The generalized triangle inequalities in thick Euclidean buildings of rank 2
aa r X i v : . [ m a t h . M G ] M a y The generalized triangle inequalities in thick Euclideanbuildings of rank 2
Carlos Ramos-Cuevas ∗ April 29, 2011
Abstract
We describe the set of possible vector valued side lengths of n -gons in thick Euclideanbuildings of rank 2. This set is determined by a finite set of homogeneous linear inequal-ities, which we call the generalized triangle inequalities . These inequalities are given interms of the combinatorics of the spherical Coxeter complex associated to the Euclideanbuilding. Let X be a symmetric space of noncompact type or a thick Euclidean building. We are inter-ested in the following geometric question: Which are the possible side lengths of polygons in X ? In this context the appropriate notion of length of an oriented geodesic segment is given bya vector in the Euclidean Weyl chamber ∆ euc associated to X . If X = G/K is a symmetricspace, the full invariant of a segment modulo the action of G is precisely this vector-valuedlength since we can identify X × X/G ∼ = ∆ euc (cf. [6]). For X a Euclidean building the samenotion of vector-valued length can be defined (cf. [7]). We denote by P n ( X ) ⊂ ∆ neuc the set ofall possible ∆ euc -valued side lengths of n -gons in X .One of the motivations for considering this geometric problem comes from the followingalgebraic question. How are the eigenvalues of two Hermitian matrices related to the eigenvalues of their sum?
This so-called
Eigenvalue Problem goes back to 1912 when it was already studied by H.Weyl. It is closely related to a special case of the geometric question above, namely, for thesymmetric space X = SL ( m, C ) /SU ( m ). We refer to [6] for more information on the relationbetween these two questions and [4] for more history on this problem.In [6] and [7] it is shown that the set P n ( X ) depends only on the spherical Coxeter complexassociated to X (i.e. on the spherical Weyl chamber △ sph ). We will therefore sometimes refer ∗ [email protected] P n ( △ sph ) as the set of side lengths of n -gons in X a symmetric space or a Euclidean buildingwith △ sph as spherical Weyl chamber.For a symmetric space X = G/K the set of possible side lengths has been completelydetermined in [6]: P n ( X ) is a finite sided convex polyhedral cone and it can be described as thesolution set of a finite set of homogeneous linear inequalities in terms of the Schubert calculusin the homology of the generalized Grassmannian manifolds associated to the symmetric space G/K . It follows, that for a Euclidean building X ′ with the same associated spherical Weylchamber △ sph as X , the set P n ( X ′ ) is also a finite sided convex polyhedral cone determined bythe same inequalities as P n ( X ) = P n ( △ sph ).As already pointed out in [7] for the case of exotic spherical Coxeter complexes (i.e. whenit is the Coxeter complex of a Euclidean building but it does not occur for a symmetric space)the structure of the set P n ( △ sph ) cannot be described with this method, since we do not havea Schubert calculus for these Coxeter complexes. Thus, the structure of P n ( △ sph ) for theseCoxeter complexes and even its convexity were unknown. It is clear that we can restrict ourattention to irreducible Coxeter complexes. By a result of Tits [12], exotic irreducible Coxetercomplexes occur only in rank 2. Our main result is the description of P n ( X ) in this case(compare with Theorem 6.14). Theorem 1.1.
For a thick Euclidean building X of rank 2, the space P n ( X ) is a finite sidedconvex polyhedral cone. The set of inequalities defining P n ( X ) can be given in terms of thecombinatorics of the spherical Coxeter complex associated to X . The inequalities given in our main theorem coincide with the so-called weak triangle inequal-ities (cf. [6, Sec. 3.8]). Moreover, our arguments also work (see Theorem 6.11) to prove theweak triangle inequalities for buildings of arbitrary rank (cf. [6, Thm. 3.34]). For symmetricspaces, these inequalities correspond to specially simple intersections of Schubert cells in thedescription of P n ( X ) given in [6]. Their description depend only in the Weyl group of X andtherefore, they can be defined for arbitrary Coxeter complexes.Consider the side length map σ : P ol n ( X ) = X n → ∆ neuc . The set P n ( X ) which we areinterested in is nothing else than the image of σ . We use a direct geometric approach todescribe this image. Our main idea is to study the singular values of σ by deforming the sidesof a given polygon in X . This strategy was already used for the case of symmetric spaces byB. Leeb in [10] to give a simple proof of the Thompson Conjecture (cf. [6, Theorem 1.1]). Inthis paper we adapt this variational method to the case of Euclidean buildings and use it todescribe the space P n ( X ). This approach is new in the sense that it does not rely in Schubertcalculus at all.Throughout this paper we state the results, whenever possible, in such a way that theyapply to Euclidean buildings of arbitrary rank. In particular, Sections 4, 5 and 6.1 (exceptLemma 6.6 and Proposition 6.7) do not use the assumption on the rank of the building. Andwhen we do use the assumption, we indicate it explicitly in the statement of the correspondingresult.The set of inequalities obtained in Theorem 1.1 constitute an irredundant system definingthe polyhedral cone P n ( X ). The inequalities given by Schubert calculus in [6] are known tobe irredundant for the cases of type A n (see [9]), however, these seem to be the only cases. A2maller set of inequalities is given in [1] by defining a new product in the cohomology of flagvarieties. The irredundancy of this set has been recently shown in [11].After a first version of this paper was written, the author learned about a recent relatedpaper of Berenstein and Kapovich [2], where the generalized triangle inequalities for rank 2 arealso determined by a different approach. Contents L n
54 Polygons 6 σ
86 The generalized triangle inequalities 11 H L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116.2 The boundary of P n ( X ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 A very good introduction to the concepts used in this paper is the work [8, ch. 2-4]. We referalso to [3] for more information on metric spaces with upper curvature bounds and to [7, ch.2-3] for the different concepts of length in Euclidean buildings.
Recall that a complete geodesic metric space X is said to be CAT (0) if the geodesic trianglesin X are not thicker that the corresponding triangles in the Euclidean space.For two points x, y ∈ X we denote with xy the geodesic segment between them. The linkΣ x X is the completion of the space of directions at x with the angle metric. −→ xy ∈ Σ x X denotes3he direction of the segment xy at x .Two complete geodesic lines γ , γ are said to be parallel if they have finite Hausdorffdistance, or equivalently, if the functions d ( · , γ i ) | γ − i are constant. The parallel set P γ is definedas the union of all geodesic lines parallel to γ . It is a closed convex set that splits as a metricproduct P γ ∼ = R × Y , where Y is also a CAT (0) space.Let ∂ T X denote the Tits boundary of X . For x ∈ X and ξ ∈ ∂ T X we write exp x ( tξ ) todenote the point in the geodesic ray xξ with distance t to x .For a polygon p , or more precisely, an n -gon in X we mean the union of n oriented geodesicsegments x x , . . . , x n − x n with x n = x . Since geodesic segments in CAT (0) spaces betweentwo given points are unique, we can also describe p by its vertices . We write p = ( x , . . . , x n − ).The union q of n oriented geodesic segments x x , . . . , x n − x n where x n = x will be called a polygonal path and we write q = ( x , . . . , x n ). We make now the convention that subindices areto be understood modulo n , whereas superindices are not. A spherical Coxeter complex is a pair ( S, W ) consisting of a unit sphere S with its usual metricand a finite group W of isometries, the Weyl group , generated by reflections on total geodesicspheres of codimension one. The set of fixed points of reflections in W are called walls of( S, W ). A half-apartment or root is a hemisphere bounded by a wall. A
Weyl chamber in S is a fundamental domain of the action W y S . The model Weyl chamber is defined as∆ sph := S/W . We say that two points in S have the same W -type (or just type ) if they belongto the same W -orbit.A Euclidean Coxeter complex is a pair (
E, W aff ) consisting of a Euclidean space E and agroup of isometries W aff , the affine Weyl group , generated by reflections on hyperplanes andsuch that its rotational part W := rot ( W aff ) is finite. The translation subgroup L ⊂ W aff isdefined as the kernel of the map rot : W aff → W . The set of fixed points of reflections in W aff are called walls of ( E, W aff ). A half-apartment or root is a half-space bounded by a wall.We define the W aff -type of a point in E as in the spherical case above. To ( E, W aff ), we canassociate the spherical Coxeter complex (
S, W ), where S := ∂ T E is the Tits boundary of E .The Euclidean model Weyl chamber ∆ euc is the complete Euclidean cone over ∆ sph .The link Σ x E of a point x ∈ E is naturally a spherical Coxeter complex with Weyl group Stab W aff ( x ).The refined length of the oriented geodesic segment xy ⊂ E is defined as the image of ( x, y )under the projection E × E → ( E × E ) /W aff . The ∆ -valued length , or just length , is the imageof the refined length under the natural forgetful map ( E × E ) /W aff → ∆ euc . We denote with σ the length map assigning to a segment its ∆-valued length.We can also define the refined length of an oriented segment xy in the spherical Coxetercomplex ( S, W ) analogously as the image of ( x, y ) under the projection S × S → ( S × S ) /W .4 .3 Buildings For an introduction to spherical and Euclidean buildings from the point of view of metricgeometry, we refer to [8].Let X be a thick Euclidean building modeled in the Euclidean Coxeter complex ( E, W aff ).The concepts of refined length and ∆-valued length of an oriented geodesic segment xy ⊂ X canbe also defined naturally by identifying an apartment containing xy with the Coxeter complex( E, W aff ).For a polygon p = ( x , . . . , x n − ) in X , we write σ ( p ) = ( σ ( x x ) , . . . , σ ( x n − x )) ∈ ∆ neuc and call σ : X n → ∆ neuc the side length map . The space P n ( X ) := σ ( X n ) is the set of possible∆-valued side lengths of n -gons in X . We say that a polygon in X is regular if all its sidesare regular, that is if their ∆-valued lengths lie in the interior of ∆ euc . The space of regularpolygons is an open dense subset of X n .We will use following result from [7] concerning the refined side lengths of polygons in X .We reproduce here its statement for the convenience of the reader. Theorem 2.1 (Transfer theorem).
