DOI 10.1515 / ADVGEOM.2009.033 de Gruyter 2010
Local recognition of the line graph of an anisotropic vector space
Kristina Altmann and Ralf Gramlich∗
(Communicated by A. Pasini)
Abstract. Letn ≥ 7 and let V be an(n+2)-dimensional vector space endowed with an anisotropic sesquilinear form. In this article we characterise the graph whose vertices are the two- dimensional subspaces ofV and in which two vertices are adjacent if and only if the corresponding two-dimensional spaces are perpendicular with respect to the sesquilinear form.
1 Introduction
A central problem in synthetic geometry is the characterisation of graphs and geometries.
The local recognition of locally homogeneous graphs forms one category of such charac- terisations, which works as follows. LetΣbe a graph. A graphΓis calledlocallyΣif for each vertexxofΓthe graphΓxis isomorphic toΣ, whereΓxis the induced subgraph of Γon the set of vertices adjacent tox. It is a natural question to ask for all isomorphism types of connected graphs which are locally some graphΣ. This classification question is called the local recognition problem for graphs that are locallyΣ, which can be found in great quantities in the literature. One of the earliest and most influential is [3].
In the present article we focus on the line graph of ann-dimensional vector space over some fieldFendowed with an anisotropic sesquilinear form. Letn∈N, letV =Vnbe ann-dimensional vector space overFand let(·, ·)be a scalar product onV ×V. For a subspaceU ⊆ V the polar ofU isUπ = {x∈ V : (x, u) = 0 for allu ∈U}. The line graphS(Vn)of the vector spaceVnis the graph on the two-dimensional subspaces of Vn, the lines ofVn, where two distinct lineslandkofVnare adjacent, in symbolsk⊥l, if and only ifl ⊆kπor, equivalently, ifk⊆lπ. For a vertexxofS(Vn)the local graph S(Vn)x =: x⊥ is the subgraph induced byS(Vn)on the set of vertices{y ∈ S(Vn) | x⊥y}, the neighbours ofxin the graphS(Vn). For a set of verticesXofS(Vn)the graph X⊥is defined asT
x∈Xx⊥.
∗The second author gratefully acknowledges a Heisenberg fellowship by the Deutsche Forschungsgemein- schaft.
The main result of this paper is the following local recognition theorem. Note that because of the size of the involved dimensions the fieldFis necessarily infinite.
Main Theorem. Let n ≥ 7 and letΓbe a connected locallyS(Vn)graph. ThenΓis isomorphic toS(Vn+2).
Forn=6 there exist counterexamples to the statement of the Main Theorem arising from anisotropic twisted forms of the semisimple algebraic groupE6(F). In the first au- thor’s PhD thesis [1] the situationn=6 is dealt with for the fieldF=C. Theorem 4.1.2 of [1] is an analogue of our main theorem for that case under a mild extra assumption.
We would like to stress that the casen =6 is much more difficult than the casen≥ 7 presented here. The proof given in [1] is based on a construction of a large automorphism group of the graphΓ, the existence of a subgraph ofΓisomorphic to the reflection graph of the Coxeter groups of typeA7, respectivelyE6(this makes serious use of [3]), and a recognition of the automorphism group via Lie theory and Phan theory (cf. [7]).
In Section 2 we discuss group-theoretic applications of our main result. Sections 3 and 4 provide some basic facts about the reconstruction ofP(Vn)fromS(Vn)and how the induced subgraphsx⊥,x∈ S(Vn), relate to codimension two subspaces ofP(Vn).
Finally, in Section 5 we prove the main result. The strategy used in that section is the same as in [4], [6].
2 Group-theoretic consequences
Assume that the automorphism group ofS(Vn) acts transitively on the set of ordered triangles ofS(Vn), as is the case forF=Cand the standard scalar product. (We refer the reader to [5] for a detailed investigation of fields admitting such a transitivity.) Then our main result implies the following group-theoretic recognition theorem.
Local Recognition Theorem. Letn≥7, letFbe a field admitting an anisotropic sesqui- linear form such that the automorphism group ofS(Vn)acts transitively on the set of or- dered triangles ofS(Vn), and letGbe a group containing an involutionxand a subgroup KCG(x)such that
1. K∼= SUn(F);
2. CG(K)contains a subgroupX ∼= SU2(F)withx=Z(X);
3. there exists an involutiong∈Gsuch thatY :=gXgis contained inK;
4. ifV is a natural module forK, then the commutator[Y, V] = {yv−v∈V |y∈Y, v∈V}hasF-dimension two;
5. G=hK, gKgi; moreover, there existsz ∈K∩gKgwhich is agKg-conjugate ofx and aK-conjugate ofgxg.
ThenG/Z(G)∼= PSUn+2(F).
Proof. Definey:=gxgandJ :=gKg. By 4. the groupXis a fundamental subgroup of Jwith centrexandY is a fundamental subgroup ofKwith centrey. Since the groupKis isomorphic toSUn(F)and sinceyandzare centres of fundamental subgroups ofKwith
[y, z] =1, we find the elementsyandzto be conjugate inKby an involution. Similarly xandzare conjugate inJ by an involution. Therefore the normaliser of{x, y, z}with respect to the conjugation action of the groupG equals the group Sym3. Hence the normaliser of{x, y}is the groupSym2.
Consider the graphΓon all conjugates ofxinGin which two verticesa,bare adjacent if there exists an elementh∈ Gsuch that(hxh−1, hyh−1) = (a, b). SinceGinduces the action ofSym3 on{x, y, z}, this definition of adjacency is completely symmetric, and we have defined an undirected graphΓin which the elementsx,y,zform a triangle.
