Gu¨nterF.Steinke AfamilyofflatMinkowskiplanesadmitting3-dimensionalsimplegroupsofautomorphisms

Advances in Geometry (de Gruyter 2004

A family of flat Minkowski planes admitting 3-dimensional simple groups of automorphisms

Gu¨nter F. Steinke

(Communicated by R. Lo¨wen)

Abstract.In this paper we construct a new family of flat Minkowski planes of group dimen- sion 3. These planes share the positive half with the classical real Minkowski plane and admit simple groups of automorphisms isomorphic to PSL₂ðRÞacting diagonally on the torus. We further determine the full automorphism groups and the Klein–Kroll types of these flat Min- kowski planes.

2000 Mathematics Subject Classification. MSC 2000: 51H15, 51B20

1 Introduction and result

Aflat Minkowski planeMis an incidence structure of points, circles and two kinds of parallel classes whose point set is the torusS¹S¹(where the 1-sphereS¹usually is represented asRUfyg), whose circles are graphs of homeomorphisms ofS¹ and whose parallel classes of points are the horizontals and verticals on the torus. We fur- thermore require that for every point pof Mthe associated incidence structure Ap

whose point setApconsists of all points ofMthat are not parallel to pand whose set of linesLp consists of all restrictions toAp of circles ofMpassing through pand of all parallel classes not passing through p is an a‰ne plane. We callAp the derived a‰ne planeat p; compare [5] or [4], Chapter 4. This implies that three mutually non- parallel points can be joined by a unique circle and that for two non-parallel points pandqand a circle KC p there is a unique circle which touchesK at pand passes throughq. Theclassical flat Minkowski planeis obtained in this way as the geometry of all graphs of fractional linear maps onS¹. Each derived a‰ne plane of the classi- cal flat Minkowski plane is Desarguesian.

When the circle sets are topologized by the Hausdor¤ metric with respect to a metric that induces the topology of the torus, then the planes aretopological in the sense that the operations of joining three mutually non-parallel points by a circle, intersecting of two circles, and touching are continuous with respect to the induced topologies on their respective domains of definition. For more information on topo- logical Minkowski planes we refer to [5] and [4], Chapter 4. The flat Minkowski planes are precisely the 2-dimensional topological Minkowski planes.

The circle spaceCof a flat Minkowski plane has two connected components; one, C^þ, consists of all circles in Cthat are graphs of orientation-preserving homeomor- phisms S¹!S¹ and the other, C, consists of all circles in C that are graphs of orientation-reversing homeomorphisms. We callC^þandCthepositiveandnegative half ofM, respectively. It turns out that these two halves are completely independent of each other, that is, we can interchange components from di¤erent flat Minkowski planes and obtain another flat Minkowski plane; see [4], 4.3.1.

Anautomorphismof a flat Minkowski plane is a homeomorphism of the torus such that parallel classes are mapped to parallel classes and circles are mapped to circles.

The collection of all automorphisms of a flat Minkowski planeMforms a group with respect to composition, the automorphism groupGofM. This group is a Lie group of dimension at most 6 with respect to the compact-open topology; see [4], 4.4. We say that a flat Minkowski plane hasgroup dimension nif its automorphism group is n-dimensional. All flat Minkowski planes of group dimension at least 4 have been classified by Schenkel [5], see also [4], 4.4.5. In particular, the classical flat Minkowski plane is the only flat Minkowski plane of group dimension at least 5 and every flat Minkowski plane of group dimension 4 fixes two parallel classes. Many flat Min- kowski planes of group dimension 3 have also been constructed, see [4], 4.3 for a summary, but no complete classification flat Minkowski planes of group dimension 3 has yet been achieved.

In this paper we contribute to the eventual classification by constructing a new family of flat Minkowski planes of group dimension 3. These planes admit simple groups of automorphisms isomorphic to PSL2ðRÞ. They are obtained from the clas- sical flat Minkowski plane by replacing the circles in the negative half in such a way that PSL2ðRÞ acts diagonally. Thus these planes are not isomorphic to the well- known flat Minkowski planes that admit PSL2ðRÞas a group of automorphisms in one of the kernels.

Main Theorem.Each incidence structureMðkÞfor k>1,see the beginning of Section 2,is a flat Minkowski plane.Furthermore, these planes are mutually non-isomorphic and the full automorphism group of each such plane is isomorphic toPGL₂ðRÞand acts diagonally on the torus.EachMðkÞis of Klein–Kroll typeIV.A.1.

The author wishes to thank the referee for his suggestions and comments in the preparation of the final version of this paper.

2 The incidence structuresM(k)

We construct a flat Minkowski plane MðkÞ by replacing the negative half of the classical flat Minkowski plane by the images of the generating circle

Ck ¼ fðx;xjxj^k1Þ jxARgUfðy;yÞg under the group

S¼ fðx;yÞ 7! ðdðxÞ;dðyÞÞ jdAPSL2ðRÞg:

More precisely, let k>1. Then the incidence structure MðkÞ on the torus S¹S¹ has circles of the following form.

.

x;axþb bxþd

xAS¹

wherea;b;c;dAR,adbc>0, with the obvious definitions forx¼yand when the denominator becomes 0. These circles are the same as the circles in the positive half of the classical flat Minkowski plane.

.

fkðxÞ ¼ xjxj^k1; if xAR;

y; if x¼y:

We shall show in the following thatMðkÞis indeed a flat Minkowski plane. Note that the restrictiong_k offk onR, that is, the function given by

g_kðxÞ ¼xjxj^k1

forxARis a multiplicative strictly increasing homeomorphism ofR. Moreover,gkis continuously di¤erentiable and its derivative is given byg_k⁰ðxÞ ¼kjxj^k1d0. Hence kgkðxÞ ¼xg_k⁰ðxÞfor allxAR.

The multiplicativity of gk implies that fk commutes with the transformation sAPSL2ðRÞgiven by

sðxÞ ¼ 1=x:

Hence d and ds define the same circle. In fact, this is the only instance that this happens, that is, if df_kd¹¼gf_kg¹ for g;dAPSL₂ðRÞ, then g¼dor g¼ds. More generally, we show the following.

Proposition 2.1.LetaAPSL₂ðRÞ.Then the homeomorphisma¹f_k¹af_k ofS¹ fixes at least three points ofS¹if and only ifa¼idora¼sas above.

Proof. ForaAPSL₂ðRÞlet F_aJS¹ be the set of all fixed points of a¹f_k¹af_k and let GJPSL2ðRÞbe the collection of allaAPSL2ðRÞsuch thatFa contains at least three points. By definition,aAG if and only if the cardinality ofCkVa¹ðCkÞis at least 3. FromsðCkÞ ¼Ck we infersAG. Applyingaandsto the intersection, we see that the setfa;a¹;as;sa¼ ðsa¹Þ¹;sasgis contained inGif one of its elements is.

