(de Gruyter 2002
Hyperplanes of dual polar spaces of rank 3 with no subquadrangular quad
Harm Pralle*
(Communicated by A. Pasini)
Abstract.LetDbe a thick dual polar space of rank 3, and letHbe a hyperplane ofD. Calling the elements ofDpoints,linesandquads, we call a quadaNH singularifHVa¼P?Vafor some pointP,subquadrangularifHVais a subquadrangle, andovoidalifHVais an ovoid. A pointPAHof a quadais said to bedeep with respect toaifP?VaHH, and it is calleddeep ifP?HH.
We investigate hyperplanesH ofD such that no quad is subquadrangular. We generalize a result of Shult proving that, if all quads are singular, then the polar spaceP¼Dis an or- thogonal polar spaceQ6ðKÞfor some (not necessarily finite) fieldK, and the hyperplaneHis a split Cayley hexagonHðKÞ.
If both singular and ovoidal quads exist, then one of the following holds:
1. H¼6PAOP?whereOis an ovoid of a quado.
2. There exists one deep point PAH such that all quads containingP are singular and the remaining quads are ovoidal.
3. The setPof deep points with respect to the singular quads is a locally singular hyperplane of a dual polar spaceD0.D0is the dual of an orthogonal polar spaceP0GQ6ðKÞfor some fieldKandP0is a subspace of the dualPofDwhere the lines ofP0are lines ofP. The set Ptogether with the lines ofDcontained inHform a split Cayley hexagonHðKÞ. The hy- perplaneHcontains all points ofDon lines ofHðKÞ.
1 Introduction and results
A subspace of a linear space is a subset of the point set of the linear space that con- tains all points of a lineliflcontains at least two points of the subspace. Ageometric hyperplaneor shortly, ahyperplane, of a point-line geometry is a proper subspaceH such that each line meetsH.
LetPbe a thick polar space of rank 3 (not necessarily finite) and letDbe its dual.
So,Dhas the diagram
* Supported by a scholarship of theGraduiertenfo¨rderung des Landes Hessen.
0 1 2
points lines quads
and as indicated, we call the elements of D points,linesandquads. Note that finite thick polar spaces are classical by Tits [12]. Considering a classical thick polar space, the residue ResPðPÞinPof a pointPAPis a classical generalized quadrangle. The residue ResDðaÞin D of a quada of Dis isomorphic to the generalized quadrangle dual of ResPðPÞ.
IfH is a hyperplane ofD(the latter being regarded as point-line geometry) anda a quad of D, then eithera is contained inHorHmeets the generalized quadrangle ain a hyperplane. IfaNH, the hyperplaneaVHofais the perp of a point or a full subquadrangle or an ovoid (cf. Payne and Thas [7, 2.3.1]). Accordingly, we call the quadasingular,subquadrangular, orovoidal, respectively. Ifais a singular quad, we call the point PAa with aVH ¼P?Va the deep point with respect to a, where ? denotes collinearity inDand whereP? is the set of points ofDcollinear withP, in- cluding P. Ifa is a quad contained inH andPa point ofa, thenP?VaHH. Ac- cordingly, we also say that every point of a quadaHHis adeep point with respect to a. Furthermore, we call a pointP deepifP?HH. So, if a pointPis deep, it is deep with respect to all quads containingP.
Pasini and Shpectorov [6] call a hyperplaneHof a dual polar spacelocally ovoidal (respectively locally singular and locally subquadrangular), if each quad a not con- tained inH meets the hyperplane in an ovoid (respectively in the perp of a point or in a subquadrangle). A hyperplane is calleduniformif it is locally ovoidal or locally singular, or locally subquadrangular. Their main result is the non-existence of finite locally subquadrangular hyperplanes apart from two small examples, one in Q6ð2Þ and one inH5ð4Þ.
The existence of locally ovoidal hyperplanes is an outstanding problem. Pasini and Shpectorov [6] prove that, given a locally ovoidal hyperplaneHof a dual polar space Dof rank 3, its complementDnH cannot be flag-transitive.
For locally singular hyperplanes, there are two possibilities. The points at non- maximal distance from a given pointPform a locally singular hyperplane called the singular hyperplanewithdeep point P. If all quads are singular and there is no deep point, then the hyperplane is a generalized hexagon by Shult [10]. More precisely, there is a bijection from the set of quads ofDonto the set of deep points of the (sin- gular) quads. In the finite case, Shult [10] shows that the polar space P¼D is an orthogonal spaceQ6ðqÞand that the generalized hexagonHis a split Cayley hexagon HðKÞfor a finite fieldK ¼GFðqÞ(for definition ofHðKÞ, see Section 4.1). By means of a theorem due to Ronan [8], we prove the following Theorem 1 in Section 4.2 without assuming finiteness; it has also been proved by H. Van Maldeghem in 1999 (unpublished).
Theorem 1.Let H be a hyperplane of a dual polar spaceDof rank 3such that every quad is singular and H does not admit any deep point. Then the polar spacePGD
is an orthogonal polar space Q6ðKÞand the hyperplane H is a split Cayley hexagon HðKÞ.