Let X and X ′ be thick Euclidean buildings modeled onthe same Euclidean Coxeter complex ( E, W aff ) . Let p = ( x , . . . , x n − ) be a polygon in X andlet x ′ x ′ be a segment in X ′ with the same refined length as x x . Then there exists a polygon p ′ = ( x ′ , x ′ , . . . , x ′ n − ) in X ′ with the same refined side lengths as of p . L n We fix a vertex o of ( E, W aff ) with
Stab W aff ( o ) ∼ = W . We obtain in this way a naturalidentification (modulo the action of Stab W aff ( o )) E ∼ = R dim E . By fixing o we get also anembedding W ֒ → W aff and the (coarser) structure ( E, W ) as Euclidean Coxeter complex. Wewill think of the Euclidean Weyl chamber ∆ euc ∼ = E/W as embedded in E , such that ∆ euc isa fundamental domain of the action W y E . Hence, the cone point of ∆ euc corresponds to o and ∂ T ∆ euc is a Weyl chamber of ( ∂ T E, W ).Let η ∈ ∂ T E be a vertex and let v η := exp o (1 · η ). We define the following linear functional: l η : ∆ euc → R v
7→ h ov, ov η i where h· , ·i denotes the standard scalar product on R dim E . We denote with L n the finite setof functionals on ∆ neuc of the form L ( v , . . . , v n − ) = l η ( v ) + · · · + l η n − ( v n − ) where all the η i have the same W -type. We write L = ( l η , . . . , l η n − ) for such a functional.Let H L denote the hyperplane L − (0) ∩ ∆ neuc for L ∈ L n . We call H L a wall in ∆ neuc .The set of walls H L divide ∆ neuc in finitely many convex polyhedral cones. We denote with C n the family of the interiors of these cones, i.e. C n is the set of the connected components ofint(∆ neuc ) \ S L ∈L n H L . 5 Polygons
Let p = ( x , . . . , x n − ) be an n -gon in X . We say that a n -tuple F = ( F , . . . , F n − ) ofapartments in X supports the polygon p if x i x i +1 ⊂ F i and the convex set F i ∩ F i +1 is topdimensional and contains x i +1 in its interior. Remark 4.1. If p is a regular polygon then there always exists an n -tuple F supporting p . F can be constructed as follows: Let A ∈ Σ x X be an apartment containing −−→ x x and −−−−→ x x n − andtake ξ ∈ A antipodal to −−→ x x . Extend the segment x x a little further than x in directionof ξ to a segment x ′ x . Inductively for i = 0 , . . . , n − F i ∈ X to be an apartmentcontaining x ′ i x i +1 and an initial part of x i +1 x i +2 and extend x i +1 x i +2 in F i a little further than x i +1 to a segment x ′ i +1 x i +2 . Finally choose F n − to contain x ′ n − x , an initial part of x x and x x ′ . This last step is possible because of our first choice of x ′ . The polyhedron F i ∩ F i +1 contains a regular segment with x i +1 in its interior. In particular, F i ∩ F i +1 is top dimensional.Let now p be a polygon and F an n -tuple supporting it. Notice that since the convex set F i ∩ F i +1 is top dimensional, there is a unique isomorphism of Coxeter complexes φ i : F i → F i +1 fixing F i ∩ F i +1 pointwise. We also write φ i for the induced isomorphism at the boundary: φ i : S i := ∂ T F i → S i +1 := ∂ T F i +1 . We obtain an associated holonomy map φ p : S i → S i defined as the composition φ p := φ i + n − ◦ · · · ◦ φ i +1 ◦ φ i . We introduce also the following notation: φ ki := φ i + k − ◦ · · · ◦ φ i : F i → F i + k and the analogous for the Tits boundaries. Recall our convention that subindices are takenmodulo n and superindices are not.Notice that the holonomy map φ p : S i → S i is an element of the Weyl group W . In particularthe set of fixed points of φ p is a singular sphere in ( S i , W ). We point out that the holonomymap (and therefore also its fixed points set) depends on the choice of the n -tuple F supporting p . We will make use of this flexibility later. Let p = ( x , . . . , x n − ) be an n -gon in X and let F be an n -tuple supporting it. We wantto construct a special polygonal path ¯ p = ( x , . . . , x n ) contained in the apartment F whichhas the same refined side lengths as p and looks like p at the vertices, that is, such that thesegments −−−→ x i x i − −−−→ x i x i +1 and −−−→ x i x i − −−−→ x i x i +1 have the same refined side lengths (see Fig. 1). For thiswe just take x := x and for i > x i := ( φ i ) − ( x i ) = ( φ i − ) − ( x i ) ∈ F . Recall that φ i | F i ∩ F i +1 = Id . The polygonal path ¯ p so defined has the required properties since x i x i +1 =( φ i ) − ( x i x i +1 ) and in Σ x i F we have −−−→ x i x i − = ( φ i ) − ( −−−→ x i x i − ) and −−−→ x i x i +1 = ( φ i ) − ( −−−→ x i x i +1 ).We remark that in general x n = x and ( x , . . . , x n ) is a polygonal path hence the expression“opening a polygon”. 6 x x x F F x x Figure 1: Opening a triangle in the apartment F This construction was first considered in [5, Sec. 6.1].For simplicity on the notation, suppose p = ( x , x , x ) is a triangle in X . There is apartition y = x , y , . . . , y k = x of the segment x x such that the triangles ( x , y i , y i +1 ) for i = 1 , . . . , k − A i , in particular, they are flat. We define pointsˆ y i in the apartment A inductively as follows: for i = 1 set ˆ y = y = x and suppose we havealready defined ˆ y i . Let β i : A i → A be an isomorphism of Euclidean Coxeter complexes, suchthat β i ( x y i ) = x ˆ y i . We define ˆ y i +1 := β i ( y i +1 ). We say that the polygon ˆ p = ( x , ˆ y , . . . , ˆ y k )is the result of folding the triangle p into A (see Fig. 2). We say that the points ˆ y i for i = 2 , . . . , k − break points of the folded polygon ˆ p . Notice that the segments x x and x x have the same refined side lengths as the segments x ˆ y and x ˆ y k respectively. Writeˆ y = y = x and define ζ i := −−−→ y i y i − and ξ i := −−−→ y i y i +1 , analogously ˆ ζ i := −−−→ ˆ y i ˆ y i − and ˆ ξ i := −−−→ ˆ y i ˆ y i +1 .A billiard triangle is a polygon ˆ p = ( x , ˆ y , . . . , ˆ y k ) in an apartment A such that for i = 2 , . . . , k − ζ i and ˆ ξ i are antipodal in the spherical Coxeter complex(Σ ˆ y i A , Stab W aff (ˆ y i )) modulo the action of the Weyl group Stab W aff (ˆ y i ). Clearly, a foldedtriangle is a billiard triangle. Conversely, the next condition is necessary and sufficient for abilliard triangle to be a folded triangle. For i = 2 , . . . , k − there is a triangle ( ζ ′ i , ξ ′ i , τ ′ i ) in the spherical building Σ ˆ y i X such that d ( ζ ′ i , ξ ′ i ) = π and the refined lengths of ζ ′ i τ ′ i and ξ ′ i τ ′ i are the same as of ˆ ζ i −−→ ˆ y i x and ˆ ξ i −−→ ˆ y i x respectively. We investigate now the relation between the constructions of opening and folding a polygon.Let p = ( x , x , x ) be a triangle in X and let F be a triple supporting p . Observe that we canchoose A = F , A k − = F and β k − = φ . Let ˆ p = (ˆ y , ˆ y , . . . , ˆ y k ) be the folded triangle andlet ¯ p = ( x , x , x ) be the opened triangle with ˆ y = x = x and ˆ y = x = x .Since ˆ p = (ˆ y , ˆ y , . . . , ˆ y k ) is a billiard triangle, there are isometries µ i of F in Stab W aff (ˆ y i )for i = 2 , . . . , k − −−−−−−→ ˆ y i µ i (ˆ y i +1 ) and ˆ ζ i are antipodal in Σ ˆ y i F (Fig. 2). We call the µ i the straightening isometries. It holds µ ◦ · · · ◦ µ k − ◦ µ k − ( x ) = x and µ ◦ · · · ◦ µ k − ◦ µ k − (ˆ y k ) = x . p is regular, then the µ i ’s are unique. x x µ (ˆ y ) x = ˆ y = x x µ µ ˆ y ˆ y ˆ y µ (ˆ y ) Figure 2: Folding and opening a triangleThe constructions for n -gons ( n >
3) are analogous. σ For a regular value of the side length map σ we mean a value s ∈ P n ( X ) for which there isa polygon p with σ ( p ) = s and such that σ is an open map at p . First we give a sufficientcondition in terms of the holonomy map for σ ( p ) being a regular value of σ . Proposition 5.1.
