Triangulating the Real Projective Plane
aa r X i v : . [ c s . C G ] D ec a p p o r t (cid:13)(cid:13) d e r e c h e r c h e (cid:13) I SS N - I S RN I NR I A / RR -- -- F R + E N G Thème SYM
INSTITUT NATIONAL DE RECHERCHE EN INFORMATIQUE ET EN AUTOMATIQUE
Triangulating the Real Projective Plane
Mridul Aanjaneya — Monique Teillaud
N° 6296
Septembre 2007 nité de recherche INRIA Sophia Antipolis2004, route des Lucioles, BP 93, 06902 Sophia Antipolis Cedex (France)
Téléphone : +33 4 92 38 77 77 — Télécopie : +33 4 92 38 77 65
Triangulating the Real Projective Plane
Mridul Aanjaneya ∗ , Monique Teillaud † Th`eme SYM — Syst`emes symboliquesProjets GeometricaRapport de recherche n ° Abstract:
We consider the problem of computing a triangulation of the real projective plane P , given a finite point set P = { p , p , . . . , p n } as input. We prove that a triangulation of P always exists if at least six points in P are in general position , i.e., no three of them are collinear.We also design an algorithm for triangulating P if this necessary condition holds. As far as weknow, this is the first computational result on the real projective plane. Key-words:
Computational geometry, triangulation, simplicial complex, projective geometry,algorithm ∗ Department of Computer Science and Engineering, IIT Kharagpur, 721302, India [Email:[email protected]] This problem was conceptualized and solved while the first author was invitedunder the INRIA
Internship Program for a summer visit. † rianguler le plan projectif r´eel R´esum´e :
Nous consid´erons le calcul de la triangulation d’un ensemble fini de points P = { p , p , . . . , p n } dans le plan projectif. Nous d´emontrons que la triangulation existe toujours d`eslors qu’au moins six points de P sont en position g´en´erale, c’est-`a-dire que trois d’entre eux nesont jamais align´es. Nous proposons ´egalement un algorithme pour trianguler P si cette conditionn´ecessaire est remplie. `A notre connaissance, c’est le premier r´esultat algorithmique connu pourle plan projectif r´eel. Mots-cl´es :
G´eom´etrie algorithmique, triangulation, complexe simplicial, g´eom´etrie projective,algorithme riangulating the Real Projective Plane The real projective plane P is in one-to-one correspondence with the set of lines of the vectorspace R . Formally, P is the quotient P = R −{ } / ∼ where the equivalence relation ∼ isdefined as follows: for two points p and p ′ of P , p ∼ p ′ if p = λp ′ for some λ ∈ R −{ } .Triangulations of the real projective plane P have been studied quite well in the past, thoughmainly from a graph-theoretic perspective. A contraction of and edge e in a map M removes e and identifies its two endpoints, if the graph obtained by this operation is simple. M is irreducible if none of its edges can be contracted. Barnette [1] proved that the real projective plane admitsexactly two irreducible triangulations, which are the complete graph K with six vertices and K + K (i.e., the quadrangulation by K with each face subdivided by a single vertex), which areshown in Figure 1. Note that these figures are just graphs, i.e. the horizontal and vertical linesdo not imply collinearity of the points. Figure 1: The two irreducible triangulations of P .A diagonal flip is an operation which replaces an edge e in the quadrilateral D formed by twofaces sharing e with another diagonal of D (see Figure 2). If the resulting graph is not simple,then we do not apply it. Wagner [18] proved that any two triangulations on the plane with thesame number of vertices can be transformed into each other by a sequence of diagonal flips, upto isotopy. This result has been extended to the torus [5], the real projective plane and the Kleinbottle [16]. Moreover, Negami has proved that for any closed surface F , there exists a positiveinteger N ( F ) such that any two triangulations G and G ′ on F with | V ( G ) | = | V ( G ′ ) | ≥ N ( F )can be transformed into each other by a sequence of diagonal flips, up to homeomorphism [14].Mori and Nakamoto [11] gave a linear upper bound of (8 n −
26) on the number of diagonal flipsneeded to transform one triangulation of P into another, up to isotopy. There are many papersconcerning with diagonal flips in triangulations, see [15, 7] for more references.Figure 2: A diagonal flip.In this paper, we address a different problem, which consists in computing a triangulation ofthe real projective plane, given a finite point set P = { p , p , . . . , p n } as input. Definition 1.1
Let us recall background definitions here. More extensive definitions are given forinstance in [19, 9].
RR n ° M. Aanjaneya & M. Teillaud ˆ An (abstract) simplicial complex is a set K together with a collection S of subsets of K called(abstract) simplices such that:1. For all v ∈ K , { v } ∈ S . The sets { v } are called the vertices of K .2. If τ ⊆ σ ∈ S , then τ ∈ S .Note that the property that σ, σ ′ ∈ K ⇒ σ ∩ σ ′ ≤ σ, σ ′ can be deduced from this. ˆ σ is a k -simplex if the number of its vertices is k + 1 . If τ ⊂ σ , τ is called a face of σ . ˆ A triangulation of a topological space X is a simplicial complex K such that the union of itssimplices is homeomorphic to X . All algorithms known to compute a triangulation of a set of points in the Euclidean plane usethe orientation of the space as a fundamental prerequisite. The projective plane is not orientable,thus none of these known algorithm can extend to P .We will always represent P by the sphere model where a point p is same as its diametricallyopposite “copy” (as shown in Figure 3(a)). We will refer to this sphere as the projective sphere . A triangulation of the real projective plane P is a simplicial complex such that each face is boundedby a 3-cycle, and each edge can be seen as a greater arc on the projective sphere. We will alsosometimes refer to a triangulation of the projective plane as a projective triangulation . a point p q rpr q z=0 (a) (b)V(p) V(q) V(r)
Figure 3: (a) The sphere model of P . (b) △ pqr separated from its “copy” by a distinguishingplane in R .Stolfi [17] had described a computational model for geometric computations: the orientedprojective plane , where a point p and its diametrically opposite “copy” on the projective sphere aretreated as two different points. In this model, two diametrically opposite triangles are consideredas different, so, the computed triangulations of the oriented projective plane are actually nottriangulations of P . Identifying in practice a triangle and its opposite in some data-structure isnot straightforward. Let us also mention that the oriented projective model can be pretty costlybecause it involves the duplication of every point, which can be a serious bottleneck on availablememory in practice.The reader should also note that obvious approaches like triangulating the convex hull of thepoints in P and their diametrically opposite “copies” (on the projective sphere) separately willnot work: it may happen that the resulting structure is not a simplicial complex (see Figure 4 forthe most obvious example), so, it is just not a triangulation (see definition 1.1). We assume that the positions of points and lines are stored as homogeneous coordinates in thereal projective plane. Positions of points will be represented by triples ( x, y, z ) (with z = 0) and INRIA riangulating the Real Projective Plane p q rpr qV(p) V(q) V(r)
Figure 4: The convex hull of duplicated points is not a triangulation.their coordinate vectors will be denoted by small letters like p, q, r, . . . . Positions of lines will alsobe represented by triples [ x, y, z ] but their coordinate vectors will be represented by capital letterslike
