THE GEOMETRY OF CROOKED PLANES

TODD A. DRUMM AND WILLIAM M. GOLDMAN

Abstract. Crooked planes are polyhedra used to construct fundamental polyhedra for discrete groups of Lorentz isometries acting properly on Minkowski (2+1)-space. This paper explores intersections of crooked planes. Criteria for the disjointness of crooked planes are developed. These criteria are applied to derive sufficient conditions for affine deformations of a discrete subgroup of SO(2, 1) to act properly on Minkowski space.

Contents

1. Introduction 2 1.1. Milnor’s Question 2 1.2. Fundamental Polyhedra for Schottky groups 3 1.3. Crooked Planes 3 1.4. Discrete groups 4 2. Background 6 2.1. Lorentzian vector spaces 6 2.2. Relation to hyperbolic geometry 8 2.3. Minkowski 2+1-space 9 3. Anatomy of a crooked plane 9 3.1. Wings and stems 9 3.2. Inversions and half-spaces 10 3.3. Automorphisms 11 3.4. A specific example 11 4. Intersections of wings 13 4.1. Intersection of planes 13 4.2. Intersections of wings 13 4.3. Representation of spacelike segments 15 4.4. Parallel null half-planes 16 4.5. Intersection of stems and null half-planes 17

Date: December 11, 1996. Drumm was partially supported by an NSF Postdoctoral fellowship and the

Mathematical Sciences Research Institute. Goldman was partially supported by NSF grant DMS-8902619 and the University of Maryland Institute for Advanced Computer Studies.

5. Intersection of stems 21 5.1. Pairs of spacelike vectors 21 5.2. Ultraparallel Stems 22 5.3. Asymptotic Stems 24 5.4. Crossing Stems 25 6. Intersections of crooked planes 26 6.1. Consistent orientation 26 6.2. Ultraparallel crooked planes 27 6.3. The polygon of intersection 29 6.4. Asymptotic crooked planes 30 6.5. Crossing crooked planes 31 6.6. Crooked planes with different orientations 32 7. Groups generated by inversions 32 References 33

1. Introduction

1.1. Milnor’s Question. In 1977, Milnor asked ([14]) whether a nonamenable group (for example a free group of rank 2) could act properly by affine transformations. He observed that by applying Tits’s alternative (see [15]), this question is equivalent to whether a nonabelian free group could act properly by affine transformations. Furthermore he proposed the following construction of such a group: Start with a free discrete subgroup of SO^{0}(2, 1) (for example a Schottky group acting on the hyperbolic plane) and add translation components to obtain a group of affine transformations which may act freely. However, “it seems difficult to decide whether the resulting group action is properly discontinuous.” ([14])

In 1983, Margulis showed that such properly discontinuous groups do indeed exist ([11],[12]), realizing Milnor’s suggestion. We study these actions in terms of their quotient spaces, which we call Margulis space-times. A Margulis space-time is characterized as a complete flat Lorentz 3-manifold M whose fundamental group is nonamenable (that is, not virtually solvable). If Γ is a discrete group acting properly on 𝔼 by Lorentz isometries, then 𝔼/Γ is a complete flat Lorentz manifold. Conversely the fundamental group of a complete flat Lorentz manifold is a discrete group acting properly on 𝔼 by isometries. (Recall that a group acts properly discontinuously if and only if it acts properly when given the discrete topology.) Since Γ contains a nonabelian free subgroup of finite index, a Margulis space-time M admits a finite covering space whose fundamental group is a nonabelian free group.

1.2. Fundamental Polyhedra for Schottky groups. A model for this construction is the Klein-Maskit combination procedure applied to Schottky groups. Namely, a Schottky group is a discrete group Γ of isometries of hyperbolic space H^{n} freely generated by cyclic groups in the following way. If γ is an isometry for which there exists a half-space H ⊂ H^{n} such that γ(H) ⊂ H, then Δ = H ∩ γ(H) is a fundamental domain for the cyclic group ⟨γ⟩ generated by γ. Suppose γ_{1}, …, γ_{r} are isometries for which there exist half-spaces H_{i} such that:

• γ_{i}H_{i} ⊂ H_{i} for each i = 1, …, r;

• The half-spaces H_{1}, …, H_{r}, H^{n} − H_{1}, …, H^{n} − H_{r} are pairwise

disjoint. Then the cyclic groups ⟨γ_{i}⟩ freely generate a discrete group Γ. Furthermore the intersection of their fundamental domains Δ_{i} = H_{i} ∩ γ_{i}H_{i} is a fundamental domain

Δ = ∩_{i=1}^{r}Δ_{i}

for Γ. Furthermore this construction is robust in the sense that for any small perturbation of either the half-spaces H_{i} or the generators γ_{i} gives another such fundamental domain.

One would like to imitate this construction for groups of affine transformations. Each generator γ is an affine transformation which leaves invariant a line ℓ, acts by a hyperbolic linear map on the quotient plane ℝ^{2,1}/ℓ, and acts by a nontrivial translation along ℓ. Choose a point p ∈ ℓ; there is a unique plane P containing p such that γ(P) is disjoint (that is, parallel) to P. The region bounded by P and γ(P) is a fundamental domain for ⟨γ⟩, which we call a slab.

Slabs unfortunately do not make satisfactory fundamental domains. They fail to be robust as above, and impossible to combine. For example, two slabs never have disjoint complements, a necessary condition for the Klein-Maskit construction. In order to combine properly discontinuous cyclic groups, one needs building blocks which are more adapted to the linear part of the generators, rather than the translational part. Crooked planes serve this purpose, as described by Drumm in his 1990 doctoral dissertation [3].

1.3. Crooked Planes. In [3], complete flat Lorentz manifolds with nonabelian free fundamental group were directly constructed using Poincaré’s fundamental polyhedron technique ([3],[4]). These examples are diffeomorphic to solid handlebodies. Their fundamental groups are “affine Schottky groups” with fundamental domains derived from fundamental domains of Schottky subgroups of SO^{0}(2, 1) acting on the hyperbolic plane H^{2}. These fundamental polyhedra are bounded by certain polyhedral hypersurfaces in ℝ^{3}, called crooked planes. (See Figure 1.) This paper investigates the geometry of crooked planes. In particular we develop criteria for disjointness of crooked planes and, in general, describe intersections of crooked planes.

Conjecture 1.3.1. Every Margulis space-time has a fundamental polyhedron whose faces are crooked planes.

A crooked plane consists of three parts: two half-planes, called wings and a pair of opposite planar sectors, called its stem. The wings lie in null planes and the stem (whose interior has two connected components) lies in a time-like (indefinite) plane. Crooked planes are sufficiently flexible to build fundamental polyhedra for affine Schottky groups. One starts with a fundamental polygon Δ for the linear part 𝕃(Γ) acting on the hyperbolic plane H^{2}. The sides of Δ correspond to spacelike directions. Using the disjointness criteria developed here, we find crooked planes corresponding to the sides of Δ. Sometimes these crooked planes bound a fundamental domain for the properly discontinuous group Γ.

Here is a basic technical result:

Corollary 4.5.5. Let C_{1} and C_{2} be two crooked planes. Suppose that the three planes containing the stems and wings of C_{1} are each transverse to the three planes containing the stems and wings of C_{2}. If a wing W of C_{1} intersects the stem of C_{2}, then W intersects one of the wings of C_{2}.

1.4. Discrete groups. These results are applied to groups generated by inversions. If l is a spacelike line, there is a unique inversion, ι_{l}, an isometric involution fixing l. For each point p ∈ l, there is a unique crooked plane C with vertex p invariant under ι_{l}. Furthermore each of the two connnected components of the complement of C is a fundamental domain for ι_{l}. Let l_{1}, …, l_{n} be a family of spacelike lines and ι_{1}, …, ι_{n} be the corresponding inversions. The criterion discussed in [7] for a Schottky group to have properly discontinuous affine deformations implies a criterion for the group Γ generated by ι_{1}, …, ι_{n} to act properly. We examine these criteria in terms of the disjointness criteria for crooked planes corresponding to l_{1}, …, l_{n}.

Although [4] presents a classification of the linear holonomy representations of Margulis space-times, the full moduli space of these structures remains mysterious. Let ℝ^{2,1} denote a 3-dimensional real vector space with a nondegenerate inner product of signature (2,1). Let Γ ⊂ SO(2, 1) be a subgroup. An affine representation of Γ with the given linear part is determined by its translational part u : Γ ⟶ ℝ^{2,1}

Figure 1. A crooked plane

which is a cocycle of Γ with values in the Γ-module ℝ^{2,1}. Thus the space of affine representations of Γ with the given linear part identifies with the vector space Z^{1}(Γ, ℝ^{2,1}) of ℝ^{2,1}-valued cocycles of Γ. Explicitly, if u ∈ Z^{1}(Γ, ℝ^{2,1}) is such a cocycle, then the affine representation

ρ_{u} : Γ ⟶ Aff(𝔼)

is given by:

ρ_{u}(γ) : x ⟼ γ(x) + u(γ)

(The cocycle condition on u derives from condition

ρ_{u}(xy) = ρ_{u}(x)ρ_{u}(y).

Compare §1 of Fried-Goldman-Hirsch [9].) Conjugation by translation by v_{0} ∈ ℝ^{2,1} alters u by the coboundary

δv_{0} ∈ Z^{1}(Γ, ℝ^{2,1})

of v_{0}:

ρ_{u}(γ) : x − v_{0} ⟼ γ(x − v_{0}) + u(γ) + δv_{0}(γ)

where

δv_{0} : γ ⟼ v_{0} − γ(v_{0}).

Thus the space of translational conjugacy classes of affine representations of Γ with the given linear part identifies with the cohomology H^{1}(Γ, ℝ^{2,1}). For given Γ ⊂ SO(2, 1), a subset of H^{1}(Γ, ℝ^{2,1}) corresponds to proper actions. Using fundamental domains for Γ acting on the hyperbolic plane, we have constructed [3],[7] a certain open cone in this subset. Here we interpret the conditions defining this cone in terms of intersections of crooked planes. Understanding how crooked planes intersect is the first part of our program to analyze the moduli space of Margulis space-times.

We are grateful to Virginie Charette for carefully reading this manuscript and correcting several errors.

