Microrrelatos

However, note that we consider both the empty set and singletons in ∂H to be totally geodesic.

Remark 2.4.4. IfV ≤ Lis a closedK-module satisfying (2.4.1), then for each

a∈F_{\ {}0},V ais a closeda−1_{Ka-module satisfying (2.4.1) (withK=a}−1_Ka). Lemma 2.4.5. The intersection of any collection of totally geodesic sets is to- tally geodesic.

Proof. Suppose that (Sα)α∈Ais a collection of totally geodesic sets, and sup- pose thatS=T_αSα6=. Fix [z]∈S, and letzbe a representative of [z]. Then for eachα_∈A, there exist (cf. Remark 2.4.4) anR-algebraKαand a closedKα-subspace

Vα≤ Lsatisfying (2.4.1) (withK=Kα) such thatz∈Vα andSα= [Vα]∩bordH. LetK=T_αKα andV =T_αVα. Clearly,V is a K-module and satisfies (2.4.1).

We have [V]_∩bordH_⊆S. To complete the proof, we must show the converse direction. Fix [x]∈S\ {[z]}. By Observation 2.3.15, there exists a representative xof [x] such that BQ(z,x) = 1. Then for eachα, we may findaα ∈F\ {0}such thatxaα∈Vα. We have

aα=BQ(z,x)aα=BQ(z,xaα)∈Kα.

Since Vα is a Kα-module, this implies x∈Vα. Since αwas arbitrary, x∈V, and

so [x]_∈[V]_∩bordH.

Remark 2.4.6. Given K ⊆bordH, Lemma 2.4.5 implies that there exists a smallest totally geodesic set containingK. If we are only interested in the geometry ofK, then by Proposition 2.4.1 we can assume that this totally geodesic set is really our ambient space. In such a situation, we may without loss of generality suppose that there is no proper totally geodesic subset of bordHwhich containsK. In this case we say thatK isirreducible.

Warning. Although the intersection of any collection of totally geodesic sets is totally geodesic, it is not necessarily the case that the decreasing intersection of nontrivial totally geodesic sets is nontrivial; cf. Remark 11.2.19.

The main reason that totally geodesic sets are relevant to our development is their relationship with the group of isometries. Specifically, we have the following:

Theorem 2.4.7. Let (gn)∞1 be a sequence in Isom(H), and let

(2.4.2) S =n[x]_∈bordH:gn([x])−→ n [x]

o .

Remark 2.4.8. An important example is the case where the sequence (gn)∞1

is constant, saygn =g for alln. ThenS is precisely thefixed point set ofg:

S= Fix(g) :={[x]∈bordH:g([x]) = [x]}.

IfHis finite-dimensional, then it is possible to reduce Theorem 2.4.7 to this special case by a compactness argument.

Proof of Theorem 2.4.7. IfS =, then the statement is trivial. Suppose

thatS ₆=_{, and fix [z]∈S.}

Step 1: Choosing representatives Tn. From the proof of Theorem 2.3.3, we see that each gn may be written in the form [Tn] for some Tn ∈O∗F(L;Q). We have some freedom in choosing the representativesTn; specifically, given an ∈S(F) we may replaceTn byTnTan, whereTanis defined by (2.3.10).

Sincegn([z])→[z], there exist representativeszn ofgn([z]) such thatzn →z. For eachn, there is a unique representativeTn ofgn such that

(Tnz)cn =zn for somecn∈R\ {0}. Then

(Tnz)cn→z.

Remark 2.4.9. If F = Q, it may be necessary to choose Tn ∈ O∗F(L;Q)\ OF(L;Q), despite the fact that each gn can be represented by an element of OF(L;Q).

Step 2: A totally geodesic set. Writeσn =σTn, and let

K=_{a_∈F:σn(a)→a}

V =nx_{∈ L}:Tnx−→ n x

o .

ThenKis anR-subalgebra ofF, andV is aK-module. Givenx,y_∈V, by Obser- vation 2.3.6 we have

σn(BQ(x,y)) =BQ(Tnx, Tny)−→

n BQ(x,y),

soB(x,y)∈K. ThusV satisfies (2.4.1). IfV is closed, then the above observations show that [V]_∩bordHis totally geodesic. However, this issue is a bit delicate:

Claim 2.4.10. If #([V]∩bordH)≥2, thenV is closed.

Proof. Suppose that #([V]_∩bordH) _≥ 2. The proof of Proposition 2.4.1 shows that [V]_∩H₆=. Thus, there existsx∈V for which [x]∈H. In particular, gn([x])→[x]. Lettingo= [(1,0)], we have

dH(o, gn(o))≤2dH(o,[x]) +dH([x], gn([x]))−→

2.4. TOTALLY GEODESIC SUBSETS OF ALGEBRAIC HYPERBOLIC SPACES 17

In particulardH(o, gn(o)) is bounded, saydH(o, gn(o))≤C.

