðPÞÞp! ðLðQÞÞp;
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:¼L0 and N?VM:¼N0 are distinct. Moreover, the intersection of the perps of L0 and Nshare the line L and so their convex closure S:¼hhN;L0iiis a symplecton which contains planePand by Lemma 7 meetsMat a planeQcontaining lineL0. 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 lineL0oppositeLinR. ThenL0¼ 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 f1ðmÞis either a single point or is a line of S.
Proof. By convexity ofS, the fibre f1ðmÞis 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 f1ðEÞis 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.Supposea0andb0are two distinct points ofF. LetAandBbe their respective fibres. ThenAandBare opposite lines ofS. Then for each pointxofA, there is a unique line Txonxwhich intersectsBat the unique pointbðxÞofx?VB. Then this transverse line maps by f to the unique line E¼a0b0 of M. Thus, for each point mAL0, the fibre DðmÞ:¼ f1ðmÞmust intersect each transverse line Tx. This gives L0JF.
On the other hand ifzADðmÞ, then eitherDðmÞ ¼AandzATz, 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 f1ðmÞ cannot contain a point not in6
xAATx.
Thus we see that the fibres of points of L0 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 ¼a0. Since f1ða0Þ:¼a0?VS is a clique with at least two points, and since lines are maximal singular subspaces ofS, we see that the fibre f1ða0Þis a lineA. Then any other line Lon pointais mapped bijectively to a line fðLÞ:¼L0ofM. Next supposeBis a line meetingLat a pointbdistinct fromA, chosen so that its image fðBÞ:¼B0 is also a line of M. Then B0 meetsL0 at the pointb0¼ fðbÞ. ThenhL0;B0i
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 pointa0, so in factpJfðSÞ. This situation is illustrated in Figure 1.
Let R:¼6ff1ðpÞ 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 Ti be a transversal from AtoBwhich is distinct from (and hence opposite to) L. Thenu is collinear with a unique point v ofTi andvis not in B. It follows that line uv is opposite lineB.
Now fðuÞand fðvÞare points ofp ðB0Ufa0gÞ, on di¤erent members of the line pencil on a0. 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 lineB0¼ fðBÞat a point fðbiÞfor somebiAB. This means uv contains a pointwwith fðwÞ ¼ fðbiÞ. Sinceuv andB are opposite lines, w0biand so are distinct points belonging to the fibreC:¼ f1ðfðbiÞÞ. By Lemma 11 above,G¼ f1ða0fðbiÞÞ ¼hA;CiS is a grid.
NowRcontainshA;BiSand 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 intersectL0¼a0fðbiÞat distinct points fðxÞand fðyÞand so intersect each other at a point pApL0. But that means f1ðpÞcontains a point ofLxand a point ofLy 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 MxofMand the planepitself lies in Mx.
Proof. Suppose pVS is a line L and choose x in pL. Suppose x belongs to a singular subspace M0 of M which does not contain p. Then pVM0¼ fxg. Also, convexity of S forces M0VS¼q. But now projection on M0 forces a mapping f :S!M0 which possesses a non-trivial fibre. By the Theorem 12, Sis a grid, an absurdity. So no suchM0exists.
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 letN2ðSÞ denote the set of points xAPS for which x?VS is a line. Similarly we write N1ðSÞfor the set of pointsxAPS for whichx?VS is a single point, and finally writeN0ðSÞfor the set of pointsxAPSfor whichx?VS¼q. Then we have the following partition of points:
P¼SþN2ðSÞ þN1ðSÞ þN0ðSÞ:
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 Mybe 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,My,Ly andtbe as chosen. Then fðxÞis the unique point ofMcol- linear withx. We haveMVMy¼qsince otherwise fðLyÞis a single point, against f being an embedding. Then fðxÞis collinear to a unique vertexzinMyLy. 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 R0 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. NowMyVR¼tzand MyVR0¼yz0wherez0is the unique point ofMy 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 isomorphismMy!M.
But since the points outside either of these lines are uniqueness points, we have MyJU.
Finally, supposepis a plane not in an element ofMmeetingSat a lineN. ThenN is not anM-line. Choose distinct pointsxandtof lineNand letMtbe a memberM on pointt. ThenMtVSis a lineLtont. ClearlyLt0N. Now chooseyinLtdistinct fromt. Then yis not collinear with x. If we rename things, writing My forMt 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 ofN2ðUÞis a uniqueness point. Now suppose yAN1ðSÞUN0ðSÞ. Then by our hypothesis, ylies in a singular subspace MyAM.