2. Background

2.1. Lorentzian vector spaces. Define ℝ^{2,1} to be the 3-dimensional real vector space with the Lorentzian inner product

𝔹 : ℝ^{2,1} × ℝ^{2,1} ⟶ ℝ

defined by:

𝔹 (u, v) = u_{1}v_{1} + u_{2}v_{2} − u_{3}v_{3},

where

u = [ \begin{matrix}u_{1} \\ u_{2} \\ u_{3} \end{matrix} ]

and

v = [ \begin{matrix}v_{1} \\ v_{2} \\ v_{3} \end{matrix} ]

are written in terms of the usual basis {e_{1}, e_{2}, e_{3}}. Denote by

SO(2, 1) ⊂ O(2, 1)

the group of unimodular linear automorphisms of ℝ^{2,1}, and SO^{0}(2, 1) its identity component. The Lorentzian cross-product ⊠ : ℝ^{2,1}×ℝ^{2,1} ⟶ ℝ^{2,1} is the skew-symmetric bilinear mapping

u ⊠ v = [ \begin{matrix}u_{2}v_{3} − u_{3}v_{2} \\ u_{3}v_{1} − u_{1}v_{3} \\ u_{2}v_{1} − u_{1}v_{2} \end{matrix} ].

which satisfies 𝔹 (x, x ⊠ y) = 0. For more details, the reader is referred to [7]. (Minkowski space ℝ^{2,1} identifies with the Lie algebra 𝔰𝔬(2, 1) of SO(2, 1) by:

ℝ^{2,1} ⟶ 𝔰𝔬(2, 1)
v ⟼ (u ↦ v ⊠ u).

Under this identification cross-product ⊠ corresponds to the Lie bracket and the inner product 𝔹 corresponds to (-1/2) times the Killing form.)

Let |v| denote the Euclidean norm of the vector v. Denote by ⟨v_{1}, v_{2}, …, v_{n}⟩ the subspace spanned by v_{1}, …, v_{n}. (For a single vector v we also reserve the notations ℝv = ⟨v⟩ for the line containing v and ℝ_{+}v for the ray containing v.) For any vector v ∈ ℝ^{2,1}, its Lorentzperpendicular plane is defined to be

P(v) = {u ∈ ℝ^{2,1} | 𝔹 (u, v) = 0}

The light cone

𝕎 = {x ∈ ℝ^{2,1} | 𝔹 (x, x) = 0}

consists of all null vectors, and its two components form the timeorientations of ℝ^{2,1}. (Each component is SO^{0}(2, 1)-invariant.) We refer to the upper nappe

𝕎^{+} = {x ∈ 𝕎 | x_{3} > 0}, (2.1.1)

as the positive time-orientation. Similarly the set of timelike vectors

𝕌 = {x ∈ ℝ^{2,1} | 𝔹 (x, x) < 0}

has a component

𝕌^{+} = {x ∈ 𝕌 | x_{3} > 0}, (2.1.2)

also defining the positive time-orientation.

If a, b ∈ 𝕎^{+} are nonzero null vectors with positive time-orientation, then v = a ⊠ b is spacelike and x^{−}(v) = a and x^{+}(v) = b, correcting a misprint in [7], p.152 (3). If v_{1}, v_{2} are two vectors, then v_{1} ⊠ v_{2} is a vector parallel to the intersection P(v_{1}) ∩ P(v_{2}) (which is typically a line).

Let v ∈ 𝕎^{+} be a nonzero null vector. The plane P(v) equals the plane tangent to the light cone containing the line ℝv and the null line ℝv divides it into two components. The mapping x ↦ x ⊠ v defines a diffeomorphism of 𝕎^{+} − ℝ_{+}v onto one component. We denote the closure of this component by

P^{+}(v) = Cl ((𝕎^{+} − ℝ_{+}v) ⊠ v)

The other component equals

P^{−}(v) = Cl (v ⊠ (𝕎^{+} − ℝ_{+}v))

A nonzero vector v ∈ ℝ^{2,1} is called • spacelike if 𝔹 (v, v) > 0;

• null (or lightlike) if 𝔹 (v, v) = 0;

• timelike if 𝔹 (v, v) < 0; A vector v is called unit-spacelike if and only if 𝔹 (v, v) = 1 and unittimelike if and only if 𝔹 (v, v) = −1. If v is a spacelike vector, P(v)∩𝕎 is the union of 2 null lines. There exists a unique pair

x^{−}(v), x^{+}(v) ∈ P(v) ∩ 𝕎^{+}

such that

|x^{−}(v)| = |x^{+}(v)| = 1 (2.1.3)

and

𝔹 (v, x^{−}(v) ⊠ x^{+}(v)) > 0 (2.1.4)

Condition (2.1.3) is a Euclidean normalization and condition (2.1.4) means that the ordered triple (v, x^{−}(v), x^{+}(v)) is a positively-oriented basis. If v is spacelike, then

x^{−}(v) ⊠ x^{+}(v) = \frac{2𝔹 (v, v)}{𝔹 (v, v) + |v|^{2}} v

and

v ⊠ x^{+}(v) = −𝔹 (v, v)^{1/2}x^{+}(v)

Note that ⟨v, x^{±}(v)⟩ = P(x^{±}(v)) is the plane tangent to the null cone at x^{±}(v). For x ∈ 𝕎^{+} the interior of P^{+}(x) is

{v ∈ P(x) | x^{+}(v) = x/|x|}.

2.2. Relation to hyperbolic geometry. It will be useful to have a concrete model for the hyperbolic plane in this geometry as we will want to pass from fundamental polygons for discrete subgroups of SO^{0}(2, 1) acting on H^{2} to fundamental polyhedra for the corresponding actions on Minkowski space. The hyperboloid defined by 𝔹 (v, v) = −1 has an induced Riemannian metric of constant curvature -1. Its two components correspond to the two time-orientations; we denote its positively time-oriented component by H^{2}. Equivalently, we may consider H^{2} as the image in the projective space ℙ(ℝ^{2,1}) of the open cone 𝕌^{+} of timelike vectors defined in (2.1.2) . Points of H^{2} correspond to timelike lines and geodesics in H^{2} to spacelike planes (polar to spacelike vectors). If v_{1}, v_{2} are timelike vectors in ℝ^{2,1}, then the distance ρ between the corresponding points in H^{2} is given by:

cosh(ρ) = \frac{|𝔹 (v_{1}, v_{2})|}{\sqrt{𝔹 (v_{1}, v_{1})𝔹 (v_{2}, v_{2})}}

Spacelike lines in ℝ^{2,1} correspond to geodesics in H^{2}. Suppose that v ∈ ℝ^{2,1} is spacelike. Then the plane P(v) ⊂ ℝ^{2,1} meets Ω in timelike lines determining a geodesic in H^{2}.

2.3. Minkowski 2+1-space. Consider (2+1)-dimensional Minkowski space, the affine space 𝔼 corresponding to ℝ^{2,1}. Minkowski space is a simply connected complete flat Lorentzian manifold of dimension 2+1. Conversely any simply connected complete flat Lorentzian manifold of dimension 2+1 is isometric to Minkowski space. For p, q ∈ 𝔼 the vector v = q − p corresponds to the translation carrying p to q. We also write q = p + v. Denote by \overline{pq} the line segment from p to q. Identify the vector space of translations of 𝔼 with the original inner product space ℝ^{2,1}. (In particular the tangent space T_{p}𝔼 canonically identifies with ℝ^{2,1}.) Denote by

H^{0} = SO^{0}(2, 1) ⋉ ℝ^{2,1}

the identity component of the group of isometries of 𝔼 and

H = SO(2, 1) ⋉ ℝ^{2,1}

its (two-component) double cover. Denote by

𝕃 : H → SO(2, 1)

the homomorphism assigning to h ∈ H its linear part 𝕃(h).

A geodesic in 𝔼 is just a Euclidean straight line, parametrized by constant speed. Geodesics are described in the usual parametric form

l = p + ℝv = {p + tv | t ∈ ℝ} (2.3.1)

where p is a point on the line, t ∈ ℝ a real parameter, and v ∈ ℝ^{2,1} a vector. l is spacelike, timelike, or null if and only if the direction vector v is spacelike, timelike, or null respectively.

If l = p + ℝv is a line where v is spacelike or timelike, then there is a unique isometry ι_{l} of order two whose fixed-point set equals l. Explicitly,

ι_{l}(p + x) = p − x + 2\frac{𝔹 (x, v)}{𝔹 (v, v)}v.

We call ι_{l} the inversion in l.

3. Anatomy of a crooked plane

3.1. Wings and stems. Let p ∈ 𝔼 be a point and v ∈ ℝ^{2,1} a spacelike vector. Define the positively oriented crooked plane C(v, p) ⊂ 𝔼 with vertex p and direction vector v to be the union of two wings

W^{+}(v, p) = p + P^{+}(x^{+}(v)), W^{−}(v, p) = p + P^{+}(x^{−}(v))

and a stem

S(v, p) = p + {x ∈ ℝ^{2,1} | 𝔹 (v, x) = 0, 𝔹 (x, x) ≤ 0}.

Each wing is a half-plane, and the stem is the union of two quadrants in a spacelike plane. The positively oriented crooked plane itself is a piecewise linear submanifold, which stratifies into four connected open subsets of planes (two wings and the two components of the interior of the stem), four null rays, and a vertex. (See Figure 1.)

We also define the negatively oriented crooked plane K(v, p) ⊂ 𝔼 with vertex p and direction vector v to be the union of two wings

M^{+}(v, p) = p + P^{−}(x^{+}(v)), M^{−}(v, p) = p + P^{−}(x^{−}(v))

and a stem

S(v, p) = p + {x ∈ ℝ^{2,1} | 𝔹 (v, x) = 0, 𝔹 (x, x) ≤ 0}.

This paper contains calculations only for positively oriented half planes. Furthermore, an unspecified orientation for a crooked plane will assumed to be the positive one.

3.2. Inversions and half-spaces. A crooked plane of either orientation is a polyhedral ruled surface: its wings are foliated by parallel null lines and its stem decomposes into timelike lines all passing through the vertex. Furthermore the line l through the vertex orthogonal to the stem is the unique spacelike line contained in the crooked plane. Reminiscent of the structure of bisectors in complex hyperbolic space([10]), we call l the spine of the crooked plane and the null and timelike lines ruling the crooked plane its slices.

The spine of the positively oriented crooked plane C(v, p) is

l = p + ℝv.

Given a spacelike line ℓ and a point p ∈ ℓ, there is a unique positively oriented crooked plane with spine ℓ and vertex p. Inversion ι_{l} in l preserves C(v, p), leaving invariant each wing, but permuting the two components of W^{+}(v, p) − l (respectively, W^{−}(v, p) − l). Similarly, ι_{l} permutes the two components of S(v, p) − {p}.

Furthermore each component of the complement 𝔼 − C(v, p) is a fundamental domain (a crooked half-space) for the group generated by ι_{l}. Given p ∈ 𝔼 and a unit-spacelike vector v, we define the crooked half-space H(v, p) to be the component of 𝔼 − C(v, p) containing

p + {u ∈ 𝕎^{+} | 𝔹 (v, u) > 0}.

Note that 𝔼 is the disjoint union

H(v, p) ∪ C(v, p) ∪ H(−v, p).

