Circuito general del proyecto

The div0(X)of a Hadamard manifoldXis the same as the “rate of divergence” of geodesics. Therefore for rank 1 symmetric spaces it is exponential since they have pinched negative curvature.

In this section we calculate divkwherek≥ 1for any rank one symmetric space. First we start with a lemma which will be needed in the proof of Theorem3.2.3.

Lemma 3.2.1. Given a Lipschitz functionf: Sk ₋→

Rn, where1 ≤ k < n, there exists a

constantc = c(n)such that for any ballBr(u0), we can find a mapg: Sk −→ Rnwhich

satisfies the following conditions: 1. fis homotopic tog. 2. volk(g)≤cvolk(f). 3. g(Sk)∩B◦_r(u0) =∅.

4. fandgagree outsideBr(u0).

5. The(k+1)-volume of the homotopy is bounded bycrvolk(f).

Proof. The proof follows the proof of Theorem10.3.3in [16]. The idea is to project the part inside the ballBr(u0)to the sphereSr(u0). We will project from a point within the ball Br/2(u0), but since projecting from the wrong center might increase the volume by a huge factor, we average over all possible projections and prove that the average is under control, and therefore we have plenty of centers from which we can project and still have the volume of the new function under control.

Let ωn−1 be the volume of the unit (n−1)-sphere in Rn. Let |Dxf|represent the Jacobian offat the pointx. Letπube the projection map from the pointu∈Br/2(u0)to the sphereSr(u0). volk(πu◦f)≤ Z f−1_(B r(u0)) |Dxf|(2r)k kf(x) −ukkcosθdx+volk(f), (3.2.1)

whereθis the angle between the ray connectingutof(x)and the outward normal vector to the sphere at the point of intersection of the ray and the sphereSr(u0). Ifu∈Br/2(u0) then it is easy to see thatθ≤π/6and therefore1/cosθ≤2/√3≤2.

By integrating (3.2.1) over the ballBr/2(u0)we get,

Z Br/2(u0) volk(πu◦f)du≤ Z Br/2(u0) Z f−1_(B r(u0)) |Dxf|(2r)k kf(x) −ukk_cosθdx du +vol(Br/2(u0)) volk(f) ≤ Z f−1_(B r(u0)) |Dxf| Z Br/2(u0) 2(2r)k kf(x) −ukk du dx +vol(Br/2(u0)) volk(f) ≤2k+1rk Z f−1_(B r(u0)) |Dxf| Z B3r/2(f(x)) 1 kukk du dx +vol(Br/2(u0)) volk(f) =2k+1rk Z f−1_(B r(u0)) |Dxf| Z Sn−1 Z3r/2 0 1 tkt n−1_{dt dµ dx} +vol(Br/2(u0)) volk(f) =2 k+1₃n−k_rn_ωn−1 2n−k_(n−k) Z f−1_(B r(u0)) |Dxf|dx +vol(Br/2(u0)) volk(f) ≤ 22k+1₃n−k_n n−k +1 vol(Br/2(u0))volk(f) ≤ 22n−13n−1n+1 vol(Br/2(u0))volk(f). (3.2.2)

The(k+1)-volume of the homotopy is bounded by2r[22n−13n−1n+1]volk(f). We take c=2[22n−1₃n−1_{n+1], which now satisfies all the requirements of the lemma.}

ferential equationx00(t) +kx(t) = 0with the initial conditionsx(0) = 0andx0(0) = 1 respectivelyx(0) =1andx0(0) =0. We also set ctk =csk/snk.

Now we use Lemma3.2.1to prove a similar result for any Riemannian manifold with bounded geometry which we will call the “Deformation Lemma”.

Lemma 3.2.2 (Deformation Lemma). Given a Riemannian manifold M, with sectional curvature L ≤ K ≤ H and injectivity radius inj(M) ≥ , then there exist constants

δ = δ(L, K, ) > 0andc = c(n, L, K, ), which do not depend on Msuch that for any Lipschitz mapf: Sk _{−→Mand any ballB}

r(p)⊂ Mwith radiusr≤δ, we can find a map g: Sk −→Mwhich satisfies the following conditions:

1. fis homotopic tog.

2. volk(g)≤cvolk(f).

3. g(Sk_)∩B◦

r(p) =∅.

4. fandgagree outsideBr(p).

5. The(k+1)-volume of the homotopy is bounded bycvolk(f).

Proof. Takeδ = min{/2, π/2√H}. By comparison with the spaces of constant curva- tureLandH, we have the following bounds withinBδ(p).