Our aim is to determine the non-uniform hyperplanes Hof Dsuch that no quad is subquadrangular, i.e. quads are either singular or ovoidal or contained inH. We prove the following:
Theorem 2.Let H be a hyperplane of a thick dual polar spaceDof rank3such that no subquadrangular,but both singular and ovoidal quads exist.Then one of the following holds:
1. H ¼6
PAOP?where O is an ovoid of a quado.
2. There exists one deep point PAH such that all quads containing P are singular and the remaining quads are ovoidal.
3. The set P of deep points with respect to the singular quads is a locally singular hyperplane of a dual polar space D0. D0 is the dual of an orthogonal polar space P0GQ6ðKÞfor some field K andP0 is a subspace of the dual Pof D where the lines ofP0are lines ofP.The setPtogether with the lines ofDcontained in H form a split Cayley hexagon HðKÞ.The hyperplane H contains all points ofDon lines of HðKÞ.
Remarks.1. The class of hyperplanes of the form6
PAOP?whereOis an ovoid of a quad has not yet been determined since ovoids of generalized quadrangles are not yet classified.
2. If we do not require the existence of ovoidal quads, Case 3 includes the possi- bilityD0¼D. Then we are back to Theorem 1.
Case 2 of Theorem 2 is non-empty. In Section 2, we describe examples for Cases 1 and 2 of Theorem 2.
No example as in Case 3 is known to the author in whichP0 is not a hyperplane ofP. Thus it is an open problem whether the existence of a hyperplane as in Case 3 forcesP0 to be a hyperplane ofP. This is of great interest because of
Corollary 1.Assume the hypotheses of Theorem2 and suppose that the hyperplane H is as in Case3of Theorem2.If P0is a hyperplane of P,then H contains exactly the points ofDon lines of HðKÞand we havePGQ7ðKÞ.
In particular,ifDis finite and the hyperplane H is as in Case3of Theorem2,thenP is isomorphic toQ7ðqÞ.
Remarks. 1. Clearly, the situation described in Corollary 1 occurs whenever P¼ Q7ðKÞ: just takeP0¼Q6ðKÞHP,HðKÞinsideP0 and defineHas in Corollary 1.
2. In the finite case, ifP¼Q6ðqÞwithqodd, neither ovoidal nor subquadrangular quads exist, whence hyperplanes of its dual are locally singular.
If P¼Q7ðqÞor H6ðq2Þ, then D admits no subquadrangular quad. The example of Case 2 of Theorem 2 exposed in Section 2 is infinite. No finite example is known
to the author. If Case 2 of Theorem 2 could be shown to be impossible in the finite case, then by means of Theorem 2 and the known results on locally uniform hyper- planes, the hyperplanes of the duals ofQ7ðqÞandH6ðq2Þwould be classified.
After a few preliminaries in Section 3, we introduce the split Cayley hexagon in Section 4.1 and prove Theorem 1 in Section 4.2.
The main part of the proof of Theorem 2 characterizes the hyperplaneHwhenH has no deep point, which leads to Case 3 of Theorem 2. One step of this proof is a result on the extension of hyperplanes of dual polar spaces stated below in Theorem 3 and proved in Section 5.1. The proof of Theorem 2 will be finished in Section 5.2.
In the following, P0 is a non-degenerate polar subspace of rank 3 of a non- degenerate polar space P of rank 3, such that the lines of P0 are lines of P. We denote their duals byD0 andD, respectively. Furthermore,s (respectivelys0) is the shadow operator of D (respectively D0), i.e. given a line l of D0, sðlÞ (respectively s0ðlÞ) is the set of points of D(respectivelyD0) onl. IfXis a point subset ofD0, we denote bysðXÞthe set of points ofDthat are on some linemwiths0ðmÞJX. Then the following holds.
Theorem 3.Let H0 be a locally singular hyperplane of the dualD0 of the polar space P0 of rank3.ThenP0 is a hyperplane ofP(and its lines are lines ofP)if and only if sðH0Þis a hyperplane of the dualDofP.In that case and ifsðH0Þhas no deep point, then all quads ofDnot inD0are ovoidal with respect tosðH0Þ.
Acknowledgment. The author wishes to thank the referee for some remarkable im- provements.
2 Examples
The points collinear with those of a hyperplane of a quad. The following proposition gives an easy method to construct hyperplanes of dual polar spaces of finite rank.
Proposition 2.Ifois a quad of a dual polar spaceDof finite rank nd3and Hois a hyperplane ofo,then the set H of points at distance at most n2from some point of Hois a hyperplane ofD.
We omit the straightforward proof. As mentioned in the introduction, the hyper- planeHois one of the three kinds of hyperplanes of generalized quadrangles, namely an ovoid, a subquadrangle or the perp of a point.
We focus again on a rank three dual polar spaceD. Ifa is a quad meetingoin a linel, thenP?VaJaVH for allPAlVHo. So, iflmeetsHoin a single pointX,a is singular with deep pointX. IflHHo, thenais contained inH.
Finally, if a quadamisseso, thenaVH is the image ofHoin the projection ofo ontoa(for definition of the projection, see Section 3). In particular, ifHois an ovoid, thenais ovoidal.
An example of Case 2 of Theorem 2.We state a lemma first.