GivenaAPSL2ðRÞone finds

a as sa sas

x7!^axþb_cxþd x7!^bxa_dxc x7!^cxd_axþb x7!^dxþc_bxa

From the above table we see that ifa:x7!^axþb_cxþd belongs toGwe may assume that either all coe‰cientsa;b;c;d ARare nonzero or thatc¼0.

We first assume that c¼0. Then a can be written in the form a:x7!rðxþsÞ where r;sAR, r>0. Since a¹ AG too, we may further assume that sd0. More- over,a¹f_k¹af_k fixesy AS¹and has at least two fixed pointsx₁<x₂inR. For these fixed pointsx_ione then findsaf_kðxÞ ¼ f_kaðxÞ, that is,jrj^k1g_kðxiþsÞ ¼g_kðxiÞ sfor i¼1;2. Eliminatingrfrom these two equations, we obtainhðsÞ ¼0 where

hðsÞ ¼g_kðsþx₂Þðsg_kðx1ÞÞ g_kðsþx₁Þðsg_kðx2ÞÞ

¼sðgkðsþx₂Þ g_kðsþx₁ÞÞ þg_kðx2ðsþx₁ÞÞ g_kðx1ðsþx₂ÞÞ:

Since x1 <x2 and sd0, the first term sðgkðsþx2Þ gkðsþx1ÞÞ is nonnegative.

The second term gkðx2ðsþx1ÞÞ gkðx1ðsþx2ÞÞ is 0 if and only if x2ðsþx1Þ ¼ x₁ðsþx₂Þ, that is, if and only if s¼0. Since g_kðx2ðsþx₁ÞÞ g_kðx1ðsþx₂ÞÞ ¼ gkðsÞ gkx2þ^x1_s^x2

gkx1þ^x1_s^x2

>0 for large s>0, it follows by continuity thatgkðx2ðsþx1ÞÞ gkðx1ðsþx2ÞÞ>0 for alls>0. HencehðsÞdgkðx2ðsþx1ÞÞ gkðx1ðsþx2ÞÞ>0 for s>0. This shows that we must haves¼0. But then r¼1.

Thereforea¼id in this case.

We now show that the second case wherea;b;c;d00 is not possible. We writea in the formaðxÞ ¼r^xþs_xþtwherer;s;tAR,rðtsÞ>0. By passing over toa¹;as;. . .; if necessary, we may further assume that 0<s<t or s<0<t. This then implies that r>0. Note that in this case neithery nora¹ðyÞ can be fixed bya¹f_k¹af_k. Then f_kaðxÞ ¼af_kðxÞis equivalent toh_r;s;tðxÞ ¼0 wherexARand

h_r;s;tðxÞ ¼ jrj^k1g_kðxþsÞðgkðxÞ tÞ þg_kðxþtÞðgkðxÞ sÞ

We show that h_r;s;t has at most two real zeros. By looking at where the factors g_kðxþsÞ;g_kðxÞ t;g_kðxþtÞ;g_kðxÞ soccurring inhr;s;tðxÞare positive or negative we find thathr;s;tðxÞ>0 forx>maxfs;g¹_k ðtÞ;g¹_k ðsÞgorx<minfs;t;g¹_k ðsÞg (note thatt>0).

UsingkgkðxÞ ¼xg_k⁰ðxÞone finds for the derivative ofhr;s;tthat

xh_r;s;t⁰ ðxÞ khr;s;tðxÞ ¼ ðjrj^k1g_k⁰ðxþsÞ þg_k⁰ðxþtÞÞðxgkðxÞ þstÞ: ð*Þ The first factor jrj^k1g_k⁰ðxþsÞ þg_k⁰ðxþtÞ on the right-hand side is always posi- tive. We now assume that 0<s<t. Then the second factor xg_kðxÞ þst in (*) is also positive. This implies that h_r;⁰_s;tðx0Þ>0 for every positive zero x₀ of h_r;s;t and h_r;⁰_s;tðx0Þ<0 for every negative zero x₀ of h_r;s;t. Hence there can be at most one positive and at most one negative zero ofh_r;s;t. Since

h_r;s;tð0Þ ¼ tjrj^k1g_kðsÞ sg_kðtÞ ¼ stðjrsj^k1þ jtj^k1Þ<0;

we see thath_r;s;t has precisely two zeros in case 0<s<t.

We finally assume thats<0<t. In this case one further finds thath_r;s;tðxÞ>0 for maxft;g¹_k ðsÞgcxcminfs;g¹_k ðtÞg; see also Table 1 below.

Thus every zero ofh_r;s;t must be in the open intervals

I¼ ðminft;g¹_k ðsÞg;maxft;g¹_k ðsÞgÞ and Iþ¼ ðminfs;g¹_k ðtÞg;maxfs;g¹_k ðtÞgÞ:

Note that for g¹_k ðsÞ ¼ t we have I¼q and h_r;s;tðxÞ>0 for all x<0. Like- wise, g¹_k ðtÞ ¼ simpliesIþ¼q andh_r;s;tðxÞ>0 for all x>0. We will see below that h_r;s;t can have at most two zeros in IG. Therefore in each of the above two cases where one of the intervals is empty we obtain the desired result. In order to avoid unnecessary special cases in the following, we now assume that g¹_k ðsÞ0t andg¹_k ðtÞ0s, that is, that both intervalsIGare nonempty.

The map x7!xgkðxÞ þst has precisely two zeros x0¼ jstj^1=ðkþ1Þ andx0. Since sþgkðtÞ and tþg¹_k ðsÞ have the same sign, we see from Table 1 that xgkðxÞ þst takes on opposite signs at the boundary points of I. We similarly obtain that xgkðxÞ þst takes on opposite signs at the boundary points of Iþ. This shows that x0AIþ andx0AI.

Ifx0x0 is a zero ofhr;s;t, then we obtain from Equation (*) thath_r;⁰_s;tðxÞ>0 for x>x0 or x0<x<0 andh_r;⁰_s;tðxÞ<0 for 0<x<x0 or x<x0. As before this implies thathr;s;thas at most one zero in each of the intervalsðminft;g¹_k ðsÞg;x0Þ, ðx0;maxft;g¹_k ðsÞgÞ, ðminfs;g¹_k ðtÞg;x0Þandðx0;maxfs;g¹_k ðtÞgÞ. Thus hr;s;t

has at most two zeros inIGunless perhapshr;s;tðx0Þ ¼0 orhr;s;tðx0Þ ¼0. Suppose thathr;s;tðx0Þ ¼0. Thenh_r;s;⁰ _tðx0Þ ¼0 too by (*) and di¤erentiating (*) atx0 we ob- tain

x0h_r;⁰⁰_s;tðx0Þ ¼ ðjrj^k1g_k⁰ðx0þsÞ þg_k⁰ðx0þtÞÞðgkðx0Þ þx0g_k⁰ðx0ÞÞÞ

¼ ðjrj^k1g_k⁰ðx0þsÞ þg_k⁰ðx0þtÞÞðkþ1ÞÞgkðx0Þ:

Henceh_r;⁰⁰_s;tðx0Þ>0 and it then follows thatx0is the only zero ofhr;s;tinIþ. The case h_r;s;tðx0Þ ¼0 is dealt with similarly and results in only one zero ofh_r;s;tinI. So in any caseh_r;s;t has at most two zeros inIG.

