Compact hyperbolic tetrahedra with non-obtuse dihedral angles

Publ. Mat. 50 (2006), 211–227 COMPACT HYPERBOLIC TETRAHEDRA WITH NON-OBTUSE DIHEDRAL ANGLES Roland K. W. Roeder Abstract Given a combinatorial description Cof a polyhedron having Eedges, the space of dihedral angles of all compact hyperbolic polyhedra that realize Cis generally not a convex subset of RE[9]. If Chas five or more faces, Andreev’s Theorem states that the corresponding space of dihedral angles ACobtained by restricting to non-obtuse angles is a convex polytope. In this paper we explain why Andreev did not consider tetrahedra, the only polyhedra having fewer than five faces, by demonstrating that the space of dihedral angles of compact hyperbolic tetrahedra, after restricting to non-obtuse angles, is non-convex. Our proof provides a simple example of the “method of continuity”, the technique used in classification theorems on polyhedra by Alexandrow [4], Andreev [5], and Rivin-Hodgson [18]. Given a combinatorial description Cof a polyhedron having Eedges, the space of dihedral angles of all compact hyperbolic polyhedra that realize Cis generally not a convex subset of RE. This is proved in a nice paper by D´ıaz [9]. However, Andreev’s Theorem [5], [13], [19], [20] shows that by restricting to compact hyperbolic polyhedra with nonobtuse dihedral angles, the space of dihedral angles is a convex polytope, which we label AC⊂RE. It is interesting to note that the statement of Andreev’s Theorem requires that Chave five or more faces, ruling out the tetrahedron which is the only polyhedron having fewer than five faces. In this paper, we explain why hyperbolic tetrahedra are a special case that is not covered by Andreev’s Theorem. We provide an explicit description of the space of dihedral angles, A∆, corresponding to compact hyperbolic tetrahedra with non-obtuse dihedral angles, finding that A∆is a non-convex, path-connected subset of R6. 2000 Mathematics Subject Classification. 52B10, 52A55, 51M09. Key words. Hyperbolic geometry, polyhedra, tetrahedra. 212 R. K. W. Roeder A description of the space of Gram matrices (and hence indirectly of the space of dihedral angles) corresponding to compact hyperbolic tetrahedra having arbitrary dihedral angles is available in Milnor’s collected works [16]. Our description of the space of dihedral angles A∆ can be derived from the result in [16], using the assumption that the dihedral angles are non-obtuse. However, we use the “method of continuity,” providing the reader with a simple example of a method that plays an important role in the classification theorems on polyhedra by Alexandrow [4], Andreev [5], and Rivin-Hodgson [18]. Let E3,1be R4with the indefinite metric kxk2=−x2 0+x2 1+x2 2+x2 3. In this paper, we work in the hyperbolic space H3given by the component of the subset of E3,1given by kxk2=−x2 0+x2 1+x2 2+x2 3=−1 having x0>0, with the Riemannian metric induced by the indefinite metric −dx2 0+dx2 1+dx2 2+dx2 3. There is a natural compactification of the hyperbolic space obtained by adding the set of rays asymptotic to the hyperboloid. We refer to these points as the points at infinity. There is no natural extension of the Riemannian structure of H3to these points at infinity, however, there is a natural way to extend the conformal structure on H3to these points at infinity. One can check that the hyper-plane orthogonal to a vector v∈E3,1 intersects H3if and only if hv,vi>0. Let v∈E3,1be a vector with hv,vi>0, and define Pv={w∈H3| hw,vi= 0} to be the hyperbolic plane orthogonal to v; and the corresponding closed half space: H+ v={w∈H3| hw,vi ≥ 0}. Notice that given two planes Pvand Pwin H3with hv,vi= 1 and hw,wi= 1, they: •Intersect in a line if and only if hv,wi2<1, in which case their dihedral angle is arccos(−hv,wi). •Intersect in a single point at infinity if and only if hv,wi2= 1, in this case their dihedral angle is 0. Compact Hyperbolic Tetrahedra 213 Ahyperbolic polyhedron is an intersection P= n \ i=0 H+ vi having non-empty interior. There are many papers on hyperbolic polyhedra, including [5], [6], [9], [13], [16], [18], [19], [20], [21], [22], [24], [26], and particularly on the groups of reflections generated by them [3], [24], [25], [27]. A hyperbolic tetrahedron is therefore a hyperbolic polyhedron having the combinatorial type of a tetrahedron. There are also many papers on hyperbolic tetrahedra including [8], [11], [12], [15], [17], [23], many of these studying volume and symmetries. If we normalize the vectors vithat are orthogonal to the faces of a polyhedron P, the Gram