On the hyperbolic unitary geometry

DOI 10.1007/s10801-009-0200-5

On the hyperbolic unitary geometry

Kristina Altmann·Ralf Gramlich

Received: 18 October 2006 / Accepted: 11 August 2009 / Published online: 1 September 2009

© Springer Science+Business Media, LLC 2009

Abstract Hans Cuypers (Preprint) describes a characterisation of the geometry on singular points and hyperbolic lines of a finite unitary space—the hyperbolic unitary geometry—using information about the planes. In the present article we describe an alternative local characterisation based on Cuypers’ work and on a local recognition of the graph of hyperbolic lines with perpendicularity as adjacency. This paper can be viewed as the unitary analogue of the second author’s article (J. Comb. Theory Ser.

A 105:97–110,2004) on the hyperbolic symplectic geometry.

Keywords Hyperbolic unitary geometry·Root group geometry·Local recognition graphs·Centralisers of involutions

1 Introduction

The geometry on the points and hyperbolic lines of a non-degenerate finite unitary polar space (or, short, hyperbolic unitary geometry) is interesting for a number of reasons.

The second author gratefully acknowledges a Heisenberg fellowship by the Deutsche Forschungsgemeinschaft.

K. Altmann

TU Darmstadt, Fachbereich Mathematik, c/o Holger Grothe, Schloßgartenstraße 7, 64289 Darmstadt, Germany

R. Gramlich (

TU Darmstadt, Fachbereich Mathematik, AGF, Schloßgartenstraße 7, 64289 Darmstadt, Germany e-mail:gramlich@mathematik.tu-darmstadt.de

R. Gramlich

University of Birmingham, School of Mathematics, Edgbaston, Birmingham B15 2TT, UK e-mail:ralfg@maths.bham.ac.uk

One reason is the fact that it belongs to the family of (partially) linear geometries that are characterised by their planes (Cuypers [6]; restated in part as Theorem4.6 of the present article). The most famous of such geometries is the projective space, which by a classical result is characterised as a linear geometry each of whose planes are projective (Veblen and Young [14], [15]). Another geometry characterised by its planes is the hyperbolic symplectic geometry (Cuypers [4], Hall [9]). It is closely related to the hyperbolic unitary geometry: while each plane of the former is dual affine (also called symplectic), a plane of the latter is either dual affine or linear. (The linear ones are, in fact, related to classical unitals, cf. [10], [11].)

A second reason why the hyperbolic unitary geometry is an interesting object to study is the 1-1 correspondence between the set of long root subgroups, resp.

fundamentalSL2’s ofSU_n(q²)on one hand and the points, resp. hyperbolic lines of the corresponding unitary geometry on the other hand via the map that assigns the respective groups to their commutator in the module. This correspondence is well- known, see e.g. [13, Chapter 2]. Cuypers’ article [5] underscores that root group geometries are highly interesting objects.

This paper can be viewed as a sister paper of [8] (where the root group geometry of Sp2n(F)is studied for arbitrary fields) and of [1] (where the authors study the line graph of a complex vector space endowed with an anisotropic unitary form).

However, the situations covered by the sister papers [1], [8] of this paper are much more behaved and a lot easier to handle than the situation in this paper. The increased difficulty compared to [8] originates from the fact that we prove Theorem1forn≥7 instead ofn≥8 (odd-dimensional non-degenerate symplectic forms do not exist), while the increased difficulty compared to [1] comes from the fact that subspaces of non-degenerate subspaces can be very far from being non-degenerate, whereas subspaces of anisotropic subspaces are anisotropic.

The first result of this paper focuses on the hyperbolic lines and their relative posi- tions. More precisely, letU_ndenote ann-dimensional vector space overFq²endowed with a non-degenerate hermitian form. The hyperbolic line graph G(Un)is the graph on the hyperbolic lines, i.e., the non-degenerate two-dimensional subspaces ofU_n, in which hyperbolic linesl andmare adjacent (in symbolsl⊥m) if and only ifl is perpendicular tomwith respect to the unitary form. Equivalently,l⊥mif and only if the corresponding fundamentalSL2’s commute.

A graph is locally homogeneous if and only if for any pairx,y of vertices of, the induced subgraphs(x)and(y)on the set of neighbours ofx, resp.y are isomorphic. Such a locally homogeneous graphis called locally, for some graph, if (x)∼= for some, whence all, verticesx of . It is easily seen (cf.

Proposition3.3) that the graph G(Un)is locally G(Un−2). Conversely, this property is characteristic for this graph for sufficiently largen:

Theorem 1 Letn≥7, letq be a prime power, and letbe a connected graph that is locally G(Un). Thenis isomorphic to G(Un+2), unless(n, q)=(7,2).

The requirement in the preceding theorem thatbe connected comes from the fact that a graph is locallyif and only if each of its connected components is locally. So its primary role is to provide irreducibility. We do not know whether the case (n, q)=(7,2)provides an actual counter example.

Forn≥8 this result has been stated without proof in the second author’s PhD thesis [7, Theorem 4.5.3]. Since counter examples to the local recognition are only known for n=6—they come from the exceptional groups of type ²E₆(q²), see [13]—publication of this result was deferred until the casen=7 could be proved.