L, M, N, . . . . We shall also state beforehand whether a given coordinate vector is that of a lineor a point to avoid ambiguity. Point p and line M are incident if and only if the dot product oftheir coordinate vectors p · M = 0. If p and q are two points then the line L = pq can be computedas the cross product p × q of their coordinate vectors. Similarly the intersections of two lines M and N can be computed as the cross product M × N of their coordinate vectors.We denote the line in R corresponding to a point p in P by V ( p ). A plane in R whichseparates △ abc from its diametrically opposite “duplicate copy” on the projective sphere willbe referred to as a distinguishing plane for the given triangle (see Figure 3(b)). Note that adistinguishing plane is not unique for a given triangle. Also note that such a plane is defined onlyfor non-degenerate triangles on the real projective plane. We first prove a necessary condition for the existence of a triangulation of the set P = { p , p , . . . , p n } of P . More precisely, we show that such a triangulation always exists if at least six points in P are in general position , i.e., no three of them are collinear. So if the number of points n in P isvery large, the probability of such a set of six points to exist is high, implying that it is almostalways possible to triangulate P from a point set.We design an algorithm for computing a projective triangulation of P if the above conditionholds. The efficiency of the algorithm is not our main concern in this paper. The existence of analgorithm for computing a triangulation directly in P is our main goal. As far as we know, thisis the first computational result on the real projective plane.The paper is organized as follows. In section 2, we devise an “in-triangle” test for checkingwhether a point p lies inside a given △ abc . In section 3, we first prove that a triangulation of P always exists if at least six points in the given point set P are in general position. We thendescribe our algorithm for triangulating P from points in P . Finally, in section 4, we presentsome open problems and future directions of research in this area. It is well-known that the real projective plane is a non-orientable surface. However, the notion of“interior” of a closed curve exists because the projective plane with a cell (any figure topologicallyequivalent to a disk) cut out is topologically equivalent to a M¨obius band [9]. For a given triangleon the projective plane, we observe that its interior can be defined unambiguously if we associatea distinguishing plane with it. The procedure for associating such a plane with any given trianglewill be described in Section 3. For now we will assume that we have been given △ pqr along with RR n ° M. Aanjaneya & M. Teillaud its distinguishing plane in R . We further assume for simplicity that this plane is z = 0 for thegiven △ pqr (as shown in Figure 3(b)). Consider the three lines V ( p ) , V ( q ) and V ( r ) in R . Theselines give rise to four double cones , three of which are cut by the distinguishing plane. We definethe interior of △ pqr as the double cone in R which is not cut by its distinguishing plane. Basedon the above definition, we define a many-one mapping s : P → R from points in P to pointsin R as follows: s ( p ) = s ( x, y, z ) = (1 , xz , yz ) z = 0;(0 , , yx ) z = 0 , x = 0;(0 , , z = 0 , x = 0 . Given three points a = ( x , y , z ) , b = ( x , y , z ) , c = ( x , y , z ) and a point p = ( x, y, z ), p liesinside △ abc if sign (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) s s s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + sign (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) s s s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + sign (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) s s s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ± △ abc if sign (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) s s s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + sign (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) s s s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + sign (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) s s s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ± s i = s ( x i , y i , z i ) for i = 0 , ,
2, and s = s ( x, y, z ). The function sign ( m ) returns 1 if m ispositive, 0 if m is zero, and − m is negative. The reader should note that similar to the orientedprojective model [17], there is no notion of interior when all a, b, c and p are at infinity. We nowconsider the case when the distinguishing plane in R is αx + βy + γz = 0, where α, β, γ arearbitrary constants. We use a linear transformation matrix M for transforming the given planeinto the plane z = 0 according to the equation M · p ′ = p , where orientation is preserved. Thistransformation takes the coordinate vector p of a point to the vector p ′ . Now the s -mapping ofequation (1) can be used for the “in-triangle” test with the new coordinate vectors, as describedabove. For the case when γ = 0, we have M = − ( β + γ ) √ ( β + γ )( α + β + γ ) α √ ( α + β + γ ) γ √ β + γ αβ √ ( β + γ )( α + β + γ ) β √ ( α + β + γ ) − β √ β + γ αγ √ ( β + γ )( α + β + γ ) γ √ ( α + β + γ ) (3)For the case when γ = 0 , β = 0, we have M = β √ α + β α √ α + β − α √ α + β β √ α + β − (4)Finally, we have the case when γ = β = 0. In this case, we simply make the X -axis the new Y ′ -axis, the Y -axis the new Z ′ -axis, and the Z axis the new X ′ -axis. So our tranformation matrix M is as follows: M = (5)Note that all the transformation matrices given by equations (2), (3) and (4) are orthogonal matrices, i.e., M − = M T . INRIA riangulating the Real Projective Plane We now proceed to discuss our algorithm for triangulating the real projective plane given a pointset P = { p , p , ..., p n } as input. We number the points in this section for diagramatic clarity. Wefirst prove the following simple result for point sets in P : (a) 1 33 1 2 4(b) 1 33 1 (c)4 21 2 35 4 L M ppr q pp q r
Figure 5: (a) Every point set P has a K -quadrangulation , unless ( n −
1) points are collinear. (b)and (c) A K -quadrangulation is sufficient for triangulating the real projective plane. Lemma 3.1
If among every set of four points in the point set P at least three points are collinear,then at least ( n − points in P are collinear. Proof:
It is easy to see that the lemma holds if all points in P are collinear. So we may safelyassume that this is not the case. We prove the above lemma by the method of contradiction.Assuming that no set of ( n −
1) points are collinear. Consider a set of four points { , , , } asshown in Figure 5(a). Since among every set of four points, at least three are collinear, so weassume that 1 , L = 12. From the given condition, it must lie on the line M = 14. Butnow no three points among the set { , , , } are collinear, a contradiction! Corollary 3.1 If ( n − points are not collinear in the given point set P , then there exists a setof four points in P , no three of which are collinear. We call such a set of four points a K -quadrangulation . Corollary 3.1 states that every point set P in which no ( n −
1) points are collinear contains a K -quadrangulation. We now make thefollowing important observation that such a point set can be used to construct a triangulation ofthe projective plane (see Figure 5(b,c)). We have the following lemma: Lemma 3.2 A K -quadrangulation can be used to construct a projective triangulation. Proof:
Consider the points of the K -quadrangulation on the projective sphere and construct thelines (great circles) { , , , , , } (see Figure 5(b,c)). The intersection of these six linesdefine three more points { p, q, r } . We call these points pseudo-points because these may or maynot be points in P . It is now easy to see that the resulting triangulation is a simplicial complexand is isomorphic to the projective triangulation shown in Figure 1(a).The reader should observe that every triangle in the above triangulation has precisely twocopies on the projective sphere which are diametrically opposite (see Figure 5(b,c)). So it nowbecomes possible to associate a distinguishing plane with each triangle in the above triangulationunambiguously. For every △ abc in the projective triangulation, we can take the plane throughthe center O of the projective sphere and parallel to the plane passing through the end-points a, b, c of one copy of △ abc . Given a query point u , we can now determine the triangle inside whichit lies. We will use this fact quite extensively in our algorithm. The procedure described aboveis incomplete in the sense that we triangulate the real projective plane with the help of somepseudo-points. We now give a necessary condition for computing a projective triangulation froma point set P . The reader should observe that every triangle in Figure 5(b,c) is incident withexactly one pseudo-point. We will refer to the set of triangles incident to one pseudo-point as a RR n ° M. Aanjaneya & M. Teillaudregion . Note that any two regions have the same set of vertices. For constructing a projectivetriangulation from P we will initially take help of pseudo-points, but we will go on deleting themas their use is over. We now present the following lemma: pp qr pp q r
11 33 2 45 5 6(c)
Figure 6: (a) and (b) Symmetric cases for constructing a projective triangulation. (c) A canonicalset always exists when six points in P are in general position . Lemma 3.3
If there exists a set of six points ( say, { , , , , , } ) in a given point set P suchthat four of them ( say, S = { , , , } ) form a K -quadrangulation and the other two ( say, { , } ) are in different regions of the projective triangulation formed by S , then it is possible to triangulatethe projective plane using these six points, unless ( n − points in P are collinear. Proof:
We give a constructive proof of the above statement. We first construct a projectivetriangulation with the set { , , , } (as described above). Suppose points 5 and 6 lie in theregions associated with the pseudo-points p and q respectively (see Figure 6(a,b)). We now addpoint 5 and make it adjacent to the vertices of its bounding region, deleting the pseudo-point p and the edges it was incident with. The newly added edges are shown by dashed lines. Thepseudo-points have also been kept for better understanding. We now add point 6 and delete thecorresponding pseudo-point q and the edges it was incident with. Now we intend to delete thepseudo-point r and construct a valid projective triangulation using only points in P . Here wemake the important observation that either the edge 12 or 34 can be flipped. To see this, notethat if flipping of neither of these edges was possible, then 6 must lie to the “left” (as shown inFigure 6(a)) or “right” (as shown in Figure 6(b)) of both the lines 52 and 54, in which case flippingof edges would induce crossings. (Note that we refer to a point being on the “left” or “right” of aline only locally with respect to front half of the projective sphere.) However, the edge 24 lies inbetween these two lines and 6 cannot lie to its left (resp. right). Thus, our claim holds.Suppose the edge 12 can be flipped. We then construct a valid projective triangulation byflipping 12, deleting the pseudo-point r and adding the edge 12 in that region. Observe that theprojective triangulation constructed is isomorphic to that shown in Figure 1(b). In the event that INRIA riangulating the Real Projective Plane , , , P , they must all lie on the line 524, implying that( n −
2) of the points in P are collinear.We will refer to such a K -quadrangulation which has two points of P in different regions as a canonical set . We now have almost all the basic tools required for triangulating the real projectiveplane from a point set P . All that we need to characterize is the existence of a canonical set. Sofar we have not used anywhere the assumption that at least six points in P are in general position .It turns out that there always exists a canonical set in P in this case. We have the followinglemma: Lemma 3.4
If at least six points in P are in general position, then there exists a canonical set. Proof:
We prove this lemma by the method of contradiction. We assume that the lemma doesnot hold, so for every K -quadrangulation in P , all other points of P are in the same region.Consider a K -quadrangulation { , , , } in P . Suppose we add two more points 5 and 6, andthey lie in the same region (as shown in Figure 6(c)). Now consider the K -quadrangulationformed by { , , , } . If this is to satisfy the property that all points in P lie in exactly one of itsregions, then it is easy to see that 2 must lie on or to the right of the line 54. But now 2 and 6 liein different regions of the K -quadrangulation { , , , } , a contradiction!So we now have a procedure for triangulating the real projective plane given a point set P with at least six points in general position. We summarize our results in the following theorem: Theorem 3.1
Given a point set P = { p , p , . . . , p n } with at least six points in general position,it is always possible to construct a projective triangulation. We now present our algorithm which outputs a triangulation of the real projective plane given apoint set P = { p , p , . . . , p n } with at least six points in general position.1. Find a set S = { , , , , , } of six points such that no three points in S are collinear.2. Construct a projective triangulation with the set S . Associate distinguishing planes withevery triangle of the triangulation.3. for all points p ∈ P\S do
4. Identify the triangle △ abc in which p lies.5. Make p adjacent to the vertices a, b and c . Make the distinguishingplane of △ apb, △ bpc , and △ cpa the same as that for △ abc .6. end for return (triangulation of P ).There are two possible approaches for finding the set S in step 1. In the first approach, wearbitrarily choose a starting point q and initialize our set S = { q } . For any point p ∈ P\S , we add p in S if p is not collinear with any two points in S . We stop when S contains six points. It mayhappen that we are not able to find such a set S of six points if we start with any random startingpoint q . So we iterate over all points in P for choosing the starting point. This approach has aworst-case time complexity of O ( n ). A slightly better approach can be adopted for performingstep 1, which works in O ( n ) time if we assume that the minimum line cover of the point set P is greater than 4. In this approach, we first choose any two points 1 and 2. Let the line definedby them be L . We delete all other points in P on L . We now choose two more points 3 and4. Let the line defined by them be M . We delete all other points in P on M . We also deleteall other points on N (the line defined by 1 and 3) and T (the line defined by 2 and 4). Nowchoose two more points 5 and 6. We now have the required set S = { , , , , , } . It is easyto see that this approach takes O ( n ) time if the minimum line cover of P is greater than 4. The RR n ° M. Aanjaneya & M. Teillaud above two approaches work reasonably well for most point sets. However, for certain point sets,it may happen that both these approaches fail to find such a set S . We are currently unaware ofan optimal method for finding such a set which works in all cases. We believe that some approachsimilar to that used for solving the “ordinary line” problem can be adopted for finding the same(see for instance [10, 12, 3]). After having found such a set S , we find a canonical set within S bya procedure similar to that described in Lemma 3.4.Once we have a canonical set, constructing the projective triangulation in step 2 takes O (1)time. We store the triangulation in a DCEL so that addition and deletion of edges and verticestakes O (1) time. The loop in steps 3-6 runs once for every point p ∈ P\S . Inside the loop, weuse our “in-triangle” test (as described in Section 2) for testing whether a point lies inside a giventriangle. We use a procedure similar to that described by Devillers et al. in [4] for identifying △ abc inside which p lies. We first choose any arbitrary vertex t of the current projective triangulation.We then identify the triangle whose interior is intersected by the line L = tp . This test is performedby checking for all edges E of all triangles sharing vertex t whether the intersection of L and the line described by E lies inside the given triangle (see Figure 7(a)). After having identified thestarting triangle, we move to its neighbor sharing the edge E . In this way, we “walk” in thetriangulation along the line L . We stop when p lies inside the current triangle. Although thismethod of “walking” in a triangulation has a worst-case time complexity of O ( n ), it is reasonablyfast for most practical purposes. So the loop takes a total of O ( n ) steps. Thus, our algorithmcomputes a projective triangulation from a given point set P in O ( n ) steps.As mentioned in the introduction, the complexity of the algorithm is not our main concern inthe present paper. Still, note that our algorithm is incremental, which is an important propertyin practice. O ( n ) is a standard worst-case complexity for incremental algorithms computingtriangulations in the Euclidean plane. After step 2, instead of inserting the points incrementally,we could do the following : for each point, find the triangular face of the initial triangulationcontaining it. Then, in each of these faces, triangulate the set of points using the usual affinemethod. This can be done since the convex hulls of subsets of points in a triangular face of theinitial triangulation can be defined with the help of distinguishing planes. This yields an optimal O ( n log n ) worst-case time (non-incremental) algorithm. p tE L (a) p pq q (b) Figure 7: (a) “Walking” in a projective triangulation. (b) Two copies of the edge pq , only one iscut by the distinguishing plane. It woud be interesting to check whether the metric on P allows to define a triangulation of theprojective plane that would extend the notion of Delaunay triangulation, which is well-known in as suggested by an anonymous reviewer INRIA riangulating the Real Projective Plane Minimum Weight Triangulation [13],
Minmax Length Triangulation [6],etc., may have meaning even on the real projective plane. The
Minimum Weight Triangulation problem was neither known to be NP-Hard nor solvable in polynomial time for a long time [8].This open problem was recently solved and was shown to be NP-Hard by Mulzer and Rote [13].The
Minmax Length Triangulation problem asks about minimizing the maximum edge lengthin a triangulation of a point set P . This problem was shown to be solvable in time O ( n ) byEdelsbrunner and Tan [6]. It would be interesting to analyze the complexity of these problems onthe real projective plane P . Acknowledgements
The authors would like to thank Olivier Devillers for his valuable suggestions and helpful discus-sions.
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Topology for Computing . Cambridge University Press, 2005.