Lemma 3.Given a dual polar spaceDof rank3and a hyperplane H ofD,suppose that H admits a deep point P.Then the following are equivalent:
(i) H is as in Case2of Theorem2,namely P is the unique deep point of H and no two points of HnP?are collinear;
(ii) Q?VH ¼ fQgfor every point QAH at distance3from P;
(iii) Q?VH ¼ fQgfor some point QAH at distance3from P.
Proof.Clearly, (i) implies (ii) and (ii) implies (iii). We shall prove that (iii) implies (i).
Assume (iii) and suppose that (i) is false, to get a contradiction. So, there exist distinct collinear pointsX;YinHnP?. AsX;Y AH, the linel¼XY is contained inH. L et P0be the point oflat minimal distance fromP. IfdðP;P0Þ ¼1, thenlandPbelong to a quada. Otherwise,dðP;P0Þ ¼2 and there exist a quadacontaining bothPand P0. In any case, we get a quadacontainingPand a pointZAHat distance 2 fromP (take Z¼X or P0 according to whether dðP;P0Þ ¼1 or 2). As P?HH, aVHK ðP?VaÞUfZgIP?Va. Therefore,aJH. Hence the pointQ0Aacollinear withQ belongs toH, which forcesQQ0JH, contrary to (iii). r We are now ready to describe an example for Case 2 of Theorem 2. Given a field K, the dual S5ðKÞ of the symplectic polar spaceS5ðKÞ is embeddable into a pro- jective spacePGPGð13;KÞby means of the mappinge:Gr3ðVÞ ! 53V from the Grassmannian of planes of PGðVÞ(whereV¼Vð6;KÞ) into the 3-fold wedge prod- uct 53V (cf. Cooperstein [4, Proposition 5.1]; note that no hypotheses on the under- lying field is needed in Proposition 5.1 of [4]).
Explicitly, suppose the polar spaceS5ðKÞis represented by the symplectic form fðx;yÞ ¼x0y3x3y0þx1y4x4y1þx2y5x5y2
forx¼P5
i¼0xiei andy¼P5
i¼0yiei withe0;. . .;e5 a basis of V. The vectors eijk :¼ ei5ej5ek for 0ci< j<kc5 form a basis of 53V. The coordinates of a vector
X
0ci<j<kc5
xijkeijk
of 53V representing a totally isotropic plane hx;y;zi of PGð5;KÞ satisfy the six equations
0¼x013x125¼x014þx025¼x023þx124
¼x034x245¼x035þx145¼x134þx235: ð1Þ where, for 1ci< j<kc5,xijk is the determinant of the matrix obtained from the 36 matrix formed by the three vectors x, y, zby picking the i-th, j-th and k-th column. Equations (1) define a 13-dimensional subspaceP<PGð53VÞ. The image ofS5ðqÞviaespansP(Cooperstein [4, Proposition 5.1]; see also [5, Teorema 4.15]).
Also, if H is a hyperplane of PGð53VÞ not containing P, then the point set fXAS5ðKÞ:eðXÞAHgis a hyperplane ofS5ðqÞ. So, in view of Lemma 3, we only need to find a hyperplaneH of PGð53VÞsatisfying the following:
eðP?ÞHH; ð2:1Þ
eðQ?ÞVH ¼ feðQÞgfor some pointQat distance 3 fromP: ð2:2Þ (Note that (2.2) impliesHQP). We chooseP¼he0;e1;e2iandQ¼he3;e4;e5i. So, the points ofP?nfPg correspond to the following (singular) planes ofS5ðKÞwhere r;s;tAK:
he0þse1;e1þte2;ste3te4þe5þre0i ð3:1Þ he0þse1;e2;se3þe4þre0i ð3:2Þ he1;e2;e0þre3i ð3:3Þ
Their coordinates in PGð13;KÞare as follows:
for (3.1): x012¼rst, x013¼x125¼st,x014¼ x025¼ t,x015¼1,x023¼ x124¼st2, x024¼ t2,x123¼s2t2andxijk ¼0 ifid2
for (3.2): x012¼rs,x023¼ x124¼ s,x024¼1,x123¼ s2, andxijk ¼0 for all re- maining choices offi;j;kg
for (3.3): x012¼1;x123¼randxijk¼0 forfi;j;kg0f0;1;2g;f1;2;3g So, considering that the above scalarsr,sandtvary arbitrarily inK, if
X
1ci<j<kc5
aijkxijk ¼0
is the equation ofH, condition (2.1) forces
a012¼a015¼a024¼a123¼0
a025a014¼a124a023¼a125þa013¼0 ð4:1Þ In its turn, the conditioneðQÞAH forces
a345¼0: ð4:2Þ
However, according to (2.2), we also want thateðQ?nfQgÞVH¼q. The points of Q?nfQgcorrespond to the following (singular) planes ofS5ðKÞwherer;s;tAK
he4þse5;e3þte4;ste0se1þe2þre3i ð5:1Þ he5;e3þte4;te0þe1þre3i ð5:2Þ he5;e4;e0þre3i ð5:3Þ
and their coordinates in PGð13;KÞare as follows:
for (5.1): x034¼x245¼ st, x035¼ x145¼ s2t, x045¼ s2t2, x134¼ x235¼s, x135¼s2,x234¼ 1,x345¼ rstandxijk ¼0 ifjc2;
for (5.2): x035¼ x145¼t, x045¼t2, x135¼ 1, x345¼rst and xijk¼0 for all re- maining choices offi;j;kg;
for (5.3): x045¼ 1,x345¼randxijk¼0 forfi;j;kg0f0;4;5g;f3;4;5g;
In view of (4.1) and (4.2), none of the above points belongs to H if and only if all of the following hold for anys;tAK:
ða034þa245Þstþ ða035a145Þs2tþa045s2t2
ða134a235Þsa135s2þa23400 ð6:1Þ a045t2þ ða035a145Þta13500 ð6:2Þ a345¼0 and a04500 ð6:3Þ Needless to say, the existence of solutions for the inequalities (6.1)–(6.3) depends on the field we consider. In order to describe some of them, we may assume some additional conditions on the scalars aijk involved in (6.1)–(6.3). Note first that, as a23400 by (6.1), we may always assume a234¼1. Suppose furthermore that a145a035¼a034þa245¼a235a134¼0 (which is allowed by (6.1)–(6.3)) and put a¼a045 (which is non-zero by (6.3)) and b¼a135=a (which is non-zero by (6.2)).