This has finally been achieved during the preparation of the first author’s PhD thesis.

Comparing the proofs of Lemmata5.5,5.6and5.7with the proof of Lemma5.8, the reader will understand why the casen=7 is so much more difficult than the case n≥8.

As mentioned before, the motivation of our research was of group-theoretic na- ture. If the fieldFhas characteristic distinct from 2, translating Theorem1into the language of group theory yields the following.

Theorem 2 Letn≥7 and let q be an odd prime power. Let G be a group with subgroupsAandBisomorphic toSL₂(q), and denote the central involution ofAby xand the central involution ofBbyy. Furthermore, assume the following holds:

• C_G(x)=A×KwithK∼=SU_n(q²);

• C_G(y)=B×J withJ∼=SU_n(q²);

• Ais a fundamentalSL2ofJ;

• Bis a fundamentalSL2ofK;

• there exists an involution inJ∩Kthat is the central involution of a fundamental SL2of bothJ andK.

IfG= J, K, thenG/Z(G)∼=P SU_n+2(q²).

This article is organised as follows: In Sections2 and3we study properties of the hyperbolic line graph G(Un)for n≥5. Section4deals with the relation of the graph G(Un)with the hyperbolic unitary geometry. In that section we also study em- beddings of G(Un−2)in G(Un), which provides us with valuable information for the proof of Theorem1that we give in Section5. Most of our arguments are based on counting in subspaces ofU_n of various dimensions and ranks. For the convenience of the reader we include a collection of results on the number of subspaces of var- ious types in AppendixA. For quick reference we also give some tables containing the necessary information in Table1. A proof of Theorem 2is not included in this article, because the problem of how to deduce a result like Theorem2 from a re- sult like Theorem1has been thoroughly studied in [3, Section 6], [7] and, thus, is well-understood.

Acknowledgement The authors would like to express their gratitude to Hans Cuypers for helpful dis- cussions on the topic. They also thank two anonymous referees for a wealth of extremely helpful remarks, comments, and suggestions.

2 The hyperbolic line graph ofU₅

Letq≥3 be a prime power and letU₅ be a five-dimensional non-degenerate uni- tary vector space overF_q² with polarityπ. Define the graph G(U5)with the set of non-degenerate two-dimensional subspaces ofU₅as the set of vertices in which two

verticeslandmare adjacent if and only ifl⊂m^π. The aim of this section is to recon- struct the unitary vector spaceU5from the graph G(U5). To this end we will define a point-line geometryG=(I,L,⊃)using intrinsic properties of the graph G(U5) and establish an isomorphism betweenGand the geometry on arbitrary points and hyperbolic lines ofU₅. From thereU₅is easily recovered.

We first determine the diameter of G(U5).

Lemma 2.1 Letlandmbe distinct hyperbolic lines ofU5. Thenlandmhave dis- tance two in G(U5)if and only if the subspacel, m is a non-degenerate plane in U₅.

Proof Letlandmbe two hyperbolic lines ofU₅which have distance two in G(U5).

That is, the graph G(U5)contains a vertexz, which is a hyperbolic line inU5, adjacent to the verticesl andm. Its perpendicular spacez^π, a non-degenerate plane ofU5, contains the distinct hyperbolic linesl andm. Hence the hyperbolic linesl andm span the non-degenerate planez^π.

Conversely, suppose thatl, mis a non-degenerate three-dimensional subspace of U₅. SinceU₅is a five-dimensional non-degenerate unitary vector space, the pole of l, mis a hyperbolic lineh= l, m^π ofU₅. By definition the vertexhis adjacent to the verticeslandmin G(U5). Since the hyperbolic lineslandmintersect inU5, it follows thatl⊥m. Therefore the verticeslandmhave distance two in G(U5).

Lemma 2.2 Letlandmbe distinct hyperbolic lines ofU₅. Thenlandmhave dis- tance three in G(U5)if and only iflandmare two non-intersecting hyperbolic lines such thatl^π∩mis a one-dimensional subspace ofU5.

Proof Suppose the verticeslandmhave distance three in the graph G(U5). Then by Lemma2.1we find a vertexzin the graph G(U5)adjacent tol such thatz, m is a non-degenerate plane ofU₅. The intersectionp:=m∩zis a one-dimensional. As z⊆l^π, the hyperbolic linemintersects the subspacel^π in at least the pointp. Since the verticeslandmare not adjacent in G(U5), we havem⊆l^π, som∩l^π=p.

In order to prove the first implication of the statement it is left to show that the hyperbolic lines l andm do not intersect in U₅. By way of contradiction we as- sume thatm, lis a three-dimensional subspace. The planel, mis degenerate by Lemma2.1, thusl^π∩m^π is a singular two-dimensional subspace ofU5. Sincep, the intersection point ofm and l^π, is incident to the hyperbolic line m, we have p∈rad(m, l)andm^π ⊆p^π, whencem^π∩l^π⊆p^π∩l^π. Of course,p is either singular or non-degenerate. Furthermore dim(m^π∩l^π)=2=dim(p^π∩l^π). Conse- quentlym^π∩l^π=p^π∩l^π.