Dexp−1_p ≤A=max{1, δ/sn_H(δ)}, (3.2.3) Dexp_p ≤B=max{1,sn_L(δ)/δ}. (3.2.4)

See Theorem2.3and Corollary2.4in chapter6of [32] for details but the reader should be aware that the estimate forDexp−1_p

is stated incorrectly there. Note thatAandBonly

depend onL,Kandbut not onpnorM. Now we use exp−1

p to lift the mapflocally, i.e. inBδ(p), toTpMthen use Lemma3.2.1 to deform this lifted map to a map that lands outside the ball Bδ(0) ⊂ TpMand then project back using exp_pto the manifoldM. Thek-volume of the new map and the(k+

1)-volume of the homotopy will be controlled because of the bounds onDexp_p

and

Dexp−1_p

given by equations (3.2.3) and (3.2.4) and the estimates given in Lemma3.2.1.

This finishes the proof of the lemma.

Theorem 3.2.3. IfX is a rank 1symmetric space of noncompact type, thendivk(X) has a polynomial growth of degree at mostkfor everyk≥1.

Proof. Letx0 be any point inX,Sr(x0)the distance sphere of radiusrcentered atx0, and dSr(x0)the Riemannian distance function on the sphere.

We will show that there exists 0 < ρ < 1such that there is a filling of any admissible k-sphere onSr(x0)outsideB_ρr◦ (x0)which grows polynomially of degree at mostkasr→

∞.

Let πR

r: SR(x0) −→ Sr(x0) be the radial projection from SR(x0) to Sr(x0), and let λR_r: Sr(x0) −→ SR(x0) be its inverse. Assume that the metric onXis normalized such that the sectional curvature is bounded between −4 and −1. By comparison with the spaces of constant curvatures−1and−4, it is easy to see that the mapπR

r decreases dis- tance by at least a factor of sinhR/sinhr, andλR_r increases distance at most by a factor of

sinh2R/sinh2r.

FixA > 0and letf: Sk ₋→_S

r(x0)be a Lipschitz map such that volk(f)≤Ark. Then volk(πr

ρr◦f)≤Ark(sinhρr/sinhr) k

.

Since horospheres are Lie groups with left invariant metrics, the curvature is bounded above and below and there is a lower bound on the injectivity radius. This puts a uniform bound on the curvature as well as the injectivity radius of distance spheresSr(x0)as long as ris big enough. Because of that and by using theDeformation Lemma, we could deform πr_ρr ◦ f on Sρr(x0) to a new function g which misses a ball, on the sphere Sρr(x0), of radius δand has k-volume ≤ cvolk(πr

ρr ◦f) where δ andc are the constants given by the Deformation Lemma. And such that the homotopy betweenπr_ρr◦fandghas(k+

1)-volume≤cvolk(πrρr◦f).

Let p ∈ Sρr(x0) be the center of this ball and let q be its antipodal point. We will coneg fromq, inside the sphere, and then project the cone from x0 back to the sphere Sρr(x0). Because the curvature is less than−1the(k+1)-volume of the cone is smaller than volk(g)≤cvolk(πr

ρr◦f)(see Remark3.1.3).

To estimate the volume of the projection of the cone, we need to find a lower bound on the distance fromx0to the image of the cone. Letx ∈ Sρr(x0)such thatdSρr(x0)(x, p) =

δ. Let x0 be the intersection point of S2ρr(q) and the ray −qx. For large values of→ r, x and x0 are close, and therefore we have dS2ρr(q)(p, x

0_{) ≥ δ/2. By comparison to the}

space of constant curvature −4 we see that θ = _∠q(p, x) ≥ δ/(2sinh4ρr). Now by comparison with Euclidian space we haved(x0,−qx→) ≥ρrsinθ≥ ρrsin(δ/(2sinh4ρr)).

So the cone from qmisses a ball of radius ρrsin(δ/(2sinh4ρr))around x0. Projecting the cone which lies inside Sρr(x0) from x0 to Sρr(x0) gives us a filling forg of volume

≤cArk(sinhρr/sinhr)k_sinh(2ρrsinsinh_(δ/22ρr_sinh4ρr))

k+1 .

Notice that sinht behaves like et/2 ast → ∞ and like t as t → 0, while sint be- haves like tas t → 0. Because of that the above estimate of the filling ofgbehaves like cArk_(eρr_/er₎k e2ρr_/2

2δρr/e4ρr

k+1

= _(4δρ)cAk+1_re((7k+6)ρ−k)r. Clearly we can chooseρ small

enough to make(7k+6)ρ−knegative. Therefore there exists r0 > 0such that for all r≥ r0 the(k+1)-volume of the filling ofgwill be smaller than1. Now the filling of the original mapfconsists of three parts.

1. The radial projection offto the sphereSρr(x0), which has(k+1)-volume≤Ark.

2. The homotopy used to deform the mapπr

ρr◦fto the new mapg, which has(k+ 1)-volume≤cvolk(πr

ρr◦f)≤cvolk(f)≤cArk. 3. The filling ofg, which has(k+1)-volume< 1.

This filling of fis Lipschitz and lies outside B◦_ρr(x0), and is therefore ρ-admissible, and it has (k+1)-volume smaller than A(c+1)rk _{+1. This finishes the proof of the} theorem.