We still have to exclude the case that h_r;s;t has more than two zeros in IþUI. Let

x h_r;s;tðxÞ xg_kðxÞ þst

t jrj^k1tðjtj^k1þ1Þg_kðtsÞ>0 tðsþg_kðtÞÞ g¹_k ðsÞ ðtsÞjrj^k1g_kðsÞgkðjsj^1=k1þ1Þ>0 sðtþg¹_k ðsÞÞ g¹_k ðtÞ ðtsÞg_kðtÞg_kðjtj^1=k1þ1Þ>0 tðsþg¹_k ðtÞÞ s sðjsj^k1þ1ÞgkðtsÞ>0 sðtþg_kðsÞÞ

Table 1.

r_þ¼ g_kðx0þtÞðgkðx0Þ sÞ gkðx0þsÞðgkðx0Þ tÞ

1=ðk1Þ

r¼ g_kðx0þtÞðgkðx0Þ sÞ gkðx0þsÞðgkðx0Þ tÞ

1=ðk1Þ

that is,rGare such thath_rþ;s;tðx0Þ ¼h_r;s;tðx0Þ ¼0. Then r_þ^k1r^k1¼s

t

x0t x₀s

k1

x0þt x₀þs

k1!

>0:

Hence rþ>r>0. Since gkðx0þsÞðgkðx0Þ tÞ<0 on Iþ, we obtain that hr;s;tðx0Þ<0, ¼0, >0 for r>rþ, r¼rþ, r<rþ, respectively. Similarly, gkðx0þsÞðgkðx0Þ tÞ>0 onIimplies thathr;s;tðx0Þ<0,¼0,>0 forr<r, r¼r,r>r, respectively.

If h_r;s;t has two zeros in I_þ, then h_r;s;tðx0Þ<0 and thus r>r_þ from above. But then r>r and h_r;s;tðx0Þ>0. From what we have seen before, this then implies thath_r;s;t has no zeros inI. The case thath_r;s;t has two zeros inIis dealt with simi- larly. This concludes the proof thath_r;s;thas at most two real zeros and the statement

of the proposition is established. r

Corollary 2.2.Two di¤erent circles ofMðkÞintersect in at most two points.Hence two points in a derived geometry at a point ofMðkÞare on at most one line.

Proof. The circles of MðkÞ are the graphs of b and dfkd¹ for all b;dAPSL2ðRÞ.

Since the first kind of homeomorphism is orientation-preserving and the latter kind is orientation-reversing, we obtain that any two such associated circles intersect in at most two points. The same is true for any two circles of the first kind because we are essentially in the classical flat Minkowski plane.

If the circles associated with gf_kg¹ anddf_kd¹ forg;dAPSL₂ðRÞhave three dis- tinct points in common, thenðd¹gÞ¹f_k¹ðd¹gÞf_k fixes three points so thatd¹g¼id ord¹g¼sby Proposition 2.1 and the circles are the same. This shows that ifgfkg¹ anddfkd¹ describe di¤erent circles inMðkÞ, that is,g0d;ds, then these circles can

have at most two points in common. r

From the definition of circles it is obvious that circles are described by homeo- morphisms ofS¹. Hence, in order to verify thatMðkÞis a flat Minkowski plane, we only have to make sure that each derived incidence geometry is an a‰ne plane.

Since the group S is a group of automorphisms of the classical flat Minkowski plane, and, by construction, also acts on the negative half ofMðkÞ, we see thatSis a group of automorphisms ofMðkÞ. Furthermore,Shas two orbits on the torus, the circle

D¼ fðx;xÞ jxAS¹g

in the positive half and its complementðS¹S¹ÞnD. It therefore su‰ces to show that the derived incidence geometries at the pointsðy;yÞandðy;0Þare a‰ne planes.

Note moreover thatSis even doubly transitive on the points ofDand thatDis the only circle fixed byS.

3 The derived geometryAat (T,T)

The lines of the derived geometryAofMðkÞatðy;yÞare the horizontal and ver- tical Euclidean lines (coming from parallel classes of MðkÞ), all Euclidean lines of positive slope (coming from circles in the positive half that pass through ðy;yÞ), and the lines

x;rf_k xt r

þt

xAR

for r;tAR, r>0 (coming from circles in the negative half that pass through ðy;yÞ). The latter circles are the images of the generating circleC_k under the sta- bilizerL¼S_ðy;yÞofðy;yÞ, that is, the group

L¼ fðx;yÞ 7! ðrxþt;ryþtÞ jr;tAR;r>0gGL₂:

Note that the transformation ss^:ðx;yÞ 7! ð1=x;1=yÞ in S leaves C_k invariant.

Therefore the cosetL^ssgives rise to the same set of circles.

Using the restrictiongk offk onR, the lines of the latter kind in Acan then be rewritten as

y¼sgkðxtÞ þt

fors;tAR, s<0ðs¼ r^1kÞ. Hence we obtain the following description of the the lines inA.

The geometryA.The lines ofAare the verticalsfcg RforcARand Ls;t¼ fðx;sxþtÞg jxARg; fors;tAR;sd0;

fðx;sg_kðxtÞ þtÞg jxARg; fors;tAR;s<0:

Proposition 3.1.The derived geometryAofMðkÞatðy;yÞis an a‰ne plane.

Proof. We first show that two distinct points ofR² can be joined by a unique line in A. Letðxi;yiÞ,i¼1;2, be two such points. Ifðy2y1Þðx2x1Þd0, there is a unique Euclidean line of nonnegative slope or a vertical line through these points.

Moreover, no lineLs;t fors<0 can pass throughðx1;y1Þandðx2;y2Þ. Ifðy2y1Þ ðx2x1Þ<0, no such Euclidean line with s>0 can exist and we have to find a unique lineLs;twhere s<0 through both points. Without loss of generality we may assume thatx1<x2. From the system of equations

y1t¼sgkðx1tÞ y2t¼sgkðx2tÞ

we obtain ðy2tÞ=ðy1tÞ ¼gkððx2tÞ=ðx1tÞÞ. Taking the inverse of the frac- tional linear map t7! ðy2tÞ=ðy1tÞ on both sides and using that gk is multi- plicative, we obtain

y1gkðx2tÞ y2gkðx1tÞ g_kðx2tÞ g_kðx1tÞ ¼t:

The left-hand side defines a strictly decreasing homeomorphismhofR. Hencehhas a unique fixed pointt₀. Sincex₁0x₂, we havet₀0x_ifor at least onei¼1;2. Then s₀¼ ðy_it₀Þ=gkðxit₀Þ is well defined and L_s0;t0 is the unique line in A through ðx1;y₁Þandðx2;y₂Þ.