Ifpis a non-degenerate point, thenp^π∩l^πis a non-degenerate line, contradicting the fact thatm^π∩l^πis degenerate. Ifpis a singular point, then of coursep^π∩l^πis a singular two-dimensional subspacesof rank one containing the pointpitself and the radical ofp^π∩l^π. Thereforep=rad(p^π∩l^π)=rad(m^π∩l^π)=rad(m, l)=p, a contradiction. Thusm, lhas to be a four-dimensional space and the two hyperbolic lineslandmhave a trivial intersection inU5.

Now for the other implication. If l andm are two non-intersecting hyperbolic lines inU₅such thatl^π∩mis a one-dimensional subspacep, then, by Lemma2.1,

the verticesl andmhave not distance one or two in the graph G(U5). To prove the statement, we construct a hyperbolic linezin the subspacel^π with the property that the subspacem, zis a non-degenerate plane inU5, implying thatl⊥zand that the distance between the verticeszandmin G(U5)is two, by Lemma2.1.

Consider the subspacel^π perpendicular to land two pointsp∈l^π∩mandx∈ l^π∩m^π. Note thatpandxare uniquely determined by the assumptions that dim(l^π∩ m)=1 and dim(l, m)=4 inU5. Moreoverp∈x^π sincep∈mandx∈m^π.

If both the pointpand the pointxare non-degenerate, thenz= p, xis a hyper- bolic line contained inl^π, sincepandxare perpendicular to each other as noted be- fore. Furthermorem, xis a non-degenerate plane ofU₅due to the fact thatx∈m^π, proving the statement in this special case.

Ifpis singular andxis non-degenerate, thenp, xis a singular line of rank one, becausep∈x^π as mentioned above. We consider the q²−q hyperbolic lines h_i, 1≤i≤q²−q, inl^π incident tox, cf. Table1, entryn+l=2,n=2,r=1. Any two different hyperbolic linesh_i andh_j span the planel^π and each subspaceh^π_i is a non-degenerate plane inx^π for 1≤i < j≤q²−q. Moreover, the intersection of h^π_i with the hyperbolic line mis a point r_i =h^π_i ∩min x^π withr_i =r_j for 1≤ i < j≤q²−q. Indeed,m⊆h^π_i , becauseh_i∩m^π is the one-dimensional subspace x for each hyperbolic line h_i. Ifr_i =r_j fori=j, then we obtainr_i=h^π_i ∩m= r_j=h^π_j ∩m=h^π_i ∩h^π_j ∩m= h_i, h_j^π∩m=l∩m= {0}, a contradiction. Thus we haveq²−q different one-dimensional subspaces ri on the hyperbolic linem.

Hence the linemcontains a non-degenerate pointr=rkfor somek∈ {1, . . . , q²−q}, becauseq²−q > q+1 forq≥3, whereq+1 is the number of singular points on a hyperbolic line (cf. Table1on page581). Note that the pointsrandpspan together the hyperbolic linem. Note also thatr^π∩l^π=h_k. For,r_i^π=(h^π_i ∩m)^π= h_i, m^π, sor^πcontainsh_k; sincel^πcontainsh_kas well and sincer^π∩l^πis two-dimensional, we haver^π∩l^π=h_k. As follows from Table1all points on the hyperbolic lineh_k different from the pointx generate together with the pointpa non-degenerate two- dimensional subspace ofl^π. Therefore the hyperbolic line h_k containsq²−q−1 different non-degenerate pointsyi such thatyi, pis a hyperbolic line. Furthermore the span of the two hyperbolic linesmandhk is a four-dimensional space of rank at least three, sincer, hk ⊆ m, hkand rk(r, hk)=rk(r, r^π∩l^π)=3.

If the four-dimensional spacem, h_kis non-degenerate, Table1implies that the hyperbolic lineh_kcontains at leastq²−2q−2>0 (recall thatq≥3) different non- degenerate pointsz_i such thatz_i, p =zis a hyperbolic line andm, z = r, p, z_i is a non-degenerate plane. Alternatively, if the rank of the four-dimensional space m, h_kis three then, by the information from Table1, the hyperbolic lineh_kcontains at leastq²−q−2>0 different non-degenerate pointsz_i, which satisfy the conditions thatz_i, p =zis a hyperbolic line andm, z = r, p, z_iis a non-degenerate plane and we are done in this case.

Next we assume the pointp to be non-degenerate and the pointxto be singular.

Then the hyperbolic lineh=l^π∩p^π is incident to the singular pointx=l^π∩m^π, becausepis incident tom. Moreover the non-degenerate pointr=p^π∩mand the hyperbolic linehspan a planeP of rank two or three. As follows from the information from Table1the planeP contains at leastq²−qdifferent hyperbolic lines incident to the pointr. Certainly, the intersections of these q²−q hyperbolic lines withh

are pairwise distinct as follows by arguments similar to the ones used above. At least q²−2q−1 of those intersection points are non-degenerate. Choosing one of those, saya, the linez= a, p ⊂l^π is a hyperbolic line, asa∈p^π. The planem, z = r, p, ahas a Gram matrix (with respect to some suitably chosen basis inr,p, and a) of the form

_{1 0 0}

0 1γ 0γ1

. This matrix has a non-zero determinant as follows from the fact thatzis a hyperbolic line, som, zis non-degenerate. Again, by Lemma2.1we are finished in this case.

