j. differential geometry 60 (2002) 315-344

ON THE ZARISKI CLOSURE OF THE LINEAR PART

OF A PROPERLY DISCONTINUOUS GROUP OF AFFINE TRANSFORMATIONS

H. ABELS, G.A. MARGULIS & G.A. SOIFER

Abstract

Let Γ be a subgroup of the group of affine transformations of the affine space ℝ^{2n+1}. Suppose Γ acts properly discontinuously on ℝ^{2n+1}. The paper deals with the question which subgroups of GL(2n + 1, ℝ) occur as Zariski closure \overline{ℓ(Γ)} of the linear part of such a group Γ. The two main results of the paper say that SO(n + 1, n) does occur as \overline{ℓ(Γ)} of such a group Γ if n is odd, but does not if n is even.

1. Introduction

A well known classical theorem due to Bieberbach says that every discrete group Γ of isometries of the n-dimensional Euclidean space ℝ^{n} with compact quotient Γ \ ℝ^{n} contains a subgroup of finite index consisting of translations. Hence such a group Γ is virtually abelian, i.e., Γ contains an abelian subgroup of finite index.

Let us now look at the group of affine transformations instead of the group of isometries of ℝ^{n}. More precisely, let G_{n} = Aff(ℝ^{n}) denote the group of affine transformations of ℝ^{n}. The group G_{n} is the semidirect product GL_{n}(ℝ) ⋉ ℝ^{n} where ℝ^{n} is identified with its group of translations. A subgroup Γ of G_{n} is said to act properly discontinuously on ℝ^{n} if for every compact subset K of ℝ^{n} the set {γ ∈ Γ | γK ∩ K ≠ ∅} of recurrencies is finite. If a discrete group Γ consists of isometries then Γ acts properly on ℝ^{n}. But this is not true for all discrete subgroups of G_{n}, e.g., for an infinite discrete subgroup of GL_{n}(ℝ).

Received December 13, 2001.

A subgroup Γ of G_{n} will be called crystallographic if Γ acts properly discontinuously on ℝ^{n} and the orbit space Γ \ ℝ^{n} is compact. In [4] Auslander conjectured that every crystallographic subgroup Γ of G_{n} is virtually solvable, i.e., contains a solvable subgroup of finite index. In [19] Milnor asked whether Auslander’s conjecture is true without the assumption that the orbit space Γ \ ℝ^{n} be compact. Milnor’s question has a positive answer for n ≤ 2 (easy for n = 2, trivial for n = 1). But for n = 3 the answer to Milnor’s question is negative. In fact, the second named author proved that there is a nonabelian free subgroup Γ of Aff(ℝ^{3}) acting properly discontinuously on ℝ^{3} ([15, 16]). On the other hand, D. Fried and W. Goldman [11] proved Auslander’s conjecture for n = 3 using cohomological arguments. For higher dimensions the existing results confirming the Auslander conjecture are proved under the assumption that the linear part l(Γ) of Γ belongs to some special subgroup of GL_{n}(ℝ) where l : G_{n} → GL_{n}(ℝ) denotes the natural homomorphism ([12, 14]). For a survey on existing results see [1].

The two main results of the present paper are:

Theorem A. For n even there is no properly discontinuous subgroup Γ of G_{2n+1} with linear part l(Γ) Zariski dense in SO(n + 1, n).

Theorem B. For n odd there are properly discontinuous free subgroups Γ of G_{2n+1} with linear part l(Γ) Zariski dense in SO(n + 1, n).

Theorem A, its proof and the statement were contained in the preprint [17] of Margulis circulated in 1991. The proof of Theorem B is based on our results [2] and the strategy is essentially the same as in [15, 16]. The results of the present paper and more were announced in [3].

The authors would like to thank different institutions and foundations for their support during the preparation of this paper: the MSRI Berkeley, the SFB 343 Bielefeld, the MPI Bonn, Yale University, Columbia University, the NSF under grant DMS 9800607 and the German–Israeli foundation for Scientific Research and Development under grant number G–454–213.06/95. Without their help this paper whose authors live on three different continents would not have seen the light of day.

2. A sufficient condition for failure of proper discontinuity

In this section we prove a general criterion for certain pairs of affine transformations to generate a subgroup of the affine group which is not properly discontinuous. Recall that G_{n} = Aff (ℝ^{n}) and that l : G_{n} ⟶ GL_{n}(ℝ) is the natural epimorphism. For any g ∈ G_{n} let us consider the decomposition of ℝ^{n} into a direct sum of the l(g)-invariant linear subspaces

(2.1)  ℝ^{n} = A^{−}(g) ⊕ A^{0}(g) ⊕ A^{+}(g),

where A^{−}(g), A^{0}(g) and A^{+}(g) are determined by the condition that their sum is ℝ^{n} and all eigenvalues of the restriction l(g) | A^{−}(g) (resp. l(g) | A^{0}(g), l(g) | A^{+}(g)) are of modulus less than 1 (resp. equal to 1, greater than 1). We set D^{−}(g) = A^{−}(g) ⊕ A^{0}(g) and D^{+}(g) = A^{+}(g) ⊕ A^{0}(g). It is clear that A^{0}(g) = D^{−}(g) ∩ D^{+}(g). Let us call an element g ∈ G_{n} pseudohyperbolic if dim A^{0}(g) = 1 and the only eigenvalue of l(g) | A^{0}(g) is 1 (not −1). Let Ω be the set of pseudohyperbolic elements of G_{n} and let Ω_{0} = {g ∈ Ω | g x ≠ x for every x ∈ ℝ^{n}} be the subset of pseudohyperbolic fixed–point–free elements of G.

For g ∈ Ω there is exactly one g–invariant line C_{g} and C_{g} is parallel to A^{0}(g). We call C_{g} the axis of g. The restriction of g to C_{g} is a parallel translation by a vector τ(g) ∈ A^{0}(g). The vector τ(g) is the A^{0}(g)-component with respect to the decomposition (2.1) of g x − x for any x ∈ ℝ^{n}. We call τ(g) the translational part of g. We have g ∈ Ω_{0} if and only if τ(g) ≠ 0.It is easy to see that

C_{g^{m}} = C_{g} and τ (g^{m}) = m τ(g) for m ≠ 0.

If g ∈ Ω_{0} then every x ∈ D^{+}(g) has a unique decomposition x = λ(x)τ(g) + a(x) where λ(x) ∈ ℝ and a(x) ∈ A^{+}(g). We say that a vector x ∈ D^{+}(g) is positive with respect to g ∈ Ω_{0} if λ(x) > 0.

Two elements g_{1}, g_{2} ∈ Ω_{0} will be called transversal if

ℝ^{n} = A^{+} (g_{1}) ⊕ D^{+} (g_{2}) = D^{+} (g_{1}) ⊕ A^{+} (g_{2})

or, equivalently, if dim (D^{+} (g_{1}) ∩ D^{+} (g_{2})) = 1 and

ℝ^{n} = A^{+} (g_{1}) ⊕ A^{+} (g_{2}) ⊕ (D^{+} (g_{1}) ∩ D^{+} (g_{2})) .

For two transversal elements g_{1}, g_{2} ∈ Ω_{0}, we say that g_{1} and g_{2} form a positive pair if the semiline {x ∈ D^{+} (g_{1}) ∩ D^{+} (g_{2}) | x is positive with respect to g_{1}} coincides with the semiline {x ∈ D^{+} (g_{1}) ∩ D^{+} (g_{2}) | x is positive with respect to g_{2}}.

We shall denote by E_{g}^{−} and E_{g}^{+} the affine subspaces containing C_{g} and parallel to D^{−}(g) and D^{+}(g), respectively. We will use throughout the whole paper the following notational conventions: affine subspaces corresponding to g will be denoted by attaching g as lower index, like C_{g}, E_{g}^{+} etc., whereas vector subspaces corresponding to g will be denoted by putting g into brackets, like A^{+}(g), D^{−}(g), etc. Let us introduce an order on E_{g}^{+} by saying that u ∈ E_{g}^{+} is greater than v ∈ E_{g}^{+} if u − v is positive with respect to g. The condition that g_{1} and g_{2} form a positive pair is equivalent to the condition that the orders on E_{g_{1}}^{+} ∩ E_{g_{2}}^{+} induced from E_{g_{1}}^{+} and E_{g_{2}}^{+} coincide.

Lemma 2.2. Let g_{1}, g_{2} ∈ Ω_{0} be two transversal elements which form a positive pair. Then there exist a compact set K ⊂ ℝ^{n} and two sequences {m_{i}} and {n_{i}} of positive integers such that

m_{i} → ∞ and n_{i} → ∞ as i → ∞ and
(g_{1}^{−m_{i}}g_{2}^{n_{i}}K) ∩ K ≠ ∅.

Proof. For x ∈ ℝ^{n} and g ∈ Ω_{0}, let B_{g}^{+}(x) denote the affine subspace containing x and parallel to A^{+}(g). Let us define the projection P_{g} : E_{g}^{+} → C_{g} by the equality

P_{g}(x) = C_{g} ∩ B_{g}^{+}(x) for x ∈ E^{+}(g).

C_{g} ∩ B_{g}^{+}(x) is a point since D^{+}(g) = A^{+}(g) ⊕ A^{0}(g) and C_{g} is parallel to A^{0}(g). The subspace A^{+}(g) is l(g)-invariant. Therefore,

(1)  P_{g}(g x) = P_{g}(x) + τ(g) for x ∈ E_{g}^{+}.

Let d denote the Euclidean metric on ℝ^{n}. The absolute values of all eigenvalues of the restriction l (g^{−1}) | A^{+}(g) are less than 1. This easily implies that there exist positive constants c and b such that for any x ∈ E_{g}^{+} and n ∈ ℕ^{+}

(2)  d (g^{−n}x, P_{g} (g^{−n}x)) ≤ ce^{−bn}d(x, P_{g}(x)).

Let us fix a point r(g) ∈ C_{g} and let R(g) denote the interval [r(g), r(g)+ τ_{g}) ⊂ C_{g}. It follows from (1) that, for every x ∈ E^{+}(g), there is a unique integer k(x, g) such that

P_{g} (g^{k(x,g)}x) ∈ R(g).

Note that if x is greater than r(g) in E^{+}(g) then k(x, g) ≤ 0.