For the parallel axiom note thatL_s;tfors>0 ands<0 are graphs of orientation- preserving and orientation-reversing homeomorphisms of R. We therefore see that the parallel axiom is clearly satisfied for horizontal or vertical lines and that the pa- rameters⁰of any parallelLs⁰;t⁰of a lineLs;tinAmust have the same sign ass. Hence there is a unique parallel inAto a lineLs;t,s>0 (that is, a Euclidean line of positive slope) through a given point. We thus only consider the cases<0.

We first verify that two linesLs;tandLs⁰;t⁰wheres;s⁰<0 are parallel if and only if s¼s⁰. Straightforward computation shows that the automorphismðx;yÞ 7! ðaxþb;

ayþbÞ, wherea;bAR,a>0, takesLs⁰;t⁰ to the lineL_a^1k_s⁰_;at⁰_þb. Using the groupL we may therefore assume thats⁰¼ 1 andt⁰¼0. Then

gkðxtÞ þt

>gkðxÞ þt>gkðxÞ; if t>0;

¼ gkðxÞ; if t¼0;

<gkðxÞ þt<gkðxÞ; if t<0:

8<

:

This shows thatL1;t is parallel toL1;0. Ifs01, thenx7!g_kðxÞ þsg_kðxtÞ þt is a continuous function onRthat tends toGyasxgoes toGyif1<s<0 and toHyforx!Gyif s<1. Therefore this function is surjective in any case and the value 0 is attained. This shows thatL_s;t intersectsL1;0 in a point ifs01.

Now given a pointðx0;y₀Þ, a line parallel toL1;0 that passes through this point must be of the formL1;t. To findtjust note thatg_kðtx₀Þ ¼ gkðx0tÞis strictly increasing intand y₀tis strictly decreasing int. Furthermore, both functions are unbounded. Hence there is a unique t0ARsuch that gkðx0tÞ ¼ y0t, that is, L1;t0 is the unique line parallel toL1;0that passes throughðx0;y0Þ.

Hence the axioms of an a‰ne plane are satisfied. r

Note that Proposition 3.1 also follows from [3], Theorem 2.7. Rotation through 45 brings the geometryAin the form used in [3]. In the new coordinates the group Lacts onR²asðx;yÞ 7! ðrx;ryþtÞand the distinguished line fixed underLis they- axis. Straightforward computation shows that the tripleðfF1g;fF2g;jÞ, whereF₁and

F₂ are functions fromR^þtoRdefined byF₂ðxÞ ¼ F1ðxÞ,F₁ðxÞ ¼ ffiffiffi p2

ux, where u is the unique solution of uf_kðuÞ ¼x, andj is given byjðxÞ ¼ x, satisfies the conditions (F1)–(F3) of [3], p. 7, so thatAis an a‰ne plane by [3], Theorem 2.7.

The transitivity of S on Dimplies that Proposition 3.1 carries over to any point onD.

Corollary 3.2.Each derived geometry ofMðkÞat a point of D is an a‰ne plane.

Note thatk¼1 does not yield an a‰ne plane because we do not get enough lines inA. This of course means that we cannot extend the definition ofMðkÞtok¼1.

Indeed, the orbit of the generating circle C1 under S only yields a 2-dimensional family of circles so that we do not obtain enough circles in the negative half in this case. (However,k¼1 results in the Desarguesian a‰ne plane for the derived geom- etry atðy;0Þ, see the following section for this geometry.)

For later, when we determine isomorphism classes, we conclude this section by showing thatAis not an a‰ne plane that occurs as a derivation of the classical flat Minkowski plane.

Lemma 3.3.Ais not Desarguesian.

Proof. We consider the triangles with verticesp₁¼ ð0;0Þ,p₂¼ ð1;1Þ, p₃¼ ð1;3Þ, and q1¼ ð2;0Þ, q2¼ ð1;1Þ, q3¼ ð1;3Þ, respectively. The lines through pi

andqi are horizontals and thus are parallel fori¼1;2;3. Furthermore, correspond- ing lines through p2 andq2 are also parallel. (The lines p2p3 and q2q3 are verticals and the lines p1p2 andq1q2areL1;0 andL1;1, respectively.) Finally, the line p1p3

isL2;0 and the line throughq1andq3 isL3gkð2Þ=2;3=2. Butk>1 impliesgkð2Þ>2 and thus ³₂gkð2Þ03. Hence p1p3 andq1q3 are not parallel, compare the proof of Proposition 3.1, and Desargues’ configuration does not close for the above six

points. r

Note that the proof of Lemma 3.3 only uses horizontals, verticals and lines L_s;t withs<0, that is, parallel classes and circles in the negative half ofMðkÞ.

4 The derived geometryBat (T, 0)

For a description of the lines in the derived incidence geometryBofMðkÞatðy;0Þ we use the coordinate transformation ðx;yÞ 7! ðx;1=yÞ. A circle through ðy;0Þis the graph of a fractional linear mapx7!b=ðcxþdÞwhere bc¼ 1 or the graph of dfkd¹ wheredAPSL2ðRÞ,dfkd¹ðyÞ ¼0. Under the above coordinate transforma- tion the former circles give rise to the lines y¼mxþt where m;tAR, m<0. As for the latter circles, note that each such circle intersects the distinguished circle D in two points, sayðu;uÞandðv;vÞwhere u;vARnf0g,u0v. Furthermore, because the derived geometry at ðu;uÞis an a‰ne plane by Corollary 3.2, the pointsðy;0Þ, ðu;uÞ andðv;vÞdetermine a unique circle. We must even have uv<0. This follows from the fact thatdf_kd¹is strictly decreasing onRnfwgwhere w¼df_k¹d¹ðyÞ<0

so that the graph must intersect D in a point with negative coordinates and one with positive coordinates. Then d¹ðuÞ and d¹ðvÞ are both fixed points of f_k so that fd¹ðuÞ;d¹ðvÞg ¼ fy;0g, that is, fu;vg ¼ fdðyÞ;dð0Þg. Let dðxÞ ¼ ðaxþbÞ=

ðcxþdÞ, adbc¼1. Then u¼dðyÞ ¼a=c, v¼dð0Þ ¼b=d, or v¼a=c, u¼b=d. In the first case we obtain a¼uc, b¼vd, and 1¼adbc¼ ðuvÞcd. Therefore dðxÞ ¼ ðucxþvdÞ=ðcxþdÞ and d¹ðxÞ ¼ ðdxvdÞ=ðcxþucÞ ¼ ðd=cÞðxvÞ= ðxuÞ. Furthermore, ðvdÞ=ðucÞ ¼d¹ð0Þ ¼ fkd¹ðyÞ ¼ fkðd=cÞ ¼gkðd=cÞ.

