AIV.3.4 DIMENSIONAMIENTO DEL HERVIDOR TIPO CALDERÍN

Through out this section, with the exception of Proposition4.2.3,Xwill denote the univer- sal cover of a closed irreducible4-dimensional nonpositively curved real analytic manifold which has no nonstandard component in∂TX.

IfXcontains any higher rank submanifold, then it is not Gromovδ-hyperbolic space for anyδ≥ 0. Easy examples of triangles which are notδ-thin are triangles which lie in a flat in a higher rank submanifold. The goal of this section is to show that all “fat” triangles have to lie near a maximal higher rank submanifold, see Theorem4.2.2. We begin with the following definition.

Definition 4.2.1. Letδ > 0and letFbe a maximal higher rank submanifold inX, a triangle ∆(p, q, r)is calledδ-thin relative toF, if every side is contained in aδ-neighborhood of the union of the other two sides andF.

This definition resembles the definition of thin triangles relative to flats used by Hruska in [25].

Theorem 4.2.2. There exists some constantδ > 0such that any triangle inXisδ-thin relative to some maximal higher rank submanifold.

Most of the proofs in this section depends on the following observation.

Proposition 4.2.3. LetXn_{be ann-dimensional Hadamard manifold, not necessarily ana-}

lytic. Letp ∈ Xand letxnynbe a sequence of geodesic segments such thatxn, ynconverge respectively toξx, ξy∈X(∞). Ifd p, xnyn

goes to infinity thenTd ξx, ξy

≤π.

Proof. The proof is easy and resembles the proof of Lemma 2.1 in [2]. Assume that the proposition is not true. IfTd ξx, ξy > πthen there exists a geodesic σinXsuch that σ(∞) = ξxandσ(−∞) = ξy. Notice that for any two pointsp, q ∈ Xthe metrics∠p and∠qdefine the same topology onX = X∪∂∞X. So without loss of generality and to simplify the notation, we assume thatp=σ(0).

Letσnbe the complete geodesic extendingxnynparameterized such thatσn(0)is the closest point toσ(0). It is easy to see that for large values ofn,xnandyn have to be on opposite sides of σn(0). Therefored σn(0), p

goes to infinity. Fix R > 0 and letqn be the point onpσn(0)such thatd(p, qn) = R. By passing to a subsequence we assume that qn converges to q and that the rays −−−qnx→n and −−−qny→n converge respectively to two rays starting atqand asymptotic to−pξ−→x and

−−→

pξy. In fact these two rays form a complete geodesic, see [2] for details. Thereforeσis contained in a flat strip of widthR. SinceRis arbitrary number, the geodesicσis contained in flat strips of arbitrary width. By a standard compactness argument, it is easy to show that a subsequence of the flat strips will converge to a flat half plane, which implies thatTd ξx, ξy

=π, and this is a contradiction.

For any convex subsetKof a Hadamard manifoldX, we denote byπK the projection map fromXtoK.

Proposition 4.2.4. There exists a constant δ > 0 which depends only on the analytic 4- manifold X, such that for any maximal higher rank submanifold F, any q ∈ F, and any

p /∈F,d πF(p), pq

≤δ. In particular the triangle∆ p, q, πF(p)

isδ-thin.

Proof. Modulo the action of the fundamental groupΓ, there are only a finite number of maximal higher rank submanifolds. If the proposition is false, then there exists a maxi- mal higher rank submanifoldFand a sequence of triangles∆ pn, qn, πF(pn)

such that d πF(pn), pnqn

≥ n. SinceFis closed, we could assume thatπF(pn)is contained in a compact subset ofF. By passing to a subsequence we assume thatpn,qn, andπF(pn)con-

verge respectively toξp ∈ ∂∞X,ξq ∈ ∂∞F, andc∈ F. By Proposition4.2.3,ξp ∈ ∂∞F, which is a contradiction. This finishes the proof.

Lemma 4.2.5. There exists a constant D1 > 0, which depends only on the space X such

that if F is a maximal higher rank submanifold andpq is a geodesic segment, then either

d πF(p), πF(q)

≤D1ord pq, F

≤D1.

Proof. Assume that the statement is false, then there is a maximal higher rank submanifold Fsuch that the following is true: for everyn ∈ _N there exists a geodesic segmentpnqn such that d πF(pn), πF(qn)

≥ n and d pnqn, F

≥ n. Since F is closed, we could assume that πF(pn)is contained in a compact subset of F. By passing to a subsequence we assume thatπF(pn), πF(qn), pn, and qn converge respectively to c ∈ F, η ∈ ∂_∞F, and ξp, ξq ∈ ∂∞X. It is not hard to see that since F is a totally geodesic submanifold, ξp∈/ ∂F. Sinced c, πF(qn)qn

≥n−1for large values ofnandd c, pnqn

≥n, using Proposition4.2.3 we see that ξp is path connected toη ∈ ∂_∞F, and therefore ξp ∈ ∂F which is a contradiction. This finishes the proof of the lemma.