matrix of Pis the matrix with terms Mij = hvi,vji. By construction, a Gram matrix is symmetric and unidiagonal (i.e. has 1s on the diagonal). The following theorem appears in [16]: Theorem 1. A symmetric unidiagonal matrix Mis the Gram matrix of a compact hyperbolic tetrahedron if and only if det(M)<0and each principal minor is positive definite. Although the hyperboloid model of hyperbolic space is very natural, it is not easy to visualize, since the ambient space is four-dimensional. We will often use the Poincar´e ball model of hyperbolic space, given by the unit ball in R3with the metric 4dx2 1+dx2 2+dx2 3 (1 −kxk2)2 and the upper half-space model of hyperbolic space, given by the subset of R3with x3>0 equipped with the metric dx2 1+dx2 2+dx2 3 x2 3 . Both of these models are isometric to H3. The points at infinity in the Poincar´e Ball model correspond to points on the unit sphere, and the points at infinity in the upper half-space model correspond to the points in the plane x3= 0. More background is available on hyperbolic geometry in [7]. Hyperbolic planes in these models correspond to portions of Euclidean spheres and Euclidean planes that intersect the boundary perpendicularly. Furthermore, these models are conformally correct, that is, the hyperbolic angle between a pair of such intersecting hyperbolic planes is exactly the Euclidean angle between the corresponding spheres or planes. 214 R. K. W. Roeder See below for an image of a compact hyperbolic tetrahedron depicted in the Poincar´e ball model using Geomview [2]. The sphere at infinity is shown for reference. The following two lemmas will be necessary when discussing compact hyperbolic polyhedra having non-obtuse dihedral angles. They are well known results and appear in many of the works on hyperbolic polyhedra mentioned above, including [5]. Lemma 2. Suppose that three planes Pv1,Pv2,Pv3intersect pairwise in H3with non-obtuse dihedral angles α,β, and γ. Then, Pv1,Pv2,Pv3 intersect at a vertex in H3if and only if α+β+γ≥π. The planes intersect in H3if and only if the inequality is strict. Proof: The planes intersect in a point of H3if and only if the subspace spanned by v1,v2,v3is positive semi-definite, so that the orthogonal is a negative semi-definite line of E3,1. If the inner product on this line is negative, the line defines a point of intersection with the hyperboloid model. Otherwise, the inner product on the line is zero and this line corresponds to a point in ∂H3, since the line will then lie in the cone to which the hyperboloid is asymptotic. The symmetric matrix defining the inner product on the span of v1, v2, and v3is   1hv1,v2i hv1,v3i hv1,v2i1hv2,v3i hv1,v3i hv2,v3i1  =  1−cos α−cos β −cos α1−cos γ −cos β−cos γ1   Compact Hyperbolic Tetrahedra 215 where α,β, and γare the dihedral angles between the pairs of faces (Pv1, Pv2), (Pv1, Pv3),and (Pv2, Pv3), respectively. Since the principal minor is positive definite for 0 < α ≤π/2, it is enough to find out when the determinant 1−2 cos αcos βcos γ−cos2α−cos2β−cos2γ is non-negative. A bit of trigonometric trickery (we used complex exponentials) shows that the expression above is equal to (1) −4 cosα+β+γ 2cosα−β+γ 2cosα+β−γ 2cos−α+β+γ 2. Let δ=α+β+γ. When δ < π, (1) is strictly negative; when δ=π, (1) is clearly zero; and when δ > π, (1) is strictly positive. Hence the inner product on the space spanned by v1,v2,v3is positive semidefinite if and only if δ≥π. It is positive definite if and only if δ > π. Therefore, the three planes Pv1, Pv2, Pv3⊂H3intersect at a point in H3if and only if they intersect pairwise in H3and the sum of the dihedral angles δ≥π. It is also clear that they intersect at a finite point if and only if the inequality is strict. Lemma 3. Given a trivalent vertex of a hyperbolic polyhedron, we can compute the angles of the faces in terms of the dihedral angles. If the dihedral angles are non-obtuse, these angles are also ≤π/2. Proof: Let vbe a finite trivalent vertex of P. After an appropriate isometry, we can assume that vis the origin in the Poincar´e ball model, so that the faces at vare subsets of Euclidean planes through the origin. A small sphere centered at the origin will intersect Pin a spherical triangle Qwhose angles are the dihedral angles between faces. Call these angles α1,α2,α3. The edge lengths of Qare precisely the angles in the faces at the origin. Supposing that Qhas edge lengths (β1, β2, β3) with the edge βi opposite of angle αifor each i= 1,2,3, The law of cosines in spherical geometry states that: (2) cos(βi) = cos(αi) + cos(αj) cos(αk) sin(αj) sin(αk). Hence, the