Since g_{1} and g_{2} form a positive pair one can find a vector v ∈ D^{+} (g_{1}) ∩ D^{+} (g_{2}) such that v is positive with respect to both g_{1} and g_{2}. Let us fix x_{0} ∈ E_{g_{1}}^{+} ∩ E_{g_{2}}^{+} and set x_{i} = x_{0} + i v, i ∈ ℕ^{+}. We have x_{i} ∈ E_{g_{1}}^{+} ∩ E_{g_{2}}^{+}. Let

m_{i} = −k(x_{i}, g_{1}) and n_{i} = −k(x_{i}, g_{2}).

Let v_{1} ∈ A^{0} (g_{1}) and v_{2} ∈ A^{0} (g_{2}) be such that v ∈ v_{1} + A^{+} (g_{1}) and v ∈ v_{2} + A^{+} (g_{2}). Since v is positive with respect to both g_{1} and g_{2} we have that v_{1} = λ_{1}τ (g_{1}) and v_{2} = λ_{2}τ (g_{2}) where λ_{1} > 0 and λ_{2} > 0. Then it is easy to see that

(3)  lim_{i→∞} \frac{i}{m_{i}} = λ_{1} and lim_{i→∞} \frac{i}{n_{i}} = λ_{2}.

It is clear that the growth of the functions f_{j}(i) = d (x_{i}, P_{g_{j}} (x_{i})) , j = 1, 2, is asymptotically linear. Thus (1), (2) and (3) imply that

(4)  lim_{i→∞} d (g_{1}^{−m_{i}}x_{i}, R (g_{1})) = lim_{i→∞} d (g_{2}^{−n_{i}}x_{i}, R (g_{2})) = 0.

We get from (4) that there exists a compact set K ⊂ ℝ^{n} such that for all i

(5)  g_{1}^{−m_{i}}x_{i} ∈ K and g_{2}^{−n_{i}}x_{i} ∈ K.

We have g_{1}^{−m_{i}}x_{i} = (g_{1}^{−m_{i}}g_{2}^{n_{i}}) g_{2}^{−n_{i}}x_{i}. This and (5) imply that K is a compact set as claimed. q.e.d.

Corollary 2.3. Let g_{1}, g_{2} ∈ Ω_{0} be two transversal elements which form a positive pair. Then the subgroup of G_{n} generated by g_{1} and g_{2} is not properly discontinuous.

Proof. Let {m_{i}} and {n_{i}} be as in Lemma 2.2. Then it is enough to check that the set {g_{1}^{−m_{i}}g_{2}^{n_{i}} | i ∈ ℕ^{+}} is infinite. We may assume that m_{i} < m_{i+1} and n_{i} < n_{i+1} for every i since m_{i} → +∞ and n_{i} → +∞ as i → ∞. Then for i < j we have g_{1}^{−m_{i}}g_{2}^{n_{i}} ≠ g_{1}^{−m_{j}}g_{2}^{n_{j}} because otherwise g_{1}^{m_{j}−m_{i}} = g_{2}^{n_{j}−n_{i}}. But g_{1}^{m} = g_{2}^{n} for positive m and n implies A^{+}(g_{1}) = A^{+}(g_{1}^{m}) = A^{+}(g_{2}^{n}) = A^{+}(g_{2}) contradicting transversality. q.e.d.

3. Orientations on subspaces of ℝ^{2n+1}

In this section we introduce orientations on certain lines. In the next section we shall compare these orientations with the ones defined in the last section.

We fix a positive natural number n. Let X = ℝ^{n+1} and Y = ℝ^{n} and let B be the quadratic form on X × Y = ℝ^{2n+1} given by

B (x_{1}, …, x_{n+1}, y_{1}, …, y_{n}) = x_{1}^{2} + ⋯ + x_{n+1}^{2} − y_{1}^{2} − ⋯ − y_{n}^{2},

where (x_{1}, …, x_{n+1}) ∈ X and (y_{1}, … y_{n}) ∈ Y. Consider the set Ψ of all maximal B-isotropic subspaces V of ℝ^{2n+1}. We have the two projections

p : ℝ^{2n+1} ⟶ X and q : ℝ^{2n+1} ⟶ Y.

The restriction of q to V ∈ Ψ is a linear isomorphism V ⟶ Y. Hence if we fix an orientation on Y we have also fixed an orientation on each V ∈ Ψ. For V ∈ Ψ let us denote the B-orthogonal complement of V by V^{⊥} = {z ∈ ℝ^{2n+1}; B(z, V) = 0}. We have V ⊂ V^{⊥} since V is B-isotropic. We also have

dim V^{⊥} = 1 + dim V = n + 1.

The restriction of p to V^{⊥} is a linear isomorphism V^{⊥} ⟶ X. Hence if we fix an orientation on X we have also fixed an orientation on V^{⊥} for each V ∈ Ψ. Thus we have orientations on both V and V^{⊥} and we can single out one of the two halfspaces of V^{⊥} ∖ V as positive by the following definition: if (v_{1}, …, v_{n}) is a positively oriented basis of V then a vector v in V^{⊥} not in V is positive (with respect to V) if the basis (v_{1}, …, v_{n}, v) of V^{⊥} is positively oriented. Accordingly, every affine line L in V^{⊥} transversal to V is oriented: For two points x_{0}, x_{1} of L we have x_{0} < x_{1} if v = x_{1} − x_{0} belongs to the positive half of V^{⊥} ∖ V.

If V_{1} ∈ Ψ and V_{2} ∈ Ψ are transversal then V_{1}^{⊥} ∩ V_{2}^{⊥} is a line which is transversal to both V_{1} and V_{2}. So there are two orientations ω_{1} and ω_{2} on L, where ω_{i} is defined if we consider L as a line in V_{i}^{⊥}. We have:

Lemma 3.1. The orientations defined above on L are the same if n is even and are opposite if n is odd.

Proof. One can easily show that there exists a basis (e_{1}, …, e_{2n+1}) in ℝ^{2n+1} such that V_{1} (resp. V_{2}) is the linear span of (e_{1}, …, e_{n}) (resp. of (e_{n+1}, …, e_{2n})) and the form B with respect to this basis is given by the equation

B (x_{1}, …, x_{2n+1}) = −x_{1}x_{n+1} ⋯ − x_{n}x_{2n} + x_{2n+1}^{2}.

Let us note that V_{1}^{⊥} (resp. V_{2}^{⊥}) is the linear span of (e_{1}, …, e_{n}, e_{2n+1}) (resp. of (e_{n+1}, …, e_{2n+1})).

Let us denote by H_{B} the group of elements of GL_{2n+1}(ℝ) preserving the form B, and let H_{B}^{0} denote the connected component (in the Euclidean topology) of the identity in H_{B}. Direct calculations show that, for every j, 1 ≤ j ≤ n, and every t ∈ ℝ, the transformation h_{j}(t) defined by

e_{i} → e_{i}, e_{n+i} → e_{n+i} if 1 ≤ i ≤ n, i ≠ j

e_{j} → \frac{e_{j} + e_{n+j}}{2} + \frac{cos t(e_{j} − e_{n+j})}{2} + \frac{sin te_{2n+1}}{2}

e_{n+j} → \frac{e_{j} + e_{n+j}}{2} − \frac{cos t(e_{j} − e_{n+j})}{2} − \frac{sin te_{2n+1}}{2}

e_{2n+1} → − sin t(e_{j} − e_{n+j}) + cos te_{2n+1}

belongs to H_{B}. We have

h_{j}(π)e_{i}, = e_{i}, h_{j}(π)e_{n+i} = e_{n+i} if 1 ≤ i ≤ n, i ≠ j,
h_{j}(π)e_{j} = e_{n+j}, h_{j}(π)e_{n+j} = e_{j},
h_{j}(π)e_{2n+1} = −e_{2n+1}.

Therefore, if we denote h_{1}(π) … h_{n}(π) by h_{π}, we have

(1) h_{π}(e_{i}) = e_{n+i}, h_{π}(e_{n+i}) = e_{i} and h_{π}(e_{2n+1}) = (−1)^{n}e_{2n+1}.

Since h_{j}(0) = Id and h_{j}(t) ∈ H_{B} depends continuously on t we have

(2)  h_{π} ∈ H_{B}^{0}.

The orientations on V and V^{⊥} defined above depend continuously on V ∈ Ψ. Hence these orientations are invariant under the action of the group H_{B}^{0}. Now assuming that the basis (e_{1}, …, e_{n}) (resp. (e_{1}, …, e_{n}, e_{2n+1})) is positively oriented in V_{1} (resp. in V_{1}^{⊥}) we get from (1) and (2) that the basis (e_{n+1}, …, e_{2n}) (resp. (e_{n+1}, …, e_{2n}, (−1)^{n}e_{2n+1})) is positively oriented in V_{2} (resp. in V_{2}^{⊥}). This immediately implies the desired statement. q.e.d.

4. The sign of an affine transformation

In this section we define the notion of sign for certain affine transformations g. It tells us whether the translational part of g is positive or negative with respect to the orientation of the preceding section. We translate this into the orientations of the second section thus obtaining a criterion for (non–) proper discontinuity. The proof of Theorem A will follow.

Let H_{B} = SO(n + 1, n) be the special orthogonal group of the quadratic form B on ℝ^{2n+1} and let G_{B} be the subgroup of those elements of G_{2n+1} whose linear part is in H_{B}. Let g be a pseudohyperbolic element of G_{B}, i.e., g ∈ G_{B} ∩ Ω, then dim A^{0}(g) = 1 and dim A^{−}(g) = dim A^{+}(g) = n. The subspaces A^{−}(g) and A^{+}(g) are isotropic, thus A^{−}(g) and A^{+}(g) are in Ψ. In the preceding section we fixed orientations on A^{+}(g) and on D^{+}(g) = A^{+}(g)⊕A^{0}(g) = (A^{+}(g))^{⊥} and similarly on A^{−}(g) and on D^{−}(g) = A^{−}(g) ⊕ A^{0}(g) = (A^{−}(g))^{⊥}. We define the sign σ(g) of g ∈ G_{B} ∩ Ω as +1 or −1 or 0 according to whether the translational part τ(g) of g is a positive or a negative vector of D^{+}(g) ∖ A^{+}(g) or is zero. Recall that g ∈ Ω is fixed point free, i.e., g ∈ Ω_{0}, iff τ(g) = 0. For g ∈ G_{B} ∩ Ω and m > 0 we have

(4.1)  σ (g^{m}) = σ(g)

since τ (g^{m}) = m τ(g) and

A^{∘}(g) = A^{∘} (g^{m})
A^{+}(g) = A^{+} (g^{m})
A^{−}(g) = A^{−} (g^{m}) .