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ISSN 0249-6399 eometri realization of a triangulation on thepro je tive plane with one fa e removedC. Paul Bonnington(cid:3), Atsuhiro Nakamotoyand Kyoji OhbazAbstra tLet M be a map on a surfa e F 2. A geometri realization of M is an embeddingof F 2 into a Eu lidean 3-spa e R3 su h that ea h fa e of M is a (cid:13)at polygon. Weshall prove that every triangulation G on the proje tive plane has a fa e f su h thatthe triangulation of the M(cid:127)obius band obtained from G by removing the interior off has a geometri realization.Keywords: triangulation, geometri realization, M(cid:127)obius band, proje tive plane1 Introdu tionLet F 2 be a surfa e with at most one boundary omponent and let G be a map on F 2.If F 2 has a boundary, we suppose that some y le of G oin ides with the boundary ofF 2. Su h a y le of G is alled the boundary of G and denoted by G. A k- y le meansa y le of length k. A triangulation on F 2 is a map on F 2 su h that ea h fa e is boundedby a 3- y le. A vertex of G not on G is alled an inner vertex. (An inner vertex of apath means a vertex not on the ends.) Throughout this paper, we suppose that the graphof a map is simple, i.e., with no multiple edges and no loops. For a y le or path C in G,a hord of C means an edge xy of G su h that x; y 2 V (C) but xy =2 E(C). A k-wheel isa triangulation on the disk su h that its boundary is a k- y le, alled a rim, and there isonly one inner vertex adja ent to the k verti es on the boundary. We often use symbolswith subs ripts to express verti es and so on, in whi h the subs ripts are taken by suitablemodulus.A geometri realization of a map G on a surfa e F 2 is an embedding of F 2 into aEu lidean 3-spa e R3 su h that ea h fa e of G is a (cid:13)at polygon. We denote a geometri realization of G by ^G. Steinitz's theorem states that a spheri al map has a geometri realization if and only if its graph is 3- onne ted [10℄. Moreover, Ar hdea on et al. provedthat every toroidal triangulation has a geometri realization [1℄. In general, Gr(cid:127)unbaum(cid:3)Department of Mathemati s, University of Au kland, Au kland, New Zealand Email:p.bonningtonau kland.a .nzyDepartment of Mathemati s, Yokohama National University, Yokohama 240-8501, Japan Email:nakamotoedhs.ynu.a .jp This resear h was partially supported by the Ministry of Edu ation, S ien e,Sports and Culture, Grant-in-Aid for Young S ientists (B), 18740045, 2005zYonago National College of Te hnology, Yonago 683-8502, Japan Email: oobayonago-k.a .jp1 2 33 1475 4758 96 887 79 946 51 23Figure 1: A M(cid:127)obius triangulation with no geometri realization onje tured that every triangulation on any orientable losed surfa e has a geometri realization [6℄, but Bokowski et al. re ently showed that a triangulation by K12 on theorientable losed surfa e of genus 6 has no geometri realization [3℄. (See also a goodsurvey paper [5℄.)Let us onsider nonorientable surfa es, in parti ular, the proje tive plane. Sin e theproje tive plane itself is not embeddable in R3 , no map on the proje tive plane hasa geometri realization. However, the surfa e obtained from the proje tive plane byremoving a disk, whi h is a M(cid:127)obius band, is embeddable in R3 , and hen e we an expe tthat a triangulation on the M(cid:127)obius band has a geometri realization. For simple notations,we all a triangulation on the proje tive plane and that of the M(cid:127)obius band a proje tivetriangulation and a M(cid:127)obius triangulation, respe tively.In this paper, we onsider geometri realizations of M(cid:127)obius triangulations. However,Brehm [4℄ has already found a M(cid:127)obius triangulation with no geometri realization shownin Figure 1, in whi h both express the same M(cid:127)obius triangulation. (In Figure 1, identifythe verti es with the same label. In the right side, the shaded part means the hole.)Can we get an aÆrmative result for geometri realizations of M(cid:127)obius triangulations?In this paper, answering su h a question, we prove the following.Let G be a proje tive triangulation and let f be a fa e of G. Let G (cid:0) f denote theM(cid:127)obius triangulation obtained from G by removing the interior of f .THEOREM 1 Every proje tive triangulation G has a fa e f su h that the M(cid:127)obius trian-gulation G (cid:0) f has a geometri realization.Why does Brehm's example have no geometri realization? We an prove that for ea hof its spatial embedding, the two disjoint 3- y les 123 and 456 have a linking number atleast 2. (See [9℄ for the de(cid:12)nition of the linking number.) However, two 3- y les withstraight segments have linking number at most one, a ontradi tion. Hen e we have thefollowing fa t.FACT 2 If a M(cid:127)obius triangulation G has a boundary y le C of length 3 and a 3- y leC0 disjoint from C whi h forms an annular region with C0, then G never has a geometri realization. 2 graph G is said to be y li ally k- onne ted if G has no separating set S (cid:26) V (G) ofG with jSj (cid:20) k (cid:0) 1 su h that ea h onne ted omponent of G (cid:0) S has a y le.Let G be a proje tive triangulation. By Fa t 2, the y li ally 4- onne tedness of G isne essary for a geometri realization of G (cid:0) f for any fa e f of G. We onje ture thatthe ondition is also suÆ ient, as follows.CONJECTURE 3 Let G be a proje tive triangulation. Then G is y li ally 4- onne tedif and only if G (cid:0) f has a geometri realization for any fa e f of G.2 Irredu ible Pro je tive triangulationsLet G be a triangulation on a surfa e and let e be an edge of G. Contra tion of e (or ontra ting e) is to remove e and identify the two endpoints of e. Furthermore, if thisyields a digonal fa e bounded by a 2- y le, then we repla e the pair of parallel edges witha single edge. Let [e℄ denote the vertex in the resulting graph obtained from e by its ontra tion. The inverse operation of the ontra tion of e is a splitting of [e℄. We say thate is ontra tible if the map obtained from G by ontra ting e has no multiple edges andloops, and that G is ontra tible to T if G an be transformed into T by ontra tions ofedges. An irredu ible triangulation is one with no ontra tible edge.See two proje tive triangulations P1 and P2 in Figure 2, in whi h we identify ea hpair of antipodal points of the hexagon. The graphs of P1 and P2 are isomorphi to K6and K4 + (cid:22)K3, respe tively, where (cid:22)Kn denotes a graph with n verti es and no edges. LetM1 be the M(cid:127)obius triangulation obtained from P1 by removing a vertex and (cid:12)ve fa esin ident to it, and let M2 be the one obtained from P2 by removing a vertex of degree 4and four fa es in ident to it. The graph of M1 is isomorphi to K5, and so M1 is alleda K5-triangulation. On the other hand, the graph of M2 in ludes a quadrangulation byK4 with vertex set f1; 2; 3; 4g, whi h is alled a K4-quadrangulation.P1 P2001 1 122 22 13 344 5 a b 1 122 34a bM212 213 54M1Figure 2: Irredu ible proje tive triangulations P1; P2 and their submaps M1; M2LEMMA 4 (Barnette [2℄) The proje tive plane admits exa tly two irredu ible triangu-lations, whi h are P1 and P2 shown in Figure 2.3et us onstru t a geometri realization of an irredu ible proje tive triangulation withone fa e removed. For