The stabiliser of{x, y}permutesxandy and therefore interchangesCG(x)K and CG(y)J. Hence the stabiliser ofxtogether with the stabiliser of{x, y}generatesG, asG=hJ, Ki ≤ hCG(x), CG(y)i. Consequently, the graphΓis connected.
Moreover,Γis a locally isomorphic to a line graph by construction. To prove this, it is enough to show that any triangle inΓis a conjugate of(x, y, z). Let(a, b, c)be a triangle and leth ∈ Gwith(hxh−1, hyh−1) = (a, b). Notice that the elementsb = hyh−1, d:=hzh−1ofhKh−1commute. The edges(a, b)and(a, c)are both conjugate inGto (x, y), so that they are conjugate to each other by an element ofCG(a)hKh−1. Since b ∈ hKh−1, we havec ∈ hKh−1 as well. We have proved thatb,c,dare centres of fundamental subgroups ofhKh−1. Since[b, c] = 1 = [b, d], the elementsc anddare conjugate by an element ofChKh−1(b). Therefore(a, b, c)and(x, y, z)are conjugate in G. Hence by the Main TheoremΓis isomorphic toS(Vn+2)and the claim follows. 2 There is an analogue of the above theorem for finite fields. See [2] for the casen≥7 and [8, Section 7.2.1] for the casen=6.
3 Reconstruction of projective space
In this section we collect some basic properties of the line graphs of the unitary vector spacesVnforn≥5, such as their diameters, cf. Proposition 3.2 and Proposition 3.3. We will also construct fromS(Vn)a point-line geometryG= (P,L,⊃)which is isomorphic to the projective space given by the points and lines ofVn.
Since the behaviours ofV5 andVn, n ≥ 6, are slightly different, we will consider these graphs separately where appropriate.
Lemma 3.1. Letlandmbe distinct lines ofV5. Then the distance oflandminS(V5)is
• one if and only ifl⊆mπorm⊆lπ;
• two if and only ifhl, miis a three-dimensional subspace ofV5;
• three if and only ifl andmdo not intersect, are not perpendicular and satisfy that the intersectionlπ∩mis one-dimensional;
• four if and only iflandmdo not intersect and satisfy that the intersectionlπ∩mis trivial.
Proof. These claims are straightforward to prove. We leave this as an exercise to the
reader. 2
Proposition 3.2. The graphS(V5)is connected of diameter four. Moreover, the graph S(V5)is locallyS(V3).
Proof. This is immediate from Lemma 3.1, because any pair of distinct linesl andm
belongs to one of the four cases considered in 3.1. 2
Proposition 3.3. Letn ≥6. Any two elements ofS(Vn)have a common neighbour. In particular, the graphS(Vn)is connected of diameter two.
Proof. Let l and m be two different two-dimensional subspaces of Vn, n ≥ 6. The dimension formula impliesdim(lπ∩mπ) = dim(lπ) + dim(mπ)−dim(hlπ, mπi) ≥ n−4≥2. Hence, inVnthere exists a two-dimensional subspaceh⊆lπ∩mπ. 2 Proposition 3.4. Forn≥6, the graphS(Vn)is locallyS(Vn−2).
Proof. For each vertexlofS(Vn), the spacelπ ⊆Vnis an(n−2)-dimensional vector space endowed with an anisotropic sesquilinear form(·,·)|lπ×lπ. The claim follows. 2 Definition 3.5. LetUbe a subspace of the unitary vector spaceVn. The set of all lines in Uis denoted byL(U).
Lemma 3.6. Letlandmbe vertices ofS(Vn)with{l, m}⊥ 6=∅. Then the vertex set of the graph{l, m}⊥⊥equals the line setL(hl, mi).
Proof. Since the vertex set of z⊥ equalsL(zπ), we have{l, m}⊥⊥ = ({l, m}⊥)⊥ = T
z∈{l,m}⊥z⊥=T
z∈{l,m}⊥L(zπ). Moreover, for everyz∈ {l, m}⊥we haveL(hl, mi)
⊆L(zπ). Therefore the setL(hl, mi)is a subset of the vertex set of{l, m}⊥⊥.
Conversely, letpbe a one-dimensional subspace not contained inhl, mi. This implies hl, miπ 6⊆pπ andhl, mi ⊆pπ. Lettbe a one-dimensional subspace ofhl, miπwhich is not contained inpπ.
By assumption 3 ≤ dim(hl, mi) ≤ 4 if n ≥ 6 and dim(hl, mi) = 3 if n = 5, thus we havedim(hl, miπ∩pπ) = dim(hl, miπ) + dim(pπ)−dim(hpπ,hl, miπi) = dim(hl, miπ)−1≥1. Hence there exists a one-dimensional subspaces⊆ hl, miπ∩pπ. The two-dimensional subspaceh=hs, tiis a vertex of the graph{l, m}⊥, so{l, m}⊥⊥⊆ h⊥ =L(hπ)⊆L(pπ). Thus{l, m}⊥⊥cannot contain two-dimensional subspaces con- tainingpas no line ofL(pπ)contains the pointp.
The claim follows becausepis an arbitrary one-dimensional subspace not contained
inhl, mi. 2
A similar statement can be proved about three different vertices under some additional assumptions.
Lemma 3.7. Let k, l and m be three distinct vertices of S(Vn). Suppose k∩l ∩m contains a one-dimensional subspace of Vn and suppose that {k, l, m}⊥ 6= ∅. Then L(hk, l, mi) ={k, l, m}⊥⊥.
Proof. Since the linesk,landmintersect in a common point inVn, the dimension of hk, l, miequals three or four.
Ifhk, l, miis a three-dimensional subspace, thenmis contained inhk, liandhk, l, mi
= hk, li. From Lemma 3.6 it follows that L(hk, l, mi) = L(hk, li) = {k, l}⊥⊥ = {k, l, m}⊥⊥as∅ 6={k, l, m}⊥⊆ {k, l}⊥.