For g ∈ G_{B} ∩ Ω we have

(4.2)  σ (g^{−1}) = (−1)^{n+1}σ(g).

This follows from τ (g^{−1}) = −τ(g) ∈ A^{0}(g) = A^{0} (g^{−1}) and Lemma 3.1 since A^{+}(g) and A^{−}(g) = A^{+} (g^{−1}) are transversal.

We shall need to compute σ(g). Let g ∈ G_{B} ∩ Ω. Note that B restricts to a positive definite form on A^{0}(g). Let e^{0}(g) be the unique vector of A^{0}(g) with B (e^{0}(g), e^{0}(g)) = 1 and positive with respect to the orientations of A^{+}(g) and D^{+}(g). Define α(g) ∈ ℝ by

(4.3)  τ(g) = α(g)e^{0}(g).

Then

(4.4)  σ(g) = sgn α(g).

We have

(4.5)  α(g) = B (e^{0}(g), g x − x)

for every point x of the affine space. To see this recall the decomposition (2.1):

(4.6)  ℝ^{n} = A^{−}(g) ⊕ A^{0}(g) ⊕ A^{+}(g).

Now (4.5) follows from (4.3) since the translational part τ(g) is the A^{0}(g)–component of g x − x for every point x and A^{0}(g) is orthogonal to both A^{+}(g) and A^{−}(g). Note that

(4.7)  α (g^{m}) = m α(g) for m > 0

and

(4.8)  α (g^{−1}) = (−1)^{n+1}α(g)  ,

which is proved like (4.2).

The following proposition is basic for our approach.

Proposition 4.9. Let g_{1}, g_{2} be transversal pseudohyperbolic elements of G_{B}. If σ (g_{1}) σ (g_{2}^{−1}) < 0 then the group generated by g_{1} and g_{2} is not properly discontinuous.

Proof. Note that g_{1}, g_{2} ∈ G_{B} ∩Ω_{0} since their signs are not zero. The proposition now follows from Corollary 2.3 and the following lemma.

Lemma 4.10. Let g_{1}, g_{2} be transversal elements of G_{B} ∩ Ω_{0}. If σ (g_{1}) σ (g_{2}^{−1}) < 0 then g_{1} and g_{2} form a positive pair.

Proof. Given g_{i} ∈ G_{B} ∩ Ω_{0} we have defined orientations on V_{i} = A^{+}(g_{i}) and V_{i}^{⊥} = D^{+}(g_{i}) and thus can talk about the positive half space of V_{i}^{⊥} ∖ V_{i}. The vector e^{0}(g_{i}) defined before (4.3) is positive with respect to V_{i}. Let L be the line V_{1}^{⊥} ∩ V_{2}^{⊥}. An element ℓ ≠ 0 in L is positive with respect to V_{1} iff (−1)^{n}ℓ is positive with respect to V_{2}, by Lemma 3.1. Write

ℓ = v_{1} + β_{1}τ(g_{1})
= v_{2} + β_{2}τ(g_{2}),

v_{i} ∈ V_{i}, β_{i} ∈ ℝ. Then g_{1}, g_{2} form a positive pair iff β_{1}β_{2} > 0. But

τ(g_{i}) = α(g_{i})e^{0}(g_{i}).

Thus, ℓ is positive for V_{1} iff β_{1}α(g_{1}) > 0 iff (−1)^{n}ℓ = (−1)^{n}v_{2} + (−1)^{n}β_{2}α(g_{2})e^{0}(g_{2}) is positive for V_{2} iff β_{2}(−1)^{n}α(g_{2}) = −β_{2}α(g_{2}^{−1}) > 0, by (4.8). Thus g_{1}, g_{2} form a positive pair iff α(g_{1})α(g_{2}^{−1}) < 0, which proves our claim in view of (4.4). q.e.d.

Proof of Theorem A. Let Γ be a subgroup of G_{n} and let H(Γ) be the Zariski closure of its linear part ℓ(Γ). Suppose we have H(Γ)^{∘} = H_{B}^{∘} for the connected components with respect to the Euclidean topology. It then follows from the main algebraic result of [13] that Γ contains an element γ_{0} such that ℓ(γ_{0}) is pseudohyperbolic. To see this look at the natural representation ρ of H_{B} on ⋀^{n} ℝ^{2n+1}. The representation ρ restricted to H_{B}^{∘} is irreducible. Then ρ(h) is proximal in the sense of [2] iff there is only one eigenvalue of h of modulus 1, counting multiplicities, see [2, 5.1]. The main algebraic result of [13] implies that ρ(ℓ(Γ)) contains a proximal element since ρ(H_{B}^{∘}) does and the representation ρ | H_{B}^{∘} is irreducible, see [2, Theorem 4.1].

We have γ_{0} ∈ Ω_{0} if Γ is properly discontinuous because otherwise γ_{0} has a fixed point and hence the cyclic subgroup of Γ generated by γ_{0} does not act properly discontinuously on ℝ^{2n+1}. We can find an element γ ∈ Γ such that

(4.11)  (A^{+}(γ_{0}) ∪ A^{−}(γ_{0})) ∩ ℓ(γ)(A^{+}(γ_{0}) ∪ A^{−}(γ_{0})) = 0

since H(Γ)^{0} = H_{B}^{∘}. Put γ_{1} = γγ_{0}γ^{−1}. Then A^{+}(γ_{1}) = γA^{+}(γ_{0}) and A^{−}(γ_{1}) = γA^{−}(γ_{0}). Then (4.11) implies that γ_{1} is transversal to both γ_{0} and γ_{0}^{−1}. On the other hand we see from (4.2) for n even that the signs of γ_{0} and γ_{0}^{−1} are different. It now remains to apply Proposition 4.9 either to the pair (g_{1} = γ_{0}, g_{2} = γ_{1}) or to the pair (g_{1} = γ_{0}^{−1}, g_{2} = γ_{1}).

q.e.d.

5. Pseudohyperbolicity of products

We are now heading for a proof of Theorem B. In this section we give a sufficient condition for when the product of two linear maps is pseudohyperbolic. In the next section we prove a generalization of the basic additivity lemma.

We need the following concept of pseudohyperbolicity which is both more general and more precise.

Before we give the actual definition which is somewhat technical we give some explanation. Let V be a real vector space of dimension m.

We fix a natural number p with 1 ≤ p ≤ m − 1. Let g ∈ GL(V). We order the complex eigenvalues of g according to their moduli |λ_{1}| ≥ |λ_{2}| ≥ ⋯ ≥ |λ_{m}|. The condition of (p, ε)-pseudohyperbolicity stipulates among other things that |λ_{p}| > |λ_{p+1}|. We thus can talk about the p eigenvalues {λ_{1}, …, λ_{p}} of maximal modulus and also of the m − p eigenvalues of minimal modulus {λ_{p+1}, …, λ_{m}}. Then V is the direct sum of two g–invariant subspaces V^{+} and V^{−} such that every eigenvalue of g | V^{+} resp. g|V^{−} is a λ_{i} with i ≤ p resp. i > p. In particular V^{+} and V^{−} have dimension p and m − p, resp. The number ε in the following definition of pseudohyperbolicity denotes the distance between these two subspaces. The number s is related to the size of the gap between |λ_{p}| and |λ_{p+1}|, more precisely, s is an upper bound for |λ_{p+1}| · |λ_{p}|^{−1}, see (5.3). The actual definition of pseudohyperbolicity is formulated not in terms of eigenvalues but in terms of norms. The reason is that we want to have pseudohyperbolicity of products of pseudohyperbolic elements, see Lemma 5.6.

Here is the definition. Let V be a real vector space of dimension m. Fix a norm ∥ · ∥ on V and let d be the corresponding metric. Let S(V) be the sphere {x ∈ V; ∥x∥ = 1}. For p ∈ {1, …, m − 1}, 0 < s < 1, ε > 0 let Ω(p, s, ε) be the set of elements g ∈ GL(V) with the following properties: There is a p-dimensional linear subspace V^{+} of V and a complementary linear subspace V^{−} of V, both g-invariant, such that

(5.1)  ∥g|V^{−}∥ ≤ s · ∥g^{−1}|V^{+}∥^{−1}

and d(v^{−}, v^{+}) ≥ ε for any two vectors v^{−} ∈ S(V^{−}), v^{+} ∈ S(V^{+}), where, of course, S(V^{−}) = S(V) ∩ V^{−} etc. Note that

(5.2)  ∥g^{−1}|V^{+}∥^{−1} = min{∥gv∥; v ∈ S(V^{+})}.

We have for every eigenvalue λ^{−} of g on V^{−} and λ^{+} of g on V^{+}:

(5.3)  |λ^{−}| ≤ ∥g|V^{−}∥ ≤ s∥g^{−1}|V^{+}∥^{−1} ≤ s · |λ^{+}|.

Since s < 1, it follows that the eigenvalues of g | V^{+} are the p eigenvalues of g of maximal modulus and the eigenvalues of g | V^{−} are the m − p eigenvalues of g of minimal modulus. In particular, the g-invariant subspaces V^{+} and V^{−} are uniquely determined by this condition.

5.4 (Inverses of pseudohyperbolic elements). Note that g ∈ Ω(p, s, ε) iff g^{−1} ∈ Ω(n − p, s, ε) with V^{−}(g^{−1}) = V^{+}(g) and V^{+}(g^{−1}) = V^{−}(g).

An element g ∈ ⋃_{s<1}Ω(p, s, ε) will be called (p, ε)-pseudohyperbolic

and we put for such g

(5.5)  s_{p}(g) := ∥g|V^{−}∥ · ∥g^{−1}|V^{+}∥.

So s_{p}(g) is the smallest s such that g ∈ Ω(p, s, ε) for some ε > 0. Note that g is (r, ε)-proximal in the terminology of [2] iff g is (1, ε)- pseudohyperbolic with s_{1}(g) ≤ r^{−1}. Two (p, ε)-pseudohyperbolic elements g, h are called (p, ε)-transversal if d(v, w) ≥ ε whenever v ∈ S(V^{−}(g)) ∪ S(V^{−}(h)) and w ∈ S(V^{+}(g)) ∪ S(V^{+}(h)). In the following lemma we will use the Hausdorff distance associated with ∥ · ∥, also denoted by d. Thus, for any two compact subsets A, B of V put

d(A, B) = max {max_{b∈B} min_{a∈A} d(a, b), max_{a∈A} min_{b∈B} d(a, b)} .