Lemma2.4.11. FixT _∈O∗F(L;Q), and letkTkdenote the operator norm of T. Then

kT_k=edH(o,[T](o))_.

Proof. WriteT=Tj,t(A⊕I), whereTj,t is a Lorentz boost (cf. (2.3.3)) and

A_∈O∗F(H;E). Then

[T](o) = [Tj,t](o) = [(cosh(t),sinh(t),0)]. Here the second entry represents the jth coordinate. In particular,

coshdH(o,[T](o)) = | BQ((1,0),(cosh(t),sinh(t),0))| p |Q(1,0)_{| · |Q}(cosh(t),sinh(t),0)_| = cosh(t) 1 = cosh(t). On the other hand,

kT_k=_kTj,tk=    cosh(t) sinh(t) sinh(t) cosh(t) I    =et.

This completes the proof. ⊳

Thus_kTnk ≤eC for alln, and so the sequence (Tn)∞1 is equicontinuous. It follows

thatV is closed. ⊳

Since #([V]∩bordH)≤1 implies that [V]∩bordHis totally geodesic, we conclude that [V]_∩bordHis totally geodesic, regardless of whether or notV is closed.

Remark 2.4.12. When #([V]∩bordH)≤1, there seems to be no reason to think thatV should be closed.

Step 3: Relating S to [V]_∩bordH. The object of this step is to show that

S = [V]∩bordH unless S ⊆∂H and #(S) ≤ 2. For each [x] ∈ S\ {[z]}, let x be a representative of [x] such thatBQ(z,x) = 1; this is possible by Observation

2.3.15. It is possible to choose a sequence of scalars (a([nx]))∞n=1 inF\ {0}such that

(Tnx)a([

x])

n →x. Leta([nz])=cn. For [x],[y]∈S, we have

a([nx])σTn(BQ(x,y))a ([y]) n =a ([x]) n BQ(Tnx, Tny)a([ny]) (by Observation 2.3.6) =BQ((Tnx)an([x]),(Tny)a([ny])) −→_n BQ(x,y). (2.4.3) In particular, (2.4.4) _|a([x]) n | · |a([ny])| −→_n 1 wheneverBQ(x,y)6= 0.

Claim 2.4.13. Unless S⊆∂Hand#(S)≤2, then for all [x]∈S we have

(2.4.5) _|a([x])

n | −→_n 1.

Proof. We first observe that it suffices to demonstrate (2.4.5) for one value ofx; if (2.4.5) holds for xand [y]₆= [x], thenBQ(x,y)6= 0 by Observation 2.3.15

and so (2.4.4) implies _|a([ny])| →1.

Now suppose thatS *∂H, and choose [x]_∈S_∩H. Then BQ(x,x)6= 0, and

so (2.4.4) implies (2.4.5).

Finally, suppose that #(S)≥3, and choose [x],[y],[z]∈Sdistinct. By (2.4.4) together with Observation 2.3.15, we have_|a([nx])| · |a([ny])| →1,|a([nx])| · |a([nz])| →1, and |a([ny])| · |a([nz])| → 1. Multiplying the first two formulas and dividing by the

third, we see that_|a([nx])| →1. ⊳

For the remainder of the proof we assume that eitherS*∂Hor #(S)_≥3. Pluggingz=xinto (2.4.5), we see thatcn→1. In particular, [z]∈[V]∩bordH. Now fix [x]∈S\ {[z]}. Sincecn→1 andBQ(z,x) = 1, (2.4.3) becomes

a([x])

n →1. Thusx_∈V, and so [x]_∈[V]_∩bordH.

2.5. Other models of hyperbolic geometry

FixF_{∈ {}R,C,Q_}and a set J, and let H=HJ_F. The pair (H,bordH) is known as thehyperboloid model of hyperbolic geometry (over the division algebraF and in dimension #(J)). In this section we discuss two other important models of hyperbolic geometry. Note that the Poincar´e ball model, which many of the figures of later chapters are drawn in, is not discussed here. References for this section include [45, 78].

2.5.1. The (Klein) ball model. Let B=BJ_F =_{x_{∈ H}:=_HJ

F:kxk<1}, and let bordBdenote the closure ofBrelative toH.

Observation2.5.1. The mapeB,H: bordB→bordHdefined by the equation

eB,H(x) = [(1,x)] is a homeomorphism, andeB,H(B) =H. Thus if we let

(2.5.1) coshdB(x,y) = coshdH(eB,H(x), eB,H(y)) = p |1−BE(x,y)| 1_{− k}x_k2p_{1− kyk}2,