In case dim(hk, l, mi) = 4 pick a line sinhk, l, mi. Forn ≥ 6 we consider the two-dimensional orthogonal subspacet =sπ∩ hk, l, mi. By Lemma 3.6 again we have L(hk, l, mi) =L(hs, ti) ={s, t}⊥⊥. Hence the claim follows by
{k, l, m}⊥⊥= ({k, l, m}⊥)⊥= \
z∈{k,l,m}⊥
z⊥ = \
z∈L(hk,l,miπ)
z⊥= \
z∈L(hs,tiπ)
z⊥
= \
z∈{s,t}⊥
z⊥= ({s, t}⊥)⊥={s, t}⊥⊥.
Ifn = 5 anddim(hk, l, mi) = 4, then the orthogonal subspacehk, l, miπis of di- mension one. This implies that{k, l, m}⊥is the empty graph, so we are done. 2 Remark 3.8. Supposelandmare distinct lines which intersect in a common point inV5. Thendim(hl, mi) =3, which implies that{l, m}⊥6=∅using Lemma 3.1. Moreover by Lemma 3.6 the vertex set of the induced subgraph{l, m}⊥⊥inS(V5)equals the line set L(hl, mi)and therefore the double perp{l, m}⊥⊥ is minimal with respect to inclusion (i.e. for any distinct verticess1, s2 in the double perp {l, m}⊥⊥ we have the equality {s1, s2}⊥⊥={l, m}⊥⊥).
Lemma 3.9. Two different lineslandmofVn, forn≥6, intersect in a common point if and only if the double perp{l, m}⊥⊥inS(Vn)is minimal with respect to inclusion(i.e.
for any distinct verticess1, s2∈ {l, m}⊥⊥we have{s1, s2}⊥⊥={l, m}⊥⊥).
Proof. Certainly, iflandmare two different intersecting lines inVn, thendim(hl, mi) = 3. For any two distinct elementss1, s2∈ {l, m}⊥⊥=L(hl, mi), the subspacehs1, s2ihas dimension three implying thaths1, s2i=hl, mi. By Lemma 3.6,L(hl, mi) ={l, m}⊥⊥
andL(hs1, s2i) ={s1, s2}⊥⊥, thus{s1, s2}⊥⊥=L(hs1, s2i) =L(hl, mi) ={l, m}⊥⊥. Supposedim(hl, mi) = 4 thenl andm are two non-intersecting lines inVn. Let h∈L(hl, mi)such thatdim(h∩l) =1. Then the span ofhandlis of dimension three and{h, l}⊥⊥=L(hh, li)$L(hl, mi) ={l, m}⊥⊥by Lemma 3.6, which shows that the double perphl, mi⊥⊥is not minimal with respect to inclusion. 2 As mentioned in the beginning of this section, our strategy is to reconstruct the projec- tive space on the one-dimensional and two-dimensional subspaces ofVnfrom the graph S(Vn). The two-dimensional subspaces are easily recovered; we can simply take the vertex set ofS(Vn)as the line set of the geometry. In a projective space of sufficient dimension each point is uniquely determined by the set of lines incident with it. We call such a set a line pencil. A successful description of line pencils essentially depends on the understanding when three lines intersect in a common point. Again we have to distinguish the casesn=5 andn≥6.
Three different pairwise intersecting linesk1,k2,k3of the vector spaceV5intersect in one point if we can find a linesinV5such that
• the linesintersects the lineki, ifs6=ki, for 1≤i≤3,
• hs, k1, k2iis a four-dimensional subspace inV5.
This observation motivates the following definition in terms of double perps.
Definition 3.10. LetΓbe a graph isomorphic toS(V5). Two different verticeskandlof Γare defined tointersect if the induced subgraph{k, l}⊥ is not empty. Three different pairwise intersecting verticesk1, k2andk3ofΓare defined tointersect in one pointif the graphΓcontains a vertexswith the following properties:
• the vertexsintersects each vertexki, ifs6=ki, for 1≤i≤3,
• the graph{k1, k2, s}⊥is empty.
Aninterior point of the graphΓ is a maximal set S of different pairwise intersecting vertices ofΓsuch that any three elements ofSintersect in one point. These maximal sets exist by Zorn’s lemma. Moreover, it is easily seen that every point ofV5can in fact be realised by three pairwise intersecting lines. We denote the set of all interior points ofΓ byP. Moreover, aninterior lineof the graphΓis a vertex of the graphΓ. The set of all interior lines ofΓis denoted byL.
Forn≥6 the geometric properties of three different pairwise intersecting lines have a slightly different meaning in terms of double perps.
Definition 3.11. Letn ≥ 6 andΓ be graph isomorphic to the line graphS(Vn). Two different vertices k andl of Γ are defined to intersect if the double perp {k, l}⊥⊥ is minimal inΓwith respect to inclusion (among double perps of distinct vertices). Three distinct pairwise intersecting verticesk1, k2 andk3 of Γare defined tointersect in one pointif there is a vertexsinΓsatisfying the following conditions:
• the vertexsintersectski, ifs6=ki, for 1≤i≤3,
• the graph{k1, k2, s}⊥is non-empty and{k1, k2}⊥⊥=L(hk1, k2i)$L(hk1, k1, si)
={k1, k2, s}⊥⊥.
Aninterior pointof the graphΓis a maximal setS of distinct pairwise intersecting vertices ofΓsuch that any three elements ofSintersect in one point. We denote the set of all interior points ofΓbyP. Moreover, aninterior lineof the graphΓis a vertex of the graphΓ. The set of all interior lines ofΓis denoted byL.