a point x and a straight segment e not interse ting x in R3 , let(cid:1)(x; e) denote the unique triangular disk in R3 ontaining x and the endpoints of e asits orners. Let M be a map and let ^M denote a geometri realization of M . Fix a pointx in R3 not ontained in ^M . We say that an edge e of ^M is ( ompletely) seen from x if(cid:1)(x; e) interse ts ^M only at e, and that an edge e of ^M is half seen from x if an intervale0 of e ontaining one of the endpoints of e is ompletely seen from x.PROPOSITION 5 Let P be an irredu ible proje tive triangulation. Then P (cid:0) f has ageometri realization for any fa e f of P .Proof. For ea h Pi, we (cid:12)rst onstru t a geometri realization of the M(cid:127)obius triangulationMi ontained in Pi. Figure 3 shows how to onstru t geometri realizations ^M1 and ^M2.In parti ular, we take the following 3-dimensional oordinates in R3 :(cid:15) For ^M1, a = (0; 0; 8), b = ((cid:0)4; 0; 0), = (0; 4; 4), d = (0; (cid:0)4; 4), e = (4; 0; 0)(cid:15) For ^M2, a = ((cid:0)4; 0; 0), b = (0; 0; 2), = (4; 0; 0), d = (0; 0; 4), e = (4; 0; 0)p = (0; (cid:0)4; 1) and q = (0; 4; 1).A AB BC CD D DEa add d d da aE B [ C add A; D add EAb e e e eb b b A BC D E FG Hd dbb a ab bD E B GC F A HD [ E add B; C; F; G add A; Ha d da p q p p pq q qFigure 3: How to onstru t M1 and M2In order to omplete a geometri realization of Pi with one fa e removed, we have onlyto add four triangular disks in ident to ^M1 and three triangular disks to ^M2. Observethat for i = 1; 2, there is only one edge on ^Mi whi h an not be ompletely seen from usin Figure 3. Su h an edge in M1 is d, and the edge in M2 is ab. Therefore, for ^M1,4xing a point x lose to our eyes in R3 (for example, x = (1; (cid:0)1; 20)), we an add fourtriangular disks (cid:1)(x; ab), (cid:1)(x; b ), (cid:1)(x; de) and (cid:1)(x; ea) with no internal ollision of anytwo triangular disks. For ^M2, (cid:12)xing a (cid:12)xed point x = (1; (cid:0)1; 20) in R3 , for example, we an add three triangular disks (cid:1)(x; b ), (cid:1)(x; d) and (cid:1)(x; da). Thus, ea h Pi with somesingle fa e removed has a geometri realization.It is easy to see that for any two fa es f and f 0 of Pi (i = 1; 2), there exists ahomeomorphism over the proje tive plane arrying f into f 0, and hen e ea h of P1 andP2 with any single fa e removed has a geometri realization.REMARK 6 In the geometri realization of M1 shown in Figure 3, b and are ontainedin a half-spa e of R3 with respe t to a plane ontaining a; d and e.3 Planar pat hesA plane graph G with boundary walk C of length m (cid:21) 3 is said to be a near triangulationand denoted by (G; C) if C is a y le and if ea h inner fa e is a triangle. Let v0; : : : ; vk(cid:0)1be k distin t verti es, alled the nodes, spe i(cid:12)ed on G in this y li order, where vi andvi+1 are not ne essarily adja ent in C, for ea h i. Let Pi be the path, alled a segment , onC joining vi and vi+1 but not ontaining vi+2, for i = 0; : : : ; k (cid:0) 1. A near triangulation(G; C) is said to be good if no Pi has a hord.LEMMA 7 Let (G; C) be a near triangulation. If C has no hord, then G (cid:0) V (C) is onne ted or empty. Moreover, ea h vertex on C is in ident to G (cid:0) V (C).Proof. We use indu tion on jV (G)j. If jV (G)j = 3, then the lemma learly holdssin e G (cid:0) V (C) is empty. Hen e we suppose that jV (G)j (cid:21) 4. Let v 2 V (C) and letP = v1 (cid:1) (cid:1) (cid:1) vk be the path of G onsisting of the neighbors of v, where v1; vk 2 V (C).Sin e G has no hord, we have k (cid:21) 3. Let G0 = G (cid:0) v, whi h is a near triangulationsin e G is hordless. Observe that G0 might have hords in ident to v2; : : : ; vk(cid:0)1. LetB1; : : : ; Bm be the hordless near triangulations ontained in G0 sharing these hords oneanother, where m (cid:21) 1. By indu tion hypothesis, the interior (cid:22)Bi of ea h Bi is onne tedor empty. Moreover, every Bi has some vj on its boundary for j 2 f2; : : : ; k (cid:0) 1g, andhen e the vertex vj is in ident to (cid:22)Bi. Therefore, (cid:22)G = G (cid:0) V (C) is onne ted, sin eV ( (cid:22)G) = fv2; : : : ; vk(cid:0)1g [ V ( (cid:22)B1) [ (cid:1) (cid:1) (cid:1) [ V ( (cid:22)Bm). Clearly, every vertex on C is in ident to(cid:22)G, sin e C has no hord.We often use the following lemma to (cid:12)nd a suitable subgraph in a near triangulationor a M(cid:127)obius triangulation. Let (G; C) be a near triangulation and let x; y 2 V (C). Apath joining x and y and interse ting C only at x and y is alled an internal (x; y)-path.The following immediately follows from Lemma 7.LEMMA 8 Let (G; C) be a near triangulation and let x; y 2 V (C) with xy =2 E(C).There is an internal (x; y)-path in (G; C) if and only if there is no hord pq for anyp; q 2 V (C) (cid:0) fx; yg su h that x; p; y; q appear on C in this y li order.5n embedding f : C ! R3 is alled an exhibition if f (Pi) is a straight segment inR3 for i = 0; : : : ; k (cid:0) 1, and if f (C) is proje ted to some hyperplane as a onvex k-gon.Note that ea h vertex of C whi h is not a node is also (cid:12)xed by f . This notion was (cid:12)rstintrodu ed in [1℄.LEMMA 9 Let (G; C) be a good near triangulation with nodes v0; : : : ; vk(cid:0)1 lying on C inthis y li order, where k (cid:21) 3, and let f : C ! R3 be an exhibition. Then f (C) extendsto a geometri realization of (G; C) ontained in the interior of the onvex hull of f (C).Proof. We use indu tion on jV (G)j. If jV (G)j = jV (C)j = 3, then we have nothing todo. If jV (C)j = 3 and jV (G)j (cid:21) 4, then (G; C) is a maximal plane graph, and f (C) is ontained in a plane in R3 . It has already been proved in [11℄ that a maximal plane graphhas a straight-line embedding on the plane with a given outer triangle.So we suppose that jV (C)j (cid:21) 4. Then we an take a hordless internal (x; y)-pathP for some verti es x; y 2 V (C) belonging to distin t segments of C. (Su h a P an betaken, by Lemma 8.) Consider two good near triangulations (G1; C1) and (G2; C2) su hthat V (G1) [ V (G2) = V (G) and V (G1) \ V (G2) = V (P ). Now, regarding x and y asnew nodes and joining x and y of f (C) by a straight-line segment orresponding to P inR3 , we an naturally obtain the exhibitions of C1 and C2, say f (C1) and f (C2). Thenthe onvex hulls of f (C1) and f (C2) interse t only at P . By indu tion hypothesis, G1and G2 have geometri realizations ^G1 and ^G2 extended from f (C1) and f (C2), whi h are ontained in the onvex hulls of f (C1) and f (C2), respe tively. So, ^G1 [ ^G2 is a geometri realization of G ontaining f (C) and ontained in the onvex hull of f (C).Figure 4 shows an example of a near triangulation D and its geometri realizationextended from an exhibition of D.v0 v1v2 x yz w v0 v1v2x w yzFigure 4: A geometri realization of DLEMMA 10 Let (G; C) be a good near triangulation with nodes v0; v1; v2, where v0v1 2E(C). Suppose that G has a fa e bounded by vv0v1, where v 2 V (G (cid:0) C). Let D beany triangular disk with orners u0; u1; u2. Then, for any inner point p of D, there isa straight-line embedding of G on D with ea h segment of C straight su h that ea h vi oin ides with ui, for i = 0; 1; 2, and that v oin ides with p.6roof. Observe that G has a hordless path, say P , from v to some u 2 V (C) (cid:0) V (P0),sin e G has no hord