Explicitly the half-space is defined as follows. Choose a point p ∈ ℝ^{2,1} and a (spacelike) vector v ∈ ℝ^{2,1} such that ⟨v, v⟩ = 1. Then there exist null vectors u_{+}, u_{−} in the positive light-cone defined by

x^{2} + y^{2} − z^{2} = 0 and z > 0

such that ⟨u_{±}, v⟩ = 0 and {v, u_{−}, u_{+}} is a positively oriented basis. Then the crooked half-space with vertex p and direction vector v is defined by:

H(v, p) = {p + v ∈ ℝ^{2,1} | ⟨u_{−}, v⟩ > 0 if ⟨v, v⟩ > 0,
⟨u_{−}, v⟩⟨u_{+}, v⟩ > 0 if ⟨v, v⟩ = 0,
⟨u_{+}, v⟩ > 0 if ⟨v, v⟩ < 0}

and the corresponding crooked plane is its boundary ∂H(v, p).

We have developed criteria for the intersection of these polyhedra H(v, p) in terms of the parameters p, v. In [3], fundamental polyhedra for discontinuous groups Γ acting on ℝ^{2,1} are derived from fundamental polygons for the action of the linear group L(Γ) ⊂ SO(2, 1) on the hyperbolic plane. (The edges of the fundamental polygon are geodesics polar to the spacelike vectors v.) The spacelike geodesic

l = p + ℝv

lies on the crooked plane ∂H(v, p). The crooked half-space H(v, p) is a fundamental polyhedron for Lorentz-inversion in l.

Similarly, for each timelike slice s of C(v, p), the inversion ι_{s} preserves C(v, p). This inversion permutes the two wings but leaves invariant each component of the stem. The crooked half-spaces comprising the complement of C(v, p) are fundamental domains for the cyclic group generated by the inversion. These properties hold as well for negatively oriented crooked planes and half-spaces.

3.3. Automorphisms. Let C = C(v, p) be a crooked plane. Let ϕ be an automorphism of ℝ^{2,1} which preserves C. The vertex p of C is fixed by ϕ. The differentiable structure on C is singular at p and along the two null-lines p + ℝx^{±}(v). However p is the only singular point for which the singularity is nonremovable (it has a nonsingular intrinsic differentiable structure). Thus ϕ(p) = p. The spine l of C is the unique spacelike line contained in C; hence ϕ(l) = l. Conversely every isometry of ℝ^{2,1} which preserves l preserves C.

3.4. A specific example. Let p be the origin and let

v = [\begin{matrix}1 \\ 0 \\ 0 \end{matrix}] .

Then

x^{−}(v) = \frac{1}{\sqrt{2}} [\begin{matrix}0 \\ 1 \\ 1 \end{matrix}] , x^{+}(v) = \frac{1}{\sqrt{2}} [\begin{matrix}0 \\ −1 \\ 1 \end{matrix}]

and

W^{−}(v, p) = \{[\begin{matrix}x \\ y \\ y \end{matrix}]| x ≤ 0\} ,
S(v, p) = \{[\begin{matrix}0 \\ y \\ z \end{matrix}]| |y| ≤ |z|\} ,
W^{+}(v, p) = \{[\begin{matrix}x \\ −y \\ y \end{matrix}]| x ≥ 0\} .

Also,

M^{−}(v, p) = \{[\begin{matrix}x \\ y \\ y \end{matrix}]| x ≥ 0\} ,
M^{+}(v, p) = \{[\begin{matrix}x \\ −y \\ y \end{matrix}]| x ≤ 0\} .

The spine of the crooked planes C(v, p) and K(v, p) is the line:

σ = \{[\begin{matrix}x \\ 0 \\ 0 \end{matrix}]| x ∈ ℝ\}

with inversion:

[\begin{matrix}x \\ y \\ z \end{matrix}] ⟼ [\begin{matrix}x \\ −y \\ −z \end{matrix}] .

A timelike line contained in S(v, p) is given by:

τ_{t} = \{[\begin{matrix}0 \\ s sinh(t) \\ s cosh(t) \end{matrix}]| s ∈ ℝ\}

and the corresponding inversion given by the matrix:

[\begin{matrix}−1 & 0 & 0 \\ 0 & − cosh(2 t) & sinh(2 t) \\ 0 & − sinh(2 t) & cosh(2 t) \end{matrix}] .

The isometry group of C is the image of the embedding

O(1, 1) ⟶ SO(2, 1) ⊂ H
A ↦ [\begin{matrix}A & 0 \\ 0 & det(A) \end{matrix}] .

4. Intersections of wings

4.1. Intersection of planes. This section describes the intersection of two wings, and a wing and a stem. The following section describes intersections of stems. From these results follow necessary and sufficient conditions for the intersection of two crooked planes. Since every stem and wing lies in a plane in 𝔼, every intersection of a stem or wing with a stem or wing will lie in the intersection of the planes which contain them.

A plane in 𝔼 may be written as p + P(v), for p ∈ 𝔼 and v ∈ ℝ^{2,1}. Suppose that p_{1}, p_{2} ∈ 𝔼 and v_{1}, v_{2} ∈ ℝ^{2,1} satisfy 𝔹 (v_{1}, v_{2}) ≠ 0. The intersection 𝔓_{1} ∩ 𝔓_{2} of the two planes 𝔓_{i} = p_{i} + P(v_{i}) is a line which can be parametrized as

p + ℝ(v_{1} ⊠ v_{2}) = {p + kv_{1} ⊠ v_{2} | k ∈ ℝ}

for some p ∈ 𝔓_{1} ∩ 𝔓_{2}. If 𝔹 (v_{1}, v_{2}) = 0 and v_{1} and v_{2} are linearly independent then 𝔓_{1} ∩ 𝔓_{2} can be parametrized as p + ℝv_{1} ⊠ v_{2}.

4.2. Intersections of wings. We now describe the intersection of two null half-planes that are the wings of crooked planes. We need only consider positively oriented null half-planes. The generic case — nonparallel null half-planes — is treated first. Lemma 4.2.1 asserts that the intersection is either empty, a point, or a spacelike line segment. Conversely, every spacelike line segment arises uniquely as the intersection of positively oriented null half-planes (Theorem 4.3.1). The degenerate case of parallel null half-planes is treated in Lemma 4.4.1, where it is shown that either the intersection is either empty or one half-plane contains the other. §4.5 treats intersections of stems and null half-planes.

Lemma 4.2.1. Let x_{1}, x_{2} ∈ 𝕎^{+} be linearly independent null vectors and p_{1}, p_{2} ∈ 𝔼 be points. The corresponding null half-planes 𝔭_{1}^{+} = p_{1} + P^{+}(x_{1}) and 𝔭_{2}^{+} = p_{2} + P^{+}(x_{2}) are disjoint if and only if 𝔹 (p_{2} − p_{1}, x_{1} ⊠ x_{2}) > 0. In case that 𝔭_{1}^{+} ∩ 𝔭_{2}^{+} ≠ ∅:

• 𝔭_{1}^{+} ∩ 𝔭_{2}^{+} is a point if and only if 𝔹 (p_{2} − p_{1}, x_{1} ⊠ x_{2}) = 0.

• 𝔭_{1}^{+} ∩ 𝔭_{2}^{+} is a spacelike line segment if and only if 𝔹 (p_{2} − p_{1}, x_{1} ⊠ x_{2}) < 0, with one endpoint on ∂𝔭_{1}^{+} and the other endpoint on ∂𝔭_{2}^{+}.

Proof. Let ℓ be the intersection of the planes that contain 𝔭_{1}^{+} and 𝔭_{2}^{+}. Then 𝔭_{1}^{+} ∩ 𝔭_{2}^{+} ⊂ ℓ. Let w = x_{1} ⊠ x_{2} and

p = ℓ ∩ (p_{1} + P (w)

so that

ℓ = p + ℝw.

The subsets ℓ ∩ 𝔭_{1}^{+} and ℓ ∩ 𝔭_{2}^{+} of ℓ are characterized as the set of all p + kw (where k ∈ ℝ) satisfying the inequalities

𝔹 ((p + kw) − p_{1}, −w) ≥ 0,
𝔹 ((p + kw) − p_{2}, w) ≥ 0,

respectively. p was chosen so that 𝔹 (p − p_{1}, w) = 0 and for the given parametrization of ℓ

ℓ ∩ 𝔭_{1}^{+} ⟷ (−∞, 0]

ℓ ∩ 𝔭_{2}^{+} ⟷ [\frac{𝔹 (p_{2} − p_{1}, w)}{𝔹 (w, w)}, ∞) .

Then:

• 𝔭_{1}^{+} ∩ 𝔭_{2}^{+} is empty if and only if 𝔹 (p_{2} − p_{1}, w) > 0;

• 𝔭_{1}^{+} ∩ 𝔭_{2}^{+} is a point if and only if 𝔹 (p_{2} − p_{1}, w) = 0;

• 𝔭_{1}^{+} ∩ 𝔭_{2}^{+} is a line segment if and only if 𝔹 (p_{2} − p_{1}, w) < 0. The proof is complete.

Here are the corresponding lemmas for pairs of negatively oriented half-planes and pairs of half-planes with opposite orientations:

Lemma 4.2.2. Let x_{1}, x_{2} ∈ 𝕎^{+} be linearly independent null vectors and p_{1}, p_{2} ∈ 𝔼 be points. The half-planes

𝔭_{1}^{−} = p_{1} + P^{−}(x_{1}), 𝔭_{2}^{−} = p_{2} + P^{−}(x_{2})

are disjoint if and only if 𝔹 (p_{2} − p_{1}, x_{1} ⊠ x_{2}) < 0.

Lemma 4.2.3. Let x_{1}, x_{2} ∈ 𝕎^{+} be linearly independent null vectors and p_{1}, p_{2} ∈ 𝔼 be points. The intersection of the half-planes 𝔭^{+} = p_{1} + P^{+}(x_{1}) and 𝔭^{−} = p_{2} + P^{−}(x_{2}) is never empty.

Figure 2. An intersection of two null half-planes

Proof. The line ℓ equals the intersection of the planes that contain 𝔭^{+} and 𝔭^{−}. Let w = x_{1} ⊠ x_{2} and p = ℓ ∩ (p_{1} + P (w) so that ℓ = p + ℝw. The subsets ℓ ∩ 𝔭^{+} and ℓ ∩ 𝔭^{−} of ℓ consist of all p + kw (where k ∈ ℝ) satisfying the inequalities

𝔹 ((p + kw) − p_{1}, −w) ≥ 0,
𝔹 ((p + kw) − p_{2}, w) ≤ 0,

respectively. The point p was chosen so that 𝔹 (p − p_{1}, w) = 0 and for the given parametrization of ℓ

ℓ ∩ 𝔭^{+} ⟷ (−∞, 0]

ℓ ∩ 𝔭^{−} ⟷ (−∞, \frac{𝔹 (p_{2} − p_{1}, w)}{𝔹 (w, w)}] .

Thus 𝔭^{+} ∩ 𝔭^{−} is never empty.