By construction we obtain the following:
Proposition 3.12. Letn≥ 5and letΓbe a graph isomorphic toS(Vn). The geometry G= (P,L,⊃)on the interior points and interior lines ofΓis isomorphic to the projective space of the vector spaceVn.
For a graphΓ∼=S(Vn)we call the spaceG = (P,L)theinterior spaceonΓ. Note that this interior space carries a canonical polarityπgiven byl 7→ l⊥. Hence we also obtain information about the automorphism group ofS(Vn).
Corollary 3.13. Let n ≥ 5. The automorphism group ofS(Vn)is isomorphic to the automorphism group of the projective space of(Vn,(·, ·)).
4 Subspaces arising from local graphs
Letn ≥ 7 and let Γ ∼= S(Vn). Proposition 3.12 allows us to reconstruct a projective spaceG= (P,L)isomorphic toP(Vn)fromΓ. By Proposition 3.4 the graphΓis locally S(Vn−2). Hence for a vertex x of Γ Proposition 3.12 also allows us to reconstruct a projective spaceGx= (Px,Lx)isomorphic toP(Vn−2)from the induced subgraphx⊥. The purpose of this section is to show that there is a canonical way to consider interior spaceGxas a subspace ofG. We will spell out the details for the more complicated case n=7 and leave the easier casen≥8 as an exercise to the reader.
Notation. Every object of the interior space Gx will be indexed by the vertex x. In particular, for vertices l, k, mof the subgraph x⊥ we use the notation {k, l, m}⊥x :=
{k, m, l}⊥∩x⊥and{k, l, m}⊥⊥x := ({k, l, m}⊥x)⊥x = ({k, l, m}⊥x)⊥∩x⊥.
Obviously the line setLxis properly contained in the line setL. Therefore again we only have to understand the behaviour of the interior points.
Lemma 4.1. Letpbe a point ofGand letl, m∈p∩ Lx. Then the lineslandmintersect each other inGx.
Proof. The claim follows from the second item of Lemma 3.1 once we have shown that {l, m}⊥x 6=∅.
Since l, m ∈ x⊥ and {l, m}⊥⊥ is minimal with respect to inclusion, the three- dimensional subspacehl, miis contained inxπ. Thusdim(xπ∩ hl, miπ) =2 implying
that{l, m}⊥x ={l, m, x}⊥6=∅. 2
Lemma 4.2. Letpbe a point inGand letk1, k2, k3 ∈ p ∩ Lxbe pairwise distinct ele- ments. Then the corresponding interior lines intersect in a common point ofGx.
Proof. By Definition 3.10 and Lemma 4.1 it suffices to find a vertexs∈x⊥such that
• for eachi∈ {1,2,3}, the graph{s, ki}⊥x is not empty,
• the graph{s, k1, k2}⊥x is empty.
The verticesk1, k2andk3are adjacent to the vertexx, thushk1, k2, k3i ⊆xπinG.
Ifhk1, k2, k3iis a four-dimensional subspace ofG, then one can chooses=k3. Then hk1, k2, siπ∩xπis a pointzand any linelofhk1, k2, siπintersectsxπ in the pointzor not at all. Thus{k1, k2, s}⊥x =∅.
Hence suppose hk1, k2, k3i is three-dimensional. By assumptionk1, k2 andk3 are elements of the pointp ∈ P. Since each of theki is a neighbour ofx, the pointk1∩ k2∩k3 =din fact lies inxπ. Fix a pointyinhk1, k2, k3iπ∩xπ and consider the line s = hd, yi. The linesis contained inxπ and intersects each lineki, i ∈ {1,2,3}, in the pointd. By Lemma 3.1, the graph{s, ki}⊥x is not empty. Sincey 6⊆ hk1, k2i, the
subspacehs, k1, k2ihas dimension four, and it follows, using the arguments from above,
that{s, k1, k2}⊥x is empty. 2
Proposition 4.3. Letpbe a point inGwithp∩ Lx6=∅. Thenp∩ Lxis a point ofGx. Proof. By Lemma 4.1 and Lemma 4.2, the intersectionp∩ Lxis a subset of some interior pointpxif|p∩ Lx| ≥2.
Letl ∈ p∩ Lx, thenl∩m ⊆ xπ for each elementm ∈ p. Asdim(xπ) = 5 we can find a two-dimensional subspacenproperly contained inxπsuch thatl∩n=l∩m.
Certainlynis an element ofpimplying that|p∩ Lx| ≥2.
To complete the proof of the statement, it remains to show that eachk∈pxsatisfies k∈ p. The vertexkcorresponds to a line in the subspacexπ ofGincident to the point d=l∩mfor two different elementsl, m∈p.
Furthermore, for each n ∈ p different from k, the space spanned by k and n is three-dimensional asn intersects kin the point d. Lemma 3.9 implies that {k, n}⊥⊥
is minimal with respect to inclusion, which is the first condition of Definition 3.11. To establish the second condition of this definition, letn, t∈ pbe distinct lines and differ- ent fromk. The linesk, n,t mutually intersect, by the arguments from above, so the subspacehk, n, tihas dimension three or four. Ifdim(hk, n, ti) =4, then{k, n}⊥⊥ = L(hk, ni) $ L(hk, n, ti) = {k, n, t}⊥⊥by Lemma 3.6 and Lemma 3.7. If hk, n, tiis three-dimensional, thenhk, n, ti=hk, ni. Therefore there certainly exists a linesinter- secting the linesk,n,tin the pointdsuch thathk, n, siis a four-dimensional space inG and{k, n}⊥⊥=L(hk, ni)$L(hk, n, si) ={k, n, s}⊥⊥by Lemma 3.6 and Lemma 3.7.