The case that both pointsxandpare singular does not occur, as otherwise the non- degenerate planel^π would contain the totally singular linex, p, a contradiction.

Lemma 2.3 Letlandmbe two different hyperbolic lines ofU5. Thenlandmhave distance four in G(U5)if and only if either

• landmare two non-intersecting lines such thatl^π∩mis trivial inU₅, or

• landmare two intersecting lines spanning a degenerate plane inU₅.

Proof Letl andm be two vertices in the graph G(U5)at mutual distance four. If the subspacel, m is a non-degenerate plane, thenl andmhave distance two by Lemma2.1. Therefore, if l, mis a plane, then l, mis a degenerate subspace of U₅. Alternatively, ifl, mis a four-dimensional subspace inU₅andl^π∩m= {0}, thenl andmhave distance one in G(U5)by definition or distance three in G(U5) by Lemma2.2, a contradiction again. It follows that, if the subspace l, m is of dimension four, thenl^π∩mis trivial.

In order to show the converse implication of the statement, letl, m be either a degenerate plane or a four-dimensional subspace such thatl^π ∩m= {0}. By the Lemma2.1and Lemma2.2the verticesl andmdo not have distance one, two, or three in G(U5). Therefore it is enough to find a path of length four in G(U5)between the verticeslandmto finish the proof of this lemma.

We choose a hyperbolic linezinl^π intersecting the spacem^π in a point. Such a choice is possible, becausel^π is non-degenerate and the intersectionl^π∩m^π is non- trivial and properly contained inl^π. By construction the verticeslandzare adjacent in G(U5). By the abovem∩l^πis trivial. Hencemandzdo not intersect, but satisfy the condition dim(z∩m^π)=1. Somandzhave distance three in the hyperbolic line graph G(U5)by Lemma2.2and, thus, the distance between the verticeslandmis

four in G(U5).

Proposition 2.4 The graph G(U5)is a connected locally G(U3)graph of diameter four.

Proof For any singular pointpin the orthogonal spacel^πof a hyperbolic linelinU5, the subspacel, pis of dimension three and rank two. As follows from LemmaA.1 it is possible to choose a hyperbolic linemdifferent fromlin the planel, p. Thusl andmspan the degenerate planel, pand hence the verticeslandmhave distance four in G(U5)by Lemma2.3. The statement about the diameter follows now from the fact that two hyperbolic lines cannot form a configuration other than adjacency and the ones described in2.1to2.3. The local property is obvious.

Remark 2.5 Letlandmbe two arbitrary vertices of the hyperbolic line graph G(U5).

An important induced subgraph of G(U5)is the common perp of the verticeslandm.

If the induced subgraph{l, m}^⊥is not empty, then the subspacel, m^πofU₅contains some hyperbolic line. We observe that{l, m}^⊥= {0}in G(U5)if and only ifl and mhave distance two in G(U5). Indeed if l and m are at distance two in G(U5), then the hyperbolic linesl andm span a non-degenerate plane inU₅ and l, m^π is a hyperbolic line, by Lemma2.1. In all other casesl, mis a four-dimensional subspace (and thereforel, m^π is a single point ofU₅) or the hyperbolic linesland mspan a degenerate planes (which implies thatl, m^π is a rank one line). Thus in these cases the subgraph{l, m}^⊥is the empty graph. Of course, if the verticeslandm have distance one in G(U5), thenl, mis a four dimensional non-degenerate space in the unitary vector spaceU₅ andl^π∩m^π= l, m^π is a non-degenerate point of U₅.

Definition 2.6 LetW be a subspace ofU5. The set of all hyperbolic lines ofW is denoted by L(W ).

Lemma 2.7 Letl andmbe two distinct vertices of G(U5)with{l, m}^⊥= ∅. Then {l, m}^⊥⊥=L(l, m).

Proof Let l and m be two distinct vertices in G(U5) such that {l, m}^⊥ is not empty. Due to Remark 2.5the vertices l and m have distance two in G(U5)and it follows that the graph {l, m}^⊥ is the single vertex l, m^π. Thus we obtain the equalities {l, m}^⊥⊥=({l, m}^⊥)^⊥=

z∈{m,l}^⊥z^⊥=(l, m^π)^⊥=L((l, m^π)^π)=

L(l, m).

It will prove useful to know whether two hyperbolic lines intersect in the projective space (i.e., the two hyperbolic lines span a plane in the projective space) or not (i.e., they span a four-dimensional space in the projective space). Lemmas2.1to2.3show that in order to distinguish the above two cases, we have to study vertices of G(U5) at mutual distance three or four more thoroughly.