v0v1. We may suppose that u 2 V (P1) (cid:0) fv1g. Then the three pathsvv0; vv1; P de ompose G into three good near triangulations. In D, even if we spe ifyany inner point p as the position of v, we an hoose a point q on the segment u1u2 ofD as the position of u so that the three straight segments pu0; pu1; pq de ompose D intothree onvex polygons. Sin e ea h boundary of the three polygons is an exhibition, we an apply Lemma 9 to them.Note that Lemma 10 does not ne essarily hold if an embedding of C into D is (cid:12)xed.However, moving the positions of inner verti es of the segments v1v2 and v0v2 on Csuitably, we an obtain a required embedding of G into D.4 Pro je tive triangulations with K4-quadrangulationsWe lassify the proje tive triangulations into those with a K4-quadrangulation as a sub-graph and those not ontaining it. In this se tion, we deal with the former.We (cid:12)rst put an important lemma.LEMMA 11 (Menger [7℄) Let G be a graph and let v; v1; : : : ; vk be distin t verti es ofG. Then G has k disjoint paths from v to vi for i = 1; : : : ; k, if and only if there is noS (cid:26) V (G) (cid:0) fv; v1; : : : ; vkg separating v and fv1; : : : ; vkg in G su h that jSj < k.LEMMA 12 If a proje tive triangulation G has a K4-quadrangulation K as a subgraph,then G has a subdivision of P2 ontaining K su h that ea h path of G orresponding toan edge of P2 is hordless.Proof. Let K be a K4-quadrangulation ontained in G and let A1; A2 and A3 be thethree fa ial 4- y les of K. Let A1 = ab d and let B1 be the plane subgraph of G withboundary A1.We take an inner vertex v1 of B1 not ontained in the interior of any separating 3- y leof B1, as follows. Sin e G is simple, we have a ; bd =2 E(B1), and hen e B1 has an innervertex. If B1 has no separating 3- y le, then let v1 be any inner vertex. Otherwise, letC be a separating 3- y le of B1 ontaining a maximal number of verti es in the interior.Sin e a ; bd =2 E(B1), at least one vertex of C, say v, is not ontained in A1. By themaximality of C, v is not surrounded by any separating 3- y le. Therefore, let v1 = v.Then, by Lemma 11, there are four disjoint paths Pa; Pb; P ; Pd from v1 to a; b; ; d inB1. Note that ea h of the four paths an be taken to be hordless if they are hosen tobe shortest. Similarly, we an (cid:12)nd a vertex vi in the interior of Ai whi h has four disjointpaths to the four verti es of Ai, for i = 2; 3. Therefore, G ontains a subdivision of P2 ontaining K.PROPOSITION 13 If a proje tive triangulation G has a K4-quadrangulation as a sub-graph, then G (cid:0) f has a geometri realization for some fa e f of G.7roof. By Lemma 12, G has a subdivision of P2 ontaining a K4-quadrangulation K.Let V (K) = fa; b; ; dg and let fv1; v2; v3g be the verti es of G orresponding to (cid:22)K3, asin the above lemma. Suppose that v1 is in the interior of the 4- y le ab d. Let D denotethe plane subgraph of G bounded by Pa; Pb; ab, where Pa and Pb are the hordless pathsfrom v1 to a and b, respe tively. Let f be the fa e of D in ident to ab. Let G0 be theM(cid:127)obius triangulation obtained from G by removing the interior of D. By Proposition 5and Lemma 9, G0 has a geometri realization ^G0, sin e ea h disk orresponding to a fa eof P2 is a good near triangulation. Even if D onsists of several fa es of G, we an putD (cid:0) f in ^G0 avoiding a ollision of fa es, by Lemma 10, moving the positions of innerverti es of Pa and Pb suitably.LEMMA 14 (Mukae and Nakamoto [8℄) If a proje tive triangulation G has no K4-quadrangulation as a subgraph, then G is ontra tible to P1.5 Split-K5's in pro je tive triangulationsConsider the irredu ible proje tive triangulation P1 isomorphi to K6. Let v be a vertex ofP1 and let Lv be the link of v. Let G be a proje tive triangulation ontra tible to P1. ThenG has a y le, say C, whi h is ontra ted to Lv. We all C a rim in G. Clearly, uttingG along a rim C, we an obtain a M(cid:127)obius triangulation GM and a near triangulation GDboth of whose boundaries are C su h that GM is ontra tible to a K5-triangulation M1,and GD to a 5-wheel W5.Let M1 be a K5-triangulation with verti es v0; v1; v2; v3; v4, where M1 = v0v1v2v3v4is alled the boundary. Consider a splitting of a vertex vi of M1. Then there are twoways of splitting vi into two verti es vi and v0i of degree 3. The splitting of vi is alleda boundary splitting when both vi and v0i lie on the boundary, where we always supposethat vi; v0i; vi+1 lie on M1 in this order. Otherwise, it is alled an inner splitting , wherewe always suppose that vi lies on M1. A split-K5 is a map H obtained from M1 byapplying either a boundary splitting, an inner splitting, or no splitting to ea h vertex ofM1, and subdividing edges. We all a vertex of H whose degree is greater than 2 is a nodeand a path in H ontaining only two nodes as its two endpoints a segment. Clearly, GM ontains a split-K5 H as a subgraph so that H = GM . We say that a vertex v of GMis alled a boundary split node (resp., an inner split node) if v is a node of H arisen by aboundary (resp., inner) splitting. Moreover, fvi; v0ig is alled a boundary split pair (resp.,an inner split node) if they arose by a boundary (resp., inner) splitting. (See Figure 5.)Throughout this paper, let P ((cid:11); (cid:12)) denote the path of H ontained in a segmentjoining two verti es (cid:11) and (cid:12). The following is the main result in this se tion.LEMMA 15 Let G be a proje tive triangulation ontra tible to P1. Then G ontains arim C separating G into a M(cid:127)obius triangulation GM and a near triangulation GD both ofwhose boundaries are C su h that(i) GM has a split-K5 H with either none or a single boundary split pair su h thatH = C (in the former ase, the nodes of H lying on C are v0; v1; v2; v3; v4, and inthe latter ase, those are v0; v00; v1; v2; v3; v4, where fv0; v00g is a boundary split pair),and 80 v1 v2 v3v3 v4 v0 v0 v1 v2 v3v3 v4 v0K5 Split-K5v01 v02v03 v03Inner split pair Boundary split pairFigure 5: K5 and split-K5(ii) in (GD; C), for any (cid:12)xed k 2 f0; 1; 2; 3; 4g, there exists a vertex v 2 V (GD (cid:0)C) whi hhas (cid:12)ve hordless disjoint paths Qi to vi interse ting C only at vi, for i = 0; 1; 2; 3; 4,su h that Qk has just an edge.Proof. We (cid:12)rst prove (i). Suppose that fa0; a00g is a boundary split pair arisen froma node a, as shown in Figure 6. Then a triangular region ab of M1 orresponds to aregion Q with nodes a0; a00; b; . Note that if we take P (a0; a00) to be shortest in G, then Qhas no edge yz with y 2 V (P (a0; )) (cid:0) fa0g and z 2 V (P (a0; a00)) (cid:0) fa0g. (For otherwise,taking z instead of a0, we an take a smaller region orresponding to Q.) Therefore, byLemma 8, we an take an internal (a0; x)-path P for some x on either P (a00; b) (cid:0) fa00gor P (b; ) (cid:0) fb; g. In the former ase, putting H0 = H (cid:0) P (a00; x) [ P , we an de reasethe number of boundary split pairs. In the latter ase, onsider H0 = H (cid:0) P (a00; b) [ P ,in whi h b an be regarded as a boundary split node but a is not. Hen e we an gatherseveral boundary split nodes into at most one node.ab a00b a0x xz yFigure 6: Move the position of a boundary split nodeNext we onsider (ii) in whi h we may suppose that k = 0, and that C is hosen inG so that