Hencek∈p, which proves the claim. 2
Next we show that each pointpx of the interior spaceGx on x⊥ is contained in a unique pointpof the interior spaceGofΓ.
Lemma 4.4. Letpxbe a point inGxand letk, lbe two distinct elements ofpx. Thenk andlintersect inG.
Proof. By Definition 3.11 we have to check that {k, l}⊥⊥ is minimal with respect to inclusion. By Lemma 3.9 this is the case if and only if the span ofk andl is three- dimensional subspace. By way of contradiction assume thatkandlspan a four-dimen- sional subspace. Sincek, l ∈ px, thushk, li ∩xhas dimension zero and therefore the subspacehk, l, xiis of dimension six. Hence dim(hk, l, xiπ) = 1, which means that {k, l, x}⊥ ={k, l}⊥∩x⊥=∅, a contradiction tok, l∈px, cf. Definition 3.10. 2 Lemma 4.5. Letpxbe a point inGxand letk1, k2, k3three different interior lines ofpx. Then the corresponding interior lines ofGintersect in a common point.
Proof. By Lemma 4.4, it remains to show, cf. Definition 3.11, that there is a vertexsof Γ∼=S(V7)such that
• fori ∈ {1,2,3}, ifs 6= ki, the double perp{s, ki}⊥⊥is minimal with respect to inclusion,
• {k1, k2}⊥⊥=L(hk1, k2i)$L(hk1, k2, si) ={k1, k2, s}⊥⊥.
Ask1, k2, k3 ∈ px, we find a line sinGx such that {ki, s}⊥x 6= ∅ ifs 6= ki, for eachi∈ {1,2,3}and{k1, k2, s}⊥x =∅. By Remark 3.8 and Lemma 3.6 the double perp {s, ki}⊥⊥is minimal with respect to inclusion, ifs6=ki, fori∈ {1,2,3}.
Furthermore∅ ={k1, k2, s}⊥x ${k1, k2}⊥x implying that{k1, k2}⊥⊥ =L(hk1, k2i)
={k1, k2}⊥⊥x ⊆ {k1, k2, s}⊥⊥x ⊆ {k1, k2, s}⊥⊥. However, as{k1, k2, s}⊥x is the empty graph,dim(hk1, k2, siπ ∩xπ) < 2 and thushk1, k2, siis a subspace of xπ of dimen- sion at least four. Hencehk1, k2i $ hk1, k2, si and s 6j hk1, k2i, so {k1, k2}⊥⊥ $
{k1, k2, s}⊥⊥. 2
Proposition 4.6. LetΓ∼=S(V7)and letx∈Γ. Any pointpxof the interior spaceGxon x⊥is contained in a unique pointpof the interior spaceGofΓ. The resulting embedding ofGxintoGturnsGxinto a codimension two subspace ofG.
Proof. It remains to prove the claim about the codimension. This, however, follows from
the fact thatGx∼=P(V5)andG ∼=P(V7). 2
The analogue holds for arbitraryn≥8.
Proposition 4.7. Letn≥8, letΓ ∼=S(Vn), and letx∈Γ. Any pointpxof the interior spaceGxonx⊥is contained in a unique pointpof the interior spaceGofΓ. The resulting embedding ofGxintoGturnsGxinto a codimension two subspace ofG.
Proof. Left to the reader as an exercise. 2
5 The global space
In this section we prove the main result, i.e., we prove that ifΓ is a connected graph which is locally isomorphic toS(Vn)forn ≥7, thenΓis isomorphic toS(Vn+2). For every vertexxofΓProposition 3.12 allows us to construct the interior spaceGx onx⊥ isomorphic toP(Vn). Our approach towards a proof of the main theorem is via a con- struction of aglobal geometryonΓ using the family(Gx)x∈Γ oflocal interior spaces.
This global geometry will turn out to be a projective space. Observe that so far geometric objects (points, lines, planes, etc.) only exist in the local interior spacesGx forx ∈ Γ.
For emphasis we call these objects local. Moreover, we index every local object by the vertexxwhose interior space it belongs to. One task will be to show that there exists a well-defined notion of global objects in order to define our global geometry.
Notation. In this section we use the following convention for the fonts of variables. Each vertexxof a graph is denoted in boldface. Ifxandyare adjacent vertices inΓ, then the line ofGxcorresponding to the vertexyis denoted byyx. Moreover, we will use the notationyx, if we want to emphasise that we consider the vertexyas an element of the induced subgraphx⊥rather thanΓ. So, for example, the symboly⊥x denotes the subgraph {x,y}⊥. The interior space obtained from the graphy⊥x will be denoted byGyx.
Definition 5.1. Aglobal lineofΓis a vertex of the graphΓ. The set of all global lines of Γis denoted byLΓ.
Letx,y,z,w ∈ Γsuch thatz⊥x⊥w⊥y. ThenUzx
x,wx := hzx, wxiπ = zπx ∩wπx is a subspace of the interior spaceGx and of the interior space Gxz. Proposition 3.12 impliesGxz = Gzx, because both Gxz andGzx are interior spaces of the same graph x⊥∩z⊥. Hence by Propositions 4.6 and 4.7 this spaceUzxx,wx can be considered as a subspace of the interior spaceGz, in fact ofxπz. For emphasis we denote Uzxx,wx by Uzzx,wx when considering it as a subspace ofGz. Altogether any local object contained in Uzxx,wx∩Uxxw,yw is a local object of each of the interior spacesGz,Gx,Gy,Gw.
Proposition 5.2. Letn≥7and letΓbe a connected graph which is locally isomorphic toS(Vn). ThenΓhas diameter two. Moreover, ifz⊥x⊥y⊥wis a chain of vertices inΓ, then there existsl∈ {z,y,w}⊥.
