A continuity property for the map + +∞

As d_σ(a, x)≤d a, ρ_σ(x)

+d ρ_σ(x), x

we deduce that

dξ(a, x) =d−(a, x) +d+(a, x)≤(4`|u|+ 2) dξ(a, ρ−(x)) +dξ(a, ρ+(x)) .

Corollary 49. Let ξ ∈ Ξ. Equip I with d= d_ξ. Then, for X ⊂ I the following assertions are equivalent:

1. X is bounded.

2. For every retraction ρ centered at a sector-germ of I, ρ(X) is bounded.

3. There exist two opposite sector-germss₊ ands− such that ifρ_s− andρ_s+ are retractions centered at s₋ and s₊, ρ_s−(X) and ρ_s+(X) are bounded.

Moreover every bounded subset of I₀ is finite.

Proof. By Theorem 47, (1) implies (2) which clearly implies (3). The fact that (3) implies (1) is a consequence of Lemma 49. The last assertion is a consequence of (3) and of Theorem 5.6 of [Héb17].

Corollary 50. The setsρ⁻¹₊ (V)∩ρ⁻¹₋ (V)such that V is an open set of an apartment A and ρ−, ρ₊ are retractions onto A centered at opposite sector-germs of A form a basis of Tm. In particular Tm is the initial topology with respect to the retractions centered at sector-germs.

Proof. This is a consequence of Lemma 49.

5.4 A continuity property for the map +

The aim of this subsection is to prove the theorem below, which will be useful to prove the contractibility of I for Tm. To simplify the notation, we write + instead of++∞.

Theorem 51. Let ξ∈ Ξ and u∈C_f^v. Equip I with dξ. Then the map I ×R⁺ → I defined by (x, t)7→x+tu is continuous.

To prove this theorem we prove that if a sequence (x_n),(t_n)

∈ (I ×R+)^Z≥0 converges to some (x, t) ∈ I ×R⁺, then (x_n+t_nu) converges to x+tu. We first treat the case where t 6= 0.

Fixξ ∈Ξand write ξ = (θ₊, θ−). Fix a norm | . |on A.

Lemma 52. Let u∈C_f^v. ThenT_u :I →R⁺ andy_u :I →(A,| .|) are Lipschitz ford_θ+ and d_ξ.

Proof. By Theorem 37 and Theorem 47, we can assume θ₊ = (+∞,| . |). Let ` ∈ R>0 be such that for all a ∈ A, `|a|u−a ∈ C_f^v. Let x, x⁰ ∈ I and (u, u⁰) ∈ U+∞(x, x⁰) be such that dθ+(x, x⁰) = |u| +|u⁰|, which exists by Lemma 29. Then x+Tu(x)u ∈ A and thus x⁰+u⁰+T_u(x)u=x+u+T_u(x)u∈A. Therefore

x⁰+u⁰+T_u(x)u+ (`|u<sup>0</sup>|u−u<sup>0</sup>) = x<sup>0</sup>+ (T<sub>u</sub>(x) +`|u⁰|)u∈A.

Hence Tu(x⁰) ≤ Tu(x) + `|u<sup>0</sup>| ≤ Tu(x) + `dθ+(x, x⁰). By symmetry we deduce that Tu is

`-Lipschitz for d_θ+. The fact that y_u is Lipschitz for d_θ+ is a consequence of the continuity of the map + (Lemma 28) and of the fact that y_u =ρ+∞+T_u.u. As d_θ+ ≤d_ξ, the lemma is proved.

Lemma 53. Let u∈C_f^v and (x_n, t_n)∈ I ×R>0 be such that x_n→x for d_ξ and t_n →t, for some x∈ I and t ∈R^>0. Then x_n+t_nu→x+tu for d_ξ.

Proof. First assume x∈ A. By Lemma 53, Tu(xn)→Tu(x) = 0. Consequently, for n ∈Z^≥0 large enough, x_n +t_nu ∈ A. Write ξ = (θ₊, θ−). By the continuity of the map + for d_θ+ (Lemma 28), x_n+t_nu → x+tu for d_θ+. As the topologies induced by d_θ+ and d_ξ on A agree with the topology induced by its structure of a finite-dimensional real vector-space (by Corollary 38 and Theorem 47 (3)), we deduce that x_n+u→x+u for d_ξ.