Let p be an n -gon in X and F an n -tuple supporting p . Suppose that theholonomy map φ p : S i → S i has no fixed points, then the space P n ( X ) is a neighborhood of σ ( p ) in ∆ neuc .Proof. Choose ǫ > B x i ( nǫ ) ∩ F i − = B x i ( nǫ ) ∩ F i ⊂ F i − ∩ F i for all i . Recall thatfor x ∈ X and ξ ∈ ∂ T X , exp x ( tξ ) denotes the point in the geodesic ray xξ with distance t to x . Notice that for 0 ≤ t < nǫ and ξ ∈ S i , exp x i ( tξ ) ∈ F i − ∩ F i . The segments x i exp x i ( tξ ) and x i +1 exp x i +1 ( tφ i ( ξ )) are two parallel segments of the same length in the apartment F i .Fix a 0 ≤ j ≤ n − ξ ∈ S j . We want to move the polygon p along the direction ξ ∈ S j in the following sense. Let t < ǫ and set x j + k ( t ) := exp x j + k ( t φ kj ( ξ )) for k ≥
0. Considerthe polygon p ( ξ, t ) := ( x j ( t ) , . . . , x j + n − ( t )) := ( x j ( t ) , . . . , x j + n − ( t )). Notice that for i = j the segment x i − ( t ) x i ( t ) is just a translation in the apartment F i of the segment x i − x i , inparticular, they have the same ∆-valued length. But since the holonomy map has no fixedpoints, then φ nj ( ξ ) = φ p ( ξ ) = ξ and we get (see Fig. 3) σ ( x j − ( t ) x j ( t )) = σ ( x j − ( t ) x j + n ( t ) + x j + n ( t ) x j ( t )) . If we think of x j + n ( t ) x j ( t ) as vectors in the Euclidean space F j − , then the set { x j + n ( t ) x j ( t ) | ξ ∈ S j , ≤ t < nǫ } is a neighborhood of the origin, because φ p has no fixed points. Hence, theset { σ ( x j − ( t ) x j ( t )) | ξ ∈ S j , ≤ t < nǫ } is a neighborhood of σ ( x j − x j ) in ∆ euc . This means8 j − x j x j +1 ( t ) x j + n ( t ) x j − ( t ) x j ( t ) x j +1 ξφ p ( ξ ) Figure 3: Variation of the side x j − x j that we can deform the ∆-valued length of every side of p independently, thus P n ( X ) is aneighborhood of σ ( p ) in ∆ neuc .The next proposition says that for a building with only one vertex the critical values of σ must lie in the walls H L . Proposition 5.2.
Let p be an n -gon in a thick Euclidean building X which has only one vertex.Let F be an n -tuple supporting p . Suppose that the holonomy map φ p : S i → S i fixes a vertexof ( S i , W ) . Then there exists a functional L ∈ L n , such that L ( σ ( p )) = 0 .Proof. Let o ∈ X be the only vertex. We have that W aff ∼ = W and W aff y F fixes o ∈ F for any apartment F ⊂ X . Let η ∈ S be a vertex fixed by φ p : S → S . Now open thepolygon p = ( x , . . . , x n ) in the apartment F to the polygonal path ¯ p = ( x , . . . , x n ). Recallthat x = φ n ( x n ) and φ n = φ p ∈ W aff . Thus, φ p ( o ) = o .Let v := exp o (1 · η ) and let η i ∈ ∂ T E for i = 0 , . . . , n − W -type as η , such that for v i := exp o (1 · η i ) holds l η i ( σ ( x i x i +1 )) = h x i x i +1 , ov i . Set L = ( l η , . . . , l η n ), then L ( σ ( p )) = Z ¯ p h o · , ov i = h ox n , ov i − h ox , ov i = h ox n , ov i − h φ p ( ox n ) , φ p ( ov ) i = 0 . Remark 5.3.
The Figure 4 shows an example of a folded triangle in a thick Euclidean buildingwith more than one vertex, which comes from a genuine triangle p (cf. [5, Sec. 6.1]). Theholonomy map φ p : S → S fixes the direction η but φ p : F → F has no fixed points. Itfollows that for the associated functional as in the proof of Proposition 5.2 holds L ( σ ( p )) = 0.Moreover, if the vertices of p are in a sufficiently general position, one can achieve that σ ( p )does not lie in any of the walls H L . But since the vertices lie in the interior of Weyl alcovesany small variation of p will leave L ◦ σ constant. Hence, σ cannot be open at p . Nevertheless,we have Corollary 5.4 below.We use the result in [7] that P n ( X ) depends only on the spherical Coxeter complex totransfer the result above to arbitrary buildings.9 y ˆ y ˆ y η ˆ y ˆ y x Figure 4: A triangle with a direction fixed by the holonomy map but with L ( σ ( p )) = 0 Corollary 5.4.
Let s ∈ P n ( X ) ∩ int ∆ neuc and suppose that L ( s ) = 0 for all functionals L ∈ L n .Then P n ( X ) is a neighborhood of s in ∆ neuc .Proof. By [7] we may assume that X has only one vertex. Let p be a regular polygon with σ ( p ) = s and let F be an n -tuple supporting p . By Proposition 5.2 the holonomy map has nofixed points. The result now follows from Proposition 5.1. Lemma 5.5.
Let p k be a sequence of regular n -gons in X such that σ ( p k ) → s in ∆ neuc , thenthere exists an n -gon p in X such that σ ( p ) = s .Proof. We assume again that X has only one vertex o ∈ X . Let p k = ( x k , . . . , x kn − ) andlet F k = ( F k , . . . , F kn − ) be n -tuples supporting p k . After transferring the polygons p k (cf.Theorem 2.1) we may assume that the sides x k x k lie in the same apartment F = F k . and that x k lie in the same Euclidean Weyl chamber ∆ euc ⊂ F . We open now the polygons p k in theapartment F to polygonal paths ¯ p k = ( x k, , . . . , x k,n ).If x k, → ∞ in F , then after taking a subsequence the segments ox k, converge to a geodesicray ρ . Let τ ⊂ ∆ euc be the smallest face of ∆ euc containing τ . Then for k big enough, foldingthe polygon p k into F will have break points only on open faces of F containing τ in theirclosure. This implies that p k is contained in the parallel set P f τ of the flat f τ ⊂ F spannedby τ . For instance, if τ = ∆ euc , then p k is contained in the apartment F for k big enough.Thus, after translating the polygons p k in the parallel set P f τ we may assume that x k, stay ina bounded region.Now we can take a subsequence and assume that the polygonal paths ¯ p k converge to apolygonal path ¯ p = ( x , . . . , x n ) with ∆-valued side lengths s . We want now to lift this polygonalpath near the polygons p k . Write ϕ k,i : F → F ki for the maps φ i : F k → F ki defined inSection 4.1. For i = 1 , . . . , n the convex set ( ϕ k,i ) − ( F ki − ∩ F ki ) = ( ϕ k,i − ) − ( F ki − ∩ F ki ) is aunion of Euclidean Weyl chamber with x k,i in its interior. Hence, for k big enough we have x i ∈ ( ϕ k,i ) − ( F ki − ∩ F ki ) = ( ϕ k,i − ) − ( F ki − ∩ F ki ) and we can define the points z k,i := ϕ k,i ( x i ) = ϕ k,i − ( x i ) ∈ F ki − ∩ F ki for i = 1 , . . . , n and z k, := x . Then q k := ( z k, , . . . , z k,n ) is a polygonalpath with the same side lengths as ¯ p , i.e. σ ( q k ) = s . However q k may still not be a closedpolygon.Notice that d ( x k, , z k,n ) = d ( ϕ k,n ( x k,n ) , ϕ k,n ( x n )) = d ( x k,n , x n ) → d ( x k, , x ) → d ( z k,n , z k, ) = d ( z k,n , x ) →
0. On the other hand, observe that x k,n and x k, = ϕ k,n ( x k,n )10ave the same W aff -type and therefore also their limits x = z k, and x n = ( ϕ ) − ( z k,n ). Hence, z k, and z k,n have the same type. But W aff is finite, so d ( z k, , z k,n ) can only take finitely manyvalues. It follows that for k big enough z k, = z k,n and q k is a closed polygon in X with ∆-valuedside lengths s . Corollary 5.6.