Then (6.1) and (6.2) read as follows:
s2ðt2bÞ01
a for anys;tAK ð7:1Þ
t20b for anytAK ð7:2Þ
Condition (7.2) just says thatbis a non-square. So, suppose thatKis the field of real numbers and chooseb¼ 1. Then (7.2) is satisfied and (7.1) says that1=ais not the product of two positive numbers, which is true whenevera>0.
Note. The trick we have exploited above whenK is the field of real numbers does not work ifKis a finite field. Indeed, whenK¼GFðqÞ, for (7.2) to hold we only need to choose a non-square for b (which can always be done, provided that q is odd).
However, even so, the equations2ðt2bÞ ¼ 1aalways has a solution in GFðqÞ, con- trary to (7.1).
Problem.Is there any example for Case 2 of Theorem 2 in the finite case?
3 Notations and preliminaries
For basic properties of dual polar spaces, we refer to Cameron [2]. Ifais a quad of a dual polar spaceDandPis a point not ina, there exists exactly one point inanearest toP. If the rank ofDis 3, there exists exactly one point inacollinear withP.
Definition 4. Leta be a quad of the dual polar space Dof rank 3. We call the map sending every point PADto the point of a nearest to P the projection onto a and denote it bypa.
It is well known thatpa is a surjective morphism fromD, regarded as a point-line geometry, toa. It is easy to see that, ifa0is a quad disjoint froma,pa induces an iso- morphism froma0ontoa.
Given a hyperplaneHof a dual polar spaceDof rank 3, we introduce the follow- ing notations. We call a line (quad) a-line(respectively-quad) if it is contained in Hand aþ-line(respectivelyþ-quad) if it is not contained inH.
4 On locally singular hyperplanes
4.1 The split Cayley hexagon. We introduce the split Cayley hexagon HðKÞ for a fieldKfollowing Van Maldeghem [13, 2.4] for notation. Most of the results are due to Tits [11].
LetPð0Þdenote the points of the quadricQþ7ðKÞfor a fieldK, letLdenote the set of lines ofQþ7ðKÞ, and letPð1ÞandPð2Þdenote the two classes of 3-spaces ofQþ7ðKÞ.
TheD4-geometryWðKÞassociated toQþ7ðKÞconsists of the elements ofL and the elements ofPðiÞfori¼0;1;2 where incidence is symmetrized inclusion fromQþ7ðKÞ and two elements ofPð1ÞandPð2Þare incident if the corresponding 3-spaces ofQþ7ðKÞ intersect in a plane.
For the following, the most important property of WðKÞ is that the geometry arising from WðKÞby any permutation of fPð0Þ;Pð1Þ;Pð2Þg is isomorphic toWðKÞ.
Hence for each iAf0;1;2g, the set of elements of PðiÞ may be identified with the points ofQþ7ðKÞ.
AtrialityofWðKÞis an incidence preserving map y:L!L,PðiÞ!Pðiþ1Þmod 3 fori¼0;1;2 such thaty3¼1. An elementxofPðiÞ is calledabsoluteifxandyðxÞ are incident. A line isabsoluteif it is fixed byy.
Now, identifying the elements of PðiÞ with the points of Qþ7ðKÞ for each i¼0;1;2, there is an algebraic description of WðKÞ available by a trilinear form T:VVV!K where V is the 8-dimensional vector space over K in which Qþ7ðKÞis defined by a quadratic form. Two elements xAPð0Þ andyAPð1Þ are inci- dent inWðKÞif and only ifTðx;y;zÞ ¼0 for allzAV. The other incidences follow similarly.
Given a vector space representation of Qþ7ðKÞ as above, denote by tthe triality sendingðx;y;zÞontoðz;x;yÞinduced onWðKÞby the identity onV. Thesplit Cayley hexagon HðKÞconsists of the absolute points and lines of t. The absolute points of tare the singular points of a non-degenerate hyperplane PGð6;KÞof the embedding projective space PGð7;KÞofQþ7ðKÞ. For the direct embedding ofHðKÞin the para- bolic quadricQ6ðKÞdue to Tits [11], see Van Maldeghem [13, 2.4.13].
4.2 Proof of Theorem 1. Let L denote the set of lines contained in H. By Shult [10, Theorem 1],H¼ ðH;LÞis a generalized hexagon, and there exists a bijection d fromHonto the set of quads ofDthat maps every pointX AHonto the quaddðXÞ of whichXis the deep point.