We no longer assume thatx∈A. LetA be an apartment containingx+∞. Letφ:A^+∞→ A. Letg ∈Gbe an automorphism inducingφ. By Lemma 41,xn+tnu=g⁻¹. g.(xn+tnu)

= g⁻¹.(g.x_n+t_nu)for alln∈Z^≥0. As g.x∈A, we deduce thatg.x_n+t_nu→g.x+tu ford_ξ. By the continuity of g⁻¹ : (I, d_ξ)→(I, d_ξ) (by Theorem 47 (2)), x_n+u→x+u for d_ξ.

It remains to prove that if a sequence (x_n, t_n) ∈ (I ×R+) converges to (x,0), for some x ∈ I, then (xn +tnu) converges to x. In order to prove this we first study the map t 7→ρ−∞(x+tu).

Tits preorder on I, vectorial distance on I and paths

Recall the definition of the Tits preorder ≤ on A from Subsection 2.2. As ≤ is invariant under the action of the Weyl groupW^v,≤induces a preorder≤_Aon every apartmentA. Let A be an apartment andx, y ∈A be such that x≤_A y. Then by Proposition 5.4 of [Rou11], if A⁰ is an apartment containing x, y, x ≤_A⁰ y. This enables to define the following relation

≤ on I: if x, y ∈ I, one says thatx ≤ y if there exists an apartment A containing x, y and such that x≤_A y. By Théorème 5.9 of [Rou11], this defines a preorder on I and one calls ≤ the Tits preorder.

Letx, x⁰ ∈ I be such thatx≤x⁰. Let Abe an apartment containingx, x⁰ andf :A→A be an isomorphism of apartments. Then f(x⁰)−f(x)is in the Tits cone T. Therefore there exists a unique d^v(x, x⁰)in C_f^v∩W^v. f(x⁰)−f(x)

. One calls d^v the vectorial distance. Let u ∈ C_f^v. A u-path is a piecewise linear continuous map π : [0,1] → A such that each (existing) tangent vector π⁰(t) belongs to W^v.u. Let x, x⁰ ∈ I be such that x ≤ x⁰, A be an apartment containing them and f : A → A. We define π_x,x⁰ : [0,1] → I by t 7→f (1−t)f⁻¹(x) +tf⁻¹(x)

. By Proposition 5.4 of [Rou11],π_x,x⁰ does not depend on the choice of A.

Letx∈ I and u∈C_f^v. Then d^v(x, x+u) = u.

Lemma 54. Let x∈ I and u∈C_f^v. Then ρ−∞◦π_x,x+u is a u-path.

Proof. This is a weak version of Theorem 6.2 of [GR08] (a Hecke path of the shape u is a u-path satisfying some conditions, see Section 5 of [GR08] for the definition).

Recall that the α^∨_i, for i ∈ I, denote the simple roots. Let Q^∨

R+ = {P

i∈Ix_iα^∨_i|(x_i) ∈ (R+)^I} ⊂A.

Lemma 55. Let u∈C_f^v and π: [0,1]→A be a u-path. Then π(1)−π(0)−u∈Q^∨_R+. Proof. Let w∈W^v. Then by Proposition 3.12 d) of [Kac94],w.u−u∈ −Q^∨_R+. Thus for all t such that π⁰(t) is defined, π⁰(t)−u∈ −Q^∨_R+ and the lemma follows.

Let| . |₀ be a norm onA such that for all q=P

i∈iq_iα^∨_i ∈L

i∈IRα^∨_i, |q|₀ =P

i∈I|q_i|. Lemma 56. Let x∈ I, u∈C_f^v and t, t⁰ ∈R⁺ be such that t≤t⁰. Then

|ρ−∞(x+tu)−ρ−∞(x)|₀ ≤(t+t⁰)|u|₀+|ρ−∞(x+t⁰u)−ρ−∞(x)|₀.