For any open cone C ∈ C n the intersection P n ( X ) ∩ C is empty or C . Moreover,if C ⊂ P n ( X ) , then ¯ C ⊂ P n ( X ) .Proof. The intersection P n ( X ) ∩ C is open by Corollary 5.4 and closed by Lemma 5.5. Thesecond assertion also follows from Lemma 5.5. H L Suppose p is a polygon in X with σ ( p ) = s ∈ H L for some functional L ∈ L n . ConsideringCorollary 5.6 the natural question is if there is a cone C ∈ C n such that s ∈ ¯ C ⊂ P n ( X ). Wewould also like to describe all cones in C n with this property. With this in mind we investigatein this section following question. When can we find polygons p ′ with ∆-valued side lengthsnear s and such that L ◦ σ ( p ) > < p . However since a Euclidean building has dimension equal to his rank,we do not have much flexibility to perturbate the polygon (compare with Remark 5.3). Thuswe must be more compliant with the variations of p that we want to admit. Therefore we willoften have to translate the polygon to other place in X where we can perform the perturbations.Let L = ( l η , . . . , l η n − ) be a functional in L n . For the rest of this section p = ( x , . . . , x n − )will be always a regular n -gon such that σ ( p ) ∈ H L , that is, L ( σ ( p )) = 0.Let F be an n -tuple of apartments supporting p . Let ξ i ∈ S i be a vertex such that if v i := exp x i (1 · ξ i ), then l η i ( σ ( x i x i +1 )) = h x i x i +1 , x i v i i . Observe that v i is of the same W -type as η , . . . , η n − . We will therefore sometimes write l η i ◦ σ = h· , x i v i i . Lemma 6.1.
If in the notation above −→ x i ξ i = −−−→ x i ξ i − for some i , then for any neighborhood U of σ ( p ) in ∆ neuc there exist n -gons p , p in X with σ ( p ) , σ ( p ) ∈ U and L ◦ σ ( p ) > > L ◦ σ ( p ) .Proof. The proof is similar to the one of Proposition 5.1. Notice that for small ǫ > x ′ i :=exp x i ( ǫξ i − ) , x ′′ i := exp x i ( ǫξ i ) ∈ F i − ∩ F i and by the hypothesis x ′ i = x ′′ i . Let θ := ∠ x i ( ξ i − , ξ i ).Consider first the polygon p := ( x , . . . , x ′ i , . . . , x n − ), then L ( σ ( p )) = l η ( σ ( x x )) + · · · + h x i − x i + x i x ′ i , x i − v i − i + h x i x i +1 − x i x ′ i , x i v i i + · · · + l η n − ( σ ( x n − x ))= l η ( σ ( x x )) + · · · + ( l η i − ( σ ( x i − x i )) + ǫ ) + ( l η i ( σ ( x i x i +1 )) − ǫ cos θ ) + · · · + l η n − ( σ ( x n − x ))= L ( σ ( p )) + ǫ (1 − cos θ ) > L ( σ ( p )) = 0 . Analogously for the polygon p := ( x , . . . , x ′′ i , . . . , x n − ) we have L ( σ ( p )) = L ( σ ( p )) + ǫ (cos θ − < L ( σ ( p )) = 0. 11uppose now that −→ x i ξ i = −−−→ x i ξ i − for all i . In particular, φ i − ( ξ i − ) = ξ i and the holonomymap φ p : S i → S i has the fixed point ξ i . Let ξ ′ i ∈ S i be the antipodal point to ξ i ∈ S i . If all ξ i and ξ ′ i coincide, then the polygon p is contained in a parallel set, namely the set P ξ ,ξ ′ of alllines connecting ξ with ξ ′ . Lemma 6.2.
Suppose p is not contained in any parallel set P ξ,ξ ′ , where ξ, ξ ′ ∈ ∂ T X are an-tipodal points such that −→ x i ξ i = −→ x i ξ for all i . Then for any neighborhood U of σ ( p ) in ∆ neuc thereexist n -gons p , p in X with σ ( p ) , σ ( p ) ∈ U and L ◦ σ ( p ) > > L ◦ σ ( p ) .Proof. Let P = ( ν , . . . , ν n − ) be an n -tuple of geodesic segments ν i : [ s − , s + ] → X with s − < < s + , ν i (0) = x i , ˙ ν i (0) = −→ x i ξ i and such that the convex hull CH ( ν i , ν i +1 ) is a (2-dimensional) flat quadrilateral. Notice that CH ( ν i , ν i +1 ) is actually a parallelogram because −→ x i ξ i = −−−→ x i ξ i − . Such a P exists, just take small parts of the lines through x i connecting ξ i with ξ ′ i .Suppose now that P is maximal with these properties, i.e. the segments ν i cannot be extended.If | s ± | = ∞ , then the ν i are parallel geodesic lines and p ⊂ P ν . Hence at least one of s + or − s − must be < ∞ . Suppose s = s + < ∞ (the other case is analogous).Now we want to displace p along ν i to the region, where it does not look locally like aparallel set anymore: set p ′ = ( x ′ , . . . , x ′ n − ) = ( ν ( s ) , . . . , ν n − ( s )). Then p ′ is an n -gon with σ ( p ′ ) = σ ( p ). Choose apartments A i containing the convex sets CH ( ν i , ν i +1 ). Let ζ i := − ˙ ν i ( s ) ∈ Σ x ′ i ( A i − ∩ A i ) and let α i ∈ Σ x ′ i A i − , β i ∈ Σ x ′ i A i be the antipodes of ζ i in Σ x ′ i A i − and Σ x ′ i A i respectively.If α i = β i for all i , then we can extend the ν i inside A i − ∩ A i contradicting the maximalityof P . Hence, there is a j such that α j = β j . Actually more is true: if it holds for all i that d ( −−−→ x ′ i x ′ i +1 , α i ) = d ( −−−→ x ′ i x ′ i +1 , β i ), then ζ i −−−→ x ′ i x ′ i +1 α i is a geodesic segment in Σ x ′ i X of length π . Let z i +1 ∈ A i be a point near x ′ i +1 with −−−−−→ x ′ i +1 z i +1 = α i +1 for i = 0 , . . . , n −
1. We can choose z i +1 close enough to x ′ i +1 , so that −−−→ x ′ i z i +1 is a regular point in the same Weyl chamber as −−−→ x ′ i x ′ i +1 because p is a regular polygon. It follows that −−−→ x ′ i z i +1 lies in the intersection of the segments ζ i −−−→ x ′ i x ′ i +1 α i and ζ i −−−→ x ′ i x ′ i +1 β i . Thus ζ i −−−→ x ′ i z i +1 α i is a geodesic segment of length π . After perhapstaking z i ∈ A i closer to x ′ i we can accomplish that CH ( x ′ i , z i , z i +1 ) is a flat triangle. It followsthat the union of the (2-dimensional) flat convex sets CH ( x i , x i +1 , x ′ i +1 , x ′ i ), CH ( x ′ i , x ′ i +1 , z i +1 )and CH ( x ′ i , z i +1 , z i ) is a flat convex quadrilateral. (See Figure 5.) Notice also that the segments ν i ( s − ) z i are extensions of the geodesic segments ν i ( s − ) ν i ( s + ). Thus this contradicts as well themaximality of P . Hence, there is a j such that d ( −−−−→ x ′ j x ′ j +1 , α j ) > d ( −−−−→ x ′ j x ′ j +1 , β j ). Analogously,there is a k such that d ( −−−−→ x ′ k x ′ k − , β k ) > d ( −−−−→ x ′ k x ′ k − , α k ).For some small ǫ > x j ∈ A j − be a point such that d ( x ′ j , ˜ x j ) = ǫ and −−→ x ′ j ˜ x j = α j . Then σ (˜ x j x ′ j +1 ) = σ ( x ′ j x ′ j +1 ) − ǫ · ov ˜ η = σ ( x j x j +1 ) − ǫ · ov ˜ η for some vertex ˜ η ∈ ∂ T E of the same typeas η j (compare with Section 3). By the above consideration we must have ˜ η = η j , otherwise d ( −−−−→ x ′ j x ′ j +1 , α j ) = d ( −−−−→ x ′ j x ′ j +1 , β j ) (recall that l η j ( σ ( x j x j +1 )) = h x j x j +1 , x j exp x j (1 · ξ j ) i and ˙ ν j (0) = −−→ x j ξ j ). It follows that l η j ( σ (˜ x j x ′ j +1 )) = h σ ( x j x j +1 ) − ǫ · ov ˜ η , ov η j i = l η j ( σ ( x j x j +1 )) − ǫ h ov ˜ η , ov η j i >l η j ( σ ( x j x j +1 )) − ǫ . On the other hand, σ ( x ′ j − ˜ x j ) = σ ( x ′ j − x ′ j + x ′ j ˜ x j ) = σ ( x j − x j ) + ǫ · ov η j − and this implies l η j − ( σ ( x ′ j − ˜ x j )) = h σ ( x j − x j ) + ǫ · ov η j − , ov η j − i = l η j ( σ ( x j − x j )) + ǫ . Thus,for p := ( x ′ , . . . , ˜ x j , . . . , x ′ n − ) we have L ( σ ( p )) > L ( σ ( p )) = 0.12 i +1 x ′ i +1 β i x ′ i α i z i +1 ζ i +1 ζ i x i z i α i +1 Figure 5: Extending the geodesic segments ν i Let now ˜ x k ∈ A k be a point such that d ( x ′ k , ˜ x k ) = ǫ and −−→ x ′ k ˜ x k = β k . It follows analogouslyfor p := ( x ′ , . . . , ˜ x k , . . . , x ′ n − ) that L ( σ ( p )) < L ( σ ( p )) = 0.The next question is what happens when p is contained in such a parallel set P ξ,ξ ′ . Inthis last situation we cannot always get the same conclusion as in Lemmata 6.1 and 6.2. Forinstance, if the wall H L lies in the boundary of P n ( X ), then we can cross H L in one directionbut not in the opposite one. Remark 6.3.