Proof. Letz⊥x⊥y⊥wbe a chain of vertices inΓ. We will distinguish between the cases thatn≥8 andn=7.
• n ≥ 8: The dimension of the intersection ofUzy
x,yx andUxy
y,wy is at least two as dim(Uzy
x,yx) = dim(Uxy
y,wy) = dim(xπy) + dim(wyπ)−dim(hxπy, wyπi)≥n−4 and bothUzyx,yx andUxyy,wy are subspaces ofxπy. Therefore we can choose an interior linelyinUzyx,yx∩Uxyy,wy. This interior linelycorresponds to a vertexlofΓand is adjacent to all ofx,y,z,w.
• n = 7: In view of the preceding paragraph it remains to deal with the case where Uzyx,yx∩Uxyy,wy =pyis an interior point ofGy, whence also of each of the interior spacesGz,Gx,Gw. Fix an interior linenxof the at least three-dimensional subspace Uzxx,yx incident to the interior pointpx. Hencenxcorresponds to a vertex n∈ Γ adjacent tox,yandz. Therefore we can consider the chainz⊥n⊥y⊥winstead of z⊥x⊥y⊥w. Again there is an interior pointqn=Uznn,yn∩Unny,wyinGn. The inte- rior pointsqyandpyare different, becausepylies on the two-dimensional subspace nyandqyis orthogonal tony. Hence the spanly =hpy, qyiis a line contained in wπy⊆ Gy. This linelycorresponds to a vertexladjacent toyandw.
We have constructed the chainz⊥n⊥y⊥linΓwith the property that the dimension of the spanhny, lyiis three as the interior linesny andly intersect in the interior pointpyinGy. Thusdim(Uny
y,ly) =n−3 and thereforedim(Uzy
n,yn ∩Uny
y,ly)≥ n−4+n−3−n+2=n−5=2. Hence there is a vertext∈Γadjacent to the verticesz,n,y,l. The verticesyandzcorrespond to interior linesytandztinGt
and the intersection ofyπt andztπ is of dimension three. Furthermore the subspace ytπ∩ztπcontains the interior pointsptandqtinduced bypy, respectivelyqy. Hence the interior lineltis incident toUztt,yt. Thereforelandzare adjacent inΓ, so thatz andwhave distance at most two.
In both cases it follows by induction that the connected graphΓhas diameter 2. 2 Our next goal is to define a notion of global points forΓ. The following observation is the reason why such a definition is possible.
Letx,y,zbe three mutually adjacent vertices of the graphΓand letpxbe an interior point ofGxcontained in the subspaceyxπ∩zπx. Thenpyx =px∩ Lyis an interior point ofGyxandpzx=px∩ Lzis an interior point ofGzxby Section 4. Letpyrespectivelypz
be the unique interior points ofGy, respectivelyGzcontainingpyx, respectivelypyx. As the intersectionyxπ∩zπxis at least three-dimensional inGx∼=P(Vn)and the point px is contained in this intersection, the subspaceUyxx,zx contains distinct interior lines g1x andgx2 incident to the interior pointpx. Hence the verticesg1 andg2 are elements of the interior pointspyx andpzx. Thusg1,g2 ∈ py andg1,g2 ∈ pz. Consequently, g1,g2∈py∩ Lz=pyzandg1,g2∈pz∩ Ly=pzy. Hencepyz =pzy, as bothpyz and pzy are interior points of the interior spaceGyz =Gzy of the induced subgraphy⊥∩z⊥ by Proposition 4.3 which have more than one element in common. Thus we have proved the following statement.
Lemma 5.3. Letx,y,zbe mutually adjacent vertices ofΓand letpxbe an interior point ofGxperpendicular toyxandzx. Then the interior pointspyandpzcorrespond to each other, i.e.,pyz =pzy.
Definition 5.4. Aglobal pointpofΓis a set of vertices of the graphΓsatisfying p=px∪ [
l∈x⊥
{pl∈ Pl|pxl ⊆pl}
for some vertexxofΓand some interior pointpxof the interior spaceGx. The set of all global points ofΓis denoted byPΓ.
A global point is stable under iteration of the above process. Indeed, letx ∈ Γand px∈ Pxand define
p0 :=px∪ [
l∈x⊥
{pl∈ Pl|pxl ⊆pl} and p1:=px∪ [
l∈x⊥
{pl∈ Pl|pxl ⊆pl} ∪ [
k⊥l⊥x
{pk∈ Pk|pxl⊆pl, plk ⊆pk}.
Certainlyp0 ⊆p1. Ifp1 6=p0, then there exists a vertexw ∈p1 such thatw 6∈ p0. By construction there exist a chain of verticesx ⊥l ⊥ k ⊥w inΓand interior pointspl
andpkwithpxl ⊆pl,plk ⊆pkandw ∈pk. Due to the proof of Proposition 5.2 there are a vertexz∈ {x,l,w}⊥ and a pathk ⊥c1 ⊥ · · · ⊥ cn ⊥zin{l,w}⊥. Since the interior linec1kis perpendicular towk, there is a interior pointpc1⊇pk
c1 incident towc1. By Lemma 5.3 we havepl
c1 ⊆pc1. Arguing along the pathk⊥c1 ⊥ · · · ⊥cn⊥z, we obtainplz ⊆pz,wz∈pz. Thereforepxz ⊆pz, sow∈p0, a contradiction.
This consideration has two immediate consequences.
Proposition 5.5. Letpbe a global point andxbe vertex of Γ.
• The intersectionLx∩pis either empty or an interior point ofGx.
• The global pointpdoes not depend on the starting interior pointpx⊆p.
The triple GΓ = (PΓ,LΓ,⊃) is a point-line geometry, called theglobal geometry onΓ.