We follow Van Maldeghem [13, 1.9] for the notation: For a point P of H, the perp-geometry in Phas point set the points ofP?VH. There are two di¤erent kinds of lines in the perp-geometry inP, firstly the lines ofHthroughP, and secondly the point setsG2ðPÞVG4ðXÞfor everyXAHat distance 3 fromP, whereG2nðXÞdenotes the set of points ofHat distancenfromXin the collinearity graph ofH. A pointP is calledprojectiveif the perp-geometry inPis a projective plane. A pointPis called distance-2-regularif two lines of the perp-geometry inPintersect in at most one point.
By Ronan [8], the generalized hexagonHis a split Cayley hexagonHðKÞif all its points are distance-2-regular and if there is at least one projective point inH(we use Ronan’s result as it is stated in Van Maldeghem [13, Theorem 6.3.1]). We prove that every point of H is projective. Then every point is distance-2-regular and the con- ditions of [13, 6.3.1] are satisfied.
For a pointPAH, letG2ðPÞVG4ðXÞbe a line of the perp-geometry inPwhereX is a point ofHat distance 3 fromP. The lineG2ðPÞVG4ðXÞcontains the points ofH collinear withPand withpdðPÞðXÞinD. Hence it is the tracefP;pdðPÞðXÞg?ofPand pdðPÞðXÞin the generalized quadrangle ResðdðPÞÞ. For every pointTAG2ðPÞVG4ðXÞ, the quaddðTÞcontains theþ-linehðP;XÞ:¼PpdðXÞðPÞ. Indeed, ifT0AP?VdðPÞand hðP;XÞNdðT0Þ, then the quaddðT0Þis disjoint fromdðXÞ, whenceT0is at distance 3 fromXandT0BG2ðPÞVG4ðXÞ. Hence the lineG2ðPÞVG4ðXÞof the perp-geometry inPis uniquely determined by the linehðP;XÞofD, and the points ofG2ðPÞVG4ðXÞ are the pointsd1ðtÞfor the quadstofDon the linehðP;XÞ.
Hence the points of the perp-geometry inPcorrespond to the lines of the residue ResDðPÞofPinD(these are the quads having points ofP?VH as deep points), and the lines of the perp-geometry inPcorrespond to the points of ResDðPÞ(these are the lines ofDthroughP). Since ResDðPÞis a projective plane, the perp-geometry inPis a projective plane isomorphic to the dual of ResDðPÞ.
Hence the generalized hexagonHis a split Cayley hexagonHðKÞembedded in an orthogonal polar spacePP~GQ6ðKÞ.
It remains to show that P¼D is isomorphic to the orthogonal polar space P~
PGQ6ðKÞ. The points ofPare the (singular) quads of D. These correspond to the points ofHvia the bijectiond, whence the quads ofDcorrespond to the points of the orthogonal polar spacePP~ ¼Q6ðKÞwhich is the ambient space ofHðKÞ.
The lines of PP~ through a point PBH are the lines of the perp-geometry in P.
By the above, the perp-geometry in Pis a projective plane, in fact a singular plane ofPP~GQ6ðKÞ. On the other hand, we have seen that the lines of the perp-geometry in P correspond to the lines ofDthrough P. So, the lines of D are the lines of the polar space PP~ GQ6ðKÞand we have proved that the dualP of Dis isomorphic to P~
P¼Q6ðKÞ. Hence we have proved Theorem 1. r
5 Non-uniform hyperplanes with no subquadrangular quad
5.1 Proof of Theorem 3.We remind the reader of the notations in Theorem 3:H0 is a locally singular hyperplane of the dualD0 of a polar spaceP0 of rank 3 that is a
hyperplane of a polar spacePof rank 3. The shadow operator ofD(respectivelyD0) iss(respectivelys0). Remark that a linelofDbelongs toD0if and only ifs0ðlÞ0 q. IfXis a point subset ofD0,sðXÞdenotes the set of points ofDthat are on some line mwiths0ðmÞJX.
SupposeP0is a hyperplane ofP. We first showsðH0Þis a subspace ofD. L etX;Y be two points ofsðH0Þon a linel. We have to showlHH0. SinceP0is a hyperplane, there is a pointbAP0onl, the latter being regarded as a line ofP. AsX;Y AsðH0Þ, there are linesh;mHH0 throughXandY, respectively. Ifh does not belong to the quadbofD0, then the pointX¼hVbbelongs toD0, thusX AH0. Ifmdoes neither belong tob, it follows similarlyY AH0. ThenlHH0 and we are done. So, assume hNbandmHb. Then, asXandmbelong toD0, the line throughXconcurrent with m, namely the line l, belongs to D0, and we are done. Finally, assume h;mHb. If bHH0, thenlHbHH0 is contained inH0. Ifbis singular,handmare concurrent in the deep point ofb. Thus eitherh;l;mform a triangle, which is impossible, orl¼ h¼mHH0. HencesðH0Þis a subspace ofD.
It remains to showsðlÞVsðH0Þ0 qfor every linelofD. SinceP0is a hyperplane ofP, every linelofPcontains a pointaofP0. Regarded inD,ais a singular quad with deep pointP, andlcontains a pointP0collinear withP. Then the linel0 :¼PP0
(possiblyl¼l0) belongs toH0since it is contained in the quadaofD0.