Suppose that p is contained in P ξ,ξ ′ with ξ, ξ ′ as in Lemma 6.2. Let b ξ ′ : X → R be a Busemann function associated to ξ ′ (see e.g. [7, Sec. 2.2] for a definition). Thenby considering an apartment in P ξ,ξ ′ containing the side x i x i +1 , we see that l η i ( σ ( x i x i +1 )) = b ξ ′ ( x i +1 ) − b ξ ′ ( x i ). In particular L ( σ ( p )) = l η ( σ ( x x )) + · · · + l η n − ( σ ( x n − x )) = n − X i =0 ( b ξ ′ ( x i +1 ) − b ξ ′ ( x i )) = 0 . Thus, if p ′ is the result of a variation of the polygon p within the parallel set P ξ,ξ ′ , it stillholds L ( σ ( p ′ )) = 0.The next lemma gives a condition that let us cross the wall H L in the positive direction.Suppose p is contained in P ξ,ξ ′ where ξ, ξ ′ ∈ ∂ T X are antipodal points such that −→ x i ξ i = −→ x i ξ for all i . Assume also that there are vertices x i , x j , x j +1 of p with the following property. Let A , A be apartments in P ξ,ξ ′ containing the segment x j x j +1 and an initial part of the segment x j x i and x j +1 x i respectively. Let y k ∈ A k for k = 0 , x j x i and x j +1 x i respectively. Thus x j x j +1 y k are flat triangles in A k . Suppose thatfor some k = 0 , α k ⊂ ∂ T A k such that the directions ξ , −−−−→ x j x j +1 and ( − k −−−−→ y k x j + k lie in the interior of α k (after the natural identification of ∂ T A k and Σ x A k for x ∈ A k ). By −−−−−→ y x j +1 we mean −−−−→ x j +1 y . (See Figure 6.) Lemma 6.4.
Under the assumptions above, for any neighborhood U of σ ( p ) in ∆ neuc there isan n -gon p ′ in X with σ ( p ′ ) ∈ U and L ◦ σ ( p ′ ) > .Proof. We show the lemma when the root α k ⊂ ∂ T A k exists for k = 0. The other case k = 1 isanalogous. 13 x j x j +1 ξ A α w Figure 6: Setting of Lemma 6.4Let w ⊂ A be a wall, such that α is bounded by ∂ T w . Let h ± ⊂ A be the roots boundedby w with ideal boundary α and its antipodal, respectively. Let ǫ > p in P ξ,ξ ′ along ξξ ′ such that x j lies in h + and d ( x j , w ) < ǫ . By thehypothesis we can take ǫ small enough so that y ∈ h − . Let A ′ be an apartment in X suchthat A ∩ A ′ = h − . Let x ′ j ∈ A ′ be the point such that d ( y , x ′ j ) = d ( y , x j ) and −−→ y x ′ j = −−→ y x j .Let z ∈ A be the reflection of x j in the hyperplane w .Observe that x ′ j / ∈ A and x j +1 / ∈ A ′ . It follows that σ ( x ′ j x j +1 ) = σ ( zx j +1 ). In particular l η j ( σ ( x ′ j x j +1 )) = l η j ( σ ( zx j +1 )) = h zx j + x j x j +1 , z exp z (1 · ξ ) i = h zx j , z exp z (1 · ξ ) i + l η j ( σ ( x j x j +1 )) > l η j ( σ ( x j x j +1 )) . Notice that the refined length of x ′ j x i is the same as of x j x i . Hence, by Theorem 2.1 wecan transfer the polygon ( x i , x i +1 , . . . , x j ) to a polygon ( x ′ i , x ′ i +1 , . . . , x ′ j ) with x ′ i = x i and withthe same ∆-valued side lengths. The n -gon p ′ = ( x ′ i , x ′ i +1 , . . . , x ′ j , x j +1 , . . . , x i − ) satisfies theconclusion of the lemma.If the polygon p is completely contained in an apartment in P ξ,ξ ′ , then the condition for thelemma above can be stated more easily. Corollary 6.5.
Suppose p is contained in an apartment A ⊂ P ξ,ξ ′ . Suppose that there are twosides x i x i +1 , x j x j +1 of p and a root α ⊂ A , such that the directions ξ , −−−→ x i x i +1 and −−−−→ x j x j +1 lie inthe interior of α . Then for any neighborhood U of σ ( p ) in ∆ neuc there is an n -gon p ′ in X with σ ( p ′ ) ∈ U and L ◦ σ ( p ′ ) > .Proof. Consider the oriented segments d = x i x j and d = x j x i . After a small variation of thepolygon p inside of the apartment A , we may assume that d (and therefore also d ) is regular.Then for one k = 1 ,
2, it must hold, that −→ d k lies in the interior of α . If k = 1, then Lemma 6.4applies for the vertices x i , x j , x j +1 and if k = 2, then it applies for the vertices x j , x i , x i +1 .Let us assume now that the building X has rank 2. We explain another method special forthis case to cross the wall H L .Let p = ( x , x , x ) be a regular triangle contained in P ξ,ξ ′ but not contained in any apart-ment. It is easy to see, that when we fold p into an apartment A , it has exactly one breakpoint. After relabeling the vertices we can assume that the break point y lies in the side x x and that the sides of the folded triangle ˆ p = (ˆ x = x , ˆ x = x , y, ˆ x ) do not intersect in their14nteriors (see Figure 7). After displacing ˆ p in P ξ,ξ ′ along ξξ ′ we can assume that y is a vertexof X . Let γ ⊂ A be the singular line through y connecting ξ and ξ ′ . Lemma 6.6.