Proposition 5.6. The point-line geometryGΓ= (PΓ,LΓ,⊃)is a projective space.
Proof. We have to show that the geometryGΓsatisfies the axioms of a projective space, i.e.,GΓis linear and satisfies the axiom of Veblen and Young. Note thatGΓ is automati- cally thick, because the spacesGx,x∈Γ, are.
We start with proving linearity. Letpandqbe two distinct global points ofGΓand fix two global linesl∈pandm∈q. By Proposition 5.2 there exists a vertexz∈Γadjacent tolandm. Hence Proposition 5.5 implies thatp∩ Lz=pzandq∩ Lz=qzare interior points ofGz. By Proposition 3.12, i.e., since the projective spaceGz is linear, there is a unique interior linekzofGzincident to bothpzandqz. Thus we have found a global line kjoining the global pointspandq. Iflwere another global line joining the global points pandq, Proposition 5.2 would give us a vertexm ∈ {k,l}and a contradiction to the linearity ofGm. SoGΓis linear.
We now turn to the axiom of Veblen and Young. Leta,b,c,d,ebe global points of GΓ such that the global pointsa,b,canda,d,eare collinear triples on distinct global lines, sayla,b,candka,d,e. We need to prove that the linemb,djoiningbanddintersects the linenc,ejoiningcande. By Proposition 5.2 there existsz∈Γsatisfyingmb,d⊥z⊥ nc,e. The claim follows from local analysis of the interior spaceGzand Propositions 3.12 and 5.5. Indeed the interior spaceGz contains the interior linesmb,dz andnc,ez and the interior pointsbz,cz,dz,ez. Moreover, the spaceGzcontains the lines joining the pair bz,cz and the pairdz,ez. This means thatlza,b,candka,d,ez are lines ofGz. Therefore the interior pointaz =a∩ Lzexists inGz. By the validity of the axiom of Veblen and Young in the projective spaceGz the interior linesmb,dz andnc,ez intersect in a pointfz. Extending this local point to a global pointffinishes the proof. 2 Definition 5.7. Letx∈Γ. Thenhx⊥idenotes the pair(Px,Lx)⊆(PΓ,LΓ)wherePx contains the global points which have a non-empty intersection withLx, i.e.p∈ Pxif and only ifp∩ Lx 6= ∅. For a global pointpwe denote byhp⊥ithe pair(Pp,Lp) :=
(S
x∈pPx,S
x∈pLx)⊆(PΓ,LΓ).
It turns out that for eachhx⊥iis a hyperline and eachhp⊥iis a hyperplane ofGΓ, cf.
Propositions 5.9 and 5.11.
Lemma 5.8. Letn≥7. The geometrieshx⊥iandhp⊥iare subspaces ofGΓ. Moreover, hx⊥i ∼=Gx.
Proof. We start with proving thathx⊥iis a subspace. We have to show that for distinct p, q ∈ Px the linel ∈ LΓ joiningpandq lies inLx. Proposition 5.5 yields interior pointspx = p∩ Lx andqx = q∩ Lx of the interior spaceGx. By Proposition 3.12 there exists a unique interior linelx incident to bothpx andqx. Hence l ∈ Lx. The maphx⊥i → Gx :p7→ p∩ Lxandl 7→lfor everyp∈ Px andl∈ Lxis the desired isomorphism.
Now we turn tohp⊥i. We have to show that for distinct global pointsa, b∈ Pp, the global lineljoiningaandbis an element of the line setLp. Letx,y∈psuch thata∈ Px andb∈ Py. Fix two interior lineshax ∈axandhby ∈by. By Lemma 5.2 there exists a vertexz∈Γadjacent toxandy. This yields the chain of verticesha⊥x⊥z⊥y⊥hb in the graphΓ. The subspaceUxxz,yz∩(hax)πhas at least dimension three and therefore this intersection contains an interior linegx. Thus we obtain the chain of verticesg ⊥ ha ⊥ x ⊥g ⊥y ⊥ hb inΓ. Therefore, by passing fromha ⊥ x⊥ z⊥ y ⊥hb to g⊥ha ⊥x⊥g⊥y⊥hbif necessary, we can assume thata∩ Lzis an interior point ofGzand thatha is adjacent tozinΓ. The intersectionaπz ∩Uxz
z,yz =: Hxz
z,yz,az has dimension at leastn−4 and is a subspace of the hyperlineyzπ. HenceHxy
z,yz,az∩bπyhas dimension at leastn−5, and thus contains a interior linety. Considering the chain of verticest⊥ ha ⊥ x ⊥t ⊥y ⊥hb ⊥ tinstead ofz ⊥ha ⊥ x ⊥z ⊥y ⊥ hb, if necessary, we can additionally assume thatb∩ Lzis non-empty and thathb ⊥z. Form the pathz ⊥ ha ⊥ x ⊥ z ⊥ y ⊥ hb ⊥ zinΓ we obtain that both interior lineshaz andhbzare contained in the polar ofpz. Thus the interior linela,bz joiningaz andbzis contained inpπz, too. Consequently, there is a vertexw∈psuch thatla,bz ⊆wπz implying
la,b ∈ Lw⊆ Lp. Hencela,b∈ hp⊥i. 2
Proposition 5.9. Letn≥7and letx∈Γ. Thenhx⊥iis a hyperline(i.e., a codimension two subspace)ofGΓ.
Proof. The subspacehx⊥ihas codimension at least two, because none of the global lines h ∈ Lx intersects the global line x. It remains to show that an arbitrary planeE of GΓ intersects the subspace hx⊥i nontrivially. Let kand lbe two distinct lines in E.
