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Real hyperbolic spaces in .NET Generating 3 of 9 in .NET Real hyperbolic spaces

2.6 Real hyperbolic spaces use none none generating tocreate none for none .NET CF is open in G none for none , since 0 is open in Hn (R). This shows that { : x } is an open subset of . The same argument shows that { : y } is an / open subset of .

Therefore x,y is open in . It remains to show that x,y is not empty. The space Hn (R) is two-point homogeneous, that is, G acts transitively on pairs of equidistant points in Hn (R); see [BriHa 99, Part I, Proposition 2.

17]. It therefore suf ces to prove that x,y = for x = [(r, 0, . .

. , 0, 1)] and y = [( r, 0, . .

. , 0, 1)],. x,y . for some 0 < r < 1. But this is clear since 0 The followin g formula for half-spaces is a variant of the Crofton formula for hyperplanes; see [Roben 98, Corollary 2.5] and [TaKuU 81, Proof of Theorem 1]. Proposition 2.

6.4 (Crofton formula) There is a constant k > 0 such that ( for all x, y Hn (R). Proof By the previous lemma, ( claim that the function.

x y) x y). = kd (x, y),. is nite for all x, y Hn (R). We Hn (R) Hn (R) R+ ,. (x, y) (. is continuous. Indeed, let x, x , y, y Hn (R). Since z) (. y ),. the function (x, y) (. x y). satis es the none for none triangle inequality. In particular, for x, y, x , y Hn (R), we have (. x y ) ( x y ) (. ) + (. Property (FH). Hence, it suf ces to show tha t, for x Hn (R),. x x lim (. ) = 0.. This follows from (i) and (ii) of Lemma 2.6.2, since by regularity of the measure , we have x x lim (. ) = lim {H S : H [x, x none none ] = }. x x = ({H S : x H }) . More none for none over the function (x, y) ( (. gx gy ) x y). is G-invariant, that is,. = (. for all x, y Hn (R), g O( none for none n, 1).. Since Hn (R) is two-point hom ogeneous, it follows that there exists a continuous function : R+ R+ such that (. x y). = (d (x, y)). for all x, y Hn (R). Fix no none for none w r1 , r2 > 0. Choose three collinear points x, y, z in Hn (R) such that d (x, y) = r1 , d (y, z) = r2 , and d (x, z) = r1 + r2 .

The set x z of half-spaces separating x and z is the disjoint union of three sets: the set x y of half-spaces separating x and y, the set y z of halfspaces separating y and z, and the set of all half-spaces containing y on their boundaries. Since, by the previous lemma, the latter set has measure 0, we have (r1 + r2 ) = (r1 ) + (r2 ), that is, is additive. Extend to a continuous group homomorphism : R R by setting (x) = ( x) for x < 0.

It is well known that such a homomorphism must be linear (Exercise C.6.12), that is, there exists k 0 such that (r) = kr for all r R+ .

By (iii) of the previous lemma, = 0. Hence, k > 0 and this concludes the proof. Let now L be a topological group, and let : L O(n, 1) be a continuous homomorphism.

Then L acts by isometries on Hn (R) and hence on , preserving the measure . This gives rise to an orthogonal representation of L on. 2.6 Real hyperbolic spaces L2 ( , ), the real Hilbert s none for none pace of real valued square integrable functions on R (Proposition A.6.1).

Proposition 2.6.5 Let L be a topological group, and let : L O(n, 1) be a continuous homomorphism.

Let be the associated orthogonal representation of L on L2 ( , ). If (L) is not relatively compact in O(n, 1), then H 1 (L, ) = R 0. In particular, L does not have Property (FH).

Proof For x Hn (R), let x denote the characteristic function of the set of half-spaces containing x. Let x, y Hn (R). By the Crofton formula, .

x ( ) y ( ). 2 d ( ) = (.
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