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 P¼SþN2ðSÞ þN1ðSÞ þN0ðSÞJU;
which we were to prove.
R
S Ly
x
f(x)
M z My
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 Ap¼Np? andNp is the unique line of the gridSon psuch thatNp is distinct from the lineL. Clearly eachAp 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, ifL0were such a line, then the symplectonhhxpðxÞ;L0iiwould 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 spaceApðxÞ, and that line is oppositeL.
Thus all lines ofGwhich are not in one of the singular subspacesApform a spread of lines transverse to the componentsAp of the partition in equation (1). Every sym- plecton on such a transverse line is a grid with its intersections with the Ap form- ing a spread. It follows that for any two distinct points x and y of the line L, the point-bijection Ax!Ay induced by the system of transverse lines takes lines ofAx
to linesAy. 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 M1 andM2 are two distinct members ofMwhich meet at a point pin the symplectonS.
If either Mi 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 zaz,zm1andzm2on pointzare all distinct and belong toR. It follows thatRis not a grid.
Now choose any pointwASp?and letMwbe a member ofMcontaining point w. IfMwVR¼q, then Corollary 16 (applied with the non-gridRand singular space My replacingS andMrespectively) would not allow M1 and M2 to intersect at p.
Thus we must conclude thatMwVRis 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 symplectonR0:¼hhN1;N20ii whereN20is another line ofM2on pdistinct from bothN2andL2. ThenR0VR¼N1. But just as we argued forRabove,R0is not a grid, and meets Sat a line. We note that the two lines RVS and R0VS are distinct since RVR0¼N1. 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 R0VS at a point sdistinct fromt. The gamma space property ofSthen forcesLwJp?, which is absurd since wALwp?.
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 lineLy. But if My is any element ofM on y, then by assumption,MyVSis a line, and so must be Ly. Thus any element ofMon ycontains the planehy;Lyi, and this forces it to be My. 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 y2 of lineL, since all of this occurs within the gridG.
Now supposeywere collinear with another point y20 ofAðpÞ. Then y20 would lie in y?Vp?JG, and so would lie inGVAðpÞ ¼L. But that forces y20 ¼y2. Thus y2 is the unique point ofAðpÞwhich is collinear with such a pointy.
Now ylies in some member, sayMy, ofM, and so applying Step 1 withMyreplac- 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, y1and y2be as in(Step2).If a point y0is collinear with both y1 and y2,it is either y or p.
Clearly, any such y0lies 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 ðy1;y2Þ 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 y1 while the right-hand coordinates range over the line py2 at which the grid G0¼hhy1;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 L0 of M. Choose point y in L, and let My be the (now unique) member of Mcontaining y. SinceLis not anM-line,MyVL¼ fyg. Since MyVM¼q, there is a bijection f :My!Mtaking each point ofMyto the unique point of M collinear with it. This mapping f is an isomorphism of linear spaces.
Note that fðyÞAL0. Thus there is a line Nin My such that fðNÞ ¼L0. Now let R¼hhy;L0ii, the symplecton on yandL0. Now each point of NUL fygis col- linear with yand a point ofL0 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 subspaceMxbelonging to the special classM. The other, we shall callAx, which may or may not belong toM. Thusx?¼MxUAx.
Now let ybe any point ofGnot inx?. We claim that yis collinear with a unique point y2 of Ax. By hypothesis (E), y is collinear with a point y1 of Mx, 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 y20 were any point of y?VAx. Now y20?contains the 2-coclique fx;yg, and so belongs toS, and in factSVAx¼N. Thus y20 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 ResGðpÞ:¼ ðLp;PpÞis 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. X1: 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 Xi is a union of connected components of the point-collinearity graph D¼ ðP;@Þ.
Proof.Suppose pAX1andqis another point collinear withp. plies on a lineLwith every maximal singular subspace containing La plane. Now ResGðpÞsatisfies con- clusion (C) with a ‘‘point’’Lhaving each maximal singular subspace containing it a
‘‘line’’. By Theorem 20, ResGðpÞis 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 pAX3 andqis another point collinear with p. plies on a lineLlying in a unique singular subspace ML 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. ThusqAX3.
Thus for all i, ifpAXiandq is a point collinear with p, thenqAXi. 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 MLofMwhich 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.P¼X3. Every point-residue ResGðpÞ 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 Pp) 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 ResGðpÞ. 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.
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