4.3. Representation of spacelike segments. Every spacelike line segment arises as the intersection of two positively (or negatively) oriented half-planes, and this representation is unique up to the choice of orientation. A null half-plane P^{+} or P^{−} is uniquely determined by its orientation and boundary l = ∂P^{+} = ∂P^{−}, which is a null line.

Corollary 4.3.1. Let σ be a spacelike line segment. Then there exist pairs of disjoint null lines (l_{1}, l_{2}) and (k_{1}, k_{2}) such that

σ = 𝔭_{1}^{+} ∩ 𝔭_{2}^{+} = 𝔭_{1}^{−} ∩ 𝔭_{2}^{−}

where 𝔭_{i}^{+} is the unique positively oriented null half-plane such that l_{i} = ∂P^{+} and 𝔭_{i}^{−} is the unique negatively oriented null half-plane such that k_{i} = ∂P^{−}. Furthermore (l_{1}, l_{2}) and (k_{1}, k_{2}) are uniquely determined by σ.

Proof. We shall only consider positively oriented half-planes. Let the endpoints of σ be p_{1} and p_{2} and let v = p_{2} −p_{1} . Take l_{1} = p_{1} +ℝx^{+}(v) and l_{2} = p_{2} + ℝx^{−}(v). Since v ∈ P^{+}(x^{+}(v)) and −v ∈ P^{+}(x^{−}(v)) ,

{p_{1} + kv|0 ≤ k ≤ 1}

is the line segment σ.

Conversely, if 𝔭_{1}^{+} ∩ 𝔭_{2}^{+} = σ, then ∂𝔭_{1}^{+} is parallel to x^{+}(v) and ∂𝔭_{2}^{+} is parallel to x^{−}(v). By Lemma 4.2.1, the endpoints of σ lie on ∂𝔭_{1}^{+} and ∂𝔭_{2}^{+} respectively. This completes the proof of Corollary 4.3.1.

Given a pair l_{1}, l_{2} of null lines, the segment of intersection of the corresponding null half-planes has the following geometric interpretation. The function

B_{l_{1},l_{2}} : l_{1} × l_{2} ⟶ ℝ
(q_{1}, q_{2}) ⟼ 𝔹 (q_{1} − q_{2}, q_{1} − q_{2})

is a quadratic function whose leading term (its Hessian) is indefinite. Let (p_{1}, p_{2}) ∈ l_{1} × l_{2} be its unique critical point. Then p_{1} and p_{2} are the endpoints of σ = 𝔭_{1}^{+} ∩ 𝔭_{2}^{+}.

Here is a formula for these endpoints. Let q_{i} ∈ l_{i} be points and v_{i} be null vectors parallel to l_{i}. (Thus l_{i} = q_{i} + ℝv_{i}.) The endpoints p_{1}, p_{2} of σ are given by:

p_{1} = q_{1} + \frac{𝔹 (q_{2} − q_{1}, v_{2})}{𝔹 (v_{1}, v_{2})} v_{1}
p_{2} = q_{2} + \frac{𝔹 (q_{1} − q_{2}, v_{1})}{𝔹 (v_{2}, v_{1})} v_{2}.

4.4. Parallel null half-planes. Now consider the intersection of null half-planes for the degenerate case when the null half-planes are parallel:

Lemma 4.4.1. Let x ∈ 𝕎^{+} be a null vector and p_{1}, p_{2} ∈ 𝔼 be points. Let 𝔭_{i}^{+} denote the half-plane p_{i} + P^{+}(x) and 𝔭_{i}^{−} denote the half-plane p_{i} + P^{−}(x), respectively, for i = 1, 2. Then:

• 𝔭_{1}^{+} and 𝔭_{2}^{+} are disjoint if and only if 𝔹 (p_{2} − p_{1}, x) ≠ 0;

• 𝔭_{1}^{−} and 𝔭_{2}^{−} are disjoint if and only if 𝔹 (p_{2} − p_{1}, x) ≠ 0; Otherwise, one of the half-planes contains the similarly oriented halfplane and their intersection is a null half-plane.

Proof. We shall only consider positively oriented null half-planes. The intersection 𝔭_{1}^{+} ∩ 𝔭_{2}^{+} lies in

(P(x) + p_{1}) ∩ (P(x) + p_{2}),

which is empty if and only if 𝔹 (p_{2} − p_{1}, x) ≠ 0.

Suppose that 𝔹 (p_{2} −p_{1}, x) = 0, that is, p_{2} −p_{1} lies in P^{+}(x). Choose any spacelike v ∈ P(x). If 𝔹 (p_{2} − p_{1}, v) ≥ 0 then 𝔭_{1}^{+} ⊃ 𝔭_{2}^{+}. If 𝔹 (p_{2} − p_{1}, v) ≤ 0 then 𝔭_{1}^{+} ⊂ 𝔭_{2}^{+}.

4.5. Intersection of stems and null half-planes. Next we analyze when a positively oriented null half-plane intersects a stem:

Lemma 4.5.1. Let x be a positively oriented null vector and v a unitspacelike vector such that 𝔹 (x, v) < 0. Let p_{1}, p_{2} ∈ 𝔼 be points. The stem S_{1} = S(v, p_{1}) and the positively oriented null half-plane 𝔭_{2}^{+} = p_{2} + P^{+}(x) are disjoint if and only if

𝔹 (p_{2} − p_{1}, v ⊠ x) > |𝔹 (p_{2} − p_{1}, x)|. (4.5.1)

In the case that S_{1} ∩ 𝔭_{2}^{+} ≠ ∅ then:

• S_{1} ∩ 𝔭_{2}^{+} consists of a point (which lies on ∂S_{1} but not on ∂𝔭_{2}^{+}) if and only if:

𝔹 (p_{2} − p_{1}, v ⊠ x) < 0 = 𝔹 (p_{2} − p_{1}, x). (4.5.2)

• S_{1} ∩ 𝔭_{2}^{+} consists of a point (which lies on ∂S_{1} and on ∂𝔭_{2}^{+}) if and only if:

𝔹 (p_{2} − p_{1}, v ⊠ x) = |𝔹 (p_{2} − p_{1}, x)|. (4.5.3)

• S_{1}∩𝔭_{2}^{+} consists of a spacelike line segment (with at least one vertex on ∂S_{1}) if and only if:

𝔹 (p_{2} − p_{1}, v ⊠ x) < |𝔹 (p_{2} − p_{1}, x)|, (4.5.4)

where the other endpoint lies on ∂𝔭_{2}^{+} if and only if:

𝔹 (p_{2} − p_{1}, v ⊠ x) ≥ −|𝔹 (p_{2} − p_{1}, x)|. (4.5.5)

and on ∂S_{1} if and only if

𝔹 (p_{2} − p_{1}, v ⊠ x) ≤ −|𝔹 (p_{2} − p_{1}, x)|. (4.5.6)

Figure 3. Wing intersecting a stem (one endpoint on stem)

Figure 4. Wing intersecting a stem (both endpoints on stem)

Figure 3 shows a null half-plane intersecting a stem of a crooked plane where the segment of intersection has one endpoint on ∂S_{1} and the other on ∂𝔭_{2}^{+}, corresponding to (4.5.5) above. Figure 4 shows a null half-plane intersecting a stem of a crooked plane where the segment of intersection has both endpoints on ∂𝔭_{2}^{+}, corresponding to (4.5.6) above.

Proof. Let ℓ be the intersection of the plane containing S_{1} and 𝔭_{2}^{+}. Let w = v ⊠ x and

p = ℓ ∩ (p_{1} + P(w)).

Parametrize ℓ as p + ℝw. The subsets ℓ ∩ S_{1} and ℓ ∩ 𝔭_{2}^{+} of ℓ consist of all p + kw (where k ∈ ℝ) satisfying the inequalities

𝔹 ((p + kw) − p_{1}, (p + kw) − p_{1}) ≤ 0, (4.5.7)

𝔹 ((p + kw) − p_{2}, w) ≥ 0, (4.5.8)

respectively. (4.5.7) is quadratic in k where the coefficient of k^{2} is 𝔹 (w, w) > 0, so that ℓ ∩ S_{1} corresponds to the interval with endpoints

\frac{±\sqrt{−𝔹 (w, w)𝔹 (p − p_{1}, p − p_{1})}}{𝔹 (w, w)}.

Because 𝔹 (p − p_{1}, w) = 0, the vector p − p_{1} is a linear combination of v and x . Solving 𝔹 (p − p_{1}, v) = 0 and 𝔹 (p − p_{2}, x) = 0 for the coefficients of v and x,

(4.5.9) p = p_{1} + \frac{−𝔹 (p_{2} − p_{1}, x)}{𝔹 (v, x)^{2}}x + \frac{𝔹 (p_{2} − p_{1}, x)}{𝔹 (v, x)}v.

Using (4.5.9) and 𝔹 (w, w) = 𝔹 (v, x)^{2},

ℓ ∩ S_{1} ⟷ [\frac{−|𝔹 (p_{2} − p_{1}, x)|}{𝔹 (w, w)}, \frac{|𝔹 (p_{2} − p_{1}, x)|}{𝔹 (w, w)}] .

Solving (4.5.8) for k and using the assumption that 𝔹 (p − p_{1}, w) = 0,

ℓ ∩ 𝔭_{2}^{+} ⟷ [\frac{𝔹 (p_{2} − p_{1}, w)}{𝔹 (w, w)}, ∞) .

The subset S_{1} ∩ 𝔭_{2}^{+} of ℓ is empty if and only if

𝔹 (p_{2} − p_{1}, w) > |𝔹 (p_{2} − p_{1}, x)|.

Otherwise S_{1} ∩ 𝔭_{2}^{+} corresponds to the interval

[max (\frac{𝔹 (p_{2} − p_{1}, w)}{𝔹 (w, w)}, \frac{−|𝔹 (p_{2} − p_{1}, x)|}{𝔹 (w, w)}) , \frac{|𝔹 (p_{2} − p_{1}, x)|}{𝔹 (w, w)}] .

This concludes the proof of Lemma 4.5.1.

Corresponding to the intersection of a stem and a negatively oriented wing is:

Lemma 4.5.2. Let x be a positively oriented null vector and v a unitspacelike vector such that 𝔹 (x, v) < 0. Let p_{1}, p_{2} ∈ 𝔼 be points. The stem S_{1} = S(v, p_{1}) and the negatively oriented null half-plane 𝔭_{2}^{−} = p_{2} + P^{−}(x) are disjoint if and only if

𝔹 (p_{2} − p_{1}, v ⊠ x) < |𝔹 (p_{2} − p_{1}, x)| (4.5.10)

The final case occurs when the null half-plane is orthogonal to the stem:

Lemma 4.5.3. Let x ∈ 𝕎^{+} and v be a unit-spacelike vector such that 𝔹 (x, v) = 0. Suppose p_{1}, p_{2} ∈ 𝔼. The positively oriented null half-plane 𝔭_{1}^{+} = p_{1} + P^{+}(x) and the stem S_{2} = S(v, p_{2}) are disjoint if and only if:

• 𝔹 (p_{2} − p_{1}, v) > 0 when x is a multiple of x^{−}(v)

• 𝔹 (p_{2} − p_{1}, v) < 0 when x is a multiple of x^{+}(v). Otherwise S_{1} ∩ 𝔭_{2}^{+} is a ray with endpoint on ∂𝔭_{2}^{+} (except if 𝔹 (p_{2} − p_{1}, x) = 0, in which case it is an entire line).