Thus

dfkd¹ðxÞ ¼d gk d c

xv xu

¼d gk

d c gk

xv xu

¼d vd ucgk

xv xu

¼d vdgkðxvÞ ucg_kðxuÞ

¼vd^g_g^kðxvÞ

kðxuÞþvd ^vdg_ug^kðxvÞ

kðxuÞþd

¼uv gkðxvÞ þg_kðxuÞ vgkðxvÞ þug_kðxuÞ

Under the above coordinate transformation we obtain the lines L_u;v given by y¼Fðu;v;xÞwhereu;vAR,uv<0, and

Fðu;v;xÞ ¼ 1 uv

ugkðxuÞ vgkðxvÞ g_kðxuÞ g_kðxvÞ :

Note that the above denominator is never 0 so that the right-hand side is defined for allxAR.

In the second case the roles of uandvare interchanged and we obtain the same equation. Note that the above equation is symmetric inuandv. In particular, we can always assume thatu<0<v.

Sincegk is multiplicative, it follows thatFðu;v;xÞ ¼_uv^{1 ug}_g^{kð1u=xÞvgkð1v=xÞ}

kð1u=xÞgkð1v=xÞ forx00 and thus limx!GyFðu;v;xÞ ¼Gy. Furthermore, ^qF_qxðu;v;xÞ ¼ ^{kðvuÞ2jxujk1jxvjk1}

uvðgkðxuÞgkðxvÞÞ²

>0 for all x0u;v so that Fðu;v;xÞ is strictly increasing in x. This verifies that x7!Fðu;v;xÞis indeed a homeomorphism ofRfor all admissibleuandv.

In summary, we have found the following description of the lines of the derived geometryB.

Figure 1. E_1;1=2andL_1;1=2fork¼3

The geometryB.The lines ofBare

.

.

.

Lu;v¼ fðx;Fðu;v;xÞÞg jxARg foru;vAR,u<0<v.

Lemma 4.1. The line L_u;v has the Euclidean line E_u;v given by y¼ _kuv¹ x^kþ1₂ ðuþvÞ

as an oblique asymptote. Furthermore, L_u;v and E_u;v have precisely the point ^uþv₂ ;^uþv_2uv

in common and Eu;v is below Lu;v to the right of that point and above Lu;vto the left.

Proof. Sincek>1, the functiongk is continuously di¤erentiable and even twice con- tinuously di¤erentiable for all x00. The respective derivatives areg_k⁰ðxÞ ¼kjxj^k1 and g_k⁰⁰ðxÞ ¼kðk1Þxjxj^k3. If zAR such that x and xz are in the same open intervalðy;0Þorð0;þyÞ, Taylor’s formula then yields

gkðxzÞ ¼gkðxÞ zg_k⁰ðxÞ þz² 2 g_k⁰⁰ðzÞ

wherezis betweenxzandx. Note that if zis fixed andxtends toGywe obtain thatg_k⁰⁰ðzÞjxj^1k tends to 0 and thatg_k⁰⁰ðzÞxjxj^1k ¼g_k⁰⁰ðzÞ¹_xjxj^3k tends tokðk1Þ.

Letu<0<vso thatxv<v<x<u<xu. Then

uvFðu;v;xÞ ¼ðuvÞgkðxÞ ðu²v²Þg_k⁰ðxÞ þ¹₂ðu³g_k⁰⁰ðuÞ v³g_k⁰⁰ðvÞÞ ðuvÞg_k⁰ðxÞ þ¹₂ðu²g_k⁰⁰ðuÞ v²g_k⁰⁰ðvÞÞ

¼ðuvÞx ðu²v²Þkþ¹₂ðu³g_k⁰⁰ðuÞ v³g_k⁰⁰ðvÞÞjxj^1k ðuvÞkþ¹₂ðu²g_k⁰⁰ðuÞ v²g_k⁰⁰ðvÞÞjxj^1k and

uvFðu;v;xÞ þ1 kx

¼ðu²v²Þkþ¹₂ðu³g_k⁰⁰ðuÞ v³g_k⁰⁰ðvÞÞjxj^1kþ_2k¹ ðu²g_k⁰⁰ðuÞ v²g_k⁰⁰ðvÞÞxjxj^1k ðuvÞkþ¹₂ðu²g_k⁰⁰ðuÞ v²g_k⁰⁰ðvÞÞjxj^1k : The numerator and denominator on the right-hand side tend to ðu²v²Þkþ

1

2kðu²v²Þkðk1Þ ¼ ^kþ1₂ ðu²v²ÞandðuvÞk, respectively, asxgoes toGy. Thus limx!GyFðu;v;xÞ þ¹_kx¼^kþ1_2k ðuþvÞ. This shows that the E_u;v is an oblique asymptote ofL_u;v.

Let

Eðu;v;xÞ ¼ 1

kuv xkþ1 2 ðuþvÞ

foru;v;xAR,u<0<v. ThenE u; v;^uþv₂

¼F u; v;^uþv₂

¼^uþv_2uvso that^uþv₂ ;^uþv_2uv is on E_u;vVL_u;v. Now consider the equationEðu;v;xÞ ¼Fðu;v;xÞfor fixedu<0<v. We writex¼^vu₂ zþ^vþu₂ forzAR. Then

0¼Fðu;v;xÞ Eðu;v;xÞ

¼ 1 uv

ugkðxuÞ vgkðxvÞ g_kðxuÞ g_kðxvÞ þ1

k x ðkþ1Þvþu 2

¼ 1 uv

ugkðzþ1Þ vgkðz1Þ g_kðzþ1Þ g_kðz1Þ þ1

k vu

2 zkvþu 2

¼ðvuÞððzkÞgkðzþ1Þ ðzþkÞgkðz1ÞÞ 2kuvðgkðzþ1Þ g_kðz1ÞÞ :

Hence ðzkÞgkðzþ1Þ ðzþkÞgkðz1Þ ¼0 or ðzþkÞ=ðzkÞ ¼g_kðzþ1Þ= gkðz1Þ ¼gkððzþ1Þ=ðz1ÞÞ. Let t¼ ðzþ1Þ=ðz1Þ so that z¼ ðtþ1Þ=ðt1Þ.