Lemma 2.8 Ifl and mare two non-intersecting hyperbolic lines ofU5 such that l^π∩mis a pointp, then in the graph G(U5)the number of different paths of length three between l and mis at most q². On the other hand, this number is at least q²−q−1 ifpis a singular point and at leastq²−2q−1 ifpis a non-degenerate point.

Proof Lethbe an arbitrary neighbor oflin G(U5), i.e.,h⊂l^π. By Lemma2.1there exists a common neighbork ofhandm(and, thus, a path of length three fromlto mthroughh) if and only ifh, m is a non-degenerate plane. In fact, ifh, mis a non-degenerate plane, thenkis uniquely determined ash, m^π. Therefore it suffices to study all non-degenerate planesE withm⊆E⊆ m, l^π such thatE∩l^π is a non-degenerate line.

Let us first deduce the upper bound in the statement of the lemma from the ob- servations made in the above paragraph. Ifp=l^π ∩mis a singular point, thenq²

different hyperbolic lines and exactly one singular line of the orthogonal spacel^πrun through the pointpby Table1, entryn+l=2,n=1,r=1 andr=0. Ifp=l^π∩m is a non-degenerate point, thenq²−q different hyperbolic lines andq+1 distinct singular lines are incident to the pointpin the subspacel^π. Hence there are at most q²paths fromltom.

Next we want to establish the respective lower bounds. Regard the four- dimensional subspaceW= m, l^π, which is of rank three or four. In the subspaceW the hyperbolic linemis contained inq²+1 different planesE_iby LemmaA.1. Each plane E_i of W intersects the non-degenerate plane l^π in a line, by the dimension formula and becausem⊆l^π. Sincep∈l^π is incident to each planeEi, every line h_i=E_i∩l^πruns throughp. Moreover the linesh_i are mutually distinct, because the identityh_i=h_jimpliesE_i= h_i, m = h_j, m =E_j.

If the subspaceWis of rank four, then the hyperbolic linemlies onq²−qdifferent non-degenerate planes E_i^m by Table 1. Therefore we obtain q²−q different lines E_i^m∩l^π=h^m_i incident to the pointpin the subspacel^π. At leastq²−q−1 lines of theq²−qlinesh^m_i are hyperbolic lines, ifpis a singular point, due to Table1. On the other hand, ifpis a non-degenerate point, then at leastq²−q−(q+1)=q²−2q−1 lines of theq²−qlinesh^m_i are hyperbolic lines by Table1again. Alternatively ifWis of rank three, then exactlyq²different non-degenerate planesE^m_i are incident to the hyperbolic linemby Table1. Hence we obtainq²different linesE_i^m∩l^π=h^m_i in the non-degenerate planel^π containing the pointp. By Table1, at leastq²−1 of these q²linesh^m_i are non-degenerate, ifpis a singular point and at leastq²−(q+1)= q²−q−1 lines are non-degenerate, ifpis a non-degenerate point.

Lemma 2.9 Iflandmare two non-intersecting hyperbolic lines ofU5which are at distance four in the graph G(U5), then there are at mostq⁴different paths of length four fromltom.

Proof By Lemma2.3we have dim(l, m)=4 withl^π∩m= {0}. A neighborhof lin G(U5)is at distance three from the vertexmif and only if dim(h, m)=4 and dim(h∩m^π)=1 by Lemma2.2. Thushis a hyperbolic line inl^π passing through the pointx:= l, m^π. If the one-dimensional subspacexis singular, thenl^πcontains q² different hyperbolic linesh^l_i incident withx by Table1. Ifx is non-degenerate point, then, by Table1again, there areq²−q hyperbolic lines throughx inl^π. By Lemma2.2the verticesmandh^l_iare at distance three in G(U5). Combining the above numbers with Lemma2.8we obtain at mostq²·q²=q⁴paths fromltom.

Lemma 2.10 Iflandmare two intersecting hyperbolic lines spanning a degenerate plane, then the hyperbolic line graph G(U5)contains at leastq⁶−3q⁵+2q⁴−q² different paths of length four fromltom.

Proof Ifhis a neighbor ofl, then the vertexhis at distance three frommin G(U5) if and only if dim(h, m)=4 and dim(h∩m^π)=1 inU5by Lemma2.2. Conse- quentlyhis a hyperbolic line in the polar space l^π ofl such that l, m^π ∩h is a one-dimensional subspace. Since the rank one linel, m^π contains exactly one sin- gular pointxandq²non-degenerate pointsp_i(see Table1), the non-degenerate plane

l^πcontainsq²hyperbolic linesh_xiincident to the pointx. Each non-degenerate point p_iadmitsq²−qincident hyperbolic linesh_pi,j ofl^π. Certainly, all those hyperbolic linesh_xiandh_pi,jare pairwise distinct as otherwise they would coincide with the line l, m^π. By Lemma2.8we have at leastq²−q−1 different paths of length three in G(U5) between each vertexhx_i and the vertex mand not less than q²−2q −1 different paths of length three in G(U5) from each vertex hp_i,j to the vertex m, again. Accordingly we obtain at leastq²(q²−q−1)+q²(q²−q)(q²−2q−1)= q⁶−3q⁵+2q⁴−q²different paths of length four fromlandmin the graph G(U5).