We use the setting above (in particular, rank ( X ) = 2 ). Suppose that the Weylchamber containing −→ yx is not adjacent to Σ y γ . Then for any neighborhood U of σ ( p ) in ∆ euc there are triangles p , p in X with σ ( p ) , σ ( p ) ∈ U and L ◦ σ ( p ) > > L ◦ σ ( p ) .Proof. Let ℓ ⊂ A be the singular line through y such that Σ y ℓ is adjacent to the simplicialconvex hull of −→ yx −→ y ˆ x and the directions ξ , −→ yx and −→ y ˆ x are contained in the interior of thesame root bounded by ∂ T ℓ . It exists by the assumptions of the lemma. Let ℓ ′ ⊂ A be thereflection of ℓ in γ . Let h − γ ⊂ A be the root bounded by γ containing x and let h + γ ⊂ A bethe antipodal root. Similarly, let h + ℓ , h + ℓ ′ ⊂ A be the roots bounded by ℓ, ℓ ′ containing ξ and let h − ℓ , h − ℓ ′ ⊂ A be the antipodal roots. Then the simplicial convex hull of −→ yx −→ y ˆ x is Σ y ( h + ℓ ∩ h − ℓ ′ ).(See Figure 7.) ℓ γ ℓ ′ yx ˆ x x A ∩ A Figure 7: The folded triangle ˆ p Let A be an apartment in X such that A ∩ A = h − γ ∩ h − ℓ . Let x ′ ∈ A be the point sothat d ( x , x ′ ) = d ( x , ˆ x ) and −−→ x x ′ = −−→ x ˆ x . Notice that x ′ / ∈ A , thus, x ′ = ˆ x . Observe alsothat x ′ / ∈ P ξ,ξ ′ , hence, x ′ = x .The concatenation of the segments −→ yx −→ yξ ′ ∈ Σ y A and −→ yξ ′ −→ yx ′ ∈ Σ y A gives a segmentin Σ y X of length π (see Figure 8). Therefore x yx is a geodesic segment and the triangle p ′ = ( x , x , x ′ ) =: ( z , z , z ) has the same side lengths as p . Set A := A and let A be anapartment in X containing the segment z z .Let ν i be the geodesic rays with ν i (0) = z i and ν i ( −∞ ) = ξ ′ . Then CH ( ν i , ν i +1 ) are (2-dimensional) flat stripes. We want to see that the ν i cannot be extended to parallel geodesiclines. Suppose then the contrary: there are parallel geodesic lines ν ′ i containing ν i . Set ζ := ν ′ i ( ∞ ). Then p ′ ⊂ Y := P ζ,ξ ′ and in particular, −→ yξ ′ , −→ yz i ∈ Σ y Y . Since −→ yz , −→ yz ∈ Σ y A areantipodal regular points, the apartment containing them is unique. Therefore Σ y A ⊂ Σ y Y and in particular, −→ yζ ∈ Σ y A .Let k ∈ { , } be so that the Weyl chamber containing −→ y ˆ x k is adjacent to Σ y ℓ ′ . Let σ k ⊂ Σ y A k − be the Weyl chamber containing −→ yz k and let ˆ σ k ∈ Σ y ( A ∩ A ) be the antipodal15hamber to σ k . (See Figure 8 for k = 2.) Notice that −→ yξ ′ −→ yz intersects ˆ σ k in its interior. Inparticular ˆ σ k ⊂ Σ y Y . It follows that the unique apartment containing σ k and ˆ σ k is containedin Σ y Y , i.e. Σ y A k − ⊂ Σ y Y . (cid:0)(cid:1)(cid:0)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1) −→ yξ ′ −→ yz −→ yξσ ˆ σ k Σ y ℓ ′ Σ y ℓ −→ yz −→ yz Σ y ℓ Σ y ℓ ′ −→ yζ Figure 8: Σ y X Let σ ⊂ Σ y ( A ∩ A ) ⊂ Σ y A k − ⊂ Σ y Y be the Weyl chamber adjacent to ℓ . The Weylchamber containing −−−→ yz − k is antipodal to σ . Hence, the unique apartment containing σ and −−−→ yz − k is contained in Σ y Y , i.e. Σ y A − k ⊂ Σ y Y .We have conclude that A , A ⊂ Σ y Y = Σ y P ζ,ξ ′ , but this is not possible because of theconstruction of A . Therefore the geodesic rays ν i cannot be extended to complete parallelgeodesic lines. The lemma now follows from Lemma 6.2 and its proof.We can show now that for rank 2 the space P n ( X ) is a polyhedral cone. Its convexity willbe shown in the next section. Proposition 6.7. If X has rank 2, then P n ( X ) is a union of the closures of polyhedral conesin C n .Proof. We have already seen in Corollary 5.6 that if for C ∈ C n holds P n ( X ) ∩ C = ∅ , then C ⊂ P n ( X ). Now let p = ( x , . . . , x n − ) be a polygon in X . We want to show that σ ( p ) iscontained in C for some C ∈ C n with C ⊂ P n ( X ). Since any polygon can be approximatedby regular polygons, we may assume that p is regular. Suppose now s := σ ( p ) ∈ H L with L = ( l η , . . . , l η n − ). If for any neighborhood U of s we can find polygons with side lengthsin U \ H L , then we are done. Indeed, in this case, there is an open cone C ∈ C n such that P n ( X ) ∩ C = ∅ and s ∈ C .Suppose then that for some neighborhood U of σ ( p ) we cannot find polygons p ′ with sidelengths in U and L ◦ σ ( p ′ ) = 0. Lemmata 6.1 and 6.2 implies that p lies in a parallel set P ξ,ξ ′ andthe functional L is given in p by taking scalar product with a unit vector in the direction of ξ .Suppose first that the triangle t = ( x , x , x ) lies completely in an apartment in P ξ,ξ ′ . Then itis easy to see that Corollary 6.5 must apply for one of the functionals L ′ := ( l η , l η , l η ′ ) or − L ′ ,where η ′ is so that l η ′ ( σ ( x x )) = h x x , x exp x (1 · ξ ) i . If t is not contained in an apartment,then we fold it into an apartment as in the setting of Lemma 6.6. Then, either Lemma 6.6applies or the Weyl chamber containing the direction −−−→ x i x i +1 of the side of t with the breakpoint must be adjacent to −→ x i ξ or −→ x i ξ ′ . If the last occurs, it is again easy to see, that Lemma 6.416ust apply for L ′ or − L ′ . In any case, we find a triangle t ′ = ( x ′ , x ′ , x ′ ) with L ′ ( σ ( t )) = 0and such that (modulo displacement in P ξ,ξ ′ along ξξ ′ ) the refined side lengths of t ′ are as nearas we want to the ones of t . After a small variation of the polygon ( x , x , . . . , x n − ) insidethe parallel set P ξ,ξ ′ and displacing it along ξξ ′ , we obtain a polygon q = ( x ′′ , x ′′ , . . . , x ′′ n − )so that the refined side length of x ′ x ′ is the same as of x ′′ x ′′ . Since q is still contained inthe parallel set P ξ,ξ ′ , it holds L ′′ ( σ ( q )) = 0 for L ′′ = ( l η ′′ , l η , . . . , l η n − ) where η ′′ is so that l η ′′ ( σ ( x ′′ x ′′ )) = h x ′′ x ′′ , x ′′ exp x ′′ (1 · ξ ) i . In particular, l η ′ ( σ ( x ′ x ′ )) = − l η ′′ ( σ ( x ′′ x ′′ )). Then by theTransfer Theorem 2.1 we can glue t ′ and q along x ′ x ′ and x ′′ x ′′ to a polygon p ′ with ∆-valuedside lengths near s and L ( σ ( p ′ )) = L ′ ( σ ( t ′ )) + L ′′ ( σ ( q )) = 0. Remark 6.8.
Proposition 6.7 is also true in rank > P n ( X ) We have seen in the previous section different methods which allows to cross certain walls H L within the space P n ( X ). We will show in this section that for the case of buildings of rank 2 thewalls where this method cannot be applied are precisely the walls that determine the boundaryof P n ( X ). That is, if a wall cannot be crossed with the polygon variations of Section 6.1, it isbecause that wall cannot be crossed at all.First we characterize the walls H L that cannot be crossed with the methods above in termsof the combinatorics of the associated spherical Coxeter complex ( S, W ). We use the samenotation as in Section 3. Let ∆ := ∂ T ∆ euc ⊂ ∂ T E = S be the spherical Weyl chamber. Let δ ∈ W be the element of the Weyl group such that ∆ and δ ∆ are antipodal. Notice that δ = Id . We say that a root α ⊂ S is a positive root if ∆ ⊂ α . Let Λ + denote the set of positiveroots of ( S, W ). A root which is not a positive root is called a negative root .For each element ω ∈ W we define a subset of Λ + : T ω := { α ∈ Λ + | ω − ∆ ⊂ α } = Λ + ∩ ω − Λ + . Let η ∈ ∆ be a vertex of the spherical Weyl chamber and let Λ + η be the set of positive rootscontaining η in their boundaries. Let B n ( η ) be the set of n -tuples ( ω , . . . , ω n − ) ∈ W n suchthat( ∗ ) T ω i ∩ T ω j ⊂ Λ + η for all i = j ,( ∗∗ ) n − S i =0 T ω i = Λ + .For ¯ ω = ( ω , . . . , ω n − ) ∈ W n we write ¯ ωη := ( ω η, . . . , ω n − η ) ∈ ( W η ) n . and for ¯ η =( η , . . . , η n − ) ∈ ( W η ) n set L ¯ η = ( l η , . . . , l η n − ). Let B n ⊂ L n be the union of the sets { L ¯ η | ¯ η =¯ ωη, ¯ ω ∈ B n ( η ) } for all vertices η ∈ ∆.We will see in Lemma 6.13 below that the walls H L that cannot be crossed in the positivedirection with our previous methods are precisely the ones of the form L ¯ η with ¯ η = ¯ ωη and¯ ω = ( ω , . . . , ω n − ) satisfying the property ( ∗ ). A motivation for this property ( ∗ ) can already17e seen in Corollary 6.5. Namely, let p = ( x , . . . , x n − ) be a regular polygon contained in anapartment A and let η ′ be a vertex of ∂ T A . Let L be the functional that in p corresponds totaking scalar product with a unit vector in the direction of η ′ . In particular, L ( σ ( p )) = 0. Fixa Weyl chamber τ ⊂ ∂ T A containing η ′ and let τ i ⊂ ∂ T A be the Weyl chamber containing thedirection −−−→ x i x i +1 after identifying Σ x i A ∼ = ∂ T A . Let ω i ∈ W be such that τ = ω i τ i . Let η ∈ ∆ bethe vertex of the same type as η ′ . Then L = ( l ω η , . . . , l ω n − η ). If T ω i ∩ T ω j Λ + η for some i = j ,then Corollary 6.5 applies and we can increase the functional L near σ ( p ). The property ( ∗∗ )is introduced to avoid later obvious redundancies in the set of generalized triangle inequalities.This can be seen in the Lemma 6.12. Lemma 6.9.