Conversely, supposesðH0Þis a hyperplane ofD. Then every linelofDcontains a pointPofsðH0Þ. The pointPlies on some linel0ofD0. The quadhl;l0ibelongs to D0since it contains the linel0ofD0. InP, the quadhl;l0iis a point ofP0on the line l. So, every line ofPmeetsP0, andP0 is a hyperplane ofP.
In the remainder, supposeP0 is a hyperplane ofPandsðH0Þhas no deep point.
Given a quadaofDnot inD0, we have to show thataVsðH0Þis an ovoid. Assume to the contrary thata contains two points X;Y ofsðH0Þon a line l. SinceP0 is a hyperplane, there is a pointbAP0 onl regarded as line ofP. SinceX;Y AsðH0Þ, there are linesh;mHH0 throughXandY, respectively. Ifh does not belong to the quadb ofD0, then the intersection pointXof the linehofD0 withb belongs toD0. RegardingXas plane ofP0, all of its points belong toP0. This forcesato belong to D0 in contradiction to the hypothesis. ThushHband similarly mHb. Sinceh;mH H0Vbandbis singular,handmare concurrent in a pointPofD0. Thus eitherh;l;m form a triangle, which is impossible, or h¼l¼m. However, the second case is im- possible, too, since it forcesato belong toD0.
Hence Theorem 3 is proved.
5.2 Proof of Theorem 2. Let H be a hyperplane such that þ-quads are either singular or ovoidal and assume that there exist both singular and ovoidal quads. In Lemmata 5–9, we investigate some properties of the hyperplaneH.
Lemma 5.Given two quadsa;a0with common deep point PAH,P is deep with respect to H.
Proof. Let b be a quad containing P and not containing the line aVa0. Then bVa0bVa0andbVaandbVa0are-lines concurrent inP. HencePis deep with respect tob. It follows that all lines throughPare-lines. So,Pis deep itself. r
Lemma 6. Given a quad aHH and an ovoidal quad o, setting O:¼oVH, the set paðOÞis an ovoid and all its points are deep.
Proof.The quadsaandoare disjoint since an ovoidal quad has no-line. LetPAO and denote the pointpaðPÞby P0. The linel:¼PP0 is contained in H. L et b be a quad onl. Thenbcontains the two-lineslandbVathrough the pointP0. HenceP0
is deep with respect tob. By Lemma 5,P0is deep. r
Lemma 7.If there exists an ovoidal quad,any two-quads are disjoint.
Proof. Assumea;a0 are-quads on a -line l¼aVa0. By Lemma 5, every point PAl is deep. Ifois an ovoidal quad, it is disjoint froma anda0, hence froml. So, poðlÞis a line ino. Since all points oflare deep, all points ofpoðlÞare contained in
H. This is impossible sinceois ovoidal. r
Lemma 8.If there exists an ovoidal quad,no two deep points of H are collinear.
Proof. AssumeP;P0 are collinear deep points. The line l:¼PP0 is contained in H.
Given a quadaonl, it is contained inHsinceP;P0are two deep points with respect toaVH. Hence all quads onlare-quads in contradiction to Lemma 7. r Lemma 9. If a is a -quad and o1;o2 are ovoidal quads with ovoids Oi:¼oiVH, i¼1;2,thenpaðO1Þ ¼paðO2Þ.
Proof. By Lemma 6,paðO1Þand paðO2Þconsist of deep points. Since by Lemma 8 no two deep points are collinear,paðO1ÞUpaðO2Þis an ovoid ofQ, hencepaðO1Þ ¼
paðO2Þ. r
Propositions 10 and 11 establish Cases 1 and 2 of Theorem 2.
Proposition 10.The hyperplane H contains at most one quad.If H contains a quad a, then H ¼6
PAOP? where O is an ovoid ofa.
Proof.AssumeHcontains a quada. By Lemma 7 and since no ovoidal quad meets a -quad, all quads meetingaare singular.
Since there exists an ovoidal quad o, by Lemma 6, the points of the ovoid O:¼paðoVHÞof the quadaare deep. LetPAO. Given a pointX AP?na, the line l:¼PX is contained inHand the quads onlare singular with deep pointP. Hencel is the only-line throughX, and the quads throughXnot containinglare ovoidal.
Since a quad a0 disjoint from a has exactly one point Y in P? (indeed the point pa0ðPÞ) and since a0 does not contain the line YP, according to the previous para- graph,a0is ovoidal. Hence all quads disjoint fromaare ovoidal.
By Lemma 9, there is one and only one ovoid Oof deep points in a. So, the hy- perplane consists of all points collinear with points ofO. r
Proposition 11.If H does not contain any quad,H has at most one deep point.If it has a deep point P,the quads through P are singular and the remaining quads are ovoidal.
Proof. LetPbe a deep point ofH. L et RAP?nfPg and denote the-line PRbyl.
Since no quad is contained inH, all quads onlare singular with deep pointP. Hence the line lis the only line throughRcontained inH. So, a quad through Rnot con- taininglis ovoidal and meetsP?in the unique pointR.