Proof. We only discuss the case when x is a multiple of x^{+}(v) . Let ℓ be the intersection of the planes containing 𝔭_{1}^{+} and S_{2}. Let

p = ℓ ∩ (p_{1} + P(x^{−}(v))).

Then ℓ = p + ℝx. The subsets ℓ ∩ 𝔭_{1}^{+} and ℓ ∩ S_{2} of ℓ consist of all p + kx (where k ∈ ℝ) satisfying the inequalities

𝔹 ((p + kx) − p_{1}, v) ≥ 0, (4.5.11)

𝔹 ((p + kx) − p_{2}, (p + kx) − p_{2}) ≤ 0, (4.5.12)

respectively. (4.5.11) and (4.5.12) simplify to

𝔹 (p − p_{1}, v) ≥ 0,
2k𝔹 (p − p_{2}, x) + 𝔹 (p − p_{2}, p − p_{2}) ≤ 0.

Write p − p_{1} and −p − p_{2} as linear combinations of x^{−}(v), x^{+}(v), and v. Solving

\begin{cases}𝔹 (p − p_{1}, x^{−}(v)) = 0 \\ 𝔹 (p − p_{1}, x) = 0 \\ 𝔹 (p − p_{2}, v) = 0 \end{cases}

for the coefficients of x^{−}(v), x^{+}(v), and v yields

p = p_{1} + 𝔹 (p_{2} − p_{1}, v)v, (4.5.13)

(4.5.14) p = p_{2} + \frac{𝔹 (p_{1} − p_{2}, x^{−}(v))}{𝔹 (x, x^{−}(v))}x + \frac{𝔹 (p_{1} − p_{2}, x)}{𝔹 (x, x^{−}(v))}x^{−}(v).

Using (4.5.13) and (4.5.14), inequalities (4.5.11) and (4.5.12) further simplify to

𝔹 (p_{2} − p_{1}, v) ≥ 0,

2k𝔹 (p_{1} − p_{2}, x^{−}(v)) ≤ \frac{−2𝔹 (p_{1} − p_{2}, x)𝔹 (p_{1} − p_{2}, x^{−}(v)}{𝔹 (x, x^{−}(v))},

respectively.

ℓ ∩ 𝔭_{1}^{+} ⟷ \begin{cases}ℝ & if 𝔹 (p_{2} − p_{1}, v) ≥ 0, \\ ∅ & if 𝔹 (p_{2} − p_{1}, v) < 0 \end{cases}

ℓ ∩ S_{2} ⟷ \begin{cases}ℝ & if 𝔹 (p_{1} − p_{2}, x^{−}(v)) = 0, \\ [−2\frac{𝔹 (p_{1}−p_{2},x)}{𝔹 (x,x^{−}(v))}, ∞) & if 𝔹 (p_{1} − p_{2}, x^{−}(v)) < 0, \\ (−∞, −2\frac{𝔹 (p_{1}−p_{2},x)}{𝔹 (x,x^{−}(v))}] & if 𝔹 (p_{1} − p_{2}, x^{−}(v)) > 0. \end{cases}

S_{2} ∩ ℓ is always nonempty, so that 𝔭_{1}^{+} ∩ S_{2} is empty if and only if ℓ ∩ 𝔭_{1}^{+} is empty. If 𝔭_{1}^{+} ∩ S_{2} is nonempty, then 𝔭_{1}^{+} ∩ S_{2} = S_{2} ∩ ℓ.

The corresponding lemma for negatively oriented null half-planes is:

Lemma 4.5.4. Let x ∈ 𝕎^{+} and v be a unit-spacelike vector such that 𝔹 (x, v) = 0. Suppose p_{1}, p_{2} ∈ 𝔼. The negatively oriented null halfplane 𝔭_{1}^{−} = p_{1} + P^{−}(x) and the stem S_{2} = S(v, p_{2}) are disjoint if and only if:

• 𝔹 (p_{2} − p_{1}, v) < 0 when x is a multiple of x^{−}(v) ;

• 𝔹 (p_{2} − p_{1}, v) > 0 when x is a multiple of x^{+}(v).

The preceding lemmas imply the following qualitative result about intersections of crooked planes:

Corollary 4.5.5. Let C_{1} and C_{2} be two crooked planes. Suppose that the three planes containing the stems and wings of C_{1} are each transverse to the three planes containing the stems and wings of C_{2}. If a wing W of C_{1} intersects the stem of C_{2}, then W intersects one of the wings of C_{2}.

5. Intersection of stems

5.1. Pairs of spacelike vectors. Consider crooked planes C(v_{1}, p_{1}) and C(v_{2}, p_{2}), where p_{1}, p_{2} ∈ 𝔼 are points and v_{1}, v_{2} ∈ ℝ^{2,1} are unitspacelike vectors. If the vectors v_{1}, v_{2} satisfy one of the following equivalent conditions, they are said to be ultraparallel:

• P(v_{1}) and P(v_{2}) correspond to ultraparallel geodesics in H^{2}; • ⟨v_{1}, v_{2}⟩ ⊂ ℝ^{2,1} is an indefinite plane (a secant plane to 𝕎);

• 𝔹 (v_{1}, v_{2})^{2} > 𝔹 (v_{1}, v_{1})𝔹 (v_{2}, v_{2});

• P(v_{1}) ∩ P(v_{2}) is a spacelike line;

• There exists a unique spacelike line orthogonal to v_{1} and v_{2};

• 𝔹 (v_{1} ⊠ v_{2}, v_{1} ⊠ v_{2}) > 0 If vectors v_{1}, v_{2} satisfy one of the following equivalent conditions, they are said to be asymptotic:

• P(v_{1}) and P(v_{2}) correspond to asymptotic geodesics in H^{2};

• ⟨v_{1}, v_{2}⟩ ⊂ ℝ^{2,1} is a degenerate (null) plane (a plane tangent to 𝕎);

• 𝔹 (v_{1}, v_{2})^{2} = 𝔹 (v_{1}, v_{1})𝔹 (v_{2}, v_{2});

• P(v_{1}) ∩ P(v_{2}) is a null line;

• 𝔹 (v_{1} ⊠ v_{2}, v_{1} ⊠ v_{2}) = 0 If v_{1}, v_{2} satisfy one of the following equivalent conditions, we say that v_{1}, v_{2} cross:

• P(v_{1}) and P(v_{2}) correspond to intersecting geodesics in H^{2};

• ⟨v_{1}, v_{2}⟩ ⊂ ℝ^{2,1} is a definite plane (a plane disjoint from 𝕎);

• 𝔹 (v_{1}, v_{2})^{2} < 𝔹 (v_{1}, v_{1})𝔹 (v_{2}, v_{2});

• P(v_{1}) ∩ P(v_{2}) is a timelike line;

• There exists a unique timelike line orthogonal to v_{1} and v_{2};

• 𝔹 (v_{1} ⊠ v_{2}, vv_{1} ⊠ v_{2}) < 0.

5.2. Ultraparallel Stems.

Lemma 5.2.1. Let v_{1} and v_{2} be ultraparallel unit-spacelike vectors. The stems S_{1} = S(v_{1}, p_{1}) and S_{2} = S(v_{2}, p_{2}) are disjoint if and only if:

|𝔹 (p_{2} − p_{1}, v_{1} ⊠ v_{2})| > |𝔹 (p_{2} − p_{1}, v_{2})| + |𝔹 (p_{2} − p_{1}, v_{1})|. (5.2.1)

Otherwise S_{1} ∩ S_{2} is a line segment whose endpoints lie on two distinct lines contained in ∂S_{1} ∪ ∂S_{2}.

Proof. Let ℓ be the intersection of the planes that contain S_{1} and S_{2}. Let

w = v_{1} ⊠ v_{2}

and

p = ℓ ∩ (p_{1} + P (w))

and parametrize ℓ as p + ℝw. The intersection ℓ ∩ S_{i} consists of all p + kw (where k ∈ ℝ) satisfying the inequality:

𝔹 ((p + kw) − p_{i}, (p + kw) − p_{i}) ≤ 0. (5.2.2)

which is equivalent to the quadratic inequality

k^{2}𝔹 (w, w) + 2k𝔹 (p − p_{i}, w) + 𝔹 (p − p_{i}, p − p_{i}) ≤ 0.

Since w is spacelike, ℓ ∩ S_{i} correspond to those k in the interval with endpoints

\frac{−𝔹 (p − p_{i}, w) ± \sqrt{(𝔹 (p − p_{i}, w))^{2} − 𝔹 (w, w)𝔹 (p − p_{i}, p − p_{i})}}{𝔹 (w, w)}. (5.2.3)

The point p may be expressed in terms of p_{1} or p_{2} by writing p − p_{i} as a linear combination of v_{1}, v_{2}, and w. Solving

\begin{cases}𝔹 (p − p_{1}, w) = 0 \\ 𝔹 (p − p_{1}, v_{1}) = 0 \\ 𝔹 (p − p_{2}, v_{2}) = 0 \end{cases}

for the coefficients for v_{1}, v_{2}, and w,

(5.2.4) p = p_{1} + \frac{𝔹 (p_{2} − p_{1}, v_{2})𝔹 (v_{1}, v_{2})}{(𝔹 (v_{1}, v_{2}))^{2} − 1}v_{1} + \frac{−𝔹 (p_{2} − p_{1}, v_{2})}{(𝔹 (v_{1}, v_{2}))^{2} − 1}v_{2},

p = p_{2} + \frac{𝔹 (p_{2} − p_{1}, v_{1})}{(𝔹 (v_{1}, v_{2}))^{2} − 1}v_{1} + \frac{−𝔹 (p_{2} − p_{1}, v_{1})𝔹 (v_{1}, v_{2})}{(𝔹 (v_{1}, v_{2}))^{2} − 1}v_{2} + \frac{−𝔹 (p_{2} − p_{1}, w)}{𝔹 (w, w)}w. (5.2.5)

Substituting (5.2.4) and (5.2.5) into (5.2.3), and using

𝔹 (v_{1}, v_{2})^{2} − 1 = 𝔹 (w, w),

S_{1} ∩ ℓ ⟷ [\frac{−|𝔹 (p_{2} − p_{1}, v_{2})|}{𝔹 (w, w)}, \frac{|𝔹 (p_{2} − p_{1}, v_{2})|}{𝔹 (w, w)}] ,

S_{2} ∩ ℓ ⟷ [\frac{𝔹 (p_{2} − p_{1}, w) − |𝔹 (p_{2} − p_{1}, v_{1})|}{𝔹 (w, w)}, \frac{𝔹 (p_{2} − p_{1}, w) + |𝔹 (p_{2} − p_{1}, v_{1})|}{𝔹 (w, w)}] .