Then the above equation becomes

g_kðtÞ ¼ðkþ1Þt ðk1Þ ðkþ1Þ ðk1Þt

and thus

hðtÞ ¼ ðk1Þjtj^kþ1 ðkþ1Þtjtj^k1þ ðkþ1Þt ðk1Þ ¼0:

The function h is di¤erentiable and even twice di¤erentiable for t00. For the de- rivatives one finds

h⁰ðtÞ ¼ ðk²1Þtjtj^k1 ðkþ1Þkjtj^k1þkþ1;

h⁰⁰ðtÞ ¼kðk²1Þtjtj^k3ðt1Þ:

Hence h⁰⁰ðtÞ>0 for t<0 ort>1 and h⁰⁰ðtÞ<0 for 0<t<1. Consequently, h⁰ is strictly increasing onðy;0Þandh⁰ð1Þ ¼0 is a relative minimum ofh⁰onð0;þyÞ.

The latter implies thathis strictly increasing onð0;þyÞand thus 1 is the only posi- tive zero ofh. The former and the fact thath⁰ð0Þ ¼kþ1>0,h⁰ð1Þ ¼ 2ðk²1Þ

<0 imply thath⁰has precisely one negative zerotfor which we have1<t<0.

Furthermore,h is strictly decreasing on the intervalðy;tÞand strictly increasing on ðt;0Þ. But hð0Þ ¼ ðk1Þ<0 and hð1Þ ¼0. This shows that1 is the only negative zero ofh.

In summary we have found that h has precisely two zeros, namely t¼1 and t¼ 1. This in turn yields the only solution z¼0, that is, x¼ ðuþvÞ=2, of our original equation. (Note that t¼ 1 corresponds to z¼0 and that t¼1 yields z¼yand thus does not contribute to a solution inR.) This proves that ^uþv₂ ;^uþv_2uv is the only point of intersection of E_u;v and L_u;v. The remaining statements on the relative positions of E_u;v and L_u;v readily follow from Eðu;v;uÞ Fðu;v;uÞ ¼ ^ðk1ÞðvuÞ_2kuv >0 andEðu;v;vÞ Fðu;v;vÞ ¼^ðk1ÞðvuÞ_2kuv <0. r Note that every Euclidean line of positive slope occurs precisely once as an asymptote Eu;v for someu<0<v. Indeed, if m;tAR,m>0, thenu¼ _ðkþ1Þm^t

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1 kmþ ^t2

ðkþ1Þ²m²

q <0 and v¼ _ðkþ1Þm^t þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1 kmþ ^t2

ðkþ1Þ²m²

q >0 are such that E_u;v is the Euclidean line given by y¼mxþt.

In the coordinates ofB, the distinguished circleDinduces the Euclidean hyperbola H ¼ fðx;yÞAR²jxy¼1g:

The stabilizer C¼S_ðy;0Þ ofðy;0Þ also fixes the points ð0;yÞ, ðy;yÞ andð0;0Þ.

Hence

C¼ fðx;yÞ 7! ðrx;ryÞ jr>0g:

This group induces a groupFof collineations ofB. In the new coordinates ofBone obtains

F¼ fðx;yÞ 7! ðrx;y=rÞ jr>0g:

C

H C

I I

0 +

+

)

1

, y

- -

( x

1

C

Figure 2.

Lemma 4.2.The derived geometryBof MðkÞatðy;0Þis a linear space,that is,any two distinct points can be uniquely joined by a line.

Proof. Given two distinct pointsðx1;y1Þandðx2;y2ÞinR² we have to find a line of Bthat passes through them. Clearly, ifðx2x₁Þðy2y₁Þc0, then these points are on a vertical line or a Euclidean line of nonpositive slope. Furthermore, such a line is unique.

We now assume that ðx2x₁Þðy2y₁Þ>0. Without loss of generality we may further assume that x₁<x₂. Since the derived geometry at each point of D is an a‰ne plane by Corollary 3.2, we may moreover assume that none of the points is on H, that is,xiyi01 fori¼1;2. A line through the two points must then be of the formLu;vwhereu<0<v. By Corollary 2.2 a joining line will be unique and we only have to verify the existence of such a line.

Since each lineLu;vis the graph of a strictly increasing homeomorphism ofRand becauseLu;vpasses throughðu;1=uÞandðv;1=vÞ, we see thatðx1uÞy1¹_u

>0 is a necessary condition, that is,

ðx1uÞðuy11Þ<0:

We similarly find that

ðx1vÞðvy11Þ>0;

compare Figure 2.

Depending on the position ofðx1;y1Þrelative to the coordinate axes and toHone obtains from the above two inequalities certain restrictions for u andvwhere IJ

ðy;0ÞandI_þJð0;þyÞdenote the maximal (open) intervals we can chooseuand vfrom; see Table 2 below.

ForLu;vto pass throughðx1;y1Þwe find the condition

y1uvðgkðx1uÞ gkðx1vÞÞ ¼ugkðx1uÞ vgkðx1vÞ which yields

ugkðx1uÞ

y₁u1 ¼vgkðx1vÞ y₁v1 that is,GðuÞ ¼GðvÞwhere

GðzÞ ¼zgkðzx1Þ y1z1

forzAR, y1z01. We denote byGGthe restriction of Gto the open intervalIG.G is di¤erentiable and has derivative

G⁰ðzÞ ¼jzx₁j^k1

ðy1z1Þ²ðky1z² ðkþ1Þzþx₁Þ:

The first term jzx1j^k1=ðy1z1Þ² in G⁰ðzÞ above is always positive on IG and it readily follows that the last factor qðzÞ ¼ky1z² ðkþ1Þzþx1 has no zero inIG. (Note that qðzÞ ¼kzðy1z1Þ þ ðx1zÞ and that zðy1z1Þ and x1z have the same sign onIG.) In the above table the sign ofqG, that is, the restriction ofqtoIG, is indicated in the columns labelledqandqþ. HenceGGis strictly increasing or strictly decreasing onIG. In Table 2 this is indicated by an arrow up"or an arrow down#, respectively.