Lemma 2.11 Two different verticeslandmof distance four in G(U5)intersect in a point in the vector spaceU5if and only if the number of different paths of length four betweenlandmin G(U5)is greater thanq⁴.

Proof Sinceq ≥3, we haveq⁶−3q⁵+2q⁴−q²> q⁴, so the claim follows from

Lemma2.9and Lemma2.10.

Lemma 2.12 Two distinct verticeslandmof the hyperbolic line graph G(U5)in- tersect in a common point inU₅if and only if either

• the subgraph{l, m}^⊥is not empty, or

• the vertices l and m have distance four in G(U5) and there are more than q⁴ different paths of length four fromltom.

Proof This is an immediate consequence of Lemma2.1to Lemma2.3together with

statements of Lemma2.11, Lemma2.7and Remark2.5.

In the next step we want to recover all points of the spaceU5as pencils of hyper- bolic lines. Therefore we need a construction to check in the graph G(U5)whether three distinct lines ofU5intersect in one point or not. Therefore take the following characterisation: three different hyperbolic linesk1, k2andk3ofU5intersect in one point if we can find a hyperbolic linesinU₅such that

• the planes, k_iis non-degenerate for 1≤i≤3 ands=k_i,

• s, k1, k2is a four-dimensional space inU5.

The same statement in terms of graph language is that three different verticesk1, k2andk3of G(U5)intersect in one point if we can find a vertexsof G(U5)with the following properties:

• the induced subgraph{s, k_i}^⊥is not empty fori∈ {1,2,3}ands=k_i,

• {s, k₁, k₂}^⊥is the empty graph.

To verify the claim that every one-dimensional subspace of theU5can be detected by three pairwise intersecting distinct verticesk1, k2andk3of G(U5)as stated above, we have to show that we can find a vertexsin G(U5)such that{s, k1, k2}^⊥= ∅and {s, ki}^⊥= ∅fori=1,2,3 ands=ki. This will be proved in the next lemma.

Lemma 2.13 Letk₁,k₂andk₃be three distinct hyperbolic lines ofU₅, which inter- sect in a one-dimensional subspacep. Then the unitary polar spaceU₅contains a

hyperbolic linelwith the properties thatk₁, k₂, lis a four-dimensional space and thatl, k_iis a non-degenerate plane fori=1,2,3 andl=k_i.

Proof In the unitary polar spaceU₅every hyperbolic linekis incident toq⁴−q³+q² different non-degenerate planesE_j^k and toq³+1 different singular planesS_i^k inU₅ as follows from LemmaA.1. For the hyperbolic linek₁we obtain theq⁴−q³+q² different non-degenerate planesE_j^k1and consider in each of these the hyperbolic lines h^E

k1 j

r containing the pointp. Ifpis a singular point then in each planeE^k_j¹ there are q²−1 different hyperbolic linesh^E

k1 j

r incident topand different from the hyperbolic linek₁by Table1. Alternatively, ifpis a non-degenerate point, then in each plane E_j^k1 we find q²−q−1 different hyperbolic lines h^E

k1 j

r passing through the point p, which are different from the hyperbolic linek₁, using Table1again. Recall that E_i^k1∩E_j^k1=k1if and only if the planes are different, which leads to the fact that a hyperbolic linesh^E

k1 j

r is not incident to the non-degenerate planeE_i^k1 if and only if i=j. Therefore, ifp is non-degenerate point, then in the unitary vector spaceU5

there are(q⁴−q³+q²−1)(q²−q−1)=q⁶−2q⁵+q⁴−2q²+q+1 different hyperbolic linesh^E

k1 j

r incident to the pointp, different from the hyperbolic linek₁and not lines of the planek1, k2. Alternatively, if pis a singular point, then the polar spaceU5contains(q⁴−q³+q²−1)(q²−1)=q⁶−q⁵+q³−2q²−1 different hyperbolic linesh^E

k1 j

r with the same properties as above.

Next we consider the singular planesS_j^k2 andS_i^k3 inU5. The pointpis not con- tained in the radicals of the planesS_j^ki, because the hyperbolic linesk₂andk₃ are passing through the pointp, and thus in each rank two plane S_j^ki are q²−1 dif- ferent hyperbolic linesl^S

ki j

r incident top and different from the hyperbolic linek_i. Therefore in the planesSk₂,j andSk₃,j are together at most 2(q²−1)(q³+1)= 2q⁵−2q³+2q²−2 different hyperbolic linesl^S

ki j

r with the assumed properties.

Ifpis a non-degenerate point, thenq⁶−2q⁵+q⁴−2q²+q+1−(2q⁵−2q³+ 2q²−2)=q⁶−4q⁵+q⁴+2q³−4q²+3>0. This implies thatU5contains a hyper- bolic lineswhich intersects each hyperbolic lineki fori=1,2,3 and such that the planess, k_iare non-degenerate fori=1,2,3 ands, k₁, k₂is a four-dimensional space.In the other case, ifpis singular, thenq⁶−q⁵+q³−2q²−1−(2q⁵−2q³+ 2q²−2)=q⁶−3q⁵+3q³−4q²+1>0, and such a hyperbolic line s exists as

well.