S_{1} ∩ S_{2} is empty if and only if either

𝔹 (p_{2} − p_{1}, w) > |𝔹 (p_{2} − p_{1}, v_{2})| + |𝔹 (p_{2} − p_{1}, v_{1})|,

or

𝔹 (p_{2} − p_{1}, w) < −|𝔹 (p_{2} − p_{1}, v_{2})| − |𝔹 (p_{2} − p_{1}, v_{1})|

as desired.

5.3. Asymptotic Stems.

Lemma 5.3.1. Let v_{1} and v_{2} be asymptotic unit-spacelike vectors such that x^{−}(v_{1}) = x^{+}(v_{2}). Let S_{i} = S(v_{i}, p_{i}) for i = 1, 2. S_{1} and S_{2} are disjoint if and only if

𝔹 (p_{2} − p_{1}, x^{+}(v_{1}) ⊠ x^{−}(v_{2}))

has a different sign from 𝔹 (p_{2} − p_{1}, v_{1}) and 𝔹 (p_{2} − p_{1}, v_{2}). Otherwise,

• S_{1} ∩ S_{2} is a point if and only if 𝔹 (p_{2} − p_{1}, x^{+}(v_{1}) ⊠ x^{−}(v_{2})) = 0 and 𝔹 (p_{2} − p_{1}, v_{1}) and 𝔹 (p_{2} − p_{1}, v_{2}) have the same signs;

• S_{1} ∩S_{2} is a line segment if and only if 𝔹 (p_{2} −p_{1}, x^{+}(v_{1})⊠x^{−}(v_{2})), 𝔹 (p_{2} − p_{1}, v_{1}), and 𝔹 (p_{2} − p_{1}, v_{2}) have the same signs;

• S_{1} ∩ S_{2} is a ray if and only if either 𝔹 (p_{2} − p_{1}, v_{i}) = 0 for only one i or 𝔹 (p_{2} − p_{1}, v_{1}) and 𝔹 (p_{2} − p_{1}, v_{2}) have opposite signs;

• S_{1} ∩ S_{2} is a line if and only if 𝔹 (p_{2} − p_{1}, v_{i}) = 0 for both i.

Proof. Let ℓ be the intersection of the planes that contain S_{1} and S_{2}. Let w = x^{+}(v_{1}) ⊠ x^{−}(v_{2}) and p = ℓ ∩ (p_{1} + P(w)) and parametrize ℓ as p + ℝx where

x = x^{−}(v_{1}) = x^{+}(v_{2}).

The intersection ℓ ∩ S_{i} consists of all p + kx ∈ ℓ satisfying:

𝔹 ((p + kx) − p_{i}, (p + kx) − p_{i}) ≤ 0,

which simplifies to

2k𝔹 (x, p − p_{i}) + 𝔹 (p − p_{i}, p − p_{i}) ≤ 0. (5.3.1)

Now express p in terms of p_{1} or p_{2} by writing p−p_{i} as a linear combination of x^{+}(v_{1}), x^{−}(v_{2}), and x. As p was chosen so that 𝔹 (p−p_{1}, w) = 0, 𝔹 (p − p_{1}, v_{1}) = 0, and 𝔹 (p − p_{2}, v_{2}) = 0, write

p = p_{1} + \frac{𝔹 (p_{2} − p_{1}, v_{2})}{𝔹 (x^{+}(v_{1}), v_{2})}x^{+}(v_{1}), (5.3.2)

(5.3.3) p = p_{2} + \frac{−𝔹 (p_{2} − p_{1}, v_{1})}{𝔹 (v_{1}, x^{−}(v_{2}))}x^{−}(v_{2}) + \frac{−𝔹 (p_{2} − p_{1}, w)}{𝔹 (x, w)}x.

Substituting (5.3.2) and (5.3.3) into (5.3.1),

k𝔹 (p_{2} − p_{1}, v_{2}) ≤ 0,

k𝔹 (p_{2} − p_{1}, v_{1}) ≥ \frac{𝔹 (p_{2} − p_{1}, v_{1})𝔹 (p_{2} − p_{1}, w)}{𝔹 (x, w)}.

S_{1} ∩ ℓ and S_{2} ∩ ℓ correspond to the following subsets of ℝ:

S_{1} ∩ ℓ ⟷ \begin{cases}ℝ & if 𝔹 (p_{2} − p_{1}, v_{2}) = 0, \\ [0, ∞) & if 𝔹 (p_{2} − p_{1}, v_{2}) < 0, \\ (−∞, 0] & if 𝔹 (p_{2} − p_{1}, v_{2}) > 0. \end{cases}

S_{2} ∩ ℓ ⟷ \begin{cases}ℝ & if 𝔹 (p_{2} − p_{1}, v_{1}) = 0, \\ [\frac{𝔹 (p_{2}−p_{1},w)}{𝔹 (x,w)}, ∞) & if 𝔹 (p_{2} − p_{1}, v_{1}) > 0, \\ (−∞, \frac{𝔹 (p_{2}−p_{1},w)}{𝔹 (x,w)}] & if 𝔹 (p_{2} − p_{1}, v_{1}) < 0. \end{cases}

If 𝔹 (p_{2} − p_{1}, v_{i}) = 0 for both i then S_{1} ∩ S_{2} is a line. If either • 𝔹 (p_{2} − p_{1}, v_{i}) = 0 for only one i = 1, 2, or

• 𝔹 (p_{2} − p_{1}, v_{1}) and 𝔹 (p_{2} − p_{1}, v_{1}) have opposite signs, then S_{1} ∩ S_{2} is a ray.

Suppose that 𝔹 (p_{2} − p_{1}, v_{1}) and 𝔹 (p_{2} − p_{1}, v_{2}) have the same sign. If 𝔹 (p_{2} − p_{1}, w) = 0 then S_{1} ∩ S_{2} is a point. If 𝔹 (p_{2} − p_{1}, w) has the same sign as 𝔹 (p_{2} − p_{1}, v_{1}) and 𝔹 (p_{2} − p_{1}, v_{2}), then S_{1} ∩ S_{2} is an interval. If 𝔹 (p_{2} − p_{1}, w) has a different sign than 𝔹 (p_{2} − p_{1}, v_{1}) and 𝔹 (p_{2} − p_{1}, v_{2}), then S_{1} ∩ S_{2} is empty.

5.4. Crossing Stems.

Lemma 5.4.1. Let v_{1} and v_{2} be unit-spacelike vectors which cross. The intersection of the stems S_{1} = S(v_{1}, p_{1}) and S_{2} = S(v_{2}, p_{2}) lies on a line and is either a line, the union of two disjoint rays, or two rays and a line segment, where all endpoints lie on the set ∂S_{1} ∪ ∂S_{2}.

Proof. Use the same calculations as in Lemma 5.2.1 except that

𝔹 (w, w) < 0.

Then S_{i} ∩ ℓ corresponds to two infinite intervals. The endpoints of the infinite intervals are given by (5.2.3). The intersection S_{1} ∩ ℓ corresponds to the union

(−∞, \frac{| − 𝔹 (p_{2} − p_{1}, v_{2})|}{𝔹 (w, w)}] ∪ [\frac{|𝔹 (p_{2} − p_{1}, v_{2})|}{𝔹 (w, w)}, ∞) ,

and S_{2} ∩ ℓ corresponds to the union

(−∞, \frac{𝔹 (p_{2} − p_{1}, w) − |𝔹 (p_{2} − p_{1}, v_{1})|}{𝔹 (w, w)}] ∪

[\frac{𝔹 (p_{2} − p_{1}, w) + |𝔹 (p_{2} − p_{1}, v_{1})|}{𝔹 (w, w)}, ∞) .

as desired.

6. Intersections of crooked planes

6.1. Consistent orientation. When considering the intersection of crooked planes, some terminology regarding the orientation of the vectors defining the crooked planes is useful. Say that spacelike vectors v_{1}, v_{2}, …, v_{n} are consistently oriented if 𝔹 (v_{i}, v_{j}) ≤ 0 and 𝔹 (v_{i}, x^{±}(v_{j})) ≤ 0 for i ≠ j.

Here is the geometric meaning of consistent orientation. A unitspacelike v defines a half-plane in the hyperbolic plane H^{2} as the projective image H(v) of

{u ∈ 𝕌^{+} | 𝔹 (u, v) < 0}. (6.1.1)

A pair v_{1}, v_{2} ∈ ℝ^{2,1} of ultraparallel unit-spacelike vectors is consistently oriented if the half-planes H(v_{1}), H(v_{2}) intersect and neither one contains the other. In that case H(v_{1}) ∩ H(v_{2}) is a strip with two ideal boundary components and boundary components ∂H(v_{1}), ∂H(v_{2}).

Lemma 6.1.1. If v_{1} and v_{2} are consistently oriented unit-spacelike vectors, then

• x^{+}(v_{1}) ⊠ x^{+}(v_{2}) is a positive scalar multiple of (v_{1} ⊠ v_{2}) − v_{1} + v_{2}.

• x^{+}(v_{1}) ⊠ x^{−}(v_{2}) is a positive scalar multiple of (v_{1} ⊠ v_{2}) + v_{1} + v_{2}.

• x^{−}(v_{1}) ⊠ x^{+}(v_{2}) is a positive scalar multiple of (v_{1} ⊠ v_{2}) − v_{1} − v_{2}.

• x^{−}(v_{1}) ⊠ x^{−}(v_{2}) is a positive scalar multiple of (v_{1} ⊠ v_{2}) + v_{1} − v_{2}.

Proof. SO^{0}(2, 1) acts transitively on the set of Lorentz unit vectors so write

v_{1} = [\begin{matrix}1 \\ 0 \\ 0 \end{matrix}]

and

x^{±}(v_{1}) = \frac{1}{\sqrt{2}} [\begin{matrix}0 \\ ∓1 \\ 1 \end{matrix}] .

Applying a hyperbolic element with v_{1} as a fixed eigenvector,

v_{2} = [\begin{matrix}−a \\ 0 \\ c \end{matrix}] ,

where |c| = \sqrt{a^{2} − 1}. Furthermore

x^{±}(v_{2}) = \frac{1}{\sqrt{2}} [\begin{matrix}−c/a \\ ±1/a \\ 1 \end{matrix}] .