Clearly,Gð0Þ ¼Gðx1Þ ¼0,

z!limGyGðzÞ ¼

Gy; if y₁>0;

y; if y1¼0;

Hy; if y1<0;

8<

:

x₁ y₁ x₁y₁ I I_þ q q_þ G G_þ a

d0 0 ðy;0Þ ðx1;þyÞ >0 <0 " # #

<0 0 ðy;x₁Þ ð0;þyÞ >0 <0 " # # c0 >0 ðy;x₁Þ ð0;1=y₁Þ >0 <0 " # #

>0 >0 >1 ðy;0Þ ð1=y1;x1Þ >0 >0 " " "

>0 >0 <1 ðy;0Þ ðx1;1=y₁Þ >0 <0 " # # d0 <0 ð1=y1;0Þ ðx1;þyÞ >0 <0 " # #

<0 <0 >1 ðx₁;1=y₁Þ ð0;þyÞ <0 <0 # # "

<0 <0 <1 ð1=y1;x₁Þ ð0;þyÞ >0 <0 " # # Table 2.

and lim_z!1=y1GðzÞ ¼Gydepending on the relative position of y₁00 to 0 and x₁, but the sign changes in any case whenzapproaches 1=y₁from opposite sides. It then follows that in any caseGGtakesIGonto the negative real numbersðy;0Þ. In par- ticular, this shows that for each uAIthere is a unique vAIþ such thatL_u;vpasses through ðx1;y₁Þ. In fact, there is a homeomorphism a₁:I!Iþ such that L_u;a1ðuÞ passes throughðx1;y₁Þ. Clearly,a₁ ¼G_þ¹G, anda₁is strictly decreasing if and only ifx1y1 <1 (that is,ðx1;y1Þis between the two branches of the Euclidean hyperbola H) and strictly increasing if and only ifx1y1>1 (that is,ðx1;y1Þis above or below H); see Table 2. Moreover,a1 is di¤erentiable and its derivative is given bya₁⁰ðuÞ ¼ G⁰ðuÞ=G_þ⁰ða1ðuÞÞ.

(Note that there is no explicit formula for a1 except for special cases. For exam- ple, in the casex1¼ y1¼0 one hasa1ðuÞ ¼ u. It then readily follows that there is a unique line Lu;u through a point ðx2;y2Þ where x2y2>0. Hence ð0;0Þ can be uniquely joined to any other point inB.)

One similarly obtains a homeomorphism a₂: ~II!II~_þ such that L_u;a2ðuÞ passes throughðx2;y₂ÞwhereII~Jðy;0ÞandII~_þJð0;þyÞare open intervals defined in a similar fashion as the intervalsIandIþfora₁.

We consider the three connected components ofR²nH; more precisely, let Cþ¼ fðx;yÞAR²jxy>1;x>0g;

C0 ¼ fðx;yÞAR²jxy<1g;

C¼ fðx;yÞAR²jxy>1;x<0g;

see Figure 2. The reflectionr about the origin ofR² given by rðx;yÞ ¼ ðx;yÞis an automorphism of the incidence structureB. (Note thatFðu;v;xÞ ¼Fðv;u;

xÞ.) Furthermore,r interchangesCþ andC and leavesC0 invariant. Using r and perhaps relabelling the points, if necessary, we can assume that x1<x2 and we can restrict ourselves to the four cases ðx1;y1ÞAC0, ðx2;y2ÞACþ or ðx1;y1Þ;ðx2;y2ÞA Cþ orðx1;y1Þ;ðx2;y2ÞAC0 orðx1;y1ÞAC,ðx2;y2ÞACþ for the relative positions of the two pointsðx1;y1Þandðx2;y2Þ. In each of theses cases we are looking at either a¼a₁a¹₂ ora¼a₂a¹₁ and verify thata fixes a point. Such a fixed pointvleads to u¼a¹₂ ðvÞ ¼a¹₁ ðvÞso thatL_u;vis a line throughðx1;y₁Þandðx2;y₂Þ.

We encounter essentially two situations. In the first one a:I!J is a strictly increasing homeomorphism and I andJ are two open intervals inRsuch that J is finite and its closure J is contained in I. If J¼ ðc;dÞ, we define v_n inductively by v₀¼c and vnþ1¼aðvnÞ fornd0, that is, v_n¼aⁿðcÞ. Then the v_n’s are increasing and bounded from above by d. Thus v¼limn!yvn exists and by continuity ofa it follows thata fixesv. In the other situationa:I!J is a strictly decreasing homeo- morphism andIandJ are two open intervals inRsuch thatIVJ is nonempty and J is finite. If I ¼ ða;bÞ, J ¼ ðc;dÞ and wAIVJ, we find that limx!aaðxÞ x¼ da>ww¼0 and limx!baðxÞ x¼cb<ww¼0. By continuity of a it follows that there is avAI such thataðvÞ v¼0, that is,afixesv.

For example, if we assume that ðx1;y1ÞAC0, ðx2;y2ÞACþ, then IJII~¼ ðy;0Þ,II~þ¼ ð1=y2;x2Þis finite andIþVII~þ0 qbecause this intersection contains

the first coordinate of the point of intersection of the positive branch of H and the line segment from ðx1;y₁Þ toðx2;y₂Þ. Moreover,a₁ is strictly decreasing and a₂ is strictly increasing so that a¼a₂a¹₁ :Iþ!a₂ðIÞ is a strictly decreasing homeo- morphism. Since a₂ðIÞJII~þ, we obtain thata₂ðIÞis finite. In order to show that IþVa₂ðIÞis nonempty we distinguish several cases.

If x₁;y₁d0, then I¼II~ and thus a₂ðIÞ ¼II~þ. In case x₁;y₁c0 we have Iþ¼ ð0;þyÞso thata2ðIÞHIþ. Ifx1<0<y1, we haveI¼ ðy;x1ÞandIþ¼ ð0;1=y1Þ; see Table 2. But thena2ðIÞ ¼ ð1=y2;a2ðx1ÞÞand because 0<y1< y2 we obtain thatIþVa2ðIÞ ¼ ð1=y2;minf1=y1;a2ðx1ÞgÞ. Finally, ifx1>0>y1, we have I¼ ð1=y1;0ÞandIþ¼ ðx1;þyÞ; see Table 2. But thena2ðIÞ ¼ ða2ð1=y1Þ;x2Þand because 0<x1<x2we haveIþVa2ðIÞ ¼ ðmaxfx1;a2ð1=y1Þg;x2Þ.

The other cases are dealt with in a similar fashion. In any case one finds thatahas a fixed point.

This finally shows thatðx1;y1Þandðx2;y2Þcan be joined by a line inB. r In order to show thatMðkÞis a flat Minkowski plane we still have to verify that the parallel axiom is satisfied inB, that is, thatBis an a‰ne plane. As a first step in that direction we characterize parallelity inB.

Lemma 4.3.Two lines L_u;vand L_u⁰_;v⁰ are parallel if and only if uv¼u⁰v⁰. Proof. We first assume thatuv0u⁰v⁰. Then the Euclidean lines given by

y¼ 1

kuv xkþ1 2 ðuþvÞ

and y¼ 1

ku⁰v⁰ xkþ1

2 ðu⁰þv⁰Þ

have di¤erent slopes and intersect transversally in a point. Since these Euclidean lines are oblique asymptotes to the linesL_u;vandL_u⁰_;v⁰ by Lemma 4.1 we see thatL_u;vand L_u⁰;v⁰ must also intersect in a point.