Definition 2.14 Letbe graph isomorphic to G(U5). Two different verticeskandl ofare defined to intersect if

• either the induced subgraph{k, l}^⊥is not empty, or

• the verticeskandl have distance four inand the number of different paths of length four betweenlandmin G(U5)is greater thanq⁴.

Three distinct pairwise intersecting verticesk₁, k₂andk₃ofare defined to intersect in one point if there is a vertexsofwith the following properties:

• the induced subgraph{s, k_i}^⊥is not empty fori∈ {1,2,3}ands=k_i,

• {s, k₁, k₂}^⊥is the empty graph.

An interior point of the graphis a maximal setpof distinct pairwise intersect- ing vertices ofsuch that any three elements ofpintersect in one point. We denote the set of all interior points ofbyI. Moreover, an interior line of the graphis a vertex of the graph. The set of all interior lines ofis denoted byL.

The discussions in this section imply the following result.

Proposition 2.15 Let be a graph isomorphic to G(U5). Then the geometry (I,L,⊃)is isomorphic to the geometry on arbitrary one-dimensional subspaces and non-degenerate two-dimensional subspaces of the unitary polar spaceU5.

3 The hyperbolic line graph ofUn,n≥6

Letq be a prime power, letn≥6, and let U_n be ann-dimensional non-degenerate unitary vector space overF_q² with polarityπ. Let G(Un)be the graph with the set of non-degenerate two-dimensional subspaces ofU_n as set of vertices in which two vertices l andm are adjacent if and only if l⊂m^π. In anology to the preceding section, the aim of this section is to reconstruct the unitary vector spaceU_nfrom the hyperbolic line graph G(Un).

Proposition 3.1 Letn≥8. Then G(Un)is a connected graph of diameter two.

Proof Letlandkbe two distinct vertices of the graph G(Un). The spaceH= l, k has dimension three or four. Since it contains the hyperbolic lineslandm, the rank of His at least two. Hence the radical ofHhas dimension at most two. The spaceH^π has dimension at least four and rank at least two, since rad(H^π)=rad(H ). Therefore H^π= k, l^π=k^π∩l^πcontains a hyperbolic lineh, so that the distance between the verticeslandkis at most two. As G(Un)obviously admits non-adjacent vertices, the

diameter of G(Un)is two.

Proposition 3.2 The graphs G(U6)and G(U7)are connected of diameter three.

Proof We first study the graph G(U6). Let l andmbe distinct vertices of G(U6).

ThenP = l, mis a three- or four-dimensional subspace ofU₆.

Assume first thatP = l, mis a plane. Then the planesP andP^π have rank two or three, because the hyperbolic linel andmare proper subspaces ofP. Therefore the planeP^π contains a hyperbolic linehand thus the verticeslandmhave distance two in G(U6).

IfP = l, mis a four-dimensional subspace ofU₆, thenP is of rank two, three, or four. In the case that P is a non-degenerate subspace, then of course P^π is a

hyperbolic line and the verticesl andmhave distance two. Finally, we assume that the four-dimensional spaceP is a singular subspace ofU₆. We fix a pointx in the radical ofP. Thenxis incident to the perpendicular spacel^πof the hyperbolic linel, which is a non-degenerate four-dimensional subspace ofU₆. We choose a hyperbolic linehinl^π passing through the pointxinl^πand certainly the vertexhis adjacent to lin the hyperbolic line graph G(U6). Ifh, mis a plane, then there exists a common neighbor ofh andmby the above, yielding a path of length three froml tomin G(U6). Hence we can assume that subspace ofh, mis of dimension four. The rank of this space is four as well. Indeed, the Gram matrix ofm, hisG=

_{0 1 0α}

1 0 0β 0 0 0δ α β δ γ

with respect to a basisv₁^m,v₂^m,x^h,v₂^hofm, hsuch that the pair of vectorsv₁^m,v₂^mis a hyperbolic pair of is linem, the vectorx^his some non-trivial vector of the pointxand v₂^his a non-trivial vector of a non-degenerate point of the lineh. So(x^h, v₂^h)=δ=0.

But that implies that the Gram matrix has determinantδδ=0 and, hence,m, his of dimension four. By the abovehandmhave distance two, so the verticeslandmare at mutual distance at most three in G(U6).

We now turn our attention to the graph G(U7). Letlandmbe distinct vertices of G(U7). Since the subspacel, mhas dimension at most four and rank at least two, there exists a non-degenerate six-dimensional subspaceW ofU₇containinglandm.

By the above, the verticeslandmhave distance at most three in the hyperbolic line graph G(W ), which is a subgraph of G(U7). Whence the diameter of G(U7)is at most three.