Let ¯ η = ( η , . . . , η n − ) ∈ ( W η ) n . If there exist j, j ′ ∈ { , . . . , n − } , j = j ′ with η j ′ = δη j and δη i = η ∈ ∆ for i = j, j ′ , then ¯ η ∈ B n ( η ) · η . If ( E, W ) has rank 2, then theconverse is also true.Proof. To prove the first statement choose ω j ∈ W with ω j η = η j and set ω j ′ := δω j . Then ω j ′ η = δω j η = δη j = η j ′ . It follows that ω − j ′ ∆ is antipodal to ω − j ∆ and T ω j ∩ T ω j ′ = ∅ , T ω j ∪ T ω j ′ = Λ + . Let ω i = δ for i = j, j ′ , then η i = ω i η and T ω i = ∅ . Hence, ¯ ω = ( ω , . . . , ω n − ) ∈ B n ( η ) and ¯ η = ¯ ωη .Now suppose that ( E, W ) has rank 2 and let ¯ ω ∈ B n ( η ) with ¯ η = ¯ ωη . By property ( ∗∗ ) thereis a j with T ω j Λ + η , that is, − η / ∈ ω − j ∆. If η ∈ ω − j ∆ (or equivalently, ω j ∈ Stab W ( η )) then η j = ω j η = η and Λ + \ T ω j ⊂ Λ + η . This and properties ( ∗ ) and ( ∗∗ ) imply T ω i ⊂ Λ + η for i = j .It follows that − η ∈ ω − i ∆, or equivalently, δω i ∈ Stab W ( η ). Hence, η i = ω i η = δδω i η = δη for i = j and the assertion follows. Suppose now that − η, η / ∈ ω − j ∆. Then there are two positiveroots α , α / ∈ Λ + η with ω − j ∆ = α ∩ − α . Then α ∈ Λ + \ T ω j . Let j ′ be such that α ∈ T ω j ′ .By property ( ∗ ) it follows that j ′ is unique and α / ∈ T ω j ′ . Hence, ω − j ′ ∆ = − α ∩ α and ω − j ′ ∆is antipodal to ω − j ∆. This implies that ω j ′ = δω j and η j = δη j ′ . Since Λ + = T ω j ∪ T ω j ′ , byproperty ( ∗ ) we deduce that T ω i ⊂ Λ + η for i = j, j ′ and this in turn implies that η i = δη for i = j, j ′ (Fig. 9). Remark 6.10.
Let B wn ⊂ L n be the set of functionals L ¯ η for ¯ η satisfying the hypothesis of theprevious Lemma. The inequalities L ≤ L ∈ B wn are the so-called weak triangle inequalities (cf. [6, Section 3.8]). Thus, Lemma 6.9 just states that B wn ⊂ B n and that for rank 2 it actuallyholds B n = B wn (Fig. 9). ∆ ω − j ∆ T ω j ω − j ′ ∆ T ω j ′ η j η j ′ ηη i ω − i ∆ Figure 9: B wn : weak triangle inequalities18 heorem 6.11 (Weak Triangle Inequalities, [6, Theorem 3.34]). For any n -gon p in X and any functional L ∈ B wn holds L ◦ σ ( p ) ≤ . That is, P n ( X ) ⊂ \ L ∈B wn { L ≤ } . Proof.
Let p = ( x , . . . , x n − ) be an n -gon in X . For the functional L = ( l η , . . . , l η n − ) ∈ B wn ,let j, j ′ be such that η j = δη j ′ and δη i = η ∈ ∆ for i = j, j ′ . Notice that l δη ≤ l η ′ in ∆ euc for all η ′ of the same type as η . That is, l η i is the smallest functional of the same type as η .After shifting the subindices of the polygon and the functional we can assume that j ′ = 0. Let ω ∈ W be so that ω η = η .Suppose first that j = n −
1. Fold the polygon p into an apartment A , so that the brokensides are x x , . . . , x n − x n − . Let ρ : A → E be an isometry that sends x to the vertex o of ∆ euc ⊂ E , induces an isomorphism of the Coxeter complexes ( ∂ T A, W ) and (
E, W ) and sothat ρ ( x x ) ⊂ ω − ∆ euc . Notice that ρ is not necessarily an isomorphism of Euclidean Coxetercomplexes with the affine Weyl group W aff . Denote with q the image under ρ of the foldedpolygon. By folding E onto the Euclidean Weyl chamber ω − ∆ euc with the natural “accordion”map, we obtain a further folded polygon q ′ = ( y , . . . , y k ) where y is the vertex of ∆ euc andthe ∆-valued side lengths of the segments y y , y y k ⊂ ω − ∆ euc are the same as for x x and x x n − respectively. Observe that q ′ is not necessarily a billiard polygon in ( E, W aff ), but ifthe side x r x r +1 of p is broken in q ′ to the sides y s y s +1 , y s +1 y s +2 , . . . , y t − y t , then the vectors σ ( y s y s +1 ) , . . . , σ ( y t − y t ) are just multiples of σ ( x r x r +1 ). This means, that if W ′ aff is the groupgenerated by W aff and the whole translation group of E , then q ′ is a billiard polygon in( E, W ′ aff ). Notice also that for i = 0 , n − l η i ( σ ( y l y l +1 )) ≤ h y l y l +1 , ov η i because of theobservation at the beginning of the proof. It follows that l η ( σ ( x x )) + · · · + l η n − ( σ ( x n − x n − )) ≤ h y y , ov η i + · · · + h y k − y k , ov η i = h y y k , ov η i . On the other hand, since y y , y y k ⊂ ω − ∆ euc and η n − = δη = δω η it follows that l η ( σ ( x x )) = l η ( σ ( y y )) = h σ ( y y ) , ov η i = h y y , ov η i and l η n − ( σ ( x n − x )) = l η n − ( σ ( y k y )) = h σ ( y k y ) , ov η n − i = h y k y , ov η i . Hence, L ( σ ( p )) ≤ h y y , ov η i + h y y k , ov η i + h y k y , ov η i = 0.The general case (i.e. j ∈ { , . . . , n − } ) now follows from the special case above by con-sidering the polygons p = ( x , x , . . . , x j ), and p = ( x , x j , x j +1 , . . . , x n − ) with the function-als L = ( l η , l η , . . . , l η j ) respectively L = ( l η , l η j , l η j +1 , . . . , l η n − ). Indeed, notice that since η = δη j , it holds l η j ( σ ( x j x )) = h σ ( x j x ) , ov η j i = h− σ ( x x j ) , ov η i = − l η ( σ ( x x j )). Hence, L ( σ ( p )) = L ( σ ( p )) + L ( σ ( p )) ≤ Lemma 6.12. If X has rank 2 and ¯ ω ∈ ( W ) n satisfies the property ( ∗ ) but not the property( ∗∗ ) for some vertex η ∈ ∆ , then there is a ¯ ω ′ ∈ B n ( η ) so that L ¯ ωη ◦ σ ( p ) ≤ L ¯ ω ′ η ◦ σ ( p ) for all n -gons p in X . If p is regular, then the strict inequality holds, in particular, L ¯ ωη ◦ σ ( p ) < .Proof. Let ¯ ω = ( ω , . . . , ω n − ) satisfy the property ( ∗ ). It is easy to see that in rank 2 at mostfor two indices i can hold T ω i Λ + η . Then we can find j = j ′ so that T ω i ⊂ Λ + η for all i = j, j ′ .Let ˆ ω j := δω j , then ˆ ω − j ∆ euc is antipodal to ω − j ∆ euc . If ¯ ω does not satisfy the property ( ∗∗ ),then ˆ ω j = ω j ′ , moreover, T ω j ′ \ Λ + η ( T ˆ ω j = Λ + \ T ω j . This implies that l ω j ′ η ≤ l ˆ ω j η in ∆ euc andsince ˆ ω j = ω j ′ the strict inequality holds for regular segments. Thus we obtain ¯ ω ′ ∈ B n ( η ) byreplacing ω j ′ by ˆ ω j in ¯ ω . 19 emma 6.13. Suppose X has rank 2 and let p be a regular n -gon in X . Suppose that σ ( p ) ∈ H L for some functional L with L, − L ∈ L n \ B n . Then for any neighborhood U of σ ( p ) in ∆ neuc there exist n -gons p , p in X with σ ( p i ) ∈ U and L ◦ σ ( p ) > > L ◦ σ ( p ) .Proof. Suppose that for a neighborhood U of σ ( p ) in ∆ neuc , we cannot find a polygon p in X with σ ( p ) ∈ U and L ◦ σ ( p ) >