Since every quad has a point of P?, it follows that the quads through P are sin- gular with deep pointPand the quads not containingPare ovoidal. r In the remainder, we investigate the hyperplaneH when it has no deep point. We prove the assertions of Case 3 of the theorem in several lemmata and propositions.
LetQbe the set of all singular quads ofD, letPbe the set of points ofHdeep with respect to some singular quad, and letLbe the set of lines ofDthat meet the point setP. Denote byP0the substructureðQ;LÞof the polar spacePdual ofD.
Lemma 12.Every quad ofDcontaining a line ofLis singular.
Proof. Let lAL. If lHH, the assertion is obvious. Assume lVH is a point X. By definition ofL,Xis deep for some singular quada. Hence every quad containingl
meetsain a-line and is singular. r
Lemma 13.P0is a proper subspace ofP.The lines ofP0are lines ofP,and two points a;bAP0 are collinear whenever they are collinear inP.
Proof. Let a;b be two quads of Q that have a line l in common. We claim lAL. Indeed, iflHH, all quads onlare singular and have their deep points onl, whence lAL. Iflis aþ-line that meetsHin a pointP, both quadsaandbare singular and have distinct-linesh;h0, respectively, throughP. Hence the quadhh;h0iis singular with deep point P. So,PAPandlAL. By Lemma 12, all quads on lbelong toQ.
Hence Q which is the point set ofP0, is a subspace of P, and the lines of P0 are precisely the lines ofPcontained inQ.
It is a proper subspace since the points ofPcorresponding to ovoidal quads ofD
do not belong toP0. r
Lemma 14.Given a point P ofD,either none,one or all lines through P belong toL. Proof.IfPAP, there is nothing to prove. Accordingly, assume in the following that Pis not deep for any quad ofQ, i.e.PBP.
We first show that there is no-line ofLthroughP. Assume to the contrary that l is a -line through P. If l0AL is a second line through P, l0 is a þ-line with l0VH¼P. It follows PAP since otherwise l0 would not belong to L—a contra- diction.
So, the two linesl;l0 ofLthrough Pareþ-lines. Since by Lemma 12, the quad hl;l0i is singular, it followsPBH since otherwise there would be a-line through P in contradiction to the previous paragraph. Hence given a line h through P not
inhl;l0i,hVH is a pointY0P. Since every quad onl(respectivelyl0) is singular, the quad hh;li (respectively hh;l0i) is singular. Since Y A hh;li VH (respectively A hh;l0i VH), there is a-line throughYinhh;li(respectivelyhh;l0i). Hence there are two-lines throughY. So,YAP, thus implyinghAL.
Given a linemthroughXinhl;l0i, the quadhh;miis singular by Lemma 12 since hAL. Hence there are the two singular quads hl;l0i andhh;mi on the þ-line m.
It follows that there are two -lines on the point mVH. Hence mVH AP, which
forcesmAL. r
Proposition 15.P0is a polar space of rank3.
Proof. By Buekenhout and Shult [1], P0 is a polar space of rank 3 if it satisfies the one-or-all axiom and if its maximal subspaces are planes. We start verifying the one-or-all axiom for polar spaces.
Let aAQbe a singular quad with deep pointZ, and let lALbe a line ofLnot contained ina. All quads onlare singular by Lemma 12. If a quada0onlmeetsain a linehAL,a0is a point ofP0 onlcollinear witha.
We show that the quads onl meetinga intersectain lines of L. This proves the one-or-all-axiom since on the one hand, all quads onlmeetaiflmeetsa, and on the other hand, there is exactly one quad onlmeetingaiflis disjoint froma.
Firstly, assumelVa¼q. Denote the line paðlÞbyh. By Lemma 12, the unique quadhh;licontainingland meetingais singular, and it remains to showhAL.
Ifh goes throughZ,hbelongs toLsinceZAP. Ifhdoes not go throughZ, de- note the pointhVH¼hVZ?byX. Sincehh;liis singular by Lemma 12, there is a -linegthroughXinhh;lisuch that the pointX Ahis deep for the quadhg;XZi. Hencehbelongs toL.
Secondly, assumelintersectsain a pointX. The quads onlare the points ofP0on the linelcollinear with the pointaby the lines that are the lines in the quadathrough the pointX. Since all quads onlare singular, it remains to show that all lines through Xinabelong toL.
Iflis a-line,Xis collinear withZand is deep with respect to the quadhl;XZi. ThusX APand all lines throughXbelong toLand we are done. Iflis aþ-line, the pointlVH is deep for a singular quadbsincelAL. We investigate the caseslVH ¼ X ð¼PVlÞandlVH0X separately.
AssumelVH0X. Then the quadbis disjoint fromasince it does not contain the linel. The pointXis not collinear with the deep pointZwith respect toa. L etgbe a quad onl. So,gis singular with deep pointDon the linem:¼gVbwhereD0mVl.
Furthermore, gmeetsain a line hthat is not contained inH sinceDBh. It follows hVH¼hVD?. The pointY:¼hVHis collinear withZsinceZis the deep point of a. So,YAPsince the linesDY andYZare distinct-lines throughY. Thus,Yis the deep point of the singular quadhD;Zi. HencehAL.