Proof. Writeρ_−∞(x+tu)−ρ_−∞(x) =tu−q₁ andρ_−∞(x+t⁰u)−ρ_−∞(x+tu) = (t⁰−t)u−q₂, with q₁, q₂ ∈ Q^∨

R+, which is possible by Lemma 55 and Lemma 56. Then ρ−∞(x+t⁰u)− ρ−∞(x) = t⁰u−q₁−q₂. One has |ρ−∞(x+tu)−ρ−∞(x)|₀ ≤t|u|₀+|q₁|₀. By choice of| . |₀,

|q₁|₀ ≤ |q₁+q₂|₀ =|ρ_−∞(x+t⁰u)−ρ_−∞(x)−t⁰u|₀, and the lemma follows.

The following lemma completes the proof of Theorem 52.

Lemma 57. Let u ∈ C_f^v. Let (x_n) ∈ I^Z≥0 and (t_n) ∈ R^Z+^≥0 be such that (x_n) converges for d_ξ and (t_n) converges to 0. Then (x_n+t_nu) converges to limx_n for d_ξ.

Proof. By Theorem 47, we can assume ξ = (| . |₀,+∞),(| . |₀,−∞). Let x = limx_n. By the same reasoning as in the proof of Lemma 54:

- we can assume x∈A, - x_n+ (T_u(x_n) +t_n)u→x. By Lemma 57, for alln ∈Z^≥0,

|ρ−∞(x_n+t_nu)−ρ−∞(x_n)|₀ ≤ |ρ−∞(x_n+ (T_u(x_n) +t_n)u)−ρ−∞(x)|₀+ (T_u(x_n) + 2t_n)|u|₀, and thus ρ−∞(xn+tnu)→ρ−∞(x).

By continuity of the map + (Lemma 28) and the continuity of ρ+∞ (Corollary 39) for d_θ+, ρ+∞(x_n+t_nu)→ρ+∞(x). Using Lemma 49 we deduce that (x_n+t_nu) converges to x, which is the desired conclusion.

6 Contractibility of I

In this section we prove the contractibility of I for T+, T− and Tm.

Let | . | be a norm on A, θ = (| . |,+∞) and ξ = (| . |,+∞),(| . |,−∞)

. To simplify the notation we write + instead of ++∞.

Proposition 58. Let u∈C_f^v. We define χ_u :I ×[0,1]→ I by







χ_u(x, t) =x+ t

1−tu if t

1−t < T_u(x) χ_u(x, t) =y_u(x) if t

1−t ≥T_u(x),

where we set ¹₀ = +∞ > t for all t ∈ R. Then χ_u is a strong deformation retract on A for d_θ and d_ξ.

Proof. Let x ∈ A and t ∈ [0,1]. Then T_u(x) = 0 and thus χ_u(x, t) = y_u(x) = x. Let x ∈ I. Then χ_u(x,0) = x and χ_u(x,1) = y_u(x) ∈ A. It remains to show that χ_u is continuous for dθ and dξ. Let (xn, tn)∈(I ×[0,1])^Z≥0 be a converging sequence for dθ or dξ

and (x, t) = lim(x_n, t_n). Suppose for example that _1−t^t < T_u(x) (the case _1−t^t = T_u(x) and

t

1−t > Tu(x) are analogous). Then by the continuity of Tu (Lemma 53), _1−t^tn_n < Tu(xn) for n large enough and thus by the continuity of the map + (Lemma 28 for d_θ and Theorem 52 for d_ξ), χ_u(x_n, t_n) = x_n+ _1−t^tn

nu → x+ _1−t^t u =χ_u(x, t). Therefore, χ_u is continuous, which concludes the proof.

Corollary 59. The masure I is contractible for T+, T⁻ and Tm. Proof. Let u∈C_f^v. We defineΥ_u :I ×[0,1]→ I by







Υ_u(x, t) = χ_u(x,2t) if t≤ 1 2 Υ_u(x, t) = 2(1−t)y_u(x) if t > 1

2.

Then Υ_u is a strong deformation retract on{0}ford_θ andd_ξ, which proves that(I,T+)and (I,Tm) are contractible. By symmetry, (I,T−) is contractible.

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