Since v_{1}, v_{2} are consistently oriented, a, c ≥ 0. Thus a > 1 and c = \sqrt{a^{2} − 1}. Write x^{+1}(v_{i}) = x^{+}(v_{i}) and x^{−1}(v_{i}) = x^{−}(v_{i}). For j, k ∈ {−1, +1}

v_{1} ⊠ v_{2} − kv_{1} + jv_{2} = [\begin{matrix}−k − ja \\ −c \\ jc \end{matrix}] ,

x^{j}(v_{1}) ⊠ x^{k}(v_{2}) = [\begin{matrix}−j − k/a \\ −c/a \\ jc/a \end{matrix}] ,

x^{j}(v_{1}) ⊠ x^{k}(v_{2}) is a multiple of v_{1} ⊠ v_{2} − kv_{1} + jv_{2} since

𝔹 (x^{j}(v_{1}), v_{1} ⊠ v_{2} ± v_{1} + jv_{2}) = 0,
𝔹 (x^{k}(v_{2}), v_{1} ⊠ v_{2} − kv_{1} ± v_{2}) = 0.

Furthermore, x^{j}(v_{1})⊠x^{k}(v_{2}) is a positive multiple of v_{1} ⊠v_{2} −kv_{1} +jv_{2} since

𝔹 (x^{j}(v_{1}) ⊠ x^{k}(v_{2}), v_{1} ⊠ v_{2} − kv_{1} + jv_{2}) = \frac{(a ± 1)^{2}}{a} > 0

as desired.

6.2. Ultraparallel crooked planes.

Theorem 6.2.1. Let v_{1} and v_{2} be consistently oriented spacelike Lorentzian unit vectors and p_{1}, p_{2} ∈ 𝔼. Let C_{i} = C(v_{i}, p_{i}) for i = 1, 2. The positively oriented crooked planes C_{1} and C_{2} are disjoint if and only if

𝔹 (p_{2} − p_{1}, v_{1} ⊠ v_{2}) > |𝔹 (p_{2} − p_{1}, v_{2})| + |𝔹 (p_{2} − p_{1}, v_{1})|. (6.2.1)

If C_{1} ∩ C_{2} is nonempty, then C_{1} ∩ C_{2} is either a single point or a polygon with 4, 5, 6, 7 or 8 edges.

Proof. Assume that (6.2.1) is satisfied. (5.2.1) is also satisfied and the stems of C_{1} and C_{2} are disjoint.

Rewrite (6.2.1) as four separate inequalities:

(6.2.2)  𝔹 (p_{2} − p_{1}, v_{1} ⊠ v_{2} − v_{1} + v_{2}) > 0
(6.2.3)  𝔹 (p_{2} − p_{1}, v_{1} ⊠ v_{2} + v_{1} + v_{2}) > 0
(6.2.4)  𝔹 (p_{2} − p_{1}, v_{1} ⊠ v_{2} − v_{1} − v_{2}) > 0
(6.2.5)  𝔹 (p_{2} − p_{1}, v_{1} ⊠ v_{2} + v_{1} − v_{2}) > 0.

By Lemma 6.1.1, the inequalities (6.2.2), (6.2.3), (6.2.4), (6.2.5) are respectively equivalent to:

(6.2.6)  𝔹 (p_{2} − p_{1}, x^{+}(v_{1}) ⊠ x^{+}(v_{2})) > 0
(6.2.7)  𝔹 (p_{2} − p_{1}, x^{+}(v_{1}) ⊠ x^{−}(v_{2})) > 0
(6.2.8)  𝔹 (p_{2} − p_{1}, x^{−}(v_{1}) ⊠ x^{+}(v_{2})) > 0
(6.2.9)  𝔹 (p_{2} − p_{1}, x^{−}(v_{1}) ⊠ x^{−}(v_{2})) > 0.

The wings W^{+}(v_{1}, p_{1}) and W^{+}(v_{2}, p_{2}) of the crooked planes are disjoint by applying Lemma 4.2.1 to (6.2.6). Similar applications of Lemma 4.2.1 to (6.2.7),(6.2.8),(6.2.9) imply that

W^{+}(v_{1}, p_{1}) ∩ W^{−}(v_{2}, p_{2}) = ∅,
W^{−}(v_{1}, p_{1}) ∩ W^{+}(v_{2}, p_{2}) = ∅,
W^{−}(v_{1}, p_{1}) ∩ W^{−}(v_{2}, p_{2}) = ∅

respectively. Since the pairs of wings and pairs of stems are disjoint, Corollary 4.5.5 implies the wings and stem of C_{1} are disjoint from the wings and stem of C_{2}. Thus C_{1} and C_{2} are disjoint.

Conversely, assume that (6.2.1) does not hold. If

|𝔹 (p_{2} − p_{1}, v_{1} ⊠ v_{2})| ≤ |𝔹 (p_{2} − p_{1}, v_{2})| + |𝔹 (p_{2} − p_{1}, v_{1})|,

Lemma 5.2.1 implies that the stems intersect. Otherwise

𝔹 (p_{2} − p_{1}, v_{1} ⊠ v_{2}) < −|𝔹 (p_{2} − p_{1}, v_{2})| − |𝔹 (p_{2} − p_{1}, v_{1})| (6.2.10)

For C(v_{1}, p_{1}) and C(v_{2}, p_{2}) to intersect it suffices that one wing of C(v_{1}, p_{1}) intersect a wing of C(v_{2}, p_{2}). The inequality (6.2.10) is equivalent to 4 inequalities, one of which is

𝔹 (p_{2} − p_{1}, v_{1} ⊠ v_{2} + v_{2} + v_{1}) < 0,

which, by Lemma 5.2.1, is equivalent to:

𝔹 (p_{2} − p_{1}, x^{+}(v_{1}) ⊠ x^{−}(v_{2})) < 0.

Applying Lemma 4.2.1, the wings W^{+}(v_{1}, p_{1}) and W^{−}(v_{2}, p_{2}) intersect. It follows that C(v_{1}, p_{1}) ∩ C(v_{2}, p_{2}) ≠ ∅ as desired.

The corresponding theorem for ultraparallel negatively oriented crooked planes is:

Theorem 6.2.2. Let v_{1} and v_{2} be consistently oriented unit-spacelike vectors. The negatively oriented crooked planes K(v_{1}, p_{1}) and K(v_{2}, p_{2}) are disjoint if and only if

𝔹 (p_{2} − p_{1}, v_{1} ⊠ v_{2}) < |𝔹 (p_{2} − p_{1}, v_{2})| + |𝔹 (p_{2} − p_{1}, v_{1})|. (6.2.11)

6.3. The polygon of intersection. Assume that C(v_{1}, p_{1}) and C(v_{2}, p_{2}) intersect, that is, (6.2.10) does not hold. If

𝔹 (p_{2} − p_{1}, v_{1} ⊠ v_{2}) = |𝔹 (p_{2} − p_{1}, v_{2})| + |𝔹 (p_{2} − p_{1}, v_{1})| (6.3.1)

then C(v_{1}, p_{1}) ∩ C(v_{2}, p_{2}) is a single point.

If

𝔹 (p_{2} − p_{1}, v_{1} ⊠ v_{2}) < |𝔹 (p_{2} − p_{1}, v_{2})| + |𝔹 (p_{2} − p_{1}, v_{1})|, (6.3.2)

then we claim that C(v_{1}, p_{1})∩C(v_{2}, p_{2}) is a polygon, that is, a piecewise linear 1-dimensional manifold without boundary. C(v_{1}, p_{1})∩C(v_{2}, p_{2}) is the union of intersections of wings and stems. There are four possible wing-wing intersections, four possible wing-stem intersections, and one stem-stem intersection, all of which are either empty, a point or a finite spacelike line segment (Lemmas 4.2.1, 4.5.1, 5.2.1).

Let σ ⊂ C(v_{1}, p_{1}) ∩ C(v_{2}, p_{2}) be a spacelike line segment with one endpoint q. Assume that q lies on a null line which is the boundary of a wing and the stem of C(v_{1}, p_{1}). If q lies in the interior of a wing or stem of C(v_{2}, p_{2}) then it is clear that q is also the endpoint of exactly one other line segment of C(v_{1}, p_{1}) ∩ C(v_{2}, p_{2}). Even if q lies on the boundary of a wing and the stem of C(v_{2}, p_{2}), it can be shown that q is the endpoint of exactly one other line segment of C(v_{1}, p_{1}) ∩ C(v_{2}, p_{2}). Thus, C(v_{1}, p_{1}) ∩ C(v_{2}, p_{2}) is a polygon.

We determine the possible number of sides of the polygon

C(v_{1}, p_{1}) ∩ C(v_{2}, p_{2}).

First consider the case where the interiors of the stems S(v_{1}, p_{1}) and S(v_{2}, p_{2}) of C(v_{1}, p_{1}) and C(v_{2}, p_{2}), respectively, are disjoint. Each wing will generically intersect the wings and stem of the other crooked plane in a line segment. The generic intersection of the crooked planes in this case will be polygon with 8 edges. It is possible that for each crooked plane there is one wing such that the intersection of the wing with a stem will be a single point. In this case, the intersection is a polygon with 6, 7, or 8 edges.

Next consider the case where S(v_{1}, p_{1}) ∩ S(v_{2}, p_{2}) is a line segment with one endpoint on ∂S(v_{1}, p_{1}) and one endpoint on ∂S(v_{2}, p_{2}). Each crooked plane has one wing which does not intersect the other crooked plane at more than a point. Each crooked plane also has one wing which intersects a wing and the stem of the other crooked plane in a line segment. (There is no possible degeneration of the wing-stem intersections in this case.) The intersection of the crooked planes in this case is a polygon with 4 edges.

Finally, consider the case where S(v_{1}, p_{1})∩S(v_{2}, p_{2}) is a line segment with both endpoints on either ∂S(v_{1}, p_{1}) or ∂S(v_{2}, p_{2}). Assume the endpoints of the line segment lie on ∂S(v_{1}, p_{1}). Both wings of C(v_{1}, p_{1}) will intersect S(v_{2}, p_{2}) and one wing of C(v_{2}, p_{2}) in a line segment. Generically, one wing of C(v_{2}, p_{2}) will intersect the wings of C(v_{1}, p_{1}) and S(v_{1}, p_{1}) in a line segment. It is possible that this intersection of a wing of C(v_{2}, p_{2}) and S(v_{1}, p_{1}). In this case, the intersection is a polygon with 5 or 6 edges.

6.4. Asymptotic crooked planes.

Theorem 6.4.1. Let v_{1} and v_{2} be consistently oriented asymptotic unit-spacelike vectors. Let C_{i} = C(v_{i}, p_{i}) for i = 1, 2. Suppose that x^{−}(v_{1}) = x^{+}(v_{2}). Then C_{1} and C_{2} are disjoint if and only if

• 𝔹 (p_{2} − p_{1}, v_{1}) < 0,

• 𝔹 (p_{2} − p_{1}, v_{2}) < 0

• 𝔹 (p_{2} − p_{1}, x^{+}(v_{1}) ⊠ x^{−}(v_{2})) > 0 . If x^{+}(v_{1}) = x^{−}(v_{2}) then C_{1} and C_{2} are disjoint if and only if

• 𝔹 (p_{2} − p_{1}, v_{1}) > 0,

• 𝔹 (p_{2} − p_{1}, v_{2}) > 0

• 𝔹 (p_{2} − p_{1}, x^{−}(v_{1}) ⊠ x^{+}(v_{2})) > 0 .