Finally, assume l is a þ-line with lVH ¼X. The line XZHa is contained in H since Z is the deep point ofa and XZ belongs toL. So, l;XZ are two lines of LthroughX, and by Lemma 14, all lines throughXbelong toL. Therefore, every quad onlmeetsain a linehAL.
HenceP0satisfies the one-or-all axiom. It remains to show that the maximal sub- spaces ofP0 are planes. On the one hand, sinceP0 is a subspace of the polar space P, the rank ofP0 is at most 3. On the other hand, by Lemma 14, the maximal sub- spaces ofP0 are maximal subspaces ofP. Consequently,P0has rank 3. r Lemma 16. A point P ofDbelongs to the dual D0 of P0 if and only if all lines ofD through P belong toL.
Proof. By Lemma 13, the polar spaceP0 is a subspace of the polar space P¼D. Since the set of all quads through a point ofDis the point set of a maximal subspace ofP, the quads through a point ofD0 are the points of a maximal subspace ofP0. If Xis a point ofD0, the quads ofD0throughXregarded as points ofP0 are points of a plane of P0. By Proposition 15, all points and lines of that plane belong to P0. Hence all lines ofDthrough Xbelong toL. Conversely, if all lines of DthroughX belong toL, then all quads onXare singular by Lemma 12. Thus regarding Xas a plane ofP, all its points belongP0, whence it is a plane of P0, namely a point of
D0. r
Proposition 17.The point setPis a hyperplane of D0.It is the hyperplaneD0VH in- duced by H inD0.
Proof. SinceP consists of the points deep with respect to some singular quad ofD, all lines through a pointX APbelong toL, andXis a point ofD0 by Lemma 16.
HencePis a subset of the point set ofD0.
To show that P is a subspace ofD0, letX;X0AP be distinct points on a linel.
SincePHH andHis a subspace ofD, it followslHH. It remains to show that ifl is a line ofDcontained inH, every point ofD0 onl belongs toP. L etVAl belong toD0. By Lemma 16, all lines throughVbelong toL. If ais a quad ofD0 through V not containing l, then a is singular by Lemma 12. So, since VAH, there exists a -line mð0lÞ of D0 through V in a. Hence V is deep with respect to the quad hl;miofD0, and it followsV AP.
Since every quad ofD0 meetsPin the perp of a point by construction ofD0,Pis a proper subspace ofD0.Pis a hyperplane ofD0since the lines ofD0 are the lines in Lthat meetPby definition.
Furthermore, we have seen in the previous paragraph that all points ofD0on a line contained inHbelong toP. Given a pointPAD0VH and a quadaofD0 onP,ais singular andPAaVH belongs to some linelUH. ThusP¼D0VH. r We denote the set of lines ofDcontained inHbyL. An immediate consequence is LJLandLPL. L etHdenote the incidence structureðP;LÞ.
Proposition 18. H¼ ðP;LÞis a split Cayley hexagon HðKÞ.The polar spaceP0¼ ðQ;LÞis isomorphic to the orthogonal polar spaceQ6ðKÞin whichHis embedded.
Proof. Since the points of P0¼D0 are the singular quads ofD, each quad of D0 is singular with respect to the hyperplaneP¼D0VH(cf. Proposition 17). By Theorem
1, H¼ ðP;LÞ is a split Cayley hexagon HðKÞ embedded in PP~GQ6ðKÞ and the
polar spaceP0is isomorphic toPP.~ r
Hence we have proved the assertions of Theorem 2. The following corollary and proposition establish Corollary 1.
Corollary 19.SupposeP0is a hyperplane ofP.Then the hyperplane H consists of the points on lines of L.
Proof. Denote byH0the set of points ofD on lines ofL. SupposingP0 is a hyper- plane ofP, we apply Theorem 3. Then H0¼sðPÞis a hyperplane ofD. We show H ¼H0. The inclusionH0JH follows from the definition ofH0.
Assume there exists a point PAHnH0. Since H0 is a hyperplane ofD, each line throughPmeetsH0in a point distinct fromPsincePBH0. So, each line throughP
is contained inHimplyingPAH0—a contradiction. r
Remark that Corollary 19 is a special case of a result of Shult [9, Lemma 6.1] ac- cording to which every hyperplane of a dual polar space is a maximal subspace.
Corollary 19 together with the following proposition prove the second assertion of Corollary 1.
Proposition 20. If D is finite, the polar space P¼D is an orthogonal polar space Q7ðqÞ.In particular,P0GQ6ðqÞis a hyperplane ofP.
Proof. By Theorem 2, the dual P0 of D0 is a polar space isomorphic to Q6ðKÞ for some finite fieldK¼GFðqÞ.
If a is a point of P0, its residue ResP0ðaÞ in P0 is a generalized quadrangle of orderðq;qÞwhich is a proper subquadrangle of the generalized quadrangle ResPðaÞ of orderðt;qÞ. Hencet>q.
Regardingaas a quad ofD, the generalized quadrangle ResDðaÞadmits an ovoid.
Thus by Payne and Thas [7, 3.4.1] and sincet>q, ResDðaÞis isomorphic toH3ðqÞ.
HencePGQ7ðqÞ. r
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Received 15 January, 2001; revised 12 February, 2001
H. Pralle, Mathematisches Institut der Justus-Liebig-Universita¨t Giessen, Arndtstr. 2, 35392 Giessen, Germany
Email: [email protected]