So d(A, B) is the minimum of the numbers r such that A ⊂ B_{r}(B) and

B ⊂ B_{r}(A), where B_{r}(X) = ⋃_{x∈X}B_{r}(x) and B_{r}(x) = {y ∈ V; d(x, y) ≤

r} is the ball of radius r with center x. For two linear subspaces V_{1}, V_{2} of V put

d(V_{1}, V_{2}) = d(S(V_{1}), S(V_{2})).

Assume now, that g ∈ H_{B}, where H_{B} = O(B) and B is the quadratic form of the signature (n + 1, n). Let g be (n, ε)-pseudohyperbolic, then g^{−1} is also (n, ε)-pseudohyperbolic and V^{+}(g) = A^{+}(g), V^{+}(g^{−1}) = A^{−}(g), V^{−}(g) = D^{−}(g), V^{−}(g^{−1}) = D^{+}(g). In that case we will define s(g) as s(g) = max{s_{n}(g), s_{n}(g^{−1})}.

Lemma 5.6. For any ε > 0 there are two real numbers a(ε) and s(ε), s(ε) > 1 such that if g and h are (p, ε)-transversal, s_{p}(g) < s(ε)^{−1} and s_{p}(h) < s(ε)^{−1} then:

(1) gh is (p, \frac{ε}{2})-pseudohyperbolic.

(2) a(ε)^{−1}s_{p}(g)s_{p}(h) < s_{p}(gh) < a(ε)s_{p}(g)s_{p}(h).

(3) d(V^{+}(gh), V^{+}(g)) < a(ε)s_{p}(g).

(4) d(V^{−}(gh), V^{−}(h)) < a(ε)s_{n−p}(h).

We often call real numbers like a(ε) constants because, although they do depend on ε, they do not depend on the other variables, in particular not on g, V^{+}(g), . . ..

Proof. We start with the case p = 1. We only show the claims (1),

(3) and the second inequality of (2). The remaining inequality of (2) and also (4) will be obtained later by passing to inverses, using 5.4, but for a different dimension. If g is (1, ε)-pseudohyperbolic then V^{+}(g) is a g-invariant line corresponding to a real eigenvalue of g which we call λ(g), so gv = λ(g)v for v ∈ V^{+}(g).

Lemma 5.7. For every ε > 0 there is a constant a(ε) > 1 with the following property. For any two (1, ε)-transversal elements g, h in GL(V) with s_{1}(g) < a(ε)^{−1} and s_{1}(h) < a(ε)^{−1} we have:

(1) gh is (1, \frac{ε}{2})-proximal.

(2) s_{1}(gh) < a(ε)s_{1}(g)s_{1}(h).

(3) a(ε)^{−1}|λ(g)λ(h)| < |λ(gh)| < a(ε)|λ(g)λ(h)|.

(4) d(V^{+}(gh), V^{+}(g)) < a(ε)s_{1}(g).

Proof. Let B(x, r) = {y ∈ V; d(x, y) ≤ r} be the ball of radius r with center x. For any (1, ε)-pseudohyperbolic element g fix one vector x^{+}(g) in S(V^{+}(g)). Put

g_{1}(x) = \frac{g(x)}{∥g(x)∥} for x ≠ 0

and

U(g) = B (x^{+}(g), \frac{ε}{2}) .

It is easy to see that there is a constant a(ε) > 0 for every ε > 0 such that if g and h are (1, ε)-transversal with s_{1}(g) ≤ a(ε)^{−1} and s_{1}(h) ≤ a(ε)^{−1} then

(5.8) h_{1} U(g) ⊂ B(x^{+}(g), r) or h_{1} U(g) ⊂ B(−x^{+}(g), r)

with r = min (a(ε)s_{1}(h), \frac{ε}{2})

(5.9) a(ε)^{− \frac{1}{2}} |λ(h)| · ∥x∥ < ∥h(x)∥ < a(ε)^{\frac{1}{2}} |λ(h)| · ∥x∥ for x ∈ U(g)

(5.10)  ∥h_{1}(x) − h_{1}(y)∥ < a(ε)^{\frac{1}{2}} s_{1}(h)∥x − y∥ for x, y ∈ U(g).

The same holds with g und h interchanged. So we have by (5.8)

(5.11)  h_{1}U(g) ⊂ ± U^{∘}(h)

and

(5.12)  g_{1}U(h) ⊂ ± U^{∘}(g).

Thus g_{1} ∘ h_{1} : U(g) → ± U^{∘}(g) and

(5.13)  ∥g_{1}h_{1}(x) − g_{1}h_{1}(y)∥ < a(ε)s_{1}(g)s_{1}(h)∥x − y∥

for x, y ∈ U(g). We then obtain the following facts from Tits’s lemma [2, 2.1] and thus basically from the Banch fixed point theorem: gh is prox-

imal, V^{−}(g h) does not intersect U^{∘}(g) and the projective map induced by g h on the projective space ℙ has a unique fixed point in the image of U(g) in ℙ and this fixed point is V^{+}(g h). This proves Claim (1). Claim (4) follows from inequality (5.8) and the corresponding inequality with g and h interchanged, and Claim (3) follows from inequality (5.9) for g and h. Finally inequality (5.13) implies that s_{1}(gh) < b(ε)s_{1}(g)s_{1}(h) for some constant b(ε) depending only on ε and of course the given norm. We may assume that b(ε) ≥ a(ε) and thus replace a(ε) by b(ε) to retain everything and also obtain Claim (2). q.e.d.

5.14. We shall constantly use the following notion. Fix p. Let k ≥ 1 be a natural number and let two functions f_{1}, f_{2} be given on some subset E of the direct product of k copies of GL(V). We say that f_{1} is dominated by f_{2} and write f_{1} ≪ f_{2} if for every ε > 0 there is a c(ε) > 0 such that f_{1}(g_{1}, …, g_{k}) ≤ c(ε)f_{2}(g_{1}, …, g_{k}) whenever (g_{1}, …, g_{k}) ∈ E and the elements g_{i}, i = 1, …, k, are pairwise (p, ε)- transversal. If f_{1} ≪ f_{2} and f_{2} ≪ f_{1} then we write f_{1} ∼ f_{2} and say that f_{1} and f_{2} are equivalent.

Proof of Lemma 5.6 for arbitrary p. Note that if g ∈ Ω(p, s, ε) then ⋀^{p} g : ⋀^{p} V ⟶ ⋀^{p} V is proximal, i.e., has a unique eigenvalue of maximal modulus and this eigenvalue has multiplicity one. The corresponding line V^{+}(⋀^{p} g) is ⋀^{p} V^{+}(g), the hyperplane V^{<}(⋀^{p} g) is {x ∈ ⋀^{p} V; x ∧ ⋀^{n−p} V^{−}(g) = 0}, the eigenvalue λ^{+}(⋀^{p} g) of maximal modulus is the product of the p eigenvalues of g of largest modulus and all the eigenvalues of ⋀^{p} g on V^{<}(⋀^{p} g) are of modulus ≤ s · |λ^{+}(⋀^{p} g)|.

In order to proceed we need the quantitative specifications of proximality as follows:

(5.15)  d (x^{+} (⋀^{p}g) , S (V^{<} (⋀^{p}g))) ∼ 1
(5.16)  \frac{∥g | V^{<}(⋀^{p} g)∥}{|λ^{+}(⋀^{p} g)|} ∼ s(g).

To see this we may assume that our norm comes from a positive definite quadratic form Q on V. Now change the form Q to the form Q_{g} with the following properties: V^{−} := V^{−}(g) is orthogonal to V^{+} := V^{+}(g) with

respect to Q_{g}, Q_{g}|V^{+} = Q|V^{+}, Q_{g}|V^{−} = Q|V^{−}. Then sup_{Q(x)=1}Q_{g}(x)

and inf_{Q(x)=1}Q_{g}(x) are ∼ 1 as functions of g, i.e., there is a compact set

K = K(ε) of quadratic forms, depending on ε, such that Q_{g} ∈ K for every (p, ε)-pseudohyperbolic g. For the induced form on ⋀^{p} V, also denoted Q_{g}, and which also belongs to a compact set K′(ε) of quadratic forms, we then have V^{+}(⋀^{p} g) ⊥ V^{<}(⋀^{p} g), which implies (5.15).

For orthonormal vectors x_{1}, …, x_{r} in V^{+}, x_{r+1}, …, x_{p} in V^{−} we have for the norm ∥ · ∥ corresponding to Q_{g}:

(5.17) ∥g(x_{1} ∧ ⋯ ∧ x_{p})∥ = ∥g(x_{1} ∧ ⋯ ∧ x_{r})∥ · ∥g(x_{r+1} ∧ ⋯ ∧ x_{p})∥.

The second factor can be estimated by

(5.18) ∥gx_{r+1}∧⋯∧gx_{p}∥ ≤ ∥gx_{r+1}∥·∥gx_{r+2}∥·· · ··∥gx_{p}∥ ≤ ∥g | V^{−}∥^{p−r},

since

∥z_{1} ∧ ⋯ ∧ z_{t+1}∥ = ∥z_{1} ∧ ⋯ ∧ z_{t}∥ · ∥π(z_{t+1})∥

where π is the orthogonal projection of z_{t+1} onto the span of z_{1}, …, z_{t}.

As to the first factor in (5.17), take the Cartan decomposition of g | V^{+} = k_{1}·d·k_{2} where k_{1}, k_{2} are unitary with respect to Q_{g} | V^{+} = Q | V^{+} and d is diagonal with entries d_{1} ≥ d_{2} ≥ ⋯ ≥ d_{p} > 0 with respect to an orthonormal basis y_{1}, …, y_{p} of V^{+}. Assuming that x_{1}, …, x_{r} forms part of the basis k_{2}^{−1}y_{1}, …, k_{2}^{−1}y_{p}, as we may, we obtain

(5.19)  ∥g(x_{1} ∧ ⋯ ∧ x_{r})∥ = ∏_{i} d_{i} ≤ d_{1} … d_{r},

where the index i runs through an r–element subset of {1, … p}. We thus obtain

∥g | V^{<} (⋀^{p}g)∥ < C · s · |λ^{+} (⋀^{p}(g))|

for some constant C, since we have a corresponding inequality for every element of an orthonormal basis of V^{<}(⋀^{p} g) by (5.17)-(5.19). Changing back to the original quadratic form Q gives the inequality ≪ in (5.16). The opposite inequality is easily proved using the Cartan decompositions of g | V^{+} and g | V^{−} with respect to Q_{g}.