0. (The other inequality follows considering the functional − L .)It follows from Lemmata 6.1 and 6.2 that p lies in a parallel set P ξ,ξ ′ and the functional L in p is just given by taking scalar product with a unit vector in the direction of ξ . Fold the polygonin an apartment A ⊂ P ξ,ξ ′ so that the broken sides are x x , . . . , x n − x n − . Let ρ : A → E be an isomorphism of Euclidean Coxeter complexes that sends ξ to the vertex η ∈ ∆ euc of thesame type.Suppose X has only one vertex o and let γ ⊂ P ξ,ξ ′ be the line through o and γ ( ∞ ) = ξ, γ ( −∞ ) = ξ ′ . Then the break points of the folded polygon all lie on γ . We may assumethat the folded polygon has at most one break point because any two consecutive break pointscan be simultaneously unfolded . Let k be so that the break point y lies on the side x k x k +1 (if there is no break point we take k = n − p ′ =( x , x , . . . , x k , y, ˆ x k +1 , . . . , ˆ x n − ). Let ω i ∈ W be so that ω − i ∆ contains the direction ρ ( −−−→ x i x i +1 )for 0 ≤ i ≤ k − ρ ( −→ x k y ) for i = k , ρ ( −−−→ ˆ x i ˆ x i +1 ) for k + 1 ≤ i ≤ n −
2, and ρ ( −−−−→ ˆ x n − x ) for i = n − L is just given by ( l η , . . . , l η n ) for η i = ω i η . After a smallvariation inside the parallel set P γ we may assume that the segments x x k and x ˆ x k +1 areregular. Let ω ′ , ω ′′ ∈ W be so that ω ′− ∆ contains the direction ρ ( −−→ x x k ) and ω ′′− ∆ contains ρ ( −−−−→ ˆ x k +1 x ).Consider the regular polygon q = ( x , . . . , x k ) ⊂ A and the functional L ′ = ( l η , . . . , l η k − , l η ′ )for η ′ := δω ′ η . That is, L ′ is the functional given in q by taking scalar product with a unit vectorin the direction ξ . Hence L ′ ( σ ( q )) = 0. Set ( τ , . . . , τ k − , τ k ) := ( ω , . . . , ω k − , δω ′ ). Supposethat there are 0 ≤ i < j ≤ k such that T τ i ∩ T τ j Λ + η . Corollary 6.5 and its proof imply thatthere is a polygon q ′ = ( z , . . . , z k ) with L ′ ( σ ( q ′ )) > q modulo displacement along γ . We can then choose x ′ , x ′ k ∈ P ξ,ξ ′ near x , x k such that x ′ x ′ k has the same refined side length (again modulo displacement along γ ) as z z k .The functional ( − l η ′ , l η k , . . . , l η n − ) applied to the polygon ( x ′ , x ′ k , x k +1 , . . . , x n − ) is 0 because itis contained in the parallel set P ξ,ξ ′ . After displacing the polygon ( x ′ , x ′ k , x k +1 , . . . , x n − ) along γ we can glue it together to q ′ and obtain a polygon p with ∆-valued side lengths as nearas we want to those of p and with L ( σ ( p )) > T τ i ∩ T τ j ⊂ Λ + η for all0 ≤ i < j ≤ k , i.e. ( τ . . . . , τ k ) satisfies the property ( ∗ ). This can be rewritten as T ω i ∩ T ω j ⊂ Λ + η for all 0 ≤ i < j ≤ k − k − S i =0 T ω i \ Λ + η ⊂ T ω ′ .Analogously, considering the polygon ( x , x k +1 , . . . , x n − ) which is also contained in anapartment in P ξ,ξ ′ we obtain T ω i ∩ T ω j ⊂ Λ + η for all k +1 ≤ i < j ≤ n − n − S i = k +1 T ω i \ Λ + η ⊂ T ω ′′ .Consider now the triangle t = ( x , x k , x k +1 ) with the functional L ′′ = ( l ω ′ η , l η k , l ω ′′ η ). Let ω ′ k ∈ W be so that ω ′− k ∆ euc contains the direction ρ ( −−−→ y ˆ x k +1 ). Then η k = ω k η = ω ′ k η . We wantto show that ( ω ′ , ω k , ω ′′ ) or ( ω ′ , ω ′ k , ω ′′ ) have the property ( ∗ ). By Lemma 6.4 applied to theside x x k we get T ω ′ ∩ T ω ′′ , T ω ′ ∩ T ω k ⊂ Λ + η . Again by Lemma 6.4 now applied to the side x k +1 x
20e obtain T ω ′′ ∩ T ω ′ k ⊂ Λ + η . Therefore if T ω ′ or T ω ′′ is contained in Λ + η , then we are done, sosuppose this is not the case for both of them. Now by Lemma 6.6 one of ω ′− ∆, ω ′′− ∆ or ω − k ∆ must be adjacent to ρ ( γ ). Notice that for ω ∈ W , ω ∆ is adjacent to ρ ( γ ) if and only if T ω ⊂ Λ + η or Λ + \ Λ + η ⊂ T ω . Since T ω ′ and T ω ′′ are not contained in Λ + η and T ω ′ ∩ T ω ′′ ⊂ Λ + η ,then neither ω ′− ∆ nor ω ′′− ∆ can be adjacent to ρ ( γ ). Hence, ω − k ∆ euc must be adjacent to ρ ( γ ). T ω ′ ∩ T ω k ⊂ Λ + η implies that T ω k ⊂ Λ + η and we are also done in this case.Putting our three considerations above together we can conclude that ¯ ω = ( ω , . . . , ω k , . . . ,ω n − ) or ¯ ω ′ = ( ω , . . . , ω ′ k , . . . , ω n − ) has the property ( ∗ ) and since p is a regular polygon with L ( σ ( p )) = 0, it follows from Lemma 6.12 that L = L ¯ ωη = L ¯ ω ′ η ∈ B n .Now we are ready to prove our main theorem. Theorem 6.14.
Let X be a thick Euclidean building of rank 2. Then P n ( X ) is a convexpolyhedral cone determined by the inequalities { L ≤ } for L ∈ B n . That is, P n ( X ) = \ L ∈B n { L ≤ } . This inequalities constitute an irredundant set of inequalities.Proof.
Let Q ⊂ C n be the subset of open cones such that T L ∈B n { L ≤ } = S C ∈ Q ¯ C . Analogously,let Q ′ ⊂ C n be the subset of open cones such that P n ( X ) = S C ∈ Q ′ ¯ C (this can be done byProposition 6.7). We have shown in Lemma 6.9 and Theorem 6.11 that Q ′ ⊂ Q . Let C ∈ Q ′ and C ∈ Q . Take a chain C , C , . . . , C k = C ∈ Q such that C i ∩ C i +1 is a face of codimensionone. We prove now inductively that C i ∈ Q ′ . Suppose then that C i ∈ Q ′ and take a regularpolygon p with σ ( p ) in the interior of the face C i ∩ C i +1 . Since C i ∩ C i +1 is not in the boundaryof T L ∈B n { L ≤ } , it lies in a wall H L with neither L nor − L in B n . It follows from Lemma 6.13that P n ( X ) ∩ C i +1 is not empty and therefore C i +1 ⊂ P n ( X ). Thus C ∈ Q ′ , and Q = Q ′ .For L ∈ B n it is clear that we can find a regular polygon p in an apartment A and γ ⊂ A a maximal singular line, such that the functional L in p is given by taking scalar product withthe direction of η = γ ( ∞ ). In particular, L ( σ ( p )) = 0. It is also clear that we can find a regularpolygon p ′ in P γ but not contained in any apartment and such that the functional L in p ′ isalso given by taking scalar product with the direction of η . It follows from Lemmata 6.1 and6.2 that L is the only functional in B n for which it can hold L ( σ ( p ′ )) = 0. Thus the inequalities { L ≤ } with L ∈ B n are irredundant. Acknowledgments.
I would like to thank Bernhard Leeb for bringing this problem tomy attention and sharing his ideas in the case of symmetric spaces with me. A last version ofthis paper was completed during a stay at the Max-Planck Institute in Bonn. The author isgrateful to the MPI for its financial support and great hospitality.
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