face angles are calculable from the dihedral angles. They are non-obtuse, since the right-hand side of the equation is positive for αi, αj,αknon-obtuse. We can now state our classification of compact hyperbolic tetrahedra: 216 R. K. W. Roeder Theorem 4. Let α1,...,α6be a set of proposed non-obtuse dihedral angles and let β1(α1,...,α6), . . . , β12(α1,...,α6)be the face angles given by equation (2), corresponding to these proposed dihedral angles. There is a compact hyperbolic tetrahedron with dihedral angles α1,...,α6if and only if: (1) For each edge ei,0< αi≤π/2. (2) Whenever 3distinct edges ei,ej,ekmeet at a vertex, αi+αj+αk> π. (3) For each face the sum of the face angles satisfies βi+βj+βk< π. Furthermore this tetrahedron is unique. Recall from Lemma 3 that the face angles βiare calculable from the dihedral angles αiand are themselves non-obtuse so that condition (3) is a highly non-linear condition on the dihedral angles. We will denote the subset of R6of dihedral angles satisfying conditions (1)–(3) by A∆. We present a proof of Theorem 4 using the “method of continuity”, the classical method used by Alexandrow [4], Andreev [5], later by Rivin and Hodgson [18], and in this author’s more recent proof of Andreev’s Theorem [19], [20]. The idea of this method is to establish a bijection between two manifolds of the same dimension: one, X, consisting of the geometric objects that you want to construct, and the other, Y, a subset of Rnconsisting of various angles, lengths, etc. The space Xshould be viewed as unknown and the space Yas known. You then consider the mapping f:X→Ywhich takes your geometric object, in X, and reads off its appropriate measurements, in Y. Of course, you need to show that the image is actually in Y, namely, that the constraints that you put on the coordinates of Y(typically something like the triangle inequality for the edges of a triangle) are indeed satisfied for each geometric object of X. This map fwill always be obviously continuous, and it is not too hard to show that it is proper and injective. (Recall that a mapping is said to be proper if the pullback of a compact set is compact.) Then, the following lemma can be used to show that the image of fis a union of connected components of Y. Lemma 5. Let Xand Ymetric spaces, and let f:X→Ybe a proper local homeomorphism. Then the image of fis a union of connected components of Y. Proof of Lemma 5: It is sufficient to show that f(X) is both open and closed in Y. Because fis a local homeomorphism, it is an open mapping, so f(X) is open in Y; and since f:X→Yis proper, it immediately Compact Hyperbolic Tetrahedra 217 follows that the limit of any sequence in the image of fwhich converges in Ymust lie in the image of f, so f(X) is closed in Y. In fact, a stronger result is true: any local homeomorphism between metric spaces which is also proper will be a finite-sheeted covering map [10, p. 23] and [14, p. 127]. This gives an alternative route to proving Lemma 5. Therefore, this lemma reduces the problem to showing that Xis nonempty and that Yis connected, which are usually the hardest parts! The result of the “method of continuity” is that you have established a bijection between your geometric objects, set X, and the measurements Y. Let Cbe a cell complex on S2that describes the combinatorics of a convex polyhedron. We say that a hyperbolic polyhedron P⊂H3 realizes Cif there is a cellular homeomorphism from Cto ∂P (i.e., a homeomorphism mapping faces of Cto faces of P, edges of Cto edges of P, and vertices of Cto vertices of P). We will call each isotopy class of cellular homeomorphisms φ:C→∂P amarking on P. Let ∆ be the cell complex on S2describing the combinatorics of the tetrahedron. Throughout this paper we will call hyperbolic polyhedra realizing ∆ hyperbolic tetrahedra. We will define P∆to be the set of pairs (P, φ) so that Pis a hyperbolic tetrahedron and φis a marking on Pwith the equivalence relation that (P, φ)∼(P′, φ′) if there exists an automorphism ρ:H3→H3such that ρ(P) = P′and both φ′and ρ◦φrepresent the same marking on P′. Proposition 6. The space P∆is a manifold of dimension 6. Proof: Let Hbe the space of closed half-spaces of H3; clearly His a 3-dimensional manifold. Let O∆be the set of marked hyperbolic polyhedra realizing ∆. By forgetting this marking, an element of O∆is a 4-tuple of half-spaces that intersect in a polyhedron realizing ∆. This induces a mapping from O∆to H4whose image is an open set. We give O∆the topology that makes this mapping from O∆into H4a local homeomorphism. Since H4is a 12-dimensional manifold, O∆must be a 12-dimensional manifold as well. If ρ(P, φ) = (P, φ), we have that ρ◦φis isotopic to φthrough cellular homeomorphisms. Hence, the automorphism ρmust fix all vertices of P, and consequently restricts to the identity on all edges and faces. However, an automorphism of H3which fixes four non-coplanar points must be the identity. Therefore Aut(H3) acts freely on O∆. 