It follows now from Lemma 5.7, the case p = 1, that there is a constant s(ε) > 1 such that if g and h are (p, ε)-transversal and s(g) < s(ε)^{−1}, s(h) < s(ε)^{−1} then ⋀^{p}(gh) is proximal since (p, ε)-transversality of g and h implies (1, c(ε))-transversality of ⋀^{p} g and ⋀^{p} h for some c(ε) > 0. The proof of 5.7 and of the Banach fixed point theorem show furthermore that then V^{+}(⋀^{p}(gh)) = lim (gh)^{n}V^{+}(⋀^{p} g) which implies that the sequence (gh)^{n} V^{+}(g) converges to a p-dimensional linear subspace of V which we call V^{+}. Similarly, using inverses and 5.4, the sequence (gh)^{−n} V^{−}(h) converges to an n − p-subspace V^{−} of V. By 5.7(4) we have

d(V^{+}(g), V^{+}) < a_{2}(ε)s_{p}(g) (5.20)

and

d(V^{−}(h), V^{−}) < a_{2}(ε)s_{n−p}(h). (5.21)

Finally, for x ∈ U(g) := B(S(V^{+}(g)), ε/2) we have

a_{3}(ε)^{−1}∥π_{h}^{+}(x)∥ · ∥h^{−1} | V^{+}(h)∥^{−1} ≤ ∥hx∥ (5.22)

≤ a_{3}(ε)∥π_{h}^{+}(x)∥ · ∥h | V^{+}(h)∥

where π_{h}^{+} and π_{h}^{−} are the projections in V = V^{+}(h) ⊕ V^{−}(h), and similarly with g and h interchanged. This implies

∥(gh)(x)∥ ≥ a_{4}(ε)^{−1} · ∥g^{−1} | V^{+}(g)∥^{−1} · ∥x∥

for x ∈ V^{+} by (5.20) and that there is a point x ∈ S(V^{+}) with

∥gh(x)∥ ≤ a_{4}(ε) · ∥g^{−1} | V^{+}(g)∥^{−1} · ∥h^{−1} | V^{+}(h)∥^{−1} · ∥x∥,

since π_{h}^{+} : V^{+} ⟶ V^{+}(h) is surjective and inf_{x∈S(V^{+})} ∥π_{h}^{+}(x)∥ ≥ a_{5}(ε)∥x∥,

and similarly for g. Thus

a_{4}(ε)^{−1} · ∥g^{−1} | V^{+}(g)∥^{−1} · ∥h^{−1} | V^{+}(h)∥^{−1}

≤ ∥(gh)^{−1} | V^{+}∥^{−1} ≤ a_{4}(ε) · ∥g^{−1} | V^{+}(g)∥^{−1} · ∥h^{−1} | V^{+}(h)∥^{−1}.

Similarly for (gh)^{−1}:

a_{4}(ε)^{−1} · ∥g | V^{−}(g)∥ · ∥h | V^{−}(h)∥
≤ ∥gh | V^{−}∥
≤ a_{4}(ε) · ∥g | V^{−}(g)∥ · ∥h | V^{−}(h)∥.

Hence for a(ε) = max(2, a_{i}(ε)) and s_{p}(g) ≤ a(ε)^{−1}, s_{n−p}(h) ≤ a(ε)^{−1}, we obtain that gh is pseudohyperbolic with V^{+}(gh) = V^{+}, V^{−}(gh) = V^{−},

s_{p}(gh) ∼ s_{p}(g)s_{p}(h)

and d(V^{+}(gh), V^{+}(g)) ≪ s_{p}(g), d(V^{−}(gh), V^{−}(h)) ≪ s_{p}(h). q.e.d.

We obtain:

Corollary 5.23. Let g and h be ε-hyperbolic, ε-transversal elements then there is a constant s(ε) < 1 such that if s(g) < s(ε), s(h) < s(ε) then gh is ε/2-hyperbolic and gh and h, gh and g are ε/2-transversal.

Proof. By Lemma 5.6 there is a constant s(ε), such that for s(g) < s(ε), s(h) < s(ε) we have d(A^{+}(g), A^{+}(gh))=d(V^{+}(g), V^{+}(gh)) < ε/4 and therefore d(A^{+}(gh), D^{−}(h)) ≥ − d(A^{+}(gh), A^{+}(g)) + d(A^{+}(g), D^{−}(h)) ≥ ε/2. The same is true for the other subspaces. q.e.d.

The next lemma will be applied in the following situation. Let h_{1}, …, h_{m} be ε-hyperbolic, pairwise ε-transversal elements and let H be the group generated by h_{1}, …, h_{m}. Under certain hypotheses for the h_{i}’s we can guarantee that certain elements h of H are still hyperbolic, but h will not be ε-hyperbolic, in general. E.g., think of h_{1}^{−n} h_{2} h_{1}^{n}. The lemma shows that we can keep our elements ε-hyperbolic and εtransversal, namely by multiplying by a fixed element g_{0}.

Lemma 5.24. Let g_{0}, h_{1}, …, h_{m} be ε-hyperbolic, pairwise ε-trans-versal elements and let s = max{s(g_{0}), s(h_{1}), …, s(h_{m})}. Let g_{ℓ} = g_{0} · h_{i_{1}}^{n_{1}} … h_{i_{ℓ}}^{n_{ℓ}}, i_{k} ≠ i_{k+1}, ℓ ∈ ℤ, ℓ > 0, n_{i} ∈ ℤ, n_{i} ≠ 0 for i = 1, …, ℓ and M_{ℓ} = |n_{1}| + ⋯ + |n_{ℓ}|. Then there is a constant b(ε) > 1, such that if s < b(ε)^{−2}, then for every s_{0}, such that s < s_{0}^{2} < b(ε)^{−2}, and ℓ ∈ ℤ, ℓ > 0, we have:

(1) s(g_{ℓ}) ≤ s_{0}^{M_{ℓ}+1}.

(2) d(A^{+}(g_{ℓ−1}), A^{+}(g_{ℓ})) ≤ \frac{ε}{2}s_{0}^{M_{ℓ−1}}.

(3) d(A^{+}(g_{0}), A^{+}(g_{ℓ})) ≤ \frac{ε}{2}.

(4) d(A^{−}(g_{ℓ}), A^{+}(h_{i}) ∪ A^{−}(h_{i})) ≥ \frac{ε}{2} for i ≠ i_{ℓ}.

(5) g_{ℓ} is ε/2-hyperbolic.

Proof. Let us consider a(ε/2), given by Lemma 5.6 and put b(ε) = 4ε^{−1}a (ε/2). Let us assume that s < b(ε)^{−2} and let s_{0} be any number such that s < s_{0}^{2} < b(ε)^{−2}. Let us first to show, that the statements of the the lemma follow from the following inequalities:

s(g_{ℓ}) ≤ \frac{ε}{4}s_{0}^{M_{ℓ}+2}
d(A^{+}(g_{ℓ}), A^{+}(g_{ℓ+1})) ≤ \frac{ε}{4}s_{0}^{M_{ℓ}+1}
d(A^{−}(g_{ℓ})), A^{−}(h_{i_{ℓ}})) ≤ \frac{ε}{4}s_{0}^{|n_{ℓ}|}.

In fact,

d(A^{+}(g_{0}), A^{+}(g_{ℓ})) ≤ ∑_{i=1}^{ℓ} d(A^{+}(g_{i−1}), A^{+}(g_{i}))

≤ \frac{ε}{4}s_{0} + \frac{ε}{4}s_{0}^{M_{1}+1} + ⋯ + \frac{ε}{4}s_{0}^{M_{ℓ−1}+1} ≤ \frac{ε}{4}.

and

d(A^{−}(g_{ℓ})), A^{−}(h_{i_{ℓ}})) ≤ ε/4.

Therefore, for all ℓ, ℓ ≥ 0

d(A^{+}(g_{ℓ}), A^{−}(g_{ℓ})) ≥ d(A^{+}(g_{0}), A^{−}(h_{i_{ℓ}})) − d(A^{+}(g_{ℓ}), A^{+}(g_{0}))

− d(A^{−}(g_{ℓ}), A^{−}(h_{i_{ℓ}}))

≥ ε − \frac{ε}{4} + \frac{ε}{4} ≥ \frac{ε}{2}

which means, that for all ℓ, ℓ ≥ 0, g_{ℓ} is ε/2-hyperbolic.

Additionally, if i ≠ i_{ℓ} we have

d(A^{−}(g_{ℓ}), A^{+}(h_{i}) ∪ A^{−}(h_{i}))

≥ d(A^{−}(h_{i_{ℓ}}), A^{+}(h_{i}) ∪ A^{−}(h_{i})) − d(A^{−}(g_{ℓ}), A^{−}(h_{i_{ℓ}}))

≥ ε − \frac{ε}{4} > \frac{ε}{2},

for all ℓ and i, i_{ℓ} ≠ i. So for all ℓ and i if i_{ℓ} ≠ i elements the g_{ℓ} and h_{i} are ε/2-transversal.

We will prove the inequalities above by induction on the index ℓ. We will assume, that they are true if the index is ≤ ℓ. Therefore, the elements g_{ℓ} and h_{i} are ε/2-hyperbolic and ε/2-transversal, if i_{ℓ} ≠ i , therefore we can use the statements of Lemma 5.6. First of all, because of (2) in Lemma 5.6,

s(g_{ℓ+1}) ≤ a(ε/2)s(g_{ℓ}) · s^{|n_{ℓ+1}|} ≤ a(ε/2)s_{0}^{M_{ℓ}+2} · s_{0}^{2|n_{ℓ+1}|} ≤ \frac{ε}{4}s_{0}^{M_{ℓ+1}+2}.

Now, by (3) of Lemma 5.6,

d(A^{+}(g_{ℓ}), A^{+}(g_{ℓ+1})) ≤ a(ε/2)s(g_{ℓ}) ≤ a(ε/2)s_{0}^{M_{ℓ}+2} ≤ \frac{ε}{4}s_{0}^{M_{ℓ}+1}.

d(A^{−}(g_{ℓ}), A^{−}(h_{i_{ℓ}})) ≤ α(ε/2)s_{0}^{2|n_{ℓ}|} ≤ \frac{ε}{4}s_{0}^{|n_{ℓ}|}.

q.e.d.

6. The Basic Lemma

The main claim of the Basic Lemma states additivity of the function α of (4.3).