GD has a minimal number of verti es. Then, GD has no hord. For a simplenotation, we put GD = D.If (D; C) has only one inner vertex, then (D; C) is a wheel, by Lemma 7, and hen ethe lemma learly holds. Therefore, (D; C) has at least two inner verti es. Pi k up anyneighbor v of v0 with v =2 V (C). This is possible sin e C has no hord in (D; C). Sin e vand v0 are adja ent in D, we want to take the remaining four disjoint paths Qi from v to9i su h that Qi interse ts C only at vi, for i = 1; 2; 3; 4. If this is impossible, one of thefollowing holds for any v, by Lemma 11,(1) There is S (cid:26) V (D (cid:0) C) (cid:0) fvg with jSj (cid:20) 3 separating fvg from fv1; v2; v3; v4g in D0for any v,(2) there is S (cid:26) V (D (cid:0) C) (cid:0) fvg with jSj (cid:20) 2 separating fv; v1g from fv2; v3; v4g (orfv; v4g from fv1; v2; v3g) in D0 for any v, or(3) there is S (cid:26) V (D (cid:0) C) (cid:0) fvg with jSj (cid:20) 1 separating fv; v1; v4g from fv2; v3g,(fv; v1; v2g from fv3; v4g, or fv; v3; v4g from fv1; v2g) in D0 for any v.v0v4v3 v2 v1ab pq v v0v4v3 v2 v1abp v(1) (2) (3)r q v0v4v3 v2 v1ab p vJ J JFigure 7: Stru tures of DWe shall prove that in ea h of the above ases, we an take another rim C0 bonding afewer number of verti es, and then getting a ontradi tion to the minimality of C.(1) Let S = fp; q; rg. (The ase when jSj (cid:20) 2 is similar.) Then there is a pathL = apqrb of length exa tly four, where a 2 V (P0) (cid:0) fv0; v1g and b 2 V (P4) (cid:0) fv0; v4g.Here we hoose a and b to be nearest to v0 on P0 and P4, respe tively. See Figure 7(1).Take the unique path J joining a and b and onsisting of neighbors of p; q; r lying on theside ontaining v0 with respe t to the ut L of (D; C), and let C0 = J [ P (a; v1) [ P1 [ P2 [P3 [ P (v4; b). Then C0 has no hord, and hen e the near triangulation with boundary C0is a minor of a wheel with rim C0, by Lemma 7. Moreover, spe ifying b = v0 and addingP (v0; b) to H, we an take a split-K5 with boundary C0, ontrary to the minimality ofD. Now onsider the ase when there is v00 on P0. If v00 is on P (a; v1), then spe ify v00 asbefore, and otherwise, add the path P (v00; a) to H and spe ify a = v00.(2) Let S = fp; qg. Similarly to the ase (1), we an (cid:12)nd a path L = apqb of lengthexa tly three, where a 2 V (P1) (cid:0) fv2g and b 2 V (P4) (cid:0) fv0; v4g. See Figure 7(2). Takethe unique path J onsisting of neighbors of p and q lying on the side ontaining v0 withrespe t to L, and onsider the smaller plane graph with no hord and in luding p and q.Then we an spe ify a = v1 and b = v0 and add P (v0; b) and P (v1; a) to H. If there is v00on P0, then add P (v00; v0) to H. Another ase is similar.(3) Take S = fpg and onsider the path L = apb, where a 2 V (P1) (cid:0) fv2g andb 2 V (P3) (cid:0) fv3g. See Figure 7(3). Take the unique path J onsisting of the neighbors ofp lying on the side ontaining v0 with respe t to L, and onsider the smaller plane graphwith no hord and in luding p. Then we an spe ify a = v1 and b = v4. Moreover, sin e10 has no hord, we an (cid:12)nd a path from v0 to p interse ting C only at v0, by Lemma 8,and hen e we an spe ify one of inner verti es on J (cid:0) fa; bg as a new v0. If there is v00 onP0, then add P (v00; v0) to H. Other ases are similar.6 Lemmas for geometri realizationLet H be a split-K5 on the M(cid:127)obius band and let v be a node of degree 4 on H. Supposethat H has a geometri realization ^H. Let A; B and C be three (cid:13)at fa es of ^H in identto v in this order. In ^H, v is said to be of type I if C bends to the same side of A withrespe t to B, and of type II otherwise. (See Figure 8, where the left and the right shownodes v of type I and II, respe tively.) In ^M1 shown in Figure 5, only the vertex a is oftype II and others are of type I.v C BA v A BCFigure 8: Verti es of degree 4 of type I and IISuppose that a node v is of type I, where A \ B = e and B \ C = e0, and a = A \ Hand = C \ H. For ^H, take a half-plane (cid:25)A parallel to A with boundary the straightline a but not ontaining e, and a half-plane (cid:25)C parallel to C with boundary the straightline but not ontaining e0. It is easy to see that if (cid:25)A and (cid:25)C share a half-line, `A;C,with endpoint v, then an inner splitting of v in ^H an be realized without hanging thepositions of all nodes of ^H. Su h a ondition is alled an inner splitting ondition of vin ^H. The following gives a riterion for a node of type I satisfying the inner splitting ondition.OBSERVATION 16 A node v of type I satis(cid:12)es an inner splitting ondition if e and e0 is ontained in a half-spa e of R3 with respe t to a plane ontaining a and .See Figure 9, where (a) is an inner splitting and (b) is a boundary splitting of a vertexof type I. (On the other hand, the left of Figure 8 is a vertex of type I whi h does notsatis(cid:12)es an inner splitting ondition, sin e (cid:25)C ollides with A.)Let f be a fa e of a geometri realization ^H su h that the boundary of f ontainsexa tly k nodes. We say that f is onvex if f is a onvex k-gon whose nodes are verti esof the k-gon f . We say that ^H is onvex if ea h fa e of ^H is onvex.LEMMA 17 Suppose that H0 is obtained from H by a single inner splitting of a vertex vof degree 4, and H has a onvex geometri realization ^H. If v is of type I satisfying aninner splitting ondition in ^H, then H0 has a onvex geometri realization.11b)(a) B AC (cid:25)A(cid:25)Caee0 vv vv0v0 `A;CFigure 9: Inner and boundary splittings of a vertex of type IProof. Put v0 on `A;C lose to v, as shown in Figure 9.LEMMA 18 Suppose that H0 is obtained from H by a single boundary splitting of a vertexv of degree 4, and H has a onvex geometri realization ^H. Then H0 always has a onvexgeometri realization.Proof. Put v0 on e0, and move the position of v along e slightly, as shown in Figure 9.7 Pro je tive triangulations ontra tible to P1In this se tion, we shall onstru t a geometri realization of a proje tive triangulation G ontra tible to P1 with one fa e removed. By Lemma 15, G has a split-K5 H with bound-ary C, in whi h C has exa tly (cid:12)ve nodes v0; v1; v2; v3; v4 or six nodes v0; v00; v1; v2; v3; v4.Let GM and GD be the M(cid:127)obius triangulation and the good near triangulation obtainedfrom G by utting along C. Let Pi be the path on C joining vi and vi+1, not ontainingvi+2, for i = 0; 1; 2; 3; 4.LEMMA 19 GM has a geometri realization ^GM su h that ea h segment on C is astraight-line segment, and that from some (cid:12)xed point x in R3 , an edge, say e, on P4in ident to v0 is half seen, and all other edges on ^GM an be ompletely seen. In parti -ular, if P0 bends at v00, then the onvex hull of P0 an be ompletely seen from x.Proof. For onstru ting a geometri realization of GM , we use that of M1 shown in thetop of Figure 3, where b and e are of type I satisfying the inner splitting ondition, byRemark 6 and Observation 16. Note that a split-K5 H in GM has at most one boundarysplit pair on C, by Lemma 15.Case 1. None of v0; : : : ; v4 are inner split nodes.If C has no boundary split pair either, then H is homeomorphi to M1, and hen e Hhas a geometri realization ^H, as shown in the left of Figure 10. Note that ^H is onvexsin e ea h fa e of H is triangular. In ^H, (cid:12)x the