218 R. K. W. Roeder This quotient is P∆, hence P∆is a manifold with dimension equal to dim(O∆)−dim(Aut(H3)) = 3 ·4−6 = 6. In fact, we will restrict to the subset P0 ∆of tetrahedra with dihedral angles in (0, π/2]. Notice that P0 ∆is not, a priori, a manifold or even a manifold with boundary. All that we will need for the proof of Theorem 4 is that P∆is a manifold and that the subspace P0 ∆is a metric space. Consider the map α:P∆→R6which is obtained by measuring the dihedral angles (ordered by the marking) of an element of P∆. Using the topology on P∆that is described in the proof of Proposition 6, it is clear that αis continuous. Therefore, we will use the method of continuity to show that αrestricted to P0 ∆is a homeomorphism onto A∆, in order to prove Theorem 4. At this point it is necessary to clarify the statement of uniqueness in Theorem 4. We will show that the map αis injective, which shows that for each set of proposed dihedral angles α1,...,α6there is a unique marked tetrahedron with the dihedral angles α1,...,α6, as ordered by this marking. This is what we mean by uniqueness in Theorem 4 and in the later Theorem 10. Proof of Theorem 4: The first step is to make sure that the dihedral angles of a compact tetrahedron satisfy conditions (1)–(3). For condition (1), notice that if two adjacent faces intersect along a line segment with dihedral angle 0, they would coincide. In addition, the dihedral angle between adjacent faces is ≤π/2 by hypothesis. For condition (2), let xbe a vertex of P. The compactness of Pimplies that x∈H3, and by Lemma 2, the sum of the dihedral angles between the three planes intersecting at xmust be > π. Furthermore, each face of a hyperbolic tetrahedron is a hyperbolic triangle of non-zero area so the Gauss-Bonnet formula gives condition (3). Therefore conditions (1)–(3) are necessary. There is an elementary proof that α:P∆→REis injective: Since the face angles are uniquely determined by the dihedral angles and each face is a hyperbolic triangle, one can calculate the length of each edge using the hyperbolic law of cosines. Before proving that α:P0 ∆→A∆is proper, we will need the following lemma: Lemma 7. Given three points v1,v2,v3that form a non-obtuse, nondegenerate triangle in the Poincar´e model of H3, there is a unique orientation preserving isometry taking v1to a positive point on the xaxis, v2to a positive point on the y-axis, and v3to a positive point on the z-axis. Compact Hyperbolic Tetrahedra 219 Proof of Lemma 7: The points v1,v2,and v3form a triangle Tin a plane PT. It is sufficient to show that there is a plane QTin the Poincar´e ball model that intersects the positive octant in a triangle isomorphic to T. The isomorphism taking v1,v2, and v3to the x,y, and z-axes will then be the one that takes the plane PTto the plane QTand the triangle Tto the intersection of QTwith the positive octant. Let s1,s2,and s3be the side lengths of T. The plane QTmust intersect the x,y, and z-axes at distances a1,a2, and a3satisfying the hyperbolic Pythagorean theorem: cosh(s1) = cosh(a2) cosh(a3), cosh(s2) = cosh(a3) cosh(a1), cosh(s3) = cosh(a1) cosh(a2). These equations can be solved for (cosh2(a1),cosh2(a2),cosh2(a3)), obtaining cosh(s2) cosh(s3) cosh(s1),cosh(s3) cosh(s1) cosh(s2),cosh(s1) cosh(s2) cosh(s3). The only concern in solving for aiis that each of these terms is ≥1. However, this follows from the triangle Tbeing non-obtuse. Lemma 8. The mapping α:P0 ∆→A∆is proper. Proof: To see that α:P0 ∆→A∆is a proper mapping, suppose that there is a sequence of polyhedra Pirealizing ∆, with α(Pi) = ai∈A∆. We must show that if aiconverges to a∈A∆, then a subsequence of the Piconverges to some P∞in P0 ∆. Throughout this part of the proof, we consider each Pito be in the Poincar´e ball. Denote the vertices of Piby vi 1,vi 2,vi 3, and vi 4. According to Lemma 7, we can normalize each Piso that vi 1is on the x-axis vi 2is on the y-axis, and vi 3is on the z-axis. Because H3is a compact space (in the Euclidean metric), we can take a subsequence of the Piso that the vertices vi 1,...,vi 4converge to some points v1,...,v4in H3. 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