So in this section we return to affine maps. Recall that H_{B} = SO(n+1, n) is the orthogonal group of the form B of signature (n+1, n) on ℝ^{2n+1} and G_{B} is the subgroup of G_{2n+1} = Aff(ℝ^{2n+1}) of those elements with linear part in H_{B}. Also, Ω was the set of those g ∈ G_{n} for which dim A^{0}(g) = 1 and the eigenvalue of ℓ(g) | A^{0}(g) is +1, see paragraphs following (2.1). These elements were called pseudohyperbolic in Sections 2-4. Let us clarify how this notion of pseudohyperbolicity is related to the one of Section 5. If ℓ(g) is in Ω(n, s, ε) of Section 5 for some s < 1 and ε > 0 and g ∈ G_{B}^{∘} then g is pseudohyperbolic. Conversely, if g ∈ Ω then some power of ℓ(g) is in Ω(n, s, ε) for some s < 1 and ε > 0. Cf. the corresponding discussion for proximality in [2, Section 2].

We need some auxiliary lemmas before we can prove the Basic Lemma.

Lemma 6.1.

(a) Let g ∈ G_{B}^{∘}, ℓ(g) ∈ Ω(n, s, ε). Then ℓ(g) ∈ Ω(n + 1, s′, b(ε)) for some constant b(ε) and \frac{s′}{s} and \frac{s}{s′} are bounded, independently of ε, if s is sufficiently small.

(b) d(v, w) ≥ c(ε) for every g ∈ Ω(n, s, ε), v ∈ S(A^{0}(g)) and every isotropic vector w with ∥w∥ = 1.

(c) For the given norm ∥·∥ and the norm ∥·∥_{B} induced by B on A^{0}(g) we have ∥ · ∥ ∼ ∥ · ∥_{B}, i.e.,

c_{1}(ε) > sup_{x∈A^{0}(g)} \frac{∥x∥}{∥x∥_{B}} > inf_{x∈A^{0}(g)} \frac{∥x∥}{∥x∥_{B}} > c_{1}(ε)^{−1}

for some c_{1}(ε) and every g ∈ Ω(n, s, ε).

(d) ∥g | A^{0}(g)∥ ∼ 1 and ∥g^{−1} | A^{0}(g)∥ ∼ 1.

(e)

s(g) = ∥g | D^{−}(g)∥ · ∥g^{−1} | A^{+}(g)∥
∼ ∥g | A^{−}(g)∥ · ∥g^{−1} | D^{+}(g)∥ = s(g^{−1}).

Proof.

(a) Take a basis e_{1}, … e_{2n+1} of V such that B(e_{i}, e_{j}) = 0 for i ≠ j and B(e_{i}, e_{i}) = 1 for i ≤ n + 1, B(e_{i}, e_{i}) = −1 for i > n + 1. Suppose our norm ∥ · ∥ is given by the Euclidean form Q with respect to this basis. Then for a ∈ H_{B} we have a^{−1} = P a^{t} P^{−1}. Write a = k_{1}dk_{2} in

Cartan form, so k_{1}, k_{2} ∈ O(Q), d = (d_{1}, …, d_{2n+1}) is diagonal with d_{1} ≥ d_{2} ≥ ⋯ ≥ d_{2n+1} > 0. Then a is (n, s, ε_{1})-pseudohyperbolic iff d_{n+1} ≤ sd_{n} iff a^{t} is (n, s, ε_{2})-pseudohyperbolic iff a^{−1} = P a^{t} P^{−1} is (n, s, ε_{3})-pseudohyperbolic, since P ∈ O(Q) iff a is (n + 1, s, ε_{4})- pseudohyperbolic, by 5.4.

To see the claim concerning b(ε), look at the map ⊥ which associates to every subspace W of V its orthogonal space with respect to B. Regard it as a map from a subset of a Grassmannian to another Grassmannian. Then ⊥ is continuous and the claim follows from D^{+}(g) = A^{+}(g)^{⊥} and D^{−}(g)^{⊥} = A^{−}(g).

(b) Otherwise there are sequences g_{i} of (n, ε)-pseudohyperbolic elements such that D^{+}(g_{i}) and D^{−}(g_{i}) converge to spaces D^{+}, D^{−} containing an isotropic vector w. Then A^{+}(g_{i}) = D^{+}(g_{i})^{⊥} and A^{−}(g_{i}) = D^{−}(g_{i})^{⊥} converge to spaces A^{+} = D^{+⊥} and A^{−} = D^{−⊥} containing w, contradicting (n, ε)-pseudohyperbolicity.

(c) By part (b) of the proof the set of pairs of spaces {(A^{+}(g), A^{−}(g)), g (n, ε)-pseu-dohyperbolic} in the product of Grassmannians is compact, hence so is the set of A^{0}(g)′s since A^{0}(g) = D^{+}(g) ∩ D^{−}(g).

(d) follows from (c) since g is a unitary operator on A^{0}(g) with respect to B | A^{0}(g).

(e) is proved by the same method as (a). q.e.d.

We also need the following inequality:

(6.2)  d(x, E_{g}^{+}) ≪ s(g)d(x, g^{−1}x)

for s(g) ≪ 1.

To see this first note that d(x, E_{g}^{+}) ∼ ∥p^{−}(x − c)∥ for arbitrary c ∈ C(g) where p^{−} is the projection to A^{−}(g) in the decomposition V = A^{−}(g) ⊕ A^{0}(g) ⊕ A^{+}(g), by (n, s, ε)-pseudohyperbolicity. Thus

d(x, E_{g}^{+}) ≪ s(g)d(g^{−1}x, E_{g}^{+})

by 6.1 (c) and (a). The triangle inequality gives

s(g)^{−1}d(x, E_{g}^{+}) ≪ d(g^{−1}x, x) + d(x, E_{g}^{+})

which yields (6.2).

Lemma 6.3. Let g and h be two ε-hyperbolic ε-transversal elements, such that gh is ε/2-hyperbolic, gh and g, gh and h are ε/2transversal. Let q be any point in the affine space. Then there is a constant d(ε) such that

d(q, E_{gh}^{+}) ≤ d(q, E_{g}^{+}) + d(ε)s(g) [|α(g)|
+d(q, C_{g}) + s(h)(|α(h)| + d(q, C_{h}))] ,
d(q, E_{gh}^{−}) ≤ d(q, E_{h}^{−}) + d(ε)s(h) [|α(h)|
+d(q, C_{h}) + s(g)(|α(g)| + d(q, C_{g}))] ,
d(q, C_{gh}) ≤ d(ε)(s(h) + s(g))d(q, C_{g})

+ d(ε)(d(q, E_{g}^{+}) + d(q, C_{h})) + s(g)|α(g)| + s(h)|α(h)|.

Proof. We know that |α(g)| = |α(g^{−1})|, s(g) = s(g^{−1}), d(q, C_{g}) = d (q, C_{g^{−1}}) and E_{g^{−1}}^{+} = E_{g}^{−} for every hyperbolic element g, then the second inequality follows from the first one, and because d(q, C_{gh}) ≪ d(q, E_{gh}^{+}) + d(q, E_{gh}^{−}), the third inequality is just a corollary of the first and the second. So we have to prove just the first inequality.

336 h. abels, g.a. margulis & g.a. soifer

Let now q_{0} be a point on L = E_{h}^{+} ∩ E_{g}^{−} such that d(q, q_{0}) = d(q, L). Let y = h^{−1}q_{0} and x = gq_{0}, then d(y, E_{gh}^{+}) ≪ s(gh)d(ghy, y) = s(gh)d(x, y). We have the inequalities

d(q_{0}, y) ≪ d(q_{0}, C_{h}) + |α(h)|
d(x, q_{0}) ≪ d(q_{0}, C_{g}) + |α(g)|.

Therefore,

d(x, y) ≪ d(q_{0}, C_{h}) + α(q_{0}, C_{g}) + |α(h)| + |α(g)|.

Now,

d(q_{0}, C_{h}) ≪ d(q, C_{h}) + d(q, q_{0}) ≪ d(q, C_{h}) + d(q, C_{g}).

The same for d(q_{0}, C_{g}) i.e.,

d(q_{0}, C_{g}) ≪ d(q, C_{g}) + d(q, C_{h}).

Then

d(y, E_{gh}^{+}) ≪ s(gh) [ |α(h)| + |α(g)| + d(q, C_{g}) + d(q, C_{h})] ,

and

d(q, E_{gh}^{+}) ≪ d(q, q_{0}) + d(q_{0}, y) + d(y, E_{gh}^{+})
≪ d(q, C_{g}) + d(q, C_{h}) + α(h)
+ s(gh) · [ |α(h)| + |α(g)| + d(q, C_{g}) + d(q, C_{h})] .

An elementary geometrical fact says, that

|d(q, E_{gh}^{+}) − d(q, E_{g}^{+})| ≪ sin (∡ E_{gh}^{+}, E_{g}^{+}) [d (q, E_{gh}^{+}) + d (q, E_{g}^{+})] . Now, using Lemma 5.6, we immediately have

|d(q, E_{gh}^{+}) − d(q, E_{g}^{+})| ≪ s(g) [d(q, E_{gh}^{+}) + d(q, E_{g}^{+})]

and

s(gh) ≪ s(g) · s(h)

with d(q, E_{g}^{+}) ≤ d(q, C_{g}).

Combining these two inequalities we have

d(q, E_{gh}^{+}) ≤ d(q, E_{g}^{+}) + d(ε)s(g) [ |α(g)| + d(q, C_{g})
+s(h) · (|α(h)| + d(q, C_{h}))]

for some constant d(ε). q.e.d.

6.4. Basic Lemma. Let g and h be ε-hyperbolic, ε-transversal elements such that gh is ε/2-hyperbolic, gh and g, gh and h are ε/2transversal. Let q be any point in affine space. Then there is a constant c(ε), such that

|α(gh) − α(g) − α(h)| ≤ c(ε) (s(g)|α(g)| + s(h)|α(h)|
+d(q, C_{g}) + d(q, C_{h})) .

Proof. Let q_{0} be a point on L = E_{h}^{+} ∩ E_{g}^{−}, such that d(q, q_{0}) = d(q, L). The elements g and h are ε-transversal, so d(q, q_{0}) ≪ d(q, E_{h}^{+})+ d(q, E_{g}^{−}), but d(q, E_{h}^{+}) ≤ d(q, C_{h}), d(q, E_{g}^{−}) ≤ d(q, C_{g}), thus

d(q, q_{0}) ≪ d(q, C_{g}) + d(q, C_{h}). (i)

Let us consider the two vectors v_{g} = g q_{0} − q_{0} and v_{h} = q_{0} − h^{−1}q_{0}. By definition,

α(gh) = B(gq_{0} − h^{−1}q_{0}, e^{0}(gh)) = B(v_{g}, e^{0}(gh)) + B^{0}(v_{h}, e^{0}(gh)). (ii)

We have v_{g} ∈ D^{−}(g) and v_{g} = α(g)e^{0}(g) + w_{1}, w_{1} ∈ A^{−}(g), and for the vector v_{h} we have v_{h} = α(h)e^{0}(h) + w_{2}, w_{2} ∈ A^{+}(h).