Proof. We only discuss the case for which x^{−}(v_{1}) = x^{+}(v_{2}). Let x = x^{−}(v_{1}) = x^{+}(v_{2}).

Suppose that 𝔹 (p_{2} − p_{1}, v_{1}) and 𝔹 (p_{2} − p_{1}, v_{2}) are negative and 𝔹 (p_{2} − p_{1}, x^{+}(v_{1}) ⊠ x^{−}(v_{2})) is positive. By Lemma 5.3.1, C_{1} and C_{2} have disjoint stems.

Since 𝔹 (p_{2}−p_{1}, v_{1}) < 0 and 𝔹 (p_{2}−p_{1}, v_{2}) < 0, 𝔹 (p_{2}−p_{1}, x) ≠ 0. The parallel wings, W^{−}(v_{1}, p_{1}) and W^{+}(v_{2}, p_{2}) are disjoint by Lemma 4.4.1.

Lemma 4.2.1 implies the other pairs of wings are disjoint. Since

𝔹 (p_{2} − p_{1}, x^{+}(v_{1}) ⊠ x^{−}(v_{2})) > 0

then W^{+}(v_{1}, p_{1}) and W^{−}(v_{2}, p_{2}) are disjoint. Now

x^{+}(v_{2}) ⊠ x^{+}(v_{1}) = x^{+}(v_{1}) ⊠ x^{−}(v_{1})

is a negative multiple of v_{1}, and since

𝔹 (p_{2} − p_{1}, −v_{1}) > 0

then W^{+}(v_{2}, p_{2}) and W^{+}(v_{1}, p_{1}) are disjoint. Similarly, W^{−}(v_{2}, p_{2}) and W^{−}(v_{1}, p_{1}) are disjoint.

Because 𝔹 (p_{1}−p_{2}, v_{1}) > 0, the stem S(v_{2}, p_{2}) and the wing W^{−}(v_{1}, p_{1}) are disjoint and the stem S(v_{1}, p_{1}) and the wing W^{+}(v_{2}, p_{2}) are disjoint (Lemma 4.5.3). Assume:

𝔹 (p_{2} − p_{1}, v_{2}) > 0
𝔹 (p_{2} − p_{1}, x^{−}(v_{1}) ⊠ x^{+}(v_{2})) > 0
S(v_{1}, p_{1}) ∩ W^{−}(v_{2}, p_{2}) ≠ ∅

By Lemma 4.5.3, the intersection must include a point on ∂S(v_{1}, p_{1}).

∂S(v_{1}, p_{1}) = ∂W^{−}(v_{1}, p_{1}) ∩ ∂W^{+}(v_{1}, p_{1}),

but the wings from different crooked planes are disjoint. Therefore, S(v_{1}, p_{1}) and W^{−}(v_{2}, p_{2}) are disjoint. Similarly, S(v_{2}, p_{2}) and W^{+}(v_{1}, p_{1}) are disjoint. If 𝔹 (p_{2} − p_{1}, v_{1}) and 𝔹 (p_{2} − p_{1}, v_{2}) are negative and 𝔹 (p_{2} − p_{1}, x^{+}(v_{1}) ⊠ x^{−}(v_{2})) is positive then the crooked planes are disjoint.

Conversely, assume one of the following conditions:

𝔹 (p_{2} − p_{1}, v_{1}) < 0
𝔹 (p_{2} − p_{1}, v_{2}) < 0
𝔹 (p_{2} − p_{1}, x^{−}(v_{1}) ⊠ x^{+}(v_{2})) < 0

If 𝔹 (p_{2}−p_{1}, v_{1}) < 0 then W^{+}(v_{2}, p_{2}) intersects S(v_{1}, p_{1}) and W^{+}(v_{1}, p_{1}). If 𝔹 (p_{2}−p_{1}, v_{2}) < 0 then W^{−}(v_{1}, p_{1}) intersects S(v_{2}, p_{2}) and W^{−}(v_{1}, p_{1}). If 𝔹 (p_{2}−p_{1}, x^{−}(v_{1})⊠x^{+}(v_{2})) < 0 then W^{+}(v_{1}, p_{1}) intersects W^{−}(v_{2}, p_{2}).

The corresponding theorem for pairs of negatively oriented crooked planes is:

Theorem 6.4.2. Let v_{1} and v_{2} be consistently oriented asymptotic Lorentzian unit vectors. If x^{−}(v_{1}) = x^{+}(v_{2}) then the crooked planes K(v_{1}, p_{1}) and K(v_{2}, p_{2}) are disjoint if and only if

• 𝔹 (p_{2} − p_{1}, v_{1}) > 0,

• 𝔹 (p_{2} − p_{1}, v_{2}) > 0,

• 𝔹 (p_{2} − p_{1}, x^{+}(v_{1}) ⊠ x^{−}(v_{2})) < 0 . If x^{+}(v_{1}) = x^{−}(v_{2}) then the crooked planes K(v_{1}, p_{1}) and K(v_{2}, p_{2}) are disjoint if and only if

• 𝔹 (p_{2} − p_{1}, v_{1}) < 0,

• 𝔹 (p_{2} − p_{1}, v_{2}) < 0

• 𝔹 (p_{2} − p_{1}, x^{−}(v_{1}) ⊠ x^{+}(v_{2})) < 0 .

6.5. Crossing crooked planes.

Theorem 6.5.1. Let v_{1} and v_{2} be unit-spacelike vectors which cross. The intersection of crooked planes of indeterminate orientation with spines parallel to v_{1} and v_{2} is always nonempty.

Proof. This is an easy corollary of Lemma 5.4.1. Since the stems intersect, the orientations are irrelevant.

6.6. Crooked planes with different orientations. Now consider the intersection of two crooked planes of different orientations.

Theorem 6.6.1. Let v_{1} and v_{2} be unit-spacelike vectors which are not parallel. The intersection of the positively oriented crooked plane C(v_{1}, p_{1}) and the negatively oriented crooked plane K(v_{2}, p_{2}) is always nonempty.

Proof. This theorem is an easy corollary of Lemma 4.2.2.

7. Groups generated by inversions

We apply the preceding considerations to affine deformations of subgroups of SO(2, 1) generated by reflections in the sides of an ultra-ideal polygon. Let Δ ⊂ H^{2} be a convex n-gon whose pairs of adjacent sides S_{1}, …, S_{n} are ultraparallel. Let ρ_{i} ∈ SO(2, 1) denote the reflection in S_{i} and Γ ⊂ SO(2, 1) the group generated by ρ_{1}, …, ρ_{n}. Then Γ acts properly and discretely on H^{2} with fundamental domain Δ. The subgroup Γ∩SO(2, 1)^{0} comprising orientation-preserving isometries of H^{2} has index two in Γ and is a Schottky group (it has a fundamental domain consisting of Δ ∪ ρ_{i}Δ for any i). There is a family of consistently oriented unit-spacelike vectors v_{1}, …, v_{n} ∈ ℝ^{2,1} such that the sides S_{i} of Δ are polar to v_{i} and

Δ = H(v_{1}) ∩ … H(v_{n})

where H(v_{i}) is defined in (6.1.1).

An affine deformation of Γ is a collection l = (l_{1}, …, l_{n}) of oriented spacelike lines such that each l_{i} has direction vector v_{i}. Such an affine deformation determines an affine representation of Γ whose image Γ_{l} is generated by inversions ι_{j} in l_{j}. A line l ⊂ 𝔼 with direction vector v is parametrized by a point p ∈ l which is determined up to translation by ℝv. The affine deformations of Γ comprise the product

𝔼/ℝv_{1} × ⋯ × 𝔼/ℝv_{n}

of the quotient affine spaces 𝔼/ℝv_{j}.

To construct a fundamental domain for the action of Γ_{l}, we must find crooked planes C_{i} = C(v_{i}, q_{i}) with spine l_{i}. (Recall that l_{i} has direction vector v_{i} and contains q_{i}.) Since a crooked plane with given spine l is determined by its vertex, the family (C) = (C_{1}, …, C_{n}) is determined by the n-tuple q = (q_{1}, …, q_{n}) ∈ l_{1} × ⋯ × l_{n}.

Conjecture 7.0.2. Let Γ ⊂ SO(2, 1) be as above. An affine deformation Γ_{l} of Γ is properly discontinuous if and only if there exist q ∈ l_{1} ×⋯×l_{n} such that H(q_{1}, v_{1}), …, H(q_{n}, v_{n}) bound a fundamental domain

Sufficient conditions for proper discontinuity of an affine deformation of a Schottky group are derived in Margulis [11, 12] and our previous paper[7].

Assume that there exist mutually disjoint crooked planes

C(q_{1}, v_{1}), …, C(q_{n}, v_{n})

which bound a common region. The linear part of the index-two subgroup

κ = ⟨ι_{l_{1}}ι_{l_{2}}, ι_{l_{1}}ι_{l_{3}}, … ι_{l_{1}}ι_{l_{n}}⟩

of Γ lies in SO(2, 1). The region bounded by the disjoint crooked planes

C(q_{2}, v_{2}), … C(q_{n}, v_{n}), ι_{l_{1}}(C(q_{2}, v_{2})), … ι_{l_{1}}(C(q_{n}, v_{n}))

is a fundamental domain for the action of κ which acts freely and properly discontinuosly on ℝ^{2,1} by [3]. Thus, Γ also acts properly discontinuosly on ℝ^{2,1} and a fundamental domain for the action of Γ is the region bounded by the crooked planes (C(q_{1}, v_{1}), …, C(q_{n}, v_{n})).

References

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4. , Linear holonomy of Margulis space-times, J.Diff.Geo. 38 (1993), 679– 691

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9. and Hirsch, M. W. Affine manifolds with nilpotent holonomy, Comm. Math. Helv. 56 (1981), 487–523

10. Goldman, W., “Complex hyperbolic geometry,” Oxford Mathematical Monographs (to appear)

11. Margulis, G., Free properly discontinuous groups of affine transformations, Dokl. Akad. Nauk SSSR 272 (1983), 937–940; 12. , Complete affine locally flat manifolds with a free fundamental group,

J. Soviet Math. 134 (1987), 129–134

13. Mess, G., Lorentz spacetimes of constant curvature, (preprint)

14. Milnor, J., On fundamental groups of complete affinely flat manifolds, Adv. Math. 25 (1977), 178–187

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Department of Mathematics, Swarthmore College, Swarthmore, PA 19081 (Drumm), Department of Mathematics, University of Maryland, College Park, MD 20742 (Goldman)

  E-mail address: tdrumm1@swarthmore.edu (Drumm), wmg@math.umd.edu (Gold-

man)

Figure 5. Crooked planes through the origin invariant under a cyclic group

Figure 6. Disjoint crooked planes invariant under a cyclic group