Corollary 4.2.6. There exists a constantD2 > 0such that ifFis a maximal higher rank sub-

manifold andpqis a geodesic segment suchd πF(p), πF(q)

> D2then there are two points p0, q0 ∈pqsuch thatd p0, πF(p) ≤ D2 andd q0, πF(q) ≤ D2. In particularpqruns withinD2 distance from the path consisting of the geodesic segmentspπF(p), πF(p)πF(q),

andπF(q)q.

Proof. Let D2 = D1 + δ, where D1 is the constant in Lemma 4.2.5 and δ is the con- stant in Proposition4.2.4. Letr ∈ pqbe a point which is closest toF. By Lemma4.2.5,

d r, πF(r)

≤ D1 and therefore the two geodesic segments pr and pπF(r) are at most D1 apart. By Proposition4.2.4, there exists a point onpπF(r)which isδclose toπF(p). Therefore there is a point onprwhich isD2 close toπF(p). Similarly there exists a point onrqwhich isD2close toπF(q)and the corollary follows.

Now we start the proof of the main result in this section.

Proof of Theorem4.2.2. Assume that the statement is false, then there exists a sequence of triangles ∆(pn, qn, rn) which are not n-thin relative to any maximal higher rank sub- manifold. In particular they are notn-thin triangles. Therefore there exists a pointcnon pnqnsuch thatd cn, pnrn

≥ nandd cn, qnrn

≥ n. SinceΓ acts cocompactly on X, we assume thatcnare contained in a compact subset. By passing to a subsequence we could assume thatcnconverges to a point c ∈ X. Notice thatpn, qn, andrndiverge to infinity. By passing further to a subsequence we could assume thatpn, qn, andrn con- verge respectively to ξp, ξq, and ξr in∂∞X. By Proposition4.2.3, Td ξp, ξr

≤ πand

Td ξr, ξq ≤ πand therefore they belong to a connected component of the Tits bound- ary. Clearly ξp 6= ξq, therefore there exists a maximal higher rank submanifoldF such thatξp, ξq, ξr ∈∂TF.

We claim thatcbelongs toF. If not, then the geodesic segmentspnqnwould converge to a geodesic σpassing through c and parallel toF. Let σ0 be a geodesic in F which is parallel to σ. By the analyticity of X, σand σ0 are contained in a2-flat. If F /∈ W, i.e. isometric toR2orR3 then the parallel setPσ0 ofσ0properly containsFwhich contradicts the maximality ofF. IfF∈ W, thenσ0 can not be a singular geodesics inF, otherwise the

parallel setPσ0 is4-dimensional andXwould be reducible. So, we assume thatσ0is not a singular geodesic inF. In this casePσis3-dimensional and therefore has to be of the form Q×_Rand therefore inW. But by the assumption onXthat can not happen. Otherwise ∂TXwould have a nonstandard component.

Letp_n0 =πF(pn),q_n0 =πF(qn), andr_n0 =πF(rn). Without loss of generality assume thatd q_n0, r_n0≥d p_n0, r_n0, we might need to pass to a subsequence to guarantee that for everyn. Notice thatd p_n0, q_n0goes to infinity and therefored q_n0, r_n0goes to infinity. We need to show thatd r_n0, p_n0goes to infinity as well. If not then by passing to a subsequence we could assume thatd p_n0, r_n0≤C, for some constantC. By Corollary4.2.6, there exist four pointstn, tn0 ∈ pnqnandsn, sn0 ∈qnrnsuch thatd tn, pn0

,d t_n0, q_n0,d sn, rn0

, andd s_n0, q_n0are smaller than or equalD2. Therefored(tn, sn) ≤2D2 +C. Therefore the two geodesic segmentsqnsnandqntnare at most2D2+Capart. But this contradicts that d c, qnrn

goes to infinity. Therefored p_n0, r_n0 goes to infinity as well. Again by Corollary4.2.6, there are two pointln, l_n0 ∈ pnrnsuch thatd ln, p_n0andd l_n0, r_n0are smaller than or equalD2. Now it is easy to see that all the triangles∆(pn, qn, rn)areD2- thin relative toF, contradicting the choice of these triangles. This finishes the proof.

Remark4.2.7. The proof of Theorem4.2.2shows that the “fat” part of any triangle is close to a maximal higher rank submanifold.