Now by ε-hyperbolicity of g and h, we have ∥w_{1}∥ ≪ d(q_{0}, C_{g}) and ∥w_{2}∥ ≪ d(q_{0}, C_{h}). Let now w and \overline{w} be the projections of the vector α(g)e^{0}(g) onto D^{−}(gh) parallel to A^{+}(g) and A^{+}(gh) respectively. Then w = α(g)e^{0}(g) + u_{1}, u_{1} ∈ A^{+}(g) and \overline{w} = α(g)e^{0}(g) + u_{2}, u_{2} ∈ A^{+}(gh). Let us show that w = α(g)e^{0}(gh) + u_{3}, where u_{3} ∈ A^{−}(gh). In fact, w ∈ D^{−}(gh), and w = α e^{0}(gh) + u_{3}, u_{3} ∈ A^{−}(gh). So we have to prove that α = α(g). We have B(w, w) = B(α e^{0}(gh) + u_{e}, α e^{0}(gh) + u_{e}) = α^{2}B(e^{0}(gh), e^{0}(gh)) = α^{2} and on the other hand B(w, w) = B(α(g)e^{0}(g)+u_{1}, α(g)e^{0}(g)+u_{1}) = α(g)^{2}B(e^{0}(g), e^{0}(g)) = α(g)^{2}. Then α = α(g) or −α(g) and because of the orientation procedure α = α(g). By (3) of Lemma 5.6 d(A^{+}(gh), A^{+}(g)) ≪ s(g) therefore ∥w − \overline{w}∥ ≪

338 h. abels, g.a. margulis & g.a. soifer

s(g)(∥w∥ + ∥\overline{w}∥). It is easy to see, using ε/2-hyperbolicity of g and gh, that ∥w∥ ≪ |α(g)| and ∥\overline{w}∥ ≪ |α(g)| and so

∥w − \overline{w}∥ ≪ s(g)|α(g)|.

Then

(iii) |B(\overline{w}, e^{0}(gh)) − B(w, e^{0}(gh))| = |B(w − \overline{w}, e^{0}(gh)|
≪ ∥w − \overline{w}∥ ≪ s(g)|α(g)|.

We have, by definition of v_{g}, |α(g) − B(v_{g}, e^{0}(gh))|

≤ |α(g) − B(α(g)e^{0}(g), e^{0}(gh))| + |B(w_{1}, e^{0}(gh)|,

but

B(w, e^{0}(gh) = B(α(g)e^{0}(gh) + u_{3},

e^{0}(gh)) = α(g) and |B(w_{1}, e^{0}(gh))| ≪ ∥w_{1}∥ ≪ d(q_{0}, C_{g}). Therefore, |α(g)−B(v_{g}, e^{0}(gh))| ≪ |B(w, e^{0}(gh)−B(α(g)e^{0}(g), e^{0}(gh))|+d(q_{0}, C_{g}) = |B(w, e^{0}(gh) − B(\overline{w}, e^{0}(gh)| + d(q_{0}, C_{g}). Now by (iii)

|α(g) − B(v_{g}, e^{0}(gh)| ≪ s(g)|α(g)| + d(q_{0}, C_{g}).

The same is true for h, namely,

|α(h) − B(v_{h}, e^{0}(gh)| ≪ s(h)|α(h)| + d(q_{0}, C_{h}).

Then by (ii) |α(gh) − α(g) − α(h)| ≤ |α(g) − B(v_{g}, e^{0}(gh))| + |α(h) − B(v_{h}, e^{0}(gh))|. Now from (i)

d(q_{0}, C_{g}) ≤ d(q_{0}, q) + d(q, C_{g}) ≪ d(q, C_{h}) + d(q, C_{g})

and

d(q_{0}, C_{h}) ≪ d(q, C_{h}) + d(q, C_{g}).

Therefore, |α(gh) − α(g) − α(h)| ≪ s(g)|α(g)| + s(h)|α(h)| + d(q, C_{g}) + d(q, C_{h}). Which proves the statement of the Basic Lemma. q.e.d.

Let g_{0}, h_{1}, …, h_{m} be ε-hyperbolic ε-transversal elements. Let us now put g_{ℓ} = g_{0}h_{i_{1}}^{n_{1}} … h_{i_{ℓ}}^{n_{ℓ}}, i_{k} ≠ i_{k+1}, ℓ ∈ ℤ, ℓ > 0, M_{ℓ} = |n_{1}|+⋯+|n_{ℓ}|. We will fix a point q ∈ ℝ^{2n+1} and put a_{ℓ} = d(q, E_{g_{ℓ}}^{+}), b_{ℓ} = d(q, C_{g_{ℓ}}), C = max{d(q, C_{g_{0}}), d(q, C_{h_{1}}), …, d(q, C_{h_{m}})}. Let now t_{ℓ} = α(g_{ℓ}), T = max{|α(g_{0})|, |α(h_{1})|, ..., α(h_{m})|} and s = max{s(g_{0}), s(h_{1}), ..., s(h_{m})}. Let a(ε), b(ε), c(ε) and d(ε) be the constants given by Lemmas 5.6, 5.24, 6.3 and 6.4, respectively. Put a_{0}(ε) = max{a(ε), b(ε), c(ε), d(ε)}. We can rewrite the results of Lemmas 6.3 and 6.4 in the following form.

(6.5) a_{ℓ+1} ≤ a_{ℓ}+a_{0}(ε)s(g_{ℓ}) (|t_{ℓ}| + b_{ℓ} + s (h_{i_{ℓ+1}})^{|n_{i_{ℓ+1}}|} [|n_{i_{ℓ+1}}|T + C]) .

(6.6) b_{ℓ+1} ≤ a_{0}(ε) (s(h_{i_{ℓ+1}})^{|n_{i_{ℓ+1}}|} + s(g_{ℓ})) b_{ℓ}

+ a_{0}(ε)(a_{ℓ} + C + s(g_{ℓ})|t_{ℓ}| + s(h_{i_{ℓ+1}})^{|n_{i_{ℓ+1}}|}|n_{i_{ℓ+1}}|T).

(6.7) |t_{ℓ+1} − t_{ℓ} − n_{i_{ℓ+1}}α (h_{i_{ℓ+1}}) |

≤ a_{0}(ε) (a_{ℓ} + C + s (h_{i_{ℓ+1}})^{|n_{i_{ℓ+1}}|} |n_{i_{ℓ+1}}|T + s(g_{ℓ})|t_{ℓ}|) .

We would like to explain now the final steps in the proof of Theorem B. The goal of the first step is to show that a_{ℓ} ≪ 1, b_{ℓ} ≪ 1 and |t_{ℓ}| ≪ M_{ℓ} for all ℓ.

The value of a_{ℓ+1} depends by (6.5) on the values of a_{ℓ}, b_{ℓ}, t_{ℓ} and the same for b_{ℓ+1}, and t_{ℓ+1}. So the idea is as follows. Let r_{ℓ} =max{10a_{0}(ε)a_{ℓ}, b_{ℓ}, 10a_{0}(ε)C, 10a_{0}(ε)T, |t_{ℓ}|/10 M_{ℓ}a_{0}(ε)}. We will show, see (6.18), that

r_{ℓ+1} ≤ r_{ℓ} (1 + 2a_{0}^{2}(ε)s_{0}^{M_{ℓ}}M_{ℓ}) ,

and then, because s_{0} < 1, we will have that

Q = ∏ (1 + 2a_{0}^{2}(ε)s_{0}^{M_{ℓ}}M_{ℓ}) < ∞

and therefore r_{ℓ} ≤ r_{0}Q for all ℓ, i.e., r_{ℓ} ≪ 1. Then, from the definition of r_{ℓ}, we have a_{e} ≪ 1, b_{ℓ} ≪ 1, |t_{ℓ}| ≪ M_{ℓ}.

The fact that b_{ℓ} ≪ 1 has a clear geometrical meaning: Let H be the group generated by the hyperbolic transversal affine transformations h_{1}, …, h_{s}. Then under certain conditions, there is a universal constant C(H), such that for every h ∈ H, d(q, C_{h}) ≤ C(H). Then we will show that |t_{ℓ}| ∼ M_{ℓ}, so one can interpret the statement of the Basic Lemma as an additive property of the function α.

Lemma 6.8. Let r_{ℓ} = max{10a_{0}(ε)a_{ℓ}, b_{ℓ}, 10a_{0}(ε)C, 10a_{0}(ε)T, |t_{ℓ}|/10M_{ℓ}a_{0}(ε)}. There are constants b_{0}(ε) > 1 and Q_{0}(ε) such that if s < b_{0}(ε)^{−1}, then for all ℓ, ℓ ∈ ℤ, ℓ ≥ 1 we have a_{ℓ} ≤ Q_{0}(ε), b_{ℓ} ≤ Q_{0}(ε), |t_{ℓ}| ≤ Q_{0}(ε)M_{ℓ}, r_{ℓ} ≤ Q_{0}(ε).

Proof. Let us first take s_{0} ∈ ℝ, s_{0} < 1 such that a_{0}(ε/2)^{2}s_{0}^{m} m ≤ \frac{1}{10} for all natural numbers m. Note that in particular s_{0} < b(ε)^{−1}. Put b_{0}(ε) = s_{0}^{−2} and let s < b_{0}^{−1}(ε). Then in view of Lemma 5.24 we can assume that the elements g_{ℓ} are ε/2-hyperbolic and ε/2-transversal for all ℓ. Using again Lemma 5.24, we can rewrite inequalities (6.5)-(6.7) as follows:

(6.9) a_{ℓ+1} ≤ a_{ℓ} + a_{0}(ε/2)s_{0}^{M_{ℓ}+1} [|t_{ℓ}| + b_{ℓ} + s_{0}^{2|n_{i_{ℓ+1}}|}|n_{i_{ℓ+1}}|T + C] ,

(6.10) b_{ℓ+1} ≤ a_{0}(ε/2) (s_{0}^{2|n_{i_{ℓ+1}}|} + s_{0}^{M_{ℓ}+1}) b_{ℓ}

+ a_{0}(ε/2) [C + a_{ℓ} + s_{0}^{2|n_{i_{ℓ+1}}|}|n_{i_{ℓ+1}}|T + s_{0}^{M_{ℓ}+1}|t_{ℓ}|] ,

(6.11) |t_{ℓ+1} − t_{ℓ} − |n_{i_{ℓ+1}}|α (h_{i_{ℓ+1}})|

≤ a_{0}(ε/2) [s_{0}^{M_{ℓ}+1}|t_{ℓ}| + s^{2|n_{i_{ℓ+1}}|}|n_{i_{ℓ+1}}|T + C + b_{ℓ}] .