position of the verti es of C so that onlyone edge e of C an be half seen and all other edges of C an be ompletely seen fromsome point x = (1; (cid:0)1; 20) in R3 . Sin e the boundary of ea h fa e of H gives an exhibitionin R3 , we an get a required geometri realization of GM , by Lemma 9.121 v0v4 v3v2 v1 v00v4 v3v2 v0H H0Figure 10: Geometri realizations of H and H0Let H0 be a split-K5 obtained from the above H by a boundary splitting of v0. ByLemma 18, a boundary splitting at v0 an be applied freely so that the resulting geometri realization ^H0 is onvex. In parti ular, the onvex hull of the bent P0 in ^H0 an be ompletely seen, sin e the segment v0v2 of ^H an be seen from (1; (cid:0)1; 20). Therefore, byLemma 9, GM has a required geometri realization, as shown in the right of Figure 10.Case 2. Both v0 and v4 are inner split nodes.If ea h of v1; v2; v3 is applied no splitting, then see Figure 11, where we give oordinatesof the verti es, as follows:v0 = ((cid:0)1; 3; 3), v00 = (1; 4; 4), v1 = ((cid:0)4; 0; 0), v2 = (0; 0; 6),v3 = (4; 0; 0), v4 = (1; (cid:0)3; 3), v04 = ((cid:0)1; (cid:0)4; 4)Observe that two quadrilateral regions v1v3v4v04 and v1v3v00v0 of ^H are (cid:13)at onvex quadri-laterals in R3 , sin e their xz- oordinates are ((cid:0)4; 0), (4; 0), ((cid:6)1; 3), ((cid:7)1; 4). Let R be thepentagonal region v0v00v2v04v4 of H. Then the embedding of R in R3 is an exhibition,sin e the proje tion of R into the plane x = 0 is a onvex pentagon whose verti es haveyz- oordinates: (3; 3), (4; 4), (0; 6), ((cid:0)4; 4), ((cid:0)3; 3). Hen e, sin e other regions of H aretriangular in R3 , we get a geometri realization of GM , by Lemma 9. However, seeingthe body from (1; (cid:0)1; 20), then the onvex hull of R hides P4 ompletely, and hen e the ondition (ii) of the theorem might not be satis(cid:12)ed.Hen e we (cid:12)nd an auxiliary path of GM , as follows:Claim. We may suppose that GM has an internal (v0; x)-path for R, denoted by P , forsome vertex x 2 P (v4; v04) (cid:0) fv4g. (See the left of Figure 11.)Proof. We may suppose that H is hosen in GM so that the number of inner verti es in Ris minimal. By the symmetry, we shall prove that there is an internal (v0; x)-path for somex 2 V (P (v4; v04)) (cid:0) fv4g or an internal (v4; x0)-path for some x0 2 V (P (v0; v00)) (cid:0) fv0g. Ifneither of them exist, then we have an edge zw su h that z is an inner vertex of P4 and wis an inner vertex of P (v00; v2) [ P (v2; v04), by Lemma 8. By the symmetry, we may supposethat w 2 V (P (v00; v2)) (cid:0) fv00g. Let H0 = H (cid:0) E(P (w; v00)) [ zw, whi h is a split-K5 witha boundary split node v0. This is not the ase. (cid:5)131 v0 v4 v3v3 v2 v1v00 v04 v0v4 v3v2 v00v04v1Figure 11: H with inner split nodes v0 and v4Sin e the internal (v0; x)-path divides R into a triangle xv0v4 and a region, say R0,bounded by v0v00v2v04x, where we might have x = v04. Hen e, take x as an internal point ofthe straight segment v4v04 in R3 and apply Lemma 9 to R0. Then the onvex hull of R0no longer hides P4 ompletely, if we see the body from (1; (cid:0)1; 20). (Figure 11 shows the ase when x = v04 and P is an internal (v00; v04)-path.) Therefore, if we let H0 be H withall edges of R0 added, then H0 has a required onvex geometri realization.Let's he k that v1 and v3 satisfy the inner splitting ondition in ^H0, sin e they arenodes of type I. Sin e ^H0 is symmetri , it suÆ es to deal with only v3. For a plane (cid:25) in R3 ontaining v2; v3 and v4 (i.e., (cid:25) = f(x; y; z) 2 R3 : 3x (cid:0) y + 2z = 12g), all other nodes liesin a half-spa e with respe t to (cid:25). Therefore, by Observation 16, v1 and v3 must satisfythe inner splitting ondition, and hen e, even if v1 or v3 are inner split nodes in H0, thenthe orresponding H0 has a onvex geometri realization, by Lemma 17.Finally, we onsider an inner splitting of v2 in H0. We put v02 = (0; 0; 6 (cid:0) 110 ), forexample, and (cid:12)x all other nodes as above. Then we an he k that the regions boundedby v2v02v04v1 (or v2v02v04v01v1) and v2v02v00v3 (or v2v02v00v03v3) are exhibited, sin e they areproje ted to the plane z = 0 as a triangle or a onvex quadrilateral. Therefore GM has arequired geometri realization, by Lemma 9.Case 3. Other ases.If exa tly one of v0 and v4 is an inner split node, then we may suppose that v4 is,by symmetry. Sin e the onvex hull of v0v2v04v4 does not hide P4, we an deal with this ase similarly to Case 2. (Figure 12 shows H with v0 and v4 applied a boundary splittingand an inner splitting respe tively, and an example of its geometri realization.) The asewhen none of v0 and v4 are applied inner splittings an be dealt with similarly to Case 2,too.The following is our main result in this se tion.PROPOSITION 20 If a proje tive triangulation G is ontra tible to a K6-triangulationP1, then G (cid:0) f has a geometri realization for some fa e f of G.Proof. By Lemma 15, G ontains a split-K5 H with boundary C ontaining (cid:12)ve nodesv0; v1; v2; v3; v4 (or six nodes v0; v00; v1; v2; v3; v4), in whi h the near triangulation, denoted141 v4 v3v1v3 v2 v04v00 v1 v00v0v2 v3v4v04v0Figure 12: H with a boundary split node v0 and an inner split node v4by (GD; C), admits (cid:12)ve hordless disjoint paths Qi (i = 0; 1; 2; 3; 4) from some inner vertexv to vi su h that Q0 is just an edge. These (cid:12)ve paths divide GD into (cid:12)ve plane graphs,denoted by F0, F1, F2, F3, F4, where ea h Fi is in ident to Qi and Qi+1, for i = 0; 1; 2; 3; 4.Here we remove a fa e f bounded by vv0p from F4 in ident to the unique edge of Q0, andlet F 04 = F4 (cid:0) f , where p =2 fv; v0g is a vertex of F4. (See the right side of Figure 13.)Let GM denote the M(cid:127)obius triangulation with boundary C. Then, by Lemma 19, GMhas a geometri realization ^GM su h that ea h Pi is a straight-line segment (only P0 mightbend), and that from some (cid:12)xed point x in R3 , an edge, say e, on P4 in ident to v0 is halfseen, and all other edges on ^GM an be ompletely seen. (See the left side of Figure 13.)v v1v0v4v3 v2 Q1Q0Q4Q3v1 v0v4 v3v2 v Q2pFigure 13: Pasting GD with one fa e removed to ^GMNow, using the (cid:12)xed point x 2 R3 , we (cid:12)rst put F 04 in ^GM , as follows. If p = v4, thenwe have F4 = f , and hen e we have nothing to do. If p is an inner vertex of F4, thenwe put F 04 in ^GM so that F 04 does not ollide with the body, moving the position of innerverti es on P4, by Lemma 10. If p is an inner point of P4, then we apply Lemma 9. Sin ethe edge e in ident to f is half seen and all other edges on P4 are ompletely seen fromx, we an put F 04 in ^GM naturally. If p is an inner point of Q4, then (cid:12)xing p on Q4 in theposition seeing v0, we put F 04 in the body, by Lemma 9.Finally, we shall put F0, F1, F2, F3 in the body of GM [ F 04, making them be (cid:13)attriangular disks by Lemma 9. (If P0 bends at v00, then the onvex hull of v0; v00; v1 an be15ompletely seen from v, by Lemma 19. Hen e we an apply Lemma 9 to F0, sin e thepositions of x; v0; v00; v1 of F0 gives an exhibition.)8 Proof of the theoremProof of Theorem 1. Let G be a proje tive triangulation. By Lemma 14, either G ontains a K4-quadrangulation as a subgraph or it is ontra tible to K6. 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