In order to establish that the diameter of the graphs G(U6)and G(U7)is three, we have to find vertices that are not at mutual distance one or two. Choose a four- dimensional rank two subspaceHofU₆respectively ofU₇. By Table1the subspace Hcontainsq⁸hyperbolic lines and any point of this space is incident toq⁴+q²+1 different lines. Sinceq⁸≥(q²+1)·(q⁴+q²+1)=q⁶+q⁴+q²+1 we find two non-intersecting hyperbolic linesl andmofU₆resp.U₇spanning the subspaceH. The polel, m^π =H^π has dimension two or three, respectively, and rank zero or one, respectively. Hence G(U6)resp. G(U7)do not contain a common neighbor ofl andm. Therefore the diameter of G(Un)with 6≤n≤7 is three.

The next proposition describes two key properties of the hyperbolic line graph G(Un)which will turn out to characterise G(Un)forn≥9 (cf. Theorem1).

Proposition 3.3 Letn≥5. The hyperbolic line graph G(Un)is connected, unless (n, q)=(5,2), and locally G(Un−2).

Proof See Propositions2.4,3.1,3.2. The local property is obvious.

Definition 3.4 LetW be a subspace ofUn. The set of all hyperbolic lines ofW is denoted by L(W ).

Lemma 3.5 Letn≥6 and letl andmbe two distinct vertices of the graph G(Un) such that{l, m}^⊥= ∅. Then{l, m}^⊥⊥=L(l, m).

Proof Since {l, m}^⊥⊥=({l, m}^⊥)^⊥=

z∈{l,m}^⊥z^⊥=

z∈{l,m}^⊥L(z^π), obviously L(l, m)⊆ {l, m}^⊥⊥.

Conversely, letk be a hyperbolic line ofU_n not incident to the subspacel, m. Then, of course, l, m^π ⊆k^π. The statement is proved, if we can find a hy- perbolic line h ⊆ l, m^π, which is not incident to the perpendicular space k^π. From the assumption that the induced subgraph {l, m}^⊥ is not empty it follows that rad(l, m^π)is properly contained in the subspacel, m^π. We claim that the unitary space U_n contains some point y in the set l, m^π\(k^π ∪rad(l, m^π)).

If rad(l, m^π)⊆k^π then by De Morgan’s laws l, m^π\(k^π ∪rad(l, m^π))= l, m^π\k^π ∩ l, m^π\rad(l, m^π)= l, m^π\rad(l, m^π) and, of course, the set l, m^π\rad(l, m^π)contains some pointy. On the other hand, if rad(l, m^π)⊆k^π, then rad(l, m^π)∪k^π is not a subspace of the vector spaceUnandl, m^π is nei- ther a subspace of rad(l, m^π) nor a subspace of k^π, thus the setl, m^π\(k^π ∪ rad(l, m^π))contains a pointy.

An arbitrary two-dimensional subspacegofUn containing the pointyintersects the setk^π∪rad(l, m^π)in at most two points by the fact that dim(k^π∩g)as well as dim(rad(l, m^π)∩g)is at most one. Therefore, we choose a hyperbolic line passing throughyinl, m^πand find a singular pointx∈ l, m^π\(k^π∪rad(l, m^π)). Using x⊆rad(l, m^π) we obtain a hyperbolic lineh in l, m^π incident to the pointx which is not contained in the subspacek^π. The lemma is now proved.

A similar conclusion can be shown for three different vertices in the graph G(Un).

Lemma 3.6 Letn≥6 andk, landmbe three distinct vertices in G(Un). Suppose the hyperbolic linesk,l,mintersect in a common point ofUnand satisfy{k, l, m}^⊥= ∅. Then L(k, l, m)= {k, l, m}^⊥⊥.

Proof By assumption the subspace spanned by the hyperbolic linesk, l, mis of dimension three or four. Denote the common intersection of the three hyperbolic lines byp.

Suppose l, k, m is a plane. Then m is a hyperbolic line of l, k and, thus, l, k, m = l, k. Using Lemma 3.5 we obtain that L(l, k, m) =L(l, k)= {l, k}^⊥⊥.

Ifl, k, mis a four-dimensional subspace, we want to find a hyperbolic linehsuch thatl, k, m = l, h. In casel, k, m has rank four, we chooseh=l^π∩ l, k, m. If the subspacel, k, mhas rank two, take ashan arbitrary line in the complement of bothl and rad(l, k, m). Indeed, we can find such a lineh inl, k, m by the fact that at most 2q⁶+4q⁴+4q²+2 of theq⁸+q⁶+2q⁴+q²+1 different lines ofl, k, m intersectlor rad(l, k, m). Certainly,his a hyperbolic line since every complement of the radical ofl, k, mis non-degenerate. Finally, ifl, k, mhas rank three, then consider the rank two planeP = k,rad(l, k, m). Since the hyperbolic lineskandl are distinct and intersect in a common point we have dim(l∩P )=1.

Moreover the radical ofP coincides with the point rad(l, k, m). In the planeP we choose the linehin the complement of both rad(l, k, m)andl∩P. Certainly the subspacehis non-degenerate. It follows from the construction thatl, h = l, P = l, k,rad(l, k, m) = k, l, mand, by Lemma 3.5, that L(l, k, m)=L(l, h)= {l, h}^⊥⊥.