Conversely assume thatuv¼u⁰v⁰. Assume thatL_u;vandL_u⁰_;v⁰have a pointðx0;y₀Þ in common. Since we already have a linear space by Lemma 4.2 the two lines are either parallel (that is, L_u;v¼L_u⁰_;v⁰) or ðx0;y₀Þ is the only common point of L_u;v and L_u⁰_;v⁰. In the latter case the asymptotes of L_u;v and L_u⁰;v⁰ are di¤erent parallel Euclidean lines so that Fðu;v;xÞ Fðu⁰;v⁰;xÞ has the same sign for large jxj. This then implies that Lu;v and Lu⁰;v⁰ touch analytically at ðx0;y0Þ, that is,

qF

qxðu;v;x0Þ ¼^qF_qxðu⁰;v⁰;x0Þ.

We now show that Lu;v is uniquely determined by the point ðx0;y0Þ on it and the slope of the Euclidean tangent line at that point, that is, the partial derivative

qF

qxðu;v;x0Þ. Then the second case is not possible and the two lines must be parallel.

By using the groupFwe may assume thatuv¼ 1. Letx0;y0;y₀⁰ AR, y₀⁰ >0. We then have to find a uniqueu<0 such that

Fðu;v;x0Þ ¼ y0

qF

qxðu;v;x₀Þ ¼ y₀⁰;

that is,

y₀¼ 1

uv uþ ðuvÞðx0vÞg_k⁰ðx0vÞ kðgkðx0uÞ gkðx0vÞÞ

¼ 1

uv vþ ðuvÞðx0uÞg_k⁰ðx0uÞ kðgkðx0uÞ gkðx0vÞÞ

y₀⁰ ¼ðvuÞðg_k⁰ðx0uÞgkðx0vÞ g_kðuÞg_k⁰ðx0vÞÞ uvðgkðx0uÞ g_kðx0vÞÞ²

¼ ðvuÞ²g_k⁰ðx0uÞg_k⁰ðx0vÞ kuvðgkðx0uÞ g_kðx0vÞÞ²: (Note thatkg_kðxÞ ¼xg_k⁰ðxÞfor allxAR.) Hence

y₀⁰ðx0uÞðx0vÞ þkðuy01Þðvy01Þ ¼0:

Butv¼ 1=u so that after multiplying through byuwe obtain the quadratic equa- tion

ðky0þx₀y₀⁰Þðu²1Þ þ ðkðy²₀1Þ ðx²₀1Þy₀⁰Þu¼0

for u. If ky₀þx₀y₀⁰00, then the above equation has precisely one positive and one negative zero (the coe‰cients of the quadratic and the constant terms have opposite signs). Thus there is at most oneu<0 that satisfies our two equations. If ky0þx0y₀⁰ ¼0, then we must also have kðy₀²1Þ ðx²₀1Þy₀⁰ ¼0. These two equations imply that y0¼ x0 and y₀⁰ ¼k. The first of these identities yields ðx0uÞgkðx0uÞ ¼ ðx0vÞgkðx0vÞ and further x0¼ ðuþvÞ=2. Thus u² 2x0u1¼0 andu¼x0 ffiffiffiffiffiffiffiffiffiffiffiffiffi

x²₀þ1 q

. Hence there is again a uniqueu<0. r Note that the proof of Lemma 4.3 further shows that there is at most one lineL_u;v with a given valueuvthrough a given point, that is, we have the following.

Corollary 4.4.Through each point there is at most one line L_u⁰;v⁰ parallel to a line L_u;v. Proposition 4.5.The derived geometryBofMðkÞatðy;0Þis an a‰ne plane.

Proof. It remains to show that the parallel axiom is satisfied inB. Since Euclidean lines of negative slope and lines of the form Lu;vare graphs of orientation-reversing and orientation-preserving homeomorphisms ofRwe see that a parallel of a Eucli- dean line or ofLu;vinBmust be of the same form. Hence there is a unique parallel in Bto a horizontal line, a vertical line or a Euclidean line of negative slope through a given point. Given a pointðx0;y0Þand a lineLu;vwe know from Lemma 4.3 that any parallel throughðx0;y0Þmust be of the form Lu⁰;v⁰ where u⁰v⁰¼uv. As in the proof of Lemma 4.3 we may assume that uv¼ 1. Then we have to find a unique u⁰<0

such thatFðu⁰;1=u⁰;x₀Þ ¼ y₀. ButhðzÞdefined byhðzÞ ¼Fðz;1=z;x₀ÞforzAR, z<0, is continuous and limz!yhðzÞ ¼ þy and limz!0hðzÞ ¼ y. This shows thath is ontoRand there is at least oneu⁰ such thathðu⁰Þ ¼ y₀. But Corollary 4.4 shows that such au⁰must be unique, that is, there is a unique parallel toL_u;vthrough

ðx0;y₀Þ. r

The transitivity ofSon points not onDimplies that Proposition 4.5 carries over to any point not onD.

Corollary 4.6.Each derived geometry ofMðkÞat a point not on D is an a‰ne plane.

Corollary 3.2 and Corollary 4.6 now imply the following.

Theorem 4.7.Each incidence geometryMðkÞfor k>1 as defined at the beginning of Section2is a flat Minkowski plane.

5 Isomorphism classes and automorphisms

Since each derived a‰ne plane of the classical flat Minkowski plane is Desarguesian, we immediately obtain the following from Lemma 3.3.

Theorem 5.1.No flat Minkowski planeMðkÞis classical.

We now turn to isomorphisms between the planesMðkÞand their automorphisms.

We want to show that, in fact, these planes are mutually non-isomorphic. As a first step in this direction we prove that any isomorphism must respect the point orbits.

Lemma 5.2. Let g:MðkÞ !MðlÞ be an isomorphism between the flat Minkowski planesMðkÞandMðlÞ.Thengtakes the distinguished circle D inMðkÞto the corre- sponding circle D inMðlÞ.

Proof. We assume thatgðDÞ0D. Then MðlÞ admits the 3-dimensional connected groupsSandgSg¹as groups of automorphisms. SinceDis the only circle fixed byS, it follows thatS0gSg¹ and hence that the automorphism groupGðlÞofMðlÞmust be at least 4-dimensional. From the classification of flat Minkowski planes of group dimension at least 4 (see [5] or [4], 4.4.5) we see thatMðlÞmust be classical or that GðlÞfixes two parallel classes. The former case is not possible by Theorem 5.1 and the latter cannot occur sinceSis already transitive on each set of allðGÞ-parallel classes.

r Theorem 5.3.Two flat Minkowski planesMðkÞandMðlÞare isomorphic if and only if k¼l.

Proof. Letg:MðkÞ !MðlÞbe an isomorphism between the flat Minkowski planes MðkÞandMðlÞ. Thengtakes the distinguished circleDinMðkÞto the distinguished