ThenE = hk,liand the two global lines k andl intersect in a global pointd, thus k,l ∈d. By Proposition 5.2 there exists a vertexz∈ Γadjacent tokandl. Moreover the intersectionkzπ∩ hkz, lzicontains an interior pointpz. Letlp,dbe the global line of GΓ joining the two global pointspanddandhbe a global line incident to the pointp and adjacent to the verticeskandz. By Proposition 5.2 there again exists a vertexysuch thath⊥y⊥x. Proposition 5.2 also yields a vertexs∈ {k,h,x}⊥. As the interior space Gscontains the two local pointsps =p∩ Lsandds =d∩ Lsthe vertexlp,dis adjacent tos, too. Furthermoredim(xπs ∩ hks, lp,ds i) ≥ n−2+3−n = 1 implying that the intersectionxπs ∩ hks, lp,ds i)contains an interior pointqsofGs. Consequently, the global planeE=hk,li=hk,lp,diintersects the subspacehx⊥iin the global pointq. 2 Corollary 5.10. The projective spaceGΓhas vector space dimensionn+2.
Proof. Indeed, the codimension two subspacehx⊥i ∼=Gxhas vector space dimensionn.
2 Proposition 5.11. Letn≥7and letpbe a global point ofΓ. Thenhp⊥iis a hyperplane ofGΓ.
Proof. Certainly the subspacehp⊥i is a proper subspace ofGΓ. Indeed,p 6∈ hp⊥i as p∩ Lx=∅for each vertexx∈p. It remains to prove that each global line meetshp⊥i.
Letl∈Γand letkbe a global line ofp. By Lemma 5.2 there exists a linez∈Γsatisfying
k⊥z⊥l. The orthogonal spacepπz ofpzis a hyperplane ofGz. Therefore the interior line lzintersectspπz in at least one interior pointqz. Hencelis incident to the global pointq.
Moreover, there existsm ∈psuch thatmπz contains the interior pointqz. This implies q∩ Lm6=∅, whenceqis a point of the subspacehp⊥iincident tol. 2 Proposition 5.12. Letn≥7and letΓbe a connected graph locallyS(Vn). The point- line geometryGΓ= (PΓ,LΓ)is isomorphic toP(Vn+2).
Proof. In view of Corollary 5.10 the only remaining question is the one for the isomor- phism type of the underlying field. That field, however, is equal to the field underlying
the projective spacesGx,x∈Γ. 2
Lemma 5.13. LetH be a hyperline of GΓ. Then there exists a unique vertexh ∈ Γ satisfyingH=hh⊥i.
Proof. Letkbe a global line in the hyperlineH. By Lemma 5.9 the subspacehk⊥iis a hyperline ofGΓ. Moreover, the intersectionL:=hk⊥i ∩His a hyperline ofH, because kandhk⊥i intersect trivially. ThereforeLis also a hyperline of hk⊥iand induces a local hyperline Lk ∼= L in the interior space Gk ∼= hk⊥i. Thus Lk = lkπ for some interior linelk. Obviously,k∈ hl⊥iandL ⊆ hl⊥i. By dimension arguments we obtain H=hk, Li=hl⊥i.
Letm∈Γanother vertex satisfyingH =hm⊥i. By Proposition 5.2 there existsz∈ Γsatisfyingl⊥z⊥m. The equalityhl⊥i=hm⊥iimplieshl⊥i ∩ hz⊥i=hm⊥i ∩ hz⊥i.
Local analysis ofGz∼=hz⊥i(cf. Lemma 5.8) yieldsl=m. 2 Lemma 5.14. LetP be a hyperplane ofGΓ. Then there exist a unique global pointpof GΓsatisfyingP =hp⊥i.
Proof. Letkbe a global line of the hyperplaneP. By Lemma 5.9 the subspacehk⊥iis hyperline ofGΓ. The intersectionL:=hk⊥i ∩Pis a hyperplane ofhk⊥i ∼=Gk. ThusL induces a unique local hyperplaneLkinGk. HenceLkis the orthogonal space of some interior pointpk. Consequently,k∈ hp⊥iandL⊆ hp⊥i. Hence by dimension arguments P = hk, Li = hp⊥i. Uniqueness ofpfollows by an argument as given in the proof of
Lemma 5.13. 2
Denote the set of all hyperplanes of the projective geometryGΓwithHΓ. Define πΓ:PΓ∪ HΓ→ PΓ∪ HΓ
p7→ hp⊥i hp⊥i 7→p.
LetS(GΓ)be the line graph ofGΓ= (PΓ,LΓ). This means the vertex set ofS(GΓ)is the set of global lines ofGΓin which two distinct global lineskandlare adjacent if and only ifk∈ hl⊥i=πΓ(l)or, equivalently, if and only ifl∈ hk⊥i=πΓ(k).
Proposition 5.15. The graphΓis isomorphic to the line graphS(GΓ)ofGΓ.
Proof. The identity mapΓ→S(GΓ)is an isomorphism. 2 The Main Theorem is proved.
Acknowledgements. The authors express their gratitude to Antonio Pasini and to an anonymous referee for several very helpful comments and suggestions.
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Eprint arXiv:0710.0034
Received 14 October, 2007
K. Altmann, c/o Holger Grothe, Fachbereich Mathematik, TU Darmstadt, Schloßgartenstraße 7, 64289 Darmstadt, Germany
Email: altmann@mathematik.tu-darmstadt.de
R. Gramlich, Fachbereich Mathematik, TU Darmstadt, Schloßgartenstraße 7, 64289 Darmstadt, Germany
Email: gramlich@mathematik.tu-darmstadt.de
Second author’s alternative address: R. Gramlich, School of Mathematics, University of Birming- ham, Edgbaston, Birmingham B15 2TT, United Kingdom
Email: ralfg@maths.bham.ac.uk