Now using the fact that a_{0}(ε/2)^{2}s_{0}^{m}m ≤ s_{0}/10 in particular a_{0}(ε/2)^{2}s_{0} ≤ 1/10, we have from (6.9)-(6.11)

a_{ℓ+1} ≤ a_{ℓ} + \frac{s_{0}^{M_{ℓ}}}{10} [|t_{ℓ}| + b_{ℓ} + \frac{T}{10a_{0}(ε/2)^{2}} + C] , (6.12)

b_{ℓ+1} ≤ \frac{b_{ℓ}}{10} + \frac{T}{10} + s_{0}^{M_{ℓ}}|t_{ℓ}| + a_{0}(ε/2)[C + a_{ℓ}], (6.13)

|t_{ℓ+1}| ≤ |t_{ℓ}| + |n_{i_{ℓ+1}}|T + s_{0}^{M_{ℓ}}|M_{ℓ}|T + \frac{s_{0}^{|n_{i_{ℓ+1}}|} · T}{10} + a_{0}(ε/2)[C + b_{ℓ}]. (6.14)

Because of our notation we have

(6.15)  a_{ℓ+1} ≤ \frac{r_{ℓ}}{10a_{0}(ε/2)} + s_{0}^{M_{ℓ}}M_{ℓ}a_{0}(ε/2)r_{ℓ} + \frac{s_{0}^{M_{ℓ}}r_{ℓ}}{1000} + \frac{s_{0}^{M_{ℓ}}r_{ℓ}}{100}

≤ \frac{r_{ℓ}}{10a_{0}(ε/2)} + 2s_{0}^{M_{ℓ}}M_{ℓ}a_{0}(ε/2)r_{ℓ} ≤

≤ \frac{r_{ℓ}}{10a_{0}(ε/2)} [1 + 2a_{0}^{2}(ε/2)s_{0}^{M_{ℓ}}M_{ℓ}] .

From (6.13) we have

(6.16)  b_{ℓ+1} ≤ r_{ℓ}

and from (6.14) we have

|t_{ℓ+1}| ≤ 10M_{ℓ}a_{0}(ε/2)r_{ℓ} + \frac{|n_{i_{ℓ+1}}|r_{ℓ}}{10d_{0}(ε/2)} + \frac{r_{ℓ}}{100d_{0}(ε/2)} + \frac{r_{ℓ}}{10} + a_{0}(ε/2)r_{ℓ} (6.17)
≤ 10a_{0}(ε/_{2})r_{ℓ}(M_{ℓ} + |n_{i_{ℓ+1}}|)
= 10a_{0}(ε/2)M_{ℓ+1}r_{ℓ}.

Therefore

(6.18)  r_{ℓ+1} ≤ r_{ℓ}(1 + 2a_{0}^{2}(ε/2)s_{0}^{M_{ℓ}}M_{ℓ}).

If Q = ∏(1 + 2a_{0}^{2}(ε/2)s_{0}^{m_{ℓ}}M_{ℓ}) < ∞, then a_{ℓ} ≤ Qr_{1}/10a_{0}(ε/2), b_{ℓ} ≤ Qr_{1}, |t_{ℓ}| ≤ 10M_{ℓ}a_{0}(ε/2)Qr. Let Q_{0} = 10M_{0}(ε/2)Qr_{1} then a_{ℓ} ≤ Q_{0}, b_{ℓ} ≤ Q_{0}|t_{ℓ}| ≤ Q_{0}M_{ℓ}. q.e.d.

Lemma 6.19. Assume that d(q, S_{g_{0}}) = α(g_{0}) = 0 and α(h_{i}) > 0 for all i = 1, …, m. There is a constant c_{0}(ε) > 1 such that if s < c_{0}(ε)^{−1} and C ≤ c_{0}(ε)^{−1}T, then t_{ℓ} ≥ \frac{2}{3}M_{ℓ}T.

Proof. Let s_{0} be a positive real number < 1 such that a_{0}(ε/2)^{2}s_{0} < 1/10 and s_{0} < b_{0}(ε)^{−1}. Put c_{0}(ε) = s_{0}^{−2}. For s < c_{0}(ε)^{−1} we have s < b_{0}(ε)^{−1} and we can therefore use the statements of Lemma 6.8. So a_{ℓ} ≤ Q_{0}(ε), b_{ℓ} ≤ Q_{0}(ε), |t_{ℓ}| ≤ M_{ℓ}Q_{0}(ε). We have from (6.11)

(6.20) t_{ℓ+1} ≥ t_{ℓ} + |n_{i_{ℓ+1}}|T − a_{0}(ε/2)C − a_{0}(ε/2)b_{ℓ}

− s_{0}^{m_{ℓ}}|t_{ℓ}| − s^{|n_{i_{ℓ+1}}|}|n_{i_{ℓ+1}}|T.

Now from (6.12) we have a_{ℓ+1} ≤ a_{ℓ} + s_{0}^{M_{ℓ}}M_{ℓ}Q_{0}, therefore for some constant Q_{1} = Q_{1}(ε), a_{ℓ+1} ≤ a_{0} + s_{0}Q_{1}(ε), but a_{0} = 0 then

(6.21)  a_{ℓ+1} ≤ s_{0}Q_{1}(ε).

Assume that c ≤ T/10a_{0}^{2}(ε/2) then from (6.10) we have

b_{ℓ+1} ≤ \frac{s_{0}b_{ℓ}}{10} + \frac{s_{0}T}{10} + a_{0}(ε/2)C + a_{0}(ε/2)s_{0}Q_{1}(ε).

Then, because b_{0} = 0, there is a constant Q_{2}(ε) such that

b_{ℓ+1} ≤ s_{0}Q_{2}(ε) + a_{0}(ε/2)C.

Let us now assume that s_{0}Q_{2}(ε)a_{0}(ε/2) ≤ \frac{T}{10}. We know that C ≤ \frac{T}{10a_{0}^{2}(ε/2)}, then from (6.20) we have

t_{ℓ+1} ≥ t_{ℓ} + |n_{i_{ℓ+1}}|T − \frac{T}{10a^{2}(ε/2)} − \frac{T}{10} − \frac{T}{10}

≥ t_{ℓ} + (|n_{i_{ℓ+1}}| − \frac{3}{10}) T

≥ t_{ℓ} + \frac{2}{3}|n_{i_{ℓ+1}}|T.

Now t_{0} = 0, hence t_{ℓ+1} ≥ \frac{2}{3}M_{ℓ}. q.e.d.

Proof of Theorem B. Using [2], we can find elements h̃_{1}, …, h̃_{m} in Aff (ℝ^{2n+1}) such that:

(1) ℓ \binom{}{h_{1}} , …, ℓ \binom{}{h_{m}} are hyperbolic and pairwise transversal.

(2) The group ⟨ℓ(h̃_{1}), …, ℓ(h̃_{m})⟩ generated by ℓ \binom{}{h_{1}} , …, ℓ \binom{}{h_{m}} is free and Zariski dense in O(n + 1, n).

(3) The Zariski closure of every group ⟨ℓ \binom{}{h_{i}}⟩, i = 1, …, m is connected.

We will assume that h̃_{i} = ℓ \binom{}{h_{i}}, i = 1, …, m, in other words, that all these elements fix the point zero. Let now g_{0} ∈ Aff (ℝ^{2n+1}) be a hyperbolic element transversal to every h̃_{i}, i = 1, …, m. We can assume that we choose g_{0} such that there is a point q in ℝ^{2n+1} such that g_{0}(q) = q. Let now ε, ε ∈ ℝ, be a positive number such that g_{0}, h̃_{1}, …, h̃_{m} are ε-transversal and ε-hyperbolic. Then there are vectors v_{1}, …, v_{m} such that if h_{i} = h̃_{i}v_{i}, i = 1, …, m then α(h_{i}) > 0, for all i = 1, …, m, see (4.3) and (4.5) for the definition of α.

It is clear that for every n,n ∈ ℤ, n > 0 and hyperbolic element h we have: s(h^{n}) = s(h)^{n}, d(q, C_{h^{n}}) = d(q, C_{h}), α(h^{n}) = nα(h). So, there is N, N ∈ ℤ, N > 0 such that the elements g_{0}^{N}, h_{1}^{N}, …, h_{m}^{N} satisfy the requirements of Lemma 6.8, but on the other hand for every n, n > N, the elements h_{1}^{n}, …, h_{m}^{n} have Property (2).

We will assume therefore that the elements g_{0}, h_{1}, …, h_{m} satisfy Lemma 6.19 and so for g_{ℓ} = g_{0}h_{i_{1}}^{n_{1}} … h_{i_{ℓ}}^{n_{ℓ}}, we have α(g_{ℓ}) ≥ \frac{2}{3}M_{ℓ}T, where M_{ℓ} = |n_{1}| + ⋯ + |h_{ℓ}| and T = max{α(h_{1}), …, α(h_{m})}.

Let us now show that group ⟨h_{1}, …, h_{m}⟩ = H acts properly discontinuously on ℝ^{2n+1}. Let K be any compact subset in ℝ^{2n+1} then {h ∈ H : h K ∩ K ≠ ∅} = {h ∈ H : g_{0}h K ∩ g_{0}K ≠ ∅}.

There is a constant c = c(K), such that if

g_{0}h K ∩ g_{0}K ≠ ∅, then d(g_{0}h x, g_{0}x) ≤ c(K)

for some point x ∈ K, hence

{h ∈ H : g_{0}h K ∩ g_{0}K ≠ ∅} ⊆ {h ∈ H : α(g_{0}h) ≪ c(K)}.

As we explained above, if h = h_{i_{1}}^{n_{1}}, …, h_{i_{ℓ}}^{n_{ℓ}} is in the latter set, then

M_{ℓ} ≤ \frac{3α(g_{0}h)}{2T} ≪ \frac{3c(K)}{2T} and therefore

#\{h ∈ H : h K ∩ K ≠ ∅\} ≤ # \{h ∈ H : M_{ℓ} ≤ \frac{3c(K)}{2T}\} < ∞.

q.e.d.

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