ANÁLISIS DE LAS DISCONTINUIDADES GENERADAS EN LAS PLACAS

In this section we prove Theorem1.2.5. The proof consists of two steps. The first step is to deform the closed curve we wish to fill to a new curve which is continuous when viewed as a curve at infinity with respect to the Tits metric. The length of the new curve and the area of the deformation will have to be under control. The proof is valid for any symmetric space of higher rank. We will establish this in Lemma3.4.1.

In the second step we will use the assumption that the rank is bigger than or equal3, to fill the new curve with a disk with controlled area. This step will be done in Proposi- tion3.4.2.

Sr(x0)to the point−x→0x(∞)∈X(∞). Notice that this map is almost never continuous.

Lemma 3.4.1. Let X be a symmetric space of noncompact type and rank k ≥ 2. There exist two constants c > 0and 0 < ρ ≤ 1 which depend only on X such that, for every Lipschitz curvef: S1 _{−→ S}

r(x0), there exists a new curveg: S1 −→ Sr(x0)which satisfies

the following conditions:

1. λ∞_r ◦gis continuous with respect to the Tits metric.

2. There exists a homotopy betweenfandgwhich lies outsideB◦_ρr(x0).

3. The area of the homotopy is bounded above bycrvol1(f).

4. vol1(g)≤cvol1(f).

Proof. Letδ = min(1, η/2), whereηis the constant given by Lemma3.3.2for =π/2. Dividefinto pieces each of lengthδr, the last piece possibly could be shorter thanδr. If that is the case we will call it “short” and all the other pieces “long”.

Letcj: [0, 1] −→Sr(x0)be one of these pieces. LetCibe a Weyl chamber containing cj(i)fori =0, 1. LetA be an apartment containingC0 andC1. Letγj: [0, 1] −→Sr(x0) be a geodesic (with respect to the Tits metric) in the apartment A connectingcj(0) to cj(1). The goal is to replace the piececj with the geodesic γj. The homotopy between them which leaves the end points cj(0) and cj(1) fixed will be through geodesics in X connectingcj(t)toγj(t)for every0 ≤ t ≤1. The length of these geodesics is no longer than2r, since they lie inside the ballBr(x0).

The new curveg will be formed by replacing each piececj with the curveγj. Notice that the length of each γj is no bigger thanπr. If vol1(f) < δr, i.e. we have no “long” pieces thengis just a point and the statement trivially follows. If vol1(f)≥δr, i.e. we have at least one long piece then it is not hard to see that vol1(g)≤2πvol1(f)/δ.

SinceXis nonpositively curved, the area of the homotopy betweencjandγjis no big- ger than2πr2_{. And the area of the homotopy betweenfandgis no bigger thancrvol1(f),} wherec=4π/δ.

To finish the proof we need to show that the homotopy lies outsideB◦_ρr(x0). So we need to show that the distance betweenx0 and the geodesic, inX, connectingcj(t)andγj(t)is at least ρrfor every 0 ≤ t ≤ 1. Assume thatt ≤ 1/2, the other case is similar. Notice that_∠(cj(0), γj(t))≤ π/2. Applying Lemma3.3.2, we get thatdX(x0, cj(0)γj(t)) ≥ηr, whereηis the constant in Lemma3.3.2.

Recall thatdX(cj(0), cj(t))≤δr. For every pointson the geodesiccj(t)γj(t)there ex- ists a points0 on the geodesicγj(t)cj(0)such thatdX(s, s0) ≤δr. ThereforedX(x0, s)≥

(η−δ)r ≥ ηr/2. By taking ρ = η/2, the image of the homotopy lies outsideB◦_ρr(x0). This finishes the proof of the lemma.

Now we proceed to the second step of the proof. We prove a more general result.

Proposition 3.4.2. Let∆be a spherical building with a re-scaled metric such that each apart- ment is isometric toSn−1(r), the round sphere of radiusr. There exists a constantc > 0which depends only on∆but notrsuch that for any Lipschitz functiong: Sk −→ ∆, wherek < n−1, we can extendgto a new functiongˆ: Bk+1 _{−→∆such thatvolk+1(gˆ)≤crvolk(g).}

Proof. Fix a pointpto be the center of a Weyl chamberC. LetAbe the collection of all opposite Weyl chambers toC, and letBbe the collection of all antipodal points top. Let > 0be the largest positive number such for anyq ∈ Bthe ballBr(q)is contained in the interior of the Weyl chamber containingq. Notice thatonly depends on∆.

The proof of Lemma3.2.1can be easily modified to deform the functiongto miss the ballBr(q)for everyq ∈ B. Now we cone the new deformed function from the pointp to obtain the desired extension.

Proof of Theorem1.2.5. The proof is immediate. We deform the functionfwe wish to fill to a new functiongusing Lemma3.4.1. We fillgwith a disc by invoking Proposition3.4.2, fork=1where∆is the Tits building on the sphereSr(x0).

Remark3.4.3. We expect a similar result to Theorem1.2.5 to hold for a larger class than Symmetric spaces. One candidate is the class of Hadamard manifolds whose boundary at infinity is simply connected with respect to the Tits metric.